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Let
denote the subgroup of G generated by P; the elements of
are the products
of points.
32
F. Bachmann
LEMMA 2: (Geom. Dedicata 4, 1975).
a,b
ula:r's if and only if ab
E
are aonneatible by a ahain of perpendia-
.
Hje1ms1ev's AKL contains a variety of surprising lemmas on double incidences.
Some of them can be proved for perpendicularity groups and some others
have a core which is valid in all perpendicularity groups. Let a,b,a,d be a chain of perpendiculars with vertices A,B,C and Ola,d; Hjelmslev says: If
b'
give two examples.
b' :";"""":---:7"""::::::=::::,.., B
is any line joining
A,B, then the fourth reflection line d'
of the concurrent lines d,a,(O,b') is incident with c.
Q
(Fig. 2.)
If a point C and a line u are given,
o
we can draw the perpendicular (C,u) and
c
d'
erect in C the perpendicular on (C,ul.
Fi g. 2
This is the line C(C,u), and it is called the C-pal'alZel of u.
I see the core of Hjelmslev's lemma in a more general lemma
3 which can be nicely proved for perpendicularity groups: LEMMA 3: 0
E
F(ABe) and A,Blu imply OIC(C,u) ,
"For points
A,B,C
holds: The c-para1lels of all lines joining
dent with any fixed point of the product of the reflections in We can say more: Under the assumption 0 bijective map of the set of lines jOining
A,B
A,B
are inci-
A,B,C".
E F{ABC), u --~ C(C,u)
is a
onto the set of lines joining o,C.
Lemma 3 and this extension are often useful, e.g. for the study of neighbour relations. Another lemma of Hjelmslev's says:
Fig. 3 a
If two lines a,b have one and only one
c
point in common, then the perpendiculars dropped from a point C to G,b have only 1
the point e in common.
b
(Fig. 3.)
lOriginal version (AKL, 1. Mitt. 62): If a,b have several points in common, then the perpendiculars dropped from a point C to a,b have several points in common.
Orthogonality in reflection geometry
33
This lemma does not hold in the group planes of Galileian groups.
But we
can replace the assumption that a,b have exactly one point in common by the stronger condition that the product of the reflections in a and b leaves exactly one point fixed, and we can prove for all perpendicularity groups: LEMMA 4: IF(ab)
I = 1 implies
F((C,a) (C,b))
= {C}.
As an application, I sketch a proof of LEMMA 5: (KnUppe1 1980).
IF(ABC)
I:::: 1,
"A product of three point reflections has at most one fixed point".
PROOF: (for perpendicularity groups). Then 0
E
Take a line f through o.
ABCf can be reduced to a product Dg of a point D and a line g
F(ABCf).
,;
(HG 2 .6), and 0
Let 0 E F(ABC).
E
F(Dg) implies that '-
Dg is a line a through 0 (HG 2*.12). b
Thus we get ABC = af
with Ola,f.
d Q
By lemma 2 (and some additional
e
reasoning), it follows that there are lines b,c,d,e such that a,b,c,d,e,f
o
rl Q
Fig. 4
is a chain of perpendiculars and b,e have a point Q in common. cd E P implies IF(cd)
then F(af)
Since a point reflection has exactly one fixed point,
I g 1; we apply lemma 4 twice and get at first
F(be) = {Q},
= {O}.
(Fig. 4), For a translation, induced by a product AB, we see from Lemma 5 that the mapping P - - + P , X ~ midpoint of
x,xA B is
injective; this is needed for
introducing "semi-translations". 5.
We should remember that the group generated by all affine line reflections of the real affine plane contains euclidean, Minkowskian and Ga1i1eian subgroups with the same point reflections.
We may ask if a perpendicularity group (G,S),
with paint set P, contains always Hje1mslev subgroups with the same point set P.
F. Bachmann
34
This is the case.
If for a
E
S, we denote by S((a)) the set of all lines which
are connectible with a by chains of perpendiculars, then «P> a Hjelmslev subgroup of (C,S) and its point set is still P.
U
a
,S(( )) is iff X crAHP'" for some \ E R, i.e. iff Xcr (a+Ak, 1 ,(a,a)+2A(a,k)). This last holds iff x = a + \kEa + in the above lemma as the point at infinity of the line a + ELand a ) . Ii S '*' () S q~
(In the subgroup, any
three concurrent lines have a fourth reflection line.) On the other hand, we can construct perpendicularity groups as extensions of Hjelmslev groups. 1974.
Such a construction theorem has been proved by M. Gerth in
Fig. 5 shows the group plane of Gerth's minimal example of a perpendicu-
larity group which is not a Hjelmslev group (In this example, we have
lsi
= 24,
ipl
= 9)1.
Ici = 216,
The perpendicularity group in question is constructed as a
semidirect product of the euclidean group over CF(3) and a cyclic group of order 3; the group plane contains a subplane which is isomorphic to the euclidean plane
over CF(3).
a
Fig. 5
1 Every line of the group plane is incident with three points. line a and three analogous lines consist of two pieces.
In Fig. 5, the
Orthogonality in reflection geometry
35
B1 BLIOGRAPHY 1.
2. 3. 4. 5. 6. 7.
8. 9.
F. Bachmann, Aufbau der Geometrie aus dem Spiege1ungsbegriff (AGS). 2. Auf1. Springer 1973. - Der H8hensatz in der Geometrie invoZutorischer GruppeneZemente. Can. J. Math. 19, 895-903 (1967). - HjeZmsZevGruppen (HG). Mathematisches Seminar, Universitat Kie1 1970/71, 2. Neudruck 1976. - Lotketten in HjeZmslevgruppen. Geom. Dedicata 4, 139-158 (1975). M. Gerth, FoZgerungen aus dem Axiomensystem des Senkrechtstehens. Dip10marbeit Kie1 1974. M. Gotzky, Uber Bewegungsgruppen affin-orthogonaZer Ebenen. Symposium: Neuere Ergebnisse Uber Projektivitatengruppen, Bad Windsheim Ju1i 1980. J. Hje1ms1ev, Einleitung in die Allgemeine Kongruenzlehre (AKL). Danske Vid. Se1sk., mat.-fys. r·ledd. 8, Nr.ll (1929); 10, Nr.1 (1929); 19, Nr.12 (1942); 22, Nr.6, Nr. 13 (1945); 25, Nr.10 (1949). B. K1otzek, Ebene aquiaffine SpiegeZungsgeometrie. ~lath. Nachr. 55, 89-131 (1973). F. KnUppe1, Homomorphismen metrischer HjeZmslevebenen, welche Geraden, die sich eindeutig schneiden, auf ebensolche abbilden. Eingereicht a1s Habi1itationsschrift Kie1 1980. R. Schnabel, 1981a, Kennzeichnungen eukZidischer Ebenen als affiner Ebenen mit SpiegeZungsoperator. Erscheint in Mitt. Math. Ges. Hamburg. 1981b, Uber die Geometrisierung gewisser Permutationsgruppen. r~anuskri pt. Y Why Tschen, AZgebraisation of pZane absoZute geometry. Am. J. r~ath. 67, 363-388 (1945). O. Veblen and J.W. Young, Projective Geometry. Boston I 1910, II 1918.
BUlow-Str. 16 D 2300 Kie1 F. R. Germany
This Page Intentionally Left Blank
Annals of Discrete Mathematics 18 (1983) 37-54 © North-Holland Publishing Company
37
ON Sm·1E TRANSLATION PLANES ADMITTING A FROBENIUS GROUP OF COLLINEATIONS Claudio Bartolone
In this note we state some results concerning with translation planes of dimension 2 over GF(q), where q
= pro From now on IT will denote such a plane.
Assume that IT has a collineation group F of order q2(q_l) satisfying the following cond it i on: there exis ts a point V E as a Frobenius group on £",-{V}.
9.", such that
F fixes V and acts (fai thfu Zly)
This situation actually occurs: (a) let TIl be a
translation plane over a semi field Q of dimension 2 over its middle nucleus Nand suppose that N - GF(q). a group U of
If a
pe~spectivities
E
Q and a
E
N, a f. 0, then (x,y) .... (ax,ax+y) defines
of order q2(q_1).
Clearly the pair (TI 1,U) satisfies our hypotheses; (b) let TI2 be a LUneburg plane [5] of order q2. IT2 possesses a
collineation group S -- Sz (q). If V E £00' it is well known that Sv has order q2(q_l) and operates as a Frobenius group (with non-commutative Frobenius kernel) 2
on £",-{V}; (c) letIT 3 be a Betten plane (see [1] for definition) of order q , q = 2r (r odd) (a Betten plane of odd characteristic is better known as Walker or Betten-Walker plane [8]).
IT3 has an abelian group L of col1ineations which fixes
the line «0,0,1,0), (0,0,0,1» lines through 0
and is sharply transitive on the remaining q2
= (0,0,0,0). L is determined by the following matrix group 0 a
0
0
0
0
b a 3 a +ab b
a
0
0
0
c
0
0
0
0
0 0 2 c 0 3 0 c
/ a,b
E
GF(q)
0
The matrix group 0 / c
E
GF( q)
";(
C. Bartolone
38
defines a co11ineation group C which fixes (only) the lines through 0 < (1,0,0,0), (0,1,0,0) > and «0,0,1,0), (0,0,0,1) > and normalizes L.
Hence LC is a Frobenius
group of order q2(q_1) acting on ioo as our hypotheses require.
(b) and (c) fill
up the class of proper (i.e. no semifie1d planes) translation planes of characteristic 2 satisfying our hypotheses.
In fact we prove (section 3) the following
theorem 1 which classifies these planes (see [2), [4), [7) for further theorems characterizing the LUneburg planes). THEOREM 1: plane.
~et
TI be of characteristic 2
and suppose that
TI is not a semifieLd
If K denotes "he Frobenius kerneL of F, then one of the following holds: (i) K is commutative and TI is a Betten pZane;
(ii) K is not commutative and TI is a LUneburg pZane. When trying to prove theorem 2 in case of characteristic
F 2, one observes
that an analogous conclusion to (ii) cannot occur: in fact a Frobenius group with even order complement has always a commutative Frobenius kernel.
Thus one could
think that the only proper translation planes of odd characteristic satisfying our hypotheses are the Betten-Walker planes.
Nevertheless this is not the case in
that the following theorem holds (see lemma 3): THEOREM 2: If the chm'acteri:stic of TI is F 2, then TI is a semifield plane. Look at the full group of collineations of a Betten-Walker plane TIw of order q2, q
= pr
= -1
(mod 6) (see (8)).
TI
w has a co11ineation group Gw in the linear
translation complement which fixes a point at infinity and acts transitively on the remaining q2 points.
G has as subgroups two p-groups Sl and S2 of order q w one of which, say Sl' is not a shear group. Gw is the split extension of SlxS 2 by a group N of order q-l, yet Gw is not a Frobenius group. However SlN is a Frobenius group with respect to N. Now G characterizes TI in the sense that we w w are specifing: assume that Hand N are two col1ineation groups of the translation plane n such that: (i) H is a p-group of order q2 and N is a group of order q-l normalizing H; (ii) G = HN fixes a point V E ioo and acts transitively on £oo-{V}; (iii) G is not a Frobenius group, nevertheless H contains a subgroup K such that: K~ G and KN is a Frobenius group with respect to N. prove (section 4):
Under these hypotheses we
On some translation planes admitting a Frobenius group
:rHEORHl 3: K has order q, moreover, if and
1T
1T
39
is not a semifield plane, q;: -1 (mod 6)
is a Betten-Walker plane.
1. BACKGROUND INFORMATION (A). plane
According to J. Andre' we can identify the points of a translation
of dimension two over GF(q) with the vectors of V (GF(q)) (vector space of 4 dimension 4 over GF(q)) and assume as 1ines of 1T through 0 = (0,0,0,0) the compo1T
nents of a spread S (a class of 2-dimensional subspaces which form a partition of the non-zero vectors of V4 (GF(q)). The group C of collineations of 1T fixing 0 is called the translation compZement of 1T: each collineation of C acts on the points of
as a semilinear map of V (GF(q)) which sends components of S in components of 4 S. Vice-versa such a semilinear map, by acting on the vectors of V4 (GF(q)), induces a collineation of C. The subgroup L(C) of collineations of C which are in1T
duced by linear maps is named the linear translation complement. suppose that A maps the point (x.) of
1T
1
on the point
(x~); 1
then the action of A 4
can be represented by a 4x4-matrix (a .. ) over GF(q), where x! lJ
(i = 1, ••• ,4).
Let A E L(C) and
1
= J=.L l
a .. x. lJ
J
Later on we will identify A with the matrix (a .. ). A collineation lJ
cr of L(C) is called a shear of axis each line parallel a semifield plane.
to~.
~
when cr fixes each point of the line
If the group of shears of axis
~
~
and
2
has order q , then
1T
is
Further information about translation planes can be found in
[ 6).
(B). We will use the following symbolism for a group G: oG
= the order of G;
ZiG) a the center of G; NG(U) = the normalizer (in G) of U. Let F be a finite group and U a proper subgroup of F.
F is called a
Frobenius group with respect to U when g-lUg n U = 1 for any g E F - U.
The ele-
ments of F which do not lie in any subgroup conjugate to U, together with 1, form a characteristic Hall subgroup K of F (the Frobenius kernel of F) and F is the split extension of K by U. Moreover oU divides !F:U!-l and if N 4 F, then N ~ K Let g E F - K; the map a ~ ag = g-lag induces a fixed-;oint free
or K ~ N.
automorphism (shortly f.p.f. automorphism) of K.
l Therefore a ~ aga- is a
bijection of K onto K and so K is inclosed in the derived group F' of F.
The
center Z(U) of U is not trivial and if oU is even, then U contains exactly one l involution j: j acts on K sending k E K to k- (whence K is commutative). A transitive permutation group is named a Frobenius group when it is a Frobenius
40
C. Bartolone
group with respect to its stabilizer, provided this is not trivial.
Hore informa-
tion about Frobenius groups and other equivalent definitions can be found in [31. In order to semplify the exposition we fix the following HYPOTHESIS A: Let p be a p-g~oup
tension of a
subg~oup
exists a lvspeet
;;0
Let (oK,oN)
p~ime
H of
G be a
and
Ol'de~ p
2r
by a
g~oup.
g~oup
Assume that G is the split ex-
N of
o~de~ p
r
-1.
K of H n01'mal -in G such that KN is a Frobenius
F~the~mo~e g~oup
there
with
N.
Rbe
the Frobenius kernel of KN.
= 1. As K is normal, K~ K or K ~
is a divisor of oK-l (1.1) K "as
Assume K
=
Rhas the same order of R. In any case R= K.
K, say pS, since The order of N
Ps -1, i.e. r divides s whence
o~del" pr o~ p2r
and it is the Frobenius kemel of KN.
H: if H is of exponent> p, then
(1.2) His cf erponent p2; moreove~ H has exactly pr_l elements of o~de~ p e:w/: of iJhie!) iJelongs to
Z(H) and is a
p-powe~
of some element in H.
PROOF: The Frobenius kernel of HN is H by (1.1), hence H admits a group of f.p.f. automorphisms of order pr_ l • As Z(H) I 1 (H is a p-group), Z(H) contains at least r
p -1 elements of order p.
\ =l
Owing to an hypothesis there exist
A,~
E
H such that
= 1: the previous arguments show that H has pr_l elements satisfying the same conditions of A. Now if s E Z(H) and sP = 1, then ('!1;)p = l when \J E H. Therefore for each p-element A E H which is a p-power of I 1 and :\p
some element in H, there are at least pr elements in H having p-power equal to A. An easy computation allows to deduce that there are in H pr (p r -1) elements of order p2 and pr -1 p-elements.
2. PRELIMINARY LEMMAS Let
TI
be a translation plane of dimension two over GF(p r ) I GF(2) and let S
be the spread which defines of
n
TI.
In this section G will denote a collineation group
in the translation complement satisfying hypothesis A of section 1.
In addi-
On some translation planes admitting a Frobenius group
41
tion we claim that G fixes a component V of S and operates transitively on the .. p2r componen t s. If U is a subgroup of G, we will denote by L(U} and remalnlng E(U}, respectively, the group of collineations of U lying in the linear translation complement and the set of shears contained in U.
The subgroup of col linea-
tions of U which fix the set X of points (lines) of n is indicated by Ux ' First of all we will state the following LH1MA 1: H acts reguZarZy on S-{V} and N fixes a further component
U of
S.
2r r PROOF: Since G operates transitively on S-{V}, oG = oG/p = p -1 where E E S-{V1 E 2r Hence from oH = p it follows H () G = 1 and so H acts regularly on S-{V}. E Moreover G = HG whence GEis a complement of H in G as well as N. As H is an Hall E subgroup of G, from a well known theorem of Zassenhaus (see, for instance, B. Huppert [31, pag. 128, theor. 18.3) it follows that Nand GE are conjugated groups in G. Therefore N = GU for a suitable component U of S. Concerning the admissible orders for K we will prove now the following lemmas 2 and 3. LH1MA 2: Let p = 2.
Then K = H or r(K}
K.
PROOF: Assume that K F H: then we see that oK = 2r by (l.l). of K.
Let
j
be a 2-element
By making use of a well known theorem of Baer (see, for instance. H.
LUneburg [61, pag. 20) and of lemma 1. we infer that j is a shear of axis V. OE(K} ~ 2r. i.e. E(K} order 2r _l. LEMr~A 3: Let p F 2.
= K. because N induces in K a f.p.f. automorphism group of
Then oK
PROOF: Assume that oK
= pr or E(K}
K = H.
F pro then (l.l) entails K = H. The order of N is pr_ l •
i.e. an even number: so N contains exactly one involution j. Moreover if .
Thus
~
E
K.
-1
~J
= ~ • By lemma 1 j fixes (besides V = V n ~oo) U = U E ~oo. Let j fix an other point at infinity WF V. then j E GU n GW' As GWis conjugate to N = G by lemma U
1.
is the only involution also for G ' Therefore, if x E K is the collineation W such that UX = W(see lemma l). it results jX = j. But we have pointed out before j that x = x- l : this means x = x- l • that is x = 1 since K is a p-group and p F 2. j
42
C. Bartolone
Hence U = Wand j is an homology of center V or of axis t
j
= x (resp.
1; E
K, ~-
pj
1
.
other words
~
by Baer's theorem.
to V (resp. point P of V).
i
Let
If
must act on the parallel lines to V (resp. points of V) as
= f;J
2
= P) for each parallel line
V
f;.
In
= 1 on the parallel lines to V (resp. points of V). As 2 does not
divide the order of
1;,
we deduce that each element of K is a perspectivity of
center V (resp. axis V), indeed, a shear because lemma 1 holds. REMARK 1: With a proper choice of the base we may assume that U = {(x,y,O,O)/x,y E
GF(pr)} and V
= {(O,O,x,y)/x,y E
GF(pr)}.
Thus if
y E
L(G),
where A,C.V are 2x2-matrices over GF(p r ) and 0 is the zero matrix: in particular 4
implies also C = 0 by lemma 1. As H is a p-group. if xi
9
j~l
hi / j (i = 1, •.•• 4) are the equations which define a collineation of H (regarded as a
'I E
L(N)
semi1inear map of V4 (GF(pr))). it is not restrictive to assume that h .. lJ i < j and, if 8 = 1, h .. = 1 for i = j.
= 0 for
lJ
The map (a ij ) ~ a (resp. (a ij ) ~ a43 ) defines an homomorphism 21 "'2) between L(H) and GF(pr)(+). Now we prove LEMMA 4: either
'p. 1
~(H)
?f~.
1
= 1.2).
(i
kern~l
(resp.
kern~2)'
(resp.
Mopeover K is contained in L(H) and
induces an homomoy?hism of K onto GF(pr)(+)
PROOF: Let x E axis V).
is the kernel
~1
OP
J:(K) = K.
then x is a perspectivity of center V (resp.
Therefore x E J:(H) by lemma 1. Conversely wE J:(H) implies VW = V and
VW = V because lemma 1 holds: hence
w E kern~ .• 1
By (1.1) K is the Frobenius kernel of KN: hence K is a subgroup of the derived group of KN and consequently K is contained in L(H). p2r by (1.1). Assume oK
= pr
K has order pr or
and oJ:(K) > 1. As N induces in K a f.p.f.
automorphism group of order pr_ l , we infer that oJ:(K) ~ pr and consequently K = Z(K).
Therefore under the hypothesis oK a pr and J:(K) F K, ~. is an iso1
morphism of K onto GF(pr)(+). Let oK = p2r; if J:(K) F K. then p = 2 by lemma 3 r and oJ:(K) = 2 by (1.2) (K = H is of exponent> 2 in view of Baer's theorem and lemma 1). This means that kern~. has order 2r whence ~. is an homomorphism of K 1
1
On some translation planes admitting a Frobenius group
REMARK 2: If L(H)
1 r(H)
(this is the case supposing r(K)
43
F K),
by lemma 4 there 0 exists (h .. ) E L(H) such that h21 101 h43 • Hence, if x~ = .il g .. x. are the 1 J= lJ J lJ equations which define a collineation of G, it is g .. = 0 for i < j because lJ vG = V and L(H) ~ G. L(N) acts completely reducible on the points of TI (since
t oL(N»: hence in case L(H) 1 r(H) we can select a base of V4(GF(pr» (in accordance with remark 1) so that the collineations of L(N) are represented by p
diagonal matrices. Lemmas 3 and 4 show us that in case p 1 2 and E(K) 1 K, E(K) is trivial. Nevertheless E(H) 1 1 in any case: in fact LEMMA 5: E(H) is aZways non-triviaZ.
Furthermore each shear of K is an eZement of
the center of H.
PROOF: The first claim is true if oK
a
p2r (lemma 4).
Let E(K) 1 K 1 H, then p 12
and oK = pr (lemmas 2 and 3): in this case E(K) = 1 as we noted above.
As
Z(H) n K 1 1 because H is a p-group and K 4 H, if we make use again of the pr_l f.p.f. automorphisms induced by N in K, we notice that K is contained in the center of H.
Therefore for each element k E GF(pr), there exists (z .. ) E Z(H) 4 lJ with z12 = k (lemma 4). Let I; E H and x~ = .E l h. j. be the equations that define 1 J= lJ J 1;: then h .. = 0 for i < j (see remark 1). By imposing the identity 1;1; = 1;1; for lJ each 1; E Z(H) we find0 = 1. Therefore L(H) = H and we can consider a collineation (a i ) E L(H) - K. By lemma 4 it is possible choose (k ij ) in K with k21 =-~l and so we find the shear (a .. )(k .. ). lJ lJ Assume E(K) = K: then in any case E(K) is contained in Z(H) (for H is an elementary abelian p-group if E(K) = K = H, while we have shown above that E(K) = K ~ Z(H) in case oK = prj. otherwise, E(K) = 1.
Let E(K) 1 K: then we may assume p = 2 because,
By lemma 2 so K = H and by (1.2) E(K)
~
Z(H) since ExpH> 2
(the 2-elements of H are shears). Evidently r(H)
~
H whence Z(H) contains some non-identical shear (s .. ) by lJ
lemma 5. Assume L(H) 1 E(H), then there exists (h .. ) E L(H) with h21 1 0 1 h43 . lJ -1 by l:~ma 4. Thus (sij)(hij)=(hij)(sij) Ylelds s32 = 0 and s31 = s42h21h43' Hence h21h43 has constant value, say k, when (h ij ) runs in L(H) - E(H).
44
C. Bartolone
REMARK 3: Let {e ,e ,e ,e } be the base used to give the coordinates to TI in l 2 3 4 accordance with remarks 1 and 2. The previous arguments show that, by replacing (if that is the case) e l by k-le • we can represent each collineation (h ij ) of l L(H) in such a way that h21 = h43 • Moreover under the hypothesis L(H) ~ I(H) (for example supposing I(K) ~ K) we can set 5 21 = s32 = s43 = 0 and 5 31 = s42 for each shear (s .. ) of the center of H (for examp 1e for the shears of K (lemma 5)). lJ
There-
31 (resp. (sij) ~ 5 42 ) defines a injective map of I(H) n Z(H) into GF(pr) since, otherwise, there would exist a shear (5 .. ) # 1 such that U(sij)nuF1. fore (sij)
~ 5
lJ
LEMMA 6: If I(K) t- K, then k32 aut,cmol"pi7.isn: of Gf(pr). a~so
=
k~l f02' any (kij)E K, UJhel'e a is an additive
Under the hypotheses L(H) = H, I(H) .;;; Z(H) and oE(H)=pr,
the remaining elements of H satisfy the above condition.
In case K is
alJt:liall, xC< = hx ].Jhe1'6 h is a non-zero element of GF(pr).
PROOF: By (1.1) either K = H or oK = pro
Let K = H, then L(H)
=
H by lemma 4.
# K entails ExpH > 2 and, by applying (1.2), we find I(H) .;;; Z(H) and
L(K)
o~(H)
By lerrrna 4 we can consider (a ij ) and (b ij ) in H such that a = -b • 2l 21 From (c ij ) = (aij)(b ij ) it follows c 2l = c 43 = 0 and c = a 32 +b : hence (c ij ) is 32 32 a shear of H. On the ground of remark 3 we may assume so c32 = 0 whence =
pro
= -b 32 • As a consequence a 21 = a43 ~ a32 defines a map a of GF(pr) into itsaf 32 (see lemma 4). Now by a simple calculation one can realize the additivity of a.
a
Suppose that there exists (h ij ) in H such that h21 t- 0 = h32 = h~l. Then, if (5 ij ) is the shear of H such that 5 42 = -h 42 (see remark 3), we find that the nonidentical collineation (h!.) = (h .. )(5 .. ) of H must fix U: a contradiction (lemma lJ
1).
This proves that If K ~ H,
a
lJ
lJ
is an automorphism of GF(pr)(+).
a
is a map and K is an abelian group because it is isomorphic to
GF(pr)(+) (see lemma 4).
Thus if (a .. ) and (b .. ) are collineations of K, from lJ
lJ
(aij)(b ij ) = (bij)(a ij ) we can infer a~lb21 = a21b~1. 50 for b21 = 1 we find a~l = ha , where h = lao Assume h = 0, then it is easy to check that a ~ a 21 21 31 defines an homomorphism B of GF(pr)(+). If there exists a non-zero element k in
ker~. then the components of 5 U and u(k ij ) must be equal in case (k .. ) is the lJ collineation of K with k21 = k. As this contradicts lemma 1, we have proved that f'
is an automorphism of GF({) under the hypothesis h
consider a shear (s .. ) 1J
k31
~
=
O.
Now by 1ell11la 5 we can
1 of the center of H and the collineation (k .. ) of K with 1J
= -5 31 • As 5 21 = 5 32 = 5 43 = 0 (see remark 3) and k32 = 0, the components of
4S
On some transla tion planes admitting a Frobenius group
Thus (k .. ) should be a shear, but E(K) = 1 in lJ this case (lemma 4): a contradiction. Therefore h must be-a non-zero element of
S
U
and U(sij )(kij) must coincide.
GF(pr) and consequently a is an additive automorphism. a To finish the proof, if K is abelian and K = H, one can prove that x = hx in the same way as in case K F H (clearly h F 0 because we have proved already that a is an automorphism of GF(pr)(+)). REMARK 4: If E(K) F K, h
=
la is a non-zero element of GF(pr). -1
Refer to the base -1
of remark 3: if we replace the vectors e l and e 2 by h el and h e2 , then we note a a that, for any (k ij ) E K, k32 = k21 = k43 where a is an automorphism of GF(pr)(+) such that la = 1. Thus a is the identity map in case K is an abelian group. We shall now denote our attention to the collineations of N.
First of all
we prove: LEl4t4A 7: Under the hypothesis E(K) F KJ eaah aolZineation v of N aan be represented by equations x~
1
= c.x.e(i 1 1
1, ••• ,4; c. F 0). 1
bijeation 1ji of N onto GF(pr);, by setting 1ji(v) =
Moreover one defines a
c c -l 2 l
PROOF: First we point out that oL(N) > 1 because oL(N) ~ oN/oAutGF(pr) = pr-l/r>l: thus there exists a non identical collineation
y
= (c lJ .. ) in L(N) (according to
remark 2 we may set c .. = 0 for i F j). Let (k .. ) E K - E(K) (whence lJ lJ k21 F 0 F k43 ), then (kij) = (kij)Y is again a collineation of K - E(K) (different from (k .. ) because Y induces a f.p.f. automorphism in K). lJ k c c -1 k' k' k -1. 21 22 ll = 21 = 43 = 43 c44 c33 ,l.e.
This yields
(2.1 ) If P F 2 (whence oK = pr by lemma 3), then k21 F k21 by lemma 4. again if p = 2 (and K = H): for if k21
= k21 ,
This is true
then by lemma 4 there should exist a
shear a in K such that (k!.) = (k . . )a. Since a E Z(H) (lemma 5) we have 2 Y2 lJ 2 lJ 2 ((k .. ) )Y = ((k .. ) ) = (k!.) = (k .. )a(k . . )a = (k .. ) • Thus Y should fix the nonlJ lJ lJ 2 lJ lJ lJ -1 identical collineation (k ij ) of K. Therefore from k21 k21c22cll it follows (2.2) If x~
1
4
E d .. xe. are the equations that define the collineation = j=l lJ J
v
of N, we
C. Bart%ne
46
may set d .. = 0 for i < j (see remark 2). N
lJ
U = U and L(N)
i >
j.
~
Besides, by imposing the conditions
N and by making use of (2.2), we notice that d .. = 0 even if lJ
Now if we set (k':.) lJ
=
(k .. ) v, we can prove, as we did above for lJ
y,
that
k21 1 k21 = k~ld22d11 -1. Thus the map v ~ k~ld22d11-l must be injective when k21 is a fixed non-zero element of GF(pr). By setting k21 1, we see that V ~
d d 22 l1
-1
r ,',
is a bijection of N onto GF(p)
r :',
since oN = oGF(p ) .
8: Under the hypothesis E(K) F K, the map Wof lemma 7 sends L(N)onto a sub-
lEt~
group D of GF(pr)*oontaining the multiplioative group of a field
GF(pn) 1 GF(2).
PROOF: In the proof of lemma 7 have pointed out that OL(N) F 1: thus one can verify easily that D = IjJ(L(N)) is a non-trivial subgroup of GF(pr/'. If we set r * =and o(N/L(N)) = m, we see that D =. If m = 1 there is nothing GF(p) to prove: so we may assume m> 1.
The automorphisms of GF(pr) used to represent
the col1ineations of N form a cyclic group A (isomorphic to N/L(N)). As oA = m, e pn the automorphi sm x .... x = x , where mn = r, generates A. let v E N be inherent to
then the coset \JL(tl) generates the factor group N/L(N).
=),
the collineation of N such that 1jJ(\1)
Therefore if
\1
is
= a, there exists a positive integer t such
n 2n (1:-1) n s q l+p +p + ••• +p q that \lE (vL(N)). If IjJ(v) = a , then we find that a = (am) (a ) for a. suitable positive integer s: hence ms+q(l+p n+p 2n +•.. +(t-1)n )=1 (mod pr -1).
t
As m divides pr_ l , we deduce (q,m) n
(a q)l+p +... +p n
k
(m-l)n
= a1+p +•.• +p
-_ (am)
z
= 1. Now from v m E
L(N) it follows
: so m must be a d"lVlsor 0 f 1+p n+••. +p (m-1)n. , l.e.
(m-l) n E
O.
But k generates the multiplicative group of GF(pn),
the fi xed fi e 1d of 0 in GF( pr).
Therefore we must prove only that e does not
generate AutGF(pr) in case p = 2.
let ~
E
Z(N), ~ F 1.
If ~ is inherent to an
automorphism a F 1, a would fix each element of D. Consequently r would have a proper divisor u such that 00 / 2u_1 and we should find the contradiction 2r_1 = = oN = oDo(N/L(N)) .;; (2 u-1)r < (2 u_l)(2 u(r/u-l)+2 u(r/u-2)+ •.. +2 u+l) = 2r _1. Hence a
= 1 and
that
j
1jJ(r,;)
E
D.
Now (W(r.))0 = w(~) 1 1 follows from r
does not generate AutGF(p ).
vI,; = r,;v
and so we note
47
On some translation planes admitting a Frobenius group
3. PROOF OF THEOREM 1 We may assume that F is contained in the translation complement C of
TI
since
F is isomorphic to a subgroup of C satisfying the same conditions of F (for each collineation of
TI
is the product of a collineation of C and of a translation and,
besides, F does not contain translations). section 1 and is transitive on too-{V}.
Clearly F verifies the hypothesis A of
Therefore we can make use of results and
In our case K = Hand E(K) f K because
symbolism of section 2.
is not a semi-
TI
field plane. K is of exponent> 2 since the 2-elements of K are shears in view of Baer's theorem and 1emma 1. Hence each shear (s .. ) of K is a square by (1. 2): (s .. ) = 2 lJ lJ = (k ij ) for a suitable collineation (k ij ) of K. Thus we see that s31 = k21k32 r r and consequently k21 ~ k21k32 is a 1 - 1 map of GF(2 ) onto itself (for OE(K) = 2 by (1.2) and (s .. ) lJ
~
s .. is an injective map as we have pointed out at the end of lJ
remark 3). Therefore the map (3.1)
a
of lemma 6 satisfies the condition
x ~ xx a is a 1 - 1 mapofGF(2 r ) onto itself
Let (k ij ) and (kij) be two co11ineations of K such that k21 = k21 f O. On the ground of remark 3and lemma 6we have also the identities k43=k21=k21 =k'43 and k32 = k32 • The co11ineation (kij) = (kij)(kij) is a shear (because k21 = 0) of the center of K (see lemma 5): according to remark 3 so k31 = k42 and consequently k31 + k31
k42 + k42 • Hence k21
k31 = k42 + k~l for such that 08 = O. (3.2)
any
~
(k ij )
k31 + k42 is a well defined map 8 and we find E
r K, where 8 is a map of GF(2 ) into itself
Let (k ij ) and (kij) be two co11ineations of K such that k21 = k21 and k42 = k42 (so k31 = k31 , k32 = k32 , k43 = k43 )· As (k ij ) and (kij) act in the same way on 23r points of TI, we infer (k .. ) = (k! .). This means that each lJ
lJ
co11ineation (k .. ) of K is different from the remaining col1ineations owing to the lJ
value of the coefficients k21 and k42 • Therefore~: (k 21 ,k42 ) bijection of GF(2 r )xGF(2 r ) onto K and it is easy to check that (3.3)
~ (x,y)q>(x'
,y' )
cp( x+x' ,y+y' +xx' a).
~
(k ij ) is a
48
C. Bartolone
The coefficient k41 of (k ij ) E K must be necessarily a function of k21 and k42 : k41 = f(k ,k 42 )· Thus from ~(x,y)~(O,y) = ~(x,O) it follows f(x,O) 21 = f(x,y)+xy+f(O,y) and, by setting f(x,O) = xY and f(O,y) = yO, we find 6 f(x,y) = xi +xy+y.
By imposing the identity
~(x,O)~(y,O)
=
~(x+y,xy
eL
) now we
deduce Y
, tfle
0
k41 = k21k42+k21+k42 for any (k. j ) E K, where Y and ':1 l ' en;:;z. t<' X y +xy i3 +y Y = ( x+y ) Y + ( xy eL)! + ( x+y ) xy CI •
(3.4)
Consider a collineation v of N.
° are maps
satisfying
If v is inherent to the automorphism 0 of
r
GF(2 ) and ,;,(-,) = c (see lemma 7), by setting e:(c) = 0 and £(0) = lone defines a r r map ( of GF(2 ) into AutGF(2 ) such that ((x) = 1 when xED u {a} (lemma 8). Let G
xi = c i Xi be the equa ti ons whi ch define v accordi ng to 1el1llla 7. Then E( c) = G , -1 where we set c = c2c l ~(v). Consider the collineation of K ~(a,O), then from ({(a,O))v E K one deduces the identities (3.5)
c 4 = c 3c;
(3.7)
c a
BE(C)
3
(3.6)
c aCl £(C)= (a£(C)c)Cl c2 ; 3
(3.8)
c aY£(C)= (a£(C)c)Y cl . 4 eL
Thus from (3.6) (by a = 1) it follows c = c c (lCl = 1 in accordance with remark 3 2 4) and consequently (3.5) gives c = CC Cl C . By substituting these values in 4 2 (3.7) and (3.8) we find respectively (a((C)c)B = as£(c)ccCl and (a£(c)c)Y = = ay£(c)c2c~. Therefore for any x, y in GF(2 r ): (3.9)
. E(X))B = xx a Y8E(X) ; \xy
2
(l
x xy
yE(x)
.
Suppose that K is commutative: on the ground of remark 4 we may assume that :1 is the identity map. From (3.9) (by Y = 1) it follows xB = lk dx ) for any r r x E GF(2 ), where k = lB. r·1oreover if x,y E GF(2 ), xY+xl+yY = (x+y)Y + (xy)o + + (x+y)xy by (3.4) and, by exchanging x with y, yY+yxB+xY = (x+y)Y + (xy)o + S S + (x+y)xy: hence xyS = yx and x = kx by Y = 1. In particular for x=z E D (see lemma 8), z ; 1, we note that k = 0, i.e. xS = 0 for any x E GF(2 r ). Furthermore 2 if we set x = y in (3.4), we deduce (x )o = 0 whence Xo = 0 for any x E GF(2 r ) because we are in even characteristic.
Now set y = 1 in (3.10): so we obtain
49
On some translation planes admitting a Frobenius group
E x = x3h (x) where h = lY.
From (3.4) it follows then
x3hE(x) +y 3hE(x) = ( x+y )3 hE(X+Y) + x2y+xy 2. r By lemma 8 there exists z E GF(2 ), z
F a.l.
such that E(Z)
E(z+l) = 1. There-
fore setting x = z and y = 1 in the above identity one infers h = 1 and conser quently x = x3 for any x E GF(2 ). By making use of lemma 1. we remark that each component W of the spread S different from V is the image of U under a collineation of K. Thus for suitable a. bE GF(2 r ), we find that W = u~(a,b) = 2
3
r
= {(x,y,(a +b)x+ay,a x+by) / x,y E GF(2 )} and we see that
Tf
belongs to the class
of translation planes defined by D. Betten in [11. map.
Now assume that K is not commutative: so by (3.3) ~ cannot be the identical r If x E GF(2 ), from ~(x.a)~(l,a) = ~(x+l,x) it follows
But on the other hand by (3.9) we have (3.12)
x13 = xx ~l i3E(X)
Now (3.11) and (3.12) together yield (3.13) By lemma 8 we can choose z in GF(2 r ) such that z
F a,l and E(Z) = E(z+l) = 1.
Since~ is additive and we suppose l~ = 1, from (3.13) it follows z+zl i3+z
zY+zlS+1Y = (z+l)Y+zo+(z+l)z and lY+zi3+zY = (l+z)Y+zo+(l+z)z whence z13 = zl13' view of (3.12) so it must be 113 = 0 and consequently (3.13) would contradict
In
~ F 1. Therefore 113 = 1 and x13 = xx~ for any x E GF(2 r ). Now (3.9) gives (xyE(X))~ = X~y~E(X) and we note that ~ induces an automorphism ~ in the field GF(2 n) of lemma 8. Moreover from the identity (3.4) it follows (by x = y) x2x~ = (xx~)o. By (3.1) the map x ~ xx~ induces a multiplicative automorphism 0 of n GF(2 ). Thus we observe that ZO = zzo-l when z E GF(2 n). So the restriction of n n over GF(2 ) is an automorphism 8 of GF(2 ) in that ~(a,a)~(a,b) = ~(a,a+b) implies
°
(a+b)o = aO+bo.
Apply again the identity (3.4), then zY+z+1Y = (z+l)Y+zo+(z+l)z
50
C. Bartolone
and lY+zzn+ zy
(l+z}y+zno+(l+z}zn whence z6+ z2 = zno+ zn. As AutGF(2 n ) is a free part of Hom(GF(2 n )}, the last identity entails that the automorphisms 6,
;6, ;
2, -6
are equal two by two: excluding the possibilities
= 2, no = n- because they are contradictory, we see that -0
from (3.10) it follows zy whence, by bearing z+l
aii z = (z) 2
~
E
6=
no, 2
= ;
and
= nand n-2 a 2.
Now
= z2z;lY.
By (3.4) we have also zY+1Y = (z+1)Y+ z;+z2 n GF(2 ) in mind, (Z;+z2}lY+z;+z2 = O. Thus lY = 1 (since
-z ) and, by (3.10), xY = x2C1 x for each x E GF(2 r ).
Now we can rewrite the identity (3.4) obtaining (xyCl)O
=
xny2.
By setting
(successively) y = 1, x = 1 and yn = u one verifies at once that n 0 is an automorphism of GF(2 r ) suc~ that (xn)n = x2• At this point we can determine in the same way as in case K commutative: W = uQl(a,b) = { (x,y,bx+any, (a a +bCl)x+by) / x,y E GF(2 r ) j(see lemma 6, (3.2) and (3.4)) and 11 is a LUneburg
WE S-{V}
2 n
plane. (In order to compare S with the spread which LUneburg determines in [5], it is enough to set a = sand b+aaCl = t).
4. PROOF OF THEORHl 3 By lemma 1 no p-element of G is a translation: hence we may suppose that G is contained in the translation complement of 11. Thus we can make use of symbolism and results of section 2. (4.1)
oK
In our case
q and K ~ Z(H)
since K ~ Hand K possesses q-l f.p.f. automorphisms. (4.2)
Now we prove
E(K) t- K.
Assume E(K) = K and let p t- 2: then N contains exactly one involution j. By Baer's theorem we see that j must fix pointwise a subplane of 11 of order q.
For
we can exclude that j is an homology, otherwise in view of lemma 1 and applying a well-known theorem of Andre (see, for instance, [6], pag. 16 lemma 3.17),11 would be a semifield plane.
If j
~
L(N), then q must be a square and, by utiliz-
ing a suitable base of V4(GF(q)) (see remark 1), j can be defined by equations x~ = x~, where Ql is the involutorial automorphism of GF(q). Consider 1 1 . -1 Ql (S;j) E E(K) = K; then (Sij)J = (sij) implies Sij = -Sij for i > j, a contradic-
On some translation planes admitting a Frobenius group
tion.
51
Thus j E L(N) and we may identify j with the diagonal matrix 0
0
0
-e
0
0
0
e'
0
0
0
e
o o o where £2
=
e,2
=
1.
follows s32 = s4l Consequently, for any k
(4.3)
In case e
= 0
E
=
(resp. s3l
-E
. e'), from (sij) j = (sij) -1 lt 0) for each co 11 i nea t ion (s .. ) E 1: ( K) lJ
-e' (resp. e s42
a
GF(q) there exist
=
(k~.) 1J
By imposing K ~ Z(H), we infer so H = L(H).
=
kit
and (k'!.) in K such that k3'l
42
lJ
K.
k
Thus 1:(H) F L(H) and, on the ground
of remarks 2 and 3, we deduce that only the case
£
= -£' can hold. Furthermore,
if (h .. ) E H - l:(H) we can set h.. = 0 for i < j and h21 = h43 F O. Let lJ '-1 lJ (gij) = (hij~ , then g2l = g43 F 0 and g32 = O. By (4.3) there exists (kij)
-g42 : so (9 ij )(kij) must fix the component U of S and this is a contradiction to lemma 1. E
K = r(K) such that k42
=
= 2 and consider an element ~ E H of order 4. If ~ ~ L(H), q must be a square and we can represent ~ by equati ons xl = xf, x = xf, 2 x' = axvG + bx~ + x~ x' = cx~ + dx~ + xv'ci (the 2-elements of H are shears of 3 1 2 3'4 1 2 4 Let p
axis V by Baer's theorem and lemma 1). As; must centralize K, we find that k.. E GF(vG) for any (k .. ) E K: so (k .. ) -+ (k ,k ) defines a 1 - 1 map of K onto 3l 32 lJ lJ lJ 2 GF(v'q)xGF(v'q) and consequently ~ E K. Suppose ~ E L(H); then L(H) F r(H) and in view of remark 3 and (4.1), we can infer again ~2
~2 v
E
E
K for any element ~
E
E
K.
Therefore in any case
H of order 4. This implies now ~v-l ~ K for any
N-{l } since, otherwise, v would fix the non-trivial element ~2 of K.
Hence H
consists of (q-l)q elements of order 4 and of q 2-elements (the elements of K). Furthermore each non-trivial element of N induces a f.p.f. automorphism in H whence G would be a Frobenius group with respect to N (see [31, pag. 497). Therefore we have r(K)
F K.
From lemma 2 it follows now p F 2 and by lemma 6 we may assume k2l k43 for each collineation (k ij )
E
= k32 =
K (see remarks 3 and 4: K is abelian because
C. Bartolone
52
i3
K ~ GF(q)(+) by lemma 4). suitable maps B,
y,
8
y
Moreover we can set k31 = k21 , k41 = k21 , k42 = k21 for 6 of K into K. By imposing that the product of two elements
of K is again in K, one gets the identities (x+y) 8 =x IS +y IS +xy; N and let x~1 = c.l. 1 1 be its equations (see lemma By lemma 4 there exists (k ij ) E K with k21 = 1. Consequently v E NG(K) Now consider an element v
E
7). implies (4.5)
-1 S i3 -1 2 -1 6 6 -1 2 -1 y y -1 3 . and (c 2c1 ) (c c, ) , (c 2c l ) (c c, ) , (c 2c, ) (c 2c1 ) • Thus 1n 2 2 view of lemma 7 and by applying (4.4), we infer that p # 3, xi3 = XO ={1/2)x 2 and x'l =(1/6)x3 for any x E GF(q).
=,
=,
=,
By lemma 5 [(H) # 1 whence Z(H) n Z(H) # 1 because r(H)
~
H.
Let
(s .. ) E t(H) n Z(H), (s .. ) # 1; then according to remark 3, we may assume 1J 1J 5 = 5 = 5 = 0 and s31 = 5 42 , By lemma 8 it is possible to select Al = (cij) 21 32 43 and '2 = (c in L(N) in such a way that Cl~lc22 = -cl~lc22' If we set (t~.) = (5 .. r"1 and (t':.) = (s .. /\2, by making use of (4.5), it is easy to verify 1J 1J 1J 1J h t t' (' ,-1)2 _ • -')2 -_ t H • Since (t~.) and (t':.) are elet a 31 - s31 c c - 5 31 (c• 22 11 31 22 c11 1J 1J ments of r(H) E Z(H) as (so .), we note so that (t~.) and (t':.) must have necessar-
1j )
1J
i1y the same coefficients.
1J
On the other hand t41 =
= -S4l(C22Cl~1)3 = -t 4l : hence s41 = 0 for any (sij)
1J
s41{C22Cl~1)3 =
L{H) n Z(H). If (s .. ) denotes again a non-trivial shear of Z(H) and (k .. ) E K, then E
1J
(5 .. )(k .. ) 1J lJ
1]
# 1 by (4.2), (4.1) and lemma 4. 2
Therefore un u(s;} (kU) = 1 by
lemma 1, whence 5 31 #-(1/12)k~1' From lerrrna 4 it follows now that -1/3 must be a non-square (hence q = -1 (mod 6»). Thus GF(p) contains both squares and nonsquares (in GF(q»).
Consequently there exist (u .. ), (v .. ) 1J
1J
E
«~so
.»
such that
1J
u31 is a square and v is a non-square. 31 Now if x~1 = c.x~ are the equations defining the collineation v of N, by 1 1 setting (u .. )') 1J
u31 = (x c 2c l
=
-1 2
(u~ .), (v .. )v = (v~ .), u31 1J 1J 1J 0
2
) and v31 =-(l/3)(Y czCl)'
=
land v = -0/3)y2, we find 31
Therefore o(r(H) n Z(H))
~
q by lemma
On some translation planes admitting a Frobenius group
53
7. Thus we note that O(KI(H}} ~ q2 since o(~(K»=l by (4.2) and lemma 4: hence H = ~(H)xK. By applying lemma 1, we see at last that, if WE S, W F V, W = {(x,y,(a-(1/2)b 2)x+by,-(1/3)b 3x+(a+(1/2)b 2)y) / x,y E GF(q)} for suitable a,b E GF(q), i.e. IT is a Betten-Walker plane. CONCLUSIVE NOTE: Let Suppose that
IT
IT
be a translation plane of dimension 2 over GF(pr) F GF(2).
admits two collineation groups Hand N such that: (i) H is a p-
group of order p2r and N is a group of order pr_l normalizing H; (ii) G ~ HN fixes a point at infinity and operates transitively on the remaining p2r points; (iii) H contains a subgroup K, K ~ G, such that F = KN is a Frobenius group with respect to N. Theorems 1,2 and 3 show that only the following cases can occur: (j) semifield plane; (ii) F
=
G, K is
F 1:1 G, K is commutative and oK
= pr and
IT
IT
not commutative and
IT
IT
is a
is a LUneburg plane; (iii)
is a Betten plane of even characteristic; (iv)
is a Betten-Walker plane.
ACKNOWLEDGH4ENT: Work supported in part by the research group G.N.S.A.G.A. of C.N.R. BIBLIOGRAPHY 1. 2. 3. 4. 5. 6. 7. 8.
D. Betten, 4-dimensionale Translationsebenen mit 8-dimensionaler Kollineationsgruppe, Geom. Ded., 2 (1973), 327-339. T. Czerwinski, Finite translation planes with collineation groups doubly transitive on the points at infinity, J. Algebra, 22 (1973), 428-411. B. Huppert, Endliehe Gruppen I. Springer-Verlag (1967). R.A. Libler, A characterization of the LUneburg planes, Math. Z., 126 (1972), 82-90. H. LUneburg, tiber projektive [benen, in denen jede Fahne von einer nichttrivialen Elation invariant gelassen wird, Abh. Math. Sem. Univ. Hamburg, 29 (1965), 37-76. H. LUneburg, Translation planes. Springer-Verlag (1979). R.H. Schulz, tiber Translationsebenen mit Kollineationsgruppen, die die Punkte der ausgezeichnete Geraden zweifach transitiv permutiren, Math. Z., 122 (1971), 246-266. M. Walker, On translation planes and their eollineation groups. Ph. D. Thesis (Univ. of London: March, 1973).
Istituto Matematico Universita di Palermo Via Archirafi, 34 90123 Palermo Italy
This Page Intentionally Left Blank
Annals of Discrete Mathematics 18 (1983) 55-60 © North-Holland Publishing Company
55
HALL TRIPLE SYSTEMS AND RELATED TOPICS Lucien Beneteau
ABSTRACT A Hall triple system (HTS) is a Steiner triple system in which every point is the unique fixed point of an automorphism. class of the corresponding Moufang loop.
Let k be the central nilpotency
In case k
a
2 (resp. 3, resp.
~
4) we
determine the possible values of the order.
Besides we prove that there exist 6 only two non isomorphic 5-dimensional HTSs of order 3 • Several open questions are stated.
1. INTRODUCTION The Hall triple systems (HTSs) are the Steiner triple systems in which any three non-collinear points generate an affine plane -or, equivalently, in which any symmetry is an automorphism (recall that a symmetry 0a with fixed point a is defined by 0a(x) = y whenever {x,a,y} is a line).
Such a HTS will be said to be
abelian in case it arises from an affine space over IF3 = GF(3).
Any HTS E has s for order a power of 3, say IEI = 3 where sis the size of E (see [8)). r·1oreover
any two minimal generator sets of such a system have the same cardinal number n+l (see [2)); we call n the dimension of E. The smallest non-abelian example of HTS contains 81 = 34 elements (see Marshall Hall Jr, [9J). It is to be denoted further by L , for it is the only non-abelian 3-dimensional HTS. 3 An exponent 3 commutative Moufang loop (or 3-CM loop) is a set E provided with a commutative binary law x.y which admits a unit (say 1), and which satisfies the two following identities: 2
2
x .(x.y) = y, and x .(y.z) = (x.y).(x.z) 3 Observe that the first identity, quoted for y=l, implies that x
= x2.x
= 1, which
is a justification of the fact that such a loop is said to have exponent 3.
L. Beneteau
S6
Loosely speaking the 3-CM loops are to the HTSs what the elementary abelian 3 groups are to the affine spaces over
~3
= GF(3). More precisely, by defining
the "lines" of a given 3-Cf4 loop as the 3-subsets of the form {X,y,(x.y)2} where x
~
y, we get a HTS.
Any HTS arises this way, and the so-defined correspondence
between 3-CM loops and HTSs is one-to-one up to isomorphism. First of all we recall a few known results about 3-CM loops.
Then relations
will be stated between the central nilpotence class k of such a loop, the size s and the dimension n.
We shall be mainly concerned by the possible values of one
of the three parameters k, sand n when the two others are given.
Besides we
determine in several speCial cases the number v( s,n, k) of pair-wise non-isomorphic 3-CM loops corresponding to given values of s,n and k. As a by-product, a complete list of the HTSs whose size is at most 6 may now be given (this list had been presented by the author in the Cogres franco-canadien de Combinatoire, 1979 141) •
We refer the reader to the literature for the classical kindship between HTSs, symmetric distributive quasi groups and centerless Fischer group. briefly that a group G is a
Fische~ g~oup
Let us say
if and only if the subset S of its
order 2 elements generates G and satisfies (Xy)3
= 1 for any x and y in S. The
group G = < S > generated by the set S of the symmetries of a HTS E is a Fischer group with Z(G)
= {l l. Here again we have a one-to-one correspondence E ~ G
between HTSs and centerless Fischer groups (see [1 I).
It is also worth mentioning
the work by Manin [91 about some unirational cubic hypersurfaces.
It is an
attempt to generalize the classical construction of an abelian group in the set of non-singular points of a cubic projective curve.
Manin considered suitable
quotients of the set of points of some cubic hypersurfaces, and he introduced structures which turn out to be direct products of 3-CM loops by elementary abelian 2-groups.
2. RELATION BETWEEN SIZE, DIMENSION AND CENTRAL NILPOTENCY CLASS Consider a finite 3-CM loop E. Recall that lEI = 3s where s is the size of E.
Define the "associator" of any three elements x,y,z by : (x,y,z) -1
=
= (x.(y.z)) .((x.y).z). The set of products of associators is called the sl
de~ived
In we have IE/D(Ell = 3n, then any minimal generator set of the
loop E involves n elements - That means (2) that any minimal generator set of the
57
Hall triple systems and related topics
corresponding HTS involves n+l elements. of E.
So n is the previously defined dimension
Let us turn now to the definition of the upper central series (Zk(E))k;;'l'
The associative center Z(E)
= Zl(E} is the set of the elements z whose behavior is,
say, "associative" (in the sense that (x,y,z) = 1 for any x and y).
The following
terms of the series are defined recursively by: Zk+l(E)
= {zl V x,y
E
E, (x,y,z)
E
Zk(E}}.
The well-known theorem of Bruck-Slaby may be expressed by: Z l(E} = E. In other nterms, there exist integers k such that Zk(E} = E, and the smallest one (the "central nilpotency class") is linked to the dimension by the inequality: k .. sup(n-l,l}. Recently this bound has been proved to be the best one (see [3] [4) and their bibliographies}, but this question is to be regarded as outside the scope of the present report.
Clearly in the abelian case, i.e. when k
= 1,
we have s
= n.
The main statement of this paper is the: THEOREM 1: If k = 2, then 3 .. n and n+l .. s .. n+(~); particularly 4 .. s. n n-l If k = 3, then 4 .. n and n+4 .. s .. n+(3}+4( 5 }; particularly 8 .. S. If k ;;. 4, then 5 .. n and n+17 .. s; particularly 22 .. s.
PROOF: The fact that k+l .. n is merely a restatement of the Bruck-Slaby theorem. Let d = s-n be the size of D(E}. 1 .. d .. (~) if k
=
It follows from a classical work by Bruck that
Besides we had proved that 4 .. d .. (~) + 4(n;1} when k
2.
(for the lower bound, see [1); for the upper bound, see [4)).
=
3
Lastly a student of
the University of Paris, Elkhouri, established that k;;. 4 implies d;;. 17 (the proof is to be published in his thesis). NOTE: In each case, all the values of
S
allowed by the given inequalities are
actually reached by at least one HTS (see [1) pp.106-107).
Besides for any n ;;. 3
(resp. 4) there exists exactly one n-dimensional HTS whose class is 2 (resp. 3)
n
n
n+l
and whose size is n+(3) (resp. n+(3)+4( 5 }).
But when k
=4
(resp. 5) the upper
bound of s remains unknown.
The table 1 represents the possible values of the class the size and the dimension for s .. 16 and n .. 12.
with respect to
The number of stars indicates
L. Beneteau
58
the number of the corresponding systems whenever this number is known. ~';
:~
For
instance in the cell corresponding to s = 8 and n = 4, we have 2 - 3,';, which means that
thel"0 al"e onZy one 3-CM Zoop with
k=2
and exactZy thl"ee 3-CM Zoops
size
~
23
djme:'i'"~L :4'
Ii,
('
4
567
l_'~1 -2~:~,.-1If- "-'- _-'- -"'I"'-
LL 1"'11....:1
.
5
9
8
I
~{'
I1,',
1 2:
1{
1e':
2
10
11
12
13
14
15
16
_ _ _ _-:-
1 2"'-3;1 .. 3 ---,-3_..0..3---'3_'·'..1-1_ _ _- . -
2'~ -31~
2
2
2-3
2-3
2-3
2-3
2-3
2-3
2
2
- ,
2
2-3
2-3
2-3
2-3
2-3 2-3 2-3
2
2
2-3
2-3
2-3 2-3 2-3
1~"
2
~.(
6
1
It
-2-~
~~:
1
7
8
!
2
1 ~
2
.';
9
-
10 11
~
2-3
2
212-3
('
2
2
2 12-3
~
2
2
~'.' 2... 2 -2-12-3
12
2 12-3
Table 1 Observe that k = 2
whenevel'
the 3-CM loops of class 2.
n < s < n+4.
Let E=F
Now it is quite easy to describe
D be a vector space over F , direct sum of 3 two subspaces F and D with respective dimensions n and d~(~). Assume that e
B = (e l ,e , ••. e } is a basis of F. Let us choose in D a generator set of (~) ele2 n ments, say {e"k!l';;;i<j
the form X = l~t~n xie i + l';;;i<J
X.V = t(xi+Yi)e i
+
i<3
turns E into a n-dimensional 3-CM loop of class 2 whose size is n+d.
Any n-di-
mensional 3-CM loop of class 2 arises this way: this derives from a classical statement by Bruck [6). ments is (X,V,Z) =
~ ~ijk
Besides in such a loop the associator of any three elee ijk where:
Hall triple systems and related topics
ll;jk
59
x. Yi Z. 1 1 det x. Yj z. J J xk Yk Zk
Therefore (X,Y,Z) is shew-symmetric and trilinear in the sense that for any a and bE lF3 = {-l,O,]}, a b (X l ·X 2,Y,Z) = (aX +bX 2,Y,Z) = a(Xl,Y,Z) + b(X 2,Y,Z). l
Notice that (e.,e.,e k) a e .. k and D(E) = D. 1 J lJ Let us turn now to the study of the case d=s-n=l.
If n=3, then E ~ L • In 3 the general case, since d F 0, at least one of the e .. k's does not vanish; assume lJ
= A F 1. One knows that
=<e ,e ,e > is then isomorphic to l 2 3 123 L • For any X in E, since D(E) == {-A, 0, + A}, there exist A, 11, v in lF3 3 such that: for instance e
'IT
(e"e 2 ,X) = AA,(e"e 3,X) = IlA and (e 2,e 3,X) = vA. By trilinearity the element u = vel -
lle
2
+ Ae 3 satisfies also:
(e l ,e 2 ,u) = AA, (e ,e 3,u) = IlA and (e ,e ,u) = vA. 2 3 l Hence X' = X-u is orthogonal to t' in
'IT.
So there is no
~oss
'IT,
of
in the sense that (t,t',X')
genera~ity
0 for any t and
in assuming that e , eS, ••• e are
n 4 In case n = 4 one gets e E Z(E) so that E ~ IL3 x F • When 4 3 n a S, consider the kernel K of the homomorphism x -~ (x,e 4,e ) from 'IT into D. S If K = 'IT, then e and e are central elements and necessarily E ';; IL3 x If 4 S
orthogona~
to
'IT.
IF;.
K F 'IT, then K is a maximal subloop of
and we can choose e 2 and e3 in K. By changing eventually e in -e 4 , one may assume then that 4 (e l , e4 , eS) = A = (e l , e 2, e3 ). We are thus led to R ';; liS where liS is the 3-CM ~oop on five generators e"e ,e ,e ,e subjeat to (e"e ,e ) = (e"e ,e ) and. for 2 3 4 S 2 3 4 S 'IT
any i < j < k suah that {l,4,S} F {i,j,k} 'f {l,2,3}, (e.,e.,e ) = O.
A direct k study of liS shows that K(ITS) = 2 and IITsl = 3. Besides Z(IT S) = D(II S)' which means that lIS may not be written as a direct product of smaller 3-CM loops (lIS is 6
"irreduaible").
1
J
We have thereby proved the:
THEOREM 2: There exist
exaat~y
two non isomorphia
5-dimensiona~
HTSs of size 6.
This completes the proof of the fact that there exaatly four pair-wise non-
60
L. Beneteau
isomorphic
HTS~
of size 6, since the unicity in the 4-dimensional case was
established in 11 J.
The next step would be the classification of the 6-dimen-
sional HTSs of size 7, which is to be dealt with in the same way, though the list of the corresponding systems is doubtless much longer.
BIBLIOGRAPHY 1. 2. 3. 4. 5. 6. 7. B.
9. 10.
L. Beneteau, Contribution d l'etude des Boucles de Moufang commutatives et des espaces apparentes, (Algebre, Combinatoire, Geometrie), These d'Etat, Univ. Marseille, 1981. L. Beneteau, Topics about Moufang Loops and Hall triple Systems, "Simon Stevin", vol. 54, n02, pp. 107-124, Avril 1980. L. Beneteau, Problemes de majorations dans les quasigroupes distributifs et Zes gl'oupes de Fischer, Actes ColI oque "A 1gebre app 1i quee et combinatoire", pp. 22-34, Univ. Sc. et Med. de Grenoble, Juin 1978. L. Beneteau, Free commutative Moufang loops and anticommutative graded rings, Journal of Algebra, Vol. 67, nOl, pp. 1-37, November 1980. L. Beneteau, Une classe particuliere de matrotdes parfaits, Proc. Colloque franco-canadien de Combinatoire, Juin 1979; Annals of discrete Math. 8, pp. 229-232, 1980. R.H. Bruck, Contribution to the theory of loops, Trans. Amer. Math. Soc. 60, pp. 245-354, 1946; t1R 8 # 134. M. Deza, Finite commutative Moufang loops, related matroids, and association schemes, to appear in Proc. Conference on Combinatorics, 1979, Arcata, California, Humboldt State Univ., Utilitus t·1ath. J. I. Hall, Note on the orders of Hall triple systems, to appear in Journ. Comb. Th. Ser. A. Marshall Hall Jr, Autcmorphisms of Steiner triple systems, 1Bt4 J. Res. Develop., pp. 460-472, 1960; MR 23 A = 1282. Yu. 1. Mani n, Cubic forms; North Ho llandp. C., Amsterdam, 1974.
Universite Paul Sabatier, Math., 118, route de Narbonne 31077 Toulouse Cedex, France
Annals of Discrete Mathematics 18 (1983) 61-76 © North-Holland Publishing Company
61
ON MAPPINGS PRESERVING A SINGLE LORENTZ-MINKOWSKI-DISTANCE. I Walter Benz
Consider a commutative field K and denote by K2 the set of all ordered pairs of elements of K.
The elements of K2 are called points of the plane K2.
The
distance PQ of the points P 1:1 (PI' P2)' Q = (ql ,q2) is defined by PQ:=
(ql-pt}(q.-P.).
The following result was proved by F. Rado, [3]:
THEOREM: Asswne char K + 2,3. "P,Q
E
K2
Given a bijection a
K2 ... K2 such that
PO = 1 if and only if pOQo = 1.
Then a must be semi linear up to a translation (considering
~
as a vector space
over K).
In case K 1:1 JR we have proved in
[1]
the sharper result
THEOREM: GiVen a mapping a of the plane JR2 into itself such that
Then a must be semi linear up to a translation and hence a Lorentz transformation.
The main question we are interested in this series of three papers I, II, I II is
PROBLEM A: Determine all mappings a of K2 into itself such that
62
W. Benz
We shall present the solution of this problem for infinitely many fields K, especially for infinitely many Galois fields by applying a result of G. Tallini, An important step in our considerations is the following generalization (for
141.
char K 4 5,7) of Rado's theorem: THEOREM 1:
Ae2wne
"P,Q ~"h;!n
E
chell" K + 2,3,5,7.
Given an injective mapping
0
K'
-->
K'
such
K'
a "lust D'] l3emiZiiwar (with respect to a monomorphism of K and not necessari-
Zy an
a,<~omorphism;
:..p to a translation.
The proof of this theorem will be given in part I of our series.
We like to
finiSh this introduction with a remark about the connection of the distance PQ with the Lorentz-Minkowski-distance d(P,Q) := (q, -P,)' - (q, -p,)' The
question we are actually interested in is
PROBLEM B: Given a cOMmUtative field K of characteristic element; k
E
K'" : = K \ {O}.
Determine all mappings
+ 2 and
given a fixed
K' .... K' such that
T :
We like to verify that problem B is solved for all fields K for which problem A can be solved.
Given now P,Q for a
T
E
Define for this purpose
K' such that PQ
= 1. Then
d(P~,Qu)
k and hence d(p~T,Q~T)
of problem B.
A consequence of the last equation is pOQo
1 for
mapping according to problem A and hence T
u·'
0
0
u.
•-
.• I
UTU
•
Thus
0
is a
k
On mappings preserving a single L-M-distance
63
1.
All occuring fields are assumed to be commutative.
Consider now a field K
of arbitrary characteristic. LEMMA 1: Assume IKI > 4.
Then every element of K2 aan be written in the form
1
1
(XI + x2 + ... + xn ' X + - + ... + - ) I x2 Xn with finitely many elements XI ,X2 , ... ,xn in K*.
(Aatually n aan be ahoosen ~ 4.)
PROOF: We notice that K2 is an abelian group with the addition (XI ,X2 )+(YI 'Y2) .- (XI + X2 'YI + y,). Observe (0,0) = (1 + (-1), + +_+). - There exists an element; that ~2+~+ 1 x, = For a
t(~
+0
because otherwise IKI ~ 4.
+ 1), X3 =
+0
Put
+= ~2+~+
1 and XI 1 1
1
-t~.
+ 0,-1
in K such
t(;2+~),
Hence (1,0) = (XI + x, + x3 ' - + - + - ). Xl x2 x3
we have 1
1
1
(a,O) = (axl + ax, + ax 3 , - + -aX) . aX+ I ax, 3 In case b
+0
we observe 1
1
(a,b) = (a - b' 0) + (b' b). 1 For the representation of (a,b) in form (XI +... + X , - + n XI at most four elements XI ,x, ,X3 ,x•.
1
+ -) we hence need xn
REMARK: In case IKI < 4 there does not exist a representation as described in lemma 1.
This will playa role later on (s. part III).
Denote by IH the set of all points (x, 1), X E K"'. X
Hence, by lemma 1
(IKI ;;'4), IH generates K2, i.e. K2 =
~ {2,3,5,7}.
GiVen points P
+ Q of
K2.
Then PQ
=0
if and
only if there exist points ft., B suah that PA = F, PB = 14', QA = 42 , QB = 16 2 ,
AB
= 122.
PROOF: For P
= (p,t), Q = (q,t), p + q, put
A = (p -
~~,
t -
~~p)'
B = (p -
j~
(q-p), t -
:~p)'
64
W. Benz
= (t,p), Q = (t,q) change the components in the definition of A,B.
In case P
Consider vice versa points P { Q and points A,B such that PA = 1', PB = 14', QA " 4', QB
=
P
Then according to the equations it turns out that Q E {(1,16),(16,1)}
= (1,1).
16', AB " 12'.
Without loss of generality we assume A " (0,0),
and hence PQ = O. REMARK: In case char K " 7 the following can be proved: Given P {Q.
Then PQ " 0
if and only if there exist points A,B,C such that PA " P, AC = ]2, CP = p,
QA
= 4', QB = 2', QC = 2', AB = 2'. 2.
Let K be a field.
Consider an injective mapping
K'
0
~
K' such that
(0,0), and
Defi ne
'P
K'~ by means of
K'" ...
:
1 0 (x, -) =:
x
.
1
(~(x),
(J
1 --(--)).
x
ip
--1
1 a
(J
(Notlce O(x, -) x = 0 (x, -) x since a is injective.
because of O(x,
,~
a. Consider>
(Xl
XI , X,
K with
E
X'I
+ x',.
x)
1.} Thus
'I'
is injective
Then
1 1 0 + X, , - + - ) XI
X,
PROOF: Denote
(Xl
+ x" - + 1 )0 by (u,v). Since 1 Xl X, 1
distance 1 from (x., -X. ) we get (u,v)(tp., -) 1 1 tp. U( (111_
'1") = V iJll '1',
«(III
-~,
),
i. e. u
=
V ~I
(Xl
+
X"
Xl
= 1 and hence
'1" 1 because
of
'PI
+ '1',.
°
+
1 ) is of X,
Now
(u -'I'Il(v -~) = 1 implies v' '1'1 '1', = V(tpl + '1''), If v = then u = v '1'1'1" = '1'1 1 1 1 1 and hence (XI + x, , -- + _)0 = 0, i.e. (XI + X, , - + - ) = 0 because of the XI x, XI X, injecti vi ty of o. Thi s contradi cts Xl' = X,' • Thus
V
'ill(jl,
= '1'1 +
tpl, i.e. (u,v) "
('1'1
+
'1'2,
1 1 -+-). '1'1
'1',
°
On mappings preserving a single L-M-distance b. Cons'Z-'del' Xl ,X2 , ••• 'X
p :=
+... +
(Xl
p:p =
K* unt . h
E
1
X , n Xl
n
1 xn
+... + -) and
(E
vfi
1
cp, v
E
1
l)
d'Z-st'Z-na. . . t
Then
cp(x i ).
:=
q>. 1
1
and P.o=
1
2 •• ' 'X 2pa'Z-1'W'Z-se " Xl 2 ,X2,
Consider n ;;. 3 and P. :=P- (x.,
PROOF: By induction. pO =: (u,v),
n
6S
l). X.
With
1
(notice induction) we get
vfi CPv
1.
Hence v
=
(1 )
E
vfi For i
f
j we get
i.e. (u If u = E cp
Ecp ) (( cp • (u
v
v
J
-
E ,I.'
JlTJ
then v = E l
CPv
cp
) - cp. Jl
1
(u -
E
.l' cp
vTl
) )
= O.
v
according to (1) and we are ready.
Assume u
f
L CPv
Hence
and thus
This implies u -
Ecpv
= - CPi - lPj for all
f j because of lPi f lPj and hence
66
W. Benz ~,
+
=r r - u
~
+ Q, ,
Q,
\i
a contradiction. e. Assume char K1 {2,3,5}. (i)
, c" ,
iii)
~!,
2
.~t2
, ••• ,Cl
+ ... +
'J.
n
n
E
K'"
with
are pairwise distinct, 1 1
'
- + ... + -
1
=
n
Then there exist e7..ements a, , a, , ... ,Ct
Ct,
Ct n
PROOF: Case char K $ {7,11,13,17,29,31,37,43,53,61,67} Put Ct,
3
=
7' Ct,
=
Ct,
:1,
=
-2
7'
= ~' as
Cl4
6
= 7' a.
Case char K E {31,37,53,61 ,67} Put
a,
10 = 19'
1, Ct, a,
(\,
15
,-g,
= -1
Cl4
a,
as
-6 = ,-g' a.
Cts
13' cx.
Cts
Case char KE (29,43}: 12 et, Put Ct, = 13'
a,
a,
-3
13'
Cl4
cx,
4
cxs
Case char K = 7 Put a,
=
5, a,
a,
Case char K = 11 Put a, = -3,
= -4, a,
(\,
2, a4
-5
Case char K = 13 : Put a, = 4, Ct, = -1 a, Case char K = 17
:
Put a, = 8, Ct, =-2, Ct, d. Assume char K ~ {3,5}.
5, a4
7
Then Q(-x)
PROOF: This is trivial for char K = 2.
- <j)(x) for aU x
E
K'"
So assume char K + 2.
Given x E K* we
get with elements cx i of step c : 1 1 (J - (0,0) ° (Q(x), -(-)) = (x, -) 4' X X
=
= (XCt l +... + XCt n ' x!, +... + X!n)O - ((-x)+ 1
- (",(-x), Q(-x)) applying step b.
v~l
xa v '
~x + v~l
x!v)O=
On mappings preserving a single L-M-distance
REMARK: d is also true in case char K = 3.
67
This will playa role in part III.
e. Asswne char K ~ {3,5.}. Consider paiY'UJise distinct eZements XI ,X, , ... ,x
n
E
Then
1
1 a
1
(XI +... + X , - +... + -) n XI xn
PROOF: Nothing is to prove for n Case n
1
(4'I+"'+4'n' - +... +-) 4'1 fjJn
1.
= 2:
For XI 1
+ x;
see a.
1
1 a
Consider XI 1
=
X, 1 and Xl
* X2.
Hence X2
=
-XI'
Now
(XI + X2, - + -) Xl X,
according to step d. (x, + X"
Thus
1 1 a l l - + -) = (fjJ(XI )+4'(X,), -(-) + -(-)) XI X, 4' Xl 4' X,
Now apply induction.
Let n be
~
3.
Nothing is to prove in case that
XI 2 , ... , X2 are pairwise distinct because of b. Otherwise suppose X2 1 X2 n nn without loss of generality and of course x _ + xn' Hence xn = - xn_ . Now n l l (XI +... + Xn , ... )
cr
=
(XI +... + Xn- 2' ... )
cr
=
(fjJI +... + 4' n- 2' ... )
LAsswne char K~ {2,3,5,7,11,13,17,19,23,31,37}. Then there exist Xl ,X2 ,X3
,~
E
K such that
1 1 1 1, - + - + - = 0, XI X2 X3 (ii) If k E {2,3,4,5} then the seven eZements _1_ _1_ - 1 . . d' . kXl, kX2, kx3, _1_ kXI' kX2' kX3' are pa'l-Y'UJ'l-se 'l-st'l-nct. (i)
XI + X2 + X3
=
PROOF: Case char K ~ {29,73,97,l03,113,157,199,239,241,349,401} 326 Put XI = 7' X2 = - 7' X3 = 7 . Case char K E {29,73,97,113,157,199,241,349,401}: 10 15 6 Put XI = 19' X2 = 19' X3 l1:li - 19 .
K'·'.
68
W. Benz
Case char K E {l03,239}: Put x,
12
13' x,
=
:l..:- Assume char
4
3
13' X3
= -
n'
Kif {2,3,5,7}.
a kl , a k2 ,···, akn(k)
E
Then for k
{2,3,4,5} then exist
K'" such that 1
(i)
E
1
a kl +· .+akn(krk=~ +... + akn(k) ,
(ii) akl, ... akn(k),-l
are pairwise distinct.
Kif {11,13,17,19,23,31,37}: Applying step
PROOF: Case char
kXI , a k2 = kx" ak3 = kX3, 1 1 1 ak4 = kX;' ak5 = kX;' a k6 = IX; . Case char K = 11 : Put ak1
=
a,l
1 , a" = 2, a23 = -3, a, 4 = -4, an
a3 I
-2, a3 ,
5 ,
a. ,
-3, a. 2
-4
1, a S 2 = -3, a, 3 -4 a" Case char K = 13: Choose a according to \1>1
1 + (-3) + 4
=2
,
(-2) + 6 + 3 + (-4)= 3 , (-2) + 6
+ (-2) + 6
Case char K = 17
=4 ,
=5
,
-5 ,
f we put
69
On mappings preserving a single L-M-distance
Case char K = 23 : 2 + (-11) + 3 + 8 = 2 , + 2 + (-11) + 3 + 8 = 3 , (-5) + 9 = 4 ,
+(-5) + 9 = 5 Case char K = 31
=2
6 + (-5)
+
,
(-11) + 14 = 3 , 1+ (-11)+14=4, 1 + 6 + (-5)+ (-11) + 14 = 5 Case char K a 37 (-3) + (-12) +
17 = 2 ,
1 + (-3) + (-12) +
17 = 3 ,
17 + (-13) = 4 , (-4) +
=5
9
!:..:.- Asswne char KEf {3,5,7}.
Given x
;,
E
K •
k a k Then (kx, x) = (krp(x), rp(x)
foI'
k E {2,3,4,5}. PROOF: This is trivial for char K = 2. 2
(2x,-) X
a-
1 a h(2) (x,-) = (1: xa v=l
X
1
= - (rp(-x), rp(-x))
,
1 xa,V
1: - - )
lV
So assume char K t 2.
a - ((-x)+l:xa lV
1 , - + -x
1
=
(rp(x), rp(x))
according to e and d.
Hence (2x,
2 a x) = (2rp(x),
2
rp(x))
The same procedure for
(vx,
~)a - ((-X)
+ vX,
~x + ~)a,
va 3,4,5,
completes the proof of h. i. Asswne char KEf {2,3,5,7}.
Given Xl, •.• ,x E K*•
n
Then
Applying g we get 1 )a = xa lV
1: - -
70
W. Benz
1 and also for n = 2 according to e and h.
PROOF: This is true for n Let n be > 3.
induction.
I{x 1"
Case A : We assume
Now apply
.. , xn} I > 2.
+ X, without loss of genera 1 ity.
Xl
Put P.1'- ( Vfi Xv' v~i l). X
Then, by induction,
v
p. ° = (vfi 1
,
Ii'
:J (u,v) p.
l).
\"
'Jfi
'J
Put P .-
(jJ
(LX ,L v
l) and Xv
pO
=: (u,v).
So flP '
1
1 implies
Ii
1, i. e.
1
(u -
Z
(j)) (v -
i:
v+i
v+i"
\i
1. v
Hence v
(2 )
= 1 and i = 2 we get
Applying (2) for (u-L4J) u .-;:",
(4'2
0 impl ies v =
v
U - Lr;J
"
+ O.
( lVl
+
(U-IID)
L:
q:tl)
(u - L W
)
V
We have
(j)l
u - L
i.e. u =
;;3
So (2) (put v = -
1
""
f
4'2
-
+
+ ... +
'<14
-
4'1
(U-Lr;J) - 4112)
=0
(3)
v
=
because of
''v =
2
l41 according to (2) and we are ready. Assume now
Hence by (3)
-
'Ill
qJl
Xl
f
qJ12 •
X,
and thus
,
I~n'
1) implies
1 1 1 + - + ... + - + = -1 + -1 +... + -1 • 1~3 IV -(jJl 413 414 4'n n
But according to induction we have
(4)
71
On mappings preserving a single L-M-distance
1
1 IT
1
1
(Xa +... + x , - +... + -) "(CPa +... + cP , - +... + -) n X3 Xn n cpa CPn This implies, since
a
is injective,
1
1
(Xa +... + x , - +... + - ) = P, i. e. x, + X2 " O. n X3 xn Hence C/l1+C/l2 = 0 by d. Thus u = L CPv by (4), contradicting the assumption u-
L
C/l v
* O. = xn = : x
Case B : x, = x, = '"
Here we have to prove (n x, ;)IT
n
(ncp, -),
cP
:"
cp(x).
x x Applying h we can assume n ;;" 6. Put y, := 2x, Y2 :" 2x, Y3 := 2' Y4 := 2' cP
Hence
Y, + Y2 + Ya + Y4 = 5 x , -1 + -1 + -1+ -1= -5 y, Y2 Y3 Y4 X Now n-6 1 1 l)IT na ( (nx,-) X " (n-6)x+YI+Yl+Ya+Y4+X, - X + -y, +... + -Y4 + -X = 1 1 1 1 1 = (x+ ... +x +Yl +Y2 +Ya +Y4 +X, -X +... + - + - +... + - +-) X Y, Y4 X
IT
' - v - - oJ
(n-6)-times -------~-----------.J
(n-1) summands
Similarly n-1 a l l 1 1 IT ((n-l)x, -x) = (x+ ... +X+YI +Y2 +Ya +Y4, - +... + - + - +... +-) X X Y, Y4 -~
(n-6)-times Since i is assumed to be true up till n-1 summands we get (nx,
~) IT
~
1 _ ( (n _1)x, n 1 ) a = (cp(x), cp(x)).
Thus (nx, ~)IT x ~Assume char
group K2.
= (ncp,
~). cp
K1 {2,3,5,7}.
Then a
~ ~ ~ is a monomorphism of the abelian
MoreOVer lH IT C IH.
PROOF: IT is injective. K1 {2,3}.
Hence P =
lemma 1. Now applying
we get
Consider P,Q E ~. We have IKI > 4 because of char 1 1 1 1 (Xl +... +X , - +... + -X ) , Q = (Y, +... +Y , - +... + - ) by n x, n m Y, Y m
72
W. Benz
(P + Q)o = (XI +... +y , ~ +... + ~)o = Y
m X,
m
pO + QO.
1 +... + -(-) 1 ) ( 'I' (X,) +... +\1 (Y) , -(-) m
X,
qJ
Ym
Since lH is the set of all points of distance 1 from 0 we get Jioe lH. ~
Assume ohar K ~ {2,3,5,7}.
* O.
assumed to be
Put X := ~. n
Consider an element n
1 + ... + 1 of K, which is
Then
Thus X
*0
and
n
Q=P+(nx,x)' Hence n) ° Qo = P° + (nx, x
by j and i.
J~
= pO
+
( n
Thus
Assume ohar K ~ {2,3,5,7}.
Then
PROOF: Nothing is to prove for P = Q.
SO assume P. Q.
ex; stence of poi nts A, B accordi ng to 1el1llla 2 such that -
OB = 16
2
-
, AB 2
QOAo = 4 since pO
2
= 12. Since 1,14,4,16,12 0
QOB = 16
,
2
,
AOBo
=
E
PO = 0 implies PA = 1', PB = 14', QA Then
~'oo
K we get P A
lz2 by applying k.
=1
2
00
, PB
the = 42
=~,
Now lerna 2 implies pOQo = 0
* QO.
3.
Consider a field K such that char K ~ {2,3,5,7}. mappi ng
uJ
:
K'
-+
K'
such that
Consider an injective
,
73
On mappings preserving a single L-M-distance
"'P,Q
E I(l
PO = 1 implies pWQw = 1
Put (O,O)w =: (al ,ad, (l,lt =: (b 1 ,bd. Because of (0,0)(1,1)
= 1
we have (O,O)w(l, l)w
= 1,
i.e. (bl -at) (b2 -a2 )
l.
Consider the distance preserving mapping (
) x' 1\ y'
a
1
= --
b1 -al
=
(x - al) ,
(b 1 -al )(y - a,)
which is bijective and linear up to a translation. Obvi ous ly, v P,Q
K'
E
and (0,0) wa
PQ
= 1 implies pwaQwa =
= (0,0), (1,1) wa = (1,1).
So section 2 applies to the mapping w a. It is {P
Especially, w a preserves distance O.
K21 (O,O)P = 0 and (T:T)P = O} = {(1,0),(0,1)}'
E
{(l ,0) , (0,1 )} wa
Hence
= {( 1 ,0) , (0,1 )} .
Define
l
x' = y
o:
,
y' = x
which is distance preserving, bijective and linear.
Define moreover
(
fl
=
~ identity
(l,O)wa = (1,0)
for (l,O)wa = (0,1)
l
Then a := wa fl is injective, preserves distance 1 and has (0,0),(1,1),(0,1),(1,0) as fixed points. The line y fact!
=
0 (i.e. {(x,o)1 x E K}) is mapped under
Given (x,O), x Ef
{D,l},
a
into the line y
O.
this point has distance 0 from (0,0),(1,0).
= (0,0), (1,0)0 = (1,0) under a into the line x = o.
(x,O)o has also distance 0 from (0,0)0 (x',O).
=
The line x = 0 is mapped
Consider a point (x,y) such that xy
* O.
The injectivity of
a
In So
and thus is of form implies
74
W. Benz
(X,O)O =: (Xl'O)
*
(0,0), (O,y)O =: (O,Yl)
Put (x,y)O =: (u,v).
Now
° imply (u - X,)V
(x,O)(x,y) = 0, (O,y)(x,y) = Hence u
= X,,
V
= y,
* (0,0).
because of (u,v)
= 0, u(v -y,) = 0.
* (0,0).
* ° it
We are now using the mapping ~ of section 2. For X 1 a 1 0 (X, -) = (,<,(X), -(-)). Thus (x,O) = (~(x),O). Also :p X
X
1
(-,y)
is
cr
y
1
* 0.
1
= (<;l(-) , ) for y y Q(.l)
Thus (O,y)0 = (0, -\-) and hence qJ(:Y)
y
(5)
for xy
* 0.
Consider X, ,X, E K'" such that X, + X,
(XI +
X, ,
1
* ° and
1 (Xl' -1-) + (x"
-1-)
-x
consider X E K \ {D,l}.
Then
1).
- -1 X
By applying j of section 2 we get
(XI +
X, ,
1
-1-)
-x
:J
=
1
(XI , -1- )
a
_ _1
+ (x"
1)
a
.
X
Now (5) implies
( <j)
1 1
(XI ) ,
) + (<;l (x, ), 1).
qJ(x - 1)
Thus
'I' (Xl
+ X,) =
qJ
(XI) +
f.l
(X, )
(6)
and
(7)
75
On mappings preserving a single L-M-distance
According to (6) we have (jJ(x + (l-x))
(jJ(x) + (jJ(l-x)
and
(jJ(l +(1 -1)) x
(jJ(l) + (jJ(x -1)
1
Thus (7) implies (jJ(x)(jJ(-!)
1 for all x'* 0,1.
=
x
But this is obviously also true for x = 1 . Define' : K ~ K by means of (
I (jJ(x)
i
,(x)
l
x '* 0 for x
0
1 For x '* 0 we have ,(x)T(x)
Thus , is injective. ,(x+y)
=0
,(x) + Tty) for all x,y
=
E
=
0 or y
O.
=
Now assume xy • 0 and x+y
,(0)
=
=
E
=
K"'. O.
It is also generally true for Thus x '* 0 and y
=
-x.
= -(jJ(x) and hence
Now d implies (jJ(y) ,(x + y)
We claim moreover that
K.
Because of (6) this is true in case x,y,x+y x
1.
0
(jJ(x) + (jJ(y)
=
=
,(x) + '(y).
Altogether we have that, is a monomorphism of the additive group of K such that 1
,(x) ,(x) = 1 for all x '* O. By Hua's theorem (s.[ 21, p. 324; the proof loc.cit. applies to the present slightly more general situation) we thus get that, is a monomorphism of K. finally prove that j the mapping (x,y) 0 for xy '* O.
0
Hence
w
=
0
is a semilinear mapping of the K-vectorspace Kl. According to
0
is a monomorphism of Kl.
= (or (x),
Because of (5) we have
,(y))
But this equation is also true in case xy = O.
(kx, ky)O
rr I
1:1
(T(kx),,(ky))
a-I
We
=
Thus for k E K
,(k)(,(x), Tty)).
is a semilinear mapping up to a translation.
the proof of theorem 1. theorem 1 are given by
This completes
It is now trivial to verify that all the mappings
0
of
W. Benz
76
(x,y)
(a
T
(x)
+
b. a
T
(y)
+
c)
(x,y) ... (a
T
(y)
+
b, 1 a
t
(x)
+
c)
where a,b,c
-+
E
K, a
and
* O.
BIBLIOGRAPHY 1. 2. 3. 4.
W. Benz, A Beckman Quarles Type Theorem for Plane Lorentz Transformations. Matn. Z., 177 (1981), 101-106. W. Benz. VOI'lesungen iweI' GeometI'ie der AZgebI'en. Grundl. Bd. 197. Springer-Verlag, New York 1973. F. Rado, On the characterization of plane affine isometries, ResuZtate d. Math., 3 (1980). 70-73. G. Tallini, On a theorem by W. Benz characterizing plane Lorentz Transformations in Jaernefelt's World. To appear in JournaZ of Geometry.
Mathematisches Seminar Bundesstrasse 55 2000 Hamburg 13 Federal Republic of Germany
Annals of Discrete Mathematics 18 (1983) 77"s6 © North-Holland Publishing Company
77
DESIGNS OBTAINED FROM AFFINE SPACES Carlo Bernasconi
We give some geometric constructions of designs (BIB-designs) and designs with parallelism obtained in affine spaces.
By these methods we get some classes
of affine designs which are not affine spaces and some special designs called calibration designs. Finally, we generalize the construction from affine spaces to designs with para 11 el ism.
INTRODUCTION We show two constructive methods of obtaining designs from affine spaces. The first method makes use of a property possessed by affine spaces - two-transitivity - and gives infinite classes of designs. 1.
This procedure is shown in Sec.
Using the second method we obtain other designs with parallelism from affine
spaces.
This procedure is outlined in Sec. 2.
The main characteristic of designs obtained is their geometrical structure. Some of them, shown in the examples of Sec. 3, have the further property of being applicable to situations of special interest; for example, all designs shown in Sec. 3 with even block size may be used as "calibration designs" (cf. (3)).
These
are designs where blocks can be partitioned into smaller blocks of a design on the same point set; that is, one obtains a refinement of the given design (which is not however a composition design (cf.
(1)
or
(5) )).
Further, we obtain some examples of affine designs with the same parameters as, but not isomorphic to, affine spaces (see Note in Sec. 3, d)). Finally, we generalize these constructions in Sec. 4. For the basic definitions and terminology on designs, parallelism and affine designs we refer to [4, Chapter 2).
78
C. Bernasconi
1. DESIGNS OBTAINED FROM AFFINE SPACES Let AG(d,q) denote a finite (desarguesian) affine space.
Let AGd_l(d,q)
denote the design (BIB-design) whose blocks are the hyperplanes of AG(d,q). Let C be any subset of points of a design D.
We call C a configuration.
Two configurations C,C' are said to be isomorphic if there exists an automorphism , of 0 such that C' = C~. The following theorem holds (cf. [21): THEOREM 1: Let AG(d,q) be a finite desarguesian affine space. figuration of AGd_l(d,q).
>'"
i; Zock
Let C be any con-
"hen the set of all configurations isomorphic to C is
set of a design en the points of AG( d ,q).
It is possible to use the above result to obtain designs from affine spaces, by means of purely geometric constructions. The most elementary class of isomorphic configurations one can imagine in an affine space AG(d,q) is the class of all t-subspaces (i.e. t-dimensional subs paces) for any given t, 1
~
t
~
d-l.
This class gives the block set of the well known
design AGt(d,q). As the most "primitive" concept in affine spaces is parallelism, our first concern will be to use Theorem 1 to obtain designs from linear configurations obtained by collecting parallel subspaces together.
Secondly, we will try to ob-
tain other designs which possess this same basic property: parallelism.
This will
be done in this Section and in Section 2, respectively. Note that the designs AGt(d,q) have this property: the parallelism of the lines of AG(d,q) induces a parallelism on AGt(d,q), for every t, 1
~
t
~
d-l.
Two
t-subspaces of AG(d,q) are parallel to each other if they are equal or if they are disjoint and contained in a same (t+l)-subspace. We begin by considering the configurations made of equicardinal sets of parallel t-subspaces of AG(d,q).
Two such configurations are not in general iso-
morphic to each other, but by means of additional properties of designs it is possible to prove (cf. [2, Sec. 41) the following: PROPOSITION 2: Let AG(d,q) be an affine (desarguesian) space.
The set of all h-
tuples of parallel t-subspaaes of AG(d,q), for any given t,h, 1
~ h .;; qd-t_ l,
~
t .;; d-l,
is the block set of a design on the points of AG(d,q).
lye denote
79
Designs obtained from affine spaces
such a design AGt(d,q,h).
NOTE 1: It is easy to check that the same statement holds for non-desarguesian pJanes A(2,q) (here the only subspaces we can take into consideration are the lines of A(2,q».
We can thus avoid restricting ourselves to the desarguesian
case of Proposition 2, which, moreover, holds for any finite affine spaces. NOTE 2: Other interesting classes of designs obtained by the same technique are the ones with the following block set: a) all sets of h parallel t-subspaces, each set contained in a (t+1)-subspaces, for any given t,h, 1
~
t
~
d-l, 1
~
h
~
q-l.
In particular: b) all sets of h parallel coplanar lines of AG(d,q), for every h, ~
h
~
q-l.
Cases a), b} are examples of designs obtained as composition designs (cf. [1) or [5 J) from the designs of Proposition 2.
2. DESIGNS WITH PARALLELISM OBTAINED FROM DESIGNS D = AGt{d,q,h) We obtain some designs with parallelism from the designs AGt(d,q,h) of the previous Section, that is from designs whose blocks are all the h-tuples of parallel t-subspaces of AG{d,q).
The procedure is as follows.
Before gOing into the details of the construction of such designs with parallelism, we observe that AGt(d,q,h) is not in general a design with parallelism, since it may be not possible to partition the pOints of AG(d,q) with h-tuples of parallel t-subspaces.
Maybe h qt does not divide qd (the size of the whole
space) at all! Our construction will consist in choosing a subset of blocks of AGt(d,q,h) in such a way that for every parallel class of t-subspaces of AG(d,q), the blocks chosen, belonging to this class, partition the class itself, so that they partition the whole point set of AG(d,q) thereby giving new parallel classes.
This
will be better understood by following the construction procedure undertaken here. Let us consider the design AGt(d,q), whose blocks are the t-subspaces of AG(d,q).
AGt(d,q) is a design with parallelism (as pointed out above).
Every
C. Bernasconi
80
parallel class C consists of qd-t t-subspaces, one parallel to another, that partition the whole affine space AG(d,q). d-t d t The idea now is that, as C has q - t-subspaces, any affine space with q points has in itself many parallelism, each one of which can be used to induce a parallelism on every class C. For every parallel class of AGt(d,q) we fix a 1-1 correspondence between the elements (t-subspaces) of that class and the points of an affine space AG(d l , pr(d-t)/d 1), where pr =q. There exists such an affine design for every integer d1 dividing r(d-t). For any given t , 1 ~ t1 ~ d -1, we consider the design AG (d ,pr(d-t)/d 1), l t l l that is the design whose blocks are all the tl-subspaces of the g~ven affine space. AG (d , pr(d-t)/d l ) is a design with parallelism. l t The '!1 correspondence we fixed between every parallel class of AGt(d,q) and the point set of AG(d" pr(d-t)/d l ) induces a correspondence between the tl-subspaces of AG(dl,p r(d-t)/d 1), i.e. blocks of AG (d l , pr(d-t)/d 1), and the k-tuples t1 of t-subspaces of AG(d,q) belonging to the same parallel class, where k is the cardinality of the tl-subspaces of AG(d , pr(d-t)/d l ), i.e. k = p(r(d-t)/d l )t l • l If we repeat this procedure for all parallel classes of AGt(d,q), we obtain a structure S on the point of AGt(d,q) whose blocks are k-tuples of parallel blocks of AGt(d,q).
The blocKs of S are not all the k-tuples of parallel blocks
of AGt(d,q). They are those k-tuples in correspondence with the blocks of . AG t (d l , pr(d-t)/d 1) by the correspondence \'we flxed above, for every parallel class of 1 . set of AG(d , pr(d-t)/d 1). AGt(d,q), between that class and the pOlnt 1 The parallelism of AG (d , pr(d-t)/d l ) induces a parallelism on the strucl t 1
ture S. S is a substructure of the design AGt(d,q,k), where q = pr, (r(d-t)/d )t k =P 1 1. It depends not only on the parameters t,d,q of AGt(d,q) but . AG (dl,p r(d-t)/d 1). also on the parameters t l ,d 1 that we can choose for the deslgn t1 In particular, we can get different substructures with the same value of k for different choices of t ,d . For this reason, for S, we will have to use the 1 l . cumbersome notatlon: S = AGt(d,q,p (r(d-t)/d 1)t 1). We call AG (d"p r(d-t)/d 1) the t1 "auxiliary" design. To finish our construction, we have to show that the structure S obtained in
81
Designs obtained from affine spaces
this way is a design.
The proof is a merely enumerative matter and proved capable
of extension to more general structures.
It will therefore be postponed and ob-
tained as a corollary of a theorem of construction given for more general structures in Section 4. We may summarize the construction given in this Section by saying that from the designs AGt(d,q,h) introduced in Section 2 we have, by use of auxiliary designs AG
t
1
(d , pr(d-t)/d1), obtained new designs with parallelism whose block set 1
is a subset of the block set of AGt(d,q,h) and which we denote AG t (d ,q, p (r(d-t}/d l )t 1) •
3. SOME CLASSES OF DESIGNS WITH PARALLELISM, AFFINE DESIGNS, CALIBRATION DESIGNS In this Section we shall need some familiar design notations - v: size of the point set; k: size of a block (number of points incident with a block); m: size of a parallel class of a design with parallelism. We start by giving two simple examples, a) and b), describing the constructions given in Section 1, 2.
We proceed by setting out a number of infinite
classes of designs obtained by applying those constructions: c), c') are designs with parallel classes made of two blocks (m
= 2); d} are affine designs; e} are
calibration designs (cf. [31) where blocks are obtained by collecting together pairs of parallel blocks of AGt(d,q}. a} Design AG 1(2, 9, 3}. With the construction given in Proposition 2 of Sec. 1 for t q
=
1, d
=
2.
= 9, h = 3 we obtain D = AG,(2,9,3}, that is the design on the points of AG(2,9)
whose blocks are all the triples of parallel lines of AG(2,9}. b} Design AG (2, 9, 3}. 1 With the construction given in Sec. 2 we obtain the result that for t d
= 1,
= 2, q = 9 the only permitted values for tl: d1 are d1 = 2, t1 = 1. Thus, we have only one kind of design AG (2,9,3), obtained by the unique 1
82
<7. Bernasconi
auxiliary design AG 1(2,3). Every parallel class of AG(2,9) contains 9 lines in 1-1 correspondence with the points of AG 1(2,3). The 12 lines of AG 1(2,3) identify the 12 triples of lines we choose out of the total Of(;)
=
84 triples of lines
existing in every parallel class of AG(2,9). c) Designs with parallelism, with m = 2: AG (d,2 r ,2 r (d-t)-1). t We use the construction of Sec. 2. If q is even then q = 2r , and we thus have AG (d,2 r ,2(r(d-t)/d l )t 1). We have m = 2, that is blocks with size t k = v/2, if and only if 2rt 2(r(d-t)/d l )t l = 2rd - l , i.e. r(~-t) (d -t ) = 1. As l 1 1 dl divides r(d-t), the last equality holds if and only if r(d-t)
= dl and
= 1. So we get designs with block size k = v/2 and with parallel classes made of pairs of blocks if and only if q = 2r, dl = r(d-t), t1 = r(d-t)-l. These dl-t l
- (d,2 r ,2 r(d-t)-l ). are the designs AG t Note that in this case we use an auxiliary design of type
AGr(d_t)_l(r(d-t), 2), whose blocks are the hyperplanes of an affine space of order 2. e') Designs with parallelism, with m = 2: AG (d,2 r ,2 r (d-t)-1). t We refer to the construction of Prop. 2, Sec. 1.
We take AGt(d,q) with q even; let q = 2r. Every parallel class of AGt(d,q) contains m = 2r (d-t) t-sub. even, m 15 . even and 2 m = 2r(d-t)-l spaces. As q 1S Take the design r r(d-t)-l D = AG t (d,2 , 2 ). It is easily to check that the blocks of D have size k = v/2 and to define a parallelism on the blocks of D in the following natural way. Let 8 ,8 2 be blocks of D. 8 ,8 are parallel if and only if either 1 1 2 81 = 82 or 82 = 81 (here 81 means the set of points non-incident with 81), Then AG (d,2 r ,2 r (d-t)-1) is a design with parallelism, with m = 2. t r r(d-t)-l NOTE 1: Recall that the block set of 0' = AG ) (the design defined t (d,2 ,2 in c)
is a subset of the block set of D = AG t (d,2 r ,2 r(d-t)-l ).
Then also the
subtraction design D - D' (cf. [1 I for terminology) is a design with parallelism with m = 2.
83
Designs obtained from affine spaces
NOTE 2: Designs c), c ' ) are examples of designs with parallelism obeying axiom (8) but not axiom (7) of [4, 2.2.6). -
r
d) Affine designs AGd_l(d,p ,p
(rid
)(d -1)
1
1
).
We refer to the construction of Sec. 2. Take t = d-l, tl = dl-l. We obr (rid )(d -11 tain D = AGd_l(d,p ,p 1 1 'J. Here dl must divide r. If we denote
rid
ql = P
r d d d-l 1, we have p = ql 1. Then D = AGd_l(d,ql l,ql 1 ).
The auxiliary
, r(d-t)/d deslgn AG t (dl,p 1) has the form AGd _l(d l ,ql). 1 1 The design D obtained in this way is an affine design: any two non-parallel blocks of D have the same number
~
of common points.
In fact, two non-parallel d
d -2
blocks of D from the same parallel class of AGd_l(d,ql 1) have a number ql 1 (d-l)-subspaces in common, hence
~
of
d d-2 d -2 d (d-l) = ql 1 qll = ql 1 ,while two nond
parallel blocks of D from different parallel classes of AGd_l(d,ql 1) have d d-2 ql 1 cOlll11on poi nts. NOTE: The construction given is nothing but a combinatorial restatement of the known representation of affine spaces AG(d,qld l ) as affine spaces AG(dld,ql). fact, AGd_l(d,qldl, qldl-l) is isomorphic to AG d d_l(dld, ql).
In
AG(dld, ql) is the
1
dl -dimensional representation on GF(ql) of the affine space AG(d,qld l ) coordinatized on the dl-th extension of GF(ql): GF(qld l ). The best known example is the classical (infinite) case given for d = 1, dl = 2 and real numbers instead of GF(ql): we get AG(2,1R) as a representation of the complex affine line. But we have two interesting exceptions that correspond to the following cases. i) Take d
= 2. Then consider any non-desarguesian plane A(2, ql dl ). Then the
structureA(2, ql dl , qldl-l) results in an affine design with the same parameters as, but non-isomorphic to, AG 2d _1(2d l ,ql). 1 ii) Take dl = 2. Consider as an auxiliary space any non-desarguesian plane A(2,ql). As A(2,ql) is non-isomorphic to AG(2,ql)' the resulting structure D = AG _ (d, q1 2,ql) is an affine design with the same parameters as, but non-isod l morphic to AG _ (2d,ql). 2d l
84
C. Bernasconi
e) Designs with parallelism, where blocks are pairs of parallel t-subspaces: r
AG (d, 2 ,2). t
r Consider the construction of Sec. 2 in the case where q is even, i.e. q = 2, r (r(d-t)/d)t and dl = r(d-t), tl = 1. Then the design AGt(d,p ,p 1 1) has the form:
r . AG t (d,2 ,2) and the auxiliary design 1S AG
t
(d1,p
r(d-t)/d
_ 1) - AG1(r(d-t), 2).
1
In this case, the tl-subspaces of the auxiliary design are the lines of AG,(r(d-t), 2) and consist of all possible pairs of points of AG1(r(d-t), 2), so r that the blocks of AG t (d,2 ,2) consist of all possible pairs of t-subspaces of - (d,2 r ,2) = AG (d,2 r ,2). AG (d,2 r ,2). Therefore we have: AG t t t We therefore have that when q is even, all designs AG t (d,q,2) are designs with parallelism.
4. GENERAl. CONSTRUCTI ON We show in this Section a general construction of designs obtained by combining designs with parallelism with other designs. Let 0 be a design with parallelism.
Let (b,v,r,k,A) be the parameters of O.
The cardinality of every parallel class of 0 is m = v/k.
The number of different
parallel classes is r. Suppose that there exists a design 0 with m = v/k paints. Then it is pos1 sible to collect the blocks of 0 in the following way. Let rl,kl,A l be the parameters of 0,.
For every parallel class C of 0 we fix a 1-1 correspondence
~
between the elements of C and the points of 0 , and we define as new blocks all 1 the kl-tuples of parallel blocks which correspond, through ~, to the kl points of a block of 01, for every block. In this way we get a structure S, on the points of 0, whose blocks are these kl-tuples of blocks of O. THEOREM: The stpuatupe S is a design. PROOF: The blocks of S have size k(S)
= k kl • In order to complete the proof we
will compute the parameter A(S). For any pair of paints P,Q of S, the blocks of S passing through P,Q are:
85
Designs obtained from affine spaces
i) the blocks coming from those parallel classes of D such that the pair P,Q belongs to a member of that class.
There are A such classes since there are A
blocks of D incident with both P and Q.
For each one of these classes the blocks
of S through P,Q are in 1-1 correspondence with those blocks of Dl incident with the point of D1 corresponding to that member of the class which contains P,Q.
We
therefore have r 1 blocks of S for every parallel class of this kind, that is we have, in all, Ar blocks of S through P,Q coming from these parallel classes; l ii) the blocks of S passing through P,Q coming from those parallel classes such that the pair P,Q belongs to distinct members of the class.
There are r-A
such classes (parallel classes are r in number, and A were considered in i)).
For
each one of these classes the blocks of S through P,Q are in 1-1 correspondence with the blocks of Dl incident with the two points of D1 corresponding through to the two members of the class incident with P,Q respectively.
~
We thus have A1
blocks of S for every such paralle1 class; that is we have, in all, (r - A)A
l
blocks of S passing through P,Q coming from these classes considered in ii). We conclude, by adding blocks of kinds i) and ii), that A(S) = Ar +(r-A)A • 1 1 This completes the proof. COROLLARY 1: The structure S of Section 2 is a design. Suppose the design 01 above is a design with parallelism. lelism can be induced on the blocks of S. Thus we get:
Then this paral-
COROLLARY 2: If 01 is a design with paraZlelism then S, too, is a design with parallelism.
This is just a generalization of the construction carried out in Section 2. We denote S; O/D l and call D1 the "auxiliary" design.
BIBLIOGRAPHY 1. 2. 3.
C. Bernasconi, Constructions of designs (1980). C. Bernasconi, Geometric designs (1981). R.C. Bose and J.M. Cameron, The Bridge Tournament Problem and Calibration Designs for Comparing Pairs of Objets, J. Res. Nat. Bur. Stand. 69B (1965), 323-332.
86
4. 5.
C. Bernasconi
P. Dembowski, Finite Geometpies. Springer-Verlag, Berlin (1968). H. Hanani, On some tactical configurations. Canad. J. Math., 15 (1963). 702722.
Istituto di Matematica Universita di Perugia Via A. Pascoli 06100 Perugia Ita ly
87
Annals of Discrete Mathematics 18 (1983) 87-94 © North-Holland Publishing Company
ON THE CODES GENERATED BY CERTAIN DIVISIBLE DESIGNS T. Beth* and D. Jungnickel
1. ABSTRACT AND INTRODUCTION Many interesting codes arise as representation modules of geometric structures, e.g. Golay codes [17], Reed-Muller codes [17J, and codes generated by planes [8] and biplanes [4].
Conversely, codes may often be used to analyse the
properties of a structure [1] which even might only exist hypothetically [18J . Among the most studied combinatorial structures the class of divisible designs plays a special role [3], [2J.
A particular example is provided by the
uniform Hjelmslev planes [14], [7] which may be defined as divisible designs by their parameters [12].
These structures are natural generalizations of projective
planes when considered as their homomorphic pre-images. In this paper we will concern ourselves with the most interesting special case of "regular uniform Hjelmslev planes" which may be described via a regular group of automorphisms.
This includes the case of the direct product of an
elementary abelian group with a cyclic group which naturally is an automorphism group of the induced code.
It is known that codes admitting such a group are very
efficient with respect to both decoding and residual error probability. 2. UNIFORM HJELMSLEV PLANES As already mentioned, Hjelmslev planes constitute a generalization of projective planes.
The precise definition of Hjelmslev planes in general [14] ,
[6J is immaterial to our present purposes as we shall only consider the special case of uniform HjeZmsZev pZanes Hq (of order q) which for q > 2 are just the symmetric divisible designs with m=q2+q+l groops of n=q2 points each, such that neighbor points (i.e. points in the same groop) are joined by precisely Al=q
lines whereas non-neighbor points have A2=1 lines in common; the term "symmetric" means that the dual conditions are likewise satisfied.
In the case q=2 one has to
add the further condition that each line meets each point class in 0 or 2(=q)
88
T. Beth and D. Jungnickei
points [12J.
Note that the neighbor classes of points and lines form an ordinary
projective plane in a natural way. Hq will be called regular (10) with respect to the abelian group G ~ Z X N provided that N acts regularly on each neighbor class of points (resp. lines) and
Z operates regularly on the set of neighbor classes. Then Hq may be represented in the following way [10,11) : The points of Hq are the elements of G and the set of lines of H ;s the orbit OG where 0 can be written in the form q
with {d , ... ,d } being a Singer difference set in Z (for the projective image o
q
plane H' of H ) whereas the set {S.: q
called a
q spread)
1
i~O,
... q} (in abuse of language henceforth
is a system of distinct representatives for the parallel classes
of an affine translation plane of order q (not necessarily related to H'). q
particular, q has to be a prime power in this case [16]. cyclic and H' is desarguesian. q
In
In all known cases, Z is
In the sequel we restrict ourselves to the case
where q=p is a prime, H is regular and Z cyclic. p
It may be mentioned that the
"desarguesian" unifonn Hjelmslev planes (i.e. those unifonn Hjelmslev planes that may be coordinatized over a suitable local ring [15]) are in fact regular [g). We finally require the following result of Bose and Connor (3): For the incidence matrix A of H we have q
(§)
Idet A I
432 (q+l)2. q2q +2q +3q +q
3. CODES
Considering the incidence matrix A (with its rows corresponding to lines) we define the code eIF generated by H to be the row space of A over the field lF • p
p
From formula (§) it is clear that elF is of no special interest unless char IF p
equals p or divides p+l.
These two cases will be considered more closely in the
sequel. In order to provide a powerful tool to work with we observe that eIF is an p
ideal in the group ring R = IF [Z x N)
89
On the codes generated by certain divisible designs
where Nl
~
N2
~ ~
p (the cyclic group of order pl.
Clearly this ring R is iso-
morphic to the quotient ring 2
F [ x ,y , zJ / ( xp- 1 ,y P- 1 ,Z P +p+ 1- 1)'
Using this language, the code elF is the principal ideal generated by the p
"polynomial" g(x,y,Z) where the polynomial p_ 1
_
i
m. i +b . J
Sj(x,y) - Li=O x Y J
is the indicator function of the j-th spread component S. J
2
{(u,v) E GF(p) : v=m.u+b.}. J
J
4. THE CASE P ;: -1 mod char IF
In this section we will see that the code generated by H here is of no p
particular interest.
In fact a similar but more simpler set-up (than that of
section 3) will prove the following result which is even independent of the uniform H-plane Hp considered (and in particular does not require the assumption of regularity) . THEOREM 1: Let H be any uniform Hjelmslev plane of prime order p. p
Then the
corresponding code elF has codimension 1 whenever char IF divides p+ 1 . p
PROOF: Let A be the incidence matrix of H
p
t
It clearly suffices to show that AAt
has corank lover IF (where A denotes the transpose of A). assume IF to be of the form IF
=
W. 1. O.g. we may
GF(r) where r is a prime with p ;: 1 mod r.
the definition of Hp it is clear that over GF(r)
From
90
T. Beth and D. Jungnicke/
B J J
J
J
B J
J
AAt = J
J
J
B
J
where J is the p2xp2-all-one-matrix and B = J-J. The set-up of section 3 now enables us to compute the rank of AAt avoiding rather messy matrix manipulations.
To this end we observe that the GF(r)-row
space of AAt is the ideal M generated by the "polynomial" 2
m(x,y,z) = (l-~(x.y» where
p-l
~'(x.y)
j k
L. k-O x y J, -
i
+ L~=~P z ~(x.y) in the ring 2
GF(r) [x,y,z) / (xp-l,l-l,zP +P+l_1). We now invoke the theorem that the GF(r)-codimension of the ideal Mequals the number of zeros of the polynomial m(x,y,z) where x,y range over the group of p'th roots of unity over GF(r) while z ranges over the (p2+p+l) 'st roots of unity Evaluating ~
~(x,y)
we note that
= n = 1 in which case
~(l,
1)
~(~,n)
2
=p
p-l
= (Ei=O
=1 mod
~
r.
m(x,y,z) in the admissable range are those with
i
p- 1 j
)(Ej=O
~
) vanishes unless
So the only zeros ~
(5).
= n = 1 where
~
(~,n,~)
of
is an
admissable zero of 2+ -1
h (z) = r.~ oP 1=
.
z1 •
Since the only root of h(z) within the set of (p2+p+l)lst roots of unity is ~=l, the assertion follows.
0
5. THE CASE char F = P We now concern ourse 1ves with the case char F = P where the above methods
91
On the codes generated by certain divisible designs
cannot be applied with exactly the same ease as the group ring IF[ Z x Nl x N J is 2 no longer semi-simple.
THEOREM 2: Let Hp be a reguZar uniform HjeZmsZev pZane of prime order p. the aode CGF(p) satisfies
Then
p
unZess p=2; in this aase exaatZy two suah planes exist [0] induaing aodes of dimension 12 and 13. respeativeZy.
PROOF: From Section 3 we recall that C= cGF(p) is the ideal generated by p
g(x,y,z)
P );i=O z
d.
1
si(x,y)
in the ring 2
GF(p) [x,y,z] / (x p-l,l-l,l +p+1_1)' It is readi ly seen that the "po 1ynomi a1s" s.1 (x ,y) generate a nil potent ideal of index p in the ring T = GF (p) [x ,y] I (xp- 1 ,y p- 1) . We first assume that the S. contain a common pOint, say 1. 1
2
1
Writing the po1y-
nomia1 g(x,y,z) as c(z) in the ring T [z] I (zp +p+ -1) we show that cia} does not vanish for every (p2+p+l ) ' st root of unity a by observing that the (almost disjoint) lines S.1 are linearly independent over GF(p).
Reasoning analogously to
Section 4, we conclude from the non-vanishing of the p2+ P+l function values c (a) that the ideal generated by c{z) in T[ Z]/{zP2+p+1_1) has GF(p)-dimension p2+ p+1. Omitting c{l) = ~(x,y) = );~-~-O xiyj we see that by successive multiplication with 1,J-
suitable nilpotent elements of degree 1, ... ,p-l we obtain p.p(p+1) GF(p)-linearly independent elements which together with the polynomial L~=O ~(x,y)zdi form a GF(p)-basis of C. If the Si are not copunctual, a sufficient condition for cia) not to vanish
92
T. Beth and D. Jungnickei
,
is that there be at least one point (a,b) which is on exactly one of the lines S.. This condition is violated only if each point is covered by exactly 0 or 2 of the lines Si'
This may in fact happen for p=2; in this case the corresponding H2
induces a code of dimension 12, whereas the other uniform Hjelmslev plane of order 2 (there are exactly two such planes by a result of Bacon, see [OJ) produces a code of dimension 13 by the above argument.
Now assume that such a configuration
would arise for p ~ 3 (though we doubt that this is possible) and let (a,b) be any one of the points covered twice, by S. in g(x,y,a) is
a
d·
11 +
a
d·
'2.
and Si
2'1
say.
Then the coefficient of xayb
2
As p +p+l is odd this coefficient can never vanish.
Using again successive multiplication with nilpotent elements we derive the desired bound p.p(p+l) + 1 ~ dim C ~ p(p2+ p+1) depending on the nilpotency index of c(l).
0
COROLLARY: Let Hp be a regular uniform HjelmsZev plane for which the spread {Si: ;=0, ... ,p} contains a common point.
Then the lower bound for dim Cis
attained.
BIBLIOGRAPHY O. 1. 2. 3. 4. 5. 6. 7. 8. 9.
P.Y. Bacon, On Hjelmslev planes with small invariants. M.A. thesis, University of Florida (1971). T. Beth and D. Jungnickel, Mathieu groups, Witt designs, and Golay codes. In: Geometries and Groups. Proc. Berlin 1981, Springer Lecture Notes 893, p. 157-179. T. Beth, D. Jungnicke1 and H. Lenz, Design Theory I. to appear. R.C. Bose and W.S. Connor, On the combinatorial properties of group divisible incomplete block designs, Annals Math. Stat. 23, (1952), 367-383. P.J. Cameron, Biplanes, Math. Z. 131, (1973), 85-101. P. Camion, Abelian codes. Math. Res. Center Univ. Wisconsin Report 1059 (1970). P. Dembowski, Finite Geometries, Srpinger 1968. D.A. Drake, Projective extensions of uniform affine Hjelms1ev planes, Math. Z. 105, (1968), 196-207. R.L. Graham and F.J. MacWilliams, On the number of information symbols in difference set cyclic codes, Bell Systems Techn. J. 45 (1966),1057-1070. M.P. Hale, Jr. and D. Jungnicke1, A generalization of Singer's theorem, PY-cc. Amer. Math. Soc. 71 (1978), 280-284.
On the codes generated by certain divisible designs
10. 11. 12. 13. 14. 15. 16. 17. 18.
93
O. Jungnickel, Regular Hjelmslev planes, J. Comb. Th. A 26, (1979), 20-37. O. Jungnickel, On balanced regular Hjelmslev planes, Geom. Ded. 8 (1979), 445-462. O. Jungnickel, On an assertion of Dembowski, J. Geom. 12, (1979), 168-174. E. Kleinfeld, Finite Hjelmslev planes, Illinois J. Math. 3, (1959), 403-407. W. Klingenberg, Projektive und affine Ebenen mit Nachbarelementen, Math. Z. 60, (1954), 384-406. W. Klingenberg, Oesarguessche Ebenen mit Nachbarelementen, Abh. Math. Sem. Hamburg 20, (1955),97-111. H. LUneburg, Translation planes, Springer 1980. F.J. MacWilliams and N.J.A. Sloane, The theory of error-correcting eades, North Holland 1978. F.J. MacWilliams, N.J.A. Sloane and J.G. Thompson, On the existence of a projective plane of order 10, J. Comb. Th. A 14, (1973), 66-78.
Institut fUr Mathematische Maschinen und Oatenverarbeitung I Universitat Erlangen-NUrnberg Martensstr. 3 0-8520 Erlangen F.R. Germany
Mathematisches Institut Justus-Liebig-Universitat Giesen Arndtstr. 2 0-6300 Gi essen F.R. Germany
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Annals of Discrete Mathematics 18 (1983) 95-112 © North-Holland Publishing Company
95
ON A CHARACTERIZATION OF THE GRASSMANN SPACES ASSOCIATED WITH AN AFFINE SPACE A. Bichara and F. Mazzocca *
ABSTRACT Affine Grassmann spaces are defined and investigated.
Under suitable as-
sumptions any such space is proved to be isomorphic to the incidence structure (Gh(IA), Fh(llI. )), where TA is an affine space, Gh(IA) ;s the family of its h-dimensional subspaces and Fh(IA) is the family formed by pencils of h-dimensional subspaces, the incidence relation being the usual set-theoretic inclusion.
1. INTRODUCTION Let
L
be an affine (projective) - either finite or infinite - space whose
dimension is at least three; for all integers h, 1 ~ h < dim denote the family of h-dimensional subspaces in E.
L -
1 let Gh(E)
The h-th Grassmann space asso-
ciated with E is defined to be the incidence structure rh(E) = (Gh(E), Fh(E)), where Fh(E) is the family of all pencils of h-dimensional subspaces in E, the incidence relation being usual set-theoretic inclusion.
Such structures will turn
out to be partial line spaces. REMARK: When
L
is a finite dimension projective space over a field, rh(E) is iso-
morphic to Grassmann's manifold representing the h-dimensional subspaces of E. Grassmann spaces associated with a projective space were completely characterized as partial line spaces in [9), [2), [3); furthermore, another
characteriz~
tion of Grassmann manifold representing the lines in a Galois projective space can be found in [ 7). In
[1)
the authors characterized the Grassmann space of index one associated
with an affine space and they will use their results and an induction argument (see sect. 4) to characterize the Grassmann space of index h associated with the affine space TA, for all h's.
A. Bichara and F. Mazzocca
96
A partial line space (PSL) is a pair (S,R), where S is a non-void set and R is a covering of S all whose elements have size at least two and such that any two elements in S belong to at most one element of R. and lines the elements in R. (p
Call points the elements in S
Two points p and q in S are said to be collinear
q) if they belong to an element of R, (p,q).
~
If this is not the case, p and
q are said non-collinear (p t q). A subspace of (S,R) is a subset T in S consisting of pairwise collinear pOints and containing the line through any two of its points. is a subspace T
A maximal subspace
S which is maximal with respect to set-theoretic inclusion.
~
{S,R} is a proper PL5 (PPL5) is there exist two non-collinear points; when this is not the case. (5,R) is a line space.
Finally. (S,R) is connected if for any two
distinct points p and q there exists a finite sequence of points a l ,a 2, .••• an such that p~al-a2- ..• -an-q. A connected PPL5 (G.F) is called an affine Grassmann space if the following axioms hold: Al
Any three pairwise collinear points belong to a subspace.
A2 - No 1i ne is a maxima 1 subspace and three non-empty co 11 ecti ons of maximal subspaces. say Sl' S2' T, exist such that any maximal subspace belongs to precisely one of them.
Furthermore, setting S
(i) 5, 5' E S2' S
(ii) S E S. T E T (iii)
If
f
F.
E
~
S'
~
5 n S'
= Sl
U
S2'
=0 ;
either 5 n T = 0 or S n T E F;
~
3! S E S. T E T: f c 5 and f
~
(iv) (Veblen axiom in S) Given Sl' 52 in S, 51 point or belonging to S2' if Sl and
52
T; ~
52' either meeting at a
are distinct elements in S both meeting Sl
and S2 at distinct points. then either Sl and
52 meet
or they both belong to S2;
A3 - S2 is a covering of G. From (iii) it follows that the set F of the lines in (G,F) can be partitioned into disjoint subsets Fl and F2 ; namely: f E F, f E Fl ¢ = the unique S in S containing f belongs to Sl; f
E
F, f
E
F2
¢=
the unique 5 in S containing f belongs to S2'
A line f will be called of the first (second) kindif it belongs to Sl (S2)' Let (G,F) and (G',F') be two affine Grassmann spaces; an isomorphism between them is obviously defined as a bijection
"i'p(S;)
= Si'
i
= 1.2,
~(T)
~
: G ~ G' such that
= T.
(G,F) is said to be of finite index if there exist a line fEF, a subspace T
97
On Grassmann spaces associated with an affine space
in T containing f and a saturated finite chain of subspaces C(f,T) with minimal f If (G,F) has a finite index, then the least integer h such that a
and maximal T.
saturated chain C(f,T) exists and IC(f,T)1
= h + 1 will
be called the index of
(G,F) and, by axiom A2, h ~ 1. When TA is a finite (or infinite) affine space of dimension at least three, it is quite easy to prove that for any integer h, 1 .;;; h < dim 1A, the h-th h Grassmann space r (1A) is an affine Grassmann space of index h. In such a space the elements of F~(1A) (F~(TA)) are the proper (improper) pencils of h-dimensional subspaces of 1A, and Sl
E
Sl(TA) -- Sl is the family of an h-dimensional subs paces of J!. through a
fixed S2
E
S2(J!. )
(h-l)~dimensional
subspaces;
~ S2 is a complete family of pairwise parallel h-dimensional subspaces of TA ;
T E T(TA)
~
T is the family of all h-dimensional subspaces of 7A belonging to a fixed (h+1)-dimensional subspaces.
After thoroughly investigating the structure of an affine Grassmann space, it will be proved that such a space - under the assumption it has a finite index h - is isomorphic to the h-th Grassmann space of an affine space TA.
2. SOME PROPERTIES OF THE AFFINE GRASSMANN SPACES The notations introduced in section 1 will be used. LEMMA 1: If S E Sl' S'
E
S2' T, T'
E
T, T ~ T',are maximat subspaces, then
IS n S' I .;;; 1
(2.1)
I .;;; 1
(2.2)
IT n T'
PROOF: Assume Sand S' have in common two distinct points p and q; therefore, the line (p,q) would belong to S n S', a contradiction (see (iii)).
In a similar way
(2.2) is proved. LEMMA 2: Any maximat subspace T E T is partitioned by the tines of second kind belonging to it.
PROOF: Let r, s be two distinct lines of second kind belonging to T.
If r
n
s
~
0,
A. Bichara and F. Mazzocca
98
then the maximal subspaces of S through rand s, respectively, would belong to $2' be distinct and have one point in common, a contradiction.
On the other hand, by
A , through any point p in T there is at least one maximal subspaces 5p E $2; 3 thus, by (ii), 5 nTis a line of second kind through p which belongs to T. The p
statement follows. LEf-lMA 3: If S, 5' E S, T E T aPe thI'ee distinct maximal subspaces such that
S n 5'
= pEG, 5 n T = f E F, 5' n T = f' E F,
then T contains p and, theI'efoI'e,
f and f' meet ac p.
PROOF: By (iii), the lines f and f' are distinct, so it is possible to choose distinct points q and q' on f and f', and q, q'
~
p.
The points p, q, q', are
pairwise collinear and - by A, - there exists a maxima' subspaces M through them and M f S, 5'.
By (iii), M belongs to T and since the line (q,q') belongs both to
M and to T, M = T.
50 pET and the statement is proved.
PROP05ITION I: Any T'lc:.xi-rruL subspace T E T is a pI'ojective space with one point leleted.
PROOF: Consider the set T' = T U {pl, where p is an element not in T. Rl
{f E F, : f
~
=
{f'
{pi: f' E F2 and f'
T}, R = Rl
R2 ; by Lemma 2, (T' ,R) is a line space and Veblen axiom holds (see (iv), A , and Lemma 3). 3 Therefore, (T' ,R) is a projective space. 5ince any line in (T' ,R) through p has =
Tl, R2
5et
U
~
U
at least three distinct points, all elements in R have this property, i.e. (T' ,R) is irreducible. REMARK: By prop. I, three distinct non-collinear points in T belong to a plane contained in T.
TI
TI
will be a projective plane with one point deleted if it con-
tains a line of second kind, otherwise it will be a projective plane.
In what
follows punctured plane will stand for projective plane with one pOint deleted. PROPOS ITI ON II: Let T, T' E T be
p talk
a)
5;
maximal subspaces such that TnT'
= pEG •
be t!zl'ee pairuise distinct maximal subspaces in $ thI'ough p
.:et Si' i = 1,2,3 <:uer; tl:at Tn
tUJO
=
fi E F, and T' n
iff tiw lines f
i aI'e
5;
=
fi E F.
The lines fi aI'e cop lanaI' (on a
coplanal' (on a plane
a').
Under these assumptions,
99
On Grassmann spaces associated with an affine space
a is a punctured projective plane iff a' is.
PROOF: Assume f , f2' f3 are coplanar on a. Thus there exist both a line f of l the first kind meeting the lines f l , f2' f3 at the distinct points Pl' P2' P3 and a line f' of the first kind meeting fi and f2 at the distinct pOints pi and P2. If Sand S' are the unique elements in Sl containing f and f', then Sand S' intersect in a point q.
(Since Sl n S2
= p, S2 n S = P2' S n Sl = Pl' Sl' S2 and
S pairwise meet at distinct points; since S' belongs to Sl and intersects Sl and S2 in the points pi and P2' respectively, by (iv), it will meet S at a point q). Therefore, also S, S' and Sl pairwise meet at distinct points, S n S'
= q, S'ns l =
= pi, Sl n S = Pl. From (iv) it follows that S3 meets S' at a point t, since S3 is in Sl and S3 n S T' n S'
= P3 and S3 n Sl = p. Moreover, S3 n S' = t, T' n S3 = f~,
= f'; by Lemma 3, f3 and f' meet at t; therefore, f l , f2' f3 belong to a
plane a'
C
T'.
If
a
is a punctured plane, then there is a line f of second kind through p
and belonging to T.
Sn
Let
S be
the unique element in S2 containing
f;
T' = f',f' being a line of second kind through p contained in T'.
then, By the
previous argument, f' belongs to the plane a'; therefore, a' is a punctured plane. Let pEG; for every maximal subspace T E T through p and for every plane a in T which is incident with p, define 1 = {maximal subspaces S* in Sl S* n a is a line through pl. p,a Thus a non-proper family of subsets in Sl is obtained. L will denote the proper family associated with {1 p,a } and the incidence structure (Sl,L) (incidence being inclusion) will now be investigated. LEI-114ft. 4:
11 p,a I;;. 2;
(2.3)
11 p,a n1,,1;;'2=>1 p,a p,a
£
I
P ,a
I·
PROOF: Since a is irreducible (i.e. through any of its points there are two distinct lines of the first kind belonging to a). (2.3) follows from (ii). If S, S'
E
Sl' S f S' and S, S'
S n S'; thus. by Lemma 1. p = p'.
1p,a n 1p,a , " then p and p' belong to Therefore, when a = a', (2.4) is trivial. E
(2.4)
100
A. Bichara and F. Mazzocca
As sume
0.
"#
0.'.
If T and T' are the maximal subs paces in T containing
respectively, then T "# T'.
Otherwise, since
0.
and
0.'
0.
and
0.
,,
have at most one line in
common, at least one of the subspaces 5 and 5' would intersect
a U a'
in two
distinct lines so that it would have these two lines in common with T, a contradiction.
Consequently, TnT'
= (pl, and - prop. I - £p,a
£
p,a
,and the
statement is proved. As a corollary to Lemma 4 and axiom A2 , the following is true. PROP05ITION III: 1"he pail' (Sl,L) is a P5L and tlJo distinct coZZineul' in such space iff 5 n 5' = pEG.
e~ements
5, 5'
E
S are
MOr'eover', L can be partitioned in two
distinc& subsets Ll und L2 as follows. £
E
p,a
L, ,_
p,e<
E
Ll
is a pr'ojective plane;
<= 0.
E L, t E L2 => 0. is a punctured projective plane. p,o. p,a In (Sl,L) the Zines belonging to Ll (L ) will be called lines of the first 2
<:
(second) kind. LEMMA 5: Lel- 5 and 5' be
1;1,.'0
distinct coZlinear points in (Sl,L); the following
c1r'e equivalent: (a)
the Zine (5,5') is of second kind;
(B)
thel'e exist;; 5"
(v)
tlwough any point p
E
$2 meeting 5 and 5' at distirict points; E
5 \ 5' there is S E $2 such that S meets 5 and 5'
at distinct points.
PROOF: Firstly, the implications (8)
~
(a)
and (y)
~
(8)
are straightforward.
If
= £q,a , denote by T E T the maximal subspace in (G,F) containing a; note that in (G,F) both f = 5 n 0. and f' = 5' n a are lines of the first kind. There-
(5,5')
fore, if
E L , then a is a punctured projective plane, so it contains a line 2 q,a of the second kind f meeting f and f' at distinct points. Thus, the unique £
maximal subspace in T containing f meets 5 and 5' at distinct points and (a)
=>
(B).
Next, let 5" be a maximal subspace in S2 meeting 5 and 5' at the distinct points q' and q", respectively.
5ince q, q', q" are pairwise collinear and don't
belong to the same line, by Al and (iii), they belong to a maximal subspace T'
E
T and, therefore, to a punctured plane
a'
contained in T'.
Let p a point in
5 \ 5' and T" the unique maximal subspace in T containing the line (p,q); let
a"
On Grassmann spaces associated wi th an affine space
101
be the plane of Til defined by the intersecting lines (p,q) and Til n S'.
If
a' F aH, then, by prop. II, all is a punctured projective plane (since
is).
Ct '
Therefore, through p there is a line of the second kind which intersects the line Til n S'; thus, the unique maximal subspace in S2 containing this line meets Sand S' at distinct points.
So (s)
~
(y) and the statement is proved.
Let pEG; the set Sp consisting of all maximal subspaces in Sl containing p is obviously a subspace of (Sl,L).
For such a subspace the following proposition
holds. PROPOSITION IV: If T is a maximal subspace of T and pET, then the subspace S p of (Sl,L) is isomorphic to the star of lines of first kind through pin T, F(p,T). Since T is a projective space with one point deleted (see prop. I), the same is true for S ; thus, it has a finite dimension n iff T has a finite dimension n + 1. p
PROOF:
By (ii), every S E S, having the point p in common with T, intersects T in
a line of first kind. ~
: SE S
p
--~
The mapping S n T E F(p,T}
is a bijection by (iii); moreover, it is a collineation (recall the definition of
L). Proposition IV guarantees that any two maximal subspaces belonging to T and meeting at a point are isomorphic.
Since (G,F) is connected, the following is
true. PROPOSITION V: TWo maximal subspaces of (G,F) belonging to the family T are isomorphic.
Furthe1'/Tlore, (G,F) has a finite index h iff every maximal subspace
T E T has dimension n + 1.
3. ON THE SUBSPACES OF (Sl,L)
Since the PLS (Sl,L) is canonically associated with the affine Grassmann space (G,F), it is worth-while to investigate its subspaces. PROPOSITION VI: If a is a plane in T E T, then the set consisting of all elements in Sl intersecting
(l
in a line is a subspace of
(Sl' L).
102
A. Bichara and F. Mazzocca
PROOF: Let 5, 5' be elements in Sl such that 5 n T = f and f' being lines in (G,F).
~ a
and S' n T = f'
f
~ a,
By definition, f and f' are lines of the first kind;
thus, belonging to the plane
a,
they meet at the point p, i.e. 5 n 5' = p.
prop. III, Sand 5' are collinear in (Sl,L) and the line (5,5') = t elements in Sl each of them intersecting
a
p,a
By
consists of
in a line.
PROPOSITION VII: Any three distinct elements S, 5', 5" of Sl which are paiY'UJise C:J
LZ ine,_lr in (S I ' L) 'J",long to a subspace.
PROOF: If 5
n
assume 5 n 5'
5' n 5" =
= pEG, then
p, S' n 5"
= q, 5"
5, 5' 5" belong to the subspace Sp.
n S
=
t and p -I q F t F p.
Therefor~
Thus, p, q, tare
three pairwise collinear points in G; hence - by Al and (iii) - they belong to a unique maximal subspace T E T of (G,F).
So the plane
a
in T through p, q, and t
defines the subspace of (SI,L) consisting of all elements in Sl intersecting a line (see prop. VI).
a
in
Since 5,5', and 5" belong to this subspace, the statement
is proved. PROPOSITION VIII: 0et p be a point of G and S an element of S not through p. set of aU elementsi'l Sl tin 0ugh p and meeting 5 either is the nun set 7
01'
The is a
line in (Sl,L).
PROOF: If Sl is an element in Sl through p and meeting 5 at a point q, then p and q are collinear in (G,F).
If T is the unique maximal subspace in T containing the
line (p,q), then - by (ii) - Tn S they form a plane
=
f E F.
Since the lines (p,q) and f meet,
in T and, therefore, the linet of (Sl,L). This line p,a consists of elements in Sl each of which is incident with S (since it is incident cY.
with f).
Now, let S2 be an element in Sl through p and meeting S at a point
q' F q.
The points p, q, q', being pairwise collinear in (G,F), belong to a
unique maximal subspace T' in T (by Al and (iii)).
So, (p,q)
~
TnT', and - by
Lemma 1 - T = T' and q' E f, and the proof is complete. PROPOSITION IX: For any pEG, if the index of (G,F) is not h = 1, then S is a p
maximal subspace in (Sl,L).
Fu:rtherrnore, T' = {Sp : pEG} is a proper family and
any we, :Ustinct elements in T' have at most one point of (5 ,L) in common. 1
On Grassmann spaces associated with an affine space
103
PROOF: If the index of (G,F) is not one, then every maximal subspace T E T has a dimension greater than two; therefore, for any pEG, the subspace Sp of (Sl,L) is not a line. p
~
By prop. VIII, Sp is a maximal subspace in (Sl,L).
If p, q E G and
q, then Sp E Sq consists of the maximal subspaces in Sl through both p and q.
Therefore, S n S P
q
= 0,
unless p - q and (p,q) E F in which case
Is P n
S
q
I = 1.
LEMMA 6: Let S, S' E Sl be two collinear points of (Sl,L). i.e. S n S' = pEG. Let Sl' S2 be two distinct elements in Sl both meeting Sand S' at distinct points. Under these hypotheses. Sl and S2 are collinear in (Sl,L) and any S3 on the line
(Sl,S2) either meets Sand S' at distinct points or belongs to the line (S,S'). PROOF: By (iv), Sl n S2 = q E G; so, Sl and S2 are collinear in (Sl,L).
The point
q,"being distinct from p, belongs either to S or to S' or to G \ (S US').
If
q E S, then - by prop. VIII - every element on the line (Sl,S2) meets S' at a point distinct from q.
A similar argument holds when q E S'.
again by prop. VIII - any S3 E (Sl,S2) meets both Sand S'.
If q
~
SUS' -
If there exists
S3 E (Sl,S2) through p, then S3 lies on the line consisting of the elements in Sl through p and incident with Sl and this line is (S,S'). For all S, S' E Sl such that S n S' o(S,S')
* = {S E Sl
f:
= pEG, set:
: either S E (S,S') or S meets Sand S' at distinct pOints}.
From prop. IX and Lemma 6 it follows that o(S,S') is a subspace of (Sl,L) and that the line (S,S') belongs to it. PROPOSITION X: Let S, S' E Sl be distinct points which are colZinear in (Sl,L) • i.e. S n S'
= pEG.
Then any subspace containing the Zine (S,S') is contained
either in S or in o(S,S'). Furthermore. S n o(S,S') = (S,S'), o(S,S') is a p p maximaZ subspace of (Sl,L) and for any distinct Sl' S2 E o(S,S'), 0(S,5') =
0(Sl'S2) • PROOF: If H is a subspace of (Sl,L) containing (5,S'), then any element in H meets both 5 and 5'; therefore, if H -~ Sp , there exists at least one S3 E H not through p meeting Sand 5' at the distinct points q and q', respectively.
By prop. VIII,
the elements in Sp collinear with S3 are precisely the ones belonging to the line (5,5'); thus, Sp n H = (5,S'). Consequently, each element in H, not belonging to
A. Bichara and F. Mazzocca
104
the line (S,S') doesn't contain p; since it intersects both Sand S', the intersection points are distinct. subspace in (Sl,L),
Therefore, H ~ 0(5,5') and 0(5,5') is a maximal
Furthermore, if Sl' S2
maxima1ity of such subspaces, o(S,S')
~
E
o(S,S') and Sl f S2' because of the
0(Sl,S2)
~
o(S,S'); it follows
a(S,S') = a(S, ,S2)' In the sequel, S' will denote the family of maximal subs paces in (Sl,L) of type o(S,S'}, where 5, 5' are distinct collinear points of Sl' LEMMA 7: ~et S, S' E Sl' S f S', be two collinear points of (Sl,L). "1,'et;:- Sand S' at distinct points, then there exists
5'
S2'
E
5'
n
If
5E
S = 0,
52
meeting
S anJ S' at ,Hstinct points,
PROOF: The points p, q
= S n 5,
and q'
= S'
n
5 are
pairwise collinear and don't
lie on the same line; therefore, they belong to a plane subspace T E T through them (see Al and (iii)),
Since
in the unique maximal
a a
contains a line of second
kind, nam1ey (q,q'), it is a punctured projective plane. q1
E
Thus, through any point
(p,q), ql ~ p, q, there is a line of second kind meeting (p,q), and so
a point q' F p, q.
5,
at
Each of these lines belongs to a unique element of S2 which
has the required properties. PROPOSITION XI: ,:et S, S'
E
Sl' S F S', be two distinct collinear points in
If the line (S,S') belongs to L , then also all the lines in the subspace 2 o(S,S') belong to L2 , (Sl,L),
PROOF: If the line (S,S') is of second kind, then - by Lemma 5 - there exists
5 E S2 incident with Sand S', respectively, at the distinct points a and b. Let (5 ,5 ) be a line other than (S,S') contained in o(S,S') and set q = Sl n S2' By 1 2 Lemma 5, we may assume q doesn't belong to 5. Therefore, one of the following holds: q E S, q E S', q intersect
5;
~
SuS'.
By (ii), in any of these cases both Sl and S2
thus, by Lemma 5, the line (Sl,S2) is of second kind.
From prop. XI it follows that the maximal subs paces of the family S' can be partitioned in two disjoint subfamilies Si and S2 according to their lines being of the first kind or of the second one. PROPOSITION XII: If (G,F) has not index one, then the
PLS
(Sl,L), together with
105
On Grassmann spaces associated with an affine space
the three families
Si. SZ'
and T' of maximal subspaces. satisfies axioms A , A •
l
3
the first part of axiom A • and properties (i). (ii) and (iii), 2
PROOF: Axiom A , the first part of A , (ii) and (iii) follow from propositions 2 l VII, IX, X, XI. If a and a' are two maximal subspaces of
S2
a = a(S,S') and a' = a(S,S"), where S', S" E Sl'
intersecting in S E Sl ' then 5et 5 n S' = pEG and
5 n S" = q E G; by Lemma 5, two maximal subspaces 5, 5 E S2 can be found such that
5n
= a E G, 5' n S = 5 n S = bEG, 5 n 5' = c; moreover, the points a,
5"
b, c can be assumed to be distinct from p and q. Now, if P X, a(S,S')
=
By (i), since
5 n 5'
F 0,5 = 5'.
q, then S" belongs to the line (S,5') E a(5,S'); therefore, by prop.
= a(S,S"),
If p F q, then - by (iv) - IS" n S'I
S" E a(S,S'), i.e. a(S,S')
=
a(S,S").
=
1.
It follows that
Thus, (Sl,L) satisfies (i).
Next, we prove axiom A3 • Let S be an element of Sl' pES and T a maximal subspace of the family T which contains p. Let f be the line S n T and a the unique punctured projective plane in T through f.
The line ~ p,a of (Sl,L) is of second kind; therefore, it lies in a subspace belonging to the family S2' Such a subspace, obviously, contains 5 and A3 follows. Under the hypotheses in prop. XII, the next Lemmas will now be proved. LEMt1A 8: Let
a. a'. a" be three distinct elements in
in distinct elements of Sl'
Set 5
exists a point pEG such that p
= a' nail.
5'
S
I
= a" n
which pairwise intersect a. S"
=a n
a'; then there
5 n S' n S", i.e. S n S' n S" E S p'
PROOF: Since S, S', S" are pairwise collinear in (Sl,L), there exists a subspace
H through them (see prop. XII and axiom Al ). The maximal subspace containing H intersects a, a' and a" in lines; therefore, it doesn't belong to S' (see prop. XII and Lemma 1), so it is an element of T'.
Consequently, there exists a point
pEG such that 5, 5' 5" belong to Sp and this proves the statement. LEMMA 9: Let a. a'. a" be three distinct elements of S I which pairwise meet at
distinct elements of Sl'
If; belongs to S I and intersects a' and a" in distinct
points. then ; and a either have one element of Sl in common or they both belong to S' • 2
106
A. Bichara and F. Mazzocca
PROOF:
= G' n u", 5' = uno", 5" = 0 n 0',
Set 5
by Lemma 8, there exist two points p, q
point q'
E
~
;no"andS"
G such that 5, S', S"
E
5, 5', S" E Sq; therefore, either p ~ q or p When p
~,
E
= q.
;no';
S , P
q, since S' and S' belong to 0", they are collinear and have a
G in common.
Analogously, a point q"
E
G exists such that q"
=
S n 5".
The points p, q,q' are pairwise collinear in (G,F) and there exists a subspace T E T through them. thus, being TnT'
r~oreover,
~
there exists a subspace T'
Ip,q}, with p
are collinear and the line f
E
q, T
~
E
T
containing p, q, q";
= T'. Hence, the points q' and q" of T
F through them is contained in precisely one maximru
Fl , then 5 E Sl' ~ E 0(5',5") = 0, S E 0(5',5") =;; thus, ~ = 0 n~. If f E F , then 5 E S2 and the lines (S',S"), (5',5") belong to S2; 2 therefore, both 0(5' ,SO) = a and o(~' ,~") = ; belong to S~. subspace 5 E S.
= q,
When p J"
If f
E
S, 5', S", 5',~"
n S :) {S,~',S'}.
E
~
and so o'ns p ~ IS,5",S"} and
From prop. XII and (ii) it follows that 5
p -
E
(S',~')n
(S",5").
Thus, the subspace of S p spanned by S, S', So, ~', 5" has dimension two, i.e. it is a projective plane, possibly with one point deleted (see prop. IV).
Therefore,
either the two lines (S',S") and (5,S") on this plane have a common point S, or they both belong to L2 . The statement follows. LEMMA 10: Let a = a(S' ,SOl) and 0"
=
o(S' ,S') crad
both beiong to
~(S"
0'
a(S',5") be wo distinct elements of S~.
,S") exist, then either' they shar'e an element of Sl Or' they
S~.
PROOF: From the hypotheses it follows p =
S2
=
S' n SOl, q
SOl n SOl, p, q, p', q' being points of G. E
If
=
S' n 5", p'
S' n S', q'
=
t10reover (by Lemma 5), both an element
52 through p and incident with S.. and an element 52
dent wi th SOl exi s t.
=
E
S2 through p' and i nci-
By prop. XII and (1), then S2 = ~2 and either q'
E
S2 or
q' 'I S2.
When the former holds, the three points p, p', q' are pairwise collinear and belong to a maximal subspace T E T of (G,F).
If T' is the subspace of T through
p', q' and q, which in turn are pairwise collinear, then TnT' ? {p',q'}.
There-
fore, T = T' (see Lemma 1). Thus the points p and q belong to T and the line (p,q)
=
then S E
f
E
F exists.
0 (S '
If f
E
Fl and S is the maximal subspace of Sl through it,
,~) = at! and 5 E 0(5",5) = a.
If, on the other hand, f
E
F , 2
the
lines (5',5) and (5" ,5") both belong to L2 ; therefore, both a" and ;; be long to S' 2"
107
On Grassmann spaces associated with an affine space
When q' , S2' setting a = 5" n 52' b = ~ n 52' the three points a, b, q' of G are pairwise collinear and belong to a subspace T E T of (G,F).
If
c E (a,b), c F a, b, and dE (a,q'), d F a, q', then the points c, d are collinear and the line (c,d) lies in a maximal subspace 5 of S. with both S" and 5' and, therefore, with ~'. 5 E a(5',S) n a(5",S").
It
On the other hand, 5 E
L2 and so both a" and a belong to
By (iv), 5 must be incident
follows that if 5 E Sl' then
S2
implies (5',5'), (5",5") are in
S2'
By lemmas 8, 9, 10, under the assumptions of prop. XII, the PL5 (Sl,L) satisfies (iv). LEMt·1A 11: Let 5 and 5' be two maximal subspaces of (G,F) belonging to the family If P E 5 and q E 5' ape any two collineaP points such that (p,q) E F , then
Sl'
2
thepe exists a point q' E 5' which is coUineap with p and (p,q') E Fl' quently, if 5, S', S"
S, 5' E
5" E
Sl'
such that 15 n
ape three distinct elements of S with
IS n s"l
S2'
Conse-
51 = 1 and IS
=
1 and Is" n 5' I = 1, then thepe exists 5 E
Sl
= 1,
n S'I
PROOF: The line (p,q) E F2 belongs to a maximal subspace T E T which intersects S' in a line f through q,
If q' E f and q' F q, then p and q' are collinear and
the line (p,q') belongs to F1, since it cannot be kong to F2 by prop. LEMMA 12: If S and S' aPe any two maximal subspaces of (G,F) belonging to the family Sl' then there exists a finite sequence ll' 12"",ln_1 such that Ii, n li+11
= 1, S Eli
and ' E In_l'
of lines of L
Thepefore, (Sl,L) is a connected
space.
PROOF: If S n 5' = pEG, then Sand 5' are collinear in (Sl,L) and the statement follows with n
=
1. Assume S n S'
S', respectively.
= ~
and let p and q be distinct points in Sand
Since (G,F) is connected, the points p and q of G are joined by
a polygonal path {f 1,f 2, •.. ,f n }, fi E F, i = 1, 2, , •• , n, and each fi is contained in a maximal subspace S.1 E S. Clearly, 15.1 n S.1 +1 I ~ 1 and w.1.0.g. it may be assumed that i F j implies S. F S .• J
1
It follows that IS. n S. 11 1+
1
subsequent elements of the sequence {Si} both belong to
S2
= 1 and no two
(see (i)).
By Lemma
11, it is possible to find a sequence {Si} of elements in Sl with S'1
=
S' n S'
=
S' and
IS~ n S~ 1 1+1
I
= 1.
The elements
S~
1
and
S~ 1 1+
are collinear in
\08
A. Bichara and F. Mazzocca
= (S~, 1
(Sl,L) and setting t.
1
S~
1+
1)' the statement follows.
As a consequence of prop. XII and lemmas 8 - 12, we have the following result. PROPOSITION XIII: If (G,F) has index at least two, then the PLS (Sl,L), together
with the thloee fOJ71ilies S', S', T' of maximal subspaces, is an affine Grassmann space.
4. A CHARACTERIZATION OF THE GRASSt·1ANN SPACES ASSOCIATED WITH AN AFFINE
SPACE
In this section the space (G,F) will be assumed to be of finite index h. For any integer t, 1
~
t
h - 1, (G(t), F(t)) will denote the affine Grassmann
~
space obtained dOing h - t times the construction which in the previous section In this notation (Sl,L) = (G(h-l), F(h-1)) and for any
lead to the space (Sl,L). t
the finite index of (G(t), F(t)) is precisely t (see prop. IV and XII).
The
families of maximal subspaces of (G(t), F(t)) which are mentioned in axiom A2 will be denoted by S(t)
= Sl(t)
u
S2(t) and T(t).
Let (G,F) and (G' ,F') be two affine Grassmann space with F F' = Fi
2.
F
U
(G,F) and (G' ,F') is a bijection mapPlng
~
-1
u F2 and
As usual, Sl' S2' T and Si' S2' T' will denote the families of
maximal subspaces of (G,F) and (G' ,F'), respectively. .
= F]
A co11ineation
~
between
: G ~ G' which, together with its inverse
~
,maps pairs and triples of collinear pOints on lines of the first
(second) kind onto pairs and triples of collinear points on lines of the first (second) kind. onto
F~
1
(i
(G' ,F'). kind,
It is straightforward to check that any co11ineation
~
maps F;
= 1,2) and maximal subspaces of (G,F) onto maximal subspaces of Since the subs paces of thr family S2 contain just lines of the second
~(S2) =
S2; thus, by (iii),
~(Sl) =
s;
and
~(T) =
T'.
The next statement
then follows. PROPOSITION XIV: Any collineation between
two
affine Grassmann spaces is an iso-
rnorphisr1.
LEt-1t·1A 13: In (G(l), F(l)), if S, S', S" are pairuise distinct elements of S such
that S E Sl(l) and
15
n
5'1
=
IS'
n
s"l
=
1, then
Is
n
S"I
=
1.
109
On Grassmann spaces associated with an affine space
PROOF: 5et 5 n 5'
=
p E G(l) and 5' n 5"
=
q E G(l); then p - q and the unique ele-
ment T E T(l) containing the line (p,q) is a punctured projective plane (see prop. V).
This plane T meets 5 and 5" in the lines fl and f2' respectively.
5ince the
line fl is contained in S E $1(1). it belongs to Fl(l); therefore, fl and f2 have a point p', on the plane T, in common, i.e. S n S" LEt~MA
14: In (G(l), F(l)), if 5 E Sl(l), 5'
PROOF: If P E 5 and q
E
E
= p'.
S(l) and 5
~
5', then 15
E
5'1
=
1.
S', (G(l). F(l)) being connected. then it is possible to
find a polygonal path {f .f ••••• f } consisting of lines in F(l) such that l 2 n p E fl' q E f and If. n f. 11 = 1. i = 1.2 ..... n-1. For all i = 1.2 ..... n-l.
,'+
n
set f., n f.1+ 1 = q.; let 5., be the maximal subspace of S(l) through f,.• W.l.o.g. , it may be assumed S., ~ S.J when i ~ j; thus, S., n S.1+ 1 S 1 = 5'. the statement follows from Lemma 13.
= q,..
Writing So
=5
and
n+
From Lemma 14, taking into account [1], the next result follows. PROP05ITION XV: The Grassmann affine spaae (G(l), F(l)) is isomorphia to the first 1
Grassmann spaae r (1A) of an affine space 1A.
PROPOSITION XVI: Let TA be an affine spaae suah that r\1A) and (G(l), F(l)) are isomorphia.
If for t> 1 rt-l(TA) is isomorphic to (G(t-l). F(t-l)). r\TA) is
isomorphia to (G(t). F(t)).
PROOF: Recall that Sl(t)
= G(t-l) and let
be the bijection which maps any point
p E G(t) onto the element in T(t-l) consisting of all those elements in Sl(t) which contain p. Next. let w be the bijection from Tt - l (IA) onto Gt(IA) which maps the collection of (t-l)-dimensional subspaces of IA all of which are contained in a t-dimensional subspace TAt onto IAt • Finally, for any isomorphism Q) between (G(t-l). F(t-l)) and (G t-l (IA), Ft-l (IA)), write lji = Q) w. By construction,
lji
is a bijection.
Let P. q, m be three points of G(t). two of them at least being distinct. on a line f of the first kind.
Let Sl and T be the unique elements of Sl(t) and
T(t), respectively, through f.
Denote by a a projective plane through f contained
in T.
Since t> 1. such a plane exists.
Under these hypotheses,
A. Bichara and F. Mazzocca
110
n tq,a n t m,a = Sl and, by prop. VI, the three lines of p n Sq n Sm = Sl' tp,a the first kind i ,{ , t in (G(t-l), F(t-l)) belong to a subspace ~ which. t-l t-l p,a p,a m,a actually. turns out to be a projective plane. Thus, in (G (IA),F (IA)), t-l (IA) containing the point qJ(Sl). {,IS ), {I(S ), and qJ(S ) are elements of T p q m Moreover, ~(l p,a ), ~(tq,a ) and ,(fm,a ) are three lines of the first kind through
S
qJ(Sl)' contained in projective plane.
) and .(Sm) respectively, and belonging to a It follows that ,(S p ), ,(S q ) and qJ{S m) are contained in a ~(S
p
),
~(S
q
(t+l)-dimensional subspace of IA; therefore, ljJ(p)
= (qJw)(S),
p
ljJ(q)
= (qJw)(S),
q
tP(m) = «(llw)(S ) are three points of (Gt(IA), Ft(IA)) on a line of the first kind. m Since the same is true for w- 1 , both IjJ and w- l preserve pairs and triples of
points on lines of the first kind.
A similar argument proves that the same is
true for pairs and triples of points on lines of the second kind.
Therefore,
IjJ
is
a collineation and from prop. XIV the statement follows. Finally, an induction argument on the sequence {(G(t), F(t)) : t
= 1,2, ...• h}
of affine Grassmann spaces, taking into account prop. XV and XVI, yields our main result, as follows. THEOREI4: Any affine Gmssmann space of finite index h is isomorphic to the h-th Grassmann space of an affine space.
ACKNOWLEDGEI1ENT: Work supported in part by the research group G.N.S.A.G.A. of C.N.R ••
BI BLI OGRAPHY
1.
2. 3.
4. 5.
A. Bichara and F. I·\azzocca, On a characterization of Grassmann space representing the lines in an affine space, to appear, Simon Stevin, Belgium. A. Bichara and G. Ta11ini, On a characterization of the Grassmann manifold representing the planes in a projectiVe space, to appear. A. Bichara and G. Ta11ini, On a characterization of the Grassmann space representing the h-dimensiona1 subspaces in a projective space, Froc. of Inter. Conf. on Combinatorial Geometries and their Appl., Rome, 1981. P. Oembowski, Finite geometries, Springer, Berlin, 1968. B. Segre, Lectures on modern geometry, Cremonese, Roma, 1961.
On Grassmann spaces associated with an affine space
6. 7. 8. 9.
10.
111
B. Segre, Istituzioni di geometY'ia supeY'ioY'e J voL I, I I, I I I, 1st. l1at. Univ. Roma, 1965. P.t·1. Lo Re and D. Olanda, Grassmann spaces, to appear, Journal of Geometry. G. Tallini, Graphic characterization of algebraic varieties in a Galois space, Atti dei Convegni Lincei J Colloquio internazionale sulle teorie combinatorie, Torno II, 153-166, Roma, 1976. G. Tallini, On a characterization of the Grassmann manifold representing the lines in a projective space, Proc. of the Second IsZe of Thorns ConfeY'ence 1980, London Math. Soc. Lecture Notes, series 49, Cambridge University Press, 1981. G. Tallini, Spazi parziali di Y'ette spazi poZaY'i. Geometrie subimmerse J Quaderni sem. geom. comb., 1st. mat. Univ. Roma, 14, 1979. J
Istituto t-1atematico "G. Castelnuovo" Citta Universitaria 00185 Roma Italy
Istituto di ~latematica Faco1ta di Ingegneria Universita di Napoli Via Claudio 21 80125 Napoli Italy
This Page Intentionally Left Blank
113
Annals of Discrete Mathematics 18 (1983) 113-132 © North-Holland Publishing Company
ON A CHARACTERIZATION OF GRASSMANN SPACE REPRESENTING THE h-DH1ENSIONAL SUBSPACES IN A PROJECTIVE SPACE A. Bichari' and G. Tall ini
ABSTRACT The Grassmann space of index h associated with a projective space will be defined and characterized.
1. INTRODUCTION Let G be a non-empty set whose elements will be called points and let F be a proper (non-empty) collection of subsets of G, which will be called lines.
The
pair (G,F) is said to be a partial line space (PLS) if the following hold: (i) (ii) (iii)
fl{Ff=G, f, f' f
E
F
E
F ;>
;>
Ifnf'l';;;l,
If I .. 2.
Two distinct points p and q of a PLS (G,F) are said to be collinear, p - q, if the line (p,q) through them exists in F.
If this is not the case, p and q will
be called non-collinear, p f q. A PLS (G,F) is a proper PLS (PPLS) if two non-collinear points exist in G. When any two points of G are collinear, (G,F) is a line space. A PPLS is connected if given any two distinct points p, q in G, a polygonal path joining them exists in (G,F). A PLS is irreducible if any line has at least three points. A subset H of a PLS is a subspace if any two of its points are collinear and the line through them belongs to H. A subspace H of (G,F) is a maximal subspace ~if
no subspace exists in (G,F) which properly contains it. Let IP be a projective space of dimension at least three; denote by Gh(IP)
114
A. Bichara and G. Tallini
the collection of h-dimensional subspaces of 11' and by Fh(JP) the family of pencils of h-dimensional subspaces of JP.
(Such a pencil consists of all elements in
Gh(IP) through an (h-l)-dimensional subspace of 11' and belonging to an (h+l)-dimensional subspace of IP).
The pair Th(JP) = (Gh(IP), Fh(JP)) will be called the
h-th Grassmann space associated with the projective space IP.
Of course, when IP
is finitely generated and coordinatized over a field, then rh(F) is isomorphic to Grassmann's manifold representing the h-dimensional subspaces of JP. PLS which turns out to be both proper and connected. spaces.
rh(JP) is a
rh(JP) contains maximal sub-
Indeed, the collect ion of all h-dimensi ona 1 subspaces of 11' through an
(h-l)-dimensional subspace of 11' is a maximal subspace in rh(JP).
Denote by
Sh ( IP) the family of a 11 such maxima 1 subs paces of rh (JP ) • Finally, for any (h+ 1 i-dimensional subspace of JP, the collection of all its h-dimensional subspaces is a maximal subspace in rh(JP).
The family of such
maximal subspaces will be denoted by rh(JP). It is easy to prove that the pair rh(F)
(Gh(JP), Fh(F)) is a connected
PPLS for which the following hold:
A • If three points in Gh(F) are pairwise collinear then a subspace of l rh(JP) containing them exists.
A . No line of Fh(JP) is a maximal subspace. There exist two families of 2 maximal subspaces in rh(JP), namely Sh(JP) and rh(JP) and any maximal subspace in rh(IP) belongs to either Sh(JP) or rh(JP).
1.
II.
SESh(p),TE
hr(JP)
Furthermore:
> eithersnT=~orsnTEFh(IP);
If f E Fh(JP), 3! S E Sh(IP), 3! T E rh(lP) such that f
~
S, f
C
T.
In.
Let S, S', and SOl be any three elements of Sh(lP) pairwise meeting at h distinct points. Then any element of S (JP) other than SOl meeting Sand S' at distinct points meets S".
A • There exists a complete finite chain of subs paces such that: its length 3 is h+l; its minimum (minimal element) is a line f E Fh(JP); its maximum is a subh space T E r (IP) (and T ~ f).
On a characterization of Grassmann spaces
Let (G,F) be a connected PPL5.
115
It will be said to be a projective Grassmann
space of finite index h if the following axioms hold.
A • If three points in G are pairwise collinear, then a subspace of (G,F) l containing them exists. A2. No line in F is a maximal subspace. There exist two collections of maximal subspaces in (G,F), say Sand T, and any maximal subspace of (G,F) belongs to either S or T.
Furthermore:
I. 5 E S, T E T
> either 5 n T = 0 or 5 n T E F;
II. V f E F, 3! 5 E S, 3! T E T : f
~
5, f
~
T;
III. If 5, 5', 5" E S pairwise meet at distinct points, then any element in S
other than 5" meeting 5 and 5' at distinct points meets 5". A3 • There exist a line f E F and aTE T, with f ~ T, and a complete, finite chain of subspaces of (G,F) such that its length is h+l, its minimum is f and its maximum is
T.
Let (G,F) and (G',F') be two projective Grassmann spaces having finite index
hand h', respectively.
Then the collections Sand T (S' and T') of
maximal subs paces in (G,F) ((G',F')) are defined.
Any mapping F : G --~ G' which
is one-to-one and onto will be called an isomorphism between (G,F) and (G' ,F') if the following hold: (a) F' is the set of all images under F of the elements in F',
s' is the set of all images under F of the elements in S; (y) T' is the set of all images under F of the elements in T. (s)
In this paper the following result will be proved (see sect. 8).
THEOREM: If (G,F) is a connected projective Grassmann space of finite index h, then a projective space IP exists and (G,F) is isomorphic to the h-th Grassmann space (Gh(IP), Fh(!p)) associated with !P.
116
A. Bichara and G. Tallini
2. SOME PROPERTIES OF A PROJECTIVE
SPACE HAVING A FINITE INDEX
GRASS~~NN
h ;;. 2. Let (G,F) be a connected PPLS satisfying axioms A , A2, and A3 in sect. 1. 1 Then: PROPOSITION I: ::Ohe cdlections Sand T ape skew. i.1.1G
Let
t~
and W (with t·l
tl') be
~
ma:l.'ima: subspacec having two distinct common points p and p'; then M and M'
belong t:o d.iffe-pent coUections of maximal subspaces.
Thus, two distinct maximal
aubsracC!s }xdonging to the same coUection have at most one common point.
PROOF: I f a max i ma 1 subspace tr' were conta i ned inS nT, then M" n rl"
W; a
contradiction to A I, as f.1" properly contains a line. 2 Since p and p' belong to t~, they are collinear; let f be the line joining them; obviously, f
~
Mand f
~
tl', the two maximal subspaces belong to different
collections and the statement is proved. PROPOSITION II: :"et Sand S be tUJO distinct m=imal subspaces in S, having a I
If T is a maxirrul subspace in T, meeting Sand S' at the lines f
common ?oint p.
and f', respectively, then f and f' meet at p, so T passes through p.
PROOF: f and f' are distinct lines (otherwise, two distinct maximal subs paces S and S' in S would meet at the line f = f', which is impossible by prop. I). any line in F has at least two points, there exist points p' that p, p' and p. are pairwise distinct. (indeed, p, p'
E
S', and p', p"
f and p"
E
f' such
These points are pairwise collinear
T); thus, by A , there exists a sub1 space N through them and N is contained in a maximal subspace M. If M belonged to E
S, p, p"
E
Since
E
S, then 11 = S (by prop. I, H n S' :: {p,p'} implies
(since 11 n S' :: {p,p"}).
~1
= S); similarly,
Therefore, if 11 belonged to S, then S
=
I~
=
S'
S' a contradic-
tion.
Hence '·1 f/. S and, by A , 11 E 1. Since t~ belongs to T and contains the 2 distinct points p' and p" in T E T, by prop. I,ll = T. Moreover, p E 11 and M T imply pET.
The point p in T belongs to S; thus, {p}
~
Tn S
=
f and p E f;
similarly, p E f'; therefore, the distinct lines f and f' meet at the point p and the statement is proved.
117
On a characterization of Grassmann spaces
PROPOSITION III: Any Tin T is a projective space. PROOF: It is enough to prove that in the line space T Veblen-Wedderburn axiom holds: Let fl and f2 be two lines in T meeting at the points P3; if f3 and f4 are distinct lines in T, each of them meeting both fl and f2 at points distinct from P3' then f3 and f4 meet at a point. Set {Pl}
=
f2 n f3' {P2}
=
fl n f3' {ql}
= fl
n f4' {q2}
f2 n f 4; then
(2.1) Through fi (i = 1, ••. ,4) there is exactly one maximal subspaces Sl
S (see A2II). Such maximal subspaces are pairwise distinct (if i 1 j and S. = S., then the J
1
maximal subspace S. would share with T E T the points in f. 1
1
U
E
f., which is J
impossible by A2I). Now, fi C Si' i = 1,2 and {P3} = fl n f2 imply {P3}':: Sl n S2'. since Sl 1 S2' by prop. I, {P3} = Sl n S2' By the same argument, {Pl}· = S2 n S3 and {P2}
= Sl
tively.
The three maximal subspaces Sl' S2' and S3 pairwise meet at distinct
n S3; moreover, S4 meets Sl and S2 at the points ql and q2' respec-
points (see (2.1)) and S4 meets Sl
and S2 at distinct points; therefore, by
A2III, S4 meets S3 at a point q: {q}
= S3
n S4'
lines f3 and f4' respectively, and S3 n S4
Since T meets S3 and S4 at the
= {q}, by prop. II, f3 and f4 meet at
the point q and the statement is proved. The projective space which are members of T, contain projective planes forming a collection
IT
of subsets of
Clearly, any element in
G.
IT
is a subspace
in (G,F) which is contained in a maximal subspace belonging to T. PROPOSITION IV: Let T and T' be two elements in T through a point p in G. S1'
S2~
Assume
and S3 are three pairwise distinct elements in S through p and such that
Tn S.1 = f.1
E
F, and T' n S.1 = f~1
E
F.
same plane a in T, then also the lines
If the lines f., i 1
f~ 1
PROOF: Let a' be the plane in T' through
= 1,2,3,
belong to the
belong to a unique plane in T'.
fi
and
fZ'
It is enough the prove that
f3 belongs to a'. Let f and f' .be two lines not through p, the former in a, the latter in
,
a •
Since a and a' are projective planes, f meets f , f2 and f , and f' meets fl and l 3
118
A. Bichara and G. Tallini
f '. 5et {q.} = f () f., i = 1,2,3, and {q~} = f' () f~, j = 1,2. It is easy to 2 1 1 1 1 check that the five points q., q~ are pairwise distinct; moreover, if 5 and 5' are 1
1
the maxima1 subspaces in S through f and f', respective1y, then the five subspaces 5,5',5 ,5 2,5 are pairwise distinct and {qi } = 5 () Si' {qj} = 5' () 5j" 5ince 51' 1 3 52' and 53 pairwise meet at distinct points and 5' (f S) meets Sl and 52 at distinct points, Sand 5' have a common point q, which is obviously distinct from
q;.
ql and
Thus, the three maximal subspaces S, 51 and 5' pairwise meet at dis-
tinct points and S3 (f S') meets 5 and Sl at the distinct points q3 and p (respectively).
Therefore, S3 and S' have a common point q'.
f3' S' nT' = f' and and f3 are coplanar. thus, it is
(I'
{q'}
Since S3 () T' =
= S3 n S', by prop. II, f' and f3 meet at q.
Hence, f'
The plane through them contains f' and the point p on f3;
and f3 belongs to a'.
3. THE PARTIAL LINE 5PACE (S,R) ASSOCIATED WITH (G,F) Take pEG, a E IT, with pEa; then the subset r
of S, consisting of p,a those maximal subspaces in S meeting a at lines in F through p, i.e. r
p,a
={SES:S3p
and
snaEF},
is uniquely defined. PROP05ITION V: If
a,
a'E
IT
and pEa, p' E a', then
Ir p,a I;;.
(3.1) (3.2)
Ir p,a
PROOF: Since
a
2;
r, ,1;;.2=>r p,a p ,a
r, p
,a
"
is a projective plane, through the point p in
two distinct lines fl and f2 of F.
a
there are at least
The maximal subspace 51 and 52 in S through
fl and f2 (respectively) belong to r p,a and now it will be shown that Sl f 52' If 51 = 52' then this member of S would share with a maximal subspace T in T a set I containing the distinct lines fl and f2' which is impossible by A I.
2
Therefore,
E rp,a and Sl f $2; (3.1) follows. Let $1 and 52 be two distinct elements inr p,a nr, p ,a " SlandS 2 meet at lines in F through p and (I' at lines in F through p'; thus p and p' belong to
Sl' S2
slnsZ'
Byprop.I,p=p'.
119
On a characterization of Grassmann spaces
When a = a
,,
(3.2) is obviously true. Assume a f a'.
Let T and T' be the
maximal subspaces in T through a and a' respectively; then T f T' (otherwise, Sl would meet T at two lines in F through p, one belonging to a, the other to a', which contradicts A2I) and TnT' = {p}. Furthermore, by prop. IV, any element S3 in rp,a' distinct from Sl and S2' meets T' at a line in F, which - being coplanar with Sl nT' and S2 n T' - belongs to r p,a ,; hence, r p,a -C r, p ,a " the same argument, r, (3.2) follows. p ,a ,C - rand p,a
By
Thus, the collection {r p,a : a E n, pEa} of subsets of S is defined; since it is not a proper collection (see prop. V), let R be the proper collection associated with it. PROPOSITION VI: The pair (S,R) is a PLS.
Moreover, two distinct elements in S are
collinear in (S,R) iff they have a common point in G.
PROOF: Let Sand S' be two distinct elements in S.
If S n S' = {p}, then let f be
a line in S through p; through f there is a maximal subspace T E T meeting S' at one point al least; thus, it meets S' at a line f' E F.
In the projective space
T, the distinct lines f and f' are joined by a plane a E IT; therefore, Sand S' belong to r p,a If S n 5' = ~, no element in R through 5 and S' exists. Let f be a line in S
E
S.
Through f there is a maximal subspace T E T.
P E f, let f' be a -line in T through p, distinct from f.
If
Through f' there is a
maximal subspace 5' E S, which is obviously distinct from S and meets S at p; thus, through S there is an element of R (joining 5 and 5'). of S.
Hence, R is a cover
11oreover, for any r E R, j rj ;;;. 2 (see (3.1)) and any two elements in R have
at most one common point (by (3.2)); it follows that (S,R) is a PLS. Let p be a point in G.
The collections S of all elements in S through p is p
a subspace of (S,R).
(Indeed, any two distinct elements in Sp are collinear in (S,R) and the line through them is completely contained in S ). p
PROPOSITION VII: Let p be a point in G and T an element in T through p.
Then the
subspace S in (S,R) is isomorphic to the star F T consisting of the lines in F p p, through p and belonging to the projective space T. Thus, S is a projective space p
and is of finite dimension h iff T is of finite dimension h+l.
120
A. Bichara and G. Tallini
PROOF: Any element in $ meets T at a line in F T (see A I). p p, 2 Let ~ be the mapping defined by :SE$-+snTEF T p,
4'
-1
is one-to-one and onto (see A2II). Moreover, ~ maps pencils onto lines, in R, belonging to $. (Indeed, any pencil in F T consists
Clearly, in F
p, T
~
P
p,
of all lines in T through p, belonging to a plane a E IT; such a pencil is the image under
of the line r p,a ). Now, it will be proved that
r
P.!l
Q
maps lines in $p onto pencils in F r Let p, , be a line in S and Sand S' two distinct points on it; they meet T at ~
P
(distinct) lines f and f', which are joined by a plane a through p in T; then, r p,a ,'" r p,a (see prop. V); furthermore, fjJ(r p,a ) '" q>(r p,a ,) is the pencil consisting of the lines in Fp, T through p and belonging to a. It follows that
Q
is an isomorphism between $p and the star of lines through p in the projective space T. From prop. VII, it follows immediately: PROPOSITION VIII: Le" T and T' be any two distinct elements in T through a point p in G.
Then T is of finite dimension h+l iff T' is of finite dimension h+l.
Next, we prove: PROPOSITION IX: Any T in T is an (h+ 1 )-dimensional projective space. the coZlection $
p
Furthermore,
oj all elements in $ through a point p in G is an h-dimensional
projective space, which is a subspace of ($ ,R).
PROOF: If T is the space Assume
T in
T, T f
I;
T in
A , then T is an (h+l)-dimensional projective space. 3 let ql be a point in f and q2 a point in T. Since (G,F) ;s
connected, there exist both a finite subset {Pl, .•• ,Pn} of pOints in G and a finite subset {fl, .•• ,fn_ll of fi
~
lin~s
{Pi' Pi+ll, i = 1. 2, .... n-1.
in F such that Pl
= ql'
Pn
= q2
and
Through any line fi there is exactly one Ti
in T (see A2II). The subspaces T and Tl have the common point Pl '" ql; since has finite dimension equal to h+l, by prop. VIII, Tl is an (h+l)-dimensional projective space.
T.1 n T.1+ 1
::l {po
-
1+
1) (i '" 1, ... ,n-2) and T 1 n T ::l {p }, all nn
T
On a characterization of Grassmann spaces
T.'s (i 1
= l, ••• ,n-l)
121
and Tare (h+l)-dimensional projective spaces.
Finally, to prove that Sp is an h-dimensional projective space, it is enough to recall that there exists an element T in T through p (through p there is at least one line f in F which is contained in a maximal subspace in T); T is an (h+l)-dimensional projective space and from prop. VII the statement follows.
4. THE SUBSPACES OF (S,R) Now, we prove the following: PROPOSITION X: Let Sl' S2' S3 be any three pairwise collinear in (S,R) points of S, and meeting at a same point p of G, as subspaces of (G,F); then there exists a projective plane in (S,R) through them.
PROOF: The family Sp consisting of all elements in S through p turns out to be a projective space of dimension h
~
2 (see proposition IX) in (S,R), which contains
Sl' S2 and S3' PROPOSITION XI: Let Sl' S2' S3 be any three pairwise collinear in (S,R) and independent points of S, which do not meet at a same point, as subspaces of (G,FJ; then a projective plane in (S,R) through them exists.
PROOF: Set Pl
= S2
S3 ' P2
n
= Sl
n
S3 ' P3
= Sl
n
S2
(and such points exist); they turn out to be pairwise distinct (otherwise, if Pl = P2 then Pl E S2 ' Pl E Sl' that is P3 = Pl = Sl n S2' and thus Pl = P2 = P3)· Then there exist in F three lines (P l ' P2), (P l ' P3), (p 2' P3), which are pairwise distinct (otherwise, Sl' S2 and S3 contain the line (P l ' P2) = (P l ' P3) and this cannot occur because of A ,II). Hence, a subspace of (G,F) (see A ) containing 2 l Pl ' P2 and P3 exists. Such a subspace, say a, belongs to IT: a meets Sl at a line, hence it is contained in a maximal subspace T of T and turns out to be the projective plane a
= {S
E
a
joining Pl ' P2 and P3 in T.
S : S n'a
E
F}
Next, we prove that
122
A. Bichara and G. Tallini
is a projective plane.
Let S, and S2 be any two distinct elements of
meet the projective plane
a
~':
a ;
they
at two distinct lines fl and f2 of F, respectively,
which intersect in a point p of
Thus, Sl' and S2' belong to the line r in R. ,., p,a p') be any two (distinct) lines in a . Since the line a.
Let rand r, ,(p tp,a p ,a (p,p') in F on ex exists, the maximal subspace S E S through it is the only element belonging to r
p,a
n
r, ,. p ,a
The statement follows.
As a corollary to prop. X and XI: PROPOSITION XII:
Ar,y subspace of (S,R) is a projective space.
5. THE COLLECTION P OF MAXH1AL SUBSPACES IN (S,R)
PROPOSITION XIII: Let; p be a point in G and S an element of S through q (c G) and
q not th!'ougiz p fsince (G,F) is a PPLS, such a subspace exists).
Then, the set
P ad S n S t- 0 } is either a Zine in R 01' the empty set. By q proF. IX, the t3Zements in S :;h!'ough p [m'm an h-dimensional (h ;;. 2) projective L = {S
E
S: S
::J
spaoe in (S,R); tin.s, there is always some element in S through p which is skew with Sq; consequently, (S,R) is a propel' partial space (see VI).
PROOF: Assume that, through p, there are two distinct elements of S, say S' and S", both meeting Sq at a point: S' n Sq q', S" n Sq = q". Then, q' t- q". (Indeed, q' = q" implies that through the line (p,q') (p,q") there are two distinct elements of S, a contradiction).
Since the three points, p, q', and q" are pairwise
collinear, there exists a subspace
in
in (S,R) p,a consists of those elements in S through p, meeting S at a point on (q',q"); a
IT
through them.
The line r
q
moreover, any point on (q' ,q") is joined to p by a line belonging to some element in r
p,a Now, assume S'" is an element in S through p, neither skew with S , nor q
belonging to r p,a ; then, S"' meets Sq at a point q"' not on (q',q"). Since the points <]' ,q'", and p are pairwise collinear, they are contained in a plane a' of E,
which is distinct from a (otherwise, an S = a' n Sand qlll
contradiction); these two planes in
q
~
q
meet at the line (p,q').
the maximal subspace in T through a and a', respectively.
E
(q',q"), a
Let T and T' be
Then T
= T' (by prop.
I,
On a characterization of Grassmann spaces
taking into account that TnT' contains the line (p,q') points
123
=a n
a').
The three
q", and q"l belong to Tn Sq; thus, the projective plane through them
in T is completely contained in Sq' which is impossible (see A2I). Finally, it must be proved that if an element Sp in S through p and incident with S exists, then there exists a line r
, any element of which is an element p,a Set Sp n Sq -::J {q'}; through the line (p,q') of S through p and incident with S. q q
there is a maximal subspace T of T, meeting Sq at a line through q', all whose points are collinear with p; and this line and p span a plane belonging to
IT.
The
statement follows. PROPOSITION XIV: Let Sp be the set of all elements in S through a point p in G. If S belongs to S and doesn't pass through p, then the subset
L = {S' E Sp' S'
'V
S}
S in (S,R) means S' is incident with S in (G,F)) of Sp is either empty or a line in R. Furthermore, Sp is a maximal subspace in (S,R). (S'
'V
PROOF: The first part of the statement follows from prop. XIII.
Again by prop.
XIII, Sp is a subspace in (S,R), containing some element which is non-collinear with S.
Therefore, no subspace containing both Sand S exists in (S,R) and S is p P a maximal subspace. Thus, the collection P = {S
p
: pEG} of maximal subspaces in (S,R), any of
them being a projective space of dimension h (see prop. IX), arises. PROPOSITION XV: The collection P = {Sp : pEG} is proper.
Furthermore, any two
distinct elements in P share at most one point of S.
PROOF: Let p and q be any two distinct points in G.
The maximal subspace Sand
Sq in (S,R) share all the elements in S through both p and q. collinear in (G,F), then there is no element of S through them.
p
If P and q are nonIf p and q are
joined by a line f E P, then (in (G,F)) there exists exactly one maximal subspace belonging to S which passes through f and so through p and q. IS p n Sq I
~
1, and the statement follows.
In both cases,
124
A. Bichara and G. Tallini
6. THE COLLECTION
l:
OF tIAXU1AL SUBSPACES IN (S,R)
First we prove: PROPOSITION XVI: Let Sand 5' be in G.
p~o
distinct elements in S meeting at a point p
If Sl and 52 are distinct elements in S both meeting Sand S' at distinct
poillt8 in G, then Sl and S2 are collinear in (S,R) and any element S3 -~ng
E
S belong-
(5 ,5 2) in R eithe1' meets both 5 and 5' at distinct points of G or 1
to the line
;'Jelongs to tht' line (5, S ') in R.
PROOF: From A2III it follows that Sl and S2 meet: {ql = Sl n S2' and thus they are collinear in (S,R) (see prop. VI). q
~
p.
If q
E
Since neither Sl nor S2 contains p, then
S, any element S I 5 on the line (Sl,S2) in R, passing through q,
meets S at the point.
If q
~
S, the set of all elements in S through
q
and inci-
dent with S is either a line in R or the empty set (see prop. XIV); since both Sl and S2 pass through q and are incident with S, any element on the line (Sl' S2) in
R is incident with S.
By the same argument, any element on (Sl' S2) is proved to
be incident with S'; thus, it is only to be proved that if S3 is an element on (Sl' S2) passing through p, then it belongs to the 1ine (S, S').
Under these
assumptions, 53 contains both p and q, and, thus, meets Sl at q.
Hence, S3 E L',
where L'
=
{S'"
hypothesis).
E
S:
S'·'
3
P and Sf, n Sl I 0l; also Sand S' belong to L' (by
By prop. XIV, 53' Sand 5' belong to a line in R; so, S3
E
(5,S')
and the statement is proved. Let Sand 5' be two distinct elements in S, which are collinear in (S,R); then 5 and 5' share a point p in G. elements in
S
Denote by o(S,S') the set consisting of all
that either belong to the line (5,5') in R, or meet both 5 and S' at
points in G distinct from p. PROPOSITION XVII: If S and 5' are distinct elements in
S which are collinear in
(S,R), then the set 0(5,S') is a subspace in (S,R)and it properly contains the lin6 (S,S') in R. PROOF: let T be a member of T through the point p, where {pl meets Sand 5' at the lines f and f' in F, respectively. q, q'
~
p,thenq
~
= S n 5'; by A2I, T
If q
q' and q and q' belong to a line f" in T.
E
f, q'
E
f',
Through f" there is
On a characterization oj Grassmann spaces
125
an element SOl E S meeting 5 and 5' at q and q', respectively.
Thus, SOl belongs to
0(5,S') \ (5,5'); therefore, the line (5,5') is properly contained in 0(5,5'). To prove that 0(5,S') is a subspace in (S,R) (i.e. any two distinct points 51' 52 E 0(5,5') are collinear in (S,R) and the line joining them is completely contained in 0(5,S')), three cases will be considered. (i) If 51 and 52 both belong to the line (5,5'), then there is nothing to prove. (ii) If 51 and 52 meet 5 and 5' at distinct pOints, then the statement follows from prop. XVI. (iii) If 51 meets both 5 and 5' at distinct points and 52 belongs to the line (5,S'), then - by an argument similar to that in prop. XVI - it is easy to prove that S1 and 52 are collinear and any element on (S1,S2) belongs to o(S,S'). PROPOSITION XVIII: Let 5 and 5' be two distinct elements in S which are collinear in (S,R), i.e. they meet at a point p in G.
Then there exist exactly two maximal
subspaces of (S,R) through S and 5'; the first one is Sp (i.e. it consists of all elements in S through p) and belongs to P; the second one is o(S,S').
These two
subspaces share exactly the elements on the line (5,5').
PROOF: Let H be a subspace of (S,R) containing both 5 and S' and so the line (S,S') in R.
Then: either H consists of elements in S all of them through p and H
is a subspace of Sp' or there exists some SOl E S contained in H and not passing through p.
Since H is a subspace and SOl, S, and S' belong to H, SOl is collinear
with both Sand S', i.e. meets Sand S' at distinct points.
By prop. XIV, the
elements in S which are collinear with SOl are exactly the elements on the line p
(S,S') in
; hence, H () Sp
11:1
(S,S').
Therefore, any element in H \ (S,S'), being
collinear with both Sand 5', meets Sand S' at points in G distinct from p (and from each other).
Thus H ~ o(S,S') and the statement follows.
PROPOSITION XIX: Let Sand S' be two distinct elements in S, which are collinear in (S,R).
0(5 1,5 2)
If 51 and S2 are distinct elements in 0(S,5'), then they meet and =
0(5,S').
PROOF: Since 51 and 52 belong to the subspace 0(5,S'}, they are collinear and thus have a common point q
E
G.
By prop. XVIII, the subspace 0(S,5'), containing 51
A. Bichara and G. Tallini
126
and $2' is contained in either the maximal subspace Sq' or the maximal subspace 0($1,5 ), 5ince the elements in 0(5,$') don't all pass through a same point, 2 0(5,5') ~ 0(5 ,5 ), Equality follows from 0(5,5') being a maximal subspace in 1 2 (S,R).
7. FURTHER PROPERTIES OF P AND
L
In the PLS (S,R) two collection of maximal subs paces have been defined (see prop. XIV and XVIII): P
= is p
=
: pEG), a proper collection;
(0(S,S'): S, 5' E S, S F S', S n 5' F 0}.
PROPOSITION XX: Fa!' any r in R, the!'e exist a unique Sp in P and a unique a in such tr...'lt res
-
l:
p and rea.
PROOF: 5ince any two distinct elements on r are incident, the statement follows from prop. XVIII and XIX. PROPOSITION XXI: If Sp E P and a n
then S n a is eithe!' the empty set a!' a
L.
Line -in R.
PROOF: The set Sp n
a
consists of all elements in S through p which are collinear
with any element in a. (see prop. XIX) a
Take S E S n a and let S' be any other element in a; then p
= a(S,5') and 5 ESp' If (pl = S n 5', then the statement
follows from prop. XVIII.
If P ~ 5', then - by prop. XIV - the set of all ele-
ments in Sp incident with 5' is a line r in R, containing 5. on r, distinct from S.
Let S" be a point
Since S" meets 5 and S' at distinct points, 5" E a(S,S');
therefore, (by prop. XIX) a(S,S')
=
a(S,S") and the previous argument proves the
statement. PROPOSITION XXII: If a'.
0"
and a'" C1!'e pai!'Wise distinct elements in 1: such that
a' n orr = {5' l. a'ne'" = {S"l, a"na'" = {S", l, I,)he!'e S', S", S"· ES and
S' F S" (so that S' F SOlI and SI! f 5'11 l"'ommon.
,
then 5', 5" and 5'" have a point of Gin
127
On a characterization of Grassmann spaces
PROOF: Since 5',5" Eo"', 5, 5' E a' and 5,5" Eo", 5, 5' and 5" are pairwise collinear in (S,R).
Thus, a subspace H of (S,R) through them exists (see X, XI);
H is contained in a maximal subspace not belonging to L (indeed, it meets (5,5') and a" in (5,5"»; therefore, H belongs to P. an element Sp in P containing 5, 5' and 5".
a'
in
Consequently, there exists
It follows that a point pEG exists
through which 5, 5' and 5" pass, and the proof is complete. PROP05ITION XXIII: If 0 , O , 0 are distinct elements of L which pairwise meet 1 2 3 at distinct points of S, then any 0 E L, other than 0 and meeting 0 and 02 at 4 3 1 distinct points of S, meets
0
3
•
PROOF: Set 01 n O2 = 5,° 1 n 03 51' O2 n 03 = 52' 01 n 04 = $1 and O2 n 04 = $2; by proposition XXII, two paints p and q exist in G such that 5,5 1,5 2 E Sp and 5, 51' $2 ESq.
Assume p = q.
5ince (see XXI) 01 meets Sp either in the empty set or in a line of R, the elements 5, 51' and $1 of S are collinear; similarly, 5, 52 and $2 are collinear (as 02 meets Sp in a line).
It follows that 5, 51' 52' $1 and $2
span, in the projective space, the plane through the lines (5 1,5 1) and (52' 52) whose common point is 5 E S; thus, the line (51' 52) and (51' 52) on this plane meet at a paint 5
E
S which clearly belongs to both 0(Sl,S2)
= 03 and
0(5 1,$2) = 04. Therefore, when p = q, there exists an $ E S such that 03
n
0
4 = {5} • Assume p r q.
that 51 n 51 = {p'}.
5ince Sl' 51
1,5 1 - 5, and a pOint p' E G exists such In a similar way, since 52' $2 E O 2 , a point q' E G
exists such that 52 n $2 p'
r
{q'}.
E 0
It
is straightforward that Sl F 52 implies
q'.
The three points p, p' and q are pairwise collinear (p, p' E 51' p', q E $1' p, q
E
5); hence, a subspace of (G,F) exists through them which is contained in a
maximal subspace T E T.
Similarly, p, q' and q being pairwise collinear in (G,F),
a maximal subspace T' E T exists through them. (p
r
q), therefore T = T' (see I).
T and T' share the points p and q
5ince a subspace T E T containing the four
paints p, p', q and q' exists, p' and q' are collinear and a maximal subspace 5E
S
exists through them.
5 meets Sl and 52 at the distinct points p' and q',
respectively, so that $ E 0(5 1,5 2) = 0 3 • ~oreover, 5 meets 51 and $2 at the distinct points p' and q', respectively, so that 5 E 0(5 1,$2) = 0 4 • Thus,
A. Bichara and G. Tallini
128
S = 03
n 04 and the statement follows.
PROPOSITION XXIV: The space (S, R) is connected. PROOF: Let Sand S' be any two distinct elements of S and take pES and q E S'. The points p and q belong to G; since (G,F) is connected, it exists a polygonal path (fl, •.. ,f n } in (G,F) joining them.
Through any line in F of this path there
is an element of S (see A2.II). The elements (Sl"",Sn} (and w.1 .o.g. it can be assumed S.1 ~ S.1+ l' i = 1, .•• ,n) of S are such that S.1 and S.1+ 1 share a point, i S'
=
n
1, ... ,n; therefore, they are collinear in (S,R). Sn 2
t10reover, S n 51 2 {p} and
It follows that the lines (S,Sl)' (Sl,S2)' ... , (Sn_1,5 n ), and
{q).
(Sn'S') form a polygonal path in (S,R) joining Sand s'; thus, (S,R) is connected and the statement is proved.
8. THE CHARACTERIZATION OF THE GRASSI·1ANN SPACE OF INDEX h ASSOCIATED WITH A PROJECTIVE SPACE In this section the main theorem will be proved (see n. 1). Let (G,F) be a Grassmann space of finite index h
~
2.
When h
2, the
theorem is proved in [41. Assume h
~
3.
The statement will be proved by induction; therefore, assume
it is true when the index is h-l.
By the results in the previous sections (3 - 7), a connected PPLS (S,R) is associated with (G,F) (see XIII and XXIV); furthermore, the following properties ho 1din (S , R) : Al ) Given any three pairwise collinear points, a subspace containing them exists (see X and XI). A2) No line is a maximal subspace; furthermore, there exist two collections - say I and P - of maximal subspaces in (S,R) such that any maximal subspace belongs either to E or to P (see XIV, XVIII, and XX) and
="
I • ° E E, Sp E P ~ either ° n Sp or ° () S P E R (s ee XX I ) • II. For any r E R, 3 ! ° E E, 3 ! S E P such that reo, res (see XX). p
III.
-
-
P
If 01'02 and 03 are elements of E which pairwise meet at distinct points
129
On a characterization of Grassmann spaces
in S, then any element of E, other then 03 and incident 01 and 02 at distinct points of S, is incident with 03 (see XXIII). A3) A complete finite chain of subspaces exists such that its length is h, its minimum subspace is a line r E R and its maximum subspace is an S p E P and res
p
(see IX). Therefore, (S,R) is a projective Grassmann space having a finite index h-l.
By the induction hypothesis, a projective space IP exists and (S,R) is isomorphic to the (h-l) -th Grassmann space (G( h-l ) (IP ), F( h-l ) (IP)) associ ated with F.
Thus,
a bijection C from G(h-l)(IP) onto S exists such that: (8.1) R is the set consisting of the images under C of all elements in G( h-l ) (IP );
(8.2) E is the set consisting of the images under C of all elements in s(h-l)(IP ); (8.3) P is the set consisting of the images under C of all elements in r(h-l)(IP ). Let Sh
E
h
G (IP) be an h-dimensional subspace of IP and Sh the set fo all
(h-l)-dimensional subspaces of Sh'
Clearly, Sh belongs to r(h-l)(IP).
under C of Sh is an element of P, that is, it is an Sp' h F : G ( IP) ---+ G defi ned by
The image
Consider the mapping
It is straightforward to prove that:
(8.4)
F is one-to-one and onto. Next, the following result will be proved:
(8.5)
i
h
Let Sh' i = 1, 2, 3, be three elements of G (IP) (two of them at least
G their images under F. Then, S~, S~, S~ pass through (h-l ) . the same (h-l)-dimensional subspace Sh_l E G (IP) 1ff Pl' P2' and P3 belong to
being distinct) and p.
E
1
C(Sh_l) as E S. To prove (8.5), note that
130
A. Biehara and G. Ta/lini
S1 n S2 n S3 - S ~ ... l n 52 n -S3 - S h h h - h- 1 \ h h - h-l
> C(S~) n
<
C(~)
n
C(~~)
From (8.5) it follows that
<,
= C(Sh_l)
s~, S~
>
<
<====>
E Gh(JP) belong to the same set
S(Sh_l) E Sh(IP) of all h-dimensional subspaces through Sh_l iff Pl' P2' P3 E = C(\_l)'
(8.6)
s
S =
Thus,
is the set of the images under F of all elements in Sh(JP).
Furthermore, since two elements of Gh(IP) (of G) are collinear in (Gh(JP), Fh(lP)) (in (G,F)) iff an element of Sh(JP) (of S) exists through them, from (8.6) it follows
(8.7)
Two elements of Gh(JP) are collinear in (Gh(IP), Fh(JP)) iff their images
under F are collinear in (G,F). Next,
(8.8)
T is the set of images under F of all elements in Th(IP).
PROOF: Take T E T and let Pl' P2' P3 be three points of T (two of them at least being distinct) not on a same line.
Since Pl' P2 and P are pairwise collinear, 3 their inverse images S~ = F-l(P ), i=1,2,3, are pairwise collinear in Gh(IP) (see i (8.7)). Thus, a maximal subspace 11 of the Grassmann space (Gh(JP), Fh(JP)) exists containing them.
1,1 cannot belong to Sh(JP), otherwise (see (8.6)) F(M) would
belong to S and meet T E T in the three points P , P2 and P which are not on a l 3 line (and this cannot happen because of A I). Therefore, ME Th(JP). 110reover, 2 the inverse image under F of any element P in T, not on the line P'P ' is an ele2 ment of M. Similarly. the inverse image under F of any element in T, not on the line P'P ' is an element of M. 3 Bya similar argument, r~
It follows that
~
M~ F-l(T).
F-l(T); thus, M = F-l(T), i.e. the inverse image
under F of any element in Tis an element of Th (JP ).
Ana 1ogous ly, the image under
F of any element in Th(IP) is proved to be an element of T and (8.8) follows. Fi na 11y.
On a characterization of Grassmann spaces
(8.9)
131
F is the set consisting of the images under F of all elements in Fh(IP).
PROOF: Take f E F and let S E Sand T E T be the two maximal subspaces through f. Since f = S n T and F is a bijection, F-l(f) = F-l(S) n F-l(T). -1 h -1 h -1 By (8.6) and (8.8), F (S) E S (IP) and F (T) E T (1P); therefore, F (f) l is an element of Fh(IP); indeed, F- (f) is the meet of two maximal subspaces of (Gh(P), Fh(IP)) which belong to distinct families.
Bya similar argument, the
image under F of any element in Fh(IP) is an element of and (8.9) is proved. From (8.4), (8.9), (8.6) and (8.8) it follows that F is an isomorphism between the Grassmann spaces (G,F) and (Gh(IP), Fh(IP )); thus, the theorem is completely proved.
BIBLIOGRAPHY 1. 2. 3.
4.
B. Segre, Lectures on modern geometry, Cremonese Ed. Roma (1961). G. Tallini, Spazi parziali di rette, spazi polari, Geometrie subimmerse, Quaderni Sem. Geom. Combinatorie, 1st. r·lat. Univ. Roma, n. 14 (gennaio 1979). G. Tallini, On a characterization of the Grassmann manifold representing the lines in a projective space, in: P.J. Cameron et al. (eds.) Finite geometries and designs. London I·lath. Soc. Lect. Notes Series n. 49. Cambridge University Press (1981), 354-358. A. Bichara and G. Tallini, On a characterization of the Grassmann manifold representing the planes in a projective space, to appear.
Istituto Matematico "G. Castelnuovo" Citta Universitaria 00185 Roma Italy
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133
Annals of Discrete Mathematics 18 (1983) 133-168 © North-Holland Publishing Company
ENRICHED CLIQUES WITH SIX VERTICES R.H. Bruck
SUMMARY Let S be a clique with 6 vertices. A spread is an unordered set of three mutually disjoint edges of S; and there are exactly 15 spreads. Let S3 be any 15 one of the 2 = 32 786 spread-sets. Create a set S2 of unordered pairs of disjoint edges of S as follows: such a pair is in S2 if and only if the unique spread containing the pair is not in S3'
Let "0 be the partial projective plane
with the elements of S2 U S3 and the vertices of S as points and with the edges of S as lines, under the natural incidence relation.
Then (in infinitely many
ways) rrO generates a projective plane rr and S appears as a 6-arc in".
If, how-
ever, we require that rr have a specified finite order (and perhaps other properties relative to "0' suggested by experience, expedience or intuition) there may be severe restrictions on the choice of S3' some of which are graph-theoretical but not necessarily well understood.
Consequently, there is a need for a com-
plete, compact and informative catalogue of spread-sets.
The present paper pro-
vides this (in a space of about 6 pages) but there is a price to be paid for brevity.
To use the catalogue to best effect one must understand the theory (here
presented) of E-cliques (enriched cliques with 6 vertices) and their decodingtrees. 1. INTRODUCTION A clique, remember, is a symmetric graph which is complete in that every two distinct vertices are adjacent (joined by an edge.) Throughout this paper we shall be concerned with a clique having six vertices.
Accordingly, we are free
to speak of "the clique" or "our clique" without fear of ambiguity. Towards the end of this section we shall explain how to "enrich" our clique (or turn it into an E-clique) by assigning a rank (subject to certain rules) to each of the 15 edges.
First, however, we wish to discuss some concepts
134
R.H. Bruck
and problems which make the study of E-c1iques seem both natural and worthwhile. Let the six vertices of the clique be denoted by the integers 1,2,3,4,5,6 and let the edge joining two distinct edges i,j be denoted by the transposition (i,j).
An unordered set of three mutually disjoint edges of the clique will be
called a spread (there are other names: this one comes from projective geometry) and will be denoted by the product of the corresponding three transpositions.
As
is easily checked, there are exactly 15 such spreads (or, equivalently, exactly 15 involutions without fixed points in the syrTllletric group of degree 6.)
For reasons coming from projective geometry (and discussed superficially in §
2) we are interested in the 15 2
32 768
subsets of the set of 15 spreads. For each integer k in the range
~k~15,
the
number of these sets of size (or cardinal number) k and the number of size 15-k is the same, namely
II
~511
and each set of size k has a complement (in the set of 15 spreads) of size l5-k. Hence it is reasonable to restrict attention either to sizes at most 7 or to sizes at least 8; we shall make the latter choice in this paper. Naturally we may classify the spread-set into equivalence classes (or orbits) under the action of the syrTllletric group on 1,2,3,4,5,6, and retain only one representative of each orbit.
However, along with each such representative, we
need to list some additional information: First (in order to make it easy to verify that no orbit has been overlooked) we need the length (or cardinal number) of the orbit.
Next - for our purposes - we need the degree of the graph of the rep-
resentative spread-set (the graph and its degree have yet to be defined.)
And
finally - since the complement of the representative represents an orbit which we are deliberately omitting - we need to list what we call the anti-degree of the representative (this is the degree of the graph of the complement of the representative.) At this point let us list some facts about spread-sets of size a1 least 8.
135
Enriched cliques with six vertices
SIZE SETS
15 14
13
12
10
11
9
8
TOTALS
(1.1)
15 105 455 1365 3003 5005 6435 16 384
ORBITS
2
5
15
9
21
24
78
The total number of orbits enumerated in (1.1) is certainly large enough to raise a serious question as to whether such a list as I have described is worth the cost of publishing it.
Luckily, as I shall explain, there is a much more
compact listing in terms of E-cliques.
This appears at the end of the paper.
(See CATALOGUE OF SPREAD-SETS.) Now let us consider the following spread-set of size 9: (12) (34) (56),
(12) (35) (46),
(12) (36) (45),
(13) (24) (56),
(13) (25) (46),
(13) (26) (45),
(14) (23) (56),
(14) (25) (36),
(1. 2)
( 15) ( 23) (46).
(Note that the first row of the list in (1.2) consists of the three spreads containing the edge (12); the second, of the three containing (13).
On the other
hand, there are only 2,1 and 0 spreads in (1.2) containing the edges (13), (15) and (16), respectively.
It is less easy, however, to see how many of the
spreads in the list contain the edge (24), for example.) By the graph of the spread-set (1.2) of size 9 we mean the symmetric graph with 9 vertices defined as follows: Each spread in (1.2) is a vertex.
Two dis-
tinct vertices are joined by an edge if and only the two spreads in question have an edge of our clique in common. With this definition understood, the degrees of the nine vertices of the graph, taken in the order written in (1.2), are found by careful inspection to be 4, 4,4,4,5,3,4,3,3. We sum up this information by saying that the set (1.2) has 1 5 3
degree 5 4 3 .
(l. 3)
We also define the anti-degree of (1.2), namely, the degree of the spread-set of
136
R.H. Bruck
size 6 complementary to (1.2).
By inspection of this complement, we find that
(1.2) has (1.4 )
-- It is important to note that the anti-degree of (1.2) is quite different from the degree of the complement of the graph of (1.2); the latter can be read off directly from (1.3) (whereas the anti-degree cannot be) and is
Now we are ready to define the [-clique associated with the spread-set (1.2).
The vertices and the edges of the [-clique are simply the vertices and
edges of our clique.
In addition, to each edge of the clique we assign a non-
negative integer called its rank, namely the number of spreads in (1.2) containing the edge.
That's all!
The [-clique of (1.2) has an incidence array as
follows: X 3 3 2 1 0
(1.5 )
3 X2 1 2 1
32Xl12 2 1 1 X 2 3
l212X3 01233X Here, for each ordered pair of distinct vertices i,j, the entry in row i, column j of the array is the rank of the edge (ij). -- By contrast, the usual adjacency matrix (or incidence array) of our clique could be obtained from (1.5) by replacing each rank by 1 (and probably by also replacing X in the main diagonal by
o --
which would be less than wise for the case of the [-clique.) There are three obvious facts about (1.5) which are equally true about the
incidence array of the [-clique associated with any spread-set. First of all ,the array is symmetric (about its main diagonal) -- since our clique is symmetric. Secondly, each rank is 0,1,2 or 3.
This is because the total number of spreads
containing a given edge is precisa1y 3. and not all of these need be in the
Enriched cliques with six vertices
spread-set in question.
137
And, finally. the sum of the five ranks in each row (or
column) of the array is equal to the size of the spread-set.
This is because, for
each vertex of our clique, say vertex i, each spread in the set contains exactly one of the five edges through i. At this pOint we are ready to make a formal definition of an E-clique, namely an enrichment of the clique obtained by assigning a rank to each edge of the clique, subject to the following two conditions: (a) each rank must be one of the integers 0,1,2,3;
(b) the sum of the ranks of the 5 edges through a vertex must
have a fixed value k, independent of the choice of vertex.
(The common sum, k,
will be called the size of the E-clique.) If we make the conventions that the symbol X is neutral with respect to addition and subtraction:
X + X = X,
X- X= X ,
then we may add and subtract incidence arrays of E-cliques componentwise, provided that we watch out for two possibilities: In the case of addition, some entry in the sum may exceed 3; in the case of subtraction, some entry in the difference may be negative.
As long as the result (in either case) has all off-diagonal
entries in the range 0,1,2,3, it is the array of an E-clique. The array corresponding to the empty spread-set is the "zero" array, with all off-diagonal entries O.
The array corresponding to the set of all spreads is
the "all" array with all off-diagonal elements 3.
If we subtract (1.3) (for
exampl e) from the "a 11" array, we get the array associ ated with the complement of the spread-set (1.2). We shall consider in
§
3 the problem of "decoding" an E-clique so as to find
all spread-sets with it as E-clique.
(There may be several, or there may be none.)
For the moment, we wish to show how to use the E-clique given by (1.5) to determine the length of the orbit of the spread-set (1.2). First of all we must contrast and compare two definitions.
On the one
hand, the stability group of a spread-set is the set of all permutations of the vertices of our clique which map the spread-set into (and hence upon) itself. (Of course, the index of the stability group in the symmetric group on the vertices is the length of the orbit of the spread-set.)
On the other hand, the
automorphism group of an E-clique is the set of all permutations of the vertices
138
R.H. Bruck
which preserve rank.
Clearly the stability group of a spread-set is a subgroup of
the automorphism group of the associated E-clique.
Hence it is worthwhile - and
generally easier - to determine the latter group first. The automorphism group of the E-clique given by (1.5) can be determined very rapidly as follows: The edges of rank 1 form a unique open path (15), (53), (34), (42), (26) from 1 to 6.
Hence the only possible automorphism of the E-clique, apart from
the identity, is the permutation (16) (25) (34). We verify easily that this permutation preserves rank and hence generates the automorphism group of the E-clique. spread set (1.2).
It also generates the stability group of the
(Note that this must be checked; it could conceivably happen
that the permutation mapped (1.2) into a different spread-set with the same Eclique.)
Thus we can say that (1.2) has orbit length 360.
( 1.6)
Apart from one important consideration (which cannot be emphasized too often) it would be appropriate to replace the spread set (1.2) and the information about it exhibited in (1.3), (1.4), (1.5), (1.6) by a single compact listing as follows: X33210 First Orbit -3 X 2 1 2 1 Orbit length: 360 1 53 3 2 X1 1 2 Degree: 5 4 3 1 2 2 1 1 X 2 3 Anti-degree: 4 3 23
(1. 7)
1 2 1 2 X3
o1 2
33 X
The consideration is that we must know how to decode the E-clique.
In addition,
in case the E-clique represents spread-sets belonging to more than one orbit, we
Enriched cliques with six vertices
139
must add further information under headings such as "Second Orbit" and so on. Luckily, the multiple columns are rarely necessary, and when they are unnecessary we can also omit the heading "First Orbit". And one last small point.
We shall introduce a normal form for E-cliques
and their incidence-arrays. The array in (1.7) is not in normal form and hence is replaced (in CATALOGUE at the end of the paper) by an equivalent array in normal form.
The actual entry is as follows: X02l33 OX3321
#
37
(1.8 )
OL 360
23X2ll 132X12
1 53 D5 4 3
32l1X2 31122X AD 41i 23
2. ARCS IN PROJECTIVE PLANES Let
~
be a projective plane (of finite or infinite order) and let k be a
positive integer.
By a k-are of
three of which be on a line
of~.
~
we mean a set of k distinct points of
~
no
Our purpose here is to show the connection
between E-cliques and 6-arcs, but first we shall consider k-arcs in general. Note that every projective plane has quadrangles, and that these are the 4-arcs. A k-arc, A, of
~
partitions the lines of
~
into three classes.
First there
are the (~) secants of A, each containing two distinct points of A. Each point of A lies on exactly k-l distinct secants of A. of
~
must be at least k-2.)
exactly one point of A. k-l.)
(Hence, for A to exist, the order
Next there are the tangents of A, each containing
(For trangents to exist, the order of
~
must be at least
And finally, there are the non-secants of A, each containing no point of A.
(These always exist if A exists.) For planes of infinite order, the situation in regard to k-arcs is very simple. LEMMA 2.1: If
~
is a projeetive plane of infinite order, then" has k-ares for
every positive integer k.
R.H. Bruck
140
Certainly
PROOF:
rr
has 4-arcs, and hence
rr
also has 1-arcs, 2-arcs and 3-arcs.
Now assume, for purposes of induction, that, for some integer k > 4,
A.
Since
has infinitely many lines and only
rr
line L which is not a secant of A. and there can be at most A.
(~) secants
of
A,-rr
rr
has a k-arc
certainly has a
Each secant of A meets l in a unique point,
(~)distinct
points of l lying on one or more secants of
Since l has infinitely many points, there certainly exists a point· P of l
lying on no secant of A. (k+l)-arc of
In particular, P is not in A.
Clearly Au {P} is a
This proves Lemma 2.1.
rr.
By the remark in the second paragraph of this section, applied with k = 6, a projective plane containing a 6-arc must have order at least 4.
This condition
is also sufficient, as follows from Lemma 2.1 and the next lemma: LEMMA 2.2: (i)
be a projective plane of finite order n, 0here n > 4.
~et rr
Then:
quadrangle of rr is contained in precisely (n-2)(n-3) distinct 5-arcs
Ev~ry
of rr. (ii) Every 5-arc of rr is contained in precisely 1 + (n-4)(n-5) distinct 6-arc of rr.
REMARK: In view of Lemma 2.2, the numbers of 5-arcs and 6-arcs in a projective plane of order n can be given as explicit polynomials in n. The proof of Lemma 2.2 is very simple and need not concern us here.
We
wish, instead, to point out that 5-arcs and 6-arcs have played important roles in the theory of Pascalian projective planes (i.e., projective planes coordinatized by fields.) k-arc.
Such a plane has conics.
A set of k distinct points of a conic is a
Every 5-arc lies in exactly one conic.
Pascal's theorem on conics, taken
with its converse, characterizes the 6-arcs (call them Pascalian 6-arcs) which lie in a conic, and the "constructive" theory of conics is essentially the theory of Pascalian 6-arcs. There are also non-Pascal ian 6-arcs, and these have received little attention. Now let us focus our attention on a 6-arc S of a projective plane (About
rr
we assume only that its order is at least 4, possibily infinite.)
is a pOint of
rr
which is not in S, then, for some integer i
i secants and 6-2i tangents of S. points of
rr.
rr
Clearly
°
~
i
~
3.
~
0, P lies on exactly
Thus S partitions the
which are not in S into four disjoint subsets
If P
141
Enriched cliques with six vertices
where, for 0 < i < 3, S. is the set of all points of 1
act1y i secants of S.
rr
not in S which lie on ex-
(Some of these sets may be empty.)
We may regard S as a clique on 6 vertices, which we denote by 1,2,3,4,5,6, and we may identify each edge (ij), i line of
rr
~
j, of the clique with the unique secant
containing the points i,j of S.
say (12), (34).
As lines of
rr,
Now consider two disjoint edges of S,
these must meet in a unique point P of
not in S and hence must be in S2 or S3'
rr
which is
If P is in S2' we denote it by (12) (34).
If P is in S3' then P must also lie on the unique secant of S disjoint from (12), (34), namely (56); in this case we denote P by (12) (34) (56). Thus each unordered triple, such as
(12), (34), (56)
of 3 mutually disjoint secants of 5,
must either define a unique point, (12) (34) (56), of 53' or three distinct points (12)(34), (12)(56), (34)(56) of 52'
In other words, 53 may be identified with
some (possibility empty) subset of the set of all 15 spreads of our clique, and then 52 may be constructed from the complementary spread-set, each spread of the latter set contributing three points of 53 forming a triangle with the edges of the spread as sides.
Conversely, if we know 52' we also know 53'
Notice that if we denote the cardinal number of 5.1 by f(i): f(i)
=
15il
(2.1)
then f(2), f(3) are non-negative integers related by the equation f(2)
3 { 15 - f(3) }.
(Of course, frO) and f(l) will be infinite unless
(2.2) rr
has finite order.)
Next suppose that we start from our clique 5 on six vertices, choose a subset, S3' of the set of all spreads of S, and define 52 as indicated above in terms of the complementary spread-set.
Then we have a set of
6 + f(2) + f(3)
51 - 2f(3)
142
R.H. Bruck
distinct points, namely 5 u 52 u 53' together with a set of 15 distinct lines (the edges or secants of 5) and a natural criterion for whether or not a given point lies on a given line. The result is a non-degenerate partial plane in the sense of Marshall Han [1] and hence can be imbedded (perhaps only by free adjunction) in at least one prOjective plane.
That is:
LEI1MA 2.3: EVe!'y spI'ead-set of oW' clique on 6 vertices can be identified wi th the 53 of a ,'-arc 5
·~n
at least one (possibly finite) projective plane.
Next suppose that 53 has been chosen (we don't claim to know how) so that the above-described partial plane can be imbedded in at least one finiteprojective plane lines.
We know the point-sets 5, 52' 53 and their relations to secant
rr.
We wish to describe the tangent and non-secant lines in terms of what
points of 5 u 52 u 53 they contain and then define the points of 50 and 51 in terms of these lines and the secants.
Can this be done? The next lemma gives a
partial answer.
LEMMA 2.4: Let
rr
be a projective pLane (of order at least
4, possibily infinite)
which contains no proper pY'ojective subplanes of oY'deY' gY'eateY' than 3.
Then
e,'el'Y t-arc of rr generates rr.
COROLLARY 2.5: Let and n I 16.
rr be a pY'ojective plane of finite oY'deY'
n, wheY'e 4
~
n
~
19
Then every 6-aro of rr geneY'ates rr.
PROOF OF LEMMA 2.4: If S is a 6-arc of w which does not generate IT. then S generates a proper projective subplane, ITO' of
IT.
However, ITO must have order at
least 4. a contradiction. PROOF OF COROLLARY 2.5: By a well-known lemma, if a projective plane of finite 2 order n has a proper subplane of order m, then either (a) n = m or (b) n
~
2
m
+
m.
In view of the inequality n
~
19, (b) cannot occur with m 2 4. and
(a) cannot occur with m ~ 4 unless m = 4, n n = 16. it follows that
rr
=
16.
Since the hypothesis rules out
has no proper subplanes of order greater than 3.
the result follows from Lemma 2.4.
Hence
143
Enriched cliques with six vertices
Finally, let us consider a 6-arc S in a finite projective plane of order n ~ 4 and define the non-negative integers f(i) by (2.1) not only for i = 2,3 but also for i
= 0, 1. Then, in addition to (2.2), we may easily establish the equa-
tions f(O) + f(l) + f(2) + f(3) f(l) + 2f(2) + 3f(3)
= n2 + n - 5 ,
(2.3)
15 (n-1).
(2.4)
We may use these three equations to solve for f(O), f{l) in terms of f(3) as follows: f(O) f(l)
(n - 7)
=
2
+ 6 -
f(3) ,
(2.5)
3 {5(n-7) + f(3) }.
(2.6)
We also notice that 3f(0) If n
+
f(l)
= 3(n - 4) (n - 5).
= 4 or 5, we deduce from (2.7) that f(O)
(2.7)
= O. Thus we get spe-
f(l)
cific values for the f(i) in these cases: If n
= 4, then f{O) = f(l) = f(2) = 0, f(3) = 5.
If n
= 5, than f(O) = f(l)
We ignore the case n
=
= 0, f(2)
= 15, f(3)
6, since there is no plane.
For n
(2.8) 10. =
(2.9)
7,8 or 9 we get the
upper bound 6,7 or 10, respectively, for f(3) (and the lower bound 0.) For n ~ 10, there are no obvious restrictions in f(3) except the inequalities
o :s; f(3)
::;; 15.
In the previous context, a difficult result of Denniston (2) (and an equivalent, point-line dual of this by Thompson et al (3) may be stated as follows:
R.H. Bruck
144
10, then f(3) < 15.
If n
(2.10)
By an elementary use of linear coding, once (2.10) is known, we can improve it to If n
10, then f(3)
~
12.
(2.11 )
-- I will leave (2.11) and other facts about arcs in the (perhaps non-existent) projective plane of order 10 for another paper. Results like (2.10), (2.11) apply to aZZ 6-arcs (in a given projective plane).
We wish to end this section with inequalities applying to some 6-arcs.
First we need two lemmas. LEMMA 2.6: Let Then
1T
7r
be a
has 7-arcs.
7-a:r.Y! of
(possibly infinite) projective plane of order at least ?
Indeed, every 5-arc of rr can be imbedded in at least one
1r.
PROOF: For
1T
of infinite order, the result follows from the proof of Lemma 2.1.
Hence we may assume that The proof for n
1T
has finite order n, where n> 7.
= 7 is difficult. We need the fact that a projective plane
of order 7 is Desarguesian.
From this it follows that every 5-arc can be
imbedded in a conic, which is an 8-arc. The proof for n o
~
i
~
~
8 is quite elementary.
2, let Fi be the set of all points of
secants of F.
Consider a 5-arc F and for 1T,
not in F, lying on exactly
The content of Lemma 2.2 (ii) is that jFol
= 1 + (n-4)(n-5) = n2 - 9n
+ 21
(2.12)
Now suppose that there exist two distinct points P,Q of FO such that the line PQ is a non-secant line of F.
It follows at once that the set F U {P,Q} is a 7-arc
containing F. Finally, suppose that no non-secant line of F contains more than one point of FO'
If P is in FO' than P lies on exactly 5 tangents and n + 1 - 5
n - 4
Enriched cliques with six vertices
non-secants of F.
We conclude from this and (2.12) that there are at least (n - 4) (n
distinct non-secants of F. 5
2
- 9n + 21)
(2.13)
Now we count the non-secants in another way. There are
10 secants of F.
exactly (2)
145
Each point of F lies on exactly 4 secants and
n + 1 - 4 = n - 3 tangents of F; hence there are exactly 5(n - 3) distinct tangents of F.
5n - 15
Since
2
(n + n + 1) - 10 - (5n - 15)
n
2
- 4n + 6
there are exactly n2 - 4n + 6 distinct non-secant lines of F.
(2.14 )
The result of subtracting (2.13) from (2.14) is
6 - (n - 4) { (n
2
- 9n + 21) - n}
2
6 - (n - 4)
(n - lOn + 21)
(n - 4)
(n - 3)(n -7) ,
= 6 -
and this is negative for n ~ 8.
Hence we have a contradiction which proves Lemma
2.6. It follows from Lemma 2.2 that a projective plane of order less than 6 has no 7-arcs.
Since there is no projective plane of order 6, Lemma 2.6 is best
possible. Now let T be a 7-arc of a projective
plane~.
A point P of
will be said
to be in T.1 provided (a) P is not in T and (b) P lies on exactly i distinct secants of T.
Clearly T. is empty for i > 3. 1
We define (2.15 )
146
R.H. Bruck
for i = 2,3.
(Clearly g(O) and g(l) will be infinite if
rr
is infinite.)
We wish
to establish the formula g(2) + 3g(3)
105.
(2.16)
First, however, let us contrast the theory of 7-arcs with that of 6-arcs. Certainly we may regard a 7-arc T of a projective plane" as a clique with 7 vertices which we call 1,2,3,4,5,6,7, and we may identify an edge (ij) i 1 j, of the clique with a secant line of T. (34).
Now consider two disjoint edges of T, say (12),
As secants, these meet in a point P of
rr
which is not in T, and must be in
exactly one of T , T . If P is in T , we may denote it by (12)(34). However, if 2 3 2 P is in T3 , P lies on a third edge, disjoint from (12), (34), and this must be exactly one of the three edges (56), (57), (67), so P may be identified with exactly one of (12)(34)(56),
(12)(34)(57),
(12)(34)(67).
Thus, if we identify T3 with a set of spreads (or unordered triples of mutually disjoint edges of T) there is a condition to be satisfied, namely: Two disjoint edges of T lie
in at most one of the spreads in T . 3
(2.17)
When we bear (2.17) in mind, we see that (2.18) counts in two ways the numbers of unordered pairs of disjoint edges of T. Now we are ready for a lemma. LEMMA 2.7: Let T be a
7-a~c
of a
p~ojective
plane rr.
seven t:-arcs 5 contained in T partition the set T3.
Then the sets 53 for the In particular, there is at
least one 6-arc 5 contained in T for which the corresponding f(3) satisfies
f(3)
~ ~ g(3) S 5 .
(2.18)
PROOF: Let 5 be one of the seven 6-arcs contained in T, and let P be a point of 53'
Then P lies on exactly 3 disjoint edges of S, and these are edges of T.
Hence P is in T . 3
Thus
147
Enriched cliques with six vertices
(2.19) for every 6-arc S contained in T.
Conversely, let P be a point in T3 . Then Plies The 6-vertices of T contained in these edges
on exactly 3 disjoint edges of T.
constitute a unique 6-arc S contained in T, and P is in S3.
Hence the S3's for
the seven 6-arcs S contained in T partition T3 . (We note that some or all of these sets may be empty.) Therefore the average value of f(3), taken for the seven 6-arcs, is precisely g(3)/7. f(3) for at least one S in T.
Thus
:;;:t g(3)
(2.20)
Since g(2), g(3) are non-negative integers, we see from
(2.17) that g(3)
~
35.
(2.21)
Inequalities (2.20), (2.21) prove (2.18) and complete the proof of Lemma 2.7. Let us go a little further for the case that rr has finite order n (with
= 0,1,2,3, are non-negative
n ~ 7.) Then the g{i), defined by (2.15) for i integers.
In addition to (2.16). we have the easily proved equations 2
g(O) + g(l) + g(2) + g(3) = n + n - 6,
(2.22)
g(l) +2g(2) +3g(3) = 21(n-l). From (2.16) and (2.22) we may solve for g(O), g(l), g(2) in terms of g(3) as follows: g(O)
(n-10)2 + 20 - g(3)
g(l)
3 { 7(n-11) + g(3)}
g(2)
(2.23)
= 3 (35 - g(O)).
In particular, if 7 ~ n S 13, then g(3) < 35, and hence (by Lemma 2.7) every 7arc must contain a 6-arc with f(3)
~
4.
Again, if 8
inequalities are improved to g(3) < 28, f(3) become g(3)
~
3.
~
n ~ 12, the corresponding
Finally, if n = 10, they
S 20, f(3) S 2. (There are also lower bounds for g(3), and hence for
148
R.H. Bruck
a suitably chosen f(3), if 7
~
n
~
10.)
It is my belief - as yet, largely untested - that in any projective plane of suitably high (possibly infinite) order every spread-set (of any size from
o to
15) of our clique on 6 vertices, can be realized as the S3 of some 6-arc S The following question concerns a possible plane in which this may not be
of~.
tl'ue. ;~oes
b:g
C:JO
(a)
~ 'JaS
t>,e1'e exiBt
C!
(possibly infinite) projective plane
7r
with the foHOUJ-
vmperti es? 6-aI'CS ani e<Jery 6-arc of
~
~.
can be imbedded in a 7-arc of
(b) 7'hD inequalit!J f(3) 2 5 holds for every 6-arc of
~.
When we compare these properties with Lemma 2.7, we deduce that f(3) = .5 for every 6-arc of
~
but hard to believe.
and g(3) = 35 for every 7-arc
I find this intriguing
of~.
(Note that, for (a), we need only require that
at least 11; the conclusion that g(3)
= 35 would then require that
~
~
have order
have order
at least 14 - and hence at least 15.) 3. E-CLIQUES OF RANK 1 Let rank
p
p
be one of the integers 0,1,2,3.
We shall say that an E-clique has
provided that some edge of the E-c1ique has rank
p
and no edge has higher
rank. In the next section we shall present a decoding algorithm for E-cliques which may be described in terms of a tree with E-c1iquesas nodes.
The root of
the tree is an E-clique which we wish to decode (by finding all spread-sets represented by it, or showing that there are none.) a spread-set attached to it.
Each node except the root has
Each spread-set represented by a node other than
the root (if the node represents a spread-set) is disjoint from the spread-set attached to the node and, in union with the attached spread-set, forms a spreadset represented by the root.
Only the root can have rank 3.
If the tree
branches at a node (which may be the root) then the node has rank 2 and there are precisely three branches.
The nodes along any path from the root have decreasing
size and non-increasing rank; the corresponding attached spread-sets have increasing size and are simply ordered by inclusion. at the end of a branch) can have rank 1.
Only an end-node (or node
An end-node is one of the following:
149
Enriched cliques with six vertices
(a) an E-clique which does not represent a spread-set (according to a simple test); (b) an E-clique which represents a simple spread-set (possibly the empty spread-set) known by means of the algorithm; (c) an E-clique of rank 1.
The
algorithm allows us to write down all spread-sets (if any) represented by the root, provided we can decode the end-nodes of rank 1. The present section is devoted to an analysis of the E-clique of rank 1 with the aid of the theory of 6-arcs in a projective plane of order 4. Let 5 be a 6-arc projective plane n of order 4.
The points of 5 are denoted
by 1,2,3,4,5,6; the secant, or edge, through points i,j of 5 (iFj) is denoted by (ij).
Each point of
* lies
on exactly 5 distinct lines of n, and each line of n
contains exactly 5 distinct points of distinct secants of 5.
n.
Each point of 5 lies on exactly 5
Hence there are no tangents of 5, and the 21 lines of 5
consist of 15 secants and 6 non-secants of 5. points of 5 and 5 points of 53'
The 21 points of 5 consist of 6
Each point of 53 lies on exactly 3 distinct
secants of 5 (disjoint as edges of 5) and is defined by the spread consisting of the 3 disjoint edges.
Each point of 53 also lies on exactly 2 distinct non-
secant lines.
Each secant line contains 2 distinct points of 5 and 3 distinct
points of 53'
Each non-secant line contains no point of 5 and 5 distinct points
of 53' Call two distinct paints of 53 disjoint if they do lie on a common secant line.
Thus they lie on a unique common non-secant line.
Through each of the two
points there is a unique second non-secant line, and the two latter non-secants meet in a third point of 53 disjoint from the first two.
Thus 3 mutually disjoint
points of 53 either lie on a common non-secant line or form the vertices of a triangle with non-secants as sides.
5uch triangles go in "polar" pairs.
Indeed,
let L , L , ... , L6 be the six distinct non-secants in some order, let T be the l 2 triangle with sides Ll , L , L , and let T' be the triangle with sides L , L , L . 2 3 4 5 6 The vertices of T are three mutually disjoint points of 53' and so are the vertices of T'.
Consider the 9 points of 53 which are neither vertices of T nor
vertices of T'.
These 9 points must lie in threes on the three sides of T and
also in threes on the three sides of T'.
Each such point, in fact, must lie on
exactly one side of T and exactly on side of T'. through a vertex of T.
Now consider a secant line
This line contains exactly three points of 53'
these is a vertex of T (the one under consideration.) site side of T and hence on a unique side of T'.
One of
A second lies on the oppo-
The third must be the opposite
150
R.H. Bruck
vertex of T'.
In this way we see that the 9 secants passing in threes through
the 3 vertices of T also pass in threes through the three vertices of T'; the two sets of vertices are "covered" by the same nine edges. that there are exactly
t (~)
=
We note, incidentally,
10 such "polar" pairs.
From the above discussion it should be clear that a set of 4 or 5 distinct, mutually disjoint points of 53 must lie on a common non-secant. Now we have enough to discuss the [-cliques of rank 1 in simple terms.
(We
could also consider all [-cliques in the same vein, but at a high cost in complexity. ) As it turns out, every [-clique of rank 1 is equivalent (under permutation of the vertices of the clique) to exactly one of the [-cliques specified by their incidence arrays as follows (listed in order of descending size): X 1
X0 1
X 0 1 1
X
o X1
oX0 1 oX0 ox
X
X 0 1
o X1
X
X0
1 1 X 1 1 (a)
o1
X0 0
0 1
o X0 o0 X
1 lOX
X 1 100 0
1 X 1 000
X
o1
1 X 000
X 1 0 0
1
(d)
(c)
(b)
X 1 000
o X0 o0 X 1
lOX 0
oX
X
X 0 0 1
0
000
X 1 000 0 1 X 0 000
o0
X 1 0 0
001 X 1 0
000 X 1
001 X 0 0
OOOlXl
000 1 X
o0
000 1 1 X
OOOOlX
o0
0 1 X
(c' )
(d' )
0 0 X 1
(b' )
Notice that if we add the "null" array (the incidence-array of the [-clique of rank 0) and denote it by (a'), the operation of interchanging 0 and 1 induces the permutation (a,a') (b,b') (c,c') (d,d') on the eight arrays and reolaces an array of size k by one of size 5-k for
lSI
Enriched cliques with six vertices
o~
k ~ 5.
A sufficient description of each of the 7 E-cliques follows.
(Orbit
and orbit length refer to spread-sets rather than E-cliques). (a) Size 5.
Represents 6 spread-sets.
One orbit, length 120.
description: A clique (with 6 vertices) on the edges of length 1.
Graphic
Geometric de-
scription: The six spread-sets are the points of the six non-secsnt lines.
(b) Size 4.
Represents 2 spread-sets.
One orbit, length 30.
Graphic de-
scription: The 3 edges of rank 0 form a spread (here the spread (12) (34) (56).) Geometric description: The 2 spread-sets consist of four points other than the
common point (here (12)(34)(56» (c) Size 3.
of one of two distinct non-secant lines.
Represents one spread-set.
One orbit, length 60.
Graphic de-
scription: The 6 edges of rank 0 form a closed cycle of length 6 (here (12), (23),
(34), (45), (56), (61» on the 6 vertices of the clique.
Geometric description:
There exist exactly two spreads all of whose edges have rank 0 (here (12)(34)(56) and (16)(23)(45»
and these are disjoint.
The unique spread-set consists of the
remaining 3 points of the non-secant line through these points. (d) Size 3.
Represents 2 spread-sets.
One orbit, length 10. Graphic de-
scription: The 6 edges of rank 0 partition the 6 vertices of the clique into two
closed cycles of length 3 (here (12), (23), (31) and (45), (56), (64).)
Geomet-
ric description: The 6 paints of the 6-arcare partitioned into 2 triangles (here
1,2,3 and 4,5,6).
The 2 spread-sets consist of the 3 vertices of each of two
"polar" triangles with non-secant sides, covered by the 9 secants containing one vertex of each triangle of the 6-arc. (c') Size 2.
Represents one spread-set.
One orbit, length 60.
Graphic de-
scription: The 6 edges of rank 1 form a closed cycle of length 6 (here (12), (23),
(34), (45), (56), (61»
on the 6 vertices of the clique.
Geometric description:
The spread-set consists of the two spreads (here (12) (34) (56) and (16) (23) (45»
all of whose edges have rank 1. (d') Size 2.
These are disjoint.
Does not represent a spread-set.
No orbit.
Graphic descrip-
tion: The 6 edges of rank 1 partition the 6 vertices of the clique into two closed
152
R.H. Bruck
cycles of length 3 (here (12). (23), (31) and (45). (56), (64)). exists no spread whatever all of whose edges have rank 1.
Hence there
Geometvic descviption:
None. (b') Size 1.
Represents one spread-set.
One orbit, length 15.
Gvaphic
desC1>iption: The 3 edges of rank 1 are mutually di sjoi nt and fonn a spread (here
(12) (34) (56).)
The spread-set consists of this spread.
Geometvic description:
Essentially identical. One final remark.
If the list at the end of the paper were complete
(rather than limited to spread-sets of size at least 8) each of the 6 incidencearrays (a), (b), (c), (d), (c'), (b') would appear (and (d') would not.) Instead, these 6 arrays are represented by arrays equivalent to the complementary arrays and of sizes 10, 11, 12, 13, 14 respectively. 4. DECODING DETAILS In
§
3 we gave a general description of our decoding algorithm in terms of
a tree E-cliques as nodes. decode.
The root of the tree is the E-clique we are trying to
We find all spread-sets represented by the root by examining all end-
nodes of the tree (together with their attached spread-sets) and also decoding the end-nodes of rank 1 with the help of the discussion in
§
3.
In the present section we will give the details for forming the tree. (This is not always unique.) In some cases the tree consists of a single node, serving as both root and end-node.
This will be true of the tree for the E-clique of rank 0 (which re-
presents the empty spread-set) and of the tree of an E-clique of rank 1 (which may represent 0,1,2 or 6 spread-sets according to the case; see
§
3.)
Other
cases will be clear from the discussion which follows. It will be convenient to separate the discussion into two parts, one concerned with E-c1iques of rank 3 and the other with E-c1iques of rank 2. E-CLIQUES OF RANK 3 (Recall the convention that an E-clique of rank 3 occurs in a decoding tree only if it is the root.)
Consider an E-clique of rank 3 and suppose, first, that
we have before us a spread-set of which it is the E-clique.
By definition (See
§
153
Enriched cliques with six vertices
1) an edge of the E-clique has rank i only because the spread-set contains exactly i of the 3 spreads containing that edge.
In particular, an E-clique cannot
have two disjoint edges, one of rank 3 and the other of rank 0, if it is to represent a spread. Thus, if an E-clique of rank 3 has an edge of rank 0 disjoint from an edge of rank 3, the E-clique does not represent a spread.
(This is one case where an
E-clique of rank 3 is the only node of its decoding tree.) Next consider an E-clique of rank 3 with the property that, for every edge of rank 3, each of the 6 edges disjoint from this edge has positive rank. the spread-set
~
as follows: A spread is in
edges of the spread has rank 3.
~
We form
if and only if at least one of the
(It should be clear that a spread-set represented
by the E-clique, if any exists, must contain
~.)
Now let A,S be the incidence
arrays of the given E-clique of rank 3 (the root) and of the E-clique associated with~.
(The latter E-clique also has rank 3.) Consider the difference array ,~
A
=A
- S •
It can occur that A" has a negative off-diagonal entry.
In this case, our
given E-clique (with incidence array A) does not represent a spread, and is the only node of its decoding tree. In the contrary case, A* is the incidence array of an E-clique, and this will be the unique second node of the decoding tree of our original E-clique. thi~
second node we attach the spread-set
~.
To
We denote some obvious properties.
First of all, every edge of our clique which had rank 3 for the root has rank 0 for the second node.
In addition, no edge has greater rank for the second node
than it has for the root.
Finally, the size of the second node is at least 3 less
than the size of the root. Since every spread in
o for
~
contains an edge having rank 3 for the root and rank
the second node, it follows that no spread in
one exists) represented by the second node. disjoint) of
~
~
can be in a spread-set (if
In fact the union (necessarily
and a spread-set represented by the second node will be a spread-
set represented by the root; and every spread-set (if any) represented by the root will have this form. The second node cannot have rank 3 but it might have rank 0 or 1 (in which case it is the unique end-node) or rank 2 (in which case we must examine it
154
R.H. Bruck
further.} E-CLIQUES OF RANK 2 (Such an E-clique may either be the root of a decoding tree which we are just beginning to construct or a node other than the root (but not necessarily adjacent to the root) of a decoding tree in course of construction. latter case, we assume that it has an attached spread-set
~O
In the
with the following
two properties: (a) LO has no spread in common with any spread-set represented by the E-clique; (b) the union of
~O
with any spread-set represented by the E-clique
is a spread-set represented by the root.} Consider an E-clique of rank 2, with incidence-array A.
Choose an edge of
the E-clique of rank 2 and consider the 3 distinct spreads containing this edge. There are three possibilities of interest: (a) At least two of the three spreads contain one or two edges of rank O. (b) Exactly one of the three spreads contains one or two edges of rank O. (c) None of the three spreads contains an edge of rank O. We may explain our interest as follows. before us with the E-clique as its E-clique.
Suppose that we have a spread-set The chosen edge has rank 2 only
because the spread-set contains exactly two of the three spreads containing the edge.
Another edge has rank 0 (if such an edge exists)
only because the spread-
set contains none of the three spreads containing that other edge.
Hence, if the
spread-set exists, we must either have case (b) (in which case we know the two spreads in the spread-set containing the chosen edge of rank 2) or case (c) (in which case there are three possibilities for the two spreads.) If, for the chosen edge of rank 2, case (a) occurs, the E-clique represents no spread and may be taken as an end-node of the decoding tree we are constructing.
If, however, case (b) or (c) occurs, there may be another edge of rank 2
for which case (a) occurs.
For this reason, there may be more than one way of
constructing the decoding-chain. Note however that if the E-clique either has no edge of rank 0 or has at least one edge of rank 0 but no edge of rank 0 disjoint from an edge of rank 2, then (c) is the only possibility, regardless of the choice of edge of rank 2. Again, if the E-clique has an edge of rank 0 disjoint from one or more edges of rank 2, each of those edges of rank 2 must yield case (a) or case (b). may be worth while to examine the edges of rank 0 (if any.)
Hence it
155
Enriched cliques with six vertices
Now let us assume that we have decided to base our construction procedure on a particular edge of rank 2. If case (a) occurs, we take the E-clique as an end-node of the decoding chain. If case (b) occurs, let Il be the unique spread-set consisting of the two spreads with the chosen edge of rank 2 as one edge and with no edges of rank O. Let B be the incidence-array of the E-clique associated with I l , and consider the difference array
(Remember, our given E-clique had A as incidence-array.) This time A": has no negative off-diagonal elements; it is the incidence-array of an E-clique which is the unique successor-node of our E-clique in the decoding chain we are constructing.
To this successor-node we attach a spread-set I, where I = Il if our E-
clique (with incidence matrix A) is the root of the tree, and E-clique is not the root and has attached spread set IO' tively if our E-clique is not the root) that
~
~
= ~O
u
~l
if our
We may verify induc-
has the requisite two properties.
If case (c) occurs, the situation is much the same except there are three different successor-nodes; the decoding tree has three branches at our given E-clique. In either case a successor-node has length 2 less than the given node, at least one less edge of rank 2 and at least one more node of rank 0, and rank at most 2. We shall stop here with the general discussion and consider two specific examples. If our root is given by (1.5) and so has rank 3, then the set
~
(defined in
the discussion of rank 3) consists of every element of the spread-set (1.2) except the spread (14)(25)(36).
The unique successor-node of the root has rank
and size 1 and represents only the l-element spread set consisting of (14)(25) (36).
Hence the decoding tree is a chain with two nodes, and (1.2) is the only
spread-set represented by (1.5). -- In particular, our remarks about (1.7) were unnecessarily elaborate, °a lthough they were appropri ate in the context of the general problem.
156
R.H. Bruck
For our second example we take as root one of the E-cliques covered by the catalogue at the end of the paper, namely the E-clique with incidence-array X 3 1 222
(4.1 )
3 X2 1 2 2 1 2 X3 2 2 213X22
222 2 X 2 22222 X This has rank 3.
Hence our first step is to subtract from (4.1) the array of the
spread-set (4.2)
(12)(34)(56), (12)(35)(46), (12)(36)(45), (15)(26)(34), ( 16 )( 2 5)( 34 ) . Thi s 1eaves
(4.3)
X0 1 2 1
o X2 1 1 1 2 X0 1
2 lOX 1
X1 1 1 X
The array (4.3) has exactly two edges of rank 2, namely (14) and (23).
Hence any
spread-set represented by the array must contain exactly one of the following sets k1' k2' l:3 and also exactly one of the following sets ~4' l:5' ~6:
(14)(23)(56), (14)(25)(36).
l:4: (14)(23)(56), (15)(23)(46).
l:2: (14)(23)(56), (14)(26)(35).
l:5: (14)(23)(56), (16)(23)(45).
(14)(25)(36), (14)(26)(35),
Z6: (15)(23)(46), (16)(23)(45).
~1:
~3:
We wark with the edge (14) and subtract from (4.3) the arrays of Zl' Z2' l:3'
(4.4)
Enriched cliques with six vertices
157
getting, respectively, C = X 0 1 0 1 1, l o X1 1 0 1
C2= X 0 1 0 1 1, o X1 1 1 0
C= X0 1 0 1 1 3 o X2 1 0 0
1 1 X0 1 0
1 1 X0 0 1
12XOOO
o 1 o X1 o 1 1 X0
o 1 oX1 1 o 1 X0 o1 1 oX
o lOX o0 1
1 0 lOX
(4.5)
1 1
X1
001 1 X
At this point in our construction of the decoding tree we have 5 nodes, including the root (4.1), the node (4.3) with associated spread-set (4.2), and for i
= 1,2,3, the node C.1 for which the associated spread-set is the union of (4.2)
and
The nodes C1 , C2 are end-nodes. the fo 11 owi ng; ~i'
Oecoding by the method of
§
3, we find
C1 represents only the set {(13)(26)(45), (15)(23)(46), (16)(24)(35)}. (4.6) C represents only the set {(13)(25)(46), (15)(24)(36), (16)(23)(45)}. 2 Combining with
~1
or
~2
as appropriate, we get two spread-sets represented by
(4.3), namely (13) (26) (45),
(4.7)
(14)(23)(56), (14)(25)(36), ( 15 )( 23 )( 46 ) , (16)(24)(35) ; (13)(25)(46) ,
(4.8)
(14)(23)(56), (14)(26)(35), (15 )(24 )(36), (16)(23)(45) . To get spread-sets represented by (4.1), we must combine each of these with (4.2) - but it is convenient not to do this at present.
Note that (4.7), (4.8) contain
~4' ~5'
respectively. The node represented by C3 is not an end-node, since it has rank 2.
In
158
R.H. Bruck
fact, (23) is the unique edge of rank 2.
Since (14) has rank 0, it follows that
any spread-set represented by C must contain ~6' On subtracting the array of ~6 3 from C we get the array which represents (13)(24)(56). Hence there is a third 3 spread-set represented by (4.3), namely (13)(24)(56) ,
(4.9)
(14)(25)(36), (14)(26}(35), (15)(23)(46), (16)(23) (45). Thus, using a coding-tree with 6 nodes, we have found that (4.1) represents exactly three distinct spread-sets, obtained by combining (4.2) with each of (4.7), (4.8), (4.9). It seems worthwhile to go a little further. E-clique defined by (4.1)
By making a sketch of the
or otherwise, we may verify that the automorphism
group, G, of the E-clique has order 8 and is generated by the permutations (12) (34), (13)(24), (46).
Obviously G maps the spread-set (4.2) upon itself.
addition G maps (4.9) upon itself.
In
Hence the union of (4.2), (4.9) is a spread-
set represented by (4.1) with stability group G, orbit length 720/8 = 90.
The
permutation (12)(34) fixes each of (4.7), (4.8), but each of (13)(24), (56) interchanges (4.7), (4.8).
Hence (4.7), (4.8) are equivalent and each has stabil-
ity group H (in G) of order 4 generated by (12}(34), (13)(24)(56).
Thus the
union of (4.2), (4.7) (or of (4.2), (4.8)) is a spread-set represented by (4.1) with stability group H, orbit length 720/4 a 180. At this point we have given c complete discussion - aside from the matter of degrees and anti-degrees - of entries numbered 31, 32 in the catalogue at the end of the paper. Although this second example was worth examining, it is far from typical. Consider, for example, the 21 orbits of size 9 listed in the catalogue. Sixteen of these are represented by E-cliques which represent a unique spread-set.
Two
others (numbered 44, 50) are each represented by an E-clique which represents two equivalent spread-sets.
Another (#52) is represented by an E-clique which repre-
sents 4 equivalent spread-sets.
Finally, orbits numbered 53, 54 are represented by
the same E-clique and this, in turn, represents a total of 5 spread-sets,two belonging tooneorbitand three to the other. A similar situation holds for sizes 10 and 8.
159
Enriched cliques with six vertices
5.
NORr~AL
FORM
Returning to our example (1.5), suppose we omit the X in each row and arrange the entries in each row in decreasing order.
The result is (5.1)
3 3 2 1 0 3 2 2 1 3 2 2 1
322 1 3 2 2 1
33210 This tells us at a glance that vertices 1;6 of the corresponding E-clique have the same type, and that vertices 2,3,4,5 also have the same type, but the types are different.
The same process, applied to (1.8), yields 33210
(5.2)
33210 322 322 1 3 2 2 1
3 2 2 1
Comparison of (5.1), (5.2) tells us that (1.5), (1.8) have the same type, with vertices differently arranged.
We prefer the arrangement (5.2), and call this
norma 1. Since (5.1) or (5.2) is very bulky, we prefer to denote each vertex type by a single letter, chosen from as many as are needed of a,b,c,d,e,f.
For example,
the type 33210 of a vertex for an E-clique of size 9 is denoted by b. 3 3 3 2 2
However, the type
R.H. Bruck
160
of a vertex for an E-clique of size 13 is also denoted by b; the same letters are used with different meanings for different sizes. two tables at the end of the paper. of the letters as vertex types.
The exact details are given in
One table gives, for each size, the meaning
The other lists the type of each E-clique dis-
played in the catalogue of spread-sets.
This type appears as a six-letter "word",
for example, abcddd.
(5.3)
This word, like all the others, has its letters in alphabetical order. An E-clique of type (5.3) is said to be normal, or in normal form, provided that, for 1 < i < 6, the type of the ith vertex is given by the ;th letter of the word.
That is, vertices 1,2,3,4,5,6 have types a,b,c,d,d,d respectively.
Simi-
larly in general. It is often more convenient to write a type such as (5.3) in a form such as
abcd
3
- but I wanted to avoid exponents in the table.
2 4 Both (1.5) and (1.8) have type b d , but (1.8) is in normal form, whereas (1.5) is not.
2 4
Consider the orbits of size 10, type c d three of these, numbered 26, 27, 28. is in normal form.
According to the table there are
In each case the E-clique in the catalogue
However, this does not specify the E-clique uniquely; it can
have 2
4
/ g
48 / g
different forms, where g is the order of the automorphism group of the E-clique. (For orbits 26,27,28, g
= 2,
2 and 8, respectively; the automorphism group coin-
cides with the stability group in each case.)
Thus given, a spread-set of size
2 4
10, type cd, we cannot identify its orbit immediately by reducing its E-clique to an equivalent normal form.
Nevertheless, the concept of normal form simplifies
the problem of equivalence a great deal. One final remark.
In constructing the catalogue I used a variety of methods
which are probably not worth discussing.
Common to all of them was the necessity
of handling some orbits a dozen times or more before arriving at a complete set.
Enriched cliques with six vertices
161
I did the work by hand; I believe that a machine program which left much less for me to do would have been very expensive.
Now that the catalogue exists, however,
it should prove useful as part of a machine program in further exploration.
162
R.H. Brock
CATALOGUE OF SPREAD-SETS ABBREVIATIONS: If = ordinal number. OL = orbit length. D = degree. AD = anti-degree. For definitions. see section I . SUGGESTION: In most cases, a well-designed sketch of the E-clique will reveal a great deal. SIZE 15--1 ORBIT
SIZE 14--1 ORBIT
X33333 If 01 3X3333 OL 001 33X333 15 333X33 D 6 3333X3 33333X AD --
X23333 If 02 2X3333 OL 015 33X233 8 6 332X33 D 6 5 3333X2 33332X AD --
SIZE 13--2 ORBITS X13333 # 03 lX3333 OL 045 33X232 4 8 1 332X23 D 6 5 4 3332X2 33232X AD 12
X23332 # 04 2X2333 OL 060 32X233 4 6 3 332X23 D 6 5 4 3332X2 2 23332X AD 0
SIZE 12-- 5 ORBITS X03333 If OX3333 OL 33X222 332X22 D 3322X2 33222XAD
05 015 5 2
12 3
X3222.3 # 08 3X3222 OL 060 23X322 2 3 6 1 223X32. D 6 5 4 3 2223X3 3 32223X AD 0
X 13323 # 06 IX3332 OL 180 33X 123 2 6 4 331X32 D 6 5 4 2323X2 32322X AD 2112
X 12333 # 07 lX3233 OL 180 23X322 2 5 4 1 323X22 D 6 5 4 3 3322X2 2 1 33222X AD 1 0
X33222 II 09 3X3222 OL 020 33X222 3 222X33 D 5 9 3 2223X3 3 22233X AD 0
SIZE 11--9 ORBITS X03332 II 10 OX3323 ot 180 33X122 6 5 331X22 D 5 4 3222X2 23222X AD 312211
X31313 # 11 3X31310t 090 13X232 1 4 6 312X23 D 6 5 4 1332X2 31232X AD 24
X11333 If 12 IX3232 OL 360 13X322 1 4 4 323XI2 D 6 5 4 3 2 3321X2 32222X AD 14
Enriched cliques with six vertices
SIZE ll--CONTINUED X32123 # 13 3X3212 OL 060 23X321 4 123X32 D 6 47 2123X3 1 32123X AD 3 13
X23132 # 14 lX2313 OL 060 3lX231 6 3 2 13lX23 D 5 4 3 313lX2 2313lX AD 31 13
X33122 # 15 3X1322 OL 360 31X232 1 3 4 3 13lX23 D 6 5 4 3 223lX2 2ZZ3lX AD Zll Z 01
X133ZZ # 16 1X33ZZ OL 045 33Xl ZZ Z 8 1 331X2Z D 6 4 2 2ZZlX3 2 ZZZ23X AD 1 0Z
X133Z2 # 17 1XZZ33 OL 180 3lX2Z2 4 Z 4 3ZlXZZ D 5 4 3 21 23ZlX2 2 2 Z3ZZlX AD 1 0
X3Z2Z2 # 18 3X2ZZ2 OL 030 2lX3Z2 1 6 4 2Z3XZZ D 6 4 3 ZZ2lX3 4 2ZZZ3X AD 0
X03133 # 19 OX332Z OL 360 33X211 13lX2Z D 53 4 6 3 1 3Z1lXZ 2 3Z1ZlX AD 3 Z3
X033Z2 # 20 OX33ZZ OL 045 33XOZZ Z 8 33OXZ2 D 5 4 ZZZlX2 1 ZZZZlX AD 4 Z4
X0332Z # Zl OX3Z3Z OL 360 33Xll Z 3Z1XZZ D 53 4 4 3 3 Z31lX2 Z 1 2 2Z22lX AD 3 2 1
X023Z3 # ZZ OX3232 OL 180 23X221 2 6 Z 32lX1Z D 5 4 3 2321XZ 1 2 Z 3Z1ZlX AD 4 Z 1
X33112 # 23 3X3Z11 OL 060 33X1 Zl lZ1X33 D 53 4 6 3 1 llZ3X3 Z 3 Z1133X AD 3 Z
X13312 # Z4 1X133Z OL 072 31X132 5 5 331X12 D 5 3 1331X2 5 ZZZ2lX AD 2
X3311Z # 25 3XllZ3 OL 360 31X3Z1 1 1 5 3 113X3Z D 6 5 4 3 1223XZ 1 3 1 23122X AD 3 2 1
X33112 # Z6 3X1321 OL 360 31X22Z 1 1 4 3 13lX2Z D 6 5 4 3 21 122lX3 3 2 21223X AD 2 1
X1133Z # Z7 lX3123 OL 360 13XZ2Z 3 3 3 1 31lXZZ D 5 4 3 Z 322lX1 1 13 Z3Z21X AD 3 Z 1
X23113 # 28 lX1331 OL 090 31X222 4 6 13lX22 D 5 3 132lX2 4 1 3122lX AD 2 0
X33112 # 29 3X1222 OL 360 31X222 2 4 2 2 12lX32 D 5 4 3 2 1223X2 2 1 2Z22lX AD 221 0
X32221 # 30 3X1222 OL 120 21X322 1 3 6 223Xl2 D 6 4 3 2221X3 1 1Z223X AD 3 13 01
SIZE 10--15 ORBITS
163
164
R.II. Bruck
SIZE lQ--CONTINUED X31222 /I 31 3X2122 OL 090 12X322 213X22 D 514 4 34 11 2222X2 4 1 2Z22lX AD 1 0
211 2 0 2
X222Z2 /I 33 lX2222 OL 006 22X222 10 222X22 D 3 222lX2 5 22222X AD 0
X00333 /I 34 OX3222 OL 060 03XZ22 3ZlXll D 56 3Z21Xl 6 32211X AD 3
X0313Z /I 35 OX331 Z OL 360 33Xl11 2 3 4 131X2Z D 5 4 3 311lXZ 4 2 Z21 ZlX AD 3 Z
X30312 /I 36 3X30Z1 OL 360 03XZZ2 1 5 3 302XZ2 D 5 4 3 1222X2 21222X AD 4 13 2 2 3
X02133 /I 37 OX3321 OL 360 Z3X211 1 5 3 13lX12 D 5 4 3 nl1X2 1 Z 3 3112lX AD 4 3 2
X0322Z II 38 OX232Z OL 360 32XllZ 1 Z 5 1 231XZ1 D 5 4 3 2 ZZI2XZ ZZ212X AD 4 1312113
X02331 /I 39 OX3Z22 OL 360 23X1l2 321X12 D 5 2 4 2 34 21 3211X2 12Z22X AD 33 2 2 11
X03132 II 40 OXZ322 OL no 3lX211 4 3 13lX12 D 514 3 21 3Z11X2 ZZ122X AD 4 1312 3 11
X33111 II 41 3X3111 OL 010 33Xl11 lllX33 D4 9 1113X3 6 11133X AD 1
X33111 /I 4Z 3Xll13 OL 180 31XZ21 112X32 D 614 3 34 21 1123X2 2 4 13122X AD 3 Z
Xl3311 II 43 1Xl133 OL 180 3lXZ21 Z Z 4 1 312X12 D 5 4 3 Z 132IX2 3 Z1 1312lX AD 3 2 1
X33111 II 44 3XOZZ2 OL 120 3OX222 6 3 122X22 D 5 2 1222X2 3 1222lX AD 3 3 1
X13311 II 45 1Xl Z3Z OL 360 31X122 3Z1X12 D 5 2 4 2 3 2 2 3 13Z1XZ lZZ22X AD 3 12411
X03ZZ2 /I 46 OX23Z2 OL 090 3lXOZ2 23OX2Z 22Z2Xl 2Z221X
X0322Z II 47 OX23Z2 OL 360 32XllZ 5 Z31XZI D 514 2 3 21 Z212X2 Z2212X AD 4 1312113
X32211 /I 48 3X1122 OL 180 ZlX321 Z13X12 D 614 13 6 21 12Z1X3 2 Z2 lZ123X AD 3 2 1
32 180 1 4 3 2 5 4 3 Z
SIZE 9--21 ORBITS
Enriched cliques with six vertices
SIZE 9--CONTINUED X32112 # 49 3X1221 OL 060 21X321 1 3 2 123Xl2 D 6 4 3 2 3 1221X3 21123X AD 312411
X31122 # 50 3Xl122 OL 090 llX322 113X22 D 45 3 2 2 2 222ZXl 22221X AD 3 2 14
X32112 # 52 3X1221 OL 360 21X222 3 3 2 12ZX22 D 4 3 2 11 122ZX2 2122ZX AD 23 12 01
X12222 # 53 lX2222 OL 015 2ZX122 8 221X22 D 3 01 222ZXl 6 22221X AD 1
X31122 # 51 3X1212 OL 360 l1X322 D 5 14 2 3 3 2 3 123X21 212ZX2 2221ZX AD 31231101 54 060 6 3 3 2 3 113 0 2
SIZE 8--24 ORBITS X00332 # 55 OX3212 OL 360 03X122 D 443 4 321XlI 3121Xl 22211X AD 4 2 3 4 21
X33002 # 56 3X0221 OL 180 30X221 4 02ZX22 D 443 022ZX2 4 2112ZX AD 4 2 3 21
X30311 # 57 3X3011 OL 180 03X122 1 2 5 301X22 D 5 4 3 112ZX2 1 6 1122ZX AD 4 3
X01133 # 58 OX3311 OL 090 13X211 4 13ZXll D 443 3111X2 4 3111ZX AD 4 2 3 21
X33011 # 59 3X0212 OL 360 3OX221 2 2 02ZX22 D 443 2 112ZX2 1212ZX AD 4 2 3 2 2 3
X03131 # 60 OX2312 OL 720 32Xll1 1 2 3 2 131X12 D 5 4 3 2 3111X2 1 4 2 1212ZX AD 4 3 2
X20312 # 61 ZX3021 OL 360 03X122 301X22 D 43 3 421 122ZXl 21221X AD 5 13 3 2 3
X20312 # 62 ZX3021 OL 180 03X212 1 2 3 2 D 5 4 3 2 30ZX21 121 ZX2 1 4 2 2121 ZX AD 4 3 2
X20312 # 63 ZX30210L 180 03X221 1 7 30ZX12 D 5 3 1221X2 4 21I2ZX AD 4 3 2
X03122 # 64 OX2312 OL 720 3ZXlli D 514 13 3 2 3 131X21 211 ZX2 1 3 2 2211ZX AD 4 3 2 11
X03122 # 65 OX1322 OL 360 31X121 D 4 3 322 3 131XI2 2221Xl 2 1 22121X AD 4 3 2 311
X03122 if 66 OX1322 OL 180 31X211 1 61 13ZXll D 5 3 1 2211X2 2 2211ZX AD 4 2 5
165
R.H. Bruck
166
SIZE 8--CONTINUEO X30212 # 67 3XZOZl OL 360 OZX2ZZ ZOZX2Z 0 4 2 3 4 22 12ZOO ZI Z21X AD 513 2 2 3 11
X30Z21
X31121 # 70 3XI112 OL 180 11X31 Z 113X21 051422411 211 ZX2 12Z1 ZX AD 3 2 2 5
X31121 # 11 3Xl1 ZI OL 180 11 X312 113Xl Z D 5134 2211 2211X2 1122ZX AD 3 4 2112
X31112 # 72 3X1l21 OL 180 11X222 2 4 llZX22 04 31Z 11 122ZXl 2 4 1 21Z21X AD 3 Z 0 XOZ222 # 75 OXZ22Z OL 090 ZZXI1Z 4 221X21 D 3 24 221ZXl 22Z11X AD 4 Zz114
#
3X2111 OL
68
no
OZXZ22
ZI 00 2 D 4 2 3 32211 2121X2 112ZZX AD 413 2 2 212
73 360 2 4 2 3 2 2 12 Z Z 3 2 31 X1l222 # 76 lXl Z22 OL 060 llX222 22ZXli 2221Xl 22211X 78 120 3 4 2 3 01
1 3 3 3 2 1
X31112 # 69 3XZlll OL 120 1 ZX311 113X21 D 613 4 2 3 l11ZX3 4 21113X AD 3 23
X02222 # 74 OXZ222 OL 015 2ZX022 22OX22 222ZXO 22220X
Enriched cliques with six vertices
167
SIZE AGAINST LETTER TYPES OF VERTICES
15 14 13 12 11
10 9 8
a
b
33333 33332 33331 33330 33320 33310 33300 33200
33322 33321 33311 33220 33210 3311 0
c
e
d
33222 33221 33211 33111 32210
32222 32221 32220 32111
22222 32211 22220
f
22221 22211
TYPES OF CATALOGUED E-CUQUES SIZE
#
TYPE
15
01
aaaaaa
14
02 aaaaaa
13
03 aabbbb 04 bbbbbb
12
05 06 07 08 09
aacccc bbbbcc bbcccc cccccc cccccc
10 11 12 13 14 15 16 17 18
aaccdd bbcccc bccccd cccccc cccccc ccccdd ccccdd ccdddd dddddd
11
SIZE
#
10
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
abcddd bbbbee bbcdde bbdddd cccccc ccccce ccccdd ccdddd ccdddd ccdddd cdddde dddddd ddddee ddddee eeeeee
9
34 35 36 37 38 39
addeee bbceef bbddff bbeeee bbeeff bdeeef
8
9
TYPE
SIZE
# TYPE 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
bdeeef cccccc cceeee cceeee cddfff ceeeef ddddff ddeeff eeeeee eeeeee eeeeff eeeeff eeffff ffffff ffffff
55 56 57 58 59 60
accddf acceef bbccff bbdddd bcceff bcdddf
SIZE
#
TYPE
8
61 62 63 64 65 66 67 68 69 70
ccccff ccccff ccccff ccddff ccddff ccddff cceeff cdefff dddddd ddddff ddddff ddffff ddffff eeeeee eeffff ffffff ffffff ffffff
71 7Z
73 74 75 76 77
78
R.H. Bruck
168
BIBLIOGRAPHY 1. 2. 3.
M. Hall, Jr., Projective planes, Trans. Amer. Math. Soc. 54 (1943), 229-277. R.H.F. Denniston, Non-existence of a certain projective plane, J. Austral. Math. Soc. 10 (1969), 214-218. F.J. MacWilliams, N.J.A. Sloane and J.G. Thompson, On the existence of a projective plane of order 10, J. Combinatorial Theory Ser. A, Z4 (1973), 66-78. Van Vleck Hall University of Wisconsin Madison, WIS. 53706, U.S.A.
Annals of Discrete Mathematics 18 (1983) 169-176 © North-Holland Publishing Company
169
BLOCKING SETS IN AFFINE PLANES :,{
A.A. Bruen and M.J. de Resmini
SECTION 1 As usual AG(2,q) (PG(2,q»
denotes the affine (projective) plane over GF(q)
the finite field of order q. Let n be a finite projective plane of order n. in
n
is a subset S of the points of
n
Recall that a blocking set
satisfying the following two conditions:
(i) Each line of
n
contains at least one point in S;
(ii) Each line of
n
contains at least one point not in S.
We say that S is irreducible if no proper subset of S is also a blocking set.
This is equivalent to saying that each pOint P of S lies on at least one
tangent to S. The following result is contained in [2) and [6]. THEOREt~
1.1: If S is any blocking set in n then
n+
v'n + 1 .;;;
IS I .;;; n2 - y'i\:
Moreover, if S is irreducible then
IS I .;;; n \I'll +
1.
Let n be a finite affine plane of order n.
A (intersection set) in
We define a blocking set
to be a subset Wof the points of n A such that each line A of n A contains at least one point of W. (Note that, unlike the definition in the projective plane, Wmay contain a n
1i ne. )
As in the case of a projective plane Wis irreducible if each point lies on at least one tangent to W. The following is shown in [5] and [7]. THEOREM 1.2: If Wis any blocking set in an affine plane
n
A
of order n then
170
A.A. Bruen andM.J. de Resmini
lWI;;. n
+ vh+
MoY'eover, if W is irreducible then IWI .;; n vh~
2.
In [101 Jamison obtains a surprising and substantial improvement on the lower bound for classical planes, as follows. THEOREM 1.3: Let
rr
A
= AG(2,q) and let Wbe a blocking set in
rr
. A
Then
II>I! ;;. 2q - 1. We should mention here that a somewhat shorter proof of Theorem 1.3 is given in [1 ). On page 254 of his paper Jamison remarks that despite "strenuous efforts" no combinatorial proof of Theorem 1.3 could be found. In r 4) we obtain new bounds on the size of irreducible blocking sets in the classical projective planes which supplement the lower bound of Theorem 1.1. method of proof uses Jamison's result.
The
The brazen idea then occurred to us that
Jamison's bound might not in fact be valid for arbitrary affine planes and that, moreover, the methods in [4) might yield some information on this. In fact this turns out to be the case! Our main result, then, is as follows. THEORH4 1.4: TheY'e exist finite affine planes lIA of order nand bloeking sets W in
rr
A
with IWI .;; 2n - 2.
SECTION 2 To justify Theorem 1.4 we offer the following two constructions. :ONSTRUCTION 1: This construction is based on the simplest case of Theorem 2.2 in 4), i.e. the case when
= l.
~
The construction can be generalized for all 'e only discuss the case
~
= 1.
~
> 1, but for simplicity here
The setting is this.
S is an irreducible
locking set in a finite projective plane of order n, with lSI = n + k. Since 5 is a blocking set no line of et us assume that some line
~
of
11
rr
can contain more than k points of S.
contains exactly k - 1 points of S.
Next, let P be any point of £ not in 5, that is P is in £ \ S. ines of
11
through P other than
£
Of the n
exactly one of them, say x(P), contains pre-
Blocking sets in affine planes
171
cisely two points of 5 while the others contain just one point of 5. We now have the following result. THEOREM 2.1: (aJ If no two of the lines x(P) meet in a point of 5, then 151 ~ n + (n + 3)/2. (bJ In particular, if rr = PG(2,q), then 151 ~ q + (q + 3)/2. (cJ Suppose that for some two points P ,. Q of £, \ S the lines x(P) and x(Q) intersect in a point of 5. Then there exist an affine plane rrA of order n and a ~
blocking set Win rrA with IWI
2n - 2.
PROOF: We consider the two possibilities. Possibility 1.
The line x(P) never meets x(Q) in a pOint of S.
I£,\ S I = (n + 1) - (k - 1) Thus lSI is 151
~
~
=
n - k + 2, it follows that IS I
2n - k + 3.
By definition lSI
n + (n + 3)/2.
This proves (a).
Possibility 2.
~
k - 1 + 2(n - k + 2~
= n + k. Therefore, k ~ (n
For some two points P ,. Q on
9., \
Then, since +
3)/2, that
S the lines x(P), x(Q)
intersect in a point Z of S. Now, S being irreducible, there exists a tangent line z to S at Z.
Let J
denote the set of those lines of rr other than z which pass through Z and which are tangent to S at Z. The lines ZP IJI
~
= x(P)
and ZQ
= x(Q)
are not tangents to S.
Thus
(n + 1) - 1 - 2 - (k - 1) = n - k - l. Let rrA be the affine plane obtained by removing z from rr.
We construct a
set Wof points in rrA as follows. On each line a of J choose anyone point R = R(a) with R(a) ,. Z. H the set of all such points R(a). Finally, set W= (5 \ {Z})
U
tha t Wintersects all 1i nes in rr A' That is IWI
~
2n - 2.
Thus IHI H.
= IJI
~
Denote by
n - k + 1.
Since S is a blocking set in rr it follows
r~oreover,
IWI
~
(n +
k) -
1 + (n - k - 1).
This proves part (c).
It follows from Jamison's result that Possibility 2 cannot occur if n
= PG(2,q). Part (b) now follows. In fact, equality can occur in part (c) of Theorem 2.1.
This will come out
in the proof of the next result. Let K be a k-arc in a finite projective plane of order n. of k points no three collinear. (k + J)-arc.
to K.
A point P of
rr
Thus K is a set
K is complete if K is not contained in any
is said to be exterior to K if P lies on no tangent
172
A.A. Bruen and M.J. de Resminl
THEOREM 2.2: Let in
71.
Let
R.
(a) If
71
b", a proojective pZane of ol'dero 9 and Zet K be a compZete 6-aroc
bl!; a secant of K. R.
Then
contains exactZy one (zero) exterior point, then theroe exist an
affille pZane of ordero 9, "A' and a bZocking set
in 7IA with I WI
= 16 (15).
(b) AZZ known non-Desaroguesian proojective pZanes " of oroder 9 contain a o-aroc Kand a secant
R.
of K which contains exaatZy one exteroior point.
PROOF: Part (b) follows from Dennistons's paper [9) which contains a very useful and detailed discussion of arcs in planes of order nine. Since K is complete. each point P Moreover, P 1ies on at most -k < n + 1
To see Part (a) we proceed as follows. of
71
lies on at least one secant of K.
secants to K.
It follows that in the plane
tn
dual to " the set of = 15 secants to K yields a blocking set S with IS I = 15. Since each point P of K 71,
=9 +
lies on exactly 5 secants it follows that lSI containing exactly 6 - 1
6 with some line of",
= 5 points. It is easy to show that S is irreducible.
Thus we are in the situation of Theorem 2.1. Elementary ad hoc counting arguments show that the hypotheses on the exterior pOints in part (a) guarantee that Possibility 2 of Theorem 2.1 occurs, and yield the results in Part (a). A specific example for the Hughes plane of order nine will be given in Section 3. CONSTRUCTION 2: A celebrated conjecture of Hanna Neumann [11) is that all finite non-Desarguesian projective planes of order n, say, contain projective subplanes of order two.
Of course, if
71
= PG(2.q)
then
subplane of order 2 when q is a power of 2.
71
only contains a projective
However the conjecture is true for
all the known non-Desarguesian planes of order nine, all of which contain projective subplanes of order two. We have exploited this fact in the case of the Hughes plane" of order nine to obtain a blocking set V in an affine plane "A of order 9 obtained by deleting a certain line of THEOREI~
71.
In fact. we have
2.3: The bZocking set V has 16 points.
It contains a Fano subpZane and
Blocking sets in affine planes
has seven points on a line.
173
Moreover, if we construct the blocking set Wof
Theorem 2.2 using for rr the Hughes plane of order nine, then Wcan be chosen so that at most five points of Ware on any line.
SECTION 3 Let rr be the Hughes plane of order nine.
We use Denniston's notations in
[ a ] and [9] for poi nts and 1i nes in rr. Firstly, we construct the blocking set Wof Theorem 2.2 starting from the "non -Desarguesian" complete 6-arc K (wl in [9]) whose points are: F, G, H, N , 3 03' Y4· Then the 15 secants to this arc are the pOints of the blocking set S in the projective plane n, dual to rr.
Namely they are
s, r, fa' f6' f4' t, 96, 9a , 94 , h3' h6' y, i 3 , k7, d7· Next, we choose for the five points on a line
2
(namely the line F of rr,)
the set s, r, fa' f6' f 4· The two lines Pa and W6 throu9h the points p and w, respectively, on F have two points in common with S and meet at the point 9 in S. 4 Now we delete 94 and take Na as the line at infinity. The tangents to S at 94 other than Na are U7 and 07 and we adjoin the point a4 on 07 and the point e on 3 U . Therefore the blockin9 set Wconsists of the followin9 16 pOints of rr, \ Na: 7 s, r, fa' f6' f4' t, 96' ga' h3' h6' y, i 3 , k7' d7, a4 , e3 · Col linearity relations' amon9 the points of Ware displayed below.
At most five
points of Wlie on any line. s, r, fa' f6' f4 r, t, h3' h6' Y f6' 9a , h6' i 3 , d7 fS' t, d7 , e3
on F;
f4' 9S ' e3 s, h3 ' d 7 r, 9 ' k7 S f6' t, k7
on G;
s, t, 96 , 9S fS' 96, h3,i 3 , k7 s, h6' k7' a 4
on N3 ; on S4;
on Za; on S7;
f6' y, 96' a4 , e3 f4' h3' a4 r, 96 , d7
on Y6; on Q6; on R4 ;
on R5; on T5;
fS' 9S' Y f4' t, i 3
on Y ; a on T7 .
on H; on °3; on T3;
Next we describe the set V of Theorem 2.3 in the HU9hes plane nine. ing:
We start from the Fano subp 1ane
C1
11
of order
in [S] whose incidences are the follow-
174
A.A. Bruen and M.J. de Resmini
n
A D N , p: D L P , x: A L X , a : A P 53' 3 6 3 3 6 D 53 X6 , £6: L S3 N , i 6 : P3 N6 X6 • 6
d3
The blocking set V will contain the seven points of lines in n. -A
=- \
0
Choosing the line
0
a,
which cover 56
as line at infinity, the blocking set V in
consists of the following 16 points:
These points lie on the following lines of -A: A, D, N , N6 on n; 4 D, L, P3' P on p; 5 H, X , N4 on h4 ; 6 A, l, X , X , X6 on X; 3 4 L, N , 53' on £6; 6 \ H, W ' Z6' X , N , V , P 6 4 6 5 5
A, P , 53' W6 3 D, 53' X6 , Z6 N6 , X , P 6 3 l, N4 , W6
on a ; 3 on d ; 3 on i6; on £4;
T , P , X , N , V5 5 3 3 4 on h . 6
on m ; 7
Note that the line h6' containing seven points of V, is tangent to the subp1ane
a
at the point N . 6 CONCLUDING REMARKS Despite the examples in sections 2 and 3, it is still not clear how good the lower bound of Theorem 1. 4 rea lly is.
It may be poss i b1e to buil d up "small"
blocking sets in affine planes by starting from subp1anes by analogy with Construction 2.
We are also examining the possibility of making more use of
Theorem 2.1 to construct smaller blocking sets. ACKNOWLEDGEr~ENT:
Research supported in part by the N. S. E. R. C. of Canada and the
C.N.R. of Italy.
A portion of this work was carried out in conjunction with the
Combinatorial Geometries Seminar of Prof. G. Ta11ini in Rome.
We wish to thank
Prof. G. Tallini and the other participants for stimulating discussions. BIBLIOGRAPHY 1. 2.
A.E. Brouwer and A. Schrijver, The blocking number of an affine space" J. Comb. Th. A, 24 (1978), 251-253. A.A. Bruen, Baer subplanes and blocking sets, BuZl. Amer. Math. Soc., 76 (1970), 342-244.
Blocking sets in affine planes
3. 4. 5. 6. 7. 8. 9. 10. 11.
175
A.A. Bruen, Blocking sets in finite projective planes, SIAM J. Appl. Math., 21 (1971), 380-392. A.A. Bruen and R. Silverman, Arcs and blocking sets, in "Finite Geometries and Designs", London Math. Soc. Lecture Note Series 49, Cambridge U.P., 1981, 52-59. A.A. Bruen and R. Silverman, Arcs and blocking sets II, preprint 1981. A.A. Bruen and J.A. Thas, Blocking sets, Geom. Dediaata, 6 (1977), 193-203. A.A. Bruen and J.A. Thas, Irreducible blocking sets, preprint 1981. R.H.F. Denniston, Subplanes of the Hughes plane of order 9, Proa. Cambridge Phil. Soa., 64 (1968), 589-598. R.H.F. Denniston, On arcs in projective planes of order 9, Manusaripta Math., 4 (1971), 61-89. R.E. Jamison, Covering finite fields with cosets of subspaces, J. Comb. Th. A, 22 (1977), 253-266. H. Neumann, On some finite non-Desarguesian planes, Arah. Math., 6 (1955), 36-40. Department of Mathematics Middlesex College The University of Western Ontario London, Ontario, Canada N6A 5B7
Istituto Matematico "G. Castelnuovo" Universita di Roma 00185 Rome, Italy
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Annals of Discrete Mathematics IS (1983) 177-lS0 © North-Holland Publishing Company
177
SOME BUILDING GEOMETRIES OF SPHERICAL TYPE AS SETS OF POINTS AND LINES F. Buekenhout
My starting point is the following list of Dynkin diagrams or rather Coxeter diagrams of spherical type. A @--o--o
0-----0---0
n
Bn,C
n
@--o---o
...
(}---{)==O
Dn
~
...
E3
~
~
E4
n vertices, n
~ E5
~
1
n vertices, n ,;;; 2
n vertices, n';; 4
®---L--o--o
E6
@-~
E7~E8~ F3 @==o--o
F4 @-----o==o-----
Each of these diagrams determines a class of buildings (and weak buildings) which has been completely classified to a large extent (see Tits (5)). such a building then
f
If r is
and the circled dot of the diagram determine uniquely a
geometry consisting of points and lines in f.
My purpose is to give a set of
axioms characterizing these geometries. A great deal of inspiration, motivation and ideas expressed here are due to various people, mainly J. Tits [5) and references given there), B. Cooperstein [3) and for his study of geometries of type En based on points and lines, A. Cohen (2) for his characterization of geometries of type F4 and Cn, P. Cameron [1) for his characterization of dual polar spaces and G. Tallini (4) for his characterization of Grassmanians of lines of projective spaces.
The author is especially grateful
to J. Tits and W.M. Kantor for useful discussions on exceptional geometries. LIST OF AXIOMS (0) f
= (P,L)
is an incidence structure consisting of a set of points and a
F. Buekenhout
178
family L of subsets of P called lines each of which contains at least two points. Points which are together on some line are called collinear. (1) if L is a line and p is a point then p is collinear with 0,1 or all points of L.
(2) any pair of points is contained in at most one line. DEFINITION: A subspace of r is a set of points S such that every line with two points in S is included in S. A singular
sl~space
of r is a subspace all of whose points are pairwise collinear.
If p,q are points a path from p to q is a sequence of points p=X ,Xl , ... ,X =q in o n which any two consecutive elements are collinear. A convex subspace S of r is a subspace such that for any pair of points p,q of S all shortest paths from p to q in r are contained in S. (3) Every singular subspace is a projective space. (4) If p,q are points of r at distance 2 then either
~
2 is the same integer
for all such pairs (and
Some bUilding geometries of spherical type
179
(9) If A,B,C are singular subs paces of rank n-l such that An Band B n C have rank n-2 and if A is a maximal singular subspace then C is a maximal singular subspace. (10) If Hand K are distinct hyperlines having a singular subspace of rank n-2 in common then they have a singular subspace of rank n-l in common. THEOREM: If r is a geometry satisfying axioms (0)-(10) then r provided with its singular subspaces and hyper lines belongs to one of the diagrams listed at the beginning with one additional case namely all projective spaces of any rank possibly infinite).
A proof will appear elsewhere. BIBLIOGRAPHY 1. 2. 3. 4. 5.
P. Cameron, Dual polar spaces. Ceo. Ded. (to appear). A. Cohen, On the points and lines of metasymplectic spaces (to appear). B. Cooperstein, A characterization of some Lie incidence structures. Ceo. Ded. 6 (1977) 205-258. G. Tallini, On a characterization of the Grassman manifold representing the lines in a projective space "Finite Geometries and Designs". Ed. By Cameron, Hirschfeld, Hughes. Cambridge U.P. (1981) 354-358. J. Tits, Buildings of spherical type and finite BN-pairs. Lecture Notes in Math., 386, Springer-Verlag, Berlin, 1974.
Departement de Mathematique, C.P. 216 Universite Libre de Bruxelles Boulevard du Triomphe 1050 Bruxelles Belgium
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181
Annals of Discrete Mathematics 18 (1983) 181-192 © North-Holland Publishing Company
"CYCLIC" SOLUTIONS FOR FINITE PROJECTIVE PLANES K.A. Bush
ABSTRACT It is well-known that if the order s of a finite projective plane is a prime or a power of a prime, then there is always a cyclic solution for the (0,1) incidence matrix of such a plane.
We raise a more difficult question.
Excluding some
type of border, it is possible that the incidence matrix for the plane can be partitioned into subsets of permutatiqn matrices.
Are there solutions which are, ex-
cluding the border, "cyclic" in the permutation matrices themselves? The permutation matrices themselves need not be cyclic.
A minor advantage of this approach
is that a specific solution can be obtained very rapidly.
We show that with the
standard border (defined later), there is always a "cyclic" solution if the order is prime.
With a Baer subplane border, we have no new results involving "cyclic"
solutions, but the fragmentary information here is given in section 3. We say we have a "semi-cyclic" solution if, exclusive of a border, the incidence matrix remaining can be partitioned into permutation matrices such that there is a further partitioning into larger subdivisions each of which is "cyclic" in the permutation matrices.
In section 3 we give a solution of this type for the
plane of order 9. With this type of approach, the isomorphism question is complex.
In the
last section we give a generalization of Hall's multiplier theorem that is useful along with some details from a communication from R.H.F. Denniston received after the conclusion of the conference in which he shows that the two types of planes of order 9 just described are inequivalent. 1. INTRODUCTION In a series of papers, the use of global multiplication (GM) was introduced by the author in the construction of partially balanced incomplete block designs which were either regular or semi-regular.
This simple idea is to partition the
182
K.A. Bush
incidence matrix into (say) n x n submatrices.
Thus two "rows" of the incidence
matrix might have the appearance Al A2
Am
Bl B2
B. m
~ AkB~. If it should happen that ~ AkB~ = A2J where J is a matrix of order n all of whose elements are 1 whereas ~Ak = ~Bk = A1J, then we have a We define GM as
start on some type of design.
In the cases considered in the past, always A FA . 1 2 Now since the structure of finite projective planes admits a natural parti-
tioning into submatrices of order s (and there are other cases as well), it seemed reasonable to investigate the suitability of this method as a tool for studying finite projective planes, particularly since this subject has been of substantial interest to the Italian school with many notable contributions by Segre, his students, and their associates.
In our lectures at the Universita di
Bologna, this approach was one of the focal points studied in some detail. 2. THE "CYCLICAL" CASE It is well-known that there is a type of canonical form for a plane of order s.
We consider the affine case.
We let Ej denote an s x s matrix whose
jth column consists of the element 1 and otherwise the entries are O.
The affine
plane can then be written in the compact form ... I P
11. .. Pl,s_1
E s
Ps- 1 , 1'" Ps- 1 ,s- 1
where I is the identity and P.. a permutation matrix of order s. lJ
We call
Pll .•• P1 ,s-1 Ps-, 1 1'" Ps-l,s-l
the kernel.
If we have in fact a solution to the plane of order s, then GM should
yield J-I for any two "rows" ("row" of course means a row of submatrices of order s) of the kernel.
We need merely note that we a lways have in the "border" I
"Cyclic" solutions for finite projective planes
183
versus I with global product I yielding J, i.e. A = 1 for the inner product of any two rows from different subdivisions whereas Ei yields inner product 1 for any two rows in the same subdivision.
We are of course further assuming that
~
k
P k=J-I u,
for all u. The first interesting question is whether there is a "cyclic" solution using As an elementary example, consider the case s = 4.
only s-l permutation matrices.
Here we can use the abbreviations 1,2,3 for the permutation matrices choosing, for example,
a1aa 1 a aa aaa1 aa1a
aaa1 aa1a a1aa
001 0
o0 a 1 100 a a1aa
2
3
1
aa0
Then a "cyclic" solution is given by (say) 2 3 3 2
2
1.
3
In this simple case, each matrix is its own inverse (it is of course the Klein
= k, i,j,k all different. Thus GM
four group), and, for nonzero values,
. j
for the first two "rows" yields 2+3+1
J - I, and it is easily verified that we
have a solution to the plane of order 4 in compact form. It is interesting to ask whether every plane, exclusive of its border, has a kernel "cyclic" in our sense where kernel means the set of s(s-l) rows and columns which can be partitioned into s - 1 permutation matrices of order s. ways possible if s is a prime.
We use the notations
a is
This is al-
the identity matrix of
order s, and 1,2, etc. are the permutation matrices of order s
a a a a
a
a a
0
0
a a a
a a
a
a
0
2
0
1
a
, etc.
Now let w be a primitive element of the group of order s such that wa a = b if
a < a,
b < s-1.
Let the first row of the kernel be 2
w w
s-l
• w
=
b
w implies
184
K.A. Bush
and advance cyclically.
Thus for s = 7 we have that 3 is a primitive element, and 2 we can write the first row as 3, 3 , ... , 36 mod 7 and then advance cyclically. This yields 3 2 6 4 5 3 2 6 4 5 3 2 6 4
5
3 2 6
4 5 6
4
3 2
5
3
2 6 4 5 3 2 6 4 5
If we take two rows such as the first and the fifth, GM yields (mod 7) 4 + 5 +1+3+2+ 6 so that every permutation matrix occurs once (except for
the border identity matrix) showing that these could be lines of the plane. More generally, if the uth row is s-u+l w
s-u+2 w
s-u w
then, with the matrix identifications we have made, it is only necessary to subtract corresponding elements under GM mod s from the first row. w(O-w
s-u
2
) + w (O-w
s-u
This yields
s-1 s-u ) + ... + w (O-w )
where 0 is the identity of the multiplicative group.
S
Since w -
u
is never the
identity, it follows that each of these matrices is different and none is the identity.
Therefore, since the identity matrix arises from the border, we have
a solution to the geometry.
(Note that + is not a group operation; it is of
course the sum of the matrices. the ith and
j
Note also that is is not necessary to introduce
th "rows", the cycl i c advance guarantees that it is suffi ci ent to
compare only the first "row" against the remaining "rows"). 3. THE CASES 8 AND 9 For the case 8, we use the same notations as in the case s initial row 254637 where x is the null matrix of order 4 and
4.
Choose as
185
"Cyclic" solutions for finite projective planes
5=
I : ~~,
2=
~ ~ :11,
3=
II: :~,
4=
~: ~~
,,~ :11,
6=
II: ~II'
7=
II: :11,
7=
II: :~
These matrices are all symmetric.
Under GM we therefore obtain:
rows (1,2):
(1,7)=6; (2,1)=3; (5,2)=7; (4,5)=1; (6,4)=2; (3,6)=5; (7,3)=4
rows (1,3):
(1,3)=2; (2,7)=5; (5,1)=4; (4,2)=6; (6,5)=3; (3,4)=7; (7,6)=1
and so on. The case s
9 is the first case of interest.
We use the notations for
3 x 3 matrices 000
100
x = 000,0= 0 1 0, 1 001 000
010
001
o0 1 , 2
100
100
010
To denote the permutation matrices of order 9 in convenient form, we use a simple group where, unfortunately, it is necessary to use 0,1,2 with two different meanings.
The permutation matrices are
oxx 0= xOx,
1 xx
1
x 1 x,
2
2xx
x0 x
x 2 x, 3
xx0
xx0
xx 1
xx2
oxx
x1x
x2x
xx
xx
4=xxl,5 1 x x
x x 2, 2 x x
6
0
o x x, x0x
7
xx2
1
1 x x,
8= 2xx
xlx
Here the inverse pairs are (1,2); (3,6); (4,7); (5,8). THEOREM: The ptane of order 9 admits a "ayaUa" kernel-. PROOF: Routinely we can compute using GM the inner products for
x2x
186
K.A. Bush
S 7 6
4
256
8
3
4 2 5 6
8
3 4
7
5 6 2
3
2
5
8 342
7
5 6
7
834
7
For the first 4 "rows", we find: versus 2:
4
5, 8
2
= 6, 3
5
= 7, 4
6
=1
versus 3:
3
4, 8
4
2, 3
2
5, 4
5
6
versus 4:
8
= 7, 8
3
1,3
4
8,4
2
3
Remarkably, no further computation is necessary, for the omitted calculations above are the inverse (e.g. in 1 versus 2, the first omitted pair is 2.7 and yields 8 since (1,2) and (4.7) are inverses).
Furthermore, the first and fifth
row are inverse pairs always and, under GM, yield the diagonal elements of the group table which are all distinct. It is possible to construct a number of solutions which are "cyclic" in our sense, but they are all easily shown to be equivalent and, in fact, Desarguesian. Another possibility is to partition 8 into two groups of 4 which each cyclic.
As an example, we give: 4 2 7
5
3 8 6
4 2
6
5
3 8
4
8
6
5
7
2 7
3
4 2 7
3 8 6 5
5 6 8 3
2
3 5 6 8
7 2 4
3 5 6
7 2
4
7
2
8
6 8
3 5
4
4
7
If we look at the dual, we see that 4 and 7 are interchanged so that is intuitively clear that this is the Hall translation plane inasmuch as it is most likely not a truly self-dual solution. verify that we have a solution using GM.
This is indeed the case, but we first Here we need compute only the global
products between "columns" 1,2 and "columns" (5,6) because of the presence of inverse pairs elsewhere.
We find:
187
"Cyclic" solu tions for finite projec five planes
(1,2) yields 1
4
= 5, 4 . 2 = 3, 5
3
= 7, 3
8
EI
1
inverses
(1,3)
= 6, 8 = 7, 6 = 8, 5 = 3, 3 = 4,
(1,4 )
7
(1,5) (1,6) (1,7) (1,8)
4
5, 5
4
3
8, 5
4
8
4
6
4
= 2, 5 = 1, 5 5 = 6, 5
= 2, 1 = 3, 4 = 6, 2 = 4, 7 = 1,
6
3 3 3 3 3
=7 7 =2 1 =4 4 =8 2 =5 5
The remaining known plane of order 9 also possesses a "cyclic" solution but with a different border.
In fact the procedure can be generalized for a plane of
2
order s , but for general s it appears to be a formidable problem.
The general
case is: find a solution to the group divisible partially balanced symmetric design with parameters b
= v = s(s 3-1), r = k = s 2, m = 5 2 + s + 1, n = 5 2
Let P(s) be a solution to the plane of order
5,
-
s, Al
= 0, A2 = 1.
and let O(s) be a solution to the
design above where first associates are 1,2, ... ,s; s+l, s+2, ... ,2s; etc.
Then if
E is the vector of order s2-s with each elemento 1, 0 is the null vector of order
s2-s, we write for the solution: P(s)
E 0
o
o o
E
o
0
E
o
E 0
o
E
o
o
0
E
While the case s
O(s)
= 2 is of no interest, we indicate the way in which a "cyclic"
0(2) can be generated.
Set
The plane of order 2 admits a skew solution on cyclic advance with initial row
o1 10
1 0 0, (symbols, not matrices).
Skew means that a
ij
+a
ji
= 1, i
r j.
Replace the diagonal term by the matrix 0,
188
K.A. Bush
replace every other 0 symbol by the matrix 1, and replace the symbol 1 by matrix x.
The skewness means that we have the "pair" (0,1) once, the "pair" (1,1) also
occurs exactly one time between any two rows showing that we have a "cyclic" solution in this case. When s = 3, we need to construct the symmetric design 0(3) with parameters v
= b = 78, r = k = 9, m = 13, n = 6, Al = 0, A2 = 1.
Here we let
r"I
o0 0
o0
010
1 ,
'I
100 , 00
2
100
001
010
100 3
o0
1
001 4
010
010 100
010 5
1 0 0
o0
1
and with x the null matrix of order 3, our 6 matrices of order 6 are
We will also now use X in what follows as the null matrix of order 6.
The group
table is
o
o
2
3
o
2
345
0
4
2
5
5
3
534
2
2
3
354
4
4
5
543
0
3
4
o
2
o
5 2
2
o
As in the preceding example where we sought a cyclic solution for a plane with 7 lines, we here seek a cyclic solution with 13 lines, 4 points on a line. choose 0, 4, 10, 12 for the points and replace them with the matrix X. remaining places could constitute the complementary design with r and we wish the sum of our 6 matrices to be J.
=k
Let us The 9, A
6,
One possibility is to use the
matrix 0 4 times with the pair (0,0) occurring once (i.e. the plane of order 3 once more but where the X positions remain fixed) and 1,2,3,4,5 occur once each. No solution of this type was found even using different matrices from our present
189
"Cyclic" solutions for finite projective planes
choice.
The search was, however, not exhaustive.
However, since 3,4,5 are self-inverse, we could use the matrix 3,4,5 twice each.
° 3 times and
This was the approach used by Glynn (unpublished thesis pp.
132-3) although he gave no hint as to how he constructed his solution.
It is
impossible to have a difference of 2d between two places occupied by the matrix unless the midposition is filled by X.
°
Thus
0-3-0 ... is impossible because we have a pair of "rows" that contain the pairs (0,3) and (3,0) which are equal.
Because of this restriction and the fact that 3,4,5 can be
interchanged under permutations such as (3 4 5) or (3 5 4), it is quite unlikely that two inequivalent solutions using this method exist. X 4 03
Glynn's solution is
X 00 5 4 5 X 3 X.
It is quite easy to verify that this is a "cyclic" solution.
Thus the differences
are: for 0,
1,3,4; for 3,5;
for 4,6;
for 5,2.
(Note that we have only included, mod 13, differences in range 1, ... ,6. difference of 8 is replaced by 5).
Thus a
By equally tedious methods, one can show that
the pair (0,4) occurs in "rows" with differences 1,4,5,7,10,11 with no two differences summing to 13. Thus (4,0) occurs in the remaining 6 differences.
Similarly (0,5) occurs in "rows" whose differences are
8,11,12,6,9,10 and (0,3) in "rows" whose differences are 12,2,3,4,7,8.
By quite
similar reasoning, taking account of the group table, one shows that between two "rows" 1 and 2 each occur precisely once completing the proof in a very direct and simple fashion. 4. ISOMORPHISM QUESTIONS Some of the preceding work suggests that a theorem on the equivalence of two circulants could be useful, for if two solutions have equivalent kernels, then they must in turn be equivalent. THEOREM: Let aO al, ... ,an·be the first row of a circulant, and Zet ao ai ... a be i · Zant were h . of a1 · .. a n. th e f~rst row of anot h er · c~rcu ai ... a i . ~s a permutat~on l n These two cirauZants are equivaZent if and JnZy iJ i =2i , i =3i , ... ,mod(n+l)
2
l
3
l
190
K.A. Bush
where (; 2' n+') = 1.
PROOF: If two circulants are equivalent, then any specific choice will also have this property. root of unity.
k
In particular we could set a
w where w is a primitive (n+l)th k In a circulant, the eigenvalues are given by \.
J
so that with our choice A. =
J
. 1) k
r w( J+ k
= 0, j
I n
so that the matrix has rank one, and equivalent matrices always have the same rank.
Considering two rows of the circulant i
w n
i w n
w
i
n-1
we have that all 2 x 2 determinants vanish so that
Thus ;2=21" i 3=3i l , ... ,1 k=ki l , mod (n+l~. Note that if w was a primitive root, , a 1so pr1ml . . t 1ve; . ' k11 d 1 ' b1e va 1ues. th en w11 1S 0th erWlse woes no t assume a1POSSl We can think of the circu1ants as permutations.
As such they are in the l same equivalence class so that a permutation P exists that PC P- =C where C is 1 2 l given by the permutation (0 1 . . . n) and C as (0 il".ni ). The effect of this 2 l is to set up a correspondence k ~ i k so that the circulant property is preserved. (By noting that we kept the initial element aD fixed, it is clear that this is a generalization of Hall's multiplier theorem for difference sets but with a much simpler proof). Unfortunately the theorem gives only a sufficient condition for the isomorphism of two planes and is therefore of limited usefulness, and, given two incidence matrices, there seems to be no easy way to answer the isomorphism question with perhaps only the possibility of coordinatization a feasible process. R.H.F. Denniston (personal cOl1111unication) has used this technique. plane the group tables are (using numbers, not our matrices):
For the Hall
191
"Cyclic" solutions for finite projective planes
Additi on
Multiplication
o1 2
345
678
12
34
56
78
120
453
786
21
68
73
54
201
534
867
36
25
81
47
34 5
678
012
48
72
35
61
453
786
120
57
46
28
13
534
867
201
63
17
42
85
678
012
345
75
83
14
26
78 6
120
453
84
51
67
32
867
201
534
Usi ng the homogeneous system
~x
+
my + nz = 0 and choosing
~=l,
z=l, we will let
x assume the values 1,4,2,8,5,3,7.6 each repeated 9 times while m assumes the values 2.8,1,4,5.3,7,6 each repeated 9 times. cycle 0,1, ...• 8. cidence structure.
The values of y will repeat the
It remains to choose appropriate values of n to produce our inThese turn out to be
0,1.2; 3,4,5; 6,7,8 followed by 0,4,8; 7,2,3; 5.6.1 followed by 0.2.1; 6,8,7; 3.5,4 followed by 0.8.4; 5.1.6; 7.3.2 followed by 0.7,5; 8.3.1; 4,2,6 followed by 0.6,3; 1.7.4; 2.8.5 followed by 0.5.7; 4.6.2; 8,1.3 and finally 0.3.6; 2,5,8; 1.4.7.
The details are onerous to
verify. An interesting question is: does a "cyclic" solution always exist for n
p • n> 1. and. if so, is it always Desarguesian?
Department of Pure and Applied Mathematics Washington State University Pullman, Washington 99163. U.S.A.
This Page Intentionally Left Blank
193
Annals of Discrete Mathematics 18 (1983) 193-196 © North-Holland Publishing Company
POINTS AND LINES IN METASYMPLECTIC SPACES Arjeh M. Cohen
Metasymplectic spaces were introduced by Freudenthal and axiomatised by Tits [3] in connection with the study of buildings of type F4. them, we need the following concepts.
In order to define
An inaidenae system (P, k) is a set P of points together with a collection of subsets of P of cardinality> 1, called Zines.
~
Its aoZZineaPity graph is the
graph whose vertex set is P and whose edges consist of the pairs of distinct collinear points.
Notions like aonneatedness, distanae, aZique, when applied to an
incidence system, are meant to hold for its col linearity graph. For x E P the subset xl of P consists of all points collinear with x (thus l x E xl). For X ~ P, we write ~ =xQxx . A subset X of P is called a subspaae of (P,~)
whenever each point of P on a line bearing two distinct points of X is
itself in X.
A subspace X of (P,k) is called singuZaP whenever it is a clique.
The length k of a longest chain X ~ X X.(O 1
~
i
~
o
1
~ ... ~
X = X of nonempty singular subspaces k
k) is called the rank of X.
A poZaP space of rank k is an incidence system (P,k) such that: (Pl) No point of P is collinear with all of P. (P2) For each point x of P and line L of
k with xi L, x is collinear with either
one point or all points of L. (P3) There is a singular subspace of rank k-l and none of higher rank. Polar spaces of rank 2 are also called generaZised quadrangles. We are now ready to define metasymplectic spaces. A connected incidence system (P,k) together with two distinguished collections
~,~
of subspaces, re-
ferred to as planes and sympZeata respectively, is called a metasympleatia space (see [3]) provided the following axioms are satisfied: (Ml) If S, T E
~
wi th I S n TJ > 1 and S # T, then S n TEL
U
V.
194
A.M. Cohen
E ~IL ~
(M2) Given a symplecton S, the incidence system (S,{L
S}) is a polar space
of rank 3 whose maximal singular subspaces are precisely the planes contained in
~.
(M3) Taking the set x
of
x"
of all symplecta on a given point x of P and calling lines
the subsets of /' consisting of all symplecta containing a plane on x,
one obtains a polar space of rank 3 whose maximal singular subs paces are precisely the subsets of x* containing a line on x. The purpose of this contribution is to state the following characterisation of metasymplectic spaces in terms of points and lines. THEOREM: Let reints.
(P,~)
be a connected incidence system
~hose
lines have at least 3
;1; consists of the points and lines of a metasymplectic space or a polar
space cf ran":
;3
if «ad only if the following five axioms hold: 1
1
(Fl) For each point x and line L with Ix n LI ~ 2, we have L ~ x . · . h x E y, 1 th e gT'ap h on x1 n y1 . 7 • W'l-t ,,13 not a c,-,,,que (F2) FoT' each pair x,y 0 f pO'l-nts . Ix1 n y1 I ~, 2 then x1 n y1 .'l-S a geneT'a~7 ( F3 ) EI,ppose x E P ana, y E P\ X1 sat"sfy ised quadrangle.
(F4) Tr.ere al'e no minimal 5-circuits; i.e. given Xl' x2 ' x3 ' x4 ' x5 E P with 1 1 Xi E xi+l\ (xi+2 U {x i +l }) for each i (indices modulo 5), there is an index i for LJhich Xi is collinear LJith a point on a line through xi+2 and x + '
(F5) If x,y
E
1
1
P aT'e sucr; that 1x n y I
~
i 3 1 1 1 2, then x n z # 0 for any Z E Y .
REMARKS: 1) According to [3] the finite examples of metasymplectic spaces, whose lines have three or more points and are in at least three planes and whose planes are in at least three symplecta, are provided by root group geometries of the 2
groups F4(q) and E (q). 6 By use of the above axioms, subdegrees and other parameters are easily derived; they are depicted in the diagrams below.
These diagrams should be read
from left to right and are to be interpreted as follows. resents a point of the geometry.
The leftmost node rep-
The next node represents its neighbor.
number inside this node stands for their cardinality.
The
The number 1 on the bond
between these two nodes and nearest to the second node stands for the neighbors in the first node of a given point in the second node.
Ad so on.
lity of P is the sum of the numbers inscribed in the nodes.
Thus, the cardina-
As is to be expected,
195
Points and lines in metasympiectic spaces
the parameters turn into the corresponding numbers for the Weyl group of type F4 upon substitution of q=l. DIAGRAMS q
4
2
15
2
q (q +q+l)
q(q +q+l)
2
E6(q). short root groups:
j
2 2 3 4 11 q (q +l)(q +l)(q +1)
J..
13 2 3 4 11 2 3 4 124\4 q (q +l)(q +l)(q +l)q (q +l)(q +l)(q +1) ~
q7 (q 2+q+ 1)
q2 (q 2+q+l) 6
3
2
q (q +l)(q +1) 2
E6(q). long root groups: 11 3 5 10 3 5 ,}--_--\ q (q+l)(q +l)(q +l)R (q+l)(q +l)(q +1)
542 q (q +q +1)
q(q 4+q 2+1) 6 3
q (q +1)( q+ 1) 2) The proof of the theorem uses Shult-Yanushka and Cameron's characterization (see [1 I ) of dual polar spaces and some of Cooperstein's techniques found in [2].
It will be published elsewhere. 3) Using axioms (Fl). (F2). (F3). (F4), (F5) many more properties may be de-
duced.
To name one: Given two pOints x.y of mutual distance 3. the subset of xl
consisting of points of distance 2 to y is a subspace isomorphic to a dual polar space. 4) The axioms (Fl), (F2). (F5) also hold for root group geometries of groups
196
A.M. Cohen
of Lie type E , E7, E8 while (F3) becomes true if 'generalised quadrangle' is re6 placed by 'polar space of rank k' for k=3,4,6 in the respective cases. Also (F4) fails whereas the weaker version 'If xl' x2 , xi' X4' X5 are the (consecutive) pOints of a minimal 5-circuit,then Ix~1 n x.1+ 2 I ~ 2 for each i' still holds.
BIBLIOGRAPHY 1. 2. 3.
P.J. Cameron, Flat embeddings of near 2n-gons, in Finite Geometries and designs, Proc. of the 2nd Isle of Thorns Conference 1980, LMS, LNS 49, Cambridge Univ. Press, 1981. B.N. Cooperstein, A characterization of some Lie incidence structures, Geometriae Dedicata, 6 (1977), 205-258. J. Tits, Buildings of spherical type and finite BN-pairs, Lecture Notes in Mathematics 386, Springer Verlag, Berlin, 1974.
Mathematical Centre Kruislaan 413 l098SJ Amsterdam The Netherlands
Annals of Discrete Mathematics 18 (1983) 197-200 © North-Holland Publishing Company
197
ON GENERALIZED PERFECT CODES AND STEINER SYSTEMS G. Cohen * and P. Frankl
I. INTRODUCTION Let (Z2)n = A be the set of n-sequences of integers mod 2.
For any x and y
in A, the Hamming distance H(x,y) counts the number of their different coordinates: H(x,y) = I{i, 1 x. ~ y.}1 and is the weight of x-y : H(x,y) = H(x-y, 0) = w(x-y), 1 where 0 is the all zero element of A.
= n, A with P{l , 2,
{i, x.1
this case H(x,y)
=
Ix
t::.
..• n
We identify x with its support supp(x)
} the collection of subsets of
y I and w(x)
=
I x I,
t::.
IB (x)1 =.r r
1 =0
(~). 1
2, ... n}.
being the symmetric difference.
Hamming sphere of radius r centered on x is B (x) = {y r
{l,
r
g
A, H(x,y)
~
In The
r} with
We denote by P (n, {r.}, {N.}, i = 1, 2, ... s) a partition of 1
A with N. Hamming spheres of radius r ..
1
When s = 1 the centers of the spheres
1 1 k
form a perfect r l error-correcting code (n, r ,2 ) [1]. We also denote by.([5]) l GS(t, K = {kl ... ks }' n) a generalized t-Steiner system, i.e. a set of n points and a set of blocks with cardinalities in K s.t. any t-tuple of points belongs to exactly one block.
For s = 1, this is the classical t-Steiner system, written
S(t, k, n). II. SOME CONSTRUCTIONS The following proposition is from [2]. PROPOSITION 1: If there exists a P(n, {r i }, {N i }, i = 1, 2, ...s)then there exist s GS ( r. + 1, {r. + r. + l} s. l' n), for i = 1, 2, ... s . 1
1
J
J=
PROOF: By translation, it is always possible to center on
Q a sphere of radius rio
We still have a partition of A, so the elements of weight r i + 1 in A must be contained in some spheres. Taking the centers of these spheres as blocks gives the result.
198
C. Cohen and P. Frankl
PROPOSITION 2: There exist P(15, {3,1} , {32,896}) GS(4, {5,7},
15)
GS(5, {6,8},
16).
PROOF: the P(15, {3,1}, {32,896}) is constructed in [2 I, using 32 spheres of radius 3 centered on the codewords of the first order Reed-Muller code (15,5), and 896 spheres of radius 1 centered on the "Bent" functions (or equivalently codewords of weight 5,6,9, 10 in the (15,11) Hamming code).
By proposition 1,
this yields a GS(4, {5,7}, 15) with 168 blocks of cardinality 5 (codewords of weight 5 in the (15,11) Hamming code) and 15 blocks of cardinality 7 (codewords of weight 7 in the Reed-Muller code).
This Steiner system is extendable to a 5-
Steiner system GS(5, {6,8}, 16) with 448 blocks of cardinality 6 and 30 blocks of cardinality 8.
The problem of the unicity of these GS, raised by Assmus [3J, is
sti 11 open.
III. ON THE SIt, t+l, 2t+3) SIt, t+1, 2t+3) can only exist when t+3 is prime [41. exist, for t =8 they don't.
For t
= 2, 4 they
The question is unsettled for t > 10. The follow-
ing proposition relates their existence to partitions of A with spheres of two different radii. PROPOSITION 3: For t > 4 a P(2t+3, N
1
= 2, N2 = 2 (2t+3)/(t+l) t
{t-l,l), {N , N }) exists onZy if l 2
and is then equivaZent to the existence of a
SIt, t+l, 2t+3). PROOF: Suppose there exists a P(2t+3, {t-1,1} , {N , N }). One can always center l 2 a sphere of radius t-1 on Q. Suppose N1 ~ 3. Then there would be at least two other spheres Bt_l(x) and B _1(y). t Now B _1(O) II Bt_l(x) = 0, Bt _ (O) n B _ (y) = 0 implies w(x) > 2t-1, w(y) > 2t-1. t 1 t 1 For t > 4 B _ (x) and B _ (y) would intersect in ~, the all "1" vector of weight t 1 t 1 2t+3, which is impossible. Hence N1 ~ 2. Suppose now Nl = 1, so the only sphere of radius t-1 is centered on O.
Call W. the number of spheres whose centers have 1
weight i. Elements of weight t in A are contained in spheres with centers of weight
On generalized perfect codes and Steiner systems
199
t+l.
Every such sphere contains t+l elements of weight t so W+ t l The same method then shows that Wt+2 = Wt+l , Wt+3 = Wt+4 = o.
(2~+3)/(t+l).
Now spheres with centers of weight t+5 must contain all elements of weight t+4 and at most the total number of elements of weight t+6, hence W (2t+3)/(t+5) t+5 = t-l and W ~ (2t+3)/(t_2) = !:l (2t+3)/(t+5) t+5 t-3 t+6 t- 1 ' a contradiction.
Finally Nl
= 2. Now the sphere packing condition
2t+3 . 2t+3 Nl IBt_l(x)1 + N2 IB1(x)1 = 2 glVes N2 = 2( t )/(t+l). Suppose now such a P(2t+3, {t-l, l} , {2, 2(2~+3)/(t+l)}) exists. sition 1, it yields a GS(t, {t+l, 2t-l}, 2t+3).
We want to prove that there is no
block of cardinality 2t-l, so this GS is a S(t, t+l, 2t+3). From Nl
By Propo-
Suppose the contrary.
= 2 we know that there is .at most one such block c of cardinality
2t-l : w(c) = 2t-l.
Denote by {a,1 } the set of elements in A having a "1" in the b first i components and b "1" in the last n-i components. Of course i i i i n-i 2t-l. . V X E { b}' w(x) = a+b, and I{ b} = ( ) ( b)· Suppose c E{2t-l OJ, Wh1Ch 1S a, a, a 2t-l' always possible by permutation of components. Let V = {t-l,l}. V n Bt_l(O) =
~
and V v E V, H(v,c) = t+l.
of radius 1 containing the elements of V. H(c,e) = t+2 (H(c,e)
Let E be the set of centers of spheres For any e in E, w(e) = t+l and
tor t+2 by triangular inequality between c, e, v, and 2t-l H(c,e) = t would imply Bt_l(c) n Bl(e) F ~.) So E c {t-l 2}. Now for every e in 2t-l 2t-l ' 4 E,Bl(e) contains t-l elements of {t-2 2 }, so lEI ~(t-2 ) (2)/(t-l). But for , 2t-l every e, Bl(e) contains only 2 elements of V and 2 lEI < 12 (t-2 )/(t-l)
=
This contradiction proves there is no such c.
In fact the same meth-
od would prove that c cannot have weight 2t, 2t+l or 2t+2.
Hence a P(2t+3,{t-l ,l}
{N , N }) has necessarily Nl=2 spheres of radius t-l centered on complementary ell 2 ements (e.g. Q and .1J. For the converse now, suppose there exists a S(t,t+l, 2t+3). Let B be the . 2t+3 blocks, w1th lsi = b = ( t )/(t+l). From [4], V Bl , B2ES, IBl n B21 ~ ~, so H(B ,B ) ~ 2t. Of course H(B ,B ) ~ 4, because S is a t-Steiner system. Consider l 2 l 2
200
G. Cohen and P. Frankl
&the
set of complements of blocks.
H(B.,B.J J
1
band
~
3.
Bspheres
Then V S., B.,E 1
6,
4 ~ H(B.,B.)~ 2t and 1
J
Now, centering on 0 and 1 spheres of radius t-1, and on elements of -
-
of radius 1 we get a P(2t+3, {t-l,l} {2,2(2~+3)/(t+l)}) (it is
easy to check the sphere packing condition 2
"i
J
~~: (2t~3)
+ (2t+4) 2 (2t;3)/(t+l)
t 3 + ).
REMARK: Cases t
~
4 are easy: for t = 2, there is a P(7, 1, 16} (Hamming code)
and a S(2, 3, 7).
For t
=
4 there is a P(ll, {3,1}, {2, 132}) [2] and a unique
S{4, 5, 11). Let us give a condition on K for the existence of a GS (t, K, 2t + 3). PROPOSITION 4: a GS(t, K, 2t + 3) exists
on~y
if t + 1
£
K.
PROOF: immediate consequence of the following bound [2J n;:' (t+l). inf k
£
((k - t + l)} K
To conclude, we can state the following PROBLEM: Is there a GS(t, K, 2t + 3) with K # {t + l} (i.e. a GS which is not a S(t, t t l , 2t
+ 3)?
BIBLIOGRAPHY 1.
2. 3. 4. 5.
F.J. Mac Williams and N.J.A. Sloane, The Theory or Error-Correcting Codes, North-Holland Math. Lib. Vol 16 (1977). B. Montaron and G. Cohen, Codes parfaits binaires a plusieurs rayons, Revue du CETHEDEC - NS 79-2, (1979), 35-58. E. Assmus, Personal communication. B.H. Gross, Intersection Triangles and Block Intersection Numbers of Steiner systems, Math. Z., 139, 87-104 (1974). I.F. Blake and M. Rahman, A Note on Generalized Steiner Systems, utilitas Math. Vol. 9,339-346 (1976). ENST 46 Rue Barrau 1t 75013 Paris France
CNRS 15 Quai Anatole France 75007 Paris France
Annals of Discrete Mathematics 18 (1983) 201-208 © North-Holland Publishing Company
201
GREEDY COLOURINGS OF STEINER TRIPLE SYSTEMS i,
C.J. Col bourn and M.J. Col bourn
ABSTRACT Three greedy algorithms for colouring the blocks of Steiner triple systems are described; worst-case colourings are studied for each, and their practical use is examined using the eighty Steiner triple systems on fifteen elements. 1. INTRODUCTION A Steiner triple system (STS) is a pair (V,B).
V is a v-set of elements,
and B is a collection of 3-subsets of V called bloaks; each 2 subset of V appears in precisely one block. 1847.
STS were apparently first studied by Kirkman [11] in
In an STS(v), a parallel alass is a set of v/3 pairwise disjoint blocks.
resolution of an STS is a partitioning of its blocks into parallel classes.
A
A
Kirkman triple system (KTS) is an STS, together with a resolution of its blocks. A substanti a1 amount of research has been carri ed out on STS (see [5] ), and much of it concerns KTS.
Ray-Chaudhuri and Wilson [16] have shown that there
exists a KTS(v) for every v=3 (mod 6); much related work is still underway, including the study of resolvable block designs and t-designs [9], nearly Kirkman systems {13], and doubly resolvable designs [17].
In addition, the size of
maximal colour classes in STS has been studied by Lindner and Phelps [14], Woolbright [ 19] • and Brouwer. Each of these studies generalizes work on KTS; we consider another generalization here.
A aolour alass in an STS is a set of pairwise disjoint blocks.
k-bloak aoZouring of an STS (V,B) is a partition of B into k colour classes. ahromatia index of a STS S is the least k for which S has a k-b1ock colouring.
A The A
KTS(v) is a (v-l)/2 -block colouring of an STS(v). Kirkman [12] , Moore [15], and others studied resolvable designs as solutions to scheduling problems for tournaments, etc.
In scheduling a tournament based on
a given design, one might ask how many rounds are required.
Viewed appropriately
c.J. Co/bourn and M.J. Col bourn
202
this is simply asking for the design's chromatic index. There are two primary questions.
First, what is the range of possible val-
ues of chromatic indices of STS? Of course, the chromatic index of an STS(v} is always at least (v-1}/2.
Moreover, applying Brooks' theorem [2] to its block
intersection graph, we obtain that it does not exceed 3(v-1}/2.
The actual range
of chromatic indices is unknown for v > 15, which suggests the second question. Can one easily compute the chromatic index of a STS? Of course, an affirmative answer to the second question might significantly simplify the first. We focus on the second question, developing heuristic methods for computing the chromatic index of STS via greedy methods and hill-climbing.
Testing these
methods on the eighty STS(15) suggests that they provide useful approximations to the chromatic index. 2. EXACT COMPUTATION In the case of graphs, Vizing's theorem [18] guarantees that a graph with maximum degree d has chromatic index d or d+1.
Arjomandi [1] gives a clever
polynomial-time method for constructing a (d+1}-co10uring.
Nonetheless, deciding
whether the chromatic index is d or d+l has recently been shown to be NP-complete [10] .
Similar results are completely lacking for STS.
The complexity of computing
the chromatic index is unknown, and the best current method involves (possibly exponential) backtracking.
An alternate direction is to consider algorithms which
always run in polynomial time, but only approximate the answer.
This is a reason-
able compromise in the context of the original tournament scheduling problems. We investigate two general classes of algorithms for approximating the chromatic index: greedy methods and hill-climbing methods. parison is the optimal co10urings of the eighty STS(15}.
Our standard of comFour have chromatic
index 7, thirteen have 8, and the remaining sixty-three have 9 [3]. 3. ACHROMATIC INDEX A simple greedy colouring technique operates as follows. block is a different colour.
Initially, each
Until there are no two disjoint colour classes,
join together two disjoint colour classes (eliminating a colour).
Although easy
to implement, this method may require many colours; we term the maximum number of colours the achromatic index of the STS.
(This use of "achromatic" is from
203
Greedy colouring of Steiner triple systems
Harary and Hedetniemi [8].)
The worst-case performance of this greedy method is
described in two lemmas. LEMMA 3.1: The achromatic index of an STS(v) is at most cv 3/ 2, for c a fixed constant.
PROOF: Suppose we have a t-block colouring; the i'th colour class contains k(i) blocks.
For each i, 3k(i)(v-l)/2
intersect.
Thus ifl k(i)
~
= t-l, since each pair of the t classes
~ = 2t(t-l)/(3(v-l)).
Since iIl k(i)
= v(v-l)/6,
t(t-l) ~ = v(v-l)2/2, which establishes the lemma. LEMMA 3.2: There exist infinitely many STS whose achromatic index is at least cv 3/2, J+'or C a f·,/-xed constant. PROOF: Take a projective plane of order 2(2t+l) __ that is, with block size 2(2t+l) + 1.
Desarguesian planes of these orders exist [7]; moreover, the plane
properties ensure that every two blocks intersect.
Now replace each block of the
plane by a KTS on those elements; these exist since the block size is 3 (mod 6). The resulting system is coloured by assigning each parallel class of each KTS a unique colour. 4. BLOCK-BY-BLOCK GREEDY METHOD The achromatic index is an indication of how badly one can do. sensible "block-by-block" method proceeds as follows. resent colours. Initially, the blocks are not coloured.
A more
We employ integers to repWe colour the blocks one
at a time, assigning a block the least integer so that the resulting colour class contains disjoint blocks.
It is immediately clear that this improves on the ear-
l i er method. LEMMA 4.1: The block-by-block greedy method uses at most 3v/2 coZours for an STS(v). PROOF: Suppose to the contrary that the number of colours exceeds 3v/2. Thus, at some interim stage, a block {x,y,z} could not be assigned to any of the first 3(v-l)/2J colours.
Then {x,y,z} intersects each of these colour classes; but
204
C.J. Co/bourn and M.J. Co/bourn
each of {x,y.z} appear in only (v-l)l2blocks.
This contradiction establishes the
desired result. At first sight. one might expect that an improvement on this crude argument might give a better upper bound. but this is not so. LEMMA 4.2: The block-by-block gl'eedy method may l'equil'e 3[ (v-l)/2J -2 coloUr's. PROOF: Select a KTS(n) with elements v(l) •...• v(n).
Construct an STS(2n+l) with
elements {x(l) •... ,x(n).y(l) •...• y(n).@} as follows.
For each block {v(i).v(j).
v(k)} of the KTS, we have blocks (x(i).x(j).x(k)}. (x(i).y(j).y(k)} • {y(i).x(j),y(k)} and {y(i),y(j).x(k)}.
In addition. we have b10cks{@.x(i).y(i)}
for each i. To colour the STS, we first order each block of the KTS arbitrarily; for example. the block {x.y.z} may be represented as<x.y.z>.
Let r
=
(n-1)/2 and let
V(l), ... ,Vir) be the parallel classes of the KTS. in which the blocks are For each ordered block
oI'de2>ed.
{x(i),x(j),x(k)J colour h. assign {x(i).x(j),y(k)} colour r+h. assign (x(i).y(i),x(k)} colour 2r+h, and assign {y(i).x(j).x(k)} colour 3r+h. assign {@,x(i),y(i)} colour 4r+i. block-by-block method produces.
Finally.
This is a (6r+l)-block colouring, which the Thus an STS(2n+l) may use 3n-2 colours.
Despite this worst-case performance of the block-by-block greedy method. one might ask now useful it is practice. showed the following:
Our computational results for order 15
GREEDY COLOURING
7
8
9
10
11
12
13
0
0
0
1
1
2
0
0
0
1
8
4
0
0
0
30
26
7
x
UJ
o z
7
B
9
5. COLOUR-BY-COLOUR GREEDY METHOD We also studied an alternate greedy method which proceeds colour-by-colour. Having selected i colour classes. we select the (i+l)st taking a maximal colour class among the remaining (uncoloured) blocks.
We do not necessarily find the
largest possible class; in this context. by maximal we mean that
205
Greedy colouring of Steiner triple systems
(1) every uncoloured block intersects the colour class (2) there is no block whose deletion from the colour class enables the simultaneous addition of two uncoloured blocks. have any results on worst-case performance.
For this greedy method, we do not It is different from the block-by-
block greedy method, however, since in this method the first colour class must contain at least (v-l)/4 blocks [14J. For order 15, this method seemed more effective than the previous: GREEDY COLOURING 13 10 7 8 9 11 12 7 u
.......
I-
8
ex:
9
~ o :J:
0
1
1
0
2
0
0
0
2
9
2
0
0
2
27
30
4
0
u
6. HILL-CLIMBING Although quite efficient, both greedy methods apparently produce some poor colourings.
We investigate here the use of some simple heuristics, for improving
eXisting colourings.
The first is quite simple: if the blocks of a colour class
can be distributed among the other colour classes, we do so.
The goal of this
heuristic is to overcome the high reliance of the greedy methods on the edge ordering. A second heuristic operates as follows.
If there are two blocks, b with
colour i and b' colour j, which can be recoloured so that b has colour j and b' has colour k, and if in so doing the vector of colour class sizes is lexicographically increased, we do so. The effects of these two heuristic on the colourings of the 5T5(15) are given here.
The block-by-block greedy colouring, after applying these heuristics,
yields:
IMPROVED GREEDY COLOURING 8 10 9 11 7
X Lo.J C
z:
u .......
i o
ex:
:J:
u
7 8
9
1
l2
13
0
1
2
0
0
0
0
12
1
0
0
0
39
24
0
0
0
206
C.J. Colbourn and M.J. Col bourn
The colour-by colour greedy method with these heuristics gives IMPROVED GREEDY COLOURING 11 7 10 12 13 9 8 x w
Cl
z
7
1
8
0
3
0
0
0
0
0
13
0
0
0
0
36
27
0
0
0
9 -
Combining the best colouring for the two methods, we obtain BEST COLOURING OBTAINED 13 7 8 9 10 11 12 x
w Cl Z
7
2
0
2
0
0
0
0
0
13
0
0
0
0
50
13
0
0
0
u
r-
eo:
8
::E:
o
c::: ::c u
9
With these heuristics improvement techniques, the greedy colouring methods are not only efficient but also apparently quite accurate. 7. FUTURE RESEARCH The worst-case performance of the greedy methods together with the heuristic improvement techniques may supply a good theoretical upper bound on the chromatic index.
In particular, our results for order 15 suggest that the upper bound
3[ {v-l )/2J is much too large.
In fact, an upper bound of v would follow from a
conjecture of Erdos, Faber and Lovasz [6J.
In the case of cyclic STS, we can
establish this upper bound of v [4], but in general, the problem is open. Whether or not improved theoretical results are obtained, our work suggests that quite simple greedy methods can provide reasonably good colourings.
Impor-
tant future work involves determining the computational complexity of computing the chromatic index of an STS exactly. ACKNOWLEDGEMENTS: Research supported by NSERC Canada Grant A5047. The results of section 3 were obtained jointly with Kevin Phelps.
Greedy colouring of Steiner triple systems
207
BIBLIOGRAPHY
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.
13.
14. 15. 16.
17. 18. 19.
L Arjomandi, "An efficient algorithm for colouring the edges of a graph with d+ 1 colours", Infor., to appear. R. L. Brooks, "On co louri ng the nodes of a network", Proc. Cambridge Philos Soc., 37 (1941), 194-197. C.J. Colbourn, "Computing the chromatic index of Steiner triple systems", Computer JournaZ, to appear. C.J. Colbourn and M.J. Colbourn, "The chromatic index of cyclic Steiner 2-designs", Int. J. Math. Sai., to appear. J. Doyen and A. Rosa, "An updated bibliography and survey of Steiner systems", AnnaZs of Discrete Math., 7 (1980), 317-349. P. ErdOs, "On the combinatorial problems I would most like to see solved", Combinatorica, 1 (1981), 25-42. M. Hall Jr., CombinatoriaZ Theory ~lad'lillan, New York, 1967. F. Harary and S. Hedetniemi, "The achromatic number of a qraph", J. Comb. Theory, 8 (1970), 154-161. A. Hartman, "Resolvable Designs", M.Sc. Thesis, Israel Institute of Technology, Haifa, Israel, June 1978. I. Holyer, "The NP-completeness of edge-colouring", SIAM J. Computing, 10 (1981), 718-720. T.P. Kirkman, "On a problem in combinations", Cambridge and Dublin Math. J., 2 (1847), 191-204. T.P. Kirkman, query, Lady's and Gentleman's Diary, 48 (1850). A. Kotzig and A. Rosa, "Nearly Kirkman systems", Proc. 5th South-Eastern Conf. Combinatorics, Graph Theory and Computing; Congressus Numerantium, 10 (1974), 607-614. C.C. Lindner and K. Phelps, "A note on partial parallel classes in Steiner systems", Discrete Math., 24 (1978), 109-112. LH. Moore, "Tactical memoranda I-III", Amer. J. Math., 18 (1896), 264-303. O.K. Ray-Chaudhuri and R.t4. Wilson, "Solution of Kirkman's schoolgirl problem", Proc. Symp. in Pure Math., 19 (Amer. Math. Soc., Providence, R.I.), 1971. S.A. Vanstone, "Doubly resolvable designs", Discrete Math., 29 (1980), 7786. V.G. Vizing, "On an estimate of the chromatic class of a p-graph", (Russian), Diskret. Analiz., 3 (1964), 25-30. D.E. Woolbright, "On the size of partial parallel classes in Steiner systems", Annals of Discrete Math., 7 (1980), 203-211.
Dept. of Computational Science University of Saskatchewan Saskatoon, CANADA
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209
Annals of Discrete Ma thematics 18 (1983) 209-224 © North-Holland Publishing Company
CONCURRENCE GEOMETRIES Henry Crapo
It was during the Colloquio Internazionale sulle Teorie Combinatorie at the Accademia dei Lincei in 1973 that I had the good fortune mino Segre.
to meet Professor Benia-
I take pleasure in dedicating this lecture to his memory.
It was
also at that meeting that I first spoke about combinatorial properties of the geometry of lines, a subject which has continued to be central to my work and has continued to fascinate me during years of research devoted to the applications of combinatorial geometry in architecture and structural engineering.
This research,
undertaken by Janos Baracs, Ethan Bolker, Thorkell Helgason, Tiong Seng Tay, Walter Whiteley, myself and others during the years 1974-1978, led us to the pioneering work of James Clerk Maxwell and Luigi Cremona on graphicaZ statics. Beginning in 1978, the Structural Topology research group, then established at the Universite de Montreal, has piloted this research activity. In today's talk, I shall deal only with the most recent aspect of this work, a problem in pure projective geometry whose solution will enable us to associate with any plane structure a combinatoriaZ object descriptive of the infinitesimal motions of that structure. For purposes of exposition, we will restrict ourselves to plane configurations of lines.
The natural generalization of our theorem to configurations of
hyperplanes in higher dimensional spaces is the subject of an article, to appear elsewhere. 1. INTRODUCTION Imagine a finite set of variable lines in a plane, each line running in a particular fixed direction (not all parallel), but otherwise free to move in the plane.
Using these lines, a variety of combinatorially distinct plane configura-
tions can be formed.
This paper is devoted to a combinatorial investigation of
this variety of configurations.
210
H. Crapo
We shall refer to plane configurations simply as drawings.
For each drawing
we pay attention only to which subsets of its of lines are concuppent. natural order on the set 0 of possible drawings.
There is a
We say a drawing F is a special-
ization of a drawing E, and write E ~ F, if and only if every set of lines concur-
rent in E are also concurrent in F. preorder
~
We shall shown that the set D, with the
of specialization, is a geometPic lattice.
It will follow that the set
D of possible drawings is itself a real projective configuration of the same dimension, the conCUI'l",mce geometpy of the given set of fixed directions.
It fol-
lows also that every drawing, being a flat of some dimension in this concurrence geometry, will have a well-defined panko
Ranks, circuits, bonds, dependence,
bases, strong and weak maps, indeed all the ideas and theorems of matroid theory can thus be brought to bear on our analysis of configurations. 2. EXANPLE In order to fix these ideas, let's start with an example.
Consider five
lines L , ... , L in the plane, such that Land L are parallel, as are Land 1
L4 ·
5
1
Fix the three directions of these five lines.
drawings in Figure 1*.
2
3
We arrange all the possible
There is a most general sort of drawing "0", in which no
three lines are concurrent: all other drawings are specializations of it.
A typi-
cal elementary specialization of "0" involves making three lines, such as 1, 3 and 5, concurrent.
Another type of elementary specialization is to make lines 1, 2
and 3 concurrent.
This can only be done by making lines 1 and 2 coincident,
forcing 1, 2,4 and 1, 2, 5 also to be concurrent.
There are six distinct elemen-
tary drawings (elementary specializations of "0"). Each elementary drawing can be further specialized in a number of ways, to yield drawings of rank 2.
Each drawing of rank 2 could have been obtained by ele-
mentary specialization of a number of different drawings of rank 1.
The connect-
ing lines between drawings in Figure 1 indicate all such elementary specializations.
Finally, the most extreme specialization occurs when all five lines pass
through a point.
This drawing has rank 3.
During the 16th OSU-Denison Mathematics Conference in May 1981, I conjectured that drawings with given edge directions, ordered by specialization, form a geometric lattice. Following the problem session at which this problem was posed, Denis Higgs, Neil Robertson, Paul Seymour, Neil White, Tom Brylawski, Tom Dowling, Torn Zaslavsky and others engaged in a lively discussion of the subject, and made a number of very helpful suggestions. It was the example in Figure 1, worked through in detail with Denis Higgs, which convinced me of the truth of the conjecture.
12
Figure 1
6 elementary drawings 7 drawings of rank 2
212
H. Crapo
The order of specialization linking these six elementary drawings and seven rank 2 drawings agrees with the order of incidence linking six points and seven lines in a certain plane projective configuration, the concurrence geometl'Y shown in Figure 2.
Three types of lines in the concurrence geometry are there singled
out, and their associated drawings are indicated.
3. CONCURRENCE GEOMETRIES: SOME BASIC FACTS A concurrence geometry has certain underlying elements, namely the triples of lines, which are combined into flats, each flat being the set of triples of
lines which are concurrent in an associated drawing.
These flats are called "0",
pointe, lines, p lane3, etc., accordi ng to thei r rank.
By extens i on every set of
triples has a rank, equal to the rank of the most general drawing in which those triples are concurrent. If the directions of the variable lines are fixed in such a way that no two lines are parallel, then the points of the concurrence geometry are simply all the single triples of lines.
This is the case in Figure 3, which the reader can eas-
ily verify is the concurrence geometry for five lines in fixed directions (no two parallel) in the plane. We have labelled the major lines of this geometry N , ••• ,N. 1
5
Notice that each line N. contains exactly those triples not containing the index 1-
i.
Note also how the concurrence geometry in Figure 2 can be obtained from this
geometry simply by moving the line N until the two shaded triangles are reduced s to single points. This is best accomplished in two steps. To obtain the concurrence geometry for the situation in which lines 1 and 2 are parallel, make N , 3
N , Ns concurrent. 4
Then, if lines 3 and 4 are also to be parallel, make Nt' N2
and N pass through a point.
s
Geometric dependence in the concurrence geometry is a result of logical implication concerning concurrence of lines in drawings.
For example, to say in
Figure 1 that the element 145 depends on the pair of elements 123 and 245 is to say that 145 is on the smallest flat containing 123 and 245 (namely the line {123, 124, 125, 145, 245)).
In tenns of concurrences, thi s means that in every drawi ng
in which 123 and 245 are concurrent triples (1 and 2 being parallel, 3 and 4 also), then 145 is also a concurrent triple.
The three triples 123, 245, 145 here form a
drcuit (a minimal dependent set of elements); any two of these three concurrences
implies the third.
Note that 123 and 124 also fonn a circuit, each implying the
other, both being equivalent to the condition that lines 1 and 2 coincide.
134,234,345
aLl" 34
123, 124, 125
/
'2IZ 3
--
~
;,:
'~"'
;::
'"
;,:
'"'"'
~ 0
~
'"
~
'I2J2
4
3 2
" Figure 2
N VJ
H. Crapo
214
134
~~~~--------~>---------------~D 235
124
Figure 3 A basis of the concurrence geometry is any minimal set of triples such that only the trivial drawing "1" (all lines passing through a single point) makes all those triples concurrent.
If 1 and 2 are the indices of two lines which are not
parallel, then for n variable lines in fixed directions in the plane, 123, 124, ... , 12n form a basis: no triple in this list is a consequence of its predecessors, and all told, they imply that all lines pass through the point of intersection of lines 1 and 2 (wherever that may be).
Consequently, the concurrence geometry for n vari-
able lines in fixed directions in the plane, not all parallel, invariably has overall rank n-2, independent of the choice of fixed directions. The basic property of geometric lattices (and matroids), which is crucial
Concurrence geometries
for understanding concurrences, is the following.
215
Any elementary specialization
(of any drawing) is generated by a single triple, and a single triple can generate at most an elementary specialization of any drawing. related exchange property. drawing D.
There is also a closely
Consider two triples sand t not concurrent in a
If in all specializations of D in which s is concurrent, t is also
concurrent, then the same is true with sand t interchanged: if t is concurrent, so is s. 4. PROJECTIVE RELATIONS So far, we have given the reader no reason to believe that concurrence geometries depend on anything more than the topology of the given set of line directions.
For instance, one might expect that every set of n distinct direc-
tions for n lines would give the same concurrence geometry. true.
This is far from
Projective relations between the fixed directions of a set of lines are
strongly reflected in the resulting concurrence geometry. a plane drawing of a tetrahedral graph, Figure 4.
Consider, for instance,
We cut the six lines of the
tetrahedral figure with a single line, which we think of as the line at infinity, so the six points of intersection are the "directions" of the lines of the tetrahedron.
These six directions form what is called a quadrilateral set.
Any
one direction can be computed from the other five, by simply drawing a tetrahedron with five edges in the correct directions, then drawing in the last edge to connect two known points.
That the resulting direction is independent of the
choice of tetrahedron (Figure 4b) is a theorem of projective geometry (see Baker 1922, vol. I, p. 15). A related theorem (ibid., p. 61), also illustrated in Figure 4b, states that the dual tetrahedron can also be drawn with the same edge directions, as a reciprocal figure. It is clear that the set of four triples 136, 145, 234, 256 will be geometrically different in the two concurrence geometries C and C , that for six lines 6
Q
in general directions, and that for six lines whose directions form a quadrilateral set permitting the drawing of a tetrahedron with those four vertices. is the difference?
What
In C , the first three triples form a flat of rank 3 (a 6
plane): the triple 256 is not on that plane, and 256 is not a concurrence in a drawing such as Figure 4c.
Only the trivial drawing, with all lines through one
point, has all four of these triples concurrent. are coplanar, forming a circuit of rank 3.
In C , however, the four triples Q
The concurrence of any three of these
216
JI.(:rapo
triples implies concurrence of the fourth. Figure 5 shows the concurrence geometry C for six lines in fixed but gen6
eral directions, not forming a quadrilateral set. typical lines in C are indicated.
Two drawings associated with
Figure 4c is the drawing corresponding to the
6
plane 136, 145, 234, which does not contain any other point of the configuration. The overall geometry has a very simple structure, not sufficiently emphasized in this sort of drawing.
The major planes N , ••• , N6 are just six planes in gener1
al position in 3-dimensional space.
As before, the plane N. contains precisely ~
those points whose labels do not contain the index "in.
The point 136, for
instance, lies at the intersection of planes N , Nand N ; in this way the com245
plete configuration is determined. another source.
This geometry, by the way, is also known from
It is the Dilworth completion 02(B
6
)
of the Boolean algebra B
6
,
the latter being truncated from below so as to remove its one-and two-element subsets. The concurrence geometry C , for six lines in directions forming a quadriQ
lateral set, is shown in Figure 6.
Here, the points 136, 145, 234, 256 are
coplanar, so the tetrahedral drawing with those vertices is of rank 3.
By the
theorem of reciprocal figures, quoted above, it is also possible to draw the dual tetrahedron using the same edge directions, so its vertices 134, 156, 236, 245 must consequently also be coplanar as points in C . Q
a whole is not hard to describe. derived planes N
7
=
The six planes N , 1
136, 145, 234, 256 and N
planes of a Mobius pair of tetrahedra.
s
=
This concurrence geometry as N together with the two 6
134, 156, 236, 245 form the face
The vertices of the first tetrahedron lie
on the faces of the second, the vertices of the second on the faces of the first. (To be precise, the vertices 136, 156, 134, 236 of one tetrahedron lie on faces N7 , N3 , N , Nt of the other, while the faces N ' N ' N4 , N2 contain the vertices 6 s s 245, 234, 256, 145 of the other, respectively.)
145
145
145
3
6 (a)
Figure 4
.
(c)
I • I
Figure 5
136
236 Figure 6
123
126
220
H. Crapo
5. COORDINATION, AND THE MAIN THEOREM We have yet to prove that drawings, with given directions for their lines, form a geometric lattice when ordered by specialization.
We will prove directly
that the elements of a concurrence geometry (that is, triples ijk of lines in the drawings) may be represented as vectors [ijkj in an n-dimensional real space (n being the number of lines).
In fact we shall show that each drawing corresponds
to an n-vector c, which is orthogonal to a vector [ijkj (c . [ijkj
= 0)
if and
only if the lines i, j, k are concurrent in the drawing indicated by the vector c. It then follows that the rank of any set of triples equals the vector space rank of the corresponding set of vectors (Crapo 1976). Assume the set of fixed directions of lines for a drawing are given by points A1 ,
...•
An on the line at infinity.
We may write the projective coordi-
nates of these points in a 2 by n matrix A = a a 1
a
2
b b 1
n
2
where each line i has slope b./a..
(There is at least one non-zero entry in each
1- 1-
column of thematrix A.)
n
b
For each 2 by 2 submatrix of A, compute the corresponding
Pluekep eoopdinate J a. a'l 1-
d .. 1-J
b.
1-
b. J
Finally, for each triple ijk of directions, we make up an n-vector [ijkj whose only non-zero components are the following: d th
component and d .. as k 1-J
jk
component.
as the ith component, -d
we have twenty vectors in 6-dimensiona1 vector space:
[ 123]
d
[ 124]
d
[ 456)
23 24
o
-d -d
13
14
o
as jth
These, we shall see, are the correct vector
representations of the elements of the concurrence geometry.
2
ik
3 d
12
4
5
6
0
0
0
0
0
0
d
o
d
12
S6
-d
46
d
4S
Thus for six lines
Concurrence geometries E~uations
221
of a 1ine L. with slope b./a. may be written in the form 1-
1-
for any finite va1ue of c .. 1-
1-
Such a line has x-intercept -c./b .. 1-
lines parallel to the x-axis) it has y-intercept c./a .. 1-
a.
1-
1-
1-
If b. = 0 (for 1-
The coordinates
= b.1- = 0 do not occur, so for each direction, and each finite value of c.1- there
is a well-defined line (not a infinity) in the plane. Any three lines L., L., Lk have a point in common if and only if the determinant 1-
J
c. c. c 1-
J
k
a. a. a 1-
J
k
b. b. b 1-
J
k
is equal to zero; that is, if and only if (c., c., c )· (d.k,-d. , d .. ) 1-
J
k
J
1- k
1-J
0,
if and only if c . [ij kj
O.
We have a Galois connection between triples and drawings: each set of triples are concurrent in certain drawings, and each set of drawings have certain common concurrences.
For each drawing, regard its vector c as a real-valued func-
tion defined on the set T of triples, the value c([ ijkj) being the inner product c . [ijkj.
The zero-set of such function c is the set of all triples on which it
takes the value 0 (ie: the set of triples concurrent in that drawing).
Since the
vectors c form a vector space, we know (Crapo 1976) that the set of intersections of zero-sets form a geometric lattice, this being a function-space geometry.
Since, moreover, the vectors c form a real finite-dimensional space (~ Rn ), any intersection of zero sets is also the zero-set of a single vector c.
To see this,
say we were given a finite list of vectors c with their various zero-sets.
A
linear combination of those vectors, using independent trascendentals as coeffi-
222
H. Crapo
cients, will produce a single vector which has as zero-set exactly the intersection of the zero-sets of the given vectors. already form a geometric lattice.
Thus, the zero-sets themselves
Since the order of inclusion on zero sets is
isomorphic to the order of specialization on drawings, we have our main theorem. (We found the rank of the concurrence geometry in section 3, above.) THEOREM 1: Given a co ~lection of n variahle lines in fixed directions (not aU paraZle:) in tr;e plane. the set of all dralJings which can be made with these n lines, Cl'dei'ed oy specialization. forms a geometric lattice of rank n-2.
All the n-vectors [ijk] are orthogonal to two fixed vectors, the n-vectors a and b, the two rows of the matrix A of projective coordinates of the fixed directions for our variable lines.
The concurrence geometry is thus a subgeome-
try of the orthogonal complement of the 2-dimensiona1 space spanned by a and b. The principal hyperplanes N of the concurrence geometry have equations x.
"
and rank n-3 in n-space.
0, and a . b
b· x
0
The reader may now wish to check that the one-dimen-
sional subspace generated by the vector [ijkJ is exactly the intersection of all hyperplanes N. for h f i,j or k, as we claimed above. r1
For a discussion of the basic properties of concurrence geometries, as reflected in this coordinatization, also for the extension of Theorem 1 to higher dimensions and for the application of concurrence geometries to the study of infinitesimal motions of plane structures, we refer the interested reader to our article "ConcU1'l'ence Geometries" (to appear in the Advances of Mathematics), of which this talk has been an extract.
BIBLIOGRAPHY 1. 2.
H.F. Baker, Pl'inciples of Geometry. Cambridge, University Press, 1922 (2nd ed. 1929, reprinted 1954). A.L.C. Cheung, "Adjoint of a geometry", Canad. Math. Bulletin 17 (1974), 363-365.
Concurrence geometries
3. 4. 5.
223
A. Cheung and H. Crapo, "A Combinatorial Perspective on Algebraic Geometry", Advances in Math. 20, (1976), 388-414. H. Crapo, "Structural Rigidity", Structural Topology 1, (1979). W.V.D. Hodge and D. Pedoe, Methods of Algebraic Geometry, Cambridge, University Press, 1947 (1952,1954).
Ecole d'Architecture Universite de Montreal Montreal H3C 3J7 Canada
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225
Annals of Discrete Mathematics 18 (1983) 225-228 © North-Holland Publishing Company
(q2+q+l )-SETS OF TYPE (O,1,2,q+l) IN TRANSLATION PLANES OF ORDER q2 Michele Crismale
A new method of constructing (q2+q+l )-sets of type (O,1,2,q+l) in the desarguesian plane of order q2 as well as in any translation plane of order q2 with kernel of order q is discussed.
l.
In an earlier paper [3] the author gives the first example of (q2+q+l )-sets of type (O,1,2,q+l) in a projective plane ~(q2) of order q2 by constructing some sets with such properties (i.e. [6,5] sets containing q2+q+l points of ~(q2) which have O-,1-,2-,(q+l)-secant lines and no i-secant line for i ! O,1,2,q+l) in any Hall plane H(2,q2) of even order. 3
This example has the characters (cf. (6,5]) 3
to = q (q-l)/2, tl = q(q-l), t2 = q (q+l)/2,tq+l = q+l (to,tl,t2,tq+l denote the numbers of O-,1-,2-,(q+l)-secant lines respectively). In this paper, we discuss a general method of constructing (q2+q+l )-sets of 2 2 type (O,1,2,q+l) and characters to = q (q-l) /2, tl = (q+2)q(q-l), t2
= q2(q2_1)/2,t +1 = 2q+l, in the desarguesian plane of order q2 as well as in
q 2 any translation plane of order q with kernel of order q.
Our construction uses the well known four dimensional linear representation of such projective planes.
In section 2, we recall briefly this representation
(for more details, see [2,1,4]).
Section 3 deals with the construction.
2.
Let S4 be the four dimensional projective space over the Galois field GF(q) of order q, where q is a prime power.
Let
L
be a fixed 3-space of 54' A spread S
is defined as a set Of"q2+1 lines in~, such that every point of ~ is contained in exactly one line of S.
Any plane in
~
contains exactly one line of S.
226
M. Crismale
The projective plane to be constructed will be denoted by T and its points and lines will be called T-points and T-lines, to avoid confusing them with elements of 54 which represent them. T-points are of two types. points of A4 = 54 \
~.
T-points of type (1) are represented by the
T-points of type (2) are represented by the lines of S.
There are q4 T-points of type (1) and q2+1 T-points of type (2). two types.
T-lines are of
T-lines of type (1) are represented by planes of 54' passing through
the lines of S and not contained in Z; there are q4+q2 T-lines of type (1).
There
is only one line of type (2), which is represented by Z. A T-line and a T-point are incident if and only if the element of 54 representing the T-line contains the element of 54 representing the T-point.
The
T-points and T-lines, with the incidence relation, form either the desarguesian 2
2
plane of order q or a translation plane of order q with kernel of order q. Conversely, any projective plane which belongs to one of these classes, can be represented by a triple (54' z, S) in the above way.
3.
Let us consider a projective plane T of order q2 which can be represented by a triple (54' z, S).
Let
y
be a plane (q+l)-arc of 54 (i.e. a set of q+l
coplanar points of 54' no three of which are collinear) and denote by V a point of 54' not in the plane of distinct from Z.
y,
such that the 3-space 5 containing V and
y
is
Consider the point set r c 54' which consists of the lines
joining V to the paints of
y,
and suppose that the unique line s(l
E
S on the plane
= 5 n L is not a bisecant of r and does not pass through V. Putting r l = r n A4 , f2 = {s E sl s n r F ~}, K = r l U f2' we have the following result.
(l
K represents in 54 a (q2+q+l )-set of T, of type (O,l,2,q+l) and characters 2 2 2 2 to = q (q-l /2, tl = (q+2)q(q-l), t2 = q (q -1)/2, tq+l = 2q+l. A complete description of (the set represented by) K and of its properties will be published elsewhere. Now, in order to prove the above result, it is convenient to use the following notation. P.
For any plane
p
For any point P E Z, sp is the unique line of S through
of 54' passing through a line of S and not contained in L, sp
is the unique line of S contained in p and pT is the T-line represented by p. and ~T are the point set and the T-line represented by K and z.
KT
Now, P --~ sp is
227
(q2+q+1J-sets o/type (O,1,2,q+lJ in translation planes ~
an injective map (because of /sa n r/
1) of r n tonto r 2• Therefore
T
2
/K /=/K/=lr 1 /+lr21=/r n A41+lr n tl=lr n s4 1=/r/= q +q+l Then lET n KTI=lr n tl=lr n al is equal to 1 if (i) r n E ={V};
or to q+ 1 if (ii) rna is a line on r; (iii) (respectively (iv)) rna is a (q+1)-arc on rand s is a O-secant a
(respectively l-secant) of rna. For any T-line pT of type (1), /pT n KTI=lp n rl cannot be different from O,1,2,q+1, because p n r = (p n S) n rand p n S is a plane or a line. Now, we show that t q+ 1 = 2q+l in all four cases above. (i) Ip n rl= q+l if and only if either a f. pes passes through s or p contains
Sv
a
and a line on r.
Hence tq+1 = q+(q+1).
(ii) jp n rl= q+1 if and only if either a f. peS passes through s or p contains
Sv
a
and a line on r different from rna.
Hence tq+1
l+q+q.
(iii) Ip n rl= q+1 if and only if either a f. peS passes through sand a
v¢
p or sp
E
f2 and V E p.
Hence tq+1 = 1+(q-1)+(q+1).
(iv) Ip n rl= q+1 if and only if either a f. pes passes through s or a
sa f. s p E r 2 and V E p. Hence t q+ 1 = l+q+q. Then, in order to complete the proof, it is enough to derive the characters t o,t ,t from the following equations [6J. 1 2 q4+q 2+1, t 1+2t +(q+1)(2q+1) 2
a
2
2
(q +q+1)(q +1),
2t 2+q(q+1)(2q+1) = (q 2+q+1)(q 2+q). REI1ARK: We observe that the above construction can be employed to get (q 2+1)- and (q+1)2-sets of type (O,1,2,q+1) in T, by using an ova1oid and a hyperbolic quadric respectively, instead of r.
We shall study these cases in a next paper.
ACKNOWLEDGEMENT: The author thanks G. Korchmaros, visiting professor at the University of Bari, for the helpful discussion. research group G.N.S.A.G.A. of C.N.R ••
Work supported in part by the
228
M. Crismale
BIBLIOGRAPHY 1. 2. 3. 4. 5. 6.
R.C. Bose and A. Barlotti, Linear representation of a class of projective planes in a four dimensional projective space, Ann. Mat. Pura Appl., (4) 88 (1971), 9-32. R.H. Bruck and R.C. Bose, The construction of translation planes from projective spaces, J. Algebra, 1 (1964), 85-102. M. Crismale, (q2+q+l)-insiemi di tipo (O,l,2,q+l) e ovali ne1 piano di Hall di ordine pari q2, Note di Matematiaa, (to appear). P. Dembowski, Finite geometries, Springer-Verlag (1968). G. Ta11ini, Problemi e risultati Bulle Geometrie di GaloiB, Re1azione n. 30, 1st. t~at. Univer. Napoli (1973), 1-30. M. Ta11ini Scafati, {k,n~archi di un piano grafico finito, con partico1are riguardo a que11i con due caratteri, Nota I, Rend. Aaa. Naz. Linaei, (8)40 (1966), 812-818.
Istituto di Geometria Universita di Bari Via Nicolai 2 70121 Bari Ita 1y
Annals of Discrete Mathematics 18 (1983) 229-240
229
© North-Holland Publishing Company
THE EMBEDDING OF (O,a)-GEOMETRIES IN PG(n,q).
PART I
F. De Clerck* and J.A. Thas
ABSTRACT A (finite) (O,a)-geometry is an incidence structure S=(P,B,I) of points and lines with a symmetric incidence relation satisfying: (a) S is connected, (b) each line is incident with s+l (s
~
1) points and two distinct lines are incident with
at most one point, and (c) if a point x and a line L are not incident, then there are 0 or a (a
~
1) points which are collinear with x and incident with L.
paper the (O,a)-geometries embeddable in PG(n,s) are considered.
In the
As a particular
case all dual semi partial geometries embeddable in PG(n,s) are determined. 1. INTRODUCTION 1.1. A(finite) semi partial geometry [6] is an incidence structure S=(P,B,I) with a symmetric incidence relation satisfying the following axioms. (i) each point is incident with t+l (t
~
1) lines and two distinct points
~
1) points and two distinct lines
are incident with at most one line. (ii) each line is incident with s+l (s are incident with at most one pOint. (iii) if a point x and a line L are not incident, then there are 0 or a (a
~
1) points which are collinear with x and incident with L. (iv) if two points are not collinear, then there are
~
(~>
0) points
collinear with both. REMARKS: (a) If in axiom (iii) 0 does not occur, then (iv) is an immediate corollary of the other axioms and
~
= (t+l)a.
In such a case S is called a par-
tial geometry [1]. (b) One proves that the dual structure of a semi partial geometry is again a semi partial geometry iff s=t or S is a partial geometry [3].
230
F. De Clerck and I.A. Thas
1.2.
A (semi) partial geometry S is said to be embedded in a projective space
PG(n,q) if the line set of S is a subset of the line set of the projective space and if the point set of S is the set of all points of PG(n,q) on these lines (then s=q). (n
~
All partial geometries embeddable in a projective space PG(n,q)
2) are known: see [2J for the case a = 1 and [9J for the case a> 1.
We
remark that also all partial geometries embeddable in an affine space are known [15J.
However, up to now only the semi partial geometries embeddable in PG(2,q),
PG(3,q), AG(2,q), AG(3,q) are classified [4J [5).
The main difficulty here is
that by axiom (iv) one cannot use induction for the embedding problem in higher dimensions. 1.3.
A (finite) (O,a)-geometry (a> 1) is a connected incidence structure
S=(P,B,I) with a symmetric incidence relation satisfying (ii) and (iii). REMARK: Let x and x', x # x', be collinear points of S, and denote by t+l (resp. t'+l) the number of lines of B incident with x (resp. x').
Then counting in
different ways the number of ordered pairs (U,U'), xIU, x'IU', xfU', x'{U, with U and U' concurrent lines of B, we obtain t(a - 1) = t' (a - 1).
Hence t=t'.
By
the connectedness of S we now see that every point of S is incident with t+l lines of B. 1.4 EXAMPLES OF (O,a}-GEOMETRIES EMBEDDABLE IN PG(n,q) Clearly a (O,a)-geometry S is said to be embedded in PG(n,q) if the line set of S is a subset of the line set of the projective space and if the point set of S is the set of all points of PG(n,q) on these lines. (a) Let U be a set with m elements, U2 = {TcU I iTi=2}, U3 = {TcU II ITi=3}, I the inclusion relation. Then U2 ,3(m)=(u ,U ,I) is a semi partial geometry with 2 3 s=a=2, t=m-3 and u=4.
For certain values of m, these geometries are embeddable
in a PG(n,q).
For example U2 ,3(5) is isomorphic to the Desargues configuration in PG(3,2}, and the geometry U2,3(7) is embeddable in PG(4,2) (8). (b) In PG(n,q) we take a symplectic polarity Wwith rank 2k, i.e. rad W= = II =PG(n-2k,q).
Let P be the set of points of PG(n,q)\ rad W.
Let B be the set of
lines of PG(n,q) which are not totally isotropic in the polarity Wand let I be the natural incidence. t=q
0-1
- 1.
Then W(n,2k,q)=(P,B,I) is a (O,a)-geometry with S=a=q,
231
The embedding afr O,aJ-geametries in PG( n,q)
If k=l, then the geometry W(n,2,q) is a partial geometry and will be denoted by Hqn . If 2k=n+l, so n is odd, then the symplectic polarity is non-singular.
In
this case the geometry W(n,n+l,q) is a semi partial geometry with ~=qn-l(q_l) and will shortly be denoted by ~) [6]. In all other cases W(n,2k,q) is a "proper" (O,a)-geometry. (c) Take a (possibly singular) quadric 0 in PG(n,2), n
~
3.
Let B be the set of
non-intersecting lines of 0, let P be the set of all points incident in PG(n,q) with the elements of B, and let I be the incidence of PG(n,2).
Then (P,B,I) is a
(O,2)-geometry, unless 0 consists of one or two hyperplanes, or 0 is the hyperbolic quadric 0+(3,2) in PG(3,2), or n ~ 4 and 0 is the cone with vertex PG(n-4,2) which projects a hyperbolic quadric of a threespace skew to PG(n-4,2) (in the last two cases the geometry is not connected). +
If n=2d-l and 0 is the non-singular quadric 0-(2d-l,2), then (P,B,I) is a ±
semi partial geometry, denoted by NO (2d-l,2), with parameters s=a=2, 2d-3 d-2 2d-3 d-l + t+1=2 -£2 , ~=2 -£2 , where £=+1 for the hyperbolic quadric 0 (2d-l,2) and £=-1 for the elliptic quadric 0-(2d-l ,2) [16]. If n=2d and Q is non-singular, then (P,B,I) is a semi partial geometry, denoted by NQ(2d,2), which is isomorphic to W(2d-l,2). In all the other cases the geometry is a "proper" (O,2)-geometry. (d) In PG(3,2 h ) (h ~ 2) we take the hyperbolic quadric. Then the same construction as in (c) gives us a (O,2 h- l )-geometry NQ+(3,2 h) with S=2h, t+l=2 h- l (2 h_l). This geometry is never a semi partial geometry. 1.5. We conjecture that the (O,a)-geometries embedded in PG(n,q), but not in PG(nl,q) with nl < n, are the design of points and lines of PG(n,q), or one of the examples in 1.4.
In the paper, part of this conjecture is proved.
2. (O,a)-GEOMETRIES EMBEDDED IN PG(n,q) 2.1.
Suppose S=(P,B,I) is a (O,a)-geometry, embedded in a projective space
PG(n,q), n
~
3, but not in a PG(nl ,q), with nl < n.
Suppose p is a point incident with a line L of S. L containing at least one other line N of S with pIN. not lying in the
plane~.
Let
~
be a plane through
Suppose q is a point of S
By the connectedness of S there is a chain (zo,zl' ...
232
F. De Clerck and l.A. Thas
... ,z u ), with z0 =p, zu =q, and z.1 collinear (in S) with z.1+ l' i=O, ... ,u-1. that
zo""'z~
are in
rr
and that
z~+l
is not in
rr,
£
E
{O, ... ,u-1}.
Assume
Then the line
z£z£+l is denoted by M. Let S' be the connected component of snPG(3,q) containing L,M and N. S' is a (O,a)-geometry embedded in PG(3,q), but not in a plane.
Then
We denote by e
the number of lines of S' through a point of S'. The points and lines of S' in points on a line and paints.
a
IT
constitute a dual Steiner system with q+1
lines through a point, together with a set of m "isolated"
We note that a point x of S' in
of 8' through x is contained in
IT.
rr
is isolated iff no one of the e lines
Hence S' (and also S) has qa-q+a lines in
rr.
It follows that S' has
~ = (qa-q+a)~(e-a}+qa-q+a+me (l) lines. a If
rr'
then by (1)
is another plane defined by two intersecting lines L' and N' of S', IT'
also has exactly m isolated points.
The following lemma is proved by counting the number of lines of S'. LEMMA 2.2: With respect to S', there are three types of planes in the space PG(3,qpS'. (aJ Planes of type (aJ containing qa-q+a lines and p =(q+l)(q-~l)+m a a
points
of S'. (bJ Planes of type (bJ containing exactly one line of 8' and
Pb=Q+1+ 9(q+l!i s - a )(a-1)+m points of P'. (c) Planes of type
(0)
containing no line of 8' and pc_(qa-q+:l(qs+s-a q ) +m
points of P' . Further e(Pb-pc)=q and the number of points v' of S' equals (q+l)pc'
COROLLARY: If there is at least one plane of type (b) and at least one plane of type (c) then slq.
REMARK 2.3: From 2.1 follows that the solution of the general embedding problem depends entirely on the classification of all (O,a}-geometries which are embedded in PG( 3,q).
233
The embedding of (O,a).geometries in PG(n,q)
3. THE EMBEDDING OF (O,a)-GEOMETRIES WITH m = 0 IN PG(3,g) THEOREM 1: Let S=(P,B,I) be a (O,a)-geometry embedded in a PG(3,q), but not in a p~ane.
If for some
p~ane
of type (a) we have m=O, then the
fo~~owing
cases are
possib~e.
a=q+l and S is the design of points and
(1)
of PG(3,q);
~ines
a=q and S=H 3 . q' a=q=2 and S=NQ-(3,2).
(2) (3)
PROOF: Suppose that all lines of some plane of type (a) are elements of B.
This
is equivalent to a=q+l, and then the dual Steiner system defined by any two 2
intersecting lines of S is a projective plane. lines of S equals (q2+1)(q2+q+l ). (q+l)(q2+q+l ).
So the number of points of S equals
Consequently S is the design of points and lines of PG(3,q).
Suppose now that in each plane which is not in B. type (a).
Hence e=q +q+l and the number of
TI
of type (a) there is at least one line L
For brevity we shall say that L is a secant of the plane
Such a secant evidently contains q- 3+ 1 points of S. a
TI
of
If T is the
number of planes of type (a) through the secant L, then there are (q- 3 +l)(e-Ta) planes of type (b) through L (implying e-T~). After some calculations there follows that e=Ta+l. First suppose e=Ta. 3 S=H (e=q2).
a
Hence (q- 3 +l)(e-Ta)~+l-T. a
e-Ta~l+ a~l - (a-l)~q+l) and so e=Ta or
Then S is a partial geometry and by [9] a=S=q,
q
If e=Ta+l, then e
Let
TI
be a plane of type (a) and let p be any point
We count in two different ways the ordered pairs (TI' ,L') with TI' a
plane of type (b) having its line of S through p and with L'=wnTI' a secant of TI. e-l e-l There are ~ planes of type (a) through a line of Sand q+l- ~ planes of type (b) through that line.
Hence
e-l
q+l-a=(e-a)(q+l-~)
(2).
Because of
e~+l,
we
2
a-a;g
have q+l-~(e-a)(q+l- ~). This is equivalent to e-a~l+ 1 Since a2 aq+a- qe~+l we have a_a2+q~, and one readily proves that a-a;q l~l if and only if aq+a- qa=2 and q>3. This means that e=a+l whenever aF2. But for any pOint p of S, the lines of S through p together with the planes of type (a) through p, form a 2-(e,a,1) design, implying e~2-a+l. always have a=2.
Hence aF2 yields a contradiction.
So we
234
F. De Clerck and l.A. Thas
For
~=2
(2) reduces to (e-3)(q+l-e)=0.
Consequently if a=2 then e=q+l or
8=3. As m=O, S is the dual structure of a semi partial geometry S'" with
. ....~ ,', a(e-l) parameters s"=a-l, t-=q, a"=a and u = l' a-
Suppose a=2 and e=q+ 1. geometry.
Then S'" is symmetri c and so Sal so is a semi parti a1
Hence S has to be NQ-(3,2) (4).
Suppose now a=2 and 8=3 (T=l). then L is contained in
¥+1
If
TI
is a plane of type (a) with secant L,
planes of type (b).
If there exists at least one
plane of type (c), then 31q (since 3(P b-p )=q). On the other hand 21q which gives c a contradiction. If there is no plane of type (c), then ~ +2=q+l, and hence q=2 and again S=NQ-(3,2). 4. THE EMBEDDING OF (O,a)-GEOMETRIES WITH m ~ 0 IN PG(3,q) THEOREM 2: Let S=(P,B,I) be a (O,a)-geometry embedded in a PG(3,q), but not in a pZan8.
If
m~O>
then there is no pZane of type
(b).
PROOF: Suppose that there is at least one plane of each type.
The total number of
planes of type (a) and (b) is less or equal than (q+l)(q2+1).
If we put qa-q+a=6
this means that g:;2 ( a 6
(8-~)+6+
me)(q+l-
e-l
~
+
8-1
6(a-l))~(q+l)(q
2 +1)
(3)
This implies that (qa-Q+a-8+1)(sq - q+ ~)+me (1- ~)~2+1. a a qa-q+a We remark that ale. Indeed, elq (since e(Pb-pc)=q) and alq (the dual Steiner system in any plane prime power.
TI
of type (a) has q-~l points on any secant in a
TI)
with q a
Since there is at least one plane of type (b) we have
a
Hence
(qa-q+a-e+l) (Eq _ q+ ~)
and this implies that a+aq-s It follows that
(4)
a
a~i~~i)
+
(q+~~~e+q-l( O.
235
The embedding of (O,a)-geometries in PG(n,q)
8 + aq(a+q»aq or (e-l)e+q{e+a 2-ae+a» 8+q-l 2
If aF2 then e+a -ae+a
0
(5)
Indeed, e+a 2-ae+a
the other hand e~ 2-a+l (see proof of the preceding theorem) and a2-a+l>a+l+ a+l ifa ~3. Hence, if aF2 then (5) implies q< e(e 21) . Because of a- l ae-a -e-a 2 q~e, this means that (a-2)e
Then by (4) q<3, a contradiction.
Finally, suppose a=2 and e>4. Since elq (q=2h) we know that e~. Putting u q=te (t=2 ) formula (5) implies (t-l)(e-6)<5, and hence t equals 1 or 2. If t=l then (4) implies (e-l)(e-2)<4 which contradicts e
e~.
If t=2 then
(5)
implies
In this case (4) gives again a contradiction.
Consequently, if mFO then the three types of planes cannot occur together. Suppose that there are no planes of type (c) but that there is at least one plane of type (b).
Then e
type (b) equals (q+l)(q2+1).
The total number of planes of type (a) and
On the other hand this number is
e-l 1 e-l . - - +q+l- -1) with I'! the number of lines of S (see (3)). aqa-q+a ae-l 2+1, which implies 1I(1- --)=q qa-q+a
I'! (-1
Hence
2
me = (e-l)(q +1) +1- (q+l)q{e-a)+{q+l)e(.9.-1) qa-q+a+l-e a
(7)
If all the points of PG{3,q) are elements of S then m={q+l-a).9.· a
Since
v={q+l){q2+1)={q+l)pc (see 2.2), there follows e=qa-q +a, a contradiction.
Hence
there is at least one exterior point (i.e. a point of PG(3,q)\P). Suppose x is an exterior point and denote by 'b the number of planes of type (b) through x.
Then
1I ='b+(q2+q+l -'b){qa-q+a), and this implies m8=-'b{q+l)(a-l)+(qa-q+a)(q+l){q-
e
~
+1)
(8)
2
Hence (q+l) Ime and by formula (7) (q+l) I (e-l)(q +1) + 1, qa-q+a+l-e which implies (q+l)le.
Hence e=(q+l)(a-l) ore"'{q+l)(a-2).
236
F. De Clerck and 1.A. Thas
Let y be a point of S and count the number of planes of S having a line of S through y.
Then we obtain
~
8-1 2 ::.(a-l) +e(q+l- ~).;;q +q+l.
(remark that we have inequality because for some planes of type (a), y can be isolated). 2
6 -8
This inequality can be written as 2
(aq+a+l)+(q +q+l)a
~
0,
the discriminant of the quadratic form being O=(aq+a+l) 2-4a(q 2+q+l)=(aq+a-l) 2-4aq 2 . One easily verifies that D
we have 0>0.
For a=q=3 and
We also note that the case a=q+l=3 cannot occur.
Now consider several cases. Case 1:
Then
e<;;(q+l )(a-2), 0>0 and 8~ a9+a\ 1+v1)
qa-2q+a-2~8;;'
aq+a+l-tvrr 2
(9)
Hence (qa-4q+a-5) 2~(aq+a+l) 2-4a(q 2+q+l) and so (4q 2-aq 2)+(10q-4aq)+(6-2a)~. ~,
2
then 4q -aq
2
~,
10q-4aq
yields a contradiction. Case 2:
Then
0>0, e.,;;
If
Also for q=a=3 (9) .-Hence D>O implies 8=(q+l)(a-l) or e~ aq+a+l-viJ 2 .
aq+a+l-vtJ 2 and
~2(q+l)
4(q+l)~";;(aq+a+l)-v1J
(10)
Hence (aq+a-4q-3) 2~(aq+a+l) 2-4a(q 2+q+l) 2 2 or (16q -4aq )+(24q-12aq)+(8-4a)~. 8-4a
If~,
2 2 then l6q -4aq ~, 24q-12aq
Also for q=a=3 (10) yields a contradiction.
The cases
which are left, are: (case 3) 8=(q+l)(a-l); (case 4) e=q+l; (case 5) a=2; (case 6) a=3 and q>4. Case 3:
e=(q+l )(0.-1)
Counting the number of planes of type (a) through a line of S we deduce that (a-l)I(8-1).
Hence 0.=2 and e=q+l.
237
The embedding o[(O,cx)-geometries in PG(n,q)
Case 4: e=q+ 1
Then by formula (7) m(q+l)= q(q:+l) +1-q(q+l)(q+l-a)+(q+l)2(1 -1). qa- q+a a One easily verifies that this is equivalent to (qa-2q+a)am=q(a-2)(a-l)(q-a)(-q-l)-aq2(a-2). If a=2, then m=O, a contradiction.
If a>2 then
(qa-2q+a)am>O and
q(a-2)(a-l)(q-a)(-q-l)-aq 2(a-2)
From e
Then q+l~<2q+3.
251 Hence e=2q+2. And by (7) 2m=-(~ tf4 2)' a contradiction.
Hence the theorem is proved. 5. CONJECTURE Let S be a (O,a)-geometry embedded in a PG(3,q), but not in a plane.
If
mfO, then there are two possibilities: (a) there are no planes of type (c) and then S=W (q) (q=a, m=l, t+l=q2); 3 (b) there is at least one plane of type (c) and then S=NQ +(3,2 h) (h~2) (a=2 h- l , m=l, t+l=2 h- l (2 h-l)). REMARK: Since there are no planes of type (b), we know that t+l=qa-q+a.
If
moreover there are no planes of type (c) we have m= 1(q-a+l) and v=(q+l)(q2+1). a
6. THE EMBEDDING OF (O,a)-GEOMETRIES IN PG(n,q), n ~ 3 THEOREM 3: Let S be a (O,a)-geometry embedded in a PG(n,q), in a PG(n' ,q), n'
~3
and q>2, but not
If for some plane oontaining two interseoting lines of S,
n
the substruoture of S induoed in that plane is oonneoted, then S=H or S is the q
design of points and lines of PG(n,q).
F. De Clerck and I.A. Thos
238
PROOF: Suppose that for the plane PG(2,q) containing the intersecting lines Land N of S. the substructure of S induced in that plane is connected. the space and S' the substructure as defined in 2.1. of type (a) do not have isolated points.
Let PG(3.q) be
Then in PG(3.q) the planes
Hence by Theorem 1 we have a=q+1 or
a=q. First suppose that a=q+1. S through p.
Let p be a point of S and let U,V be two lines of
Since a=q+l every line of the plane UV is a line of S.
There fol-
lows that the union of all lines of S through p is a subspace PG(n' ,q) of PG(n,q), n'-l n'-2 and that every line of PG(n' ,q) belongs to S. So t+l=q +q +... +1 and the lines of S through any point p' of PG(n' ,q) are all the lines of PG(n' ,q) through p'.
Since S is connected, S is the design of points and lines of PG(n' ,q).
Now
it is immediate that n'=n. Next let a=q.
Then any two intersecting lines of S generate a dual affine n
plane.
By [10] S=H q .
REMARKS: 1. Let S be a (O.a)-geometry embedded in a PG(n,q), PG(n' ,q), n'
~3,
but not in a
If a=q+l. then by the same argument as above it is immediately
clear that S is the design of points and lines of PG(n,q).
If a=q>2, then by
( 10] , S=W(n,2k,q) for some k. In [8] we prove that there are no other possibiZities for a if q>2 and
~.
Consequently the only semi partial geometries, with 0>1, embedded in a PG(n,q), ~3>
q>3, but not in a PG(n',q), n'
PG(n,q),
fl, q
and if n is odd W(n,q).
2. If a=q=2, then as mentioned in 1.4., there are more examples.
In [11] M.P.
Hale Jr. announced that there are no other possibilities than these known examples.
However, the proof is ommitted. Moreover, he did not give the explicit
representations of the embeddings.
In (8) we prove that the only (0,2)-geometries
embeddable in PG(4,2), but not in PG(3,2), are NQ(4,2) (here v=15, t=3),NQO(4,2) where QO(4,2) is the cone which projects Q-(3,2) from a point (here v=20, t=5), 4
U2 ,3(7) (here v=21, t=4), H2 (here v=24, t=7), W(4,4,2) (here v=30, t=7) and the design of points and lines of PG(4,2). Up to a projectivity of PG(4,2), each of these embeddings is unique. 3. Finally we notice that the graphs of all (O,q)-geometries embeddable in PG(n,q) were determined by J.I. Hall [12].
This is a generalization of a result due to
LL Shult [14] (see also J.J. Seidel f 13] ).
239
The embedding of(O,cxJ-geometries in PG(n,q)
7. THE H1BEDDING OF DUAL SEMI PARTIAL GEOMETRIES IN PG(n,q) THEOREM 4: If S* is the dual of a semi partial geometry S with ~l, and if S'l is embedded in a projeative spaae PG(n,q),
n~3,
but not in a PG(nl ,q), nl
n=3 and S* is the design of points and lines in PG(3,q), S*=H~, or S*=NQ-(3,2). PROOF: As S* satisfies the axiom of Pasch, there exists a partial geometry -~'; -1~ -* -* 1t-* S = (P ,B ,I ) with B the set of lines of S and with P the set of dual designs -1,
~.
defined by two intersecting lines of S [7].
This immediately implies that a line
of S* not lying in a plane containing two intersecting lines of S*, intersects point set of the dual design in this plane.
Hence n=3 and m=o.
4',
By Theorem 1, S
.is the design of points and lines in PG(3,q), S*=H 3q , or St=NQ (3,2).
BIBLIOGRAPHY 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
R.C. Bose, Strongly regular graphs, partial geometries and partially balanced designs, Faa. J. Math., 13 (1963), 389-419. F. Buekenhout and C. Lef~vre, Generalized quadrangles in projective spaces, Arah. Math., 25 (1974), 540-552. I. Debroey, Semi partiele meetkunden, thesis, Ghent (1978). I. Debroey and J.A. Thas, Semi partial geometries in PG(2,q) and PG(3,q), Rend. Aaaad. Naz. Linaei, 64 (1978), 147-151. I. Debroey and J.A. Thas, Semi partial geometries in AG(2,q) and AG(3,q), Simon Stevin, 51 (1978), 195-209. I. Debroey and J.A. Thas, On semi partial geometries, J. Comb. Th., Ser. A, 25-3 (1978), 242-250. I. Debroey, Semi partial geometries satisfying the diagonal-axiom, J. of Geometry, 13 (1979), 171-190. I. Debroey, F. De Clerck and J.A. Thas, The embedding of (o,a)-geometries in PG(n,q), Part II (to appear). F. De Clerck and J.A. Thas, Partial geometries in finite projective spaces, Arah. Math., 30 (1978), 537-540. K.B. Farmer and M.P. Hale Jr., Dual affine geometries and alternative bilinear forms, Lin. Alg. and Appl., to appear. M.P. Hale Jr., Locally dual affine geometries, in "Proceedings of the Conf. of finite groups" (ed. W.R. Scott-F. Gross), (1976), 513-518. J.I. Hall, Classifying copolar spaces and graphs (preprint). J.J. Seidel, On two-graphs and Shultis characterization of symplectic and orthogonal geometries over GF(2), T.H.-Report 73 -WSK-02, Tech. Univ. Eindhoven (1973). E.E. Shult, Groups~ polar spaces and related structures, in "Proceedings of the Advanced Study Institute on Combinatorics, Breukelen" (ed. M. Hall Jr. - J.H. van Lint) (1975). Math. Centre Tracts nr. 55, 130-161.
240
F. De Clerck and I.A. Thas
15.
J.A. Thas, Partial geometries in finite affine spaces, Math. Z., 158 (1978),
16.
H. Wi1brink, Private communication.
1-13.
Seminar of Geometry and Combinatorics State University of Ghent Krijgs1aan 281, B-9000 GENT, Belgium
241
Annals of Discrete Mathematics 18 (1983) 241-278 © North-Holland Publishing Company
ON CUBIC SURFACES OVER A FIELD OF CHARACTERISTIC 3 M. de Finis and M.J. de Resmini *
Cubic surfaces in a 3-dimensional projective space over an algebraically closed field of characteristic 3 are investigated and classified according to possible configurations of their nuclei, and a comparison with the classification based on singular points is made. A special combinatorial configuration, consisting of 10 points, 10 lines, and 5 planes and leading to the definition - that seems new - of a composed PBIBD, is found; this configuration motivates the study of some cubic surfaces in PG(3,3) .
1. INTRODUCTION Let K be an algebraically closed field of characteristic 3 and P a 3-dimensional projective space over K.
A cubic surface in P will be denoted by F; thus,
the most .general equation for F is Hxy 2 + 222 222 Jy 2 z +2 Ky t + Lxz + Myz + Nz t + Pxt + Qyt + Rzt +
f(x,y,z,t)
(1.1 )
333222 = Ax3 + By + Cz + Dt + Ex y + Fx z + Gx t
+
Sxyz + Txzt + Vxyt + Wyzt = 0, x,y,z,t being homogeneous projective coordinates in P. A line in P will be called a tri-tangent line (for short a tri-tangent) of F if it does not belong to F and meets F at three coincident points.
A plane in P
will be called a tri-tangent plane of F if it is tangent to F at three (distinct) pOints. F uniquely defines the family E (depending on three parameters) of the polar quadrics (with respect to F) of all points in P; E is determined by the four quadrics:
242
M. de Finis and M.J. de Resmini
af = - Exy - Fxz - Gxt + Hy 2 + Lz 2 + Pt 2 + Syz + Tzt + Vyt ax af -ax = Ex 2 - Hxy - Jyz - Kyt (1. 2)
af
-az = Fl H
G/
lat
Ml + Qt2 + Sxz + Vxt + Wzt
0
2 + Ji - Lxz - Myz - Nzt + Rt + Sxy + Txt + Wyt
0
+ Ky
2
Conversely, when base point of
L
0
+
+ Nl - Pxt - Qyt - Rzt + Txz + Vxy + Wyz
=0
is given, a family of cubic surfaces is defined.
L
Any
(i .e. any solution to equations (1.2)) wil be called a nucleus of
F. Simple computations prove the following statements [4]. PROPOSITION 1.3: The
quadPic (with respect to F) of a point p,
po~ar
p (and in the most genera~ case has rank
tangent
p~ane
po~ar
quadPic
Of~;
(i) p is the vertex
~
have the same
~
of the point p is a cone, then one of the
occurs;
are tri-tangent Zinee. and
~
contains
at p.
PROPOSITION 1.4: If the lu~lowing
4); if P E F, then F and
~,
then p is a
nuc~eus
of F and
a~Z
generators of Qp
In case p belongs to F, p is a singular point (conic node)
is the tangent cone to F at P.
(ii) The vertex of
~
is a point q # p; then q is not a nucleus and
~
is a
aone having p as its vertex. Conversely, if p is a nuaZeus of F, then the polar quadPic of p is a cone, having p as its vertex, whose generators are tri-tangent lines to F.
COROLLARY 1.5: If the polar quadPia
~
of the point p splits into two planes, then
p is a nucleus iff p belongs to both planes.
PROPOSITION 1.6: The polar quadPia (with respect to
F) of every point in
P van-
ishes iff F is a pZane counted three times.
PROPOSITION 1.7: If three non-aolZinear points exist whose polar quadrics vanish, then the same is true for an points in P and F is a plane aounted three times.
243
On cubic surfaces over a field of characteristic 3
In what follows it will be assumed F is not a plane counted three times. All proofs will be only sketched; more details can be found in (4).
2. ON SURFACES FOR WHICH POINTS EXIST WHOSE POLAR QUADRICS VANISH
PROPOSITION 2.1: Let F be a cubic surface such that two points p and q exist whose polar quadrics vanish; then (i) the polar quadric of any point on the line through p and q, (ii) a plane
00
exists through
~
any point on which (not belonging to
nucleus and all polar quadrics degenerate into two planes: (iii) F is a cone whose vertex belongs to through
~,
~
00
vanishes; ~)
is a
and a plane through
~;
and all whose tangent planes pass
any non-degenerate plane section of F is a cubic curve with a cusp.
~;
PROOF: Take coordinates so that p =(0,0,0,1) and
q
(0,0,1,0); then
l:
is defined
by (2.2)
y(- Ex
+
Hy) = 0
x(Ex - Hy)
1:11
0,
where E and H are not simultaneously equal to zero. It is straightforward to check (i) and (ii). F is a cone whose vertex is the point (0,0, ~,~). F degenerates into three planes through
~:
If C = D = 0, then
x=y=O, and conversely.
Assume F is a
proper cone; w.l.o.g. take its vertex as 04 = (0,0,0,1). If a is a plane not through 0 , then the cubic curve a n F has a cusp at the pOint g n a, where 4 g = 00 n F, the tangent at the cusp being the line a n 00, and all tangents to this cubic curve are tri-tangents and pass through the point
t
n a.
Thus the canonical
form for F is 3 2 z - x y =0 .
(2.3)
A similar argument proves (4) the converse of prop. 2.1; namely: PROPOSITION 2.4: If a plane (i) vanish;
00
00
exists any point on which is a nucleus of F, then
contains a tine t such that the polar quadrics of alt its points
244
M. de Finis and M.J. de Resmini
(ii) aU thr·ough
non-var;'~shing
polar quadl'ics split into two planes:
w
and a plane
~;
(iii) eithe1' F is a cone with vertex on t, an whose tangent planes pass through
t
and aU whose plane sections have a cusp and tangent lines forming a
p,mcil,
01'
F degener'ates into th1'ee planes through
to
if F is a cone as in (iii), then there exist both a line t the
~onverselYJ
poLO"£' quadric of any point on which vanishes and a pLane to aU quad1'ias in
~,
w
through
~
-which beLongs
and the locus of nuclei is w.
PROPOSITION 2.5: Fo1' a cubic surface F the following are equivalent: (iJ there exists exactly one point w whose polar qUad1'ic vanishes; (ii) aU polO"£' quadrics are cones with the same vertex w; (iii) a point w exists such that any line through it is a tri-tangent line. If eithe1' (i), 01' (ii),
01'
(iii) holds, then:
F;
(a) w is a nucleus of
(b) aU cones in 1: have three common generators and any point on these lines 1-$
a nucLeus;
,'c) either' F is a cone with Ve1'tex
W,
01'
F has three (generaUy distinct)
tina.kG, and aU lines through w O"£'e tri-tangent to F. -leD,)
(When the binodes coincide,
,jus't; one uninode is possible.)
PROOF: Taking the point 0 (
4
= (O,O,O,l) as w,
2
2
- Exy - Fxz + Hy + Lz + Syz Ex 2 - Hxy - Jyz
+ Mz2 (2.6) ) 2 ) Fx 2 + Jy - Lxz - Myz
(1.2)
become
0
+
Sxz
0
+
Sxy
0
Iat vams es af
. h
l
then it is easy to check that (i), (ii), and (iii) are equivalent and (a) is true [4 J.
To prove (b), consider the three conics (2.6) on the plane t = 0; they have
three common points; indeed, solving the simultaneous equations (2.6) one gets (2.7)
y3(FH 2
+
E2J + EHS) + y2 z (-EFJ - EHL _ E2M + ES2) +
YZ2(F2J + FHL + EFM - FS2) + z3(_ EL2 - FLS - F2M)
= O.
245
On cubic surfaces over a field of characteristic 3
Thus, all cones in
L
have three common generators, and any point on these
three lines is a nucleus.
°
Taking as 1 , 02' 03 the points common to the conics (2.6), the equation of F becomes 1. Eqn. (2.7) has three distinct roots. Ax 3 + By 3 + Cz 3 + Dt 3 + Sxyz
(2.8)
(1.i)
F.
WE
F can contain at most one of the lines 0i04 (i
otherwise it degenerates. plane
= 0.
Assume F contains g: x
inflexional point. 3
1,2,3),
= y = 0, then any section by a
B w is a cubic curve having a node at the point g n
0
=
0
and exactly one
The canonical form for F is
3
x + y + xyz = 0.
(2.9)
When F contains none of the lines 0i04' any section by a plane
cr ~
04 is a cubic
curve having exactly three (collinear) inflexional points and the canonical form for F is (2.10)
333 x + y + z + xyz = 0. Let gi (i = 1,2,3) be the lines 0i04'
(l.ii) w ¢ F.
Each of the three
planes through two of these lines intersects F in a line counted three times; the latter lines (which are the only lines on F) belong to the plane (2.11)
~A
x +
~B
y +
~c
z +
~D
t
= 0.
Since all lines through ware tri-tangent lines, F contains exactly three of its nuclei; therefore, it has three binodes and the biplanes do not intersect on F.
(When the tangent cone at a singular point degenerates into two planes, any of
them will be called a biplane [1) ; similarly, a unip1ane is defined.)
Coordinates
can be chosen so that the canonical form for F is (2.12)
t
3
+ xyz = 0.
Since all tangent planes pass through w, F has class three. 2. Eqn. (2.7) has two coincident (and one distinct) roots; taking as y
=
°
the double root and as (H,E,O,O) the remaining common point, the generators belonging to all cones in
L
are g: x
°
°
= y = and h: Ex - Hy = z = and the
equation of F becomes (2.13)
a 3 3 32 2 2 Ax + By + Cz + Dt + Ex y + Hxy + Sxyz - (HS/E)y z = 0, (2.i) 04
E
F.
ES
r 0.
Then F is a cone and, provided neither g nor h belongs to F,
246
M. de Finis and M.J. de Resmini
both planes y = 0 and Ex - Hy = 0 meet F at a line counted thrice. any section by a plane a
~
0
When 9
is a cubic curve with a node at g n o.
4
E
F,
When hE F,
such a curve has a cusp and three inflexional points. (2.ii) 0
F.
~
4
Then F has two binodes and the biplanes do not intersect on
F.
The reference simplex can be chosen so that the equation of F is t
(2.14 )
3
xy(x
+
+
= O.
z)
Eqn. (2.7) has three coincident roots. Thus its left hand side is a
3.
cube and the following must hold: - E(FJ
(2.15 )
Taking E = F =
= 0 and F(FJ
+
HL
EM - S2)
0
as a solution, the line g: y
+
and coordinates can be chosen so that
= 0, z(- Jy
Syz
(2.16 )
+
Sx
= 0, y(Jy
where J and M are not simultaneously naught. Ax
(2.17)
3
By
+
3
+
Cz
3
+
Dt
3
+
EM - S2)
+
z =
0
=0
.
is the locus of nuclei
is defined by [41:
L
Mz)
+
HL
+
+
Sx - Mz)
= 0, S #
0,
Then the eqn. of F becomes
yz(Jy + Mz
+
Sx)
= 0,
and there are three planes meeting F at a line counted three times. (3.i) 0 E F, then F is a cone. 4
plane a
~
0
4
If g belongs to F, then the section by a
has a node at g n a and just one inflexional point.
The following
canonical form can be obtained [4J: (2.18)
y3
+ Z
3
+
yz(Sx
+
Mz)
0,
S, M #
o.
If g does not belong to F, then three cases may occur, according to the distribution of inflexional points on plane sections, namely: three collinear inflexional points: (2.19)
x3
+
yz(Jy
+
Mz
+
Sx)
= 0,
J, M, S # 0
three distinct non-collinear inflexional points: (2.20)
3
x
+
y
3
+
yz(Jy
+
Sx)
= 0 ,
J, S # 0 ;
two inflexional points: (2.21 )
x3 + yz(Mz + Sx) (3.ii) 04
~
F.
counted three times.
0,
M, S # 0 •
Then F has just a binode and each biplane meets F at a line F contains just three lines and the equation of F is [4J:
247
On cubic surfaces over a field of characteristic 3
t3
(2.22)
+
yz(Jy
+
Mz
+
= 0,
Sx)
J, M, S F
°.
REMARK: There is no point whose polar quadric has rank 1; the only points whose polar quadrics have rank 2 are the points on the planes x Solving (2.15)'s by FJ
+
HL
=
0, y = 0, Z = O.
EM - 52 = 0, one gets the same results [4].
+
However, case 3. can be obtained directly under the assumptions that the conics (2.6) have exactly one common point and the same tangent line at it. Taking such a pOint as 0 3 (
I I
)
(2.23)
)
- Exy - Sxz
+
2
= (0,0,1,0), ~
Hy 2
Ex - Hxy - 5yz 2 5(x - y) ,,0,
+ +
is defined by
° 5xz = ° Syz
=
5 F
\
°,
and E, H not simultaneously equal to zero.
°
(3.i ') 04 E F; then F is a cone. If the line g: x = y = (locus of nuclei) belongs to F, then the cubic curve cr n F on a plane cr ~ 04 has a cusp at the point g n cr and no inflexional point and the canonical form for F is [4]: 322 x + Ex y + 5(x - y) z
(2.24)
= 0,
E, 5 F 0 .
If g i F, then the cubic curve cr n F has just one inflexional point and the canonical form for F is z
(2.25)
3
+
2
Ex Y + 5z(x - y)
2
= 0,
E, S F
°
(3.ii') 04 ~ F. Then F has a uninode at the pOint g n F, the uniplane being x - y = 0, and the following canonical equation for F can be obtained [4]: (2.26)
t To
~
3
2
+ Ex Y + 5z(x - y)
2
= O.
defined by equations (2.23) there belong quadrics having rank 1.
REMARK: When all polar quadrics are cones (with the same vertex w), all nuclei on the same common generator have the same polar quadric degenerating into two fixed planes. In the next sections only cubic surfaces for which no pOint exists whose polar quadric vanishes will be considered.
248
M. de Finis and M.J. de Resmini
3. ON CUBIC SURFACES WITH THREE COLLINEAR NUCLEI PROPOSIiION 3.1: If tt,r'ee eJollinear' nuclei exist, then any point on the tine r througr. them is a
nu~leus
and either'
(i) no other' nUeJleus exists; Or'
(ii) theN} exi3ts a nUeJleus w not on r; then there exists a tine s inoident ~ith
r all whose points aPe nuclei and the plane
TI
joining r and s meets F in a
Zin-? acur.ted thr'ee times.
When
(7:)
holds, either' F is a r'uled sUPface (a sCr'oU)
01'
it has just one
singular' point, whieJn can be either' a uninode, with the uniplane intersecting
F
a: ;h!>ee distinat lines, or a birwde. ;"hen (ii) holds, eithel' F is Cayley's r'uled sUPfaeJe, Or' it has two binodes, Or' it has a uninode and the uniplane meets F at thr'ee coincident lines.
PROOF: Take coordinates so that the collinear nuclei are: 04 03 = (0,0,1,0) and (0,0,1,1); then equations (1.2) become: - Exy - Fxz - Gxt + Hy 2 + Syz + Yyt 2
Ex - Hxy - Jyz - Kyt + Sxz + Yxt (3.2)
2
Fl + Jy + Sxy 2
Gx + K/ + Vxy
(0,0,0,1),
0 0
a 0
Thus, the line r: x = y = a belongs to all quadrics in t, i.e. any point on it is a nucleus and its polar quadric splits into two planes through r. Assume -F = -J = -S does not hold [4]. On r there are two points whose polar G K Y quadrics have rank 1; their coordinates are given by 2 y2 (FJ - 52) + yo (GJ + KF + SV) + 6 (KG - y2)
(3.3)
=0 .
Two cases must be considered: (i) eqn. (3.3) has distinct roots; (ii) eqn. (3.3) has coincident roots. W.l.o.g. take the points 03 and 04 as solutions to equation (3.3) and 2 their polar quadrics as x o and y2 = 0, resp.; thus eqn.s (3.2) become: (i)
- Exy - Fxz + Hy 2 (3.4)
El
- Hxy - Kyt
fx2
0,
K/
0,
0, 0,
F, K ~
a,
249
On cubic surfaces over a field of characteristic 3
and the only solutions are all points on r.
The locus of points whose polar
= 0 and y = 0; besides r, there are Fz = x = 0, all whose points (distinct
quadrics are cones consists of the planes x two lines, namely Hx + Kt
= Y = 0 and Ey
+
from 0 and 04) have rank 2 polar quadrics.
The equation of the surface can be
3
written as
[4]:
3 333 2 2 Ax + By + Cz + Dt + Fx z + Ky t = 0 .
(3.5)
(a) The line r belongs to F, then F is a general ruled surface and completing the choice of the coordinates [4], the equation for F becomes: 2
2
x z + Y t = O.
(3.6)
(b) The line r does not belong to F.
Since r is a tri-tangent line, F
contains exactly one of its nuclei; thus, it has just one singular point. (b.l) The point r n F is a solution to equation (3.3).
Take this point as
04; then 04 is a uninode and the uniplane y = 0 meets F in three distinct lines
(if two of them coincide, then (a) holds) and the equation of F becomes [4]: 2 2 2 z(x - z ) + Ky t = 0,
(3.7)
K~ 0 .
(b.2) The polar quadric of the point r n F
= P = (0,0, V3-D,
3-
- vC) has rank
2; then p is a binode and the biplanes do not intersect on F, whose eqn. is [4]: 3
z
(3.8)
+
t
3
+
2
xz
+
2
Yt
= O.
(ii) W.l.o.g. take the unique solution to equation (3.3) as 04 and the polar quadric of this points as y2
( (3.9) J
'I
y (-
O.
Thus E is defined by
Ex + Hy + Sz) = 0
EX: - Hxy - Jyz - Kyt + Sxz
=0
Jy 2 + Sxy = 0 Ky = 0,
K, S
~
0 ,
l
and any point on the line s: Ex
+
Sz a y
= 0 is a nucleus, and taking 01 on s,
E = O.
The plane
times.
The polar quadric of any point not on
on
TI
TI
joining rand s (y
=
0) meets F at a line g counted three TI
(distinct from 04) splits into two planes,
has rank 4, while that of a point TI
and
(3.10)
a(Hy + Sz) + y(Jy + Sx) + oKy = 0
through 04'
Coordinates can be chosen [4] so that equations (3.9) become
250
M. de Finis and M.J. de Resmini
Syz = 0, - Kyt + Sxz = 0, Sxy = 0, Ky2
(3.11)
0.
The locus of nuclei is the degenerate conic C xz
(3.12)
=
yO,
and no nucleus exists off the plane
The equation of F is
IT.
3 3 3 3 2 Ax + By + Cz + Dt + Ky t + Sxyz =
(3.13)
Since both g and C belong to
°
(K, S
~
0)
•
one of the following occurs: (a) g meets C
IT,
at two distinct points; (b) g passes through the double point of C; (c) g belongs to
C.
(a) The two nuclei g n C belong to F and are two binodes. they are the paints 03 and 01'
W.l.o.g. assume
Each biplane distinct from the common biplane
meets r at three distinct lines.
IT
With a suitable choice of the unit point and
taking 02 on F, the equation for F is (3.14 )
t(t
2
2 - y ) + Sxyz =
°
(S
(b) 04 is a uninode and the uniplane three times. (3.15 )
All lines on 3
3
IT
0).
intersects F at the line g counted
are tri-tangent lines and the equation of F is [4):
2
x + z + Ky t + Sxyz (c) Assume 9 = r.
IT
~
=
°
(K, S
~
0).
F is Cayley's ruled cubic surface (any point on r but 04
being a binode) and coordinates can be chosen (4) so that the equation of F is (3.16)
3
x
+
y(yt
+
xz) = 0.
The line s is a tri-tangent and the section by any plane (z
= AY)
through it
is a cubic curve having a cusp at 04 (the tangent being s) and a nucleus at the point (-1,0,0,.<). When (ii) holds, no pOint exists whose polar quadric has exactly rank 3. REMARK: Case (ii) in the statement of prop. 3.1 can also be proved under the assumption that r has a nucleus not belonging to r (4). Ruled cubic surfaces are characterized by prop. 3.1; indeed, the following can be proved (see [4], where also more details are given). PROPOSITION 3.17: The Locus of nucLei of the ('l'cciBPly of aU the points on a 'line.
Fo~
gene~aL ~uled
Cayley 's
~uled
cubic surface consists cubic surface the locus
251
On cubic surfaces over a field of characteristic 3
of nuclei consists exactly of all the points on a degenerate conic on a plane
IT
(one component of the conic being the double line of the surface); moreover, the
polar quadric of any point not on
IT
has rank 4 (while that of any point on
TI
but
the double point of the conic has rank 2).
4. ON CUBIC SURFACES WITH FOUR COPLANAR NUCLEI PROPOSITION 4.1: If F has four coplanar nuclei, on a plane collinear, and no point whose polar quadric
IT,
no three of them
vanishes exists, then
IT
contains a
conic C all whose points are nuclei and if C is non-degenerate, then (a) there exists a nucleus p
~ TI;
(b) the polar quadric of any point q on C splits into
and the plane
TI
joining p with the tangent to C at q; (c)
TI
intersects F at a line counted three times.
When C degenerates, all its points are nuclei and no nucleus exists off ConVersely, if a plane
TI.
exists meeting F at a line counted three times (and
TI
no point whose polar quadric vanishes exists), then (a') the polar quadric of any point on
IT
splits into
and a plane through a
TI
fixed point p; (b') p is a nucleus; (c') on
TI
there exists a conic C all whose points are nuclei, and C degener-
ates iff p belongs to
TI;
provided C is a non-degenerate conic, (d') the polar quadric of any point q E C splits into
TI
and the plane
joining p with the tangent to C at q. Under these assumptions, C being a non-degenerate conic, one of the following holds for the surface F: (i) F has exactly two binodes joined by the line 9 = F n biplanes share the plane
TI,
TI;
the two pairs of
the other biplanes meeting F at two distinct lines
(both pairs Of biplanes do not intersect on F); moreover, through 9 there are three planes (beside
IT)
meeting F at two distinct lines (# g).
(ii) F has two binodes (as in (i)) and a conic node; there is just one plane (~
IT)
through 9 =
IT
n F intersecting F at two lines
to the biplanes distinct from
TI);
(~
g, and these lines belong
the cone tangent to F at its conic node inter-
252
M. de Finis and M.J. de Resmini
sects F exactly at the lines joining the binodes with the conic node. (iii) F has one binode, the biplanes (one of them being
F has
(iv)
'IT)
meet on F at g.
one binode and one conic node; the biplanes intersect on F at g
ana the biplane distinct from
meets F at the line joining the singular points
11
cowzted twice.
When C degenerates> prop . .3. J (ii) is obtained.
,° ,°
PROOF: Assume the existing nuclei are 0 and U = (0,1,1,1) (then 01 will 4 3 2 l be a point whose polar quadric does not vanish); thus ~ is defined by
=
- Exy - Fxz - Gxt + Syz + Vyt - (S + V)zt
°
2
Ex + Sxz + Vxt = 2 Fx + Sxy - (5 + V)xt = Gx 2 - (5 + V)xz+ Vxy = 0,
(4.2)
° 5,V~0[4).
The polar quadrics of the given nuclei split into Ex + 5z + Vt Gx - (5
+
V)z
+
Vy
° °
(E + F + G)x + (S + V)y - Vz - St respectively.
p
(4.4)
11
=
°,
The planes (4.3) have exactly one common point:
x
- SV(5 + V)
y
- E(5 + V)2 - GS(S + V) - FV(5 + V) G5V - FV 2 - EV(S + V)
z
and the planes
0
Fx + 5y - (5 + V)t (4.3)
°
t = S(FV - E{5 + V) - GS). P is a nucleus [4) and the polar quadric of any point belonging to into (4.5)
11
and a plane through p. Syz + Vyt - (5 + V)zt
are nuclei.
+
V#
splits
Furthermore, all the pOints on the conic C
=x =
°
C is a degenerate conic iff p E
avoid this case, S
'IT
11
°will be assumed [4).
and then prop. 3.1(ii) holds.
To
Therefore, the locus of nuclei
consists of p and the points on C, and it is easy to check statement (b) [4}.
°
Taking p as a point 1 , the equation of F is 3 3 3 3 (4.6) Ax + By + Cz + Dt + Sxyz + Vxyt - {S
+
V)xzt
0,
253
On cubic surfaces over a field of characteristic 3
and
meets F at a line counted thrice.
TI
All lines on
are tri-tangent lines and
TI
the same holds for the generators of the cone r projecting C from p.
The polar
quadric of a point on r \ C has rank 3; the polar quadric of any point neither on r nor on
TI
has rank 4.
Conversely, assume a plane
TI
exists meeting F at a line g counted three
times (and no point exists whose polar quadric vanishes). that
= 0,
L
is defined by
+
Tzt
+
+
Vxt
+
Vyt '" 0 Wzt '" 0
is the plane t
TI
Pt 2
(
- Gxt
+
I
- Kyt
+
I
- Nzt + 2 2 Gx + Ky
(4.7) )J
\
Qt2 Rt 2 +
Txt + Wyt '" 0 2 Nz - Pxt - Qyt - Rzt
Taking coordinates so
+
+
Therefore, the polar quadric of any point on
Txz
+
Vxy
+
Wyz '" O.
splits into
TI
TI
and a plane through
the point p whose coordinates are
( I
(4.8)
i
l
2
x '" P(KN - W ) - Q(- VN - TW)
+
R(VW
KT)
+
Y '" - P(- NV - TW) + Q(GN - T2) - R(- GW - TV) z '" P(VW + KT) - Q(- GW - VT) + R(GK _ V2) t = GKN + TVW - KT2 - NV 2 - GW 2
It is easy to check p is a nucleus; moreover, all points on the conic
(4.9)
22 Gx 2 + Ky + Nz
are nuclei.
+
Txz
+
Vxy
+
Wyz '" t
=0
C is a degenerate conic iff p is on
Assume C is non-degenerate.
TI
[4] (then prop. 3.1 (ii) holds).
Taking the canonical form for C and the point p as
0 , the equation of F becomes [4]: 4 (4.10)
Ax
3333 +
By
+
Cz
+
Dt
+
t(Gx
222 + Ky + Nz ) '"
o.
All generators of the cone r projecting C from p and all lines on
TI
are tri-
tangent 1i nes. Since g and C have two common pOints, F always contains two of its nuclei, and either (1) g meets C at two distinct points, or (2) g is tangent to C. (1) Take suitable coordinates on (4.11 )
Cz 3
+
Dt 3
+
t(Nz 2
+
TI
[4]; then the equation of F is
Vxy) = 0 ,
C, N, V
0 and O are binodes sharing the biplane
1 2 and y = 0, taO do not intersect on F.
TI
#0 .
and the biplanes x '" 0, t
Two cases are to be considered.
=0
M. de Finis and M.J. de Resmini
254
(i) The nucleus p is not on F; F has only the two binodes 0 and O2 , 1 (ii) P belongs to F; then F has a conic node besides the two binodes. (i)
Through the line g
=Fn
there are exactly three planes, beside TI,
TI
intersecting F at two lines other than g and each of them contains a pair of lines on the biplanes (F TI); the canonical form for F is [41: 332 t + t(xy + z )
z +
(4.12 )
All the lines on
TI
=0
.
and all the generators of the cone xy + z2
0 are tri-
tangent lines and no other tri-tangent line exists.
= 0 in (4.11). The tangent cone at the conic node p meets F exactly
(ii) 0
at the lines joining p with the binodes. (beside
Through 9 there is just one plane
and the plane through the singular points) meeting F at g and at the
~
lines Cz
(4.13 )
+
Nt
=
x
=
0, and Cz
Nt = Y = 0 ,
+
each of them belonging to one biplane.
F contains no other line.
(2) The point at which g is tangent to C is a nucleus belonging to F and is a binode.
On
(4.9) 141.
Ky2
+
~
coordinates can be chosen to get the canonical form for the conic
The polar quadric of any point on TI has rank 2, that of a pOint on r:
Txz = 0, but not on C, has rank 3.
rank 4.
The polar quadric of any other point has
Two cases have to be considered.
(iii) F has just a binode at the pOint g n C = 0, the biplanes intersect on F (at g) and the biplane other than
TI
meets F at two distinct lines (beside g).
F contains no other lines and the canonical form for F is 3
(4.14 )
z - t
3
+ tty
2
+
xz)
= O.
(iv) F has also a conic node at the point p
= 04 ,
the tangent cone being r.
The only lines on Fare g and the line joining p with the binode.
The canonical
form for F is [4]: (4.15 )
z3 + t(y2 + xz)
= O.
As a corollary to prop. 3.1(ii) and 4.1, the following holds. PROPOSITION 4.16: If foul' coplanar nuclei exist, on a plane TI, three of them !,c;'ng ,":Uinear (on a line
9.)
and no point exists whose polar quadric van1:shes,
On cubic surfaces over a field of characteristic 3
(i) the polar quadric of any point on
TI
splits into
TI
255
and a plane through a
point p on £; (ii) on
TI
there exists a degenerate conic, containing £, all whose points are
nuclei, and no nucleus exists off (iii)
TI
TI;
meets F at a line counted three times and F is one of the surfaces
considered in prop. 3.1(ii).
5. ON CUBIC SURFACES HAVING THREE DISTINCT NON-COLLINEAR NUCLEI Beside characterizing some cubic surfaces, the results in this section will show the role the rank of polar quadrics of nuclei plays. PROPOSITION 5.1: If F has three distinct non-collinear nuclei whose polar quadrics have exactly rank 3 and the plane
TI
joining them does not intersect F at a line
counted three times, then two other nuclei not on (a) the nuclei not belonging to
TI
TI
exist, and either
are distinct, or
(b) they coincide and there is a plane, distinct from
TI,
meeting
F at a line
counted three times so that prop. 4.1 holds. When (a) is true, F has exactly five nuclei (no four Of them being coplanar) and one of the following holds: (i) F contains none of its nuclei (therefore, it has no singular point); (ii) F contains one of its nuclei and so it has a conic node; (iii) F contains two of its nuclei and has two conic nodes; (iv) F contains three of its nuclei; thus it has three conic nodes; (v) F contains four of its nuclei and has four conic nodes. In all these cases the numbers and configurations of lines on the surface are as in the classical case (projective 3-space over the complex number field).
PROOF: Assume the given nuclei are 0 , 0 and O , Thus, 4 3 2 (
- Exy - Fxz - Gxt + Syz + Tzt + Vyt
I I,
Ei + Sxz + Vxt + Wzt Fx 2 + Sxy + Txt + Wyt 2 Gx + Txz + Vxy + Wyz
(5.2) {
a
L
is defined by
=0
0 0 0,
W# 0 [41.
Solving the simultaneous equations (5.2), the following one is obtained:
256
M. de Finis and M.J. de Resmini
2 2 2 t 2 (GW - TVW) + tx(STV - GSW) + x (GS - EST - FSV + EFW)
(5.3)
Therefore, two nuclei not belonging to (a)
The two nuclei not on
TI
TI
=
°
exist, and two cases are to be considered.
are distinct.
Take one of them as 01' then the
other one is (W,T,V,S), and by a suitable choice of the unit point, the equation of F becomes (5.4)
3
3
Ax + By + Cz
3
3
+ Dt + xyz + xyt + xzt + yzt
=
°.
(b) The two roots of equation (5.3) coincide; then there exists a plane (# TI) meeting F at a line counted three times and prop. 4.1 holds.
Thus, assume (a) holds and the equation of F is (5.4).
All possible cases
for F will now be considered (for more details, see [4]). (i) When A, B, C, D # 0, A + B + C + D + 1 # 0, none of the nuclei belongs to F; there are exactly five non-degenerate cones all of whose generators are tritangent to F. (ii) F contains exactly one of its nuclei, thus is has a conic node; assume it is 04; then the eqn. to F is (5.4) where D = 0, and A, B, C # 0, A+B+C+l #0.
there are four cones all whose generators are tri-tangent to F. (iii) Two of the five nuclei belong to F; therefore, there are two conic nodes and three cones all whose generators are tri-tangent to F. (iv) F contains three of its five nuclei; thus it has three conic nodes and there are exactly two cones all whose generators are tri-tangent to F. (v) All nuclei but one belong to F; therefore F has four conic nodes and there is just one cone all whose generators are tri-tangent to F. nuclei on F are the points 0i (i (5.5)
xyz + xyt + xzt + yzt
Assume the
= 1,2,3,4); then the equation of F is =
°.
In all cases (i) - (v) the numbers and configurations of the lines belonging to F are as in the classical case [1 I, [3 I (see also (4)). The canonical forms for the surfaces obtained from (5.4) in cases (i) - (v) are different from the classical ones; the latter will be discussed in sect. 7. REMARK: When (v) holds, in the dual space the planes tangent to F form - as in the classical case - Steiner's Roman surface. The next result shows the assumption on rank in prop. 5.1 cannot be dropped.
257
On cubic surfaces over a field of characteristic 3
PROPOSITION 5.6: If F has
th~ee
non-collinear nuclei, their plane does not inter-
sect F at a line counted three times and the polar othe~
given nuclei has rank 2 (the nuclei.
If the polar
q~ics
qua~ic
of exactly one of the
two having rank 3), then F has just three
of two nuclei have rank 2, then there exists a
conic all of whose points are nuclei and a nucleus not belonging to the plane of the conic, and prop. 4.1 holdS.
Finally, if the polar
nuclei have rank 2. then there exists a point whose
qua~ics
of all three
pola~ qua~ic
vanishes and
prop. 2.5 holdS. Under the assumption just one of the polar nuclei has
~ank
qua~ics
of the three given
2, one of the following occurs:
(i) F has two conic nodes and one binode. and either (a) the line
s at whioh the biplanes intersect belongs to F. or
(bJ s does not belong to F; (ii) F has two conic nodes and there are two pencils of tri-tangent lines through the same point on two distinct planes; (iii) F has a conic node and a binode and there is exactly one cone all whose generators are tri-tangent; either (aJ or (b) in (i) holds; (iv) F has just one oonic node and both a cone and two pencils of tri-tangent lines (as in (ii)) exist; (v) F has just a binode and either (a) or (b) in (i) holds; oones all whose
are two
are tri-tangent to F;
generato~s
(vi) F has no
the~e
singula~
point; all the generators of two distinot oanes and
all the lines of two pencils
(as
in (ii)) are tri-tangent.
PROOF: Take coordinates as in the proof of prop. 5.1; r is defined by (5.2). where Wf O.
Under the further assumption the polar quadric of the point 04 has
rank 2 and splits into the planes y
= 0 and z = 0, r is defined by
Exy - Fxz + Syz = 0
f - Ei
(5.7)
) )
l
+ Sxz + Wzt 2 Fx + Sxy + Wyt
Wyz
0
If either F
1:1
0
°
Wf O.
= 0, or E = 0, then there exist a conic all of whose points are
nuclei and a nucleus not on its plane, and prop. 4.1 holds. E
If both F
0 and
0, then a pOint whose polar quadric vanishes exists and prop. 2.5 holds [4].
M. de Finis and M.J. de Resmini
258
Thus, E, FrO will be assumed. Let s be the line y
Take as 01 the point (W,O,O,-S); then S =
(W,O,O,-S) has rank 3. in eqn.s (5.7).
= z = 0; the polar quadric of any pOint on s but 04 and
The polar quadric of any point on
nuclei and not on the line Ey rank 4.
°
not on a line joining two
= x = (which contains just one nucleus) has
Fz
+
TI
°and E,F,W F °
°3, 04;
The only solutions to equations (5.7) are the given nuclei 02'
therefore. all surfaces for which E is defined by (5.7) (with S
= 0) have exactly
three nuclei, the equation of such an F being Ax
(5.8)
3
+
By
3
+
Cz
3
+
Dt
3
+
2
2
Ex Y + Fx z
+
Wyzt =
°.
(No plane meeting F at a line counted three times exists.) (i) All three nuclei belong to F; then F has two conic nodes (02 and 03) and a binode (04)' (b) s
~
The biplanes intersect at the line s. and either (a) s
E
F, or
F.
(a) F contains exactly five distinct lines [4] (as in the classical case), no proper tri-tangent exists. and the canonical form for F is 2
x (y
(5.9)
+
z)
+
yzt =
°.
(b) Since s does not belong to F. A F
° in (5.8).
Each biplane meets F in
the line joining the binode to the conic node belonging to it (counted twice) and at another line.
The configuration of lines on F is as in the classical case \ 4)
and the canonical form for F is 2
x (x
(5.10)
+
y
+
z)
+
yzt
0.
(ii) F has two conic nodes.
The number and configuration of the lines on
the surface are as in prop. 5.1(iii), but now the tri-tangents to F are exactly all lines through 04 belonging either to the plane y \4].
= 0. or to the plane z =
°
The canonical form for F is 2 x (x
(5.11)
+
y + z) + t(yz
+
2 t ) =
°.
(iii) F contains the nucleus whose polar quadric has rank 2 and one of the other nuclei.
There is exactly one cone all of whose generators are tri-tangent
to F and no other tri-tangent line exists. (a) s
E
F (then A = 0).
One biplane meets F at s counted twice and at
another line, the other biplane meets F at s and at two distinct lines.
Besides
the biplanes. there are three planes meeting F at lines (one of them through s)
259
On cubic surfaces over a field of characteristic 3
[4].
With a suitable choice of the unit point, the equation of F becomes 2
(5.12)
2
2
y(y - z ) + z(Fx + Wyt)
=0
F, W-F 0 .
,
One biplane meets F at the line 0 0 g counted twice and at 34 the line r: Ax + Fz = Y = 0, the other biplane meets F at three distinct lines. (b) s
~
F.
Through g there are three other planes meeting F at lines; also through r there are three other planes intersecting F at lines [4]. 3
(5.13)
3
2
2
x + y + x y + z (x + yt)
The equation for F is
= 0 .
(iv) F contains one of its nuclei having a rank 3 polar quadric; thus, F has just a conic node.
The situation is as in prop. 5.1(i), but now the tri-tangent
lines are all generators of a cone (the polar quadric of O2 ) and all the lines through 0 belonging either to the plane y = 0, or to the plane z = 0 [4]. The 4
canonical form for F can be obtained from eqn. (5.8) (and it is not the classical one) . (v) F contains its nucleus whose polar quadric has rank 2. a binode. (a) s
Then F has just
The tri-tangents to F are exactly the generators of two cones. E
F.
Each biplane meets F at s and at two distinct lines and there is
exactly one other line of F (not through 0 ) incident with s [4].
F contains four
4
other lines and the (classical) canonical form (5.14)
(y + z)(y + z + x)(y + z - x) + Wyzt (b) s
~
F.
is obtained
[4]:
W-F 0
= 0 ,
Then each biplane meets F at three distinct lines [4].
are as in the classical case.) (5.15)
[1]
(Lines
The canonical form for F is
(- x + y + z)((- x + y + z) (vi) F contains none of its nuclei.
2
2 + x ) + Wyzt
=0
All generators of two cones and all the
lines through the same point (0 4 ) on two distinct planes are tri-tangent to F and no other tri-tangent line exists. F has 27 lines. For instance, all the required conditions are satisfied by [4]: (5.16)
3 3 2 3 y - z + x (y - z) + t - yzt = 0
6. ON CUBIC SURFACES WITH EITHER TWO OR ONE NUCLEI
PROPOSITION 6.1: If a cubic surface F has two (distinct) nuclei and no point on
260
M. de Finis and M.J. de Resmini
the Zine joining them is a nucleus, then one of the following holds: the polar> quadJ"ics of both nuclei have rank 3; then there are exactly
(-i)
three other nuclei and either they are distinct and prop. 5.1 holds,
01'
they
coincide at a nucleus whose polar quadric has rank 2 and prop. 5.6 holds; ~he
(ii)
poZar> quadrics of the two nuclei have rank 3 and 2, respectively;
then either 1. they-e is a third nucleus whose polar quadric has rank 3, and prop. 5.6
2. no other nucleus exists;
01'
3. there exists a conic all whose points are nuclei and prop. 4.1 holds. When (ii)2. holds, one of the following occurs:
F contains a binode and a conic node; no tri-tangent line exists not
(aJ
passing through the singular points; the biplanes meet at the line s, which may belong
F
:;0
01'
nor;.
(b) F contains just a binode (and either
S E
F,
01' S
i F) and the tri-
tangent,,: are exactly ,;he gener-ators of a cone. (c) F conr;ains Just a conic node and its tri-tangents ar>e exactly the lines of
~-:,)o
pencils 1.Jith trze same centre on two distinct planes.
F has no
(1)
sil~gular
point and the locus of tri-tangents consists of the
"el1eratcrs of a cone and the lines through a point on two distinct planes. (;:ii) The polar> quadrics of both nuclei have rank 2; then either 1. r;he pair's of planes into which they split are distinct (then a line all 1.Jhose points are nuclei exists and prop. ;).1 ho Ids); or 2. the pairs of planes share a plane (then a conic all whose points are exis~s
nucZei
and prop. 4.1 holds).
PROOF: (i) Coordinates can be chosen so that 03 and 04 are the existing nuclei, their polar quadrics both have rank 3, and no point on the line a nucleus. (
(6.2)
I
) ) I
l
joining them is
Then. with a suitable choice of points 01 and 02 [4]. E is defined by
2 - Exy + Hy + Syz + Tzt + Vyt
0
E/ - Hxy + Sxz + Vxt + Wzt
0
Sxy + Txt + Wyt
° ° The polar quadric of any point on I
t
T, S, V. WF 0 •
Txz + Vxy + Wyz
t but q
(O.O,V.-S) has rank 3.
The
261
On cubic surfaces over a field of characteristic 3
polar quadric of q splits into distinct planes not through
~.
Solving the simultaneous equations (6.2) the following one is obtained: 23 ES x
(6.3)
+
(ESW
+
22 HTS - VS )x t
23 VW t = 0 ;
+
eqn. (6.3) has either three distinct or three coincident roots according to whether (6.4)
ESW
+
HTS - VS 2 I 0 , or
(6.5)
RSW
+
HTS - VS 2
=
0
When (6.4) holds, F has five distinct nuclei (the nuclei given by eqn. (6.3) are non-collinear and belong to the plane Sz - Vt = 0), and prop. 5.l(a) holds. When (6.5) holds, the third nucleus is the point V ~E2W2(EW + HT),
x
= -
z
= EV 2 , t = E(EW
(6.6)
+
Y=
-
ETV
~ HT{EW
= EW +V HT
HT) , where S
+ HT) ,
I 0 ;
its polar quadric has rank 2 and prop. 5.6 holds. (ii) Take as 04 the nucleus having a rank 2 polar quadric, let it be yz = 0, and as 0 the nucleus having a rank 3 polar quadric; then E is defined by 3
(6.7)
( - Exy - Fxz + Hy 2 + Syz = 0 J Ex 2 - Hxy - Jyz + Sxz + Wzt Fx 2 + Jy2 + Sxy + Wyt = °
l
Wyz
=
=
0
0,
F, W I
°,
and no nucleus on the line 03°4 exists. When E I 0, the equations (6.7) have, beside 03 and 04' another solution, namely the point (6.8)
x
= - EHW, Y = - E2W, z = 0, t = FH 2
+
JE
2
+
EHS ,
whose polar quadric has rank 3, and prop. 5.6 holds (see [4]). Now, assume E = 0; then the simultaneous equations (6.8) have just the pOints 03 and 04 as solutions. The choice of the reference simplex can be completed in order to get a canonical form for the polar quadric of 03 [4]. The polar quadric of any point but 04 on the lines xz = y = 0 has rank 3. tion of F becomes [4]:
(6.9)
Ax
3
+
By
3
+
Gz
3
+
Dt
3
and one of the following occurs:
+
2 Fx z
+
Hxy
2
+
Wyzt = 0 ,
The equa-
F,H,WIO,
262
M. de Finis and M.J. de Resmini
(a) both nuclei belong to F; (b) the nucleus 04 belongs to F', (c) the nucleus 0 belongs to F', 3
(a) F has a conic node and a binode; the line s: y = z
°mayor not
belong to F. (a.l) s E F.
One biplane meets F at the line
t
= 0 04 (counted twice) and 3
at S; the other at s (counted twice) and at By + Hx = z = 0, and no other plane through s exists meeting F at lines. lines: FBx - HWt = Hx + By =
°and
The tangent cone at 0
3
intersects F at two
F contains no other line and by a suitable
t.
choice of the unit point, the equation of F is 2
2
=
y (y + x) + z(x + yt)
(6.10)
(a.2) s
F.
~
°
One biplane meets F at t (counted twice) and at the line
Ax + Fz = Y = 0; the other at three distinct lines.
The equation of F becomes
[ 4\: (6.11)
x(x
2
2 - Y + Fxz) + Wyzt =
°.
(b) F has just a binode and the tri-tangents to F are exactly the generators of a cone (the polar quadric of 0 ), 3
(b.1) s E F. One biplane meets F at s and at the two distinct lines 2 2 Cz + Fx = Y = 0, while the other meets F at s counted twice and at By + Hx = z = O. These three lines belong to two planes, namely Hv'Tx+B
VTy~H
each of them containing another line on F. 2
Y (x + y) + z(x
(6.12)
(b.2) s
~
F.
2
~z=O,
The canonical form for F is
2
- z ) + Wyzt =
°.
Each biplane meets F at three distinct lines and F contains 9
other 1i nes [4\. (c) F has just a conic node, 03'
The tri-tangents are exactly the lines of
two pencils with the same centre 04 and belonging to the distinct planes y = z = O.
°and
The lines on F are as in prop. 5.1(ii). (d) F does not contain any of its nuclei.
The locus of tri-tangents consists
of the generators of a cone and the lines of two pencils as in (c). lines on F.
An example of such a surface is given by [41:
There are 27
263
On cubic surfaces over a field of characteristic 3
Cx
(6.13)
3
+
(C
+
W)y
3
+
Cz
3
+
(C
+
3
2
W)t - Wx z
+
Wxy
2
+
Wyzt
=
°.
(iii) Take coordinates so that 04 and 03 are the nuclei and xy = 0, Fx +
2
Jy =
°their polar quadrics, respectively.
Writing down
L
(where F,J,V
2 ~
+
0)
it is straightforward that a line all of whose pOints are nuclei exists, and prop. 3.1 holds
[4].
When the polar quadrics of the nuclei 03 and 04 share a plane, taking coordinates so that these quadrics are yt
=
°and yz
=
0, resp .• equations (1.2)
become - Exy
(6.14)
+
Hy 2
= 0, Ex 2 - Hxy
+
Wzt
= 0, Wyt = 0, Wyz = 0,
where at least one between E and H is not zero.
Then - Ex
2
+
Wzt
=Y
W~ 0,
°
is a non-degenerate conic all of whose points are nuclei and prop. 4.1 holds [4].
PROPOSITION 6.15: If a point p exists whose polar quadric splits into two planes, both of them through p, then p is a nucleus and one of the following holds: (i) there are two other nuclei and both their polar quadrics have rank 3
(then prop. 5.6 holds); (ii) there exists but one other nucleus and its polar quadric has rank 3
(then prop. 6.1(ii}2 holds); (iii) there is a line all whose points are nuclei and prop. 3.1 holds.
PROOF: Take 04 as p and yz = p is a nucleus. y
0, or to z
°as its polar quadric; then it is easy to check that
If another nucleus exists, then it must belong either to
= 0, and one of the following holds [4]:
(i) there are two nuclei, distinct from p, one on y = 0, the other on z
0;
(ii) one of the nuclei in (i) coincides with p; (iii) both nuclei in (i) coincide with p. The statement follows. PROPOSITION 6.16: If a point p exists such that its polar quadric has rank 1, being the plane w counted twice, then p is a nucleus, and either
1.
w intersects
F at a line counted three times, in which case w contains a
degenerate conic all of whose points are nuclei and prop. 4.1 holds; or
2.
w does not meet
F at a line counted three times; then one of the fol-
lowing occurs: (i) p is the unique nucleus; (ii) w contains a line t through pall
264
M. de Finis and M.J. de Resmini
b.'huse points Clr8 nuc:2i and no nucleus off R. exists, and prop. :3·1 (i) holds.
(:,) F has a unique singular point, namely the uninode p, the uniplane being (,; (a'ld eitht.:r w meet.3 F at three distinct lines, or
w
intersects it at two
coincident (mld one c','stinct) Zines); or' Ii,}
F has no s:ngular' point and its tn-tangents are exactly the lines on
PROOF: Taking p as 04 and
as x
w
0,
1F/
(6.17)
i\
Gx2
Thus P
J =
+ Jy
2
is defined by
L
- Exy - Fxz - Gxt + Hy2 + lz2 2 Ex - Hxy - Jyz + Mz2 + Sxz - lxz - Myz + Sxy
w
+
° °
Syz =
°
o,
G 'f 0 .
04 is a nucleus.
If w meets F at a line counted three times, then 2 2 N = 0 in eqn.s (6.17) and the degenerate conic Hy + Lz + Syz = x = 0 is the
locus of nuclei and prop. 4.1 holds. iherefore, assume J and M are not simultaneously equal to zero.
Then,
either p is the unique solution to equations (6.17), or HM2
(6.18)
+
lJ 2
SMJ = 0 ,
+
and the line Jy - Mz = x = 0 belongs to all quadrics in I. 2 (i) HM2 + LJ + SMJ 'f O. The polar quadric of any point on w is a cone with vertex at p and w contains a cubic curve the polar quadrics of all whose points have rank at most 2, and this curve has a cusp at p (the tangent being Jy - Mz
x = OJ.
Taking the following canonical form for this curve [4]: 2 3
2
H Y + GMz t
(6.19)
= x =0
,
H, G, M
and as yz = 0 the polar quadric of 0 ,
I
3
(6.20)
- Gxt + Hy
2
= 0,
- Hxy + Mz
2
'f 0 ,
is defined by
= 0, - Myz = 0, Gx 2 = 0 •
All points in P have a rank 4 polar quadric but the points on the quartic surface (6.21 )
2 2 x (Hxy - Mz ) = 0 ;
the plane w contains the curve (6.19) all whose points but the cusp p have a rank
265
On cubic surfaces over a field of characteristic 3
2 polar quadric and p is the unique point whose polar quadric has rank 1. The equation of F is Ax 3 + By3 + Cz 3 + Dt 3 + Gx 2t + Hxy2 + MYZ2
(6.22)
0,
G,H,M
-f 0 .
All lines through p on ware tri-tangent to F and no other tri-tangent exists. (a) p E F; P lines.
= 04 is a uninode, the uniplane is wand meets F at three
Since M -f 0, either these lines are all distinct and the following equa-
tion for F is obtained [4): By3 + Hxy 2 + Myz 2 + Gx 2t = 0 ,
(6.23)
B, H, M, G -f 0 ,
or two of them coincide and the equation of F is 2 2 2 3 Cz + Hxy + Myz + Gx t
(6.24)
o,
H, M, G -f 0 •
(b) P It F. Since all lines through p on ware tri-tangent to F, coordinates can be chosen so that the equation for F becomes [4): (6.25) All tri-tangents to F are exactly all tri-tangents to the cubic curve F n w. surface contains 27 lines [4). (ii) HM2 + LJ 2 + SMJ = O.
Then all points on the line Jy - Mz = x
The
0 are
nuclei and it is easy to check that prop. 3.1(i) holds [4). PROPOSITION 6.26: If a point exists whose polar quadric has rank 4 and if F has a nucleus whose polar quadric has rank 3 (and not less), then one of the following is true: (i) F has exactly five nuclei (whose polar quadrics all have rank 3) and prop. 5.1{a) holds; (ii) F has another nucleus whose polar quadric has rank 2 and prop. 6•. 1(ii) holds; (iii) there are two other nuclei with rank 2 polar quadrics; then a nondegenerate conic all whose points are nuclei exists and prop. 4.1 holds.
PROOF: No pOint exists whose polar quadric vanishes.
Assume 04 is the point with a rank 4 polar quadric and take the canonical form Vxy - Rzt = 0 (V, R -f 0) for this quadric [4).
W.l.o.g. assume 03 is the nucleus whose polar quadric has rank 3; then equations (1.2) become
266
M. de Finis and M.J. de Resmini
r - Exy2 Ex
)
(6.27)
Fx 2
1l
Fxz
+
Hy 2
- Hxy - Jyz 2 + Jy2 + Rt
Vxy - Rzt
+
Syz
+
Vyt = 0
+
Sxz
+
Vxt = 0
+
Sxy
=0 V, RIO, FJ - S2 I 0 .
=0 ,
Three cases must be considered [4]: 2 2 (i ) JV + RH2 I o and FV + E2R I 0 2 (i i) Jv2 + RH2 o and FV + E2R I 0 (or conversely) ; 2 (iii) JV 2 + RH2 o and FV + E2R = 0 (i) With suitable coordinates, S
= E = H = 0 can be assumed in equations
(6.27), while F, J, R, V I 0; then the four nuclei beside 0 are: 3 x = v'JR, y = v'FR, z = V, t = v'FJ; x = - v'JR, y = v'FR, z = V, t = -v'FJ; x = v'JR, Y = - v'FR, z = V, t = - v'FJ;
(6.28)
x
=
- v'JR, y
=
- v'FR, z
=
V, t =
v'FJ.
Since they are independent, F has five nuclei, and prop. 5.1(a) holds [4]. 2 2 2 2 (ii) Suppose JV + RH = 0 and FV + E RIO. Then a nucleus exists on the line x
= z = 0 and its polar quadric has rank 2; so, prop. 6.1(ii) holds. (iii) In this case another nucleus exists on the plane z = 0, thus, there
are two nuclei, both with a rank 2 polar quadric, and a non-degenerate conic all whose points are nuclei exists, and prop. 4.1 holds [4] (see also prop. 6.1).
7. ON CUBIC SURFACES WITH FIVE NUCLEI AND ON CUBIC SURFACES WITH FIVE NUCLEI AND 27 LINES It was already shown that cubic surfaces with five nuclei exist; now it will be proved that this is the general case. PROPOSITION 7.1: If a point exists whose polar quadria has rank 4, then the most general aubia surfaae has five nudei, no four of them aoplanar.
PROOF: Choosing coordinates as in the proof of prop. 6.26,
~
is defined by
267
On cubic surfaces over a field of characteristic 3
r -
(7.2)
)
"\
l
2 2 Exy - Fxz + Hy + Lz + Syz + Vyt El - Hxy - Jyz + Mz2 + Sxz + Vxt Fl + Jy 2 - Lxz - Myz + Rt 2 + Sxy Vxy - Rtz
=0 =0 =0
o,
V, R t-
o.
Under the assumptions HM2 + LJ 2 + SMJ
(7.3)
t-
0 and EL2 + MF2 + FLS
t-
0 ,
solving the simultaneous equations (7.2), the following is obtained [4]: x5(FLSV 3 + EL 2V3 + F2MV 3) + x4(FRS2V 2 _ FHLRV 2 - F2JRV 2 + L2V4 _ EFMRV 2) + x3(EFJR 2V + EHLR 2V + E2R2MV + FMRV 3 + LRSV 3 - ER 2S2V) + x2(_ FH2 R3_ E2 JR 3 _ FJR 2V2 + EMR 2V2 - EHR 3S) + x(MR 2V3 - HR3SV - EJR 3V) - JR3V2 - H2 R4 = 0 , t =1
(7.4)
Therefore, five (generally distinct) nuclei exist and by prop. 4.1 no four of them are coplanar. 2 When HM2 + LJ + SMJ = 0, the second condition in (7.3) being satisfied, 2 2 prop. 6.26 holds. When also EL + MF + FLS = 0, prop. 6.1 holds [4]. It is not difficult to verify that the surfaces for which E is given by (7.2)(and (7.3) hold) have 27 lines [4]. Cubic surfaces containing 27 lines have already been considered. Now it will be shown that also over a field of characteristic 3 a cubic surface containing 27 lines and having the same canonical form as in the classical case can be constructed; furthermore, such a surface has five distinct nuclei (no four of them being coplanar). Starting from equation (1.1), assume t = 0 is a tri-tangent plane intersecting F at the degenerate cubic curve xyz
= t = 0; then the equation of F
becomes: (7.5)
3222 222 Dt + Gx t + Ky t + Nz t + Pxt + Qyt + Rzt + Sxyz + Txzt + Vxyt + Wyzt
=0 •
= A z meeting F at a degenerate cubic curve consisting of three lines [4], assume one of them is z = 0 and the cubic curve on Since there are five planes t
it contains 04. So, D = 0 in eqn. (7.5) and 2 (7.6) PQV - p2K - Q G = 0
268
M. de Finis and M.J. de Resmini
(and the other three planes through t
= z = 0 intersecting F at lines are distinct
if F does not contain any of its nuclei (41). Bya similar argument (4), there are five planes through the line y meeting F at lines; assume one of them is y
=t =0
= 0; then
2 2 PRT - P N - R G = 0
(7.7)
must hold.
Also tht'ough x
t
=
° there are five planes intersecting F at lines;
=
take one of these planes as x = 0; then
(7.8) must hold. Taking into account (7.6), (7.7), and (7.8), being D (7.9)
G
K
0, eqn. (7.5) becomes
N
5xyz +t(pX + oy + RZ + t)(Px + Qy + Rz) = 0 ,
(5 f. 0).
This surface contains 27 lines and its equation is the classical one [1]. The cubic surface defined by equation (7.9) has five distinct nuclei (none of them on it), provided that G, K, N, P, Q, R f. 0[41. As special cases (for more details see (4)), since
(7.10) the following are obtained: (i) G = 0 and TV
P5 f.
°:
F
has a conic node (and four nuclei not on it).
(ii) G = P = 0 : F has a binode, the biplanes intersect on F; (7.9) becomes (7.11)
2 2 5xyz + t(Ky + Nz + Wyz + Qyt + Rzt) =
°;
the binode takes into account three nuclei, and two other nuclei exist (prop. 5.6 (l)(a) holds). (iii) G = 0, P f. 0, S
=
PNK OR
=
TV
p : F has a binode and in the most general
case the biplanes do not intersect on F (prop. 5.6(i)(b) holds). (iv) G = K = 0 (N f. 0), P, Q f.
° : F has two conic nodes and there are three
other nuclei (prop. 6.1 (i) holds).
°:
(v) G = K = 0, P = 0, Q f. F has a binode and a conic node (prop. 6.1(ii) holds). (When also Q = 0, F splits into a plane and a quadric.) (vi) G = K = N = 0, P, Q, R f. 0 : F has three conic nodes and two other nuc 1ei not on it. (vii) G = K
N
P
0, Q, R f. 0
F has two conic nodes and a binode, the
On cubic surfaces over a field of characteristic 3
269
biplanes intersecting on F, and no other nucleus exists (prop. (5.6)(i)(a) holds). No other non-degenerate case can be obtained from equation (7.9).
8. ON CLASSIFICATION OF CUBIC SURFACES In this section the classification of cubic surfaces according to the numbers and types of singularities (i.e. the classification over the complex number field) will be compared with the classification based on possible configurations of nuclei, taking into account the loci of tri-tangent lines and the planes intersecting F at a line counted three times. In this comparison, neither cones nor ruled cubic surfaces will be considered.
(The former can be classified according to the classification of plane
cubic curves [6], the latter were completely characterized in sect 3.) First, recall some classical notation for singular pOints [1]: C2 B3
conic node;
B4
binode with biplanes intersecting on F;
B5
binode with biplanes intersecting on F at s, one biplane meeting F at s
binode with biplanes not intersecting on F;
counted twice and at another line; B6
binode with biplanes intersecting on F and one biplane meeting F at the line s counted three times;
U6 U 7
uninode such that the uniplane meets F at three distinct lines; uninode such that the uniplane meets F at two coincident (and one distinct) lines;
U ~ uninode such that the uniplane meets F at three coincident lines. 8 In the following table the 35 different types of cubic surface other than cones and ruled surfaces over an algebraically closed field of characteristic 3 are listed and the notation here below is used: C+ P
~
the locus of nuclei consists of a non-degenerate conic and a point not on its plane;
n(v)
number of points whose polar quadric vanishes;
n(n)
number of planes meeting F at a line counted three times.
270 #
M. de Finis and M.J. de Resmini
and type of nuclei
n(v)
singularities configuration
locus of
ref.
n(1I)
tri-tangents
5 distinct
0
5 cones
0
5.1(a)(i) & (7.9)
3 distinct
0
2 cones & 2 noncoplanar pencils
0
5.6(vi)
2 distinct
0
1 cone & 2 noncoplanar pencils
0
6.1(ii)2.(d)
1 only
0
1 penci 1
0
6.16.2.(i)(b)
5 distinct
0
4 cones
0
5.1 (a)(ii)
3 distinct
0
1 cone & 2 noncoplanar pencils
0
5.6( i v)
2 distinct
0
2 non-coplanar penci 1s
0
6.1(ii)2.(c)
1 line
0
special 1ine complex
0
3.1 (i )(b.2)
1 1i ne
0
all 1i nes through a point
3
2.5(b)(3ii),(2.23)
3 distinct
0
2 cones
0
5.6(v)(b)
2 distinct
0
1 cone
0
6.1 (i i )2. (b.2)
B4
3 distinct
0
2 cones
0
5.6(v)(a)
B5
2 distinct
0
1 cone
0
6.1(ii)2.(b.l)
B6
C+ P
0
1 cone &all lines on a plane
1
4.1(iii)
U6
1 line
0
special line complex
0
3.1(i)(b.l)
1 only
0
~
0
6.16,2(i)(a),(6.23)
U7
1 only
0
~
0
6.16,2(i )(a) ,(6.24)
Us
2 incident 1ines
0
all 1i nes on a plane
1
3.1(ii)(b)
1 1i ne
1
all lines through 3 a point
0
C2
B3
2. 5( b)( 3. iii)
On cubic surfaces over a field of characteristic 3
#
and type of nuclei
n(v)
singularities configuration
locus of
n(1f)
271
ref.
tri-tangents
5 distinct
0
3 cones
0
5.1(a)(iii)
3 distinct
0
2 non-coplanar pencils
0
5.6(ii)
3C 2
5 distinct
0
2 cones
0
5.1 (a)(iv)
4C 2
5 distinct
0
1 cone
0
5.1 (a)(v)
2B3
2 incident 1ines
0
all 1i nes on a plane
1
3.1 (ii )(a)
C+ P
0
1 cone and all lines on a plane
1
4.1 (i)
2 incident 1i nes
1
all 1i nes th rough a point
3
2.5(b)(2. ii)
3 concurrent non-coplanar 1ines
1
all 1i nes through a point
3
2.5(b)(1. ii)
3 distinct
0
1 cone
0
5.6(iii)(b)
2 distinct
0
~
0
6.1 (ii )2. (a.2)
B4 + C2
3 distinct
0
1 cone
0
5.6(iii)(a)
B5 + C2
2 distinct
0
~
0
6.1(ii)2.(a.l)
B6 + C2
C+ P
0
1
4.1(iv)
B3 + 2C 2
3 distinct
0
0
5.6(i )(b)
2B3 + C2
C+ P
0
1
4.1(ii)
B4 + 2C 2
3 distinct
0
0
5.6(i )(a)
2C 2
3B3
B3 + C2
-
(Remark.
all 1i nes on a plane ~
a 11 1i nes on a plane ~
Of course, only proper tri-tangent lines, i.e. not through any
singular point, were considered.)
M. de Finis and M.J. de Resmini
272
9. A COMBINATORIAL CONFIGURATION AND THE DESIGN ASSOCIATED WITH IT In this section only cubic surfaces having exactly five nuclei and no singular point will be considered. Any tri-tangent line through two nuclei will be called a special tri-tangent; thus, for such an F there are exactly 10 special tritangents; therefore 10 points on F are determined and in this set of points there are exactly 10 collinear triples and 5 coplanar hexads. The set conSisting of the 10 points on F belonging to its special tritangents, the 10 triples of collinear points (i.e. the 10 lines containing these triples) and the 5 planes containing the hexads will be called configuration N (N for short).
If (5.4) is the equation of F, then the 10 pOints in N are: 1
= (0,0,
3-
v D, - ~C)
3-
6 '"
3-
3-
3- 3- 3-
= (0, v D,O, - v B) 333 = (0, v c, - v B,O) 34 = (v D,O,O, - ~A) 35 = (v C,O, - ~A,O)
2
3-
(v B, - v A,O,O) 3
7 '" (v D, v D, v D, - v A + B + C + 1) 3- 3-
8 = (v
c, V C,
3-
3
3
3-
- v A + B + D + 1, v C) 3- 3-
9 '" (v B, - v A + C + D + 1, v B, v B) 10
= (-
3
3- 3- 3-
vB + C + D + 1, vA, vA, vA).
The tangent plane at each of these points intersects F in a cubic curve having a cusp at that point (the tangent being the special tri-tangent) and the polar quadric of any of these points contains the special tri-tangent through it. When A '" B = C = 0 = 1 the polar quadrics of all points in N have rank 2, one of the planes in which they split being tangent to F (the pairs of planes do not intersect at the special tri-tangents). The planes of N are: vAx+vBy
+VCz
3-
+ ~O t
=D
33'A : V A+l x + V By
3+v Cz
+ qO t
=
0
3-
+ qO t
=
0
=
0
'IT
3-
3-
"
3-
3"B ::;;Ax
+ v B+1 y + v C z 3-
3-
+vBy
3-
3-
'ITC :vAx
3-
3-
3-
3-
+ V C+ 1 z + V' D t
lTD :V'Ax + vB Y + v C z + V' D+ 1 t = 0 and the hexads are the following ones: 2
3
4
5
6
3
5
6
8
9
10
2
4
6
7
9 10
4
5
7
8
TI
"D TIC: '"B:
10
273
On cubic surfaces over a field of characteristic 3
2
3
7
8
9.
Finally, the lines of N (i.e. the lines containing the triples of collinear points) are: 1TD n 1TB
5
8
10
5
1TD n 1TA
3
8
9
1T n 1TA
2
3.
1T n 1TB
4
1T n 1TC
2
4
6
1TC n 1TB
4
7
10
1T n 1TD
3
5
6
1TC n 1TA
2
7
9
1T Dn 1TC
6
9 10
1TB n 1TA
7
8
Generally, the planes of N are not tri-tangent planes to
F',
however, in the
family of all cubic surfaces having the same five nuclei there is a surface
~~
F
to
which all planes in N are tri-tangent planes (but they are not the only tritangent planes).
i:
The equation to F is:
3 3 3 3 x + y + z + t + xyz + xyt + xzt + yzt
(9.1 )
=0
An easy but tedious computation shows [4) that the five tri-tangent planes ,~
of N contain exactly 15 lines belonging to F and these lines are the only lines on F": over the prime field of K. Now a design to describe Nwill be defined. First, remark that any hexad of coplanar points of N, taking as blocks the triples of collinear pOints, is a PBIBD with two association classes and parameters (v,b,r,k,A ,A ,n l ,n 2) = (6,4,2,3,1,0,4,1), two points being first associates l 2 if they belong to a block and second associates if they do not [2). The definition of a composed PBIBD (which up to the authors' knowledge is new) is the following one. A set of p PBIBD's with the same parameters (b,v,r,k,A ,A ,n ,n ) and two l 2 l 2 ~Iith parameters (p,B,r,b,A ,A , l 2 Nl ,N ) and two association classes if the set consisting of the B blocks of the p 2 given PBIBD's is a PBIBD (whose points are exactly the blocks of the p PBIBD's)
association classes is a CPBIBD (Composed PBIBD)
containing p blocks of size b, such that through any point there are r blocks. Consequently, the parameters of a CPBIBD must satisfy the following equalities: (9.2)
vr
= bk,
nl + n2
=v -
(9.3)
Br
= pb,
Nl + N2
= B-1,
1, nlA l + n2A2 N1Al + N2A2
= r(k
- 1) ,
= r(b
- 1)
274
M. de Finis and M.J. de R esmini
(Of course, starting from p PBIBD's with m association classes, and the same parameters, a CPBIBD with m association classes can be defined.) When
N
is considered, the p PBIBD's are the already defined 5 PBIBD's
consisting of the coplanar hexads; now, call points the blocks (triples of collinear points) of N, first associates two points on the same block (i .e. two coplanar triples of collinear points) and second associates two points through which no block exists (i.e. two non-coplanar triples of collinear points); thus, a CPBIBD with parameters p
5,B
10,b
4,r=2.A l =1,A 2 =0,N l
(v
6, k
3, ;'1
1, "2
= 0,
n
l
= 4,
n
2
6,N 2
3
= 1)
is obtained, and (9.2), (9.3) hold.
10. ON CUBIC SURFACES WITH FIVE NUCLEI IN PG(3,3) Usually, when cubic surfaces are investigated in PG(3,q), the assumption q >3 is made [5],
[7),
since lines on the surface are looked for. On the other
hand, the existence of configuration N motivates the investigation of cubic surfaces in PG(3,3). PROPOSITION 10.1: In PG(3,3) there are but two projectively inequivalent cubic surfaces having five nuclei, none of them on F.
The surfaces of the first type
contain exactly 25 points over GF(3); namely, the 10 points of N and the 15 points in PG(3.3) not belonging to any tine joining two nuclei.
The planes of N are the
t!'i-tangent planes to F and together they contain aU 15 points of F, 9 on each plane (two planes sharing exactly 3 points on F).
The lines of N do not belong
to F.
The cubic surfaces of the second type contain exactly 16 points; these points belong to three out of the five planes of N; these three planes have a unique common point and each of them intersects F at a line and an irreducible conic.
The three lines (one on each plane) are coplanar, their plane not being a
pZane of N, and have a conmon point. It; points on F.
Together these tines contain 10 out of the
The remaining 6 points be long to the three conics.
The planes
of these conics pairwise meet at secant lines of the conics; thus, two planes and the conics on them (which have two conmon points) determine the third plane (the
On cubic surfaces over a field of characteristic 3
latter being the plane through the unshared pairs of points). belong to the same elliptic quadric. just six points of N.
275
The three conics
Each of the other two planes of N contains
Therefore. F consists of the points on three plane
section~
by three planes through the same point p. of an elliptic quadric (the planes pairwise meeting at secant lines of the quadric) and of the points on three concurrent coplanar lines on a pZane through p which is tangent to the quadric at a point not in N.
PROOF: Since the proof consists of tedious computations and checking [4), here only the canonical forms for the surfaces will be given. Taking as nuclei the vertices of the reference simplex and the unit point, the equation of the 25point surface is ( 10.2)
3 3 3 3 x + Y + z + t + xyz + xyt + xzt + yzt
0,
while the equation of the 16-point surface is either (10.3)
3 x3 + y3 - z
- t 3 + xyz
3 3 x3 - y - z
- t 3 + xyz + xyt + xzt + yzt = 0 •
+
xyt + xzt + yzt = o
,
or (10.4 )
For the sake of completeness, the main results on cubic surfaces in PG(3,3) with five nuclei and singular points will now be stated (for more details see [ 4 J).
(i) Four of the five nuclei belong to F. All such surfaces are projectively equivalent and the equation of F (choosing coordinates as in prop. 10.1) is (10.5 )
xyz + xyt + xzt + yzt
=0 .
F has exactly 19 points and 9 lines; N degenerates, its points being the 4 conic nodes and its planes the 4 planes joining them by threes. (ii) Three of the five nuclei belong to F. All such surfaces have three conic nodes and are projectively equivalent to (10.6)
x3 + xyz + xyt + xzt + yzt = 0 ,
whose conic nodes are O2 , 0 , 0 , 3 4 F contains exactly 16 points and these together belong to two planes.
The
plane joining the three conic nodes contains 9 points (on the lines joining pairs
276
M. de Finis and M.J. de Resmini
of conic nodes); the other 7 points belong to another plane (namely, y
z
+
t
+
=O~
and their configuration is shown in the picture at the left.
(The two planes meet at a
line not belonging to F.l
N degenerates; its points are the three conic nodes and the point p on the unique special tri-tangent line (not through any singular point); its planes are the plane through the three singular points and the three planes projecting from p pairs of conic nodes. (iii) Two of the five nuclei belong to the surface and are conic nodes.
All
such surfaces are projectively equivalent to 3
3
x - y
(l0.7)
+
xyz
+
xyt
+
xzt
+
yzt
0,
having its conic nodes at the points 03 and 04' F has just 12 points, four of them on the line
t
through 03 and 04'
The
other points belong to two planes through a point p on t; these planes meet at a line tangent to F at p and intersecting F at q # p. through
t
Two of the four planes
contain two points of F not on t; the remaining two just one point on F.
N degenerates: there are just 5 points of N, but all five planes exist; the three points of N distinct from the conic nodes are collinear. (iv) The surface contains just one of its five nuclei. therefore this point is a conic node. inequivalent types of such surfaces.
Assume it is 04;
There are exactly two projectively The only possible equations of a cubic
surface wi th five nuclei, one of them on the surface, are (l0.8)
x3
(10.9)
3
x
(10.10)
+ +
y3 3
+
3
z
3
y - z
3 3 3 - x - y - z
+
xyz
+
xyt
+
xzt
+
+
xyz
+
xyt
+
xzt
+
+
xyz
+
xyt
+
xzt
+
°, yzt = ° yzt = ° yzt =
Equations (10.8) and (10.9) yield two projectively equivalent surfaces, while equation (10.10) yields another type. The surface (10.8) contains exactly 10 points. points coinciding in 04; there are only 4 planes.
N degenerates, four of its
The 9 points on F distinct
from its conic node belong to two planes; each of them contains a triple of collinear points - the same for both planes - and a triple of non-collinear One of these planes contains three of the five nuclei, the other none.
point~
(The same
On cubic surfaces over a field of characteristic 3
277
is true for the surface (10.9).) Also the surface (10.10) contains exactly 10 points, but the points distinct from its conic node are precisely the points on three non-concurrent coplanar lines, which are the only lines on the surface.
BI BLIOGRAPHY 1. 2. 3. 4.
5. 6. 7.
W.H. Blytbe, liOn modeZs of mAhia surfaaes", Cambridge U.P. 1905. R.C. Bose, Graphs and Designs, in "Finite Geometria Struatures and their AppZiaations", CIME (1972), Ed. Cremonese, Roma, 1973. W. Burau, AZgebraisahe Kurven und FZaeahen, Band II, W. De Gruyten and Co., Berl in 1962. M. de Finis and M.J. de Resmini, Sulle superficie cubiche in uno spazio proiettivo tridimensionale sopra un campo a caratteristica 3, Quad. Sem. Geom. Comb. n. 36, Maggio 1981, 1st. Mat. "G. Castelnuovo", Univ. Roma. J.W.P. Hirschfeld, Classical configurations over finite fields: I. The double-six and the cubic surface with 27 lines, Rend. Mat. e AppZ., 26 (1967), 115-152. J.W.P. Hirschfeld, ''Projeative Geometries over Finite FieZds", Clarendon Press, Oxford, 1979. J.W.P. Hirschfeld, Cubic surfaces whose points all lie on their 27 lines, in P.J. Cameron et al. (ed.), Finite Geometries and Designs, Cambridge 1981, pp. 169-171.
Istituto Matematico "G. Castelnuovo" Citta Universitaria 00185 Roma Italy
This Page Intentionally Left Blank
Annals of Discrete Mathematics 18 (1983) 279-282 © North-Holland Publishing Company
279
SOME CHARACTERIZATIONS OF FINITE 3-DIMENSIONAL PROJECTIVE SPACES AND AFFINOPROJECTIVE PLANES A. Delandtsheer
A lineaP space is a non-empty set of elements called points together with a family of distinguished subsets called lines such that any two distinct points are contained in exactly one line, each line containing at least two points. A linear space is said finite if it has a finite number of points. A transversal of two lines Land L' will be any line intersecting LUL' in two distinct points. The generalized projective spaces may be defined as the linear spaces in which for any two disjoint lines Land L', any point p 4 LUL' is on at most one transversal of Land L' (the dimension being';; 3 iff "at most one" is replaced by "exactly one").
The finite affine planes of order n have a similar property: for
any two disjoint lines Land L', any point p 4 LUL' is on exactly n transversals of Land L
I.
It is natural to try to classify all finite linear spaces having such a property. THEOREt1 (A. Beutelspacher and A. Delandtsheer, [1]): Let S be a finite linear space.
If there is a non-negative integer t such that for any two disjoint lines
L, L of S, any point p outside L and L is on exactly t transversals of Land L 1
1
I,
then one of the following possibilities occurs: (i) S is a generalized projective space (moreover, if the dimension of S is
at least 4, any line has exactly two points)
(ii) S is an affine plane, an affine plane with one point at infinity, or a punctured projective plane
(iii) S is the Fano quasi-plane (i.e. the linear space obtained by breaking one line of PG(2,2) into three lines of two points).
Note that in case (ii), S is always an affino-projective plane, that is a linear space obtained by deleting any number m of collinear points from a projec-
280
A. Delandtsheer
tive plane of order n.
S is an affine plane if m = n+1, an affine plane with one
point at infinity if m = n, and a punctured projective plane if m = 1.
A similar problem is to classify the finite linear spaces satisfying the following condition (1,) for any two intersecting lines Land L', any point outside Land L' lies on
exactly t transversals of Land L'. PROPOSITION: If S is a finite linear space containing at least two lines and for JcJ,:/ic;: -:;her"! is all hlteger t such that ("') is satisfied, then
(i) S i3 a degenerate projective plane
(ii) S is a Steiner system S(2,k,v)(k;;' t+l)
such that
(1) ti(k-1)(k-2) (2) kil (k-1)(k-2)/t+21l (k-1)(k-2)/t+1J (3)
t(2k-2-t)ik(k-1)2(k-2)
hi particuZai>. S is a projective plane if t
k-l and S is an affine plane
Conditions (1) and (2) are obtained by easy counting arguments.
Condition
(3) is a consequence of the existence of a partial geometry (k,k-1,t) defined as follows: given a line L of S, the points of the partial geometry are the points of S which are not on L and the lines are the restrictions to S-L of the lines of S intersecting L. Brouwer has observed that a symmetric 2-design is obtained by taking two points p and q of S, by call ing "points" the lines through p distinct from the line pq and "blocks" the lines through q distinct from pq and by saying that a point and a block are incident iff the corresponding lines intersect.
If we apply
the Bruck-Ryser-Chow1a theorem to this symmetric 2-design St(2,k-1,(k-1)(k-2)/t+1) we get a fourth condition on the parameters t and k. These four conditions together rule out many values of t and k but there remain infinitely many values which satisfy all those conditions, for example t
= gn and k = (t+1)(t+2)/2. In order to get not only affine and projective planes but also affine and
projective spaces of any dimension, it seems natural to replace in ("') "t trans-
Characterizations of finite 3-spaces and planes
versals" by "0 or t transversals".
281
There is no hope that this condition charac-
terizes affine and projective spaces.
However, it is possible to prove:
PROPOSITION: Let S be a finite linear spaae suah that (a) for any two interseating lines Land L', the set of points which are on
L U L' or on at least one transversal of Land L' is a linear subspaae of S (b) there is an integer t suah that for any two intersecting lines Land L', any point outside L and L' is on exaatly 0 or t transversals of L and L' Then
(i) if t
~
2, S is a planar spaae all of whose lines have the same number
of points and in which all planes satisfy condition (*) In partiaular, if t = k-l S is a projeative space,if t = k-2 S is an affine spaae
(ii) if t = 1, S aonsists of disjoint pieaes joined by lines of two points, these pieces being - planar S(2,3,v.) all of whose planes are affine 1
- lines having more than two points - points
(iii) if t
0, S is a S(2,2,v).
Another similar problem is to classify the finite linear spaces for which there is a positive integer s such that for any two disjoint lines Land L', any point outside Land L' is on exactly s lines disjoint from both Land L'.
THEOREM: If S is a finite linear space for whiah there is a positive integer s such that for any two disjoint lines Land L', any point outside Land L' is on exaatly s lines disjoint from both L and L', then one of the following oaaurs:
(i) S is a projeative spaae of dimension d
~
3
(ii) S is a degenerate projeative plane, a generalized projeative spaae aonsisting of two disjoint lines of the same size or a trivial linear spaae
S(2,2,v) (iii) S is an affino-projeative plane (but not an affine plane with one point at infinity)
(iv) S is an affine plane from whiah either one point or one line has been deleted.
The proofs of these theorems will be published elsewhere.
282
A. Delandtsheer
BI BLIOGRAPHY
1.
A. Beutelspacher and A. Delandtsheer, A common characterization of finite projective spaces and planes, European JOUPnal of Combinatories (1981) 2, 213-219.
Universite Libre de Bruxe11es Campus Plaine, C.P. 210 Boulevard du Triomphe, B - 1050 Bruxelles Belgium
Annals of Discrete Mathematics 18 (1983) 283-290 © North-Holland Publishing Company
283
ON A PROBLEr~ OF A. KOTZIG
J. Denes
In 1975 during the Graph Theory Conference in Oberwolfach, Prof. A. Kotzig suggested to me the following problem: Let Q be a finite quasigroup, a graph TQ corresponds to Q in the following way: The vertices of TQ are labelled by the elements of Q and an edge is directed from a to b (a,b E Q) if and only if b=a -1 l l where a- denotes the local right inverse of a, i.e. ae = a, aaea' What a kind of properties of Q can be characterized by T ? Q Obviously TQ is a graph which has been called a transformation graph, see [31.
A regular transformation graph is called a permutation graph.
In several
papers transformation graphs have been studied in detail and all our results can be found in [41.
For definitions of loop, isotopy et cetera the reader is
referred to [ 51. In 1976 a computer search was set up.
The graphs corresponding to some
quasi groups of order 6 (which are the class representatives, for definitions see
[ 51) were determined. These graphs will be published in [41. No definitive solution was obtained by the computer investigation.
In the
meantime we had correspondence with Dr. A.D. Keedwell and that inspired the present author to prove the following: THEOREM 1: Let Q be a finite loop of order n, then there exist isotopiaquasigroups Q=Ql,Q2, ••• ,Q(n-l).1 with right identity element suah that TQ ,TQ , ... TQ 1 2 (n-l)! are different permutation graphs with n labelled vertiaes. (The isotopes are obtained using (n-l)! aolumn permutations).
PROOF: Let us consider a latin square which is the multiplication table of loop Ql of order n. The isotopes Q2 ••• Q(n-l)! of Ql will be quasi groups with the same right identity element if we permute n-l columns and fix the column where the border
el~
284
J. Denes
ment is the right identity element.
(See Figure 1. where the case when n=3 is
Let Li i=2.3 ••••• (n-1)! denote the multiplication table of 0;.
exhibited).
If
the cells which are occupied by the right identity element are replaced by 1 and all others by O. then to Li there corresponds a permutation matrix P.1 and P1'P 2 •••• 'P(n-1)! are all different. It is easy to see that Pi is the adjacency matrix of TO. and this completes the proof. 1
'1
2
3
2
3
0
0
1- 1=1. 2- 1=3. 3- 1=2
2231---+00 3
2
3
o
o
o
() 3
2 3 3
o2 =
2
2
2 3 --
0
o
0
o
332
0
0
123 T
O2
000
=
Figure 1. REMARK: It is easy to find an example of a quasi group 0 such that TO is not a permutation graph (see Figure 2.).
I
1 2 3 '
2 3
0
2 3
2 3
T • Q
3
2
Figure 2.
V
3
285
On a problem of A. Kotzig
As we have seen by theorem 1. isotopism does not keep the Kotzig graph of a quasigroup. Theorem 2 will show us that this is not the case if isomorphism is considered instead of isotopism. THEOREM 2: Let Q and Q2 be two isomorphic quasigroups, then TQ and TQ are isol 1 2 morphic graphs, (i.e. graphs which are equal, apart from labelling).
On the con-
trary if Q and Q are not isomorphic, TQ and TQ might be isomorphic.
l
PROOF: Clearly if
2
1
2
denotes the isomorphism between Ql and Q2 then for every a E Ql' ~(a) = ~(a e ) = ~(a) ~(e ) holds. This implies that e ( ) = ~(e ) -1 a_ l -1 -1 cp a a a = ~(a a ) = ~(a) cp(a ) as a consequence ~(a ) = cp(a) holds and TQ and TQ ~
1
2
are isomorphic.
It is obvious that TQ and TQ are isomorphic if Q and Q2 are l 2 two non isomorphic groups of order n ifl n is a composite odd integer. and this
completes the proof. Theorem 3 will show us that there are other types of TQ graphes than implied by theorem 1. To formulate theorem 3 we shall need some definitions. (see also [51): a transversal of a latin square of order n is a set of n cells one in each row. one in each column. and such that no two of the cells contain the same symbol. A transversal is called cyclic if it has the form
THEOREt·l 3: If n is odd, n
~ 0
mod 3 then there exists at least one quasigroup Q
of order n such that TQ is a single cycle.
PROOF: For n odd and n ~ 0 mod 3 such a quasigroup (Q •. ) always exist. h.j = 3h-2j+4(mod n).
Then h·(h+l)
a
We define
h+2 for h = 0.1 •...• n-l defines a cyclic -1
transversal and we have h·(h+2)=h. so eh = h+2 and h
= h+l.
For fUrther results on cyclic transversals see [51. especially Theorem 9.1 .7. of [ 51.
It is easy to show that there exist quasi groups whose Kotzig graph is different from the ones given in Theorem 1 and 3. One can use (to obtain such
286
J. Denes
quasigroups) some results of K. Heinrich see
(7).
(9).
In [7) K. Heinrich gave a
partial solution of a problem of L. Fuchs who has posed the following problem (see [5J problem 1.9.): If n is any positive integer and n=n +n +••• +n
any fixed parl 2 k tition of n. is it possible to find a quasigroup Q of order n which contains sub-
quasigroups Ql' Q2'···.Qk of orders nl , n2 •• ··, nk respectively. whose set theoretical union is Q? K. Heinrich gave necessary and sufficient condition for the affirmative reply in the case when k
~
which is not necessary in the case k > 4. the same subject very recently see (9).
4.
She gave a sufficient condition
Also she gave some further results on In [9] she wrote the following: "Denote
by RP(a,b;s,t), a < b, a latin square of order n=as+bt with s disjoint subsquares of order a and t of order b.
We show that RP(a,b;s,t) exist for all s
that RP(a,b;l,t) exist if and only if t s
~
3 and (5-1) a
that for all t
~
~
~
~
3, t
~
3;
3; that RP(a.b;s.l) exist if and only if
b; and that there exist integers T=T(a.b) and S=S(a.b) such
T and s
~
S, RP(a.b;2,t) and RP(a.b;s.2) exist".
It is worth remarking that the fact that a latin square has no kxk latin subsquares is not equivalent to the corresponding quasi group having no subquasigroups of order k.
While the quasi group corresponding to a latin square with no
kxk latin subsquare has no subquasigroup of order k, it is quite easy to construct quasigl'oups having no subquasigroups of order k whose multipl ication tables contain kxk latin subsquares. In (5) (p. 486) there is a problem which asks for all sufficiently large n there exist quasi groups of order n which contain no proper subquasigroups.
This
problem (which has a trivial affirmative solution see (12). [17] and (18). has been attributed erroneously to A.J.W. Hilton). wanted to be proved.
In [10] one can find what Hilton
He wrote: "It is probably true for most orders n there is a
latin square with no proper sub-latin square of order greater than 1; however, Dr. A.D. Keedwell has pOinted out to me that this is not true for n=4.
A partial
solution of Hilton's problem can be found in [8) where K. Heinrich proved that for orders n=pq where p and q are distinct primes and n
F 6. there exist latin squares
which contain no proper latin subsquare. A restricted version of Hilton's problem has been suggested and investigated in 113). Namely is it true that for every n there exists a latin square L with no latin subsquare of order 2? The reply is in the affirmative, if n F 2,4. \61,
(7),
[13], [14J, 115}. [16J).
(see
There exists an obvious connection between
latin squares without subsquares of order 2 and the Kotzig graph.
If L represents
287
On a problem of A. Kotzig
the multiplication table of a quasigroup 0, then the necessary and sufficient condition for L have no latin subsquare of order 2 is that every isotope of 0 is without subquasigroup of order 2.
With the aid of Theorem lone can prove Theorem
4. THEOREr~
4: If L denotes a Zoop of order n. then there exist quasigroups with Zeft
identity isotopic to L such that they have no proper subquasigroup.
PROOF: A sufficient condition for a quasi group 0 not to have proper quasigroup, is that TO has a cycle of length m(m >[ a
E
%]).
To prove this let us observe that if
0 is contained in a subquasigroup so does its left inverse. Further it is
well known: if n is any positive integer then an upper bound for the order of a subquasigroup of a quasigroup of order n is
[%]
(the greatest integer in
%),
(see
e.g. Lemma 8 of [18] and Theorem 1.5.3. of [5]). By the proof of our Theorem 1 it is immediate that for any given loop L of order n there exist isotopic quasigroups such that their Kotzig graphs contain cycles of length m(m > [~]). REMARK: One can use Theorem 3 instead of Theorem 1 to prove Theorem 4. If R denotes quasigroup of order n such that its multiplication table is a latin square which has no proper sub-latinsquare then every quasigroup isotopic to R is such that it has no proper subquasigroup. A sufficient condition for the latter property is that TRW
(R* denotes an arbitrary isotope of R) contains a
n
cycle of length m(m > [2]). This sufficient condition never fullfils, this is implied by Theorem 1. A groupoid G is said to have the right inverse property if, for each x in G, there is at least one b such that (xy) b=y for every y in G. The left inverse property can be defined analogously.
When a groupoid G has the right and the left
inverse property, then is said to have the inverse property. In the case of quasi groups the definitions above can be used too.
The read-
er can find a detailed study of quasi groups and loops with the inverse property (I.P. quasigroups) in [18],
[l~.
The best known I.P. quasi groups other than
groups are those characterized by the mild associative law a(b.cb) = (ab.c)b and the existence of a unique two sided identity element.
Finite quasigroups of this
type were introduced by R. Moufang by whose name one usually refers to them. A
288
J. Dl!nes
Moufang loop of odd order satisfies both Lagrange's theorem and Cauchy's theorem (see e.g. (5) p. 63}.
Consequently, the orders of all its elements and subloops
divide the order of the loops. A loop which satisfies the identity [(ab).c] .b=a[ (bc).b] is called a
BoZ
loor and such a loop satisfies the weak form of Lagrange theorem (namely that the
order of every element divides the order of the loop) (see e.g. [5) p. 32).
Since
Bol loops have the one sided inverse property, every Moufang loop is also a Bol loop. It is easy to see that if Q is a quasigroup with the right inverse property then TQ is a permutation graph.
Since every loop isotope of a Moufang quasigroup
is Moufang (see e.g. Corollary of Theorem 9 in [18J). if L is a loop isotope
R. Artzy independently from A. Kotzig defined and investigated the graph which exhibits the mapping which maps every elequasigroup, then TL is a permutation graph.
ment of a loop into its right inverse (see
[1
J).
Since he investigated loops with
one sided inverse property (he called them weak-inverse (WI) loops) his graphs are permutation graphs.
(It is a special case of the Kotzig graph).
Artzy proved
that if L is WI loop then \ consists of one sling (cycle of length 1) and m cycles of length k, such that k divides 2m, (see [11, [21).
WI loops which are
not Moufang are scare in the literature. One can find construction of non-Moufang WI loops in [ 11). Another special class of loops which have been studied by Artzy is the crossed inverse (Cl) loop.
Here we assume that to each x in G there corre-
spond a, b in G such that (xy)a=y=b(yx) for every y in G.
Artzy proved that a
necessary condition for a WI loop L not to be CI that TL consists a single cycle of length 1 and m cycles of length k such that k is a divisor of 2m. We would like to suggest a problem which can be consirlered as a semi group theoretical analogue of Kotzig's problem.
Let S be a finite semi group.
Since
every abstract semigroup of order n can be represented isomorphically as an appropriate subsemigroup of the symmetric semigroup of degree n+1 (F n+ 1)' without loss of generality we may suppose that S is a transformation semigroup. Every element of S could be represented as a transformation graph.
The restricted transforma-
tions to the elements in the cycles of the transformation graph will be called main permutation (really the graph representation of the main permutation is a permutation graph).
The main permutation of
a E
Fn will be denoted by f{a}.
1 2 3 4 5 6)
example, if a = ( 2 2 3 5 6 5
its graph representation is
For
On a problem of A. Kotzig
1
289
3
~O
Since the main permutation of is f(a)
_ (2 356) ,. ts 236 5
,,'ph "p""ot,t;oo ;, "follow,
c3
~
\5 J
6
The quasi inverse of a is the least power a k such that f(a)-1
k = f(a).
This notion
has been introduced in [3] and for further results see [4]. PROBLEM: Let S be a finite semigroup of order n, a graph TS corresponds to S in the following way; the vertices are labelled by the elements of S and an edge directed from a to b (a,b
E
S) if and only if b is the quasiinverse of a.
What
kind of properties of S can be characterized by TS? REMARK: Clearly TS is a transformation graph, but no further results could be obtained by the author. My sincere thanks to Mr. P. Ablonczy, Mr. B. Biro, Dr. A.D. Keedwell who kindly read the manuscript and made many helpful comments.
BI BLIOGRAPHY 1. 2. 3.
4. 5. 6. 7.
R. Artzy, Inverse cycles in weak inverse loops, Froc. Am. Math. Soc., 68 (1978), 132-134. R. Artzy, A functional equation solved by loops. Lecture given at the Functional Equations Conference at Graz, September 1978. For summary see Aequationes r4ath. 19 (1979), J. Denes, Connections between transformation semigroups and graphs, Theorie des graphes. Journee internationales d'etude. Rome, Juillet 1966. Dunod, Paris, Gordon and Breach, New York, 1967, 93-101. J. Denes, Transformations and transformation semi groups. To appear. J. Denes and A.D. Keedwell, Latin squares and their applications. Akademiai Kiado Budapest, Academic Press New York, English Universities Press, London 1974. R.H.F. Denniston, Remarks on Latin squares with no subsquares of order two, Utilitas Math., 13 (1978), 299-302. K. Heinrich, Latin squares composed of four disjoint subsquares, Combinatorial mathematics V. (Proc. Fifth Austral. Conf. Roy. Melbourne Inst. Tech., Melbourne 1976). pp. 118-127. Lecture Notes in Math. Vol. 622. Springer, Berlin, 1977.
255.
290
8. 9. 10.
11. 12. 13. 14. 15. 16. 17. 18. 10
J. Denes
K. Heinrich, Latin squares with no proper subsquares. Journal of Combinatorial Iheory A. 29 (1980), 346-353. K. Heinrich, Disjoint subquasigroups. To appear in Proc. London Math. Soc. A.J.W. Hilton, On the Szamkolowicz-Doyen classification of Steiner triple systems, Proc. London Math. Soc., 34 (1977), 102-116. K.W. Johnson and B.L. Sharma, Construction of weak inverse property loops, Rocky Mountain Journal of Mathematics, 11 (1981), 1-8. T. Kepka, A note on simple quasigroups, Acta Univ. Carolin. - Math. Phys., 19 (1978) no. 2, 59-60. A. Kotzig, C.C. Lindner and A. Rosa, Latin squares with no subsquares of order two and disjoint Steiner triple systems, Utilitas Math., 7 (1975), 287-294. A. Kotzig and J. Turgeon, On certain constructions for latin squares with no latin subsquares of order two, Discrete Mathematics, 16 (1976), 263270. M. Mc. Leish, On the existence of latin squares with no subsquares of order two, Utilitas Math., 8 (1975), 41-53. M. Mc. Leish, A general construction of latin squares with no subsquares of order two, Ars Combinatoria, 10 (1980), 179-186. N.S. Mendelsohn, Letter to the author dated October 12, 1976. R.H. Bruck, Some results in the theory of quasigroups, Trans. Amer. Math. Soc., 55 (1944), 19-52. R.H. Bruck, Contributions to the theory of loops, Trans. Amer. Math. Soc., 60 (1946), 245-354.
Csaba utca 10. 1122 Budapest Hungary
291
Annals of Discrete Mathematics 18 (1983) 291-294 © North-Holland Publishing Company
A SMALL 4-DESIGN R.H.F. Denniston
What is exhibited here is a t-design (repeated blocks not allowed) for which t=4,
k=5, v=12, A=4.
Infinitely many 4-designs are known, and a single one would
not as a rule deserve to be published; but this construction is so easy to grasp intuitively that is seemed to be worth talking about at the conference.
If a 4-
design with these parameters could be extended to a 6-design, that would be more interesting - but in fact this particular one cannot be extended. Our set of 12 points is divided into two hexads, which are represented in the tables as above and below a line. the hexads; where one
x
There is a one-one correspondence between
appears directly above another, it signifies that two
corresponding points belong to the subset with which we are concerned.
Each
hexad is unordered, but has a structure imposed on it which may conveniently be described in the language of Sylvester [2].
He used the term duad for an unor-
dered pair, and syntheme for a partition of the hexad into three duads; and he observed that a famity of five synthemes can be found which partitions the collection of 15 duads.
So, on our upper hexad, let us choose and fix one family (out
of the collection of six families that in fact exists): and let us use the one-one correspondence to carryover that structure to the lower hexad.
If we describe
two duads as "parallel" when they belong to the same syntheme of the chosen family, we have an axiom of a familiar kind:- Given a duad and a point not in it, we can find just one parallel duad that includes the point. We can now partition the whole collection of 5-subsets of our 12-set, as shown in Table 1, into two subcollections Land R.
This is done mainly by asking
how many pOints of a given 5-subset have been chosen from the upper hexad, and how many pairs of corresponding points are contained it it.
If, for instance, there
are two upper points and two pairs, the 5-subset goes into the subcollection Rand is said to be of type R4 ; two upper points, with no pair, give a 5-subset of type L . When, however, we come to two upper points with one pair, we need to discrim7
292
R.H.F. Denniston
inate by means of our fixed family of synthemes.
Namely, if the duad of upper
pOints and the duad of non-corresponding lower points are parallel, we have a 5subset of type L5 , otherwise of type R . Table 1 puts this in evidence by means 6 of vertical lines separating the duads of one syntheme. We proceed to show that, of the eight 5-subsets in which any given 4-subset is contained, four will be found in L and four in R.
In fact, the details, for
all the types of 4-subset that we can distinguish, are exhibited in Table 2. Discrimination between types L5 and R6 is easy in the second case on the right of Table 2. where the 4-subset includes a single upper point and also includes the corresponding lower point.
Namely, considering the two non-corresponding lower
points as a duad, we assert that just one parallel duad includes the single upper point, and therefore that our 4-subset can be embedded, in just one way, in a 5subset of type L5. In the next case down, where the single upper point has its corresponding lower point outside the 4-subset, we look at the duad of upper points with the property that the subset excludes them and also excludes their images below.
For
a 5-subset of type L , such a duad exists, and is parallel to the duad of chosen 5 upper points - but not parallel for type R6. And so we can make a similar assertion to the one at the end of the last paragraph. The other cases in Table 2 are easy to establish. that the ccllection L of 5-subsets
And the conclusion is
(and likewise, of course, the collection R) is
a 4-design lJi ttl the I'equiY'ed parameteY's.
The question remains whether we can be sure that L is not a contraction of a 5-design (k=6. v=13. A=4).
We can, however, apply a theorem proved by Alltop [1.
p. 184] about a (hypothetical) 5-design (Q,O) with these parameters, a point a of the point set
and a block
Q,
~
the six 5-subsets of the 6-set
(not including a) of the block set ~,
o.
Namely, of
either two or three are blocks of the contrac-
tion with respect to a, a 4-design which Alltop denotes by (Q ,Oa). a
But now (I go on from his theorem) suppose 6 is a 6-subset of not a block of O.
Q
a
which is
Then we observe that the complement C of 0 (in the collection
of all 6-subsets of Q) is another 5-design with the same parameters, and the a contraction C of C with respect to a is the complement of Va in the collection of 5-subsets of of
6,
Applying the theorem to C, we have that, of the six 5-subsets either two or three belong to Ca , and either four or three to Oa. Q .
a
293
A small 4-design
TABLE 1 L
R
x x x x x x x x x x
LZ
x
•
RZ
x x x x
x x x x
x • •
x x x x
• x
x
x x x x
x x x
x x
x x
x x x
RS
LS LS
x x x
x x
x x x
x x
TABLE 2 x x x x x x x x x x x x
2Ll. 2L3. 4RZ
x x x x
4LZ. 2Rl. 2R3
2Ll+. 2LS. 3Rz. IRs
x • x x x
3Lz. lLS. 2Rl+. 2Rs
2L3. 2L6. lRz. IRs. 2R7
• x x x x
ILz, IL S' 2L 7• 2R 3• 2R6
x X x x
4LlIJ 4Rl+
x x x x
ILl+. lLS. 2L 6 • 1RlIJ IRs. 2RS
t
x xl' • x x •
2LS. 2L7. 2RS. 2R7
x x I' '1' • x • x
2L6 • 2L7. 2R6. 2R7
294
R.H.F. Denniston
So we have a corollary to Alltop's theorem.
Suppose, if possible, that a
gl:ven 4-design (k=5, v=12, 1.=4) is a eontX'aetion of a S-design.
Then the nwnbeX'
of bloeksaontained in any given hexad must be two, thX'ee, oX' foUX'.
Since our L
in Table 1 includes all six 5-subsets of the upper hexad (and R of the lower), this condition is not satisfied.
BIBLIOGRAPHY 1. 2.
W.O. Alltop, Extending t-designs, J. CombinatoX'ial TheoX'y (A) 18 (1975), 177-186. J.J. Sylvester, Note on the historical origin of the unsymmetrical sixvalued function of six letters, London, EdinbuX'gh and Dublin Philos. Mag. and J. Sei. (21) (1861), 369-377.
Department of Mathematics University of Leicester Leicester LE1 7RH, U.K.
295
Annals of Discrete Mathematics 18 (1983) 295-314
© North-Holland Publishing Company
SOME NEW GENERALIZATIONS OF SHARPLY t-TRANSITIVE GROUPS AND SETS M. Deza
Recently the author tried to obtain permutation analogs of some of his previous results on largest set-intersection systems.
This gave rise to a series of
joint works dealing with the three following new concepts: a) permutation geometry (an analog of a matroid) which generalizes sharply t-transitive sets and some Jordan groups; b) (L,n)-sharp groups which generalize sharply t-transitive groups (the case L
{O,1,2, •.• ,t-l}); c) sharply edge-transitive digraphs which approximate a finite projective
plane over near-field, i.e. sharply t-transitive groups. This paper is a short up-to-date survey of results in this direction (without proofs but with some new observations and problems).
1. SET INTERSECTION MOTIVATION At first, I wanted to find a permutation analog of the following THEOREM 1.1 ([6), (7): Let A = A(L,k,v) be a family of k-subsets A. (i.e. 1
IA.I ,
=
k) ofa given v-set V such that any IA., () A.I J
E
L (A.,
~
A.) where J
= P o, 2 , ••• ,2r- l}' 0';; 20 < 21 < ... < Jl r- 1 < k. Then there exists a largest 1 nwnber Vo (L,k) (depending only on L and k) such that v> Vo (L,k) impZies: L
a)
([
6]) I A(L,k,v) I .;;
v - Jl. 11 - - ' ,
,
k - R..
b) ([ 7]) the equality holds above if and only if A(L,k,v) is the famiZy of hyperplanes of a perfect matroid-design (i.e. a matroid such that aZl flats of same rank have same cardinality); (see [11) and references to it for perfect matroid-designs).
296
M. Deza
The three most interesting (and most studied special cases of A(L,k,v) are: 1) L = {O,l, ... ,t-l}, a code with all distances dH(A.jA.) = IA. 1
- 2t
+
is a
S~;eineY'
J
1
l!.
A.I > 2kJ
lor, in other words, t-packing; the corresponding perfect matroid-design system 5(t,k,v) (which is also the smallest t-covering);
2) L = ~t,t+l , ... ,k-l },an anticode, considered by the famous Erdos-Ko-Rado's theorem; 3}
L = It}, (r,A)-design (r
= k,A
t) or, in other words, equidistant code
or weak D.-system. The first series of papers on permutation analogs concerned analogs of the three cases above : namely [8] for the case 1); [8] and [12] for 2); and [9] and most of its references for the case 3). The permutation analog of A(L,k,v) will be the set B = B(L,n) of permuta,
-1
tlons b.1
E
5n such that any f(b.1
o ~ £o <
~1
b.) J
E
L (b.1 ; b.) where L = {~'~l""'~ l}' J 0 r-
1 < n. Here the number f(b) of fixed points of the rpermutation b is an analog of the cardinality IAI of the set A. Each permutation < ... <
~
5n can be represented by its permutation matrix, i.e. (O,l)-matrix (a c d)nn such that a cd = 1 if and only if b moves c to d. Thus, b can be seen as an n-subset of l the n2-set of all ordered pairs (c,d) of integers (1 ~ c, d ~ n) and f(b: b.) is 1 J the size of the intersection of corresponding n-subsets of this n2-set. 50 any
bE
2 B(L,n) is a special case of a A(L,n,n ).
2 (More exactly, a A(L,n,n ) comes from a
B(l,n) if and only if it has two orthogonal resolutions, see [10]). The function f(b) of argument b caned the ;::'ermutation chaY'actel'. -1
group of 5 we have all f(b. n
f(b)
E
L (b
1
E
b.) J
E
5n is the trace of the permutation matrix,
In the special case when B = B(L,n) is a subE
L (b.,b. 1
J
E
B, b. ; b .) if and only if all 1
J
B, b ; < 1»; we will call any B(L,n) which is a group a B(L,n)-
gl'OUp or 3ubgl'oup of Sn of type L.
The program was to find permutation analogs of Theorem 1.1, i.e. "good" upper bound for \B(L,n)\ such that the case of equality corresponds to some natural subset of S. The group structure of Sn and specificity of B(L,n) as n 2 A(L,n,n ) were expected to give tools (impossible for general A(L,k,v)) and to give tighter characterizations (at least, for the groups of type L). (partially) came about.
This
297
New generalizations of sharply t-transitive groups
v-~.
2. NEW INEQUALITIES AND EXTREMAL CASES
It turns out that the upper bound ~~L k-~~ (for IA(L,k,v)1 in Theorem 1.1) is just a polynomial P(v) such that P(v) ~ 0 iff ~ ELand P(v) = 1 iff v = k; it is the Lagrange interpolation formula for parabolic interpolation. (n-~.)
polynomial Q(n) = RIkL •
I expected the
to be "good" upper bound for some IB(L,n)l.
1
This
happened for two im~ortant special cases (Theorems 2.1, 2.2 below) but it is not
true for the case L = {l}.
(The permutation analog of the Lagrange formula is the
following function 0(b) of argument bEG on a group G
e (b)
=
~
lkL
.
1
(f (b) - L1 1G)
where f(b) = e is the permutation character and lG is the identity (or principal) character of G.
8 (b)
is a generalized character of G because 0(b) = 0 for
bEG, b ~ <1>; 0 «1» =IkL (n-~.).) ~ . 1 1
THEOREM 2.1 ([16] ): If B(L,n) is a group G then IB(L,n)I';;;~lL (n-~i)· 1
In fact, IGI = ~lL (n-~i) / (8, lG)G 1
where ~
(o,
lG)G is the multiplicity of lG in
e (it
is an integer because 8 is a
-linear combination of irreducible characters of G).
We will call a B(L,n)-
group G (L,n)-sharp group if it realizes equality in Theorem 2.1 (i.e. if 0 is the regular character of G).
An (L,n)-sharp group with L = {O,l, ••• ,t-l) is
exactly a sharply t-transitive group. sharp groups in
§
We will collect known information on (L,n)-
3.
THEOREM 2.2 ( [8]): If L = {O,l, ••. ,t-l} then n!
IB(L,n) I ~ ~iIkL (n-q =-t) ( 1 n .I with equality if and only if B(L,n) is a sharply t-transitive subset of S • n
A sharply t-transitive subset of Sn (if it exists) is not only the largest permutation t-packing (i.e. a set B({O,l, ••• ,t-l},n) but it is also the smallest
M. Deza
298
permutation t-covering containing the <1> (i.e. a t-transitive subset of S ). n
This is another analogy to Steiner systems S(t,k,v); we remark that both are Tdesigns (in semi-lattices) of De1sarte. For a general B(l,n) we mention only two following inequalities. THEOREM 2.3 ([10]): a) jB(l,n)j.jB({1,2, •.• ,n-1}\ l,n).;; n! (duality); 2 b) IB(l,n)j .;; n if for some integers k,m (0';; k < m), n f k (mod m) but i; == k (mod m) faY' any g,i
E
l.
If a sharply t-transitive subset of Sn exists, then for l we have max jB(l,n)j
= n!
/ (n-t)
equality; but already for n
=
{O,l} we have max <
6
let us define (by iteration) a "(A
+
of B(l,n). a,b
6, l
, ..• ,b
{0,1, ••• ,t-1}
and the inequality a) above becomes an
= 4! and 18.4!
max IB({2,3,4,5}, 6)1
(1)
=
=
B({0,1},6
=
18,
l)-wise intersection", a generalization
let Be S ; we call it BA(l,n} if for any A + 1 its different elements (A)
E
n
B we have where F(c} is the set of fixed points for any c
Of course a B (l,n) is just a B(l,n). 1
E
S. n
For BA(l,n) we have a generalization of
Theorem 2.2, but not of Theorem 2.1. THEORH4 2.4 ([3]): If l = {0,1, ... ,t-1} then JSA(l,n)j .;; g,~~l (n-g,i) 1
with equality if and only if BA (l,n) is a A-w'lifor'fTl t-transitive set (i.e. for any two t-tY'iples of distinot elements of {1.2 •.•• ,n} there are exaotly A members of this set which oarry the first t-triple to the seoond).
Any t-transitive group is an example of A-uniform t-transitive set. let r
=
(V,R) be a directed graph with vertex set V,jvj
=
n.
let us use
short expressions SETG (corresp. SET) for sharply edge transitive group (corresp. set).
We say that digraph r admits a SETG G (corresp. a SET)
if G is a subgroup
(corresp. a subset) of Aut rand G is regular (i.e. sharply l-transitive) on the set R of edges of r.
Now we give an obvious
299
New generalizations of sharply t-transitive groups
PROPOSITION 2.5: If r = (V.R) admits a SETG (corresp. a SET) G then IRI = IGI
~
n(n-l), with equality if and only if there exists a sharply 2-transi-
tive group (corresp. set).
In fact, IRI = IGI by the definition of SET. means that r
The equality IRI = n(n-l)
n (the complete digraph). i.e. Aut kn = Sn contains a subgroup (corresp. subset) G regular on R. i.e. on the set of all ordered pairs. In other a k
words. G is a sharply 2-transitive subgroup (corresp. subset). We will finish this section with a table of objects which will be considered in this paper: 1) permutation geometry Perm G (L.n) (see L
§ 4
below); in the special case
{0.1 •••• ,t-1} it corresponds to the equality in Theorem 2.2; 2) (L.n)-sharp group (see
§ 3
below); it is the case of equality in Theorem
2.1; 3) digraph r = (V,R) admitting a SETG (or SET); the max IRI will be an IVI=n approximation of sharply 2-transitive group (or set) because of Proposition 2.5. These three objects are treated in So, most results in
§§
§§
3-5 but they are still too general.
3-5 will concern two more special objects:
a) geometric groups (i.e. groups which are both permutation geometries and (L,n)-sharp); b) complete 1-partite digraphs admitting (or not) a SETG. Most of the results of [ 1].
§
3 are in [101. [16], [3], [13], of
§
4 in [10], [3] and of
§
5 in
300
M. Deza
Table 1
~
~--------~=-,
,-~-------,
B(L.n) of size
I
9,
subgroup of
~L (n- 9,.
i
Sn of type L
)
1
~--~ ---:-::"'--:-="-:----; Perm. geometry I' (L.n)-sharp
I Perm
G( L.n)
!
! I
digraph admitting SDG
group
-----------~
I geometric I
9,
group
k • n m
= R.
m,
admitting SETG
case L={O.l ••••• t-i 1
~ 2- t ra ns it i ve T-g-e-om'::"e-:t-r7ic--g-r-ou-p--o-::-f-;
i.e.sh.t-trans. set
geometri c group
type L= {Q ,m }
sharply t-transitive
sharply 2-transitive
group
group
3. (L,n)-SHARP GROUPS In this section we let G be a (L,n)-sharp subgroup of Sn' i.e. it has type L = {i ,ll, ... ,1 1 I (i.e. the number of fixed points f(g) E L for any g E G, o r9 f' < 1 >and 0 .;; .1.0 < £1 < ... < R. _ ) and the order IGI = R. ~L (n-R. ). We suppose r 1 i 1
that G acts in the set Q, I QI = n, and denote by F(G) the set of points fixed by any element of G.
By
~
we denote permutation isomorphism.
It is easy to check that a) G is t-transitive iff (O,l, ... ,t-l}E L;
b) G is sharply t-transitive iff {O,l, ... t-l} = L; c) the stabilizerGaof a point a LI
oE
E
Q is a (L',n-l)-sharp group with
P',-l, 9.2-1' .... l r_l-l} •
The conjecture is that G has 9, o+1 orbits, it is true in the three cases: L (i.e. G transitive). ILl = 1. and G is a geometric group (i.e. G generates a
301
New generalizations of sharply t-transitive groups
permutation geometry, see next
§
4 for the definitions).
We want to classify all (L,n)-sharp groups.
Below in
§
3 we give all known
results; the classification is obtained only for the three following cases: I)
L = p" H 1, ... , Ht-l };
II)
L
III} L
U,},i=l,2,3; {i,H2}, L = {i, H3}.
The subcase i=O of the case I corresponds exactly to sharply t-transitive groups; they were classified by Jordan in about 1870.
They are: Sn (t . = nand
t = n-l), An (t = n-2). t,lathieu groups M12 (t = 5), Mll (t = 4) for t> 3. All sharply 2 (corresp. 3) transitive groups are known and they are exist iff n = pa (corresp. n = pa+l ).
The regular (i.e. sharply l-transitive group) L corresponds
exactly to a Latin square for which the rows form a group. Any cyclic Latin square and, for example, the Klein four-group provide examples but not all such Latin squares are known.
(Sharply t-transitive sets which are not groups we know
for t=l (any n), t = 2 (n = pal t
3 (n = pa + 1); but for larger t the existence
a
of such sets is an open problem).
It was proved in [13] that in the general case
I) we have IF(G)I = i and so G is exactly sharply t-transitive on the remaining n-ipoints of
fl.
In the case II (L = {i}) we know that G has a representation as a ({i},n)sharp group,i > 0, iff G has a G-invariant proper partition (and all such nonsolvable groups were classified).
For i = 1 we have IF(G) I = 1 and so G is
regular on the remaining n-l points.
The case i=2 and i=3 are classified in [14] •
•
For example, all ({2}.n)-sharp groups are A4 , S4. A , the generalized dihedral 5 group and. of course. the case IF(G)I = 2 with G regular on the n-2 remaining points. Let us consider the case III.
For the subcase L = {0.2} the classification
(see [17) is: D8 (n=4). Sn (n=6 with two different representations). GL(2.3) (n=8) and PSL(2.7) (n=14); moreover. G is transitive of rank 3 and G is a geometric group (see
§
4 for the definition).
(In [lJ, [3J many examples of ({O.l }.n)-
sharp groups. geometric or not are given; they are given implicitly in
§§
4-5).
The crucial reduction lemma in [13] says that for any ({i.£+s}.n)-sharp group IF(G) I ;;.
i-l+s~-(s-l)sl. where s,= max (1.[ (s-1)/21).
302
M. Deza
and the equality occurs for s s ;. 5).
= 1,2,3,4 (it is conjectured to be impossible for
Without lost of generality we can suppose (until the end of this section) that F(G) a)
=
0.
The classification of (U,£+s},n)-sharp groups given in
for L
{£,£~2}:
is:
either L = {0,2} (those 5 groups are given above) or
L = {1,3}, G has two orbits (n
=~
U
~I) and G ~,S4 (G~ = S3' G~I = S4' n=7) or
G;?; PSL(2,7) (G~ is 2-transitive of degree 7, G~ n a
[13]
is 2 transitive of degree 8,
15); b) for L = {£,
£~3}
: either L = {O,3} (and n = 6,9,15,24,27) or L = {2,5},
G has 3 orbits and G ;?; (71 3 x Z3) )() Z2 (n = 8). The 5 corresponding ({0,3},n)-sharp groups are: (Z3 x Z3 (Z3 x Z3)
~
~
Z2 (n = 6),
S7 (it has two different representations on 9 points), Z3 x PSL(2,4),
Z3 x PSL(2,7), and (Z3 x Z3 x Z3)
~
S4'
4. PERMUTATION GEOMETRY AND GEOMETRIC GROUPS Let us consider the set (denoted by Sin) of all bijections (in other words, partial permutations, subpermutations) between subsets of the set N = {1,2, ••• ,n}. Si we will denote by ran a its range and by a= dom a its n domain; the inverse a- l of a is the inverse map from ran a to dom a. The com-
For any bijection a
E
position aob of bijections a,b is defined (on its doman a~) by aob(i) = a(b(i)). The set Si with this composition aob (generalizing the composition of permutan
tions) becames an inverse semigroup.
It is called the symmetric inverse semi-
group and any inverse semigroup can be embedded (up to isomorphism) in it. Let a be a bijection (for example a = fl,3,4}
~
{2,3,1}).
We will
represent a in following ways: 1) the vector form, by the vector; (for example,
(a(l), a(2), ..• ,a(n)) with a(i) = 0 if
¢
a
a = (2,0,3,1));
2) the matrix form, by (O,l)-matrix «a .. ))n with a .. =1 if and only if a(i)=j (for lJ
n
lJ
example,
o1
0 0
000 0 001 0 1 000
);
3) in subset form; by the lal-subset A of the n2-set NxN of all ordered pairs
303
New generalizations of sharply t-transitive groups
(i,j) of integers I
~
i,j
~
n, with (i,j) E A if and only if a(i)
a
j (for
exampl~
A= {{1,2}, {3,3}, {4,l}}).
For any two bijections a,b we define now the new operation, meet C = a A b. In vector form, c(i)
=
b(i) if a(i)
=
b(i) and c(i)
=
0 otherwise.
In matrix form,
all Co 0 = min (ao o,bo 0) = ao 0 bo 0' i.e. «co 0)) is Hadamard product of matrices lJ lJ lJ lJ lJ lJ corresponding to a and b. In subset form C = A n B. The set Si with respect of n
this operation becames a semilattice (i.e. commutative idempotent semigroup). Now we define the new partial operation join c
= a v b for some a, b E Si n .
In matrix form, all Co 0 = max (ao o,bo 0) and c is defined if and only if «co 0)) is lJ lJ lJ lJ the matrix form of some bijection, i.e. if any column and any row of «co 0)) has lJ at most one element 1; in subset form C = AU B. We obtain partial the semilattice (Si n ,V) and the partial bisemilattice (S; n ,AlV), let us denote by <0> the bijection (0,0, ..• ,0) (in vector form), i.e. with We call two bijections a,b disjoint if a A b = <0> and
empty range and domain.
a v b is defined (if, moreover, a vb E S , we denote it by a
1
bo; also denote by
allc parallelism, i.e. the existence of b such that alb, c
1
b).
n
bijection a with
I§I= 1, i.e. of a form {I}
~
J
let us consider the following partial order a
~
b on Sin'
We say that a
(i.e. a covered by b) if a is the restriction of b to a subset of subset form, a
~
b means that A ~ B.
We have a
We call any
{j} a point (or singleton).
~
6a
~
b
dom bo; in J
b if and only if a A b
= a and
so the semilattice (Sin,A) with this partial order is a meet-semi lattice. (Actually, the partial bisemilattice (Sin,A,v) is meet distributive, i.e. the meet operation A distributes over the join partial operation v). The partial order a
~
b is a natural partial order in the inverse semigroup (also a-l';;;b- l );
(Si ,0), i.e., a ~ b implies coa ~ cob, aoc ~ boc for any c E Si n
we have a';;; b if and only if a
n
= eob for some idempotent e in (Si n ,0).
Now using these constructions A, V, .;;; on Si n we can define a permutation geometry; denote it by Perm G (l,n). let G be a subset of S ; denote by T(G) the n
meet-semi lattice of bijections generated by G; i.e. T(G) o
= {a A b:
a,b E 6}.
let
-1
b): a,bE G, a -; b} = l = {~'~l'''''~ o r- I}' (In the special case when G is a group l = {f(a): o ~ ~ o < ~l < ... < ~ r- I < n. a E G, a -; < I >} holds, i.e. G is a group of type l). It is evident that for any G be a B(l,n), l.e. {f(a
c = a ~b the set of fixed points of the permutation a-lb is exactly of the meet a A b. To ,Tl, .•• ,T r- 1,T r
c,
the domain
So the semi lattice T(G) is partitioned by the classes
= G where
for 0';;; i .;;; r-l,
M. Deza
304
T.1
{a
1\
b: a, bEG, f(a
-1
b)
= t .1} = {c E T(G): lel=
L}. 1
r
DEFINITION: Let G C Sn and G be a B(L,n); let T(G) = i~oTi be the above-described partition of the meet-semi lattice T(G) generated by G.
Then we call T(G) permuta-
~'ion geometry if for any a E Ti (0 " i < r) and for any point b(i .e.
where a and b are disjoint (i .e. a 1\ b
= <0>
Ibl
=
1)
and a v b defined) there exists a
unique c E T. 1 such that a vb" c. 1+
The above definition gives a permutation (more exactly, bijection) analog of perfect matroid-design.
For example, any sharply t-transitive set G, G c S , r n generates a semilattice T(G) = .u T. where T. = {c E Si : lei = i} for all 1 =0 1
o~
n
1
t-1, T = G. So, this T(G) is a Perm G(L,n) with L = {0,1, ••• ,t-1}; the t blocks of S(t,k,v) work as hyperplanes in the lattice of matroids in a similar
way.
i
~
The definition of Perm G(L,n) given in [31 was more general; but (because of
Proposition 3.4 in 131 ) the above definition (which is much simpler) excludes only the trivial permutation geometry T(G
= Sn ).
In [31, [101 bijection (and transformation) analogs of general matroids were defined and given their first elementary properties; we will develop this subject in another paper. It is easy to check the following elementary properties of a permutation r
geometry T(G)
T. which is Perm G(L,n):
.U
1=0 1
1 (from now we will suppose, withiut lost of generality, that o T = {
T
°
3) any a E T(G) is the least upper bound (in T(G» 4) for any a
E
G define f~a
=
{ii:<
0
of elements of T ; l >"b "a, bE T(G)} , then Ma is the
lattice of flats of perfect matroid-desiqn; 2
5)
IT.1 I
=
i =1
II j=o
(n- ~. ) _--><J_ if 0 " i " r- 1 , c - ~. 1 J
The equalities 5) (which follow from 4»
IGI
IT r I
show that any subset G c S
n
generating a permutation geometry is a B(L,n) of the size same as the order of
(L,n)-sharp group.
The first cases of Perm G(L,n) to study are L
~IIEL(n-~.),
i
{O,.q and L
i.e. the
1
{O,l,t}; i.e.
New generalizations of sharply t-transitive groups
305
an analog od the matroid of rank 4, i.e. of planar space. From now on we consider in this section only special case of geometric group G, i.e. a subgroup G of S , such that T(G) is a permutation geometry, say n
Perm G(L,n).
Of course, any geometric group generating Perm G(L,n) is an (L,n)-
sharp group (see
3).
§
(An example of ({O, q}, q2)-sharp but not gemetric group
is the group {(x,y)
--+
(x+c, ax + by + d)}
of permutations of 2-dimensional vector space over GF(q) with a,b,c,d b
E
GF(q),
"i 0).
For geometric group G, all matroids M , a E G (defined in 4) of the above a proposition) are equal. We denote this matroid by M(G); matroid M(G) is geometric iff 0,1
L, i.e. iff G is 2-transitive group.
E
the fixed-point sets of subgroups of G.
The flats of M(G) are exactly
G is a subgroup of Aut M(G).
Further, G is a Jordan group, i.e. a transitive group with a subgroup fixing some points and
tra~sitive
on remaining points; moreover, the pointwise
stabilizer of any hyperplane of M(G) is sharply transitive on its complement.
In
fact, any Jordan group acts as a group of automorphism of some matroid such that the pointwise stabilizer of any flat is transitive on its complement. Using the above observations one can reformulate a theorem of Kantor ([ 15) ) as the following classification of all 2-transitive geometric groups (Le. with 0,1
E
L); 2r -1, L = { 0, 1,3 ,7 , ••• ,2 r-l - l} ; AGL (r-l,q), n = qr-l , L = {O,l,q,q 2, ••• ,q r-2 };
1) G = GL ( r , 2), n 2) G = 3) G 4) G
=
15, L = {0,1,3}; G = V16 .A , n = 16, L = {0,1,2,4}; 7 .harply t-transitive, La {O,l, ••• ,r-l}. , n
=
ere is a striking similarity of above list to the list of all known perf
matroid-designs (with 0,1
L):
E
1) truncations of projective geometry PG(d,q), d-i -1 1 L = {O " 1 q+l , • • • q q-l -} ' k
=
qd-i -1 d+ 1 1 v = -q--- . q-l' q-l'
2) truncations of affine geometry AG(d,q), L = {O,l ,q,q 2, ... ,q d-i-2 }, k
=
qd-i-l
v
=
qd
3) finite commutative Moufang loops of exponent 3,
306
M.Deza
m L = {O) 1)3 }, k = 9, v = 3 ; 4) Steiner systems S(t,k,v), L = {O, 1 , ••• , r- 1 }.
So, for example, in the first interesting case, L = {O, 1,3} (and so both n, k are
= 1,3
(mod 6)) a geometric group of the above list exists iff n = 7,9,15
(cases 1), 2), 3)) while a perfect matroid-design exists for k = 7,9 and the first unknown cases are k = 13 (v = 133, 183, ... ), k = 15 (v = 171,183, ... ,), k = 19 (V = 307, ... ).
The problem of classification of 1- but not 2-transitive geometric group 0.e.
oE
L, 1 ¢ L)) is open; for permutation geometry,we do not have the same
procedure of geometrization as for matroids.
Concerning the simplest such case of
geometric group of type L = {O,t}, we know only that it is exactly a group of order n (n-t) having TI-subgroup H (i.e. for any g
E
G, either Hg=H or H II H =<1»
such that IHI = n- t and the order of its norma 1i zer is (n- t) t.
r1any examp1 es of t m
such groups are given in 11]; they are given implicitly (as SETG of digraph k ; n = tm) in the list of examples of k£ admitting SETG at the end of m
§
5.
In [3] a "H1-wise generalization" Penn). G(L,n) of permutation geometry T(G) was also considered, where T(G) is defined about as the set of all A+1-wise meets of elements of G, see Theorem 2.4 for related ideas.
The corresponding
generalization of a geometric group is always a Jordan group (but the converse is false) and all such 2-transitive groups are either r-transitive with L ={O,l, ... , r-l} or a group of automorphisms of a projective geometry over GF(2) or of an affine geometry.
The (>.+l)-wise generalization of Theorem 2.1 is unknown.
Finally, let us come back to the pennutation geometry T(G) generated by the geometric group G.
Then(T(G),o) is an inverse semigroup whose semi lattice of
idempotents is (semi1attice-) isomorphic to the lattice of flats of the matroid M(G).
Moreover, the semi group (T(G),o) is fundamental (i.e. equality is the only
idempotent-separating congruence) iff 0,1
E
L. The structure of T(G) as a
permutation geometry can be recovered from its structure as inverse semi group and also, it is possible to decide, for a given inverse semi group, whether it arises from a geometric group in this way.
307
New generalizations of sharply t-transitive groups
5. DIGRAPHS ADMITTING SETG (or SET) As was defined at the end of
§
2 a digraph r
= (V,R) admits a SETG (or SET)
G if G is a subgroup (or subset) of (transitive) Aut r which is regular (i.e. sharply l-transitive) on R.
Being both vertex- and edge-transitive, the digraph r
is: a) regular (i.e. the indegree and outdegree of any vertex are equal to the same number d(r), called the degree of r; b) IRI = Ivl. d(r); c) r is either an undirected or an oriented (partial tournament) graph; d) in the case d(r) > Ivl / 2, the digraph r is undirected and, moreover, forodd IVI the number d(r) must be even. An example of digraph r with SETG is a l-regular graph (in terms of [18], i. e. G = Aut r, and, in particular, an elementary circuit of degree 1 and an elementary cycle of degree 2.
(In fact, we will classify below all digraphs of
degree more than Ivl -3 with SETG). The direct product r of digraph r and r 2 admitting a SETG (corresp. SET) l admits a SETG (corresp. SET) and d(r) = d(r l ) d(r 2).
= (V,R) be a digraph admitting a SETG G. Then:
Let r
a) if d(r)
~
Ivl / 2 then r is undirected (i.e. equal to its converse);
b) conjecture; if d(r)
~
Ivl / 2 and G is primitive then either r
=k \
lV (and G is sharply 2-transitive) or G is A5 acting on unordered pairs for L(k 5), i. e. the complement of Petersen's graph (IVI g 10, d(r) = 6); c) if d(r) > Ivl - 1 - ~~ then r is a compete multipartite graph
k~, Ivl = R.m; d) the digraph r -
(~n-l)
2
=n -
-
2~n +
= (V,R) = k~11 x k~11 has (if IVI = n = p2a) d(r)
g
1; it is the largest known degree of digraph which is not
a kR. but which admits a SETG. m
We define d(n) (corresp. d'(n)) admitting a SETG (corresp. SET)}. 2 ~ d(n)
~
d'(n)
~
=
max {d(r): r is digraph (V,R), IVI
=
n,
Because of Proposition 2.5 we have
n-l with d'(n) = n-l if and only if there exists a sharply 2-
transitive subset of Sn (and hence a projective plane PG(2,n)) and with d(n)
=
d'(n)
=
n-l if and only if there exists a sharply 2-transitive subgroup of
S (and hence a projective plane PG(2,n) over a near field which is possible if n
and only if n
= pa,.
308
ftI. L>eza
The known lower bounds on den) are: a) d{n 1nZ) ? d(n l ) d{n Z) (it comes from the direct product fl x fZ; compare with r~cNeish's lower bound min (N(n l ), N(n ) for the maximal number N(n) of Z pairwise orthogonal Latin squares); S ai S a. b) den = i g1 Pi )? i~l(Pi 1 - 1) (it comes from a)); it becames equality a for all n = p and as the smallest other n we have d(Z8) = 18) and then den) > Cn/log log n for any n and some constant C; d) conjecture: den) ? Cn for any n and some constant C; d) if conjecture c) is true then C ~ l/Z since d(Zp) = P for any prime p
> 7;
2 e) den) ? n/2 if n is even (because any km admits a SETG) and den) ? Zn/3 3 if n i5 divisible by 3 because any km admits a SETG); f) den) > n/2 for all n < 5.7.11.17.19.23.29 = 82944785; d(llO) > 72 (from the Mathieu group Mll acting on unordered pairs); d(q(q+l)(q2+q+q+l )
?
q2(q_1)2
for prime powers q (from PGL(3,q) acting on ordered pairs of points of the projective plane). The known upper bounds and exact values of den) are:
= n - 1 iff n = pa; den) 2a a for any n = (2 _l)(2 +l) with odd a; a) den)
~ n - Vi1 if n # pa but den) > n - ~i1
= n-2 iff n = 6 or 14 (and d'(6) = 4, d'(14) = 12); c) den) = n-3 iff n = 15 or 24 (b) and c) come from k~ and k~ admitting b) den)
geometriC ({O,2},2 t )- and ({O,3},3t)-sharp groups as SETG; to compare b) and c) with similar "extention" results in nets and orthogonal Latin squares); d) d(2p) = p for any prime p > 7; d(63)
= 56; d(78) = 72, d(455) = 448 etc.
Table 2 The values of den) for n < 30, n # pa • n
6
10
den)
4
6
digraph realizing den)
a
12 14 15 18 20
21
22
24 26
28
8 12 12 12 16 14
11
21
18
2 k3 L(k ) k3 k7 k5 k3 k5 k2 kll 4 2 3 6 4 7 2 5
13
8 k k2 k4 x k7 3 13
309
New generalizations of sharply t-transitive groups
(The examples of SETG G in above table are the group of rotations of the octahedron for k~, the alternating group A5 for the line graph L(k 5) of k5 (the complement of the Petersen's graph) etc.). 1 From now on we consider complete multipartite digraphs k admitting SETG. m
Some examples are: a) ~ = 2,m-arbitrary; b) ~ an odd prime, m = S(~-1)/2 for some S> 1 (in particular, ~
= 3, m
arbitrary) ; c) ~ = 4, m = S2 where S is a product of prime powers congruent to 1 (mod 3); d)
~
=ma
e)
~
= q+1, m = q-l where q is a prime power;
f)
~
=
g)
~
h)
~
prime power;
q+l, m = (q-1)/2 where q is a prime power with q
4);
q + q + 1, m = q(q-l) where q is a prime power; = 22a + 1, m = 2a - 1 for a odd. =
For all (except the case q> 3 of g)) pairs SETG G.
=3 (mod
2
(~,m)
above we have a geometric
For the case a) we have also a nongeometric SETG G.
~
on k ) has IGI = n(n-m), n = m
~m.
Any SETG G (acting
If, in addition, G has type {O,m} then G is
{O,m}-sharp and, moreover, geometric (see
§
4).
G is geometric for all prime m which are less or equal to the smallest prime divisor of
~-l.
But for any geometric ({O,m}, n
= ~m)-sharp group G the
digraph (V,R) with V = {1,2, ..• ,n}, R = {(a,b): Ga does not fix by b} is k~ and G acts on (V,R) as a SETG. The problem is to classify all pairs (~,m) for which k~ admits a SETG. If m 2 3 m is fixed then we know that k , k always admit a SETG but we conjecture that m m for ~ > 3 the set {m: km admits a SETG} has density zero.
For the other case of fixed 1 the basic result was ([ 17)) that k~ admits a
SETG iff
~
= 2,3,4 or 7. Below we give (from (1) ) two generalizations of it.
THEOREr~ 5.1: For any m;;' 2 the set {~: k~ admits a SETG} is finite. m
THEOR£I.1 5.2: For any prime m;;' 2. the digraph k~ admits a SETG iff R, 1:11 2,3, m+2 m 2 a (with m = 2 - 1), 2m + 2 (with odd m and 2m+1 = pal, m + 2m + 2 (with m+1 an odd power of 2), mt (with t exeept
R,
=
mt for t > 1
and
~ m) or ~ = 7, m = 2 (examples are known for all eases m odd).
310
M. Deza
T. Ito (personnal communication) classified all groups G acting as SETG on £
k and {(O,m}, n = £m}-sharp (we remind,that such (O,m}-sharp groups are, morern over, geometric) for all p~ime m> 2. (For m = 2 it is just the list of groups given by Tsuzuku in [17J ). 1) (Zm x Zrn)
l'l
There are exactly five groups:
Z2 (£ = 2, any m);
2) (Zm x Zm) '" S3 (£ = 3, any m); 3) (Zm x Zm)
1 0 )0
{(":
U = m), the right factor is a subgroup of GL(2,m)
,.,)}
which acts naturally on Z x Z considered as vector space of dimension 2 over m m this field; a
4) Zm x PSL(2,q) (t = q+l, m = (q-l)/g.c.d.(2,q-l), q is a power p of a prime pl. 2 2a+l 5)Zm xS(q)(£=q +l,m=q-l,q=2 ). z For all parameters l,m (except pending case £ = mt, t> 1, m> 2) there are a group of above list but there are also non-{O,mrsharp SETG G on k£. m
Table 3 k£ admitting SETG with t,m ~ 10 m
the sign :', denotes that d(k£) = Rm-m=d(£m)). m
"'-,
~9.i
m~1 2 3
4
2
3
I +
+
I+ I I I
+
6
I+ I+
7
,,, +
5
I I
I
4 :',
+
7
9
10
+
:'~
'O',
+ +
8
'O':
+
+ ..+:
6
5
+
+
?
?
?
?
?
:',
+
+
+
+ ~';
+
?
?
?
?
+
~':
+
+
+
!
8 9 10
I+ +
I
!
i+
+
?
?
?
+
+
?
+
?
+
?
+
?
?
+
?
+
?
?
?
?
+
?
?
:':
I+ I
+
We remark now that the problem of recognizing digraphs admitting a SETG can be formulated in the language of group theory.
Let G be a group with proper sub-
311
New generalizations of sharply t-transitive groups
-1
group H and element g E G such that H n g Hg
= <1>. Define digraph (V,R g ) where V is the set of all left cosets of H in G and Rg = {(xH, xhgH): hE H, xH E V}, then digraph (V,R g ) admits G as a SETG (with G acting on V by left multiplication) and the subgroup H of G is the stabilizer of the vertex HE V. Any digraph admitting SETG is isomorphic to the digraph constructed as above.
As corollaries
we have: a) d(n) = max{IHI: H is a subgroup of index n of G and H intersects some G-conjugate of itself trivially}; b) the digraph admitting a SETG G is k~ if and only if for any x E G there -1
m
exist hl , h2 E H such that x = g hl gh 2 or x = hl g h2. Described above the sugroup H (of the group G) having H n g H g-l =<1> (i. e. hlg
~
g h2 for any hl f h2 E H) for some g E G, can be seen as a permutation
group-code correcting the error consisting of possible multiplication of its elements (from the left or from the right) by the given element g E G.
If g is a
cycle we can see H as a synchronization code. Finally, a generalization of d(n) is dt(n) = max d(r) where r = (V,R) is a uniform directed hypergraph on IVI = n vertices such that Aut r contains a subgroup G which is regular on R.
Then dt(n)
~
n-l with equality iff there exists a
sharply t-transitive subgroup of Sn'
6. DIVERS RELATED PROBLEMS A) One could try to approximate sharply 2-transitive groups by the numbers max IGl/n where H is a Frobenius subgroup of Sn, i.e. a B({O,l},n)-subgroup. But a· S a. it is easy to see that max IGr/n = g.c.d. (Pi 1 - 1) where n = i~l Pi 1 is the prime factorization of n.
So max IGl/n ~ min (p~i -1), but the well-known lower i
1
bound for the largest number of pairwise orthogonal Latin squares is exactly McNeish's bound min(p~i -1). i
1
B) One could try to approximate sharply 2-transitive subsets of S for small n
n f pa (i.e. 6,10,12 .... ) by the numbers IP(n)l/n (where P(n) is a largest 2packing. i.e. a B( {O.l }.n)) and IC(n) lin (where C(n) is a smallest 2-covering containg <1> i.e. a 2-transitive subset of S). n n = 6 IP(6)1/6 = 3 (<5<) 7
~
IC(6) 1/6
~
We know only ([ 10)) that for
10 and that IP(lO)1
C) Let B be a B({t.t+l •...• n-l}.n). i.e. an anticode. the exact value IBI for n ~ no (n-t) and the easy bounds IBI
~
32.
Then we know (181) ~
(n-t+log n)! (if
M. Deza
312
t < n/2) and IBI > (n-t)!
We have IBI
= (n-t)!
sitive subset of Sn and we conjecture IBI with respect of t.
if there exists a sharply t-tran-
= (n-t)! for all n ~ n0 (t), i.e. n large
D) Sharply t-transitive groups G are (see [2] ) asymptotically good for coding (i .e. din
~
l-R where d
n-t is the minimal distance of code G and R =
=
= log iGi/iS n I is its rate) and they admit good (majority type) decoding procedures.
Two other related coding type problems are to study B(L,n)-schemes (i.e. 2 B(L,n) such that the corresponding A(L,n,n ) is a subscheme of Johnson scheme 2
2
J(n ,n) of all n-subsets of an n -set), see [10J) and to study other distances on S (see [4]). n
E)
The Hamming distance on Sn is dH(a,b)
=
n - f(a-
l
b)
=
t- b(i)}.
{i: a(i)
The volume V of a sphere of radii g (in this metric space (S ,d H)) is IV 1= 9 g i· n g = 1 + . 'i: 2 (~) D. where D. = i! .l: ( - 1) J 1 Ij! - i! lei s the number of 1= 1 1 1 J=o derangements c of degree i (i.e. c E S.1 such that f(c)
= 0). As in any metric
space we have here the sphere-packing bound for a code B correcting g errors (i.e. having all distances dH ~ 29
+
1)
In the case of equality we will call B a perfect code; in this case Sn is partitionned by the spheres Vg having as center an element ov B. V = (~) + 1 and 2 the smallest n for which it divides n! is n = 11; does there exist n perfect permutation code B({0,1,2,3,4,5,6}, 11)7 Similar problems concerning the possibility of partition by equal spheres can be interesting for the metric space (A n , dH) where An is the alternating group. F) It would be desirable to get a unified theory for the best upper bounds of the size of family of objects having pairwisely prescribed "inner products" L=
{t , ••• .0
,t
r-
l} (and with a finite beuty realizing the equality) for the follow-
ing three cases: 1) Theorem 1.1; the objects are k-subsets A of a large v-set, L
{IA n BI},
the beuty is a perfect matroid-design; 2) Theorem 2.1; the family is a group consisting of permutations a L
E
Sn'
= {f(a-lb)}, the beuty is an (L,n)-sharp group; 3) The special bound in r 5]; the objects are points a on the unit sphere in
IR" (i. e. 1i nes through the ori gi n), L = {( a, b,z) (i. e. (a, b)
2 = cos 2ex where
ex
the angle between lines and sin 2ex is the distance), the beuty is a spherical t-
is
313
New generalizations a/sharply t-transitive groups
design which is an association scheme. The bound for case 3) contains (under the name annihilator) the Lagrange polynomial of the same type as for 1), 2).
The similarity between the cases 2)
and 3) is that we have in both cases some orthogonal basis (of irreducible characters for the permutation case 2)), we expand the Lagrange polynomial in this basis and multiply both sides by the identity element of the basis. The information on a (the coefficient of the identity element in the above expansion) gives the o bound (a is an integer for the case 2)). o It would be interesting to derive the bound in the case 1) from fUrther study of the case 3). Actually, any A(L,k,v) can be seen as a set of points a of a unit sphere in IRvn such that the vector a has l/-/-k on some k positions and zeroes on the other v-k positions. Any (a,b) is a v-~. J
£~EL k-£. 1
1
v/k - L/k
= £~EL l-£./k 1 1
1
=
for some
~. 1
ELand
(Y..) e: F(Y..)
k
k
where F(x) in the annihilator polynomial of [5]
x
~./k 1
we remark that Y.. = (x,x) where k
(1,1, ... ,1) Nk.
The case
was recently generalized by A. Neumaier (Combinatorial configu-
3)
rations in terms of distances, Lecture notes, Eindhoven University of Technology, 1980) on what he call Delsart metric space (by analogy with the Q-polynomial
association schemes of Delsarte). over IF
= IR, It,
The special cases are projective spaces FP n-l
JH, ID (reals, complex numbers, quaternions, octonions).
As the
models for the first three spaces one can take lines through the origin in IRn, Itn, ~ with distance sin 2a. Here also there appear annihilators in the upper bounds and generalized t-designs in the case of equality; (see, especially for the cases IF
It, JH, ID the paper of S.G. Hoggar, t-designs in projeetive spaees, to
appear).
BIBLIOGRAPHY 1. 2. 3.
L. Babai, P.J. Cameron, M. Deza and N. Singhi, On sharply edge-transitive permutation group, J. of Algebra, 73 (1981), 573-585. I. Blake, G. Cohen and M. Deza, Coding with permutations, Information and Control, 43 (1) (1979), 1-19. P. Cameron and M. Deza, On permutation geometries, J. London Math. Soe., 20 (3) (1979), 373-386.
314
4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
15. 16. 17. 18.
M. Deza
G. Cohen and M. Deza, Some metrical problems on S. Proc. of France-Canada Meeting. Montreal (1979), in Annals of Dise~ete Math., 8 (1980), 211219. P. Delsarte, J.M. Goethals and J.J. Seidel, Bounds for systems of lines and Jacobi polynomials, Philips Res. Repts, 30 (1975), 91-105. M. Deza, P. Erdos and P. Frankl, On intersection properties of the system of finite sets, ppoe. London Math. Soe., (3) 360978), 369-384. M. Deza, Perfect generalized packing as a generalization of simple t-design and matroid-design, Proc. de colloque into du CNRS sur Ie combinatoir~ Paris-Orsay (1976), 97-100. M. Deza and P. Frankl, On the maximum number of permutations with given maximal or minimal distance, oT. Combin. Theopy, 22(A) (1977), 352-360. 11. Deza and S. Vanstone, Bounds for permutation arrays, J. Statist. Planning and Infepe~we, 2 (1978), 197-209. M. Deza, On permutation cliques, Annals of Diserete Math., 6 (1980),41-55. 11. Deza and N. Singhi, Some properties of perfect matroid-designs, Annals of Disepf-te Math., 6 (1980), 57-76. M. Deza, On maximal permutation anticodes, Proc. of 10 S.E. Conference on Combinatorics, Boca Raton (1979), 381-392. T. Ito and 1·1. Kiyota, Sharp permutation group, Ppoc. Math. Soe. Japan, 33(3) (1981), 435-444. N. Iwahori, On a property of a finite group. Part I. J. Fae. Sei. Univ. Tokyo, X (1964), 46-64. N. Iwahori and T. Kondo, On a property of a finite group. Part 2, ,r. Fae. Sei. Univ. Tokyo, XI (1965), 113-147. W.M. Kantor, 2-transitive designs, Combinatopies (ed. M. Hall Jr. and J.H. van Lint) D. Reidel, Dordrecht (1975), 365-418. M. Kyota, An inequality for finite permutation groups, Seminar on permutation group and related topiCS, Kyoto (1978), 78-83. T. Tsuzuku, Transitive extensions of certain permutation groups of rank 3, Nagoya Math. J., 31 (1967), 31-36. t1.T. Tutte, A family of cubical graphs, Ppoe. Cambpidge Philos. Soe., 43 (1947), 459-474.
U.E.R. de Mathematique Universite Paris VII Tour 45-55 5me Etage 2, Place Jussieu 75221 Paris Cedex 05 France
315
Annals of Discrete Mathematics 18 (1983) 315-318 © North-Holland Publishing Company
DISSECTIONS OF POLYGONS :,'(
J. Doyen and M. Landuyt
1. INTRODUCTION
A polygon P in the Euclidean plane E2 is said to be dissected into the polygons P ,p ""P (which we write P k l 2
= Pl
+
P2 + ... + Pk ) if
k
(i )
P
=1,u=l
P.
1
(ii) P. n P. (i f j) has no interior point. 1
J
2 Let G be any subgroup of the group I of all isometries of E.
Two polygons P and
Q are said to be equivaZent by dissection with respect to G (which we write P
G = Q)
if there is a positive integer k such that
and, for every
P = Pl + P + 2
+
P
Q=Q + Q + l 2
+
Q
k
k
1,2, ... ,k, there is an isometry gi
E
G mapping Pi onto Qi'
Intuitively, G is the group of isometries used to rearrange the pieces of the dissection of P in order to construct Q from P. Bolyai proved in 1832 that area P a area Q ~ P
I = Q
A stronger result was obtained in 1951 by Hadwiger and Glur [3) who proved that
area P
= area Q ~ PTH =Q
where TH denotes the group of translations and half-turns in E2. In some sense, this result is best possible: indeed, two polygons with the same area are not necessarily equivalent by dissection with respect to the group T of translations (for example, one can prove that an equilateral triangle and a square with the same area are never equivalent by dissection with respect to T).
316
J. Doyen and M. Landuyt
2. MINIMAL DISSECTIONS Given two polygons P and Q having the same area, we denote by I1(P,Q) the minimum number of pieces in a dissection of P from which Q can be reconstructed by rearranging the pieces with the full group I of isometries of E2.
We
write 11
C
(P,Q)
if the pieces of the dissection are supposed to be convex. Let P and Q denote respectively a regular convex m-gon and a regular m n convex n-gon having the same area. We can prove that, if 3 3 ~
~
m < n, then
I1(P ,Q ) m n
~
(2m + 4)(n + 1)
In some cases, the upper bound is rather far from the exact value: for example, the dissections in figure 1, discovered by Dudeney in 1902, show that
Figure 1 We can prove that
~
c
(P ,Q4) = 4. but the value of I1(P 3 ,Q4) does not seem to be 3
known at present. The dissections in figure 2, due to Goldberg [2], show that
Dissections of polygons
317
Fi gure 2
We can prove that ~ c (P 3 ,Q6) = 5 or 6. (Lindgren [~l has produced a dissection using only 5 pieces, one of which is non convex). The details of the proofs will be published elsewhere. 3. REPLICATING POLYGONS A polygon P in E2 is said to be rep~icating of order k (briefly rep-k) if there is an integer k ~ 2 such that
where P ,P "",P are pairwise isometric and similar to P. For example the nonk l 2 convex hexagon in figure 3 i~ rep-4 as is shown by the dissection on the right.
p
Figure 3
318
J. Doyen and M. Landuyt
It is easy to show that the only rep-2 polygons are the isosceles right triangles and the parallelograms whose lengths of sides are in the ratio ~: 1. The only convex rep-3 polygons are (up to a similarity) the right triangles with lengths of sides 1, ~, y3-, and the parallelograms with lengths of sides 1,
/:3
(it is not known whether there exists a non convex example). In 1974, Va1ette and Zamfirescu (5) have completely classified the convex
rep-4 polygons (here several non convex examples are known, including the hexagon in figure 3). We have proved that the only convex rep-5 polygons are (up to a similarity) the right triangles with lengths of sides 1,2, \!!f, and the parallelograms with lengths of sides 1, v'!f . Betke [ 1 J has proved that if a convex n-gon is rep-k, then n = 3 or 4, and it has been conjectured that every convex rep-k polygon is either a triangle or a trapezoid.
BIBLIOGRAPHY 1. 2. 3.
4. 5.
U. Betke, Zerlegungen konvexer Polygone (to appear). M. Goldberg, Solution of problem E 400, Amer>. Math. Monthly, 47 (1940),
490-491. H. Hadwiger und P. G1ur, Zer1egungsg1eichheit ebener Po1ygone, El. Math., 6 (1951), 97-106. H. Lindgren, Ge:Jmetr'ia dissections, Van Nostrand, Princeton, 1964. G. Valette and T. Zamfirescu, Les partages dlun po1ygone convexe en 4 polygones semblab1es au premier, J. CombinatOI'ial Theor'Y, 16 (1974), 1-16.
Departement de Mathematique C.P. 216 Universite Libre de Bruxel1es B - 1050 Bruxel1es, Belgique
319
Annals of Discrete Mathematics 18 (1983) 319-332 © North-Holland Publishing Company
GROUP MODIFICATIONS OF SOME PARTIAL GROUPOIDS A. Drapal and T. Kepka
~'¢
1. INTRODUCTION Let G (0) and G (*) be twogroupoids with the same underlying set. We put dist(G(o),G(*)) = card {(a,b) E G2; a b ~ a * b}. For an integer n ~ 2, let 0
gdist(n)
= min dist(Q,Q(o)) where Q runs through all groups of order nand Q(o)
through all quasigroups with the same underlying set such that Q ~ Q(o).
If 2 < m
divides n then clearly gdist(n) < gdist(m), and therefore gdist(n) < gdist(p), p being the least prime number dividing n. gdist(n) < gdist(p) for some n
~
2.
It seems to be an open problem whether
In the present paper, a technique which might
be useful in solving this problem is developed.
In particular, the modifications
(i.e. relfexions) of certain partial groupoids into the category of groups are investigated.
Let us note here that the numbers gdist(n) play important role in the
enumeration questions concerning the number of associative triples of elements in fi ni te quas i groups (see [ 1 1) . 2. PARTIAL GROUPOIDS By a partial groupoid we mean a non-empty set together with a partial binary operation (possibly empty).
This operation will be denoted multiplicatively in
this section. Let K be a partial groupoid. For every a E K we put
= {(b,c) E K2; a = bc}, so that the operation of K is defined just for ordered pairs from the set M = M(K) = U M(K,a), a E K. Further, let
M(K,a) B(K)
{b E K; (b,c) EM}, C(K)
D(K)
{be; (b,c)
E
M}.
=
{c E K, (b,c) EM}, A(K)
Finally, let m(K)
1:11
=
B(K)
U
C(K) and
card M.
Let K and L be partial groupoids. By a homomorphism of K into L we mean any mapping f of K into L satisfying the following condition: (1) If (a,b) E M(K) then (f(a),f(b))
E
M(L) and f(ab) = f(a)f(b).
phism f is said to be strong if the following is true:
The homomor-
A. Drapa/ and T. Kepka
320
(2) If (x,y) E M(L) and x,y E f(K} then there exists a pair (a,b) E M(K) such that f(a) = x and f(b) = y. Partial groupoids with homomorphisms form a category which will be denoted by
P. Let f be a homomorphism of a partial groupoid K into a partial groupoid L. Put f(M(K)) = {(f(a),f(b)); (a,b)
E
M(K)}
~
M(L}.
groupoid I(o} =f[K las follows: 1= f(K} and u iff there are a,b (then (u,v)
E
K such that (a,b)
E
M(L) and
Z
= uv}.
E
0
Further, define a partial v
M(K), f(a}
Z
is defined for
u, f(b) = v and f(ab) =
It is easy to see that f:K
homomorphism and the identity mapping id: I(o}
~
U,V,Z E
~
Z
I(o} is a strong
L is a homomorphism.
By a pseudoimmersion (resp. immersion) of a partial groupoid K into a partial groupoidL we mean any homomorphism f of K into L such that the restrictions f
i B(K) and f I C(K) (resp. f I B(K), f I C(K) and f I O(K)) are injective mappings. Let K be a partial groupoid. By a congruence of K we mean any equivalence
relation r defined on K such that: (3) If (a,b),(c,d)
E
rand (a,c),(b,d)
E
M(K} then (ac,bd) E r.
In this case, we
can define a structure of a partial groupoid on the factorset K/r by xy = z for x,y,z E Klr iff there are a E x and bEy such that (a,b) E M(K) and ab E z.
The
natural projection of K onto K/r is then a strong homomorphism. An element e of a partial groupoid K is said to be a partial neutral element of K if the following three conditions are satisfied: (4) If (e,a) E M(K) then ea
a.
(5) If (a,e) E M(K} then ae
a.
(6) (e,e)
E
r"(K).
A partial groupoid K is said to be reduced if K = A(K)
U
O(K).
We denote by
R the full subcategory of P formed by reduced partial groupoids. Obviously, every partial groupoid K can be expressed uniquely as the disjoint union K = L u J where M(K) c L2 and L is reduced. A partial groupoid K is said to be balanced if the sets B(K), C(K} and O(K) are pair-wise disjoint. A partial groupoid K is said to be cancellative if the following two conditions are satisfied: (7) If (a,b),(a,c)
E
M(K) and ab = ac then b = c.
(8) If (b,a),(c,a) E M(K) and ba = ca then b = c. A group is a semi group with (unique) division.
Thus a group is a (reduced
321
Group modifcations of some partial groupoids
cancellative partial) groupoid and we denote by 9 the full subcategory of groups in the category R.
Further, we denote by S (T) the full subcategory of R consisting
of reduced balanced (cancellative) partial groupoids. LEMMA 2.1: Let K E R. of L onto K.
Then there exist L E S and a surjective strong immersion k is cancettative, provided K is so.
Moreover,L
PROOF: Let B,C,D be pair-wise disjoint sets and f: B ~ B(K), g: C ~ C(K), h: D ~ D(K) bijective mappings.
Put L = B u CuD and define bc = h-l(f(b)9(C))
for all bE B, c E C such that (f(b),g(c)) E M(K).
Now, it suffices to put
k = f u 9 u h.
Let K be a partial groupoid and x = (a,b) E M= M(K).
We denote by s = Sx
the least congruence of K such that (a,b) E sand a/s is a partial neutral element of K/s (it is easy to check that there exists such a congruence).
Further, define
a relation s'
s'x on K as follows: (9) (c,c) E s' for every c E K.
(10)
If (a,c) EM then (ac,c),(c,ac) E s'.
(11)
If (d,b) EM then (db,d),(d,db) E s'.
(12) If (a,c),(d,b) EM and ac
= db then (c,d),(d,c) E s'.
LEMMA 2.2: Let K E T and x E M(K).
Then s
x
s~
and the naturat homomorphism of
K onto K/s is an immersion. PROOF: One may verify easily that
s~
is a congruence of K and the rest is clear.
LEMMA 2.3: Let h be a pseudoimmersion of a partiatgroupoid K into a canceltative partial groupoid L.
Then K is canceUative.
PROOF: Obvious. LEMMA 2.4: Let I,J,K be a partiatgroupoids and g: J such that J E S, g(M(L))
~
K, f: I
f(M(I)) and f is an immersion.
homomorphism h of J into I such that fh
PROOF: Easy.
~
=
g.
~
K homomorphisms
:I'hen there exists a
322
A. Drapa/ and T. Kepka
An element a of a partial groupoidK is said to be idempotent if (a,a)
E
M(K)
and aa = a. 3. GROUP MODIFICATIONS OF REDUCED PARTIAL GROUPOIDS
In this section, the operations of general partial groupoids are denoted by
0
and the operations of groups multiplicatively. Let K E R and let F(K) designate the free group of words over the set A(K). -1 -1
Let N(K) be the normal subgroup of F(K) generated by all the words bce d and uvz- l where (b,c),(d,e) E M(K,a), a E D(K), and (u,v) E M(K,z), z E A(K) n D(K). Put G(K) = F(K)/N(K) and define a mapping g and g(b
0
c) = bcN(K) for all a
PROPOSITION 3.1: Let KE R.
E
A(K) and (b,c)
Then g
= gK
G(K) and it Is Just ,;he modification of g(e)
= gK of K into G(K) by g(a) = aN(K)
is a
E
~ell
M(K). defined homomorphism of K into
K into the category 9 of gl'OUpS.
= 1 jor "very pcn,tial neutral element
e
Moreove1}
K.
E
PROOF: Using the definition of N = N(K) and the fact that K is reduced, it is easy to show that g is a homomorphism.
H. a
Now, let h be a homomorphism of K into a group
= h(a) for every A = A(K). Clearly, N ~ Ker p, and hence we have a homomorphism f of G = G(K) There is a homomorphism p of F(K) into H such that pta)
E
into H such that fraN)
=
h(a) for each a
E
A.
Then h
=
fg and the rest follows
from the fact that G is generated by g(A). Let K,L
E
R and let h be a homomorphism of K into L.
According to 3.1,
there exists a unique homomorphism G(h) of G(K) into G(L) such that G(h)9 K = gLh. This homomorphism is defined by G(h)(aN(K)) = h(a)N(L) for every a E A(K). Further, gK(a)gK(b)
-1
E
Ker G(h) whenever a,b
E
K and h(a) = h(b).
Finally, G(h)
is surjective, provided h is so. LEMMA 3.2: L"t K,L
Rand Zet h be a surjective strong homomorphism of K onto L. -1 Ker G(H) ts jU3t ~he Iwrmal. subgroup of G(K) generated by aU 9 (a)9 K(b) , K
~'hel;
a,b
E
E
K, h(a) = h(b).
PROOF: Denote by P that normal subgroup.
Then P ~ Ker G(h).
Put H = G(K)/P,
denote by p the natural homomorphism of G(K) onto H and define a mapping f of L into H by fh(a) = P9 K(a) for every a
E
K.
Then f is a homomorphism, and hence
323
Group modifcations of some partial groupoids
f
= k9 L for a homomorphism
P = kG(h) and Ker G(h)
~
k of G(L) into H.
Ker p
Let K E R, 9 = gK and x = (a,b) G = G(K) generated by all g(a)
= kgLh
=
kG(h)gK'
P.
=
-1
Now, pgK = fh
E
M(K).
g(c), g(d)g(b)
Denote by H(K,x) the subgroup of -1
where (c,d)
E
P(K,x) be the normal subgroup of G generated by g(a) and g(b).
M(K).
Further, let
Denote by p the x
natural homomorphism of G onto G/P(K,x). LEMMA 3.3: Let K E Rand x,y px(H(K» H(K)
= G(K)/P(K,x).
E
M(K).
Then H(K,x)
= H(K,y) = H(K)
and
Moreover, if K has at least one idempotent element then
= G(K).
PROOF: Easy. LEMMA 3.4: Let K,L G(h)(H(K»
~
H(L).
R and let h be a homomorphism of K into L.
E
Moreover, G(h)(H(K»
= H(L),
Then
provided h is surjeative and
strong.
PROOF: Easy. Let KE R.
A normal subgroup R of G(K) is said to be pseudoregular (resp.
regular) if the homomorphism f9 of K into G(K)/R, f being the natural homomorK phism of G(K) onto G(K)/R, is a pseudoimmersion (resp. an immersion). The groupoidK is said to be pseudoregular (resp. regular) if the unit subgroup of G(K) is pseudoregular (resp. regular). LEMMA 3.5: Let K E R be pseudoregular.
Then K is aanaellative.
PROOF: Every group is cancellative and the result follows from 2.3. LEMMA 3.6: Let K E R.
Then K is (pseudo)regular iff there exist a group G and a
(pseudo)immersion of K into G.
PROOF: Obvious. LEMMA 3.7: Let K,L E R and let h be an (pseudo)immersion of K into L.
If L is
324
A. Drapal and T. Kepka
(pseudoJreguUlr then K is so.
PROOF: Use 3.6.
4. GROUP MODIFICATIONS OF REDUCED BALANCED PARTIAL GROUPOIDS LEMMA 4.1: :'et KE S be such that ccu>d K =3.
Then G{K) is a free group of rank 2
and H( K) = 1.
PROOF: Obvious. Let K E S.
Denote by V{K) the normal subgroup of G{K) generated by H(K).
LEMMA 4.2: !.et KE gK(K)
n
S. 'I'hen
G{K)/V{K) is a free group of rank 2, H{K)
~
V{K) and
V{K) = 0.
PROOF: As one may see easily, there exist L E S and a surjective homomorphism h of K onto L such that card L = 3 and h is strong.
The rest is clear: from 3.2 and
4.1. LEMMA 4.3: :'et K E S and x E M{K).
Then there exists an isomorphism fx of
G(K)/P(K,x) onto H{K) such that f xpx I H{K)
=
id.
= Px I H, G = G(K,x) and P = P{K,x). There is a surjective homomorphism q of F{K) onto H = H{K) such that q{c) = g{a)-l g{c) and -1 q(d) = g{d)g{b) for each (c,d) E M= M{K). It is easy to check that PROOF: Let x = (a,b), 9 = gK' P
N = N(K)
~
Ker q and a,b
E
Ker q.
homomorphism f = fx of G/P onto H. and fp(g(d)g(b)
-1
) = g(d)g(b)
PROPOSITION 4.4: Let KE I:';
-1
Hence P ~ Ker q/N and q induces a surjective' For (c,d)
E
M, fp(g{a)
-1
g{c)) = g(a)
-1
g(c)
,so that fp = id W Similarly, pf = id G/ p'
$, X E I~(K) and
U(K,x)
=
H(K) is iso"1orphie to G(K)/P(K,x), H(K)
of G{K) and G(K) is gener>ated by H{K)
U
P{K,x).
(i:) P(K,x)/U{K,x) is a free group of rank 2.
P(K,x) (\ V(K). n
Then:
P(K,x) = 1, H(K) is a retract
Group modifcations of some partial groupoids
325
(iii) G(K)/U(K,x) is isomorphic to the direct product H(K) X F where F is a free group of rank 2.
PROOF: Apply 4.2 and 4.3. Let K E S,
XE
M(K) and t
=
tx
phism of G = G(K) onto H = H(K) and t and R = t
-1
(S).
=
f xp. x
By 4.3, t is a surjective homomor-
I H = id. Let S be a normal subgroup of H
We shall say that S is an x-(pseudo)regular subgroup of H if R is
(pseudo)regular in G.
Thus S. is x-(pseudo)regular iff qtg K is a (pseudo)immersion, q being the natural homomorphism of H onto HIS. LEMMA 4.5: Let KE S and let S be a normal subgroup of H(K). x,y
E
M(K),S is x-(pseudo)regular iff it is y-(pseudo)regular.
PROOF: Suppose that S is x-pseudoregular. u,v
E
Then, for all
Let x
= (a,b),
y = (c,d) and let
B(K) be such that qt g(u) = qt g(v) where 9 = gK and q is the natural y
-1
y
-1
homomorphism of H = H(K) onto HIS. Then g(c) g(u)g(v) g(c) E S, and hence g(a)-lg(u)g(v)-lg(a) = g(a)-lg(c)g(C)-lg(u)g(v)-lg(c)g(c)-lg(a) E S, since g(a)-lg(c)
E
Hand S is normal in H.
S being x-pseudoregular. for u,v
E
Consequently, qtxg(u) = qtx9(v) and u = v,
Similarly for u,v
E
C(K) and, provided S is x-regular,
D(K).
Let K E S and let S be a normal subgroup of H(K). (pseudo)regular if it is x-(pseudo)regular for some x LEMMA 4.6: Let K E S be regular.
We shall say that S is
E t~(K).
Then gK is injective.
PROOF: Easy.
5. GROUP tl0DIFICATIONS OF REDUCED BALANCED CANCELLATIVE PARTIAL GROUPOIDS Let K E T and x
E
M(K).
Consider the congruence s
denote by qx the natural homomorphism of K onto Kx'
=
Then Kx
neutral element and qx is a strong immersion. LEMMA 5.1: Let KE T and x E M(K).
Then Ker G(qx)
K/s and R, Kx has a partial
sx' put Kx
P(K,x).
E
=
326
A. Drapai and T. Kepka
PROOF: The assertion follows easily from 2.2 and 3.2. LEr+1A 5.2:
K E T and x E M(K). that a
HOl'nai
Then there is an isomorphism h of G(K ) onto
x
subgroup S of G(K lis
(pseudo)regu~aJ'
x
x
iff the subgroup
PROOF: According to 5.1, there is an isomorphism kx of G(Kx) onto G(K)/P(K,x) such that p
x
=
k G(q). x x
LEMMA 5.3:
Now, it suffices to put hx
K E T Cl>!d
x,y
E
M(K).
=
f k . x x
Then there exists an isomorphism h of G(K )
x G(Ky } suc:; tiwr a normd 2ubgroup S of G(Kx) is (pseudo)regulm' iff the suoP'c:A.? h(S) cf G(K ) :'$ so. y ;'),-;:0
PROOF: This is an immediate consequence of 5.2. PROPOSITION 5.4:
(ii)
:"-;;S'
x
is (psed.oh'egular for evel'Y
PROOF: (i) implies (ii). (pseudo)regu1ar.
M(K).
X E
We must show that the subgroup P = P(K,x) of G(K) is
For, let x = (a,b) and c,d
pxg(c} = pxg(d), 9 = gK' g(c)g(d)
for K E T:
Kx is (pse-..do)regular for some x E M(K).
(iii) K
-1
equiva~ent
foUowing conditions are
E
Then fxPx(g(C)9(d)-1)
B(K) be such that
= 1, g(a)-lg(c)g(d)-lg(a)
1,
= 1, g(c)=g(d) and c = d, since K is (pseudo)regular. Similarly the
rest. (ii) implies (iii). (iii) implies (i).
This is an immediate consequence of 5.3. Use R.7 and the fact that qx is an immersion.
COROLLARY 5.5: Let K E T and x 't:'~8~iic)-:'mmtn~sion
E
M(K).
Then K is (pseudoh'egular iff txg
K
is a
of K into H(K).
6. COUPLES OF COMPANIONS In this section, we shall use various symbols for partial binary operations. Using another notation than the multiplicative one, we put the operation symbol
327
Group modifcations of some partial groupoids
into the brackets behind the symbol for the underlying set. Let K(o) and K("') be partial groupoids with the same underlying set. ~all
say that the pair (K(o), K(*»
We
is a couple of companions if both the partial
g.roupoids are reduced and cancellative, M(K(o» (a,b) EM there exist c,d,e,f E Msuch that c
= ~
M(KU'» b
~
e, d
=
Mand for all
~
a
~
f,
(a,c),(d,b),(a,e),(f,b) EM, a ° b = a * c = d * b and a * b = a ° e = fob. LEMMA 6.1: Let (K(o),K(*» be a couple of companions. of companions (L(o),L(*»
Then there exist a couple
and a surjective mapping k of L onto K such that both
L(o) and L(*) are balanced, k is a strong immersion of L(o) onto K(o) and k is a strong immersion of L(*) onto K(*).
PROOF: Easy (use the same technique as in the proof of 2.1). A partial groupoid K is said to be exchangeable if it is reduced and cancellative and if it possesses at least one companion. LEMMA 6.2: Let K be a partial grupoid, L(o) an exchangeable partial groupoid and f a pseudoimmersion of L(o) into K. canceZZative.
Suppose that either f is an immersion or K is
Then the partial groupoid 1(0) = f[ L(o)] is exchangeable.
PROOF: Let L(*) be a compani on of L(0) . Oefi ne a parti a1 bi nary opera ti on ,', on 1= fill by u ,', v = z iff there are a,b E L such that (a,b) E f4(L("'», f(a) = u, fib) = v and f(a {, b) = z.
It is easy to verify that the pair (1(0),1("'»
is a
couple of companions. A partial groupoid K is said to be (strongly) primary if it is exchangeable and every pseudoimmersion of an exchangeable partial groupoid into K is a surjective strong pseudoimmersion (immersion). LEMMA 6.3: Let f be a pseudoimmersion of an exchangeable partial groupoid L into a primary partial groupoid K. (i)
Then:
f(M(L)) = M(K), f(A(L)) = A(K), f(B(L)
f(O(L»
=
= B(K), f(C(L»
C(K) and
O(K).
(ii) The partiaZgroupoid L is primary.
Moreover, L is strongly primary (resp.
balanced, provided K is so. (iii) If K is balanced and f is an immersion then f is an isomoprhism.
328
A. Drapal and T. Kepka
PROOF: Easy. LEMMA 6.4: :.oet K be a (strongly) primary partial-groupoid. Then there exist a balance,:! (strongly) primary partial groupoid L and a surjective strong il71l1ersion k of L on;;o K.
PROOF: Use 6.1 and 6.3 (ii). LEMMA 6.5: Let K be a finite exchangeable partial groupoid. Then there exist a ?rimw'" partial groupoid L and an injeotive homomorphism f of
PROOF: We shall proceed by induction on m(K).
L into K.
Suppose that K is not primary.
Then there are an exchangeable partial groupoid J and a pseudoimmersion g of J into K such that g(M(J)) f M(K). of I into K.
Put 1(0) = g[J) and denote by i the identity mapping
By 6.2, 1(0) is exchangeable.
Moreover, M(I(o))
and i is an injective homomorphism of 1(0) into K.
g(t4(J))
=
~
M(K)
We have m(I(o)) < m(K), and
hence there are a primary partial groupoid l and an injective homomorphism h of L into 1(0).
Now, it suffices to put f
= ih.
COROLLARY 6.6: ;:'e t K be a fini te exohangeab Ze partia l groupoid. Then there exis t a i;',;!ianced primw'y partial groupoid L and a immersion f of L into K.
PROOF: Apply 6.4 and 6.5. PROPOSITION 6.7: Let K be a finite exohangeable partial- groupoid. Then there exist ;;: balanced stmngly primary partial groupoid L and a pseudoimmersion f of L into K.
PROOF: Put m = card B(K) and n
=
card C(K).
If I is a reduced partial groupoid and
9 is a pseudoimmersion of I into K then I is finite and cardI
~
m+n+mn.
Hence
there are a balanced primary partial groupoid L and a pseudoimmersion f of L into K such that card
Dill
~
card D(J) whenever J is a balanced primary partial groupoid
with a pseudoimmersion h of J into K (use 6.7).
LEMMA 6.8: Let K,L
E
R and let
Obviously, l is strongly primary.
f be a surjective strong immersion of
:;'h.:n K is exclumgeabZe (resp. primary, strongly primary) iff L is so.
L onto K.
Group modifcations of some partial groupoids
PROOF: If L is exchangeable then K is exchangeable by 6.2.
329
Conversely, if K is
exchangeable (resp. primary, strongly primary) then it is easy to check that L is so (see the proof of 6.1 and 6.3(ii)).
Now, suppose that L is (strongly) primary
and g is a pseudoimmersion of an exchangeable partial groupoid I into K.
Then K is
exchangeable and there exist a balanced exchangeable partial groupoid J and a surjective strong immersion h of J onto I. of J into L.
By 2.4, gh
= fk for a homomorphism k
It is easy to see that k is a pseudoimmersion, and hence k is
surjective and strong.
The rest is clear.
7. GROUP DISTANCES OF FINITE QUASIGROUPS Let G be a groupoid and Ma non-empty subset of G2 C = {c; (b,c) E M} and D = {ab; (a,b) EM}. K(o)
=
Put B = {b; (b,c) EM},
We define a partial groupoid
K[G.M las follows: K = B U CUD and for a,b E K, a
0
b is defined iff
= abo It is easy to see that K(o) is reduced and it is cancellative, provided G is so. Moreover, M(K(o))) = M. (a,b) E M; in that case, a
b
0
PROPOSITION 7.1: Let Q(o) and Q("') be 2
and M= {(a,b) E Q ; a
0
b -f a", b}.
quasigroups with the same underZying set Then the partiaZgroupoidsK[Q(o),M] and
K[Q(*),M] have the same underZying set and form a aoupZe of companions. PROOF: Easy. Let G be a grupoi d, t4 a non-empty subset of G2 and K( 0) = K[ G,M].
Let K(>,')
be a partial groupoid with the same underlying set as K(o) and let M~ M(K(*)). define
agroupoidG('~) =
(a,b) , M; a * b
G[M,KU')] as follows: ai, b
=
ab for all a,b E G,
= a * b for all a,b E G, (a,b) E M.
PROPOSITION 7.2: Let Q(o) be a quasigroup and Ma non-empty subset of Q2. that the partiaZ groupoid K( 0)
Q("')
=
Q(o)[M,K("')]
KU') = K[Q('·'),M]. PROOF: Easy.
We
Then
= K[ Q( 0) ,M]
Q('~)
Suppose
has a companion K("') and put
is a quasigroup, M= ({a,b); a
0
b -f
a'~
b}and
A. Drapal and T. Kepka
330
PROPOSITION 7.3: Let Q be a finite group, Q("') a quasigroup, M = {(a,b); ab# a", b} ani K(o) = K[ Q,M\.
::hen:
U) K(o) is a regular exchangeahle partial groupoid and M(K(o))) = M. (ii) 'TiJe!'e exist a regular balanced primary partial groupoid L and an irrrnersio'l f of Linto K( 0). (iii) There exiar a pseudoregular balanced strongly primary partial groupoid and a pseuQoimT1n'sion 9 of I into K( 0).
PROOF: Apply 3.6, 6.6 and 6.7. LEMMA 7.4: '-ee Q ce a group of a finite ol"del" n;;' 2 and ~hat
gdist(n)
~'hen
K(o) ia a l"egular pl"imary partial groupoid and gdist(n)
=
dist(Q,Q("')).
Q('~)
Put M = {(a,b); ab # a", b}
a quasigl"oup such and K(o) = K[Q,Mj. =
m(K(o)).
PROOF: The result is an easy consequence of 7.1,7.2 and 6.2. PROPOSITION 7.5: For every n ;;. 2, gdist(n) = min m(K) where K runs through all (ps"udo,'pegulal' balanced (strongly) primary partial groupoids such that thel"e exist Q of ol"der n and a (psew1o)immersion of K into Q.
'-' gl'OUp
PROOF: Apply 3.7, 7.3 and 7.4. Let Q be a group, u,v M ((a,b) x
0
y
E
R2; ab t a
= xy if x,y
a,bE R.
E
J.
E
b).
Q, R a subgroup of Q, R(*) a quasi group and Define a new operation
Q and either x , uR or y , Rv; (ua)
We shall use the notation Q(o)
=
E
Q, xy # x
0
y iff x
=
ua and y
=
on Q as follows: 0
(bv) = uta * b)v for all
Q[R(,"),u,v).
LEMMA 7.6: Q(o) is a quasigroup, dist(Q,Q(o)) x,y
0
= dist(R,R(*))
bv for (a, b)
card Mand, for
E r~.
PROOF: Easy. Let Q be a group, Q(o) a quasigroup, M = {(x,y)
E
Q2; xy # x
B
{x; (x,y) EM}, C = {y; (x,y) E M} and D = {xy; (x,y) E M}.
D
{x
0
y; (x,y) EM}.
0
y},
Then
Further, let (u,v) E M and let R be the subgroup of Q
331
Group modifcations of some partial groupoids
-1
1
generated by all u x, yv- , X E B, y E C. a i, b
=
-1
ab if a,b E Rand (a,b) t- (u x,yv
-1
= u (x
y)v
0
-1
* on
Define an operation -1
R as follows:
-1
) for all (x,y) EM; (u X)i'(yv
-1
) =
for all (x,y) E M.
LEMMA 7.7: R(",) is a quasigroup,{(a,b) E R2; ab t- a
i,
b}
-1
{(u x,yv
-1
);(X,y)E M}
= Q[ R(,,) ,u ,v] .
and Q( 0)
PROOF: Easy. Further, let K E T and let f be a pseudoimmersion of K into Q such that f(M(K)) = M. f(a) h(d)
Clearly, m(K) = card M.
u and f(b) = v.
a =
f(d)v
-1
Let x = (a,b) E M(K) be such that
Define a mapping h of K into R by h(c) = u-lf(c),
and h(cd)
=
-1-1
u f(c)f(d)v
for every (c,d) E M(K).
pseudoimmersion and it is an immersion, provided f is so.
Then h is a
We have h(a) = 1 = h(b)
and sx = ker qx -c ker h. Hence h induces a pseudoimmersion k of Kx into R such that h = kqx' If f is an immersion then k is so. Moreover, there is a unique homomorphism g of G(Kx) into R such that k = ggK' and S
=
Ker g is a pseudoregul ar subgroup of G(
<).
Obviously, g is surjective If f is an immers i on then S
is a regular subgroup.
The group R is isomorphic to the factorgroup G(K liS, x Now, consider the isomorphism hx of G(Kx) onto H(K) (see 5.2). Then T = hx(S) is
a normal subgroup of H(K), T is pseudoregular and it is regular, provided f is an immersion.
Moreover, the groups Rand H(K)/T are isomorphic.
If Q is finite then
H(K)/T is finite and card H(K)/T divides card Q. LEMMA 7.8: For every n
~
2 there exist a finite {pseudo)regular (strongly) primary
balanced partial groupoid K and a (pseudo)regular subgroup T of (H(K) such that
m(K) = gdist(n), H(K)/T is finite and card H(K)/T divides n. PROOF: The assertion is an easy consequence of 7.3, 7.4 and the preceding considerations. LEt-1MA 7.9.: Let n
~
2 be an integer and Kan exchangeable balanced partial groupoid
such that there exists a pseudoregular subgroup T of H(K) such that H(K)/T is finite and card H(K)/T divides n.
Then gdist(n)
~
m(K).
332
A. Drapai and T. Kepka
PROOF: There is a group Q of order n such that H(K)/T is isomorphic to a subgroup R of Q.
Let x E M(K), q be the natural homomorphism of H(K) onto H(K)/T and p an
isomorphism of H(K)/T onto R. into R (see 5.5).
Put f
= pqtxg K so that f is a pseudoimmersion of K
Let L(o) = f[ KJ = K[Q,MJ, M= f(M(K)).
exchangeable, and hence gdist(n) THEOREM 7.10: For every n
~
~
By 6.2, L(o) is
m(L(o)) = m(K).
2, gdist(n)
= min m(K)
where K runs through all finite
(psewioJregular balanced (strongly) primary partial groupo ids such that H(K)/T is fiiJhe and card H(K)/T divides n for a (pseudo)regular subgroup T of H(K).
PROOF: Apply 7.8 and 7.9.
BI BLI OGRAPHY 1.
A. Drapal, Quasigroups rich in associative triples (to appear).
HFF UK 186 00 PRAHA 8 - Karl{n Sokolovsk~ 83 Czechoslovakia
333
Annals of Discrete Mathematics 18 (1983) 333-334 © North-Holland Publishing Company
BIREFLECTIONALITY Erich W. Ellers
Let G be any group and IT
2
1.
IT
an element in G.
Then
IT
is called an involution if
A group G is called bireflectional if every element in G is a product of
two involutions.
Many groups are bireflectional. We mention some of them, espe-
cially those with geometric significance: The motion group of a Euclideen, an elliptic, and a hyperbolic plane; the group of motions of the real Euclidean geometry of any finite dimension; the group of projectivities of a line, the projective general linear group PGL 2(K); the symmetric group Sn; the dihedral group D ; the orthogonal group. 2n The bireflectionality of the orthogonal group has been proved by M.J. Wonenburger [11] for the case that the fi e1d of scalars has characteri sti c di stinct from 2.
For the case that the field of scalars has characteristic 2 the
problem has recently been solved by E.W. Ellers and W. Nolte [8].
In the same
paper it is also shown that the symplectic group is bireflectional if the field of scalars has characteristic 2. The symplectic group is not bireflectional if the characteristic of the field of scalars is distinct from 2.
The reason is that in any regular metric
vector space a simple mapping (i.e. a mapping that keeps every vector of a certain hyperplane fixed) is an involution if it is a product of two involutions (see E.W. Ellers [6]).
Now any symplectic simple mapping is a transvection and if the
field of scalars has characteristic distinct from 2, then a transvection is not i nvo 1utory. Let Pl and P2 be involutions in G and IT = P1P2' Then , , ' t t 't ' S 1 2 l = P2P l = P1ITP l = P2ITP 2 , 1.e. IT 1S conJuga e 0 1 S 1nverse. evera authors [2], [4], [10], [11] showed that if IT is an invertible linear transformaLet G be any group.
IT -1
P-1 P-1
tion of a finite dimensional vector space over a commutative field and if conjugate to its inverse, then
IT
IT
is
is a product of two involutory transformations.
This theorem does not hold any more if the field of scalars is not commutative.
334
E.W. Ellers
counterexamples exist for a vector space over the quaternions (see E.W. Ellers {7J).
The same paper contains a criterion which turns out to be very useful for
detecting if a certain subgroup of GL(V) is not bireflectional.
Namely, if a
simple mapping
0
transvection.
As a consequence of this lemma we obtain easily that the general
is a product of two involutions, then
G
is an involution or a
linear group GL(V) is not biref1ectional since it clearly contains simple mappings that are neither transvections nor involutions. W.H. Gustafson, P.R. Ha1mos, and H. Radjavi [9] show that over a commutative field every square matrix with determinant involutions.
±
1 is a product of not more than four
While the above stated result gives a criterion when a transforma-
tion is a product of two reflections, no criterion is known for products of three reflections.
C.S. Ballantine [3] gives some conditions under which a mapping is
the product of three reflections. BIBLIOGRAPHY 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
F. Bachmann, "Aufbau der Geometrie aus dem Spiege1ungsbegriff", 2nd edition. Springer Verlag, New York-Heidelberg-Ber1in (1973). C.S. Ballantine, Some involutory similarities, Lin./Multilirz. Alg., 3 (1975), 19-23. C.S. Ballantine, Products of involutory matrices, Lirz./Multilirz. Alg., 5 (1977), 53-62. D.Z. Djokovic, Product of two involutions, Al'ch. Math., 18 (1967), 582-584. E.W. Ellers, Decomposition of orthogonal, symplectic, and unitary isometries into simple isometries, Abh. Math. Sem. Univ. HambU1'g, 46 (1977), 97-127. E.W. Ellers, Bireflectionality in classical groups, Canad. J. Math., XXIX (1977), 1157-1162. E.W. Ellers, Products of two involutory matrices over skewfields, Lin. Alg/Appl., 26 (1979), 59-63. E.W. Ellers and W. Nolte, Bireflectionality of the orthogonal group. (To appear) . W.H. Gustafson, P.R. Ha1mos and H. Radjavi, Products of involutions, Lin. Alg/AppZ., 13 (1976), 157-162. F. Hoffman and E.C. Paige, Products of two involutions in the general linear group, hdiana Univ. Math. J., 20 (1971), 1017-1020. M.J. Wonenburger, Transformations which are products of two involutions, .T. f.iat/;. Mech., 16 (1966), 327-338.
Department of Mathematics University of Toronto Toronto, Canada M5S lA1
335
Annals of Discrete Mathematics 18 (1983) 335-342 © North-Holland Publishing Company
A GRAPHIC CHARACTERIZATION OF HERMITIAN CURVES 1r
G. Faina and G. Korchmaros
1. INTRODUCTION In PG(2,q), q square, a set U of points is called a Hermitian arc (or a unital) if it is a
(qv'q + 1, v'q + l)-arc
of type (1,
v'q + 1).
In particular, U is
called a Hermitian curve if there is a Hermitian polarity for which U is the set of self-conjugate points. It had been conjectured that a Hermitian arc is always a Hermitian curve. In fact this is not true.
There is a counter-example for every q > 4, see
Buekenhout [ 1), Metz [ 10) • In PG(2,q), q square, a set
r of v'q + 1 collinear
subline if there is a Baer subplane B containing
r.
points is called a Baer
The notion of Baer subline
forms the starting point of a classification of Hermitian arcs, see Lefevre-Percsy [7, 8). The following theorem is noteworthy in this context. THEOREM (Lefevre-Percsy [7)): A Hermitian arc U is a Hermitian curve if and only if U has the following property: (a) every (v'q+l)-secant r of U meets U in a Baer subline.
The technique used by Lefevre-Percsy depends on the characterization of the sets of type (0,1 ,n,n+l) of PG(4,n) given also in [9). In this paper we obtain the above theorem in another way.
Our proof uses a
procedure originally due to Segre [11) and a Tallini-Scafati theorem on reciprocal Hermitian arcs [14).
Furthermore, it involves the application of Hering's theorem
[4) on group spaces and finite permutation groups. A discussion of Hermitian arcs and curves may be found in [2), [5), [6) and [ 13). For a thorough discussion of projective planes the reader may refer to the books [3), [5), [6), [12) . It is well known that every Hermitian curve has the above property (a).
336
G. Faina and G. Korchmaros
Furthermore, the only Hermitian arcs of PG(2,4) are the Hermitian curves, see [5, p. 3451.
We assume from now on that q > 4 and U is a Hermitian arc having property
(u).
In order to prove that U is then a Hermitian curve, the first step consists
in proving that U is "quasi"-reciprocal. group leaving U invariant.
Next we shall determine a collineation
In the final step we shall prove that U is reciprocal.
As the reciprocal Hermitian arcs are Hermitian curves (Tallini-Scafati [14, Prop. XI] see also [5, 12.3.10]), it will follow the Theorem.
2. "QUASI"-RECIPROCAL HERMITIAN ARCS We say that a Hermiti an arc K is "quas i "- reciproca 1 if it has one of the following equivalent properties (B) F01' every point Q not on K, the y'q+ 1 unisecants to K through Q meet K in
y'q+ 1 c,;;Uineal' points. (13') For eVe!');; (y'q+l)-secant r of K, the y'q+l unisecants to K at the points
cr r
n K, meet at a point.
By a theorem of Tallini-Scafati ([ 14], [5, 12.3.9)), the reciprocal Hermitian arcs are also "quasi "-reciprocal. Given a "quasi"-reciprocal Hermitian arc K, we can define a bijective correspondence w between points that are not on K and (y'q+l)-secants to K as follows: If Q ~ K, then w(Q) is the line containing the y'q+l points of contact of the yq+l unisecants to K through Q.
If r is a (yq~l)-secant to K, then w(r) is the
common point of the yq+l unisecants to K at the points of r n K. For the next results, and in various places later on, the following two simple but useful facts are needed. PROPOSITION 1: If B is a Bael' subp~ane and r is a (y'q+ 1 )-secant to U such that B n U has more than
~o
points of r, then r n U=r n B.
PROOF: It is well known that whenever three points are collinear they are contained in a unique Baer subplane.
As U has property (a) this proves Prop. 1.
337
A graphic characterization of Hermitian curves
PROPOSITION 2: If a Baer subpZane B contains two Baer subZines of U then B n U consists onZy of these Baer subZines.
PROOF: Assume by way of contradiction that B n U contains a further point P.
Then,
from P there is a line r belonging to B and meeting the two common Baer sublines in two distinct points.
By Prop. 1 it follows that r n U=r n B. This means that
B n U contains three Baer sublines which form a triangle.
As every point T of B
lies on a line belonging to B and meeting this triangle in three distinct points, we have by Prop. 1 that T belongs also to U. this is not possible.
Therefore, B is contained in U.
But
In fact, as lUI> IBI, from B C U it follows that there is
a point S of U which does not belong to B.
In view of B C U, the unisecant to U
at S has no common point with B, contradicting the well known Baer's theorem that each line of PG(2,q), q square, has at least one common point with every Baer subplane. PROPOSITION 3: U is Iquasi"-reciprocaZ. PROOF: We prove that U has property (S). Assume by way of contradiction that from Q there are three unisecants t ,t 2,t to U such that l 3
are three non collinear points. As each two Baer sublines having a common point are contained in a unique Baer subplane, it follows from (a) that there is a Baer subplane B for which (1 )
By Prop. 2, we have then that (2)
Take A1A2A3 as the triangle of reference for a homogeneous coordinatization of PG(2,q).
Thus the triangle is
Al = (1,0,0),
A2
g
(0,1,0).
A3 = (0,0,1).
Any line r.1 through A.1 and distinct from A.A. 1 and A.A. 1 can be written as 1 11 1+
338
G. Faina and G. Korchmtiros
where o.1 is a non-zero element of the coordinatizing field GF(q) of PG(2,q). call
We
the slope of r .•
p. 1
1
Through A. there are v'q-1 lines 1 .. which intersect A. 1A. 1 n U and are 1 1J 1- 1+ distinct from A.A. 1 and A.A. l' Denote by A•• the slope of 1 ..• As A. lAo 1 n U 1 11 1+ 1J 1J 1- 1+ is a Baer sub1ine containing A.1- 1 and A.1+ l' there is a fixed element y.1 of GF(q) for which yq..1 U
j=l
Put '\ =
A..
1J
I),
{AY1.
E
GF(vlq-)-{O}}.
(3)
..rq:1 j~l \r
Since the product of the non-zero elements of GF(v'q) is -1, from (3) we have v'q:l
'\ = (- 1)
Yi
.
(4)
Without loss of generality, we can assume that (1,1,1) belongs to B. by (1) Yl'Y2 belong to GF(v'q) but by (2) Y 3 implies that 1\
E
GF(q)-GF(vCi).
Then,
Therefore, (4)
= - 1,
2
(5)
Through Ai there are q-1 lines different from Ai Ai _ and AiAi+1; they 1 consist of q-2 (v'q+l)-secants s .. and the unisecant t .• Put 1 1J q-2 Li
where
D .. 1J
j g1 Dij'
denotes the slope of s ..• 1J
Moreover, denote by
T. 1
the slope of t .. 1
Since the product of the non-zero elements of GF(q) is -1, we have (6) t~oreover, if
slope of RA i , then
R is a point of U' = U - {A1A2,A1A3,A2A3} and \1i denotes the J
l \12\13 = 1.
Thus, (7)
It is easy to see that
339
A graphic characterization of Hermitian curves
(R ~ u' \.I1\.12\.13)('l"2A:3)
=
(~1~2l:3)
v'q
•
By (5), (6) and (7), this implies that (8)
As, by assumption, t 1,t 2 and t3 have a common point, we have '1'2'3 = 1. From this and (8) it follows that A3 = -1, contradicting (5). This proves Prop. 3.
3. COLLINEATION GROUP LEAVING U INVARIANT
The next proposition that will be useful in the sequel is a consequence of Prop. 3. PROPOSITION 4: If two Baer sub lines belonging to U have a common point P on U then the Baer subplane containing these Baer sublines contains also the unisecant to U at P.
PROOF: Let r 1 and r 2 denote two (v'q+1)-secants to U such that r 1 n r 2 belongs to U. Let B be the (unique) Baer subp1ane containing both r l n U and r n U. Denote by 2 t the unisecant to U at P. We have to prove that t belongs to B. Let w(r 1) be as in Section 2. It is clear that w(r l ) lies on t. Let us consider any line r which belongs to B but does not pass through P. Then r n r f r n r , r n r l E U, r n r 2 E U. Thus r is a (v'q+1)-secant to U. In 2 l particular, r cannot pass through w(r 1). As there is at least one line which contains w(r ) and belongs to B, by Baer's theorem, it follows that t must belong 1 to B. This proposition enables us to find nontrivial homologies leaving U invariant. PROPOSITION 6: If T is any point of U and t denotes the unisecant to U at T, then T is the center and t is the axis ofv'Ci elations which form a group 6(T,t) having the following properties
(i) 6(T,t) Zeaves U invariant; (ii) 6(T,t) is an elementary abelian group of order q;
340
G. Faina and G. Korchmtiros
(iii) 6(T,t) acts regularZy on r'-{T}, where r' is any Baer subline containing T and belong1:ng to U.
PROOF: Let f(T,t) be the full group of elations with center T and axis t. and L' be two points of U which are collinear with T. ~(L,L')
which maps L onto L'.
Let L
Then there is an elation
We prove that ,(L,L') leaves U invariant.
Let S be
any point of U. Assume first that S ~ TL. LS
n
Putting S' = S·
U.
lies on t.
(L L')
,
Let B be the Baer subplane containing LT n U and
,we have that the common point H of LS with L'S'
As, by Prop. 5, t belongs to B, H E B.
of two lines belonging to B.
It follows that S'
and 5' EST, this implies that S' Now we assume that S E TL.
E
Thus, S' is the common point B.
As B n U contains ST n B
U.
E
We can repeat the above argument by replacing L
by a point J i 5T and L' by J' = J' (L , L') . We obtain S' S E TL.
E
U
also for the case
So we have (i).
It is clear that 6(T,t) is the subgroup of r(T,t) which leaves U invariant. It is well known that r(T,t) is an elementary abelian group which acts regularly on r - {T}, where r is any line through T and distinct from t. r' -
{T}
As each set
has exactly v'q points, we have also (ii) and (iii).
PROPOSITION 7: Every point Q not on U is the center of a nontrivial homology with axis w(Q) "-'hich leaves U invariant.
PROOF: Let G be the co11ineation group generated by all elations contained in ~(T,t)
with T E w(Q).
By Prop. 6 each point T of w(Q) n U is the center of an
elation of G which fixes Q.
This and (ii) of Prop. 6 enable us to apply Hering's
theorem on group spaces [4,3.1].
By this theorem it follows that G is isomorphic
to SL(2,v'q). We prove that G acts regularly on U - w(Q). point 0 E U - w(Q) for which g(O)=O. unisecant to U at O.
Then I
~
Oenote by I the common point of w(Q) and the
w(Q) n U and I F Q.
triangle and QI is a (v'q+1)-secant to U. also g(w(QI))=w(QI).
Let g E G such that there is a Furthermore, I,O,Q form a
Clearly g(I)=I.
As g leaves U invarian~
We prove that I,O,Q,w(QI) form a proper quadrangle.
to verify that w(QI) does not lie on any side of QOI.
We have
As Qw(QI) is a unisecant to
U, w(QI) ~ QO and w(QI) ~ QI. As Iw(QI) is a (yq~l)-secant to U and as Dr is a
A graphic characterization of Hermitian curves
unisecant to U, w(QI) proper quadrangle.
~
DI.
341
It follows that 9 fixes four points which form a
As G is generated by elations, this implies that g is the
identity collineation of PG(2,q).
Since [SL(2,~[ =[ U - w(Q)[, this proves that
G acts regularly on U - w(Q). Let f(Q,w(Q)) be the full group of homologies with center Q and axis w(Q). By [3,3.1.12] for every h E f(Q,w(Q)) and 9 E G hg=hg. Let J and J' be two collinear points with Q. Then there is a homology h such that h(J)=J'. U - w(Q). g(J)=M.
We prove that h leaves U invariant. Let Mbe any point of
Then, as G acts regularly on U - w(Q), there is a 9 in G such that As gh=hg,
M'=h(M)=h(g(J))=hg(J)=gh(J)=g(h(J))=g(J'). Therefore, M' belongs to U - w( Q) • U - w(Q).
r~oreover,
h fi xes every poi nt of
It follows our assertion which proves also Prop. 7.
PROPOSITION 8: Let Ql and Q2 be two points not on U. Let hl E f(O ,w(Q )) and h2 E f(Q2,W(Q2)) be two homoZogies which leave U invariant. If QllE W(Q2) then also Q2 E w(Ql ).
PROOF: Ql E w(Q2) implies that h2(Ql)=Ql' As h2 leaves U invariant, it follows that h2 maps w(Ql) into itself. As w(Ql) # w(Q2)' this implies that Q2 E w(Ql)' 4. U IS A RECIPROCAL HERMITIAN ARC A Hermitian arc K is called reciprocal if any four of its points Pl ,P 2 ,P 3 , P , three of which are never collinear, with respective unisecants Pl,P2,P3,P4' 4 satisfy the condition that when Pl n P2 lies on P3P4 then P3 n P4 lies on P1P2• By a theorem of Tallini-Scafati ([ 14], [5, 12.3.10]), any reciprocal Hermitian arc is a Hermitian curve. In order to prove that U is reciprocal let us consider any four points P ,P ,P ,P of U, three of which are never collinear. Assume that Pl n P2 lies on l 2 3 4 P3P4• Put Ql = Pl n P2 and Q2 = P3 n P4' Then w(Ol) = Pl n P2 and w(Q2)=P3 n P4' By Prop. 7, there are two homologies hl E f(Ql,W(Ql)) and h2 E f(Q2,W(Q2)) which leave U invariant. Since Ql E w(Q2)' by Prop. 8, we have now also Q2 E w(Ql)' This means that P3 n P4 lies on P1P2 •
G. Faina and G. Korchmaros
342
BIBLIOGRAPHY 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
F. Buekenhout, Existence of unita1s in finite translation planes of order q2 with kernel q, Geom. Dedicata, 5 (1976), 184-194. F. Buekenhout, Characterizations of semi quadrics, Atti dei Convegni Lincei, 17 vol. I (1976), 393-421. P. Dembowski, linite Geometpies, Springer Verlag (1968). Ch. Hering, A theorem on group spaces, Hokkaido Math. Journal, 8 (1979), 115-120. J.W.P. Hirschfeld, Ppojective Geometpies ovep Finite Fields, Calderon Press, Oxford (1979). D.R. Hughes and F.C. Piper, Projective PZanes, Springer Verlag (1973). C. Lefevre-Percsy, Characterization of Hermitian curves. Apchiv. dep Math., (Basel) (to appear). C. Lefevre-percsy. Semi quadriques en tant que sous-ensembles des espaces projectifs, BuZl. Soc. Math. BeZg., Ser. B 29 (1977), 175-183. C. Lefevre-percsy. Ensembles de classe (O.l,q,q+l) de PG(d,q), Joupnal of Geometpy, 15 (1980). 93-98. R. Metz, On a class of unitals, Geom. Dedicata, 8 (1979). 125-126. B. Segre. Ovals in a finite projective plane, Ganad. JouPnal of Math., 7 (1955), 414-416. B. Segre. Lectupes on modePn geometry, Edizioni Cremonese (Roma) (1959). B. Segre, Forme e geometrie Hermitiane, con partico1are riguardo a1 caso finito, AnaaZi Mat. Pupa AppZ., 70 (1965). 1-201. M. Ta11ini-Scafati, {k.n}-archi di un piano grafico finito, con partico1are riguardo a quel1i con due caratteri. Nota I e II, Rend. Accad. Naz. ~incei, (8)40 (1966) 812-818, 1020-1025.
Istituto di ~~atematica Universita di Perugia Vi a Vanvite 11 i 06100 Perugia Italy
Istituto di Matematica Universita di Bari Via Nicolai 2 70121 Bari Italy
343
Annals of Discrete Mathematics 18 (1983) 343-354 North-Holland Publishing Company
©
ON THE CONNECTIVITY OF MAXIMAL PLANAR GRAPHS WITH mNIMUM DEGREE 5
Stefano Fanelli
ABSTRACT
Let Fn, k be the classes of maximal planar graphs Gn (without loops or multiple edges) having k vertices of degree 5 and n-k vertices of degree more than 5. Hakimi and Schmeichel proved that if Gn E Fn, k and k=12, 13, then Gn is 5connected. In this paper the author gives some general conditions in terms of the integers nand k for a graph Gn E Fn, k to be p-connected, with p=4,5.
1. I NTRODUCTI ON
k
dpP denote the vertex degree sequence of an undirected graph Gn on n vertices (without loops or multiple edges) having k, vertices of degree d , k2 vertices of degree d2.....• kp vertices of d~gree dp' with l kl +k 2+ •.•.• kp = nand dl > d2 > ..... > dp' kl k2 k The notation Gn dl d2 ••••.. dpP will be used throughout this paper to specify briefly the vertex degree sequence of a graph Gn with the above mentioned
=
properties. Let Fn, k be the classes of graphs GR following conditions:
=dkll
k
d P which satisfy the p
(i) G are maximal planar, n
(ii) G have k vertices of degree 5 and n-k vertices of degree more than 5. n
In [7J Schmeichel and Hakimi proved an existence theorem related to the kl k2 k graphs Gn E Fn, k such that : Gn 7 6 5. This theorem, which was extended by the author in [lJ, reads as follows:
=
344
S. Fanelli
THEOREr~
1.1
kl
G n
=7
(see [7) Theorem 2, Theorem 3; (1) Theorem 3): The classes
1<2 I< 6 5
F k are non empty, n, 513 , G16 71 62 513 ' E
=
and "possibly" the following undecided cases:
h=O,l,2.
In [51 Hakimi and Schmeichel proved some fundamental results on the connectivity of maximal planar graphs.
Using the same terminology of [51, such results
can be referred to the classes Fn, k as follows: THEOREM 1.2 (see [51 Lel1l11a 1, Lemma 2): If G E F k contains a separating 4-cycle n n, 01'
a separating 3-cycle diffe2'ent from those bounding the faces, then both the
i>lter'ior' an.d the exwr'ior of this cycle must each contain at least 7 ver'tices of degr'ee 5.
Gn
=
F 12 1'ff Gn = 6n- 12 512 and Gn E Fn, 13 l'ff S1· nce a grap h Gn En, 1 n 14 7 6 - 513 , from Theorem 1.1 and Theorem 1.2 one derives the following re-
su1t which gives a complete answer to the existence and the connectivity of the classes Fn,12 and Fn ,13: THEOREM 1.3: Ther'e exists a graph Gn E Fn, 12' V n ~ 12, n ~ 13. There exists a gr'aph G E F ,13' V n ~ 17. Moreover' in the classes F • n n n 12 and Fn ,13 all the graphs a2'e 5-connected.
The latter connectivity-result cannot be extended to the graphs Gn E Fn, 14' since Hakimi and Schmeichel showed that in such a class there exists a graph having connectivity equal to 3 (see [5] Table 1,
a =
0 S = 14).
The aim of the present paper is to extend Theorem 1.3, giving sharper sufficient conditions in terms of the integers nand k for a graph Gn E Fn, k to be pconnected, with p=4 or 5. In this work, we prove in particular: a) sufficient conditions, in terms of the integer n, for a graph Gn E Fn, k to be 5-connected, when k is fixed, 14 ~ k ~ 16. b) necessary conditions, in terms of the integer n, for a graph Gn E Fn, k to
345
On the connectivity of maximal planar graphs
have connectivity 3. when k is fixed. 14 < k < 18 (hence. sufficient conditions for a graph G
n
E
F k to be 4-connected). n.
2. 5-CONNECTED GRAPHS In [6] Ruscitti showed that: E F 2 iff n ;;'14, n n.nFurthermore if n +12+10h (h=1,2 •••. ), then 3! G E F 2 and the latter n n.n-
THEOREt1 2.1 (see [6] Theorem 6.1. Corollary 6.2): 3G even.
graph is 5-connected.
THEORH12.2 (see [6] Theorem 7.1): 3G
n
E
F 3 iff n=15+10h, n=19+10h. n=21+10h. n.n-
n=24+2h (h=O.l ••••.. ). We first prove the following existence theorem: THEOREM 2.3: I} 3 Gn E II} 3 Gn E I II} 3 Gn E IV} 3 Gn E V} 3 Gn E
F n.14 F n.15 F n,16 F n.ll F n.18
If
n ;;. 16 • n n ;;. 19
If
n ;;. 18
If
n ;;. 21
• If
n ;;. 20
• If
+ 17;
PROOF: It is well known that the classes Flare empty (see [4] p. 272-275). n.nThus there are no graphs G15 E F15 •14 • G16 E F16 •15 • G17 E F17 •16 • G18 E F18 •17 G19 E F19.18' Moreover. by Theorem 2.1 and Theorem 2.2 it follows that the classes F17 ,15 • F19 •17 ' F20 • 17 are also empty. Since Theorem 1.1 implies that n 73 6n- 18 515 3G 72 6 - 16 514 E F If n;;' 16 with n + 17 and 3G n n.14 n If n ;;. 19, one can immediately prove I} and II}.
=
=
E
F n.15
To prove III}. IV} and V}. it is sufficient to notice that: i} the classes F18 • 16 • F19 •16 • F20 •18 • F21 • 18 are non empty by Theorem 2.1 and Theorem 2.2; ii} the sequences:
346
S. Fanelli
are planar (see (8) Theorem 4 n 16 iii) 3 Gn '" 7 6 - 20 5 E Fn,16 17 3G 75 6n- 22 5 E F n n,17 6 n 24 18 3G '" 7 6 5 E F n n,18
=
by virtue of Theorem 1.1.
4.6, Theorem 4.7); V n ;;;. 20; V n;;;' 23; V n ;;;. 24,
0
In [3) the author proved a lemma on the structure of vertices in the interior (or equivalently in the exterior) of a separating 4-cycle in a graph Gn E Fn,k' This lemma, which makes use of some results announced in [2], reads as follows: LEMMA 2.1 (see [3] Lemma 3.3): Let w(k) be the nwnbe1' of ve1'tices of deg1'ee k in in~erior
the
I(X) (in the exterio1' E(X)) of a sepa1'ating 4-cycle induced by a
vertex set X = {xl' x ' x ' x } in a g1'aph G
Asswne that there does not
edst,:n G a
C
2
XI
C
n Xu E(X)). ~f
3
n E Fn, k'
4
4-cycZe induced by any set XI
sepa!'a~ing
X u I (X) (by any set
;he,;;
w(5) = 7
w(6);;;' 4 , w(k)
~
0
V k ;;;. 7.
From 1heorem 2.3 and Lemma 2.1. we may state the following result which is an extension of Theorem 1.3: THEORErl 2.4:
SUPPOSE-
G
E
t
13;
n
F
n,k
with 12 .;; k .;; 16.
Consider the following hypoth-
esie: a)
k = 12 n ;;;. 12. n
bJ k c)
k
d) k 2)
k
13
n :;;. 17;
14 16 .;; n .;; 25. n 15 19 .;; n .;; 22;
t
17;
16 18 .;; n .;; 19.
1f one of the conditions a), b), c), d), e) is satisfied, then G is 5-connected. n
PROOF: By virtue of Theorem 1.3 and Theorem 2.3 each of the hypothesis stated in aI, b), C), d). e) implies the existence of graphs Gn E Fn, k' If k=12 or k=13 there is nothing to prove. Suppose that Gn contains a separating 4-cycle C. If k=14. from Theorem 1.2 and Lemma 2.1 there must be exactly 7 vertices of degree 5
On the connectivity ofmaximal planar graphs
347
and at least 4 vertices of degree 6 both in the interior and in the exterior of C. Hence n ~ 26, which proves the theorem under the hypothesis c). If k = lS, consider the graph drawn in Fig. 1; since in the interior of the 4-cycle xl x2x3x4 there are 8 vertices of degree S and no other vertices, from Theorem 1.2 and Lemma 2.1 we deduce that the minimum number n of vertices which may exist in a graph Gn E Fn, 15 having connectivity 4 is equal to 23. the theorem under the hypothesis d). The last case is trivial and is left to the reader.
This proves
0
We conclude this paragraph, giving a list of planar sequences, whose planar realizations are 5-connected. (n-2/2) 2 Sn-2 n ~ 16 , 6h S12 h ~ 0 , h f 1 8 1 6h S14 4 ~ h ~ 10 1 1 h 1S 8 7 6 5 2 ~ h ~ 5;
n even, n f 12+10h (h=1,2 ..• ); 71 6h 513 h ~ 3; 72 6h 514 2 ~ h ~ 9; 73 6h SlS 1 ~ h ~ 4; 91 66 SlS; 1 2 16 2 1 8 7 5 ; 8 6 516 .
3. 4-CONNECTED GRAPHS Let I(X) denote the set of vertices in the interior of a separating 3-cycle induced by a vertex set X = {xl' x2 ' x3 }. Let di(x j ) and wi(k) be respectively the number of vertices in I(X) to which x. is adjacent for j = 1,2,3 and the J
number of vertices of degree k in I(X). LEMI·1A 3.1: If G
We state some lemmas.
F aontains a separating 3-ayaZe n,k set X = {xl ' x2 ' x3 }, then: dl n
3 j~l
E
di(x j ) = witS)
C, induaed by a vertex
- h~7 (h-6) with)
where d denotes the maximum degree of the vertices of I(x). l
PROOF: Let G' be the maximal planar graph induced by Xu I(X). q
formula it follows:
Therefore one derives that:
By Euler's
s. Fanelli
348
d1
~
J
.Zl d.(x.) = 6q - 13 - 5 w.(S)
J"
J
1
6 wi (6) - h~7 h with)
1
=
d1 6 [&J.(S) + ",.(6) + hh with) + 3) - 18 - 5 witS) 1 1 d
dl
- h~7 h with) = witS) -
1
The following result was proved in [5) (see Lemma 1): LEW-1A 3.2: Sl
n
£;;;iucc;L:.
[';!
F
n, k'
E
n
VCl"t2X set X = {xl' x
a
l"c:wing 3-ayeleinduc"U: by
- d.(x.) ;;. 2
• x } and that thel"e does not exist a sepa3 2 any set X' c X u I(X). Then:
j=1,2,3
J
1
Assume that G contains a separating 3-cycle C
- 3j : di(x );;. 3. j
LEMMA 3.3: Let the assumptions of Lemma 3.2 be satisfied.
Then:
V h ;;. 8
PROOF: Suppose witS) ,,7.
By Lemma 3.1 and Lemma 3.2 it follows:
~
J
.Zl d.(x.) ,,7
J=
J
1
w.(h)
,
1
=
V h ;;. 7.
0
Therefore, applying Theorem 1.2, it is easy to check that the situation is as shown in Fig. 2. The vertices a,b,c,d in this figure induce a separating 4-cycle.
By Lemma
2.1, there must be at least 4 vertices of degree 6 in the interior of the cycle abed.
This proves a). Suppose w.(5) = 8. 1
Again from Lemma 3.1 and Lemma 3.2 we deduce that one of
the following two situations may occur: 3
i)
j~l d.1 (x.) J 3
i i ) .I
J= 1
d (x j i
)
=
7 ,
w. (7) = 1
= 8 , u)h)
Thus in both cases: w.(7) 1 that "'i(6) .;; 2.
~
1
= 0
• w.1 (h) = 0
V h ;;. 8;
V h;;' 7.
1 and w.(h) 1
= 0 V h ;;. 8. Now, suppose ab absurdo
349
On the connectivity of maximal planar graphs
The case i) is illustrated in Fig. 3; in such figure at least one of the four vertices of the separating 4-cyc1e abcd must have degree 5.
This is in
contrast to Lemma 2.1 or Theorem 1.2. The case ii) is illustrated in Fig. 4 and in Fig. 5; in both figures at least one among the three vertices a, c, d must have degree 5.
Suppose, for
instance, that the vertex a satisfies this condition; then, applying once again Theorem 1.2 and Lemma 2.1 to the 4-cycle xl ax x , we derive a contradiction. 3 2 proof is analogous for the vertices c and d. C ~Je
The
are able to prove now the following result, which gives necessary condi-
tions for a graph Gn E Fn, k to have connectivity 3. THEORH1 3.1: Suppose Gn E Fn, k' with 14 .;; k .;; lB. If G has connectivity 3, then one of the following conditions must be satisfied: n
a) k=14, n
~
33;
e) k=lB, n
~
21.
b) k=15, n ~ 29; c) k=16, n
~
25;
d) k=17, n
PROOF: The result follows from Theorem 1.2 and Lemma 3.3.
~
23;
Let C be a separating
3-cycle contained in Gn; denote by wi(k) (we(k)) the number of vertices of degree k in the interior (in the exterior) of C. If k=14, then: w.(5) = 7, 1 Hence n
~
33.
e (5) = 7, w.(6) 1
~
B,
e (6)
~
W
B.
If k=15, then one of the following two situations may occur:
I) wi (5) = 7, we(5) II) wi (5) = B, we(5) Therefore n
~
W
29.
7, wi (6) 7, wi (6)
~ ~
8, we (6) 3, we (6)
Suppose k=16.
~
8;
~
8.
Since the l-skeleton of the icosahedron is an
example of a separating 3-cycle having exactly 9 vertices of degree 5 and no other verti ces in its interior, the following four cases may occur: 7, w.(6)
~
7, w.(6)
~
III) w.(5) = 9, w (5) 7, w.(6) 1 1 e IV) w.1 (5) = 8, we (5) = 8, w.(6) 1
~
I) w.1 (5) = 7, we (5) = 8, w (5) II) w.(5) e 1
Hence n
~
25.
1
1
~
B, we (6) 3, w (6) e 0, w (6) e 3, w (6) e
~
8;
~
8;
~
8;
~
3.
If k=17 and k=lB the proof is trivial.
Using Theorem 2.3, we state:
0
350
S, Fanelli
COROLLARY 3.1: Suppode G
n
E
F
n,
k' with 14 .;; k .;; 18.
Consider the following hy-
;xl:.hesis:
a' ) k=14,
16 .;; n .;; 32,
b' ) k=15,
19 .;; n .;; 28;
c' ) k=16,
18 .;; n .;; 24;
d' ) k=17 ,
21 .;; n .;; 22;
e' ) k=18,
n=20.
n
t
17;
If one of the condih'ons a'), b'), c'), d'), e ') is satisfied, then G is n 4-conneeted.
•
F IG.1
5-vertices
On the connectivity a/maximal planar graphs
•
6-vertices
FIG.2
F I G.3
351
s. Fanelli
352
FIG.4
FIG.5
On the connectivity of maximal planar graphs
353
BIBLIOGRAPHY 1.
2. 3.
4. 5.
6.
7. 8.
S. Fanelli, On a conjecture on maximal planar sequences, J. Graph Theory, 4 (1980), 371-375. S. Fanelli, An unresolved conjecture on non maximal planar graphical sequences, Discrete Math., 36 (1981), 109-112. S. Fanelli, On the existence and connectivity of a class of maximal planar graphs. Calcolo, to appear. B. GrUnbaum, Convex polytopes. Wiley, New York (1967). S.L. Hakimi and E.F. Schmeichel, On the connectivity of maximal planar graphs, J. Graph Theory, 2 (1978), 307-314. A. Ruscitti, Sulla struttura delle classi F k di grafi planari massimali, Rend. Mat., Serie VI, 11 (1978), 243-~~9. E.F. Schmeichel ans S.L. Hakimi, On planar graphical degree sequences. SIAM J. Appl. Math., 32 (1977); 598-609. M. Tersigni, Problemi di esistenza, cardinalita e connessita in grafi planari massimali. Tesi di laurea, Universita dell'Aquila (1981).
Istituto t1atematico Universita dell'Aquila Via Roma 33 67100 L'Aquila Italy
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Annals of Discrete Mathematics 18 (1983) 355-376 © North-Holland Publishing Company
355
AFFINE GEOMETRIES OBTAINED FROM PROJECTIVE PLANES AND SKEW RESOLUTIONS ON AG(3,q) R. Fuji-Hara and S.A. Vanstone
{,
The main result of this paper is to give a construction of AG(n+l,q) from the projective plane PG(2,qn).
This construction can be applied to construct what
we call skew resolutions of the lines in AG(n,q) and pairs of orthogonal resolutions.
Denniston has shown that there exists a skew resolution of the lines in
AG(3,q) by constructing a resolution of the lines in PG(3,q). As an application of the main result, we give a direct proof of this and show that there is a one to one correspondence between skew resolutions of lines in AG(3,q) and resolution of lines in PG(3,q).
A skew resolution in AG(n,q) along with the natural resolution
of lines in AG(n,q) obtained from parallelism form a pair of orthogonal resolutions.
Orthogonal resolutions have recently attracted the interest of various
authors.
1. INTRODUCTION Although we are primarily interested in BIBDs, we begin by defining a slightly more general design.
An (r,l)-design D is a collection B of subsets
(blocks) from a finite set V (varieties) such that (a) every element of V is contained in precisely r blocks of B. (b) every pair of distinct elements of V is contained in exactly one block. If all blocks of D have the same cardinality (block size), then D is said to be a balanced incomplete block design (BIBD) with parameters (v.k,l). where v
= Ivi. and k is the block size. A (u,k,l)-BIBD with block set B' and variety
set V' is called a subdesign of a (v,k,l)-BIBD with block and variety set B and V respectively if V'
~
V and B' c B.
An (r,l)-design D is said to be resolvable if the blocks of B can be partitioned into classes (resolution classes) Rl , R2 , •.•• Rr such that every variety of V is contained in precisely one block of each class. Such a partitioning is
356
R. Fuii-Hara and S.A. Vanstone
called a resolution of D. The finite affine geometry AG(n.q) is usually obtained from a vector space over a Galois field.
Here, we state combinatorial axioms for AG(n,q) n
~
3, taken
from Lenz { 10J , as they are convenient for proofs of isomorphism which are discussed later. Let L be a collection of subsets (blocks or lines) of a finite set of points P. tion
A paralZelism of the system
L
= (P,L) is defined to be an equivalence rela-
II satisfying
(i') For any line
£
and point, there is a unique line 9,' such that
Let r be a system with a parallelism
tIlt' and
PE 9,'.
II satisfying the following conditions:
(i) Any two points Pl ' P are incident with (contained in) exactly one line 2 (ii) If P P II P P and P is a point on P P distinct from P and P , then 1 2 3 4 1 3 1 3 there is a pOint P' on PP 2 and P P 3 4 (iii) If no line contains more than two points and P , P ' P are distinct l 2 3 points, then the line 9,2 and 9,3 such as P2 E £2 II P1P3' P3 E £3 II P1P2 have a point in common. (iv) Some lines contains exactly q
~
2 points.
(v) There exist two lines neither parallel nor with a common point. AG(n,q) for some n If
I
~
Then
3.
is a BIBD, then it is easily seen that every parallel class, that is,
every class of the equivalence relation
II, contains equally many blocks. But
parallelism of BIBD's may not be unique, since any resolution is a parallelism. A finite affine plane is more simply defined to be a (q2,q,l)-BIBD.
It is
well known that an affine plane has a unique resolution (parallelism). If II is a parallelism of an affine geometry satisfying the axioms, then disjoint lines which are not parallel are called skew lines.
A resolution of
lines in an affine geometry, in which each class consists of skew lines, is called a skew resolution.
We display a construction for a skew resolution of AG(3,q) in
Section 4.
2. CONSTRUCTION OF RESOLVABLE «gn+1- l1/ (q-l),1)-DESIGNS 2 A projective plane of order m is an (m +m+l,m+l,1)-BIBD, varieties and block blocks in the design are called points and lines of the projective plane, respec-
357
Afline geometries obtained from projective planes
tively.
Let P be a projective plane of order qn, q a prime power.
be two lines in P intersecting at a point z. 2~
Let R. 00 and R. 0 U is a set of points on R.oo - {z},
lUI = u ~ qn. Let V(a), a E U, be a set of lines in
P-R."" passing through a
point a. We show construction for ((qn+l_l)/(q-l),l)-deSigns on the variety set
v=
U
V(a).
aEU Since every line of V meets the line R. o at a point of £0 - {z}, any element of V(a) can be represented by a pai r <x,a>, x E R. - {z}. o
Therefore,
V = {<X,a>; x E R.o - {z}, a E U}. Let AG(n,q) be an affine geometry of dimension n and order q.
Since £o -{z}
contains qn points, the qn points in AG(n,q) can be put into 1 - 1 correspondence with the points on R. -{z}. o
Let L be a collection of q-subsets in R. -{ti·which o
corresponds to lines of AG(n,q).
Hence, let G = (R. -{Z}, L) be an affine geometry o
on £o-{z} isomorphic to AG(n,q). Now, we define two types of blocks on V. BI(y) where
ya
For each point y E P-R. , o
tYa; a E U},
is a line passing through points y and a.
For every line R. ELand every
a E U
BII(£,a) =
{<x ,a>;
xE
£}.
These two types of blocks are said to be blocks of type I and blocks of type II, respectively. By generalizing Lemma 1 in Dickey and Fuji-Hara [3], we have (r,l)-designs on V. LEMMA 2.1: {BI(y); y E P-£o}
U
{BII(£,a); £ E L, a E U} is the set of bZ-oaks of a
((qn+l_l)/(q-l),l)-design D on V with bZ-oak sizes q and u.
PROOF: Let £1 and R.2 be any two distinct lines of V.
The common pOint of £1 and
£2 is in U or in P-R. oo • If £1 n £2 = a E U, R., and £2 meet £0 at distinct points, x and y.
There is a unique line xy of G which contains x and y.
BII(xy,a) is a unique block containing £1 and £2·
Then, the block
If £1 n £2= a E p- £00' then
R. Fu;i-Hara and S.A. Vanstone
358
2• 0 We show that the design D has a resolution if
BI(a) is the unique block containing
and
~1
t
lUI
~ qn.
Since AG(n,q) has
(qn_l)/(q_l) parallel class of lines, G also has parallel classes, denoted by t Let U = {O,1, ••• ,u-1'.
= (q n-1)/(q-1).
A collection of blocks of type II,
is a resolution class for 0
~
i
~
t-1.
Every block of type II is contained in one
of those classes.
T ,T , ••• ,T _ is a partition of the set of type II blocks. O 1 t 1 Let c be a point on koo which is not in U, and let M be the set of lines in P-~oo
passing through the point c. the lines of M.
Every point in P-t oo is contained in exactly one of
Any line of V meets each line of M at a point.
Therefore, for
any line mE t1,
is a resolution class. Hence, R =
{R: mE M} forms a partition of the set of type I blocks. m
i .;; t-1}
{T.; 0.;; 1
U
{R ; mE M} is a resolution of D. m
We can state the following result: THEOREM: The «q
n+1.
-1)/(q-1),1)-des~gn
D of Lemma 1 has a pesoZution if
lui.;; qn•
(Note: The type I blocks form a transversal design).
3. ISOMORPHISM RESULTS If u
= q, the design D constructed in the previous section has the same
parameters as a BIBD obtained from a finite affine geometry of dimension n+1 and order q.
In this section, we show certain conditions under which D is isomorphic
to the affine geometry. Let u
= q. For each pOint i
E U, the set of lines V(i) is in 1 - 1 corre-
spondence with the set of points in G = =
(V(i),{BII(£,i);
~
(~
o
-{z}, L).
E L}) is isomorphic to G.
Therefore H(i)
=
Then, H(O),H(1), ••• ,H(q-1) are
disjoint n dimensional affine geometries in D, where U = {O,l, ••• ,q-1}. every block of type I meets any H(i), 0
~
i
~
Since
q-1, at one element of V(i), then
any (q2,q,1)-subdesign (plane) in D containing a block of type I meets H(i) in a block of type II for any i E U.
359
Affine geometries obtained from projective planes
Let c(x) by a set of points on to-{Z} at which lines of Br(x) meet to' for example, c(x) = {a o .a 2 •••• ,a q- l} if Br(x)
=
{
THEOREM 3.1: If D is isomorphic to a n + 1 dimensional affine geometry, the, for any point x of P-(t
'"
U £ I. c(x) is a line of G or no three points in c(x) are 0
collinear in G.
PROOF: Suppose that. for some x E P-t",' c(x) has three points a l .a 2.a 3 on a line and one point b not on the line. Let Br(x) ~ {.
0
(2) For all x E P-(t u t ), c(x) has no three collinear points. '"
0
(3) P has two types of points x, x'. c(x) is a line of G. C(X') has no three collinear points. Here we need to define a transversal design for the following lemma.
A
transversal design TD(m,t) is a pair (X.B) where X is a union of mutually disjoint sets G., 1 ~ i 1
~
t (groups) of elements (varieties) of cardinality m and B is a
set of t-subsets (blocks) of X with the property that any B E B contains precisely one element from each Gi • 1 ~ i ~ t, and two elements from distinct groups are contained in exactly one block of B. rt is well known that a transversal design TD(q,q)
= (X.B) is isomorphic to an affine plane with pOint set X and line set
l';;;;iU~q{Gi}UB.
rn the next lemma and theorem. we show that the case (1) is a sufficient condition for isomorphism. when q
~
3.
360
R. Fuji-Hara and S.A. Vanstone
LEMMA 3.1: When 0 izaD tize pl'opel'ty that c(x) is a line of G fol' any xE P-U u
£ ),
'"
0
theil, fOl' rno blocks Brr(m,i) and BII(m',j), i F j, thel'e exists a (q2,q-l)-subJesign (~lane) Pin 0 containing BIr(m,i) and BII(m' ,j), if and only if m and m' al'e paralleZ lines iYi G.
PROOF: Let F be a set of points in P-£", at which lines
B1r(m,i) and BIr(m',j) are disjoint blocks belonging to H(i) and
H(j), respectively.
If P is a plane containing BII(m,i) and Brr(m' ,j) in 0, a
block of type r, Br(x), must be a line of P for any point x E F.
From the rela-
tion between an affine plane and a transversal design TD(q,q), we can say that there exist a plane in 0 containing BrI(m,i) and BII(m' ,j) if and only if T = {Br(x); x E F} forms a transversal design with groups BIr(mi,i), i
= a,l, •.. ,
q-l where mo ,ml, .•• ,mq- 1 are lines of G including m and m'. Suppose that m is parallel to m' in G but m F m'. There is a plane P' in G which contains m and m'.
Let S be a parallel class of P' containing m and m'.
{c(x); x E F) is the set of lines of P' excluding S. i E U.
Let Si={
If Si has two elements <x,i> and
of S, then line xy in G meets m and m', that is, xy is a c(a) for some a E F. a block containing <X,i> and
Si = BIr(mi,i), i E U.
with groups B1I(mi,i), mi E S. BrI(m,i), i = a, ... ,q-l. of G such that £ and
£'
So any Si' for i E U, is a
Therefore, T is a transversal design
When m = m', T is a transversal design with groups
Suppose that m and m' are skew lines. 9.'
But
There are no two distinct lines £ and
both meet m and m', and £ and £' have a common point
not on m nor m'.
Therefore, there are no two distinct block of type I, Br(x l ), Br (x 2) such that Br(x l ) and Br (x 2 ) have a common point in H(k), k F i,j, and Br(x l ) and Br (x 2) both intersects BrI(m,i) and Brr(m',j). So, there is no affine subdesign in 0 containing BI1(m,i) and B11(m' ,j). Suppose m and m' have a point c in common.
Let P be a plane in 0 containing
Brr(m,i) and BrI(m' ,j), i # j. Let xa and Yb be points in P-£", at which
a
E
iZ;
Yb
~
Si nce xa # xa , if a -; a', and Yb, if b F b'. then Ak = m-{c} and Bk = m'-{c}. Let d be a point in P-£", at
Affine geometries obtained from projective planes
which
<e,k>
E
m-{c}, b E m'-{c}.
E
Bl(d), is not in mUm'.
ments in H(k), k BII(m',j).
F i,j.
Then point e of G, where
Therefore, P has at least 2(q - 1) + 2> q ele-
So there is no plane of 0 containing BII(m,i) and
D
LEMMA 3.2: Suppose 0 has the ppopepty (1) fop c(x). vapiety e
361
F B of
Fop any block B and any
0, thepe exist a unique (q2,q-l)-subdesign (plane) of 0 contain-
ing B and e.
PROOF: First, we show that if a (q2,q-l)-subdesign (plane) S in 0 contains a type I block B = {
H(i), i
E
U.
E
U}, then S contains exactly one type II block from each
There is a variety in S but not on B, say e
II block C = BII(bka,k) must be a block in S. containing e and disjoint to B.
=
There is a unique block B' in S
If B' is a block of type II, which is a block in
H(k), then two distinct blocks B', C having a common variety e are contained in S. B' and C generate a plane in H(k), which implies that S is in H(k).
But S has
points not in H(k). Therefore B' must be a block of type I, say B'
= {
i
E
U}.
Then S has at least one type II block BII(bibi,i) in each H(i), i
By the same manner above, S cannot have two type II blocks from a H(i).
E
U.
Further,
Lemma 3.1 says that there is a plane containing BII(£,i) and BII (£' ,j) i F j if and only if £ is parallel to £' in G. Then, blocks of type II BII(£i,i), i E U contained in S have the property that the £.'s are parallel in G. 1 Suppose B is a block of type I {
j
E
U, there is unique line
£~
J
E
Uland e
=
containing b. and parallel to Bka in G.
From the
J
proof of Lemma 3.1, a plane can be constructed from those parallel lines. Suppose B = BII(£,i) and e
=
i
F j. If there is a plane containing
Band e, it must contain a block of type I, say C, such that e,
E
C, where b
There is a unique line £' in G which contains a and parallel to
By Lemma 3.1, there is a plane containing Band BII(£',j), which is the unique
plane by the structure of S mantioned above. Suppose B = BII(£,i) and e =
We denote para 11 eli sm in G by I IG.
a relation liD on blocks of 0 using II G•
Here, we define
R. Fu;i-Hara and S.A. Vanstone
362
For any blocks B,B' of 0, BIIOB' if and only if (a) Band B' are both type II blocks, say B = BII(i,i), B = BI I ( 9' . , E U, S uc h th a till G £ ' • . ,1") , 1.,1
or (b)
blocks, say B = {; i U}, satisfying:
Band B' are both type B'
= {
1
,i>; i
E
U}
E
and
(blJ B = B', or (b 2 ) B n B'
0 such that bib'i IIG b}'j for any i,j
=
E
U.
From Lemma 3.1 and 3.2, it is clear that the design 0 with the relations
110 satisfies the axioms of an affine geometry if lID is a parallelism. next two lemmas, we show that lID is a parallelism on blocks of O. Let A be an affine plane in G.
In the
For any A in G, a subset Z = {<X,i>;
x E A, i E U} of V and the blocks of 0 contained in Z form a (q3,q-l)-subdesign of 0, say SIAl.
This follows from the proof of Lemma 2.1.
We require the
following result. LEMMA 3.3: SIAl is
isomo~phic
to AG(3,q)
fo~
any affine ptane A in G, q
~
3.
PROOF: Let n be the set of (q2,q,1)-subdesigns (plane) of 0 contained in Z.
As a
step of the proof, we show that a pair E(A) = (Z,n) is a (q3,q2,q + l)-BIBO.
It
is clear that,ZI= q3 and every plane of n contains q2 varieties of V.
Let us show
that, for any arbitrary pair of distinct varieties v ,v of Z, there are q+l l 2 planes in n containing v v • There is a unique block B containing v ,v and a l 2 l 2 plane containing v ,v must contain B. l 2 Suppose B is a block of type II, say BII(i,k), where t is a line of A and k E U.
From the proof of Lemma 3.2, A plane containing B is in H(k) or contains a
type I block.
Any plane of the first type consists of {
is uniquely determined by k. II block from each H(i), i containing Band B'
E
A}, say Pk. It Any plane of second type contains exactly one type
U.
= BII(~',k'),
k' Uk), there are q blocks
E
From Lemma 3.1 and 3.2, there is unique plane k' # k if and only
BII(.~',k')
IIG
if.~
such that i' IIG
L
i'.
By fixing
Hence, in total,
there are q + 1 planes containing B in Z. Suppose B is a type I block of 0, and
E
B.
Any plane in 0 containing There are q + 1 lines in A
363
Affine geometries obtained from projective planes
which contain the point a.
By Lemma 3.2, for each line t such that a E t, Band
BII(t,i) generate a unique plane in 0 which is contained in Z. that E(A) = (Z,IT) forms a (q3,q2,q
+
Hence, we can say
l)-BIBD.
Next, we show that E(A) has a parallelism on IT. t E C, i E U}, where C is a parallel class of A.
Let Xc = {BII(t,i);
Xc can be partitioned into
G = {BII(t,i); t E C} for i = O,l, •.• ,q-l. Let Bc be the set of planes in E(A) i which have at least one block of Xc and at least one block of type I. Then, from Lemma 3.1, any plane of Bc has exactly one block from each Gi • Further, for any two blocks BE G , B' E Gj , i t j, there is a unique plane in Bc ' which implies i (X ,B ) forms a transversal design T(q,q). It follows that there are q parallel c c
classes in Bc for each parallel class C of A, which we take as parallel classes of E(A}. q(q
+
Also, {P.; i E U} is another parallel class of E(A). 1
1)
+
1 disjoint parallel classes in IT.
There are, in total,
It follows that E(A) has a parallel-
ism. We now appeal to a result due to Dembowski [11 (pp. 74). The following properties (a), (b) of a
~IBD
F with a parallelism such that
m ~ 3 (m is the number of blocks in a parallel class of F) are equivalent: (a) F is isomorphic to the system of points and hyperplanes of an affine geometry. (b) Every line consists of m points, where a line in F is defined to be all points incident with every block through two distinct points of F. In E(A), lines of F will be a block of 0 contained in Z, and a block of F, of course, is a plane of IT.
Therefore, by the Dembowski result above, S(A) is
isomorphic to the affine space AG(3,q). 0 LEMMA 3.4: The relation //D on blocks of 0 is a parallelism of 0, when q PROOF: From Lemma 3.2, //0 satisfies the condition (*) of parallelism. show that the relation //0 is an equivalence relation. tive property of the relation //0 are easily seen.
~
3.
Here, we
The symmetric and reflec-
We show that //0 possesses
the transitive property. Let A, B, C be blocks such that A //0 B and A //D C.
If all of these
blocks are type II blocks, then it is clearly true that B //0 C.
If A, B, Care
distinct blocks (of A = B or A = C, then trivially B //0 C) of type I, then ab / / G a'b' and
ac //G a'C',
where
<])'
,i '> E Band
The following two cases must be considered.
364
R. Fuji-Hara and S.A. Vanstone
(1) ab f a'b' and ac f a'c', (2) at least one of the two pairs is the same line, say ab Consider the case (1). ~ by ac and
d'C'.
= aotl'.
Let r be a plane in 0 generated by ab and-aob', and
r containd the blocks A and B, and L contains A and C.
From
the proof of Lemma 3.1, the set of points in G at which varieties contained in
r(t respectively) intersect to induces an affine plane of G, say rG (LG respectively).
respectively) naturally correspond to blocks in r(L reG Let £a' l , tc be lines of G corresponding to A, B, C, respectively. b
Lines in 'G
spectively).
(6
a b
c a' c'
b' 9-
a t
£b
c
(Figure 3.4) 9b IIG £c' ab IIG aotl', ac IIG d'C' in G. Since G is the Desarguesian affine geometry AG(n,q), then bc IIG b'C', which implies B IIG C.
Now, we have xa IIG
Consider, next, the case (2); ab = a't)I and ab = a'b' planes r,
in
ab
=
a'b' implies
= bb' in G. Let A be a plane of G containing aa' and ac. The
aa' :l
ac IIG aoc:r.
0
are both contained in the affine space S(A) defined in Lemma 3.3.
It follows that the same configuration as Figure 3.4 occurs in the affine space
SiAl, that is B liD C.
o
Now, we can state the following result: THEOREM 3.2: If c(x) is a line of G for any point x in P-(t u t ), then the n+1 . 0 (q ,q,l)-BIBD 0 isomorphic to AG(n+l,q), when n ~ 2, q ~ 3.
.3
00
Next, we show that the design 0 which is isomorphic to an affine geometry is constructable on a particular plane. n
Consider the desarguesian projective plane of order q ,P
= PG(2,q n ), q a
Affine geometries obtained from projective planes
prime power, q > 2.
We use a method in Hughes and Piper [9] for coordinatization.
The points in p-~oo are assigned coordinate from {(x,y); x,y on
~oo
365
E
GF(qn)}.
The points
are coordinatized as
= {( oo) } u
~ oo
{( a);
a
E
n
GF (q )}
such that line (O,a)(l.O) meets (a) and the line (0,0)(0,1) meets (oo). The points in p-~oo can be thought of as elements of GF(q2n) provided a suitable primitive irreducible polynomial over GF(qn) is selected. . element of GF *(q 2n ) = GF(q 2n )-{O}. Then. tlve GF"'(q2n) Let
~o
{Xi; i
=
=
Let x be a primi-
0.1, .... q2n_ 2 }
be
~
o
{oo} u {X j {(O,a); a
where'"
=
(0.0).
; j EI E
O(mod qn+l)}
GF(qn)}.
£0 is incident with £oo at (oo).
We define an affine geometry G
of dimension n and order q by: (i) a point of G is a point of £0' (ii) a line xy of G passing through x and y. X,y
E
£0' is defined by
xy = {Ax + (l-A)Y; AE GF(q) • Consider two types of collineations a,S on n+ ) a: x ... X(q l • X E 'P-l oo •
p-~oo
defined by
where a E ~ • o n C = {a o, a l ••••• a q -l} and C2 = {S ; a E £ } form collineation groups on P-£ . l a 0 "" Any j E C fixes lines of P-£"" passing through oo, and any Sa E C2 maps a line £ l passing through", to its coset. Therefore. any collineation of these two types fixes £oo pointwise, and fixes £. These collineations also act on G. S. 0 a for any a E £o • fixes the points on ~ oo pointwise. Let a point x E P-£ oo be on a line £ of P-~ incident to point z = £ n £ • then {S (x); a E £ } is the set of points on "" 0 oo a 0 £.
Let point x E P-£oo be on a line m of P-£oo passing through oo, then
{ai(x); i = 0,1, .... qn_ 2} is the set of points of m-{oo}. point x E P-(£ oo {a
i
Sa(x);
U
a
i
£0 ), Sa
E
Cl
x
C2 }
So, for an arbitrary
366
R. Fuji-Hara and S.A. Vanstone
is the set of points of P-(£ u '"
t
)_
0
Let BI(x) = (, •••• ,J be a block of type I of D. a(BI(x)) is also a block of type I of D; '"I(BI(x)) Let BlIP,.i)
BI(cl(x))
=
Then,
{
(,, •••• J be a block of type II of D, then l a(Brr(t.i))iS also a block of type II. =
,:.(BII(t.i))
BII (a{9,),i) {
Similarly, any co1lineation Sa of C2 also acts on D. state the following results: LEf4MA 3.5: If a co:Zinearion a on P which fixes
£",
Therefore, we can
pointUJise and preserves the
a.ffine geometry G cwfined on £0' then a induces an automorphism of D.
Let x be a point on P-(£", a.1
E
m meets x at u1.•
£0) and let m be a line of G.
U
A line a.x, 1
Let D be a BIBD constructed by the previous method with
00
U = {u o .u 1 '···,u q- 1}'
Then any block of type I, BI(y). where y i
be generated from BI(x) by some a Sa
E
C1
x
E
P-(£",
U £),
can
C2 ,
i
Br(Y) = BI(a ea(x)). This implies. for every point x E P-(£
o
U
£ ), c(x) is a line of G. '"
Therefore, we
can state the following result. THEOREM 3.3: If there is a set of collineation C on P of which any collineation of C sat'!:3fie3 Lemma 3.;;, and for any two points x.y
E
P-(£o
U
tJ
there is an a
E
C
c:uch that y = a(x), then the BIBD D constructed on P is isomorphic to AG(n+l,q), ",hel: n
> 2, q > 2.
4. SKEW RESOLUTIONS OF AG(3,g)
Let D be an (r,l)-design having two resolutions Rand R'. orthogonal to R' provided iR n
SI .;;
1
We say that R is
Affine geometries obtained from projective planes
367
R'. Orthogonal resolutions in block designs have recently been of interest to various authors. (see for example [7). [11)). If we can construct a skew resolution of the lines in AG(3,q) then this resolution and the natural resolution from parallelism provide a for each resolution class R E R and each resolution class S
pair of orthogonal resolutions.
E
In the remainder of this section, we construct a
skew resolution of the lines in AG(3.q). In R. Fuji-Hara and S.A. Vanstone [4). orthogonally resolvable (q2 + q+ 1.1) -designs 0 are constructed on a finite projective plane P of order q2 where P-t is coordinatized by two dimensional vectors over GF(q2).
In the plane. we can
choose q points U on too such that the design 0 is isomorphic to AG(3,q).
But any
two of the resolutions constructed by the method of [4) do not necessarily form a skew resolution.
In this section. we show a construction for a skew resolution of
AG(3,q) obtained from PG(2,q2) in Section 3. LEMMA 4.1: Two blocks of type I, Bl(x l ) and Bl (x 2), of 0 are parallel if and only if line x x is incident at the point Z = too n £0' l 2 PROOF: Let Bl(x l ) }.
=
{
=
{
Two disjoint lines are parallel in an affine geometry if and only if they are disjoint and there exists a unique plane containing the two lines.
From Lemma
3.1. Bl(x l ) and Bl (x 2) are parallel if and only if lines aib i and ajb j , i G are parallel or the same line.
~
j, in
a. can be represented as a. = a. + (d. - d.) where d., and d.J are points at J J _J_' _,_ which zX l (a coset of £0) meets (i)oo and (j)oo. respectively. bj also can be represented as b.J = b._,_ + (d~ - d~). Then d. dj = di dj if and only if aib i is J J , parallel or equal to a.b .• J J
, d.J , d., - , d. J
d.
d~
d~
,-
dO. and d. d. J J -( j )00 is 00. So d., -
d~
,
- dO.J
= d.J -
d~
J
d~
J
are points of (i)oo and (j )00.
d~
J
The common point of (i)oo and
= O.
Therefore ajb j = aibi~di - dj parallel or equal to aib i . 0
).
Since (d i - dj ) is a point on to' ajb j is
R. Fuii-Hara and S.A. Vanstone
368
If we choose a point c from
U U) and construct resolution classes of
~oo-{Z}
blocks of type I for the set of lines passing through c. then any of these resolution classes are skew classes of D.
But. from the definition of Ti in section 2.
every T. obviously contains parallel lines. 1
P-~oo
Let Mc be a set of lines of
passing through c in
~oo-U,
Since Rm={B1(x);
x Em}. for mE Mc' is a resolution class. every element of V appears precisely once in Rm for each mE
r\.
x ~ m BI(x) = V = i ~ U Vii) H(i) is an affine geometry isomorphic to G =
(~o-{z};
L). having Vii) as its point
set and {Bn(£.i); x E L} as lines. Define a 1 -
mappi ng
m 1J
y .. :
V(i)
V(j) by
-+
y~.(
if and only if there exist a block containing
r\,
m
Y
ij is an isomorphism from H(i) to H(j).
PROOF: Let moo be a line of Mc which passes through moo = {x
j
; j ==
k(mod qn+l)}U
00
moo can be represented by
{oo}.
where x is a primitive element of GF*(q2n). k is some integer.
10
is defined in
section 3 as
Let
~
be the col lineation on P-l oo defined in section 3;
Using .:'. elements of Vii). for any i E U. can represented as V(i)
{<) (a). i>; j = O.1 .... qn- 2} U {<",.i>}
where a is a generator in ~o (arbitrary point. not 00). If a line of Vii) meets ~o d xet+k • respect1ve . 1y. were h . t eger. Th en an d m at xana t = qn+1 • e some . 1n a '" 0 t et+k et+k t . . ~(x ) = x .X E I and a(X ) = X ·x Em are also on a llne of V(l). That o '" et k is. a line I of Vii) meets 1 o at y if and only if the line 1 meets m00 at YX + Similarly. a line ~ of V(j) meet 10 at y if and only if the line ~ meets m", at y'X ft+k • f some integer. Let a line 1 of Vii) meet 10 and moo at a and b respectively. and a line
~'
of V(j) meets
1
a and m at d and b. respectively. 00
The
Affine geometries obtained from projective planes
b -_ a·x et+k
369
d'X ft+k
d = a'x(et+k) - (ft+k)= a.x(e-f)t d = a· x( e- f ) t 1. s a co 11 1. nea t·lon 1. n cl'
So, Y~j is an isomorphism which carries
Let a line
line
of V(j) meet
-
~'
b = d + P. = 1
~
of V(i) meet
~
0
m is an additive coset
, m and m at a,b and d, respectively. '"
, m and m at g, band h, respectively. o '" h + P., ~
Let a
Then
J
where Pi and Pk are points at which m meets (i)'" and (j)"', respectively. h = d + (Pi - Pj ) We know that d = a'x et+k and g = h·x -(ft+k) • Therefore, (a.x et +k+ (P. _ P.))x-(ft+k)
9
J
1
a.x(e-f)t+ (P. _ p.)x-(ft+k) J
1
P. - P. 1· S a po 1. nt.1 n m, hence (P . - P) . X- (ft+k) 1. S a po 1. nt 1. n ~ . 1 J m . '"e-f 1 J 0 So y .. :
1
J
Let E = {X l ,x 2' ... ,x q } c m, Fl(E) = {Bl(xl),Bl(x2), .•. ,Bl(Xq)} is a set of skew blocks such that F (E) = {f(E,i); i E U} is a set of skew blocks of type II, 2 where f(E,i) = {
LEMMA 4.3: Suppose D is the design obtained from PG(2,q2) and c E ~ -(U u {z}). o For any block B of type II in H(i), B and y~.(B) E H(j) are skew for any mE M , . lJ C i#j,i,jEU. PROOF: E is a set of q points on m such that f(E,i) = B. k E U, f(E,k) is a block of type II in H(k). blocks in D.
From Lemma 4.2, for any
Now, {Bl(x), x E E} is a set of skew
If f(E,i) and f(E,j), i # j, are parallel in D, then Bk(x) and
Bl(x'), x,x' E E, x # x', must be parallel.
This is a contradiction.
So, for any
i,j E U, i # j, f(E,i) and f(E,j) are skew; that is, e-f in Lemma 4.2 cannot be divided by (qn_l)/(q_l).
0
R. Fuji-Hara and SA. Vanstone
370
LEMMA 4.4: Suppose D is the design obtained from PG(2,q2) and c E £",-(U
U
{z}) and
E induces a net, then Rm - Fl(E) + F (E) is aZso a skew class of D faY' any mE Mc. 2 PROOF: F {E) is a set of skew blocks of D from Lemma 4.3, and any block of 2 Rm-Fl(E) does not meet any of F2(E). Disjoint blocks of type I and type II are skew, since any parallel block of a type II block is also type II. Therefore, Rm - Fl(E) + F2 (E) is a skew class of D. 0 Next, we show a construction of skew resolutions of AG(3,q), q > 2.
Let let D
be a projective plane of order q2 coordinatized by the method of section 3. 3
be a (q ,q,l)-BIBD isomorphic to AG{3,q) and constructed on P by the method of section 3. Step 1. Let a a a
q+l
,
where
a
is the collineation on P defined in section 3.
On a line mE Mc -{m o }, takeo q-subsets E ,E ,El, ••• ,E q- 2 such that fiE ,0), ", 0 fiE o ,0), .•. ,f(E q- 2,0) is a parallel class in H{O). Then, of course, for any i E U, fiE ,i), fiE ,i), ••. ,f(E 00
0
E. induces a skew net.) 1
q-
2,i) is a parallel class in H(i).
(That is, any
Suppose that E contains a point on £0. '"
Step 2. Let M be a set of lines in P-t", passing through the point
z :
£00
n £0.
z Define Ni for each i E Q
{oo,
0, 1, •••• q-2} to be a set of lines of
Mz such that any line of Ni meets m at a point of Ei • N",' N , ••• ,N q_ is a o 2 partition of Mz • a(E ) : (a{x); i
some
j
E Q.
X
E E. }, i E Q is the point set on a (m) and lines of N. for
cyclically represented.
J
1
But points of a(E",) are also on lines of Noo •
Points of Pare
Therefore, we can assume. without loss of generality
that points of a(E i ), i E Q-{oo}, are on lines of Ni+l(mod. q+l).
Affine geometries obtained from projective planes
371
z
t co
Step
3. In
{a
i
(E j );
E
Q,
0,1, •.• ,q-2}, we choose q - 1 sets under the
following rule; i (1 ) one set from each 1ine a (m) for 0
~
i
~
q-2,
(2) one set from each N , i
E Q, i (3) There are no two sets E and E' in chosen q-l sets such that
f(E,i) = f(E' ,i) for some i E U. If E is a q-subset on m-t o ' which induces a skew net, such that any of f(E,O), f(E,l) •••• ,f(E,q-l) do not contain
E, a(E), ••. ,aq-2 (E) obviously satisfies the above conditions. We prove the existence of such a q-subset when q
~
3.
PROOF: Let xa ,xl; •.• ,x q- 1 be points in m-t a such that
E
U.
Suppose that, for a set of three points Y ~ X-Ix}, f(Y,O) consists of collinear three points of H(O).
Then, for any i
E
U, f(y,i) consists of collinear
We assume without loss of generality that Y = {xo 'X l ,x 2 }. Then, <"',0>, <00,1> and<~2> are elements of BI(x o )' BI(x l ) and BI (x 2), respectivethree points of H(i).
ly.
The blocks BI(oo) contains those elements, B(oo)
=
{
312
R. Fuji-Hara and S.A. Vanstone
Br(w} meets Br (x 2 } at an element <00,2> of H(2). Let Ao and Al be blocks of H(O) and H(l} having f(Y,O} and f(Y,l), respectively. Br(oo) and Br (x 2 ) both meets Ao in distinct points, and meet Al too. This means Ao and Al are parallel in D. But Ao and Al = Y~l(Ao} are skew from Lemma 4.3.
Contradiction.
Suppose that Y is a 3-subset of X including x.
Block Br(x) forms
We assume Y = {X,Xo,X }. Then l Br(x o } and Br\X l } contains <00,0> and <00,1>, respectively. Let Co be a block of H(O) containing <x,D> and <00,0>, and let C be a block containing <X,l> and <00,1>. l Co ~ BIr(xoo,O) and Cl = BrI(Xoo,l). That is, Co and Cl are parallel from Lemma {<x,D>, <X,l>, ••• ,<X,q-l>} since x is on
~o.
But Cl = Y~l (Co) and Co must be skew from Lemma 4.3, contradiction. Therefore, for any i E U, f(X,i} is a set of q + 1 element in H(i) in which
3.1.
no three elements are collinear.
In a H(i}, the number of blocks which has at
least one element of f(X,i) is (q;l) + q + 1
= (q2
+ 3q + 2}/2.
the number of blocks, q2 + q, in an affine plane H(i) when q ~ 3.
This is less than So, there
exists q - points set E on m where f(E.i) is a line of H(i) and does not contain <"',i> for any i Step 4.
in step 3.
slJ.. for 0
~
i
E U.
0
Let Ko
atE. ) ••••• K 2
q-
11
q 2 a - (E.
1
Then = R i j( ) a a m ~
q, 0
Step 5.
~
j
} be subsets chosen q-2
- Fl(a(K.)) + F (a(K .)) J 2 J ~
q-2 is a skew class from Lemma 4.4.
f(KO,i), f(K l ,i) •••• ,f(Kq_ ,i) are q-l parallel blocks in H(i) for 2
each i E U. PROOF: From the definition in step 2. any two blocks f(ai(Ej),k) and f(a i ' (Ej'),k), where k E U. i 1 i'. i,i'
E
Q and 0
~
j,
j ~
q-2, are parallel or the same block.
rf they are same line (j = j'), then f(oi(E.),k) contains an element
n
J
step 3 includes those cases.
0
There is exactly one line in H(i) which is parallel to f(K.,i), j J
Denote the line by gi'
\
=
0
<;
~~
Then
q-2 Fl(ak(K j )) + {/(gi); i
is a skew class for 0
~
k
~
q.
E
U}
0.1 •••. ,q-2.
373
Affine geometries obtainedfrom projective planes
Step 6. {5 .. ; O..;;i .;;;q. 0.;;;j";;q-2}U{5.; O..;;i ';;;q}UR lJ 1 moo
is a skew resolution of D. PROOF: We know that each class of these is a skew class. and no block of type I appears in two of these classes. classes. i
E
k
f(K o i). f(Kl.i) ..... f(K q- 2.i), g.1 are parallel classes in H(i) for each
U. from step 5.
a (g.)
Sk'
E
1
We prove that no block of type II appears in two
f(K .• i)
5 . and kJ The set fa (f(K .• i)); 0';;; k.;; q} U {a (g.); 0.;;; k.;;; q} is the set of ~
E
5 . and g. oJ
E
5.
10
Hence. ak(f(K.i)) J
k
J
lines in H(i). for any i
E
U.
E
1
So, no block of type II appears in two classes.
This finishes the construction of a skew resolution in AG(3.q).
0
A skew res-
olution in AG(3.q) can be used to induce a resolution of the lines in PG(3.q). Resolution of lines in PG(3.q) was first done by Denniston [2].
We include an
alternate proof here for completeness. LEMMA 4.5: Let R be a skew
reso~ution
of AG(3.q).
Then, for each class S of R,
there is a unique parallel class of planes C of AG(3.q) such that there is no
s
line of S contained in any plane of Cs '
Moreover, C
PROOF: Suppose that S is a skew class of R. tains at most one line of S. S is q3
+
q2
s
Cs ' if and
if S
Let P be a plane of AG(3.q).
= S'.
P con-
The number of planes of AG(3.q) containing a line of
But the total number of planes in AG(3.q) is q3
there are q planes which do not contain any line of S. containing any line of S.
on~y
+
q2
+
q. Therefor~
Suppose Po is a plane not
Then. the lines of 5. which meet Po at distinct pOints.
meet any parallel plane to Po at distinct points.
That is. any parallel plane to
Po is also a plane of AG(3.q) which does not contain any line of 5.
So. we can
say that. if S is a skew class of AG(3.q). there is exactly one parallel class of planes in AG(3.q) where any plane of the class does not contain any line of S. Let R be a skew resolution. and let 5 and 5' be distinct classes of R. here are parallel classes of planes Cs and Cs , which do not contain any line of S and S', respectively. Suppose now that Cs and Cs ' be the same class. Let P be a plane of Cs (= Cs I)' Each lioe of P must be contained in distinct skew classes of R.
The number of lines in P is q2 + q, but the number of skew classes of R
excluding 5 and 5' is q2
+
q - 1. That is. some skew class of R must contain two
lines of p. contradiction. 0
374
R. Fuii-Hara and S.A. Vanstone
Here we define a packing of PG(3,q) to be a resolution with respect to the lines in PG(3,q). THEOREM 4.1: Any skew resolution in PG(3,q) induces a packing of PG(3,q) and oonveraely, any packing of PG(3,q) induces a skew resolution in AG(3,q).
PROOF: Suppose that R is a skew resolution of AG(3,q).
Let
space obtained by attaching an infinity plane Poo to AG(3,q). of AG{3,q) are incident at the same point on Poo ' there is a unique line
~s
in Poo where any point on
L
be the projective In L, parallel lines
For each skew class S of R, ~s
is not incident with any
line of S, because planes of Cs are incident with a line of Poo and only the lines in those planes are incident with the line of Poo ' Therefore if S f S' E R then i
s
So {S U t ; S E R} is a packing of projective space L. s s Suppose Q is a packing of PG(3,q). Let Po be an arbitrary plane of PG(3,q).
f 1 ,.
Each class A of Q contains precisely one line 2A of Po'
Since if A contains two
lines from a plane then those lines are intersecting, and each class of Q must contain at most one line of Po'
The lines of A-{i A} meet Po at distinct points, which means lines of A-{tA} are skew in the affine space PG(3,q)-P ' 0 o From the proof of Theorem 4.1 it is easy to extend the result to show that there exists a skew resolution in AG(n,q) if there exists a packing in PG(n,q). The converse of this may be false.
The authors believe that there exists a skew
resolution in AG(4,q) but it is well known that there is no packing in PG(4,q). Several recursive constructions for skew resolutions exist ([6]).
As yet, there
is no skew resolution in any AG(n,q), n even. ACKNOWLEDGEMENT: The authors would like to thank Professor U.S.R. Murty for valuable comments made during the preparation of this paper.
BIBL IOGRAPHY 1. 2.
P. Dembowski, Finite Geometries, Springer, 1968. R.H.F. Denniston, Some Packing of Projective Spaces, Rend. Accad. Naz. Lin-
3.
L.J. Dickey and R. Fuji-Hara, A Geometrical Construction of Doubly Resolvable (n 2+n+1,1)-designs, Ars Combinatoria vol. 8, 1979.
oei, 52 (1972).
Affine geometries obtained from projective planes
4. 5. 6. 7. 8. 9. 10. 11.
375
R. Fuji-Hara and S.A. Vanstone, On Automorphism of Doub~y Reso~vab~e Designs. Proceedings of the 7th Australian Conference on Combinatorics, 1979. Springer-Verlag Lecture Note 829 (1980). R. Fuji-Hara and S.A. Vanstone, Transversal Designs and Doubly Resolvable Designs. Europ. J. of Comb., 1 (1980), 219-223. R. Fuji-Hara and S.A. Vanstone, Recursive Constructions for Skew Resolutions in Affine Geometries, Aequationes Mathematicae (to appear). R. Fuji-Hara and S.A. Vanstone, On the spectrum of doubly resolvable Kirkman systems, Congressus Numerantium, 28 (1980), 399-407. J.W.P. Hirschfeld, Projective Geometries over Finite FieZds, Clarendon Prss, Oxford, 1979. D.R. Hughes and R.C. Piper, Projective PZanes, Springer-Verlag, New York, Heidelberg, Berlin. H. Lenz, Zur BegrUndung der analytischen Geometrie, Bayer Akad. Wiss.Math Natur. K~, S.-B., 17-72, 1954. A. Rosa, Room squares generalised, Anna~s of Discrete Math., 8 (1980), 4557.
Department of Combinatorics and Optimization University of Waterloo Waterloo, Ontario, Canada N2L 3Gl
This Page Intentionally Left Blank
Annals of Discrete Mathematics 18 (1983) 377-400 © North-Holland Publishing Company
AUTOMORPHISt~S
377
AND GENERALIZED INC IDENCE MATRICES OF POINT-DIVISIBLE DESIGNS
Dina Ghinelli Smit
I NTRODUCTI ON A point-divisible design is a tactical configuration for which the v points are split into m classes of size c, such that points have joining number A' A) if and only if they are in the same (resp.: in different) classes.
(resp~
We consider
square (i.e. with b=v blocks) point-divisible designs, with a nonsingular incidence matrix (i.e. k 1 A' and k2 1 vA), and we write simply p-divisible for them. These were introduced by Bose and Connor [3] as Symmetric Regular Group Divisible Designs (SRGDD, for short).
They include many interesting structures such as
divisible semisymmetric designs (A' = 0, A> 0), divisible semiplanes (A' = 0, A = 1) and semibiplanes (A' = 0, A = 2) and, when c=l, symmetric 2-designs.
Generalizing to p-divisible designs well known results by Hughes (see [16] on automorphisms of symmetric 2-designs, we show the existence of certain matrix equations which must be satisfied if a p-divisible design is to possess a standard automorphism group G (i.e. an automorphism group with orbits of length either 1 or iGi).
These equations involve matrices with rational entries, and the Hasse-Min-
kowski theory of rational congruence is applied to give numerical conditions on the parameters v=mc, k, A', A, the order u of G and the numbers N of fixed points and
~
of fixed classes. In section 1 we review the relevant points of the Hasse-Minkowski theory.
In section 2 we prove some general lemmas on matrices of a special type and on their Hasse invariants.
In section 3 we establish a rational congruence for a
general incidence structure and we deduce from it a different proof of a theorem by Hughes ([ 16] , 2.1) for symmetri c 2-des i gns. In section 4 we specialize the above rational congruence to p-divisible designs with a standard automorphism group.
We then apply (section 4 and 5) the
Hasse-Minkowski theory to obtain nonexistence theorems which generalize, simultaneously, the well known Bose and Connor, Hughes, and Bruck-Ryser-Chowla theorems.
378
D. Ghinelli Smit
If the dual structure of a p-divisible design is also a p-divisible design (i.e. the design is both point and block-divisible) we call it simply a divisible design.
We also show, in section 4 and 5, that for divisible designs all the
nonexistence theorems can be improved.
Most of the proofs in section 4 and 5 are
simply sketched, but it appears clearly how crucially they depend upon the lemmas proved in section 2 (see [10], for complete proofs). We point out that all the results in this paper concern standard tactical decompositions of p-divisible designs and we do not need to assume the decomposition to be the orbit decomposition of some automorphism group.
If we make use of
the actual group action, it is possible to get stronger results.
In this case we
need a different approach (due, for synllletric designs, to E. Lander [22] (1) which uses algebraic coding theory and modular representation theory.
We note, howev-
er, that these methods do not seem to be applicable for standard tactical decompositions that do not arise from the orbit decomposition of a group.
1. THE LEGENDRE SYMBOL. THE HILBERT NORM RESIDUE SYMBOL AND THE HASSE INVARIANT In this section we give a description of the relevant points of the Theory of congruence of quadratic forms (over the field of rationals) due to Minkowski [26] and Hasse [ 14]. fa
L
Instead of considering invariants of quadratic forms
a .. x.x., we consider invariants to refer to the corresponding matrices lJ 1 J
((a .. )) (for a more detailed account see Jones [19), chapter 2). lJ
Let p be an odd prime.
The integers a 1 0 (mod p) are divided into two classes called quad1~tic residues and quadratic nonresidues according as x2 a
=
(mod p) does or does not have a solution x (mod pl. The Legendre symboZ is defined by (1)
If a
+1
if a is a quadratic residue modulo p;
-1
if a is a quadratic nonresidue modulo p.
(!) p
=0
(mod p) we may write (!) P
= O.
The following theorems give some of the main properties of quadratic resi-
-----
(1) Added in proof: see also E.F. Assmus, Jr. and D.P. Maher, Nonexistence proofs for projective designs, Amer. Math. Monthly, 85 (1978), 110-112, and C. Hering, On codes and projective designs, Kyoto University Mathematics Research Institute Seminar Notes 344 (1979). 26-60.
379
On divisible designs
dues and nonresidues (for proofs see Hardy and Wright [13], chapter VI; Theorem (1.2) is due to Gauss and (6) and (7) are known as the taw of quadPatia reaiproaity) •
THEOREt~
(2) (3) (4)
1.1: If p is an odd prime, then b (mod p) implies (~) p a(P-l)/2=,(~) (mod p) ,
a
='
(E.) p
(~) (E.) p p
(ab) p
THEOREM 1.2: If p and q are two odd primes, then (5)
(.::!. ) p
(6)
(.?)
(7)
(E.) (51.) = (-1)[ (p-l)/2][ (q-l)/2]. q p
p
A slight generalization of the Legendre symbol is the Hilbert p-norm residue symbol (a,b)p (Hilbert symbol, for short).
Where no ambiguity is possible, we
shall omit the subscript p on the Hilbert symbol.
For any two nonzero integers a,
b, the Hilbert symbol is defined to be 1 or -1 according as the (8)
a x2 + b
i
='
z2
congrue~ce
(mod ph)
does or does not have solutions in integers x,y,z not all multiples of p, for arbitrary h. (9)
We also include p = ~ to mean that 2
2
ax+by=z
2
does or does not have solutions in real numbers x,y,z,not all zero. (10)
(a,b)~
Thus
= + 1, unless both a and b are negative.
THEOREM 1.3: The equation ax 2 + by2 zero if and only if (a,b)
p
=+
z2 has solutions in integers x,y,z, not all
1 for all primes p (inaluding p
= ~).
Pall [29] has shown that if a=a'pCl and b=b'pf3, ~/ith a' and b' prime to p, then
380
D. Ghinelli Smit
that is, by (5) (12)
(a,b) p = (_1)cxS(p-l)/2 (r)s (%')cx, if p> 2;
(13 )
(a,b)2
=
If p = 2, then
(-1)[ (a'-1)(b'-1)/4)+I s(a,2- 1)/S)+[cx(b,2_ 1 )/S).
The following theorem gives further computational properties of the Hilbert symbol. THEOREM 1.4: The Hilhert symbol has the following properties, where a,b,c,r,s aPe integers
= 1 the product being over all primes p (including p = 00),
(14)
IT(a,b)
(15 )
(a,b)p = (b,a)p ,
(16 )
(ar ,bs) p = (a,b) p ,
( 17)
(a,-a)p = 1,
(lS)
(a,b)p (a,c)p
( 19)
(a,a)p = (a,-l)p ,
(20)
(ar,br)p = (a,b)p (r,-ab)p)
p
2
2
2
(a ,b)p
1,
= (a,bc)p ,
If P is an odd prime,
= 1 if a
(21)
(a,b) p
(22)
(a , p) = (~ ) if a '1 0 (mod p)
(23)
if a1
p
and b are prime to p (see (11)).
P
= a2 '1
0
(mod pi, then (al,b)p = {a 2,b)p'
If x and yare nonzero rational numbers the Hilbert symbol (x,y) p is also defined and has similar properties; in particular, it has the property (24) where
(x ,y)
2
p
2
= (xp ,yo )
p ~~
i:
p, 0
£
Q = Q - {a}. Therefore, if x=a/b and y=c/d, with a,b,c,d
£
Z =Z-{O},
then (2S)
(x,y) p
= (xb 2,yd 2) = (ab,cd)
and actually we only have to consider Hilbert symbols of integers. Now, let A,B be two rational symmetric and nonsingular matrices of the same
381
On divisible designs
order and let A - B mean that A and B are rationally congruent matrices: that is
U A UT = B
(26)
for some nonsingular matrix U with rational entries (throughout the paper we shall let MT denote the transpose of the matrix M). If x and yare nonzero rational numbers such that x/y is a rational square then we write x .:: y. Let u be the order of A and let D. (i=l, ..• ,u) denote the leading principal 1
minor determinant of order i in A.
For our purposes we can assume that all of the
D. are nonzero(2).
Then for each prime p the Hasse invariant H (A) is defined by u-l P Hp(A) = (-1 ,-Du)p i~l (Di ,-D;+l )p'
1
(27)
The major theorem that we shall use is THEOREM 1.5 (Hasse-Minkowski): The necessary and sufficient conditions for two positive definite, rational and symmetric matrices A and B of the same order to be rationally congruent are that (28)
(3)
IAI.:: IBI
and (29)
H (A)
P
= H (B) P
for all primes p (including p a
oo).
By property (14) of the Hilbert symbol, we are able to exclude p = 2 from consideration since H2(A) = H2(B) is a consequence of Hp(A) = Hp(B) for all odd primes p and p = 00.
The invariant Hp may be expressed in a more symmetrical form as follows:
-----
(2) The matrices we need to consider will be all positive definite. Otherwise Pall's generalization of the Hasse symbol can be used, that is: if Dk=O the symbols (Dk,-Dk+l) and (Dk-l,D k) are interpreted to be (±l,-Dk+l) and (Dk-l,+l) respectively with the ± sign chosen arbitrarily. Actually, lemmas 1 and 2 in Jones [191 show that A must be nonsingular and Dk = 0 implies Dk+l Dk- l ~ O. (3) We denote bYIMlthe determinant of a square matrix M. In what if x and yare positive integers, Ix will denote the identity matrix of x, J(x,y) the (x,y)-matrix all of whose entries are 1, and J x = J(x,x). matrlx M (not necessarily square) we define a(M) to be the total sum of entries.
follows, order For any its
382
D. Ghinelli Smit
u-1 H (A) p
u-1 = (-1,-1)
°
where o
a
II
;=0
(Oi+1,D i ) (Oi+l,-l) ,
1; that is u-l
If Au- 1 denotes the matrix obtained by leaving out the last row and column of A, we get from (3~}, (31 )
Hp (A) = Hp (A u- 1) (IAI,-IA u- 1 1 ).
From this and the properties of the Hilbert symbol we can at once deduce the follow; ng theorems (see Jones [ 19J , and Pa 11 [29J).
THEOREM 1.6: If A ,A , .•• ,A are mtionaZ, nonsinguZar and syrrmetric matrices and m l 2 if A = e.A. = diag(AI, •.• ,A ) , then 11m (32)
Hp(A) = (-1,-1)
m-l
m
m Hp(Ai)} {. ~ (IAil,IAjl)}. i=l l,J=l ;<j
{II
As a particular case we have the following COROLLARY 1.7: The Hasse invariant of A = 1m x B (where x is the KPonecker product of matric:es) is
(_l,_l)m-l {H (B)}m (IBI._1)m(m-l)!2.
(33)
p
THEOREM 1.8: For A = a1 we have m
(34)
H (A) = (-1,-1) (_1,a)m(m+l)/2. p
THEOREM 1.9: If
p
is a nonzero rationaZ number
383
On divisible designs
We shall state another lemma on a property of the Hilbert symbol which we shall use later (see Bose and Connor [3] , lemma III). LEMMA 1.10: For any two nonzero rational numbers x,y and any prime p
(X,y) p = (-xy,x+y) p •
(36)
2. SOME USEFUL LEMt-1AS
In this section we prove some general lemmas about matrices.
The notations
are in footnote (3) (see section 1). LEM~1A
2.1: Let 0 be a nonsingular matrix of order u with entries in a field F, and
let d
E
F.
Then
110 + dJ u I
(1)
=
101(1+do(0
-1
)).
If (l+dO(O-lr f 0, then the matrix Z = O+dJ
is invertible and
u Z-l = 0- 1 _ d (O-lJ 0- 1). l+do(O-l) u
(2)
If the above matrices have rational entries, p is any prime (including p
= 00),
and we can consider the Haase invariants (see section 1. footnote (2)),
then (3 )
H (Z) p
a
H (D) (dIOI, l+do(O-l)) p p
or, equivalently (4)
PROOF: We give the proof for matrices with rational entries since the extension of
(1) and (2) to matrices over any field is obvious. matrix
(5)
Clearly
Z'
We consider the ((m+l)x(m+l))-
384
D. Ghinelli Smit
(6)
Iz'l
and by (31) 1
-dlzl,
=
(4)
Hp (Z')
(7)
=
Hp (Z) (IZ'I,-IZI).
Now we consider the ((m+1)x(m+1))-matrix
Z" =
(8)
\ I
0
l-dJ(l,u)
-dJ(U,l)l -d )
Let B the square matrix of order m+1 with all entries 1 in the diagonal and in the last column, and with entries 0 elsewhere. B Z' BT
(9)
Since
= Z",
we have (see (6) and (7)) (10)
IZ" I
=
IZ'I
=
-diZI ,
and Z' -::- Z", that is, by the Hasse-Minkowski theorem. Hp(Z")
(11 )
= Hp(l') = Hp(Z) (lZ'I. -IZI).
On the other hand (12) and, by (31)1' (13)
Hp (l")
= Hp (0)
(ll"I,-IOI).
Compari ng (10). (11) and (12), (13) we get (1) and, since Iz' 1= IZ" I ( 14)
1 H ( l ) (! Z' I , -I 0 I ( 1+do (0- )) = H ( 0 ) (I l' I ,-I 0 I ) • p
P
By (18)1 we can cancel (Il' I .-IDI) in both sides and from (6) we obtain (4).
From
(1). (18)1' (19)1 we get the equivalence between (3) and (4). The proof of (2) is an exercise. Let a, b be ra ti ona 1 numbers, u a pos iti ve integer and, fo 11 owi ng Hughes! 16] • define an (a,b,u)-matrix to be a square matrix of order u with a+b on its main
(4) Here the subscript denotes that we refer to the formula (31) of section 1. We shall use this notation throughout all the paper, when referringto formulae of different sections.
385
On divisible designs
diagonal and b elsewhere.
In other words, an (a,b,u)-matrix is defined by
A = aI + bJ ,
(15 )
u
u
and we can apply our lemma (2.1) in this particular case -1
-1
-1
D =a I u , a(D )
D 1:1 aI , d=b,
(16)
u
=
ua
-1
•
we thus get COROLLARY 2.2: Let A = aI u+bJ u be an (a,b,u)-matrix, with a (17)
IAI
If a+ub
~
=
a u-
l
Then
(a+ub).
A- l = l I _ a u
-1
O.
0, then A is invertible and
( 18)
(i.e. A
~
is an (a
-1
b
a(a+ub)
,-bl a(a+ub)J
-1
J
u
,u)-matrix).
We aZso note that
(19)
a(A)
=
u(a+ub).
Hence
u(a+ub)
(20)
-1
•
The following lemma is an easy consequence of lemma (1.10) and of the 2 property (a ,b) = 1 of the Hilbert symbol. LEMMA 2.3: Let u, a, b be the nonzero rational numbers. (21)
(-uab,a(a+ub))p
Then for any prime p
= 1.
From the Lemmas (2.1) and (2.3) we can at once deduce the following LEMMA 2.4: Let A = aI u+bJ u be a nonsingular (a,b,u)-matrix (i.e. a a+ub
~
0).
Then for any prime p
(22)
H (A) = (-1,-1) (-l,a)
Furthermore,
(22)
p
is equivalent to
u(u+l)/2
u (ba ,a(a+ub)).
~
0 and
386
(23)
D. Ghinelli Smit
H (A) p
(-1.-1) (_1.a)u(u-l)/2 (-uau-l.a+ub)(u.a).
PROOF: By (1.8), (22) is lel1Jl1a (2.1) in this particular case (see (16)).
From
(21), and (16)1' (18)1' (19)1 it follows that (24)
u (ba ,a(a+ub))
«-uab)(-ua
u-l
).a(a+ub))
(-uab.a(a+ub))(-ua (-l.a)(u.a)(a.a)
u-l
u-l
.a)(-ua
(-ua
u-l
u-l
.a+ub)
.a+ub)
u u-l ( -1 .a) (u.a)( -ua • a+ub) • But ( 25)
~:~ (mo d 2) 2 +u- 2
and. therefore, (22) is equivalent to (23). The formulae (17) and. for odd primes. (23) were proved by Hughes (see [16], lemmas 2.1 and 3.1) in a different way.
Formula (22) seems to us more
interestin~
as it gives at once (see (25)) the proof of the Bruck-Ryser-Chowla theorem. In order to calculate the Hasse invariants of our matrices in section 5. we shan need, essentially. only the previous Lemmas. (25) and the following elementary property of congruences modulo 2: (26)
ab{a+b) : ab(ab+l) 2
-
2
(mod 2).
3. A RATIONAL CONGRUENCE FOR TACTICAL DECOMPOSITIONS.
THE ORBIT DECOMPOSI-
TION Let S=(P.B.I) be a finite incidence structure with v =Ipi points and b = IBI blocks. A tacticaZ decomposition of S is a partition of the points and the blocks of
S into classes Pl •.•.• P ,. and Bl ••••• B ,. respectively such that b v (i) the number of points of P. incident with a block of B. depends only on
J
1
and j and is denoted by (P .• B.). and. dually 1
J
(ii) the number of blocks of B. incident with a point of P. is a constant J
(Bj.P i ) depending only on the choice of classes.
1
On divisible designs
Given a tactical decomposition
~
387
of a structure S we consider the diagonal
matrices
and
Obviously tr D(P)
(3)
1:1
v,
Let X be the (v'
w.1 I of
with
its copies.
tr D(B) x
=
b.
v)-matrix obtained from I v , replacing the i-th column Clearly
T X X = D(p).
(4)
Similarly, if Y denotes the (b' column with (5)
IB.I
b)-matrix obtained from I b, replacing the j-th of its copies, then x
J T Y Y = D(B).
Let A be an incidence matrix for S associated with the given tactical rows of A correspond to the points of Pl , the next iP2i rows to P2' etc. and similarly for the columns). If we denote by L the
decomposition (i.e. the first (v'
IPl I
b')-matrix whose (i,j)-th entry is (P.,B.), we have
x
1
J
X A = L Y,
(6)
and, by transposition of matrices
Hence:
Equation (8) is particularly interesting when L is a non singular square matrix. In this case, which is always valid when A is nonsingular (see Block [1), cor. (2.1)), (8) establishes a rational congruence between the diagonal matrix D(B)
= yyT and the matrix X(AAT)XT. Thus (8) permits the application of the
theory of Hasse-Minkowski outlined in section 1. The most important example of tactical decomposition of a structure S is
388
D. Ghinelli Smit
given by the orbits of any automorphism group G. We say that S is a square nonsingular struature if it has a square nonsingular incidence matrix.
In this case, it follows from a result of Brauer [5]
that any automorphism group of S has equally many point and block orbits, thus v' = b'
(9)
and L is a square matrix (see also Parker
[30] ,
Hughes [ 16]
,
Dembowski
(6),
LUneburg [ 241 ) • We choose in each point-orbit base bloak x. (i,j=l, ••. ,v').
p.
1
J
J
1
IG p . I = r., 1
IG
1
1
x.
1
which fixes P., and, respectively, x .• ( 10)
a base point P. and in each block-orbit a
Let Gp . and respectively G ,be the subgroup of G
I =
Xj
J
If
5.,
J
then each point in P. has a stabilizer of order r. and each block in B. has a 1
stabilizer of order
J
1
5..
J
We consider the diagonal matrices of order v'
(11 )
and we put IGI (12)
=~.
D(P) =
With the notation introduced in (1)-(8), we have ).l
-1 R ,
D(B)
= ).l
-1 S ,
and thus ( 13)
For our purposes it will be better to consider, instead of the matrix l = (( (Pi,B ) )) the generalized inaidenae matrix, that is the (v'
Ha
j ((h ij )) where hij = I{g
E
G I Pig
x
v')-matrix
Xjl!.
Clearly (14)
H=RL
and (8) becomes (15)
H S-l HT
= ).l-1 R X (AAT) XT R.
In spite of its appearance, (15) is, in our case, simpler than (8).
For
example, when the structure is a symmetric 2-design 2-(v,k,A) of order n = k-A, then (16 )
A AT
=n
I + AJ v
v
389
On divisible designs
Hence
But it is easy to see that (18)
so that, from (13), we get T
(19)
T
-1
-1
X (AA ) X =)J R (nR+)JAJ v ') R
and (15) gives
-1 HT = nR + )JAJ ,.
HS
(20)
v
This is a different and very easy proof of theorem 2.1 of Hughes [ 16).
Equation
(20) establishes a rational congruence between the diagonal matrix S-l and the matrix nR + )JAJ v , (which is of the type considered in Lemma (2.1». Hughes in [16) applies the Hasse-Minkowski theorem to (20) under the hypothesis that G is a standard automorphism group.
We note that, using lemma (2.1), it is not difficult
to write the Hasse-Minkowski conditions for more general
automorphism groups (see
(10) and compare with Dembowski [8) , theorem (2.1.11), [6) and [7), Lenz (23), p. 117, Block [lJ, theorem 6.1, Goldhaber [111, p. 391). An interesting generalization of symmetric 2-designs are the structures introduced by Bose and Connor [31, as Symmetric Regular Group Divisible designs. The main result in [31, consists in considering the rational congruence given by (8) when the tactical decomposition is trivial.
In the next sections we consider
those structures and apply the Hasse-Minkowski theory to (15), to give numerical conditions which must be satisfied if such a design is to possess a standard automorphism group.
4. RATIONAL CONGRUENCES FOR P-DIVISIBLE DESIGNS.
FIRST NONEXISTENCE
CONDITIONS DEFINITION 4.1: A point-divisibZe design '(p-divisible for short) is a tactical configuration for which the v points are split into m classes of size c, such that points have joining classes.
nu~ber
A' (resp.: A) if they are in the same (resp.:
differen~
390
D. Ghinelli Smit
The study of these designs commenced with the work of Bose and Connor [3) who introduced them as an important subclass
of the Partially Balanced Incom-
plete Block Designs with two associate classes of Bose and Nair [4).
Bose and
Connor use the term "group" instead of "classes" and the term "group divisible design" (shortly GD-design).
We avoid this terminology as there is no connection
with the algebraic group notion and as we will deal with automorphism groups of the designs. Since the appearance of (3), much research has been done in connection with point-divisible designs.
For a survey of known results see Raghavarao [32),
Dembowski [8) (pp. 286-289, where the designs are called "divisible partial designs"), Jungnickel [20) and C. Mitchell [27). For sake of brevity, we will assume, in this paper, the following: DEFINITION 4.2: A p-divisible design D=D(m,c,k,A',A) is an incidence structure which satisfies (4.1) and is: (i) square (i.e. with b=v=mc blocks), (ii)
(i.e. it has a nonsingular incidence matrix, or, what is the same, k f A' and k2 f vA).
nonsingu~al'
When the dual structure
D'·:
is also a p-divisible design (i .e.
D
is both
pOint and block-divisible) we will say that D is a divisible design. Necessary and sufficient conditions for a p-divisible design to be divisible are given by C. Mitchell in [28); in particular he shows that if
D
is divisible,
then D* is a p-divisible design with the same parameters. We note that in Bose and Connor's terminology our "p-divisible designs" are called "Symmetric Regular Group Divisible Designs" (SRGDD, for short), and, by the above mentioned result by Mitchell, our "divisible designs" are precisely the structures that Bose in [21 calls "SRGDD with the dual property". For divisible designs we shall use the following result, stated by Bose ([ 21) in equi va 1ent form THEOREM 4.1: Let D = D(m,c,k,A' ,A) be a point classes
p] •..•• ?m
divisib~e
design.
and the divisibility block cZasses
tacticaZ decomposition of D and the matrix
Then the divisibiZity
B1••••• Bm form
a
391
On divisible designs
(1 )
« (P. ,B.) »
L
J.
1
satisifies (2)
L LT = n1
(3 )
L J m = J m L = kJ m,
m
+
CAJ , m
where
(4)
n
= (k~A')
+ C(AI -A).
Note that (2) establishes a rational congruence between the (n,cA,m)-matrix LLT and the identity matrix 1m' From our (2.2), (2.4), we get easily, applying the Hasse-Minkowski theorem, the following THEOREM 4.2: If a divisible design D=D(m,c,k,A',A) exists, then (i) if m is even. n is a square; (ii) if m is odd, the Diophantine equation
n x2 + (1)(m-l)/2 c", 2 y
(5)
= Z
2
has a solution in integers x,y,z not all zero.
If m and c are both odd, Theorem (4.2) appears to be an improvement of Bose and Connor [3J, theorem 9. REMARK 4.3: An interesting special class of divisible designs is that with AI
= 0,
A> O.
In this case p-divisible is equivalent to divisible.
These de-
signs are also called Divisible Semisymmetria designs or, when A = 1, Divisible semiplanes, when A = 2, Divisible Semibiplanes, (see [9] and [10], or Hughes [18),
Jungnickel [20J, Wild [33) ).
SemisYlTllletric designs with c
=
1 are symmetric 2-
designs. For semisYlTllletric designs L is a (O,l)-matrix and (2)-(4) show that L is an incidence matrix for a symmetric 2-(m,k,cA). Then, it is natural to call this 2design the quotient design(with respect to the divisibility relation).
In this
case, theorem (4.2) is the Bruck-Ryser-Chowla theorem for the quotient design. Mielants gives in [25) a direct construction of the dual of the complement of the quotient design 2-(m,k,cA), and applies the Bruck-Ryser-Chowla theorem to
392
D. Ghinelli Smit
get nonexistence conditions. the particular case
~'
= 0
Therefore the conditions in (25) are equivalent to
of (4.2) (see, also, (21) and Payne [31), theorem
(5.1) ).
REMARK 4.4: In general l is not a (O,l)-matrix but, by abuse of terminology we will consider l as a sort of pseudoinoidenoe matrix and, also in general, we shall speak of the quotient symmetrio pseudodesign which is a 2-(m,kH'(c-l),o).
(6)
Going back to p-divisible designs (not necessarily divisible), let
Abe
an
incidence matrix for D=D(m,c,k,A' ,A) associated with the division (i.e. the first crows of
Acorrespond
to the points of the first class
poi nts of the second class
PZ'
etc.).
(7)
A AT
(8)
A Jv = Jv A = k J v
where
x
= I
m
x
C + AJ
Pl ,
the next c rows to the
Then
v
is the Kronecker product of matrices and C is the (k-A',A'-A,c)-matrix
(9)
C = (k-A')l c
+
(X'-X) J c •
We will call order of the design the integer (10)
n = (k-t.') + CP'-A).
It is easy to prove (see (3.1) in (3) (11 )
that
2 n = k - >v.
Now, let G be an automorphism group of D, and let A be an incidence matrix associated with the orbit tactical decomposition.
With the notation of section 3,
we have
Since A and
Aare
both incidence matrices for D, there exist permutation matrices
P and Q such that (13)
A = P AQ.
Hence, from (7), we get
393
On divisible designs
( 14)
and, for our purposes, the nature of the permutation Q on the blocks is irrelevant. As in section 3, we have
X J XT = ~2 R- l v
R- l ,
J '
v
and, therefore -1
T
H =~
HS
(15)
-1
(RX) P (1m
T
C) P (RX)
x
T
+ ~A J
v'
From now on G will be a standard automorphism group. points fixed by G.
Since D is nonsingular, v'
Let N be the number of
= b' which implies that N is also
the number of fixed blocks. Let
be the number of fixed classes, and let i be an integer, 0
~
~
i
~
c.
We define
orbits of length
(c-l)/~
We note that
~).
~).
Clearly C
C
(16 )
~
=
N=
l:
i=O
l:
i
i=O
Counting orbits of non fixed points contained in fixed classes, we get v.
(17)
where
=
1
h={m~)/~
(18 )
When ( 19)
Also, if
v'-N-ch
c~-N
= -~
is the number of orbits of non fixed classes.
= v'-ch,
l:
~
> c the situation is much simpler; then
~
=
~ ~
1
1
c
l:
Hence
= N/c.
c, then
=
0 for every i F O,c and
= N/c.
We prove in [10] the following theorem concerning the generalized incidence matrices of a p-divisible design THEOREf.l 4.5: Let D = D{m,c,k,A',A) be a p-divisible design. automorphism group of order
~.
'bJhich fixes N points and
then D has a generalized incidence matrix H satisfying (20)
-1 T
HS
H
= (I h
x
C)
e
K+
~AJV'
,
~
If D has a standard
divisibility classes,
394
D. Ghinelli Smit
;)hel'e C is a ((k-\'),
(l.. '-A),
fixed a[asses, v'={(v-N)/ll c
(21 )
K=
(1
i=O
~i
x
c)-matl'ix, h=(m- )/ll is the nwnber of orbits of non-
~N
is the number of point orbits (and bZoak orbits) and (5)
K.) 1
with
(v.=(c-i}/ll, i=O,.,c)(5) •
( 22)
1
The matrix 5 is defined in seation 3.
(Sinae G is standard S-1 is a diagonal
matrix '.iii th v'-N entT'ies 1 and N entries II -1, corresponding to the N fixed b loaks) •
From the last observation in (4.5) and (1.8), (1.6), it follows at once (23)
IS-ll = ll-N,
and, for every prime p (24)
H (S-l)
(_1,_1)(_l,Il)N(N+l)/2.
p
Hence, if (25)
Z
(I
h
x
C)
K + llAJ
V
'
,
then (26) must be a square. We note that Z is a matrix of the type considered in section 2 with
D = (I h x C)
Izi
= k 2uN (k-A') v'-(h+
)n(hi4> )-1.
From (26) and (27) we derive the following nonexistence theorem: THEOREM 4.6: A neaessary condition fol' the existence of a p-divisibZe design D=D{m,c,k,A ',A) with a standard automorphism group of order u, having fixed a[asses, N fixed points, is that
(5) If x=O we mean that the term containing Ix must be suppressed in a direct sum of matrices.
395
On divisible designs
(28)
(k-A')
v'-u
u-l n is a square. 2
Here n = (k-A')+C(A'-A)=k -AV is the order of D, ber of point orbits (v=mc) and u=h-i4>
={(m-~)/~}H
v'={(v-N)/~}+N
is the num-
is the number of orbits on
cZasses.
REMARK 4.7: As it should be, when
~=l
(~v'=v=mc,
u=m)
(28) gives, as a particu-
lar case, that (k_A,)m(c-l)(k 2_VA)m-l must be a square,
(29)
that is theorem 7 of Bose and Connor [3]. If c=l, that is the design is a symmetric 2-(v,k,A), then v=m,
~
=N, and
thus v'=u and (28) asserts that v'-l n is a square
(30)
that is the condition of Hughes [ 16] • If
~=c=l
we have, of course, the first condition of the Bruck-Ryser-Chowla
theorem (see
~1.
Ha 11 [12] ):
(31)
v-l n is a square.
I
We can now use condition (29) to simplify, when m is even or m is odd but c is even, theorem (4.6).
Thus we get, for instance, the following
THEOREM 4.8: If a p-divisibZe design D=D(m,c,k,A' ,A) with a standavd automorphism group G having v' orbits on points and u orbits on cZasses exists then (i) If m is even then n=k2-vA=(k-A')+C(A'-A) is a square and either k-A' is aZso a square or v'=u (mod 2). (ii) If m is odd and c is even, then (k-A') is a square and either n is aZso a square or u is odd (iii) If m and c are both odd we have condition (28).
For divisible designs the previous results can be improved, as we can consider the quotient symmetric pseudodesign 2-(m,k+x'(c-l),cA) (see (4.4». The group G acts on it standardly and has
~
fixed "points" and u orbits.
Since
theorem (4.6) actually concerns matrices and not incidence structures, we can u-l apply it to this pseudodesign and we get that n also must be a square. Hence
396
D. Ghinelli Smit
we have THEOREM 4.9: fieaessaY'Y aondition foY' the existenae of a divisible design U=D(m,c,k,A',A) with a standard automoY'phism gY'oup of oY'der pOl:nts,
1>
having N fixed
fixed point alasses is that
(k-A')
(32)
~
v'-u
u-1 and n are both perfeat squares.
Hei'e v' and u are defined as before Y'elatively to the points, but it aan be shown that they, as well as 4> and N, have the same value relatively to the bloaks.
5.
THE HAW NONEXISTENCE THEOREMS
Consider a p-divisib1e design D=D(m,c,k,A',A) with a standard automorphism group G of order u. ( 1)
Let 4>,N,h=(m-4»/u, v' ,n as in (4.5) and let
l = (I h xC) e K + u AJ v I
,
where C is a (k-A ' ,A'-A,c)-matrix and K is defined in (21)4' We have seen in section 4 that the matrix Z is rationally congruent to the matrix S-l defined by (11)3 (see theorem (4.5»).
Hence a necessary condition for
the design to exist is that (2)
(-1,-1) (_1,~)N(N+1)/2
H (Z) = H (S-l) p
p
for every prime p. From lemma (2.1), we have (3)
Therefore, in order to calculate H (Z), we have to calculate H «I x C) e K) p P h (that is, by (1.6) Hp (C), Hp (K) and (iClh,IKI) and 1+ ~Ao«Ih x C-l) e K- l ». Applying the lemmas of section 2 we can show that (4)
-1 l+lJAo«Ih xC)
e
-1 2 K ) = kin.
2 From this, as k /n ~ n and III = ~-NIHI2 ~ ~N, we get (5)
Hp(Z) = Hp«Ih x C) e K) (-lJ
N+l
A,n).
Comparing (5) and (2) we can derive the following THEOREM 5.1: If a p-divisibZe design D=D(m,c,k,A',A) with a standard automoY'phism
397
On divisible designs
group G of order jl having N fixed points,
v'={(v-N)/jl}+N orbits on points and
1>
u={(m~
fixed classes, (and thus
)/]1
J~~
orbits on classes) exists, then
for every prime p
(6)
(k-A' ,(_1)(V'-u)(v'-u+l)/2 c U ]1N+1»p (n,(_1)U(U-l)/2 cUAjl¢+1)p = (n,_l)u-l. The proof (see [10]) is extremely tedious and long, but is rather straight-
forward, relying highly on the lemmas proved in section 2 and on (16)4-(18)4' This should be clear from the definition of K and from the properties (1.6)-(1.9) of the Hasse invariant.
We observe that if ]1 > c the proof is very simple and
could be given directly as then (7)
K = ]1(I /
Nc
x
C)
REI4ARK 5.2: (i) If]1 = 1 (=> v'=v, u=m, v'-u=m(c-l» (8)
then (6) gives, as (n,-l)
u-l
=1
(k_A',(_1)m(C-l){m(C-l)+l}/2 Cm)p (n,(_1)m(m-l)/2 cm A)p = 1
which, by (26)2' is exactly the condition in Bose and Connor [31, and combined ~/ith
(29)4 gives the well-known Bose and Connor Theorem. ( v'-u)!v'-u+1)/2 u N+q, (i i) If c= 1 (=> v=m, 1> =N, v' =u) the (-1) \ c i s a square u-1 and (6) becomes, as (n,-l) ='1, (9)
(n,(_1)v'(v'-1)/2 A]1 N+1)p = 1,
that is the condition given by Hughes in [16] , which, together with (30)4 gives the Hughes theorem. (iii) If ]1=c=l we have, of course, the second condition of the Bruck-RyserChow1a theorem. Hence, combining (4.6) and (5.1) we have a simultaneous generalization and, a 1so, a unifi ed proof of Bose ant Conn:)t', Hughes, and Bruck-Ryser-Chow1a Theorems. From (1,.3) we can rephrase the conditions in terms of Diophantine equations, and state the following nonexistence theorem. THEOREM 5.3: A necessary condition for the existence of a point-divisible design D=D{m,c,k,A',A) with a standard automorphism group G of order ]1 which fixes N points andq, classes (and so has v'={(v-N)/]1}+N orbits on points and
u={(m-1»/]1}+q, orbits on cZasses) is that (i) if v' and U both even, or m is even, then n is a square and the equation
398 ( 10)
D. Ghinelli Smit
(k-\') x2 + (_1)(V'-U)(V'-U+l)/2 CU ~N+~ y2
z
2
possesses a nontrivial solution in integers. (ii) if v' is even and u is odd, or m is odd and C is even, then k-\' is a square and the equation (11 )
n x2 + (1)u(u-l)/2 -
u
C
Hl Y2
z
\~
2
possesses a non trivial solution in integers. (iii) if v', u,m and C are all odd, then (10) has a nontrivial solution in
integers if and only if (11) has suah a solution. (iv) ifv',m and
( 12)
C
are odd but u is even then
(k-\')n is a square and thus
\ N+ 1 2 = z2 n x2 + (1 - ) ( v' -1 ) /2 ,,~y has a nontrivial solution in integers.
For divisible designs the nonexistence theorem (5.3) can be improved applying the condition (9) to the quotient symmetric pseudodesign 2-(m,k+\'(c-l), CA)
(see also theorem (4.9)). More precisely, then
( 13)
(k-\')
v'-u
andn
u-l
are both squares,
and (14 )
( n, ( - 1) u(u-l)/2
Hl)
CA~
p
. p = 1 f or every prlme
are necessary conditions for the existence.
Hence we can state the following
THEOREM 5.4: If a divisibiZe design D=D(m,c,k,\',\) exists with a standard automOI'phism group of order
~,
and if ~ , N, v·, u, n are defined
as
above, then we have
the aondition (13) and the Diophantine equations (10) and (11) are equivalent and both possess nontriv1:al solutions in integers.
Applications of the nonexistence theorems given here can be found in [10], especially in the case of semi symmetric designs. We conclude by pointing out that all the results in this paper concern standard tactical decompositions of p-divisible designs and we do not need to
On divisible designs
399
assume the decomposition to be the orbit decomposition of some automorphism group. ACKNOWLEDGEMENTS: The Author acknowledges with great pleasure the guidance of Prof. D.R. Hughes, her research supervisor at the University of London. She would also like to thank the British Council and the University of Rome for support during the time of this research.
Bl BL IOGRAPHY 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.
R.E. Block, On the orbits of collineation groups, Math. Z., 96 (1967), 3349. R.C. Bose, Symmetric Group Divisible Oesigns with the dual property, J. Stat. planning Inf., 1 (1977),87-101. R.C. Bose and W.S. Connor, Combinatorial properties of group divisible incomplete block designs, Ann. Math. Statist., 23 (1952), 367-383. R.C. Bose and K.R. Nair, Partially balanced incomplete block designs, Sankhya, 4 (1939), 337-372. R. Brauer, On the connections between the ordinary and the modular characters of groups of finite order, Ann. Math., 42 (1941), 926-935. P. Dembowski, Verallgemeinerungen von Transitivitatsklassen endlicher projektiver Ebenen, Math. Z., 69 (1958), 59-89. P. Dembowski, Tactical decompositions of A-spaces, Proc. Colloq. Found. Geometry Utrecht 1959; Pergamon (1962), 15-23. P. Dembowski, Finite Geometries, Springer-Verlag, New York (1968). D. Ghinelli Smit, On semisymmetric designs, (Westfield College, University of London, 1980). D. Ghinelli Smit, Nonexistence theorems for automorphism groups of divisiblE square designs, (Ph. D. Thesis, University of London, to be submitted~ J.K. Goldhaber, A note concerning subspaces invariant under an inCidence matrix, J. Algebra, 7 (1967), 389-393. M. Hall Jr., Combinatorial theory, Blaisdell, Waltham (1967). G.H. Hardy and E.r~. Wright, An introduction to the theory of numbers, Oxford Clarendon Press, (1938). H. Hasse, tiber die Aquivalenz quadrati scher Formen in Korper der rationalen Zahalen, J. reine angew. Math., 152 (1923), 205-224. D.R. Hughes, Regular collineation groups, Proa. Amer. Math. Soa., 8 (1957), 165-168. D.R. Hughes, Collineations and generalized incidence matrices, Trans. Amer. Math. Soa., 86 (1957), 284-296. D.R. Hughes, Generalized incidence matrices over group algebras, Ill. J. Math., 1 (1957), 545-551. D.R. Hughes, Biplanes and semibiplanes, in Combinatorial Mathematias, Springer Lecture Notes 686 (1978), 55-58. B.W. Jones, The arithmetic theory of quadratic forms, MAA Carus Math. Monographs 10, John Wiley and Sons Distr. (1950).
400
20. 21. 22. 23. 24. 25. 26.
27. 28.
D. Ghinelli Smit
D. Jungnickel, On automorphism groups of divisible designs (to appear). C.W.H. Lam and S.E. Payne, Erratum to "Generalized relative difference sets", PI·oo. Amer. Math. Sao., 56 (1976), 392. E.S. Lander, Topics in algebraic coding theory, Ph. D. Thesis, University of Oxford, October 1980. H. Lenz, Quadratische Formen und Kollineationsgruppen, Aroh. Math., 13 (1962), 110-119. H. LUneburg, Gruppentheoretische Methoden in der Geometrie, Ein Berioht. Jahresber. Deutsohe Math. Ver., 70 (1967), 16-51. W. Mielants, On the nonexistence of a class of symmetric group divisible partial designs, J. Geom., 12 (1979), 89-98. H. Minkowski. Uber die Bedingungen. unter welchen zwei Quadratische Formen mit rationalen Koeffizienten ineinander rational transformiert werden konnen, J. reine angew. Math., 106 (1980). 5-26; see also Ges. Abh. I. Teubner. Berlin and Leipzig 1911, 219-239. C.J. Mitchell. On divisions and decompositions of l-designs, Ph. O. Thesis, University of London (1979). C.J. Mitchell, Group divisible designs with dual properties, (to appear in Disorete Math.).
29. 30. 31. 32. 33.
G. Pall, The arithmetical invariants of quadratic forms, BuZZ Amer. Math. Sao., 51 (1945), 185-197. E.T. Parker, On collineations of symmetric designs, Proa. Amer. Math. Sao., 8 (1957), 350-351. S.E. Payne, Generalized relative difference sets. Proa. Arner. Math. Sao., 25 (1970), 46-50. D. Raghavarao, Construction and combinatorial problems in design of experiments, John Wiley and Sons, Inc. New York (1971). P.R. Wild, On semibiplanes, Ph. D. Thesis, University of London (1980).
Westfield College (University of London) Kidderpore Avenue London NW3 7ST, U.K. permanent address: Istituto Matematico "G. Castelnuovo" Citta Universitaria 00100 Roma, Italy
401
Annals of Discrete Mathematics 18 (1983) 401-408 © North-Holland Publishing Company
SOME RESULTS ON PARTIAL STEINER QUADRUPLE SYSTEMS Mario Gionfriddo
ABSTRACT In this paper we prove some results concerning DMB PQSs having at least an element of degree seven and with m = 14 or 15 blocks.
In a previous paper we
have determined, to within isomorphism, all Dr·1B PQSs having 12, 14, 15 blocks and satisfying some particular conditions.
By means of the results contained in this
paper, we purpose to construct all DMB PQSs having 12, 14, 15 blocks.
1. INTRODUCTION A Partial Steiner QuadrupZe System (PQS) is a pair (P,q) where P is a finite non-empty set and q is a collection of 4-subsets (called bZocks) such that every 3-subset of Q is contained in at most one block of q.
Using graph theoretic ter-
minology, we will say that an element x of P has degree d(x)=h if x belongs to exactly h blocks of q.
Clearly Lpd(x)=4.lql. XE
We will call the degree-set (DS)
of a PQS (P,q) the n-up1e DS = [d(x), d(y) •••. ] where x,y, •• are the elements of P.
If there are r., elements of P having degree h., for i=1,2, •• ,s, we will write 1
DS=[(h l ) ,(h ) , ••• ,(h)] ,wherer +r 2+... +r =IPI. 2r l rl s rs s 2 then we will write (h')l=h .• 1 1
Ifr.=l,forsomei, 1
Two partial quadruple systems (P,q1) and (P,q2) are said to be disjoint and mutually balanced (DMB) if they have not blocks in common and any given triple of
distinct elements of P is contained in a block of ql if and only if it is contained in a block of q2'
It is easy to see for any two DMB PQS (P,q1) and (P,q2)
that Iql I = Iq21. A Steiner quadruple system (SQS) is a partial quadruple system such that every 3-subset of Q is contained in exactly one block of q. In a previous paper [4] we have determined, to within isomorphism, all DMB
402
M. Gionfriddo
PQSs having 12, 14, or 15 blocks (observe that OMS PQSs with 13 blocks do not exist, see (2)) and satisfying some particular conditions (blocks with at least two elements of degree 4 or blocks with one element of degree 7) along with their admissible degree-sets.
Our aim is to construct all OMS PQSs with 12, 14, 15
blocks and after, by means of these OMS PQSs, to find pairs of SQS(v) with qv-12, qv-14, qv-15 blocks in common, for qv = v(v-l)(v-2)/24. In this paper, we prove some results concerning OMS PQSs having at least an element of degree 7 and with m = 14 or 15 blocks.
In (4) we have constructed
a~~
OMS PQSs with 12, 14, or 15 blocks and such that there exists in them a block with at least two elements of degree 4.
Therefore, in what follows we shall always
consider OMS PQSs having blocks with at most one element of degree 4.
(P,ql) and
(P,q2) will be two OMS PQSs with P = {O,l, •• ,n-l} and Iqll = Iq21 = m.
Further,
for i=1,2, we will put Li = x
E
P, then JK(x,i) = {b
E
~q.
P3(b), where P3(b) = {X : Xc b, IXI = 3}. If q. : ~ E b}, further'" = {x E P : d(x) = r}. Finalr
1
ly, if x,y E P, x F y, we will write (x,y)
r
to indicate that {x,y} is contained in
exactly r blocks of ql or q2"
The following results on OMS PQSs are known (2):
1) for every x E P, d(x) > 4
further n > 8, m ;:. 8 ;
2) if k = MAX {d(x): x E P}, then m ;:. 2 k ; 3)
if d(x) > 4, then d(x) ;:. 6 ;
4) if m> 8, then m = 12 or m ;:. 14.
2. RESULTS From now on (P,ql) and (P,q2) will be two OMS PQSs with m = 14 or 15 blocks.
Further we shall always suppose that IM4
n
bl
~
1 for every bE ql
U
q2.
The results are call obtained to within isomorphism. LEMMA 1: Let (P,ql)' (P,q2) be two OMS PQSs with M = 14 or 15 that 1M4 n bl ~ 1 for every b E ql U q2. ;chat d(x) = 7 and
(x'Y)2~
then
{x,y,a,b} {x,y ,c ,d}
{x,c,f,h}
{x,e,f,g} E ql {x,e,b,d} {x,d,g,h}
{x,e,a,c}
b~ocks
and such
If there are two e~ements x,y E P such
403
Some results on partial Steiner quadruple systems
{x,y,a,c} {x,y,b,d} {x,e,a,b}
{x, h ,c, d}
{x,c,e,f} {x,d,e,g}E q2 {x,f,g,h}
PROOF: Let {x,y,a,b}, {x,y,c,d} E ql' {x,y,a,c}, {x,y,b,d} E q2'
If {U,x,a,c},
{V,x,b,d} E ql and {W,x,a,b}, {Z,x,c,d} E q2' then it is easy to see that it is not U = V
W = Z and that U '* V or W '* Z implies {U,V} n {W,Z}
that if U
V (resp, W = l) then {U,V} n {W,Z}
{U,V} n {W,Z}
a~,
~~,
let {W,x,i,j}E ql'
We wove
In fact, let U = V and let
Since W ~ Z implies {U,V} n {W,Z}
Necessarily {W,x,a,d} or {W,x,b,c} E ql'
~~,
~ ~,
it follows W = Z,
Let {(i,j),(i',j')}= {(a,d),(b,c)} and
Since (x,j)3 in ql' then {U,x,i,j}E q2'
Observe that
(since (x, i')3 in ql) {W,x, i' ,j' } E ql if and only if {U,x, i' ,j' } E q2'
Moreover,
if {W,x,i'}, {W,x,j'} are contained in two distinct blocks of ql and {U,x,i'}, {U,x,j'} are contained in two distinct blocks of q2' then we have (W,x)3 and (U,x)3'
In every case d(x)
Suppose U = W.
~
Necessarily, it follows that {U,V} n {W,Z}
7.
We have
{x,y,a,b}
{x,y ,a,c} {x,y,b,d}
{x,y,c,d} {U,x,a,c}
~ ~.
E q
1
{U,x,a,b}
{V,x,b,d}
E q
2
{Z,x,c,d}
where it is not V=U=Z.
Let Z
~
Suppose V ~ U.
U.
hence {x,c,Z,V} and {x,b,U,Z} E ql' Therefore, it is V = Z.
If V
~
Z, then {x,b,V,Z} E q2'
It follows {U,x,c,Z} E q2 with {d,X,C,Z}E q2'
Since (x,b)3 in ql' then {x,b,V,e} E q2'
Necessarily
{x,c,V,e} E ql and {x,c,U,e} E q2' with d(x) < 7
At this point, we can say that necessariZy U Z
h.
We have {x,y,a,c}
{x,y,a,b}
{x,y ,b,d}
{x,y,c,d} {x,a,c,e}
E q
1
{x,b,d,e}
{x,a,b,e}
E q
2
{x,c,d,h}
Hence, necessarily {x,c,f,h}
{x,c,e,f}
{x,d,g,h} E ql
{x,d,e,g} E q2
{x,e,f,g}
{x,f,g,h}
V
W F Z.
Let U
V
W
e,
404
M. Gionfriddo
The statement is so proved. 0
THEOREM 2.1: Let (P,q1)' (P,q2) be two OMS PQSs with m = 14 or 15 bZocks and such ~hat 1M4 Ii
bl "
1 fei' every bE q1
that d(x) = d(y) = 7 and I
IK( x, i)
Ii
IK(y, i)
IK(x,i)
U
Ii
q2'
IK(y,i)F0 (for i=1,2), then
F 2.
1
PROOF: Let x = 1, y = 2, and let II«l,i)
p =
{D, 1,2, •••
If there are two eZ-ements x,y E P suah
,n-1}.
Ii
IK(2,i)
1
2, for i
1,2,
From lel11lla 1, we have
{l
,4,6,7}
{1
,3,4,7}
(l
,3,5,7)
{l,5,6,O}
; 1,6 ,9 ,0 }
{1
(1,7,8,9) E ql
{l,6,7,9}E q2
{l,5,8,O}
{l,5,7,8}
n,2,3,4}
{l,2,3,5}
{l,2,5,6}
{l,2,4,6}
Observe that (2,7,8,9) E q2'
,8,9,0}
Otherwise, we have that {3,5,7}, {4,6,7},
U,8,9} are contained in exactly three blocks b. such that b. Ef 1«2,2), hence 1
IK(2,2)
Ii
contained in blocks b. E IK(2,1) , it follows 1
1M4 d(6)
Ii
~
1
1<(7,2) = 0 and, since {3,4,7}, {5,6,O}, {5,7 ,8}, {6,7,9} are not m;;'
16.
At this point, since
bl " 1 for every b E ql U Q2' we have d(8) + d(9) + d(O) ;;. 16, d(5) ~ 6, 6, d(3) + d(4)
follows m = 15 (i.e.
~
10, d(l) = d(2) = d(7) = 7, hence
~p
d(x) = 60).
OS = [(7)4' (6)4' (4)2] with {{2,5,8,O), {2,6,9,0}}
Ii
{3,4}
x~p
d(x)
~
59.
It
Further, the only admissible degree-set is Ii
M4 F 0 and {8,9,0}
Ii
IK(2,2) F 0, it follows (2,0);.;. 2'
possible that {2,0,5,6} E Q1' hence d(O) ;.;. 6.
M4 F 0.
Since
Further, it is not
It follows {2,8,9,O} E q1'
Otherwise we have (8,9)3 in Q1' hence d(8) ;.;. 6, d(9) ;.;. 6.
Observe that, since
m = 15 and {5,7,8}, {8,9,O}, {6,7,9}, {3,4,7l,{5,6,O} E fl. ' we have {2,7,8,5} or 1 {2,7,9,6} E Ql' Let {2,7,i,i+3} E Q1' for i=5 or 6. Since {2,8,O}, {2,9,O}, {2,7,iJ, {2,i,i+3}
E
fl. ,
2 (2,5,6,7} E Q2 and that {2,3,5,O }
{2,3,4,0}
{2,4,6,O} {2,5,7,8} {2,6,7,9}
it is a routine matter to see that necessarily
{2,5,6,7} E
Q 1
{2,5,8,O} {2,6,9,O}
E
Q2 •
Some results on partial Steiner quadruple systems
Since
{3,4} n M4 1
I
follows {3,4} n M6 "I 0, {2,0,u,u+2}
4'
q,!
405
then we have (3,4)2' hence {3,4,7,0}
.,; "
But {5,6,0,u}
E
q"
for u
E
E
ql'
It
{3,4}, implies
[JJ
LEMMA 2: Let (T,t,)and (T,t ) be two DMB PTSs (partiaZ tripZe systems) with m = 7 2 Nooks, It foUows that: I) If x,y II) If
E
T, d(x) '" dey) = 3, then IK(x,i) n !K(y,i) "10, for i=',2.
ITI = n = 7.
then d(x)
=3
for every X E
T;
further it is (to within
isomorphism) {2,3,4} {4,5,7}
{2,3,5} {5,6,7}
{2,5,6} {4,6,8} E t"
{2,4,7} {3,7,8} E t , 2 {2.6,8} {4,5.8} {3.4,6 }
{2,7,8} {3,6,7} {3,5,8 } III) It is n
IV) If
It '"
= 7 or n = 9,
9. then we have (to within isomorphism)
{2.3.4} {4,7,9} {2,5,6} {6,8,O}
{2,3.5} {5.6,0} E
{2.4.7} {l,8,0}
t"
{2,7,8} {5,7,0} {3,5,9 }
{2.6,8} {5,?9}
E
t , 2
{3.4.9 }
PROOF: I) Let 2 and 3 be two elements of P having degree three and such that for every b', b" E ti (i'"' or 2) if 2 E b', 3 E b" then b' n b" "0. then d(4)
If {4.5.6} E t • l
= d(5) = d(6) = 3.
hence {4,5.6} E t , 2 II) If T = {2.3, .... 8}. it is easy to prove the statement, III) It is a routine matter to see that for n < 7, n
= 8,
n
~
'0, there are
U
t , then we 2
not pairs of OMB PTSs with m = 7. IV) Let T
= {2.3 .... ,9.0},
We have
{2.3.4}
{2,3,5}
{2.5,6}Et,.
{2,4.7}Et2.
{2.7,8}
{2,6.8}
Necessari'y d(9)
= d(O)
have (to within isomorphim)
2,
If {9.0} 1. b for every bE tl
406
M. Gionfriddo
{3,4,9 }
t3,5,9 } {4, 7,9 ) {6,8,0 }
{5, 7,0
{5,6,0 } t ,
E
{7 ,8,0} E t2'
l
(5, 7,9 )
j
If {y,9,0} E t , it follows x -f y and 2 {x,9}, {x,O} are contained in two distinct blocks of t2' This implies m ~ 8. Therefore, suppose {x,9,O} E tl'
0
COROLLARY: Let (P,G ) and (P,q2) be two DMB PQSs. If x,y,z E P, X E M]' (x'Y)3' l (x,z)3' then there c-xiets a bZock bE D«x,i) such that {y,z}::. b, for i=1,2.
It is immediate to see that the statement is equivalent to lemma 2,I).
THEORHt 2.2: Let (P,ql) and (P,q2) be
tJ~o
DMB PQSs vith m = 14 or 15 blocks.
x E M , Y E t~6 sucr ~ilOt IIK(x,i) n IK(y,i) 1= 3 (i=1,2).
7 e,}f!l",f b
E
ql
U
~;16n ~e
Q2'
If 1~14 n bi .;; 1 for
have (to with-in isomoI'phism):
{x,y,3,4}, (9,y,3 ,5 )
{x,y,3,5 },
{9,y,3,4 }
( x ,y , 5 ,6 ), {9 ,y ,4 , 7 )
{x,y,4,7 },
{9,y,5,6 }
{x,y,7,8}, {9,y,6,8}
{x,y,6,S },
{9,y,7,S}
{x,3,5,81, {9,3,4,6) E ql
{x,3,4,6 },
{9,3,5,S} E q2
{x,4,5,7}, {9,5,6,7}
{x,5,6,7 },
{9,4,5,7 }
(x,4,6,8), {9,3,7,S}
{x,3,7,S },
{9,4,6,S}
{x,3,6,7}, {9,4,5,S}
{x,4,5,S },
{9,3,6,7 }
The para'1leters are m = 14, n
PROOF: Let x
1, y
2.
Let
9, OS = [ (7) 2' (6) 71 '
USing lemma 2 and corollary, we have the following
two cases, 1) ibE~(1,i~ - (1
II
7.
i=l or 2 We have i 1,2,3,4)
(1,2,5,6 ) (1,2,7,8 ) : 1 ,3,5,S} Let further
(1
{1 ,2,3,5} {1 ,4,5, 7l
{l ,2,4,7}
{1,4,6,S}E ql { l,3,6,71
(2,3,5,~)
E ql'
{l ,2,6,S} {1,3,4,6}
{1,5,6,7} {l,3,7,8}E q2 {1,4,5,8}
Necessarily a = 9 (it is easy to see that a -f 4,6,S,
= 7 implies {2,3,4,7l or {2,3,7,8}E q2 with {1,2,4,7l, (l,3,7,8}E q2)'
407
Some results on partial Steiner quadruple systems
It follows that {2,3,5,9}, {2,4,7,9}, {2,6,B,9} E ql and {2,3,4,9}, {2,5,6,9},
{2,7,B,9} E q2'
At this point, observe that there are in q2 exactly four blocks
containing {3,5,9}, {4,7,9}, {6,8,9}, {3,6,7} respectively. it is easy to see that S
F 1,2,4,5,B.
Further, it is S
F
If {3,6,7,S} E q2'
°:
in fact, since
{3,5,8}, {4,5,7}, {4,6,B} E lL2 and d(O) ;;. 4, if 6 " 0, then m > 15.
Necessarily
6 = 9.
Hence {3,4,6,9}, {5,6,7,9}, {3,7,B,9} E ql and, since {3,5,9}, {3,B,9}ElL , 2 it follows {9,2,3,5}
{9,2,3,4}
{9,5,6,7}
{9,2,4,7}
9 3 7 B}
{
{9,2,6,B}
,
"
E
{9,2,5,6}
ql
{9,2,7,B}
{9,4,5,8}
{9,3,4,6 } 2) IbE~( 1,
{9 {
,4 , 5 , 7 }
9 4 6 8 ,
,
,
} E
q2'
{ 9 ,3 ,6 ,7 }
{9,3,5,B}
if - {l } I = 9.
i=l or 2 We have {l,2,3,5}
{l,2,3,4} {l,4,7,9}
{l,2,5,6}
{124 , , ,7} {l,2,6,8}
{l,6,B,0} Eq l'
{l,2,7,B} {l,3,5,9}
{l,5,7,O}
{1
, 3 ,4 , 9 }
n,5,6,O} n,7 ,8,0} E q2' {l,5,7,9}
Since {l,3,5,9} E ql and {1,5,6,O} E q2' it is a routine matter to see that
=0
{9,O} n b
for every bE JK(2,i).
FUrther, d(O) < 7 and d(9} < 7.
Otherwise,
for d(O) = 7 (resp. d(9) = 7}, {5,7 ,9,O} E ql (resp. E q2)' with n ,5,7,0} E ql Finally, d(O) < 6 and d(9) < 6.
(resp. {l,5,7,9} E q2)' d(O)
=6
(resp. d(9)
= 6),
Otherwise, for
{3,4,9,0} E ql (resp. {6,B,9,0} E q2) with {3,5,9,0},
{4,7 ,9,0} E q2' {3,4,0} E lL2 (resp. {5,6,0,9}, {7 ,B,0,9} E ql' {6,B,9} E lL ), l hence d(O) = 7 (resp. d(9) = 7). Necessarily d(O) = d(9) = 4. It follows ~
M6 u M • 7 This implies that
{3,4,5,6,7,B}
Hence
{5,6,0,v}
{4,7 ,9,u}
E
ql
{6,B,0,v}
{7,B,O,v} with u
=6 0
= 10.
or B, v
E
q2
{5,7 ,O,v}
=3
or 4.
two distinct blocks of q2' theorem.
d(x) ;;. 57 and, therefore, m = 15, n
{3,5,9,u}
{3,4,9,u} {5,7 ,9,u}
~p
It follows that {3,4,u} and {5,7,u} are contained in Hence d(u) ;;. B.
This completes the proof of the
408
M. Gionfriddo
BIBLIOGRAPHY 1. 2. 3. 4. 5. 6. 7. 8.
J. Doyen, Recent developments in the theory of Steiner systems, Teorie Combinatorie, Colloq. Roma 1973, Atti dei Convegni Lincei 17, Roma (1976), 277-285. M. Gionfriddo and C.C. Lindner, Construction of Steiner quadruple systems having a prescribed number of blocks in common, Discrete Math., 34 (1981), 31-42. M. Gionfriddo and C.C. Lindner, On the block intersection problem for Steiner quadruple systems, Combinatorica. M. Gionfriddo, On some particular disjoint and mutually balanced partial quadruple systems, Ars Combinatoria, 12 (1981), 123-134. C.C. Lindner and A. Rosa, Finite embedding theorems for partial Steiner quadruple systems, Bull. Soo. Math. Belg. 27 (1975), 315-323. C.C. Lindner and A. Rosa, Steiner quadruple systems, A survey, Discrete Math. 21 (1978), 147-18l. C.C. Lindner, Two finite embedding theorems for partial 3-quasigroups, Discrete Math. 16 (1976), 271-277. A. Rosa, Intersection properties of Steiner quadruple systems, Annals of Discrete Math. 7 (1980), 115-128.
Seminario Matematico del1'Universita Citta Universitaria Viale A. Doria 6 95125 Catania Italy
Annals of Discrete Mathematics 18 (1983) 409-418 © North-Holland Publishing Company
409
SOME RECENT RESULTS ON CYCLIC STEINER QUADRUPLE SYSTEMS - A SURVEY M.J. Grannell and T.S. Griggs *
1. INTRODUCTION AND ENUMERATION RESULTS The necessary and sufficient condition for the existence of a Steiner quadruple system of order v (SQS(v)), namely v
=2 or 4 (mod 6), was established by
Hanani (14) in 1960 and is now well known.
Such a v will be called admissible.
In comparison, the problem of determining those values of v for which cyclic Steiner quadruple systems exist is still unresolved.
A survey of results on
Stei ner quadruple systems was gi ven by Lindner and Rosa (18) in 1978.
However, at
that time no infinite families of cyclic Steiner quadruple systems were known.
A
great deal of the work on the subject has been developed recently, much of it in response to some of the questions posed in Lindner and Rosa's paper. It is an aim of this paper to provide a unified account of this work, present some of the conjectures which have been made, and discuss some of the problems which remain.
We will denote a cyclic Steiner quadruple system of order v
by CSQS(v) and represent it in the usual way as the union of 4-block orbits Qnder the action of the cyclic group Cv = < i ~ i+l(mod v) >. Orbits will be called full, half or quarter respectively according as the number of distinct 4-blocks which they contain is v, v/2 or v/4. Firstly, enumeration results have been obtained for all admissible v
~
22.
In the table below n(v) denotes the number of distinct CSQS(v) and N(v) the number of non-isomorphic CSQS(v). ~
8
~
0
~
0
Reference Guregova and Rosa (13) Barrau [ 1)
10 16
o o
o o
20
152
29
Phelps [20]
22
210
21
Diener (6)
14
Guregova and Rosa [13) Guregova and Rosa (13)
M.J. Grannell and T.S. Griggs
410
~.
METHODS OF CONSTRUCTION FOR CSQS(v)
The first infinite families of CSQS(v) were constructed by Phelps [19).
By
exploiting the structure of the groups PGL(Z,qk) the following theorems were obtained. PHELPS THEOREM 1:
If theI'e exists an
exists a CSQS(qZ+l)
THEOREM Z:
~ontaining
If there exists a
SQS( q+ 1) with q a p1'ime POWeI' then ther>e
the SQS(q+l) as a subdesign.
CSQS(q+l) with q a prime power> then ther>e exists a
k
CSQS (q +1) fOI' a Zl k > O. In [Z), Cho gave a doubling construction for certain CSQS(v). CHO THEOREM 1:
If there exists a
exists a CSQS(Zv). a.t leaBt
CSQS (v) with v == Z or 10 (mod lZ) then ther>e
.'foreoveI' if the CSQS(v) compr>ises n orbits then there exists
Zn pail'<Jis;; distinct CSQS(Zv).
The basis of Cho's construction is as follows: (a) From each full orbit in the CSQS(v), choose any block, say {w,x,y,zl, 0
w< x < y <
< v.
Form the orbits under C of the blocks Zv (w,x,y,z},{w,x,y+v,z+v},{w,y,x+v,z+v) and {w,z,x+v,y+v}. ~
Z
(b) From each half orbit in the CSQS(v), choose any block, say {w,x,W+v/Z,x+v/Z},
0
Form the orbits under C of the blocks Zv {w,x,w+v/Z,x+v/Z} and {w,x+v/Z,x+v,w+3v/Z}. ~
w < x < w+v/Z.
(c) The orbits formed in (a) and (b) are full and the CSQS(Zv) is completed by adjoining all half orbits and the quarter orbit of 4-blocks under C ' Zv A problem with Cho's construction is that it can not be re-applied because the CSQS(2v) contains the quarter orbit.
In [4), Colbourn and Colbourn devised
a method which in certain circumstances overcomes the problem.
The basic idea is
to omit part of the cyclic Steiner quadruple system, including the quarter orbit, apply Cho's construction to the remainder and then complete the doubled system
Some recent results on cyclic SQS's
411
with another appropriate cyclic Steiner quadruple system.
To formalize the dis-
cussion we need some further definitions. From any given 4-block {w,x,y,z}, 0
~
w < x < y < z < v, a cyclically
ordered difference quadruple < x-w,y-x,z-y,w-z+v > may be formed which is characteristic of the orbit from which the 4-block is drawn. ordered difference triple may be formed from a 3-b1ock.
Similarly a cyclically Then, since each 4-block
contains four 3-b1ocks, each difference quadruple gives rise to four difference triples.
If mlv, define an m-beheaded CSQS(v), denoted by CSQS(v,-m), as a col-
lection of 4-b1ock orbits with the property that any 3-block whose difference triple entries have a common factor vim does not occur in the system but all other 3-b1ocks occur exactly once.
We note that in this definition it is not in
fact necessary for v to be admissible.
However, the significance of the defini-
tion is that if there exists a CSQS(v,-m) and a CSQS(m) then there exists a CSQS(v) . The main result in [4] is: COLBOURN AND COLBOURN THEOREM 1: If there exists a CSQS(2v,-2m) with v
=m(mod
2)
then there exists a CSQS(4v,-4m).
The theorem is of great importance and has wide ranging applications which are dealt with in Section 4. Finally in this section we announce that we have recently obtained product constructions for cyclic Steiner quadruple systems which with an appropriate parameter and minimal modification yield the constructions of Cho and of Col bourn and Col bourn. In [12], the following results are proved: GRANNELL AND GRIGGS THEOREM 1: If there exists a CSQS(u) and a CSQS(v), the latter being composed entirely of fuZl orbits then there exists a CSQS(uv,-2u).
THEOREM 2: If there exists a CSQS(u) and a CSQS(v,-m) with v
=m(mod
2) and the
latter being composed entirely of fulZ orbits then there exists a CSQS(uv,-um).
THEOREM 3: If there exists a CSQS (u), a CSQS (2u) and a CSQS (v) wi th v = 2 lO(mod 12) then there exists a CSQS(uv).
0)"
M.J. Grannell and T.S. Griggs
412
THEOREM 4: If there exists a CSQS(u), a CSQS(2u) and a CSQS(v,-m) with v,m
=O(mod 2)
and tne latter being composed entirely of fuZl and half orbits then
there exists a CSQS(uv,-um).
Theorem 3 and 4 with u
2 in effect give Cho's and Col bourn and Colbourn's
constructions respectively. 3. R-CYClIC AND S-CYClIC STEINER QUADRUPLE SYSTEMS CSQS(v) which admit additional automorphisms are also of interest particularly those which are stabilized by the mapping i reverse Steiner quadruple systems.
~
v-i, i.e. which are also
We will call these R-cyclic Steiner quadruple
systems and denote them by RCSQS(v).
An easy argument shows that no RCSQS(v) may
contain a half orbit and hence that v
=2,4,10 or 20(mod
tion.
24) is a necessary condi-
It has been conjectured that this is also sufficient, but little work ap-
pears to have been done on RCSQS(v). Of more importance both from an historical point of view and because of their use with Colbourn and Colbourn's construction are symmetric cyclic or S-cyclic Steiner quadruple systems, denoted by SCSQS(v).
These are defined as
CSQ5(v) in which each orbit contributing to the system is stabilized by the mapping i
~
v-i.
The present authors [9) and Diener [7) independently obtained
the following results on the structure of SCSQS(v). GRANNELL AND GRIGGS/DIENER THEOREM 1: If there exists an SC5QS(v) then v 4n whel'e the prime factors of n are aU ~i
= 1 or
5(mod 12).
= 2n
or
Moreover, if m is even
mlv then the SCSQS(v) contains an SCSQS(m) as a subdesign. We conjecture that the necessary condition given in the theorem is also
sufficient.
Until recently, most known CSQS(v) were also S-cyc1ic.
However, a
non 5-cyc1ic C5Q5(26) is given (although with a misprint) in a paper by Fitting [8)
of 1915. Constructional techniques for 5CSQS(v) have been investigated in two papers
(16J, [17] by Kohler.
The basis for this work is as follows:
For a given v, not necessarily admissible, define a graph H"'(v) where the set of vertices
= {{x,y,z}:x,y,z
E
{1,2, ... v-3}\{v/2},x,y,z unequal, x+y+z
v}
and the set of
Some recent results on cyclic SQS's
413
edges is defined by the relation (which it is easily verified is both irreflexive and symmetric) that {x,y,z} is joined to {x' ,y' ,z'} if there is some ordering X,Y,l of x,y,z such that {x',y',z'} = {X,X+Y,v-2X-Y}.
v
=2,4,10 or
In the case where
20(mod 24) careful analysis of the difference quadruples and the
difference triples they contain leads to the graph H'\(v) where each vertex reppresents two difference triples and each edge represents a difference quadruple. Hence KOHLER THEOREM 1: There exists an SCSQS(v) iff H*(v) aontains a l-faator. (This result is in fact slightly stronger than that given originally by Kohler [16] and ·is due to Diener [7].
The basic idea of the work goes back to
Fitting [8]). Kohler proceeds to prove various properties of the graph H*(v). ular it is not necessarily connected.
In the case where v
or 5(mod 12), H*(v) consists of two components.
In partic-
= 2p with p a prime
=
In one of these components,
H *(V), the elements of each set which comprises a vertex are all even and in the 2 other component, H,*(v), at least one element (and therefore precisely two elements) of each set comprising a vertex are odd.
Now it can be verified that a
l-factor of H,*(v) may be obtained by selecting the set of difference quadruples
< a,b,a,v-{2a+b)> with a
=
1,3, ... (v-8)/2 and b
=
2,4, ... (v-2a-4)/2.
In addi-
tion the graph H *(V), which is also denoted by H(v) in [16], can easily be seen 2 to be isomorphic to the graph H*(p) which leads to: THEOREM 2: There exists an SCSQS(2p) with p a prime
= 1 or
S(mod 12) iff H*(p)
contains a l-faator.
When p group < i
~
=5(mod
12), Kohler proceeds further with the analysis.
ai+b(mod p) > with a,b
E
GF(p), a
f
By using the
0, in order to reduce the number
of orbits required, another graph B(p) is obtained as follows: Let F = GF(p)\{0,1,(p-l)/2, p-2, p-l}.
If
a E
F define
= {a,l/a,-a/{a+l),-a-l,-l/a-l,a/(a+l)-l}. Then it follows that for a,S E F either ~ = $ or ~ n S =~. Take as the set of vertices of B(p), the set ~
{~ a
: a
= 8+1
E
F} and join ~ and or
a
= 8-1.
e by
an edge if there exists
a E ~
The following two theorems are proved.
and S E $ with
414
M.J. Grannell and T.S. Griggs
THEOREM 3: If Sip) co"Ztains a l-factoT' then theT'e exists an SCSQS(2p). THEOREM 4: If Sip) is oT'idgeless with p
= 53 OT'
77(mod 120) then theT'e exists an
SCSQ5(2p) . These results enabled Kohler to construct SCSQS(v) for v = 26,34,50,58,74, 82,106,178,202,226,274,298,346,394,466,586 and 634.
Using similar techniques Cho
[3) has constructed SCSQS{v) for other values of v not necessarily covered by the theorems. Enumeration results for R-cyclic and S-cyclic Steiner quadruple systems are few.
The unique CSQS(lO) is also S-cyclic.
For v
= 20, there exist 16 distinct
RCSQS(v) which partition into 4 isomorphism classes (Phelps [20]).
Of these, 4,
all within a single isomorphism class, are also S-cycl ic (Jain [15] ). authors [11) enumerated SCSQS(26).
The present
There are 87 distinct SCSQS(26) in 18
isomorphism classes. 4. EXISTENCE RESULTS FOR CSQS(v) In this section we apply the known methods of construction, particularly those dev; sed by Cho, Col bourn and Col bourn, and ourselves, to vari ous "small" systems constructed by hand or with the aid of a computer search to give an account of the known spectrum of CSQS(v). (a) In [4], Colbourn and Col bourn give a CSQS{16,-8).
Together with a
CSQS(32), two examples of which were constructed by the present authors [ 10) , all n
CSQS(2 ) for n
~
5 may be obtained using the following constructional scheme.
The
details appear in [4). CSQS(16,-8)--~--
C5Q5(32,-16)
~
CSQS/64,-32)
CSQS/128,-64)
-~-
CSQS(32)---+-- CSQS(64)---- CSQS(128}------n THEOREM A: CSQS(2 ) exist foT' n = 2 and n ~ 5.
CSQS(2 3 ) and CSQS(2 4 ) do not
(b) Colbourn and Phelps [5] constructed a CSQS(40,-20).
Again, together
with one of the known CSQS(20}'s a similar constructional scheme to the one above may be adopted.
Some recent results on cyclic SQS's
CSQSt40,-20)
-~--
CSQS!80,-40)
415
-~- CSQSf160,-80)--~
csQs(2o)-------csQs(4o)---------- CSQS(80)-------- CSQS(160)-------n THEOREM B: CSQS (2 . 5) exist for n;;' 1 . (c) It is clear that what is really required are appropriate "starter" systems to which Colbourn and Colbourn's construction may be applied. provided by the S-cyclic Steiner quadruple systems.
These are
From the theorem on the
structure of these systems, if m is even and mlv then by removing all those orbits which contribute blocks of the SCSQS(m) subdesign from the SCSQS(v), a CSQS(v,-m) is obtained.
In particular if m = 10 and u = vim is odd we have the
following: SCSQS(lOu)-----CSQ1(lOu,-10)-----CSQSf20u'-20)--~--CSQSf40u'-40)-~-CSQ~ (lOu)
CSQS (20u)
CSQ5(lO) CSQJ(20) n using the CSQS(2 .5) from the·previous theorem.
CSQSI( 40u) CSQst(40) An SCSQS(50) is given by Kohler
[16], and Cho [3] has constructed SCSQS(lOu) for u
=
13,17,25,29 and 37.
n THEOREM C: CSQS(2 .5u) exist for n;;' 1 and u = 5,13,17,5 2,29 and 37. (d) The results given in Theorems A, Band C appear to be the only known infinite families, the orders of whose members increase by a factor of 2, which are totally determined. In many others there are gaps. Using our Theorem 3, let u = 2nw for some n ;;. 1 and w = 5 or 5u for any of the values of u given in Theorem Then for any v = 2x, with x odd, for which a CSQS(v) exists, it follows that a CSQS(2 n+ l wx) exists i.e. only the existence of a CSQS(2wx) is left undetermined. C.
Values of x for which CSQS(2x) are known are x = 5,11,13,17,19 (Colbourn and Phelps [51),5 2 ,29,37,41 and 72 (Cho [31) as well as other values greater than 50 constructed by Kohler [161 and listed towards the end of Section 3. THEOREM D: If there exists a CSQS(2x) with x odd then there exist CSQS(2 n.5x) and n
2
CSQS(2 .5ux) for u = 5,13,17,5 ,29 and 37 and n;;' 2. (e) The existence of a CSQS(2x) with x odd also implies the existence of another infinite family.
Immediately from Cho's construction there exists a
Some recent results on cyclic SQS's
416
CSQS(4x) and also, by removing the quarter orbit, a CSQS(4x,-4).
Applying n n Colbourn and Colbourn's construction one can then produce CSQS(2 x,-2 ) for all n n ~ 2 and when n ~ 5 these systems can be completed using the CSQS(2 ) from Theorem A. Again a gap appears i.e. the existence of a CSQS(2 3x) and a CSQS(2 4x) is left unresolved. THEOREM E: If there exists a CSQS(2x} tJith x odd then there exists CSQS(2 n.x) for n
=2
and n
~
5.
(f) There are other values of v for which CSQS(v) are known to exist. These include v = 2n.7 for n = 2 and n ~ 5, constructed as in (e) from a CSOS(28). Phelps' theorems enable further systems to be constructed and there are applications of our Theorem 3 in addition to that given above such as the construction of n
a CSQS(2 .5xy) for n ~ 3 from a CSQS(2x) and CSQS(2y) with x and y odd. [3J has constructed CSQS(v) for v
=
Also Cho
88,92 and 124.
5. CONCLUDING REMARKS The evidence of the results in the previous section would seem to support the conjecture that CSQS(v) exist for all admissible v apart from v = 8,14 and 16. The only values less than 100 for which the existence of a CSQS(v) is unresolved are v
= 46,56,62,70,86 and 94. A major problem with the methods known at present appears to be that the
non-existence of CSQS(8), CSQS(14) and CSQS(16) frustrate the construction of certain systems of higher order.
To overcome this, it would be useful to have a
generalisation of Cho's construction which in effect "doubles" the quarter orbit. If v
= 4m then starting from a CSQS(4m,-4), a CSQS(8m,-8) can be constructed but
not completed.
However, it seems not unlikely that the removal of a relatively
small number of orbits from the CSQS(8m,-8) may result in completion being possible. Other major lines of investigation would appear to be the construction of CSQS(2x) with x odd and the development of more general recursive constructions. Finally, the question of whether the known necessary conditions for R-cyclic and S-cyclic Steiner quadruple systems are also sufficient remains open.
M.J. Grannell and T.S. Griggs
417
BIBLIOGRAPHY 1.
2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
J.A. Barrau, Over de combinatorische opgave van Steiner, Kon. Akad. Wetensah. Amst. Verlag Wis-en Natuurk. Afd., 17 (1908), 318-326. (= On the combinatory problem of Steiner, Kon. Akad. Wetensah. Amst. Proa. Seat. Sai., 11 (1908), 352-360). C.J. Cho, On cyclic Steiner quadruple systems, Ars Combinatoria, 10 (1980), 123-130. C.J. Cho, private communication. C.J. Col bourn and M.J. Co1bourn, A recursive construction for infinite fami1es of cyclic SQS, Ars Combinatoria, 10 (1980), 95-102. C.J. Co1bourn and K.T. Phelps, Three new Steiner quadruple systems, Utilitas Mathematiaa, 18 (1980), 35-40. 1. Diener, On cyclic Steiner systems S(3,4,22), Annals of Disarete Math., 7 (1980), 301-313. I. Diener, On S-cyc1ic Steiner systems, unpublished. F. Fitting, Zyk1ische Losungen des Steiner'schen Problems, Nieuw Arah. wisk~ (2) 11 (1915), 140-148. M.J. Granne11 and T.S. Griggs, On the structure of S-cyc1ic Steiner quadruple systems, Ars Combinatoria, 9 (1980), 51-58. M.J. Granne11 and T.S. Griggs, A cyclic Steiner quadruple system of order 32, Disarete Math., 38 (1982), 109-111. M.J. Granne11 and T.S. Griggs, An enumeration of S-cyclic SQS(26), Utilitas Mathematiaa, 20 (1981), 249-259. M.J. Grannel1 and T.S. Griggs, Product constructions for cyclic block designs -I, Steiner quadruple systems, (to appear). M. Guregova and A. Rosa, Using the computer to investigate cyclic Steiner quadruple systems, Math. Casopis Sloven. Akad. Vied., 18 (1968), 229239. H. Hanani, On quadruple systems, Canad. J. Math., 12 (1960), 145-157. R.K. Jain, On cyclic Steiner quadruple systems, M. Sc. thesis, McMaster University, Hamilton, Ontario (1971). E. Kohler, Zyk1ische Quadrupe1systeme, Abh. Math. Sem. Hamburg, 48 (1979), 1-24. E. Kohler, Numerische Existenzkriterien in der Kombinatorik, Numerische Methoden bei graphentheoretischen und kombinatorischen Problemen, Birkhauser, Basel (1975), 99-108. C.C. Lindner and A. Rosa, Steiner quadruple systems - a survey, Disarete Math., 21 (1978), 147-181. K.T. Phelps, Infinite classes of cyclic Steiner quadruple systems, Annals of Disarete Math., 8 (1980), 177-181. K.T. Phelps, On cyclic Steiner systems S(3,4,20), Annals of Disarete Math., 7 (1980), 277-300.
Division of Mathematics and Statistics, Preston Polytechnic, Corporation Street, Preston, Lancashire, PR1 2TQ. ENGLAND
This Page Intentionally Left Blank
Annals of Discrete Mathematics 18 (1983) 419-426
419
© North-Holland Publishing Company
ON A CONJECTURE OF S. ILKKA L. Guerra and E. Ughi
,'t
I NTRODUCTI ON In this paper we give a new proof of a well known conjecture of S. I1kka (1972) [31, which concerns regular and pseudoregu1ar points with respect to an affine conic C defined over a finite field F , and which claims that such points, q
apart from the centre of the conic, may only exist for small values of q.
This
conjecture was demonstrated in 1973 by B. Segre [71 albeit through a very laborious and detailed study. In a recent work, we were concerned with the distribution of squares and non-squares in a Galois field, emphasizing that it could have useful applications to combinatorial geometry.
The aim of this paper is to show a much shorter proof
of the conjecture of S. I1kka than B. Segre's may be obtained, as one of these applications.
Substantially, this proof relies only on the Hasse-Weil theorem on
the number of points of a curve defined over a finite field. We refer to the work of B. Segre [71 for everything concerning general properties and definitions and for a detailed description of the exceptional cases. There exists another exceptional case pointed out by F. Karteszi [4]; see also the last sentence of
§
2.
After a section containing preliminary material, in the following two sections we give our proof of the conjecture of S. I1kka separating the case of an ellipse from that of a hyperbola. In the latter case, we obtain explicit
t
formulas for the number of regular and pseudoregu1ar secants through a point P not belonging to the hyperbola (and different from the centre), relating those numbers to the number of points of an elliptic plane curve defined over F. q
We hope
that these formulas will allow us to answer in a future paper, at least in this case, the general question proposed by B. Segre at the end of the introduction to his fundamental paper. We extend our heartiest thanks to prof. G. Korchmaros for his valuable help
420
L. Gue"a and E. Ughi
during this work.
1. PRELIMINARY Let Fq be a Galois field of order q, which is assumed to be always odd in this paper; let
be an affine plane over Fq ; let C be a non singular conic of 11. Then, it is well known that C contains exactly q+l pOints of 11, no three points of 11
which lie on a line (and which include possible points at infinity); moreover the points of are
11
exterr~Z
not belonging to C fall into two classes, according to whether they or internal with respect to C.
This distribution may be related to the concept of an internal or external point with respect to a segment of line AB (A,B
E 11,
A ~ B).
Indeed, recalling that a point P belonging to the line AB is said to be external or internal to the segment AB according to whether the cross-ratio (A B P (AB),,,) is a square or a non-square if Fq ( (AB)oo = the point at infinity of the line AB), then we have (cf. [7], LEMMA 1: If A,B
E
§
10):
C, A ~ B, then the set of the (q-l)/2 external (internal) points
to the segment AB of the line AB is the set either of the external or of the internal points with l'espect to the conic C (including (AB)oo).
In other words, for any secant line AB, either the two classes of the external and the internal points remain fixed or they are exchanged, according to whether the point at infinity (AB)oo is external or internal with respect to the conic C. This allows us to divide the affine secants to C into two classes, corresponding to each one of the two given possibilities; we call any secant line which does not exchange the character of points regular, and any other we call pseudoregular.
Naturally, there arises at this point the problem of classifying any point pE
11,
P ~ C, according to the behaviour of the secants to C through P.
These
lines are divided into two classes, that of regular and that of pseudoregular ones, whose numbers of elements we call respectively Rp' Sp.
On a conjecture of S. Ilkka
421
As we pointed out in the introduction, the questions of calculating the integers Rp' Sp for any point PE IT and of describing the outline of regular and pseudoregular secants through P or, more generally, the configuration of the set of points P E
having given values of the characters Rp' SP' are even today still But, if we confine ourselves only to the extreme cases Rp=O, Sp=O,
open ones.
IT
then the question may be answered. Call any point P for which Sp=O regular, and any P for which Rp=O pseudoregular.
Then it can be proved that:
CONJECTURE OF SEPPO ILKKA: if the conic C has a centre, then regular or pseudoregular points, different from the centre, may only exist for a finite number of values of q.
We shall give our proof of the conjecture in the following sections.
For
the moment we point out that the case of a parabola is exceptional, because of the following (cf. [7),
§ 8}:
LEMMA 2: Any secant to a parabola C of
TI
is regular.
Moreover, the centre of C , when C is an ellipse or a hyperbola, is always a regular or a pseudoregular point, according to the following (cf. [7),
§ 14):
LEMMA 3: The centre of an ellipse is a regular or a pseudoregular point according to whether - 1 =f}.or - 1 =0, i.e. q == - 1 (4) or q == 1 (4); while the centre of a hyperbola is a regular or a pseudoregular point, according to whether - 1
=C
or
- 1 = /)..
(0 denotes, as usual, a non zero square in Fq , and /). a non-square in Fq}. NOTE: We emphasize that our definition of a regular or a pseudoregular point is slightly different from that of [7), as we prefer not to include the limiting case of a tangent line. We end this section with some algebraic preliminaries concerning the distribution of squares and non-squares in Fq and the Hasse-Weil theorem.
L. Guerra and E. Ugh;
422
As usual, (~) is the Legendre symbol of x in F , which equals x(q-l)/2, so q
q
that it is 0,1-1, according to whether x is respectively zero, a square or a nonsquare in F. q
The key result is the following:
LEMMA 4: :'et t'zere be given r polynomiaZs
fl(X), ... ,fr(X)
FJ X], of positive
E
degree, with no multiple zevos, and such that no two have a common zero.
Then
"i:ere exists an integer C, depending only on the degvees of the polynomials fi(X), and such that, for any fixed r-tuple of quadratic chal'actevs £i = ± 1, i =1, ... , r ,
Ci~X) )
=
E.
1
c.Z1J:J.ys has a solution in F , fov q ;;. C.
q
PROOF: It is clear that the question depends only on the case
Ci~X))= -
Indeed, for any equation
1 we may substitute
1, i = l, ... ,r.
E. 1
(gi~X))
1, where
g.(X) = s f.(X), s being a non-square in Fq , which is equivalent to the former, so 1 1 that the general assertion that holds in Fq follows from the case i = 1, ... ,r.
= 1,
Now, in this case, call the number of solutions of the given system
of equations N.
N=
E. 1
Clearly we have, (cf. [ 2),
>JFq [1
+
Cl~X))]
§ 1)
[1 + C~(X))] +
N ,
o
where, if we denote by Z (f.) the set of zeros of f.(X) in F , then q
1
1
q
Distributing the various terms and collecting those which contain the same number of factors fi' we obtain N
.9.
2r
+
2r
r i ~1
[J,J:(X)) l
1 L Cl:X') ... C<X)) + N0 . 2r xEF q
+-
Then we can write
On a conjecture of S. Ilkka
423
where Ih
't'f
= lEF E
q
(f..(g) q
and f is one of those divisors of the polynomial fl(X) ... fr(X) obtained by multiplying the various factors f.(X). 1 Now observe that, denoting the set of zeros of f. in the algebraic closure 1
of Fq by Z(f.), then 1
I
N
a
1
I.;;; -2 I
r
.u
1=1
Moreover, repeating the arguments used in [2], lemma 1.5, one can easily prove that
= f(X)
less q, so
that
whence the conclusion follows. NOTE: This lemma might be proved directly by considering the curve y~ ~ f.(X) i
= 1, ... ,r
1
1
in (r+l)-space, but we preferred the given proof as it explicitely
relates the number N to the numbers of points of certain algebraic plane curves. The lemma in fact holds under a slightly more general hypothesis but obviously not for an arbitrary system of polynomials.
2. Finally, we come to the proof of the conjecture of S. Ilkka. First, we consider the case of an ellipse C C
n
and choose a coordinate system so that the
equation of C is X2 _ s y2
1, wi th s
/::'.
424
L. Guerra and E. Ughi
Then, let P=(L,M) be a point of (0,0)
the centre of
C.
not belonging to C and different from
rr
It is easy to see that a secant line AB, A,B
E
C, A ~ B,
through P parallel to lY = mX is regular or pseudoregular according to whether s m2 _ 12
is a non-square or a square in Fq , respectively. Moreover, a line through P parallel to 1Y = mX is a secant to C if and only if s (1 M - m L)2
+
(1 2 - sm 2)
=
D .
Then, clearly, a regular point with respect to C lies among those points 2 2 (L,M) E n such that, if ~ E F satisfies s (~ M - L) + (~ - s) = 0, then s - ~2 =
q
6
holds too, while a pseudoregu1ar point with respect to C lies among
those points (L,M) E s -
~
2
c
such that, if
~
E
Fq satisfies the given condition, then
= D.
Now, from lemma 4 it follows that for q»O there then exist
~1 E
Fq such
that s (1;1 M - L)
and
~2 E
2
+ (~1
2
- s)
=0
,
- s)
=0
, s - ~2
s - 1;1
2
=0 ,
Fq such that
s (<:2 M - L)
2
+
(~2
2
2
=~ ,
2 2 2 whence the conclusion (since the polynomials X (s M + 1) - 2 X L Ms + (L s - s) 2
and - X +
S
never have a multiple zero, and they have a common zero only when
L=M=O) . Using the expression of the number N which we stated in the proof of lemma 4, one actually obtains that the two systems always have a solution for q Segre [71 obtains q
~
~
17 (B.
7 in the pseudoregular case, asserting moreover that there
does not exist any regular point, different from the centre, for every value of q; but it is true that q
= 7 is still an exceptional case, as F. Karteszi shows in
[4); afterwards I. Oebroey [ 1) proved the assertion holds without conditions for q :;. 9).
3.
As we pointed out in the introduction, in the case of a hyperbola calculations are easier, and we really are able to find an explicit formula
for the
On a conjecture of S. llkka
425
integers Rp'Sp depending on the number of points of an elliptic plane curve associated with P. In fact, if we choose a coordinate system in such a way that the equation of the hyperbola C is X Y = 1, then we have at once that the line joining a point (t,l/t) E C to another point (T, liT) E C is regular if and only if t T = C. Now, consider a point P=(L,M} E
n
different from (O,O) = the centre of the
hyperbola; the line joining P to (t, lit) meets the hyperbola C again at the point ((t - L}/(t M - l) , (t M - l}/(t - L)} (which possibly may be equal to (t.l/t) itself). naturally when such a line is not a parallel to anyone of the two asymptotes of C. Now it follows that the number of regular secants through P is half the number of solutions of the equation x (x - L) (x M - l) = 0 , excluding any solution corresponding to an eventual tangent line through P. The case when P belongs to either one of the two asymptotes may be solved at once, since supposing for example M=O, then we have Rp = (N - 1}/2, where N is now the number of solutions of the equation x (x - L)
= [] ;
whence Rp (cf. also [71,
1
"4
=
(q
26).
§
Moreover in this case Rp + Sp = (q - 3}/2 , so that S
p
If. instead,
LM~O,
= l4
(q - 2 +
(:1)) q
we obtain at once N = (E - 3}/2 •
where E is the number of pOints of the elliptic plane CUrve defined by the equation 2 Y
=X
(X - L) (X M - 1) ,
426
L. Guerra alld E. Ughi
so that
(we put Nl
~
Nl
~
Rp
= N
Sp
=
t (2q - E - 3)
=1
+
(1 ~ LM)
= E
=
3
-"2
[1 + (1
~
LM) ]
'
the number of tangent lines through P, and bear in
mind that, con5equently, R +S P P
=!l...:...l. _ 1.2 2
[1 + (1 -q LM) ]
Finally, we observe that the conclusion follows for q case, and for q
~
~
11 in the regular
17 in the pseudoregular case (as in B. Segre [7 I).
ACKNOWLEDGEMENT: Work carried out under the aegis of the G.N.S.A.G.A., while the authors were receiving a study grant from the C.N.R.
BI BLI OGRAPH Y 1.
1. Debroey, Note: A remark on a result of B. Segre concerning pseudoregular points of an elliptic quadric of AG(2,q), q odd, Journal of Geometry,
2.
L. Guerra and
3.
S.
4.
F.
5. 6.
B. B.
7.
B.
8.
A.
vol. 14, (1980), 159- 163. E. Ughi, On the distribution of Legendre symbols in Galois fields, to appear in Discrete Mathematics. Ilkka, On the inner altd outer points of conics and betwennes-relations in finite linear planes of odd characteristics, Report of the Helsinki University of Technology, Mat. A 16 (1972). Karteszi, Su una congettura di Seppo Ilkka, Ann. Univ. Sci. Budapest Rolando Eotvos, Sect. Math., 20 (1977), 167-175. Segre, Lectures on modern geometry, Cremonese, Roma (1961). Segre, Introduction to Galois geometries, Mem. Acc. Naz. Lincei, (8) 8, fasc. 5 (1967), 133-236. Segre, Proprieta elementari relative ai segmenti ed alle coniche sopra un campo qualsiasi ed una congettura di Seppo Ilkka per il caso dei campi di Galois, Annali di Matematica, 96 (1973), 290-337. Weil, Sur les courbes algebriques et les varietes qui s'en deduisent, Hermann, Paris 1948.
Istituto di Matematica Universita di Perugia Via Vanvitelli 1 06100 Perugia - Italy
427
Annals of Discrete Mathematics 18 (1983) 427-432 © North-Holland Publishing Company
ON HOMOLOGIES IN FIXED-POINT-FREE AFFINE GROUPS Christoph Hering
Let (P,B) be a projective plane, G a group of automorphisms of (P,B), P the subgroup generated by all perspectivities in G and S the subgroup generated by all elations in G.
We denote by B(P) the lattice consisting of all substructures of
(P,B) which are left invariant by P and by Bl (P) the set of atoms of B(P).
By
Theorem 5.10 of [5] there are, up to duality, 16 possibilities for the structure of Bl (P).
We assume here that P is of type (11) according to the notation intro-
duced in this theorem, i.e. that P leaves invariant a line a but no point. we assume that P is finite.
duced.
Also,
In [7] a classification for this situation was intro-
This investigation clarifies pretty much the situation as far as elations
in G are concerned.
Like in the Lenz-Barlotti classification [2] and
[11] it
would now be nice to be able to handle homologies also, which at this time seems to be somewhat more difficult.
The object of this paper is to show that in 5 of
the cases given in the classification G does not contain any non-trivial homology at all. We use the same notation as in 1i s ted i n
[8].
(11,
are 2r ITI = q and SIT ~ SL(2,q) trans.) 4r ITI = q and SIT "" Sz(q) trans.) 4r ITI = 3 and SIT ~ SL{2,5))
(11,
ITI
(11,
4r ITI = q and SIT "" Sz(q) intrans.)
(11,
ITI = lr and 21IS/TI)
(11,
S=T+l)
(11,
S ~ PSL(2,q))
(11, (11,
[8]
= q2r and SIT ~ SL(2,q) intrans.)
(11, S ~ Sz(q))
In particular, T = G(a,a).
The cases
G. Hering
428
(11,
S~PSU(3,q))
(11,
S = 1).
Here always q is a prime power larger than 2, and rEIN. in the first type. unless S
We hope that this list is complete.
Also, q is even except It can be shown to be,
+ T = 1.
(11, S ~ PSL(2,q)} In this case (r,B) ~ PG(2,q) and S acts as an orthogonal group on (P,B).
As
pointed out already in ! 8), this implies that G does not contain any non-trivial homology. q2r and SIT ~ SL(2,q) intrans.}, where q is a power of 2, q > 2
(11, ITI and rEIN.
Suppose that G contains a non-trivial homology. ~
of prime order s.
shows that
5<~>
Clearly, s
spl its over T.
+ 2.
Then G contains a homology
The argument in the proof of [4, (3.10})
Let H be the complement of T in 5
Then H con-
taines a Sylow s-subgroup of 5
All ele-
ments of order 2 in S are elations (see [4, p. 39)}, and as H n T = 1, the involutions in H are elations with affine axes (i.e. axes f a). =
S1
HT/T ~
~
SIT
H/H n T
=SL(2,q).
fine axes. (4.4)].
=H,
Also, S
and there exists a normal subgroup Sl
~H
=
such that
As we have seen, the involutions in Sl are elations with af-
We apply [4, Theorem 2.8).
If
!I,
E SI {a}, then IG{£,£}I
Therefore we have case c) in this theorem.
~
2 by [4,
But this implies, that t; acts
on a desarguesian subplane and leaves invariant a hyperoval in this subplane, which is impossible. ( 11, S
=Sz ( q) }
We can assume w.l.o.g. that (P,B) trivial homology. group S
= (P,B). Suppose that G contains a non-
Then G contains a homology n of prime order s.
Clearly, S
=
M so that CGS
=
1 by [7, Theorem 1).
isomorphic to a subgroup of the automorphism group of S. note H =
NS
<[ n,x».
Consider the Hence S
Let x E S\ Csn, and de-
We prove that
P:
Let Z be
On homologies offlXed-point-free affine groups
429
a, then IzG I > 1, so that T f 1 by Andre's Theorem [1], a contradiction. So t f a and hence Z E a. If IX a 2, then -1 -1 x [n,x] = n (x nx) E G(t), so that
If t
a
then In,x] E G(Z), and
f
Assume now, that ZX
f Z and
Then all fixed points of [n,x] = n-lx-lnx lie in auU n tX} by [5, Lemma 5.1], and also, t n t X is a fixed point. Suppose that 2 n t X Ea. Then Q,x
Q,.
[n,x] = (x-l)n XES does not fix any point in T\a and therefore must be an element of order 4.
This follows, because all other elements in S are products of
the form as, where a and S are involutions and hence elations.
If a and Shave
different centers, then the intersection of their axes is an affine fixed point of If a and S have the same center, then as is itself an elation and thus fixes
as.
pointwise an affine line even. But now [n,x]2 is an elation and hence fixes only one point on a. Thus In t X is the unique fixed point of [n,x] on a and therefore invariant under H and hence clearly also under
n
2
l
Ef a.
Then t n
l
is the unique fixed point of [n,x]
in P\ a and therefore i nvari ant under H and hence a 1so under < n ,H >. Thus
P,
which implies that
Thus n is an abstract perspectivity, as defined in [6].
Even, n has property (*1).
By [9, Theorem 3.1 and Corollary 4.3] it follows that
n is an involution, which clearly is impossible. (11, IT I
= q4r and SIT "" Sz(q) intrans.)
Here we assume in addition that SIT splits over T. Suppose that G contains
a homology n of prime order s. By the GaschUtz Theorem (see [10, p. 121, Hauptsatz 17.41), S< I» splits over T also. We now proceed as in the case (11,
I
IT = q2r and SIT ';;; SL(2,q) intrans.) to show that S contains a subgroup of type (11, S';;; Sz(q)) which contains non-trivial homologies.
But this is impossible,
as we have seen above. (11,
S ""
PSU(3,q))
Here q is a power of 2, and q > 4.
Suppose that G contains a non-trivial
430
G. Hering
We may assume that (P,B) = (P,B).
homology n.
Then S
Let Ql and Q 2 be the centers of two different Sylow 2-subgroups of Sand H =
H ~ SL(2,q), and the involutions in H are elations. (11, S
~
~
PSU(3,q).
Also, H is of type
PSL(2,q», i.e. the axes and centers of non-trivial elations in H gener-
ate a desarguesian subplane (P,B) of order q.
Here a
E
B, and the lines in
8\{a} which are not axes of non-trivial elations form a dual ovalO of (P,B), whose (dual) knoten is a (see [B] and [4, Theorem 2.B]).
Let A be a "secant
point" of 0, i.e. a point which lies on two different lines of O. dihedral group of order 2(q-l). where
v
= (3,q + 1).
Also, CSH is a cyclic group of order (q+l)/v,
Clearly, CSH acts trivially on (P,B), so that CSH
Now obviously S does not leave invariant A. group of S containing CSH CSH
x
I[ AJ
x
H (see Hartley [3).
SA = CSH
x
Then HA is a HA < SA'
So SA is contained in a maximal sub-
HA. But the only such maximal subgroup of S is By the Dedekind modular law it follows that
1
HA. From this it follows that IASI = q3(q3+ 1) and is the set of axes of non-trivial elations in G.
(SA n H) = CSH
n UI = q-l, if U
x
x
Let B be an "outside point" of 0, i.e. a point which is not incident with any line in O.
Then HB is a dihedral group of order 2(q+l} and SB contains
HB. If SB ~ CSH < H, then SB = CSH x (SB 2 then [SBi= 6(q+l) Iv and I[ ~n U 1= 3(q+l).
CSH
x
n
H) = CSH
x
HB.
If SB
1 CSH
x
H,
We now define
r
= (X E P
I I[X)
nUl
>
2}
and consider the map of the set consisting of all pairs (g,h) of two different elements of U into
defined by
I
(g,h)l-.... g
n
h.
Here each element of 'ie and AS has (q-l)(q-2) preimages.
If ISBI = 6(q+l)2/v ,
then the number of preimages of B is (3q+3)(3q+2) and IBSI = ~ q3(q2_ q+l )(q_l). 3
3
3
But then (q +l)(q-l)[ (q +l)(q-l)-ll=lul(IUI-l»(q +1)(q-l)(q-2)+
.~ q3(q3+ 1)(q_l)(q_2)+ -
2
1\1
S
i
q3(q2_q+l )(q-l)(3q+3)(3q+2), which is impossible. 1
3
2
Thus
= 2(q+l) lv, IB 1="2 q (q -q+l)(q-1) and I[BJ n ul= q+1. As IUI(lul -1) = 3 133 132 = (q +1)(q-l)(q-2)+ 2 q (q +l)(q-l)(q-2) +"2 q (q -q+l)(q-l)(q+l)q, we see that 1 = Q U AS U BS.
e
431
On homologies of fixed-paint-free affine groups
As
I
Let g
E
Finally, we can investigate the homology n. point X E Re which is not left invariant by n. As Xn
+ X,
gn
+ g,
so that Y E I.
In fact, YE 1\
Q
e
Rei> 2, there exists a [X] n U and Y = g n gn.
= AS
U
BS, and we can assume
that Y E {A,B}. Clearly, n leaves invariant
U
ry n
[y] n U
and hence the set of axes
a] n U.
yE[Y] r)U
But this set generates the desarguesian subplane (P,B).
Therefore n induces a
homology of (F,B) leaving invariant the dual ovalO, which clearly is impossible.
BIBLIOGRAPHY 1. 2. 3. 4.
5. 6.
7.
8. 9.
J. Andre, Uber Perspektivitaten in endlichen projektiven Ebenen, A~ch. Math., 6 (1955), 29-32. A. Barlotti, Le possibili configurazioni del sistema delle coppie punto-retta (A,a) per cui un piano grafico risulta (A,a)-transitivo, BoZZ. Un. Mat. ItaZ., 12 (1957), 212-226. R.W. Hartley, Determination of the ternary collineation groups whose coefficients lie in GF(2n), Ann. Math., 27 (1926), 140-158. C. Hering, On projective planes of type VI. Teorie Combinatorie, Torno II, Rome, 1976. (Coll. Intern. Teorie Combinatorie, Accad. Naz. dei Lincei, Roma 1973). Atti dei Convegni Lincei 17, Torno II, (1976), 29-53. C. Hering, On the structure of finite collineation groups of projective planes, Abh. Math. Sern. Univ. Hamburg, 49 (1979), 155-182. C. Hering, Finite collineation groups of projective planes containing nontrivial perspectivities. The Santa Cruz Conference on finite groups, Santa Cruz 1979. Proceedings of Symposia in Pure Mathematics 37 (1980), 473-477. C. Hering, On shears in fixed-poi nt-free affine groups. Finite geometries and designs. Proceedings of the Second Isle of Thorns Conference 1980, edited by P.J. Cameron, J.W.P. Hirschfeld and D.R. Hughes, London Math. Soc. Lecture Notes 49, Cambridge 1981, pp. 146-152. C. Hering, On Beweglichkeit in affine planes. To appear. C. Hering and M. Walker, Perspectivities in irreducible collineation groups of projective planes, II. J. Statist. PZann. Infe~ence, 3 (1979), 151-177.
432
10. 11.
C.Hering
B. Huppert, l:'16.Uche GT'uppen 1. Berlin, Heidelberg, New York, 1967. H. Lenz, Kleiner desarguesscher Satz und Dualitat in projektiven Ebenen, <'"hT'estep. Deutsche Math. VeT'., 57 (1954), 20-31.
Mathematisches Institut Universitat TUbingen Auf der Morgenstelle 10 7400 TUbingen 1 Federal Republic of Germany
Annals of Discrete Mathematics 18 (1983) 433-448 © North-Holland Publishing Company
433
ON PELLEGRINO'S 20-CAPS IN S4,3 R. Hi 11
1. INTRODUCTION Let Sr,q denote the projective space of dimension r over the Galois field GF(q) of q elements. A k-cap in Sr,q is a set of k points, no three of which are collinear.
The largest size of cap in Sr,q is denoted by m(r,q) and a cap of this size is called an ovaloid. For q = 3 it is known that m(2,3)
4
[2],
m(3,3)
10
[2],
m(4,3)
20
[5],
56
[3].
m(5,3)
EI
=
For q f 2 and r > 3 the above values of m(4,3) and m(5,3) are the only known values of m(r,q). Two k-caps are said to be equivaZent if there is a linear transformation mapping one into the other.
Up to equivalence, the ovaloids in S2,3 and S3,3 are
unique, being respectively conics and elliptic quadrics, while the uniqueness of the 56-cap in S5,3 was shown in [4]. In [6], Pellegrino showed that every 20-cap in S4,3 is one of two geometric types, but it is by no means clear how many inequivalent 20-caps there are of each type.
Our main aim is to show that there are in all exactly nine inequivalent 20-
caps in S4,3'
We will also show that just two of these occur as the intersection
of a 56-cap in S5,3 with a hyperplane. 2. NOTATION AND DEFINITIONS We write pOints of Sr,q as (r+l)-tuples (a l ,a 2,·· .,a r +l ), where a i £ GF(q) and not all the ai's are zero; (a ,a 2,· .. ,a + ) = (b l ,b , ... ,b + ) if and only if l 2 r l r l Ab. for all i, for some A£GF(q) with A f O. 1
A line in S which meets a cap K in exactly two points is called a secant r,q
434
R. Hill
of K.
For any point P of S not in K the degree of P with respect to K is the r,q number of secants of K passing through P. The intersection of a cap K with a
hyperplane H of Sr,q is called an i-section of K if IK n HI = i. A m;;,ri.r of a k-cap K in Sr,q is a k x (r+l) matrix over GF(q) whose rows are the points of K. Clearly, any permutation of the rows, or multiplication of a row by a non-zero scalar, yields a matrix of the same cap. Let G denote the group of non-singular linear transformations of S r,q r,q We regard the elements of Gr,q as non-singular (r+l) x (r+l) matrices which transform points of Sr,q via right multiplication. Two caps Kl and K2 are said to be
.:ql
The {g
if Klg = K2 for some g
c;L
£
Gr,q' where Kg denotes the set {Pg ; P
K}.
£
Aut K of a k-set Kin S i s defi ned to be the group r,q
Kg = K). r,q As i n I 4], if K i s a k-ca pin S r,q with matrix A, we define the code of K to be the r-dimensional subspace of Sk_l,q generated by the columns of A. E
G
The
wei9~r w(~)
of a k-tuple
~
is the number of non-zero co-ordinates of
~
and we define the :xl..'eight z(!S) of !S to be the number of zero co-ordinates of .!S' i.e.
z(~) = k - w(~).
Let K be a k-cap in Sr,q with code C generated by the columns, £1'£2"" '£r+l' of a matrix A of K. Then the coweight of the codeword r+l '~l .'.. c. (Ao E GF(q)) is just the cardinality of the section K n H where H is the 1= 1 -1 1 r+l hyperplane {(x l ,x 2 , .. ·,x r +l ) ; ih Ai xi = OJ. Let hi be the number of codewords of C of coweight i, or, equivalently, the number of hyperplanes of S which r,q intersect K in exactly i points. If i l ,i , ... ,i are the values of i, in increas2 t ing order, for which hi + 0, then the t-tuple (ilhil • i2hi2, ••• ,ithit) ia called the
cD~cight
distribution of the code of K (or just of K).
Equivalent caps
have the same coweight distribution, though the converse need not be true, as we shall see in §4. 3. THE [-CAPS OF S4,3 In
[6],
Pellegrino showed that every 20-cap in S4,3 is either a r-cap
(defined in the following theorem) or a 6-cap (defined in §4.) THEOREt~
3.1 (Pe 11 egri no
[6 J); ",et
E be a ZO-cap in a hyperplane H of S4,3'
Pl ,P 2 ,··· ,P lO be the points of E and let V be a point of S4,3' H.
Let
Then if we take
435
On Pellegrino's 20-caps in S4,3
any two of the three points different from V on each of the 10 lines VP. we obtain 1
a 20-cap.
Such a cap is called a r-cap or a cap of type r or a pseudo-cone "-'ith
base E and vertex V.
Since any 10-cap in S3,3 is an elliptic quadric and is unique up to equivalence (see e.g. Theorem 5.2.3 of Barlotti [11), it follows that any choice of E and V is equivalent to any other. However, for a given choice of E and V, there are 310 different r-caps given by Theorem 3.1. The seemingly difficult task of finding how many of these are inequivalent will be made relatively simple by using the triple transitivity of the group Aut E on the points of E, together with the uniqueness of the ternary Golay code.
First we give some further notation.
Suppose K is a r-cap with base E = {P ,P , ... ,P } and vertex V. Let Q be l 2 10 i the point on the line VP.1 other than V and the two points of K. We denote by K the 10-set {Ql ,Q2, ... ,Q10} and we call K the anti-cap of K. Gr,q : Eg = E, Vg = V}. Two r-caps Kl and K2 (or their respective anti-caps Kl and K2), having base E and vertex V, will be Let G(E,V) be the group {g
called strongly equivalent if Klg Klg
= (~) for any g
~
= K2 (or Klg = K2) for some g
~
G(E,V).
Clearly,
~ G(E,V), and so Kl and K2 are strongly equivalent if and
only if Kl and K2 are strongly equivalent. For the remainder of this section we take E to be the quadric {(xl' x2 ' x3' x4 ' 0) : x3x4 = x/ + x/} and V to be the point (0, 0, 0, 0, 1). r-cap K with base E and vertex V has a matrix of the form
A
KNill
436
,;0
0 Y1,1 0 Y1,2
0
:0
0
A = \0
0
0
Y2,1
:0 0 0
Y2,2
I
I,
0
I,
Y3 ,1
0
Y3 ,2
12 0
Y4 ,l
I
I I
!2 0
1° 1°
Y5 ,2 Y6 ,1
0 2
Y6 ,2 2 Y ,l 7 2 Y],2
1
,11
2
I
\l 2
12
2 Y ,1 S
12 11 11
2 Y8,2
I
2 Y • 9 1
2
:2 2
12 '-
Figure
Y5,1 2
)0 I
Y4,2
2
Y9,2
2 YlO,l 2
YlO.~
where y.1 , 1 ' y.1 • 2 for ; = 1. 2. . .. , 10. We will classify all such caps up to equivalence by first classifying them up to strong equivalence (this turns out to amount to the same thing).
It is more
convenient to aChieve this aim by classifying the corresponding anti-caps up to strong equivalence. Let Yi be the element of GF(3) other than Y;,1 and Y ,2' i Then the anti-cap Kof the above cap has matrix
On Pellegrino's 2O-caps in S
0
0
0 Y l
0 0 0
Y2
0
Y3 Y4
A"
2 0 0 0
437
4,3
=
[MI~J
Ys 2
Yo
2
2 Y7
2
2 Ys 2 Yg
2 2
2 Y lO
T where M is the 10 x 4 matrix consisting of the first four columns of A and X T (Y 1.Y2 ••.•• Y10). We will call X the defining veator of the r-cap K. The group G(E,V) consists of all linear transformations of the form
h
o
0 0 0
where h g Aut E, and bi e GF(3) for i g 1, ... ,4. If K is a r-cap with defining vector yT we denote by (yT)g the defining vector of the cap Kg. i.e. (yT)9 is that
ve~tor
such that
[M-(~T)gJ
is the matrix of (K)g.
For example, the
t;ansformation 93 below acts on a defining vector ~T via (Yl'Y2'" • 'YlO)9 3 = (Y 2 'Y l ·yS ,y6 'Y3 ,y4 ,2y? , 2yS, 2Y 10 ,2y 9)' Our problem of classifying all r-caps up to strong equivalence is essentially that of finding the orbits of the group G(E,V) on the set of all lO-tuples yT, future reference we list here some of the members of the group G(E,V).
For
438
R. Hill
20 0 0 0
o2
g,
000
g2
00200
2 0 000
10000
g3
00100
000 , 0
00020
000 , 0
00100
0000'
00001
o0
g5
o 1 , 10
o2
0 0 1
2 1 020
102 ° 0 12 000° g4
o 1 000
02000
, 10
1 2 1o 0 o , 1 0 2 0 1
0
o 0 001
Looool
Figure 2 Let C be the code of K and
Cthe
code of
R.
Let fl' f2' f3 and ~ be the
columns of M and let C be the subcode of C generated by them. For any given l defining vector rT, the coweight distribution of K can be found by first finding that of
K(e.g.
by writing out all the codewords of
C)
and then applying part
(ii) of the following lemma. LEMMA 3.2: , ., It)
The
(in If
" cowetg~t
d'"strt' b ' ut"on
0
f
(110 , 430 ) . hO hl hlO
C- l"s •
C has coweight distribution (0
,1
, ... ,10
)
>
then the code C
has cOLJeigi;t distribution
(0
hlO
,1
hg
,2
h8+10
h7 h6 h5 h4- 30 h3 h2+30 hl-l0 hO , 3 , 4 ,5 ,6 ,7. 8 , 9 ,10).
PROOF: (i) is a straightforward verification. (ii) Denote the columns of A (see Figure 1) by
~1' ~2' ~3' ~4'~'
Then,
the subcode of C generated by £1' £2' £3 and £4 has co~eight distribution 30 10 (2 , 8 ). On the other hand, if a codeword of the form y +.L A. c. of C has 4 - 1'" 1 1-1 coweight z, then the corresponding codeword y +.L A. d. of C has coweight 10 - z, _ 1'" 1 1-1 and so the result follows. by (i),
We now give the main result of this section. THEOREM 3.3: Any 20-cap of type r in S4,3 is equivalent to one whose defining
vector
~nd
cOLJeight distribution are one of those of Figure 3.
439
On Pellegrino's 20-caps in S 4,3
Coweight Distribution of K Type
T
Y
hO
hl
h2
h4
h3
fl
(0,0,0,1 ,0,2, 1 ,0,1,1)
f2
(O,O,O,O,O,O,O,O,O,O)
f3
(O,O,O,O,O,O,O,O,O,l)
1 10
f4
(O,O,O,O,O,O,O,O,l,l )
11
f5
(O,O,O,O,O,O,O,l, 1,1)
10
f6
(0,0,O,O,0,O,0,1,2,2)
10
f7
(0,0,O,O,0,O,l,1,2,2)
10
15
f8
(l ,0,0,0,0,0,0,1,1,2)
10
9
h5
h7
36
10
10
1
h6
h8 75
60
30 20
18 18 24 39
9
2
20 46
6
2
4 12 17 22 45
8
1
5 10 21 2 3
h lO
h9
9 21
3
21
48
3
3
60 30 9
18 18 48
6 9
Figure 3 PROOF: For a given defining vector yT of a r-cap K, we denote by d(~, minimum distance of ~ from the code where the distance d(l' which l and
~
differ.
~)
Gl ,
i.e. d(~,
Gl ) = min.
G1l
{d(~,~)::
the E
Cl },
is, as usual, the number of co-ordinate places in
(Note that d(l'
~)
= w(l -
We will find all the distinct anti caps
~».
Khaving
matrix
A= [MI~],
up to
strong equivalence, by considering separate cases for the value of d(y, ell. Case 1 d(~,
Gl )
~ 5.
minimum weigh! 6, and so_if d(~, e) ~ 5, then the O code generated by the columns of the matrix I ~J is a linear code with By Lemma 3.2 (i),
minimum weight 6.
e1 has
lO °M °
But it was shown by Pless (7] that the only such code is the
ternary Golay code.
An alternative proof of this uniqueness, given in Theorem
8.14 of [4], essentially shows that the anticap T
Kis
unique up to strong equiva-
lence and that y may be taken to be (0, 0, 0, 1,0, 2, 1,0, 1, 1). ~ . 10 36 75 coweight distribution of K 1S found to be (2 ,5 , 8 ). Case 2 d(y,
el ) ~ 3. 4
Let ~c =.L A. c. be such that 1 =l 1 -1
d(~, .-
£)
~
3.
The
440
R. Hill
=
Then taking g
° ° ° -).1-).2 ° ° ° -\'3 ° ° ° -).4 °°° 1 ° °T° °
(tT)g = (~ -
we have
vector has weight
~
in G(E,V),
£) , and so K is strongly equivalent to a cap whose defining
3.
We may therefore assume in this case that w(y)
~
3.
Now
AutE contains the orthogonal group 0-(4,3) which acts triply-transitively on the pOints of E and so there exists some member f of G(E,V) of the form
° h
f =
° ° °
°°°° T
such that (y )f has all its non-zero entries in the last three co-ordinate places. Hence it is enough to determine the strongly inequivalent caps having defining vectors of the form y
T
=
(0, 0, 0, 0, 0, 0, 0, YS' yg' YlO). T
The effect of each of gl' g2' g3 and g4 above (Figure 2) on such a y is given by T (y )gl
=
(t
=
-T
)9 2 T (y )g3 I (t: )9 4
= =
(0, 0, 0, 0, 0, 0, 0, 2yS' 2y g , 2Y10) (0, 0, 0, 0, 0, 0, 0, YS ' Y10 , Yg) (0, 0, 0, 0, 0, 0, 0, 2yS' 2y 10' 2y g) (0, 0, 0, 0, 0, 0, 0, 2y 10' 2yS' Yg)
while (0, 0, 0, 0, 0, 0, 0, 0, Yg' Y10)g5 = (0, 0, 0, 0, 0, 0, 0, 0, Yg , 2y ). 10 It is now easily shown that for each of the 27 defining vectors of the form (0, 0, 0, 0, 0, 0, 0, ys' Yg, Y ) there exists a suitable product 9 of 9 's such 10 i that (~T)g is of similar form with (YS ' Yg' Y10 ) equal to one of (0, 0, 0), (0,0,1), (0, 1, 1), (1, 1, 1) or (1,2,2). The respective coweight distribu. 30 gO 1 2 19 9 24 48 18 1 tlonsofKarefoundtobe(l , 4 , 1 0 ) , ( 0 , 1 , 2 , 3 , 4 , 5 , 9 ) , 16 2 (0 ,116,2 , 451 , 510 , 65 , Sl), (0 1 , 118, 215 , 322 , 447,5 12 ,64 ,7 2) and
lO,
On Pellegrino's 2O-caps in S4,3
3 13 lS 21 52 S 3 3 (0 ,1 ,2 ,3 ,4 ,5, 6 , 7 ).
441
.. The dlstlnctness of these shows that exact-
ly five inequivalent r-caps arise from Case 2 and by Lemma 3.2(ii) their coweight distributions are those of r 2, r , ... ,r of Figure 3. 3 6 Case 3 d(t,
Cl ) = 4.
As in Case 2, we may assume that w(t) = 4 and by the triple transitivity of AutE, we may assume that YS' Yg and Y10 are three of the four non-zero entries of y. We consider three sub-cases. T Case 3A Y
(0, 0, 0, 0, 0, 0, Y7 ' YS' Yg , Y10 ), where Y7' YS' Yg and Y1 0 are all
non-zero. By considering Kg , if necessary, we may assume that Y = 1. The table 7 below then shows that m6st possibilities for (YS' Yg , Y ) actually belong to Case 10 2. (Y8 ' Yg , YlO )
~
(1 , 1, 1), (1, 1 , 2) or (1 , 2, 1)
2£3 + £4
(2, 1, 1), (2, 1, 2) or (2, 2, 2)
£1 2£2
(2, 2, 1 )
Cl such that d(t, £) .;;; 3
e:
For the remaining possibility, yT (0,0,0,0,0,0, 1, 1,2, 2), the coweight distribution of K is (0 6 , 110, 360, 430, 615 ) and so we obtain a further r-cap, which is denoted by type r7 in Figure 3. T Case 3B Y
(Y l ' 0, 0, 0, 0, 0, 0, YS ' Yg , Y10 ), where Yl ' YS' Yg and Y10 are all non-zero. As in Case 3A, we may assume that Y = 1 and again the table below l shows that most possibilities actually belong to Case 2. £
e:
Cl
such that
d(y, £) .;;; 3
(Y8 ' Yg , YlO ) (1, 1 , 1)
2£3 + £4
(2, 2, 2) or (1 , 2, 2)
£3 + 2£4
(2,1,1), (2, 1, 2) or (2, 2, 1)
£1 + 2£2 + £3
442
R. Hill
The remaining possibilities for (yS' Yg , Y ) are (1, 1, 2) and (1, 2, 1). 10 The corresponding caps are strongly equivalent, for (1,0,0,0,0,0,0, 1, 1, 2)g2
= (1, 0, 0, 0, 0, 0, 0, 1,2, 1). We thus obtain
just one further r-cap, of type rS ' with coweight distribution as shown in Figure 3.
Case 3C w(y)
4, where yS' yg' Y10 and exactly one of Y2' Y3 ' Y4' YS and Y6 are
non-zero. Any such cap is strongly equivalent to one given by Case 3B, as may be seen by considering the effect of the transformations g3' g4' g3 g2g4' g3 g4 and g2 g4 on
tT
in the cases where Y2' Y3' Y4' YS and Y6 are non-zero respectively. This completes the proof of Theorem 3.3.
The distinctness of the coweight
distributions of the caps of types r l , r 2 , ... , rS shows that the classification of r-caps up to strong equivalence actually gives their classification up to equivalence.
The theorem also shows that the equivalence type (one of r l , r 2 ,···, rS) of a r-cap is uniquely determined by its coweight distribution, though this result does not extend to 20-caps in $4,3 in general, for as we shall see in §4, a so-called
~-cap
has the same coweight distribution as, but is inequivalent to, a
fl-cap. 4. THE A-CAP$ OF $4,3 The reader is referred to Theorem 3.4 of (6) for the proof of the following. THEOREM 4.1 (Pellegrino [6]): There exists in $4,3 a 20-cap which every point of $4,3' 6 has degree 2, 3, 4, S or 6. structed as follows.
Let C , C , C , C
l
2
3
4
6 with respect to
The cap may be con-
be the points of a 4-arc in a plane
IT
in $4,3'
Let Zl ,z2 and z3 be the points C1C2 n C C , C C n C C and C C n C C 3 4 2 3 4 l 24 1 3 respectively. Let zil,zi2 be the points on the line ZjZk(j, k + i) other than ~ skew to IT. Then 6 is the 2 3 4 ser of points C , C , C , C together with the points on the following lines AB l 2 3 4
Zj and zk'
Let Vl , V , V , V be the points of a line
other than A and B:
On Pellegrino's 20-caps in S
443
4,3
A cap given by the construction of Theorem 4.1 ;s called a type
~-cap
or of
~.
EXAMPLE 4.2: Taking C l (1,0,0,0,0), C2 = (0, 1,0, 0, 0), C3 = (0,0,1,0,0), C 4 (1,1,1,0,0), Vl = (0, 0, 0,1,0), V2 = (0,0,0,0, 1), V3 (0, 0, 0, 1, 1) and V4 a (0, 0, 0, 1, 2), we get the ~-cap with matrix
°° ° 1 °°° ° °1 °° 1 1 °° °1 1 ° 1 °1 2 ° °1 °1 °1 °2 °1 1 ° °1 2 °
°1 °1 °1 °2
100
THEOREM 4.3: All
2111 2122
°
1 211 1 022 2 21
12
21
21
° 1 212 ° 1 221 ~-caps
in 5 ,3 are equivalent.
4
PROOF: The essential uniqueness of choice at each stage of the construction in Theorem 4.1 follows from the fact that (i) The plane
in 54,3 is unique up to equivalence. (ii) A 4-arc in ~ in unique up to equivalence. ~
(iii) The subgroup the lines skew to
G~
of G4,3 which fixes the points of
TI
acts transitively on
~.
(iv) For a given line
i
skew to
~,
any permutation of the points of
achieved by some linear transformation in
i
can be
G~.
By listing the codewords of the code of the
~-cap
of Example 4.3 it is found
that: 75 LEMMA 4.4: The coweight distribution of a ~-cap is (210, 536 , 8 ).
Thus a
~-cap
and a fl-cap have the same coweight distribution.
However they
R. Hill
444
are inequivalent caps for, with respect to any r-cap, there is a point (the vertex) of degree 10, whereas by Theorem 4.1, with respect to a 6-cap, every pOint has degree at most 6. The main result of Pellegrino (6) is that every 20-cap in S4,3 is either of type
or of type r and so, combining this with Theorems 3.3 and 4.3, we have:
6
THEOREM 4.5: There ape exactly nine inequivalent 20-caps in S4,3' of which 8 are of type r and one id of type 6.
5. THE 20-SECTIONS OF A 56-CAP IN S5,3 The aim of this section is to identify those 20-caps which occur as the intersection of a 56-cap in S5,3 with a hyperplane.
The following theorem was
proved in (4). THEOREM 5.1: The Zapgest size of cap in S5 3 is 56 and, up to equivalence, there 308 ) . . . S5,3' The cowe'/.g • h' t d''/.str'/.' b ' one 56-cap 1',.' '/.n ut'/.on 0 f K' '/.s (1156 , 20 .
'I.S JUS"
PROOF: Combine Theorems 8, land 8,18 of (4). We next show that every 20-section of K is either of type 6 or of type rl' LEMMA 5.2: Let K be a 56-cap in S5,3 and Kl a 20-section of K. Kl as a cap in S4,3' every section of Kl has size 2, 5
01'
Then, regarding
8.
PROOF: Let Hl be the hyperplane (an S4,3) in S5,3 such that Kl = K n Hl , Let Kl n S be an h-section of Kl , where S is a hyperplane (an S3,3) of Hl . Let H2 , H3 and H4 be the other hyperplanes in S5,3 which contain S,
Then
4
.LlIKnH.I=56+3h 1= 1
(1)
where iK n Hl I = 20 and for each of i = 2, 3, 4, IK n Hi I = 11 or 20 by Theorem 5.1.
For (1) to give a positive value of h, the number of IK n Hi I 's equal to 11
must be 0, 1 or 2, giving respectively h = 8, 5 or 2. Inspection of the coweight distributions of the r-caps (Figure 3 of Theorem 3.3) and the 6-cap (Lemma 4.4) gives
445
On Pel/egrino's 20-caps in S 4,3
COROLLARY 5.3: Every 20-seation of a 56-aap in 5 ,3 is either a r -aap or a 6-aap. 1 5 We show in our final result that caps of both these types do occur as sections of a 56-cap in 55 ,3' THEOREM 5.4: Let K be a 56-aap in 55 ,3' 6-aaps and 56 are r -aaps.
Then of the 308 20-seations of K, 252 are
1
PROOF: We first describe the 56-cap K constructed in [31.
Consider the quadratic
form on 55 ,3 defined by 5
i~l (x i Yi+1 + x i +1Yi)
where P = (xl' x2 ' ... , x6 ) and Q = (Y1' Y2' ... , Y6)' quadric, a 112-set, of points P satisfying
Let E be the elliptic It was
shown in [31 that
the following subset K of points of E is a 56-cap. 1 220
o 202
2
2 2 0 0
2 000
02120
12220
012101
o 220
1
20 0 0 0
o1 2 1 2 1
02212
2 0 2 1 1
o 0 001 2 o0 0 1 2 0
o1 o1
000 1 1 2
o0
2 2
001 200
o1
2 2 2 0
020 1
2 1 002
001 222
012222
o 202
2
o0 o0
0 1 0
2 0
o0
1 2 1
1 2
2 1 210
1 2
o0
2 0 1
2 2
2 1 2 1 1
010202
o 1 002
200
2 1 2 2 1
010212
220
22021
222
2 2 1 0 1
2 1
200 0
2 2 2 0 0
o 1 221 o 202 0
2 1 0 1
2 2 22 0
2 1 0 2
22222
1
o1 o1
2 0
o 1 022 o1 0 2
2 2
o1
o1
2 2 2
o1
2 0 0
Figure 4 For each point P in 55 ,3' let H(P) be the hyperplane {Q : < P, Q> = a}. Let 0-(6,3) be the orthogonal group {g £ G5 ,3 :
R. Hill
446
Then (H(P))g = H(Pg) for any P
5 ,3 and any g £ 0-(6,3). It was shown in [3) 5 that AutK is a subgroup of 0-(6,3) and it is easy to show that the orbits of AutK €
on 5 ,3 are the sets K, E \ K and 5 ,3 \ E. It follows that the sections K n H(P) 5 5 fall into (at most) three equivalence classes according as P is in K, E \ K or 5 ,3 \ E. Hence it is sufficient to identify K n H(P) in each case for just one 5 such point P. Then H(P) = {(xl' x2 ' ... , x6 ): OJ and it is found immediately from Figure 4 that IK n H(P) 1= 11.
x4
(i) Consider P = (0, 0, 0, 0, I, 2)
£
(ii) Consider P = (0, 1, 2,0,1,2)
£
Xl = OJ and so K 11 4.
K.
E \ K.
Then H(P) ={(x ,x 2 , ... ,x ): 6 l H(P) consists of the first 20 points of K as listed in Figure
K n H(P) is a r-cap, for the point P is itself of degree 10 with respect to
K n H(P).
(iii) Consider P = (1, 0, 0, 0, 0, 0)
55 ,3 \ E. Then it may be shown directly that K n H(P) is a 20-cap with respect to which no point has degree 10. £
However, the following is a less laborious verification that K n H(P) is not a r-cap, and hence is a 6-cap. It is shown in [3] that every point of 55 ,3 \ K has degree 10 with respect to K. Let A(P) denote the 20-set of points of K lying on the 10 secants of K through P.
Then if P is to be a point of degree 10 with respect to some 20-sec-
tion of K it must be the case that A(P) is the intersection of K with a hyperplane. We have already seen, in (ii), that this is so if P K n H(P) in the case P P
£
E \ K.
£
E \ K.
In fact A(P)=
However, it is easily verified that if
5 ,3 \ E, say P = (1, 0, 0, 0, 0, 0), then A(P) is not contained in any 5 hyperplane. £
Thus of the 308 20-sections of K, the 56 of the form K n H(P), where PeE \ K, are of type r l , while the 252 of the form K n H(P), where P are of type 6.
£
5 ,3\ E, 5
BIBLIOGRAPHY 1. 2. 3.
A. Barlotti, "Some Topics io Finite GeOOletrical Structures", Institute of Statistics Mimeo Series No. 439, University of North Carolina, 1965. R.C. Bose, "Mathematical Theory of the Synmetrical Factorial Design", 3ankhya 8, 107-166 (1947). R. Hi 11, "00 the largest size of cap in 55 ,3", Rend. Aeead. Naz. Lined 54 (8), 378-384 (1973).
On Pellegrino's 20-caps in S
4. 5.
6.
7.
4,3
447
R. Hill, "Caps and Codes", Disarete Mathematias 22, 111-137 (1978). G. Pellegrino, "Sul massimo ordine delle ca10tte in S4,3'" Matematiehe 25, 149-157 (1971). G. Pellegrino, "Su1le ca10tte massime dello spazio S4 3'" Atti deU 'Aaaad. di Saienze lettere e Arti di PaZermo, Serie IV,'Vo1. XXXIV, 297-328 (1974-75). V. Pless, "On the uniqueness of the Golay codes", J. Comb. Theory A5, 215228 (1968).
Department of Mathematics University of Salford Salford M5 4WT England
This Page Intentionally Left Blank
Annals of Discrete Mathematics 18 (1983) 449-466 © North-Holland Publishing Company
449
CAPS IN ELLIPTIC QUADRICS J.W.P. Hirschfeld
1. INTRODUCTION In PG(n,q), projective space of n dimensions over the Galois field GF(q), a k-aap is a set of k points no three of which are collinear.
that k can take is denoted by m2(n,q). m2(n,2) = 2n
The maximum value
The only values known are the following:
rq + 1
n ;;. 2
, q odd,
m (2,q) -) 2
-1 q
m (3,q) = q
2
2
m (4,3)
2 m (5,3) 2
+ 2 , q even; + 1 , q
>
2 ;
20
56
When n = 2, a k-cap is usually called a k-ara. For q odd, a (q+l)-arc is a conic and a (q2+1)-cap is an elliptic quadric. The following three theorems were shown by Segre [4] in 1967. THEOREM 1.1: For q odd, a k-ara in PG(2,q) with k> q unique aonia.
t Iq + ~s aontained in a
0
THEOREM 1.2: For q odd and sUffiaientZy Zarge, a k-aap in PG(3,q) with 2 3/2 1 k > q - cq , where c < 4' is aontained in a unique eZZiptia quadria. THEOREM 1.3: For q odd and suffiaientZy Zarge, m2(n,q) < qn-l - cq n-3/2 where c <
41 and
n.. 4.
0
0
450
J. W.P. Hirschfeld
The object of this paper is to make Theorems 1.2 and 1.3 exact.
In fact it
only requires a slight adjustment to Segre's methods to achieve this. THEOREM 2.10: In PG(3,q) with q odd, q
~
67, a k-cap with
2 1 3/2 k> q - - q + 2q 4
is cJornained in a unique elliptic quadrie. 0
THEOREM 3.3:
1>1
PG(n,q) with q odd, q> 121, n
~
4,
n-l 1 n-3/2 n-2 m (n,q) < q - 4q + 3q 0 2 In both these theorems the respective remainder terms 2q and 3qn-2 are a little 1 3/2 d 1 n-3/2 larger than the proofs deliver. Any improvement on the terms 4 q an 4 q 1
would require an improvement on the term 4 Iq in Theorem 1.1. To contrast the case of q even, we have the following results by Segre [4]. THEOREM 1.4: For q epen, q > 2, 3 1 3/2 3 m2(4,q) < q - ~ + ~ + 1; n-l 1 n-5/2 3( n-3 n-4 m2 (n,q) < q - ~ +4q +q +••• + q) + 2, n > 4.0 This was improved for some even values of q by Thas [5] using a slight variation of Segre's methods. THEOREM 1.5: For n
~
5,
1 n-5/2 3 n-3 5 n-4 n-l m2(n,q) < q - ~ q +4q - 4 (q +••• + q) + 1, (ii) m (n,8) < 8n- l + (6 - 8/2)(8 n-4 +... + 1) + 1. 0 2
(i)
Hill 12] proved the following recurrence relation. LEMMA 1.6: For
q
> 2 and n > 4,
q
45\
Caps in elliptic quadrics
m2(n,q)
~
n-4 n-4 n-5 q m2(4,q) - q - 2(q + ... + 1) + 1.0
The estimate of m2(4,q) from Theorem 1.4, namely that, for q even and q > 2, 3 1 3/2 3 + '4 + 1 , m2(4,q) < q - 2" q gave the following improvement on Theorem 1.4 for all even q > 2.
These values
are not as good for those q for which Theorem 1.5 applies. THEOREM 1.7: For q even, q> 2 and n
~
5,
n-l 1 n-5/2 3 n-3 n-5 m2(n,q) < q - 2" q + '4 q - 2(q + ... + 1) + 1. 0 There are results due to Barlotti [11 comparable to Theorem 2.10 which are effective for small q. THEOREM 1.8: Let K be a k-cap in PG(3,q), q odd. and k ~ q2 for q
= 3 or 5,
If k
~ q2 - q
+
7 for q ~ 7
then K is contained in a unique elliptic quadric.
For further results of this type, see [11, [41, [51.
0
For a list of previous
estimates for m2(n,q), see [41, p. 166. The results and proofs in the next two sections are either the same or just slight variations on those in [41.
However, the nature of the results is such
that, without the full details, it would be difficult to follow the argument. The following notation is also used: y y
+
= GF(q), GF (q)
U {oo}
P(X) = the paint of PG(n,q) with coordinate vector X = (xO""'x n) , V(F 1' ... , Fr) = {P (X) E PG (n , q) I F1(X) ITd = subspace of dimension d.
F (X) r
O},
452
J. W.P. Hirschfeld
2. ELLIPTIC QUADRICS IN PG(3,q) For notation and background material on PG(n,q), we follow [3].
The irre-
ducible quadrics in PG(n,q) are En
= V(f(xO'x l )
+ x2x3 + •• + xn_lx n)
1
I
Hn = V(XOX l + X2X3 + ••• + xn_lx n)
n odd;
n even; Then every quadric Q in PG(n,q) can be
f is an irreducible binary quadratic form.
written ITrV, where llr is the subspace of singular points and V is a section of Q 1 skew to II ; that is, Q is a cone with vertex llr and base n-rr ducible plane quadric is a conic. by a
Jl
V.
An irre-
LEMMA 2.1: In PG(3,q). the maximum value of k for which a k-cap can lie on a quadric sUl'face is as follows.
Symbol
q even
q odd
repeated plane
IIlo
q +2
q +1
1i ne
III El
2
2
plane pair
2(q + 2)
2(q + 1 )
2(q + 1)
2(q + 1)
hyperbolic quadric
1l1Hl 1l 0P2 H3
elliptic quadric
E3
2(q + 1 ) 2 q +1
2(q + 1) 2 q +
Description
cone
LEMMA 2.2: In PG(3,q), q> 2, Then, if k;.
t
0
let K be a k-cap contained in a (q2 + l)-cap O.
(q2 + q + 4),any cap K' containing K lies entirely in O.
PROOF: Suppose there is a point Q in K' \ O. meets K in at most one point.
Each of the q+l tangents from Q to 0
Hence there are at least k - (q+l) ;.
21
(q
2
- q + 2)
lines through Q which meet K and are bisecants of O. So 0 contains at least 122 (q + 1) + 2 • 2 (q - q + 2) = q + 3 points, a contradiction. 0 LEMMA 2.3: In PG(3,q), let K be a k-cap in a quadric V.
Then, if
453
Caps in elliptic quadrics
k~
21
(q + q + 4)
2
k ~
21
(q + q + 6)
2
for q
~
5 ,
for q
= 3,4,
V is elliptic and every cap containing K lies on V.
PROOF: This follows immediately from Lemmas 2.1 and 2.2.
0
LEMMA 2.4: If A,B,C are three sets such that C ~ A U B,then IA n BI
~
IAI + IBI -
lei.
PROOF: This follows from the identity IA n BI
IAI + IBI - IA U BI. 0
LEMMA 2.5: If PG(3,q) with q odd, let C and C' be conics in the respective planes 11
and
11'
such that the line R. =
11
n
11'
meets both C and C' in the points It and P'.
Then the quadrics containing C and C' form a pencil F if n2 ,n ,n
+
-
= {Fp I
p
E y+}
such that,
3 4 and n4 are respectively the number of plane pairs, cones, hyperbolic
quadrics and elliptic quadrics in F, then
n2
= 1 • n3 = 0 or 2 ,
+
n 4
(q-l-n )/2 3
~
(q-3)/2
n4
(q+l-n 3 )/2
~
(q-l)/2
The plane pail' in F is nal to C is internal
PROOF: Let
11
11
01'
= V(x ) and 3
+ 11', and
n3 is 0
2 as a point P on
01'
R.
which is exter-
external to C'.
11'
= V(x 2 ); let P = P(l,O,O,O) and Q = P(O,l,D,D). Also
2 let C = V(X 3 ,X/ - xOx l ) = {P(t ,1,t,O) I tE /}; then C'=V(x ,bxOX + cx x + 2 3 l 3 + dX/ - xOx l ) = {P(cs+d,s2 - bs,O,s-b) I s E /L Hence, if Fp = V(Fp) is a ,
.
quadric containing C and C • then for some p ln F
p
y
+
,
J. W.P. Hirschfeld
454
So, with F( i )
aF
P
fjx. 1
-x
1
+ bx , F(l)
3
So F is a plane pair when p = '" and otherwise has a singular point where F(O)D= F(l) = F(Z) = F(3) = 0, which occurs at P(2c,2b,-p,2) when p Z = 4(bc + d). Hence n = 1, and n3 = 2 or 0 as bc + d is a square or not. Also 2 P(l,-r,O,O) is external or internal to C as r is a square or not, and is external or internal to (' as (bc + d)r is a square or not. +
-
It remains to find n and n . If Q and Q' denote respective points of ( 4 4 and (' other than P and P', there are (q_1)Z lines QQ', each of which lies on exactly one F. p
Also, every line on an irreducible Fp not through P or P' is some
QQ'.
Each cone contains q-1 lines QQ' and each hyperbolic quadric 2(q-l) lines
QQ'.
So +
n3( q- 1) + 2n 4 ( q- 1) = (q - 1) +
n3 + n4 + n4
=q
Z
,
,
whence the result. 0 Three conics (1,C -cetl"ali<JLirai:
2
and C in the respective planes 3
IT
l ,IT2 and
IT3
form a
system if
with P,P ,P ,P3 not coplanar. 1 2 The tetrahedral system is frat if the tangents
~1'~2
and
~3
to Cl 'C and C3 at P 2
are coplanar. LEMMA 2.6: The three conics of a tetrahedral system are contained in a quadric if and ,m"y if the system is [rat: the quadric is then unique and irreducible.
PROOF: If. in the above notation, the conics form a tetrahedral system, then by
455
Caps in elliptic quadrics
the previous lemma there is a pencil F of quadrics Fp through C and c • If the l 2 system is flat, each Fp has the tangent plane ~1~2~3 at P. So such an Fp is tangent to C at P and contains the points P and P2 of C • Hence a quadric F 3 l 3 p containing one further point of C3 contains C and is necessarily irreducible. As 3 two quadrics cannot intersect in a curve of degree more than four, the quadric is unique. If there is a quadric Fp containing the conics Cl 'C and C of a tetrahedral 2 3 system, then Fp has a simple point at P = Cl n C2 n C3 • So the tangent plane TI at P to Fp meets TI 1 ,TI 2 and TI3 in the tangents £1 and ~3 are coplanar. 0 LE~~A
'~2
and £3 to the conics; so
tl'~2
1
2.7: In PG(3,q) with q> 13, let K. be a k.-arc with k. > q - -4 Iq + 7/4 and 1
so contained in a conic C , i
i
= 1,2,3.
1
1
If C ,C and C form a non-flat tetrahe-
l
2
3
dral system, then there is a line meeting K ,K and K3 in distinct points. l 2 Kl U K2 u K3 is not a cap.
So
PROOF: By Lemma 2.5 there is a pencil F of quadrics Fp through C and C . 1 2 such Fp has in common with C3 the points P'P l and P2 ' where, as above,
Any
C. n C. = {P,P }. As the system is not flat, there is exactly one F tangent to k 1 J p C3 at P, one Fp tangent to C3 at P and one Fp tangent to C3 at P2• As there are l n + ~ (q-3)/2 hyperbolic quadrics in F, at least (q-3)/2-3 = (q-9)/2 hyperbolic 4 quadrics in F meet C in a point Q other than P'P and P • Let M be the set of 3 l 2 such points Q: then m = IMI ~ (q-9)/2. 1
P } and put r i = 1Ki' I; then r3 > q - "4 Iq - 5/4 and, li 2 for i = 1 and 2, r i > q - "4 Iq - 1/4. Also, let M' = M n K3' and m' = 1M' I; finally, put C ' = C\{P,P ,P } and so IC 'I = q - 2. Then, by Lemma 2.4 with 3 l 2 3 3 Let Ki'
A
= M,
B
= K3 ,
=
Ki \ {P,P
C
= C3 ', m'
~
m + r3 - (q-2) 1
> (q-9)/2 + (q - "4 Iq - 5/4) - (q-2) =
(2q - Iq - 15)/4.
If R3 is a point of M', then R3 is the meet of two generators g and g' of an H3 in F.
Also g and g' meet C2 in distinct points R2 other than P and Pl ' and meet Cl in distinct points Rl other than P and P2• If g or g' contains P3 ' it is
456
J. W.P. Hirschfeld
coplanar with the tangents to C and C at P ; these tangents are distinct and l 2 3 their plane meets C in at most two points. If R3 is not of this type, then g and 3 g' each meet each C in some paint other than P,P ,P and P . There are at least i 3 l 2 m' - 2 such R3 each giving two different R2• each of which is obtained from at most two R3• Hence the number of such R2 is ;;. m' - 2
> (2q - Iq - 23)/4, which is positive for q ;;. 14.
Hence the number of R2 also in K ' is, again by 2
Lenma 2.4,
> (2q - Iq - 23)/4 + (4q - Iq - 1)/4 - (q-2)
= (q - Iq - 8)/2 , which is positive for q ;;. 13.
Each of these R2 lies on at least one line R2R3
meeting C2 and C3 in points of K2 and K3 , and meeting Cl in a point Rl • Each such Rl arises from at most two R2 • So the number of Rl in Kl' is, by Lemma 2.4, 1
> (q - Iq - 8)/4 + (q - '4 Iq - 1/4) - (q-2) = (q - 21q - 1) /4 ,
which is positive for q ;;. 7. Anyone of the corresponding lines R1R2R3 meets K ,K and K3 in distinct l 2 points.
0
LEMMA 2.8: Let K be a k-aap in PG(3,q), q odd and q
~
67 ,with
2 1 3/2 + R(q) k > q - '4 q R(q)
and
(31q + 14/q - 53)/16 2q -
::'et .~ be any line such that
I~
n KI
(I,q -
;,
7)2 + 41/16.
1 and 'let the q+ 1 planes through ~ be
i , i=l ••••• q+l. If Ki = TIi n K is a ki-arc, and, if ki > q i = 1, •.. , S, then S > 3(/q + 9)/4.
n
I~
PROOF: If k i
~
n KI = r, then r
= 1 or
As ki q - '4 /q + 7/4 for; = S+l, ••• ,q+l, so 1
2.
~
q + 1 for
t Iq + 7/4 only for 1, ... ,S and
Caps in ellip tic quadrics
457
,
k = r + E(k. - r) .;;; Ek. - q ,
1
.;;; S(q+l) + (q+l-S)(q-"4 Iq+7/4) - q = S(/q-3)/4 + q2 - 41 q3/2 + (7q - Iq + 7)/4. If the lemma is false and S .;;; 3(/q+9)/4, then 2
k .;;; q -
41 q3/2
= q 2 - 41 q 3/2 a contradiction.
+ (31q + 14/q - 53)/16 + R(q),
0
2 1 3/2 2 NOTES: (i) It is necessary to take q ;;;. 67 only so that q -"4 q + R(q) < q + l. It
only requires q ;;;. 11 for q - 41 Iq + 7/4 < q+l, and q ;;;. 9 for 3(/q+9)/4 < q + l. (ii) The number 3(/q+9)/4 is the smallest that is large enough for both the
next lemma and the final theorem. (iii) This is the lemma that determines R(q): it is the smallest value consistent with the required lower bound for S. 2
LEMMA 2.9: If K is a k-aap in PG(3,q), q odd and q ;;;. 67,with k > q -
41 q3/2+ R(q),
then there exist aonias Cl ,C and C3 forming a fZat tetrahedraZ system suah that,
2
,
,
1
if K. = K n C.is a k.-ara, then k. > q - -4 Iq + 7/4 for i = 1,2,3.
"
PROOF: As k> 1, take two distinct points P and P3 in K. By the previous lemma, there are S > ~(/q+9)/4 planes ni through PP 3 such that K n ni Ki is a ki-arc with ki > q - "4 Iq + 7/4. Take two such planes nl and n2; then Kl and K2 contain P and P3• Also Kl is contained in a conic Cl and K2 in a conic C2 , Theorem 1.1; hence K n Cl = Kl and K n C2 = K2 • Let Pl be a point of K2 other than P and P3 • Through PP l there are more than 3(/q+9)/4 planes n.' such that K n n.' = K.' is a k.'-arc with 1 ' " , ki ' > q - "4 Iq + 7/4. One of these planes, namely n2' passes through P3 and at most one is tangent to Cl at P. Therefore there are more than 3(/q+9)/4 - 2 =
= 3(/q+19)/4 planes ni ' each meeting Cl in a distinct point P2 other than P and P3" If there are m of these points P2 , then m > (3/q+19)/4. points P2 in K1 is, by Lemma 2.4,
So the number of
J. W.P. Hirschfeld
458 ~
m + kl - (q+l)
> (3/q+19)/4 + (q- ~/q + 7/4) - (q+l) =
(/q+ 11 ) /2.
Therefore there exists P2 in Kl and so the plane n3 = PP 1P is one of the 2 planes IT; '. Put K n n3 = K3 ; then K3 is obtained in a conic C3 • The three conics C1 ,C and C constitute a tetrahedral system. If the system were not flat, then 3 2 Kl u K2 u K3 is not a cap, Lemma 2.7. However, as Kl U K2 U K3 is contained in K, it is a cap and therefore the system is flat. 0 THEOREM 2.10: :n PG(3,q) with q odd and q ~ 67, if K is a k-cap with . .. K''l-S an e II"'l-pt'l-C k > q2 - '41 q3/2 + 2q, then t he on l y comp l ete cap conta'l-m.ng JuadT'ic Q.
In fact, it sUffices to take k
R(q)
2
1 3/2
> q - '4 q
+ R( q) ,
(3lq + 14/q - 53)/16 = 2q - [ (/q-7)
2
+ 4] /16.
PROOF: By Lemma 2.9, there exists a flat tetrahedral system of conics C ,C ,C l 2 3 1 such that K n Ci = K. is a ki-arc with k > q - '4 Iq + 7/4. By Lemma 2.6, the i 1 the three conics are contained in a unique quadric Q. show that K C Q; for, as k
~
It is now sufficient to
2
(q +q+4)/2, so, by Lemma 2.3, Q is elliptic and the
only complete cap containin9 K. Suppose there is some point Q in K \ Q.
As before, C ,C and C have the l 2 3 point P in common and have the residual intersections Pl ,P 2 and P • So Q is on no 3 C and Q F P,P ,P ,P • 1 2 3 i Let m be the number of planes n ' through PQ meeting C , apart from P, in a i l 1 point Q1 of K1 with Q1 distinct from P,P 2 ,P 3 and such that K n n i ' = Ki' is a 1 k.1 '-arc wi th k.' > q - -4 Iq + 7/4. 1 Let m be the number of these m1 planes ni ' meeting C2 in a point Q of K2 2 2 with Q2 distinct from P,P ,P . Similarly, let m3 be the number of these m planes 2 1 3 meeting C3 in a point Q of K3 with Q distinct from P,P ,P 2• 3 l 3 It will be shown that m3 > O. Firstly we note that IKl \ {P,P ,P 3 }1 > q - ~ Iq - 5/4. From Lemma 2.8 the number of planes n ' with i ki ' > q - '4 Iq + 7/4 is more than 3(/q+9)/4. So we obtain in succession the fol-
t
lowing, always using Lemma 2.4:
Caps in elliptic quadrics
459
1
m1 > (q- "4 Iq-S/4) + 3(1q+9)/4 - (q+l)
(/q+9)/2,
m2 > (q- ~ Iq-S/4) + (lq+9)/2 - (q+l)
(/q+9)/4,
1
m3> (q-"4 Iq-S/4) + (/q+9)/4 - (q+l)
O.
So there exists a plane nO', say, containing P and Q, and meeting Cl in Ql' C2 in Q2 and C3 in Q3 with each Qi distinct from P,P l ,P 2 ,P 3, such that KO' = K n nO' is contained in a conic CO'. The four conics CO' ,C l ,C 2 ,C form four tetrahedral systems. As 3 CO' U Cl U C2 C K, the system of CO' ,C l ,C is flat, Lemma 2.7. The conic C ' has 2 u at Q a tangent coplanar with the tangents at Q to C, and C?; this plane is the tangent plane to Qat P. Q3'
So CO' is tangent to Qat P and meets Q in Ql,Q2 and
Therefore CO' 1i es on Q. and, as Q is on CO', so Q is on Q. IJ
COROLLARY: In PG(3,q), q odd and q
~
67, if K is a complete k-cap which is not an
elliptic quadric, then
k .;;; q 2
_l
4
q3/2 + (31q + 14/q - 53)/16
_ ~ q3/2 + 2q. 0 NOTE: This theorem gives a better result than Theorem 1.8 for q
~
139.
3. THE MAXIMUM SIZE OF A CAP Three planes in PG(4,q) lie in a pencil if their join is n3 and their intersection TIl' LEI~MA
3.1: In PG(4,q), three quadric surfaces Ql,Q2,Q3 in respective solids
l , a2 , a3 have in common the conic C in the plane n. Then Q.l'~'~ lie on a quadric primal Q if and only if at each of three points of C the tangent planes to
a
the three
~
lie in a pencil.
PROOF: (i) Suppose Q exists. a simple point of Q. to
~
at P.
Consider a point P in C.
The tangent solid
a
Then Q n a i = ~, and P is a. in the tangent plane
to Qat P meets
,
These planes have the line ann in common and lie in the solid
a.
460
J. W.P. Hirschfeld
(ii) Here we give an analytic proof of the converse as opposed to the synthetic one in [4J.
~
Let
= V(x O2 - x1x2
The tangent plane at P(Y) to Tp(y)(~)
+
~
is
= V(2x oyo - x1Y2 - x2Y1 +
x3(a i yO + bi Y1 + c i y2 )
+
Y3(a i x + bi x1 + c i x2), x - di x3 )· O 4
(t,t 2 ,1,0,0) so that P(Y)
If Y
x (a i x + b x + c x2 ), x4 - d x )· 3 i 1 i i 3 O
E
C, then
where The T. have the line 1
in common. i
=
The
lie in a solid if there exist A ,A ,A and F such that, for 1 2 3
1,2,3,
This requires that
\.1 =
A
2 2 and a;t + bi t + c i - Ad; = G = F - AX 4 - (2txO - xl - t X2) ; that is, the vectors (A ,A ,A ), (d ,d ,d ), (1,1,1) are linearly dependent, where 1 2 3 1 2 3 2
Ai = ait + bit + c i • This occurs for three values and so for all values of t if and only if (a 1,a 2 ,a 3 ), (b 1,b 2,b ), (c 1 ,c ,c ) is each linearly dependent on (d ,d ,d ) and 3 2 3 1 2 3 (1,1,1 ). Now, Q. is a solid section of a quadric Q if and only if there exist 1
e,f,g,h such that, for i
1,2,3, 2
Q = V(x O - x1x2 +
As the d. are distinct, so h 1
+
x3 (aix O + bi x1 + ci x2)
(x - di x3 ) (exO + fX 1 + gX + hx 3 ))· 4 2
= O. Also
461
Caps in elliptic quadrics
a.1 - d.e 1
1:11
a
b.-d.f=b 1
1
c. - d.g 1
1
=C
But these relations mean precisely that (a ,a ,a ), (b ,b ,b ), (c l ,c ,c ) is 1 2 3 1 2 3 2 3 each linearly dependent on (d ,d 2 ,d 3 ) and (1,1,1). 0 1 LEMMA 3.2: In PG(4,q), q> 121 and odd, let K be a k-cap and
~
n K is an s-arc with s > q -
t Iq + 7/4.
TI
a plane such tlmt
Then there do not exist three distinct
IT such that K. = a. n K is a k.-cap with 1 2 3 2 1 3 /2. 1 21 1 3/2 11 ki > q - 4 q + R(q), 1 = 1,2, and k3 > q - 4 q + /I q + 1.
solids a ,a ,a containing
PROOF: Suppose that the theorem is false. contained in a unique elliptic
quadric~.
Then, by Theorem 2.10, each Ki is Now we consider the alternatives posed
by Lemma 3.1. (i) There exists a quadric Q meeting The set M= Kl
U
K2 U
i in ~, i = 1,2,3. K3 is an m-cap contained in K with a
m = s + I(ki-s) = k1 + k2 + k3 - 2s. As s
~
q + 1, so
2 1 3/2 2 1 3/2 11 m> 2(q - 4 q + R(q)) + (q - 4 q + /I q + 1) - 2(q+1) 3(q2 -
t
q3/2) +
i (37q + 14 Iq - 61).
There are two possibilities for Q.
(3.1 )
Either Q = P4' the nonsingu1ar
quadric, or Q = TIOE3' the singular quadric with vertex TIO and base E3 • (a)
Q= P 4
P comprises (q2+1)(q+1) points on the same number of lines with q+1 lines 4 through a pOint. Through each point of a line t on P4 there pass q other lines, whence q(q+1) lines t' on P meet t. No two of these lines t' meet off t, as 4 otherwise their plane would meet P4 in a cubic curve. Also P4 contains q2(q+l) points off t. So through each point of P4 ' t there is exactly one line t'. The m-cap M has at most two points on t and on each t', and every point of M lies on t
or some t'.
Hence
462
J. W.P. Hirschfeld
2
m ~ 2 + 2q(q+l) = 2(q +q+l).
(3.2)
From (3.1) and (3.2),
*
3(q2 -
q3/2) +
i
(37q + 14/q - 61) < m ~ 2(q2+q+l ).
Hence q2 _ l q3/2 + ~ (3q + 2/q - 11) < 0 , 4
8
a contradiction.
2
Through ITO there are q + 1 generators of Q, each containing at most two poi nts of
:11.
So 2 m ~ 2(q +1).
(3.3)
From (3.1) and (3.3), 3(q2 -
*
q3/2) +
i (37q+14/q-61) <
2 m ,;; 2(q +1).
Hence q2 _
%
q3/2 + ~(37q + 14/q - 77) < 0,
a contradiction. (i i) There is a pencil .~ of quadri c prima 1s through Q and Q2' none of whi ch
l
conta ins Q3' The members of
~
cut out on 03 a pencil
ing the conic C, the unique conic through and Q rr. 3
So
1>'
~
~'
n K.
of quadric surfaces all containOne member of
cuts out on Q3' apart from C, a pencil
planes TI'; the quadric C' is either a conic or a point. planes
~'
have a common line
~
'¥
is TI repeated,
~'
of quadrics C' in
Also C E
'¥,
and the
in n.
2 1 3/2 7 k3 - (q+l) > q - 4 q +4 q
As
1 7 =q(q-4 /q + 4)'
there is some plane 1
7
k' > q - 4/q + 4'
TI'
other than
TI
meeting K3 in a k'-arc K' with
So, by Theorem 1.1, K' is contained in a conic C' = TI' n Q3'
Caps in ellip tic quadrics
463
It will now be shown that there exists a line P'P P ' where P' 1 2 and P2 E K2' Let V be the quadric of
% in
meeting
C + C',
E
K',P
l
E
Kl
Take a point P' in C' \ C,
Since it is simple, the tangent space Tp'(V) to Vat P' meets V in a cone P'P2' So there are q + 1 lines £' of V in Tp'(V),
As Tp'(V) does not contain C it meets
C in at most two points, whence at most two lines £' meet n,
The others, in num-
l in a point Pl of Ql and a 2 in a point P2 of Q2' with P ,P not in C, Also, P # P since a n a 2 = TI and P ,P ~ TI, Further, l l 2 l 2 l 2 P. # P', i = 1,2, since every point of a n C' lies on C, l 1 1 1 Let P' E K' \ C and note that IK' \ cl > q - 4 Iq - 4' For each such P', ber at least q - 1, all meet
a
there are at least q - 1 points P ' Conversely, each P is derived from at most l l two P', namely K' n Tp (V), unless Tp (V) contains TI' and hence K'. This exceptional case can
O~ly occur twice,lwhen Pl lies on the polar line of the plane
Thus each P' gives at least q - 3 points Pl ' apart from the exceptions; each P comes from at most two P', Thus, with A = {P in Ql obtainable from some P' in l l TI',
K' },
1 1 1 IAI > '2 (q - '4 Iq - '4) (q - 3)
= l q2 _ l q3/2 _ l (13q - 3/q - 3). 2
8
8
Let B = K , C = Q and Kl f' = {P in Kl obtainable from some l l l Kl ,', = A n B. So, by Lemma 2,4,
pI
in
K'}.
Then
IK/I >~q2 _ ~q3/2 _~ (13q - 3/q - 3) _ (q2 _ l q3/2 + R(q)) _ (q2+l) 4
= l q2 _ ~ q3/2 + ~ (5q + 20/q - 63) 2
8
16
'
The line P'P with P' in K' and P in Kl* meets a in a point P of Q \ C, l 2 l 2 2 Such a P2 is obtained at most twice when ITp (V) n K' 1 ~ 2, unless Tp (V) ~ TI',
which can occur for at most two points P ' w~ere the polar line of rr,2meets Q2' 2 Thus, with A = {P in Q obtainable from some P1P'}, 2 2 1A1
3 q3/2 + 32 1 (5q + 20 I q - 63), = '21 1K1f'l > '41 q2 - 16
464
p'
E
J. W.P. Hirschfeld
K', P,
E
K/'}, Lemma 2.4 gives that (5q + 20/q - 63)
+
1
=4
q
1 3/2
2
(q - 4 q
2
-
7
16
q
3/2
2
+ R(q)) - (q +1)
1 + 32 (67q + 48/q - 201)
> O. So there is a line meeting K',K 1,K in distinct points. 2 provides the desired contradiction. 0
So K is not a cap, which
NOTE: The condition that q > 121 is only necessary to ensure that both 2 1 3/2 2 1 3/2 11 2 q - '4 q + R(q) and q - 4 q + "4 q + 1 are less than q + 1. THEOREM 3.3: In PG(n,q), n
~
4, q> 121 and odd,
2(n,q) < q
m
n-1
-
41 qn-3/2
+
3q
n-2
•
In fact,
m (n,q) < qn-l ! qn-3/2 - 4 2
+
!4 (11q n-2
- 2(q
n-5
+
2qn-5/2
+ ••• +
-
l4qn-3)
1) + 1,
where there is no tem _2(qn-5 +... + l} for n = 4.
PROOF: Let K be a k-cap in PG(n,q). {il
(a) There is no plane
11
n
=4
such that
11
n K
Take a line t meeting K in two paints. through
l
1
7
is an s-arc with s > q - 4 /q + 4 . There are q2
each meeting K in an m-arc with m" q -
+
t /q + i.
q + 1 planes So
11
465
Caps In elliptic quadrics
k " 2 + (q
= q3
2
+ q + 1) (q -
1
4
Iq -
1
4)
1 5/2 1 2 3/2 - -4 q + - (3q - q + 3q - Iq - 1) 4
3 1 q5/2 3 2 < q-4 +q.
(b) There is a plane rr such that rr n K is an s-arc with s > q -
t i. Iq +
by Lemma 3.2 and Theorem 2.10, there are at most two solids through
an elliptic quadric, and, for the other q - 1 solids a through n.
la So k " 5 + 2(q
2
n
KI
"q
2
-
1 3/2
4q
+ 1 - s) + (q - 1) (q
2
11
+ "4 q + 1
-"41 q3/2
11 +"4 q + 1 - s)
2 1 5/2 1 3/2 1 7
= q3 -"41 q5/2
1 2 3/2 + 4 (llq + 2q - 14q + 4)
1 5/2 3 2 < q3 - 4 q + q.
(ii)
n>4.
The induction formula of Lemma 1.6 gives that n-4 n-4 n-5 m2 (n,q) " q m2(4.q) - q - 2(q + ••• + 1) + 1 < qn-l _
t qn-3/2 + t (11qn-2 + 2qn-5/2 _ 14qn-3)
- 2(q
n-5
+ ••• + 1) + 1
n-1 1 n-3/2 n-2
0
rr
Then
meeting K in
466
J. W.P. Hirschfeld
BIBLIOGRAPHY 1. 2. 3. 4. 5.
A. Bar1otti, SGme topics in finite geometrical structures, Institute of Statistics Mimeo Series No. 439, University of North Carolina, 1965. R. Hill, "Caps and codes", Discrete Math., 22 (1978), 111-137. J.W.P. Hirschfeld, Projective geometry over finite fields, Oxford University Press, 1979. B. Segre, "Introduction to Galois geometries", Atti Accad. Naz. Lincei Mem., 7 (1967), 133-236. J.A. Thas, "Contributions to the theory of k-caps in projective Galois spaces" (Flemish) Med. Konink. Vlaamse Acad. Wetensch. Lett. Schone Kuns7::. 2dgie Kl. i·ie tensch., 30 (1968), no. 13.
Mathematics Division, University of Sussex, Fa1mer, Brighton, BNl 9QH, England
467
Annals of Discrete Mathematics 18 (1983) 467-468 © North-Holland Publishing Company
ON COHNFUNCTIONS Herbert Hotje
In this lecture I will give a theorem which I found together with a student of mine, Winnfried Balkwitz.
In 1961 P.M. COHN showed how to characterize the
multiplicative group G of a field by an element e E G and an involutorical function E: of G'
G\
:=
namly by setting:
{l},
-1 ) E:
x-1 xE: e,
i)
II x E: G' : (x
ii)
V x,y E: G', x , y : (y-lx)E: = (y-l)E:((yE:)-l xE:)E:.
iii)
-1 E: II x ,y E: G' : x y x
=
= (x -1 y x) E: .
In the field e means -1, and xE: case. 71 n
= 1 -x. We will restrict ourselves on the finite
So we have the field GF(pn), and the multiplicative group is isomorphic to Is there a simple way to describe the cohnfunctions within
(+).
p-l 71 p~l
? 71
p~l
Case 1:
has a ringstructure, and we will try a description by polynomials.
. a prlme. . pn- 1 lS
n p -1
iUl i,j
Then 71 is a field, and for each polynomial n p-l -1 The polynomial (x-i) there exists ('P j (j) )
'P j
.'=
f
pn -1 .e: -1 . - '};l J 'P . (x) ('PJ' (j ) ) is equa 1 to J=
J
n
Case 2: p - 1 =
r' s wi th r, s
E IN \ {l}.
e:
on 71 \ {O}. n p-l
Then we have the proper ideal r . 71
and a proper partition {r 71 , r 71 + 1 , ... ,} of 71 rs mrs . rs Suppose e: is a polynomial '};O a. Xl. 1=
For z
E
71
1
rs and t Ef r 71 rs we have
rs
468
H. Hotje
t. ,.
f is bijective on II
II
/ r II
rs
\ {O}, and so f induces a bijective mapping on 1 ,which fixgs the neutral element. n
rs II We look at the meaning of ii) in the factorring rs/ r II
Because of rs
r,s ~ 1 we have two different elements x := r·ll
r·ll + 1. It is rs' rs (-y + x)£ ,. (r . II + (r-1))£, and (_y)E + (_(YE) + i(E)£ = (r . II + (r-1)) E + rs rs ii) means -eyE) = r . II , which is in contradiction to + (_(9E) + r II rs rs - (r . II + l)E ~ r II . rs rs Hence we have the
r.
THEOREr4: A cohnfunc:.-ion of GF(pn) is exactly then a polynomial on II ,when n n-1 -~8 . a pFl-Ine, . l ' f n 3 n l ' M . p-1 naN y 1- P = Ol' P - 1-S a el'senne pl'1-me. P n EXAMPLES: For P n for p n for p
3 is
E
X --->
x, 2x
4
is E
x
8
is £
X --->
E
X -->
---->
6x 4 + 4x or 4 X + 4x
REMARK: The last example showes, that there may be more than one cohnfunction for a field. function.
One can show, that there are exactly ¢(pn-1) n
ones, ¢ being the eu1er-
Is £ one of them, so [ =
-1 r.:;.
E ~,
~
E Aut(ll n )} is the set of all of them.
p-1 BIBLIOGRAPHY 1.
p.r4. Cohn, On the embedding of rings in skew fields, Pl'OC!. London. SOC!., 11
(1961), 511-530. Universitat Hannover Institut fUr Mathematik Welfengarten 1 3000 Hannover Federal Republic of Germany
Annals
of Discrete Mathematics 18 (1983) 469-472
469
© North-Holland Publishing Company
A CLASS OF STRONGLY REGULAR GRAPHS RELATED TO ORTHOGONAL GROUPS X. Hubaut* and R. Metz
INTRODUCTI ON In this paper we describe a class of strongly regular graphs related to P0 2n + {q). If q is odd, the vertices of the graphs may be seen as a set of points l off a quadric in PG{2n,q). F. Buekenhout noticed that strongly regular graphs of the same kind occur in symplectic spaces over GF{2 r ). We give a unified description of those graphs for a field or arbitrary characteristic. Let Q be a quadric in PG{2n,q).
The intersections of Q with non tangent ±
hyperplanes are hyperbolic or elliptic subquadrics Si' elliptic) sections may be tangent or secant. If
±
Two hyperbolic (resp.
or + is written. the upper
(lower) sign denotes the hyperbolic (elliptic) case.
THEOREr~: Let G be a graph with S~ (resp. S~) as vertic:es.
Two vertic:es S~, S;
(resp. S:, S~) are adiaaent if and onZy if the two c:orresponding subquadric:s are 1
tangent.
J
Then G is strongZy regular with parameters V
= 21
n n q (q ±
1).
k={q n +l){q n-l ±1), 1
£=2{q-2)q
n-l
n_ (q +1),
A
= q - 2 + q2n-2 + q{q n-l + l){q n-2
jl
-
±
1),
_ 2qn-l{ qn-l -+ 1) .
COMPUTATION OF THE PARAMETERS Let us recall the following facts:
_ g2n _ 1 the number of points of a quadric in PG{2n,q) - q _ 1
470
X. Hubaut and R. Metz
the number of ~oints of a hyperbolic (elliptic) quadric in PG(2n - l,q) _ (qn + l)(qn- ± 1) -
q - 1
1) /iurnbe!1
or vtErtices
of G : v.
From now on we shall restrict ourselves to the case of hyperbolic sections. The other case can be handled in exactly the same way. We proceed by induction.
If n = 1 the hyperbolic sections of a conic are
sections by secants; hence v = (q ; 1) =
t q(q
+ 1).
t qn-l (qn-l+
the number of hyperbolic sections is
Suppose that in PG(2n - 2,q)
1).
In PG(2n,q), take S, any hyperbolic section of Q, and let p be one of its points; then the tangent hyperplane Hp intersects S along a hyperbolic cone. Conversely, given a hyperbolic section of the cone of the generators at p, any hyperplane (beside the tangent H ) will intersect Q along a hyperbolic section. p
We compute the number of ordered pairs (S,p), where S is a hyperbolic section and p a point of S, in two different ways. n
Beginningwith 5 we get v (9
-
n-
1
~)~ql
+ 1); on the other hand, through any
point p of Q, the number of hyperbolic sections is equal to the number of hyperbolic sections in the cone of the
hyperplanes in the pencil; thus we get the two results we obtain v
times the number of non tangent - 1 1 n-l n-l -2 q (q + 1) . q. Comparing
genera~grs
= 21 q n (q n
q
q - 1 + 1).
2) VaZency k of G.
Take any hyperbolic section S of Q. pencil of q
+ 1
Through any point p of S there is a
hyperplanes intersecting 5 along the tangent cone at p; one of
those hyperplanes is obviously the hyperplane containing 5, another is the tangent hyperplane Hp at Q.
The remaining q - 1 hyperplanes intersect Q along hyperbolic
sections tangent to S at p.
Thus the number of tangent sections is equal to q - 1
times the number of points of 5. n n-1 So k = (q - 1) (q - 1) (q + 1) q - 1
=
(qn _l)(qn-1 + 1).
3) Determinati,'i'l of >..
Let Sl and S2 be two hyperbolic sections of Q tangent at some point p, and let Hl and H2 be the two corresponding hyperplanes. planes of the pencil generated by Hl and H2
Then all the other hyper-
intersect Q along a hyperbolic
section except the tangent hyperplane H ; so there are q - 2 of them. p
Now let S
471
A class of strongly regular graphs
be a hyperbolic section tangent to Sl at ql and to S2 at q2; then the two tangent hyperplanes in Hl at ql and in H2 at q2 meet in Hl n H2 · Thus if ql is given, the intersection in Hl n H2 is known and the tangent hyperplane in H2 is determined; so S is known. It follows that the number of such S is equal to the number of f Sl not 1. n H1 n H pOl. nt s O 2' ·l. e. q2n-2 Finally there are hyperbolic sections tangent to Sl and S2 at a point lying in Hl n H2 (and different from pl. quadrics.
Given such a point, there are q - 1 such (qn-l_l)(qn-2+1) On the other hand the number of pOints in Sl n S2 is q q: I
Gathering those three types of hyperbolic sections we get the value of A: A=q-2+q 4) computation of
2n-2
+q(q
n-l
-l)(q
n-2
+1)
~
Again let Sl and S2 be two intersecting hyperbolic sections of Q lying in 1-\1 and H2 •
First note that if S is tangent to both Sl and S2 the touching point cannot belong to Sl n S2. So let S be a hyperbolic section tangent to Sl at PI and to S2 at P2.
The
tangent hyperplane in HI at P, intersects S, n S2 into a subquadric contained in a subspace K of codimension 2 in H2; this subquadric is contained in exactly two cones of S2' the vertices of these cones being the intersection of the polar line of K in H2 with respect to S2.
Thus to any given point PI of Sl not on S2' there
corresponds two points of S2 and therefore two quadrics S.
We thus obtain
~ = 2 [(gn - l)(qn-l + ') _ g2n-2 - 1J = 2qn-l(qn-l + 1) q-l
q-l
RANK OF THE GRAPH The graph G obviously has P0 + (q) as an automorphism group. A straight2n l forward verification shows that P0 + (q) acts on G as a rank ~ group if q is 2n l odd; if one fixes a section, all the tangent sections lie in the same orbit, the conjugate sections in another orbit and all the others form q ; 3 distinct orbits. r If q = 2r , there are no conjugate sections; so it follows that P0 1(2 ) 2n+ r-l acts on G as a rank (2 + 1) group. Rank 3 graphs are obtained only when q = 3 or 4.
X. Hubaut and R. Metz
472
BIBLIOGRAPHY 1. 2.
X. Hubaut, Strongly regular graphs, Discr. Math., 13 (1975), 357-381. A. Rudvalis, (v,k,A)-graphs and polarities of (v,k,A)-designs, Math. Z., 120 (1971), 224-230. Universite Libre de Bruxel1es Departement de Mathematique C.P. 216 B-1050 Bruxe11es Belgium
Geislaterstrasse 22 0-5205 St. Augustin Federal Republic of Germany
473
Annals of Discrete Mathematics 18 (1983) 473-480 © North-Holland Publishing Company
ON THE NON-EXISTENCE OF A SEMI-SYMMETRIC 3-DESIGN WITH 78 POINTS Daniel Hughes
A semi-symmetric 3-design (SS3D) for A is a connected structure in which any 3 points are on
a or
A blocks and any 3 blocks meet in
a or
A points.
If A > 1
then such a structure is square, with k points on every block and k blocks through every point.
Furthermore, the existence of an SS3D for A > 1 implies the exis-
tence of an extended symmetric design.
Except for hadamard 2-designs (which are
always associated with SS3D's, although not in a very interesting way), the only symmetric design known to have an extension is the projective plane P of order 4, which has a unique extension to M, a 3-(22, 6, 1). There are 2 possible SS3D's associated with P: (1) a design with 100 paints and k = 22, which is unique, and (2) a "special" symmetric design S for (78, 22, 6). are shown in [2].
These and other related facts
In this paper we prove that there is no SS3D S which is also a
symmetric design for (78, 22, 6). If S exists, then it has the following properties (see [2] for (i), (ii), (i i i ) ) ; (i) if Y is
a~y
block of S, the structure Sy of all points on y and all
blocks of S except y is isomorphic to M (see above). (ii) for any anti-flag (P, y) of S, the structure Sp ,y of all points on y and all blocks on P is a semi-biplane for (22, 6, (2)) (that is, a structure of 22 points and 22 blocks, with 6 blocks on every point and 6 points on every block, such that 2 points (blocks) are on
a or
2 blocks (points)).
(We abbreviate semi-biplane by SSP.) (iii) if P and Q are distinct points, neither on the block y, then the SSP's Sp,y and SQ,y have 6 blocks (of Sy) in common. We let B represent the unique SSP for (22, 6, (2)), and let unique hadamard 2-design for (11, 5,2) (note that [ 3],)
(iv) if A is an incidence matrix for L, then
L
L
is also an SSP).
represent the (See Wild
474
D. Hughes
A
B
is an incidence matrix for B. (v) a maximal set of points of points (and is called an
ovaZ);
L
such that no 3 are on a block contains 3
there are 55 ovals in
L
and there is a unique
"tangent block" on each point of an oval (see Assmus and van Lint [l)}. A1so in [ 1 lis:
(vi) let T be a 22 x 55 matrix whose rows are indexed by the points and blocks of L and whose columns are indexed by the ovals of L, with T = (t (
t
I
a,O
=
1 if a is a point of oval
o or
u,
O) and
a tangent
)
1o
block to oval 0 otherwi se;
then associating the point and block rows in the natural way, letting B be the matrix of (iv), M=
I BTl
is an incidence matrix for M. This demonstrates: (vii) isomorphic copies of B exist in M. From (iv) we can make morea"geometric" construction of B: for each point P and block y of L, we let (P) and (y) be points and [PI and [y) be blocks, where (a) (P) is on [PI, (b) (P) is on [yl if P is on y, (c) (y) is on [y], (d) (y) is on [P J if Pis on y.
a
Then:
(viii) B contains two isomorphic copies of L, say Ll and L , and a bijection 2 2 from Ll to L2 which is an anti-automorphism and such that a = 1; any block y'
of B contains the 5 points of a block y of Li and the one point (y}a of Lj' and dually for points; if 0 is an oval in Li' then a maps the 3 points of 0 onto the tangent blocks of an oval of Lj • and letting Oa be the oval so determined in L • j we call 0 u Or). an oual in B; the 55 ovals of B are the remaining 55 blocks of M (i.e., besides the 22 blocks of B). These observations allow us to carry out the search for S within the much simpler design L.
On the non-existence of a semi-symmetric 3.( 78,k)J
LEMMA 1: If B is anSBP for
(22,
6,
(2))
are, as point sets, the 55 ovals of B_
475
in Mthen the remaining 55 blocks of M (So B is embedded in M in essentially only
one way.)
PROOF: If y is a block of M not in B, then y meets one of Ll or L2 in at least 3 If y meets Ll in 4 points then these 4 points have a subset of 3 paints
points.
on a block of Ll hence on a block of B (see (v)); this would force two blocks of M to contain the three pOints, which is impossible.
So y meets each of Ll and
L2 in 3 points, and similarly these point sets are easily seen to be ovals and their union an oval of B. D We fix an anti-flag (P, y), identify Mwith Sy and B with Sp,y in what follows. LEMMA 2: If Q is a point not on y, Q ~ P, then the 6 blocks of B n SQ
,y
covel' 18
points of B, each in 2 blocks.
PROOF: A point on one of the 6 blocks is certainly in 2 of the 6 blocks, since S is an SS3D for A = 2. blocks: then 2t
= 6·6, since each point is on 2 blocks and each block contains 6
points, and so t = 18. LE~~~~A
Let t be the number of points in B which are on the 6 0
3: If Q is chosen as in Lemma 2 then the 6 blocks of B n SQ ,y are the blocks
of an oval of B, and there is a one-to-one correspondence between the 55 ovals of B and the 55 points Q.
PROOF: We say that a block of B is in Ll (or L2 ) according as it has 5 of its pOints in Ll (or L2 ); see (viii). If 4 of the blocks of B n SQ,y were in Ll , then the dual of (v) asserts that some 3 of them are on a point, which violates Lemma So the 6 blocks are 3 each in Ll and L2 , and it is similarly easy to see that they are the tangent blocks to an oval of B. Also, it is straight-forward that
2.
different choices of Q lead to different ovals of B.
0
The rest of our proof will consist in showing that we cannot choose another SBP in M to play the role of SQ .
oval.
,y
,that is, to meet B in the six blocks of an
Let us write Bl for such an SBP SQ,y' 0 for the oval of B that it deter-
mines, and 0 for the oval in Ll determined by 0; 6 blocks of Bl are the 3 tangent 1
476
D. Hughes
blocks of 01 and the 3 blocks of L2 which are a-images of the 3 points of 01' Its remaining 16 blocks are 16 ovals of 8; i.e., 16 ovals of Ll and their in L2 • The automorphism group of Ll is transitive on its ovals, and
~-images
clearly the automorphism group of Ll extends to an automorphism group of Band hence to M (see the construction of Min (vi)), so we may take 01 to be an arbitrary oval of Ll . (Or note that every oval of Ll must be used.) We represent the points of 01 by Xl' X2 , X3 , and their respective tangent blocks by xl' x2' x3. Then the two points in xl and x2 are J , i = 1,2, and so 3i on; the two points which are not X's or J' S are W ' W2 . A precise representation l of Ll is given by a difference set: the points of Ll are the elements 0,1,2, ... , 10 of the additive group ell' and its blocks are the sets D + i, where D = {l, 3, 4, 5, 9 ).
= 0, X2 = 1, X3 7}, {W.} = {2,5}. 1
If we choose the oval Xl (9, 10},
{J
.} 31
= {4,
We give below all the ovals of Ll , according to types. Type TO
XXX
o1
Tl
XXJ
o14 o1 7
36
039
38
o3
068
1 9 10
347
046
098
078
o6
1 4 10
1 6 10
179
189
346
379
378
3 4 10
045
095
072
o 10 2
142
165
175
18 2
362
392
385
3 10 5
3
10
XJJ
T2
(T' ) X.J.J. 1 1 1 2 (T 2) X.J.J. 1 1 J
T3
XJW
T4
XWW
025
125
325
T5
JJJ
469
48 9
4 8 10
7 8 10
769
7 6 10
472
682
9102
475
685
9105
JJW
T6 (T' )
6
J.J.W 1 1
10
=
3, then
477
On the non-existence of a semi-symmetric 3.( 7B,k)..J
(T ") 6
J .J. W 1
J
485
762
8 10 2
492
7105
6 9 5
We let ni be the number of ovals of type Ti in B , and also nj, ni' for l = 2, 6, the number of ovals of the appropriate types Tj and Ti' so n = n + n ' 2 2 Z for instance.
PROOF: The oval of type TO cannot be used, since the 3 points Xi are each joined to the a-images of Xi in L2 by two blocks.
Similarly 2 points J
il
and J
i2
are
already on 2 blocks. 0
LEMMA 5: The n.1 satisfy: (1)
(2) (3) (4) (5) (6)
nl + n2 + n3 + n4 + n5 + n6 = 16 2n 1 + n + n3 + n4 = 12 2 nl + 2n 2 + n3 + 3n 5 + 2n 6 = 24 n3 + 2n 4 + n6 = 12 2n + 6n + 2n = 24 2 5 6 2n +n +n =12. 2 3 l
PROOF: (1) counts the total number of ovals used; (2), (3), (4) count respectively the number of ovals that must contain points of type X, J, W. Points of type i and J j , where i F j, must be joined one more time and this is counted in (5). Finally points of type X. and J., where i F j, must also be joined one more time
J
J
1
and this gives (6).
0
LEMMA 6: The equations of Lemma other n
i
=0
5 have the unique solution
(under the assumption n
PROOF: (6) and (2) give n4 = O. n6
i
~
n3
12. n5 =4 and all
0 for all i).
Then from (4) and (5) we have
= 12 - n3
n2 = n3 - 3n 5 · Then (1) gives n1 = 4 - n3 + 2n 5 and substituting in (3) we find n5 = 4, so n2 = n3 - 12. This, plus the equation for n6 above,yields n3 = 12 and all other n. must be zero. 1
0
478
D. Hughes
THEOREM 7: There does not exist an SS3D for A
PROOF :
2 which is a symmetric design for
Using the notation preceding the theorem, 4 and 6 must occur together
again in some oval, and this must be of type T ; so 4, 6, 9 must be used in 8 , 1 5 Similar considerations for 7 and 8 implies that the oval 7, 8, 10 is used. Again, considering the pairs 4, 10 and 7, 9, we find that the 4 ovals of type T5 are the 2 already given plus 4, 8, 10, and 7, 6, 9.
But this gives two ovals on 6, 9, and
since 6, 9 are also on the block x , we have a contradiction. 0 3 This in fact implies; COROLLARY 8: If V is a geometry for the Buekenhout diag1'0Jn
c 0 _ _0 _ _ 0 _ _ 0
then 0 is (a) the) tl'ivia: design 3-(5, 4, 2); (io)
the compLer",ent (i.e., blocks are point-compLements of hyperpLanes) of
th.e P'oJ ecti ve geome tr'y PG (3, 2); (a) the biaffi"Je geometry of dimension -1 over GF(2) (i.e., the 16 points of
PG( 4, 2) not un a f,:xed hyperplane H and the 16 hypeI'Planes not on a fixed point
(d.)
the "Higman-SirrJs geometry" of 100 points, 100 blocks, constructed in
[ 2];
(,,) o'!e of a J7:nite nwnber of possible SS3D's for A = 2, with 112 points on a block and 11;; blocks on a point, constructed from a projective pLane of order it)
and its extended Z-design.
The "finite" in (e) follows from a private cOl1111unication from Douglas leonard, who has shown that such a geometry has bounded diameter.
It might be
possible to show that (e) does not occur, independently of a possible determination of the existence or otherwise of a projective plane of order 10. The author would like to express his appreciation to A. Brouwer, who made some valuable observations that were crucial to the results in this paper.
On the non-existence of a semi-symmetric 3.(78,k)J
479
BIBLIOGRAPHY 1. 2. 3.
E.F. Assmuss Jr. and J.H. van Lint. Ovals in projective designs. J. Cambin. Th. (AJ, vol. 27 (1979). 307-324. D.R. Hughes. Semi-symmetric 3-designs. Proc. Pullman Symposium. April 1981 (to appear). P. Wild. On semibiplanes. Ph.D. Thesis. University of London. (1980). Westfield College (University of London) London NW3 7ST. U.K.
This Page Intentionally Left Blank
Annals of Discrete Mathematics 18 (1983) 481-492 © North-Holland Publishing Company
481
ON CERTAI N LI NEAR CONGRUENCE CLASS GEOMETRI ES Thomas Ihringer
SUMMARY A subset of the points of a finite affine desarguesian geometry, equipped with the induced subspace structure and the induced parallelism, ;s called a rep res e n tab 1 e geometry.
The question is answered as to which repre-
sentable geometries are congruence class geometries in the sense of Wille, i.e. allow "enough" dilatations to generate the induced subspace structure and the parallelism.
Each geometry induced on a subset of the points of an affine desar-
guesian geometry satisfies the Exchange Axiom for subspaces, and hence is linear, i.e. every line is determined uniquely by any two of its points.
Thus, using
results of Wille, Pasini and Herzer on finite linear congruence class geometries, one easily obtains an answer to the above question (Theorem 3.2.5).
Moreover, the
problem is solved as to which finite congruence class geometries have the same points and subspaces (but possibly a different parallelism) as a representable geometry (Theorem 3.2.6). Corollary 3.2.7 is a consequence of this theorem, stating that a finite congruence class geometry satisfies the Exchange Axiom, if all its subplanes do.
1. INTRODUCTION 1.1.
In [14] Wille introduced the concept of congruence class geometry in
order to relate algebraic and geometric theories: In accordance with Werner [ 13] a pair (A,n) is denoted an equivalenae alass geometry (henceforth simply a geometry), if A is a set and n : A x P(A)
satisfies: n(xI0) = {x}, n(xln(yIX)) ~ n(xIX), y E n(xlx,y),
~
P(A) a map, which, for all x,y
E
A, X~ A,
482
T.lhringer
n(xIX) = U{n(xIE) IE::: X finite}. Consider the sets (0):= 0 and [XI:= TI(xIX) for every nonempty set X::: A and some x E X as subspaces of the geometry.
[ J is a closure operator on A.
Each element
of A is called a POillt and, for all x E A, X::: A, rr{xIX) the subspace through x paraZZe[ to [Xl.
to be of
mnk
Let IXI denote the cardinality of X.
A subspace i X) is called
lXI, if [XI is not generated by less than IXI points.
A subspace of
rank 2 is called a line, a subspace of rank 3 a plane. A geometry
(A,~)
is called a congruence class geometry if there exists a
(possibly infinite) set F of finitary operations on A, defining a (universal) algebra (A,F) such that, for all x
E
A, X::: A, rr(xlX) is the congruence class
[xla(X) containing x of the smallest congruence relation (A,F) identifying all elements of X. Universal Algebra).
8
(X) of the algebra
(Cf. Gratzer [61 for the notations of
Thus every congruence class geometry arises from an algebra
(A,F) by simply taking the elements of A as the points of the geometry, the congruence classes as subspaces, and the parallelism map n as defined above. In a geometry (A,TI) a map f: A ~ A which, for all x,y fix)
E
r:(f(y)lx,y) is denoted a dilatation of (A,rr).
If
E
A, satisfies
one is interested in the
question as to whether a geometry (A,TI) can be regarded as a congruence class geometry with respect to some algebra, it is sufficient to examine the set of all nonconstant dilatations of (A,TI), since the unary admissible operations of an algebra, which completely determine the congruence relations of the algebra, are exactly those maps satisfying fix)
E
w(f(y)lx,y) for all x,y
E
A in the associated
congruence class geometry. 1.2.
In [14) Wille examines linear congruence class geometries: In accor-
dance with Libois
(10)
a geometry is called linear if every line is uniquely
determined by any two of its points or, more formally, if y always implies Xl
E
E
[xl,x l
and ytx2
[y,x21.
General Assumrtion: In order to eliminate trivial exceptional cases from the
succeeding results, every geometry is henceforth assumed to contain a plane. If a linear congruence class geometry has only finitely many points, then, by
Hilfssatz 4.3 in Wille [14), its nonconstant dilatations form a permutation
group on the set of points which is henceforth called the dilatation group. Essenti a lly by work of IIi 11 e, Pas i ni and Herzer (i n [ 141 , [111 and [81) the fin i t e linear congruence class geometries can be characterized algebraic1y,
On certain linear congruence class geometries
mainly by use of group theoretic methods.
483
Omitting all proofs, a summary of this
characterization is given in chapter 2. 1.3. The well-known equivalence between vector spaces and affine des arguesian geometries can, however, be formulated in terms of congruence class geometries: Let V be a vector space, x E V, x'
E
X ~ V, and n(xIX) the coset through
x of of the vector subspace generated by -x'+X. (V,n) is then denoted by A(V).
The affine desarguesian geometry
On the other hand, every affine desarguesian
geometry can be regarded as an A(V) with respect to a suitable vector space V, def; ned on the poi nts of the geometry.
(Cf. Dembowsk i [51, Pi ckert [121 and Wi 11 e
[141 for the geometric notations left undefined in this paper). For a geometry (B,n) and a subset A ~ B, the induced parallelism map nA on A is defined by nA(xIX) ~ n(xIX) n A for all x E A, X ~ A. (A,nA) is then denoted the induced geometry of (B,n) on A.
The induced subspaces of this geometry are
the sets [XI A g [XI n A for all X ~ A. A geometry (A,n) is called representable if, by an injective map from A into a vector space V, A can be identified with a subset of V such that (A,n) is the induced geometry of A(V) on A.
2. THE FINITE LINEAR CONGRUENCE CLASS GEor·1ETRIES The finite linear congruence class geometries can be characterized by an examination of their dilatation groups. the group theoretic notations).
(Cf. Hall, Jr. [7] and Huppert [9J for
Every dilatation of a linear congruence class
geometry is uniquely determined by the image of any two pOints. As a consequence, every nonconstant nonidentity dilatation leaves at most one point fixed (cf. Hilfssatz 4.6 in Wille [14]).
Fixed point free dilatations, and the identity, are
denoted translations, and nonconstant dilatations with exactly one fixed point proper dilatations.
If the dilatation group of a finite linear congruence class
geometry operates transitively on the set of points, then, by a well-known theorem of Frobenius (cf. Theorem 16.8.8 in Hall, Jr. [7]), the translation subgroup operates transitively on the set of points. Thus in this case the points of the geometry can be identified with the elements of the translation group, and the geometry can be regarded as a "translation structure" as defined by Andre in [21. Now consider a finite linear congruence class geometry with a transitive translation group and a proper dilatation.
If the translation group is abelian,
484
T.lhringer
then by PropoSition A of Pasini list.
(11),
the geometry is of type I in the following
If the translation group is not abelian, then, by use of Satz 4.2 in Andre
t2J and Theorema 3.2 in Biliotti [4], the geometry proves to be of type II.
If the translation group of a finite linear congruence class geometry is transitive, but the geometry has no proper dilatation, then it is of type III. If the dilatation group of a finite linear congruence class geometry does not operate transitively on the points of the geometry, then by Satz 4.11 of Wille 1141, every nonconstant dilatation is a translation, and the geometry is of type
IV.
There are the following types of finite linear congruence class geometries: l/pf3 I
consists of all finite affine desarguesian geometries.
I,iep II: In [8] Herzer proves that every finite nonabe1ian group G, equipped
with a nontrivial partition p and a p-admissible fixed-point-free automorphism ¢, is of nilpotence class 2 and of prime exponent, and he gives a method of ing all such groups. partition
P,
If G and ¢ have the above properties with respect to some
then consider G as the group of left translations on G: For every
g E G let g: G ~ G be the map with g(x):= g·x. of the algebra (G,G cosets of
construc~
U
~-admissib1e
{¢}) is linear.
subgroups of G.
Then the congruence class geometry
The subspaces of this geometry are the left In chapter 3 the fact will be used that
every such congruence class geometry contains a plane, which does not satisfy the Exchange Axiom. l'ype III:
prime order.
Let G be a finite group, in which every non-identity element is of
Consider again G as the group of left translations on G.
Then the
congruence class geometry of the algebra (G,G) is linear (and contains a plane, if G is noncyclic), because the subspaces of this geometry are the left cosets of subgroups of G.
By Theorem 3 of Pasini (11) there are three different types of
such groups: Type IlIa: G is a group of prime exponent for some prime p.
fupe fIfb: If in G exactly two different prime numbers occur as orders of
elements, then G can be regarded as a semi direct product of a finite group H of prime exponent p with the cyclic group F of order q, for two different primes p and q, with respect to a fixed-poi nt-free automorphism ¢ on H of order q: If F is the addition group modulo q on the integers O,l •••• ,q-l, the group operation on H x F is defined by (x,m)·(y,n):= (x'<j>JT1(y),m+n) for all X,jEH,m.nE{O,l, ... ,q-l}. On the other hand. every semidirect product with the above properties gives rise
On certain linear congruence class geometries
485
to a group of type IIIb. Type IIIc: If exactly three different prime numbers occur in G as orders of
elements. then G is an isomorphic copy of the alternating group A5 on 5 symbols. There is no finite group of type III with more than three different prime numbers occurring as orders of elements of the group. Type IV: Let G again be a (possibly cyclic) finite group. in which each non-
identity element is of prime order.
Consider. for some integer k ~ 2 (k
~
3. if G
is trivial). pairwise disjoint copies Gi of G. i = 1.2 •.••• k. Let G operate simultaneously as left translation group on each group Gi . The congruence class geometry of the algebra (·-1 2() k G1.• G) obtained in this way is linear. with G 1- , , •.• , as dilatation group. and consists of k sets of transitivity under G.
3. THE REPRESENTABILITY PROBLEr1 3.1. A geometry is said to satisfy the Exchange Axiom. if y E [XI and y Ej: [X \ {x}) always implies x E r (X \ {x}) u {y}). As every affine desarguesian geometry satisfies the Exchange Axiom. the following lemma is valid. LEMt4A 3.1.1: If a geometry is representaNe oVe:!' an affine desarguesian geometry, then it satisfies the Exchange Axiom.
PROOF: Let I.
(V.~)
be an affine desarguesian geometry with subspace closure operator
Let A be a subset of V. equipped with the induced subspace closure operator
I A and parallelism map
and x.y EA. X:: A such that y E [XI A and y 1 [X \ {xll A• [XI A g [XI nA and [X \ {x}I A = [X \ {xlI n A implies y E [XI and y 1 rx \ {x}). whence. by the Exchange Axiom in (V.~). x E [(X \ {x}) u {y}] n A = = [ (X \ {x}) u {y}] A•
~A'
o
By the Lemma. every representable finite congruence class geometry trivially occurs among those geometries of the list in chapter 2 which satisfy the Exchange Axiom in particular for sets X with
Ixi = 3. i.e. for all their subplanes. There-
fore in the rest of this section those geometries are selected from the list in chapter 2 which satisfy this condition. The geometries of type I trivially have this property. but the geometries of type II have not. as already mentioned in chapter 2.
If a geometry is of type
486
T.lhringer
IlIa. and the group G is abelian and of exponent P. then the congruence relations of the algebra (G.G) are identical to those of a vector space over GF(p) having G as additive group tcf. Huppert [9]. p. 80. or Wille (14). p. 36).
Therefore the
geometry ;s affine and desarguesian. and thus was already listed under type I.
If
G is not abelian. then the geometry contains a subplane which does not satisfy the Exchange Axiom: Consider two noncommuting elements x and y of G. and denote the identi ty of G bye.
Then the plane [e.x,y] is just the subgroup <x,y> of G
generated by x and y.
As<x.y>is a finite group of prime exponent. by Hauptsatz
2.3 of Huppert [9J it contains a central element z
ing x·z=z·x and y·z=z·y. z
~
[e.xJ.
Thus z
E
[e.x.YJ. z
~
~
e. i.e. an element z satisfy-
[e.xJ or z
~
[e.yJ. say w.l.o.g.
In contradiction to the Exchange Axiom for subplanes y 1! e.x.zl.
because [e.x.zJ is an abelian subgroup of G. namely the group generated by x and z. Now assume for the rest of this section that every geometry satisfies the Exchange Axiom for all its subplanes. If the geometry is of type IIIb. then. by the same argument as for type IlIa. the normal subgroup H of G is abelian.
If the identity of H is denoted bye.
then. for every x E H. (4)(x) .0) = (e.l) • (x.O) • (e.l)
-1
• Therefore the element
(}(x).O) is always contained in the subgroup generated by (x.O) and (e.l). hence in the subspace [(e.O).(x,O),(e.l)J. Exchange Axiom yields (1)(x).O)
E
As (e,l)
[(e.O),(x.O)].
~
[(e.O).(x.O).(4>(x).O)l. the Thus. for every x
E
H. 4>(x) is
contained in the subgroup of H generated by x. i.e. 4>(x) = xn for some n
E
{1.2 •.•.• p-l 1.
As
4>
is a fixed-point-free automorphism of order q. it is an
easy calculation that n does not depend on x. if x
e.
~
Thus. up to this point.
exactly those groups of type IIIb remain. where H is abelian. and 4> is of the form x - xn for some (fixed) n E {2.3 •..•• p-l} satisfying nq=l (mod p).
These pro-
perties are sufficient to yield the Exchange Axiom for all subplanes of the geometry. because. by the proof of the forthcoming Theorem 3.2.5. every such geometry is representable. The geometry of type IIIc contains a subplane which does not satisfy the Exchange Axiom: Consider the permutations x = (123). y = (12)(34). z = (13)(24) and the identity id on the set n.2.3,4.5}. x
~
Then y
E
[id.x.zJ and y 1 [id.z]. but
[id.y,zJ, contradicting the Exchange Axiom.
If the geometry is of type IV. then the subspace Gl of the geometry is isomorphic to the congruence class geometry of the algebra (G.G). All subplanes of this subspace satisfy the Exchange Axiom. and G thus is type IlIa or IIIb and has
On certain linear congruence class geometries
the properties described for these groups in this section.
487
These properties for G
are again sufficient to yield the Exchange Axiom for all subplanes of the geometry: The only nontrivial case of the Exchange Axiom for
IXI
~
3 is when
IXI = 3,
E G , xl ,x E G , i ~ j, Y E [x,x ,x ] and y E/: [xl ,x ], i.e. j l 2 2 2 l 2 i y E x·U., where U. is the copy in G. of the subgroup U. of G. generated by -1 1 .. 1 .1 J J Xl ·x 2• TrlV1ally then x E y.U , 1.e. x E [y,x l ,x ]. i 2 Thus the characterization of those finite congruence class geometries which
X = [x,x ,x ] , x
satisfy the Exchange Axiom for all their subplanes is complete. 3.2.
For each of the geometries described in section 3.1 either a represen-
tation is given (Theorem 3.2.5), or a representable geometry is found, having the same set of points and the same subspaces, but a different parallelism map (Theorem 3.2.6).
To this end some preliminaries are necessary.
The following lemma is essentially Theorem 11.2.3 in Artin [3]. LEMt·1A 3.2.1: In a representable geometry every dilatation is uniquely determined by the image of two distinct points.
PROOF: Let f be a dilatation of the representable geometry (A,rr), x,y,z E A, x
~
y and z E/:[x,y].
Such x,y,z exist by the General Assumption of n. 1.2.
f(x} and fey) are given, then fez)
E
If
rr(f(x}lx,z} n rr(f(y)ly,z), which is the
intersection of two subspaces, each of them being a point or a line, because the geometry is representable.
Thus fez) is uniquely determined, except in the case
that rr(f(x}lx,z} and rr(f(y)ly,z) are both lines and are equal. [x,z]
But then
= [y,z], whence y E [x,z], and thus z E [x,y] by the linearity of the
geometry, contradicting the choice of z.
If w is an arbitrary element of A, then
there are two elements in {x,y,z} which generate a line not containing w. fore, by the same argument as for fez), few) is uniquely determined. LEt4t·1A 3.2.2: Let V be a vector space of dimension;;' 2 over the field F.
There-
0
Then
every dilatation of the associated affine des argues ian geometry A(V) is of the form x .... rx+v for some
V E
V and rEF.
PROOF: Every map of the form x .... rx+v clearly is a dilatation of the geometry. Now let f be a dilatation, and y, z
E
V such that fey) E/: [y,z}.
If
488
f(z)
T. lhringer
E
TI(z!y,f(y», then the map 9 with x
~
rx+v, r=l and v=f(y)-y, satisfies
f(y)=g(y) and f(z)=g(z), thus by Lemma 3.2.1 f=g.
If f(z)
~
TI(ziy,f(y», then the
map 9 with x ~ rx+(l-r)c satisfies f(y)=g(y) and f(z)=g(z), if c
E
I y,f(y)l n I z,f(z)] and r is such that r(y-c)=f(y)-c.
3.2.1 f=g, and the proof is complete.
Therefore by Lemma
0
The next two lemmas are of central importance for the results of this chapter. LEt~r,lA
3.2.3: If V id a vector space, and (A,TI) is a geometry which is represent-
,_:':,Zd over A(V), tiler, a map f: A ~ A is a dilatation of (A,TI) iff it is the !'(:sL!'ictior, of a dnutation of A(V) to A.
PROOF: Because of Lemma 3.2.1 a dilatation of (A,n) is uniquely determined by the image of two distinct points y and z of A.
Using Lemma 3.2.2 one can easily prove
that there is a dilatation g on A(V) satisfying g(y)=f(y) and g(z)=f(z). proof of Lemma 3.2.1 then yields that f is the restriction of 9 to A. direction of the lemma is trivial.
The
The lIifll_
0
If f is a dilatation of the geometry (A,TI), and x E A such that f(x) # x, then the 1i ne [ x, f (x)] is ca 11 ed a trace of f. LEf.1I-1A 3.2.4: Ler; V Ie a vector space and f a dilatation of a geometry which is representable over A(V).
Then f is the restriction of a translation in V, i.e. of
a map cf r;he form x .... x+a, iff f has wo distinct parallel traces.
In this case
').;;: :.mces of f QJ:'e paralleL
PROOF: The assertion of the lemma is an immediate consequence of the and 3.2.3.
Lemma~
3.2.2
0
Now the main result of this paper can be proved. THEOREI1 3.2.5:
.4 finite congruence class geometry which satiSfies the Exchange
Axiol"! foT' every subpZane is representable over an affine desarguesian geometry iff ':tu dilar;aLion grouF is either cyclic or transitive on the points of the geometry.
PROOF: One has to check which geometries of the list in section 3.1 are representable.
On certain linear congruence class geometries
489
The theorem holds trivially for geometries of type I and type IlIa.
Let the
geometry be of type IIIb, and the group H have pm elements for some positive integer m.
Consider a vector space V of dimension m+l over GF(p), a vector sub-
space U ~ V of dimension m, and a vector
WE
V\ U.
H can then be regarded as the
vector subspace U, equipped with the vector addition as group multiplication. Since the group multiplication in H is written additively now, the group automorphism cp of H maps u to nu. Consider the subset A:= {u+n i wlu E U,i=O,1,2, ... ,q-l} of V, and identify the element (u,i) of H x F with the element u+niw
E
A.
The
congruence class geometry under consideration may now be denoted by (A,TI).
It is
sufficient to show that the geometry (A,TI) and the induced geometry (A,TI A) of A(V) on A are equal. The left translation (u,i) of H x F corresponds to the map f
.: A ~ A with v+njw ~ (niv+u)+ni+jw, which is the restriction of the dilatation
U,l
.
of A(V) mapping x to n1x+u.
Thus by Lemma 3.2.3 every dilatation of (A,TI) is a
dilatation of (A,TI A), whence, for every x E A, X ~ A, TI(xIX) ~ TIA(xIX), because (A,TI) is a congruence class geometry. If [ 1, [ 1A denotes the subspace closure operator of (A,TI), (A,TI A) respectively, then, by transitivity of the dilatations of each geometry, ITI(xIX) I = I[ xli, ITIA(xIX) I = I[ X1AI, and [Xl ~ [X1 A• Let {x l .x 2' ••• ,x t } be a minimal subset of A such that [xl.x2'···.x~ A = [Xl A• then the intersection of a subspace of rank t of A(V) with A, whence t-l t-2 . I[XlAI = p or q'P • For each 1 E {l,2, .... t-l}, xi+l Ej: [X l ,x 2' .... xf cause otherwise xi+l
E
[xl,x2, ... ,xil
~
[~A
'
is
be-
[xl,x2, ... ,xi1A' contradicting the choice
of {x l ,x 2 '· .. ,x t }· Therefore 0 J [xl1 J [x l , x21 J ... J [x l ,x 2, .. ·, xt 1. Every nonempty subspace of (A,TI) obviously has pr or q'pr elements for some nonnegative integer r. Using induction over t, it is now easy to prove that I[X l ,x 2, .•• ,X l I t :;;, t-2 t-l . ~ q.p or p ,whence [X l .x 2, ... ,xt l [xl,x2, ... ,xtlA only lf t-2 t-l . . I[X l ,x 2, ... ,x t ll = q'P and I[X l ,x 2, ... ,x t 1AI = p • ThlS would lmply qlp, contradicting nq=l (mod p): As n can be regarded as primitive q-th root of unity
J
in the field GF(p) of integers modulo p, qlp-l.
Thus TI(xIX) =TIA(xIX), which
proves the theorem for geometries of type IIIb.
(Cf. Theorem 5 in Pasini [111 to
this part of the prOOf). Now let the geometry be of type IV, and G be cyclic of prime order p. Consider a prime number q such that plq-l, a vector space V of dimension k over GF(q), a basis
el~e2,
••• ,ek of V, a primitive p-th root of unity r of GF(q) and
the subset A = {rJeili=1,2, ... ,k; j=O.1.2 ..... p-l} of V.
Then the geometry (A.TI A) is a representation of the congruence class geometry of the algebra
490
T.lhringer
('-1 2(1 1- "
••• ,
k G.,G): Choose an element g. of each set G., which generates G., 1
.
.
1
identify the element g.J with rJ e . 1
1
E
1
1
A, and the dilatation group G with the
dilatation subgroup
,
restricted to A.
,
Therefore the congruence class geometry of ('--1 "2().. , k G.• G) may now denoted by (A,n). and for every x E A, X ~ A 1T(xIX) ~ 1T A(xIX). Let Ai for i=l,Z, ••• ,k be the subset of A corresponding to G.: A. = {rje. Ij=O,l,2, ••• ,p-l 1 1
1
1
For each X ~ A there are subsets Kl and KZ of {1,2 •••. ,k} such that IX n Ai I = 1 iff i E Kl and ix n \ I ;;. 2 iff i E KZ' Let x E Aj • If j E K \ (Kl U K2), and U is the vector subspace of V generated by {e Ii E Kl U K }, then i 2 TIA(xiX) ~ (x+U) n A = (xl, proving n(xIX) = nA(xIX). If j E Kl U K2 and KZ = 0, xi
E
X n Ai for
{x;-\,Ii,i'
E
Kl , and U is the vector subspace of V genera ted by Kl I, then 1T A(xIX) = (x+U) n A, x+U is of rank IK11 in A(V), and ever~1
i
E
contains one element of each Ai' i E Kl • Thus x+U cannot contain an element of A., i ~ Kl , or two different elements of one set A., because then x+U would be of 1 1 rank> IKll.
Therefore I11A(x!X)1 = !K l !. ,(xiX) = TIA(xIX).
Clearly, In(x!X)1 = IK1I, and thus
KZ and KZ f 0, and U is the vector subspace of V generated by lei Ii E Kl U K21, then x E U and nA(x!X) = UnA = i E K?U K Ai' which is Z trivially equal to ,(xix). If
j
E
Kl
U
Now let the geometry be of type IV, and G consist of one element.
Then the
geometry is discrete, i.e. n(X!X) = {xl and [Xl = X for every element x and subset X of . --1 1
"
z()
•• ,
kG .• and it can be represented for example on a bas i s of the vector 1
space of dimension k over an arbitary finite field, as is easily proved. Let the geometry be of type IV and the dilatation group be noncyclic and of type IlIa.
G then is elementary abelian and contains at least p2 elements.
Thus
there are elements x,y of G and g E G such that g·x ~ [x,yl (recall that each Gi l is an isomorphic copy of G on which the elements of G operate as left translation~. Then goy E n(ylx,g'x), because y=(y.x
-1
).x, g.y=(y.x
to be represented over A(V) for some vector space V.
-1
)·g·x.
Assume the geometry
By Lemma 3.2.4 g is then the
restriction of a translation in A(V), and thUS, by the same lemma, g,z2 E n(zZlzl,g'Zl) for elements zl E Gl ,z2 E G2• This contradicts 11(z2Iz1,g-Zl) = (z2 1• If the group is of type Illb, and the geometry is assumed to be represented over A(V) fora vector space V, then the element (x,D) E H
x
F, for some xE H\ {el,
491
On certain linear congruence class geometries
has two parallel traces: (x.l)=(x,O)·(e,l)
E
1T((e.l)l(e.O).(x.O)·(e.O)). if all
elements occuring are again simultaneously regarded as elements of G and Gl . This leads to a contradiction by the same argument as was used for groups of type IlIa.
0
THEOREM 3.2.6: For every finite congruence cZass geometry satisfying the Exchange Axiom for each subpZane, there exists a geometry having the same points and subspaces which is representabZe over an affine desarguesian geometry.
PROOF: This theorem is a consequence of Theorem 3.2.5. except for geometries of type IV. having a noncyclic dilatation group.
Consider such a geometry equipped
with a dilatation group G of type IlIa. such that G has pm elements for some integer m ~ 2. and a vector space V of dimension k+m over GF(p). a basis e l .e 2•.•. ,e k+m of V. the vector subspace U of V generated by {ek+l.ek+2 •...• ek+m}. and the subset A 11:1 1.-1 , 2U k (e.+U) of V. The group G can be regarded as the , ••• , , additive group on U. and an element of Gi • which is a copy of x E G. is identified with the element e.+x E A. The congruence class geometry of the algebra 1
(.--1 2U k G.• G) can thus be regarded as a geometry (A.1T) on A. and. by similar 1 " ••• , 1 arguments as were used in the proof of Theorem 3.2.5. it can be shown that the subspace closure operators [ I of (A.1T) and [ I A of the induced geometry (A.1T A) of A(V) are equal: Every dilatation of (A.1T) is a dilatation of (A.1T A). therefore [XI
~
I[XII
[XI A for every set X ~ A. 11:1
By elementary considerations one obtains
I[ XIAI, thus proving [XI = [XI A• The details of this part of the proof
are omitted, because they are easy but rather technical. Now let the geometry be of type IV and the dilatation group G of type IIIb such that the group H has pm elements for some positive integer m.
Consider a
vector space V of dimension k+m over GF(p). a basis e l .e 2•...• e k+m of V. the vector subspace U ~f V generated by {ek+l.ek+2 •..•• ek+m} and the subset k (nJe.+U). A = 1.-1 , 2U , ••• , 1
The group H can then be regarded as the additive group
on U. and an element of G.1 which is a copy of (x.j) . J
the element n ei+x.
E
G = H x F is identified with
The congruence class geometry of the algebra (·=1 2U 1
"
••• ,
kG .• G)
can thus be regarded as a geometry (A.1T) on A. and again. by the methods used above. it can be proved that (A.1T) and the induced geometry (A.1T A) of A(V) have the same subspaces. [] The following corollary could. of course. have been proved directly.
,
T.lhringer
492
COROLLARY 3.2.7: A f'inite congl'uence class geometl'Y satisfies the Exchange Axiom iff
aLZ subpZanes 0;' the geometl'Y do.
PROOF: Immediate from Theorem 3.2.6, Lemma 3.1.1 and the fact that the Exchange Axiom only involves subspaces, but not the parallelism.
0
BIBLIOGRAPHY 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
J. Andre, Uber Parallelstrukturen. Tei1 I: Grundbegriffe, Math. z., 76 (1961), 85-102. J. Andre, Uber Para11e1strukturen. Tei1 II: Translationsstrukturen, Math. Z., 76 (1961), 155-163. E. Artin, Geometric Algebm, Interscience Publishers, Inc., New York, 1957. 1·1. Bil i otti, Strutture di Andre ed S-Spazi con tras 1azi oni. Geom. Dedic. (to appear). P. Dembowski, Finite Geometl'ies. Springer-Verlag, Ber1in-Heide1berg-New York, 1968. G. Gratzer, Univel'sal AZgebl'a. D. van Nostrand, Princeton, N.J., 1968. M. Hall, Jr., The Theory of Gl'OUps. MacMillan, New York, tenth printing, 1968. A Herzer, End1iche nichtkommutative Gruppen mit Partition TI und fixpunktfreien 11 -Automorphi smus, Arch. Math., 34 (1980), 385-392. B. Huppert, Endliche Gruppen I. Springer-Verlag, Ber1in-Heide1berg-New York, 1967. P. Libois, Quelques espaces 1ineaires, Bull. Soc. Math. BeZg., 16 (1964), 13-22. A. Pasini, On the Finite Transitive Incidence Algebras, Boll. Un. Mat. Ital., (5) 17-B (1980), 373-389. G. Pickert, Pioojektive Ebenen. Springer-Verlag, Ber1in-Gottingen-Heide1berg, 1955. H. Werner, Produkte von Kongruenzk1assengeometrien universel1er A1gebren, Math. ;:-., 121 (1971), 111-140. R. Wille, Kongl'uenzklassengeometY'ien. Lecture Notes in l-lathematics 113, Springer-Verlag, Berlin-Heidelberg-New York, 1970.
Technische Hochschule FB Mathematik, AG 1 Schloagartenstr. 7 0-6100 Darmstadt Federal Republic of Germany
Annals of Discrete Mathematics 18 (1983) 493-510 © North-Holland Publishing Company
493
TRANSLATION PLANES OF CHARACTERISTIC P THAT ADMIT SL(2,pr) N.L. Johnson
1. INTRODUCTION One of the most basic problems in the study of translation planes (characteristic p) is to determine the possible collineation groups generated by collineations of order p. In this article, we explore the situations where the p-elements generate SL(2,pr) and conversely investigate the permutation group action of a collineation group isomorphic to SL(2,pr) acting on the points and lines of a translation plane.
We are also concerned with the combinatorial nature of the spaces fixed
pointwise by the Sylow p-subgroups. 2. THE KNOWN TRANSLATION PLANES ADMITTING SL(2,pr) The Desarguesian plane terized by such a group.
~
of order n admits SL(2,n) and conversely is charac-
One of the first classification results is:
2.1 LUneburg [32]: Let E denote a projective plane of order pe which admits a collineation group isomorphic to SL(2,pe). ~
is determined.
Then ~ is Desarguesian and the orbit structure of
In particular, if the group acts on an affine plane then the p-
elements are elations. The question of whether affine elations generate SL(2,pr) (in a translation plane) has been studied by Ostrom in [33) , [34) and by Hering [15) . 2.2 Hering-Ostrom: Let
~
be a translation plane of characteristic p and let E denote the
collineation group in the-translation complement generated by the affine elations of
~.
Then E is either
494
N.L. Johnson
(1) elementary abelian,
(2) SL(2,pr), (3) SL(Z,5) and p = 3, r
(4) SZ(Z ) and p = Z, or
(5) E contains a normal subgroup N of odd order and index 2. By Hering [161, if the group E is Sz(Zr) there is an invariant LUneburg-Tits Zr subplane of order Z If the group E is SL(Z,pr) there is an invariant Desarguesian subplane of order pr by Ostrom 1331 . Foulser [71 has shown that for p > Z the results of Hering-Ostrom remain valid for collineation groups generated by Baer p-elements (the group element fixes a Baer subplane pointwise) .
.?-)
Foulser ( 71:
Let
TI
be a translation plane of characteristic p.
Let B denote the collin-
eation group in the translation complement generated by all Baer p-elements.
If
p > 3 then the Baer subplanes pointwise fixed by p-elements are disjoint (as subspaces).
Furthermore, B is either elementary abelian or SL(Z,pr).
Moreover, the
Baer subplanes pOintwise fixed share the same infinite points. 2.4 The Hall Plane (of order q2 admits SL(2,q) where the p-elements (pr=q) are Baer): Z
The Desarguesian plane n of order q admits SL(2,q) and the Sylow 2-subgroups define a derivable net N in n.
The p-elements in n act as elation groups. Z
Deriving N produces the Hall plane of order q where the p-elements now fix Baer subplanes pointwise . .Z.5 The Hering Planes (( 13J , [ 141) (order q2 and admit SL(2,q)): Let P = PG(3,q), q odd> 3, and let C denote a twisted cubic in P (set of qtl points in P such that no four points are coplanar, Segre [38]). tangents T to C forms a partial spread.
Now further assume that q =
The set of mod 3 and
let S denote the subgroup isomorphic to SL(Z,q) of GL(V ) which leaves the 4 twisted cubic invariant. Let g be an element of order 3 in S acting on P.
Then
g fixes exactly two lines of P (Z-subspaces of V ) and fixes one of them pointwise. 4 1 In SL(2,q) there are zq(q-l) subgroups of order 3 and each such subgroup fixes
Translation planes that admit SL(2,pr)
two distinct lines in P.
495
The set of q+l tangents to C together with these two
sets of ~(q-l) lines form a spread (see e.g. LUneburg [31) (45.1)).
Moreover, S
leaves invariant precisely one spread of P and acts irreducibly on V4. 2.6 The Ott-Schaeffer Pl anes ([ 37] , [39)) (order q2 and admit SL (2 ,q) ): Bya theorem of Segre [38) any twisted cubic C in P may be brought into the canonical form 3
2
2 3
«s ,s t,st ,t ) is,t
E
GF(q), (s,t) f (0,0»
for q odd> 3. By using a similar curve C(a)
=<ssa,tsa,sta,tt~ for q even where a is an
automorphism which generates the automorphism group of GF(q), planes similar to the Hering planes may be constructed.
Let S2 be a Sylow 2-subgroup of the group S
isomorphic to SL(2,q) in GL(V 4 ) fixing C(a). Then S2 fixes a point P of C(a) and fixes each line of a plane in P incident with P. Moreover, the elements of S2 flare elations with axis thru P.
The two remaining lines incident with Pare
in distinct S orbits and each orbit defines a regulus in P (derivable net in V4). The trick of the 3-elements noted in the discussion of the Hering planes gives the -1
It turns out that a,a give rise to the same curve C(a) so there are ~(r) if q = 2r such curves and two remaining q(q-l) lines of a spread for either regulus.
planes (one the derived plane of the other) for each curve. Ott-Schaeffer planes. automorphism).
Thus, there are
~(r)
(The Ott planes [37) actually correspond to the Frobenius
Here, as in the Hering case, the group S
~
SL(2,q) acts irreduc-
ibly on V4 . The Ott-Schaeffer planes may be characterized in the following way: THEOREM 2.7 (Johnson [19]): Let
n
be a translation plane of even order 22r that
admits a collineation group S in the translation complement isomorphic to
r SL(2,2 ).
Assume the Sylow 2-subgroups of S fix Baer sub lines pointwise (2
fine points on a line of n).
Then n is an Ott-Scaheffer plane.
The proof of (2.7) relies on the group orbit structure established by Johnson and Os trom in [27] .
r af-
496
N.L. Johnson
THEOREM 2.8 (Johnson-Ostrom): Under> the hypothesis of (2.?J the points at infinity of
1
1!
1
aT'e hi exact Zy ;;hr>ee orbi ts of r>espective Zengths, 1+q, ~ (q- 1), ~ (q-l ) .
Once one knows where to look, it is not hard to show that the orbit of length q+l defines a derivable net V in
1!
(note that we are not assuming anything
about the dimension of the vector space). V is derivable and S leaves V invariant so by Foulser we can imbed S in
GL(4,q). Now we establish that S must act irreducibly on V . This tells us (by 4 Brauer and Nesbitt [21)) the form S can take on V and using this we can work out 4 a form for components invariant under certain key collineations. The components are then seen to be subs paces over V which implies 4 (see section 3).
1!
is an Ott-Schaeffer plane
2.9 The Walker Planes of Order 25 [41) (admits SL(2,5»: 2
Let" be a translation plane of order q and kern
~
GF(q) that admitsSL(2,q)
Assume S acts reducibly but indecomposably on V4 . 5 (Walker [41]) and there is a canonical form for S (see e.g. LUneburg
in its translation complement. Then
q =
S may be represented as generated by
[31] 46.12, p. 240). 0 s 3s s
0
2 3
s 3s
2
s
0 -1 lO '"' c· :
([ '-'y s
Let w,
~l
0
as ~ . normalizes
0
0
0
0
0
-1
0-1
:J
-1
s
~,p
to be a component of
Then
~
fixes exactly six components of
permutes these six components. 1!
so that
p
fixes (X
1!
and
The form for S forces X = 0
= 0) and permutes the remaining five.
It turns out there is exactly one translation plane of each of the three cases: fixes 1, 3 or 5 of these components.
p
One of these three planes turns out also to
admit a group acting irreducibly on V which forces it to be the Hering plane of 4 order 25.
Translation planes that admit SL(2.pr)
497
2.10 The Dempw01ff Plane of Order 16 {3] (admits SL(2,4)): Let n be a translation plane of order 16 and let X = 0, Y = 0, y = x, y = xT be the components where T is a fixed point free element of GF(4,2). GL(4,2) ~ A8 and using ~ one can determine the conjugate forms for fixed point free elements in GL(4,2).
To construct a translation plane of order 16 one needs l a set Mof 15 such elements such that if a,S E M then as- is fixed point free. In {3] Dempwolff determines (originally with the aid of a computer program) a number of such sets M.
One of these sets is left invariant by GL(2,4) which
implies GL(2,4) acts as a collineation group of the corresponding plane fixing X = 0 and Y =
o.
This plane may actually be obtained by derivation of the semifield plane of order 16 with kern GF(2) (see Johnson [22]). .
4
2.11 The Foulser-Ostrom Planes of Order u (admitting SL(2,v), GL(v)
~
GL(u)):
There is a nice way of obtaining these planes using PlUcker coordinates, but the following description is more succinct. Let M(q) denote the Miquelian-Mobius plane of odd order q.
Then if P,Q are
pOints of a circle C there is a unique circle D F C orthogonal to C such that Dnc={p,Q}. Let Pq = PG(1,q) (C - PG(1,q2) - = M(q)). Let A = {circles N E M(q)/N is orthogonal to P and N n P = {Q,Q Vq} for Q E P - PI.-(P r = PG(l,..,.iq))}. Then the q q q vqyq circles of A are disjoint and permuted transitively by PGL(2,v'q) (Ostrom l36] ). The circles correspond to derivable nets in V4(q2) and by replacing each of these derivable nets by its unique replacement net, the constructed plane of order q2 will admit SL(2,)q) as a collineation group where the p-elements (pe=q) are elations. Let B = {circles N E M(q)/ N is orthogonal to Pq and N n P = {S,T}, SF T in PG(l,y'q)}. Clearly, PGL(2,jq} permutes the circles of B. q
By selecting sets of disjoint circles in AU B which are orbits under PSL(2,v) for GF(v)
~
GF(vq), various translation planes can be constructed admitting SL(2,v) as
a collineation group where the p-elements are elations (see Foulser [81]). For example, planes of order u8 admitting SL(2,u) may be constructed which may not be obtained simply by replacing a set of disjoint derivable nets corresponding to circles in A.
498
N.L. Johnson
2 Similarly for q even, planes of order u4 admitting SL(2,v) for v f u, GF(v)
~
GF(u) may be constructed.
Note in all of these planes the order is u4 , the group is SL(2,v) for some GF(v)
~
GF(u), and the kern
~
2
GF(u ).
Now let nO denote the plane obtained by replacing all the nets corresponding to circles in A (so has order u4 , admits SL(2,u), has kern GF(u 2 ) and there are
u(u;~ replaced derivable nets). There is in "0 a derivable D that shares u+l infinite points with PG(1,u 2 ) and two infinite points with each of the U(~-l) replaced derivable A-nets.
Actually these points are conjugate under x-->x
that GL(2,u) leaves D invariant.
u
so 4
By replacing D, a translation plane of order u
admitting GL(2.u) may be constructed where the p-elements (u
= pi) in the group
are Saer collineations and the kern is GF(u) as it turns out the Baer subplanes of Dare GF(u) but
no~
2
GF(u )-subspaces (see Foulser [9)).
2r 2.12 The Johnson-Ostrom Planes of Order 3 Admitting SL(2,3) (Bruck type {l)
planes):
These planes are of particular interest as apparently they are the only known translation planes that admit both affine homologies and elations with the same center.
The planes were obtained by trying to derive generalized Andre planes to produce planes with affine central collineations. THEOREM 2.13 (Johnson-Ostrom [28]): Let
n
be a Desarguesian affine plane coordi-
'uzr-;:zeG by GF(q2), q ;;. 25 and let; be a generalized Andre plane constructed by r'cplacement of disjoint nets Nl ,N , ... ,N in " such that the slopes in all of 2 k til£se are non-sqUal'ec. :'et NO be the net in ,,(or ;) denoted by {mim E GF(q) L<:t if f:Jc' the plane obtained from; by ~eriv~ng NO'
or m= w}. (1) "
adrzits a Desarguesian decomposition
!Nli > (u+l)(q+l). o
Hel'e Nl
="
NO
U
Nl
U
N2
Then U ••• U
Nk such that
n ii.
0
(2) If NO and Nl are invariant under the full collineation group of ii and rr '--<~ ncr-
a Hall plane, then nontrivial affine central. collineations will have order
2. o
0
(n
If k > 2 in (]) then both NO and Nl are invariant.
(1)
if k = 2 and
Nl
is not invariant then
NO
is invariant.
o
':i
If NO is not invariant then k = 1 or' 2 and ii is an Andre pl.ane.
If
Translation planes that admit SL(2,pr)
k
= 2,
499
N2 consists of one or two disjoint derivabZe nets.
So if we are to obtain planes admitting central collineations of orders
F2
in this manner then we are forced to replace exactly three derivable nets in TI. Let 0 and -6 be nonsquares in GL(2,q)(so q=l mod 4). Let Ny be the net inTI q 1 whose slopes m are such that m + = y. Let TI = NO U NUN U M. Replace 6
-6
(derive) NO' No' and N_ 6 to obtain TI = NO U No U N_6 U M. Then; admits a group isomorphic to PGL(2,3) on the points at infinity of M (see Johnson-Ostrom (28] (2.6)).
And, if q
=0 mod
3, ; admits exactly four centers of affine elations of
order 3 and each elation center is also the center of an affine homology of order 2 (see also Johnson [20] for the connection with Bruck's classification). 2.14 The Kantor Planes (order q3, q
= 22e+l admitting SL(2,q), kern
GF(q),
the involutions are elations): Let V be a vector space of dimension 4m over K ~ GF(q), q even.
Let Q be a
quadratic form on V and assume there exist totally singular 2m dimensional subspaces.
That is, let (V,f) be a n+(4m,q) space (see Kantor [29]).
Let L be a spread of n+(4m,q) by totally singular 2m dimensional subs paces (that is, a partition of the associated quadric in PG(4m-l,q)). 2m-l of q +1 2m-spaces. Let y be a nonsingular vector, then /
A spread consists
n L/y ={< y,/ n F> /yl FE L} is a
spread of the 4m-2 dimensional space yl/y by totally isotropic (2m-l)-spaces (constructed by sections of the quadric by hyperplanes). So, this produces a translation plane of order q
2m-l
with kernel
~
GF(q).
In [29) Kantor defines several new n+(4m,q) spaces and finds spreads for certain of them. GU(2,q)
One of the spreads for n+{8,q) is found using
SL(2,q) X Zq+ 1 (see [29), section 6). By the choice of the "slice" y, SL(2,q) can act on the corresponding translation plane of order q3. The construc:=
tion of n+(8,q) requires q
=2 mod
3, so q
= 22e+l.
2.15 The Dye-Kantor Plane of Order 83 that Admits SL(2,8): Dye [5), section 4, constructs a spread L of a n+(8,8) space which is preserved by the group A9 . A particular slice of L is left invariant under SL(2,8) (~A9). The Dye and Kantor planes of order 8 are nonisomorphic (see Kantor [29), section 9}.
500
N.L. Johnson
2.16 The Liebler Planes (even order q3 admitting SL(2,q)): The Kantor planes mentioned above of order q3 have a S(=6L(2,q))-invariant Desarguesian subplane TIO'
Furthermore, there is a collineation group of order
q+l fixing TIO pOintwise and there are two S-orbits on too of, respectively, lengths 3
q+l and q -q. Liebler [30) uses his theory of enveloping algebras to construct planes of even order q3 possessing the above properties of the Kantor planes.
It is not
entirely clear whether the set of Liebler's planes contains those of Kantor. Liebler assumes that S is a partial spread associated with a subplane of order q and assumes the enveloping algebra £ K ~ GF(q).
Further, he assumes H ~
= £(S)
Mat ,2(K) $ Matz,z(K), 2 GL(2,q) acts on V so that the Sylow 2-sub~
groups determine S (of degree q+l) and H/Z(H) acts regularly on a set R of 3 spaces.
V = X s Y where X is the irreducible £-module consisting of points of
the subplane of order q and Y is a sum of two isomorphic irreducible £-modules. There are q+l isomorphic irreducible £-modules in Y which define a regulus M.
Q denote the locus of points on lines joining X to M (points in PG(V)).
Let
Then Q is
a degenerate quadric with vertex X. Liebler uses the quadric Q to give necessary and sufficient conditions that
SUR is an H-invariant spread and then exhibits a 3-space R satisfying the condiThe SL(Z,q)-module has the form Va $ V $ V where V 2 z z 2 a+l is the canonical representation. The condition on a F 1 is that X = 1 has no
tions (see (3.2) [30) ).
nontrivial solution in K ~ GF(q).
3. THE TRANSLATION PLANES WITH KERN
j
GF(q) OF ORDER qZ ADMITTING SL(Z,q)
Walker [41) and Schaeffer [31] have completely determined the translation planes of order q2 that admit SL(Z,q) as a collineation group in the translation complement and that have GF(q) in the kernel. 3.1 Walker [39], Schaeffer [31): Z
A translation plane TI of order q and kern
~
GF(q) admits S
~
SL(Z,q) iff TI
is Desarguesian, Hall, Hering, Ott-Schaeffer or one of the two Walker planes of
Translation planes that admit Sa2,pr)
501
25. TI is Desarguesian if and only if S is completely reducible and the p-element are elations. TI is Hall if and only if S is completely reducible and the p-elements are Baer. TI is Hering or Ott-Schaeffer if and only if S is irreducible. TI is Walker or Hering of order 25 if and only if S is reducible but not completely reducible. 4. TRANSLATION PLANES OF CHARACTERISTIC TWO IN WHICH ALL INVOLUTIONS ARE BAER In [25] Johnson and Ostrom study translation planes as in the title, adapting some of Schaeffer's arguments. 4.1 Johnson and Ostrom [25] : Let TI be a translation plane of dimension 2 over GF(q) (kern q = 2r.
~
GF(q», where
Assume S is a nonsolvable subgroup of the translation complement of TI and
contains only Baer involutions.
Let Sl denote the normal subgroup generated by s the Baer involutions. Then Sl is isomorphic to SL(2,2 ) and sir. 2s If Sl is irreducible, TI has an Ott-Schaeffer subplane TIO of order 2 and s
Moreover, for every factor So of s F 1, TI contains Ott-Schaeffer subplanes of order 22s 0 The elements of order 3 as affine homologies and determine is odd.
the component of TIO not fixed by a Sylow 2-subgroup. If Sl is reducible then TI is derived from a plane admitting SL(2,2 s ) where the involutions are elations. In the Foulser (-Ostrom) plane of order q4 admitting SL(2,q) where the pelements are Baer there are invariant Hall subplanes upon which SL(2,q) acts faithfully.
Apart from this example, I know of no plane TI that contains a non-
Desarguesian SL(2,pr)-invariant subplane TIO such that the planes TIO and TI are in different isomorphism classes. For example, if TI is a translation plane of order 22r admitting SL(2,2 s ), sir where SL(2,2 s ) leaves invariant an Ott-Schaeffer subplane TI O' is TI also Ott-Schaeffer? In [17] , Hering and Ho consider translation planes of dimension 2r, r odd (vector space has dimension 4r).
N.L. Johnson
502
4.2 Hering-Ho (5.7) [17]: Let
n
be a translation plane of even order and dim 2r, r odd.
Let S be a
co11ineation group in the translation complement containing only Baer involutions. Let S denote the subgroup generated by the involutions in S.
Let S be non-
solvable, then either (a) 5/0(S) or
=PSL(2,q),
=
q ±3 mod 8 s (b) S ~ 5L(2,2 ) for some s > 1.
Hering-Ho [17l and Johnson-Ostrom [ 26}a1so classify the co11ineation groups in even order planes when the involutions are not necessarily Baer. Oempwo1ff [4l studies translation planes of even order and arbitrary dimension that admit Baer groups of large even order. ~.3
Oempwo1ff [4]:
Let
n
be a translation plane of order 2n and S a co1lineation group in the Let E be a subgroup fixing a Baer subp1ane pointwise and
translation complement.
1et S'" denote the group generated by all subgroups that fi x a Baer subp 1ane poi nt-
wise and have order and IE; > 2n/4. (1) n
or
(2)
s*
lEI. Assume
Then one of the following possibilities occurs:
4 and S
~
S contains only Baer involutions, is nonsolvab1e
1:.
=SL(3,2)
5L(2,2 s ).
=
Note the Lorimer-Rahilly or Johnson-Walker (see e.g. [24]) are planes satisfying condition (l). Also, Oempwo1ff's results more generally include the case where S may contain elations. Although Ostrom's work [35J does not fit this section, the ideas are similar to those involved in the proof of (4.1). 4.4 Ostrom [35J (2.8): Let p > 5.
r
be a translation plane of order q2 whose kern contains GF(q), q
~
pr ,
Suppose that the linear translation complement has a subgroup S iso-
morphic to SL(2,ps).
Then either
(1) the p-e1ements are elations and S has an invariant Oesarguesian subplane of order pS (2) the p-e1ements are Baer and
n
is derived from a plane in which S acts as
503
Translation planes that admit SM2,pr)
in (1), or (3) S
2s has an invariant Hering plane of order p
The action of S on the
vector space is irreducible. The cases p = 3 or 5 are probably not much different (at least for SL(2,3 s ), s > 1) but the proof of (4.4) uses Suprunenko and Zalesski 's classification (see [40)), valid only for p > 5.
5. A CHARACTERIZATION OF THE DESARGUESIAN PLANE 5.1 Foulser-Johnson-Ostrom [12) : Let
IT
2
be a translation plane of order q , q f 3,4, that admits SL(2,q) as a
collineation group in its translation complement. (1) If the p-elements (q
= prj are elations then
IT
is Desarguesian.
(2) If the Sylow p-subgroups fix Baer subplanes pointwise then
IT
is Hall.
PROOF: (1) Sketch for q f 8. There is a Desarguesian net N of degree q+l (Ostrom [33)).
Let e be a
collineation of SL(2,q) such that lei is a prime 2-primitive divisor of q2_ 1. Then e fixes exactly two components of IT-N, acts irreducibly on each fixed component and generates over GF(p) a field isomorphic to GF(q2).
This may be done in
such a way that the (O,i )-homologies in the corresponding Desarguesian plane fix N componentwise and ;iX the two e-invariant components.
There are
~(q-l)
k
groups of SL(2,q) conjugate to <e> and to each corresponds a distinct pair of components.
Each field generated by a given 2-primitive element contains the same
field K of order q.
That is, K fixes each component so the result follows from
Walker-Schaeffer. The proof of (2) requires that the Baer subplanes pointwise fixed by Sylow p-subgroups belong to the same derivable net.
This is true by Foulser [7 ) (see
2.3)) if q f 3 and p f 2 and by Johnson [21) for p = 2, q f 2.
6. THE TRANSLATION PLANES OF ORDER q2 AND CHARACTERISTIC TWO ADMITTING GL(2,q) As seen in section 3, Walker and Schaeffer have determined the planes of the title with kern
~
GF(q).
However, without making an assumption on the size of the
N.L. Johnson
504
2
kern, the same results may be obtained for planes of even order q , q
~
4 with the
additional assumption that GL(2,q) is a co11ineation group (instead of merely SL( 2 ,q)).
By using the action of the center of GL(2,q), detailed combinational arguments yield a contradiction unless the Sylow 2-subgroup fixed point spaces are Baer sub1ines, lines or Baer subplanes.
By (2.7) and (5.1), we obtain:
6.1 Johnson [181: Let
IT
2r r be a translation of order 2 1 16 admitting GL(2,2 ) in the transla-
tion complement.
Then
IT
is Desarguesian, Hall or Ott-Schaeffer, depending on
whether the Sylow 2-subgroup fixed point spaces are lines, Baer subp1anes, or Baer sublines.
7. THE TRANSLATION PLANES OF ORDER 16 ADMITTING NONSOLVABLE GROUPS As seen in sections 5 and 6, planes of order 16 present an isolated problem. Are translation planes of order 16 that admit SL(2,4) or GL(2,4) Desarguesian or Hall?
(There are no Ott-Schaeffer planes of order 16). We mentioned the Dempwolff plane (2.10) admits GL(2,4) so we have at least
to add another plane to the list. The Lorimer-Rahilly and Johnson-Walker planes (see (241) of order 16 admit PSL(2,7) but the list stops here. 7.1 Johnson [24] : Let
IT
be a translation plane of order 16 admitting a nonso1vable col1in-
eation group.
Then
r
is Desarguesian, Hall, Dempwo1ff, Lorimer-Rahilly or
Johnson-Walker.
8. THE TRANSLATION PLANES OF ORDER
l
THAT AD/HT SL(2,q) AS A COLLINEATION
GROUP Fou1ser and Johnson [ 101 , [11] have detennined the planes of the title. Apart from the planes with kern
~
GF(q), only the Dempwolff plane is obtained.
We understand that Hering and Schaeffer have also worked on this general problem.
Translation planes that admit SU2,pr;
505
8.1 Foulser-Johnson [10] ,[ 11]: Let
IT
be a translation plane of order q2 (where pr
= q), which admits a
collineation group S isomorphic to SL(2,q) or PSL(2,q). Then IT is Desarguesian Hall or the Dempwolff plane of order 16 if and only if S is completely reducible.
S
IT
is Hering or Ott-Schaeffer if and only if S is irreducible.
IT
has order 25 and is Hering or one of the two Walker planes if and only if
is reducible but not completely reducible.
PROOF: Sketch (for q F 4) Let F
= GF(p), E = GF(pr). The irreducible representations of S ~ SL(2,pr)
have been determined by Brauer and Nesbitt [2]. Let N be an irreducible EG-module of dimension d and let K be the smallest field over which the matrices of S may be written.
Form the restriction module
NK. Assume K ~ GF(ps) and replace the elements of K in S by s X s matrices over F and replace the elements of K in a d-tuple in NK by an s-tuple over F. This makes NK into an irreducible FS-module of dimension ds.
Conversely, any irreduciple
FS-module may be obtained in this way (see Foulser-Johnson [10] (1.9)). Fong and Seitz [6] (usi ng Brauer and Nesbitt) have determi ned the i rreducible FS-modules of degree d
~
4r (= dimension of
IT
over G).
There are seven pos-
sible irreducible cases. It is shown that the only irreducible cases that support translation planes IT
are those involved in the dimension 2 situation and then shown that the plane
must also have dimension 2 in this case.
That is,
IT
is Hering or Ott-Schaeffer.
For the completely reducible cases, we may use the results of Fong and Seitz to determine the action on the invariant subspaces. structure to show the modules must be canonical.
We then use the group
This shows the plane
IT
to be
Desarguesian or Hall. Let S be reducible but indecomposable on submodule and
IT/W
Case 1: p
l
IT.
Let Wl be an irreducible FS-
= W2.
= 2.
Alperin's theorem [1] for p
= 2 on extension spaces shows that Wl ~ W2.
Dimension arguments then yield a contradiction, so the situation cannot occur. Case 2: p
F 2.
N.L. Johnson
506
=
Here W W and the possibilities for Wi given by Fong-Seitz. The module l 2 structure imposed yields impossible situations unless both Wl and W2 act canonically.
In this case, we really have a situation as in the dimension 2 case where
matrix arguments may be used (as Walker did in [41]) to show the characteristic is 5 and then that the plane have order 25. (In many of these arguments, we show the plane has dimension 2 and then use Walker-Schaeffer). Note this work reproves (6.1).
However, (6.1) may be obtained without
using representation theory.
9. PROBLEMS We have described the Kantor, Dye-Kantor and Liebler planes of even order q3 that admit SL(2,q) (possibly the Dye-Kantor and Kantor planes are also described by Liebler).
Are there others (besides the Desarguesian)?
Schaeffer [39] has described the possible modules on planes of order q3 with kern
~
GF(q) but not the planes in this situation.
In the even order case the
module is v~ ~ v~ w v~ where V is the canonical representation and a,B,o are the 2 twisting automorphisms of GF(q). In Liebler's planes, B
a+ 1
1 and x
= 6
1 has no solutions in GF(q).
PROBLEM 9.1: (a) Determine the translation planes where kern
rr ~
rr
3
of even order q admitting SL(2,q)
GF(q).
(b) Same problem for odd order. (c) Same problem for even or odd order but with no restriction on the kern. Problem
(e)
at least would require representation theory.
The planes of Fou1ser-Ostrom have order q4 and admit SL(2,q).
Some have
2
kern GF(q ) and one has kern GF(q). PROBLEM 9.2: (a) Determine the translation planes of order q4 admitting SL(2,q) with kern
~
2
GF(q }. (b) Same problem with kern
~
GF(q).
(e) Same problem with no restriction on the kernel.
Translation planes that admit SL(2.pr)
507
Obviously, the picture is clear on how one might proceed, clicking the exponent i in qi up to one at a time while the plane of order qi admits SL(2,q). Probably, (9.2) is too hard.
But, (9.1)(a) is probably within reach, at least with
some slight modification such as adding (as Liebler does) the loo-orbit structure and a particular group fixing an invariant subplane pointwise. PROBLEM 9.3: Let
TI
be a translation plane of order pr that admits S ~ SL(2,ps) and
S acts irreducibly.
(a) Is there an invariant Hering or Ott-Schaeffer subplane (if r/2 kern ~ GF(p ), see (4.1) and (4.4))? (b) If there is an invariant Hering or Ott-Schaeffer subplane, must r/2 kern ~ GF(p )? (c) If there is an invariant Hering or Ott-Schaeffer subplane, is the plane TI
Hering or Ott-Schaeffer?
ADDED NOTES - ADDED PLANES 2.17 Kantor has recently constructed a class of translation planes of order p6r admitting GL(2,p2r) where the p-elements are elations. least pr(pr+l) nonisomorphic planes of each order. 6r
There are at
2.18 There is also a plane of order 43 admitting SL(2,4) which comes from the work of Dye and Kantor. BIBLIOGRAPHY 1. 2. 3. 4. 5. 6. 7. 8.
J.L. Alperin, Projective modules for SL(2,2 n) (to appear). R. Brauer and C. Nesbitt, On the modular characters of groups, Ann. Math., 42 (1941), 556-590. V. Dempwolff, Einige Translationsebenen der Ordnung 16 und Ihre Kollineationen (to appear). V. Dempwolff, Grosse Baer-Untergruppen auf Translationsebenen Gerader Ordnung (to appear). R.H. Dye, Partitions and their stabilizers for line complexes and quadrics, AnnaZi di Mat., (4) 114 (1977), 173-194. P. Fong and G. Seitz, Groups with a (B,N)-pair of rank 2, Invent. Math., 21 (1973), 1-57. D.A. Foulser, Baer p-elements in translation planes, J. AZgebra, 3 (1974), 354-366. D.A. Foulser, Derived translation planes admitting affine elations, Math. Z., 131 (1973), 183-188.
508
9.
10. ll. 12. 13.
14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33.
N.L. Johnson
D.A. Foulser, A translation plane admitting Baer collineations of order p, Ai'clz. ,'·;.;;tri., 24 (1973), 323-326. 2 D.A. Foulser and N.L. Johnson, The translation planes of order q that admit SL(2,q) as a collineation group T, J. Algebm (submitted). D.A. Foulser, The translation planes of order q2 that admit SL(2,q) as a collineation group. II, J. Geometry (submitted). D.A. Foulser, N.L. Johnson and T.G. Ostrom, Characterization of the Desarguesian and Hall planes of order q2 by SL(2,q), Inter. J. Math. Sci., (submitted) . Ch. Hering, A new class of quasifields, Math. Z., 118 (1970), 56-57. Ch. Hering, Uber Translationsebenen, auf denen die Gruppe SL(2,q) operiert. Atti del Convegno di Geometria Combinatoria e sue Applicazioni. Perugia: 1st. t1at. Univ. Perugia, 259-261. Ch. Hering, On shears of translation planes, Abh. Math. Sem. Hamburg, 37 (1972), 258-268. Ch. Hering, On projective planes of type VI. Atti dei Convegni Lincei 17. Teorie Combinatorie (1976), 30-53. Ch. Hering and C.Y. Ho, On free involutions in linear groups and collineation groups of translation planes. Trabulho de r·latematica No. 129, Funda~ao Universidade de Brasilia (1977). N.L. Johnson, The geometry of GL(2,q) in translation planes of even order (and correction), Internat. J. Math. & Math. Sci., 1 (1978), 447-458. N.L. Johnson, The translation planes of Ott-Schaeffer, Arch. Math., 36 ( 1980), 183-192. N.L. Johnson, The translation planes of Bruck type {l }, Arch. Math., 26 (1975), 554-560. N.L. Johnson, Addendum to "The geometry of SL(2,q) in translation planes of even order", Geom. Ded. (submitted). N.L. Johnson, On the construction of the Dempwolff plane of order 16, Geom. ,'"Jed. (submi tted). N.L. Johnson, The translation planes of order 16 that admit SL(2,4), J. Cornb1:natorial Theory (submi tted). N.L. Johnson, The translation planes of order 16 that admit nonso1vable co 11 i nea ti on groups, Geom. Ded. (submitted). N.L. Johnson and T.G. Ostrom, Translation planes of characteristic two in which all involutions are Baer. J. Algebra (2) 54 (1978), 291-315. N.L. Johnson and T.G. Ostrom. Translation planes of dimension two and characteristic two, G6om. Ded. (to appear). N.L. Johnson and T.G. Ostrom. The geometry of SL(2,q) in translation planes of even order, Geom. Ded. 8 (1979), 39-60. N.L. Johnson and T.G. Ostrom, Translation planes with several homology or elation groups of order 3, Geom. Ded., 2 (1973). 65-81. W.M. Kantor, Spreads, translation planes and Kerdock sets. I. (to appear). R.A. Liebler, Combinatorial representation theory and translation planes (to appear). H. LUneburg, Translation Planes, Springer-Verlag: Berlin-Heidelberg-New York, 1980. H. LUneburg, Charakterisierungen der endlichen Desarguesschen projektiven Ebenen, Math. z., 85 (1964), 419-450. T.G. Ostrom, linear transformations and collineations of translation planes, J. Algebra, 14 (1970), 405-416.
Translation planes that admit SL(2,pr)
34. 35. 36. 37. 38. 39. 40. 41.
509
T.G. Ostrom, Elations in finite translation planes of characteristic 3, Abh. Math. Sem. Hamburg, 41 (1974), 179-184. T.G. Ostrom, Translation planes of dimension two with odd characteristic, Ganad. J. Math., (5) 32 (1980), 1114-1125. T.G. Ostrom, A class of translation planes admitting elations which are not translations, Arch. Math., 21 (1970),214-217. V. Ott, Eine neue Klasse von Trans1ationsebenen, Math. z.,143 (1975), 181185. B. Segre, On complete caps and ovaloids in three-dimensional Galois spaces of characteristic two, Acta Arith., 5 (1959), 315-322. H.J. Schaeffer, Trans1ationsebenen, auf denen die Gruppe SL(2,pn) operiert. Dip10marbeit TUbingen, 1975. 1.0. Suprunenko and A.E. Za1esskii, Classification of finite irreducible linear groups of degree 4 on fields of characteristic p> 5. Inst. Mat. Akad. Nauk. BSSR, preprint No. 13 (1976). M. Walker, On translation planes and their collineation groups, Ph. D. Thesis, Univ. London, 1973.
The University of Iowa Iowa City, Iowa 52242
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Annals of Discrete Mathematics 18 (1983) 511-518 © North-Holland Publishing Company
511
NONDESARGUESIAN PLANES, PARTIAL GEOMETRIES, STRONGLY REGULAR GRAPHS AND CODES ARISING FROM HYPERBOLIC QUADRICS William M. Kantor
ABSTRACT Spreads of hyperbolic quadrics are used to construct translation planes, partial geometries, strongly regular graphs and codes, all having a rich geometric structure. 1. INTRODUCTION A hyperbolic quadric Q+ in PG(2n-l ,q) has (qn_ l )(qn-l+ 1 )/(q_l) points, and contains n-l-spaces each having (qn_l)/(q_l) points.
A $ppead of Q+ is a family
r of qn-l+ 1 subspaces of Q+ of dimension n-l partitioning the points of Q+.
A
spread can only exist if n is even (when n is odd, Q+ does not even contain 3 pairwise disjoint n-l-spaces).
The only known examples occur when n=2, when q is
even, or when n=4 and q;: 0 or 2(mod3). In this paper, we will summarize some of the ways spreads of hyperbolic quadrics have been used recently. 2. TRANSLATION PLANES AND THE CONSTRUCTION OF SPREADS If X is any subspace of PG(2n-l,q), then Xl will denote its polar with respect to Q+.
Consider a spread r of Q+, and assume that q is even.
x ~ Q+, and form the family
i
n r = {xl n wlw
n-2-spaces partitioning the quadric xl n Q+.
E
r}; clearly, this consists of
Since q is even, x
a PG(2n-3,q) equipped with a symplectic (or null)
Fix any point
polarity.
E xl
and xl Ix is
If r(x) denotes the
projection of xl n r into xlix, then r(x) is a spread in the more usual sense: qn-l+1 subspaces of dimension n-2 which partition the points of xl/x.
Consequent-
ly r and x determine a translation plane A(r(x)) arising from a symplectic spread. Conversely, suppose that nand q are even, and that spread of PG(2n-3,q). [4], Dye [5]).
~,
is a symplectic
Then~' arises from some Q+, r and x, as follows (Dillon
We can regard PG(2n-3,q) as our former xl/x, related to Q+ as be-
512
W.M. Kantor ;.)-1
fore. Xl
(I
Each of the j'
+1 members of
~,
is the projection of a unique n-2-space of
Q+, and the resulting family ~" of n-2-spaces partitions
xl
Q+.
II
Fix one of
the two families of ~-l-spaces of Q+; each n-2-space of Q+ is in a unique member of this family.
Thus, r" lifts to a family ~ of qn-l+l of these n-l-spaces, any
two having at most a point in common.
Since no two of these n-l-spaces can have
exactly one point in common (as n is even), l consists of pairwise disjoint subspaces.
Consequently, r is a spread of Q+.
The preceding construction produced an essentially unique spread of Q+ from 1
a symplectic spread of x Ix.
In particular, l(x) essentially determines l (where
"essentially" means that the symplectic geometry on / Ix can arise from several quadrics in PG(2n-l,'7), and that we singled out one of the two families of n-lspaces of and x
1
+
~).
However, more is true.
be pOints off Q+.
Let l and
~
+
1
be spreads of Q , and let x
Then any isomorphism from A{l{x»
collineation of PG(2n-l ,q) sending Q+ to itself, ~ to l In particular, each collineation of
A(~(x»
1
to A{l (x » induces a 1
and x to x
1
1
(Kantor [6]).
is the product of a perspectivity with
axis at infinity and a collineation of PG(2n-l,q) preserving Q+, land x.
Thus,
the collineation groups of many translation planes can be found simultaneously once the group c{r) of collineations preserving Q+ and
L
is known.
Several spreads l are described in Kantor [6,7], and some of the resulting n-l
translation planes of order q
are studied in detail.
q=2, only desarguesian planes occur.
When n=2, or n=4 and
In all other cases, new planes are obtained.
Some of the planes are uninteresting, but others have collineation groups behaving in unusual manners (such as flag-transitively, or behaving as decribed in Johnson's paper in these proceedings).
Complete arcs and dual ovals in all of
these planes were found by Thas [13] . The Simplest example of a hyperbolic spread is obtained as follows.
The
desarguesian plane AG(2,qn-l) arises from the unique spread of pc(l ,qn-l), which is trivially symplectic.
This produces a symplectic spread l' in PC{2n-3,q), and
hence (if nand q are even) a spread L of a hyperbolic quadric in PC(2n-l ,q}. This spread is called desarguesian, for obvious reasons.
It was found by Dillon
[4] and Dye [4]; G{l} was determined by Dye [5] and Cohen and Wilbrink [2].
As we
will see in §6, an "affine" version of this spread was discovered much earlier by Kerdock [ 10) . Other hyperbolic spreads arise from the hermitian curve in PC(2,q2), from triality, and from field changes generalizing that of the preceding paragraph.
Geometric structures arising from hyperbolic quadrics
513
For coordinate descriptions of many of the known examples we refer to Kantor[6,7). It seems unlikely that hyperbolic spreads can exist when n > 4 and q is odd. On the other hand, if q is even and fixed, the number of inequivalent hyperbolic spreads probably
+00
as n+oo.
3. MORE TRANSLATION PLANES; OVOIDS Let Q+ be a hyperbolic quadric in PG(7,q), where q is now even or odd. Consider a spread ~ of Q+. Q+.
Then ~ belongs to one of the families of 3-spaces of
Let T be any member of that family not in~.
empty or a line.
Set T n
~
= {T n
wlw E
~
and T n
If W E ~ then T n W is either ~
is a line}.
spread of PG(3,q), and hence determines a translation plane of order q2. translation planes obtained from
~
~
Then T n
is a
(The
3
as in §2 when q is even have order q , not q 2 .)
These translation planes are studied in Kantor (8). Let, be a triality map. Then, cyclically permutes the following three sets: the points of Q+, and the two families of 3-spaces on Q+.
At the same time,
, sends lines of Q+ to lines of Q+, while preserving incidence between lines and It follows that ~' is either another spread, or
both points and 3-spaces of Q+.
else consists of q3+ l pairwise noncol1inear points of Q+. Let Q+ be a hyperbolic quadric in PG(2n-1,q). Thas (14)) to be a set n of
q
_1
An ovoid of Q+ is defined (by +
+1 pairwise noncollinear points of Q.
count shows that each n-1-space on Q+ contains a unique point of n.
A simple
It follows
that, if x E Q+ - D, then xl n n projects onto an ovoid D(x) of the obvious . 1n . x1;x. qua dr1C
If n > 4, ovoids probably do not exist, but this has only been proven in PG(2n-l,2) (Kantor [9, (4.3))).
If n=3, ovoids correspond (under the Klein corre-
spondence) to spreads in PG(3,q), and hence to translation planes of order q 2 . n=4, ~ is a spread, and ~' is not a spread, then ~' is an ovoid n.
If
Moreover, if T is as before then T' is a point x, and T n ~ and n(x) are related by the Klein correspondence. with D(x) than T n
~,
When dealing with coordinates, it is easier to work
since points of n require 8 coordinates while 3-spaces in
The translation planes A(n(x)) are studied for
are more complicated to describe. all known
~
recover n or
in Kantor (8). ~
~
Unlike the situation in §2, it is not clear how to
from A(D(x)); moreover, some co1lineations of the plane need not be
related to automorphisms of n or
~.
W.M. Kantor
514
4. STRONGLY REGULAR GRAPH Let ~ be a spread of a hyperbolic quadric Q+ in PG(2n-l,q), where n ~ 4. Let 11 be the set of all hyperplanes of members of ~, so that 1111=1~I(ql1-1)/(q-l) is the number of points of Qt.
If
X, Y E
11 and
X
F
Y,
write x-y=-iny F O.
Then
(11,-) is a strongly regular graph having the same parameters as the col linearity +
+
graph (Q ,1) of Q (Kantor [9]).
If n=4, these graphs are isomorphic (an
isomorphism being induced by triality). isomorphic.
However, if
11
> 4 they are probably never
This is known if q=2 (see [91); the proof uses the nonexistence of
ovoids of Q+ when q=2.
The graphs are also not isomorphic when ~ is the
desarguesian spread defined in §2; this was proved in [9) by brute-force calculations. If
WE
~ 1et ;/' be its set of hyperp 1anes.
partition of n into cliques. (
11-1
Then
T.'"
{lIWET.}
is a
{,
If X
E
11 - W then X is joined to exactly
{,
* Ix1
-1)/(
n
WE
Y}.
Thus, each clique
inherits from (n,-) the structure of a PG(n-l,q) in a natural manner.
Further
properties of (11,-) reminiscent of properties of (Q+,l) are found in [91. it is easy to reconstruct Q+ and
~ given (11,-) and
i',
';,':
W
While
it is not known whether ~'"
can be determined from (11,-) alone. Variations on the construction of (11,-) are found in [91.
Instead of a
spread of a hyperbolic quadric, spreads of other quadrics, of hermitian geometries, and of symplectic geometries could have been used: strongly regular graphs again arise in exactly the same manner, having the same parameters as the underlying geometries.
In the symplectic case, new symmetric designs are also obtained
having the parameters of PG(211-1,q).
5. PARTIAL GEOMETRIES In this section, ~+ and ~ will be as before, but q will be 2 or 3.
Define n
as in §4. + If q=2, let i: be the complement of Q.
If q=3, let N consist of the points
of the complement of "length" 1 (where "length" refers to the value at x of the quadratic form defining Q+). It is straightforward to check that the incidence structure (N,n,l) is a partial geometry
pg(~)
(DeClerck, Dye, Thas [3,15)).
Namely, if x
E N
then ": is perpendicular to ql1-1+1 members of 51 (one per member of ~),X dicular to
qn-i tl
members of
N
(lying in an affine space), and (if x ~
and
X E
11,
is perpen-
i)
there
Geometric structures arising from hyperbolic quadrics
515
are exactly qn-2(q_l) pairs (y,Y) with yEN, YEn. y E Xl n yl and x E yl. Let x, yE N, x ,.
Then x.
y.
l
y E
for some
X E n if
and only if q=2 and x
and yare perpendicular. or q=3 and x and yare not perpendicular. +
PG(2n-l,q) and Q can be recovered from
Consequently,
Moreover, projectively inequiva-
pg(~).
lent spreads L produce nonisomorphic partial geometries (Kantor [9. (7.3)]). n=4 then
so is
~
If
is essentially unique (cf. Patterson [12]; Kantor [6, §lO]). and hence If q=2 and n-l is composite then
pg(~).
~
is not unique (Kantor [7.(9.12)]);
presumably, the same is true for all even n > 4.
On the other hand. if q=3 and
> 4 it seems likely that no ~ exists. Let x. YEn, X ,. y. Then X, Y E X1 for some x are not adjacent in the graph (n.-) obtained in §4.
n
E N
if and only if X and
y
6. CODES
Let Q+ and ~ be as usual. introduce coordinates.
In order to define some codes. we will have to
The vector space underlying Q+ can be written
E
$
F
with
Fix a basis e •... ,e of E, and let f 1, ... ,f be the dual basis of F, 1 n n so that the underlying bilinear form has (e.,f.) = 0 ... We will write matrices
E, FEn.
1-
J
1-J
with respect to the ordered basis e 1 ,···.en , f 1 ,··· ,fn . Let
F' E
n -
{E}.
Then
F' = F(IM' 0) I for a uniquely determined skew symmetric nXn matrix M' (with zero diagonal), where
o and
I are the zero and identity nXn matrices. + Q • and is the identity on E.)
Thus,
~
and E determine a family
This provides an "affine" version and
M"
of~.
K=K(~,E)
If
are the correspondi ng matri ces. then
nonsingular.
(The above 2nX2n matrix preserves of skew symmetric nXn matrices.
F', F" E ~ -{E} , F' ,. F",
F'
n
F" = n-l
Thus, K is a Kerdock set: a set of q
0 imp 1i es that
and if
M '-M"
M'
is
skew symmetric nXn matrices
(with zero diagonal) such that the difference of any two is nonsingular. Conversely, any Kerdock set of nXn matrices defines a hyperbolic spread via (1').
Note that this spread is uniquely determined; thus, inequivalent spreads
produce different Kerdock sets. produce the same spread.
Note, however, that different Kerdock sets can
A further discussion of equivalence of Kerdock sets can
be found in Kantor [6, §5].
W.M. Kantor
516
Now let q=2, and fix a Kerdock set K of nXn matrices.
oE
We may assume that
If ME K, there is a quadratic form QM(x) associated with M:
K.
xMyt= QM(x+y) - QM(x) - QM(Y)
for all x, y
E z~.
Consider the following functions Zn.~ Z : 2
Q,/ x)
+ L (x) +
2
( ,"'~)
e
2 , There are 2n -}. 2n '2 = 22nf unctions. 2 Let; = C(X) be the set of all sets of zeros of the various functions (**).
for
MEg,
L a linear functional, and c
E
Clearly, c contains all hyperplanes of AG(n,2) as well as If x,
YE
~
n
and Z2.
n 1 C, then their symmetric difference XAY has size 0, 2n , 2 - ,
2n- 1±2!n, corresponding to whether XAY is {a}, z~n, an affine hyperplane, or a
quadric or its complement in z~n (cf. Cameron and Seidel [1 J). z2r!
are called v , .•• ,V'7 with N = 22n, then each subset x of zn can be written
2 N
X
If the vectors in
='E 1
c,.V. 7,.
with
7,.
1" a. E Z , 7,. 2
N
2
where'E refers to symmetric difference.
In this way, C
1
can be regarded as an error-correcting code (MacWilliams and Sloane [11)), having .. d'1S t ance 2n- 12!n ' 22n . It'1S ex t rema l '1n a sense , an dSlZe 1engt h 2n ,m1nlmUm di scussed on p. 667 of [ 11) . The codes
c(x)
were first discovered by Kerdock [10), with
arising from the desarguesian hyperbolic spread described in §2.
K
a Kerdock set
Since inequiva-
lent spreads exist whenever n-l is composite, inequivalent codes C(K) exist as well.
However, from a coding theoretic point of view, it is not clear how differ-
ent codes C(x) differ (for a fixed n): those corresponding to desarguesian spreads seem as if they should be "best", but it is not clear what this means.
BI BLI OGRAPHY 1.
4. 5.
P.J. Cameron and J.J. Seidel, Quadratic forms over GF(2), Ken. Ned. Ak. Wet., PY'ee. A76 (1973) 1-8. A.M. Cohen and H.A. Wi1brink, The stabilizer of Dye's spread on a hyperbolic quadric in PG(4n-l,2) within the orthogonal group (to appear). F. DeClerck, R.H. Dye and J.A. Thas, An infinite class of partial geometries associated with the hyperbolic quadric in PG(4n-l,2) (to appear). J. Dillon, On Pall partitions for quadratic forms (unpublished manuscript). R.H. Dye, Partitions and their stabilizers for line complexes and quadrics.
6.
W.M. Kantor, Spreads, translation planes and Kerdock sets. I (to appear).
2. 3.
Annal{ Ji Mat. (4) 114 (1977) 173-194.
Geometric structures arising from hyperbolic quadrics
7. 8. 9. 10.
W.M. W.M. W.M. A.M.
ll.
F.J.
12.
N.J.
13.
J .A.
14. 15. 16.
J.A. J.A. J.A.
517
Kantor, Spreads, translation planes and Kerdock sets. II (to appear). Kantor, Ovoids and translation planes (to appear). Kantor, Strongly regular graphs defined by spreads (to appear). Kerdock, A class of low-rate non-linear binary codes, Inform. ControZ 20 (1972) 182-187. MacWilliams and N.J.A. Sloane, The theory of error-correcting codes. North Holland, Amsterdam 1977. Patterson, A four-dimensional Kerdock set over GF(3), J. Comb. Theory (A)20 (1976) 365-366. Thas, Construction of maximal arcs and dual ovals in translation planes, Europ. J. Comb. 1 (1980) 189-192. Thas, Ovoids and spreads of finite classical polar spaces (to appear). Thas, Polar spaces,generalized hexagons and perfect codes (to appear). Thas, Some results on quadrics and a new class of partial geometries (to appear).
Bell Laboratories f4urray Hi 11 New Jersey 07974, U.S.A. Permanent address f4a thema tics Depa rtment University of Oregon Eugene, OR 97403, U.S.A.
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519
Annals of Discrete Mathematics 18 (1983) 519-534 © North-Holland Publishing Company
THE PROJECTIVITY GROUPS OF OVALS AND OF QUADRATIC SETS H. Karzel
;':
and M. Marchi
We know several theorems on conics: the classical theorem of Pascal, the Steiner generation of conics, the theorem of P. Kustaanheimo-B. Segre that any oval in a projective plane over a finite field F of characteristic
+2
is a conic,
the theorem of Buekenhout that any projective plane with a Pascal oval is a Pappian plane and the oval is a conic
(3), [1], [7J.
Here we will give a further characterization of conics.
For the Conference
in Bad Windsheim in 1980, H. Karzel and H.-J. Kroll got the task to report on projectivity groups in circle geometries (see [6]).
Therefore it was necessary to
look at the different types of perspectivities which one can define in a circleplane.
One type is the following: Let (a) C be a circle of a circle plane,
(b) a,b be two distinct points not on C, and
r C------+- C (c) [C,a,b,C] :
I
(.r---..
J
l
.r---..
ja,b,x n C\ {x} x-- )
if
la,b,x n CI
x
if
la,b,x n CI
I
2
.r---..
l
.r---..
where a,b,x is the circle passing through these points. generate a group r which is 3-transitive on C.
These perspectivities
Now let a be a fixed point; the
subgroup r , generated by all the perspectivities [C,a,b,C) is still 3-transitive. a If we consider the derivation of the circle-plane in the point a, we obtain an affine plane (P,B) where the circle C appears as an ovalO, and the mappings [C,a,b,C) as involutorial perspectivities ( 0----+ 0
I b: ~
Il.
(
x__~b,X n 0 \ {x} Ix
l
if
Ib,x n
if
Ib,x n
01 01
2
l.
520
H. Karze! and M. Marchi
The group r(o) generated by all perspectivities b, with bE P \ 0, will be called -eh" pmJectivit!., group of the ovalO (see § 1).
In
[6)
the following theorem is
stated and the proof is sketched: THEOREM 1: If the projectivity group r(O) of an ovalO of an affine or projective plana (P,B) is snm'pZy 3-transitive, then there exists a suhset B' of the pOIJer-
o
.wt of P such -chat
(P,B~ U
Bl
U
B ) (see § 1J is a papp-ian affine or projective 2
"lane i" which 0 is a conic.
In
§
2 we shall give a complete proof of Theorem 1.
This theorem gave rise
to the following questions. 1. Are there examples of projective planes (P,B) with ovals 0 such that r(O) is sharply 3-transitive but (P,B) is not pappian (in this case (P,B) would moreover be non-desarguesian)? 2. Which conditions do we have to add, to obtain a pappian plane? 3. Is it enough to assume that (P,B) is a finite plane? To tackle these problems we started our investigations in a more general setting by replacing the ovalO by a quadratic set Q (see have lately attracted great interest.
§
1).
Quadratic sets
So far as we know the incidence structure
of a maximal kinematic space can be obtained from a projective space (P,B) by omitting a quadratic set Q. The "trace space" is then an example of a so-called 2-punched-space.
These
spaces are characterized as incidence spaces containing only 0-, 1- or 2-lines. By the deep theorem of G.P. Kist [10) we know that with few exceptions any 2punched space of dimension> 3 is isomorphic to a trace-space given by a projective space and a quadratic set. In
§
3 we consider self-dual ovals O.
group f(O) in a dual way.
Here we can define a second oval
We shall give examples of projective planes (P,B) with
self-dual ovals 0, where the groups r(O) and r(O) are different.
By Theorem 1 any
example where one of the groups r(O) or r(O) is sharply 3-transitive and (P,B) is not pappian can be derived from a projective plane over a commutative field F,
The projectivity groups of ovals and of quadratic sets
521
where 0 is a conic, by replacing the lines which have an empty intersection with 0 by other curves. tions (see
§
To obtain such examples the field F has to fulfil suitable condi-
3).
Theorem 2 of
§
4 gives an answer to the second question and Theorem 3 of
§
5
deals with the group r(Q) of a quadratic set Q consisting of two distinct lines of a projective plane.
1. DEFINITIONS Let (P,B) be a projective space with the point set P and the line set Band Q + ~ a subset of P.
For the structure (P,B,Q) we define the following types of
lines: B:={LEB
LeQ},
B;== {L E B
IL n Q I =
i}
for i E
]'j
The set Q is called a quadratic set, if the following axiom holds: 1 B = Bo U Bl U B2 U Boo In the same way we can define a dual quadratic set
Q
Q.
restrict ourselves to the case of a projective plane.
In this paper we will Then
0 is
a subset of B
characterized by the axiom Pl U P2 U Poo If [p] := {X E B : p E X} then Poo:= {p E P: [p] eO}and Pi := {p E P:I[p] nQI=
Q l'
P = Po
U
n.
A subset Q e P is called a self-dual quadpatic set if for Q and O:=B l the axioms Q 1 and Q l' are valid respectively. Special types of quadratic sets Q are the ovals in projective planes which are characterized by the condition: Q 2 For any q E Q we havel [q] n Bl I = 1.
We shall write [P]l := [p] nB l · For an ovalO we can identify the point-set 0 with the line-set
0 .-
Bl by the
bijection
o --~
0 = Bl ; x ---+[x] n Bl , and we have Boo = By these definitions we obtain:
~
.
If 0 is a self-dual oval of a projective plane (P,B) then O:=B l is an oval in the dual projective plane (B,F) with P:={[p] : pEP}. ~
In the case of a projective plane over a commutative field with char (F)F2,
522
H. Karzel and M. Marchi
any ellipse 0 is a self-dual oval. For any projective space with a quadratic set (P,B,Q) we define now the projectivity group of Q. For any pEP \ 0 let
p be
the permutation of 0 defined
by (
I 0 --... ;
p
Q (
i x ___~x \
lx'
p,x n Q = x
if
= p,x
n
0 \ {x}
if
p,x n Q + x.
\
By p,x we denote the line L E B with p,x E L.
The permutation group
,(Q) := <{p: pEP \ O} > generated by all these mappings is called the pY'ojectivi~i!
group of the quadi'atic set Q.
If
0 is
a dual quadratic set of a projective plane then any line L E B \ Q
defines a permutation
I Q- ---... 0(
L
J
(
Ix
I
I x ---~ I
IX'
if :=
[ X n L) n 0 = X
[X n LJ n Q \ {X}
if (X n LJ n Q + X,
\
and all these maps generate the projectivity group r(o) := <{L:L E B \ OJ>.
Any
self-dual quadratic set 0 has the two projectivity group r(O) and r(o) with Q = B . In the case of a self-dual ovalO (cf. (1.1)) we will consider r(Q) as a l permutation group of O. For n
IN a permutation group (M,r) is called n-transitive if for any two n n-tuples (Pl, ... ,Pn),(ql, ... ,qn) E M with I{Pl, ... ,Pn}1 = l{ql, ... ,qn}1 = n there E
exists ayE r with Y(Pi} = qi for i E fl,2, ... ,n}.
We say that a permutation
group (M,r) has the pY'opeY'ty (P ) if a map y E r fixing n distinct points of M is n the identity. Moreover we say that a permutation group (M,r) is sharply n-transitive if it is n-transitive and has the property (P ). n
2. OVALS WITH A SHARPLY 3-TRANSITIVE PROJECTIVITY GROUP In this section let 0 be an oval of a projective plane (P,B) with IO!
~
6 and r:= r(O}. 2.1
First we show the following.
Let p,q E P \ 0 be any two distinct points; then
P+ q.
The projectivity groups of ovals and of quadratic sets
523
PROOF: Let x be any point on 0; then x' := pix) E p,x, x" := q(x) E q,x. x
f
p,q we have x'
If
f x".
From this proposition we have: 2.2 There exists at most one point pEP' 0 such that p = id.
We have to distinguish the two cases: I) for all pEP \ 0 we have p
f
i d,
II) there exists exactly one point kE P' 0 with k = id; the point is then called nueZeus of O.
Henceforth by a,b we shall mean any two distinct points of 0; we denote: ~ a, b:= {p : p E a,b, p
f
a,b} and
f(a,b):= {y E r(O) : y(a) = a, y(b) = b} .
.£.:l
For any poi nt pEP' 0 we have:
p E a,b' {a,b} ~ pia) = b
<=>
P E ~ a, b'
2.4 The permutation set (0' {a,b},
~
a, b'
~
a, b) is transitive.
PROOF: Let x,x' EO' {a,b} be any two distinct points; if yEO' {a,b,x,x'} then we denote {r}:= x,y n a,b, {s} := x' ,y n a,b.
Since r,s
~
{a,b} we have
sr(x)=x'.
By this proposition and by ~ a, b'
~
a, b ~ r( a, b) we obtain:
2.5 The permutation group (0 '{a,b}, r(a,b)) is transitive. 2.6 The projectivity group (O,r) is 3-transitive. PROOF: Let x,y,z,x',y',z' E 0 with l{x,y,z}1 = l{x',y',z'}1 = 3. assume x = x'.
Since 101
~
By (2.5) we can
6 there is an element u E 0 '{x,y,y'} and (by (2.5))
524
H. Karzel and M. Marchi
an a E '( x,u ) such that a(y) = y'. Therefore we can now suppose x = x' and y = y'. Again by (2.5) there exists a 8 E f( x,y ) such that 8(z) = z'. Let us assume from now on that (O,r) has the property (P ); then we have: 3 -2.7
il a, b' ~ a, b' ~ a, b
PROOF: Since (~a,b)3 q,r,s E a,b \ {a,bL
~a,b
.
~ ~a,b'
We have to prove (6a~b~3_~ ~a,b' Let For x E 0 \ {a,b} let x' := s r q(x) and {u}:= x,x' n a,b .......
(where x,x'
""'-'
................
if x = x'); then the projectivity u s r q fixes the points 1 ............... -- ...... a,b,x and hence, by (P ) u s r q = id. that means u = s r q. 3 :=
[xl
2.8 The permutation group (0 \ {a,b}, f(a,b)) is regular and commutative. PROOF: Since the subset ~ a, b' 6a, b of f( a, b) operates transitively on 0 \ {a,b} (by (2.4.)), f(a,b) is regu1~r_b: ~P3); hence f(a,b) = 6a ,b . ~a,b' To prove the commutativity of r( b) let p,q,r,s be involutions in 6 b; then by (2.7) we have - - -
a,
_ _ _ _
_ _ _ _
_ _
~, -1-
--1 --1 --1 -
(p ____ q r) E 6a, b and hence: (p___ q)( r s) = (p q r) s = (p_ q___ r) s -----1 = (r q p )s = r q p s. Again by (2.7) q p s E 6a, b and hence: r q p s = r(q p s) = (r
s)(p q). ~
Let a E r be any involutorial transformation (a
+ id.).
Then there
exists exactly one point pEP \0 such that p = a. PROOF: By (P 3 ) there are x,y E 0 with x + a(x),y, a(x) + y and a(y) + y. For {p} := x,a(x) n y,a(y) the map p a has the three distinct fixed points X,a(X),y
-
and hence by (P ) p a 3
=
id. i.e. -p
=
a.
By (2.1)
2.10 For a EO and p E P\ 0 we have: p(a)
P is unique. =
a" p E [all'
For a E 0 let 6 := {p : pEP \ 0 with p,a n 0 = {a}}. a Viith the same arguments as in the proof of (2.4) it follows: 2.11 - The permutation set (0 \ {a}, 6a • 6a ) is transitive •
The projectivity groups of ovals and of quadratic sets
Let us denote by J the set of all involutions in r(O).
2.12 {p {p
525
By (2.9) we have:
pEP \ O} = J if 0 has no nucleus pEP \ O} = J U {id.} if 0 has a nucleus k.
Let now 00,0,1 be three distinct points on 0 and
:= (00],
k
n
[0]1'
Since (O,r) is
sharply 3-transitive there exists a KT-field (F,+,·,a) (see [4], [8]) such that i)
O=FU{ooL
ii)
0,1 are the neutral elements of (F,+) and (F'~,.) resp. (where F'~:= F \ {O}).
iii)
For a,b E F, b
+° let
(
[a,b] : F U {oo}
---->
F U {oo}; x
x 4 oo ---->t00 + bx for for x = oo 1
Then roo = {[ a, b] : a, b E F1\ b iv)
Let.: F U {oo}
--->
+ o}.
l
F U {oo} with .(x) .- a(x) for x E F''', .(0) = 00 and
.(00) = O.
Then r = roo U JiF [c, 1] .r oo ' Let F+ := {c+ : c E F} with c+ := [c, 1] and Joo = roo () J.
v)
(k vi)
Then F+=Joo ·kU{id.} is the involution fixing 0 and oo, or the identity if 0 has a nucleus)
(F*,·) and r(O,oo) are isomorphic. By (2.8) r(O,oo) is commutative; hence (F,+,·) is distributive by (vi), and
therefore, by [9], a commutative field. -1
By the theorem of Benz-Elliger ([2]) we have moreover a(x) = x
for all
)t,
X E F and therefore r = PGL(2,F). . ax + b For a E PGL(2,F) wlth a(x) = CX + d we have ad - bc
°
aEJ ~ a + d = for char(F) a E J U {id.} ~ a + d = 0 for char(F)
+
°and:
+ 2, 2.
Hence together with (2.12) we may identify the point set P \ 0 with the set of all points F*(a,b,c) E F3*/F* of the projective coordinate plane R(F) over the commutative field F for which we have /+bc 3'~
,':
3'~
+ 0.
This bijection between P \ 0 2
and F IF":\C with C := {F (x ,x 2 ,x ) E F IF'~ : Xl + x2x3 = O} is extended to a l 3 ,': 3* ,~ 2 bijection w : P --> F IF* by w(oo) = F (0,-1,0) and w(t) = F (t,-t ,1) for t E F. Now we prove:
526
H. Karzel and M. Marchi
2.13
For L E Bl
PROOF: Let p 1. Case:
LE
pEL" b =
B 2
B2 the set w(L) is a line of
U
F (x, ,x ,x ) E P \ O. 2 3 • Let {a,b} := L n 0 with a
p( a)
x,a
+
=
x2
+
00.
IT
2
(F).
By (2.3) we have:
(, (*)x,(a+b)+x -x a b = 0, for b + 2 3
1
00
.. )
for b = x3a - xl (**)x l -x 3a = 0, Since oo(a) = F*(a,-a 2,1) and w(b) = F*(b,-b 2 ,1) or w(a) and w(oo)
00
F*(O,l,O) are
3'~
the only other points of F IF''; satisfying the linear equation ("') or (,';"') respectively, will is a line of IT 2(F). :;. C,we;: L E B . Let {a} := L n O. l pEL"
a = pta) .. J )
By (2.11) we have:
2 (*) 2x l a+x 2 - x3 a
l ('' ' ') x3 ° =
Since w(a) 3 F "'1 F'"
= FA(a,_a 2 ,1) or w(a) = w(oo)
=
°for a +
00
for a F(O,l,O) is the only remaining pOint of
satisfying U') or (""") resp., will is aline of
IT
2(F).
With these lemmas we have proved: 2.14 Let 0 be an oval of a projective plane (P,B) with a sharply 3-transitive projectivity group rIO). P such that (P,B'o By (2.14)
U
Bl
U
Then there exists a subset B'o of the power-set of
B ) is a pappian projective plane in which 0 is a conic. 2
Theorem 1 is proved for the projective case.
For the attlne case
let 'A(O) be the oval group generated by perspectivities with centers in the affine plane, and frO) the oval group generated by perspectivities with centers in the extended projective plane.
Then one shows that fA(O) is 3-transitive and that
the condition (P 3) for fA(O) implies (P 3 ) also for f(O). completely.
So Theorem 1 is proved
3. SELF-DUAL OVALS WITH DIFFERENT PROJECTIVITY GROUPS In this section (P,B,O) will always be a projective plane with a self-dual ovalO.
We will give examples of self-dual ovals 0, where one of the groups frO)
or rIO) fulfills condition (P 3 ), the other does not. By the results of § 2 we have:
The projectivity groups of ovals and of quadratic sets
527
Let r(O) fulfill condition (P 3 ) and let F be the corresponding commutative field. Then there exists a line set B' such that: 3.1
a) (P,B') is the pappian projective plane of F; and 0 an ellipse, b) For any two distinct pOints p,q E P2 u 0 we have p,q n P 2 p,qEp,q'EB'.
=
p,q' n P , where 2
+ 2 and
Now let F be a commutative field with char (F)
0 the ellipse of the 2 projective plane over F given by the quadratic form q(x) = x + x x . Let 1 23 f(x,y) = q(x+y) - q(x} - q(y) = 2x l Yl + x2Y + x3Y2 be the corresponding bilinear 3(2) (2) 2 ,', form, and F/F(2) := {x.F : x E F} with F := {x : x E F } the set of
quadratic classes.
Then the point set P is decomposed into the subsets := {F*' x: q(x) E a.F(2}}, and we obtain:
Pta} := p(a'F(2» 3.2
0 = P = P(O), P = P(l), Po =
3.3
For any L E Bl we have L C P2
l
2
{P(x) : x E F'': with x
U
U 0
3.4 For any L E B2 we have L n P(r)
= P(l)
+ 1/1
for all rEF.
*
3.5 For all L E B we have L n P2
1(F(2»}.
U P(O}.
PROOF: For {('a, F b} := L n 0, c := f(a,b) (+ 0) we have q(a + c c- l rf(a,b) = r; hence F*(a + c-lr • b) E L n P(r). 6
$
-1
r b)
+1/1.
PROOF: By (3.3) and (3.4) we have LEBo to consider.
Since for A E Bl , An L ~
1/1
we have An L c P by (3.3). 2 3
1,
3.6 Let L E Band u E F with L = {F x E P L C p -=-q(U) 2
=-1
(F(2»
f(u,x}
=
O}.
Then:
F is pythagorean. ,,:
(2),':
PROOF: By (3.5) there is a point F a E L n P2 ; hence q(a) E F . Let F bEL with f(a,b) = O. Then, by f(a,u) = f(b,u) = f(a,b) = 0 and q(a)·q(b)·q(u) + 0, we
=
have q(a).q(b).q(u)E(-1)'F(2); hence q(u) (-l)·q(b). For any point Fi'.x E L we 2 2 have x = xla + x b and q(x) = Xl q(a) + x2 q(b). Hence: L C P2 ~ V Xl ,x 2 E F with 2
528
H. Karzel and M. Marchi
(Xl ,x Z) ~ (0,0) we have both pythagorean
<=>
xl q(a) + xl q(b)
E
F(Z) and F is
q(u) '" -1.". F is pythagorean.
The following conditions are equivalent:
~
o ' ~ ALE B0 => L C PZ ' b) F is an euclidean field.
a) B i
PROOF: let l LEBo l C P
=>
z
<=>
=
{F'" x
E
P: f(u,x)
a}.
=
Then by (3.Z):
q(u) Ef F(Z) u {a} and by (3.6): q(ll) E - F(Z)", F(Z) + F(Z) C F(Z) (2)
Hence a) is equivalent to F
.'. + (', (..'. \ F(2)
i. e.
F is euclidean. For F(Z) ~
F'" we have Po =
=
if>
and Bo
= if>
and hence:
Let (P,B,O) be a projective plane with a self-dual oval 0 such that one
of the groups riO) and riO) fulfills (P ). 3 field F, then (P,B) is pappi an.
If F(Z) = F* for the corresponding
To obtain examples (P,B,O) such that (P3) is valid for riO) but not for riO) we have to start from a projective plane over a field F with char(F)
1 F"',
F(2)
~
Z and
and by (3.1) we have to replace the intersections L n Po between lines
L and the interior Po of the ellipse 0 by other subsets of Po'
If F is an
euclidean field this can be done with an easy method: With the SUbstitution
= YZ - Y3 we obtain q(x) " Yl 2+y~
Xl " Yl' Xz = Y2
+ Y '
fine plane {(x,y)
:=
F"'(x,y,l)
l
+
(1 = ,;
3
((x,y) E F2 :
x3
i
x,y
= l}.
E
- y;, and in the af-
F} the ellipse 0 has the form
The line U of infinity (spanned by ("(1,0,0) and
F (0,1,0)) belongs to Bo; hence by (3.7) U C P and 0 C P \ U. 2 Now we consider also a Moulton plane over F where the x-axis is the bending line.
For two distinct points p,q
B2
P let p,qm be the joining line of p,q in the
BZ let LZ := (P u 0) n L, {ll' lZ} L no, 2 :" Po n (1 1,1 2 ), and L' :" L~ u LZ' Then (P, Bo U B1 u B ,0) with 2 :=' [L' : L E B } is a non-desarguesian projective plane with a self-dual oval Z
Moulton plane. L~
E
For any line l
E
--m
riO). 3 Another example (P,B,O) where r(O) fulfills (P3) but not riO) is the
0, where riO) fulfills (P ) but not
projective plane given by Hilbert (1899) in the first edition of his "Grundlagen
529
The projectivity groups of ovals and of quadratic sets
der Geometrie".
This was the first example of a non-desarguesian projective plane
Hilbert chose an ellipse in the euclidean plane over the reals and replaced the segments of secants by sections of suitable circles.
Also the example of T.
Vahlen (1905) in his book "Abstrakte Geometrie" of a non-desarguesian projective plane has this property.
4. THE PROJECTIVITY GROUP OF A SELF-DUAL OVAL In this section let 0 be a self-dual oval a projective plane (P,B).
Then we
may consider the two projectivity groups (O,r(O)) and (O,r(O)), which are 3-transitive by (2.6) and (1.1).
Now let us assume that (O,r(O)) fulfills the condition
(P 3) hence that (O,r(O)) is a sharply 3-transitive permutation group.
By theorem
1 there exists a subset B; of the power set of P such that (P,B; U Bl U B ) is a
2
pappian projective plane where 0 is a conic.
Since this conic 0 is self-dual, the
tangents cannot pass through a common point.
Therefore the corresponding commu-
tative field F has a characteristic different from 2.
By
§
2 every point p of P
is represented by F,·,(Pl,_P2) with (Pl,P2,P3 + (0,0,0). The determinant P3' Pl 2 _(X + x x ) is a quadratic form and the corresponding bilinear form 2x Y + l l 23 1 + x Y3 + x Y defines a bijective polarity n: P---~ B; U Bl U B2 with n(P ) = 6 3 2 1 2 l and n(P2) =B2 . For any pEP \ 0 the involution p is induced by the projective reflection of the projective plane (P,B;
Bl
U
U
B2) with center p and axis nIp).
Therefore we have: :=<{p : p E P2 }> of rIO) is also generated by the set {L : L E B2 } i.e. ~ is a subgroup of rIO). ~
The subgroup
~
By calculation one obtains:
{F
f,
(Xl ,X 2 ) x3 ' - x1
E
PGL(2,F): x x - x 1 4
X
2 3
(2)
EF
U
(2)
(-1) F
}
•
530
H. Karzel and M. Marchi
As we saw in
2 (cf. (2.12)v)) the addition group (F.+) of the correspond-
§
ing commutative field (F,+,') is isomorphic to the group F+ = J oo . k. + tion we obtain F = {p q ; p,q E [001 1 /\ p,q + oo}. ~
Since ([ 0011\ roc))
By calcula-
~
Pz we have
C
+ ++ + 4.3 F uF 'o=F UF (-1)'
(a
+
+ aEF)u{a(-l)'
aEF)
F+ u F+ (-1)' ~ ~.
Since the polarity TI interchanges the points of the tangent with the lines containing
00
we have TI([ooll\oo) = Bl
n [001
[00 1
1 and by (4.3);
through
00
The multiplication group (F*,,) can be considered as a subgroup of
'I
);= <{p ; p E
0,00 \
(o,oo}»
F;' (01 01 ) _
.0,"',..
x- l for
of index Z as follows; If we let t(x) -
,', u 'J:'.F'"
x E Fft - that is, t = then f{o,oo}= f(o,oo) u t.r(O,l) = F ~ By (4.2) t E P and a' E ~ if a E F(Z) u - F(2). Hence we obtain;
z
4.5
f{o,oo} n ~ = (a'; a E F(Z) u - F(Z)} u t·{a·; a E F(Z) u _ F(Z)} =«j) ; p
E
0,,, n P }>' Z
By (4.2) we see
-
From now on we assume that besides (O,r(O)) also the group (O,r(O)) fulfills the condition (P3)' (F,ffi,0). F= 0,,,
Then by
§
2, to (O,r(O)) belongs also a conmutative field
If we choose for constructing F and Fthe same points 0,1,00 E 0, then
F and + = ffi by (3.4). The polarity n P with the lines [[01 n [oolf n z 1 Q
a' = a
;
TI interchanges the points of the set B • 2
Hence for a E F(2) u - F(2) we have
i.e. a·b = a 0 b always holds if a E F(2) u - F(Z) or bE F(2)
U _
F(2).
Since (F,+,') and (F,+,0) are fields, the distributivity and conmutativity gives us; a,b = a 0 b for a
E
A or bE A where A ;=
(F,+) generated by the squares and -1. x = (2
-1
(x+l))
2
- (2
-1
(x-l))
2
{-l}>
+
is the subgroup of
Since for any x E F the equation
is fulfilled we have A = F and hence'
= ~.
The projectivity groups of ovals and of quadratic sets
531
We have proved the following THEOREM 2: Let (P,B,O) be a projective plane with a self-dual ovalO such that riO) fulfills the condition (P 3). a) (P,B) is pappian (and b)
riO)
z
Then the following statements are equivalent:
°a conic),
riO),
c) riO) fulfills (P 3 ), d)
5. THE PROJECTIVITY GROUP OF A PAIR OF LINES
In this section let (P,B,Q) be a projective plane with a quadratic set Q consisting of two distinct lines F,H E B. r
:=
r(Q)
r(F
=
U
H) fixes
00
:= F n H, then any element from
Therefore we can consider the affine plane obtained
00.
by omitting a line U E B with
If
00
E U and U + F,H.
For a point pEP \ (F
U
H) the
permutation p interchanges the lines F and H hence r contains the subgroup r'
:=
{y E r : y(F)
=
F} of index 2.
By dualization (P,B,Q) can be also consid-
ered as a 2-structure (P, Gl , G2 , K) in the sense of [5] or [6]; that means: Nl for each point pEP and each i E {1,2} there is exactly one generator G E G.
1
with pEG; this generator will be denoted by (p).; 1
N3
every two generators of distinct classes Gl , G intersect in exactly one point; 2 every element K of K intersects every generator in exactly one point;
N4
for any p,q E P with (P)l
N2
with p,q
E
+ (q)l
and (P)2
K
K) correspond the involutions with KE K
)
x
They generate a group r. 5.1
U
f Gl Il
there is exactly one K E
K.
To the permutations p with pEP \ (H K
+ (q)2
for X E Gl for X E G2 • From [5] , [6] we know:
The permutation groups (F,r') and (Gl,r') with
are 2- transitive.
r' .-
{y E r
y(Gl)~l)}
H. Karzel and M. Marchi
532 ~
If (F,r') or (Gl,r') fulfills condition (P 2) then F or Gl respectively can be provided with an addition "+" and a multiplication "." such that (F,+,·) or (G ,+,.) becomes a planar nearfie1d and r' 1 1"
= {[
= {[ a,b] : a,b E F /\ b
~
D} or
a,b] : a,b E G /\ b + D} respectively, where [a,b] (x) = a + bx. l
Let us now assume that (F U H,r) or (G l U G2 ,r) fulfills (P 3 ). (F,r') or (G , r') condition (P ) is valid and we obtain: l 2
Then for
Let DE F, hE H, x,y,z E D,h' {O,h}, a E F \ {O}, a' .- x y z(a) and
~
u := a,a' () D,h.
Then x y z
= -u.
PROOF: The map u x y z fixes the points 0, h and a, hence by (P 3 ) u x y z = id.; i.e. x y z = u. 5.4 The corresponding nearfield (F,+,.) is a commutative field. PROOF: By (5.3) the set {x y . ."t
transitive on F (F",·)
3'!
F\
{O},
x,y E O,h' {O,h}}is a commutative group acting
(x y :
r~=
hence
x,y E O,h' {D,h}}.
Because of
r~ the nearfield is a commutative field.
With this we have proved the
THEOREM 3: Ler (P,B) be a projective plane, Q : = F U H wi th F Q : = [ f]
U [
U {oo},
HU
h] wi th f, h E P and f
r := r(Q), [:= r(Q), r' := {y
r:
[' := {y E
E B and F
{"'}
y([ fl) = [f]}.
E
~
+ H,
h,
r : y(F)
F} and
Then:
a) If (If]' {f,h}, [-') fulfills (P2J then there exists a planar nearfield
(F,+,·) such that [-, direct product r' morphic
3'!
([a,b) : a,b E FA b ~ OJ,
e (£) with £2 = id.
and £[a,b)£
:=
{{F
isomorphic to the semi-
= [a,b] -1,
and (P,B) is iso-
to the projective closure of the affine plane (P , G U G U K)
the llearfield (F,+,') given by P := F
G2
r is
x
a} : a E F},
a 1 2 F} : a E F},
x F, G := {{a x a 1 K:= {(a,b]g : a,b E FA b ~ O} with'
over
The projectivity groups of ovals and of quadratic sets
[a,b] 9 := {(x,a + bx) :
X E
533
F}.
b) If (F,r') fulfills (P 2) then we have the dual situation to a). e) If one of the permutation groups (Q,r) or (Q,r) fulfills (P 3 ) then (P,B) is a pappian plane,
r'
e.
r' ~ {[a,b] : a,b
E
F,\ b
f
O} where (F,+,') denotes the
eoordinate field of (P,B).
BIBLIOGRAPHY 1. 2. 3. 4. 5. 6.
7. 8. 9. 10.
R. Artzy, A Pascal theorem applied to Minkowski-geometry, J. Geometry, 3 (1973), 93-105. W. Benz and S. Elliger, tiber die Funktionalgleichung f(l+x)+f(l+f(x))=l, Aequationes Math., 1 (1968), 267-274. F. Buekenhout, Etude intrinseque des ovales, Rend. Mat., 25 (1966), 333393. H. Karzel, Inzidenzgruppen. Vorlesungsausarbeitung von I. Pieper und K. Sorensen, Univ. Hamburg 1965. H. Karzel, Zusammenhange zwischen Fastbereichen, scharf zweifach transitiven Permutationsgruppen und 2-Strukturen mit Rechtecksaxiom, Abh. Math. Sem. Univ. Hamburg, 32 (1968), 191-206. H. Karzel and H.-J. Kroll, Perspectivities in Circle Geometries. Geometry von Staudt's Point of View. Edited by Peter Plaumann and Karl Strambach, Nato Advanced Study Institutes Series; Series C: Mathematical and Physical Sciences, Volume 70, 51-99. D. Reidel PUb. Compo Dordrecht 1981. H. Karzel und K. Sorensen, Die lokalen Satze von Pappus und Pascal, Mitt. der Math. Gesellseh. Hamburg, Bd. X, Heft 1, (1971), 28-39. W. Kerby, On infinite sharply multiply transitive groups, Hamb. Math. Einzelschriften 6, Gottingen 1974. W. Kerby und H. Wefelscheid, tiber eine scharf 3-fach transitiven Gruppen zugeordnete algebraische Struktur, Abh. Math. Sem. Univ. Hamburg, 37 (1972), 225-235. G.P. Kist, Projektiver AbschluS 2-gelochter Raume, Res. der Math., 3 (1980), 192-211.
ACKNOWLEDGEMENT: Supported in part by the research group G.N.S.A.G.A. of the Italian National Research Council (C.N.R.) while the first Author was Visiting Professor at the University of Brescia.
Institut fUr Mathematik Technische Universitat MUnchen Arcisstr. 21 0-8000 MUnchen 2, FRG
Istituto Matematico Universita Cattolica via Trieste 17 1-25100 Brescia, Italy
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Annals of Discrete Mathematics 18 (1983) 535-548 © North-Holland Publishing Company
535
ON R-SEQUENCEABILITY AND Rh-SEQUENCEABILITY OF GROUPS A.D. Keedwell
ABSTRACT A finite group (G,·) of order n is said to be R-sequenceable (or near-sequenceable) if its elements a, a l , .•. , an- 1 can be ordered in such a way that o the partial products bo = ao ' bl = aoa l , b2 = aoal a 2,···, ... ,b n_2 = aoala2 ••. an_2 are all different and so that the product bn- 1 = a0 a l a 2•.• an- 1 = b0 = e. We show that a finite dihedral group is R-sequenceable if and only if it is of doubly even order and that a non-abelian group of order pq such that p has 2 as a primitive root, where p and q are distinct odd primes with p < q, is R-sequenceable.
We also
discuss the generalized concept of Rh-sequenceability and its relevance to the construction of sets of mutually orthogonal latin squares.
1. INTRODUCTION We begin with some definitions. A finite group (G,') of order n is said to be sequenceable if its elements can be arranged in a sequence a
a e, a , a , •.• , a _ in such a way that the l 2 o nl partial products bo = ao ' bl = aoa l , b2 = aoala2, ••• ,bn_l = aoala2 •.. an_l are all distinct (and consequently are the elements of G in a new order). It is said to
be R-sequenceable (see [2]) or near-sequenceable (see [5]) if its elements can be ordered in such a way that the partial products bo = ao ' bl = aoa l , b2 = aoala2, .•. ,bn_2 = aoala2 •.. an_2 are all different and so that the product bn- 1 = a0 al a 2···a n- 1 = b0 = e. A one-to-one mapping 9 ~0(g) of a finite group (G,·) onto itself is said to be a complete mapping if the mapping 9 ~ $(g), where $(g) to-one mapping of G onto itself. -1
= 9.0(9) is again a one-
If (G,') is R-sequenceable, then the mapping
= bi bi +l = ai +l for i = 1,2, ... ,n-2,a(b n_l ) = a 1.0 (c) = ao = e, where cis the element of G which does not occur in the set {bo,b l ,b 2, ... ,b n_1}, is a com-
8(b i )
A.D. Keedwell
536
plete mapping, as was shown by l.J. Paige in [7).
Thus the condition that (G,·)
be R-sequenceable is a sufficient condition for G to have a complete mapping but it is not always necessary.
For example, for abelian groups, the condition
b I = e is alone sufficient (l.J. Paige [61). When and only when a group G has a ncomplete mapping, the latin square formed by the Cayley table of G has an orthogonal mate (see section 1.4 of [1 I for a proof) and, if the group is R-sequenceable, this orthogonal mate can be constructed by the procedure called the column !':ii,
[;10:1 in [4).
permutation
(See a 1so section 7.4 of [ 1 J).
The column method makes use of the
defined above which we can write in cycle form as (c)(b b2 b l 3 -1 -1 -1 -1 bn_l ) and for which the elements bl b2, b b , ••• , b _ b _ , bn_lb are the 2 3 n2 n l l ~
non-identity elements of G. a re-ordering of the non-identity elements of G, then the column method permits the construction of at least three mutually orthogonal latin squares based on the Cayley table of G. r,lore generally, we shall say that a group (G,') of order n is Rh-sequenceif n-1 of its elements can be arranged in a sequence c ' c ' ••• , c n_ in l 2 l -1 such a way that the set of elements c.1 c.1+ 1 for i = 1,2, ... ,n-l, are all distinct
able
(where arithmetic of suffices is modulo n-l), and likewise the sets of elements -1
-1
-1
c i c i +2' c i c i +3 ' .•• , c i c i +h' In particular, a group which is Rl-sequenceable is R-sequenceable with -1 -1 -1 -1 ao = e and a 1 = cn_lc l ' a 2 c l c 2' a3 = c2 c3 ' ... , .... a n_l = c n_2c n_l • If a group (G,·) is Rh-sequenceable, the column method permits the construction of at least h
+
1 mutually orthogonal latin squares based on the Cayley table of G
(as has been shown in [41 and [I J). R. Friedlander, B. Gordon and M.D. Miller have investigated the R-sequenceability of various classes of abelian groups in [21.
Here we obtain some results
on the R-sequenceability of certain non-abelian groups and make a few observations about Rh-sequenceability. In the investigation of sequenceability of groups, the concept of a quotient aequencing has played an important role.
We shall make use of the same concept in
the present paper. DEFINITION: If P(~)
is any sequence ul ' u2 ' ••• , un of elements of a group G, then will denote its sequence of partial products VI = ul ' v2 = ul u ' 2 a
537
On R-sequenceability and Rh-sequenceability of groups
v3 = u1u2u3,···,···, vn = u1u2···u n• Let G be a non-abelian group of order n which has a normal subgroup H of order w, and let G/H = gp {x 1,x 2' ••• ,x t } where t
= n/w. A sequence
a
of length n consisting of elements of G/H is called a
quotient sequencing of G if each xi' 1
The image under the mapping
~
~:G ~
i
~
t, occurs w times in both a and P(a).
G/H of either a sequencing or a near-se-
quencing of G is a quotient sequencing of G.
In the latter case, the last entry
in the quotient sequencing is necessarily the identity element of G/H.
2. R-SEQUENCEABILITY OF DIHEDRAL GROUPS 1 The dihedral group Dn = gp{a,S : an=e:, S2=e:, as = sa- } of order 2n cannot be R-sequenceab1e if n is odd since the product of all its elements takes the form sna s for some integer s, 0 ~ s < n, when the relation as = Sa- 1 is used to simplify it and this cannot be equal to the identity element unless n is an even integer. We have been able to show that this necessary condition for R-sequenceability is also sufficient. That is, the dihedral group Dn of order 2n is R-sequenceab1e if and only if n is an even integer. We exhibit a suitable R-sequencing in FIG. 1. This is based on the quotient sequencing x
x m
l's
m l's
where n = 2m, H = gp{a : an=e:}and D /H = {l,x}. n
have to be treated separately. values n
x
x
(2m-1)
x' s
The cases n = 4h and n = 4h - 2
(The R-sequenceabi1ity of Dn for the particular
= 2,4,6,8,10,14,20,22,32,34,36,38,50,64,66 and for eighteen other values
between n = 100 and n = 4084 has already been established in [21].)
3.
R-SEQUENCEABI LlTV OF NON-ABELIAN GROUPS OF ORDER pq
In a recent paper [5], we showed that the non-abelian group of order pq, where p and q are distinct primes with p < q, is sequenceab1e whenever p is a prime which has 2 as a primitive element.
We show next that these same groups are
also R-sequenceab1e. Our method is almost a copy of that described in [5] and uses the same quotient sequencing. Let H be the unique normal subgroup of order q in the given group G of order pq and suppose that the factor group G/H is generated by the element x. A neces-
A.D. Keedwell
538
sary condition for the existence of G is that q integer h.
We suppose that p has 2 as a primitive element and let 0 satisfy the
congruence 20 = oh
=1 ~
= 1 + 2ph for some positive
2h
mod p.
= 1 mod
p.
Then 0 is a primitive element of GF[ pI since The following is a quotient sequencing for G (as is
shown in [5 J): a sequence of 2ph l's, followed by x, followed by a sequence of 2ph-l 0-1 0 2_0 O3 2 op-2_ op-3 1_op-2 (p-l)-tuples x x X -0 x x , where indices are computed modulo p, followed by a sequence 2P- 2 1_op-2 2 4 B 0-1 02-0 °3 -0 2 oP- 2-oP- 3 x x x ... x 1xxx ... x x
2
;-2
The partial products are 2ph l's, followed by the sequence x X Xo x 2 p-2 p-2 p-3 p-4 o 0 0 0 a o repeated 2ph-l times, and then the sequence x x x •.• x x x x O
~
2
XV
x° x 1.
(We use the fact that 0p-2
2.)
q let G be generated by two elements a, b such that a = e, bP ab = bas, where sP
= 1 mod
q and e is the identity element of G.
= e and
Then,
x (buav)(bxa Y) = bu+xa vs +Yand G has a unique normal subgroup H = gp{a : aq = e} of The natural homomorphism G ~ G/H maps G onto the cyclic group C with 2 2 p-1 P elements 1 = H, x = bH, x b H, ... , x = bp-l H.
order q.
let r be a primitive element of GF[ q). Then (compare [5]) we seek a nearh h-l h+l h h-2 h-3 r -r r -r r-r sequencing of G in the form e, a , a , ..• , a , b, 2 2 3 boP-2-oP-3 al(p-2) bl-oP-2 al(P-l) 0-1 a1(1) 0 _0 al(2) 0 _0 a1(3) b a , b a , b a , ... , a, a , (p-l) 1 (1) 2 (2) (1) 2 (2) 1 p-2 b0-1 a:<2 ,b 0 -0 aCl2 , ... , b -0 aa2 , b0- aa3 ,b 0 -0 aa 3 , ... , ... , (p-l) p-2 p-3 (p-2) p-2 p-3 p-4 lJ bo -0 aa q- 1 ,a, b0 ,b 0 ,b 0 , •.• , b0 ,b -1 aa q-1 , where jJ
= r
h-1
-r
h-2
We make use of the facts that op-r = 2r - l , that 1_op-2 p-r p-r-l that 0 -0 = -0 p-r ,where 20 1 mod p.
-1, and, generally,
=
This sequencing gives rise to the partial products which are listed in FIG. 2.
(For ease of printing, we have represented the element bVa w by the
ordered pair v, w in FIG. 2.) i
= 1,2, •.. ,q-2 and
j
The expressions denoted by E~j) (for 1
= O,1, ..•• p-2) in FIG. 2 are as given below. We equate
these expressions to expressions in a primitive element t of G8 ql exactly as in 15\.
We also make the definitions a~:~
ta~j) for i = 1,2, •.• ,q-2 and
On R-sequenceability and Rh-sequenceability of groups
539
= 1,2, ••• ,p-l.
j
. E(O) 1
=
2 3 p-2 i 1 l-a +Q (p-l) -(t£ a.(1) S ]-a +a (2) . S l-a +a (3) . S ]-a +•.• +a (p-2) . S . - - t£-l)t - , 1
1
1
2
1
3
1
u.
4
a-l +Q.(1)1 -_ t l(t£ - t£-l)t,-l , E1(1) . = a.1(2) S a-a +a 1(3) . S a-a +Q 1(4) . S a-a +.•• +Q 1(p-l) . S 1+ 23 24 25 2 u . (2) (3) a -a (4) a -a (5) a -a (1) a -a (2) _ t 2(t£_t£-l)t,-1 E. = ex. S +ex . S +ex . S +... +ex . +1s +ex . +1 , 1
1
1
1
l '
From these equalities, we obtain
(p-2) (p-3) aP- 2_a P- 3 _ (p-2) (p-2) E] -E l S - ex -exl 2 l_aP-2 tE 1(O)_E(P-2) 1 5
= t (p-l)_ (p-l) exl
exl u Hence, ex~l) = t£-l(t 1_ sa-1), u
u
a~2)
= t£-l(t
ex~3)
= t£-l(t 3_ t
2
2_ t lsa -a),
U
3 2 2s a -a ),
U
It is easy to check, using FIG. 2, that, with these choices for the expressions E~k) and the indices ex~j) in terms of the primitive element t of GF[q], all 1
1
540
A.D. Keedwell O
j
8 (j)
the partial products of the form b
=t
~
which occurs provided that
a 1
Up_2
(-t
are different including the last one
\!,-2)
).
(This has been illustrated in [51).
Finally, it is easy to see that all but the last of the partial products which are powers of a alone are different.
We require in addition that the last
such product is equal to the identity element e.
-~
~s-2
+ a(P-1)(1_s-1) +
q-l
= 0 mod
That is we require u
Using the fact that ~ = t p-2(_t\!,-2) and
q.
u
that a(P- 1) q- 1
= t{-2(t-t p-2 S -1). this becomes
u 2 2 2 U 2 1 -1 u 2 2 2 u 2- 1 t p- t~- +t l - (t-t p- s- )(l-s ) - t p- t£- s- = 0, whence t p-1. Since is. r
u
h-2
= r h-1 -r h-2.1n ( r- 1) = t
£-1
FIG. 2, we must also have r
• since t
u -1 p-2
h-1
-r
h-2
= - 1•
We conclude that an R-sequencing (near-sequencing) of the non-abelian group of order pq, where 2 is a primitive element of GF[ pI, can be obtained by the following procedure. First choose one of the roots s of the congruence sP compute the integer
= 1 mod
q and also
such that 20 a 1 mod p.
Then choose primitive elements r Up_2- 1 and t of GF[ql and integers ul ,u ' ••• ,u -2 such that, modulo q, t = -1 and 2 U U -u 2 u -u ~ 2 U -u p-2 p-3 1 0-1 2 1 0 -0 3 2 0 -0 p-2 p-3 0 -0 t I' s ,t I' s , t I' s , ••• , t Fs (s i nce we a
require a~j) F 0 for j i
= 1,2 •••.• q-1
modulo q.
and j
= 1,2, ••• ,p-l). Compute the indices a~j) for = 1.2, .•• ,p-1. Finally choose h so that1 r h-2 (r-1)
t
£-1
This gives an R-sequencing for arbitrary choice of the index \:'.
In order to illustrate the procedure. we have carried out the necessary calculations for the case p
=5
and q
= 11
and we display the resulting near-
sequencing of the non-abelian group of order 55 in FIG. 3. Since 20
= 1 mod
p, we have 0 = 3.
We may choose s
3 (since 3
5
= 1 mod
11)
and r = t = 2 (since 2 is a primitive element of GF[ 111 ). Then 5 t = -1 mod 11 so u 6. We find that a~l), a~2). a~3) (and a~4)) are all non3 u1
u2
zero if we choose t t l .
We also put t£-
1
= 1.
Then ~ = 1
h 2
=2
- (2-1), so
On R-sequenceability and Rh-sequenceability of groups
h
(1)
2. We obtain a'1 a
(2) l
a
(3) l
a
(4) l
u t 1_ so-1
= 1 _ 32 = 3,
u u t 2-t
2
541
\0 -0 1 - 36 = 9, u u 3 2 t 3_ t 2so -0 = -2 - 318 = 4, u 1-0 3 = t( l+s -1 ) = 2(1+4) = -1, = t - t 3s
where these latter calculations are made modulo 11. We also have the group 2 29 3 generating relations ab = ba 3, ab = b a , ab 3 = b a5, ab 4 = b4a4.
4. Rh-SEQUENCEABILITY In general, the problem of determining which groups are Rh-sequenceable seems to be a difficult one.
For abelian groups G, it is certainly necessary
that the Sylow 2-subgroup of G should be either trivial or non-cyclic. However, even for cyclic groups, the problem is not trivial. group is R-sequenceable if and only if it has odd order. G. Ringel and also has been proved in [2].
(See [31.)
Certainly a cyclic
This has been stated by
The following R-sequencings for
C4m+ 1 and C4m+3 are alternative to those given in the latter reference. For C4m +l an R-sequencing is
o 4m,
2, 4m-2, 4, 4m-4, 6, 4m-6, ••• , •.. , 2m-2, 4m-(2m-2), 2m, 2m-l, 2m+3,
2m-3, 2m+5, .,., ••• , 2m-(2m-3), 2m+(2m-l), 2m-(2m-l), 2m+l. For C +3 an R-sequencing is 4m 0, 4m+2, 2, 4m, 4, 4m-2, 6, 4m-4, ••. , .•• , 2m-2, 4m-(2m-4), 2m, 2m+l, 2m+3,
2m-l, 2m+5, 2m-3, ••• , .•• , 2m-(2m-3), 2m+(2m+l), 2m-(2m-l), 2m+2. However, a computer search has shown that Cg is not R -sequenceable. A 2 similar search has shown that C15 is not R3-sequenceable but that it has 32 isomorphically distinct R2-sequencings. The cyclic group C is R2-sequenceable but 2l at the present time the question whether it is also R3-sequenceable remains undecided. The Rh-sequenceability of cyclic groups of prime order is covered by the following theorem:
THEOREM: An eZementapY abeZian group of opdep pn is Rh-sequenaeabZe fop
542
A.D. Keedwell
1, 2, ••• ,pn_ 2•
h
PROOF: If '" is a primitive element of the Galois Field GF[pn] , the following sequencing of the elements of the additive group GF[pn) (which is elementary abelian) has the required properties: 0, w-1,
2 uJ
-w,
:.J
3 2
-w ,
n n ... , wp -2 -w p-3
n
1-w
p -2
(The n-l elements c 1 ' c ' ••• , c _ given in the definition of R -sequenceabi1ity n l h 2 pn_2 2 pn_2 2 are then 0, w-1, w -1, ••• , w -lor 1, w, w, ••• , w .) This gives at once the well-known result that COROLLARY: i"porr. an elementary abelian group of order> pn a set of pn_l mutually ol'thogonal latiY!. squares all
~ased
on that gr>oup can be constructed and hence a
c.csm'Jue[J~an PpoJ£ct;:;)e plane of order pn.
2
8
In GF191, there exists a primitive element w such that w = w + 1, w = 1. 3 3 We can write C x C = {a : a = e} x {b : b = e} or, in additive notation, as 3 3 ;0,1,-1)$ {O,w,-wl.
Our R -sequencing is 7 23243546576
0, w-l, w
-~, ~ -w
or 0, w-1, 1,
w,
, ~ -w , w -w , w -w , W -00 ,
w+1, -w+1, -1,
-w,
1-00
7
-w-l.
In multiplicative notation, we have the sequencing 2 22 2 2 2 2 e, a b, a, b, ab, ab 2 , a , b , a2b with partial products e, a 2b, b, b , a, a b , 2
ab , ab, e.
Thus the sequence of n-l elements c ' c ' ••. , c _ given in the l 2 n l 2 2 2 2 2 definition of Rh-sequenceabi1ity becomes e, a b, b, b , a, a b , ab , ab for this group. The first "differences" c~l c. 1 are as listed in the sequencing above. -1 1 1+ 2 2 2 2 2 2 The second "differences" c. c. 2 are b, ab, ab , a , b , a b , a b, a; and so on. 1
1+
543
On R-sequenceability and Rh-sequenceability of groups
,
R-seguencing of D4h e
I a
2h-3 a
-(2h-4) a
a
2£-1
a
a
-(2£-2)
a
a
a
-2
a
a
a
a
S
a a
a
2h-2
a
2
a
a
h-Hl
a
a
a
h
a
Sa
2 Sa
-3 Sa
2£-2 Se.
-(2£-1 ) Sa
2h-2 Sa
a
2h-4 a
-(2h-5) a
2£-2 a
-(2£-3) a
2 a
-1 a
3h
Sa
3h-l
a
3h+l
a
3h-2
a
2
h-£ hH-2 h-£+l
h-2 h h-l
Sa Sa
-2 Sa
3 Sa
Sa
3hH-l
a
3h-£
a
-(2£-2) Sa
2£-1 Sa
- (2h-l )
~
2h-3
4h-3
3h-Hl
-1
2h-2
-(2h-3)
a
h-l
Sa a
a
h+l
-1
1 e
a
hH-l
Sa
Partial products
2h-2
h-£
Sa a
a
a
a
a
2h-l
-(2h-2)
a
R-SegUrCing of D4h - 2 e
e
2h-l
a a
Partial products
Sa a
2h-3
(continued over)
Sa
3h-2 3h-l 3h-3 3h
3hH-3 3h-£-1 3hH-2
2h 4h-3
544
A.D. Keedwell
R-sequencing of D4h 0.
0.
Partial products
-(2h-l) 8a
-2h
2h
8'1 6':1
a
2h
e·),
Cl
2h+l
Sa i3a Sa
4h-2 a
Sa 13·:1
0.
13;,.
80.
3h+l
Sa
Zh+l a
4h-5 0;
Sa
a
2h-l e
3h-3
2 0;
h-l
3h-l
So;
4h-l 2h-l
h-Z
Sa
13
3h+2
Sa
Zh
4h-4
So.
2h-2
2h-l
2h
h+l
Sa a
a
Sa
Sa
3
S
2h-l
h-l
h
2
0;
Sa
Sa
Sa
2h-l
2h-l
3h-l
4h-l
B:t
Sa
Sa
Sa
Partial products
-(2h-2)
Sa
2h+l Zh+2
a a
S
2h
R-seguenci ng of D 4h 2
h
3h
4h-3 0; Sa Sa
FIG. 1 (continued)
Zh-3 Sa
2h-2 e
2h-2
On R-sequenceability and Rh-sequenceability o[groups
0,
°
h h-l 0, r -r 0, r
0, r
a
p-2
h+ 1-r h
h-2 h-l -r
,-~s
1,
a P-
-~s+ (
a
=- ~
2
(1) aP- 2_a
+a 1
5
(2) aP- 2_a 2 (p-3) a P- 2_a P- 3 (p-2) +a 1 5 +•.• +a 1 5 +a 1
2 2 . (p-2) 1_a P(p-l)) (1) 1-a (2) 1_a 5 +a 5 +••• +a 1 5 +0: 1 1 1
2 3 a (1) ( (2) a_a +0: 1 + a 1 5 +0: 1(3) 5 a-a +••• +0: 1(p-l) 5 a-l +0:2(1)) 2 2 2 3 2 2 2 a (1) a _a (2) ( (3) a _a (p-l) a _1 (1) a -a+a (2)) a ,-~s +0: 5 +0: 1 + (Xl 5 +••• +0: 1 5 +0:2 5 2 1 a,
-~s
FIG. 2 (contd. on next page)
545
A.D. Keedwell
546
a
J
p-2
, -us
o
p-2 , --'s o ~
p-2 p-2
p-2 p-2 p-3 q-2 (1) 0 -0 (p-3) 0 -0 (p-2) E(p-2) +Ct 1 5 +•. ·+'-'1 s +a 1 +i~l; (1) C +u 1 s
p-3 ,
-\15
c
+a
(1) ~ S'
p-3
-0
p-2 p-3 ~ q-2 (p-3) 0 -0 (p-L) (p-2) +•.• +a 1 5 +a 1 +;=1; Z E +"~
p-3 p-4 q-2 p-2 (p-4) 0 -0 (p-3) (p-3) (p-2) 0 5 +a +.1: E. +(a 1 +[1)5 1 1 1 1 =1 1 qp-4 p-4 p-4 p-5 q-2 p-4 , o (1) G -0 (p-5) 0 -0 (p-4) ~ E(p-4) J +a 1 5 5 +a 1 +;~1; -u5 + .•• +a 1
,1
p-3
p-2
-0
+... +a
(p-2) ) 0 +( aq- 1 +u 5
p-2
+0
p-3
(p-3) 0 +Ct q- 1 5
p-3
q-2 p-2 p-3 2 p-3 p-4 2 ) 0 +0 +••• +0 +0 (p-3) 0 +0 + •• . +0 +0 + 1 ,-us+i~l E;(O) +«p-2) aq_1 +\.1 5 +a _ 5 ... q 1 (2)
.•. +a q- 15
0
2
+0
(1)
+a q- 15
0
q-2 (0) (p-2) p-2 ( p - 3 2 -U5+.1: E. +(a 1 +U)5 1- 0 +a P- 3)5 1- 0 +••• +a(2 )5 1- 0 +a(11)5 1- O 1= 1 1 qq- 1 q- 1 q2 q-1 (1) -0 (2) _0 (p-2) _op-2 (p-1) _op-1 (p-l) _op-1 _op-2 0, -\1+,'-_):1 (a. 5 +a. S +..• +a. 5 +a.1 S )+a q- 1 (1-5 )+U5 1 1 -1 -2 (p-1) -\.1+'-' q- 1 (l-s )+\.15
FIG. 2 (contd)
On R...sequenceability and Rh-sequenceability of groups
Sequencing e+
a a a a a
2 4 B
5 10
a
9
7
a
a a
3 6
b 23 ba ba 9 3 4 ba b\lD 26 ba b/ 3B ba 4 9 ba 2 ba 3 ba b3a 5 4 b/ 2 2 ba ba 6 lO b3a 4 3 ba 24 ba ba 3 9 ba b\6 2B ba 2 ba
Partial eroducts
i
e
2
a a a a a
6 3 B 7
a
5
a a4 10
a baB 3 9 ba 43 ba 2B ba 9 ba 3 lO ba 4 4 b a 26 ba b 3 ba 4 6 ba 2 b
l
Sequencing
Partial products
(contd) 3 b/ 4 ba 25 ba 4 ba 33 ba 4 2 ba 2 lO ba baB 6 b3a 44 ba 9 b2a 5 ba 3 ba B b4a b2/
(contd) 2 ba 5 ba 36 ba 4 b 23 ba 3 ba 34 ba 4 9 ba 2 b/ balD b3 4 5 ba 24 ba 2 ba 3 3 ba b\B 29 ba 2 lO ba 4 2 ba 8 b3a
balD 3 2 ba a 2 b 4 b b3
b/
4 5 ba
e
4
ba 3 5 ba 4 lO ba 25 ba ba 3 2 ba 4 b / 2 b 6 ba 3 b/ 4 ba
FIG. 3
547
548
A.D. Keedwell
BIBLIOGRAPHY 1. 2.
3.
4. 5.
6. 7.
J. Dines and A.D. Keedwell, "Latin Squares and their Applications". (Akadimiai Kiad6, Budapest/English Universities Press, London/ Academic Press, New York, 1974.) R.J. Friedlander, B. Gordon and M.D. Miller, On a group sequencing problem of Ringel. Proc. Ninth S.E. Conf. on Combinatorics, Graph Theory and Computing. Florida Atlantic Univ., Boca Raton, 1978. (Congressus Numerantium XXI, Utilitas Math, 1978), pages 307-321. M. Hall and L.J. Paige, Complete mappings of finite groups, Pacific J. Math., 5 (1955), 541-549. A.D. Keedwell, On orthogonal latin squares and a class of neofields, Rend. Mat. (Roma) (5) 25 (1966), 519-561. A.D. Keedwell, On the sequenceability of non-abelian groups of order pq, Discrete Math., 37 (1981), 203-216. L.J. Paige, A note on finite abelian groups, Bull. Amer. Math. Soc., 53 (1947), 590-593. L.J. Paige, Complete mappings of finite groups, Paoific J. Math., 1 (1951), 111-116.
Department of Mathematics, University of Surrey, Guildford, Surrey, GU2 5XH, England
Annals of Discrete Mathematics 18 (1983) 549-552 © North-Holland Publishing Company
549
ON THE CHARACTERIZATION OF FINITE MIQUELIAN MOBIUS PLANES Hans-Joachim Kroll
1.
One of the main problems in the theory of finite inversive geometry is the question whether any Mobius plane of odd order is a Miquelian Mobius plane. Any affine derivation 1A (p) of a
t~iquelian
Mobius plane is a Pappi an affine plane, and
the circles not containing the point p are conics in 1A(p). And it is even unknown whether these properti es characteri ze the
r~i que 1ian
r·lObi us planes.
I n the
case of a finite W6bius plane of even order the mentioned properties are indeed sufficient for the theorem of Miquel, but for other Mobius planes the problem is unsolved.
However, we shall give a partial result concerning the case of odd or-
der (cf. 4.).
2.
Let us consider a Mobius plane (P,R)l) of odd order q. define R(oo) := {C lA(oo)
:=
E
R 1 00
E
(P\ {oo}, RH''').
For a point 00
E
P we
C}, R("')"':= {C\ {oo} ICE R(oo)} and
1A (00) is an affine plane.
Let 1A(oo) denote the projec-
tive closure of the affine derivation 1A (oo) and let F be the line of infinity. Any circle C E R not containing 00 is an oval of 1A (00) with C c P \ roo}.
For any
ovalO let T(O) be the subset of points of F incident with a tangent of the ovalO. By the theorem of Qvist (cf. [21 we have (1)
IT (0) I = '21 I F I = .9:'2J.
fol' GJ1y ova Zoe P\ r",}.
1) For the definition cf. [1), [3). circles.
P is the set of points and R the set of
550
H.J. Kroll
3. Now let (P,R) be a Miquelian Mobius plane of odd order q and (2)
For any
17010
ro
E P.
ci1'C!les A,B E R\ R(oo) we have T(A) = T(B) or T(A) = F \ T(B).
PROOF: Since (P,R) is a Miquelian Mobius plane of odd order there exist a commutative field K with char K P\
(~J
* 2 and
= K x K and A = f(x,y)
E
a non square n E K" := K \ {OJ such that
K x Ki
/ -
n/ = l}.
For the circle B there is a
translation, of IA(oo) and abE K": with ,(B) = {(x,y) E K x K1 / -
ni = b}.
If
there is an (xo'yo) E A and an (xl'Yl) E ,(B) such that the tangent of A through (xo'yo) and the tangent of ,(B) through (Xl'Yl),,:have the same point of infinity, then (xl'Yl) E K(x ,y ), i.e. there is an a E K with (xl'Yl) = a(x ,Y ), hence 2 2 0 0 2 2 2 O2 0 b = xl - nY = a (x - ny ) = a , and therefore the dilatation 8 : K x K --.... K x K, l -1 0 _1 0 (x,y) -~ (a x,a y) maps the circle T(B) onto A, hence T(A) = T(8 ,(B)) = T(B). Together with (1) we are finished.
By the theorem of Qvist and (2) we obtain Let A,B,C E R such that IA n BI = IB n C 1= 1 and 00 E C, 00 ~ A,B.
(3)
Then
therc::s a circZe DE R with 0 n C = {oo} and ID n AI = 1.
(4)
Let C c P \ {oo} be any oval and A E R \ R(oo) such that
i.,et q == 3 modulo 4.
T(C) = T(A).
Then C E R.
PROOF: The assumption on the order q implies -1 assume A = {(x,y) E K x K1 / + / IA (00).
C
By B. Segre [4] the oval C is a conic in
Then there is a translation T of IA (00) and r,s,c E K such that
T(C) = {(x,y) E K x KI / T(C)
= l}.
~ K(2):= {/Ix E K1'}. We may
P \ fool.
+ 2rxy + si = c}.
This implies s - r
2
2 K.
E
Since C C P \ {oo} = K x K, also ,': 2 2 Let WE K with w = s - r , w t s
and let ex
: K x K - - > K x K be the affine map of 14.(",), defined by -1 -1 ,': a(X,y) = (x,w rx + w sy). Then there is a dE K such that D :=
Cl
T(C) = {(x,y)
I/
+
i
= d} is a circle of R.
F := (F \ T(O)) n T(C). flow we show 2 (a) If r to or if r = 0 and s
Let F := T(D) n T(C) and l
none of the sets F ,F is empty. l 2 For mE K let (m) denote the point of infinity of the line K(l,m). Let t
u = (00) be the point of infinity of the line K(O,l).
For the extention ~ of the
55l
Characterization of finite Miquelian Mobius planes
--
-
affine map a onto 1A(00) we have a{u)
-
-1-1
and a{{m)) = (w r + w sm).
a U
Besides u
there is exactly another fixed point f in F. The equation T{O) F
=
T{O)
U
=
= ~(T{C))
;(T{T{C)))
implies Fl
* ~,
for otherwise
T{C) and u = ;(u) ~ T{C), T{O); but u E F.
Now let us assume F2 =~.
Then T{D) = T{C)
= ~(T{C)).
We define
F3 := F \ {u,f}, T := F3 n T(O) and S := F3 \ T. First we consider the case r (-m)
E
= (2s
-1
* O.
If (m)
-
T{C).
Since arT) = T we have (m)
r + m)
E
T.
Hence for any (m)
E
E
E
T(O)
= T(C) then -(m) :=
T if and only if - a--1 (- a(m))) -1
T the orbit F{m) := {(2s rz+m) I z
contained in T, and therefore IF{m) I = char K := p devides ITI. that p devides lSI too.
lSI
E
* 1 the
T{O) if and only if d
E
K(2), hence ITI
= m - 1 and
-1
E
T then ;({m))
Similarly follows
second fixed point of ; in F is f = (D),
= m + 1 or ITI = m + 1 and lSI = m - 1. Let a := w s, then
If (m)
E
~((m))
T, and therefore the orbit M(m) := {(azm) I z
11 of the cyclic subgroup of that 11 devides lSI too, hence 11
{,
= (am). Z} c T,
E
hence the order
K devides
way follows
devides I lSI - ITI I
a
Z} is
Hence p devides ITI + lSI = q - 1 in contradiction with
the fact that q is a power of p. In the case of r = 0, w2=s and we have u,{O)
E
ITI.
In the same
= 2, but
* 1,-1.
= T{A) = T{C) or T{O) = F \ T(C) by (2), hence Fl = ~ or F2 = ~ and therefore r = 0 and s = 1 by (a), i.e. T{C) E R, hence C E R. (b) 0 E R implies T{D)
4.
By comparing the set of circles of a given Mobius plane with the set of circles of a suitable Miquelian Mobius plane we obtain from (4) the THEOREM: Let (P,R) be a Mobius plane of order q
=3 modulo
4 and let 00
E
P be a
point such that the follohling thlo conditions hold: 1) 1A ( 00) is a Pappian affine plane.
2) Let A,B,C E R hlith IA n BI exists a 0
E
R hli th 0
nC=
{oo}
and
= IB n CI = 1 and
Ion AI
= 1.
Then (P,R) is a MiqueZian Mobius pZane.
00 ~ A,B, 00 E C, then there
H.J. Kroll
552
PROOF: The second condition implies (2) for (P,R).
Let A E R, 00
Segre [4] A is a conic in the Pappi an affine plane VI. (00).
~
A.
By B.
Hence there is a second
subset R' of the power set of P such that (P,R') is a Miquelian Mobius plane, A E R' and R(oo) = R'(oo). For any C E R, 00
~
(4), i.e. R C R'.
By (2) there is a circle BE R' with T(B) = F \ T(A).
C we have T(C) = T(B) or T(C) = T(A) by (2), hence C E R' by This implies R = R'.
BIBLIOGRAPHY 1. 2. 3. 4.
W. Benz, l'orZesungen weT' Geometrie deT' AZ.gebren. Berlin-Heidelberg-New York 1973. P. Dembowski, Finite Geometries. Berlin-Heidelberg-New York 1968. W. Heise and H. Seybold, Das Existenzproblem der t·lobius-, Laguerre- und Minkowski-Erweiterungen endlicher affiner Ebenen. Sitz. Ber. Bayer. kkad. Wiss., Math. Nat. Kl. 1975, 43-58. B. Segre, Ovals in a finite projective plane. Canadian J. Math., 7, 1955, 414-416.
Institut fUr Mathematik der Technischen Universitat Arcistrasse 21 8 MUnchen 2 Federal Republic of Germany
Annals of Discrete Mathematics 18 (1983) 553-566 © North-HoIland Publishing Company
553
COPOLAR SPACES FULLY EMBEDDED IN PROJECTIVE SPACES C. Lefevre-Percsy
1. INTRODUCTION
The non-isotropic lines structure of a symplectic polarity has inspired several authors during the last years.
Let us mention Shult (1974) with his cotrian-
gle property [ 14] , Hale (1978) with his geometries whose planes are dual affine [8], Thas and Debroey (1978) with their semi-partial geometries [3] and Higman (1979) with his A-spaces [10].
In 1980, J.I. Hall [9] introduced a notion of
copolar space, which is less general than A-spaces but covers most of the other geometries.
Recently, Thas and Declerck [5] enlarged the concept of semi-partial
geometry, in order to include all finite copolar spaces. A copolar space is essentially a semilinear space whose points and lines satisfy some analogue of the "one-all" property of polar spaces (for precise definitions, see Section 2).
J.I. Hall's work leads to a nice classification result:
under some assumption of irreducibility, a copolar space is either isomorphic to a well known substructure of a projective space, or is associated to a Moore or a triangular graph. An important problem remains open: are all these copolar spaces embeddable as points and lines of a projective space and in how many ways are they? This question leads to the following problem: classify all copolar spaces whose points and lines are points and lines of a projective space.
It turns out
that an answer may be obtained without any irreducibility condition. A first step is Farmer and Hale's result [7], which holds for finite projective spaces of order greater than 2.
Recently, using a quite different method, we proved the same
result for arbitrary projective spaces P whose lines have at least four points [ 12].
Let us say rough ly the idea of the proof: we show that the geometry of
points, lines and planes of a copo1ar space Q fully embedded in P, belongs to the Buekenhout diagram
C. Lefevre-Peres}"
554
FA
AG
0--0--0,
where AG denotes the geometry of points and lines of an affine space.
Then, it is
possible to build a polarity whose non isotropic lines are the lines of Q. We present here (Section 5) a somewhat different and extremely short proof of the first part of the proof of space has finite dimension.
[l~
(mentioned above), when the projective
We use a characterization of affine spaces as subsets
A of a projective space (whose order is at least 3) satisfying the following: each line intersecting A in at least two points, has exactly one point not in A.
This
characterization is a corollary of [11] and is not too hard to obtain. Except for low dimension (see [21), no classification of copolar spaces fully embedded in projective spaces of order 2, has been obtained.
In the order 2
case, numerous examples are known, in particular several embeddings of the geometries associated to triangular graphs (see Section 3).
However, most of them
are "embedded" in the non-isotropic lines structure of a symplectic polarity.
In
Section 6, we give a necessary and sufficient condition for a copolar space Q of PG(d,2), under which it is possible to construct a symplectic polarity such that the lines of Q are non-isotropic.
Furthermore, we mention some characterizing
properties of the classical examples in PG(d,2) and notice the relation between copolar spaces of order 2 and Fischer spaces. The author is grateful to F. Declerck and A. Neumaier for helpful discussions.
2. DEFINITIONS A semiZinear space (or partial linear space) is an incidence structure Q=(P,L), where the elements of P and L are called points and lines respectively, which satisfies the following axioms: (i) two distinct points are incident with at most one common line; (ii) any line (resp. point) is incident with at least two points (resp. lines). Each line L of a semilinear space Q may be identified with the set of points incident with L.
This allows to use a set theoritic language.
If a and b are two
distinct points on a common line of Q, we say that they are adjacent and we write a-b.
The semi linear space Q is connected if the adjacency graph on the points
Copo/ar spaces fUlly embedded in projective spaces
is connected.
555
Following the terminology of [9], we say that Q is peduced if no
distinct points are adjacent to the same set of points. A linear subspace of Q is a set S of points such that each line having two points in S is contained in S. A plane is the smallest subspace containing two intersecting lines.
A copolar space (J.I. Hall [9]) is a semi linear space O=(P,L) such that (iii) given a line L and a pOint p not on L, then p is adjacent to none or all but one point of L. We say that a copolar space Q=(P,L) is fully embedded in a projective space P if P and L are points and lines of P such that each point of P on a line of L is a point of Q.
It is easy to see that all
planes of a fully embedded copolar
space are necessarily dual affine: each of them coincides with a plane of P in which one point and the lines through it are deleted. Consequently fully embedded copolar spaces satisfy Pasch axiom.
3. EXAMPLES We shall restrict ourself to connected copolar spaces; non connected ones are disjoint union of connected ones. We list all known examples of connected and fully embedded copolar spaces.
Let us note that this list provides embeddings
in a projective space for all reduced copo1ar spaces of Hall's classification, up to one exception: the geometries associated to the Moore graphs.
The latters are
described in (f); they cannot be embedded in a projective space (except for Petersen graph). (a) Two tPivial classes of copolar spaces are mentionned in order to be complete: copolar spaces having a single line (they are embeddable) and copolar spaces whose lines have exactly two points i.e. graphs without triangles (they are not fully embeddable). (b) Let n be a (possibly degenerate) symplectic polarity of a projective space p(l).
Let Pn be the set of points p of P such that n(p) # P, and let Ln be
the set of lines which are not totally isotropic for n. connected copolar space, fully embedded in P.
Then Qn =(Pn ,Ln ) is a
We roughly call it, a symplectic
(1) A polarity of P in the sense of Tits [15] is a synunetric relation n C P x P such that n(x)={yl(x,y) E n} is a hyperplane or P itself. The polarity n 1S symplectic (or nUll) if x E n(x) for every point x E P. A line L is totally isotropic for n if L C n(x), for all x E L.
556
C. Le[evre-Percsy
()OpoZal" spae<3.
If" is non-degenerate, then Q is reduced. 11
degeneracy in a finite projective space PG(d,q), then Q1I is
If
11
has maximal
H~ described in [4].
(c) Let 0 be a (possibly degenerate) orthogonal quadric of a projective space P of order 2.
Let Po be the set of points of P not belonging to 0 and let Then QO=(PO,L O) is a copolar space, Furthermore Q is connected if and only if 0 is not a 3-
LO be the set of lines contained in PO' fully embedded in P.
O
dimensional hyperbolic quadric nor a cone projecting such a 3-dimensional quadric. We call Q an orthogonal copolal" space. O
If 0 is non-degenerate, then Q is O
Let us mention that, if P has even dimension, then Q is isomorphic (in O terms of abstract copolar space) to a symplectic copolar space. reduced.
(d) Let [be a set of arbitrary cardinality. pairs of elements of
Ii
Consider the set PQ of all and let LQ be the set of all triples of elements of Q.
Then Q;:(Pn,L n ), where incidence is inclusion, is a copolar space which is necessarily reduced. Q is called a triangular copolal" space. (d.l) The above definition of triangular copolar space does not provide an embedding of them in some projective space. j
It is well known that Qn' with
nl=n+2, may appear as a substructure of a symplectic copolar space embedded in
PG(n,2).
When Inl is odd, the following description is equivalent to a more gen-
eral embedding, due to Neumaier [ 13] , of arbitrary "reduced" graphs into a symplectic polarity.
Let (x.). with i=O,i, ••• ,n be some coordinatization of a n1 dimensional projective space P over GF(2). Let n={O,l, •.•• noo }. Each point {i,j} of Pn' with i and j different from 00, corresponds to the point a(i.j) of P, whose all but ith and jth coordinates are zero; each point {k. oo } of Pn is associated to the point b(k) of P whose all but kth coordinates are zero(2)
Then. it is easy to
verify that the points corresponding to {i,j}, {j.k} and {k,i} are on a line of p. y i j. (d.2) The following embedding ofO\! with I QI=n+3. in PG(n.2) is a substruc-
which is not totally isotropic for the polarity defined by f(x.y) = ture of a symplectic copolar space if and only if n is odd.
i~j X
In this case. i.e.
when !nl is even, our description is again equivalent to Neumaier's one (Xi) be a coordinatization of p. as above.
Let Q={O.1 •••• ,n.
[l~.
Let
'002} and consider l the same correspondance as above, between {i,j} and a(i.j}, when i and j are different from ooi and 00 , between {k,ool} and b(k}. 2
oo
Associate to the pair
{~.002}'
(2) This is also a classical projective embedding of the cycle matroid associated to the complete graph K.
557
Copo/ar spaces fully embedded in projective spaces
the point c(t), whose all but tth coordinates are 1, and to the pair {ooi,002}' the point d whose all coordinates are 1. Then, it is easy to verify that the points of a line of LQ correspond to pOints of a line of P. All these lines are not totally isotropic for the polarity defined by f(x,y) = .4. x. y., if and only if P l;OJ
has odd dimension.
1
J
When n=4, this construction provides an embedding of QQ' with
IQI=?; it turns out that it is the same as the embedding first discovered by Debroey, Declerck and Thas [2]. (d.3) The same authors also describe in [2] a special embedding of QQ' with IQI=9, in PG(5,2). (d.4) Let us also mention the following isomorphisms, which provide exceptional embeddings of triangular copolar spaces: QQ with I QI=5 is isomorphic to QO' where 0 is an ovoid in PG(3,2); QQ with I QI=6 is isomorphic to QTI , where
TI
is a
non degenerate symplectic polarity in PG(3,2); QQ with 101=8 is isomorphic to QO' where 0 is a hyperbolic quadric in PG(5,2). (e) We give a construction of a (necessarily non reduced)copolar space in P, from a copolar space Q fully embedded in a hyperplane H of P.
Let p be a point of
P_H(3). For each line L of Q, consider the dual affine plane obtained from the projective plane
Let L be the union
of all lines in these dual affine planes and let P be the set of pOints on the lines of of
Q.
L.
Then
Q=(P,L)
is a non reduced copolar space, which we call a covering
If Q belongs to one of the families (b) or (c), then
these families.
Qagain
belongs to
If Q is some embedding of a triangular copolar space, then
Qis
some new example. (f) Let Mbe one of the Moore graphs, with valency k. vertices of M. a line. space.
Let PM be the set of
The set of all vertices adjacent in M to a given vertex is called
If LM is the family of all such lines, then QM =(PM,LM) is a copolar It is known [9] that, if k> 3, then Q does not satisfying Pasch axiom.
Hence, if k > 3, Q is not embeddable in any projective space (see end of Section M 2). If k=3, it is easy to check that QM is isomorphic to an orthogonal copolar space QO' where 0 is an ovoid in PG(3,2). If k=2, QM is a graph without triangle.
(3) J.I. Hall Dointed out that actually p may belong to H, provided that, for each point of q of Q, the third point of the line pq is not on Q.
C. Le[evre-Percsy
558
4. PRELIMINARY RESULTS We first establish some results which are valid for copolar spaces in arbitrary projective spaces P. LEMMA 1: Let Q be a copoLar space, fuZZy embedded in P and generating P.
If Q is
aonneowd, then the set of Zines of Q through every point of Q generates P.
PROOF: Suppose there is some point a of Q such that the lines of Q through a generate some proper subspace S of P. point b exists, because Q generates P.
Let b be some point of Q not in S: such As Q is connected, we can refer to [ 12,
Lemma 3) which insure that there exists a point c adjacent to a and b.
Then a is
adjacent to a point of the line
If SI
Hence, the set of lines of Q through a
F P, we again apply the same argument and Lemma
generis
achieved. DEFINITION: In the same way as in [12). consider, for each point p of Q, the following residual geometries: P is the projective space of lines and planes of P through p: lines through p
p are points of P and planes through p are lines of P •
P P Qp is the incidence structure of lines and planes of Q through p: lines of Q
through p are points of Q and planes of Q through p are lines of Q . p
p
Q is a substructure of Pp which satisfies the following properties: p
(l) the points of Qp are points of Pp ' (2) the lines of Q are all non trivial intersections of the lines of P
P
p
with the set of points of Q. Indeed, let a and b be two points of Q. The p p corresponding lines A=
As this plane is the dual affine plane obtained from
the projective plane
Furthermore, this also show
that: (3)
the lines of Qp are completed into a line of Pp by adding exactly one
poi nt. (4) Furthermore, if Q is connected, Q generates P (see Lemma 1). p
Consequently, we have proved the following result.
P
Copo/ar spaces fully embedded in projective spaces LEt~MA
559
2: Let Q be a fuHy embedded and connected copotaT' space, generating P.
Then the residuat geometry Q- is a subset of points of the projective space P ,
P
P
generating P , such that each tine of P meeting Q in at teast two points has
p
p
p
exactly one point not in Q . p
We now mention a proposition which is actually the last step of the proof of
(12) :
PROPOSITION 1: Let Q be a connected copolar space, fully embedded in P.
If all
residual geometries Q aT'e affine spaces deduced from P , then Q is a symplectic p
p
aopolaT' space.
PROOF: Let us sketch the proof, which already appears in
(12).
Consider the fol-
lowing correspondance. To each point of Q, associate the hyperplane n(p)=P-Q , p
where Q is the union of all lines of Q through p; to each point p which is not a p
point of Q, associate the whole space P. symplectic polarity.
This correspondance clearly defines a
Furthermore, the lines of Q are necessarily the non iso-
tropic lines of this polarity and so the proof is finished.
5. THE CASE OF ORDER BIGGER THAN 2 Let P be any projective space whose lines have at least 4 points.
Then the
classification of copolar spaces fully embedded in P is completely solved. proved in [7 ) for finite projective spaces, and in general in
It is
[12].
THEOREM 1: Let P be a projective space whose lines have at teast 4 points.
If Q
is a connected copotar space, futty embedded in p. then Q is a symptectic copolaT' spaae.
We give here a very short proof of this result: the basic idea is the same as in
(12)
but the additionat assumption of finite dimension of P allows to use
the fo 11 owi ng known propos iti on [ 11) • PROPOSITION 2: Let P be a finite dimensional projeative spaae whose lines have at teast 4 points.
Suppose A is a subset of p. generating p. suah that each tine
c. Le!evre-Percsy
560
'!Ieet,:ng A in at least
2
points has exaetly one point not in A.
Then A is the
eOl'lplement of a hype1'plane (i.e. A is an affine spaee dedueed f1'om P).
This proposition is an immediate corollary of [11].
If P is a finite pro-
jective space of order at least 3, this means that all sets of class (O,l,q) are affine spaces deduced from P. PROOF OF THEOREM 1 IN THE FINITE DIMENSIONAL CASE: Lemma 2 shows that Qp is a subset of P satisfying the assumptions of Proposition 2.
P
Consequently, if P is p
finite dimensional (i .e. if P is), then Qp is an affine space deduced from Pp and so Proposition 1 applies.
6. THE CASE OF ORDER 2
Let Q=(P,L) be a copo1ar space, fully embedded in a projective space P of order 2.
Section 3 shows that there exists other such copo1ar spaces than the
and QO ones. We first give a characterizing property, for each of these symplectic and orthogonal copo1ar spaces.
classical
Q~
PROPOSITION 3: Let Q be a connected copolai' space, fully embedded in a projective space P of 01'de1' 2.
If each plane of P containing a line of Q eontains two lines
of Q, then Q is a syrrrpZectie eopolai' space.
PROOF: Consider the residual geometry Qp at some point p of Q (see Section 4). If each plane of P through p, containing a line of Q, contains two lines of Q, then there must be two lines of Q through p, because all plane of Q are dual affine. Then, Qp is a subset of Pp intersected by every line in 0 or 2 points. So Q is p an affine space deduced from Pp. Consequently, Proposition 1 applies and the proof is finished. PROPOSITION 4: Let Q=(P,L) be a eopolai' space fully embedded in a finite projeet-ive spaee P=PG(d,2), such that there exists a plane
TT
of L.
n P is the union of a line
If. for eaeh such plane
TT,
the intersection
TT
containing exaetly one Zine
a singZe point. then Q is an o1'thogonal eopolai' space.
Copo/ar spaces fully embedded in projective spaces
PROOF: 1) We first prove the proposition for d=3.
561
If Q is connected, we can
refer to the classification of Debroey, Declerck and Thas [2J. Q is necessarily either symplectic or orthogonal. We shall show that a non connected copolar space in PG{3,2) does not satisfy the hypothesis of Proposition 4. If Q contains (properly) a connected component Q* generating P, then Q* is symplectic or orthogonal .. If Q* is symplectic, then Q* must be associated to a degenerate polarity, because Q contains Q* properly. 1~
<;':
':t'~
Let
L
be the line exterior
~':
':to:
Then Q={P,L) is obtained from Q =(P ,L ) by adding to P some points of
to Q
and, possibly, to L* the line L.
Consequently, either there is no plane of P
containing exactly one line of L or there is a plane containing a single line of L and more than one point not on his line. Proposition 4.
Now, if Q* is orthogonal, Q is the union of Q* and some points
(but no line) not in Q*. a single line of
This contradicts the hypothesis of
L
Then it is clear that there exists a plane containing
but more than one point of Q not on the line.
Suppose the connected components of Q do not generate P. each point of Q is on at most two lines of Q.
This implies that
If Q has two lines on a point, then
Q contains a dual affine plane a'"', obtained from a plane a of P by deleting some
point a.
Every line of Q not in a* must be through a, because there are at most
two lines at each point of Q. through a.
Hence Q may be the union of
a'"'
and 0,1 or 2 lines
In every case, there exists a plane containing exactly one line L of Q
and more than one poi nt not on L.
Now, if Q has at most one 1i ne at each of its
points, then, either Q is the union of two skew lines i.e. an orthogonal copolar space associated to a ruled quadric, or Q does not satisfy the assumptions. 2) Suppose P has dimension 4. Consider a plane a of P. We shall study its intersection with Q. and let H=
If a contains some point p of Q, consider a line L of Q on p
If H contains a plane rr having a unique line in L, then, by 1),
the copolar space H n Q is orthogonal and so a n Q is the complement of a (possibly degenerate) conic.
Now, if each plane of H containing a line of Q contains
two lines of Q, then Q is symplectic or is the union of the sympletic copolar space Q* associated to a degenerate polarity (whose kernel is a line L*) together with one or two points on L. But, as there exists a plane rr of P meeting Q in the union of one line Mand one point m, there is a line N=rr n H meeting H n Q in one or two points exactly. or 2 points.
Consequently, H n Q is necessarily the union of Q'" and 0,1
If M intersects H in a point n of Q*, then m is the second point of
Q1' belonging to a line joining n to a point not in Q.
And so <M,l'"'> is a 3-di-
C. Le[evre-Percsy
562
mensional subspace H'.
If M intersects H in a point of l*, then it is clear that
c
But H' contains the
and consequently, by 1), H' n Q is an orthogonal copolar space.
When Q is
Q*, we get a contradiction with the existence of a line l* of H' having no point of Q (this would imply that H' n Q is the union of two skew lines, which is not -k
the case, because H n H' n Q contains more than two points or a point and a line); when Q contains a point £ of l*, we get a contradiction with the existence of a line through t, whose all points are in Q but which is not a line of Q. can conclude that, for every plane
a
of P,
a n
SO, we
Q is necessarily either empty or
the complement of a (possibly degenerate) conic.
Then, by a well known charac-
terization of a quadric by its plane sections, Q is the complement of an orthogonal quadric. 3) If P has dimension d
~
plane H of P containing a plane assumption, H n Q is orthogonal.
5, we use an induction on d. rr
having a single line in L.
Consider a hyperBy the induction
Now consider a hyperplane H' not on
The sub-
rr.
space H n H' is a hyperplane in H and so it surely contains a plane having exactly one line of the orthogonal copolar space H n Q.
Consequently the induc-
tion assumption also applies to H' and so all intersections H n Q, where H is a hyperplane, are orthogonal copolar spaces.
This implies that P-P is a set whose
all hyperplanes sections are orthogonal quadrics and so Proposition 4 is proved.
Except the even-dimensional examples of (d.2) and that of (d.3) in Section 3, all known fully embedded copolar spaces are substructures of a symplectic copolar space.
We shall establish some necessary and sufficient condition on Q,
for "embedding" Q=(P,L) into a symplectic copolar space Q=(P,L).
We mean L is a
subset of Land P is the union of all points of P on the lines of L.
We need the
following easy lemmas. lEMMA 3: Let P be a 3-dimensional projective space PG(3,2) of order 2.
If A is
come set generating p. uhich contains no line of p. then either A is contained in 4 a (um:que) affine space deduced from P or A is an ovoii ) of P.
(4) let us notice that an ovoid of P is a 5-uple of linearly independent points, i . e. a frame of P. As the group PGl ( 4,2) is (sha rp ly) trans iti ve on the frames, all ovoids are elliptic quadrics.
Copo/ar spaces fully embedded in projective spaces
PROOF: Consider a 4-uple of linearly independent points of P. generates Pl.
563
(This exists, for A
This 4-uple generates a (unique) affine space in P.
all points of A are in this affine space. independent pOints, i.e. an ovoid of P.
Suppose not
Then A contains a 5-uple of linearly It is clear that no set, containing no
line of P, can properly contain an ovoid.
And so the lemma is proved.
LEMMA 4: Let P be the projective space PG(d,2), d
~
3.
If A is some set
generating P, which contains no line of P and no ovoid in any 3-dimensional projective subspace of P, then A is contained in a (unique) affine space deduced from
P. PROOF: We use an induction on d.
If d=3, see Lemma 3.
(d+l)-uple U of linearly independent points in A. ates Pl.
U generates an affine space in P.
If d > 3, consider a
(This exists because A gener-
Let H=P-A.
If there is a point of A
in H, then, by the induction assumption, this point cannot belong to any subspace H n S, where S is a hyperplane generated by points of U.
But it is easy to verify
that all such H n S cover H and so we have proved that A is contained in P-H. Let us consider a copolar space Q whose lines are lines of a symplectic one
Q.
Since each residual geometry
Qp
in
Qis
an affine space, each residual geome-
try Q in Q is contained in an affine space and consequently does not contain any p ovoid. We shall prove that this condition is also sufficient for embedding Q into a symplectic copolar space
Q.
THEOREM 2: Let Q=(P,L) be a connected copolar space fully embedded in P=PG(d,2), d
~
2.
If no point of Q is adjacent to the points of an ovoid in some 3-dimen-
sional subspace of P, then there exists a symplectic copolar space Q=(P,L) such that L :::J
L
PROOF: Consider, for each pEP, the residual geometry Q. p
This is a set of
points A generating P (see Lemma 1) which does not contain any 1i ne of P (s ee p
Lemma 2).
p
p
By the assumption on Q, the set Ap does not contain any ovoid.
Hence,
Lemma 4 shows that Ap may be completed uniquely in an affine space Ap deduced from Pp ' Define the incidence structure Q=(P,L), where I is the set of all lines L which are points of some A and P is the union of all members of I. If Qis a p
564
C. Le!evre-Percsy
connected copolar space, then Proposition 3 may be applied and copolar space; so the theorem is proved.
Qis
a symplectic
The remaining problem is to show that
Q
is actually a copolar space (which will necessarily be connected, because Q is). We proceed in several steps. 1) Let pEP and eEL with pEe.
Then every plane
a
of P through
Ccon-
tains a second line of I through p.
Indeed, as the residual geometry Q is an p affine space in P , the lines of P through p not in I, are the lines of a hyperp plane on p. As a is not contained in thi s hyperplane, there is exactly one line in
~.
through p, not in
I.
2) Let pEP and eEL with p ~ L.
Then, by the above step 1, there is
exactly 0 or 2 lines of I through p, which intersect [. 3) Let a,b,c be the points of a line L E L.
Let H , Hb and H be the hypera c planes of lines of P through a,b and c, which are not in I. Then Ha n Hb = Hb n Hc = Hc n H. a Indeed, let x be the intersection with Ha of a line X of L through b. This point x must be adjacent (with respect to Q) to c, because of copo1ar axiom (iii). pOints in H. a H n H • a
c
4) Let
p,
Consequently band c are adjacent to the same set of
But, by Lemma 1 the set of points x generates H • So, H n Hb = a a
pE
P-P and L E
L
with pEL.
intersecting L at a point a.
shows that
p is
If
p is
Suppose there is a line [ in I through
not joined to a point b of L, then step 3
joined to the third point c of L.
If
Pwas
joined to all point of
L, then, by step 1,
the points band c cannot be joined by a line of I to the
third point d of [.
Then a and b would not be adjacent to the same set of points
in Hc ' contradicting step 3. 5) Let p E P-P and CE L-L.
Suppose there is a line f1 E
C. Notice that H~ L. If the and Mrespectively, are distinct,
I
through
p inter-
secting
(uniquely determined) points f and m of P,
on [
then steps 1 and 4 allow to conclude that
there exists a second line of I through Suppose now £=m. space S=
p intersecting
Consider a line L E
L
[.
through f and the 3-dimensional sub-
In the sequel, for all x L, Hx will denote the hyperplane of all
lines through x which are not line of I, and nx will be the plane S n Hx' By step 4,
p is joined by a line of L to a second point i l F f of L. Let i2 be the third point of L.
By step 2, fl (resp. f ) is joined to a point a (resp. b) of M, by 2 some line [, (resp. [2) of I. The points a and b must be distinct, by step 4.
Copo/ar spaces fully embedded in projective spaces
565
Consequently a (resp. b) is a point of IT1 (resp. IT1 ). If c (resp. d) denotes the third point of [1 (resp. [2)' this im~lies that ~ (resp. d) is joined to 12 (resp. 11)' We have to prove that p is joined by a line of I to exactly one of the paints a or b. If there is no line joining p to a nor b, then
Since <12,c>and <12,d>are lines of I, the lines <12,e>and <12,f>are also
in L (where e and f are the third points on
Then there are five
lines of I through 12 in S: the lines L,<12,c>,<12,d>,<12,e>and<12,f>. This contradict·s the existence of the plane IT1 • Suppose
I.
Then there are five lines of I through 11: the lines L,
This contradicts the existence of IT1 •
We can now conclude: steps 2, 4 and 5 show t~at copolar axiom (iii) is
Q=(P,L)
satisfied by
and the theorem is proved.
We close this section with some remarks on the relation between copolar spaces of order 2 and Fischer spaces, in the sense of Buekenhout [1]. be a copolar space of order 2, satisfying Pasch axiom.
Complete it by lines of
two points, consisting in all pairs of non adjacent points of Q. structure Q=(P,L) obtained in this way, is a linear space. every plane
a
Let Q=(P,L)
The incidence
It is trivial that for
of Q, containing a line of three points, either
a
contains a unique
such line, or the set of lines of three paints in a forms a dual affine plane of order 2.
Consequently, following a known characterization of Fischer spaces [1],
Q is a Fischer space without affine planes of order 3. Let us mention that Fischer subspaces of a symplectic Fischer space with minimal degeneracency (i.e. of a symplectic copolar space of order 2, associated to a polarity whose kernel has minimum possible dimension) have been determined in [6]: they are of type (c), (d) or (e) of Section 3. This result, together with Theorem 2, give a partial answer to a classification of connected copolar spaces fully embedded in PG(d,2).
BI BL IOGRAPHY 1.
F. Buekenhout, La geometrie des groupes de Fischer. Seminar notes U.L.B., Brussels 1972.
566
2. 3.
4. 5. 6. 7. B. 9. 10. 11. 12. 13. 14. 15.
C. Lefevre-Percsy
I. Debroey, F. De Clerck and J.A. Thas, The embedding of (O,a)-geometries in PG(n,q), Part. II. In preparation. 1. Debroey and J.A. Thas, On semi-partial geometries, J. Combin. Th. A, 25 (1978), 242-250. F. De Clerck and J.A. Thas, Partial geometries in finite projective spaces, Al"clz_ Math., 30 (1978), 537-540. F. De Clerck and J.A. Thas, The embedding of (O,a)-geometries in PG(n,q). To appear. G. Demeur, Espaces de Fischer hermitiens. Thesis U.L.B. 1979. K.B. Farmer and M.P. Hale Jr., Dual affine geometries and alternative bilinear forms, Lin. Alg. and Appl., 30 (1980), 183-199. f4.P. Hale Jr., Locally dual affine geometries, in Pl"oceedings of the Confel"elice on Finite gl"OUpS; PGI'k City 1975. Academic Press 1976, 513-518. J.I. Hal" Classifying copolar spaces and graphs. To appear. D.G. Higman, Lectures at East Lansing ans Santa Cruz 1979. C. Lefevre-Percsy, Ensembles de classe (O,l,q,q+l) de PG(d,q), J. Geom., 15 (1981), 93-98. C. Lefevre-Percsy, Geometries with dual affine planes and symplectic quadrics, Lin. Alg. and Appl., 42 (1982), 31-37. A. Neumaier, Personnal communication. E.E. Shult, Groups, polar spaces and related structures, in Pl"oceedings of the Advanced Study Institute on Combinatol"ics, Breukelen 1975. Math. Center Tracts N. 55, Amsterdam (1974) 130-161. J. Tits, Buildings and BN-pail"s of sphel"ical type. Lecture Notes, Springer Verlag, Berlin 1974.
Universite Libre de Bruxelles Campus de la Plaine C.P. 216 Boulevard du Triomphe 8-1050 Bruxelles Belgium
Annals of Discrete Mathematics 18 (1983) 567-574 © North-Holland Publishing Company.
567
A PHYSICAL CHARACTERIZATION OF CONFORMAL TRANSFORMATIONS OF MINKOWSKI SPACETIME J.A. Lester
1. INTRODUCTION
Minkowski spacetime M4 can be described as R4 equipped with a scalar product ( , ) of the form
for "events" r 1 .- (x1'Yl,zl,t 1), r 2 .- (x 2'Y2,z2,t 2) in IR4. tion, a light signal can pass between events r ,r E M4 iff 1 2 ......
In this formula-
)'(
This relation is preserved by similarities of M , i.e. by composition of Lorentz 4
transformations, translations and dilatations on M. 4
Conversely, all bijections
of M4 preserving {, in both directions must be similarities (see, e.g., [1], [4]). The situation changes when we consider transformations defined only on a proper subset of M4. The conformal inversion r ~ r/(r,r), for example, is defined only on M \ C(O) (C(O) is the cone with equation (r,r)=O) and also pre4
serves "'.
By adjoining to M4 a "null cone of points at infinity", this inversion
can be made into a bijection of the whole extended space. known as conformal Minkowski space similarities and
M , 4
This extended space is
and its transformations (generated by
inversions) as conformal transformations [5].
These transformations are of some physical interest because, while much of the physics of Minkowski spacetime is not conformally invariant, Maxwell's equations (which describe the behavior of electromagnetic fields) are (Cunningham and Bateman, 1910 [2] ).
Though the significance of this invariance has been widely
investigated, the results appear to be mainly inconclusive (cf. [3]). In this article, we characterize conformal transformations as mappings pre-
568
I.A. Lester
serving the speed of light (i.e. preserving the relation *) on a domain in M . 4
Specifically, we prove the following theorem. THEOREM: Let V be an open, connected subset of n-dimensional Minkowski space Mn(3
~
n < 00), and let f: V (x-y, x-y)
~
Mn be a function such that for all x,y
0 iff (f(x) - flY), fIx) - fly))
ihen there exists a conformal transformation
c:
r~ ~ n
E V,
O.
Mn (where Mn denotes the
conforrnal extension of Mn) such that c IV = f.
Some terminology concerning metric vector spaces is in order (see (6) for fu 11 er detail s).
A real metric vect01' space is areal vector space V together
with a metric (symmetric bilinear form) ( , ) on V. no non-zero x
E
V satisfies (x,V) = O.
We say V is non-singular iff
An isometry between real metric vector
spaces is a linear, metric-preserving bijection between them.
Any isometry be-
tween subspaces of a non-singular space V is the restriction of an isometry of V (the Witt theorem). For any basis {e
1
, .•.
,e k} of a subspace U of V, the numbers p,q,r of (resp.)
positive, negative and zero eigenvalues of the metric matrix[(e.,e.)) classify U 1
up to isometry.
J
Some special cases: U is totally isotropic if p=q=O, Minkowskian
if p= 1, q* 0, r=O, and an Artinian 2-space if p=q= 1, r=O.
(An Artinian 2-p lane is
a translated Artinian 2-space.) Finally, some notation: brackets <... > denote the span of whatever they contain. 2. THE
GE~~ETRY
AND TOPOLOGY OF M
n
For any a in a real metric vector space V, the null cone with vertex a is the subset C( a)
{x
E
V I(x-a, x-a) = a}.
When V = Mn' these null cones have
two very useful properties. First of all, tangent Minkowskian null cones intersect in a null line, i.e. for bE C(a), b * a, CIa) n C(b) is a line of the form a (here, k=b-a).
+
It follows that all Lorentz transformations (isometries of M ),
translations and dilatations of
n
r~n
preserve null lines.
Furthermore, since the
569
Conformal transformations of Minkowski spacetime
function f of our theorem preserves null cone sections of V, it also maps any null line section of V into a null line. Secondly, the interior of C(a) (consisting of those x
E
Mn with
(x-a, x-a) > 0) is divided by C(a) into two disjoint halves (the "absolute future" and "absolute past" of the event a).
These two halves may be distinguished as
follows: for an arbitrary timelike vector t (i.e. one with (t,t) > 0), a vector x in the interior of C(a) is inside the same half of C(a) as a+t iff (x-a,t) > O. This property makes it possible to define "causal intervals" (or Alexandrov neighbourhoods) on Mn. DEFINITION 2.1: For a,b
E
Mn with (a-b, a-b) > 0, the open causal interval N(a,b)
is defined to be N(a,b):={x
E
Mn l(x-a,x-a)>O,(x-b,x-b)>O,(a,b-a)«x,b-a)«b,b-a)}.
By putting t=b-a, it is easily seen that N(a,b) is the intersection of the absolute past of one of C(a) or C(b) and the absolute future of the other.
Since the
open causal intervals generate the same topology of Mn as the usual ffin c-neighbourhoods, we may assume that the domain V in our theorem is open and connected with respect to open causal intervals. We shall also have occasion to use closed causal intervals N(a,b), formed by allowing equality in all the inequalities in Definition 2.1.
3. THE GEOMETRY OF ~
With respect to some basis of Mn' the metric ( , ) has the form (x,y) for x := (Xl' ... ,x n), y := (Yl' ... 'Yn)
E
Mn.
Points of Mn may also be represented
*
by conformal coordinates X := (X 1 , ... ,X n ,H,N), given by H 0, X.1 := Hx., 1 i=l, ... ,n; N := H(x,x) for x
E
Mn. These coordinates are homogeneous (proportion-
al X's represent the same point) and satisfy [ X,X] : =
n-1 X + X2 - H N = O. " - "i=l i. n
The metri c [ , ] defi ned on IR n+2 by the above equati on makes IR n+2 into another
l.A. Lester
570
metric vector space C(M ), called the coordinate space of M.
n
n
Points of Mare
n
thus represented by the one-dimensional totally isotropic subspaces <X> of C(M ) n which satisfy H -2(X,E] 0, where E := (0, ... ,0,1) is called the antipode.
*
=
Equations of the form (X,A) = 0 for A
*0
in C(M ) represent quadrics, null cones n
or hyperplanes in M (see (5] for details); if (A,A] = 0 and (A,E) t 0, for examn
p1e, we obta in the nu 11 cone with vertex E Mn. ,~'onformal Minkowski space Mn
is obtained by allowing paints with H = 0,
i.e. by adjoining to M those <X>c C(M ) satisfying [X,X) = [X,E) = n
n
o.
Since
(E,E) = 0, we say that the added points form the antipodal null cone, or the cone (~t
infini ty .
We set e : = <E >E M and C(M ) n
n
= C(Mn ).
Note that X E C(M ) repn
presents the finite point x E M iff X cr(x,(x,x)). n
We shall be interested in the extensions of null lines in M to n
Mn ,
particu-
lar1y in those properties summarized in the following lemma.
LEMMA 3.1: a) Le;; a +
7'hen
n
U
Mn
:eotally isotropic 2-space in C(f\),E Ef
Cl
'ih:) a +
COniJel'seZy, for any totally isotropic 2-space W in C(f\) with
E Ef W,{ <X> I X E W, [X,E) t)
* O}
For n:;,[Z lines a
1
is a null line in M . n +
2
p~' .- (k ,0,2(a ,k )), p~ := (k ,0,2(a ,k )). 1
1
1
2
2
2
2
n
Then the lines are paralleZ iff
<Jl'7,P;,E> is a totai.lu isotropic 2-space in C(~)
PROOF: a) Parts i) and ii) follow from direct calculation.
For part iii),
oo
X E
A,B in W with [A,E) = [B,E)
a = E Mn' A - B = (k,O,w) for some k E Mn' wEIR. a +
t
0, and set
Then k is null and
* OJ.
b) From a)i)
1
P; = (0,O,2(a
l
If the 1ines
2
-a ,k 2
1
)) =
B E for some BE IR. i.e. p7 E
ly, if P: = a P; + B E for some u,s E IR, then kl = uk
2
;
Converse-
hence the lines are par-
Conformal transformations of Min kows ki spacetime
alle1.
571
0 oo
We may think of the point
A totally isotropic 2-space W in C(Mn ) which does contain E satisfies [W,E] = 0; in this case, the set {<X>!X E W} C Mn may be thought of as a null line at infinity containing e, i.e. as a generator of the null cone at infinPart b) of our Lemma then states that parallel null lines in Mn intersect the same generator of the null cone at infinity. ity.
The isometries of its coordinate space induce the conformal transformations Since isometries of C(M n) preserve totally isotropic 2-spaces in C(M n), conformal transformations map null lines in Mn onto null lines. Furthermore, the
of Mn
conformal transformations which fix e (and thus map Mn onto itself) induce the similarities of Mn (see [5] for detail). The two lemmas which follow will be used in the proof of our theorem. LEMMA 3.2; For p,q
E
Mn with (p-q, p-q)
tion g1 of ~ such that g1(P)
* 0,
there exists a conformal transforma-
= e, g1(q) = o.
PROOF: Set P := (p,l,(p,p)), Q:= (q,l,(q,q)), Z:= (0,1,0) and,,:= -2[P,Q] = (p-q, p-q)
* O.
The linear mapping
isometry, and is thus the restriction of an isometry G1 of C(M). n
The conformal
transformation 9 of M induced by G has the required properties.
0
n
1
LEMMA 3.3: For a,b
E
1
Mn with (a-b, a-b) > 0 and any t
E
Mn with (t,t)
exists a conformal transformation 9 of Mn such that 9 (a) 2
2
= 0,
9 (b) 2
1, there e and
PROOF: Set A:= (a,l,(a,a)), B;= (b,l,(b,b)), Z:= (0,1,0), T:= (t,l,l) and " := -2[A,B] = (a-b, a-b) > nates of (1/2)(a+b). 1
I
o.
Then C := A + B + "E gives the conformal coordi-
The linear mapping
A ..... ,,2Z, B ..... ,,2E, C ..... 2,,2T, is an isometry, and is thus the restriction of an isometry G2 of C(M). n
The corresponding conformal transformation g2 of Mn
satisfies g2 (a) = 0, g2 (b)
=
e, and g2[Ha+b)] = t. .
Now, N(a,b) is that region bounded by C(a) and C(b) which contains !(a+b). It
follows that g [N(a,b)] is that region bounded by CeO) and the null cone at 2
572
J.A. Lester
infinity which contains t, thus g [N(a,b)] n M has the stated form. n
2
0
Fix t E Mn with (t,t) = 1, and denote by S that region of Mn bounded by C(O) and the cone at infinity which contains t. The set S := S n M has the important o
n
property that all null line sections of So have their points at infinity in S.
4. PROOF OF THE THEOREM (Suppose f(a) = fIb) for some a
First of all, f must be injective.
then a-b is null, and CIa) n C(b) is the null line joining a and b.
* b;
An open set
DC V containing a contains point d on CIa) but not on C(b), thus f(d) is on C{f(a)] but not on C[f(b)), which is impossible.) LEMMA 4.1:
POP
a,b E Mn with (a-b, a-b) > 0, suppose that an injective function
9 : N(a,b) .... Mn satisfies, fop aU x,y E N(a,b), (x-y, x-y) i'::en
= 0 iff (g(x) - g(y), g(x) - g(x) - g(y)) = O.
*
9 is the pestl'iction of a confopmal tpansformation of M • n
PROOF: From Lemmas 3.2 and 3.3, there exist conformal transformations 9 ,g
M such n
1 2
that 9 1(g(a)] = 0, 9 1(g(b))
for S as in
§
= e, g 2 (a) = 0, 9 2 (b) = e, and 92[N(a,b)]
S
4. -
-1
Define h : S .... Mn by h(x) = gl[g(g2 (x)] for all XES; then h(O) hIe) = e, so from *, XES is finite iff h(x) is. sections of S into null lines, thus for So := S
= 0 and
Furthermore, h maps null line
Mn, h takes null line sections of S into null lines in M , while preserving their points at infinity. Since h o n also maps sections of S with generators of the cone at infinity into like n
generators, Lemma 3.1 yields that the function ho := hiS maps parallel null line o
sections S into parallel null lines. o Any three parallel null line sections of So lying in an Artinian 2-p1ane are all intersected by another null line sections of So'
Their parallel images are
then also intersected by a null line; they must lie in an Artinian 2-p1ane. Therefore h maps Artinian 2-p1ane sections of S into Artinian 2-p1anes, and o 0 moreover, maps parallel Artinian 2-plane sections into parallel Artinian 2-p1anes, since such planes are determined by parallel families of null lines. Two Artinian 2-p1anes which intersect at a point pESo contain between them at least three null line sections of So through p.
The corresponding image 2-
Conformal transformations of Minkowski spacetime
573
planes cannot coincide, since then two of the null line sections would be mapped into a single null line, thereby mapping some pair of points x,y with (x-y,x-y)+ 0 into points satisfying (ho(x) - ho(y), ho(x) - ho(Y))
= 0,
and thus contradicting
Since any line section of So lies in two distinct Artinian 2-planes, and
1:
indeed, parallel line sections of So lie in two pairs of parallel Artinian 2planes, ho maps line sections of So into lines and preserves their parallelism. It now followd easily that ho is the restriction of a bijective linear transformation of Mn, which, since it maps C(O) onto itself, must be a scalar times a Lorentz transformation. This multiple of a Lorentz transformation must then be the restriction of a conformal transformation s of Mn which agrees with h on all of S, since both preserve the points at infinity of null line sections of So'
It follows that 9 is the restriction of the conformal transformation
-1
gl
0
S
--
0
g2 to N(a,b).
0
Since V is open, for any x E V there exist a,b,c,d in such that x
E
N(a,b)
C
N(a,b)
C
N(c,d) c V.
a conformal transformation.
From Lemma 4.1, fIN(a,b) is the restriction of
On non-empty intersections of two such N(a,b)'s, the
corresponding conformal transformation agree; thus, since they are analytic functions, they agree on the union of the two N(a,b)'s.
Since V is connected, all
conformal transformations so obtained agree, thus there exists a single conformal transformation c of Mn such that f
=c ' iV
ACKNOWLEDGEMENTS: This work was completed with the financial assistance of the Alexander von Humboldt Foundation.
BIBLIOGRAPHY 1. 2.
3. 4.
H.J. Borchers and G.C. Hegerfeldt, The Structure of Spacetime Transformations. Comm. Math. Phys. 28, (1972), p. 259. E. Cunningham, The Principle of Relativity in Electrodynamics and an Extension Thereof. Proc. Lond. r~ath. Soc. 8 (1910), p.77, and H. Bateman, The Transformation of the Electrodynamical Equations. ibid., p. 464. T. Fulton, F. Rohrlich and L. Witten, Conformal Invariance in Physics. Rev. of Modern Phys. 34 (1962), p. 442. J.A. Lester, Cone Preserving Mappings for Quadratic Cones over Arbitrary Fields. Can. J. Math. 29 (1977), p. 1247.
574
5. 6.
l.A. Lester
J.A. Lester, ConformaZ Spaces, J. of Geomtry 14, (1980), p. 108. E. Snapper and R.J. Troyer, Metpic Affine Geometry, (Academic Press, 1971).
Mathematiches Seminar der Universitat Hamburg Bundesstr. 55 2000 Hamburg 13 Federal Republic of Germany Present address: Dept. of Pure Mathematics University of Waterloo Waterloo, Ontario Canada
Annals of Discrete Mathematics 18 (1983) 575-580 © North-Holland Publishing Company
575
THE VAN DER WAERDEN CONJECTURE J.H. van Lint
1. INTRODUCTION For the world of combinatorics the 1980's started off quite nicely with the appearance of G.P. Egoritsjev's elegant and elementary proof of the famous Van der Waerden conjecture (cf. [2], [9]).
In the meantime at least two expositions of
the proof have been written (cf. [4], [6]) but at the time of the Rome meeting these were not available yet. Therefore it seemed appropriate to show a large gathering of combinatorialists how the proof works. If A is an n
x
n matrix with entries a .. (i lJ
l, •••• n; j
l •••.• n) then
the permanent of A, which we denote by per A, is per A : = ~ ala(l) a2a (2) ... ana(n) •
(1.1)
where a runs through all permutations of 1,2 •••• ,n. co 1umns of A as vectors al , •.. ,a in lR -n
n
We usually consider the
and we therefore wri te
per A = per (a- l , -a .... , -n a ) • 2 where
~j
=
(a lj , a 2j , ... , anj)T • (j
=
1, ... , n) •
It is well known that the permanent is a linear function of a. (for each j). -J
In
fact we have per A
(1.2)
a .. per A (i Jj) • (for any j) • lJ
where A(iJj) denotes the matrix obtained from A by deleting row i and column j. Permanents play an important role in several counting problems in combinatorics.
We mention the probleme des rencontres, the probleme des menages,
systems of district representatives. l-factors of graphs, the dimer problem and we give one geometric example.
Consider all 1-1 mappings (not correlations only) of
the points of a projective plane onto the lines such that each point is incident
576
I.H. van Lint
with its image.
How many are there?
The answer is the permanent of the incidence
matrix of the plane. In design theory (O,l)-matrices with constant line sum play an important role.
After dividing by this line sum we obtain matrices which belong to the
class of so-called doubZy stochastic matrices of order n, which we denote by nn' A matrix A is called doubly stochastic if all entries are non-negative and all The set r. is a convex polyhedron with the permutation matrices
line sums are 1. as vertices.
n
The simplest matrix in the interior of n is the matrix J which has -1
n
all entries equal to n •
n
The l'an der Waerden conjecture states: ( 1.3)
If A E n and A f. J then per A > per J • n n n A matrix A =
will be called a minimizing matrix if per A= min{per SISEn }. n n There is extensive literature on the Van der Waerden conjecture for which we refer II
the reader to the book Permanents by H. Minc [7].
We mention the most important
partial results. (1.4) If A E n is a minimizing matrix and a .. > 0 then per A(i Ij) n lJ
= per A.
(1.5) If A E n is a minimizing matrix then per A(ilj) ;;;. per A for aU i and j. n
It was known that the following improvement of (1.4) and (1.5) would be sufficient to prove (1. 3) : (1.6) If A E
'h
is a minimizing matrix then per A(ilj) = perAfor aU i and j.
The proof of (1.6) given by Egoritsjev depends on an inequality of A.D.
Alexand~f
which we treat in the next section.
2. ALEXANDROFF'S INEQUALITY In 1937 A.D. A1exandroff [11 proved an inequality for so-called mixed volumes of compact convex sets (also called mixed discriminants). also been announced by W. Fenche1 [31. R. Schneider [81.
The theorem had
In 1966 the inequality was generalized by
It seems that for a long time the inequality was only known to
geometers working in convex geometry (cf. e.g. [5]).
Recently a number of
combinatoria1ists apparently independently realized that the inequality was also useful for them.
Below we shall give a direct proof of the inequality which
577
The van der Waerden conjecture
Egoritsjev derived as a special case of Alexandroff's inequality. This proof is taken from [61. THEOREM 2.1: Let 2.1,2.2"" '2.n-l be vectors in bE IRn. Then (2.2)
(per(2.1 ,2.2"" '2.n-l ,E))2
~
and equatity hotds if and onty if
n
R with positive coordinates and tet
per (2.1 , ••• '2.n-l '2.n-l) ·per(2.1'··· '2.n-2'~'~) ~
= A2.n_l
for some constant A.
REMARK 2.3: Clearly the inequality (2.2) is also true if we only require that the In that case the claim about the
coordinates of the vectors -1 a. are non-negative. consequence of equality cannot be made.
We shall prove Theorem 2.1 using the concept of a Lorentz space. n following we consider IR with the standard basis.
In the
DEFINITION 2.4: The space mn is called a Lorentz space if a symmetric inner product
~,t>
= ~TQr
has been defined such that Q has one positive eigenvalue and
n - 1 negative eigenvalues. We call a vector
~
and isotropic if
~,~
~,~
positive (resp. negative) if
~,~
is positive (resp. negative)
= O. By Sylvester's theorem there is no plane such that
is positive on this plane
(~
; 0).
The following lemma is a consequence of
this fact. LEMMA 2.5: If 2. is a positive vector in a Lorentz space and
~..!?2 ~ ~f>. ~,E.> PROOF: If
~
and equality hotds iff
=
is arbitrary, then
A2. for some contant A.
is not a multiple of 2. then the plane spanned by 2. and
isotropic vector and a negative vector. form in A.
~
~
Consider
~
+ A~.
~
contains an
2. + A..!? as a quadratic
Since this form is 0 resp. negative for suitable values of A it has a
positive discriminant. 0 Consider vectors 2.1.2.2 .... '2.n-2 in mn with positive coordinates. Let!i denote the ith basis vector (1 ~ i ~ n). We define an inner product on IR by
(2.6)
.- per (a-1.•-a 2•••••-a f I2.x.r). --
578
J.R. van Lint
i.e. where Q is given by q1'J' .- per (a- l ,a- 2 ,···,a-n- 2,e.,e.) -1-J
(2.7)
per A(i,j In - l,n}, where A is a matrix with columns -al, ... ,a. -n a 1 and a • -n-fI
Note that at this point we do not use
THEOREM 2.8: IRn with the inner product given by (2.6) is a Lorentz space. For n = 2 we have Q = (~ ~) and the assertion is true. l Now assume that the asserti on is true for IRn- • We shall fi rs t show that Q does
PROOF: We use induction.
not have the eigenvalue O.
(2.9)
Suppose Q£ = Q.
per (a- 1 .a- 2, •.. ,a-fI- 2,c,e.) = 0 for 1 --J
Then by (1.2) ~ j ~
n.
Consider the n - 1 by n - 1 matrices
and apply the induction hypothesis and Lemma 2.5.
o = per
(al,· .. ,a-n- 3,a-n- 2,c.e.}2 - -J
~
We find
e .) per (a- 1,···,a-fI- 3,a-n- 2,a-n- 2.-J per (a- 1,···.a-n- 3' -c.c.e.) --J •
Since the first factor on the right-hand side is positive it follows that for 1
<;;;
j <;;; n
per (a- 1 , ... ,a-n- 3,c.c,e.}.:;; 0 - --J and for each j equality holds iff all coordinates of c except c. are O. -
assumption Q£
g
J
The
Q, i.e.
n
o =
_j
:= (l,l, ••• ,l)T. T
~ Q~ :"
We consider
per ((1 - {I}l. +
the-inn~r
{}~-1' ...
product xTQ y defined by
,(l - {}).J. +
-
-5"-
{}2.n-2'~',t}·
For every {} in [0,1) this satisfies the condition of the theorem. not have the eigenvalue O.
Hence Q{} does
Hence Q " Ql has the same number of positive eigen-
The van der Waerden conjecture
579
values as QO and since QO is a multiple of nJ - I this number is one. 0 n The proof ·of Theorem 2.1 is nothing but the observation that the theorem is a combination of Theorem 2.8 and Lemma 2.5.
3. PROOF OF THE VAN DER WAERDEN CONJECTURE As was pointed out in Section 1 we need only prove (1.6). This is the essential part of Egoritsjev's paper. THEOREM 3.1: If A E
~n
is a minimizing matrix then per A(ilj)
per A for aZZ
and j.
PROOF: Suppose the statement is false. that per A(rls) > per A.
Then by (1.5) there is a pair r,s such
For this r there is a t such that art> O.
We now apply
Theorem 2.1 (using Remark 2.3) (per A)
2
2 = per (a-. •••. ,a-s ••..•-at •••••-t1 a )
> per (al a , ••••-s a •...•-t1 a ). per (al •.•••-at ••••• -at ••••• ~ a ) - •••••-s =
(k~l
a ks per
A(klt))(k~l
a kt per A(kIS)).
On the right-hand side every subpermanent is at least equal to per A. while per A(rls) > per A.
Furthermore per A(rls) is multiplied by art' which is positive.
Therefore the right-hand side is larger than (per A)2.a contradiction.
0
Actually the proof in [2) uses (3.1) directly instead of the fact that (3.1) had previously been shown to be a sufficient condition for (1.3) to hold.
BIBLIOGRAPHY 1. 2. 3.
A.D. Alexandroff. Zur Theorie der gemischten Volumina von konvexen Korpern IV. Math. Sbornik 3 (45). (1978). 227-251 (Russian; German summary). G.P. Egoritsjev, Solution of van der Waerden's permanent conjecture. preprint 13M of the Kirenski Institute of Physics. Krasnojarsk (1980), (Russian). W. Fenchel. Inega1ites quadratiques entre les volumes mixtes des corps convexes. C.R. Aaad? Sai. Paris, 203 (1936). 647-650.
580
4. 5. 6. 7. 8. 9.
I.H. van Lint
D.E. Knuth, A Permanent Inequality, Computer Science Dept., Stanford University (preprint). K. Leichtweisz, Konvexe Mengen, Springer-Verlag, Berlin (1980). J.H. van Lint, Notes on Egoritsjev's Proof of the van der Waerden Conjecture, Linear Algebra and its AppLications. H. Minc, Permanents, EncycZopedia of Mathematics and its AppZications Vol. 6. Addison-Wesley, Reading, Mass. (1978). R. Schneider, On A.D. Alexandroff's Inequalities for Mixed Discriminants, J. of Math. and Mech., 15 (1966), 285-290. B.l. van der Waerden, Aufgabe 45, Jber d. D.M.V., 35 (1926), 117.
Department of Mathematics Eindhoven University of Technology P.O. Box 513 5600 MB Eindhoven The Netherlands
Annals of Discrete Mathematics 18 (1983) 581-590 © North-Holland Publishing Company
581
SOME EXTREMAL RESULTS ABOUT THE GENUS OF A PAIR OF PERMUTATIONS A. Mach;* and P. Pizzi
1. INTRODUCTION Let a and a be two permutations belonging to Sn. the symmetric group on n elements. and assume that the group (a.a) they generate is transitive.
Then it
can be proved ([ 1 1. [21) that there exists a non negative integer g=g(a.a) such that -1
Z(a) + z(a) + Z(a a) = n+2-2g
(1 )
where. for a permutation y. z(y) denotes the number of cycles of y. g = g(a.a) is called the
genus
of the pair (a.a).
The integer
If s is any circular permuta-
tion of n elements, then the following "triangular inequality" holds ([2]): g(a,a) Let
~
~
g(a,s) + g(a.s)
(2)
be the maximum value attained by the sum on the right hand side of (2) as s
varies in the set of circular permutations on n elements.
In the present paper we
consider the following problems: 1. Find the value of
~
and a circular permutation
~
for which
this value is
attained; 2. Find a circular permutation
~
for which the minimum value g
g(a,a) is at-
tained; 3. If i is any integer between g and
~.
does there exist a circular permutation
~
such that the sum on the right hand side of (2) equals i? We shall solve these three problems; the answer to problem 3 will be in the affermative. provided a # a.
Note that the right hand side of (2) equals
-1
g (a,a ) +
n+z(a a)-c 2 -1
where we have set c = Z(a s) + z(a values of c.
(3)
-1
~).
We are thus led to study the possible
In section 2 we consider two permutations a and a and the sum
582
A. Machi and P. Pizzi
Z(~s) + Z(aB) as S varies in the set of aLl permutations of Sn, without restric-
tion to circular permutations.
Problems 1,2 and 3 will be solved in this case.
The results obtained will be used in section 2, where the case of circular permutations is considered.
2. MINIMUM AND MAXIMUM VALUE OVER THE SET OF ALL PERMUTATIONS
THEOREM 1: ~e t
a,
cr
E
Sn and let
m = min {z(aS) + Z(crB)} S E Sn '1'1,en: (i)
If a and a have the same parity, then m = 2.
(i i) Ii a and a have different parity, then m = 3.
The following lemma is nothing more than an observation. LEMMA 1: Let a and ~uJo
claDses.
G
be two partitions of a set A and assume a and cr have at least
T;;en there exis t two elements a ,b
classes of a and
E
A that be long to two different
0.
PROOF: If all pairs a,b belonging to different classes of class of cr, then clearly
0
belong to the same
only has one class.
LEMMA 2 (Serret): Let y ES~ and let
J z(a)+l,
a
1:
=
(ij) be a transposition.
Then:
if i and j belong to the same cycle of a,
I
l z(a)-l,
otherwise
In the first case we say that
1:
disconnects a.
PROOF OF THEOREM 1: If z(a) = z(o) z(u)
= 1,
then B
In the second case we say that
= identity
will do.
= 1 and z(o) > 1. If z(o) is odd, in which case a and
ty, let '1 be a transposition connecting o.
Then
0
1:
Assume
have the same
pa~i
Results about the genus of a pair ofpermutations
583
By Lemma 1, there exists a transposition T2 connecting aT 1 and aT • Thus, 1
= 1 and z(aT 1T2) a z(a)-2.
z(aT 1T2)
If z(aT 1T2) = 1, then S = T1T2 is the required permutation. In any case the above argument yields a set of p = z(a)-l transpositions T ,T , ••• ,T such that p l 2 Ti connects aT 1T2••• Ti _ if i is even and disconnects aT 1T2••. T _ if i is odd l i l (let TO = identity). Thus, for S = T1T2••• Tp we have Z(aS) + Z(aS) = 2. If Z(a) is even, then
a
equal to z(aS). p
and
a
have different parity, so that for no S E Sn can Z(aS) be
Thus in this case m
3.
The above argument yields a set of
= z(a)-l transpositions Tl ,T 2, ••• ,T p such that z(aT 1T2••. Tp) = 2 and
z(aT T2••• Tp) = 1 and S = T T2••• Tp is the required permutation. 1 1 Suppose now z(a) > 1, and let Z(a) ~ z(a). Let p = z(a)-l.
By Lemma 1,
there exist p transpositions Tl ,T 2 , •.• ,T p such that Ti connectsa T2••• T _ and 1 i l aT 1T2 ••• Ti _l • Thus, z(aT 1T2••• Tp) = 1 and z(aT 1T2••. Tp) a Z(a)-p = z(a)-z(a)+l. Now if a and a have the same parity, then z(a)-z(a) is even, z(a)-z(a)+l is odd and we are in the first case considered above, with aT 1T2 ••• Tp playing the role of a and aT1Tc ••• T that of a. If a and a have different parity, the we are in the p second case, q.e.d. THEOREr4 2: Let a,a
E
Sn and Zet
M = max {z(aS) + z(aS) }. S E Sn Then
-1
MEln+z(a a). Let S be such that
PROOF:
z(aS) + z(aS) If as
M.
= identity, then S = a -1 and M has the stated value. If as # identity, let
T be a transposition disconnecting as. S'
= ST
we would have Z(aS') + z(aS')
Now T cannot disconnect as, otherwise with
= M+2,
contrary to the choice of S.
Thus
there exist transpositions T ,T , .•• such that Ti disconnects aT 1T2••• Ti _l and l 2 connects aT T ••• T _ ·, and such that at each step the value of the sum is M. The 12 i l
584
A. Machi and P. Pizzi
process ends when, for some k,
~T1T2 •••
Tk = identity.
-1
Then S = 'lT 2··• k
= a
-1
and M = n+z(a a), q.e.d. Note that the integers m and Mgiven by the theorems above have the same parity (which is even if a and
a
have the same parity and odd in the other case)
and that this is also the parity of any other integer i such that i=z(aS)+z(aS) for p
E
Sn.
We now consider problem (iii) of the Introduction in the case of a
permutation S E Sn not necessarily circular. THEOREM 3: Let ~,a
E
Sn, m and Mas in Theopems 2 and 3, and Zet
be an integep
such that
m~ i
~
Mand i
=m (mod
2).
Jhen thepe exists S E Sn such that Z(~B)
+ z(aS) = i.
PROOF: It is sufficient to prove that given j > m and S E Sn such that z(aS)
+
z(aS) = 1, then there exists S'
Then starting with z(~S)
j
E
Sn such that z(~S') + z(aS')
= Mwe get the result.
j-2.
Assume
+ z(aS) = j.
If z(a0) and z(aS) are both greater than 1, then Lemma
applies, and if , is a
transposition connecting as and as, then S' = S, gives a sum equal to j-2. sume Z(aB) = 1.
Then z(aS) > 2, otherwise j = m.
and a611 each have at least two cycles. as'l'
Let '1 connect as.
As-
Then as'l
By Lemma 1 again, let 12 connect as'l and
Now with 6' = S1 11 2 we have a sum equal to j-2, q.e.d.
REMARK 1: Note that throughout this section we have not required that the group generated by
a
and
a
be transitive.
3. MINIMUM AND MAXIMUM VALUE OVER THE SET OF CIRCULAR PERMUTATIONS In this section we consider the three problems as stated in the i.e. we restrict our attention to circular permutations. THEOREM 4: Let a,a
E
Sn and let
Introductio~
Results about the genus of a pair of permutations
m = min ~ E
{Z(a~)
+
z(a~)
585
}
C
where C is the set of oiroular permutations of Sn. (i)
If a and a have different parity, then m = 3;
(ii)
If a and a are both odd, then m = 4;
Then we have:
(iii) If a and a are both even, then m = 2 or m = 4. PROOF: (i).
By Theorem 1. there exists S E Sn such that
z(aS) + z(aS)
= 3.
Assume z(aS) = 1 and z(aS)
a
2.
If S is circular we are done.
Otherwise let Tl
connect S and as (Lemma 1). Then
z(S) - 1 transpositions T1.T 2' •••• Tp such that Ti connects ST T2.•• T _ and only one of aST T ••• T _ and aST T ••• T _ • At the end of the 1 1 2 1 2 i 1 i l i 1 process we have ~ = ST 1T2.•• Tp' a circular permutation. and
Thus we can find p
z(a~)
+ z(a~)
= 3.
(ii) By Theorem 1. there exists S E Sn such that z(aS) + z(aS)
= 2.
Note that S cannot be circular. and as a matter of fact S must have an even number of cycles.
Thus if a is odd and
~
is circular. then
Starting from S we now construct a circular permutation connect S.
~
Z(a~) ~
2 so that m ~ 4.
such that m = 4.
Let Tl
Then:
Let T2 be a transposition connecting ST 1 and aT l (Lemma 1). Then either T2 connects aT 1 or it disconnects it. so that either
or
Note that ST1T2 cannot be circular. since p
= z(S)-l must be odd. Thus. a trans-
586
A. Machi and P. Pizzi
position '3 connecting ST T either disconnects the first two permutations (in the 1 2 first case) or it can be chosen as to connect ST 1'2 and 0'1'2 in the second case, In either case, the sum Z(aS") + Z(OS") = 4, where S" = s'lT 2'3' We can therefore choose p = Z(8)-1 transpositions 'l"2"""p such that, for even, either z(aB'1'2" "i)= 1,
z(oS'11 2,· "i) = 1,
Z(81 1'2"' "i)
z(B)-i,
or one of aB'1'2"'Ti and 081112""i is circular and the other one has three cycles,
Let now i
circular and
z(a~)
+
p-l.
In the first case, 'p is such that 811'2"'1p
z(o~)
= 4,
~
is
If we are in the second case, then 1p can be
chosen as to connect B'1'2 •• , l p_l and the non circular one of the other two, In any case, ~ = 611'2""P gives rise to a sum equal to 4. 2 must have an odd (iii) In this case a permutation B such that z(aB) + z(08) number of cycles, and an argument similar to the above shows that either m = 2 or m = 4, q.e.d.
REMARK 2: It is easily seen that there are pairs of even permutation a,o such that m = 2 and others such that m = 4. and
0
= (1,3)(2,4): here m
o = (1,2,3)(4), then m = 2. lem:
unde~
For instance, for n = 4, let a = (1,2)(3,4)
4; if instead we take a = (1,2)(3,4) and The above theorem leaves it open the following prob-
What conditions on a and 0 is the minimum of the sum
aU ail'aulal' permutations ~ of Sn equal to 2?
z(as)+z(o~)
over
Note that the examples gi ven above
show that the transitivity of the group generated by a and 0 is not relevant. COROLLARY: Let a,o
E
Sn and Zet the group
Then the maximum value
~
of the sum
g(a,~)
eiraular permutations of Sn is given by: (i )
if a and
0
have different parity, -1
j.J
(ii)
-if a and ~
(i i i)
= g(a,o) 0
n+z(a 0)-3 are both odd, -1
a 9 (a,o ) +
11 a
and
0
2
+
n+z(a 0)-4 2
are both even, then either
w = g(a,o) + n+z(a
-1
2
0)-4
+
g(o,~)
as
~
varies in the set of
Results about the genus of a pair ofpennutations
g(a.a) + n+z(a
-1
2
a)-2
Let now< a.a> be transitive. -1
exceed n + z(a a).
587
By (2) and (3) the sum Z(az;) + z(az;) cannot
That there exists
z;
such that this value is actually attain-
ed is shown in [2].
We sketch the proof here. As
of
-1
P = z(a)-l such that ,. p. 1
connects a'1'2""i-l and dlsconnects a a'1'2""i-l' Then, lf z; = a'1'2""p' z;' =z;-l is the required circular permutation. We state this in the form of a theorem. THEOREM 5: Let a,a
E
Sn such that
M= max {Z(az;) + z(az;) } z; E C where C is the set of circular pe~utations of Sn.
Then
-1
M=n+z(a a).
Let now m and M have the meaning of above. arises as to whether given an integer i. m ~ i
As in section 2, the question
~
M. having the same parity of m
there exists a circular permutation z; such that the sum z(az;) + z(az;) is precisely i.
The answer is given in the following theorem.
THEOREM 6: Let a,a
E
Sn, a F a.
Let
m = min {Z(az;) + z(az;) } z; E
C
and
M= max {z(az;) + z(az;)} z; E C where C is the set of circular
m~ i
~
Mand i
Then there exists z;
E
=m (mod
pe~utations
2).
C such that
z(az;) + z(az;) = i.
n
of S.
Let
be an integer such that
588
A. Macht and P. Pizzi
For the proof of this theorem we need two lemmas.
LEMMA 3: least three
Let a and
0
~lasses
and
~o distin~t
be
at least two.
0
Then there exist
the set, belonging to different alasses of both a and the w,,:on of the alasses of i and j in a and (ij) su~h
.'nent k of the set
k E (ij)
0
~o
that in
o
elements i and j of
su~h
and
0,
that a has at
that if (ij)
a
is
there exists an de-
that
(ij) \ (ij) n (ij) •
U
a
su~h
partitions of a set
a
0
0
PROOF: Let i and j be as in Lemma 1.
Assume first that (i)
a
= (i) 0 and
(j) a (j) 0 (where the notation is self-explaining). As a f 0, there exists £ such that (£) f (£). Let k E (l) \ (,I.) (or k E (.~) \ (£». Then i and £ are a
a
0
0
k ffi (i) ( U1(i) 0
(j)
U
a
(j)
U
0
and t
E
t rf: (i)
o
U
(j)
(k) ,
U 0
0
LEMMA 4: Let a,o
then
Sn, a f
E
partitions of {1 ,2, ... ,n} ~o
have at least
z(a~) + z(o~)
If t
E
(k)
E
(i) 0\ (i) a ).
If t
E
(j)
o
,then j, k and t will do.
o
t and i will do, q. e. d.
j,
0, and let indu~ed
classes.
Assume now (i) a f (i) 0 ,
(i) a\ (i) 0 (or t
just take j and k as in Lemma 1 and t.
a
0
as in Lemma 1 and k is the element we seek.
~ be a ~ir~ular permutation su~h that the
by the cyales of
a~
and
o~
are the same and
Let
= i.
Then there exists a circular permutation r;' such that
z (a~') +
Z (0 ~ ') =
i
and such that the partitions of {1,2, ••• ,n} induced by the cycles of ar;' and or;' are different.
PROOF: Since that
al;
a f
0,
we have
takes k into land
a~ 01;
f
Then there exist two elements k and £ such
01;.
takes k into t, where t f £.
not belonging to the cycle of k and
l
(11;
(l, .•. ,s)(k,l, .•• ) •••
01;
(l, ... ,w)(k,t, ... ,l, ... ) ...
and we have either:
in
a1;
and
01;.
Then:
Let 1 be an element
If
Results about the genus of a pair of permutations ~
(l ••••• u.k.v ••••• I •••. )
~ =
(l •.••• u.l.v ••••• k•..• )
589
or
(where u and v can be equal to l.k or I).
If
~
is of the first type. then by
multiplying by (l.k.!) we have a~(l.k.l)
(l ••••• s.k)(I ••.• ) •.•
a~(l.k.l)
(l, •••• w,k.t •••• )(!, ••• ) ...
~(l
,k.l)
I,; I .
Now the cycle of k in t.
Moreover. t
F k,
al,;
contains t. whereas the cycle of k in al,; does not contain
since if no such t between k and I exists in
would take k into I, contrary to assumption. z(a~') + z(a~')
and the cycles of
a~'
The case in which
1,;
al,;.
then
as
Then we have:
=i and al,;' do not induce the same partition of {1.2, •••• n}.
is of the second type is dealt with in a similar way (multiply
by (l.!.k)). q.e.d. PROOF OF THEOREM 6: It is sufficient to prove that if z(al,;) there exists a circular permutation starting from a circular permutation both equal to 1. al,;
~'
such that
Z(a~')
1,;
such that
Z(a~) + z(a~)
Thus assume i > m.
we get the result.
+
+
This implies that
z(al,;) = i > m. then
z(a~')
z(a~)
= i-2. Then
= M. the maximum. and
z(a~)
are not
By Lemma 4 we can assume that the partitions induced by as and
on n.2 ..... n} are different.
(i) al,; has at least three cycles and al,; at least two. i.j and k such that: a~
(i, ... )(j, ... )(k .... ) ...
a I,;
(i ..... k .... )(j .... ) ...
Then if
i
l
is the transposition (ij) we have
al,;i 1
(l .... ,j .... )(k .... ) ...
a~il
(i ..... k..... j .... ) ...
I,;i
l
(i .... )(j .... ).
Then by Lemma 3 there exist
590
A. Machi and P. Pizzi
If k belongs to the cycle of i of 1;'1' then (j,k) connects the first and the third permutation and disconnects the second one. ~Tl'
If k belongs to the cycle of
j
of
Then with '1 = (i,j) and '2 = (j,k) or '2 = (i,k~
the same happens for (i,k).
s' = ~'1'2 is the required circular permutation. (ii) i
z(a~)
= 1. If
Z(a~) ~
3, the either i
= m+2 and in the latter case z(a~)
Assume ing
G~T1
= m, contrary to assumption, or
is the circular permutation giving the minimum.
~'
> 3 and let '1 be a transposition connecting aT and '2 one connect-
and
~~l;
if such a '2 can be found as to connect
the required circular permutation.
~1;'1'
then
~' = ~'1'2
is
Otherwise,
and we are in case (i). (iii) z(al;) = 2.
Then we can assume
REMARK 3: If a = 0, then given sible to find
~'
~
z(o~) ~
such that the sum equals i it is clearly not pos-
such that the sum equals i-2.
s' such that the sum equals i-4 (for i COROLlIl.RY: Let ",a m=
min ~ E
3 and we are in case (i), q.e.d.
E
Sn, a "!
{g(a,~)
° and
~
It is possible, however, to find
m+4, of course).
Let
Let
+ g(o,l;) }
C
and \J
=
max {g(a,r;) +
1; E
g(o,~)
}
C
;,;here C is the set of eil'euLar pel"T7lutations of Sn. that m .;; i .;; 9(~,1;)
fl,
then thei'e exists
+ g(a,l;)
I; E
Then if i is an integer suah
C sueh that
= i.
BIBLIOGRAPHY 1. 2.
A. Jacques, Sur le genre d'une paire de substitutions, C.R. Aead. Sai. Paris, 267 (1968). R. Cor; and A. Machl. Su alcune proprieta del genere di una coppia di permutazioni, Bollettino U.M.l., (5) 18-A (1981). 84-89. Istituto Matematico "G. Castelnuovo" Citta Universitaria 00185 Roma Italy
591
Annals of Discrete Mathematics 18 (1983) 591-
PARTIAL t~.
PARALLELISt~
Marchi
~'(
SPACES AND SLIT SPACES
and S. Pianta
The notion of affine space has been generalized in many ways (see e.g. Andre [1], Karze1 [3], [4], and Sperner (P,L) a relation
[7])
by introducing in a linear space
for the set L of all lines, called parallelism.
p
The relation
p
is sometimes assumed as a fundamental relation ([4], [7]), or the para11e1ity of lines is defined by incidence conditions ([ 3] ,[4]). As a consequence of different definitions,
p
may be an equivalence relation or not.
In this paper we continue the study started in [6] of a class of generalized affine spaces (P,L) called partial parallelism spaces.
Here the parallelism II is
an equivalence relation defined in a subset L C L, satisfying Euclid's axiom. a Andre's affine parallel structures (see [1]), Sperner spaces, an important class of Karze1's incidence spaces (see [4]) are included in the class of partial parallelism spaces. In a partial parallelism space it is of interest to consider, for any point q E PJ the so called affine kernel C P consi sti ng of all poi nts pEP (to-
ghether with q) such that the lines {q,p} are in La' In this paper we shall be mainly concerned with partial parallelism spaces in which exchange axiom is fulfilled.
In theorem 1 we give a characterisation of
spaces where the affine kernels are desarguesian affine spaces. describes the spaces which are imbeddab1e in a projective space.
Theorem 2 In this way we
obtain a further characterisation of slit spaces.
1. PRELIMINARY DEFINITIONS AND RESULTS Let (P,L) be a linear space (incidence space), i.e. a pair consisting of a set P, the set of pointsJ and a subset L of the powerset of P, the set of lines, such that (cto [2]. [3], [8]): A1 - V p,q E P, p
* q,
3! L E L such that pEL. q E L; we shall write
592
M. Marchi and S. Pianta
L=: {P.q};
A2 - ILl
~
2 and V L E L: ILl
~
2.
Now we assume that a subset La of L is distinguished and an equivalence relation II caned paralletisn is given such that: ~ - V
pEP.
V
LELa •
L'E
3!
La
such that pEL I and LI II L; we sha 11
write L'=: (pilL). We shall call affine tines the lines of La'
Then the quadruple
.,.- (r.L,La.ll) is called partial pal'aUetism space (in short "p.p. space").
If La
= L. then
Z is an affine paratleZ structure (in the sense of Andre
Examples of p.p. spaces where La
[ 1 J).
* L are
the "geschlitzte RaUme" of H.
Karzel 131. [41. A tineal' variety of a p.p. space L
i) a p.p. space
E' = (P',L',L~.II)
= (P.L,L a , II ) is:
with pI ::. p. L' ::. L.
L~
.- La n L';
ii) any line and any point of I; iii) the empty set
~.
Because of this definition. the intersection of any set of linear varieties is still a linear variety. The closure of
Mof
a subset Me P is the intersection of all linear varieties
containing M.
L
The dimension of a linear variety I' C I is defined by: dim (E')
{I M I: M::' pl. M = E I } A p.p. space z is called
i nf
1.
exchange p.p. space if the exchange axiom is valid
with respect to the closure operation defined above. i.e.: ~ - V
Po
E
Po' P' .... 'Pn' p' E P with p' E {Po.Pl .... 'Pn}\ {Pl ... Pn}' we have:
{pi
,P, ... •Pn l.
For any point pEP the set
{x
E P
{P.x}
E La}
U
{p} is called affine
r.erneZ.
1.' A p.p. space E = (P, L, L • II ) whieh fuZfUs one of the foUowing
a
equivaZent conditions (cf. [6)) will be eaZZed regular: i)
every affine kernel of E is a tin ear variety;
-:-:) every affine kernel is an affine paraUe l structure; {i-i)
tht>. affine kernels f01'l71 a partition of P; •(01'1
L.t~ E La with L n M* ~, Z E L \ t·1 and m E M\ L, we have {t,m) E La •
Partial parallelism spaces and slit spaces
593
1.2 In a regular exahange p.p. spaae the following properties hold (cf. [6]): i) all affine kernels have the same dimension; ii) every 2-dimensional linear variety,
~hiah
is not aontained in any affine
kernel, aontains at most one paraZlelism alass of affine lines.
By 1.2ii) it follows: 1.3 Let
If there is a q E P ~ith R:= r
variety. R II
(f
be a regular exchange p.p. space and r a 2-dimensional linear
~
Ii
rK
q> E L , then V pE a
f: f
n
2. AFFINE KERNELS AS AFFINE SPACES We shall now characterize p.p. spaces in which the affine kernels are affine spaces. DEFINITION: a line T E L\ La will be called projeative if
'r/
pEP \ T any line of
the linear variety T U {p} meets the line T. We shall indicate the subset of all projective lines of 2. 1 For T E Lp and A E La PROOF: Let
{t}:=
Tn A; By
~
~ith
T n A '*'
I/J we have
there exist points
u
~
by Lp
C L.
ITI ;;. IAI
E T\
{t}
+ 1.
and b E
(u
II A)\
{u}.
Then, for any aEA \ {t}, b is contained in {a,u,t} = T U {a} and is different from a.
Since T E Lp the line {a,b} intersects T in a point different from t, u.
Let us now consider p.p. spaces E = (P,L,La,h ) which fulfil the following axioms: PL -
V
pEP
3
T E Lp such that pET.
Vl' - Let R,S E L \ La with that <1"> n S '*'
I/J,
then
V
Rn
S '*'
pER :
I/J;
if there exists a point
I'
E R-S such
I/J.
2.2 In a regular exahange p.p. space r, Vl' is equivaZent to the folZowing VebZen-like axiom:
Vl - Let R,S E L \ La with R n S * I/J; if there exist two distinct points
1" E R. s
E
S with {r,s} E La' then V pER: (p II {r,s}) n S ~
I/J
(and furthermore
594
M. Marchi and S. Pianta
(plI {1",s!} n S = < P > () is a point.
PROOF: i) V1' =>-~.
i.e"
Since R,S ~ La and {1",s)
La we have1" ~ Sand
E
= b',s} () S C
If dim(R
IJ
R then {p,q}
a
~ ~
ii)
If q
V1 '.
E
by V1'.
Since {1",s) , {p,q)E La' {1",s} , {p,q}C Ru S,
S}=2 and R,S ~ L ,we have {~} II
{r,q'n S = {q).
{p,q} by 1.2ii}, hence (p II {1",s}}ns =
R then q = p and hence (p II
{~}) n S = {p} = R () S.
By the definition of affine kernel, it follows
>.
{l',si) ~
(F II
La'
E
* I/J
From (2.2) we obtain:
2.3 In
Cl
1'egulal' exchange p.p. space
L:
fulfilling V1' > if all lines
X E L \ L contain at least th1"ee points, then parallel affine lines have the same
a
::'.7.p'_{i:nc
l i tu.
Let us now prove the theorem:
= (P,L,L a ,1/ )
THEOREM 1: Let L
affine i<el'ne 1s ape at least 2-dimensional;
:;.) ti;:::
; \ E.b
be a regula1" exchange p.p. space such that:
and Vl ' are
f~J..Zfi.lled;
,-'wn thr:: affine ke1"nels a1"e desarguesian affine spaces.
The proof is given by the following lemmas.
2.4
~:ndel"
the asswnptions of Theo1"em 1
the Itaffine Veblen axiom" V2 holds
in the fol"ms: ~ i) - Let R, S E La' with R n S "I/J and S
E S \
Rand pER, we have (p II b"s}) n S
by 1.1 iv), R () S
* I/J'"
{1",s}
E
IRI ;;;.
* I/J
3.
Then for r
(Tmpezaxiom in [51) (note that,
La'
i i) Let R,S E La' with R n S '" I/J, 1" E R \ Sand s E S \ R.
(Y' II S) () (sll R)
* I/J
R \ S,
E
Then we have
(Parallelog1"amm=iom in [5]) (If affine lines have at least
three points, then ii) follows from i) (see [51); vice versa ii} is indipendent from i) if affine lines have at most two points).
PROOF: Let {v}:= R n S, let V be a projective line through v (existing by let;.; to V \ {v}.
Let us project the trian!]le (v,r,s) from a pOint q
E
~)
and
V \ {v,w},
Partial parallelism spaces and slit spaces
existing by 2.1, onto <w>.
595
Since VEL ,<w> '"
Let R':= (w#R) and S':= (wIIS).
By 2.2, Vl'
is equivalent to
~
and, by
U
~,
with
respect to the lines {q,r}, {q,v} and {q,s}, {q,v}, there exist points {r'} .R' n {q,r} and {s'}:= S' n {q,s}.
Since {r,s}C
1" E {q,r} f/:. La' we have {r',s'} = (1"# {r,s})
Hence {r',s'} # {r,s}.
by~.
Now, in the case i), let pER \ {l',v}; then R' ::. V U hence {Z}:= {r',p} n V '"
because V E Lp'
Applying
~
{p}
because R' #R and
to the lines {z,r' },{z,s'}
and {~}, {z,s'} resp., there exist the points {k}:= {z,s' }n(pil {~) and {k"'}:= {z,s'}n S.
Since z f/:.
,~
we have {k} = {k } =
sn (pll
{r,s}).
In the case ii), applying
'!..l several times,
we obtain the following points: {z }:= {w,r}n (qll R), {k}:= {z,s'}n (sll R) since {z,s'}, {q,s'}E La and {q,z}= (qll R)E La' {k;'}:= {z,s'}n (I'll S) since {z,s'}, {z,w} f/:. Land S' E L Since z f/:.
E
(sll R) n (I'll S).
2.5 Under the assumptions of Theorem 1, and if att tines X E L \ La contain at teast three points, att affine tines have the same cardinatity n
PROOF: By
~,
~
2.
all affine lines through the same point have the same cardinality;
hence, by 2.3, we obtain that all affine lines have the same cardinality n
~
2.
2.6 An affine paraUet structure which futfns the affine Vebten axiom is an affine space.
(For the proof, see Lenz [5] ). PROOF OF THEOREM 1: By l.lii), the affine kernels are affine parallel structures in which, by 2.4, the affine Veblen axiom
~
holds.
Then, by 2.6, the affine
kernels are affine spaces which are desarguesian if their dimension is at least three, or even if their dimension is two and If R E La IRI = 2.
Now, let us assume
that there exist affine lines containing at least three points: in order to complete the proof of the theorem, we have only to consider the case that the dimension of affine kernels is exactly 2. If a,b,c, a',b',c'
We have to show that If v E P and
E
{v} = {a,a'} n {b,b'} , the inclusion v E {c,c'} holds.
Let V be a projective line through v; now let us project, as we made in 2.4,
596
M. Marchi and S. Pianta
the triangles (v,a,c), (v,c,b) and (v,a,b) from a point q oJ E
E
V \ {v} onto< w>, with
V \ {v,q} (existing by 2.1). Applying y..! to the lines {q,a}, {q,b} , {q,c} respectively and to {q,v} we
obtain the following points: {a"} : = {q,a} n (wll {v,a}) , {b"} : = {q,b} n n (vii {v,v J) , {a"; : = {q,c}n (w/! {v,c}). lq,c}
Applying now Vl to the 1ines {q,b} ,
resp. and to {q,a} , we have: (aliiI {a,b}) n {q,b} f. 4>, (a"/! {a,c}) n
n {q,c}
f
4>; hence by {q,b} <{.<w>, {q,c}
rf. <w>we obtain
{a",b"}/! {a,b}
and
Again by Vl applied to the lines {q,b} and {q,e} we have
{al,e"}11 {a,e}.
(IJ· II {b ,c)) n {q ,e}
* 4>
and then, because {q ,e} rf. <w> , we have {b" ,e" }II {b ,e}.
In the linear variety V U {a'} , the line {a",a'} intersects the line V in a point :; different from v, w, q. Then applying Vl to the lines {z,a"}. {z,v} resp. and {a' ,b'}
n {z,b"}
applying now n
f.
{
again
~
f ¢, {v,b'}
n {z,b"}
* 4>
and b'
E {z,b"}
{~}
we obtain
because {z,b"} rf.
to the lines {z,a"}, {z,b"} resp. and {z,e"} we obtain {a',e'} n
¢, {b' ,e' }
n {z,e"} '"
¢
and since {z,e"} rf.
{z,e"};
applying
to the lines {z,e"} and {z,v} we obtain {v,e} n {z,e"} F 4>; then since
~
{v,e}n {z,e"} =
E
{V:;}, that is
v E {e,e'}.
3. EMBEDDING OF REGULAR EXCHANGE P.P. SPACES INTO PROJECTIVE SPACES In [3] the so called slit spaces are defined: they are projective spaces from which a projective subspace, or a line, or a point, has been deleted. Let us consider exchange p.p. spaces
(P,L,L ,/!)
L
a
which are regular and
fulfil the following axiom: ~
L n M
- Let L,M
* 4>
<=>
(l U {p}) n
3.1
with dim (L U M)
=
2; it holds
3 pEP \ L U t~ such that (L U {p}) n (-M-U-{-p-)) E L. (t~
U
{p })
V2 and
- II R,S
~
E La
{r ,s]} n {r ,s2} 1 2
*¢
(Then we have:
n L U M E P).
In a regular exchange
at Least 2, ~
L\ La
E
imply
p.p. space L whose affine kernels have dimension
the axiom:
with RII S, II r ,r E R, II l 2 or {r ,51} II {r ,8 }. l 2 2
5
,5
1 2
E
S, it holds:
PROOF: Let
Partial parallelism spaces and slit spaces
*
*
597
*
Suppose now <8 1>
*
kernel has dimension 2, then it follows that there exists {q} : = Hl II H • If 2 the affine kernels have a dimension, greater than 2, then we Rave to show that Hl and H2 lie in a linear variety of dimension 2, hence by V2 and 2.6, in an affine plane. Let us consider the linear variety r : = R U {P,sl}; then dim (r)
*~
*
*
and hence r n <8 1> r (because
r n
dim (r II <8 1» < 3. have: 2 ~ dim (r n <8 »,
3,
* H2,
we
Therefore dim (r II <8 » = 2, and then, by V2 and 2.6, 1
1
r II <8 > is an affine plane; hence there exists {q} : = .Hl II H • Since 1 2 q E {rl'sl} U {p} II {r 2,s2} U {p}, and q p, we have {rl,sl'p} II {r 2 ,s2'p}= = {p,q} E L. Thus, by ~ and dim ({rl,sl } U {P ,s2}) = 2, we have 2 {rl,sl } II {r 2,s2} ~. If Pl = 1'2 or sl = s2 then respectively Pl E {Pl,sl } II
*
*
--
--
We are able now to prove the following theorem: THEORH1 2: Let
l:
= (P,L,La,11 ) be a regular exchange p.p. space of dimension n
such that: a) the affine kernels have dimension m ~ 2; b)
!:!:, Vl'
and
~
are fulfilled;
then l: is a slit space, obtained fpom an n-dimensional projective space
IT
by
deleting an (m-l)-dimensional subspace.
The proof is given by the following lemmas.
We recall that a 'pointed
plane" ("punktierte ebene" in [3]) is a 2-dimensional linear variety whose lines
are only projective or affine of a unique parallelism class. 3.2 In a regular exchange p.p. space mensional linear variety
n
l:,
Yland V3 imply that every 2 -di-
-
-
which contains exactly one class of affine parallel
lines is a pointed plane.
PROOF: Let A be an affine line of n.
i) First, let R,S be two distinct lines of
n
such that R II A *~, S n A * ~ (i.e. R,S f1. La)' R II S
n
= R:LJ:S because
l:
is an exchange space.
* t/l
and RIIS f1. A.
Then
We have to prove that V r ,P E R \ S l 2
598
and
M. Marchi and S. Pianta
\j
sl,s2
{1"2,s2}
E
$ \ R, {l"l,sl} n {1"2,s2}
E
'*
rf>
or {l"l,Sl } II {1"2,s2}'
La' then {l"l,sl} II {1"2,s2}' since
parallel lines. Applying
~
contains exactly one class of
11
rt
Now suppose {l"l,sl}' {1"2,s2}
If {l"l,Sl },
La and Al := (1"/1 A), A2:=(s211 A).
to the lines Rand $, Rand {1"2,s2}' $ and {l"l,sl} respectively,
we obtain the following points: {al}:= Al n $,
{a
} := A2 n R, {b } := Al n
1 n {1"2,8 2 } , {b 2 } := A2 n {1"1 ,8 }. We now apply V3 to the lines {l"l,Sr = {1"1 ,b 2 } 1 and {1'2,s2} = {s2,b l }, obtaining: {l"l,sl} n {1"2,s2} cjo
E
rt
La and {1"2,s2}
(°211 {l"l's,}) n R
*
{J'l,sl } n {1"2,s2}
* ¢'
lines R,$ of
11
La'
2
Applying Vl to the lines R,$ we have:
and then, by
~
with respect to Rand
{P
2 ,s2}' we obtain
$0 we have proved the Veblen axiom for any two non affine
having a common point and meeting an affine line of
11
in two
distinct points. ii) Now, let R be a non affine line of
11
such that RnA!
we have to
rf>;
R \ A and Ii a l ,a 2 E A \ R : {l"l,a l } n {1"2,a 2 } '" rf>. By ~, --with respect to the lines R = {1"1 ,1"2} and {1"2,a }, we obtain {a' }:= (1"11 A) n show that Ii i'1'1"2
E
2
n {1"2'u 2 }. ~, with respect to the lines {1"1 ,a' } II A gives {1"1 ,a l }l{r 2,a } f ¢. 2 Thus the Veblen axiom has been proved also for any two intersecting lines of 11, one of them being affine. iii) Since
L
is an exchange space, we can always assume that
by two lines R,$ as in case i).
11
is generated
By i), ii) any two lines R',$' intersecting Rand
$ in distinct points have a common point or are parallel and, moreover, they both intersect the affine line A or one of them is parallel to A and the other one intersects A. iv) For the lines R',$', intersecting R,$ in distinct points, we can now Prove that
s'} n {;;-;o} '" or l' 1'" 2 E R' \ $' and s'l' S'2 E $' \ R' ' {p'1'1 2' 2 'f' {l"l,sl} II {1"2,s2}' because for R' ,$' it is possible to repeat the arguments used I.J
1'"
for Rand $ in i) or ii).
A.
I.J T
y
Then any line intersecting R' ,$' in distinct points is
also intersecting Rand $ and, if it is non affine, by iii) it intersects the line A too. In this way, all lines of R n $ with points of A.
11
either intersect R,$ in distinct points or join
With the arguments i) or ii) all these lines are inter-
secting in one point or are parallel to A and this means that
11
is a pointed
plane. 3.3 In a regular exahange p.p. spaae r, Vl, V3 and
~
imply that the 2-di-
Partial parallelism spaces and slit spaces
mensionaL Linear varieties
TI
599
of E not containing affine Lines are projective
planes.
PROOF: Let v, r, s be any three distinct points such that show that II rl'r 2 E R:= {v,d and II sl,s2 n {r 2 ,s2} '" ~.
E
TI
= {v,r,s}.
We have to
s:= {~} it holds: {rl,sl} n
Let us denote for short Tl := {rl,sl} , T2 := {Y'2,s2}' affine line through v and p := R u A, a := SuA and
r~oreover
let A be an
= R u S; p and a are 2-di-
TI
mensional linear varieties with an affine line and a non affine one, so that they are not contained in any affine kernel.
By 1.2, p and a have exactly one
parallelism class of affine lines and then, by 3.2, they are pointed planes. q
be any point of
a\
Let
(A u S) and let us project the lines T1 and T2 from q onto
by means of the linear varieties 'i :=
{q}
p
u Ti (i=l,2).
In order to obtain the projections 'i n p (i=l,2) of T1 and T2 onto p, by and the definition of linear variety, we have to consider the intersections
~, ~,
of the lines {q,r i } and {q,si} (i=1,2) with p. Since r E p, {q,r }np= {r } i i i (i=l ,2). Moreover since A = p n a and sl ,s2,q E a, we have {q,si } n p = = {q,s. } n A. 1
Now, two cases are possible.
a) {q,si} are non affine for i=l,2; then by 3.2, since {~f {q,si} n A a : {hi} for i=l ,2.
' A C a,
Hence {hi,Y'i} = 'i n p (i=1,2) are non affine
lines. b) One of the lines {q,sf is affine; let {q,sl } be an affine line, then, by 1.2ii), {q,s2} is non affine.
Thus '1 n p is the affine line (1'1# A) because
{q,sl}11 A, and '2 n p is a non affine line.
At any rate, '1 n p and '2 n p are two lines of
p
which is a pointed plane,
so that these two lines (not both affine) must intersect, by 3.2, in a point = ('1 n p) n ('2 n pl· that the lemma is proved. {t
}:
Then '1 n'2 = {q,t} and hence,
by~,
T] n T2
*~,
By 2.4,3.1.3.2.3.3, we observe that in a p.p. space E=(P,L.La ,#) which fulfils the assumptions of Theorem 2 it holds: L = La U Lp , and moreover L is regular in the sense of Karze1 [3], [4].
Hence
L
is exactly a slit spaoe (P,L).
For any slit space it holds (see [3]. [4]): 3.4 Let (P,L) be a slit space of dimension n.
Then theY'e exists a unique
so
600
M. Marchi and S. Pianta
n-dimensional- projective space II = (P, L) and a pro.iective subspace V* V such that: i) p = P \ V , U) L = {L \ V
Lp
{l
E
By
L: [
3.4,
n V =
I
L I/>}
E
L, [~ V},
and La = {[ \ V
I[
LE L,
n
vi
is then imbedded into the projective space
3.4 we obtain that, for any a
E
P, the set
U
V
l}.
and by the proof of
II
is a projective subspace of
projective closure od the affine space ; hence dim (V)
=
dim (
=
II,
m - 1.
Thus Theorem 2 is completely proved. ACKNOWLEDGEMENTS: The Authors are deeply indebted to Professor suggestions during many stimulating discussions.
H.
Karze 1 for hi s
Work supported in part by the
research group G.N.S.A.G.A. of the Italian National Research Council (C.N.R.), while the second author was enjoying a study grant of C.N.R ••
BIBLIOGRAPHY 1.
J. Andre, Uber Parallelstrukturen Tei1 I, Teil II, Tei1 III, Math. Z., 76
2.
F. Buekenhout, Une caracterisation des espaces affins basee sur 1a notion de
(1961),85-102, 155-163. 240-256. 3.
4. 5. 6.
7. 8.
droite, Math. Z., 111 (1969). 367-371. H. Karze1 and H. Meissner. Geschlitzte Inzidenzgruppen und normale Fastmodu1n, Abh. Math. Sem. Univ. Hamburg, 31 (1967), 69-88. H. Karze1 and I. Pieper, Bericht Uber geschlitzte Inzidenzgruppen, Jber. Deutsch. Math. Verein., 72 (1970). 70-114. H. Lenz, Zur BegrUndung der analytischen Geometrie, Sitzber. Bayer. Akad. wiss., Math.-NatU!'lJ. Kl., (1954), 17-72. t1. 1·larchi, S. Pianta, Su una classe di spazi affini debo1i, Rend. Ist. Lomb. Cl. Sa., (A) 114 (1980). E. Sperner, Affine Raume mit schwacher Inzidenz und zugehorige a1gebraische Strukturen, J. Reine Angew. Math., 204 (1960), 205-215. G. Tal1ini, Spazi di rette e geometrie combinatorie, Sem. Geom. Comb., 1st. I~at. Uni'l. Roma, 3 (1977).
Istituto r~atematico Universita Catto1ica Via Trieste. 17 25100 Brescia Italy
Annals of Discrete Mathematics 18 (1983) 601-{i 16 © North-Holland Publishing Company
601
ON A CLASS OF KI NH1A TI C SPACES M. t4archi and E. Zizio1i
;,
The classical kinematia mapping introduced by Blaschke and GrUnwald (see [3]. [4]) represents the motion group of a Euclidean plane. by means of a subset of the real 3-dimensiona1 projective space.
This first idea gave rise to a wide
range of research concerning the so called kinematia spaaes (for an extensive bibliography of this subject see Hotje [5] and Karze1 [7]). A kinematic space is a group (G,·) with a suitable incidence structure (G,L) where every left and right translation: with a
E
G ~ G, x ~ ax and a r : G ~ G, x ~ xa; G, is an automorphism of (G,L) and where every line through 1 is a suba~:
group (cf. Definition 1.1). If the group (G,.) is commutative, the kinematic space is just a translation struature (see [1 J).
In a more general sense, kinematic spaces are obtained from
a suitable class of translation structures satisfying the following property (see [9]): for every line L and every point p ~ L, the orbit of p under the action of the stabilizer of L is still a line L'.
Here we obtain a further equivalence
tion in the line-set such that Euclid's axiom is fulfilled.
re1~
This second parallel-
ism makes then the translation structure to a double parallelism space (see [8]) such that the additional conditions for kinematic spaces are fulfilled. In this paper we shall be concerned with a general construction method of kinematic structures by means of a commutative kinematic space A and a suitable monomorphism semi group Mof A. constructed (see Theorem 1,
§
In this way a kinematic incidence semi group E is 2); if M is a non-trivial automorphism group, E is a
kinematic space (see Theorem 2, In
§
§
3).
4 we shall investigate properties of E concerning the lattice of linear
varieties and the exchange axiom. condition so that
L
Theorem 3 states a necessary and sufficient
is an exchange space.
602
M. Marchi and E. Zizioli
1. DEFINITIONS We recall some well known definitions (see [6]). DEFINITION 1.1: A kir.e~atic space is a triple (P,L,') where the following axioms hold: ~
(P,·) is a group (where 1 will be the identity);
1.2 L c 2P is such that: V a,b E P, a # b,
I.3
a,b E X (we shall write X=:a,b);
L :
V a E P, a ,a , defined by { t r -+ P a . P -+ P at . x -+ ax' r· x -+ xa
{p
=
are
3! XE
~ollineations
v XE
of (P,L), that is:
a \, (X), a r (X) E
L
L;
1.4
V X E L,
1 E X ~ X ~ (P,·) (i.e.: X is a subgroup of (P,·));
1.5
iLI;;'l;
VXEL
IXI;;'2.
The following property holds: 1.1 L(l):'" {X
E
L: 1
E
X} is a pa!'tition of (P,·) such that:
(i) V X E L(l) , V a E P: aXa- l E L(l); (ii)
L:=
{aX: a E P, X E L(l)}.
Hence L(l} is a normal pa!'tition (cf. [2], [8]) of (P,·).
The following
proposition characterises the kinematic spaces (see [8]): 1.2 Let (P,·) be a g!'oup and L(l) a pa!'tition of P consisting of subg!'oups such that:
v XE
L(l),
V a E P: aXa-1E
L(l}.
::hen (P,L,.), whe!'e L:= {aX: a E P , XE L(l)}, is a kinematic space.
In a kinematic space (P,L,·) we can introduce two equivalence relations II and Ilr' which we call left-and !'ight-pa"l'allelism respectively. DEFINITION 1.2: For any A,B E
L
we define:
t
On a class of kinematic spaces
A II
~
A II
B ~ A-l.A = B-l.B and B ~ A.A- l = B.B- l
-1 r
where A
603
:= {a
-1
: a E A};
A.B:= {ab
a E A, b E B}.
By definition 1.2 we have: 1.3 FoT' any A,B
E
L:
(i) A II~ B ~ 3X E L(l), 3a,b E P such that A= aX , B= bX; (ii) A Ilr B <=> :lY E L(l),
3a',b' E P such that A=Ya', B=Yb'.
Both the left-parallelism and the right-parallelism verify the condition (Euclid's axiom): P V X E L , V pEP :;)!Y E L such that p E Y and Y II X; t
V X E L , V pEP 3!Y'E L such that p E Y' and Y'II X r
A kinematic space is then an affine parallel structure (in the sense of Andre [11) either with respect to the left-parallelism or to the right-parallelism. We denote by (plltX) the line through p left-parallel to the line X; in the same way we denote the right-parallel through p by (pll X). r
DEFINITION 1.3: A kinematic space E'=(P',L',·) is a kinematic subspace of a kinematic space E=(P,L,·) if (P',·)
~
(P,.) and L'
C
L.
In the following we shall consider in particular commutative kinematic spaces Q.e. (P,·) is a commutative group) in this case the additive notation will be used. The proof of the following is straightforward: 1.4 A kinematic space (P,L,') is commutative if and onZy if II t
= Il r •
2. CONSTRUCTION OF INCIDErlCE SEMIGROUPS Let (A,L,+) be a commutative kinematic space.
By a monomorphism of (A,L,+)
we understand a monomorphism of the group (A,+) with the property:
v XE
L(O) : Il(X)
E
L.
Now let M be a semigroup of monomorphisms of (A,L,+) containing the identity 1. Then:
604
M. Marchi and E. Zizioli
A)( M := Ha,ul : a
2.1
to the multiplication
01
A, u
E •
01
E
M} is a semigroup with identity [O,11 in respect
defined by (1)
2.2 ;'ize mapping A -- A x M { a--+[a,l]
i
is a I'1cnomorpizism.
If M is an automorphism group, then (A
x
M,·) is the semidirect product of A
and M. We shall identify henceforth A with i(A):= A x 1 and X E L(O) with i(X). 2.3 V [a,ll]
A x Mone has:
E
REMARKS: I) For
[a.~)
.
Ac A·
[a,~)
E
A x M, let us consider the mapping:
{ x --.... a+u(x).
A-~
[~).
Then [~) is injective because if
11
is surjective.
11
A
is injective and [~) is surjective if and only
In this case we have:
[~) (X)
II) [a,lll
[a.ul.
E
for all X E L.
L
A x M is a unit of (A
E
x
-1
M,·) if u is a bijection and u
EM;
then [ a, ul
-1
= [11
-1
( -a ),
-1
u ).
(2)
In order to provide A x Mwith a kinematic structure, we define for any a
E
A the
following subset of Ax M: Ba . = {[ a-
2.4
For any a
E
~
(a),
~)
F,; EM}.
A. the set B is a subsemigroup of (A x M,')
a
(3)
isomorphic to M.
On a class of kinematic spaces wi th B
a
II
A
1:1 {[
605
0,1] }.
2.5 The following statements ape equivalent:
K. 1) "I 11 E M\ a) b) X E
=: M* , "I a E A \ {O} =: A* one has 11 (a) .,. a;
{l }
"I a,b E A with a "'b: Ba
II
Bb ={ [O,l]};
1,
"I 11 EM, "I a E A the equation X-11(X) = a has at most one solution
A.
PROOF: Let a .,. b. Then = b-n(b)
a-~(a)
= a-b .,. O.
~ ~(a-b)
For a E A condition a) implies Bo
=a
~ ~
II
Bb
II
= nand
~ ~
Therefore K.l implies
1,
=[ 0,1]. ~(a)
= [b-n(b),nl E Ba
[a-~(a),~]
~
= 1 and
[a-~(a),~]
Ba = {[ 0,1] }, hence
= 1 and K.l is valid. (x-y) K.l and b) are equivalent.
=
2.6 The following statements aPe equivalent:
K.2) "I 11 E a) U{ B
a
l', : a
E
"I a
E
A :Ix with X-11(X)
A}
U
A
=A
x
a;
M.
PROOF: If K.2 holds, then for any [a,11] E A x Mwith 11 .,. 1 there exists an x E A = [X-11{X),11] E Bx and for 11 = we have [a,l] EA. Hence a) Now let a) be fulfilled. For any [a,11] E A x M'" there exists at least one
such that holds. xE
A
[a,~
such that [a,11] = [X-11(X) ,11] i.e. the equation X-11(X) = a has at least one
solution. By 2.5 and 2.6 we have: 2.7 The conditions K.l and K.2 hoZd if and only if fop any a
E
A, and any jJ
the equation x-jJ(x) = a has exactly one solution x E A.
2.8 Let v E Mbe given. a)
( i)
"I 11 EM, ::I
Then ape equivalent:
EM (ii)"I"EM,:l11 E M A
b) :I a,b,c E A with [b,vj.B c)
"Ia,b
E
and
V11
"V
V]l
AV;
a
= B .[ b,v] ; c
A and c:= b+v(a) one has: [b,v] '6
a
B·[ b,vj • c
E
l'
606
M. Marchi and E. Zizioli
"lI b+v(a)-vl1(a),wj :
PROOF: Since [b,v]'B ).11)
:
A EM} 0)
a implies
a).
Now we prove
a) =>
B • [b,vj = {[ b+v(a)-:..(b)-AV{a)H(b),AVJ : A E c and by a) this set is equal [b,~· Ba'
2.9 Fo1' [b,v]
E
A
x ,\I,
X E L(O)
[ b, v]
• X
and
M} and B .[ b,vj = ([ c-,,(c)H(b), c c). For c := b+v(a) we have
11 E
M}
= {[ b+v(a)-J,v(a),Av]
Y:=v(X) we have
y. [b, v]
(4)
•
PROOF: [b,v] . X = [b,v] • {[ x,l] : x EX} = {[ v(x)+b,v] = ([ v(x) ,1]
: A EM}
X
EX} =
: x E X} • [b,v] = Y • [b,v] •
Now we can define and incidence structure (P,L) on the set (A
x
M,·) in the
following way:
P .- A )( M;
(5)
L .- ([ b,v) • Ba : [b,v) E A x M, a E A} U ([ b,v)' X
[b,v]
E
A
x
M, (6)
XEL(O)}.
: a E A} U {X : X E L(O)}. The incidence a structure (P,L) with the multiplication (1) is an incidence semigroup according to By definition we have: L([ 0,1)) = {B
the definition of Wahling [10].
We recall that a triple (P,L,') is called inci-
cience semi group if:
G.I.1
(P,·) is a semigroup;
G.I.
P L .::. 2 ;
~'
,';.I •.3
a£(x)
E
pfor all a E P, the mapping a : { t X -
P
ax
is such that V X E L
L.
Moreover a kinematic incidence semigroup is an incidence semi group in which any line through 1 (1 is the identity of (P,-)) is subsemigroup of (P,·).
THEOREM 1: Let (A, L,+) be a commutative kinematic space, M a monomorphism semi-
(I lA, L, + I such that the condi tions K. 1 and K. 2 ape utl'ucbre l: : = (P, L,·), where P, L and ". are defined in
gmup
II
kinema~ic
incidence semigroup.
fut fil red.
Then the
(5), (6) and (1), is a
607
On a class of kinematic spaces
PROOF: By 2.1, (A
x
-
-
and the definition of
(A,L,+)
Finally for any
[a,~]
E
p
- -..... P
[x,,,;]
- - - + [a,~]
[ a,~])!, :
{
-
p
M,·) is a semi group and by (6) L ~ 2. Moreover by (6), 2.4
maps certainly lines of
-
the lines of LOO,l]) are sub-semi groups of
-
(P,·).
P the mapping:
L
•
[x,~]
into lines of
L
by definition (6).
REMARK: It is possible to introduce an equivalence relation (left-parallelism:
II)!,) among the lines of an incidence semigroup E in the following way: V [al'~l] ,
[a2'~2] E
-
P, V Ba' Bb
L([O,l]):
E
[al'~l]·Ba
V [al'~l]
II)!, [a 2,1l 2] • Bb <= a=b; , [a2'~i E P, V X,Y E L(O):
[al'~l]'X
II)!,
[a2'~21
. Y <=> X=Y.
Moreover if we assume that the condition 2.8a) is valid, it is possible to introduce a further equivalence relation (right-parallelism: II r ) in a subset of the following way:
-L in
V [al'~ll , [al,lll]
[a2'~21 E
-
P, V Ba,B b E L(IO,ll):
Ba Ilr [a 2,1lZ" Bb
<==>
al + Ill(a) = a2 + 1l 2(b).
-
In order to extend this right-parallelism to all the lines of L, it is necessary, by 2.9, that: V v
E
M, V X E L(O) : v(X)
E
L(O).
If this condition holds, the elements of Mare automorphisms of (A,+) and we can define
3. CONSTRUCTION OF KINEMATIC SPACES
Now let M be an automorphism group of THEORm 2: Let
(A,L,+).
Then we have:
(A,L,+) be a aorrurrutative kinematia spaae and
group of (A, L,+j, with
IMI ;;;. 2,
Man automorphism
suah that the aonditions K.l and K.2 are fuZfUZed.
Then E:=(p,r,·) as defined in (1 ),(5),(6) is a (non aommutative) kinematia spaae.
608
M. Marchi and E. Zizioli
PROOF: By Theorem 1, E is a kinematic incidence semigroup, where (P,·) is a group because M is a group. L
In order to prove that E is a kinematic space we show that
fulfils the assumptions of 1.2.
-
LO 0, lJ) is a set of subgroups of (Pi') because, by 2.2, all X E L(O) are
subgroups and, by 2.4, all Ba E LO 0,1)) are isomorphic to the group M.
L([ 0,1))
and 2.6,
P which
is a partition of
By 2.5
is normal because [b,vj • X • [b,vJ -1=
= v(X) for X E L(O) (by 2.9), and, since in a group the conditions (i) and (ii) of 2.&1)
are valid, we have [b,vJ • Ba • (b,vJ
-1
= Bb+v(a) by 2.8b),c).
Therefore
E
is a (non commutative) kinematic space. 3. 1
L~t
r:=(p,L,') t2 a kinematic space constructed as in Theorem
:: ' :" ( P' , L ' , .)
IJi th
pI: = {[
a, 1) : a E A},,: A xl,
2 and let
LI : = {[ a, 1] • X : a E A, X E L( 0) }.
: i,cn tree fo ZlCTuYing properties ho ld: (i)
z'
(ii)
L' is a kinematic subspace of r
is a cOrmlutative kinematic space isomorphic to (A,L,+);
(iii) any tine [a,l1j'B E L(Va,c E A, VllEM) meets c [a-c+>J(c),l) •
E' exactly in the point
PROOF: (i), (ii)- By 2.2, the mapping i is the required isomorphism. [a,l1l' Bc " {[ a+\l(ch.lf;(C),\lE;1 :
E;
(iii) Since
E M} we have
[a-c+\l(c),ll ,,[ a,l1]' Bc () pl. 3.2 Let /\
oe
a kinematic subspace of
BC E L(f 0,11 ),
1;;1. th
Bc
E
=( P, L,· )with Aif A and I\' : = 1\ () A.
For any
we have /\ "I\' . Bc •
C /\,
PROOF: Every line of the partition {(p
II~
Bc) : p E /\} of/\ meets A' exactly in
one point, by 3.1, and so {(p
lit Bc) :
Since (a II £Bc' 1\ " U{ (a
=
p E I\}
= {(a
/1~Bc):
a
E I\'}.
[a, 11 • Bc we have
I I 'l.B c) : a
E/\')
= 1\'.
Bc'
Now we investigate if, in the kinematic space E introduced by Theorem 2, there are
-
-
lines L ELand points pEP \ L such that: (p
119. L) "
(p
Ilr L).
609
On a class of kinematic spaces
The following properties hold:
3.3 For any BEL ([ 0,1] ) and any [b, v] ¢ B one has
a
a
PROOF: By 1.3 and 2.8, we have: ([ b,v] II, Ba) = [b,v] • B = Bb ()[ b,v] and ([ b,v] II B) " a +v a r a
B • [b,vj. a
This implies: [b,v] . B = B .[ b,v] a a
<=>
3.4 For any X E L(O) ~
b+v(a)=a, that is b=a-v(a) and [b,v] = [a-v(a) ,v] E B . a
L([ 0,1]
) and any [b,v] ¢ X one has:
([b,v] II ,Q, X) = ([b,v] II r X)
<=>
v(X) = X.
PROOF: It is evident by 2.9.
4.
KINEr~ATIC
EXCHANGE SPACES
Let L = (P,L,.) be a kinematic space (cf. Definition 1.1) and L' =(P',L' ,.) a kinematic subspace of L.
For any pEP, the point subset P:= p.p' is called
proper tinea!' variety of L.
By a tinea!' variety we understand a proper linear
variety, the empty set and all the points of P. 4.1 For any a,b
E
PROOF: Let a' :=p
-1
therefore a,b ~ P.
P:= p.p', ·a, b':=p
a:tl ~ P. -1
-
--
·b E p', then, by axiom 1.3, a,b = p.(a' ,b') and
~
~
4.2 For any proper tinea!' variety P:= P'P', there exists a kinematic subspace
L"=(P",L",') such that P = P"·p. PROOF: With L'=(P',L',') is also L":=(P",L",·) L":={pXp
-1
\~ith
P":=P'P"p
-1
: X E L'} a kinematic subspace and we have P = P".p.
and
610
M. Marchi and E. Zizioli ~
4.3 POr' ani! a E P .- p·p'=P"·p and L ~ P, ;,,'ith L E L, we have
I I ~ L)
(a
~ 1-'
(allrL)CP.
and
PROOF: Let L = pL' = L"p and a=pa'=a"p with L'
~
P', a' E P' and L"
Since P' and pOI are kinematic subs paces we have: (a' II (a" I I
r
L")
C
-
P".
L')
~
C
-
~
p lI , a" E PII.
P' and
This gives us: ~
(a 11 9" L)
pta' II
(all r
(a"ll
L)
r
t
L')
L")p
p'P'
C
C
-
and
P ~
p
lI
p= P.
4.4 'The set-theor'etical inter'Gection of any set of lineal' var'ieties of
L 7:13
a
Unea:r ))ar'7:e ty.
PROOF: P = P • P' and
Q = q.Q' where P' and Q' are kinematic subspaces.
a E P n Q. then there exist pEP' and q ~ --1 ~ --1 have: P = ap . P' a· P' and Q =aq P n Q = a'P' n a.Q'
Q' such that a=
E
Q' = a • Q'.
P'P
Let
= q.q; hence we
Then
a(P' n Q').
Therefore P n Q is a linear variety because P' n Q' is a kinematic space.
By
the associative property of set-theoretical intersection one has the thesis.
DEFINITION 4.1: A linear variety P of a kinematic space CZC8Ul'e
if P is
of a point-set 1
containing I. set of P. min
~
-
~
{I I I
for any a.b E
4.5 For' I
the intersection of all linear varieties
-
As the dimension of a linear variety P we define the cardinal number: ~
I~P.I=P}-l.
if the exchange axiom
b
= (P,L.·) is called the
We shall write P:= 1 and the set I will be called a generating sub-
DEFINITION 4.2: A kinematic space
~
l:
E
{a} U I.
C
~
l:
= (P,L,·) is called kinematic exchange space
holds with respect to this closure operation:
P and for any subset I b
~
I
~
I.
a E {b} U
P and pEP, we have p.!
~
=
p·r.
P:
On a class of kinematic spaces
611
PROOF: The linear variety P'I contains the set p'I, hence p7I ~ p·I. Since -1 -1 --1 I =P p·I is contained in the linear variety p • p'I we have I ~ P • p·I hence p'I
4.6 E
~
p·I.
= (P,L,')
ex~hange
is a kinematic exchange space if and only if the
axiom
holds for the lattice y of aZl kinematic subspaces of E.
PROOF:
For I
C
P let be the smallest kinematic subspace of
If 1 E r then < I > =
E
containing I.
I.
First we assume that y is an exchange lattice. Let I ~ P with F~, -1 -1 -1 j E rand bE {a}U r \ I. Then j .b E j . ({a}u 1) \ j (1) = -1 -1 -1 -1 -1 -1 = {j . a } u j . 1\ j I by 4.5 and hence j . b E <{j •a } u j 1>\ < j I>. -1 -1 -1 -1 From ~ we obtain j a E<{j b}u j 1> = j ·{b}u I -hence a E {b} u I. Now let
-::-r-
E
be a kinematic exchange space.
For Ie P and b E\ <1> we have
b E < a u 1 u I> \ <1 u 1> = a u 1 u I \
4.7 Let
~
TUI
and hence by ~: a E b u 1 u I =
= (P,L,') be a kinematic exchange space of dimension n
exist a line L and a point q (q
~
~
3.
If there
L) such that:
= (q IlrL)
(q II~L)
then for all points pEP we have:
PROOF: Let M:=(qll L)=(qll L). r
~
For each point p
~
L,M we consider the linear
variety: P:= {'j)}UL n {p} u M. (i) For
p~
----= {p} U L would
{q}U L we have {p}U L., {p}U
imply p E {q} U L by
~;
dim "{j)TUL = dim {p} U M=2 by~.
{p} U L or {p} U 14 and dim
P<
we have: (pll L)=(pil M)
{j)TUL n {p} U 11 =
~
C
~-
because q E {p} U M =
~1
therefore P is a proper subset of
Besides, by 4.3,
P and
(pil r L)=(pll r M) -C ~L n {p} U M = P. Hence (pl/ ~ L)=(pll r L). (ii) Now let p E {q} U L. q'
~
{q} uLand by (i)
(pII~L)
= (pllrL) by
Since dim
(q'II~L)=(q'II/).
(i).
l: ~
3, there exists a point
Since p
~
{q'}U L, we have
612
M. March; and E. Zizioli
From now let
~
= (P,L,') be a kinematic space constructed as in Theorem 2,
We investigate the problem: Is
L
a kinematic exchange space under the assumption
that (A,L,+) is a kinematic exchange space? First we prove the following: 4.8 Let t=(P,L,')
be a kinematic space constructed witiz a kinematic exchange
spaC0 (A,L,+).
~
If
\.l
E
M,
If
If
is a kinematic exchange space then
X E L(O)
\.l(X)=X.
PROOF: For dim A = 1 i.e. L = {A} we have u(A)=A for all
dim A ~ 2; then dim P ~ 3 because A F Land
L
M.
\.l E
Now let us assume
is an exchange space.
Since (A,L,+)
is a commutative kinematic space, we have: (q//£L)=(q//rL) for all L E L(O) and any point q E A \ L, (pll~L)=((c,Ylll9.L)=
= (pll
r
and, by 4.7, {[c+Y(k),yl : \<E L1 = ([c,ylllrL) = {[c+k,yl:kE L}=
L) for all points p= [c,y]
E
A x M = P.
This gives us c+y(L)=c+L hence
y(L)=L, We now assume that for the automorphism group Mthe following condition is fulfilled: If " E M, If X E L(O)
\l(X)=X.
(7)
Then the following theorems hold: 4.9 Fol' any kinematic subspace 1\' of (A,L,+) and for any Bk
E
L([ 0,1] ) \ L(O),
the set 1\' 'B k is a kinematic subspace of L.
PROOF: By 1.2 it is sufficient to verify: (i) A'.
Bk is a subgroup of
(ii) the lines of L([O,l]) through any point of N.B k' {[O,l]} are contained in A'.B k. (i) Since: (P,.) = (A" M,·);
1\"B k=
([
a, 1]
•
[k -c;(k) ,.';] : a E 1\', .'; EM} = ([ a+k-t;(k) ,.';] : a E 1\',.'; EM}, we
have for any a1,a 2 E I\' and .';1'.';2 EM: [a 1+k-s l (k),sl] -1 -1-1 = [aj+k-s (k) ,f.: ] • [- s2 (a )+k-s (k) ,s2] = 2 2 1 1
[a2+k-sz(k),s2]
-1
613
On a class of kinematic spaces
-1
-1
-1
= [al-~1~2 ~2)+k-~1~2 (k)'~1~2 ] E A"B hence
al-~1~;1(a2)
-1-
because ~1~2 (a 2) E a,a 2
k
by (7) and
E A'. \ {[O,l]} with a E /\'. If a f. 1 then there k A with x-a(x)=a+k-a(k) by 2.7, and therefore
(ii) Let [a+k-a(k),a] E N'B is exactly one x
E
[0,1], [a+k-a(k),a] =B = {[x-x(x),x]:xEM}. x
Let a f. a.
Fora=OwehaveB
x
=B
C/\,·B. k k
From a=x-k-(a(.)-a(k))=(x-k)-a(x-k), we obtain with (7):
a,(x-k) = O,a C I\' and b :=x-k-x(x-k) E I\' for all x E M. This gives us: X [x-x(x),x]=[b +k-x(k),x] EN·B hence B cl\'·B. For a=l we have x k xk [a+k-a(k),a] =[a,l] maE;\' hence [O,l],[a,l] =O,ac/\' CA'·B. k
4.10 For any BE L([a,l])
and any qo
{ q o} u B n A E L( a) or
{q
0
}
U
E
-
P \ B we have:
B~ A
PRoaF: (i) First we assume B ~ A and qo
~
A.
Then there are h,k E A with Bh=B,
qo E B , and hf.k. For any t:= [h-~(h),~] E Bh we have k A n (til B ) = An t·B = {[h-k-~(h-k),l]} hence by (7): .e, k
A n Bh·B ~
k
k
a,(h-k)
= A n {(tll.e,Bk) : t E B } = {[h-k-~(h-k),lJ h
~
E M} ~
(8)
and
0, (h-k) • B ::l Bh·B ::l Bh U {qo}' kk-
(9)
Since l,qo E Bh U {qo} we have Bk ~ Bh U {qo} hence Bh·B ~ {(tll.e,B ) : t E B } ~ Bh U {qo}.
k
h
= {t.B : t E B } = k k h Together with (8) and IMI ~ 2 we obtain:
(la)
By 4.9, a,(h-k) • Bk is a kinematic subspace, hence O,(h-k)'B ~ Bh U {qoJ by (9) k and a,(h-k} = O,(h-k}.B n A:: Bh U {qo} n A:: a,(h-k} E L(O} by (10). k (ii) Now let be B=B ~Aand qo E A (or B ~ A and Bk=O,qo ~ A). Then we h have: {qo}U Bh n A:: By 4.9, O,qo·B is a kinematic space containing qo h --and B ; therefore {qo} U Bh = a,qo • Bh and {qo } U Bh n A = O,qo E L(O}. h (iii) For BE L(O) and qo E A we have {qo} U B ~A because A is a kinematic
o,q:-.
space.
---
M. Marchi and E. Zizioli
614
4.11 If (A,L,+) is a kinematic exchange space, then for any line BE L([O,l]) and f01'
any point- qo
E
P \ B t;he linear variety {qo}
U
B is a kinematic: exchange sub-
spae.;:. c;" dimension 2.
PROOF: (i) Let B ~ A or qo i!:=
~
Then {qo} U B n A=:A E L(O) by 4.10 and
A.
{Go} U B is a kinematic subspace because [0,1] E {qo} U B.
Hence
r;= {qo } U B = A'C for any line C= [O,l],x with x E {qo} U B \ A because of 3.2. This gives us: ii=AU{x}foranyxEn\A.
(11)
Now let L C IT be a line and p E IT \ may assume [0,1) ELand L f A by (11).
L.
Since IT is a kinematic subspace, we
Then L U {p} is a kinematic subspace
contained in n and intersecting A in a line.
Therefore L U {p} n A = A and
= A'L = n.
L U {pl
(ii) If qo E A and Be A then IT:= {qo} U B ~ A and IT is with A a kinematic exchange subspace.
By 4.6, 4.8 we have proved one part of the following:
,~et
THEOREH 3:
(A,L,+) be an exchange commutative kinematic space and Man auto-
IMI ;;.
W
lTVl'phism
group of (A,L,+) with
futfUZeL
,;le;i "i:c !:7:nematic space L :=(?,L,') defined by (1),(5),(6) is a
kinematic exchange Yu
d~ace
2 such that the condition
and K.2 are
I.lith dim L = dim A+l if and only if:
M, Y X E L(O) : u(X) = X.
E
PROOF: By 4.6 it is enough to consider a kinematic subspace A and two points x,y E P with Y E {x} U;\\;\.
Let !\,:=An A, r:= {x} U;\ and r':= rnA.
(i) For /\. = 1\ and x E A we have x E {y} u;\ because A is an exchange space. (ii) ForN=A and x
;\'.B ~ x U /\' for B := [O,l],x hence {x} u;\' = /\"B by k k k Since y E /\'B k \ /\ there are a E ;\ and bE Bk \ {( 0,1] } with y=a·b hence
{x} uA = {x} 4.9.
with y=a·b x
E' {f
E
U
I\'
~
{[O,l] ,a,x}\ [0,1] ,a.
0,1] , a ,y}
=
~'.T.
Because of 4.11 we obtain
C {y} U /\ •
(iii) let;\ ;. ;\'. also
rI. A we have
By 3.2 there is T E L([ 0,1] ) with ;\ = 1\'. T and we have
615
On a class of kinematic spaces
Since x E r there are a E r' and t E T with x=a·t and we obtain x E {a} u/\ because T ell.
This gives us r = {x}
U/\
kinematic subspace containing /\ = /I."T.
= {a}
U /\'
{y}u/\
\.Je have a'
= {a}
/\
U
/\'.T.
and t' E T with y=a'·t' implying
(A,L,+) is an exchange space and since/\' gi ves us {a'} u 1\' = {a} u
By 4.9, {a; u/\'.T is a
This implies r = {a} U
Since y E r \ A there are a' E {a} {a'}u/\ = {a'}uN·T.
U/I.
~
~
1\'
because
y~
/\=/\"T.
Since
A, a' E A the inclusion a'E {a}Jf{\ /\'
1\' •
flow we have: {x} U /\
= {a}
U!\
= {a}
U
/I.' • T
{a' } U I\' . T
{a' } U
1\
= {y}u 1\
•
ACKNOWLEDGEMENT: The authors are very grateful to Professor H. Karzel whose suggestions were valuable in improving this paper. This work is supported in part by the research group G.N.S.A.G.A. of the Italian National Research Council (C.N.R.).
BI BLI OGRAPHY 1.
2. 3. 4. 5. 6. 7. 8. 9. 10.
J. Andre, Uber Parallelstrukturen. Teil I, Teil II, Teil III, Math. z., 76 (1961) 85-102, 155-163, 240-256. R. Baer, Partitionen Endlicher Gruppen, Math. Z., 75 (1961) 333-372. W. Blaschke, Euklidische Kinematic und nicht-euklidische Geometrie, Ztsehr. f. Math. und Phys., 60 (1911) 61-91, 203-204. J. GrUnwald, Ein Abbildungsprinzip, welches die ebene Geometrie und Kinematik mit den raumlichen Geometrien verknUpft. Sitzungsber. Akad. Wien, math. nat. Klasse IIa, 120 (1911) 677-741. H. Hotje, Zur Geschichte der kinematischen Raume und 1nzidenzgruppen, Beitrage zur Geometrie und Algebra, Nr 5, TUr1 - NATH - 7916, Technische Universitat t-1Unchen, july 1979, 26-48. H. Karzel, Kinematic spaces. Ist. Naz. AZta Matem. Symp. Math., 11 (1973) 413-439. H. Karzel, Spazi cinematici e geometria di riflessione, Sem. Geometrie Combinatorie 1st. Mat. Fac. Sc. Univ. Roma, n. 30- settembre (1980). H. Karzel, H.J. Kroll and K. Sorensen, 1nvariante Gruppenpartitionen und Doppelraume. J. reine angew. Math., 262/263 (1973) 153-157. M. f1archi and C. Perelli Cippo, Su una particolare c1asse di S-spazi. Rend. Sem. Mat. di Brescia, IV (1979) 66-83. H. Wah1ing, Projective Inzidenzgruppoide und Fastalgebren. J. of Geom., 9, 1/2 (1977) 109-126. 1stituto Matematico Universita Cattolica Via Trieste 17 25100 Brescia Italy
This Page Intentionally Left Blank
617
Annals of Discrete Mathematics 18 (1983) 617-624 © North-Holland Publishing Company
A BRIEF SURVEY OF COVERING RADIUS H.F. Mattson Jr.
1,
and J.R. Schatz
The covering radius, though a fundamental geometric parameter of a code, has not been much studied until recently. It is defined as the smallest integer
p
such that the spheres of radius p
(in the Hamming metric) centered at the codewords cover the containing space. Here we restrict our attention to binary codes C, which are nonempty subsets of n
Z2· Background.
The Hamming distance between two points of Z~ is the number of
coordinate-places in which the vectors are different: If x=(xl, ... ,x n) and y=(Yl'··· 'Yn) then dist(x,y)=I{i; XifYi' i=l, ... ,n}l· its distance from O.
The weight of a vector is
The minimum distanae, often denoted by d, of a code C is the
smallest nonzero distance between two codewords.
The paaking radius of a code is
defined as the largest integer TI such that the spheres of radius TI centered at the codewords are mutually disjoint. code is called perfeat. L
~
Thus TI
~
p (the covering radius); when TI =p the
If d is the minimum distance, then
TI~(d
-
stands for the familiar greatest-integer, or "floor" function.
1)/2~,
where
The code can
correct all errors of weight at most the packing radius, but only some of those with weights between TI and p. We call n the length of the code C.
If C is a linear subspace of dimension
k of Z~ the code is called an (n,k) code, sometimes an (n,k,d) code where d is the minimum distance. The redundanay r=n-k is the codimension. Sometimes we "extend" a code C, linear or not, by the process x -+ (X,E), where x runs through all the vectors of C and E is the sum of all the coordinates of x.
Thus C is mapped to a code of length 1 + n, and the dimension remains the
same.
Thi s process is ca 11 ed "extendi ng the code by an overa 11 parity-check"; it
always increases the covering radius by 1. Sometimes we define a linear code by means of its parity-aheak matrix, a matrix of n columns of rank r=n-k for which the code is the null space of
n~ctor~
618
H.F. Mattson Jr., and J.R. Schatz
that is. the code is the set of all linear relations on the columns. A c!pliu code is one which is invariant under the cyclic permutation of Thus C is cyclic if and only if (xl,···,x n)£ C~(x2,x3, ... ,xn.xl)£C The BCH codes are cyclic. and the Reed-Muller codes are extended cyclic codes.
coordinates.
Any element of least weight in a coset of a linear code is called a z,eaJ.er-.
coset
The wei ght of a coset is defi ned to be the wei ght of its 1eader.
The
minimum distance from each element of a coset to a nearest codeword is the weight of that coset.
Thus the covering radius of a linear code is the weight of the
coset of greatest weight.
(This result extends to the nonlinear case).
The covering radius is a measure of the quality of the code, in that a code with covering radius at least as big as the minimum distance can be improved, as we shall explain later.
The covering radius is the maximum weight of errors
correctable by the code and is the maximum distortion encountered in the use of the code for data-compression.
And some interesting nonlinear codes, such as the
kerdock code, are the unions of linear codes with cosets of maximum weight. After Golay's discovery of essentially all of the perfect codes in 1949 l 41. there was no work done on covering radius until 1960, when Gorestein, Peterson, and Zierler [61 easily determined that the double-error-correcting BCH code is quasi-perfect (meaning that rr + 1 = pl.
The single-error-correcting BCH code
is the perfect Hammi ng code (d i scovered by Go 1ay c f.
[4), [5), [7]).
So the
next step in this sequence was to find the answer for the case where d = 5 (rr = 2). As we said, the answer is was more di ffi cu It.
p
ly.
The triple-error-correcting BCH code (d=7. rr=3)
Its coveri ng radi us was found to be 5 in three papers [ 10) ,
[OJ. [8) dated 1976 to 1978. dancy) 3m.
= 3.
This code has length 2m_l and codimension (redun-
The papers dealt with the cases m=4a, m odd, and m=4a + 2. respective-
Each of them directly or indirectly involved considerable effort, the last
even resting on A.
Weil's proof of the Riemann hypothesis for function fields
with finite fields of constants. Thus the difficulty of finding the covering radius of the BCH codes appears to grow rapidly with
TI.
In [9) Helleseth proved a conjecture from [6). that no BCH code with rr > 2 is quasi-perfect. The first-order Reed-Muller code is an extended BCH code.
This is the only
other BCH code of which we have even partial knowledge of the covering radius, apart from the general bounds to be mentioned later.
It is therefore natural to
A brief survey of covering radius
turn now to the Reed-Muller codes.
619
These are extensions of cyclic codes; the
m
Reed-Muller code of length 2 and order r, RM(r,m), can be defined as the set of all polynomials in m binary variables of degree at most r. m m m-r 1+(1)+···+(r),andd=2.
The code has dimension
It is not hard to see that when m is even, the covering radius of the firstorder Reed-t·1uller code RM(l ,m) is 2m- l - 2(m-2)/2 (see [13, Ch. 14]). It is known, from the supercode lemma (q.v.) and from the weight-distributions calculated by Kasami [12] or Sloane and Berlekamp [21] or McEliece [15], that when m is odd, cr(Rt4(1,m)) ;;. 2m- l - 2(m-l)/2. But the determination of the covering radius here has resisted many efforts. 1
bound is cr(Rt·1(1 ,m)).
We and Wolfmann [23] conjecture that this