Translation Generalized Quadrangles (Series in Pure Mathematics)

Chapter 3. Elation and Translation Generalized Quadrangles

52

PG(2n —l,q)= n). If for some i and at least two distinct j and f we have %j = %j> with %j regular, then S = T 2 ( O ' ) / 0 r some oval G' o/PG(2,s) with s = qn. Conversely, if S s T2(0') for some oval O' o / P G ( 2 , s ) , then %j = %ji for all i, j , j ' , with i ^ j ^ j ' ^i and all %j are regular. Proof. If S s T 2 ( C ) for some oval O' o f J P G ^ s ) , then by §3.6 and Corollary 3.5.3, 0 ( n , n , g ) is regular and so %j = %j> for all i,j,f, with i ^ j ^ j ' ^ i, and all %j are regular. Conversely, suppose that for some i and at least two distinct j and j ' we have Tij = %j< with %j regular. First of all, as %j is regular, there are precisely n lines over GF(g") which intersect the extensions of the spread elements of %j in precisely one point. Fix such a line L*. Let M* be the unique line over GY{qn) through the point of L* in the extension of £tjj, — {TTJ,TTJ') n Ti to GF(g n ), which intersects the extensions of TTJ and •Kji. Let T* be the Desarguesian projective plane over G¥(qn) which is generated by L* and M*. Now let itk be an element of O, with i ^ k =£ j and k ^ j ' . Then ^k and ^j>k are elements of %j = %y, distinct from 7TJ, r\ and ^ y . Let U* be the line of T* which is incident with the point of L* in the extension of ^jk to GF(qn), and with the point of M* in the extension of TTJ to GF(g n ). Let V* be the line of T* which is incident with the point of L* in the extension of ^k to GF(qn), and with the point of M* in the extension of 7iy to GF(g n ). Then over GF(g n ), U* n V* is a point of the extension of (ityKk) D (7iy, 7Tfc) — ^k to GF(g"). We have shown that over GF(qn), each element of O meets P \ Hence over GF(g n ), there are n planes which all meet each element of O, implying that O is regular. So by Section 3.6 we have <S £* T2(£>') for some oval of PG(2, qn). •

3.10

Eggs

In this section we prove fundamental results on eggs. Also, some interesting and useful characterizations of regular eggs will be given. Theorem 3.10.1 Let 0(n, m, q) he an egg,

son^m.

(i) Each point o / P G ( 2 n + m - 1, q) which is not contained in an element of 0(n,m,q) belongs to either 0 or qm~n + l tangent spaces of the egg. Ifm = 2n, then each point of PG(4n - 1, q) not contained in an element of 0(n, In, q) belongs to exactly qn + 1 tangent spaces of the egg. If m ^ In, then there are points contained in no tangent space of 0(n,m,q).

3.10. Eggs

53

(ii) For q even we necessarily have m = 2n. Hence for any 0(n, m, q) with q even, we have m e {n, 2n}, that is, for any TGQ S of order (s, t), with s and/or t even, we have t e {s, s2}. Proof, (i) Let 0(n,m,q) be an egg, so n / m. Take a point x not in an element of 0{n, TO, q). Firstly we will show that the number of tangent spaces containing x is either 0 or at least qm~n + l. The number of hyperplanes on x containing just one element of the egg is denoted by a; the number of hyperplanes on x containing qm~n + l elements of the egg is denoted by /?. Now we count in two ways the number of pairs (7, 7r), with 7 an hyperplane on x and -IT an element of 0(n, m, q) contained in 7. We obtain „ m + n - l _ -I

a + p(qm-n

+ l) = (qm + l)l

—

.

(3.1)

Let p be the number of tangent spaces of 0(n, m, q) through x. Now we count in two ways the number of pairs (7', it), with 7' a hyperplane on x containing just one element of the egg and n the element of the egg in 7'. We obtain an - 1

an~l - 1

So a = pqn-1 + (qm + l)qn

_ ^ .

(3.2)

From (3.1) and (3.2) it follows that pqn~l + p(qm-n

+ 1) = ^

^

>-.

(3.3)

As ma = n(a + 1), with a G N0 and a odd, we have that qm~n + 1 divides qm - 1, and so, by (3.3), qm~n + 1 divides p. It follows that either p is zero or at least qm~n + 1. Next, we will prove that p e {0, qm~n + 1}. Let Zi be any point of PG(2n + m — l,q) not in an element of the egg, and let pi be the number of tangent spaces containing zt. We count in two ways the number of pairs (zt, T), with r a tangent space containing zt. We obtain

J2Pi = (qm + i)qn(qm -1)/(« - 1 ) -

(3-4)

Chapter 3. Elation and Translation Generalized Quadrangles

54

Next, we count in two ways the number of ordered triples (zt, T, T' distinct tangent spaces containing z,. We obtain

T,T'),

YlPi(j>i-l) = (qm + l)qm(qm-l)/(q-l).

with (3.5)

i

From (3.4) and (3.5) it follows that £ > ( P i - Of 1 - " + 1)) = 0. i

As pi is either zero or at least qm-n + l, we necessarily have pt e {0, qm~n + 1}. That proves the first part of (i). Now assume that m = 2n. Counting the number of points contained in qn + 1 tangent spaces, we obtain (q 2 " + l ) ( g 3 " - g " ) ___q4n-l ( g - l ) ( g " + l) g-1

2n W

J

qn - 1 ?- l '

and so each point not contained in an element of 0(n, 2n, q), is contained in exactly qn + 1 tangent spaces. This proves the second part of (i). Finally, assume m ^ 2n. Counting the number of points contained in at least one tangent space, we see that there are points contained in no tangent space. (ii) Let q be even, and assume by way of contradiction that m ^ 2n. By (i) there is a point z of PG(2n + m — l,q) contained in no tangent space of 0(n,m, q). Let 0(n,m,q) = {iro,iri,... ,7rg™}. Then the space (z,iTi) intersects some itj\ also (Z,TTJ) intersects 7TJ. Hence we obtain a pairing in 0(n, m, q), implying that qm + 1 is even, a contradiction. We conclude that m = 2n. • Remark 3.10.2 Theorem 3.10.1 is due to Payne and Thas [128], but here we present a simpler proof. Also Theorem 3.10.3 below is due to Payne and Thas [128]. Theorem 3.10.3 Every three distinct tangent spaces of 0{n,m,q), have as intersection a space of dimension m — n — 1.

m ^ n,

Proof. By Theorem 3.10.1 the number of points contained in exactly qm~n + 1 tangent spaces is equal to (qm + l)qn(qm - 1) (qm~n + l)(q - 1) •

3.10. Eggs

55

Now we count in two ways the number of ordered 4-tuples (z,Ti,Tj,Tk), with Ti,Tj,Tk distinct tangent spaces of 0(n, m, q) and u e nllTj D Tk. If Ujk = |TJ n Tj n Tfc|, with i, j , k distinct, we obtain Ujk=

(q^

+ l)(q-l)

-{q

+l)q

^

-1)-

i,j,k

As Ujk > q q_il, and the number of ordered triples (n^^Tk), with i,j, k distinct, is equal to (qm + l)qm(qm -1), we necessarily have tijk = (qm~n l)/(g - 1) for all distinct i,j, k. • By Theorem 3.10.3 the tangent spaces of G(n, m, q), with m ^ n, axe the elements of an egg O* (n, m, q) in the dual space of PG(2n + m — l,q). Definition 3.10.4 The tangent spaces of an egg 0(n,m,q) in PG(2n + m — 1, q) form an egg O* (n, m, q) in the dual space of PG(2n + m — 1, q). So in addition to the TGQ T(n,m,q), a TGQ T*(n,m,q) arises. The egg O* (n, m, q) = O* will he called the translation dual of 0(n, m, q) = O, and T*(n, m, q) = T*(0) = T(C*) will be called the translation dual of the GQ T(n,m,q) = T(O). Remark 3.10.5 The translation dual S* of the TGQ S is (up to isomorphism) uniquely defined by S, that is, is independent of the chosen subfield of the kernel to construct 0(n, m, q). For regular eggs O, we clearly have O ~ O*, respectively T(O) = T*(C). For q odd there are examples with O ¥ O*, respectively T(O) ^ T*(C); see Chapter 10. For q even we have O = O* for all known examples (all known examples are regular). The following result is taken from Thas and K. Thas [191]. Theorem 3.10.6 The TGQ S of order (s,t), with s ^ t, and its translation dual S* have isomorphic kerneb. Proof. Let S = T(O), with O = 0{n, m, q) an egg in P G ( 2 n + m - 1, q), and assume that GF(g) is the kernel of S. Then O* = 0*(n,m,q) is also an egg in a PG(2n + m - 1, q), and so, by Theorem 3.5.1, GF(q) is a subfield of the kernel of S*. Interchanging roles of S and S* yields the result. •

Chapter 3. Elation and Translation Generalized Quadrangles

56

To end this section on eggs we give some interesting and elegant characterizations of regular and classical eggs. Characterization Theorem 3.10.7 is contained in Payne and Thas [128], but as the proof is quite short we repeat it here. Theorem 3.10.7 The egg 0(n, 2n, q) is regular if and only if one of the following holds. (i) For each point z not contained in an element of 0(n, 2n, q), the qn + 1 tangent spaces containing z have exactly {qn — l)/(q - 1) points in common. (ii) Each P G ( 3 n - 1, q) containing at least three elements of 0(n, 2n, q), contains exactly qn + 1 elements of 0{n, 2n, q). Proof, (i) If 0(n,2n,q) is regular, then it is clear that for each point z not contained in an element of 0(n, 2n, q), the qn + 1 tangent spaces containing z have exactly (qn — l)/(q - 1) points in common. Conversely, assume that for any point z not contained in an element of 0(n, 2n, q), the qn + l tangent spaces containing z have exactly (qn — l)/(q — 1) points in common. This means that for any triad {(oo), x, y} of the corresponding T(n, 2n, q) we have |{(oo), x, y}-1-1] = qn + 1. Hence the point (oo) of the GQ T(n, 2n, q) is 3-regular. So by Theorem 2.7.l(i) it follows that T(n,2n,q) ^ T3(0') for some ovoid O' of PG(3,g"). Consequently, by §3.6 and Corollary 3.5.3, 0(n, 2n, q) is regular. (ii) If 0(n,2n,q) is regular, then clearly each P G ( 3 n - l,q) containing at least three elements of 0(n, 2n, q), contains exactly qn + 1 elements of

0{n,2n,q). Conversely, assume that any P G ( 3 n — l,q) containing at least three elements of 0(n, 2n, q), contains exactly qn + 1 elements of 0(n, 2n, q). This means that the translation dual 0*(n, 2n, q) of 0(n, 2n, q) satisfies the condition in (i). So the GQ T*(n, 2n, q) is isomorphic to a T 3 (G') of Tits, with O' some ovoid in PG(3,g"). It follows that T(n,2n,q) is isomorphic to T 3 ( C * ) , with O'* the translation dual of O'. Clearly O'* is also an ovoid of some P G ( 3 , q n ) ; in fact O' and O'* are isomorphic; see Hirschfeld [72]. Consequently, by §3.6 and Corollary 3.5.3, 0(n,2n,q) is regular. We can also argue as follows. As T*(n, 2n, q) = T 3 ( C ) , the kernel has size qn. So by Theorem 3.10.6 the kernel of T(n, 2n, q) has size qn, that is, T(n, 2n, q) is isomorphic to a T3(0") of Tits, with O" an ovoid of some P G ( 3 , g n ) . Hence 0(n, 2n, q) is regular. •

3.11. The Stabilizer of the Base-Point of a Translation Generalized Quadrangle

57

3.11 The Stabilizer of the Base-Point of a Translation Generalized Quadrangle From each translation generalized quadrangle of suitable order S one can construct another translation generalized quadrangle S* with the same parameters, namely the translation dual of S. Essential for the classification of translation generalized quadrangles is the isomorphism problem between S and S*. Only two (infinite) classes of translation generalized quadrangles <S are known for which indeed S = S*, namely the Td(0) of Tits, where d e {2,3} and the characteristic is not two if d = 2, and the dual Kantor-Knuth generalized quadrangles, see Theorems 4.7.3 and 5.1.13. Using Theorem 3.7.2, we will start to investigate this problem by completely explaining the connection between the full automorphism groups of <S and S*, see Theorem 3.11.1. In fact, the only classes of translation generalized quadrangles where the sizes of these groups are different are precisely the translation generalized quadrangles which are the point-line dual of a flock generalized quadrangle S(F), F not a Kantor-Knuth semifield flock, a result which will be described in Chapter 10. (This observation was first done in work of K. Thas — see [201], the extensive paper [202], and [205], or [206] as a general reference, where it was first shown that these examples indeed have different automorphism groups than their translation dual.) We now have the following result which is an important step towards the determination of all translation generalized quadrangles. Theorem 3.11.1 Suppose S = T(0) is a TGQ of order (qn, qm) with basepoint (oo), where q is odd if n = m, and suppose that G(oo) is the stabilizer of (oo) in the automorphism group G of S. Furthermore, suppose (oo)' is the base-point of T(0*) = S*, and let G',y be the stabilizer of (oo)' in the automorphism group G' of S*. Suppose u and v are arbitrary points of T(0) and T(0*), not collinear with (oo) and (oo)', respectively. Then [G(oo)]u = [G'ioo),]v We also have that |G(oo)l =

\G[oo)'\-

Proof. Suppose O = {-K0,TTI, ... ,nqm} is contained in II = PG(2n + m - l,q), and embed IT in the (2n + m)-space IT = P G ( 2 n + m,q). Let T0, T\, ..., rqm be the tangent spaces at respectively w0, TTI, ..., wqm. Now consider the dual space LT'* of LV; then the set of tangent spaces interpreted in LT'*, say U* = {TQ, T-J*, . . . , r*m), forms a set of qm + 1 spaces of dimension n which satisfy the following properties:

Chapter 3. Elation and Translation Generalized Quadrangles

58

(1) they intersect two by two in the same fixed point z (which corresponds to IT); (2) for distinct i,j and k, T*T*T£ has dimension 3n; (3) for each i, there is an (n + m) -space n* which contains r*, and so that TT*nT* =

{z}Hi^j.

Now consider an arbitrary point x e II' \ II. Then the corresponding hyperplane n^ in the dual space II'* of II' is so that n^ n U* is an egg O* in n x , which is clearly isomorphic to the dual egg of O. Now consider an arbitrary coUineation 9 of T(O) which fixes (oo) and the point x. Then by Theorem 3.7.2, 9 induces a coUineation of n' = PG(2n + m, q) which fixes O and x. Now interpret 9 naturally as a coUineation of II'*, that is, if r\ is an r-dimensional space in II'* and if rf is the corresponding (2n + m - r — 1)dimensional space in the dual space n, then rf is the r-dimensional space of II'* which corresponds to rfe. Then 8 fixes z, Hx, and hence O*, and thus 9 induces a coUineation of T(C*) which fixes (oo)' and z, where (oo)' is the translation point of T(0*). It follows immediately that [G(oo)]u

=

[ G ^ ^ v ,

where u, respectively v, is an arbitrary point of T(O), respectively T(0*), not collinear with (oo), respectively (oo)'. For any TGQ S^ = (V,B,I) with translation point y, translation group H and full automorphism group G, it holds, by Lemma 3.7.1, that Gv = (Gy)zH = H{Gy)z, where z is arbitrary in V \ y^. As the translation groups of T(0) and T(0*) are isomorphic, the theorem now readily follows. • It may be useful for other purposes to define the following object U in PG(2n + m, q): U = { C 0 , ^ • • • iC™} is aset of qm + l spaces of dimension n which satisfy the following properties: (1) the elements of U intersect two by two in the same fixed point x; (2) for distinct i, j and k, ClCJCfc n a s dimension 3n; (3) for each i, there is an (n + m) -space -IT1 which contains £\ and so that 7r*nCJ' = {a;}ifi#j. We call U a generalized ovoid cone, respectively generalized oval cone, with vertex xifn^m, respectively n = m.

3.12. Structure of the Automorphism Group of a Translation Quadrangle

3.12

59

Structure of the Automorphism Group of a Translation Quadrangle

Let S = T(n, m, q) = T(O) be a TGQ of order (
Chapter 4

Generalized Quadrangles and Flocks Dembowski [45] used the word flock to refer to the partition of all but two points of an inversive plane into disjoint circles. The concept was naturally extended to other circle geometries, and significant applications have been discovered in each of its settings. For surveys of the existence problems and their solutions and of the different applications we refer to Fisher and Thas [58], Thas [172, 174, 178, 187], and the webpage of Cherowitzo ( h t t p : //www-math. cudenver. edu/~wcherowi).

4.1 Flocks Let O be an ovoid of PG(3, q). A partition of all but two points of O into q-l disjoint ovals is called a flock of O. The two remaining points x, y are called the carriers of the flock. If L is a line of PG(3, q) having no points in common with O, then the q-l planes through L which are not tangent to O intersect O in the elements of a flock T. Such a flock is called linear. Next, let H be a hyperbolic quadric of PG(3,g). A partition of H into q + 1 disjoint nonsingular conies is called a flock of H. If L is a line of PG(3, q) having no points in common with H, then the q+1 planes through L intersect H in the elements of a flock T. Such a flock is called linear. Finally, let O be either an oval in PG(2,q) for any q or a hyperoval in PG(2,g) for q even, and let K. be the cone with vertex a point x of PG(3, q) D PG(2, q), with x £ PG(2, q), and base O. A partition of K\{x} into q disjoint ovals or hyperovals in the respective cases is called a flock of K. If L is a line of PG(3, q) having no points in common with K, then the 61

62

Chapter 4. Generalized Quadrangles and Flocks

q planes through L but not through x intersect /C in the elements of a flock T. Such a flock is called linear.

4.2 Flocks and Translation Planes We next present a construction, discovered independently by Walker [237] and Thas [167], of a class of translation planes. Let T = {Ci, C 2 , . . . , Ca} be a flock of a quadric Q of P G ( 3 , q), with a = q — l,q+l,ox q according as Q is elliptic, hyperbolic or a cone. We now embed Q in the nonsingular hyperbolic (or Klein) quadric TL of P G ( 5 , q). Denote the plane of d by 7TJ, the polar plane of -Ki with respect to TL by TT\, and the conic IT* n TL by C*, with i = 1,2,..., a. Note that since, for i ^ j , -Ki n TTJ is a line exterior to TL, IT* U IT* lies in a 3-space that intersects TL in an elliptic quadric; in particular, no two points of C* U C* are on a common line of TL. (a) When Q is elliptic and T has carriers x and y, the set Cj" U C£ U • • • U C*_j U {x, y} = O has size q2 + 1 and no two points of this set are on a common line of TL; (b) When Q is hyperbolic, the set C* U C\ U • • • U C*+1 = O has size q2 + 1 and no two points of this set are on a common line of TL; (c) When Q is a quadratic cone with vertex x, the set C\ UQ U• • • UC* = 0 contains x, has size q2 + 1 and no two points of this set are on a common line of TL. A subset O of size q 2 +1 of TL such that no two of its points are on a common line of TL is called an ovoid of TL; see e.g. Thas [178]. Equivalently, an ovoid of TL is a subset of "H intersecting each plane of TL in exactly one point. Let P be the Klein mapping from the points of TL to the lines of P G ( 3 , q); see e.g. Chapter 5 of Hirschfeld [72]. Then in each of the three cases O13 = T is a set of q2 + 1 mutually skew lines of PG(3,
4.3. Flocks of Ovoids and Hyperbolic Quadrics

63

4.3 Flocks of Ovoids and Hyperbolic Quadrics In the context of generalized quadrangles flocks of quadratic cones are much more important than flocks of ovoids or hyperbolic quadrics; it will appear that flocks of cones play a key role. However, we emphasize that e.g. the fundamental theorem on flocks of ovoids appeared to be very useful in the proof of strong characterizations of certain classes of GQs, see e.g. Theorem 5.3.1 in Payne and Thas [128]. The following theorem was proved in the odd case by Orr [105], in the even case by Thas [159]. A short proof of Orr's result can be found in Fisher and Thas [58]. Theorem 4.3.1 Any flock of an ovoid O of P G (3, q) is linear.

•

The next theorem is due to Thas [166]. Theorem 4.3.2 For q even any flock of a hyperbolic quadric of P G ( 3 , q) is linear. •

From now on assume that q is odd. Let Q be a hyperbolic quadric of PG(3,g), with q odd. In the set-of all nonsingular conies of Q we define the following equivalence relation, see Thas [166, 169]: two conies C\ and C2 are equivalent if and only if there is a nonsingular conic C on Q which is tangent to both C\ and C2- There are two equivalence classes, denoted by I and II. Let L be a line having no point in common with Q, and let L' be the polar line of L with respect to Q. The set of all conies of Class I, respectively II, whose plane contains L is denoted by V, respectively V. For q = 1 mod 4 the set of all conies of Class II, respectively I, whose plane contains L' is denoted by W, respectively W; for q = — 1 mod 4 the set of all conies of Class I, respectively II, containing L' is denoted by W, respectively W. Then it was shown by Thas [166] that V U W and V U W are nonlinear flocks. By most authors these flocks are called Thas flocks. Further, Bader [1] showed that for q = 11,23,59 the hyperbolic quadric Q of P G ( 3 , q) has a flock which is neither linear nor a Thas flock. These flocks were independently discovered by Johnson [77], and for q = 11,23 also by Baker and Ebert [7]. Since these flocks were derived by Bader from exceptional nearfield planes, exploiting the relationship between flocks and translation planes mentioned in Section 4.2, Bader calls these flocks the exceptional flocks.

Chapter 4. Generalized Quadrangles and Flocks

64

The following theorem classifying allflocksof the hyperbolic quadric Q of PG(3,g), with q odd, is a combination of results by Bader and Lunardon [2] andThas [173]. Theorem 4.3.3 Any flock of the nonsingular hyperbolic quadric Q of PG(3, q), with q odd, either is a linear flock, a Thas flock, or an exceptional flock. U

4.4

Flocks of Cones

Let O* be a hyperoval of PG(2, q), with q = 2h, and let K* be the cone of PG(3,g) D PG(2,g) which projects O* from a point x of PG(3, ?)\PG(2, q). Further, assume that T* = {CI, C*,..., C*} is a flock of /C*. Now, let y e O*, O = 0*\{y}, and let K be the cone which projects the oval O from x. Then it is clear that the planes TTI, 1T2, • • •,7rg, with C* C -Ki, define a flock J" = {C\, C2, • • •, Cq} with Ci c C*, of /C. In the next theorem we prove the converse. Theorem 4.4.1 Let O be an oval of PG(2, q), with q = 2h, and let K. be the cone which projects O from a point x of PG(3, q) \ PG(2, q). Further, assume that T = {C\, C 2 ,..., Cq) is a flock ofK. Let n be the nucleus (kernel) of O, and let O* be the hyperoval O U {n}. If K* is the cone which projects O* from x, then the planes m, 7r 2 ,..., nq, with d C in, define a flock T* = {Ci,(%,..., C*q}, with d c CI ofK.*. Proof. Suppose that z e C* n C*, with i =fi j . Since O* is a hyperoval, the line -Ki n TCJ has exactly two distinct points y, z in common with K*. Clearly {y, z) = C* n C*. Hence \d n Cj-| > 1, a contradiction. So C* n C* = 0, for all i ^ j , that is, {C{, C|,..., C*} is a flock of /C*. • Let ^ = {Ci,C 2 ,... , C J be a flock of the quadratic cone /C in PG(3,g). Let x be the vertex of K and let 7Ti be the plane of d, with i = l,2,...,q. Dualizing in PG(3, q), the point x becomes a plane PG(2, q), the q+1 lines on AC become the q + 1 elements of a nonsingular dual conic T in PG(2,9), and the plane m becomes a point Xi not in PG(2, q), with = 1,2,..., q. Then C» n Cj = 0, with i 7^ j , is equivalent with X^XJ n PG(2, q) not lying on an element of T. We say that T = {xi,x2,...,xq} is a/iiocfc of the dual conic T; T is called linear if and only if T is linear. If q is odd, then let 2/1,2/2 j • • •, yg+i be the tangent points of the elements of T, that is, the points

4.4. Flocks of Cones

65

of PG(2, q) lying on exactly one element of T. Then {y1,y2,..., yq+i} = C is a nonsingular conic in PG(2, q). In such a case we also say that ~T is a flock of the conic C. In the remaining part of this section we will only consider flocks of quadratic cones. We shall use the following notation: the set of all elements m of GF(g), q even, for which the polynomial X2 + X + m is reducible, respectively irreducible, over GF(q) is denoted by Co, respectively C\. Theorem 4.4.2 With each flock T of the quadratic cone K, o/PG(3,g) corresponds a set of q ordered triples (a,, bj, Cj), ai: bi,Ci G GF(q), such that for q odd (c, - Cj)2 — 4(a; — a,j)(bi - bj) is a nonsquare whenever i ^ j , and for q even Ci ^ Cj and (a» + a,)(6* + bj)(a + Cj)~2 G C\ whenever i ^ j . Conversely, with each such set of q ordered triples corresponds a flock T of the quadratic cone /C. Proof. Let X0Xi = X\ be the equation of the cone K. Now we consider q planes 7r,, with i = l,2,...,q, which do not contain the vertex x(0,0,0,1) of /C. Such a plane -K% has equation aiX0 + foXi + CiX2 + X3 = 0. The plane -K^ projecting 7TJ n TTJ, with i ^ j , from the vertex x has equation (a,i - aj)Xo + (bi - bj)X\ + (CJ - Cj)X2 = 0. Let q be odd. Then the plane mj has no line in common with K. if and only if (ci - Cj)2 - 4(at - aj)(bi - bj) is a nonsquare. Hence the conies 7TJ n K. forma flock of K if and only if (c, — Cj)2 — 4(aj — aj)(6j - 6 j ) is a nonsquare whenever i 7^ j . Let g be even. Then the plane 7r„ has no line in common with K, if and only if the homogeneous quadratic polynomial X2(al + a,j)2 + XY(ci + Cj)2 + Y2(pi + bj)2 is irreducible, that is, if and only if a ^ CJ and (m + aj)2(bi + bj)2(ci + Cj)~4 e C\, if and only if CJ ^ Cj and (aj + a,j){bi + bj)(ci + Cj)"2 e C\. Hence the conies 7r, n K form a flock if and only if Cj 7^ 9 and (a, + aj)(6j + 6J)(CJ + Cj)~2 € Ci whenever i ^ j . • Remark 4.4.3 Clearly coordinates can be chosen in such a way that m has equation X 3 = 0, that is, (ai, &i, ci) = (0,0,0). Now we will classifiy all flocks T = {C1X2, • • • ,Cq} for which the planes 7Ti, 7T2,..., 7rg of the respective conies C\, C 2 ,. • •, Cq all contain a common point.

66

Chapter 4. Generalized Quadrangles and Flocks

Theorem 4.4.4 Let T = {C\,C2, • • • ,Cq) he a flock of the quadratic cone K, of P G ( 3 , q), with q even. If the planes ni, 7r 2 ,..., 7rg of the respective conies C\, C2,..., Cq all contain a common point, then the flock T is linear. Proof. For q = 2 the statement is trivial, so assume that q > 2. Let the cone AC be embedded in the Klein quadric H of P G ( 5 , q). If n* is the polar plane of Wi with respect to Ji and if -K* n H = C*, with i = 1,2,..., q, then by Section 4.2 C\ U C2 U • • • U C* = O is an ovoid of U. The planes -K\ , ir2,.. •, 7r* contain a common tangent line L of H, where L is the polar line of P G ( 3 , q) with respect to 7i. Since the planes m, 7r 2 ,..., nq contain a common point y, the planes TT^ , 7r 2 ,..., ir* are contained in the polar space PG(4, q) of y with respect to H. Also, y is the nucleus of the quadric Q = PG(4,g) n H. Notice that each line of Q has exactly one point in common with O. Now consider distinct points z\,z2 of O, and let C* be the intersection of Q and the plane yz\z2. Let C* n O = V, with \V\ = a. Now we count in different ways the number of ordered pairs (u, z), with u e 0\V, z £ C*\V, and uz a line of Q. If u e 0\V, then the tangent 3-space of Q at u contains exactly one of the lines joining y to a point of C*. Hence we obtain q2 + 1 - a = (q + 1 - a)(q + 1), and so a = 2. Consequently the plane yz\z2 only contains the points zi,z2 of O. Now we project O from y onto a P G ( 3 , q) C PG(4, q) not containing y. By the foregoing no three points of this projection O of O are collinear, and so (since q > 2) O is an ovoid of PG(3,g); see also Section 1.5. The projections of C\, C 2 , . . . , C* are conies C\, C2, • •., Cq having a common tangent line at their common point. By a theorem of Glynn [63], a short proof of which can be found in Storme and Thas [148], O is an elliptic quadric. Let C be a conic on O containing the common point of C i , C 2 , . . . ,Cq, but distinct from any of these conies. The conic C is the projection from y of a conic C* on O through the common point of C*, C2,..., C*, which is the vertex x of K. As \C* n C*\ = {x, Ui}, with x ^ ut, the conic C* belongs to the 3-space (C*,L), with i = 1,2,..., q. So O is contained in (C*,L). Hence O is an elliptic quadric, and so the planes wi, w2,..., 7r9 have a common line. We conclude that the flock is linear. •

In the next two theorems we will handle the odd case. Theorem 4.4.5 Let T = {C\, C 2 ,... ,Cq} he a flock of the quadratic cone K, of P G ( 3 , q), with q odd. If the planes m, ir2,..., nq of the respective conies Ci, C 2 ,... ,Cq all contain a common interior point x of K,, then T is linear.

4.4. Flocks of Cones

67

Proof. Let X0Xi = X\ be the equation of the cone K, and let a,iX0 + biXi + CiX2 + X3 = 0, with i = 1,2 . . . , q, be the equation of the plane 7TJ. Since the automorphism group of P G ( 3 , q) leaving K, invariant is transitive on the interior points of /C, we may assume that x = (1, - m , 0,0), with m a nonsquare of GF(g). As x is contained in -Ki, we have a» - hm = 0, with i = l,2,...,q. Let GF(gr2) be the field obtained by adjoining a root u> of X2 = m to GF(q). If a £ GF(q2) with a = a + bu, and a, b £ GF(g), then let a = a - bco. With the q planes iti we now let correspond the q elements Cj + 2biU> = a, of GF(qr2). Assume that at - OLJ, with i 7^ j , is a square in GF(q 2 ), so let ai — aj = (r + SUJ)2 with r,s £ GF(q). Then we have (ci - Cj)2 - 4(«i - aj)(bi - bf) = (c, - c^-)2 - 4(6, - 6j) 2 m = (^ - Cj)2 - 4(6, - 6j)2w2 = (^ - Cj + 2(bi - bj)uj)(ci - Cj - 2(bt 2

= (^ - aj)(cii - ctj) = (r + su) (r - su)

2

2

= (r -

2

-b3)u)

2

s m) .

Hence ( Q - Cj)2 — 4(a, - aj)(bi ~ bj) is a square in GF(g), contradicting Theorem 4.4.2. Consequently a, - a>j is a nonsquare in GF(g 2 ) whenever Let v be a given nonsquare of GF(^ 2 ). Then the set V consisting of the q elements v(pn — ct\) contains 0 and has the property that 5 - 7 is a square whenever S,j £ V. Now by a theorem of Blokhuis [15] we have V = {da || d £ GF(q)} with a some nonzero square in GF(g 2 ). Hence v{pn a±)/a £ GF(q),iori = l,2,...,q. Ifu/a = u+u'ui, with u, u' £ GF(q) and 1 (u,u ) ^ (0,0), thenu'ci + 2biU — u'ci + 2b1u,i = l,2,...,q. Consequently the q planes 7r; all contain the point (0, 2u, u', —u'ci - 2b\u). So the q planes •Ki all contain a common line, that is, the flock F is linear. • Theorem 4.4.6 Let T = \C\,Ci,... ,Cq}, with q odd, he a flock of the quadratic cone K and suppose that the planes 7Ti, 7r 2 ,..., -Tr,, with d C 7rw contain a common exterior point of K. If m is a given nonsquare of GF(q), then coordinates can be chosen in such a way that K has equation X 0 Xi = X2 and the planes 7r; have equation atX0 - mafXi + X3 = 0, with {a\, a2,..., aq} = GF(q) and a an automorphism of GF(q). Conversely, given a nonsquare m ofGF(q), the planes -Ki with equation a%XQ - mafXi + X3 = 0, with {ai, a 2 , . . . , aq} = GF(q) and a any automorphism ofGF(q), define a flock T of the cone K, with equation XQXi = X\. Also, the planes 71-j all contain the exterior point (0,0,1,0) ofK. Finally, the flock is linear if and only if a = 1.

68

Chapter 4. Generalized Quadrangles and Flocks

Proof. Consider the cone K, with equation X0Xi — X2 and the q planes 7Tj with equation aiX0 -ma^Xy + X3 = 0, with {ai, a 2 , . . . , aq} = GF(q),m a given nonsquare of GF(q), and a any automorphism of GF(q). All these planes contain the exterior point (0,0,1,0) of /C. Since - 4 ( a j aj){—ma% + ma") = 4m(ai - o,j)a+1 is a nonsquare whenever i ^ j , the planes 7ri, n2,..., nq define a flock of K. by Theorem 4.4.2. It is easy to show that the planes iri, ir2, • • •, -Kq have a line in common if and only if a = 1. This proves the second part of the theorem. Conversely, consider any flock T — {Ci,C2,... ,Cq} of a quadratic cone K. where the planes -K\, IT2, •.., irq, with d c TT*, contain a common exterior point x of K. Coordinates can be chosen in such a way that x = (0,0,1,0), that 7ri is the plane X3 = 0, and that K has equation X0Xi = X\. Then the plane Tn has an equation of the form a,iX0 + biX\ + X 3 = 0, with ai = b\ = 0. Assume, by way of contradiction, that a* = a,j for i ^ j . Then 7TJ n TTJ lies in the plane X\ — 0. Since Xi = 0 and K. have a line in common, the line 7TJ n -KJ and /C are not disjoint, a contradiction. Hence a* ^ a^ whenever i ^ j ; similarly 6, ^ bj whenever i ^ j . Consequently 6 : ai H-> 6j, with i = 1,2,..., q, is a permutation of GF(q), with 0 9 = 0. By Theorem 4.4.2 -(ai - a,j)(af - a|) is a nonsquare whenever i ^ j . Putting <2j = 0 and at = 1, we see that - l e is a nonsquare. Hence ((al/l6) — (a8j/le))/(ai - a,j) is a nonzero square whenever i ^ j . Let a : O J M (a^/161), with i = 1,2,..., q. Then a is a permutation of GF(g) for which 0CT = 0 and V7 = 1. Moreover (af - a?)/(en — a,j) is a nonzero square whenever i ^ j . Then by a theorem of Carlitz [36] a is an automorphism of the field GF(g). The plane Wi has equation a,iX0 + \ea%X\ + X3 = 0. Now let m be any given nonsquare of GF(q), and consider the following change of coordinates: x0 = x'0,xi = (—m/l6)x'1,X2 = sx'2,x3 = x'3 with —m/1 9 = s 2 . Then K, has still equation X0X\ = X% and ivi has equation aiX0 - ma°Xi + X3 = 0, with i = 1,2,..., q. Now the theorem is completely proved. •

Let T = {Ci,C2, • • • ,Cq} be a flock of the quadratic cone /C of P G ( 3 , q), q odd, and suppose that the planes iri,TT2, ..., 7rq, with d C 7Tj, contain a common exterior point of /C. By Section 4.2 the flock T defines an ovoid O of the Klein quadric H. Since all planes TTI, 7r2,..., -Kq contain a common point, the ovoid O is contained in a hyperplane PG(4, q) of the space PG(5, q) containing H. Hence O is an ovoid of the GQ Q(4, q) with point set PG(4, q)r\H. Consequently, if/3 is the Klein mapping, then Op = T is a spread of the GQ Q(4, q)13 = W(q). It is easy to check that the ovoids O cor-

4.5. Semifield Flocks

69

responding to the flocks described in Theorem 4.4.6 are ovoids described by Kantor [82], for which the translation planes defined by the corresponding spreads T are Knuth semifield planes. For this reason these flocks will be called Kantor-Knuth flocks. We conclude this section onflocksof cones with a short description of the process of derivation introduced by Bader, Lunardon and Thas [5]. Let T = {Ci,C2,...,Cq} be a flock of the quadratic cone K, with vertex x of PG(3, q), with q odd. The plane of d is denoted by iti, i — 1,2,..., q. Let K, be embedded in the nonsingular quadric Q of PG(4,g). Let the polar line of ni with respect to Q be denoted by Li and let L, n Q = {x, xt}, i = 1,2,..., q. If Hi is the tangent hyperplane of Q at xi; then put Hi flQ = K,i, Hi n Hj n Q = K,i n Hj = K,j n Hi = Cij = Cji and Cu = Ci, i,j = 1,2,...,q and ij^j. Then Bader, Lunardon and Thas [5] prove that Ti = {Cn,Ci2, • • •,Ciq} is a flock of Ki, with i = 1,2,... ,q. We say that the flocks T\,p2, • • • ,Fq are derived from the flock T. It is clear that T,T\,..., Ti-\, J^i+i ,...,Fq are derived from the flock Ti. There is an extensive literature on the subject, see e.g. Thas [178], Johnson and Payne [79], Payne [122], and the recent paper by Johnson [78]. Remark 4.4.7 Derivation for any q can be described algebraically; see Bader, Lunardon and Payne [3], Payne [121] and Section 2.9 of Cardinali and Payne [35]

4.5

Semifield Flocks

Casse, Thas and Wild [37] give the following construction of flocks. Let GF(g n ) be an extension of GF{q) and let PG(3n - 1, qn) be the corresponding extension of PG(3n - 1, q). Let is be a plane of PG(3n - 1, qn) with the property that -K and its n - 1 conjugates with respect to the extension GF(g") of GF(g), generate P G ( 3 n - l , qn). Then each point x of it, together with the rc-1 conjugates of x, define an (n-l)-dimensional subspace of PG(3n - 1, q). These (n - 1)-dimensional subspaces of PG(3n - 1, q) form an (n - l)-spread T of PG(3n - l,q). Now embed PG(3n - l,q) in PG(3n, q). Then the incidence structure with as point set PG(3n, q) \

70

Chapter 4. Generalized Quadrangles and Flocks

P G ( 3 n - l,q), with as lines the n-dimensional subspaces of PG(3n,g) which contain an element of T but are not contained in PG(3n - l,q), and with as incidence the natural one, is isomorphic to the design of points and lines of AG(3, qn). Now consider a line L of n. Then L and its n - 1 conjugates define a (2n - 1)-dimensional subspace of P G ( 3 n - 1, q). The set of these (2n - 1)-dimensional subspaces is denoted by T'. The planes of AG(3, qn) correspond to the 2n-dimensional subspaces of PG(3n, q) which contain an element of T ' but are not contained in P G ( 3 n — 1, q). See also Section 5.1 of Dembowski [45]; for an algebraic description of this representation of AG(3, qn) we refer to our Section 3.6. Next, let r be a nonsingular dual conic of IT and, for q odd, let C be the nonsingular conic consisting of the tangent points of V. With the qn + 1 lines of r correspond qn + 1 elements of T'\ this subset of T will be denoted by C 2n . For q odd, with the qn + 1 points of C correspond qn + 1 elements of T; this subset of T will be denoted by Cn. For q odd, the set Cn is a pseudo-conic 0(n, n, q) and C2„ is the set of all tangent spaces of 0(n,n,q). We remark that C2n is a pseudo-conic in the dual space of PG(3n - l,q). Let P G ( n - l,q) be an (n - 1)-dimensional subspace of P G ( 3 n - 1 , q) intersecting no element of C 2n . Further, let P G ( n , q) be an ndimensional subspace of PG(3n, q) which contains P G ( n - l,q) but which is not contained in PG(3n —l,q). If P G ( 3 , qn) is the projective completion of AG(3, qn), with n = P G ( 3 , g n )\AG(3, qn), then T defines a nonsingular dual conic r of If and, for q odd, C defines a nonsingular conic C of n. If T is the point set of A G (3, qn) which corresponds to P G ( n , q)\PG(n— l,q), then the line of P G ( 3 , qn) joining any two distinct points of T has no point in common with the union of the lines of T; for q odd this line intersects W in an interior point of the conic C. Hence J 7 is a flock of the dual conic r . This flock T is not linear if and only if P G ( n - 1, q) £ T. Assume that P G ( n - 1, q) is contained in an element -q of T'. Then the set T is contained in a plane. Dualizing, we obtain a flock T of the quadratic cone for which the planes of the conies of T all contain a common point. Hence, by Theorems 4.4.4 up to 4.4.6, either P G ( n - l,q) e T and T is linear, or q is odd and T is a Kantor-Knuth flock; if T is not linear, then r\ contains two distinct elements of Cn. Conversely, let q be odd and let T be a flock of the quadratic cone K, for which the planes 7TJ all contain a common point. Then, using the equations of the planes m given in the statement of Theorem 4.4.6, dualizing, and applying the mapping C of Section 3.6 from AG(3, q) onto AG(3n, q'), with q'n = q and GF(q') a subfield of the field of fixed elements of the

4.5. Semitield Flocks

71

automorphism a, one easily sees that we have a flock of the type described above with P G ( n - 1, q) in an element of T'. Let X0X\ — X$ be the equation of the quadratic cone K. in P G ( 3 , q), and consider a flock T = {C\,C2, • • • ,Cq) of K. Let the plane 7TJ of d have equation a,X 0 + hXi + CiX2 + X3 = 0, with i = l,2,...,q. Clearly we may put aj = t, 6j = zt, Ci = y t , with t e GF(g) and y0 = z0 = 0. As T is a flock, by Theorem 4.4.2, t <-^ zt defines a permutation of GF(q), and for q even also £ H-> yt is a permutation of GF(g). The construction of Casse, Thas and Wild yields a flock T if and only if the points (t, zt, yt) of AG(3, q) define an AG(n, q') for some subfield GF(g') of GF(g), that is, if and only if yt and zt are additive. On the other hand the translation plane defined by the flock T (see Section 4.2) is a semifield plane if and only if yt and zt are additive; see Kallaher [80]. For this reason these flocks are called semifield flocks. For more details we refer to Lundardon [100]. The following fundamental result is due to Johnson [76]. Theorem 4.5.1 A semifield flock of a quadratic cone in P G ( 3 , q) for q even is linear. • Remark on the Proof. The original proof of Theorem 4.5.1 relies on the connection between the flock and the semifield which defines the corresponding semifield plane. A geometrical proof will be given in Section 5.1 (Theorem 5.1.11).

Known Examples of Semifield Flocks With the notation introduced above, we will give for each flock the set of triples (t, zt, yt), with t e GF(g) and q odd. (a) The Kantor-Knuth flocks (t, -mta,

0),te

GF(q), q odd,

m a given nonsquare of GF(qr), a any automorphism of GF(q). All planes of the conies of the flock contain a common exterior point of the cone. The flock is linear if and only if a = 1. Derivation yields no new flock [122].

Chapter 4. Generalized Quadrangles and Flocks

72

(b) The Ganley flocks

{t,t3,~mt-m~1t9),te

GF{q),q = 3h,

m a given nonsquare of GF(q). This flock is linear if and only if q = 3; it is isomorphic to a KantorKnuth flock if and only if q e {3,9}. For q > 9 derivation yields one extra (non-semifield) flock [122]. (c) The Penttila-Williams-Bader-Lunardon-Pinneri flock (i,-i9,i27),iEGF(35). This flock is not of the previous types, and derivation yields one extra (non-semifield) flock [122]. This example was found by an interesting mix of theory and computer assistance. Penttila and Williams [131], with a slight assist of a computer, discovered a translation ovoid of the GQ Q(4,3 5 ); see also Chapter 7. In 1994 (see the proceedings of the Oberwolfach meeting on Designs and Codes in 1994) Thas [181] had shown how to interpret a translation ovoid of a GQ Q(4, q) as a semifield flock; a detailed and more complete treatment of this correspondence was given by Lunardon [100]. Using this technique Bader, Lunardon and Pinneri [4] succeeded to determine the exact form of this semifield flock over GF(3 5 ).

4.6 Generalized Quadrangles and Flocks In this section we will consider a particular class of GQs S(G, J). If u — {ui,u2),v = (vi,v2) e GF(q) x GF(q), then put uv — u\V\ + U2V2- Let G = {(a,c,/3) || a,/3 G GF(q) x GF(q), c e GF(g)}, with the group operation defined by {a,c,P)o(a,,c',(3')^{a

+ a',c + c,+(3a',(3 + (3l).

This operation makes G into a group of order q5. Put A(oo) = {(0,0, /?) e G || 0 e GF(q) x GF(q)}. For each t e GF(q), let A

*=

-n 0 zt

4.6. Generalized Quadrangles and Flocks

73

be an upper triangular 2 x 2-matrix over GF(q), with the convention that A0 be the zero matrix. For each t G GF{q), put Kt = At + Af, and let A(t) = {(a, aAtaT,

aKt) || a G GF(q) x GF{q)}.

It follows that A(t) is a commutative subgroup of G having order q2, for each t e GF(q) u {oo}. Let C = {(0, c, 0) G G || c G GF(g)}, and put A*(t) = >4(t)C, t G GF(g) U {oo}. Then A*(t) is a commutative subgroup of G having order q3, t G GF(g) U {oo}. Moreover A*(t) D A(i). Further, let J = {A{t) \\ t e GF(q) U {oo}} and J* = {A*(t) \\ t E GF{q) U {oo}}. With the foregoing notation we have the following two important theorems. Theorem 4.6.1 (Kantor [83]) If q is odd, then G, J, J* satisfy Conditions (Kl) and (K2) of Section 3.2.2, so that S(G,J) is a GQ of order (q2,q), if and only if -det(Kt

- Ku) = (yt - yu)2 - 4(xt - xu)(zt - zu)

is a nonsquare ofGF(q) whenever t,u£ GF(q), t ^ u.

•

Theorem 4.6.2 (Payne [117]) If q is even, then G, J, J* satisfy Conditions (Kl) and (K2) of Section 3.2.2, so that S(G,J) is a GQ of order (q2,q), if and only if Vt 7^ Vu and (xt + xu)(zt + zu)(yt + yu)~2 G d whenever t,u G GF(g), t ^ u.

•

A set of q upper triangular 2 x 2-matrices over GF(q) satisfying the conditions of Theorems 4.6.1 and 4.6.2 is called a q-clan. Comparing Theorems 4.6.1 and 4.6.2 with Theorem 4.4.2 yields the following fundamental result. Theorem 4.6.3 (Thas [172]) The group G and the sets J, J* satisfy Conditions (Kl) and (K2) of Section 3.2.2, that is, S(G, J) is a GQ of order (q2,q), if and only if the planes xtX0 + ztXx + ytX2 + X3 = 0, t G GF(q), define a flock T of the quadratic cone with equation X0Xi = Xf in GF(q). Each flock T of the quadratic cone of P G (3, q) gives us a GQ, and each GQ S(G, J) of the type described above gives us a flock of the quadratic cone; the GQ 5(G, J) is also denoted S(F) and is called a flock GQ. •

74

Chapter 4. Generalized Quadrangles and Flocks

Payne and Thas [128] show that if the flock T is linear then S(G, J) is isomorphic to the classical GQ H(3, q2); conversely, it follows from Payne [120, 121] that if S(F) ^ H(3, q2), then T is linear. In [129] Payne and Thas show that if the GQ S{T) is not classical, then the elation point (oo) of S (T) is fixed by the full automorphism group of «S(JF). Finally, as C is a normal subgroup of G it follows from 8.2.2(v) of Payne and Thas [128] that (oo) is a center of symmetry of S(T). Finally, by recoordinatization, every line of S (J7) incident with the elation point (oo) can be taken as line [.A(oo)]; see Payne and Rogers [125] and Payne [120, 121].

4.7 Semifield Flocks and Translation Generalized Quadrangles 4.7.1 Position of a Translation Line in a Flock Quadrangle Let L be a translation line of the flock GQ S(T) of order (q2,q), q odd, and suppose that L is not incident with (oo). Then L is incident with a point x g (oo)-1-. As x is a center of symmetry, there is a symmetry 0 with center x mapping (oo) onto (oo)' ^ (oo). So by Payne and Thas [129], S{T) = H(3, q2) and consequently T is linear. Hence when T is nonlinear, any translation line is incident with (oo). When T is linear, each line is a translation line and so also in that case a given translation line may be taken to be incident with (oo).

4.7.2 Additive g-Clans and Translation Generalized Quadrangles Suppose q is a prime power. A g-clan {At = (

V° ztJ

I || t e GF(q)} is

additive if for each t, t' e GF(g) the following property holds: At + Af

=At+v.

Let C = {At || t G GF(q)} be a q-clan, put Kt = At + Af, and define fft(A) = A.4tAT and X5t = \Kt for A e GF(q)2. Put G = {(a, c, p) || a, (3 G GF(q)2,c G GF(g)},

4.7. Semifield Flocks and Translation Generalized Quadrangles

75

and define a binary operation on G by (a, c, (3) o (c/, c', /3') = (a + a', c + c' + (3a'J', /3 + /3'). This makes G into a group. Define a family of subgroups J of G as follows ^(*) = {(a,gt(a),aSt)

\\ a G GF(q)2},

t G GF(g),

and A(oo) = { ( 0 , 0 , / ? ) | | / 3 e G F ( g ) 2 } . Define a family of subgroups J* of G by A*(t) = {(a,c,as*)

|| a £ GF(q) 2 ,c G GF(g)}, i £ GF(q),

and yl*(oc) = {(0,c,/3) || c G GF(<7),/3 G GF(g) 2 }. Then J ' is a 4-gonal family which defines a flock GQ of order (q2, q); see Section 4.6. The next theorem is due to Payne [118]. Theorem 4.7.1 A q-clan C = {At = (lVt)

II t G GF(q)} is additive if

and only if the point-line dual of the corresponding flock GQ is a TGQ with translation point corresponding to [A(oo)]. Proof. Let <S be the GQ of order (q2, q) defined by C, and let SD be its point-line dual. Suppose SD is a TGQ with translation point [.A(oo)] and translation group G. As automorphism group of S, G induces a regular permutation group on {[A(i)] || t G GF(g)}, which is an elementary abelian p-group, where q = ph for the prime p, of size q. Denote this group by G*. Let T be the flock of the quadratic cone K (with vertex v) in PG(3,g) defined by the Ats. Then G* can be interpreted as a subgroup of P r L z t ^ ) that acts regularly on the flock planes (cf. Addendum A (§4.12) to this chapter). In the dual space of P G ( 3 , q), the flock planes become q points lying in the affine 3-space obtained by deleting the plane corresponding to v, on which an elementary abelian p-subgroup of P r L 4 ( g ) acts regularly. So this point set can be considered as the point set of an affine space over some subfield of GF(q). Whence T is a semifield flock by Section 4.5, and therefore 6 :t ^ is additive.

At

Chapter 4. Generalized Quadrangles and Flocks

76

Conversely suppose 0 is additive. Then 5t+r = St + 8r. Now define, for r G GF(g), 4>r : (a,c,P) H-> (a,c + gr(a), f3 + aSr). It is routine to check that 4>r is a symmetry of the flock GQ about the point A*(oo) sending A(t) to A(t + r). So A*(oo) is a center of symmetry, and then each point of [A(oo)] is as such, in view of the fact that (oo) is an elation point and that (oo) is a center of symmetry (see Section 4.6). The theorem follows. • Theorem 4.7.2 The flock F defined by the q-clan C = {At = ( * Vt j || t e GF(q)} is a semifield flock if and only if the line A[(oo)] of the flock GQ S(F) is a translation line. Proof.

Follows from Theorem 4.7.1 and Section 4.5.

•

4.7.3 Known Cases With each semifield flock corresponds a TGQ, and hence an egg. We give an overview of some terminology on these eggs and their translation duals. KANTOR-KNUTH EGGS.

Let J 7 be a Kantor-Knuth flock. The corresponding

GQ S(T), which is a TGQ for some base-line, was first discovered by Kantor [83], and is called a Kantor-Knuth generalized quadrangle. The kernel K of the TGQ is the fixed field of a, see [133]. The following was shown by Payne in [118]. Theorem 4.7.3 (Payne [118]) Suppose a TGQ S = T(O) is the point-line dual of a flock GQ S{F), T a Kantor-Knuth semifield flock. Then T(0) is isomorphic to its translation dual T(C*). • GANLEY EGGS AND ROMAN (GANLEY-PAYNE) EGGS. Let T be a Ganley flock. The point-line dual of the corresponding GQ S{T) is a TGQ for some base-line, and so the dual S{T)D oiS{T) is isomorphic to some T(O). The GQ T(0) is called a Ganley generalized quadrangle. By [133], the kernel K is isomorphic to GF(3). Payne [118] shows the following.

Theorem 4.7.4 (Payne [118]) For q > 9, T(O) is not isomorphic to its translation dual T(0*). •

4.8. Derivation and BLT-Sets

77

Further, Payne proves in [118] that T(C*) is a TGQ which does not arise from a flock. In [118], the GQs T(0*) were called the Roman generalized quadrangles. THE PENTTILA-WILLIAMS-BADER-LUNARDON-PINNERI EGG AND ITS TRANSLATION DUAL. Let T be the Penttila-Williams-Bader-Lunardon-Pinneri

flock. The translation dual T(0) of S(F)D is referred to as the PenttilaWilliams-Bader-Lunardon-Pinneri generalized quadrangle. The kernel of this GQ is isomorphic to GF(3) [122]. For a detailed study we refer to Bader, Lunardon and Pinneri [4]. Theorem 4.7.5 (Bader, Lunardon and Pinneri [4]) T(0) is not isomorphic to its translation dual T(C*). •

4.8 Derivation and BLT-Sets Let T — {Ci, Ci,..., Cq} be a flock of the quadratic cone K. with vertex x of PG(3, q), with q odd. The plane of d is denoted by 7TJ, i = 1,2,..., q. Let AC be embedded in the nonsingular quadric Q of PG(4,
Chapter 4. Generalized Quadrangles and Flocks

78

Remark 4.8.2 Derivation for any q can be described algebraically. In the flock GQ it amounts to choose any given line incident with (oo) as the line [.A(oo)]. From this approach it is clear that the process of derivation does not produce new GQs, but, as in the odd case, also in the even case new flocks and planes arise. For details we refer to Bader, Lunardon and Payne [3], Payne [121] and Section 2.9 of Cardinali and Payne [35].

4.9

Constructions

(a) The Construction of Knarr Start with a symplectic polarity £ of PG(5, q). Let p e PG(5,g) and let P G ( 3 , q) be a 3-dimensional subspace of PG(5,
POINTS

are of three types:

(i) the points of PG(5, g) not in p^; (ii) the lines of P G ( 5 , q) not containing p but contained in one of the planes nt = pLt, with Lt e V; (iii) p• LINES are of two types:

(a) the totally isotropic planes of ( not contained in p^ and meeting some 7Tt = pLt, with Lt 6 V, in a line (not through p); (b) the planes wt = pLt, with Lt e V. • The INCIDENCE PG(5,?).

RELATION

I is just the natural incidence inherited from

Then Knarr [89] proves that S is a GQ of order (q2, q) isomorphic to one of the GQs arising from any of the q + 1 flocks defined by the BLT-set V. If T is some flock corresponding to V, then the elation point (oo) of the GQ S{T) = S corresponds to the point p of S.

4.9.

Constructions

79

The problem of finding a geometric construction of a flock GQ was open for quite some time. As the years passed after Knarr found such a construction for q odd, many researchers came to believe that no such construction could exist for q even. However, as a byproduct of the proof of a fundamental characterization of flock GQs, see Section 4.10, Thas [183] presented such a construction, with refinement of related material in Thas [182, 186].

(b) The Construction of Thas Let K. be a quadratic cone with vertex x of P G ( 3 , q). Further, let y be a point of K,\{x} and let n be a plane of PG(3,q) not containing y nor x. Now we project K\{y] from y onto TT. Let r be the tangent plane of K at the line xy and let r n TT = T. Then with the q2 points of JC\xy correspond the q2 points of the affine plane TT\T = TT', with any point of xy\{y} corresponds the intersection oo of xy and TT, with the generators of K distinct from xy correspond the lines of TT distinct from T containing oo, with the (nonsingular) conies on K passing through y correspond the affine parts of the q2 lines of TT not passing through oo, and with the (nonsingular) conies on K. not passing through y correspond the q2(q - 1) (nonsingular) conies of TT which are tangent to T at oo. Let J7 = {C%,C2, • • • ,C*} be a flock of the cone /C. Now consider the set T = {C\,C2, • • • ,Cq-i,N} consisting of the q - 1 nonsingular conies Ci,C 2 ,... ,Cg-i and the line TV of TT, which is obtained by projecting the elements of T from y onto TT. So C\, C2,..., C 9 -i are conies which are mutually tangent at 00 (with common tangent line T) and N is a line of TT not containing 00. Now we consider planes Woo ^ TT and /x ^ TT of PG(3,g), respectively containing T and TV; in /x we consider a point r, with r g TT U 7roo. Next, let Oi be the nonsingular quadric which contains C», which is tangent to TTOO at 00 and which is tangent to /1 at r, with i = 1,2,..., q - 1. As C, n N = 0, the quadric 0 , is elliptic, z = l , 2 , . . . , g - l . Next let S be the following incidence structure. •

POINTS

are of five types.

(a) The q3(q - 1) nonsingular elliptic quadrics O containing d n 2 TTOO = £ S U M ^ (over GF(q )) such that the intersection multiplicity of Oi and O at 00 is at least three (these are Oi, the

Chapter 4. Generalized Quadrangles and Flocks

80

nonsingular elliptic quadrics O ^Ot containing i S u M ^ (over GF(g2)) and intersecting £>* over GF(q) in a nonsingular conic containing oo, and the nonsingular elliptic quadrics O ^ d for which OnOi over GF(g2) is L^ U M& counted twice), with i= l,2,...,q-l. (b) The q3 points of PG(3, g)\7r00. (c) The q3 planes of PG(3, q) not containing oo. (d) The q — 1 sets Oit where O j consists of the q3 quadrics O of Type (a) corresponding to Oit with i —1,2...,q — 1. • LINES are of five types. (i) Let (w,r)) be a point-plane flag of PG(3, q), with w <£ ir^ and oo ^ 7. Then all quadrics O of Type (a) which are tangent to 7 at w, together with w and 7, form a line of Type (i). Any two distinct quadrics of such a line have exactly two points (00 and w) in common. The total number of lines of Type (i) is q5. (ii) Let O be a point of Type (a) which corresponds to the quadric ou i € {i,2,...,q-i}. if o n ^ = d nTTOO = I S U M S 2 (over GF(g )), then all points £>' of Type (a) for which O ' n O over GF(
(v) {00, -Koo, Oi, 0 2 , • • •, Oq_i} is the unique line of Type (v). • INCIDENCE is containment. Then it is proved in Thas [183] that <S is a GQ isomorphic to the point-line dual of the flock GQ S(T), We emphasize that this construction works for any prime power q.

4.10. Property (G) for Generalized Quadrangles of Order (s, s 2 )

81

4.10 Property (G) for Generalized Quadrangles of Order In Section 2.5 we introduced 3-regularity in GQs of order (s, s 2 ). Here we will weaken this definition so as to obtain a strong characterizing property for flock GQs. Let S = (V,B,l) be a GQ of order (s,s2), s ^ 1. Let x\,y\ be distinct collinear points. We say that the pair {x\,y{\ has Property (G) or that S has Property (G) at {xi,7/i}, if every triad {xi,X2,X3}, with J/! e {xi,X2,xs}-L, is 3-regular. Then also every triad {yi,y2,2/3}, with x1 e {2/1, 1/2, to}"1, is 3regular. The GQ S has Property (G) at the Zine L, or the line L has Property (G), if each pair of points {x,y}, x ^ y, and xlLly, has Property (G). If (x, L) is a flag, that is, if xlL, then we say that S has Property (G) at (a;, L), or that (x,L) has Property (G), if every pair {x,y}, x ^ y, and ylL, has Property (G). The point x of the GQ 5 of order (5, s 2 ), with s / 1, is 3-regular if and only if each flag (x, L) has Property (G). If s is even and if S contains a pair of points having Property (G), then by Theorem 2.6.2, S contains subquadrangles of order s. In Payne [118] Property (G) is defined for any GQ of order (s,t), with t > s > 1. Motivation for introducing Property (G) is the following result of Payne [118]. Theorem 4.10.1 If S(T) is the GQ of order (q2,q) arising from the flock T of the quadratic cone K o / P G ( 3 , q), then 5 (J") has Property (G) at its elation point (00). • Now we prove that if 5 is a GQ of order (0, q 2 ), q ^ 1, for which at least one pair of points {x, y}, x ~ y, has Property (G), then S defines a projective space P G ( 3 , q). This is due to Payne and Thas; see Thas [176]. Let S = (V,B,I) be a GQ of order {q,q2),q / 1, for which the pair of distinct points {x,y},x ~ y, has Property (G). We introduce the following incidence structure Sxy = (Pxy, Bxy, Ixy);

(i) Vxy =

xx\{x,y}x.

(ii) Elements of Bxy are of two types:

Chapter 4. Generalized Quadrangles and Flocks

82

(a) the sets {y, z, u}±x\{y}, and

with {y, z, u} a triad having x as center,

(b) the sets {x, w}-L\{a;}) with x ~ w / y. (iii) Ixy is containment. Then \Pxy\ = q3, Bxy contains q3(q3 ~ q)/{q{q - 1)) — q4 + q3 elements of Type (a) and q2 elements of Type (b). Lemma 4.10.2 Let Mi, M2 be different elements of Type (a) ofBxy, let L\, L2 be different elements of Type (b) of Bxy, and assume that Li n Mj ^ 9 for all i, j e {1,2}. Then each element of Type (b) of Bxy which contains a point of Mi abo contains a point of M2. Proof. Let L, n Mj = {xij}. Since Mi ^ M2, we have {xn,x2i} ^ {xi2,x22}Suppose, e.g., that x2i ^ x22. Let u e Mi,u <£ Li,u <£ L2,N = {a;,u} ± \{a;}. If N does not contain a point of {y,xii,x22}±Jthen by Lemma 2.5.1, and since the point x e {y,xn,x22}x is collinear with u, the point u is collinear with a second point x' of { y ^ n , ^ } ^ - So x' is collinear with y, xn,x22, u. Hence x' G {y, xn, u}1- and so x' ~ x 2 i, giving a triangle {x',x2i,x22}, a contradiction. Consequently N contains a point of {y,xn,a:22}"L"L- If xn ^ xi2, then a similar argument shows that JV contains a point of M2; if x n = xi2, then, as {y,xu,x22}-LJ= M2 U {y}, it is clear that N contains a point of M2. • Now we introduce the set Hxy- The elements of Hxy are of two types. (a) The sets {x, 2} x \{y}, with 1 / 2 and y e {x, z}-1. (b) Each set which is the union of all elements of Type (b) of Bxy containing a point of some line M of Type (a) of Bxy. The set Hxy contains q3 elements of Type (a), and by Lemma 4.10.2 it contains q2 + q elements of Type (b). Lemma 4.10.3 The elements ofVxy and Bxy in an element ofHxy are the points and lines of a 2-(q2,q, 1) design, that is, an affine plane of order q. Proof.

Easy, using Property (G) and Lemma 4.10.2.

•

4.10. Property (G) for Generalized Quadrangles of Order (s, s2)

83

Theorem 4.10.4 The elements of Vxy,Bxy and Tixy, respectively, are the points, lines and planes of an affine space AG(3, q). Proof. We have \Vxy\ = q3, each element of Bxy contains q elements of Vxy, and any two distinct elements of Vxy are contained in a unique element of Bxy. Hence Sxy is a 2-(q3, q, l)-design. If q = 3, then by Section 1.8 we have <S ^ Q(5,3). By Theorem 1.5.2(h) Q(5,3) ^ T 3 (C), with O an elliptic quadric of PG(3,3). Now the statement of the theorem is easily checked by taking for y the point (oo) of T 3 ( 0 ) , and for x any point distinct from (oo) but collinear with (oo). Now assume q > 3. By a theorem of Buekenhout [30] it is sufficient to show that any three points of Vxy, not contained in a common element of Bxy, are contained in a common element of Hxy. Let z,u,w be three distinct points of Vxy, not in a common element of Bxy, and not in a common element of Type (b) of Hxy. Then no two points of {z, u, w} are collinear in S (that is, no two points of {z, u, w} are incident with a common line of S through x), and {y, z, u}±± has no point in common with {x, w}^. Since w ~ x and x e {y, z,u}L, by Lemma 2.5.1 w is collinear with a second point x' e {y,z,u}^. Clearly {x,x'} contains y,z,u,w, and hence z,u,w are contained in an element of Type (a) of Hxy. • Corollary 4.10.5 If S = {V, B,l)isa GQ of order (q, q2), q ^ 1, for which at least one pair of points {x, y}, x ~ y, has Property (G), then q is a prime power. Proof.

Immediate from Theorem 4.10.4.

•

Let <S(JF) be the GQ of order (g2, q) arising from a flock T of the quadratic cone K of P G ( 3 , q). If «S(J")D is the point-line dual of S(T), then by Theorem 4.10.1 the GQ S(F)D has Property (G) at its elation line L. Let x,y be distinct points of L and consider VXy,Bxy,Hxy and the corresponding affine space AG(3, q). Then by interpreting the points and lines of S(T)D in AG(3, q), Construction (b) of Section 4.9 is obtained. Let S = (V, B, I) be a GQ of order (q, q2), g / 1, which satisfies Property (G) at the flag (x,L). Let y!L,x ^ y, and as above we consider the affine space AG(3,g). The lines of Type (b) of AG(3, q) are parallel, so define a common point at infinity oo. Let P G ( 3 , q) be the projective completion of AG(3, q), with P G ( 3 , q) \ AG(3, q) = ir^. If x / z ^ y, then ({x, zy^iu}) U {oo}, with z ~ ulL, is an ovoid Oz of P G ( 3 , q) which is

84

Chapter 4. Generalized Quadrangles and Flocks

tangent to 77^ at the point 00; see Thas [183]. Then in Thas [183] the following fundamental characterization theorem of flock GQs is obtained. Theorem 4.10.6 Let S = (P,B,I) be a GQ of order (q,q2), q > 1, and assume that S satisfies Property (G) at the flag (x,L). If q is odd, then S is the point-line dual of a flock GQ. If q is even and all ovoids Oz are elliptic quadrics, then we have the same conclusion. • If S is the point-line dual of a flock GQ and we construct the affine space AG(3, q), then all ovoids Oz are elliptic quadrics; see Thas [183]. Now consider the GQ T 3 (C), with O a Tits ovoid in PG(3,g), with q = 2 2 e + 1 and e > 1. As by Section 2.5 the point (00) of T 3 (C) is 3-regular, the GQ T 3 (C) satisfies Property (G) at any flag ((00), L). We leave it as an exercise to check that here the ovoids Oz are Tits ovoids isomorphic to O. Hence T 3 ( 0 ) is not the point-line dual of a flock GQ. This also follows from Theorem 4.5.1 and Section 4.7. So in Thas [183] it was conjectured that a GQ <S = (V, B, I) of order (q, q2), q even, is the point-line dual of a flock GQ if and only if it satisfies Property (G) at some line L. Relying on Thas [183], Barwick, Brown and Penttila [9] obtain the following generalization of Theorem 4.10.6. Theorem 4.10.7 Let S = (P,B,I) be a GQ of order (q,q2), q > I, and assume that S satisfies Property (G) at the pair {x, y}, with x and y distinct collinear points. If q is odd, then S is the point-line dual of a flock GQ. If q is even and all ovoids Oz are elliptic quadrics, then we have the same conclusion. • Finally, Brown [25] proved the conjecture of Thas in the even case. Theorem 4.10.8 Let S = {V,B,I) be a GQ of order {q,q2), q even, and assume that S satisfies Property (G) at the line L. Then S is the point-line dual of a flock GQ. • Now we show that Theorem 4.10.7 can be deduced easily from Theorem 4.10.6. Proposition 4.10.9 Let S = (V, B, I) be a GQ of order (q, q2), q > 1, and assume that S satisfies Property (G) at the pair {x, y}, with x and y distinct collinear points. If q is odd, then S satisfies Property (G) at the line L = xy. If q is even and all ovoids Oz are elliptic quadrics, then we have the same conclusion.

4.11. Flocks, Subquadrangles and Ovals

85

Proof. We use the notation Vxy, Bxy, Hxy, TT^, OO introduced in this section. Let TTxy e Hxy be a plane of Type (b) of the affine space AG(3,g). Then the point oo belongs to the projective completion w^ of irxy. All elliptic quadrics Oz contain oo and are tangent to the plane 7 ^ at oo. Now we consider the conies Oz DW^ = Cxy. Then Cxy contains oo and is tangent to £<x> = TT^ n -n-oo at oo. The number of quadrics Oz is q4 - q3, and there are exactly q3 — q2 nonsingular conies in W^ which are tangent to L^ at oo. In a GQ of order (q, q2), for each set X of size q + 1 consisting of mutually noncollinear points, we have | X -11 < q +1. Hence each of the q3-q2 conies arises from at most q quadrics Oz. Consequently each nonsingular conic of W^ which is tangent to L^ at oo, arises from exactly q quadrics Oz. It easily follows that S satisfies Property (G) at the line L = xy. • Remark 4.10.10 (a) Lavrauw and Penttila [97] give an alternative proof of Theorem 4.10.6 in the particular case that S is a TGQ. (b) From the proof of Proposition 4.10.9 follows that over GF(q2) all elliptic quadrics Oz, z being collinear with a fixed point ulL, intersect the plane 7Too in the same two lines M^ and N^. This provides a short proof of Lemma 7.1.1 inThas [183].

4.11 Flocks, Subquadrangles and Ovals If S is a GQ of order (q2, q), with q even, having Property (G) at a pair of concurrent lines, then, by Theorem 2.6.2, it contains subquadrangles of order q. So by Theorem 4.10.1 any flock GQ <S(.F) contains subquadrangles of order q containing the elation point (oo). Also each such subquadrangle is a GQ T 2 (C) of Tits, with O some oval of PG(2,g); see Payne and Maneri [124] and Payne [113, 117]. In this way Payne [117] discovered a new infinite class of ovals. By Cherowitzo, Penttila, Pinneri and Royle [40], for any flock T of the quadratic cone K, of P G ( 3 , q), with q even, the hyperovals defined by the ovals arising from S{J7), can also be described as follows. Theorem 4.11.1 The q planes xtX0 + ztX1+tX2 + X3 = 0, withte GF(q) and x0 = z0 = 0, define a flock T of the cone K, with equation X0Xi = X2 o / P G ( 3 , q), with q even, if and only iffor all (ai,a2) e GF(g) 2 \{(0,0)}, the set £(<.!,<*) = {(1, y/atxt + awt

+ alzt, t)\\t£

GF(q)} U {(0,1,0), (0,0,1)}

86

Chapter 4. Generalized Quadrangles and Flocks

is a hyperoval o / P G ( 2 , q). Such a set of hyperovals is called a herd of hyperovals. • In Storme and Thas [148] Theorem 4.11.1 is generalized to partial flocks, that is, sets of k disjoint conies on the quadratic cone of PG(3,g), and herds of (k + 2)-arcs in PG(2,g), with q even. In that paper also a short proof of Johnson's Theorem 4.5.1 is given, avoiding the classification of certain classes of semifields but relying on the connection between flocks and herds of hyperovals. In Thas [186] it is shown how in a purely geometric way the herd can be constructed directly from the flock T. For q even, the connection between flocks, spreads, g-clans, GQs of order (q2,q), subquadrangles of order q and herds of hyperovals is given in detail in the survey by Johnson and Payne [79]. In this monograph, herds of hyperovals will not be studied in detail as by Theorem 5.1.11 any flock TGQ of order (q2,q), q even, is classical, so that the corresponding flock is linear, and consequently all hyperovals of the herd are conies together with their nucleus.

4.12. The Fundamental Theorem of q-Clan Geometry, and Applications

87

Addendum A: Isomorphisms of Flock Quadrangles and Associated Geometries In this section, we describe the so-called "Fundamental Theorem of g-Clan Geometry". Starting with a natural definition for equivalent q-clans (defined below), it interprets the equivalence of g-clans C\ and C2 as an isomorphism between Q(Ci) and 5(C 2 ), where Q(Ci) is either the flock T{Ci) associated to Ci, or the flock quadrangle S(T{€{)), i = 1,2.

4.12 The Fundamental Theorem of g-Clan Geometry, and Applications Suppose q is a prime power, and suppose A = I

J and B = I

J be

2 x 2-matrices over GF(g). We write A = B to mean that x = r, w — u and y + z = s + t. It follows that A = B if and only if o>AaT = aBaT for all a e GF(q)2. If C = {At || t G GF(g)} is a g-clan, the matrices At G L are used to construct quadratic forms aAtaT', with a G GF(g) 2 , so an At G C may be replaced by an A't whenever At = A't without "effectively changing" C.

If q is odd, one could adjust each At G C to be symmetric, say, for instance,

For any q we can adjust each At G C to be upper triangular, so, for instance,

,, = ( ; - ) , «eor(,). The g-clan is normalized if T40 is the zero matrix. For any prime power g, let c={At

^(ot!tt)l|iGGF(5)}

and C' =

{A't^^^\\teGF(q)}

Chapter 4. Generalized Quadrangles and Flocks

88

be two not necessarily distinct g-clans. We say that C and C are equivalent and write C ~ C provided there exist 0 ^ \i G GF(g), 5 G GL 2 (g), cr G Aut(GF(q)), M a 2 x 2-matrix over GF(g) and a permutation n : t H-> t on GF(g) such that the following holds: Aj = u,BAatBT + M for all t G GF(g). If for a g-clan C each A t is replaced by A\ = At - A0, then there arises a normalized g-clan equivalent to C. Theorem 4.12.1 (Fundamental Theorem of q-Clan Geometry) Assume that C = {At \\ t € GF{q)} and C = {A't \\ t G GF(g)} are normalized q-clans. The following are equivalent: (i) C ~ C. (ii) The flocks F(C) and F{C) are protectively equivalent. (hi) The GQs <S(.F(C)) and S(T{C')) are isomorphic by an isomorphism mapping (oo) to (oo), [A(oo)] to [A'(oo)}, and (0,0,0) to (0,0,0). These three equivalent conditions hold if and only if the following exist: (i) a permutation n : t H-> t of the elements of GF(q); (ii) T G Aut(GF(g)); (iii) A G GF(g), A ^ 0; (iv) D G GL2{q) for which Aj - XDJA\D zero diagonal) for all t G GF(q).

- A$ is skew-symmetric (with

Given T,D,\ and the permutation ir : x f-* x satisfying Condition (iv), an isomorphism 6 = 0(T,D,X,IT) of the GQ S{F(C)) onto S{F{C')) arises, as follows: 6 =

0{T,D,\,*):(<X,C,I3) l r

~ {\- a D~T,

X~lcT +

X~2aT{D~TA^l){aT)T,pTD

+X-laTD-TK^). (We write D~T for the matrix {D-l)T

=

{DT)~l.)

4.12. The Fundamental Theorem ofq-Clan Geometry, and Applications

89

a b . Now let T{C) = T{C') = T. Define cd a projective semilinear collineation Te o / P G ( 3 , q) as follows (defined on the planes o / P G ( 3 , q), and with a general plane having equation xX0 + zX\ + yX2 + X3 = 0):

For 6 = 6(T, D, X, it), write D =

Te

I Xa2 Xab Xb2 f \ 2Xac X(ad + be) 2Xbd y^ Xc2 Xcd Xd V 0 0 0 1/

Then Te fixes the cone K, with equation X0Xi = X2, and leaves invariant T = T{C) precisely when 6 defines a collineation ofS(T) fixing (oo), [.A(oo)] and (0,0,0). • The map

T-.e^Tg is a homomorphism from the subgroup of Aut(<S(jF)) leaving (oo), [A(oo)] and (0,0,0) invariant onto the subgroup of PT04(q) leaving the flock J7 invariant. The kernel N(T) of T is N(T)

= {6a || (a, c, (3) >-> (aa, a2c, a/3), 0 ^ c £

GF(q)}.

Note that N(T) is a group of whorls about (oo) and (0,0,0). The following statement is a direct corollary of the Fundamental Theorem. Corollary 4.12.2 Let T and T1 be protectively equivalent flocks of the quadratic cone in P G ( 3 , q). Then S(T) ^ S(T'). Proof. Let C be the g-clan associated to T and C the g-clan associated to P. Then C ~ C, and S(7) = S(f(C)) =* S(T{C')) = S(T'). U Let J 7 be a flock of the quadratic cone in P G ( 3 , q), q odd. Then T together with its q derived flocks correspond to the q + 1 lines incident with (oo), cf. Remark 4.8.2. Let J"0 = T, T\,..., Tq be these q + 1 flocks. Then in this section we denote by L(i) the line corresponding to T{. The following result can be found in Payne and Rogers [125]. Theorem 4.12.3 Let T and <S(Jr), etc. be as above. Then there is an automorphism of S(T) fixing (oo) and mapping L(i) to L(j), with i,j e {0,1,...,q}, if and only if Ti is projectively equivalent to Tj. •

90

Chapter 4. Generalized Quadrangles and Flocks

The converse of Corollary 4.12.2 is not necessarily true; this is because nonequivalent g-clans can be associated to the same flock generalized quadrangle. For, let T = T0 be a flock of the quadratic cone in P G ( 3 , q), q odd, and let T\, JT 2 ,..., Tq be its derived flocks. Then S{T) = S(fi), i e { 0 , 1 , . . . , q}, according to Section 4.8. On the other hand, Ti is projectively equivalent to Fj, with i,j e { 0 , 1 , . . . , q], if and only if there is an automorphism of S(T) that fixes (oo) and maps L(i) to L(j). But many examples of flock GQs are known for which Aut(5)(oo) does not act transitively on the lines incident with (oo). Theorem 4.12.5 determines such examples for flock GQs that are TGQs. Note that, in any case, if S(J7) is a dual TGQ of order (q2,q), q then there are at most two isomorphism classes in the set {To, F1:..., as the translation group of 5(J 7 ) r i induces an automorphism group of that fixes the translation line and acts transitively on the other lines dent with (oo).

dual odd, Fq}, S(F) inci-

We first mention the following result, which can be found in [206], and which we will encounter later in Chapter 10, cf. Theorem 10.7.2 in combination with Theorem 6.4.4. Theorem 4.12.4 Let S be a TGQ of order (s,s2) which contains a line of translation points, and which is the point-line dual of a flock GQ 5(J"). Then either s is even and T is linear, or s is odd and J7 is a Kantor-Knuth flock. • Theorem 4.12.5 Let T he a nonlinear flock of the quadratic cone in PG(3,q), q odd, with <S(.F) the point-line dual of a TGQ. Then Aut(<S)(oo) acts transitively on the lines incident with (oo) if and only if J7 is of KantorKnuth type. Proof. By the assumption, each line incident with (oo) is a translation line. So SiJ7)0 contains a line of translation points. By Theorem 4.12.4, it now follows that T is of Kantor-Knuth type. • Hence a flock GQ in odd characteristic of which the point-line dual is a TGQ, but which does not arise from a Kantor-Knuth flock, admits precisely two isomorphism classes of flocks. We end this section with the following useful observation by Payne [121]:

4.12. The Fundamental Theorem ofq-Clan Geometry, and Applications

91

Theorem 4.12.6 (Payne [121]) Suppose S(F) is a nondassical flock GQ of order (q2,q), q > I, and let H be the full group of automorphisms o/<S(Jr) fixing (oo) and (0,0,0) linewise. (i) Ifq = 2h, thenH = N{T). (ii) Ifqis odd, and H* = {9(T,D,X,TT) e H \\ T = 1}, then H* = N(T), except if J7 is a Kantor-Knuth semifield flock. In that case, H = H* if a2 ^ 1, and then [H : N(T)] = 2, where a is the automorphism of GF(q) which is used to define T.Ifa2 = \ ^ a, then [H : H*] = [H* : N(T)] = 2. m

Chapter 4. Generalized Quadrangles and Flocks

92

Addendum B: Basic Questions on Elation Groups In this addendum, we pose four fundamental and old questions on (elation) generalized quadrangles, and survey tersely answers on these questions coming from work of Payne and K. Thas [130], K. Thas and Payne [214], Rostermundt [134] and K. Thas [207]. The importance of collineations that fix proper (but not necessarily thick) subGQs will be underlined, and in particular this will be clear when the Kantor-Knuth GQs will serve as counter examples for one of the conjectures. Not all proofs of [130, 214, 134] and [207] will be given. But we will give the ones which reveal the important nature of those collineations fixing subGQs pointwise. For more details on the theory sketched in this addendum, we refer the reader to the monograph "Lectures on Elation Quadrangles" [212] of K. Thas.

4.13 The Standard Conjectures and Questions on Elation Generalized Quadrangles In Chapter 8 of [128], the following is quoted: "In general it seems to be an open question as to whether or not the set of elations about a point must be a group." In the same chapter of loc. cit, the authors study TGQs, and show that all elations about the elation point are in the translation group (cf. 8.6.4 of [128]). We will call the aforementioned question "QUESTION (1)". Most of the known GQs are, up to duality, EGQs with at least one elation point. (In fact, each known GQ is as such, or is constructed from an EGQ — see [206], and in particular Chapter 3 of that book, for details.) We therefore formulate the following specialization of Question (1): QUESTION

(2). Given an EGQ S(x\ is the set of elations about x a group?

The following question makes sense if the answer is "not always".

4.13. The Standard Conjectures and Questions QUESTION

93

(2'). Given an EGQ S^x\ when is the set of elations about x a

group? Let T be a Kantor-Knuth semifield flock of the quadratic cone K, in PG(3, q), q = ph, with p an odd prime. Let S(T) be the corresponding flock GQ of order (q2,q). Recall that each flock GQ has a "special" point (oo) for which there is a group of elations K making it into in an EGQ with elation point (oo). In [214], Payne and K. Thas constructed an elation 9 about (oo) which generates a group not only consisting of elations. So 6 is not an element of K. The construction is as follows. We use the standard notation of §4.6. \,soQ = QT = Q'1. It is noted in [119] that the map

Let Q = (

T

: {a, c, 0) ^ {aQ, c, (3Q)

is an automorphism of K which induces a collineation of S that fixes (0,0,0) and (oo) linewise. Let g = (a',c',/3') be a fixed element of K to be determined later. Put 9 = T o w(g) = T o ir(a',cf,(3'). For the appropriate choice of g we will establish the following: (1) 9 is an elation about (oo) that is not in K; (2) 9P is an involution whose fixed element structure is a subquadrangle of order q, so that in particular 9P is not an elation about (oo); (3) 92 G K. Lemma 4.13.1 If j is a positive integer, let j'2 denote the element of {0,1} to which j is congruent modulo 2. Similarly, let —ji denote —lifjis odd and 0 if j is even. Also, I is the 2 x 2 identity matrix. /

+

0 + Q2 + -

+

Q^(j;1(j+°1)2).

Q + Q2 + --- + Qj = (J0_°J2)-

(4.1) (4-2)

94

Chapter 4. Generalized Quadrangles and Flocks ( (3-1)j

Qi-1 + 2Qi~2 + 3Q^- 3 + • • • + (j - l)Ql = I

^

n

u

\

\

(4.3)

In Equations (4.1) and (4.2) we take j > 1. In Equation (4.3) we must take j>2. Proof. All three equations are established by routine induction arguments (note that |_§J = i ^ ) • By definition we have 6 = Ton(a',c',p')

: (a,c,P) .-> {aQ + a',c+ J + PQoa',PQ

+ P'). (4.4)

Then by an easy calculation we have 82 : (a, c, /?) h-v (aQ 2 + a'(Q + / ) , c + 2c' + /3(Q + Q2) o a' + /3'<9oa',/3Q 2 +/3'(Q + / ) ) .

(4.5)

It now follows by a routine induction that 0i : (a,c,/?) ^ (aQ* + a ' ^ - 1 + Q i _ 2 + • • • + Q + I),

(4.6)

c + id + p(Q + Q2 + Q3 + • • • + Ql) o a1 + (3'{Ql-x + 2Q{-2 + 3 Q ' " 3 + ... + ({- l ) ^ 1 ) o a', PQ' + P\Ql~x

, (l

a

+ Q'~2 + --- + Q + I))

0 \

+a

,(i 0

= ( U(-i)v U*2

K;<-°I)0 + ^> At this stage it is convenient to have written out the image of 6l separately for odd and even i. 92k:(a,c,P)^(a

+ a'(^°Y

(4.7)

4.13. The Standard Conjectures and Questions

95

,fc 0 ^(-°)o . ^v - *H^fo"> c+ 0 +

^MT

0 ! ) > •

We now determine whether or not #l fixes some point (a, c, (3). First, #2fc fixes (a, c, /?) if and only if

(ll)2fcc+/3

UoJ°a+/? A

o

-fcJoa=0;

and

™ "•(??)-("•")It follows readily that if a' = (ai, a2) with a\ ^ 0, then we have proved the following lemma. Lemma 4.13.2 Assume that a\ ^ 0. Then 82k fixes some (a, c, j3) if and only if k = 0 mod p, in which case 92k = id. In particular, 82 is an elation for which (62) is a group of elations. • Note that it is also easy to check that 92 € K. Now we consider whether or not 92k+1 fixes some (a, c, j3). Since we are assuming that a' = (ai,a2) with a\ ± 0, it follows easily that if 92k+l , x. u /0 0\ /2/c + l 0 \ fixes some (a,c,p) it must be the case that a I I = a I I, which forces 2k + 1 = 0 mod p. So consider the fixed points of 9P. It is routine to check that 9P fixes the point (a, c, /3) if and only if (a, c, j3) = ((a,-ka2),c,(b,-kb2)), a,c,b £ GF(g), where a' = (oi,a 2 ), p" = {bi,b2)

Chapter 4. Generalized Quadrangles and Flocks

96

andp = 2fc + 1. It now follows that 8P is an involution with a subquadrangle of order q as its fixed element structure. This suggests the definition of a "standard elation" about a point x: this is an elation for which () is a group of elations ( acts freely on the points not collinear with x). The following natural question arises: QUESTION

(3). Given an EGQ S^x\ is the set of standard elations about x

a group? In the same way as for Question (2), one could now also formulate Question (3'). (3'). Given an EGQ <S(x), when is the set of standard elations about x a group? QUESTION

4.14

Some Results by Payne and K. Thas

For nonclassical flock GQs there is a complete answer to Question (3): Theorem 4.14.1 Let S(!F) be a nonclassicalflockGQ of order (q2,q). Then the set of standard elations about (oo) is a group. This group is the usual elation group K. When q is odd, the same conclusion holds in the classical case. Proof. See the proof of Theorem 6.1 of [130].

•

Remark 4.14.2 In [130], the condition that S(T) is nonclassical for q even was forgotten in the statement of the theorem (cf. Theorem 2.4 and Theorem 6.1). Generalizing Theorem 4.14.1, the following was also obtained in [130]. First define a skew translation generalized quadrangle (STGQ) to be an EGQ ((p), G) for which the elation point p is a center of symmetry about which the group of symmetries is contained in G.

4.14. Some Results by Payne and K. Thas

97

Theorem 4.14.3 Let <Sl. two possibilities:

Then we have

(a) the set of standard elations about pis a group; (b) s = t2, s is a power of 2, and there is a subGQ isomorphic to W(i) containing p which is fixed pointwise by an involution ofS^p\ Proof. See the proof of Theorem 7.2 of [130].

•

Remark 4.14.4 Examples of (b) yield counter examples to Question (3). Such STGQs will play a central role in §4.15. From the next theorem of [214], it will follow that "most of the time", the answer to Question (2) is that the set of elations about an elation point is not a group, and it also explains precisely why. SETTING. In this section, S = (V, B, I) is a GQ of order (s, t), s, t > 1, and x is an elation point for the elation group G. We let W be the group of all whorls about x.

Now let H < W be any subgroup of W that acts transitively on V \ x1-. We apply Burnside's Lemma on the permutation group (H, V \ xx) to obtain s2t-l 2

\H\ =s t+ Y^

i6(

S)i

where S(j) denotes the number of elements of H that fix precisely j points

olV\xL. Let E be the number of elations in H (not including 1) about x; then, as s2t-l

s2t-l 2

l + E+Y,

5(i) = s t+ J2

i=l

i<5

W'

i=l

it follows that the number of elations (now including 1 = id!) is given by the following: s2t-l 2

1 + E = s t+ ^ which is clearly at least s2t. The following theorem now easily follows.

(i-l)S(i),

Chapter 4. Generalized Quadrangles and Flocks

98

Theorem 4.14.5 Let («S(p\G) be an EGQ, and let W be the group of all whorls about p. Then the set of elations about p is a group if and only if there is no nontrivial element in W fixing more than one point not collinear with p if and only if W is a Frobenius group. • The Known Examples It is convenient to mention the known (classes) of examples of EGQs with elation point p for which the set of elations about p is not a group. These classes are treated in detail in [214]. • The classical and dual classical examples. H(4,q2);H(4,g2)^.

W(q) with q odd; H(3, q2);

• Flock GQs. The flock GQs with an even number of points on a line. • Dual flock GQs which are EGQs. As the order is (q, q2) for some q, it can be shown that not more than one point not collinear with the elation point can be fixed by a nonidentity collineation. So there are no examples possible. • TGQs. There are no examples possible. • Dual TGQs which are EGQs. GQs «SD, where S is a TGQ of order ( Q2), Q odd, that is the translation dual of the point-line dual of a flock GQ.

4.15 Elation Generalized Quadrangles with Nonisomorphic Elation Groups The following fundamental question (especially in construction theory for GQs) was posed by Payne [123] during the 2004 Pingree Park Conference "Finite Geometries, Groups and Computation": QUESTION (4). Let S^ morphic elation groups?

be an EGQ. Can x be an elation point for noniso-

The following theorem considers a class of GQs which do admit nonisomorphic elation groups, thus answering Payne's question affirmatively. The only known examples of this class are the GQs H(3, q2) with q even.

4.15. Elation Generalized Quadrangles with Nonisomorphic Elation Groups

99

Theorem 4.15.1 Let S = {S^P\H) be an EGQ of order (q2,q), where q is even, which contains a subGQ S' of order (s, q), s > 1, fixed pointwise by a nontrivial automorphism 9 of S. If H2 < Z{H), and if for all Lip and xILlp, with p y£ x, we have that \H(x,L,p)\ = q2, where H(x,L,p) is the group of whorls about x, L and p, then there is an automorphism group H' ofS such that H' ^ H and (S^ ,H')is an EGQ. Proof.

SeeK. Thas [207].

•

Corollary 4.15.2 Let S = (S^,!!) be an EGQ of order (q2,q), where q is even, which contains a subGQ S' of order (s, q), s > 1, which is fixed pointwise by a nontrivial automorphism 8 of S. Let z ^ p and suppose z ~ zi ~ pfor i = 0,l,...,q. If all groups H(p,pzi,Zi)nH are elementary abelian and have size q2, then there is an automorphism group H' of S such that H' ^ H and (S
100

Chapter 4. Generalized Quadrangles and Flocks

He then constructs q2 -1 distinct elation groups Kt = 1,2,..., q2 — 1 of size q5 with base-point (CXD), and shows that all Ki are mutually isomorphic. The Kis have nilpotency class 3, while K has nilpotency class 2, so that K p Ki for all i. The proofs are long and technical. For details and several other results, see Rostermundt [134].

Chapter 5

Good Eggs In this chapter we will define good eggs, explain the relationship with Property (G) and flock GQs, and give some interesting characterization theorems. There is also a section on the coordinatization of good eggs and their translation duals, and coordinates of all known examples are given.

5.1 Good Eggs and Good Translation Generalized Quadrangles Definition 5.1.1 The egg 0(n,2n,q) = O is good at its element TT if any PG(3n - 1, q) containing ir and at least two other elements of 0(n, 2n, q) contains exactly qn + 1 elements of 0(n, 2n, q); the space n is called a good element of 0(n, 2n, q). In such a case we also say that 0(n, 2n, q) is good. If O is good at ir, then we say that the TGQ T(O) is good at its line IT, and ir is a good line of T ( 0 ) ; shortly we say that the TGQ T(O) is good. We emphasize that each known egg 0(n,2n,q) good.

or its translation dual is

Now we give the relation between good and Property (G); see Thas [176]. Theorem 5.1.2 The egg 0(n,2n,q) — O satisfies Property (G) at the flag ((oo),7r) if and only if the translation dual T(G*) of the TGQ T(0) is good at its element r with T the tangent space of O at n. Proof. Suppose that T((9) satisfies Property (G) at the flag ((oo), ir), with •n e O. The tangent space of O at it is denoted by r. Let nj,irk be distinct 101

102

Chapter 5. Good Eggs

elements of 0\{n} and put P G ( n - 1, q) = r n Tj H rk, with T; the tangent space of Oatiri, I e {j,k}. By Theorem 3.10.1(i) each point of P G ( n - l , g ) is contained in exactly qn + 1 tangent spaces of O. Let PG(4n, q) be the space in which T(O) is defined, and let P G ( 4 n - l,q) be the space of O. Further, let u £ P G ( n - 1, q), let N be a line of PG(4n, q) through u not contained in PG(4n - 1, q), and let u\,v,2 be distinct points of N not contained in PG(4n - 1, q). Then each of the points (N, T), {N, TJ), (N, rk) of Type (ii) of T(0) is collinear with each of the points (oo), ui,u2 of T ( 0 ) . Clearly {(oo),u1,u2}J- is the set {(N,T), (N,Tj), {N,Tk}, • • • } with T,T0,Tk,. • • the qn + 1 tangent spaces of O through u. As T(O) satisfies Property (G) at the flag ((oo),7r), we have \{(oo),ui,U2}1"L\ = qn + 1. Hence (N, r) n (AT, T,-} n (N, rfe) n • • • = r) contains qn points of PG(4n, q) \ P G ( 4 n - 1, q). Consequently rj is an n-dimensional space, that is, TDTJ f) Tfc n • • • is (n - l)-dimensional. So the intersection of these qn + 1 tangent spaces is P G ( n - l,q). We conclude that P G ( n - l,q) is contained in exactly qn + 1 tangent spaces of O. Hence the translation dual O* of O is good at its element T. Conversely, let O* be good at its element T. SO for any two distinct elements Tj,Tk € 0*\{T} the space P G ( n - l,q) = TDTJ C\Tk is contained inexactly qn + 1 tangent spaces of C Let T, TJ, r^ be points of Type (ii) of T(O) collinear with (oo), where r c r, r,- c Tj, T^ C r^. Then T fl Tj fl Tfc is an n-dimensional space P G ( n , g ) . Clearly P G ( n — l,q) = P G ( n , q) n PG(4n - 1, g). If r, Tj, Tfc,... are the g™ + 1 tangent spaces of O containing P G ( n - l , q ) , then the ij™+ 1 spaces (PG(n,q),r) = T, (PG(n,g),Tj) = Tj, (PG(n,q),Tk) = Tfc,... are points of Type (ii) of T(O), each of which is collinear in T(O) with (oo) and the qn points of Type (i) in PG(n,q). Hence the triad {f,Tj,Tk} is 3-regular. So we conclude that T(O) satisfies Property (G) at the flag ((oo), ir). • Remark 5.1.3 As the translation group of T(O) is transitive on the points of Type (ii) of T(O) incident with n, in the statement of Theorem 5.1.2 the flag ((oo),7r) may be replaced by a point pair ((oo),7r), with n a point of Type (ii) incident with -IT. Corollary 5.1.4 If the egg 0(n, In, q) = O, with q odd, is good at its element IT, then T(C*) is the point-line dual of a flock GQ with elation point T, where T is the tangent space ofO at n. Conversely, if the flock GQ S has a translation line, then the point-line dual ofS is isomorphic to a T(O) for which O* is good at its element which corresponds to the elation point of S.

5.1. Good Eggs and Good Translation Generalized Quadrangles

Proof.

From Theorems 4.10.1, 4.10.6 and 5.1.2.

103

•

Theorem 5.1.5 Ifqis even and if the TGQ T ( 0 ) satisfies Property (G) at the flag ((oo),7r), with n G O, then the translation dual T(C*) satisfies Property (G) at the flag ((oo), T), with r the tangent space of O at TT. Proof. Suppose that T(O) satisfies Property (G) at the flag ((oo), TT), with q even. Let r, rj, TJ be points of Type (ii) of T ( 0 ) collinear with (oo), where T C T . T J C Tiy Tj c Tj, with r, r,, T,- the tangent spaces of O at the respective (distinct) elements TT, iri,itj of O. Then, by Theorem 2.6.2, {¥, Ti,Tj}± and {T, T^T,}- 1 - 1 - are contained in a subquadrangle <S' of order qn of T(0). The space 7 n f , n T., is an n-dimensional space PG(n,q), and the space P G ( n - 1, q) = P G ( n , q) n PG(4n - 1, q) = r n Tj n TJ is contained in exactly qn + 1 tangent spaces T,n,Tj,... of O. Then the points of P G (4n, g) \ P G ( 4 n - l , g ) in (PG(n,q),iv), (PG(n,g),7r i ), ( P G ( n , g ) , f , ) , " - are the points of Type (i) in S'. Now we consider a line of Type (a) concurrent with 7r and incident with a point of Type (i) in the space (PG(n, q), Wi). As this line is also a line of S', it is incident with a point of Type (i) in each of the qn - 1 spaces (PG(n, q), TVJ), • • •. Hence the spaces (PG(n, q), TTJ), • • • are contained in the 3n-dimensional space (PG(n, g),7r,7r,) = PG(3n, q). Consequently the spaces P G ( n — 1, q), IT, m, TTJ, ... are contained in a common P G ( 3 n - 1, q) = PG(3n, q) n PG(4n - 1, q). So, by Theorem 3.8.1 (i), the (3n - 1)-dimensional space (TT, m, TTJ) contains exactly qn + 1 elements of O. Now, by Theorem 5.1.2, the GQ T(C*) has Property (G) at the flag ((oo),r). • Corollary 5.1.6 If q is even and if the egg 0(n, 2n, q) is good at its element IT, then the translation dual 0*(n,2n,q) is good at the tangent space r of 0(n, 2n, q) at IT. Proof.

Directly from Theorems 5.1.2 and 5.1.5.

•

Remark 5.1.7 An alternative proof of Corollary 5.1.6, without using GQs, goes as follows. Assume that q is even and that the egg 0(n,2n,q) = O is good at its element IT. Let 7^,71-., be distinct elements of 0\{TT}, and let T,Ti,Tj be the respective tangent spaces of O at ir,Tri,iTj. Further, let TT, 7Tj, TTJ, ... be the qn + 1 elements of O in (n, TTi,TTj) = PG(3n - l,q) and let O' be the pseudo-oval of P G ( 3 n — 1,
104

Chapter 5. Good Eggs

tangent spaces of O at the elements of O' all contain 77. It follows that the translation dual O* of O is good at its element r. Assume that the egg 0(n, 2n, q) = O, with O = {IT, m,..., 7rg2n}, is good at its element n, let r be the tangent space of O at -K, and let n be the tangent space of O at 7TJ, with i = 1,2,..., q2n. Further, let C be a (3n - 1)dimensional subspace of the P G ( 4 n - 1, q) containing O, with 77 n C = 0Now we project the elements of 0\{TT} from n onto (. The projection of 7TJ from 7r onto C is denoted by 7^, that is, n[ = {71-, 7r;)n C, with i = 1,2,..., g 2 n . Then 7ri U 71-3 U • • • U 7i^2n is C\T', where r ' is the (2n - 1)-dimensional space T f l ( . The set of all intersections (7r,7rj,7rj) n ( , i ^ j , is denoted by B. Then, with respect to inclusion, the elements of V = {TT[, TT'2, . . . , ir'q2ri} and B, are the points and lines, respectively, of a 2 - (q2n, qn,l) design A, that is, of an affine plane of order qn = s. The affine plane A will be called the derived or internal plane of the egg O at the good element n. If p' and p" are distinct parallel lines of A, then p ' n p " is necessarily an (n-1)-dimensional subspace of r'. Hence all lines of A parallel to p' contain a common (n — 1)dimensional subspace of r'. These qn + 1 (distinct) subspaces of r ' are denoted by v0, v\,..., vqn. If for some j ^ j we have J/J n i/j ^ 0, say re £ ViC\Vj, then the lines p' and p" of ^l containing respectively ^ and i/j, do not intersect in a point of A, so are parallel, that is, vt = z/,-, a contradiction. Hence {^o, I/J, . . . , vq™} is a partition (an (n - 1)-spread) of T' . The elements va,vi,...,vqn and r ' extend A to a projective plane II of order qn. Now by a theorem of Segre [140] such a plane II is Desarguesian, so in C there are n planes £i> £2, • • •, £n over GF(
5.1. Good Eggs and Good Translation Generalized Quadrangles

105

Assume again that the egg O = 0(n,2n,q) is good at its element w. Let T be the tangent space of O at IT, and let r, be the tangent space of O at 7Tj, with 7Ti e 0\{TT} and i = 1,2,..., <j2n. Further, let O be contained in P G ( 4 r a - l , q), let P G ( 4 n - l , g) be ahyperplane of PG(4n, g), and let T ( 0 ) be the corresponding TGQ of order (qn, q2n). If irk, 7Tj are distinct elements of 0\{ir} then PG(3n - 1, q) = {n, iTj,nk) contains exactly qn + 1 elements of O. Let PG(3n - l,q) C PG(3n,g) c PG(4n,g), with PG(3n,g)
106

Chapter 5. Good Eggs

Theorem 5.1.10 Let 0(n,2n,q) = O be an egg in PG(4n - l,q), q even, which is good at IT e O. If the q2n + qn pseudo-ovals on O containing IT are regular, then O is regular, so the corresponding TGQ T(O) is isomorphic to a T3(0')ofTits.

Proof. Let O = {n, TTI, TT2, • • •, irq^} be an egg in P G ( 4 n - 1, q), q even, which is good at IT. Suppose that the q2n + qn pseudo-ovals on O containing IT are regular. Clearly we may assume that n > 1. Next, assume that qn = 4, so q = n = 2. As qn = 4, the q2n + qn regular pseudo-ovals on O containing IT are pseudo-conies. Now, by Brown and Lavrauw [26], cf. Corollary 7.8.4 below, the egg is classical, hence regular. From now on we assume that qn > 4. Let PG(3n - 1, q) be a subspace of P G ( 4 n — 1, q) which is skew to IT and let {7r,7Ti)n PG(3n - l,q) = 7T-, with i = 1,2,... ,q2n. Let r be the tangent space of O at IT, and let n be the tangent space of (!) at 7r„ with i = 1,2,..., q2n. If T' = r n P G ( 3 n - 1, q), then the intersections of r ' with the tangent spaces at IT of the pseudo-ovals on O containing IT, form an (n —1)-spread T* of r'. Then T* U {TT[, ir'2,... ,7r' 2 „} is a regular (n — 1)spread T of PG(3n — 1, q). This spread T is the point set of the projective completion II of the derived plane A of O at IT. If we extend the elements of T to G¥(qn), then these extensions have a point in common with each of n planes £i, & , . . . , £„ over GF(qn). Let 7f • n £j = {pij}, with 7f • the extension of TT[ to GF(g"), i = 1,2,..., q2n and j = 1,2,..., n. Then £.,- = {pij,p2j, • • • ,Pq2™j} U I/j, with L,- the intersection of £j with the extension of r' to GF(q n ), with j = 1,2,..., n. Let Mj be a line of £,-, with Lj ^ Mj. Then (MJ,TT), with 7f the extension of IT to GF(q"), intersects the extensions to GF(qn) of all elements of a pseudooval O' on O containing IT. AS 0 ' is a regular pseudo-oval, there are planes 7^ over GF(g") intersecting the extensions of all elements of O', with k = 1,2,...,n. Then {<¥,7fc) D PG(3n - 1,qn) \\ k = 1,2,...,n} = {(O1) n ^ II j = 1,2,..., n}; here we rely on the fact that a regular (n - 1)-spread in a P G ( 2 n - 1, q) uniquely defines n transversals over GF(<7n). So we may assume that (7f,7 J )nPG(3n-1,<7™) = (0')n£j = M,-,withj = 1,2, . . . , n . Clearly the extensions of the elements of O' intersect ^3 in the points of an oval G'j over GF(qn), with O) n n = {rj},j = 1,2,..., n. This oval O) is contained in (w, £j). Now let iV, be the tangent of Oj at rj. Then (A^-, W} intersects ^ in a point of Lj. If CK and O" are two such ovals, with O'j ^ O", then we have the following cases.

5.1. Good Eggs and Good Translation Generalized Quadrangles

107

• If the corresponding lines of £, intersect in a point not on Lj, then {O'j n Q'!)\K is a point. • If the corresponding lines of £,- intersect in a point of Lj, then (O' n 0")\W is empty. Let the distinct points Vij-.Vi'j-.Vv'j define a triangle of £j. The ovals defined by the lines VijVvj->VviVvjiVviVij a s above, are denoted by uf', Oj , Oj ', respectively. Then one easily sees that all q2n + qn ovals O'j are contained in the space (Oj ,Oj\ Oj). So all these ovals generate a P G ( j ) (m, qn), with 3 < m < 5.

(a) The Case m = 3 Then (TF, PG (i) (3,g™)) = {W,Q is (n + 2)-dimensional, so TfPiPGU)(3,qn) is a point u. This point u belongs to the q2n+qn ovals Oj. The union of these ovals is a set W of size g2™ + 1 . As no three points of W are collinear, this set W is an ovoid O of P G ^ ' (3, qn). Hence the extensions of the elements of O meet the n spaces P G ( 1 ) ( 3 , ? n ) , P G ( 2 ) ( 3 , g " ) , . . . , P G ( n ) ( 3 , q n ) and the egg O is regular (the TGQ T(O) is isomorphic to the TGQ T 3 ( 0 ) of Tits).

(b) The Case m = 5 Then (TF, P G ( i ) ( 5 , g n ) } = (¥,£,-) is (n +2)-dimensional, so TF n P G ( j ) ( 5 , g n ) is a plane 5j. Let Wi be the extension of 7TJ to GF(g n ) and let Wi n PG^(5,<7 n ) = {si}, with i = l , 2 , . . . , g 2 r a . Now we project from sis 2 onto a subspace P G ( 3 , g n ) of P G ^ (5, qn) which is skew to sis 2 . Let the qn ovals containing si but not s 2 be O n , O i 2 , . . . , Oig™, and let the qn ovals containing s 2 but not s\ be 021,022, • • •, 02 g »- Further, let Lu be the tangent line of On at si, with Z = 1, 2 and i = 1,2,..., qn. By projection of Ou\{si} from s\S2 onto PG(3,g"), there arise qn points of a line Tu. Let (si,s2,Lu) n P G ( 3 , g") = {in}; then fe e 7^. Assume, by way of contradiction, that Tu C\Tik ^ 0 for i ^ k. Then all ovals in PG ( j ' ) (5,g") are contained in (Tij,T( fc ,si,s 2 ) which is at most 4-dimensional, a contradiction. So Tn,Ti2,...,Tiqn are mutually disjoint, with I = 1,2. If O is the oval containing si and s 2 , then 0 \ { s i , s 2 } is projected onto a point i of P G ( 3 , g n ) . The notation is chosen in such a way that 0 2 i is the oval on s 2 which contains no point of OH\7F, with i = 1,2,..., qn.

108

Chapter 5. Good Eggs

So Tu n T2fc ^ 0 for k =£ i. As qn > 4, it follows that there is a hyperbolic quadric H in PG(3,g n ) such that {Tn,Ti2,. • . , 2 V } is a subset of a regulus ft; of H, with Z = 1,2. Let O H n W = On n 6j = {n*}, and <si,s 2 ,ni> n P G ( 3 , g " ) = {SH}. If n is the extension of n to GF(g n ), then Tt n P G ( i ) (5, qn) is a plane Pji, with j = 1,2,..., n and i = 1,2,..., q2n. So the lines Ln,Li2,..., L(g« are contained in the plane /a,-;. It follows that Ui,ti2, •••, Uq™ are points of some line T/, with I = 1,2. Clearly {t} = T^nT^ and SH = s2i = s'j, with i = 1,2,..., g n . Hence ovals which define a common point of Lj intersect 8j in a common point. The points s^s^,..., s'qn belong to (Sj,si,s2) n PG(3,g"), so are contained in a plane ip. Hence s[,s'2, ...,s'qn belong to the nonsingular conic C = ip n H. Also, we have

tec. Now we consider an oval O'j not containing si nor s 2 . By projection of O'j from S1S2 we obtain the intersection of H with a plane of PG(3,g") through t, but not containing T[ nor T 2 . Hence this intersection is a nonsingular conic, and so O'j is a conic. It follows that the corresponding pseudo-oval O is a pseudo-conic. Now, by Brown and Lavrauw [26], see also Corollary 7.8.4, the egg is classical. But in such a case m is necessarily three, a contradiction.

(c) The Case m = 4 Then (7f,PG (j) (4,g ri )) = ( ¥ , ^ ) is (n + 2)-dimensional, SOTF n P G ( i ) ( 4 , g " ) is a line Wj. Let Wi be the extension of 7TJ to GF(g") and let 7fj n P G ( i ) (4, qn) = {Si}, with i = 1,2,..., g 2 n . Let w e Lj. With the g n lines of ^ through w, but distinct from Lj, correspond qn ovals. The tangent lines of these ovals at their intersection point with Wj are contained in the plane P G ^ (4, g n ) n (TT, W). Assume, by way of contradiction, that the point u G U = W3 is on at least g™ + 1 ovals O'j. So at least two of these ovals, say Oj and Oj, have a point s^ <£ U in common. Let Oj be a third oval containing u. As all ovals generate P G ( i ) ( 4 , qn), the oval of^ cannot contain a point of both 0^\{u, st} and

of\{u,Si}. First, assume that Oj contains s,. If r* is the extension of n to GF(g n ), then T j n P G ( i ) (4, g n ) is a plane pjt, with j = 1,2 . . . , n and i = 1,2,..., g 2 n . So the tangent lines at st of the ovals Oj , Oj , of' are coplanar. It follows that these three ovals are contained in the threedimensional space (pji, u),

5.1. Good Eggs and Good Translation Generalized Quadrangles

109

and so all q2n+qn ovals O'j are contained in (pji, u), clearly a contradiction. So any oval of containing u, with of ^ of ^ of\ does not contain s^ Let of and of] be distinct ovals containing u, with \of n of\ = \of] n of\ = 2 and I e {1,2}. Then o}fc) n OJ.fc'} = {«}, as otherwise (0J 0 , Of \ of]) is 3-dimensional. That means that all ovals containing u are elements of two classes W\ and W2, where of G Wi and any two distinct ovals of Wj just have u in common, with i = 1,2. Now we project from u onto a PG(3, qn) in PG ( i ) (4, qn), with w £ PG(3, g n ). The ovals of W, are denoted by o g = of\o$,..., with Z = 1,2. Let Tfe(0 be the tangent line of ofl at u, and let T^ n PG(3,? n ) = {4°}- Then the points 4 > 4 > • • • a r e distinct, as otherwise all ovals are contained in a 3-dimensional space, with i = 1,2; as ovals of different classes W\ and W2 always have a point of PG^(4,g")\[7 in common, we clearly have 4 1} ^ 4^- L e t t h e projection from u onto PG(3,g") of ofk\{u} belong to the line Vfew. Then 4^ e Vfe(i). Now we shall prove that the lines v}l),V2(l\ ..., with I = 1,2, all belong to a common hyperbolic quadric H of PG(3, qn). Suppose, by way of contradiction, that these lines do not belong to a common hyperbolic quadric. As T{',T2 ,... define a common point on Lj, the points t\', 4 , • • • are collinear, with I = 1,2. Assume that \Wi\ < \W2\. Then the only possibility is that |Wi| = 1 and \W2\ = qn, as otherwise all lines Vk ' belong to a common hyperbolic quadric. The points tf, 4 2 ) , • • • are on a line R; let u' be the (g™ + l)-th point of that line. Then {u1} = UC\ PG(3, qn). With the q2n - 1 ovals not containing u correspond, by projection from u, q2n - 1 ovals through u'. Any such projection is a subset of V1(2) U V2(2) U • • • U V^2) U {u'}. The planes of these projections, together with the qn + 1 planes on R in PG(3, qn) and the plane («', V^ ) are the g2n + qn + 1 planes of PG(3, gn) through w'. Now consider a line V, with Vf> ^V ^R, which intersects v}2) ,V^2\v£2). Then (V, u') intersects v}2) U V2(2) U • • • U Vg(2) U {u'} in g" + 1 points which form an oval, a contradiction as this intersection contains three distinct collinear points. Consequently the lines V^ all belong to a common hyperbolic quadric H ofPG(3, g n ). Assume \WX\ < \W2\ and let U n PG(3,g n ) = {u'}; from the foregoing it follows that | W\ | > 2. Then the points u', t^p, 4 ' \ • • • a r e o n a common line Ru with I = 1,2. Assume, by way of contradiction, that we are not in the case \WX\ = \W2\ = qn. Then there exists an oval C , with £>'• n of =

Chapter 5. Good Eggs

110

0. The projection O'j of O'j from u onto PG(3,q") contains u', and the projection of the tangent line of £K at O'j nU is the line Rx. It follows that (O'j) contains Rx. As i?i c H, the oval £^ contains \W2\ > qn - 1 points of a line of H, clearly a contradiction. Hence \W\\ = \W2\ = qn. Any oval O'j not in W\ U ^ 2 is then projected onto a conic of H, and so C^ is a conic. It follows that the corresponding pseudo-oval O is a pseudo-conic. Now, by Brown and Lavrauw [26], see Corollary 7.8.4, the egg is classical. But in such a case m is necessarily three, a contradiction. So we have shown that each point u of U is contained in exactly qn of the q2n + qn ovals. Consider again distinct ovals of^of*1 assume that Oj

n Oj

containing the point u e U and

= {u, s j , with s» ^ [/. Then as before all ovals

containing u are elements of two classes W\ and W2, where Oj e Wj and any two distinct ovals of W; just have u in common, with I = 1,2. Assume that |Wi| < \W2\. We proceed as before, using similar notation. Then, if the lines Vfe ' do not belong to a common hyperbolic quadric H, we have \Wi\ = 1 and |VF2j = qn — 1. In such a case the planes of the q2n ovals not on u together with the qn + 1 planes through R in P G ( 3 , q") and the plane (u', V}1'), are the q2n + qn + l distinct planes in P G ( 3 , qn) containing u'; the unique oval O* having no point in common with the union of the elements of W2 defines a plane of PG(3,g") which contains R (the projection O* onto PG(3,g n ) of this oval is tangent to R at u'). Now consider a line V, with Vi(1) ^ V ^ R, which intersects v/ 2 ) , V2{2), V3(2). Then (V,u') intersects v / 2 ' U V2(2) U • • • U T/(„2^ U (5* U {u'} in qn + 1 points which form an oval, a contradiction as this intersection contains three distinct collinear points. So the lines Vk' belong to a common hyperbolic quadric H of P G ( 3 , qn). Also, from the foregoing reasoning it follows that \W\\ > 2. Consider an oval O'j with of] n O'j = 0. The projection O'j of O'j from u onto P G ( 3 , g") contains u', and the projection of the tangent line of O'j at O'j C\U is the line Rx containing u', tf ) , i^ 1} ,.... It follows that (O'j) contains Ri. As Ri contains at least three points of H, it belongs to H. It follows that O'j contains \W2\ > qn — 2 points of a line of H, clearly a contradiction. So if Oj ,Oj ]

C>W n of PG

(i)

are any two distinct ovals containing a point u of U, then = {u}. Let u e U and project from u onto a P G ( 3 , f ) C

B

( 4 , g ) , with u £ P G ( 3 , g n ) . Let of] ,of],...

containing u, and let Tt be the tangent line of Oj

,0{f] at u.

be the ovals Further, let

5.1. Good Eggs and Good Translation Generalized Quadrangles

111

Ti n P G ( 3 , f ) = {U} and let the projection of of\{u} belong to the line Vi, with i = l,2,...,qn. Then ij e Vi. The points ti, t2, •.., V are not necessarily distinct. Also, as Ti, T 2 , . . . , Tq™ define a common point on Lj, the points ti,t2, • • • , V are collinear. If U n PG(3,g") = {V}, then the points u',ti,t2, • •. ,tqn are on a common line 17' of P G ( 3 , qn). Let it; e U\{u}, with / = 1,2,..., qn. The tangent lines at m of the
= t2 = ••• = tq-n. Hence Ti = T2 = • • • = Tqn.

A similar conclu-

sion holds for each point m e U, with I = 1,2,..., qn. So in PG^ j ) (4, q") the nuclei of the q2n + qn ovals are on qn + 1 lines Bi,B2l..., Bqn.+1. As Bi,B2,..., Bqn+1 define all qn + 1 points of L,, these lines are mutually skew. In the tangent space r of O at 7r, the space 7r together with the nuclei of the pseudo-ovals on O containing IT form a regular (n — 1)-spread T* in T; see Remark 5.1.7 and the text preceding Lemma 5.1.8. As the nuclei of the q2n + qn pseudo-ovals are distinct, also the nuclei of the q2n + qn ovals are distinct. The extensions to G¥(qn) of the elements of T* meet n planes ££, ££> •••>££ o v e r GF( n. Now the theorem is completely proved. • As a corollary of the previous theorem we will prove in a geometric way a result which is equivalent to Johnson's Theorem 4.5.1. Theorem 5.1.11 If the TGQ T(0) of order (s, s2), with s even, is the dual of a flock GQ, then T ( 0 ) is classical.

112

Chapter 5. Good Eggs

Proof. Let T(O) be a TGQ of order (s, s2), s even, which is the dual of a flock GQ. By Theorems 4.10.1, 5.1.2 and 5.1.5 the egg O is good at some element IT = PG(n - l,q), where s = qn. Let O' be any of the pseudoovals on O through IT. By Payne [117] the point-line dual of the TGQ T ( C ) of order qn is isomorphic to a T 2 (0') of Tits, with O' some oval of PG(2, qn); the translation point of the point-line dual of T2(<5') is the line 7T of T (£>'). As the elements of O' are regular points of the point-line dual of T 2 (£>'), the GQ T(O') is a T 2 (0") of Tits, where O" is a translation oval of PG(2, s); see Payne [109] and Section 12.5 of Payne and Thas [128]. By Corollary 3.5.3 and Section. 3.6 the pseudo-oval O' is regular. Now by Theorem 5.1.10 the egg O is regular, that is, T(0) is isomorphic to a T 3 (0) of Tits, with O some ovoid of PG(3, qn). With the notation of Section 4.10, the ovoids Oz are isomorphic to the ovoid O. As T(0) is the dual of a flock GQ, all ovoids Oz are elliptic quadrics; see also the paragraph preceding Theorem 4.10.7. Hence O is an elliptic quadric, that is, T 3 (0) = T(O) is classical. • If T is a semifield flock of the quadratic cone K in PG(3, q) with q even, then, by Section 4.7, the corresponding flock GQ <S(.F) has a translation line. So by Theorem 5.1.11 S(T) is classical, so T is linear (see e.g. Payne [120, 121]). This is Johnson's Theorem 4.5.1. Theorem 5.1.11 also follows from the theorem of Johnson. Consider a flock GQ S{T) having a translation line L, with T a flock of the quadratic cone K of PG(3, q), with q even. If S(F) is not classical, then, by Payne and Thas [129], the elation point (oo) of SiJ7) is fixed by the full automorphism group of S(F). So we may assume that L contains the point (oo). By Payne [121] we have S(T) = S{T'), where in S(.T) the line L is the line [oo] (we "recoordinatize" the flock GQ). Then, by Section 4.7, T' is a semifield flock. Hence, by Theorem 4.5.1, T' is linear, and consequently, by Payne and Thas [128], the GQ S{T) = <S(.F') is classical. • We now consider TGQs T(0) of order (s, s2) for which the egg O has more than one good element. Theorem 5.1.12 If q is odd and the egg 0(n,2n,q) = O has at least two good elements, then O is classical and so the TGQ T(0) is classical. If q is even and the egg 0(n,2n,q) = O has at least four good elements, not contained in a common pseudo-oval on O, then O is regular and so T(0) is aT3{0')ofTits.

5.1. Good Eggs and Good Translation Generalized Quadrangles

113

Proof. Assume that q is odd and that the egg 0(n, 2n, q) = O has at least two good elements. Let n be a good element of O. By Corollary 5.1.4 the translation dual T(C*) of T(O) is the point-line dual of a flock GQ with elation point r, where r is the tangent space of O at TT. Let TT' e 0\{TT} be good, and let TT" e 0\{TT, TT'}. There is an elation of T(C*) with center T, mapping the tangent space r ' of O at 7r' onto the tangent space T" of O at 7r". By Theorem 3.7.2 there is an automorphism 9 of PG(4n - 1, q) D 0 , fixing 0*, fixing T and mapping r ' onto T". Clearly 0 fixes 0 , fixes 7r and maps TT' onto 7r". Hence all elements of 0 are good. By Theorem 3.10.7(ii) the egg is regular, and so by Section 3.6 the egg 0 is classical. Next, assume that q is even and that the egg 0 ( n , 2n, q) = O has at least four good elements, not contained in a common pseudo-oval on 0 . Let TT and TT' be distinct good elements of 0 . Now we consider a pseudo-oval 0 i on 0 containing TT, but not containing TT'. Let P G ( 3 n - l,q) be the subspace of P G ( 4 n - 1, q) D 0 generated by Oy. By projection of 0\{TT'} from TT' onto PG(3n - 1, q) there arises the derived plane A of 0 at TT'; see Section 5.1. Clearly 0\ is an oval of A, and so 0 i is regular. As there are four good elements, not contained in a common pseudo-oval on O, it easily follows that all pseudo-ovals on O containing TT are regular. Now by Theorem 5.1.10 the egg O is regular. • From Theorem 4.7.3 and Theorem 5.1.12, one can now deduce the following result, which is essentially contained in Bader, Lunardon and Pinneri [4]. Theorem 5.1.13 Suppose T is a Kantor-Knuth semifield flock of the quadratic cone in PG(3,g), q odd, and let S(J-)D = T(O) he the associated dual flock GQ. Let TT be a good element of O, and let TT* be a good element of O*. Then there exists an isomorphism between T(0) and T(C*) that maps TT onto TT*. If T is nonlinear, then any isomorphism between T(O) and T(C*) has this property. Moreover, in the latter situation, the good element of O* is the tangent space of O at TT. Proof. If F is linear, then T(O) ^ T(£>*) ^ Q(5,
Chapter 5. Good Eggs

114

TTS = 7r* ^ T, with r the tangent space of O at n; if IT* is tangent to 0 at ir', then, by Corollary 5.1.4, n' corresponds to the point (oo) of S(T). Let Ll(oo) correspond to the translation point of T(O). Then since S(T) is an EGQ, the stabilizer of the flag ((oo), L) in the automorphism group oiS(F) acts transitively on the points incident with L and different from (oo). It readily follows that O* contains more than one good element, contradicting the hypothesis that T is nonlinear. So -K* = T, and the theorem is proved. • The next theorem provides an interesting condition for a good TGQ to be a Kantor-Knuth GQ. Theorem 5.1.14 (Blokhuis, Lavrauw and Ball [16]) Assume that T(0) is a good TGQ of order (qn,q2n), q odd, where GF(g) is the kernel of the TGQ, with the additional condition that q > An2 - 8n + 2. Then T(O) is isomorphic to a Kantor-Knuth GQ. Sketch of Proof. Since T(£>) is good, by Corollary 5.1.4 T(0*)D is a flock GQ. By Theorem 4.7.2, T(£>*)D ^ S(T), where T is a semifield flock of the quadratic cone K, with equation X\ — X§X\ = 0 in PG(3,g"), and where the kernel of the corresponding TGQ is GF(q). In the dual space of PG(3, q), the generators of /C become a set of qn + 1 lines in the plane irv which is the dual of the vertex v of the cone. These lines are the tangents to a conic C. If the dual is taken w.r.t. the standard inner product, then C' is given by the equations X3 = X\ - 4X 0 X! = 0. In the dual space, let T be the set of qn points which corresponds to T. Since every intersection line of distinct flock planes is skew to every generator of K,, it follows that in the dual space the line joining two points of T meets nv in an internal point of C (the external points and the points of C are incident with a tangent); see also Section 4.4. Let I be the set of these internal points. With the notation of Section 4.6, and taking into account that f(t) = —zt and g(t) = yt,t e GF(qn), are additive, we have X = { ( * , - / ( * ) , 5(4), 0) || t € G F ( g n ) } .

5.1. Good Eggs and Good Translation Generalized Quadrangles

115

A nondegenerate quadratic form is either always a square on the external points of the conic it defines, and a nonsquare on the internal points, or vice versa. The quadratic form Q(X0,X1,X2)

=

X2-4X0X1

is a square on all external points of the conic C; (0,0,1) is incident with a tangent line and Q(0,0,1) = 1. So it is a nonsquare on the set of internal points I , implying that g(t)2 + 4tf(t) is a nonsquare for all t £

GF(qn).

The argument just given can be reversed, so that any pair of functions / and g which are GF(q) -linear with the property above, gives rise to a semifield flock and TGQ with kernel GF(q). There are two cases to consider now. CASE (1): J is CONTAINED IN A LINE. Then the flock planes share a common point. Theorems 4.4.5 and 4.4.6 imply that the flock is a Kantor-Knuth flock.

CASE (2): W E ARE NOT IN CASE 1. Then n > 3 and X generates the plane TTV (over GF(qn)). The set J is a P G ( n - 1, q) when interpreted over GF(g), so X contains a subplane of order q of nv which generates nv and which is contained in the set of internal points of the conic C.

The following lemma of Weil can be found in [135]. Lemma 5.1.15 (Weil) The number N of solutions in GF(q) of the hyperelliptic equation Y2 = g(X), where g e GF(q)[X] is not a square and has degree 2m > 2, satisfies \N-q+l\<

(2m-2)y/q.

BlokhuiSi Lavrauw and Ball [16] use Lemma 5.1.15 to obtain

Chapter 5. Good Eggs

116

Lemma 5.1.16 (Blokhuis, Lavrauw and Ball [16]) Let f(X) = X2 + uX+v e GF(qn)[X] be a nonzero square in GF(qn)for all X = x e GF(q), where q is odd and q > An2 — 8n + 2. At least one of the following holds: (i) f(X) is the square of a linear polynomial; (ii) n is even and f has two distinct roots in

GF(qn/2);

(iii) the roots of f are a and aa for some a £ Gal(GF(g")/GF(g)), where a e GF(qn). • Using the latter lemma, it is then shown that when q > An2 — 8n + 2, no subplanes of order q can be contained in the internal points of a conic in PG(2,g"). •

5.2

Good Eggs and Veronese Surfaces

There is a strong connection between good eggs and the Veronese surface V| of all conies of PG(2,g). So we include a short section on Veronese varieties; a good reference is Chapter 25 of Hirschfeld and Thas [74]. The Veronese variety of all quadrics of PG(n, K), n > 1 and K any commutative field, is the variety V = {(XQ, XJ,...,X1,

(x0,xi,...

X0Xi,X0X2,

. . . , X0Xn,

XxX2, • • • , XxXn, . . . , Xn-iXn)

7xn) is a point of PG(n,K)}

of PG(iV,K) with N = n(n + 3)/2. The variety V has dimension n and order 2 n [74]; for V we also write Vn or V2". It is also called the Veronesean ofquadrics of PG(n, K), or simply the quadric Veronesean of PG(n, K). It can be shown that the quadric Veronesean is absolutely irreducible and nonsingular [74]. Let PG(7V, K), with TV = n(n + 3)/2, consist of all points (yoo ) 2/11) • • • ; 2/nn) 2/01) 2/02) • • • ) 2/Ori) 2/12) • • • j 2/ln) • • • )

Vn—l,nji

for yij we also write y^. Let C, : PG(n,K) ^ PG(iV,K), with n > 1, be defined by (xQ,xi,...,xn) H^ {yoo,yn,...,yn-i,n), with y^ = XiXj. Then £ is a bijection of PG(n, K) onto the quadric Veronesean V of PG(n, K). It then follows that the variety V is rational.

\\

5.2. Good Eggs and Veronese Surfaces

117

The quadrics of PG(n, K) are mapped by £ onto all hyperplane sections of V. It follows that no hyperplane of PG(iV, K) contains the quadric Veronesean V. The lines of PG(n, K) are mapped onto irreducible conies of V; for K ^ GF(2) each irreducible conic of V is the image of a line of PG(n, K). Hence for |K| > 2 any two distinct points of V are contained in a unique irreducible conic of V. If K = GF(q), then it is clear that Vn contains 9(n) = qn + qn-1-\ \-q + l points. As no three points of Vn are collinear, the quadric Veronesean Vn is a 0(n)-cap of PG(iV, q), with N = n(n + 3)/2. Let n = 2 and assume from now on that K = GF(q). Then V is a surface of order 4 in PG(5, K). Apart from Vf, which is a conic, the variety V| is the quadric Veronesean which is most studied and characterized. With the conies (irreducible or not) of PG(2,q) correspond all hyperplanes sections of V|. The hyperplane is uniquely determined by the conic if and only if the latter is not a single point. If the conic of PG(2, q) is one line, then the corresponding hyperplane of PG(5, q) meets V| in an irreducible conic; the surface V| contains no other irreducible conies. It follows that V| contains exactly q2 + q + 1 irreducible conies, that any two distinct points of V| are contained in a unique irreducible conic of V|, and that any two distinct irreducible conies on V| meet in a unique point. If the conic C of PG(2, q) is two distinct lines, then the corresponding hyperplane PG(4, q) meets V| in two irreducible conies with exactly one point in common; if C is irreducible, then PG(4, q) meets V| in a rational quartic curve. The planes of PG(5, q) which meet V| in an irreducible conic are called the conic planes of V|. Any two distinct conic planes of V| have exactly one point in common, and this common point belongs to V|. The tangent lines of the irreducible conies of V| are called the tangents or tangent lines of V|. Since no point of the surface V| is singular, all tangent lines of V| at the point p of V| are contained in a plane ir(p). This plane ir(p) is called the tangent plane of V| at p. Since p is contained in exactly q + 1 irreducible conies of V| and since no two conic planes through p have a line in common, the tangent plane 7r(p) is the union of the q+1 tangent lines of V| through p. Also V|ri7r(p) = {p}. For any two distinct points pi and p 2 of V|, the tangent planes ir(pi) and TT(J>2) have exactly one point in common. If C is a nonsingular conic on \>2, if 7T is the plane of C and if p' e V|\C, then ir(p') n n = 0. Suppose that q is odd. Then PG(5, q) admits a polarity that maps the set of all conic planes of V| onto the set of all tangent planes of V| • Suppose that q is even. Then the nuclei of the q2 + q + 1 irreducible conies

118

Chapter 5. Good Eggs

on Vf are the points of a plane PG(2, q), called the nucleus or kernel of V|. In Thas [180, 183], the following classification of good eggs is obtained. Theorem 5.2.1 Consider a TGQ T(n,2n,q) = T(0), q odd, with O = 0(n, 2n, q) = {TT, TTI , . . . , irq2n } . If O is good at IT, then we have one of the following. (a) There exists a P G ( 3 , qn) in the extension PG(4n - 1, qn) of the space PG(4n - 1, q) of 0(n, 2n, q) which has exactly one point in common with the extension 7Fj of 7r, to GF(qn), with i = 1,2,..., q2n. The set of these q2n + 1 points is an elliptic quadric of P G ( 3 , qn) and T(O) is isomorphic to the classical GQ Q(5, qn). (b) We are not in Case (a) and there exists a PG(4,g") in PG(4n 1, qn) which intersects the extension W of IT to GF(qn) in a line M and which has exactly one point ri in common with each space TTI, with i = 1,2,..., q2n. Let W = {n \\ i = 1,2,..., q2n} and let M be the set of all common points of M and the conies which contain exactly qn points of W. Then the set W U M is the projection of a quadric Veronesean V| from a point pin a conic plane of V| onto a hyperplane PG(4, qn); the point p is an exterior point of the conic o / V | in the conic plane. Abo, the point-line dual ofT(0) is isomorphic to the flock GQ S(T) with T a Kantor-Knuth flock, that is, O is a Kantor-Knuth egg. (c) We are not in Cases (a) and (b) and there exists a P G ( 5 , g n ) in PG(4n - 1, qn) which intersects if in a plane /x and which has exactly one point n in common with each space Wi, with i — 1,2,..., q2n. Let W = {ri || i = 1,2,..., q2n} and let C be the set of all common points of \x and the conies which contain exactly qn points of W. Then the set W U C is a quadric Veronesean in P G ( 5 , qn). • Thas [180] contains a long technical proof of Theorem 5.2.1, buth with a weaker version of Case (b); Case (b) could be improved relying on Thas [183], see Section 9 of this paper. An alternative proof is contained in Lavrauw and Lunardon [96]. Remark 5.2.2 If we project the Veronesean V | in PG(5,g r a ) onto some PG(3,g") c P G ( 5 , g n ) from a line TV that intersects V| in two points of PG(5, q2n) \ P G ( 5 , qn) which are conjugate with respect to the extension GF(g 2 n ) of GF(g"), then we obtain an elliptic quadric of P G ( 3 , qn).

5.2. Good Eggs and Veronese Surfaces

119

If the egg 0(n, 2n, q) = O, with q odd, is good at its element n, then, by Corollary 5.1.4, the translation dual T(C*) of T(O) is the point-line dual of a flock GQ S {J7) with elation point T, where r is the tangent space of O at 7r. The point (oo) of T(£>*) is a translation line of S{T). We may assume that (oo) is the line [oo] of <S(JF) and then T is a semifield flock; see Section 4.7 and Payne and Rogers [125]. Conversely, if T is a semifield flock, then, by Section 4.7 and Corollary 5.1.4, S{T) = T(0) where O* is good at its element r which corresponds to the elation point of T. If J 7 is a Kantor-Knuth flock which is not linear, then S{T) = T(O) ^ T(C*) with O of Type (b) in Theorem 5.2.1; see Theorem 4.7.3. If T is a Ganley flock or the Penttila-Williams-Bader-Lunardon-Pinneri flock, then S{F) = T(0*) with O of Type (c) in Theorem 5.2.1. If the TGQ T(O) of order (s 2 , s), with s odd, is good and the kernel has either size s or size
120

Chapter 5. Good Eggs

exactly qn + 1 elements of O. If m, with i e {3,4}, does not belong to = PG(3n — 1, g) contains -K, a contradiction. So O is a Kantor-Knuth egg. Next, assume that the subspace PG(3n - 1, q) of PG(4n - 1, q) contains at least five elements of 0\{ir}, say ni, 7T2,..., 7r5. We rely again on Theorem 5.2.1. Assume, by way of contradiction, that O is not classical. Then, by the first part of the proof, O is a nonclassical Kantor-Knuth egg. Let W n 7TJ = {r,}, with i = 1,2, . . . , 5 . Clearly (ri,r 2 ,... ,r 5 ) ^ PG(4,q n ), as otherwise O is contained in PG(3n — 1,q). Now suppose that (r\,rz,...,r$) = PG(3, qn). Then, with the notation of Theorem 5.2.1, M n PG(3, gn) ^ 0, and as in the previous section we find that IT c PG(3n - l,q), a contradiction. Consequently ri, r2,..., rs are contained in a common plane PG(2,qn). Clearly PG(2,g n ) n M = 0, as otherwise TT C PG(3n - l,q). The set W U jVl is the projection of a quadric Veronesean V| from a point p in a conic plane 7 of V| onto a hyperplane PG(4,g n ); the point p is an exterior point of the conic C of V| in 7. Let n be the projection of the point r[ e V|, with i = 1,2,..., 5. If (rlr'z, ...,r'5)= PG'(2, g"), then r'1; r'2,..., r'5 are contained in a common conic of V|, hence C n PG'(2, g") ^ 0, and so PG(2,g") n M ^ 0, a contradiction. Consequently, ( r i , ^ , . . • ,7-5) = PG'(3,g n ), withp G PG'(3,g n ). Clearly PG'(3,g n ) n 7 = {p}, as otherwise PG(2,g n ) C\M ^ 0. As ir ^ PG(3n - l,g), no three points of {r'i,r'2,.. .,r'5} are contained in a conic plane of V|. Let £ be the bijection of a plane PG* (2, gn) onto V| which maps the lines of PG* (2, qn) onto the conies of V|. Further, let C'c = L, {r1^1 = pi} with i = 1,2,... ,5. Then no three points of {pi,p2, • • • ,Ps} are collinear. So pi,p 2) • • • ,Ps belong to just one conic C" of PG*(2, qn). It follows that r[,r'2,..., r'5 generate a hyperplane of PG(5, qn) D V|; see page 117. So we have a contradiction, and conclude that O is classical. • Remark 5.2.4 (a) If O is a nonclassical Kantor-Knuth egg of PG(4n - 1 , q), with q odd, then from Thas [180] it easily follows that there exist (3ra - 1)-dimensional subspaces in PG(4n - l,q) which contain exactly four elements of O. (b) Independently and in a completely different way, Lavrauw [94] also proved the last part of Theorem 5.2.3.

5.3.

Coordinatization and Applications

121

5.3 Coordinatization and Applications We now introduce a coordinatization method of eggs and their duals which was first used by Payne in [118], and later on generalized by Lavrauw and Penttila [97].

5.3.1 Coordinatization of Eggs and Dual Eggs The next theorem states the connection between additive g™-clans and eggs, and is taken from [97]. We use the notation of §4.7.2.

Theorem 5.3.1 The set C = {At = ( *

Vt

J \\ t £ GF(qn)}

is an additive

qn-clan if and only if the set O = {TT(A) || A G G F ( g n ) 2 U {oo}}, with TT(A)

= {(*, -gt(X), -A 5 ') || t G GF(qn)},

A G GF(
and TT(OO) =

{(0,*,(0,0)) I)

teGF(qn)},

where the triples defining w(X) and 7r(oo) are considered as vectors of the in-dimensional vectorspace over GF(q), is an egg o / P G ( 4 n — l,q), where gt(A) = A^4tAT and XSt = X(At + AT). The tangent space at the element n(X), X E GF(qn)2, is r(A) = {(*, PXT + A5'AT - gt(X), p)\\te

GF(qn), (3 e G F ( ^ ) 2 } ,

and the tangent space at IT{oo) is r(oo) = {(0,i,/3) || t G GF(qn),(3 G GF(g") 2 }.

• For the rest of this section, it is convenient to write F instead of

GF(qn).

If At is additive, so defines an egg in P G ( 4 n — l,q), we can write At as

A

>=t (-:)"'

Chapter 5. Good Eggs

122

with (a0, a i , . . . , a„_i) = (1,0,..., 0), for certain bt, Ci £ F. If an egg O is written in this form, we denote it by 0{b,c) where b = {bo,bi,... , 6„-i) and c = (c 0 ,ci,... ,c„_i). If O is as such, it is easy to explicitly write down the elements of the dual egg, and the tangent spaces to the latter. This was first done by Lavrauw and Penttila [97] in its full generality, based on observations of Payne in [118], who carried out the calculations for the eggs corresponding to the Roman TGQs. We start with a lemma. Lemma 5.3.2 Let tr be the trace map from F to GF(q). Then /n-l

\

tr Y; "**** for allt <E F if and only if n-l

E°r1-i=°i=0

Proof. Since the trace function is additive and tr(a;) = tr(a;'?) for x e F, we have that

5>r-> B

fn-l

tr J2 ^

}=

tr

\i=Q

vi=0

The fact that tr(ax) = 0, Vx e F implies that a = 0, ends the proof.

•

Theorem 5.3.3 The elements of the translation dual 0*(b,c), and the corresponding tangent spaces, of the egg 0(b, c) are given by TT*(A)

= {{-9*tW,t, -Ai) || t G F}, A E F 2 , 7T* (C») = {(«, 0 , ( 0 , 0 ) ) | | t 6 F } ,

r*(A) = {(/*(/?, A) +g*t (A), t,/3) \\teW,0£

F 2 }, A e F 2 ,

r * M = {(t,0,/3)||ieF,/3GF 2 }, with

n-l

2 2 1 1 9t (a, 6) = £ ( a i a + bia& + Cift ) /'** /'* i=0

5.3. Coordinatization and Applications

123

with (a 0 , ai, • • •, a„_i) = ( 1 , 0 , . . . , 0) and n-l

f*((a, b), (c, d)) = ^ ( 2 a i a c + &i(ad + 6c) + 2ci&d)1/
Proof. We calculate the vector space dual of 7r(A) and r(A) in V(4n, q) with respect to the standard inproduct ((x, y, 2;, w), (a;', ?/', 2:', to')) i-» Xx{xx' + yy' + zz' + ww'), with tr the trace map from F onto GF(g). If (x, y, z, w) is an element of the vector space dual of r( A), thentr[xt+y((3\T+\St \T -gt(\)) + (z,w)/3T] = 0, 2 for all i e F and all (3 e F . With A = (a, 6) and j3 = (c, d), we obtain tr[xi + y(ac + bd + XStXT - gt(X)) +zc + wd] = 0,

(5.1)

for all c,d,t e F. When t = 0, this equation is satisfied for all c, d e F if and only if w = -by and z = -ay. Substituting this back into Equation (5.1), we obtain that tr[a;i + y(XStXT - gt{X))} = 0 for all t e F. Using the formulas for gt and St this is equivalent to n-l

tr[(x + y{a0a2 + b0ab + c0b2))t + ^

y(aia2 + btab + Cib2)tq%] = 0,

i=l

for all t e F. By Lemma 5.3.2, it follows that (x, y, 2;, to) is of the form n-l

( - ^ ( a , a 2 + 6,06 +

Ci 6

2 1/
)

i

,t, - a t , -bt),

i=0

for some i e F . The tangent spaces are handled in the same way.

5.3.2

•

The Known Examples Revisited

Using Theorem 5.3.3, we now write down, for q odd, the known examples of good eggs, and their translation duals; for q even each known egg arises by field reduction either from an elliptic quadric of P G ( 3 , q) or from a Tits ovoid. We only provide elements of type 7r(A) and 7r*(A), A e F = GF(q"), as the elements 7r(oo) and 7r*(oo) are all of the form {(0, t, 0,0) II t e F} and {(£, 0,0,0) || t e F} in this representation.

Chapter 5. Good Eggs

124

•

In this case, the egg is isomorphic to its translation dual; see Payne [118]. Let q be odd, a an automorphism of F, and m a given nonsquare in F. Then KANTOR-KNUTH EGGS.

TT(A) = {(t, -a2t + mb2ta, -2at, 2bmta) \\ t G F} and TT*(A) = { ( - a 2 i + ( m 6 2 ) C T _ V ~ \ t , -at,-bt)

•

\\

te¥}.

GANLEY AND ROMAN EGGS. For q > 9, the Ganley eggs are not isomorphic to their translation dual; see Payne [118]. They exist for q = 3h, h > 1. Let m be a nonsquare in F. Then TI-(A)

= {(*, - ( a 2 - b2m)t - abt3 + b2m~H^, at - bt3, -at3 + m~H9))

-b(mt

\\teW}

and 7r*(A) = { ( - ( a 2 - b2m)t -

{abf'3tll3

+ (^m" 1 ) 1 / 9 * 1 / 9 , t, -at, -bt) || * G F}. • THE PENTTILA-WILLIAMS-BADER-LUNARDON-PINNERI TRANSLATION DUAL.

EGG AND ITS

The Penttila-Williams-Bader-Lunardon-Pinneri

egg is not isomorphic to its translation dual. Here q = 3 5 . Then 7r(A) = {(t, ~a2t - abt3 + b2t27, at - bt3, -at3 - bt27)

\\te¥}

and ,r*(A) = {(-a2t

- {abflh1'3

+ {b2)l'27t1'27,

t, -at, -bt) \\ t G F}.

5.3.3 A Geometric Connection between Semifield Flocks and Good Eggs Let J 7 be a semifield flock of the quadratic cone K with vertex v in P G ( 3 , qn), q odd. If one dualizes in P G ( 3 , qn), the qn flock planes become the affine points of an n-dimensional affine space AG(n,q) over GF(q). The (n — 1)-dimensional space at infinity P G ( n - 1, q) of AG(n, q) is contained in the set of internal points of the conic C, that corresponds to the

5.3. Coordinatization and Applications

125

set of generators of /C, in the plane TTV corresponding to v. Over GF(q) the conic C becomes a pseudo-conic C in the space P G ( 3 n — l,q) which corresponds to 7r„. Dualizing in P G ( 3 n - 1, q), the space P G ( n - 1, q) becomes a (2n — l)-space PG(2n — 1, q) skew to the elements of the pseudo-conic C obtained by dualizing C. Now suppose O is a good egg in P G ( 4 n — l,q) = (O), q odd. Suppose O' is one of the pseudo-conies containing the good element TT (see Theorem 5.1.9), and let TT' e O \ O'. Suppose T' is the tangent space of O atir'; then T' n (£>') = /3 is a (2n - l)-space in P G ( 3 n - 1, q) = (£>') skew to £>'. Now reverse the argument of the first paragraph to obtain a semineld flock. In [95] Lavrauw proves that such a semifield flock is isomorphic to the semifield flock defining the good egg O.

5.3.4

Further Characterizations of Good Eggs in Odd Characteristic

The following interesting theorems are due to Lavrauw [94]. Theorem 5.3.4 Let O he an egg o / P G ( 4 n - 1, q), q odd, which is good at TT. Then the following properties are equivalent. (i) O is classical. (ii) There exists a triple (ir',w",TT'"), with TT',TT",TT'" distinct elements of 0\{n} and where n' is not contained in the pseudo-conic O' of O in (IT, TT", TT'"), such that the tangent space of O at n' contains two distinct elements of the regular (n - l)-spread of (w, TT", IT'"} which contains O'. (iii) Each triple (TT', n", TT'"), with n', IT", TT'" distinct elements ofO\{ir} and where TT' is not contained in the pseudo-conic O' of O in (n, IT", TT'"), is such that the tangent space of O at TT' contains two distinct elements of the regular (n - l)-spread of (TT, TT", TT'"} which contains O'. Proof. Clearly (i) implies (iii), and (iii) implies (ii). It remains to show that (ii) implies (i). So assume that (ii) is satisfied. By Corollary 5.1.4 and Section 4.7 the translation dual O* of O defines a semifield flock T. In the (2n - 1)-dimensional space (TT, TT", TT'") n r ' = £, with T' the tangent space of O at TT', a (n - 1)spread is induced by the regular (n - 1)-spread defined by O' (the space £

126

Chapter 5. Good Eggs

is a line of the PG(2, qn) defined by the regular (n - l)-spread). It follows that the semifield flock T is linear, and so O is classical. • Theorem 5.3.5 Let O be an egg o / P G ( 4 n — \,q),q odd, which is good at TT. Then the following properties are equivalent. (i) O is a Kantor-Knuth egg. (ii) There exists a triple (TT',TT",TT'"), with TT'' ,TT"\TT'" distinct elements of 0\{TT} and where TT' is not contained in the pseudo-conic O' of O in (TT, TT", TT'"), such that the tangent space of O at TT' contains an element C of the regular (n - l)-spread of (TT, TT", TT'") which contains O'. (hi) Each triple (TT', TT", TT"'), with TT', TT", TT'" distinct elements ofO\{Tr} and where TT' is not contained in the pseudo-conic O' of O in {TT, TT",TT'"), is such that the tangent space of O at TT' contains an element ( of the regular (n - l)-spread of (TT, TT", TT'"} which contains O'. Proof. Assume that O is a Kantor-Knuth egg and let T be the corresponding flock. Let TT',TT",TT'" be distinct elements of 0\{7r} where TT' is not contained in the pseudo-conic O' of O in {TT,TT",TT'"). By Theorem 4.4.6 all planes of the elements of T contain a common point. It follows that (n, TT", TT'") n T' = £, with r ' the tangent space of O at TT', contains an element C of the regular (n - l)-spread denned by O'. Hence (i) implies (iii). Since (iii) obviously implies (ii), it remains to show that (ii) implies (i). So assume that (ii) is satisfied. The egg O* defines a semifield flock T. Also, all planes of the elements of T contain a common point. By Theorem 4.4.6 the flock T is a Kantor-Knuth flock, and so O = O* is a Kantor-Knuth flock. • Remark 5.3.6 If O is not classical, then the element ( in Theorem 5.3.5 defines an exterior point of the conic C of P G ( 2 , g n ) defined by the pseudoconic O'; see Theorem 4.4.6. Theorem 5.3.7 Let O he an egg in PG(4n - 1, q), q odd, and let r\ he the subspace generated by three distinct egg elements. (i) If O is a Kantor-Knuth egg, then any element TT' of O is either contained in r\ or intersects r\ in at most one point. (ii) If O is a Ganley-Payne egg, then any element TT' of O is either contained in rj or intersects r\ in at most a line.

5.3. Coordinatization and Applications

127

(iii) If O is the translation dual of the Penttila-Williams-Bader-LunardonPinneri egg, then any element IT' of O is either contained in n or intersects n in at most a line. Proof. Using coordinates, see Lavrauw [94].

•

Chapter 6

Generalized Quadrangles, Nets and the Axiom of Veblen Each regular point of a thick GQ <S of order (s, t) defines a dual net N® • If TVjP is not a dual affine plane and satisfies the Axiom of Veblen, then M® is isomorphic to the "classical" dual net H™, for some q and n. In such a case strong properties for the GQ S can be deduced. In this chapter it is also shown that the existence of particular automorphisms of A/jP, where the regular line L is incident with some coregular point x of S, forces S to be a TGQ.

6.1 Generalized Quadrangles and the Axiom of Veblen Consider a GQ T 3 (C) of Tits, with O an ovoid of PG(3,g). Here s = q and t = q2. Then the point (oo) is coregular, that is, each line incident with (oo) is regular. It is an easy exercise to check that for each line incident with (oo) the corresponding dual net is isomorphic to H^; see Section 2.2 for the definition of H | . Hence for each line incident with the point (oo) the corresponding dual net satisfies the Axiom of Veblen. We now prove the converse. Theorem 6.1.1 and its corollary are due to Thas and Van Maldeghem [193]. Theorem 6.1.1 Let S = {V, B, I) be a GQ of order (s, t), with s ± t, s > 1 and t > 1. If S has a coregular point x and if for each line L incident with x the corresponding dual net N® satisfies the Axiom of Veblen, then S is isomorphic to a T 3 (O) of Tits. 129

130

Chapter 6. Generalized Quadrangles, Nets and the Axiom ofVeblen

Proof. Let Li,L2, L3 be three lines no two of which are concurrent, let Mi, M 2 , M 3 be three lines no two of which are concurrent, let Li / M, if and only if {i,j} = {1,2} and assume that xlL\. By 5.3.8 of Payne and Thas [128] it is sufficient to prove that for any line L4 e {Mi,M2}± with Li Li,i = 1,2,3, there exists a line M 4 concurrent with L1,L2,L4. So let L4 e {Mi,M 2 }- L with L4 / Li} i = 1,2,3. Consider the line R containing L2 n M 2 and concurrent with L\. Further, consider the line R' containing M2 n L4 and concurrent with L\. By the regularity of Li there is a line R" e {Mi, Ms}- 1 - 1 through the point L 3 n M 2 . Clearly the lines Lx and .R" are concurrent. So the line L\ is concurrent with the lines R, R', R"; also the line M 2 is concurrent with the lines R, R', R". By the regularity of Li the line R" belongs to the line {R, R'}1-1 of the dual net A ^ denned by L1. Hence the lines {R, R'}-1-1 and {Mi,M3}xlof A ^ have the element R" in common. By the Axiom ofVeblen, also the lines {Mi,i?'} ± J - and {M3,R}±Jof A/j^ have an element M 4 in common. Consequently M 4 is concurrent with Li,L2,L4. Now fro m 5.3.8 of Payne and Thas [128] it follows that S is isomorphic to a T 3 ( 0 ) of Tits. • As a corollary the following characterization of the classical GQ Q(5, s) is obtained. Corollary 6.1.2 Let S be a GQ of order (s, t) with s ^ t, s > 1 and t > 1. (i) J/s is odd, then <S is isomorphic to the classical GQ Q(5, s) if and only if it has a coregular point x and if for each line L incident with x the corresponding dual net A/jP satisfies the Axiom of Veblen. (ii) If s is even, then S is isomorphic to the classical GQ Q(5, s) if and only if all its lines are regular and for at least one point x and all lines L incident with x the dual nets A/JP satisfy the Axiom of Veblen. Proof. Let (x, L) be a flag of the GQ Q(5, s). By 3.2.4 of Payne and Thas [128] there is an isomorphism of Q(5,s) onto T 3 (C), with O an elliptic quadric of P G ( 3 , s), which maps x onto the point (oo). It follows that A/JP satisfies the Axiom of Veblen. Conversely, assume that the GQ S of order (s, t), with s odd, s / (, s > 1 and t > 1, has a coregular point x such that for each line L incident with x the dual net A/p1 satisfies the Axiom ofVeblen. Then by Theorem 6.1.1 the GQ 5 is isomorphic to T 3 (C). By Barlotti [8] and Panella [106] each ovoid

6.2. Translation Generalized Quadrangles and the Axiom ofVeblen

131

O of P G ( 3 , s), with s odd, is an elliptic quadric. Now by Theorem 1.5.2(ii) we have S a T 3 (C) ^ Q(5, s). Finally, assume that for the GQ S of order (s, t), with s even, s / t, t > 1, all lines are regular and that for at least one point x and all lines L incident with x the dual nets MP satisfy the Axiom of Veblen. Then by Theorem 6.1.1 the GQ S is isomorphic to T 3 (C). Since all lines of S are regular, by §2.1 we finally have S =* T 3 (C) =* Q(5, s). •

6.2

Translation Generalized Quadrangles and the Axiom of Veblen

This section is also taken from Thas and Van Maldeghem [193]. It explains the relationship between the Axiom ofVeblen and good eggs. Theorem 6.2.1 Let S^ he a TGQ of order (s, s2), s / 1, with base-point x. Then the dual net MP defined by the regular line L, with xlL, satisfies the Axiom ofVeblen if and only if the egg 0(n, 2n, q) which corresponds to S^ is good at its element -Ki which corresponds to L. Proof. Assume that the dual net MP satisfies the Axiom of Veblen. Let the egg 0(n, 2n, q) correspond to S^ and let •K1 correspond to L. We have s = qn. The dual net has qn + 1 points on a line and q2n lines through a point. By Theorem 2.2.3 the dual net MP is isomorphic to H;j„. Consider the TGQ T(n, 2n, q) = 5^) and let PG(3n, q) be a subspace skew to •Ki in the projective space PG(4n,q) in which T(n,2n,q) is defined. Let 0(n,2n,q) = {7r0,7Ti,... ,Tcq2n}, let (7r,,7rj) n PG(3n, q) = TT'J for all j / i (TT'J is (n-l)-dimensional), let P G ( 4 n - l , 9 ) n P G ( 3 n , g ) = P G ( 3 n ~ l , g ) , with PG(4n - 1, q) the space of 0(n, 2n, q), and let rt n PG(3n, q) = T[, with Ti the tangent space of 0(n, 2n, q) at Hi {r[ is (2n - 1)-dimensional). Then the dual net MP is isomorphic to the following dual net MD: • POINTS of MD are the q2n spaces -K'J, j ^ i, and the q3n points of PG(3n,q) \PG(3n-l,q); • LINES of MD are the q4n subspaces of PG(3n, q) of dimension n which are not contained in PG(3n - 1, q) and contain an element it'p j ^ i, and •

INCIDENCE

is the natural one.

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Clearly the points TT'J, j ^ i, of ND form a parallel class of points. Let M be a line of ND incident with 7^ and let it'k ^ ir'j, with k ^ i ^ j . As J\fD = H 3 „ the elements -n'k and M of MD generate a dual affine plane AD in AfD', and the plane AD contains qn points ix[,l ^i. Clearly the points of AD not of type ir[ are the q2n points of the subspace (ir'k, M) of PG(3n, q) which are not contained in P G ( 3 n -l,q). Hence the qn points of AD of type -K\ are contained in (ir'k,M) n P G ( 3 n - 1, q). It follows that these qn elements 7r,' are contained in a (2n - 1)-dimensional space PG'(2n - 1, q); also, they form a partition of P G ' ( 2 n - 1, q)\r!-. Consequently for any two elements -K[^'V, I ^ i ^ I', the space (7^,71-;',) contains exactly q n elements Tv'r, r ^ i. Hence for any two spaces 717 and -KV of 0(n, 2n,q , )\{7r I }, the (3ra - 1)-dimensional space (71-4,717,717/) contains exactly <jn + 1 elements of 0(n, 2n, 5). We conclude that 0(n, 2n, q) is good at 717. Conversely, assume that 0(n, 2n, q) is good at the element 717 which corresponds to L. As in the first part of the proof we project onto a PG(3n, q) and we use the same notation. Since 0(n, 2n, q) is good at 717, for any two elements 7Tj,7r[,, with I ^ % ±V', the space (71-;, 7^,) contains exactly qn elements n'r, r ^ i; these qn elements form a partition of the points of (717', ix'v) which are not contained in T[. If M, M' are distinct concurrent lines ofAfD, then it is easily checked that M and M' generate a dual affine plane AD of order qn in AfD. As AD satisfies the Axiom of Veblen, also MD does. • Corollary 6.2.2 Let S^ be a TGQ of order (s,s2), s odd and s ^ 1, with base-point x. If the dual net N® defined by some regular line L, with xlL, satisfies the Axiom of Veblen, then S^ contains s 3 + s2 classical subquadrangles Q(4, s) containing L. Proof.

This follows immediately from Theorem 6.2.1 and Theorem 5.1.9.

6.3 Property (G) and the Axiom of Veblen Property (G) for GQs of order (s, s 2 ) was introduced in Section 4.10. If the GQ <S of order (s, s 2 ), s / 1 , satisfies Property (G) at the line L, then we shall prove that L is regular and shall consider the dual net Af® in some detail. Theorem 6.3.1 Let S = (P, B, I) be a GQ of order (s, s2), s ^ 1, satisfying Property (G) at the line L. Then L is regular.

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133

Proof. Let S = (V, B, I) be a GQ of order (s, s2), s even, satisfying Property (G) at the line L. Consider distinct points x and y incident with L, and let x be a center of the triad {y, z, u}. By Theorem 2.6.2, S has a subquadrangle S' of order s which contains {y, z, u}1- U {y, z, u}±J-. Let w be a point of <S', with w / i and w •/ y. Further, let r, r', r" be distinct points of {x, w}^' (here " 1 " ' denotes "perp" in <S'), with rlL. Then {r, r', r"}- 1 and {r, r', r"}^1- are contained in a subquadrangle of order s of 5. Clearly the lines xr,xr',xr",wr,wr',wr" are common lines of S' and 5". By 2.3.1 of Payne and Thas [128], see also Section 1.1, the common points and lines of S' and S" form a subquadrangle S' n <S" of order (s,t'), with £' > 1, of S". Now Lemma 1.1.1 readily implies that S' n S" = S' = 5". It follows that {r,^}- 1 '- 1 ' = {r,r',r"}±±, {r,^}1' = {r,r',r"}x, and so {r,r'} and {x, w} are regular in S'. Hence x is regular in S'. With the notation of the previous paragraph and interchanging the roles of x and y, we see that also y is regular in S'. Since in the previous paragraph y and r play the same role in S', also r is regular in S'. It follows immediately that all points of L are regular in <S'. By Theorem 2.1.5 also the line L is regular in <S'. Consider now a line M of S, with M ^ L. Let y ~ x'lM, x ~ y'lM, and x" e {y,y'}-L\{a;,a;'}. The subquadrangle 5 of order s defined by x, x', x" contains L and M. Since L is regular in <S, the span {L, M}-11- has size s + 1. We conclude that L is regular in S. Next assume that s is odd, s ^ 1, and that the GQ <S = (V, B, I) of order (s2, s) satisfies Property (G) at the point x. By Theorem 4.10.6 the GQ S is a flock GQ. Now we consider the construction of Knarr, see Section 4.9. If the point y is not collinear with x, that is, if y is a point of PG(5, s) not in x^, with £ the symplectic polarity used to construct S in PG(5, s), then {x, y}-11- consists of the s + 1 points of the line (x, y) of PG(5, s). As |{x, y}"1"^! — s + 1 the point x is regular. • Remark 6.3.2 In the first part of the proof, where in the even case we show that x is regular in S', it is sufficient to assume that S satisfies Property (G) at the flag (x,L). By Theorem 4.10.7 it is sufficient in the odd case to assume that <S satisfies Property (G) at some point pair {x, y}, with xlLly. In fact, for all s ^ 1, the GQ S of order (s, s2) is the point-line dual of a flock GQ. Hence L is an axis of symmetry (cf. Section 4.6), and consequently regular. The following theorems of this section are also taken from Thas and Van Maldeghem [193].

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Theorem 6.3.3 Let S = (V, B, I) be a GQ of order (s2,s), s even, satisfying Property (G) at the point x. Then the dual net Mx satisfies the Axiom of Veblen, and consequently Mx = H^. Proof. Let S = (V, B, I) be a GQ of order (s 2 , s), s even, satisfying Property (G) at the point x. By Theorem 6.3.1 the point x is regular. Let y be a point of the dual net Mx, let A\ and A2 be distinct lines of Mx containing y, let B\ and B2 be distinct lines of Mx not containing y, and let Ai n Bj ^ 0 for all i,j e {1,2}. Let {z} = Ax n Bx and let zlM, with x IM. Further, let xlL, with z IL, let u be the point of A\ on L, and let u' be the point of B\ on L. The line of S incident with u, respectively u', and concurrent with M is denoted by C, respectively C; the line incident with z and a; is denoted by N. Since <S satisfies Property (G) at x, the triple {C, C, N} is 3-regular. By Theorem 2.6.2 the lines of S concurrent with at least one line of {C, C", N}1- and one line of {C, C", A^}-1-1-, are the lines of a subquadrangle S' of order s of 5. As x is regular for S it is also regular for S'. By Theorem 2.2.1 the point x in S' defines a projective plane wx of order s. Clearly A\,A2, B\,B2 are lines of the projective plane TXX. Hence B\ and B2 intersect in irx. Consequently Nx satisfies the Axiom ofVeblen, and so N° = H^ by Theorem 2.2.3. • Theorem 6.3.4 Let S^ = T(O) be a TGQ of order (s, s2), s / 1, with base-point x. Then the dual net N® defined by the regular line L, with xlL, satisfies the Axiom ofVeblen if and only if the translation dual <S'(x -1 = T(C*) of S^ satisfies Property (G) at the flag {x' ,L'\ where L' corresponds to L, that is, where L' is the tangent space of O at L. In the even case, A/jP satisfies the Axiom ofVeblen if and only ifS^ satisfies Property (G) at the flag (x, L). Proof. By Theorem 6.2.1 the dual net Af® satisfies the Axiom ofVeblen if and only if O = 0(n, 2n, q) is good at the element L. By Theorem 5.1.2 the egg O is good at L if and only if T(0*) satisfies Property (G) at the flag (x', V). By Theorem 5.1.5, for s even, T(0*) satisfies Property (G) at the flag (x', V) if and only if T(0) satisfies Property (G) at the flag (x, L). • Theorem 6.3.5 Let S^ be a TGQ of order {s,s2), s odd and s ± I, with base-point x. If xlL and if the dual net N® satisfies the Axiom of Veblen, then all lines concurrent with L are regular. Proof. Let N be concurrent with L, x IN, and let the line M of S^ be nonconcurrent with N. From Theorem 5.1.9 easily follows that the lines

6.4. Flock Generalized Quadrangles and the Axiom ofVeblen

135

M,N are lines of a subquadrangle of S^ isomorphic to Q(4, s). Hence {M, N} is a regular pair of lines. We conclude that the line N is regular in

S(*\

*

6.4 Flock Generalized Quadrangles and the Axiom of Veblen Let T be a flock of the quadratic cone K, of P G ( 3 , q). Then, by Theorem 4.6.3, with T corresponds a GQ S{T) of order (q2,q). In Payne [118] it was shown that <S(.F) satisfies Property (G) at its point (oo); see Theorem 4.10.1. Hence by Theorem 6.3.1 we have the following result. Theorem 6.4.1 For any GQ 5 (J") of order (q2, q) arising from a flock T, the point (oo) is regular. • (Of course, we also know that (oo) is regular because it is a center of symmetry.) The following theorem is taken from Thas and Van Maldeghem [193]. Theorem 6.4.2 Consider the GQ <S(JF) of order (q2,q) arising from the flock T. If q is even, then the dual net A / ^ always satisfies the Axiom of Veblen and so AfB^ = H^. If q is odd, then the dual net M^ satisfies the Axiom of Veblen if and only if T is a Kantor-Knuth flock. Proof. Consider the GQ S(T) of order (q2,q) arising from the flock T. Then S(T) satisfies Property (G) at the point (oo). Hence for q even, by Theorem 6.3.3, the dual net NP , satisfies the Axiom ofVeblen, and consequentlyA/^^H^. Next, let q be odd. Suppose that J 7 is a Kantor-Knuth flock. Then by Theorem 4.7.2 the point-line dual of S(T) is a TGQ, so is isomorphic to some T ( 0 ) with O = 0(n, 2n, q') and q = q'n and by Theorem 4.7.3, T(O) = T ( 0 * ) . In Payne and Thas [129] it is shown that if S(T) is not classical, then the elation point (oo) of S(T) is fixed by the full automorphism group of <S(JF). Hence if T is not linear, then the point (oo) of S{T) corresponds to some line n of Type (b) of T(O), that is, to some element n of O; if J77" is linear, so S{T) is classical, then we clearly may assume that (oo) corresponds to some element n of O. Consequently T(O) satisfies Property (G) at n. By Theorem 5.1.2 the translation dual O* of O is good

Chapter 6. Generalized Quadrangles, Nets and the Axiom of Veblen

136

at T, with T the tangent space of O at n. Hence, by Theorem 6.2.1, the dual net AfjP which corresponds with the regular line r of T(0*) satisfies the Axiom of Veblen. By Theorem 5.1.13 also the dual net M£ which corresponds with the regular line -K of T(0) satisfies the Axiom of Veblen. It follows that the dual net MP-, satisfies the Axiom of Veblen. Hence A/"/^ — H^. Conversely, assume that q is odd and that the dual net A / ^ satisfies the Axiom of Veblen. Hence A/"/^-, = H^. In the representation of Knarr, see Section 4.9, this dual net looks as follows: •

POINTS of A/"^\ are the lines of P G ( 5 , Q ) not containing p = (oo) but contained in one of the planes pLt = 7rt, with Lt &V and V the BLT-set which corresponds to T,

•

of NR^ can be identified with the threedimensional subspaces of p£ not containing p (£ is a symplectic polarity of P G ( 5 , q)), and

•

INCIDENCE

LINES

is inclusion.

By point-hyperplane duality in p^, the net A / ^ ) , which is the point-line dual ofAfPy is isomorphic to the following incidence structure: •

POINTS of A/(oo) are the points of p^ \ <5, with S some threedimensional subspace of pt,

•

of A/"(OO) are the planes of p^ not contained in 8, but containing one of the lines of a BLT-set V in 5, where V ~ V', and

•

INCIDENCE

LINES

is the natural one.

As the net A / ^ ) is isomorphic to the dual of Hjj, it is easily seen to be derivable; see e.g. De Clerck and Johnson [43]. Let V = {L'0, L[,..., L'q} be contained in the GQ W(
6.4. Flock Generalized Quadrangles and die Axiom ofVeblen

137

Corollary 6.4.3 Suppose that the TGQ T(0), with O = 0(n,2n,q) and q odd, is the point-line dual of a flock GQ T, where the point (oo) of S^) corresponds to the element IT of O. Then O is good at the element TT if and only if T is a Kantor-Knuth flock. Proof. Assume that J 7 is a Kantor-Knuth flock. Then by Theorem 5.1.2 and Theorem 4.7.3 the egg O is good at TT. Conversely, assume that the egg O is good at TT. Then by Theorem 6.2.1 the dual net J\f^ = NR^s satisfies the Axiom of Veblen. By Theorem 6.4.2 the flock T is a Kantor-Knuth flock. • In Bader, Lunardon and Pinneri [4] a stronger version of Corollary 6.4.3 is obtained; the authors rely on Corollary 6.4.3, however. Theorem 6.4.4 Suppose that the TGQ T(0) with O = 0{n, 2n, q) and q odd, is the point-line dual of a flock GQ <S(.F), or, equivalently, assume that O* is good at some element. Then T(C*) is the point-line dual of a flock GQ, or, equivalently, O is good at some element if and only if T is a Kantor-Knuth flock. Proof. Assume that T is a Kantor-Knuth flock. Then by Theorem 5.1.2 and Theorem 4.7.3 the eggs O and O* are good. Conversely, assume that the egg O is good at some element IT. By Corollary 5.1.4 the GQ T(0*) is the point-line dual of a flock GQ with elation point T, where r is the tangent space of O at IT. Hence T(C*) has an elation group with base-line T. SO G acts transitively on the lines, distinct from r, incident with the point (oo) of T(C*). By Theorem 3.7.2, G is induced by a collineation group G' of PG(4n,q) (with PG(4n,g) the space in which T(0*) is constructed), which fixes O*, T, and acts transitively on 0*\{T}. Hence there is a collineation group G of (O) = PG(4n - l,q) fixing O, IT, and acting transitively on 0\{-7r}. AS T(O) is the point-line dual of a flock GQ, the GQ T(C) has an elation group with base-line IT' e O. Hence there is a collineation group G of PG(4n - 1, q) fixing O, IT', and acting transitively on 0\{IT'}. If IT ^ IT', then the collineation group (G, G) acts 2-transitively on O. Consequently, O is good at each of its elements. Now by Theorem 5.1.12, O is classical, so the flock T is linear. If IT = IT', then by Corollary 6.4.3, T is a Kantor-Knuth flock. •

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138

6.5

Subquadrangles and the Axiom of Veblen

In this section, which is also taken from Thas and Van Maldeghem [193], we describe the impact of the Axiom of Veblen on the existence of subquadrangles of order s of GQs of order (s 2 , s). Theorem 6.5.1 Let S = (V, B, I) be a GQ of order (s, t), s / 1 / t, having a regular point x.Ifx and any two points y, z, with y / x and x ~ z / y, are contained in a proper subquadrangle S' ofS of order (s',t), with s' / 1, then s' = t = y/s and the dual netMx satisfies the Axiom of Veblen. It follows that s and t are prime powers, and that for each subquadrangle S' the projective plane irx of order t defined by the regular point x of S' is Desarguesian. Conversely, if the dual net Mx satisfies the Axiom of Veblen, then either (a) s = t, or (b) s = t2, s and t are prime powers, x and any two points y, z, with y / x and x ~ z / y are contained in a subquadrangle Sf of S of order (t, t), and the projective plane irx of order t defined by the regular point x of S' is Desarguesian. Proof. Let S = (V,B,I) regular point x.

be a GQ of order (s,t), s ^ 1 ^ t, having a

First, assume that x and any two points y, z with y ^ x and x ~ z •/• y, are contained in a proper subquadrangle S' of <S of order (s', t), with s' ^ 1. As x is also regular for S', the GQ S' contains subquadrangles of order (1, t). Then Lemma 1.1.1 implies s' = t = y/s. By Theorem 2.2.1 the dual net N'x arising from the regular point x of S', is a dual affine plane of order t. Hence J\f'x satisfies the Axiom of Veblen. Now consider distinct lines Ai,A2,B1,B2 of the dual net Afx , where Ai n A2 = {z}, z <£ Bu z g B2, and AitlBj / 0 f o r a l H , j e {1,2}. Let AtnBi = {u}, A2C\B2 = {w}, and let y e {u,«;} J -\{a;}. Let S' be a subquadrangle of order t containing the points x, y, z of <S. Then Ai,A2, Bi,B2 are lines of the dual net Af'x . As Af'x satisfies the Axiom of Veblen, we have Bi n B2 / 0. It follows that the dual net J\f® satisfies the Axiom of Veblen. Consequently Mx = H^, and so s and t are prime powers. For any subquadrangle S' the dual net Af'x is a dual affine plane of order t, which is isomorphic to a dual affine plane of order t in Hf. Hence Af'%, and also the corresponding projective plane TTX, is Desarguesian.

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139

Conversely, assume that the dual net N® satisfies the Axiom of Veblen. Also, suppose that s^t, and so s > t by Theorem 2.1.3. Then, by Theorem 2.2.3 and Corollary 2.2.4, we have N® = Hf and s = t2, with s and t prime powers. Now consider any two points y, z, with y / x, x ~ z ^ y. As .A/jf = Hf it is easily seen that z and {x, y}1- generate a dual affine plane A of order t inAfJ^. LetAi,A2,..., Ati be the lines of A. Further, let V' be the point set of <S consisting of the points of Aj- U A£ U • • • U A^-2 and the points of A. Clearly V' contains z and y, and \V\ = t3 + t2 + t + 1. Further, any line of <S incident with at least one point of V' either contains a; or a point of A; the set of all these lines is denoted by B'. Also, any point incident with two distinct lines of £>' belongs to V'. Then by 2.3.1 of Payne and Thas [128], see also Section 1.1, S' = (V',B', I'), with I' the restriction of I to (V x B') U (B1 x V') is a subquadrangle of S of order t. As in the first part of the proof one now shows that for any such subquadrangle S' the projective plane itx denned by x is Desarguesian. • Remark 6.5.2 Note that in the first part of the statement of Theorem 6.5.1, x need not be a regular point a priori; one easily deduces the regularity of x from the assumption about the subquadrangles, and Lemma 1.1.1.

6.6 Nets and Characterizations of Translation Generalized Quadrangles First we give the definition of a T-net, then we prove a characterization theorem for TGQs in terms of T-nets, and finally several interesting corollaries will be obtained. This section is taken from Thas [185]. Let J\f = (V, B, I) be a net of order k and degree r. Further, let R be a line of M and let T be the parallel class of B containing R, that is, T consists of R and the k - 1 lines not concurrent with R. An automorphism 9 of N is called a transvection with axis R if either 9 = 1 or if T is the set of all fixed lines of 9 and R is the set of all fixed points of 9. The net J\f is a T-net if for any two nonparallel lines M,N e B\T there is some transvection with axis belonging to T and mapping M onto N. In particular, let M = (P, B, I) be an affine plane of order k and let V be the corresponding projective plane. Then M is a T-net if and only if the point z of V defined by T is a translation point of V; see Hughes and Piper [75].

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Chapter 6. Generalized Quadrangles, Nets and the Axiom of Veblen

If M is the dual of H™, then it is easy to check that TV is a T-net for any parallel class T. Theorem 6.6.1 Let (S^X\G) be a TGQ of order (s,t), s ^ 1 ^ t. Then for any line L incident with x, the dual net M£ defined by L is the dual of a T-net ML, with T the parallel class of ML defined by the point x. Proof. Let M, N be two distinct nonconcurrent lines of <S, concurrent with L but not incident with x. So M, N are nonparallel lines of the net ML, not belonging to the parallel class T defined by the point x. Let R e {M, iV} ± J -, with xiR, and let LimlM, LlnlN. In S there is a symmetry 9 about R with me = n. Then Me = N. The symmetry 8 induces an automorphism 9 of the net ML which fixes the parallel class T of ML elementwise, and all points of ML incident with the line R of T; it is easily checked that no other elements of ML are fixed. Also, Me = N. Hence 8 is a transvection with axis R of ML mapping M onto N. It follows that ML is a T-net. • Now we state the main theorem of this section. Theorem 6.6.2 Let S coregular point x. Iffor the dual of a T-net ML a TGQ with base-point

= (V,B,I) be a GQ of order (s,t), s ^ 1 ^ t, with at least one line L incident with x the dual net M® is with T the parallel class of ML defined by x, then S is x.

Proof. Let 9 ^ 1 be a transvection with axis R of the net ML, with ReT. Then 9 fixes all lines of S, distinct from L, incident with x and all sets {U, R}-1-1, with x IU, U ~ L. Now we prove that 9 induces a symmetry 7 about R of the GQ S. First, let r be a point of S not in x1. Put rlUlylL and rlMlzlR. The line Ue is concurrent with M; let UeIr'lM. Now we put r' = r 7 ; clearly r' ^ r and distinct points not in x1- are mapped by 7 onto distinct points not in xL. We show that r ~ u, r ^ u, r g x±, u £ xL, implies that r 7 ~ u1, r 7 ^ u~«. UrlA - L, then r^lA6 - L; iiulD ~ L, then u, V are on a line of ME, so also A 9 , D e , l / 9 = V are on a line of JVf. So W ~ £»9. The lines i?, ru, W are concurrent with V and with rr1. Consequently, as R is

6.6. Nets and Characterizations of Translation Generalized Quadrangles

141

regular and ru ~ uu1 ~ i?, we have uu1 ~ W. So a triangle with sides uu 7 , W,De arises, clearly a contradiction. It follows that u 7 l W , and so r 7 ~ u 7 . Clearly r 7 7^ u 7 . Now we define Z 7 for any line Z in 23. lfxIZ, then we define Z 7 = Z. Next, assume x IZ. Let n l Z l r 2 , n ^ r2, n 76 x •/ r 2 . Then Z 7 — 7-7r-7. From the foregoing section follows that in this way Z1 is well-defined. Clearly Z*i = Z for all lines Z ~ R. From the foregoing it also follows that for x IZ the lines Z and Z 7 are concurrent with a common line incident with x. Let Zi, Z2 be two distinct lines incident with a common point z G ar 1 . We will prove that Z 7 , Z 7 are incident with a common point z' G a;-1 with z ~ z'; also, Z7 7^ Z 7 . If at least one of the lines Z\,Z2 is incident with x, this is obvious. So assume x I Zit with i = 1,2. If zli?, then clearly Z 7 I z l Z 7 and Z 7 7^ Z 7 . Now let z IX. Then Zj and Z 2 are parallel lines of the net ML, SO also Zf and Zf are parallel in ML- Hence Z 7 = Z{, ZJJ = Zf, L are concurrent in <S; also Zj ^ Z^. So let z I i? and H I , say zlVlx. From the foregoing we have Zj ~ V — Z^. Assume, by way of contradiction, that Z7 7^ Z 7 . Let ailZj, a 2 i z ; j , ai I V I a 2 , ai ~ a 2 . Since 7 defines a permutation of the set of all pairs consisting of distinct collinear points of S not in x-1, we have a 7 ~ aT, , with aj I Z I and aT, IZ 2 , clearly a contradiction. Hence Z7, Zg, V are concurrent in S; also Z7 ^ Z7. Consider any point z e x1- and let ZiIzlZ2, with Z\ ^ Z 2 . Then, by 1 definition, z is the point incident with both Zl and Z.J. Now it is clear that 7 is an automorphism of S which fixes all lines concurrent with R. Consequently, 7 is a symmetry about R. Let li and l2 be incident with L, l\ 7^ x 7^ l2 and l\ 7^ l2. Now we choose two lines A1,A2, with hlAi, l2lA2 and R e {Ai,A2}±J-. Then there is some transvection 6 with axis R' of ML, for some R! e T, mapping Ai onto A 2 . As ({R', Ai}-11)6 = {R^A^-11= {R1, A2}-L±, it follows that R = R'. For the corresponding symmetry 7 about R we clearly have lj = l2. Consequently, R is an axis of symmetry of S. Hence any line R incident with x, but distinct from L, is an axis of symmetry of S. Now by Theorem 6.7.10 of K. Thas [206] <S is a TGQ with base-point x. • This theorem has several interesting and strong corollaries.

142

Chapter6. Generalized Quadrangles, Nets and the Axiom of"Veblen

Corollary 6.6.3 Let S = (V, B, I) be a GQ of order s, s ^ 1, with coregular point x. Iffor at least one line L incident with x the corresponding projective plane TTL is a translation plane with as translation line the set of all lines of S incident with x, then S is a TGQ with base-point x. Proof. Follows immediately from Theorem 2.2.1, the first part of Section 6.6, and Theorem 6.6.2. • Corollary 6.6.4 Let S = {V, B, I) be a GQ of order s, s ^ I, with coregular point x. If for at least one line L incident with x the projective plane -KL is Desarguesian, then S is a TGQ with base-point x. If in particular s is odd, then S is isomorphic to the classical GQ Q(4, s). Proof. As any line of a Desarguesian plane is a translation line, by Corollary 6.6.3 the GQ S is a TGQ with base-point x. Now let s be odd. Then, by Theorem 3.9.6, we have <S^Q(4,s). • Corollary 6.6.5 Let S = {V,B,I) be a GQ of order s, with s e {5,7}. If S contains a coregular point, then S = Q(4, s). Proof. As there is a unique projective plane of order s, s e {5, 7}, by Corollary 6.6.4 we have S = Q(4, s). • A GQ of order 3 is isomorphic either to Q(4,3) or to its dual W(3); see Section 1.8. Corollary 6.6.6 Let S = (V,B,I) be a GQ of order (s,s2), s ^ I, with a regular line L for which the dual net A/jP defined by L satisfies the Axiom of Veblen. If L is incident with a coregular point x, then S is a TGQ with base-point x. Proof. Immediate from Theorem 2.2.3, the first part of Section 6.6 and Theorem 6.6.2. • Corollary 6.6.7 Let S = (V, B, I) be a GQ of order (s, s2), s even, satisfying Property (G) at the line L. If Lis incident with a coregular point x, then S is a TGQ with base-point x. Proof.

Immediately from Theorem 6.3.3 and Corollary 6.6.6.

•

6.6. Nets and Characterizations of Translation Generalized Quadrangles

143

Corollary 6.6.8 Let <S(J") be the GQ of order (q, q2), q even, arising from a flock T. If the point (oo) of <S(JF) is collinear with a regular point x, with (oo) 7^ x, then S(F) is isomorphic to the classical GQ H(3, q2) arising from a nonsingular Hermitian variety of P G ( 3 , q2).

Proof. The GQ S(T) is an EGQ with elation point (oo); see Section 4.6. It follows that the corresponding elation group acts transitively on the points distinct from (oo) which are incident with the line L = (oo)a;. Hence L is coregular. By Theorem 6.4.2 and Corollary 6.6.6, S(T) is a TGQ with baseline L. Now by Johnson's Theorem 5.1.11, S^) is isomorphic to H(3, q2).

The last application of Theorem 6.6.2 is on subquadrangles. Corollary 6.6.9 Let S = (V, B, I) be a GQ of order (s, t), s + 1 ^ t, having a coregular point x. If for a fixed line L, with xlL, and any two lines M, N, with M / L and L ~ N ^ M, there is a proper subquadrangle S' of S of order (s, t'), with t' ^ 1, containing L, M, N, then t' = s = VI, S is a TGQ with base-point x, and S' is a TGQ with base-point x. For s even S satisfies Property (G) at the flag (x, L) and for s odd the translation dual S* of S satisfies Property (G) at the flag (x', L'), with x' the base-point of S* and L' the line of S* corresponding to the line L of S. For s odd the subquadrangles S' are isomorphic to Q(4, s) and S* is isomorphic to the dual of a flock GQ 5(JF). If sis even and ifS* is isomorphic to the dual of a flock GQ S(T), then 5^Q(5,s). Proof. By Theorem 6.5.1 we have that t' = s = y/i and the dual net A/jP satisfies the Axiom of Veblen. Hence by Corollary 2.2.4 A/"f = H;?. By Corollary 6.6.6 the GQ S is a TGQ with base-point x. As S' contains x, S' is also a TGQ with base-point x. Let «S = T ( 0 ) , with O = 0(n, 2n, q), and where -K e O corresponds to L. Now we show that the condition on the existence of subGQs implies that for any two spaces 7TJ,7T.,- G 0\{X}, with i ^ j , the (3n - 1)-dimensional space {-K^^-KJ) contains exactly qn + 1 elements of O. Let PG(3n, q) be any 3n-dimensional space in the PG(4n,g) containing T ( 0 ) , which contains (IT, ni,TTj) but is not contained in P G ( 4 n — l,q) = (O). Next, let /x be an n-dimensonal space in PG(3n,q) containing n, with \i % (ir,iTi,-Kf), let n be an n-dimensional space in PG(3n,q) containing wi, with -n % (TT,^,^) and fj, n r\ = 0, and let p be the n-dimensional space containing TTJ and intersecting p, and 77.

144

Chapter 6. Generalized Quadrangles, Nets and the Axiom ofVeblen

By the condition on the existence of subquadrangles, there is a subGQ of order qn = s of T(C) containing the lines it, 77, fi, p. It easily follows that this subGQ contains all points of (7r,7Tj,p)\PG(4n — l,q) = AG(3n,q). Consequently the qn + 1 lines of the subGQ incident with (00) belong to a common PG(3n - l,q), which clearly is (it,iti,itj). So O is good at it, and consequently, by Theorem 5.1.2, the translation dual S* of <S satisfies Property (G) at the flag (x', V), with x' the base-point of S* and V the line of S* corresponding to the line L of S. So, by Theorem 5.1.5, for s even the GQ S satisfies Property (G) at the flag (x,L). Now let s be odd. Then, by Theorem 5.1.9, all subquadrangles S' are isomorphic to Q(4, s). By Corollary 5.1.4 the GQ <S* is isomorphic to the point-line dual of a flock GQ <S(T). Finally, let s be even and assume that S* is isomorphic to the point-line dual of a flock GQ <S(JF). Then, by Theorem 5.1.11 the GQ S*, and consequently also S, is isomorphic to Q(5, s). •

Chapter 7

Ovoids and Subquadrangles The first part of this chapter elaborates on ovoids of the classical GQ Q(4, q). In particular subtended ovoids and translation ovoids will be considered. The second part of Chapter 7 deals with subGQs of order q of TGQs of order (q,q2). Also, there are sections on subquadrangles of flock GQs, on TGQs of order (q, q2) having at least one subGQ of order q and on EGQs admitting certain subGQs.

7.1 Ovoids of Q(4, q) An ovoid O of the GQ Q(4, q) is a set of q2 + 1 points, no two of which are collinear; equivalently, O has a point in common with each line of Q(4, q). If the GQ Q(4, q) is contained in PG(4, q) as a nonsingular quadric Q, and if the hyperplane PG(3, g) intersects the quadric Q in an elliptic quadric O, then O is an ovoid of Q(4, q); such an ovoid is called classical. For q an even power of 2 no other ovoids of Q(4, q) are known. For q = 2 2/l+1 , with h > 1, one other type of ovoid is known; its projection from the nucleus n of Q onto a hyperplane not containing n, is the Tits ovoid; see Thas [158] (see also Subsection 12.3.2 below). In fact, in Thas [158] it is shown that with any ovoid of Q(4,g) with q even, corresponds, by projection, an ovoid of PG(3, q), and, conversely, with any ovoid of PG(3, q) with q even, corresponds an ovoid of Q(4, q). For q odd, Kantor [82] constructed two types of nonclassical ovoids. Starting from a Kantor-Knuth flock T, the construction in Section 4.2 defines an ovoid O of Q(4, q). By Theorem 4.4.6 these ovoids of Q(4, q), q odd, are characterized by the property that they are the union of q conies which are mutually tangent at a common point. 145

Chapter 7. Ovoids and Subquadrangles

146

Such an ovoid is classical if and only if the corresponding automorphism a of GF(q) is the identity. An ovoid of Q(4, 1, then Kantor [82] proves that Q(4, q) has an ovoid which arises from the Ree-Tits ovoid on the nonsingular quadric in PG(6, q); such an ovoid is always nonclassical. An ovoid of Q(4, q) of this type will be called a Kantor-Ree-Tits ovoid and will be denoted by /C2. Further, Penttila and Williams [131], with a slight assist from a computer, discovered a new ovoid of Q(4,35). This ovoid will be called the Penttila-Williams ovoid. Finally we mention that for q odd one other class of ovoids of Q(4, q) is known; see Section 7.2. With each ovoid O of Q(4, q) corresponds a spread of the dual GQ W(g). And with a spread ofW(g) corresponds, by the standard argument, a translation plane n of order q2; see e.g. Section 3.1 of Dembowski [45]. The plane -n is Desarguesian if and only if O is classical. If O is of Tits type, then 7r is the Liineburg plane; see Thas [158]. To an ovoid /Ci corresponds a Knuth semifield plane; see Kantor [82].

7.2

Subquadrangles and Ovoids

Let S' be a subGQ of order (s, t') of the GQ S of order (s, t), with t' < t. If z is a point of S which is not contained in S', then the 1 + st' points of S' collinear with z form an ovoid Oz of S'; see 2.2.1 of Payne and Thas [128] (or §1.1). Such an ovoid Oz of S' is called a subtended ovoid. For example, if we consider the subGQ Q(4, q) of the GQ Q(5, q), then Oz is an elliptic quadric on the quadric Q denning Q(4, q). Consider an egg O = 0(n, 2n, q) and assume that 0(n, 2n, q) is good at its element TT. By Theorem 5.1.9 the TGQ T(0) contains exactly q3n + q2n subGQs of order qn containing the line 7r of T(C), and for q odd each of these subGQs is isomorphic to the classical GQ Q(4, qn). Fix such a subGQ S' ^ Q(4, qn). Then by Lavrauw [93, 95] and Theorem 10.6.3 of K. Thas [206], up to isomorphism the subtended ovoid Oz of S' is independent of the choice of the point z. Further, Lavrauw [93, 95] proves that up to isomorphism the ovoid Oz is independent of the choice of the subGQ S'. Let O be an ovoid of Q(4, q), let x G O and let L be a line of Q(4, q) incident with x. We call O a translation ovoid with respect to the flag (x, L) if there is an automorphism group GXIL of Q(4, q) which fixes O, which fixes x linewise and L pointwise, and which acts transitively on the points

7.3. Translation Ovoids and Semifield Flocks

147

of (O \ {x}) n y1- for each t / / i incident with L. We call O a translation ovoid with respect to the point x if O is a translation ovoid with respect to the flag (x, L), for each line L incident with x. In Bloemen, Thas and Van Maldeghem [14] it is shown that the ovoid O of Q(4, q) is a translation ovoid with respect to the point x e O if and only if Q(4, q) admits an automorphism group of size q2 which fixes O and all lines incident with x. It is easy to see that the ovoids Oz of S1 in the previous paragraph are translation ovoids of S1 with respect to the point (oo). For more on translation ovoids we refer to Bloemen, Thas and Van Maldeghem [14] and Lunardon [100]. Let 0(n, 2n, q) = O be a good egg in PG(4n— 1, q), with q odd. By Corollary 5.1.4 and Theorem 4.7.2 the TGQ T(C*), with O* the translation dual of 0, is the point-line dual of a flock GQ Stf), with T a semifield flock. Conversely, if T is a semifield flock, then S{F) is the point-line dual of a TGQ T(£>*), where O is good (see Theorem 4.7.2). So with any semifield flock corresponds an ovoid of Q(4, s), which is a translation ovoid with respect to some point. Also, with any translation ovoid with respect to some point of Q(4, s) corresponds a semifield flock. For details we refer to Lunardon [100]. In the next section we will show how to construct, in a purely geometrically way, the semifield flock from the translation ovoid, and conversely. In Thas [176] it was conjectured that the translation ovoids of Q(4,3 h ), h > 2, arising from the Ganley flocks (See Section 4.5) are new. This conjecture was proved by Thas and Payne [190], and the new ovoids are called Thas-Payne ovoids. Together with the ovoids of Q(4, q) mentioned in Section 7.1, these are the only known ovoids of the classical GQ Q(4, q).

7.3 Translation Ovoids and Semifield Flocks The following construction is taken from Thas [181], see also Lunardon [100]. The plane PG(2,qn), with q = ph and p prime, can be represented in PG(3n - l,q) in such a way that points of the plane become (n - 1)dimensional subspaces belonging to an (n - l)-spread T of PG(3n - 1, q) and lines of the plane become (2n - 1)-dimensional subspaces of PG(3n 1, q); see also Sections 3.6 and 4.5. In each of these (2n - 1)-dimensional subspaces corresponding to lines an (n - 1)-spread is induced by T. Let C be a nonsingular conic of PG(2, qn). With C corresponds a set C of qn + 1

148

Chapter 7. Ovoids and Subquadrangles

(n-l)-dimensional subspaces of PG(3n—1, q). This s e t C is a pseudo-conic ofPG(3rc-l,g). Let 7r be a (2n — 1)-dimensional subspace of P G ( 3 n - 1, q) having no point in common with the elements of C. Now we consider the GQ T 2 (C) = Q(4,9 n ) of Tits defined by C. Then T 2 (C) ^ T(C'), where C corresponds to C and the points of Type (i) of T 2 (C), which are the points not collinear with the point (oo) of T 2 (C), correspond to the points of PG(3n, q) \ PG(3n -l,q) where PG(3n, q) is the space containing T(C'); see Sections 3.6 and 4.5. Let n be a 2n-dimensional subspace of PG(3n, q) containing re, but not contained in PG(3n -l,q). Then (n \ PG(3n - 1, q)) U {(oo)} is an ovoid of the GQ T(C'), so defines an ovoid of T 2 (C), and hence defines an ovoid of Q(4, qn). Also, this ovoid of Q(4, qn) is a translation ovoid with respect to the point which corresponds to the point (oo) of T(C) (respectively, T 2 (C)). By Lunardon [100] any ovoid of Q(4, qn) which is translation with respect to some point can be obtained in this way. Now we dualize in PG(3n - 1, q); choose e.g. a polarity in P G ( 3 n - 1, q) and consider the polar spaces of the subspaces of PG(3n—1, q). With C corresponds a set C' of qn +1 (2n-1)-dimensional subspaces of PG(3n -l,q); C' can be considered as a dual conic C in a PG(2, qn). With n corresponds an (n — 1)-dimensional subspace n of PG(3n - l,q) which is skew to all elements of C'. By Section 4.5 this subspace TT defines a semifield flock, and, conversely, any semifield flock can be obtained in this way. Hence any ovoid O of Q(4, s) which is a translation ovoid with respect to some point, defines a semifield flock To, and, conversely, any semifield flock T of the quadratic cone K in P G ( 3 , s) defines an ovoid 0? of Q(4, s) which is a translation ovoid with respect to some point. Let J 7 be a semifield flock. Then the point-line dual of the GQ S(T) is a TGQ T(£>*) of order (s,s2), where the egg O is good. By Section 7.2, O defines a subtended ovoid Oz of the subquadrangle S' = Q(4, s). By Lavrauw [93, 95] this ovoid Oz is isomorphic to the translation ovoid 0? of Q(4, s), which corresponds to the flock T.

7A

Coordinates of the Known Nonclassical Ovoids of Q(4,<7)

Suppose that Q(4, q) has equation X\ = X0Xi the known ovoids of Q(4, q) in coordinates.

+ X3X4. We now describe

7.4. Coordinates of the Known Nonclassical Ovoids o/Q(4, q)

149

First suppose that q is even. [219]. Define 7 : i i-> tr+1, with t G GF(q) and g = 2 2 e + 1 , e > 1. Then the Tits ovoids Tl are described as follows (see also §12.3.2):

TITS OVOIDS

r i = {(0,0,0,1,0)} U {(x, y,x^2+l

+ y^2, xy + x^+ 2 + y\ 1) || x, y G GF(g)}.

Next suppose that q is odd. KANTOR OVOIDS /CI [82].

Let n be a nonsquare of GF(q)

and let a ^ 1

be an arbitrary automorphism of GF(g). Then the Kantor ovoids K\ are given by / d = {(0, 0,0,1,0)} U {nx°,x, y, y2 - nx°+1,1)

|| x, y G GF(g)}.

/C2 [82]. Define 4> : x ^ x3h, with x G GF(g) and q = 3 2 / l _ 1 , h > 2. Then /C2 is described as follows: KANTOR OVOIDS

/C2 = {(0, 0 , 0 , 1 , 0 ) } U { ( x ^ + 3 + / , x, y, y2-x2^-xy\ THAS-PAYNE OVOIDS T7 7 [ 1 9 0 ] .

1) || x, j, G GFfa)}.

Put q = 3h, h > 3, and let n be a non-

square of GF(q). Then T7? is given by TV = {(0,0,0,1,0)} U {{xn + (xn" 1 ) 1 7 9 + y 1 / 3 , x, y, y 2 - xixn + (xn" 1 ) 1 7 9 + t/ 1 / 3 ), 1) G Q(4, q) || x, y G GF(g)}. If ft is odd, - 1 is a nonsquare, and then the ovoid can be written as TV = {(0,0,0,1,0)} U { ( - x - x 1 / 9 + 2/1/3, x, y, y2 + x2 + x 1 0 / 9 - xy 1 / 3 ,1) GQ(4,g)||x,yGGF(g)}. THE PENTTILA-WILLIAMS OVOID

VW [131].

The Penttila-Williams ovoid

PWofQ(4,35)isgivenby P W = {(0,0,0, l,0)}U{(x 9 + y 8 \ x , y , y 2 - x 1 0 - x y 8 1 , l ) || x,y G GF(3 5 )}.

Chapter 7. Ovoids and Subquadrangles

150

7.5

Subquadrangles of T(£>), with O Good: the Even Case

First we discuss briefly the connection between GQs with a regular point and covers of the net associated with the regular point. In particular, we shall consider subGQs of order s of GQs of order (s, s2) in this context. Let TV = (V, B, I) be a net of order k and degree r. An m-fold cover of A/" is a geometry M = (V,B,I) such that there exists a surjective map

satisfying: (i) for any x e V, x1

consists of m pairwise noncollinear points;

(ii) for any pair {x, y} of collinear points of 77, {x, y}1 disjoint pairs of collinear points;

consists of m

(iii) for any pair {x,y} of noncollinear points of N, {^,y}7 pairwise noncollinear points of Af; (iv) for any line L of M, if VL = {% € V || xlL), then VI points of m disjoint lines of A/".

is a set of consists of the

The map 7 is called a covering map. A point or line of M is said to be a cover of its image under 7. Now let 5 = (V, B, I) be a GQ of order (s, t) with a regular point x, and let Afx be the associated net of order s and degree t + l. The geometry 77^ with • POINT SET V

\x^,

• LINE SET B \ {L e B |j xlL}

and

• INCIDENCE inherited from S, is a i-fold cover of A/^. The covering map 7 maps the point y e V \ x1to the point {x, y}1- of Afx. The cover has the additional property that for a nonincident point-line pair {y, L} of J7^, either y1 = z1 with zlL and z £ {x,y}~LJ~, or 2/is collinear with a unique point of 772 incident with L. In general, given a net of order s and degree t + l with such a i-fold cover,

7.5. Subquadrangles of T(0), with O Good: the Even Case

151

it is possible to construct a GQ of order (s, t), which in the above case is a reconstruction of <S. Suppose that Mx has a proper subnet J\fx of order s', with s' > 2, and degree t + 1. Then the subset V' of V consisting of x, the lines of J\f'x and j e P \ i i such that {x, y}1- is a point of M'x, is the point set of a subGQ 5 ' of order (s', t) of 5 . The lines of S' are the lines of S incident with at least one (and hence s' + 1) points of V. The point x is a regular point of S1 with associated net A^. It follows that t = s',t2 = s and so the subnet Mx is an affine plane of order t. For more details see K. Thas [200]. Let S be a GQ of order (q2, q) with a regular point x such that the associated dual net Mx is isomorphic to H^. In Theorems 5.1.2, 6.2.1, 6.3.3 and 6.3.4 examples were given. In the next results, due to Brown and Thas [28], subGQs of such GQs will be considered. Theorem 7.5.1 Let S = (P,B,I) be a GQ of order (q2,q), 9 / 1, with a regular point x such that the associated dual net N® is isomorphic to H3. Then x is contained in the maximal number q3 + q2 subGQs of order q. If S' = (V, B', I') is any subGQ of order q not containing x, then S' is isomorphic to the point-line dual ofT2(0)forsome oval O o / P G ( 2 , q). Proof. Suppose that the net (H^) D = J\fx is constructed from the line L of P G ( 3 , q); that is, by taking as POINTS the lines of P G ( 3 , q) not meeting L, as LINES the points of P G ( 3 , q) \L, and as INCIDENCE the incidence from P G ( 3 , q). Then each plane n of P G ( 3 , q) not containing L gives rise to an affine plane subnet of (H^) D , which is the dual of IT, with the point TT n L and the lines of n on n n L removed. From the foregoing we see that TT gives rise to a subGQ of order q. This gives q3 + q2 distinct subGQs of order q containing x, the maximal number by Lemma 5.1.8. Now suppose that S' = (V, B\ I') is a subGQ of order q of S, not containing x. The set 'P\a;-L is the point set of a g-fold cover of the net J\fx = (H^) with covering map 7 taking the point y e V \ x1- to the point {x^}1- of A4 = (H^) . The subGQ S' contains a unique line, say M, incident with x; see Section 2.2 of Payne and Thas [128]. The points of S incident with M, but distinct from x, define a parallel class of (H^) the elements of which are contained in a plane of P G ( 3 , q) containing L which we will denote by M 7 . The subGQ S' contains q + 1 points of M. These points define a set O of size q + 1 in the plane M 7 , and O n L = 0. Consider a line TV of S' not concurrent with M. If PN = {z e V \\ zlN}, then Vjj defines a point of P G ( 3 , q) not in the plane M 7 . Further, since no two lines of S' are incident

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Chapter 7. Ovoids and Subquadrangles

with a common point of xL not incident with M, it follows that 7 gives a one-to-one correspondence between the q3 lines of S' not concurrent with M and the q3 points of PG(3, q) \ M 7 . Each point of S' not incident with M is collinear with a unique point of M and so, under the map 7, defines a line of PG(3, q) meeting M 7 in a unique point of O. Since no two lines of S' are concurrent in a point of x1- not incident with M, it must also be the case that no two points of S', not incident with M, correspond under 7 to the same line of PG(3, q). Thus 7 gives a one-to-one correspondence between the set V \VM, with VM = {z € V \\ zlM}, and the lines of PG(3, q) not in M 7 meeting M 7 in a point of O. It is now a straightforward exercise to verify that O is an oval and that S' is isomorphic to the point-line dual of T 2 (£>). • We have an immediate corollary. Corollary 7.5.2 Let S = (V,B, I) be a GQ of order (q2,q), q odd, with a regular point x such that the associated dual net Af® is isomorphic to H3. If S' = (V, B', I') is any subGQ of order q not containing x, then S' is isomorphic to the classical GQ W(g). Proof. This follows from Theorem 7.5.1, Theorem 1.5.2(f), the remark which follows Theorem 1.5.2, and Theorem 1.4.1. • Suppose that S = (V, B, I) is a TGQ of order (s, s2), s ^ 1, with S = T(0), where O is good at its element -K. By Theorem 6.2.1 the dual net N® is isomorphic to Hf. Suppose that H^ is constructed from the line L in PG(3, s). We identify H^ and J\f®, and assume that 7 is the covering map from V \ x1- to the point set of (H^) D . Let G be the translation group of S about (00). Then G has order s4, fixes each line incident with (00) and acts regularly on the elements of V \ ar1. The group G induces an automorphism group G' of the dual net A/jf = H^. The identity mapping of J\f® is induced by all symmetries of S about n, and so G' has size s3. The group G' is induced by a collineation group of PG(3, s) fixing the line L; this collineation group will also be denoted by G'. If we denote the plane of PG(3, s) corresponding to the elements of O \ {71-} by (oo)7, then each element of G' fixes all points of (00)7, so is an axial collineation with axis (oo)7. If 9 e G fixes a line of ir1- not incident with (00), then it follows that 9 is a symmetry about n. Hence G' acts regularly on the points of PG(3, s) \ (oo)7 and must be the group of elations of PG(3, s) with axis (oo)7.

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153

By Theorem 5.1.9 <S = T(0) contains exactly s 3 + s2 subGQs of order s containing the line n. Now suppose that S' = (V',B', I') is a subGQ of order s of S not containing the point (oo); then n cannot be a line of S'. Consequently, by applying Theorem 7.5.1 it follows that S' = T2(0'), for some oval O' of PG(2, s), where PG(2, s) corresponds to the point m of S' which is incident with w; this plane PG(2, s) will be denoted by vrC. Lemma 7.5.3 Let S = (V, B, I) be a TGQ of order (s, s2), s ^ 1, with S = T(0), where O is good at its element TT. Suppose that S' — (V',B', I') is a subGQ of order s ofS not containing the point (oo). Then S' is isomorphic to the classical GQ Q(4, s). Proof. By the foregoing we have S' = T 2 (£>')• N o w w e show that O' is a conic in m 7 , with m the unique point of S' incident with the line w. Let y be a point of (oo)7 not on the line L and let

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Theorem 7.5.4 Let S = (V, B, I) be a GQ of order (s, s2), s even, having a subGQ S' isomorphic to Q(4, s). If in S' each ovoid Ox consisting of all the points collinear with a given point xofS\S'isan elliptic quadric, then S is isomorphic to Q(5, s). • Remark 7.5.5 The equivalent theorem was proved in the odd case by Brown [22] and a proof valid for both s odd and even may be found in Brouns, Thas and Van Maldeghem [19] and in Brown [24]. We are now equipped to prove the classification theorem. Theorem 7.5.6 Let S = (V,B,I) be a TGQ of order {s,s2), s even, with S = T(O), where O is good at the element IT. Suppose that S' = (V, B', I') is a subGQ of order s ofS not containing the point (oo). Then S is isomorphic to the classical GQ Q(5, s). Proof. By Lemma 7.5.3, if m is the point incident with the line IT and contained in S', then S' = T 2 (C) where C is a conic in the plane rrC of PG(3,s). Let y be a point oiT)\V' such that y ^ m, and let Oy be the ovoid of S' subtended by y. We now show that Oy corresponds to an elliptic quadric of Q(4, s) ^ T 2 (C). Let tp be a nontrivial elation of P G ( 3 , s) with axis (oo) 7 and center on the line L of P G ( 3 , s). Let Tp be an automorphism of <S that induces ip in P G ( 3 , s). By Sections 2 ^ and 2.3 of Payne and Thas [128] there are three possibilities for S' n S'v: (i) a point u, the s + 1 lines of S' and S^ incident with u and the points incident with these lines; (ii) an ovoid of S' and S'v, and (iii) a grid. Hence the subGQs S' and S^ can share either 0,1,2 or s + 1 lines incident with the point m. Consequently, C and Cv may share 0,1,2 or s + 1 points. However, ip cannot fix C so the possible intersection numbers are 0,1 and 2. As any nontrivial elation of P G ( 3 , s) has order 2, we may choose

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155

the subGQs S' and S'v share exactly two lines incident with m, and so S' n S'^ is a grid. Under the covering map 7, with the lines of the grid correspond z, z' and the 2s lines on z or z', not in m 7 , in a fixed plane 5 of PG(3, s). Since the point y of <S is not contained in 5', but is contained in S'v, it follows that y1 fi 5. Each of the s + 1 lines of <S'^ incident with y is incident with a unique point of 5 ' n S^. Looking in PG(3, s) at the T2(C) and T2(CV) models for 5' and <S'^, respectively, we see that this set V of s + 1 points must contain the s - 1 points of the projection of Cv \ {z, z'} from y7 onto 5, that is, s - 1 points of a conic (completed to a conic by adding z and z')- The remaining two points of V are the planes spanned by y1 and the tangent lines of C at z and z'\ these tangent lines are also the tangent lines of Cv at z and z'. Under the isomorphism from T2(C) to Q(4, s) the images of these s + 1 points of T2(C) are the s + 1 points of a conic, so under the isomorphism from Q(4, s) to W(s) the corresponding s + 1 points of W(s) are also the s + 1 points of a conic C; see Brown [23] for an explicit isomorphism from T2(C) to W(s). To Oy corresponds an ovoid 0'y ofW(s). By Thas [158] 0'y is also an ovoid of the projective space PG(3, s) which contains W(s). As Oy contains a conic, by Brown [23], Oy is an elliptic quadric of PG(3,s). Hence to Oy corresponds an elliptic quadric on Q(4, s). Now suppose that y e V \T" such that y ~ m with y ^ m. Again let Oy be the ovoid of S' subtended by y. Let TV be any line of S incident with y but not with m. Then N meets S' in a unique point, n say. Each of the s - 1 points incident with N, but distinct from y and n, subtends an ovoid of S' which corresponds to an elliptic quadric on Q(4, s). Since the s points incident with N, but distinct from n, subtend s ovoids that partition the points of S' not collinear with n, it follows that the ovoid Oy is determined by the s - 1 ovoids subtended by the points incident with N, but distinct from y and n. Looking in the Q(4, s) or T2(C) model of S', we see that with Oy corresponds an elliptic quadric of Q(4, s). Since each point oiV\T" subtends an ovoid in S', which defines an elliptic quadric in Q(4, s), it follows from Theorem 7.5.4 that S = Q(5, s). • Note that a similar proof may be possible in the case when s is odd. However, given that the characterizations of an elliptic quadric of Q(4, s) (as ovoid of the latter) are more involved for s odd than in the case where s is even, any such proof will be more complicated and probably much longer than that given in Theorem 7.6.2. Now we will consider subGQs of order s of TGQs of order (s, s2), for any

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s ^ 1, where S' does contain the point (oo). The following theorem is folklore. Theorem 7.5.7 Let S = {V,B,I) be a TGQ of order (s,s2), s ^ I, with O = G(n,2n,q), and assume that S' = (P',B',I') is a subGQ of order s containing the point (oo) of S. Then S' = 1(0'), with O' = 0'{n,n,q) a pseudo-oval on O. Proof. Assume S admits a subquadrangle S' = (V, B', I') of order s containing the point (oo). Let L0, Li,..., Ls be the s + 1 lines of <S incident with (oo) and contained in <S', say Li = 71-j with 7r; e O, i = 0 , 1 , . . . , s. If PG(n, q) = M is a line of S' concurrent with L 0 and not containing (oo), then (PG(n,q),iti) \ PG(4n - l,q), with (O) = P G ( 4 n - l,q) and i e { 1 , 2 , . . . , s}, is a 2n-dimensional affine space AG ( -^(2n, q) consisting entirely of points of <S'. In this way arise the s3-s2 points of S' not collinear with (oo) or with L0 n M, together with the s points of M not incident with L0. Remark that the points of A G ^ (2n, q), i E { 1 , 2 , . . . , s}, are the points not collinear with (oo) of a grid W. Now let N = P G ' ( n , q) be a line of <S' concurrent with L\, not concurrent with M and not containing (oo). As N contains a point of the grid <SW, the space PG'(n, q) has a point in common with the space A G ^ (2n, q), i = 2 , 3 , . . . , s. Letting vary P G ' ( n , q), we see that the s - 1 affine spaces AG1-1^ (2n, g), i == 2 , 3 , . . . , s, are contained in (iri, AG ( 2 ) (2n,g)) = PG(3n,g). Clearly also AG ( 1 ) (2n,q) is contained in PG(3n, q). It easily follows that the points of S' not collinear with (oo) are the points of the affine space AG(3n, q) = PG(3n,g) \ P G ( 4 n - l,q), that 7ro,7ri,... ,7rs are contained in a common PG(3n - l,q), and that S' = T(O'), with O' = {TTO.TTI, . . . ,7r s }. • Remark 7.5.8 Several (more general) versions of Theorem 7.5.7 can be found in K. Thas [197], see also [203].

7.6

Subquadrangles of T(0), with O Good: the Odd Case

The theorems of this section are due to Brown and Thas [29]. Theorem 7.6.1 Let S = (V, B, I) he a TGQ of order (s, s2), s odd and s ^ 1, with S = T(O) and O good at some element IT. If S' is a subGQ of order s of S containing the point (oo), then S' = Q(4, s) and either S' is one of the s3+s2 (classical) subGQs of order s containing the line n ofS, orS = Q(5, s).

7.7. Subquadrangles: Remaining Cases and Some Applications

157

Proof. By Theorem 5.1.9 the GQ contains exactly s 3 + s 2 subGQs of order s containing the line -K of S; each of them is isomorphic to the classical GQ Q(4,s). Assume <S admits a subGQ S' = (V, B', I') of order s containing the point (oo) but not containing the line n. If <S = T(O), with O = 0(n,2n,q), then 5 ' = T(O'), with O' = {TT0, v n , . . . , TTJ, TT, e O for » = 0 , 1 , . . . , s, and 7ro, 7ri,..., 7rs contained in a common P G ( 3 n - 1, q); see the proof of Theorem 7.5.7. Now by Theorem 5.2.3 and the uniqueness of the GQ of order (3,9) the egg O is classical, and so <S = Q(5, s). • Theorem 7.6.2 Let S = (V, B, I) be a TGQ of order (s, s2), s odd and s=£l, with S = T(O) and O good at some element z.IfS'isa subGQ of order s of S not containing the point (oo), then S = Q(5, s). Proof. By Corollary 5.1.4, T ( 0 ) is the translation dual of the point-line dual of a flock GQ. If -K is the good element of O, then, by Section 3 of K. Thas [205] (see also K. Thas [206] or Chapter 10) every point of the line 7r is a translation point of the TGQ S. One of these translation points, say m, belongs to S'. As there is an automorphism of <S mapping m onto (oo), there is a subGQ S" of order s which contains the point (oo). Hence, by Theorem 7.6.1, S ^ Q(5, s). •

7.7

Subquadrangles: Remaining Cases and Some Applications

Each known GQ of order (s, s 2 ), s ^ 1, is either the point-line dual of a flock GQ, or a TGQ T(O), with either O or O* good. In this section we will determine the subquadrangles of order s of flock GQs and of TGQs T ( 0 ) , with O* good. As most of the proofs are quite long, just a few of them will actually be given. Finally, we will mention some interesting characterization theorems, which are related to the determination of these subquadrangles. The next theorem is due to O'Keefe and Penttila [92], but we give here another and very short proof. Theorem 7.7.1 If S(F) is a flock GQ of order (q2,q), q even, with center (oo), then S(J7) contains exactly q3 + q2 subGQs of order q containing the point (oo).

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Chapter 7. Ovoids and Subquadrangles

Proof. Payne [118] proved that | { L , M , i V } ± x | = q + 1 for all triads {L, M, N}, with (oo)lL, for which there is a line U such that (oo)lU and {L,M,N} c UL; see also Theorem 4.10.1. That is, the set {L,M,N} is 3-regular. Now let {L, M, N} be such a triad. If B' is the set of lines of S {J7) incident with points of the form X nY, with X e {L, M, N}1- and Y e {L, M, N}-11-, and "P' is the set of points which are incident with at least two distinct lines of £>', then endowed with the induced incidence I', S' = (V',B', I') is a subGQ of order q of S{F), see Theorem 2.6.2. Now a standard counting argument yields q3 + q2 of such subGQs. By Lemma 5.1.8, 5(J - ) contains exactly q3 + q2 subGQs of order q containing the point (oo). • Now we mention, without proof, a strong characterization theorem for Q(5, s) on which we will rely in the proof of Theorem 7.7.3. Theorem 7.7.2 and 7.7.3 are due to Brown and Thas [28]. Theorem 7.7.2 Let S = {V, B, I) he a GQ of order (s, s2), s + 1, and assume one of the following conditions is satisfied: (i) s is odd and S satisfies Property (G) at the pair {x, y} with i ~ t / / i and L = xy, or (ii) s is even, S is the point-line dual of a flock GQ and L is the corresponding base-line of S. If at least one triad \z\,zi,z), with zlL and where {zi,z2,z}1contain a point incident with L, is 3-regular, then S is classical.

does not •

Theorem 7.7.3 Let Sbea nonclassical flock GQ of order (q2,q), q even, with base-point (oo). Then any subGQ of S of order q contains (oo). Proof. Let SD be the point-line dual of a nonclassical flock GQ of order (q2,q), q even. The base-line of SD will be denoted by L. Suppose, by way of contradiction, that S' is a subGQ of order q of SD which does not contain L. Let x be the point of <S' incident with L an let ylL, with x ^ y. Further, let Oy be the ovoid of S' consisting of the q2 + 1 points of S' collinear with y. Also, let z e Oy\ {x}, let M be a line of S' incident with x, let N be a line of S' incident with z, and let M / N. Let Mlu ~ z and Nlw ~ x. Then by Theorem 4.10.1 and Theorem 2.6.2 there is a subGQ S" of order q of SD containing y, u, w, x, z. From Lemma 1.1.1 easily follows that <S' n <S" is

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159

a grid containing the lines M, N. Hence the pair {M, N} is regular for any line AT of S' not concurrent with M. Consequently M is a regular line of <S', so in S' any line incident with x is regular. Now by Theorem 2.1.5 the point x is regular in S'. If r is a point of S' not collinear with x, then {x, r}- 1 ' and {x, r}-1-'1-' contain q + 1 points of S'. Let {x, r}1-' = A and {x, r}1-'x' = A. If r' e A\{x, r}, then {x, r, r'} is 3-regular in 5 and {x, r, r'}-1 = A does not contain a point incident with L. By Theorem 7.7.2, the GQ SD is classical, a contradiction. • Theorem 7.7.4 Let S fee a flock GQ of order (q2, q), q odd, with base-point (oo). If S' is a subGQ of order q of S containing the point (oo), then S is a Kantor-Knuth semifield flock GQ. Hence S' ^ W(g) and either S ^ H(3, q2) or S' is one of the q3 + q2 subGQs of order q containing the point (oo). • Remark 7.7.5 The proof of this theorem is quite long; the last part of the statement follows from Theorem 7.6.1. Corollary 7.7.6 Let S be a flock GQ of order (q2,q), q odd, with base-point (oo). If the net M^oo) defined by the regular point (oo) of S contains at least one proper subnet having the same degree as M^), then S is a Kantor-Knuth semifield flock GQ. Proof. In K. Thas [200] it is shown that the subnet is an affine plane of order q. From the proof of Theorem 6.5.1 then follows that <S has a subGQ of order q containing the point (oo), and so by Theorem 7.7.4 we are done.

• Theorem 7.7.7 Let S be a nonclassical flock GQ of order (q2, q), q odd, with base-point (oo). Then any subGQ ofS of order q contains (oo). • Remark 7.7.8 The proof of Theorem 1.1.1 is long and technical. Central in this proof is the construction of Knarr; see Section 4.9. As a byproduct of the proof of Theorem 1.1.1 we obtain the counterpart of Corollary 6.6.8; no proof will be given. Theorem 7.7.9 If S = (V,B,I) is a flock GQ of order (q2,q), with q odd, then the point-line dual of S is a TGQ if and only if S has a regular point x, with (oo) ^ x ~ (oo), where (oo) is the elation point of S. •

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Next we will consider subGQs of TGQs T(0*) of order (s, s2), where the translation dual O* of O is good. If s is even, then the egg 0(n, 2n, q) is good if and only if 0*(n, 2n, q) is good; see Corollary 5.1.6. If 0*(n,2n,q), q odd, is good, then, by Corollary 5.1.4 the TGQ T(C) is isomorphic to the point-line dual of a flock GQ. So from Theorems 7.5.6, 7.5.7, 7.7.4 and 7.7.7 we have the following result. Theorem 7.7.10 Let S = T(0), with O = 0{n, 2n, q), be a TGQ such that the translation dual 0*ofOis good at the tangent space TofOatit&O, and assume that S contains a subGQ S' of order qn = s. If q is even and S' does not contain the translation point (oo), then S = Q(5, s); if q is even and S' contains the translation point (oo), then S' = T(0'), with O' = 0'(n,n,q) a pseudo-oval on O. If qis odd and S' does not contain the translation point (oo), then S = Q(5, s); if q is odd and S' contains the translation point (oo), then S is a Kantor-Knuth semifield flock GQ, S' = Q(4, s) and either S = Q(5, s) or S' is one of the s3 + s2 subGQs of order s containing the point (oo). •

7.8 Translation Generalized Quadrangles with One Classical Subquadrangle Recall Brown's result on ovoids in PG(3, q) and its equivalent statement in the theory of TGQs. Theorem 7.8.1 (Brown [23]) (i) If an ovoid o/PG(3,?), q even, has a conic section, the ovoid must be an elliptic quadric. (ii) If a T 3 (C) of order (q, q2), where Oisan ovoid o/PG(3, q), q even, has a classical subGQ of order q containing the point (oo), then T 3 (O) = Q(5, q), that is, T 3 (C) arises from a nonsingular elliptic quadric in PG(5,g). Proof. See Brown [23].

•

We note that the only classical GQs of order q are the GQ Q(4,q), which arises from a nonsingular quadric in PG(4, q), and the symplectic quadrangle W(q). However, if q is even, then by Theorem 1.4.1, Q(4, q) = W(g).

7.8. Translation Generalized Quadrangles with One Classical Subquadrangle

161

Brown and Lavrauw [26] generalized Theorem 7.8.1 by obtaining a similar result for generalized ovoids. We first prove a lemma. Lemma 7.8.2 Every (2n~l)-space a in PG(3n— 1, q), q even, which is skew to all elements of a pseudo-conic, is the span of two elements of the regular (n — l)-spread induced in a by the pseudo-conic. Proof. Let a be a (2ra — l)-space skew to all elements of the pseudoconic O in w = P G ( 3 n - l,q). After dualizing, we obtain an (n - 1)space a* that is disjoint from all elements of a dual pseudo-conic O*. Embed 7r in a 3n-dimensional space PG(3n, q), and embed a* in an n-space P G ( n , q) C PG(3n, q), where P G ( n , q) ) is a TGQ of order (qn, q2n), where Oisa generalized ovoid o / P G ( 4 n —l,q), q even, which has a classical subGQ S' of order q containing the point (oo), then T(0) = Q(5, qn). Proof. Suppose O' C O is the pseudo-conic on O which corresponds to S'; see Theorem 7.5.7. For x a point of S \ S1, let Ox be the ovoid of S' subtended by x. First suppose that x is a point of Type (ii) of T(O) — so a: is a subspace of dimension 3n meeting PG(4n - l,q) in the tangent space at some egg element. Let T(0) be contained in PG(4n, q) and let PG(3n, q) C PG(4n, q) be the 3n-space containing O' that yields T(O') = S'. Then Ox consists of the point (oo) plus the q2n points contained in

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Chapter 7. Ovoids and Subquadrangles

(xnPG(3n, g ) ) \ P G ( 3 n - 1 , q), with P G ( 3 n - 1 , q) the space containing O'. The (2n-l)-dimensional space x n P G ( 3 n - l , g ) = P G ( 2 n - l , q ) is skew to the pseudo-conic O', so applying Lemma 7.8.2, we have that it is the span of two elements of the regular (n - 1)-spread of P G ( 2 n -l,q) induced by the pseudo-conic. Now represent S' over QF(qn), so as a T 2 (C) = Q(4, qn) with C a conic of P G ( 2 , q n ) . Then the ovoids of T 2 (C) consisting of the point (oo) of T 2 (C) together with the affine points of a plane skew to C correspond to the elliptic quadrics of Q(4, qn) containing a fixed point. As a subtended ovoid can be subtended by at most two distinct points (by Section 2.5), and as there are q2n(qn - l ) / 2 elliptic quadrics on Q(4,g n ) containing a given point, it follows that each of the q2n(qn - 1) points of S \ S' of Type (ii) subtends a doubly subtended classical ovoid. Now let y be a point of <S \ S' not collinear with (oo), and let Oy be the corresponding subtended ovoid of S'. We now work in the T 2 (C) model of S'; assume T 2 (C) is contained in P G ( 3 , qn). As y ^ (oo), we have that Oy = A U {irp || p £ C}, where A is a set of q2n - qn points of Type (i) of T 2 (C) and ITP is a point of Type (ii) of T 2 (C) (a plane containing p e C). If a plane •K of P G ( 3 , qn) contains no point of C, then ( T T \ P G ( 2 , qn)) U {(oo)} is a classical ovoid subtended by two points, x and x' of S \ S', by the first part of the proof. If y is collinear with x or x', then ir n A is a single point. If y is collinear with neither x nor x', then {x, x', y} is a triad of S. Hence |7rn„4| = qn + l, as this is the number of centers of {x, x', y}. Next, suppose that 7T contains a unique point p of C. If -K = ixp C Oy, then ir contains no point of A. If -K ^ irp, then the qn lines of -K incident with p and not in the plane of C are lines of T 2 (C), and so contain precisely one point of A. Whence J7r n A\ = qn. Next, suppose that n contains two distinct points, p and p', of C. Of the qn + 1 projective lines in n incident with p, one is contained in PG(2, qn), one in ixp and qn - 1 are lines of T 2 (C) containing a unique point of A Hence J7rn^| = qn - 1. Finally, if 7r = PG(2, g"), then •K contains no point of A. Consider the set of points of PG(3,g™) defined byOy = AllC. Bythe above, the plane intersections with Oy have size 1 or qn + 1, and a straightforward count shows that Oy is an ovoid of P G ( 3 , qn). Further, since Oy contains C, it is an elliptic quadric by Theorem 7.8.1. Hence Oy is a classical ovoid of S' = Q(4, qn). So every point of S outside S' subtends an elliptic quadric in S'. By Theorem 7.5.4 the theorem now follows. • Corollary 7.8.4 An egg in P G ( 4 n - l,q), q even, contains a pseudo-conic if and only if it is classical. •

7.9. Elation Generalized Quadrangles with a Subquadrangle

163

Further on in this chapter, we will improve these results by showing that an elation generalized quadrangle of order (q,q2), q even, with a subGQ Q(4,g) containing the elation point, arises from a nonsingular elliptic quadric in PG(5,q). This is taken from K. Thas [208]. The theorem is a corollary of a structural result (cf. Proposition 7.9.2 and Corollary 7.9.3) independent of the characteristic. Another corollary generalizes also a result of Brown and Lavrauw [27].

7.9 Elation Generalized Quadrangles with a Subquadrangle Recall that a center of transitivity a; of a GQ S is a point for which there is an automorphism group of S fixing x linewise and acting transitively on the points noncollinear with x. The next lemma is well-known (it can, for instance, be found in [197]). Lemma 7.9.1 Let S be a GQ of order (s, s2), s > 1, and suppose that S' is a subGQ of order s. Furthermore, assume that x e S' is a center of transitivity of S. Then x is also a center of transitivity of S', and there are s distinct subGQs of order s which all contain the same lines through x, and of which the sets of points not collinear with x partition the set of points of S not collinear with x. Proof. Let y, y', y ^ y', in S' both be not collinear with x. Let 6 be a whorl of S about x mapping y onto y'. If S'e ^ S', then S'C\S'e must be a subGQ of order (s, 1) (by Lemma 1.1.1), a contradiction. Whence S' n S'e = S' and it follows that a; is a center of transitivity for S'. The lemma now easily follows. • Proposition 7.9.2 (i) Let S^ be an EGQ of order (q, q2), q > 1, containing a subGQ S' of order q which has an axis of symmetry L incident with x. Then L is an axis of symmetry in S^x\ (ii) Let S^ be a GQ of order (q,q2), q > 1, with center of transitivity x, containing a subGQ S' of order q which has an axis of symmetry L incident with x. Then L is an axis of symmetry in S^. Proof, (i) Let S' be as in the statement of the proposition, and suppose G is the group of elations of S about x. Then there are q elements in the

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orbit S'G by Lemma 7.9.1, and by that same lemma they precisely intersect in the q + 1 lines of <S' which are incident with x, together with the points on these lines. Whence the elements of S' partition the lines of S not incident with x and concurrent with one of the q + 1 lines on x in S', and also the points of S not collinear with x. Let L be as in the statement of the proposition, and suppose M ~ L ^ M is not incident with x. Then there is precisely one element, denoted S'M, in S'G which contains M, and no other element of S'G contains a point of M\{LnM}. Take an arbitrary element 8 of GM- Then 8 fixes S'M. By Theorem 3.2.1, 8 induces a symmetry about L in S'M, so that 8 fixes all lines of S'M through each point of L. We obtain the following crucial property for G: (S) For each point ylL, y j^ x, each element of G fixes a constant number c = aq, a G N, of lines incident with y and different from L. We emphasize the fact that this constant is the same for each point on L different from x. Suppose that 8 ^ 1 is as above (8 fixes M ) . Suppose that 8 fixes some point z ~ x which is not on L. Then Theorem 3.2.1 implies that the fixed elements structure of 8 is a thick subGQ of order (q,q2), and so 8 = 1, contradiction. So 8 fixes / = q + 1 points, namely the points of L. Suppose that z is a point not collinear with x for which z9 ~ z ^ z6; then zze ~ L (otherwise z, z e ,proj L z are the points of a triangle). It follows that the number of points of S which are mapped by 8 onto a collinear point different from itself, is g — q3 + q2c. Applying Theorem 3.3.1, we obtain (g2 + l){q + l) + q3 + q3a = q3 + 1 mod q2 + q. Hence a + 1 = 0 mod q+ 1, so that c = aq > q2, hence c = q2. It follows that 8 is a symmetry about L of <S. (ii) By Bose and Shrikhande [18], see also Section 2.5, we have that for each z 1, containing a subGQ S' of order q which has at least one axis of symmetry L incident with x. Then S^ is a TGQ for the translation point x.

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165

(ii) Let <S(x) be a GQ of order {q,q2), q > 1, with center of transitivity x, containing a subGQ S' of order q which has at least one axis of symmetry L incident with x. Then S^ is a TGQ for the translation point x. Proof. By the proof of Proposition 7.9.2, we only have to obtain (i). Suppose that <S(Z) is as assumed in (i). By Lemma 7.9.1, one derives that <S' is an EGQ with elation point x (the stabilizer of S' in the elation group of S^ is an elation group of S'). Say this elation group of S' is G, and suppose Llx is an axis of symmetry in S'. Then clearly the full group of symmetries (of S') about L is a (normal) subgroup of G. By Theorem 3.3.18, it follows that S' is a TGQ with translation point x, so that each line incident with x in S' is an axis of symmetry of S'. By Proposition 7.9.2, each of these lines is also an axis of symmetry of S. Now the theorem follows from Theorem 3.3.17. • Remark 7.9.4 (a) Let <S(x) be as in (i) or (ii) of Proposition 7.9.2, and suppose there is a subGQ S' of order q which has a regular line L incident with x. Then by Theorem 3.3.13, L is an axis of symmetry of S'. So by Proposition 7.9.2, L is an axis of symmetry of S^. Whence Corollary 7.9.3 can also be restated in this way (and the same holds for some of the following results). (b) Let S^ be an EGQ of order (s, s2), s > 1, containing the s distinct subGQs Si, S2, • •., Ss of order s which share a line L incident with some point x of S, and which also share the same s +1 lines incident with x. Suppose L is regular in all these subGQs. Suppose that there is some line M ~ L ^ M which is not incident with x so that there is a group of automorphisms H of S which fixes x linewise, and acts transitively on the points of M different from L n M . Then there is an i e {1,2,..., s} so that M e <%. By Theorem 2.3.10 of [206], L is an axis of symmetry in «S», and a similar argument as in the proof of Proposition 7.9.2 leads to the fact that L is an axis of symmetry in S. Not much is known about EGQs which are also dual EGQs, and which are nonclassical. We will prove some specific results in Section 12.2. In O'Keefe and Penttila [90], some results on certain TGQs which are dual EGQs can be found. In Thas and K. Thas [192], see Chapter 8, a general theory is initiated about TGQs which are also dual TGQs. Also, if S is a thick nonclassical GQ with elation point x and elation line L, then by K. Thas

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and Van Maldeghem [215], or Van Maldeghem [231], one may assume that xlL. (If x IL, it follows easily that each point of <S is an elation point, and each line an elation line, and then [231] — see also Corollary 11.7.3 — implies that the GQ is classical or dual classical.) Here, we obtain a result on flock GQs which are also dual EGQs. Theorem 7.9.5 Let S = <S(.F) be a flock generalized quadrangle of order (q2,q)> 1 even, which is abo a dual EGQ. Then S is classical, i.e., S(T) = H(3,g 2 ). Proof. We look at the situation in the point-line dual 5(^ 7 ) r> of S (T). Then the line [oo] which corresponds to the point (oo) of S{F) is contained in q3 + q2 subGQs of order q; see Theorem 7.7.1. Suppose S' is such a subGQ. We may assume that the line [oo] is incident with an elation point x, so that by Lemma 7.9.1, <S' is an EGQ with elation point x. The line [oo] is an axis of symmetry of S(T)D, cf. Section 4.6, so that it also is an axis of symmetry of S'. Now by Corollary 7.9.3 we have that S is also a TGQ with translation point x. The conclusion follows from Theorem 4.5.1. • The following theorem is a nice corollary of Corollary 7.9.3. Theorem 7.9.6 (i) Let S^ be an EGQ of order (q,q2), q > 1, containing a subGQ S' of order q which is a TGQ with translation point x. Then is a TGQ with translation point x. (ii) Let S^ be a GQ of order (q,q2), q > 1, with center of transitivity x, containing a subGQ S' of order q which is a TGQ with translation point x. Then S^ is a TGQ with translation point x. Proof.

Each line of <S' incident with x is an axis of symmetry of S'.

•

The next theorem generalizes the result of Brown and Lavrauw to EGQs. Theorem 7.9.7 Let S^ be an EGQ of order (q,q2), q even, containing a classical subGQ S' of order q containing x. Then S^ = Q(5, q). Proof. The only classical GQ of order q, q even, is Q(4, q) = W(q), and this is a TGQ for each of its points. By Theorem 7.9.6, S^ is a TGQ. Now apply Theorem 7.8.3. •

7.9. Elation Generalized Quadrangles with a Subquadrangle

167

Let O be a generalized oval in PG(3n - l,q) = (O), q even. Then by Theorem 3.9.1 the qn + 1 tangent spaces to the elements of O meet in an (n - 1)-dimensional subspace 77 of P G ( 3 n - 1, q). It is easily seen that for any IT e O, O' = (O \ {ir}) U {77} is again a generalized oval. If O is a generalized conic, that is, if T(£>) = Q(4, qn), then O' is called a pointed generalized conic or a generalized pointed conic. We will not prove the next result in the present monograph. Theorem 7.9.8 Let S ^ = T(O) be a TGQ of order (qn,q2n), q even, containing a subGQ T ( C ) through (00), where O' is a pointed generalized conic in PG(3n - l,q). Then either qn = 4 and S ^ ^ Q(5,4), or qn = 8 and 5(00) ^ T 3 ( C " ) with O" a Suzuki-Tits ovoid o/PG(3,8). Proof.

See Brown and Lavrauw [27].

•

We now generalize to EGQs. Theorem 7.9.9 Let <S(°°) be an EGQ of order (qn,q2n), q even, containing a subGQT(0), where O is a pointed generalized conic in P G ( 3 n - l , q), having (00) as translation point. Then either qn — 4 and <S^°°) = Q(5,4), or qn = 8 and <S(°°) ^ T 3 ( 0 ' ) with O' a Suzuki-Tits ovoid o/PG(3,8). Proof. This theorem follows immediately from Theorem 7.9.6 and Theorem 7.9.8. •

Chapter 8

Translation Generalized Ovals In this chapter, we first introduce new objects called "translation generalized ovals" and "translation generalized ovoids", and make a thorough study of these objects. We then obtain several new characterizations of the T 2 (C) of Tits and the classical generalized quadrangle Q(4, q) in even characteristic, including the complete classification of 2-transitive generalized ovals for the even case. We also prove that translation generalized ovoids do not exist.

8.1 Translation Generalized Ovoids and Translation Generalized Ovals Let 6 be an involution in the projective general linear automorphism group PGLfc + i(q) of the projective space PG(/c, q), with q even. Then the set of all fixed points of 9 is a subspace PG(r, q) of PG(fc, q), called the axis of 8. The fixed hyperplanes for 9 are all hyperplanes containing some subspace PG(fc - r - 1, q) of PG(fc, q), which is called the center of 8. Also, PG(fc r — l,q) Q PG(r,q) and so A: < 2r + 1. For any point x $ PG(r,q) we have xx9 n PG(fc - r - l,q) ^ 0. If TTI and 7r2 are mutually skew (k - r + 1)-dimensional subspaces of PG(fc, q), which are skew to a given PG(r,q) C PG(k,q), then there is exactly one involution 9 of PG(k,q) with axis P G ( r , q) mapping -K\ onto 7r2; the center of 9 is the subspace (7ri,7T2) n P G ( r , q ) . Further, all involutions of PG(k,q) with given axis PG(r, q) form an elementary abelian group. For more details on involutions of finite projective spaces we refer to Chapter 16 of [139]. Notice that in [139] the term "fundamental space" is used 169

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instead of "axis", and the term "fundamental axis" instead of "center". Let O be a generalized ovoid in PG(2n + m - l,q), n ^ m, respectively a generalized oval in PG(3n - 1, q), with q even. Then O is a translation generalized ovoid, respectively a translation generalized oval, with axis the tangent space P G w ( n + m - l , ? ) at P G w ( n - l , g ) e O, if there is a group of involutions of PGL 2 n +m(g) with axis P G ^ ( n + m-l,q), fixing O and acting regularly on O \ { P G ( i ) (n-l,q)}. If n = m = 1, then a translation generalized oval is just called a translation oval; if 2n = m = 2, it is called a translation ovoid. All translation ovals of PG(2, q), q = 2h, were determined by Payne in [107]; up to recoordinatization, they are always of the form { ( l , M 2 i ) || t £ G F ( g ) } U {(0,0,1)}, where i is fixed in { 1 , 2 , . . . , h - 1} and (h, i) = 1. Suppose S = (P,B, I) is a TGQ of order q n with translation point (oo). Let x T^ (oo) be a regular point. Then each point of V \ (oo)-1- is regular. Whence each point of <S is regular, and S = W(g ra ) by Theorem 2.2.2. Theorem 8.1.1 (Structure of Translation Generalized Ovoids and Ovals) (i) A translation generalized ovoid in PG(2n + m — l,q), q even, does not exist. (ii) Let O be a nonclassical generalized oval in P G ( 3 n — 1, q), with q even. Then the following are equivalent. (a) S = T ( 0 ) contains a regular point x different from the translation point (which is necessarily collinear with (oo),); (b) the point-line dual SD of S is a TGQ (with as base-point the line x(oo) ofS); (c) O is a translation generalized oval with axis the tangent space of O at its element x(oo). Proof. Let O be a nonclassical generalized oval in P G ( 3 n - 1 , q). Suppose a; is a regular point different from (oo). Then by the observation preceding the theorem, x is collinear with (oo). Hence each point yla;(oo), y ^ (oo), is regular. By Corollary 3.3.9, also the point (oo) is regular. The translations of T(C) fixing the point x, induce elations, with as axis the set of points

8.1. Translation Generalized Ovoids and Translation Generalized Ovals

171

of x(oo), of the projective plane TTX defined by the regular point x of S. Hence TTX is a translation plane having as translation line the set of points on x(oo). Then by Thas [185], SD is a TGQ where the base-point is the line x(oo) of 5. So S contains a regular point x different from the translation point if and only x ~ (oo) and SD is a TGQ with as base-point the line x(cx>) of <S.

Suppose that SD is a TGQ with as base-point the line x(oo) of S. Then the translation group of SD has a subgroup of size qn acting regularly on the lines, distinct from a;(oo), incident with (oo). Remark that all translations of SD are involutions. Further, let x(oo) = TT and let r be the tangent space of O at TT. We have TT0 = TT, T0 = T and rf = r\, with r? the kernel of O. Clearly the axis of 9 is contained in T, as otherwise elements of O \ {TT} would be fixed by 9. If TT\ e O \ {TT}, then (TTI, TT0) PIT = TT' belongs to the center of 9, so is a subspace of the axis of 6. Let u e TT'. The unique line containing u and intersecting r\ and TT is fixed by 9, and it follows that also TT and T] belong to the axis of 9. Hence T is the axis of 9 and TT' is the center of 9. Hence O is a translation generalized oval with as axis the tangent space of O at its element x(oo). Finally, suppose that O is a translation generalized oval with axis the tangent space T of O at TT, where TT is the line x(oo). Let G(0) be the group of involutions with axis T which acts regularly on O \ {TT}. Embed P G ( 3 n — l,q) = (O) in PG(3n, q). Let 77 be an arbitrary 2n-dimensional subspace of PG(3n, q) which is not contained in PG(3n - 1, q), and which contains r. Then there is a group G' of involutions of PG(3n, q) with axis r\ which induces G(G) in PG(3n - 1, q); the corresponding automorphism group of T(O) will also be denoted by G'. The axis 7 is a point of T(O) and G' fixes 71- pointwise. As \G'\ = qn, the point 7 is a center of symmetry, and so is a regular point by Theorem 3.3.4. Part (ii) of the theorem is proved. Now suppose that O is a translation generalized ovoid in P G ( 2 n + m — 1 , q), so n < m, with axis T (T being the tangent space of O at the element IT). Then by the final part of the proof of Part (ii), each (n + m)-dimensional subspace of PG(2n + m,q) D P G ( 2 n + m — l,q) which is not contained in PG(2n + m — l,q) and which contains r, is a regular point of T(O). Hence s > t, so n > m, contradiction. The theorem is proved. • Remark 8.1.2 (On the Notion of "Translation Generalized Oval/Ovoid") Let O be a generalized ovoid, respectively generalized oval, in PG(2n + m - l,q), and let TT G O. Suppose r is the tangent space of O at TT. Suppose there is a subgroup G of P r L 2 n + m (q) that fixes r pointwise and

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that acts regularly on O \ {K}. We will call this Property (*) for the moment. Then by an argument similar to the one in the final part of the proof of Theorem 8.1.1, each (n + m)-dimensional subspace of PG(2n + m,q) 3 PG(2n + m-l,q) which is not contained in PG(2n + m — l,q) and which contains T, is a regular point of T(O). So n = m. It easily follows that also (oo) is regular, and so, by Theorem 2.1.5, q is even. (Alternatively by 1.5.2(f) of [128] it follows that q is even and so, by Theorem 2.1.5, (oo) is regular.) Let Ll(oo) be the line of regular points of T(0) that corresponds to the space r of PG(3n - 1, q). Then in the same way as in the first part of the proof of Theorem 8.1.1, the dual T(0)D of T(0) is a TGQ with as base-point the line L. It follows that G is an elementary abelian group, so that each nontrivial element of G is an involution. Hence a generalized oval, respectively ovoid, satisfies Property (*) if and only if it is a translation generalized oval, respectively ovoid. This is the reason why we only defined translation generalized ovals and ovoids in even characteristic. Corollary 8.1.3 (see also Chapter 12 of [128] and K. Thas [206]) Let O be a nonclassical oval in PG(2,g), q even. Then S = T 2 (C) contains a regular point x different from the translation point if and only if' x ~ (oo) and SD is a T 2 (C) with base-point the point corresponding to x(oo), if and only if O is a translation oval with axis the tangent space of O at its element a;(oo). • Remark 8.1.4 Tits [221] defines a translation ovoid ("ovoide a translation") O of P G ( n , I ) , n = 2,3 and K a (not necessarily finite) field, in a slightly different way; here O is a set of points such that for each point x e O the union of the tangent lines of O at x is a hyperplane of PG(n, K), called the tangent space of O at x. He demands that for each tangent space T to O, there is a group of elations with axis T, fixing O globally, and acting regularly on the points of O not incident with T. Amongst many other results, he then proves that PG(n, K), n = 2,3, contains a translation ovoid if and only if the characteristic of K is 2 and there is a subfield L of K for which (n - 1)[K : L] < [L : L2]. Suppose that K is finite, i.e., K is isomorphic to GF(2 r ) for some r. If n = 3, then clearly no such subfield exists, so PG(3,2 r ) does not contain translation ovoids in the sense of Tits. If n = 2, then the only possibility is that L = K = GF(2 r ), and in this case the translation ovoid is a conic of PG(2, K). Let S = T(0) be a TGQ of order s, with s = qn even. Let O C PG(3n 1, q), and write O = {it, Tt\,..., wqn.}. Suppose n is the nucleus of O, that is,

8.1. Translation Generalized Ovoids and Translation Generalized Ovals

173

the common (n -1)-dimensional space of all tangent spaces of O. Consider the (2n - l)-space r = (IT, 77) (that is, the tangent space of O at 71-). For each i e {1,2,..., qn}, the set { 7 r,r ? }U{{7r i ,7r i )nr||i^j} is an (n - 1)-spread of r, denoted 7^. Note that if O is a translation generalized oval with axis T, then all the %'s coincide. It is also important to note that if all the 7^'s coincide, we have that if 7 is an element of that spread distinct from it and 77, and if j G {1,2,..., qn}, then (7,itj) contains precisely one other element irr of O (and hence is disjoint from itu if k ^ j , r). Suppose that all the (n — 1)spreads 7^ coincide. We say that O is projective atr if the following property holds: Let 7 be an element of T = % (for all i), where it ^ 7 ^ 77, and let j •£ k be in {l,2,...,qn} such that (7,^} ^ (l,itk)- By the above, there are elements ity, j ' ^ j , and -K^, k! ^ k, so that 7iy C (7,^) and 7Tfc' c (7,7Tfc). Then (itj, itk) n (7Tfe/, ity) is an element of T. Remark 8.1.5 Each translation oval O of PG(2, s) with axis L, s even, has the (in that case trivial) property that the %s coincide, and also is projective at L. It should be noticed that for a generalized oval the fact that the %s coincide, does not imply that O is projective; this follows from Theorem 8.1.6 below, and the existence of ovals in PG(2, q), q even, which are not translation ovals. Suppose all the %s coincide, and that the generalized oval O is projective at (ir, 77) = r. Let 7r, and Wk be arbitrary distinct elements of O \ {it}, and suppose that (TTJ, irk) HT = r ^ . Let 6jk be the involution with axis r, center Tjk, and which maps TTJ onto irk- Then 9jk fixes O globally, and sends TTJ onto Tifc. If then G is the group consisting of all the involutions 6jk, j ^ k, and j , k € {1,2,..., qn}, together with the identity, then it is clear that G is a group of involutions with axis T, acting regularly on O \ {-K}. Whence O is a translation generalized oval. So we have the following theorem. Theorem 8.1.6 Let O be a generalized oval in PG(3n - l,q), q even, and use the above notation. Then O is a translation generalized oval with axis r if and only if all the %s coincide and O is projective at r. •

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Note that for n = 1, this is the well-known geometrical characterization of translation ovals in PG(2, q) {q even).

8.2

Note on the Definition of Translation Generalized Oval/Ovoid

In this section, we make the following useful observation. Theorem 8.2.1 A generalized oval O o / P G ( 3 n - 1, q), q even, is a translation generalized oval with axis T, where r is the tangent space ofOatireO, if and only if there is a subgroup of involutions o / P G L 3 n ( g ) which fixes n and acts regularly on the remaining elements of O. Proof. Let G be a group of involutions as stated. Let 9 be a nontrivial element of G. Suppose r\ is the kernel of O. Let n be an arbitrary ndimensional subspace of P G ( 3 n - 1, q) containing IT but not contained in T, and suppose x is a (2n - 1)-dimensional subspace of PG(3n -l,q) containing r\ and meeting T in r\. First of all, one notes that Ue ^ II. For, suppose that this is not the case. Let IT' be the unique element of O \ {IT} intersecting II. Then n' = TT', clearly a contradiction. For each p e x, put pa = (p, ir)e Hx- Then a is a linear involutory automorphism of x- As a does not fix any point of x \ V> it follows that a must fix each point of r\ (as the dimension of r\ is n - 1). But 0 and a have the same action on n, so that 9 must fix r/ pointwise. By the same argument, 9 also fixes IT pointwise, so that r is fixed pointwise by 6. Now suppose ( e 0 \ {IT}. Then (CC") n (7r,??) belongs to the center, so that the center has dimension at least n - 1. So the dimension of the axis is at most 2n - 1, from which it follows that r is the axis of 9. The theorem readily follows. • From the proof of Theorem 8.2.1 we have the following useful observation. Observation 8.2.2 Let Obea generalized oval o / P G ( 3 n — l,q),q even, and let T be the tangent space of O at the element IT G O. Suppose that there is an involution in PGL 3 n (q) which stabilizes O, which fixes n and does not fix any of the remaining elements of O. Then 9 is an involution with axis r. • Remark 8.2.3 It should be remarked that Theorem 8.2.1 is not true for generalized ovoids. For, consider the GQ Q(5, qn), q even, and suppose p

8.3. Characterizations oftheT2(0)

of Tits

175

is a point of Q(5,g ,n ). Then p is a translation point, and Aut(Q(5,g n )) p contains a subgroup H which acts naturally as PSL2(g2™) on the set of lines incident with p. In particular, for any line Lip, the stabilizer of L in H contains a normal elementary abelian 2-subgroup which acts regularly on the lines incident with p and distinct from L. So, if O is the generalized ovoid in P G ( 4 n — l,q) corresponding to the TGQ Q(5, qn) , and r is the tangent space of O at IT e O, where IT corresponds to L, then there is a subgroup G of involutions of PGL 4 „(q) which fixes IT and acts regularly on the remaining elements of O. But G can not fix r pointwise by Theorem 8.1.1.

8.3

Characterizations of the T 2 (O) of Tits

Not many characterizations of the T2(£>) of Tits, O an oval in PG(2,s) with s even, are known. It is the goal of this section to provide new characterizations of these objects relying on Theorem 3.9.7 and the theory of translation generalized ovals. Theorem 8.3.1 Let <S(°°) = T(O) be a TGQ of order s with base-point (oo). Suppose there is some line Ll(oo) for which the associated projective plane irL is Desarguesian. Suppose that x ^ (oo) is a regular point, x I L. Then O is a regular translation generalized oval, i.e. S = T2(0') of Tits for some translation oval O' o / P G ( 2 , s), and conversely. Proof. First suppose that x is not collinear with (oo). Then since <S(°°) is a TGQ, each point not collinear with (oo) is regular. It follows that each point is regular, and then <S(°°) = W(s) by Theorem 2.2.2. Also, s is even by Theorem 2.1.5. The theorem follows. Now suppose x ~ (oo). By 1.5.2(f) of [128] s is even (alternatively, all points of the line x(oo) are regular, so s is even by Theorem 2.1.5). Then Theorem 8.1.1 implies that O is a translation generalized oval with axis the tangent of O at x(oo). The result now follows from Theorems 8.1.6 and 3.9.7. The converse is obvious. The following theorem is very general.

•

Chapter 8. Translation Generalized Ovals

176

Theorem 8.3.2 Let S = <S(°°) = T(£>) be a TGQ of order s, s even, with base-point (oo). Suppose there is a line LI (oo) for which the associated projective plane TTL is Desarguesian. Suppose abo that there is a line M ^ L which is incident with (oo), and an involution 8 of Aut(<S)(oo) which fixes M pointwise and which does not fix any of the lines incident with (oo) and different from M. Then S £* T 2 (C) of Tits for some oval O' o/PG(2, s). Proof. Let it, respectively n', be the element of O corresponding to L, respectively M, and let r' be the tangent space of O at -IT'. Suppose 8 is as in the theorem. Then by Observation 8.2.2, 8 fixes r' pointwise and does not fix any element of 0\{ir'}. So n and ire induce the same spread T in T'. Moreover, since TTL is Desarguesian, we have that T is regular. Now apply Theorem 3.9.7. • Corollary 8.3.3 Let S = <S(°°) = T(£>) be a TGQ of order s, s even, with base-point (oo). Suppose there is a line LI (oo) for which the associated projective plane -KL is Desarguesian. Suppose that there is a nontrivial symmetry about some point x I L. Then S = T 2 (C) of Tits for some oval O' of PG(2,s). Proof. If x is not collinear with (oo) then one observes that each point of S is a translation point, so that each point is regular. The corollary then follows from Theorem 2.2.2 and Theorem 1.5.2(i). If x ~ (oo), we consider a nontrivial involution in the group of symmetries generated by the given one, and then the corollary follows from Theorem 8.3.2. •

8.4 A Characterization of Translation Generalized Ovals LetO = {71-, 7ri,... ,7r9n}be a generalized oval in PG(3n-l,g), with q — 2h even, and let r be the tangent of O at n. Now we define a point-line incidence structure A(0) as follows: • POINTS are the elements of O \ {n}; • LINES are the pairs {7r,,7rj}, with i, j e {1,2,... ,qn} and i ^ j ; • INCIDENCE is containment.

8.5. Classification of2-Transitive Generalized Ovals in Even Characteristic

177

Hence A(O) is the complete graph with vertex set O \ {IT}. Further, two lines {-Ki, -Kj) and {71-fc, 17} are called parallel if either {71-j, TTJ} = {-Kk, -K{\ or (•Ki.TTj) DT = (lTk,1Tl) PIT.

Theorem 8.4.1 The incidence structure A((D) provided with parallelism is isomorphic to the hn-dimensional ajfine space AG(hn, 2) over GF(2) if and only if O is a translation generalized oval with axis r. Proof. Assume that A(O) provided with parallelism is isomorphic to AG(hn, 2). If {wi,TTj} is a line of A(0) and -Kk $ {TT^TTJ}, then there is exactly one line of A(0) containing irk and parallel to {71^,7^}. Also, if the distinct lines {7^,77,} and {71-fc, 717} are parallel, then also {71^, 71-fc} and {ITJ,TTI} are parallel. Hence, by Theorem 8.1.6, O is a translation generalized oval with axis r. Conversely, assume that O is a translation generalized oval with axis r. Then, by Theorem 8.1.6, the qn (n - l)-spreads of r defined by the respective elements of O \ {-K} coincide, and further O is projective at TT. NOW by [98] the point-line incidence structure A(O) provided with parallelism is isomorphic to AG(hn, 2). • Remark 8.4.2 Let P G ( k , 2) be the projective completion of the affine space A(0). Then the points at infinity of A ( 0 ) , that is, the points of PG(/m, 2) not in A(0), can be identified with the elements of T \ {n, 77}, where T is the common (n - 1)-spread of r defined by the elements of O \ {TT} and 7? is the kernel of O. Any line of P G ( k - 1 , 2 ) = PG(hn, 2)\ A(C) is of the form {a,/?, 7}, with a = (iri,irj) n (7^,7^), (3 = (7ri,7rfc) n {TTJ,TTI}, 7 = (-Ki,-KI) n (7Tj, 7Tfc) and 7T,, 7Tj, 7Tfe, 7T; distinct elements of 0 \ {7r}.

8.5

Classification of 2-Transitive Generalized Ovals in Even Characteristic

We call a generalized oval O in P G ( 3 n -l,q) 2-transitive if there is a subgroup of PTL3n(q) which stabilizes 0 and acts 2-transitively on its elements. Note that by Thas and K. Thas [191], cf. Theorem 3.7.2, this is equivalent to asking that the automorphism group of T(0), which we may demand to fix the translation point (00) (as otherwise T(O) = Q(4,
178

Chapter 8. Translation Generalized Ovals

at the equivalent problem for the corresponding TGQs. From the theorem will follow which groups arise. Theorem 8.5.1 Let S = S ^ ) = T(O) be the TGQ of order {qn, qn), q even, which arises from a 2-transitive generalized oval O in PG(3n — l,q). Then S = Q(4, qn), and so O is classical. Proof. Since q is even, the point (oo) of T(O) is regular. The plane # , , which is the dual of TT^, is a translation plane of order qn, where the parallel classes correspond to the lines of S incident with (oo). We suppose that Aut(S) fixes (oo) for reasons of conveniency (otherwise S = Q(4, qn)). Then Aut(<S) acts 2-transitively on the parallel classes of 7I-P*. We have the following possibilities by Schulz [136] and Czerwinsky [42], see also Liineburg [101] (Chapter VI): (a) TTP, is Desarguesian, and Aut(S) has a subgroup K which induces PSL 2 ((7 n ) on the line at infinity of nPy (b) nP, is a Liineburg plane (so that n is even, and h odd, with q = 2h — cf. also the next chapter), and Aut(<S) has a subgroup K which induces the Suzuki group Sz( v /g") on the line at infinity of irPy We take K in such a way that the translation group G of S is contained in K. Let a be an involution of K which fixes the element n e O but not all elements of O; then -K is the only element of O which is fixed by a. We regard a as being an element of ~PVL5n(q). Since h is odd, a is a linear involution. By Observation 8.2.2, a is an involution with axis r, the latter being the tangent space of O at n. As PSL 2 (g n ) and Sz( v / q") both are generated by their involutions (a fact that can be proved, for instance, by using their representations given in the next chapter), we have proved that the nucleus 77 of O is fixed pointwise by K (regarded as a subgroup of PGL 3 „(g)). Let z' be a point of <S not collinear with (00). Then since S is a TGQ, it is easy to see that Kz< = K' still induces PSL 2 (g"), respectively Sz( v /g") J on the lines incident with (00). Let LI(00) be arbitrary, and consider a Sylow 2-subgroup 5 2 of K'L. As 5 2 — seen as a subgroup of PGL 3 n ((/) — fixes 77 pointwise, S2 fixes each line of S incident with projL2;' (and 5 2 acts sharply transitively on the set of lines incident with (00) and different from

8.5. ClassiScation of 2-Transitive Generalized Ovals in Even Characteristic

179

L). If we interpret 5*2 as an automorphism group of nL, it thus follows that 5*2 also must fix L pointwise (since in a finite projective plane any central collineation is also axial, and conversely). The group Gp ro j z, has order q2n and fixes L pointwise, so together with 5*2, it generates a group H such that, if N is the group of symmetries about L, H/N acts sharply transitively on the set of spans {[/, V} ±J - of nonconcurrent lines U,V e L-1, and fixes L pointwise. So H/N is a translation group of 7r£ of order q2n, and hence it is elementary abelian. In particular S2 = S2N/N also is. So PGL3n(g)e> has a subgroup of involutions that fixes 7r and acts sharply transitively on O \ {n}. By Theorem 8.2.1, we conclude that O is a translation generalized oval with axis r. So any point incident with L is regular. As L was an arbitrary line incident with (00), it follows that each point of (oo)-1- is regular, so that each point of S is regular (cf. 1.3.6(iv) of [128]). It follows that S ^ Q(4,qn) by Theorem 2.2.2, so that Case (b) cannot occur. Hence by Theorem 1.5.2 and §3.6, O is classical. •

Chapter 9

Moufang Sets and Translation Moufang Sets

In this chapter, we gather some results from group theory, that we will need later on. The central notion is the notion of a Moufang set. This can be viewed as a generalization of a one-dimensional translation group. We begin without any finiteness restriction, but later on, when dealing with classification, we will have to consider only finite Moufang sets.

9.1 Definition and General Results A Moufang set (on the set X) is a system (X, (Ux)xex) consisting of a set X,\X\ > 2, and a family of groups Ux of permutations of X indexed by X itself and satisfying the following conditions. (MSI) Ux fixes x € X and is sharply transitive on

X\{x}.

(MS2) In the full permutation group of X, each Ux normalizes the set of subgroups {Uv || y £ X} (that is, it acts on the latter by conjugation). The groups Ux will be called root groups. The elements of Ux are often called root elations. If Ux is abelian for some x G X, then for all x e X and we call the Moufang set a translation Moufang set. If (X, (Ux)x€X) is a Moufang set, and Y c X, then Y, \Y\ > 2, induces a sub Moufang set if for each y eY, the stabilizer (Uy)y acts sharply transitively on Y \ {y}. In this case (Y, ((Uy)Y)yeY) is a Moufang set. 181

182

Chapter 9. Moufang Sets and Translation Moufang Sets

The group S generated by the Ux, for all x e X, is called the little projective group of the Moufang set. A permutation of X that normalizes the set of subgroups {Uy || y e X}, is called an automorphism of the Moufang set. The set S of all automorphisms of the Moufang set, is a group G, called the full projective group of the Moufang set. Any group H, with S < H < G, is called a projective group of the Moufang set. We have the following easy lemma. Lemma 9.1.1 (i) The little projective group S of a Moufang set (X, (Ux)X£x) acts doubly transitively on the set X. (ii) A permutation group H (acting on X) is a projective group if and only if Ux
9.1. Definition and General Results

183

A group element g is said to act freely on a set S, if the group generated by g does not contain any nontrivial element that fixes x G S. A group acts freely on 5 if only the identity fixes some x e S. Proposition 9.1.2 Let (G, X) be a sharply transitive group. (i) Then we can identify X with G and the action of g G G on the element x G X = G is given by right multiplication xg. (ii) Suppose some permutation u commutes with every element of G = X. Then u acts freely on X and the action ofu is given by left multiplication with some element h G G (so u maps every x G X to hx). (iii) If a transitive permutation group H centralizes G, then H acts sharply transitively on X and there is an isomorphism ip : H —» G such that the action ofheHonG = Xis given by h : x \-> h~vx. If in particular G is abelian, then H = G. (iv) Suppose the sharply transitive permutation groups H and G normalize each other. Then either G and H have a nontrivial intersection, or G and H centralize each other. Proof, (i) Take any fixed element x e X and identify each element g e G with xs G X. Then (x9)h gets identified with gh, and (i) follows. (ii) Now put X = G. If [u, G] = {id}, then every element of (u) centralizes G, so we only have to show that u acts fixed point freely if it is nontrivial. Let e be the identity in G and suppose that u maps e to ft G G. If g G G is arbitrary, then gu = (eg)u = (eu)g = hg (the second equality uses the fact that u commutes with g), hence u maps g to hg. (iii) By (ii), H acts freely on X and hence sharply transitively, and there is a bijection 9 : H —> G such that the action of ft G H on X = G is given by left multiplication with he. If fti, ft2 G H, and g e G, then (h1h2)eg = ghlh2 = (ghi)h2 = heheg> s o (hlh2)e = he2h{. Putting y : H -» G : h .-> (ft 6 )" 1 , we see that (p is an isomorphism and so we can write he = (h~ly = h~v. (iv) If G and H intersect trivially, then from [g,ft]= g~1h~1gh e H9h C H and g~1h~1gh G g~lGh = G we deduce that [g,ft]is trivial, for all g e G and all h G H. •

184

Chapter 9. Moufang Sets and Translation Moufang Sets

By (iii) of the previous proposition, and its proof, it follows that for every sharply transitive permutation group G acting on a set X, there is a unique sharply transitive permutation group H acting on X and centralizing G in Sym(X). We say that the action of H is opposite the action of G and denote H by G°PP. Note that, if G is abelian, then the two actions are obviously the same. We now get back to the Moufang sets. A first question that can be asked is: do we really need all the root groups in the definition? This is one answer. Proposition 9.1.3 Let X (\X\ > 2) be a set and let a,b G X be distinct. Let Ua and Ub be two permutation groups acting on X such that Ua fixes a and acts sharply transitively on X \ {a}, while Ub fixes b and acts sharply transitively on X \ {b}. Then Ua and Ub are root groups of at most one common Moufang set. Also, they are root groups of a Moufang set if and only if for each u G U£, u ^ id, there exists v e Ub such that U% = Ujf, if and only if Ua is normal in Ga, Ub is normal in Gb, and Ua is conjugate to Ub in G:=(Ua,Ub). Proof. Suppose first that (X, (Ux)xex) is a Moufang set. For given u € Ua \ {id}, define v e Ub by av = bu. Then (MS2) implies that Uva = Ua« = Ub* =Ubu.By the doubly transitivity of G, it maps Ua under conjugation to Ub- Also, since G normalizes the Ux, the stabilizer Ga of a must normalize Ua, hence Ua < Ga and likewise Ub
9.1. Definition and General Results

185

and so, by assumption, U™ lg 1 = Ua, hence Uva = U%u = U%. The previous paragraph completes the proof of the proposition. • The condition that for each u e C/ax there exists v G Ub such that U% = Ug is used by Timmesfeld [216] to define rank one groups. But a rank one group has the additional condition that all root groups are nilpotent, and we do not want this restriction. For a set X, elements a, b e X (with \X\ > 2), and permutation groups Ua, Ub of X such that Ua (respectively, Ub) fixes o (respectively, b) and acts sharply transitively o n I \ {a} (respectively, X \ {&}), we call (X, Ua, Ub) a Moufang triple if, for G = (Ua, Ub), Ua is normal in Ga, Ub is normal in Gb, and Ua is conjugate to Ub in G. We can use the previous proposition to recognize easily sub Moufang sets. Corollary 9.1.4 Let M = (X,(Ux)xex) be a Moufang set. Suppose that Y C X, with \Y\ > 3, and consider two distinct elements oo,0 e Y. Let Voc and Vo be the permutation groups induced on Y by the stabilizers (£/oo) y and (UO)Y, respectively. Then (Y, V^, Vo) is a Moufang triple — and hence induces a sub Moufang set of M — if and only if Vx and Vo act transitively on Y \{oo} and Y \{0}, respectively. Proof. It is clear that the stated transitivity condition is necessary. We now show that it is also sufficient. First we observe that the group H generated by V^ and Vo acts doubly transitively on Y. Let h e H be such that oo'' = O and Oh = oo. Then h can be written as product of elements of Voo U Vo, and reading these elements as belonging to U^ U [/"o, we see that there is some element g e (f/oo, Uo) such that G/Y = h. It follows that v£ is the restriction to Y of (C/oo)y, and hence this must be equal to Vo- Hence V^ and Vo are conjugate in H. Similarly one shows that V^ and Vo are normal in Hoc and Ho, respectively. The assertion now follows directly from Proposition 9.1.3. • Proposition 9.1.3 tells us something about the possibility of two root groups to be contained in the same Moufang set. Now we want to look at the situation where we have two Moufang sets acting on the same set X and sharing at least one root group, and we want to find conditions under which these Moufang sets are the same. More precisely, we have the following result.

186

Chapter 9. Moufang Sets and Translation Moufang Sets

Theorem 9.1.5 Let (X,(Ux)xGx) be a Moufang set, and let a, b £ X be distinct. Moreover, let Wb be a permutation group acting on X, fixing b and acting sharply transitively on X\ {b}. Suppose that (X, Ua, Wb) is a Moufang triple. If UbC\Wb is nontrivial, or if Uh and Wb normalize each other, then Wb = Ub- If Ub and Wb centralize each other, then Ub = Wb and the Moufang set is a translation Moufang set. Proof. Let M be the Moufang set determined by Ua and Ub, and let Af = (X, (Wx)x€x) be the one determined by Ua and Wb, with Ua = Wa. First suppose that v e Ub n Wb is nontrivial. Then Wav = Wl = Uva = Uav is a root group in both M and Af, and since these Moufang sets are both determined by Ua and !/ 0 ., they must coincide. Hence Ub = Wb. Now suppose that Ub centralizes Wb. Then the action of Wb is opposite the action Ub. Take arbitrary elements u e Ua, v e Ub and w e Wb. Then [v,w] = id, hence [vu ,wu] = id and we see that the action of Wb* is opposite the action of Ubu. li\X\ = 3, then clearly Ub = Wb is commutative and the result follows. Hence we may assume |X| > 4, so that we can take two distinct elements c, d in X \ {a, b}. We then see that jV is determined by the permutation groups Wc and Wd the actions of which are opposite those of Uc and Ud, respectively. By the doubly transitivity of the little projective group G(M) of M, there is a permutation g e G(M) interchanging a with b. Since [Ub, Wb] = {id}, we also have [U^Wj]} = {id}. Hence, if we denote Ua by Va, then, since Ujj = Ub9 = Ua, and since W{] acts sharply transitively o n I \ { a } , w e obtain Wb9 = Va.lfc9 = c' and d9 = d!, then similarly, one easily shows 9 that W C, = Wc and W|, = Wd- Hence the conjugate Af9 of M contains the root groups Wc and Wd and hence coincides with Af. Comparing the root groups fixing b, we see that Ub = Wb, and [Ub, Wb] = {id} implies that Ub is abelian. Hence M = N is a translation Moufang set. Finally suppose that Ub and Wb normalize each other. If they share a nontrivial permutation of X, then by the first part of the proof Ub = Wh. If they intersect trivially, then by Proposition 9.1.2(iv) they centralize each other, and the second part of our proof implies again Ub = Wb. •

9.2 Finite Moufang Sets In this section, our aim is to comment on the following classification result of finite Moufang sets.

9.2. Finite Moufang Sets

187

Theorem 9.2.1 Let M = (X, (Ux)xeX) be a finite Moufang set with little projective group S and full projective group G. Then one of the following cases occurs. (2T) S acts sharply doubly transitively on X, there is a prime number p and a positive integer n such that \X\ = pn, and S contains a normal sharply transitive subgroup N, which is an elementary abelian p-group. (Ch) S is a Chevalley group, there exists a prime number p and a positive integer n such that \X\ =pn + l, and Uxis a p-group ofnilpotency class at most 3. We have one of the following cases: - S = PSL2(p"), pn > 4, is simple and X is the projective line

PG(l,p"); - nisa multiple of3,S^ is the Hermitian unital

PSU 3 (p™/ 3 ), pn > 21, is simple and X Un(pn^3);

- p = 2, n = 2n' is even, n' is odd, S = Sz(2 n ), n' > 3, is simple and X is the Suzuki-Tits ovoid C ST (2™'); -

= 3, n = 3n', n' is odd, S ^ R(3™'), n' > 1, is simple for n' > 3, and X is the Ree-Tits unital UR(3n'); if n' = 1, then R(3) = P r L 2 (8) has a simple subgroup of index 3, which coincides with the commutator subgroup of S.

p

In all cases, we have that G is the full automorphism group of S. Also, the root groups are precisely the Sylow p-subgroups of S.

We will not prove this theorem here, but the interested reader can find proofs in Shult [143] and Hering, Kantor and Seitz [69]. We will content ourselves with some comments on this theorem, and with an explicit definition of the Moufang sets appearing in (Ch) above. It should be noted that in the papers [143, 69], the term "split BN-pair of rank 1" is used instead of "Moufang set". Concerning the sharply doubly transitive case, it follows from Frobenius' Theorem that there is a normal sharply transitive subgroup N acting on X. The group Ux, for arbitrary x e X, clearly acts transitively on iV x by conjugation, hence all elements in N have the same order and N is a p-group. Since every p-group has a nontrivial center, and since Ux acts transitively on Nx ,we see that ,/V must be elementary abelian. It follows

188

Chapter 9. Moufang Sets and Translation Moufang Sets

that |X| is a prime power. The standard example here acts on a finite field GF(g), for some prime power q, and the actions are given by GF(g) -> GF(g) : x >-> ax + b, with a e GF(q) \ {0} and b e GF(q). The normal sharply transitive subgroup consists of the maps with a = 1. A second class of examples is given by a slight modification of the previous example. If q is odd and a perfect square, then we retain the previous maps with a a square in GF(q) \ {0}, and substitute the other maps with the maps GF(q) —• GF(q) : x i-> ax^ + b, with a a nonsquare in GF(g). We call it the nonstandard example related to GF(q). We now turn to the examples under (Ch), the Chevalley type groups. As a general remark, we would like to point out that in the following descriptions we are primarily interested in the finite case, but we choose the notation in such a way that also the general (infinite) case is covered, disregarding nevertheless noncommutative fields.

9.2.1 The Case PSL2(g) This is the prototype of all Moufang sets. Let K be any (finite) field, and denote by PG(1, K) the projective line over K. So we may identify PG(1,K) with the set of all 1-dimensional subspaces of a given 2-dimensional vector space I x l . The elements of PG(1, K) can be written as (a, b), with a,b e K, (a,b) ^ (0,0) and (a,b) identified with (ca,cb) for all c e Kx. We set O = (0,1) and oo = (1,0). We define U0 as the (multiplicative) group of matrices (k)0 := (

), k e K, Uoo consists of the matrices

(fc)oo := ( , 1 ), for all fc e K, and the action of a matrix M on an element (a, b) is by right multiplication (a, b)M (conceiving (a, b) as a (2 x 1)matrix). It is easy to calculate that U^}° = UQ ) O °, hence by Proposition 9.1.3, we indeed have a Moufang set acting on PG(1,K). Since l ^ is abelian, we have a translation Moufang set. We denote this Moufang set by M(PG(1,I)). The little projective group is equal to PSL 2 (K) and is simple if |K| > 4. For |K| = 2, we have the unique Moufang set on 3 points, and for |K| = 3, we obtain the unique Moufang set on 4 points. Both are improper Moufang sets, isomorphic to the standard examples related to GF(3) and GF(4), respectively.

9-2. Finite Moufang Sets

9.2.2

189

The Case PSU3(g)

Let K be a field having some quadratic Galois extension F. This means that F admits some field involution a and the elements of F fixed by a are precisely those of the subfield K. In order to treat all characteristics at the same time, we introduce the following notation. For an arbitrary i G F \ K, we can write any element k € F as

a

a

a

k"i -ki" —. :

k- ka + -. —i,

l — la

i — ia

_1

with both (ki - k i)(i - i ) and (k - ka)(i - i")"1 in K. So F can be regarded as a 2-dimensional vector space over K and the map F —> K : x i-> x + x" is K-linear, nontrivial, hence surjective with 1 -dimensional kernel. We denote the inverse image of the element k G K by K^. Now we define our set X. It is the set of all projective points (x0,xi,X2), up to a multiplicative nonzero factor, of the projective plane PG(2,F) satisfying the equation XQX^ + x^x2 = x\x\. We can write X = {(1,0,0)} U {(jfc, x, 1) j| x e F, k G K ^ " ) } , and also X = {(0,0,1)} U {(1, x, k) || x G F, k e K^*")}. We set O = (0,0,1) and oo = (1,0,0). The group of collineations Uoo = {{x,k)oo || x GF,fcGK( xx ' T )},where(x,fc) 00 isthecollineationofPG(2,F) defined by the linear transformation with matrix

acts sharply transitively on X \ {oo} (on the right!) and fixes oo. Likewise, the group Uo = {(x,k)o || x G F, k G K ^ " ) } , where (x,k)o is the collineation of PG(2, F) defined by the linear transformation with matrix

fixes O and acts sharply transitively o n I \ {O}. It is also easy to calculate that, if k G K^'K then k~x e K«x/fc)(x/fc)a>, and also that rrC^/rOoo _ O —

u

Tr(x/k,l/k)o

u

oo

190

Chapter 9. Moufang Sets and Translation Moufang Sets

By Proposition 9.1.3, we obtain a Moufang set, which we denote by M{H(2,F,a)), or when |K| = q, briefly by M(H{2,q2)), since in this case a is uniquely determined and given by a : x \-^ xq. We call it a Hermitian Moufang set. The little projective group of M(H(2,¥,a)) is the unitary group P S U 3 ( F , a), denoted just PSU 3 (g) when |K| = q. It is a simple group whenever |K| > 3 and has order q3(q3 + \){q2 - \)/{q + 1,3). Note that \X\ = q3 +1 in this case. For |K| = 2, PSU 3 (2) is isomorphic to the sharply doubly transitive nonstandard example related to GF(9), and hence is not simple (but solvable). There is a little proposition that can be proved now. Proposition 9.2.2 Let U^ = {(a;, k)x \\ x e F, k e K^ 35 ")} be, as above, a root group of the Moufang set M(H(2,¥,a)), with |F| > 4. Then U^ is nilpotent of class 2 and the following subgroups of U^, coincide: (i) the center Z(Uoo); (ii) the derived group [[7^, [7^]; (iii) the subgroup of elements in Uoo that induce a central collineation in PG(2, F) (necessarily with center (1,0,0)). If \K\ = q, then {U^ = q3 and |Z(t/oo)| = q. Proof. It is easy to calculate that all these subgroups coincide with {(0,fc)oo|| k G K(°)}. The other assertions are obvious. •

9.2.3 The Case Sz(q) Let K be a field of characteristic 2, and denote by K 2 its subfield of all squares. Suppose that K admits some Tits endomorphism 9, i.e., the endomorphism 6 is such that it maps x9 to x2, for all x e K. If |K| = 2 n , then n must be odd, say n = 2e + 1, and 8 maps x to xT , for all x e K. Let Ke denote the image of K under 9. In the finite case K6 = K. Let L be a subspace of the vector space K over Ke, such that tfcl (this implies that L \ {0} is closed under taking multiplicative inverse). In the finite case L = K. We also assume that L generates IK as a ring. We now define the Suzuki-Tits Moufang set M(Sz(K, L, 9)).

9.2. Finite Moufang Sets

191

Let X be the following set of points of PG(3,K), where the coordinates with respect to some given basis are: X = {(1,0,0,0)} U {(a2+e + aa' + a'6,1, a', a) || a, a' e L}. One now calculates that X = {(0,1,0,0)} U {(1, a2+e + aa' + a'6, a, a') || a, a' £ L}. Wesetoo = (1,0,0,0) and O = (0,1,0,0). Let (x, a;')oo be the collineation of P G ( 3 , K) determined by / (x0 xi x2 x3)

H^

(x0 xi x2 x3)

V

1 0 0 0\ x2+9 + xx' + x' 1 x' x x 0 10 1+e x + x' 0 xe 1 /

and let (x, x')o be the collineation of P G ( 3 , K) determined by

(x0

XI X2 X3) H^ (X 0 XI X 2 X 3 )

/lx2+^+xx' + x ' % x ' \ 0 1 0 0 0 x1+e + x' 1 x6 \0 x 0 1/

Define the groups )oo || x,x' G L} and C/0 = {(x,x')o

|| x,x' G L}.

Both groups [/oo and [/o act on X, as an easy computation shows (for Uo use the second description of X above), and they act sharply transitively o n I \ {(1,0,0,0)} and X \ {(0,1,0,0)}, respectively. Moreover, one can check that (U0){x'x')a° = ([/ 00 ) (!/ ' J/ ' )c, ) with _./ 2-2+61 _)_ xxi

_ _|_ xit

and y'

c2+e + xx,

+ x,b

It follows from Proposition 9.1.3 that we obtain a Moufang set M(Sz(K,L,6)), called a Suzuki-Tits Moufang set. The group Sz(K,L,0) is the Suzuki group generated by Uoo and Ifo. If |K| = g, then we denote it by Sz(g). One has |Sz(g)| = q2(q2 + l)(g - 1) and \X\ = q2 + l. All Suzuki groups are simple groups unless |K| = 2. The group Sz(2) is a sharply doubly transitive standard Moufang set related to GF(5).

Chapter 9. Moufang Sets and Translation Moufang Sets

192

In order to understand better the structure of Uoo, we can identify each point (a2+e + aa' + a'e, 1, a', a) of X \ {oo} with the ordered pair (a, a'). Then the unique element of U^ that maps (0,0) to (b, b') is given by {b,6')oo : {a,a') ^{a + b,a' + b' + abe). The root group Uoo is hence given abstractly by the set {(a,a')oo II a, a' G £} with operation (a, o')oo © {b, &)'<x> = (a + 6, a' + 6' + ab 6 )^. We can now state the following proposition. Proposition 9.2.3 Let U^ be as above (a root group of the Moufang set M{Sz(K,L,9))), and suppose \Uoo\ > 4. Then Uoo is nilpotent of class 2 and the following subgroups of Uoo coincide: (i) the center Z(Uoo); (ii) the derived group [Uoo, Uoo}; (hi) the subset of elements in U^ which fix at least two points o/PG(3, K); (iv) the subset of elements of order 1 and 2 of UooIf\K\ = q, then \Uoo\ = q2 and \Z(Uoo)\ = q. Proof. The equality of the subgroups under (i), (ii) and (iv) follows from an elementary and easy computation, using the abstract description of Uoo above. For (iii) we consider the matrix representation of U^o and one easily calculates that (0, o')oo fixes the line spanned by (1,0,0,0) and (0,0,1,0) pointwise. If a ± 0, then (a, a')oo only fixes (1,0,0,0), as the reader can verify for himself. • Remark 9.2.4 The Moufang set .M(Sz(K, L, 9)) can also be defined as the set of absolute points of a polarity in a certain generalized quadrangle. We explain this in detail further on. 9.2.4

The Case R(g)

Let K be a field of characteristic 3, and denote by K3 its subfield of all third powers. In the finite case, K3 is just K. Suppose that K admits some Tits endomorphism 9, i.e., the endomorphism 9 is such that it maps xe to x 3 , for all x G K. In the finite case, |K| is a power of 3 with odd exponent, say

9.2. Finite Moufang Sets

193

|K| = 3 2 e + 1 , and 0 maps x to a;3 . Let Ke denote the image of K under 0. In the finite case, again necessarily Ke = K. We now define the Ree-Tits Moufang set M(R(K, 0)). For a, a', a" e K, we put „"\ - _„4+20 fi(a,a ' ,a")

f2(a,a

',an"\) =- -_ ^ a3+9 ,

/ 3 (a, a', a") = - a

3+2e

aa

+a

/0

a

T •

a

+a

+a

-a a

- a a ,

2 I

•a — aa • a a , - a"6 + aea'9 + a'a" + aa'2.

Let X be the following set of points of PG(6,K), where the coordinates with respect to some given basis are: X = {(1,0,0,0,0,0,0)} U{(f1(a,a',a"),-a',-a,-a",l,f2{a,a',a"),f3{a,a',a"))

\\ a,a',a"

A tedious calculation shows that X = {(0,0,0,0,1,0,0)} U{(1, f2(a, a', a"),h{a,

a', a"), a", fx{a, a', a"), - a ' , - a ) || a, a', a" e

We set oo = (1,0,0,0,0,0,0) and O = (0,0,0,0,1,0,0). Let (x, x', x")oo be the collineation of PG(6, K) determined by (XQ XI

(

x2 x 3 x 4 x5 xe) i-> (x 0 xi x 2 x 3 x 4 x 5 x6)

1 0 0 0 1 0 —x p X6 1 x' - x 1 + e a 1 0 0 x" fi(x,x' x") -x' —x -x" x' — xi+e 0 0 0 X 0 0 0

0 0 0 \ 2 —x" — xx' 0 x r s 0 0 X -x1 1 hi* , X x")f3(x,x',x") 0 1 -xe 0 0 1 )

with p = x3+e - x'6 q = x"6 + xex

•f x'x" — xx'

r = x"~ xx' + x „ /

X

1+(

2

r

2+e

V _TV

x2+ex>

_

xwx»

„3+2e

GK}.

Chapter 9. Moufang Sets and Translation Moufang Sets

194

and let, with the same notation, (x, x', x")o be the collineation of PG(6,1 determined by (xo xi x-i X3 X4 x5 xe) "-> (xo xi x2 x3 x\ x5 XQ) ^ 1 f2(x,x',x")

V

0 0 0 0 0 0

1 0 —x 0 X2

r

f3(x,x',x")

x"

-X6

0 0 1 0

1 x' 0 —x" — xx' X i+e_ x s

fi(x,x',x") x' _xi+e X

-x" 1 p

x'

Q

—x' — X '

0 0 0 0 1 xe

0 0 0 0 0

w

For the computations, it can be interesting to note that (x,x',x")o (x, x', x")^, with g the collineation of PG(6, K) determined by (x0,Xi,X2,X3,X4,X5,Xe)

H->

=

(xi,X5,Xe,-X3,X0,Xl,X2)-

Define the groups ^oo = {{x,x'',a;")oo || x,x'7x" e K} and U0 = {(x,x',x")o

\\ x,x',x" € K}.

The groups U<x, and Uo both act on X, as an easy computation shows (for Uo use the second description of X above), and they act sharply transitively o n I \ {(1,0,0,0,0,0,0)} and X \ {(0,0,0,0,1,0,0)}, respectively. Moreover, it can be checked that (t/0)(*.*V)oo = (jjoo^v,y',y")o> w j t h

y' =

y" =

f3(x,x',x") fi{x,x',x"y f2(x,x',x") fi(x,x',x"Y x" fi(x,x',x")'

Hence, by Proposition 9.1.3 we obtain a Moufang set, which we denote by A/((R(K, 6)) and call a Ree-Tits Moufang set. Its little projective group, generated by Ux and Uo, is the Ree group R(K, 9). If |K| = q, then we denote the corresponding Ree group by R(g). In this case we have |R(g) | = q3(q3 + l)(q - 1). All Ree groups are simple groups unless q = 3, in which case R(3) = PrL 2 (8). Hence [R(3),R(3)] ^ PSL2(8) is simple, but not sharply doubly transitive, because it has order 504.

9.2. Finite Moufang Sets

195

Following 1.1.1 of [232], we can now define U^ abstractly as follows. Define (a, a', a") := (/i(a, a', a"), -a', -a, -a", 1, / 2 (a, a', a"), /3(a, a', a")). Then the unique element of Uoo that maps (0,0,0) to (x, x', x") is given by (x,x',x")oo

'• (a,a',a")

t—y (a + x, a' + x'+ axe ,a"+ x" + ax' — ax — axl+6).

The root group Uoo can now be defined as the set {(a, a', a")^ || a, a', a" e K} with operation (a,a',a")oo®{b,b',b")oo

= (a + b,a' + b' + ab0,a" + b" +

ab'-a'b-ab1+e)oo.

We have the following proposition. Proposition 9.2.5 Let U^ be as above (a root group of the Moufang set M(R(K,9))). Then Uoo is nilpotent of class 3 and the center Z(JJoo) coincides with the second central derived group [Uoo,Uoo][2]- Also the following subgroups of Uoo coincide: (i) the derived group [Uoo, Uoo]; (ii) the subset of elements in U^ which fix at least two points o / P G ( 6 , K); (iii) the subset of elements of order 3 of UooAlso, the derived group [Uoo, Uoo] is abelian, hence Uoo is solvable of length 2. Finally, if\K\ = q, then \UX\ = q3, {[U^, Uoo]\ = q2 and \Z{Uoo)\ = q. Proof. The coinciding of Z(Uoo) and [Uoo, E^»][2] is readily verified using the above abstract description of Uoo- Likewise, the equality of the subgroups under (i) and (iii) follows from an elementary and easy computation. For (ii) we consider the matrix representation of Uoo and one easily calculates that (0, x', x")<x> fixes the plane spanned by (1,0,0,0,0,0,0), (0,0,0,0,0,0,1) and (0, x'2,0, -x'x", 0, x"2 - x'l+e, 0) pointwise. If x ^ 0, then {x,x',x")oo only fixes (1,0,0,0,0,0,0), as the reader can verify for himself. • Remark 9.2.6 The Moufang set A4(R(K, 9)) can also be defined as the set of absolute points of a polarity in a certain generalized hexagon, see 1.1 of [232].

196

9.2.5

Chapter 9. Moufang Sets and Translation Moufang Sets

Sub Moufang Sets

We end this chapter with describing which finite Moufang sets are sub Moufang sets of other finite ones. First consider M(PG(1, q)). Clearly if GF(q') is a subfield of GF(q), then by considering a projective subline over GF(q') of PG(l,g), we obtain a sub Moufang set isomorphic to M(PG(l,q')). Up to isomorphism there are no other sub Moufang sets of M(PG(l,q)). Note that the improper standard Moufang sets related to any field cannot be isomorphic to sub Moufang sets of M(PG(l, q)) unless they are isomorphic to M(PG(1,2)) or M{PG(1,3)), although they are somehow related to affine lines, which can be viewed as subgeometries of projective lines. Next consider M{H(2, q2)). Again, M(H.(2, q'2)) is isomorphic to a sub Moufang set if and only if GF(g'2) is a subfield of GF(g2) but not of GF(g). But here, also M{PG(1, q)) is a sub Moufang set, together with all its sub Moufang sets mentioned in the previous paragraph. Indeed, with the notation of Subsection 9.2.2, we restrict X to the projective points with second coordinate 0, i.e., to the set {(1,0,0)} U {(fc,0,1) || k + ka = 0}. Also, we restrict [Too to the elements (0, k)^, with k + k" = 0 and with Uo to the elements (0, k)o, with k + k? = 0. There are no other sub Moufang sets up to isomorphism. The only sub Moufang sets of the Suzuki-Tits Moufang set M(Sz(q)) are isomorphic to the Suzuki-Tits Moufang sets M{Sz(q')), where GF(q') is a subfield of GF(q). No projective lines to be found here! Finally, concerning the Ree-Tits Moufang set M(R(q)), we have the sub Moufang sets over a subfield, M(R(q')), with GF(g') a subfield of GF(q), and also M(PG(l, q)), together with all its sub Moufang sets. This can be seen as follows. With the notation of Subsection 9.2.4, we restrict X to the point set Y = {(x'1+$ ,-x',0,0,1, x'e ,0) \\ x' G K} U {(1,0,0,0,0,0,0)}. Then one easily sees that the commutative groups {(0,x',0)oo || x' e K} and {(0,x',0)o || x' e K} act sharply transitively on Y \ {oo} and Y \ {O}, respectively. Corollary 9.1.4 now implies that we indeed obtain a sub Moufang set, which is readily seen to be isomorphic to M{PG(l,q)). There are no other sub Moufang sets up to isomorphism. Remark 9.2.7 Let M(R(q)) be a Ree-Tits Moufang set on the set X. Recall that |X| = q3 + 1. Let B be the collection of subsets of size q + 1 on which a sub Moufang set isomorphic to M(PG(1, q)) exists. Then the space UR = (X, B) is a unital, i.e., a system of points and blocks such that every

9.2. Finite Moufang Sets

197

two points are contained in a unique block, and, for some natural number q, the number of points is equal to q3 +1, and the number of points on each block is q + 1. This unital is called the Ree-Tits unital. The Hermitian curve H(2, q2) is also a unital, if endowed with all secants. In fact, this Hermitian unital can also be constructed as follows. Let M{H(2, q2)) be a Hermitian Moufang set on the set X. Let B be the collection of subsets of size q + 1 on which a sub Moufang set isomorphic to M(PG(l,q)) exists. ThenUH = (X,B) is that unital.

Chapter 10

Configurations of Translation Points Let S^ be a TGQ with translation point p. Then an easy exercise yields that we have the following possibilities for its configuration of translation points: (a) p is the only translation point; (b) there is a line of translation points (incident with p); (c) every point is a translation point. If we are in Case (c), then by Theorems 10.5.3 and 10.7.1 below, and Theorem 2.2.2, the TGQ is isomorphic to one of Q(4, q), Q(5, q) for some q. Often in the existing literature, one speaks of "the translation point" of a given nonclassical TGQ S. So in view of the hypothetical Class (b), the question arises whether there are nonclassical examples with a configuration of translation points as described in (b). In the papers [86, 201, 202] and the monograph [206], a more general class of GQs was studied, namely the "span-symmetric generalized quadrangles" ("SPGQs"). The defining property is that they have nonconcurrent axes of symmetry. The main result of Kantor [86] and K. Thas [201] is that an SPGQ of order s (s > 1) necessarily is isomorphic to Q(4, s) (and then s is a prime power). So since each SPGQ of order s (s > 1) is classical, each nonclassical TGQ of order s has a unique, and whence well-defined, translation point. Surprisingly there are nonclassical examples of TGQs of order (s, s2), s > 1 and s an odd prime power, with distinct translation points, so TGQs for which the notion "translation point" is not uniquely defined! This was first 199

200

Chapter 10. Configurations of Translation Points

noted in [205]. From the paper [202], a classification of all GQs in the Class (b) could then be deduced. In this chapter, we will describe several aspects of the theory of SPGQs (which have direct implications for TGQs), state the main results, along with a number of lemmas (without proofs) by which we will point out the general ideas of the proofs of these main results. Some lemmas have an elementary (combinatorial) proof, and sometimes a sketch of a proof will be given. We will then elaborate on the consequences for the theory of TGQs. Also, we will comment on the classification of TGQs which lie in (c), in particular on the proof of K. Thas [199] which does not use the deep group theory of Fong and Seitz [59, 60].

10.1 Span-Symmetric Generalized Quadrangles Suppose S is a GQ of order (s, t), s,t^\, and suppose L and M are distinct nonconcurrent axes of symmetry; then it is easy to see by transitivity that every line of {L, M}-11- is an axis of symmetry, and <S is called a spansymmetric generalized quadrangle (SPGQ) with base-span {L, M}^^. Note that \{L, M}-1-1] = s + 1, as L and M are regular lines by Theorem 3.3.4. Let 5 be a span-symmetric generalized quadrangle of order (s,t), s,t ^ 1, with base-span {L, M}-1-1. Throughout this chapter, we will use the following notation. First of all, the base-span will always be denoted by £. The group which is generated by all the symmetries about the lines of £ is G, and sometimes we will call it the base-group. This group clearly acts 2-transitively on the lines of £, and fixes every line of £-*-. The set of all the points which are on lines of {L, M} XJ - is denoted by Q (of course, Q, is also the set of points on the lines of {L, M } x ; we have that \{L, M } x | = \{L, M}^\ = s + 1). We will refer to r = (fi, £ U £-•-, I'), with I' being the restriction of I to (fi x (£ U £-L)) U ((£ U £ ± ) x n), as being the base-grid. The substructure of fixed elements of an element of the base-group is given by the following theorem. Recall that a partial spread of a GQ is a set of mutually nonconcurrent lines. Theorem 10.1.1 Let S be an SPGQ of order (s,t), s ^ 1 ^ t, with basespan £ and base-group G. If 0 / 1 is an element of G, then the substructure Se = (Pe, Be,I$) of elements fixed by 0 must be given by one of the following.

10.2. Groups Admitting a 4-Gonal Basis

201

(i) Ve = 0 and Be is a partial spread containing C1-. (ii) There is a line L e Cfor which Ve is the set of points incident with L, and M~Lfor each M e B9 (C1- C B6). (iii) Be consists of C1- together with a subset B' of C; Ve consists of those points incident with lines of B'. (iv) Se is a subGQ of order (s, t') with s < t' < t. This forces t' = s and t = s2.

Proof. See 10.7.1 of [128].

•

We also mention the next result. Theorem 10.1.2 If S is an SPGQ of order (s,t), s,t ^ 1 and t < s2, with base-group G, then G acts regularly on the set of (s + l)s(t - 1) points of S which are not on any line of C. Proof. See 10.7.2 of [128].

•

10.2 Groups Admitting a 4-Gonal Basis Let S be an SPGQ of order s ^ 1 with base-span C, and put £ — {Lo,Li,...,Ls}. The group of symmetries about Li is denoted by d, i = 0 , 1 , . . . , s. One notes the following properties (see [114] and 10.7.3 of [128]): (1) the groups Go, Gi, • • •, Gs form a complete conjugacy class in G, and are all of order s, s > 2; (2) GinNG(Gj)

=

{l}foii^j;

(3) dG-j n Gk = {1} for i,j, h distinct, and (4) \G\ =

s3-s.

We say that G is a group with a A-gonal basis 55 = {G0, G\,..., Gs} if these four conditions are satisfied. It is possible to recover the SPGQ <S of order s, s > 1, from the base-group G starting from 4-gonal bases, in the following way, see [114, 128].

202

Chapter 10. Configurations of Translation Points

Suppose G is a group of order s 3 - s with a 4-gonal basis 03 = {G 0 , Gi, • . ., Gs}, and let G* = NG(Gi) for i = 0 , 1 , . . . , s. Define a point-line incidence structure S
POINTS.

(a) Elements of G. (b) Right cosets of the G*s. •

B
of three kinds of

LINES.

(i) Right cosets of Gi, 0 < i < s. (ii) Sets Mi = {G*g \\ g e G}, 0 < i < s. (iii) Sets U = {G*g || G* n G* 3 = 0,0 < j < s,j ^ i } U {G*}, 0 < i < s. •

INCIDENCE. I«B is the natural incidence: a line dg of Type (i) is incident with the s points of Type (a) contained in it, together with that point G*g of Type (b) containing it. The lines of Types (ii) and (iii) are already described as sets of those points with which they are to be incident.

Then 5® = (V^s, 6
See 10.7.8 of [128].

•

It is also important to recall the following from [128]. Theorem 10.2.2 Let S be an SPGQ of order s =£ 1, with base-span £. Then every line o / £ x is an axis of symmetry.

10.3. SPGQs and Moufang Sets

Proof. See 10.7.9 of [128].

203

•

So for any two distinct lines U and V of £-*-, S is also an SPGQ with basespan {£/, V}-1-1. The corresponding base-group will be denoted by G-1. It should be emphasized that this property only holds for SPGQs of order s, s > 1, as will follow from later observations. Now suppose G is a group of order s3 - s, s > 1, where s is a power of the prime p, and suppose that G has a 4-gonal basis 05 = {Go, G\,..., Gs}. Since the groups Gi all have order s, all these groups are Sylow p-subgroups in G. Since 03 is a complete conjugacy class, this means that every Sylow p-subgroup of G is contained in 03, and hence G has exactly s + 1 Sylow p-subgroups. We have proved the following theorem. Theorem 10.2.3 Suppose G is a group of order s3 — s, s > 1, with s a prime power. Then G can have at most one A-gonal basis. In particular, if G has a A-gonal basis, then it is unique. • As a corollary we obtain the following fundamental result, see [206]. Theorem 10.2.4 Suppose Sis an SPGQ of order (s,t),l<s
10.3 SPGQs and Moufang Sets By the following theorem, the base-group of an SPGQ is essentially known.

Chapter 10. Configurations of Translation Points

204

Theorem 10.3.1 Suppose S is an SPGQ of order (s,i), s,t ^ I, with basespan £ and base-group G. Let N be the kernel of the action of G on the lines of £. Then G/N acts as a sharply 2-transitive group on £, or is isomorphic (as a permutation group) to one of the following: (i) PSL 2 (s); (ii) the Ree group R( ^fs); (iii) the Suzuki group Sz(y / s); (iv) the unitary group PSXJ3(
10.4 Basic Structural Lemmas Let S be an SPGQ of order (s, t), with s ^ 1 ^ t. We keep using the standard notation introduced earlier. Suppose that s
The following set of lemmas can be found in Kantor [86] and K. Thas [201]. One can also use [206] as a general reference. Lemma 10.4.1 The group G is a perfect group if (G/N,£) 2-transitive, and if G/N ¥ R(3). Proof.

See Lemma 7.6.2 of [206].

Lemma 10.4.2 N is a subgroup of the center ofG.

is not sharply

•

10.4. Basic Structural Lemmas

205

Proof. Let H be a full group of symmetries about any line L of £. Then H and N normalize each other, so N n H = {1} implies that they centralize each other. The lemma now follows as G is generated by the symmetries about the lines of C. • Lemma 10.4.3 We have that G/N either acts naturally as PSL 2 (s) on C, or sharply 2-transitively. Proof. See Lemma 7.6.5 of [206].

•

The proof of the following lemma uses the theory of universal central extensions of (perfect) groups. Lemma 10.4.4 If G/N acts as PSL 2 (s), then G ** SL2(s) and S ^ Q(4, s). Proof. See Lemma 7.6.6 of [206].

•

It remains to consider the case where G/N acts sharply 2-transitively. First suppose that G/N acts sharply 2-transitively on the lines of C, and that s = t. Every line of £-L is an axis of symmetry by Theorem 10.2.2. Since the lines of C1- are also axes of symmetry, we can assume that the base-group G1- corresponding to these lines also acts sharply 2-transitively on C1 — otherwise G1- is isomorphic to SL2(s), and then <S is classical by Theorem 10.2.4. Hence G and G1- contain normal central subgroups N and N1- which act trivially on the points of tt, both of order s - 1 (where fi is the set of points on the lines of the base-span). Also, G and G1- act regularly on the points of S not in £1 by Theorem 10.1.2. Let a; be a point and L a line of a projective plane II. Then II is said to be (x, L)-transitive if the group of all collineations of II with center x and axis L acts transitively on the points, distinct from x and not on L, of any line through x. The following theorem is a step in the Lenz-Barlotti classification of finite projective planes, see e.g. [45,145]; it states that the Lenz-Barlotti Class III.2 is empty. Theorem 10.4.5 (Yaqub [239]) Let II be a finite projective plane, containing a nonincident point-line pair (x, L) for which II is (x, L)-transitive, and

206

Chapter 10. Configurations of Translation Points

assume that II is (y, xy)-transitive for every point y on L. Desarguesian.

Then II is •

As every axis of symmetry L of a GQ is a regular line, there is a projective plane irL canonically associated with L (by Theorem 2.2.1) if the GQ has order s (s ^ 1). Using Theorem 10.4.5, one then obtains the following result. Theorem 10.4.6 Suppose that S is an SPGQ of order s, where s / 1 , with base-group G and base-span C. Also, let N be the kernel of the action of G on the lines of L, and suppose that G/N acts as a sharply 2-transitive group on the lines of C. Then S is isomorphic to Q(4,2) or Q(4,3). Proof.

See Theorem 7.7.2 of [206].

•

Now suppose that s s and t 7^ s +1, we have t = s + 2. Since there are nontrivial symmetries about lines of <S, Theorem 3.3.2 implies that s is even. The paper [86] (see Chapter 7 of [206]) contains a group theoretical treatment for the case s < t < s2, using Maschke's Theorem, cf. Gorenstein [64], which was also needed in [206] to handle the case t = s — 2 with s even. We now give a combinatorial solution also for that case, which completes the combinatorial approach already largely obtained in [206]. As G/N is a sharply 2-transitive group, it contains a normal elementary abelian p-group A of size s + 1, with s + 1 = ph for some h (this follows from the fact that (G/N, C) is a Frobenius group). Call A' the subgroup of G for which A'/N = A; as A is normal in G/N, it now follows easily that A' is normal in G. Let a; be a point of S \ O; then there are two distinct lines incident with x that do not hit T, so that the total number of such lines is s(s + 1)2. So

10.4. Basic Structural Lemmas

207

the stabilizer in G of a line M which does not contain a point of Y has size at least (s + l)/2. Since s + 1 = ph with p odd, it follows that the size of the stabilizer is exactly s + 1 (recall that G acts regularly on the points of S \ fi), and that the action of the stabilizer on the points of M is regular. As A' is a Sylow p-subgroup of G, we have that GM < A', and \MA' \ = s + 1. As N < Z(G) acts transitively on the lines of M A , we have that GM fixes each line in MA . If two distinct lines of MA' would intersect, then any element of GM fixes their common point, clearly a contradiction. Hence MA is a partial spread of the GQ. The set of all points incident with lines of MA is an orbit R of A'. If mlM and M is the second line incident with m which does not contain a point of Q,, then similarly R is the set of all points incident with lines of MA'. So R is the point set of a grid having as lines MA' U MA . It easily follows that MG is a partial spread of S and that MGUCJ- is a spread of S. Consider any i e {0,1,..., s}, take any point mlM and suppose O is the unique line on m meeting L» e £. As any orbit R' of A' is the point set of a grid, it contains at most one point of O. Hence O has a unique point in common with every orbit of A1. Take any 9 £ Gt and j3 e GM- Then Me/3 is concurrent with O13 and belongs to the spread MG U £-"-, so M9/3 = Me. It easily follows that GM fixes every line of the spread, while acting sharply transitively on the points of each of these lines. By definition, this means that C1- U MG is a spread of symmetry. Before proceeding, we need a nice construction, first observed by Payne, of GQs of order (s - 1, s + 1) from (thick) GQs of order s with a regular point.

The Payne Derivative Let s be a regular point of the GQ S = (V, B, I) of order s, s > 1. Define an incidence structure V{S, x) = S' = {V, B', I') as follows: • The POINT SET V is the set V \

x^.

• The LINES of B' are of two types: - the elements of Type (a) are the lines of B which are not incident with x; - the elements of Type (b) are the hyperbolic lines {x, y}1-1- where y i- x. • INCIDENCE I' is containment (regarding a line of S as a set of points).

208

Chapter 10. Configurations of Translation Points

Then S' is a GQ of order (s - 1, s + 1), which we call the Payne derivative of <S with respect to x. Theorem 10.4.7 (De Soete and Thas [49]) Let S be a GQ of order (s, s + 2), with s > 1, and suppose S has a spread of symmetry T. Let H be the group of automorphisms of S that fix T linewise. Then there exists a GQ S' of order s + 1 with center of symmetry x, such that S^V{S',x). Moreover, T arises from the set of hyperbolic lines containing x, and H is induced by the group of symmetries about x. • So there exists a GQ S' of order s + 1 with center of symmetry (oo) such that V{S', (oo)) = S, and £ x U MG = V arises from the hyperbolic lines containing (oo), while GM is induced by the symmetries about (oo). Moreover, as G preserves T', a result of De Winter and K. Thas [50] states that G is induced by an automorphism group of S' that fixes (oo). Now consider a nontrivial symmetry <j> of S about some fixed line Lj of the base-span C Then Lj is a line of Type (a) in S', and is induced by a collineation of S' that fixes (oo), every point on Lj (as a line of S'), and is a whorl about y with ylLj, y •/ (oo). This contradicts Theorem 3.2.1, and ends the combinatorial proof of the sharply 2-transitive case. •

10.5 Classification of SPGQs of Order (s,t),l < s
10.6. SPGQs of Order

(s,s2)

209

Theorem 10.5.2 A finite group is isomorphic to SL2(s) for some s if and only if it has a 4-gonal basis. • It also follows that the order of an SPGQ is essentially known. Theorem 10.5.3 Let S be an SPGQ of order (s, t), s,t^l.

Then t e {s, s2}.

10.6 SPGQs of Order (s,s2) In [202], the author proved that if S is a span-symmetric generalized quadrangle of order (s, s2), s / 1 and s odd, with base-span £, then <S contains s + 1 subquadrangles, all isomorphic to the classical GQ Q(4, s), which mutually intersect in the base-grid T = (Q,C U £-L, I'). Also, the basegroup G acts freely on the points of S \ £1 and G = SL2(s). (Note that |G| = |SL 2 (s)| = s 3 - S . ) Already since the beginning of the study of SPGQs (late 1970's), it was believed (and posed as a problem to prove) that SPGQs of order (s, s2), s > 1, always have ("many") classical subGQs of order s, all passing through the base-grid. The solution of that problem in the odd case by K. Thas [202] involved a proof that was largely valid in the even case. In Chapter 7 of [206], the problem was completely solved for the general case. The fact that an SPGQ of order (s, s2) has s +1 subGQs isomorphic to Q(4, s) in this position endows the GQ with a lot of structure, and this observation is crucial for many applications in theory of SPGQs. For instance, Kantor used this to classify what he calls "grid-symmetric generalized quadrangles" — these are SPGQs for which the perp of the base-span also is a base-span, see [87]. Other applications will be obtained further on. Here, we give a brief account of the proof. The reader is referred to [206] for further details. For the rest of this section, let S be an SPGQ of order (s, t), s ^ 1 ^ t, with base-grid T = (fi,£ U £-*-, I') and base-group G, and let Li,Gj, etc. be as before. If p is a point which is not an element of f2, and U is a line through p which meets fl in a certain point q!Lk of Y, then clearly every point on U which

Chapter 10. Configurations of Translation Points

210

is different from q is a point of the G-orbit which contains p. We will use this observation freely throughout this chapter. The proof of the next lemma was first obtained in K. Thas [202]. Lemma 10.6.1 Suppose S is an SPGQ of order (s,t), s / 1 / t, with basegrid T and base-group G. Then G has size at least s 3 - s. Proof. If s = t, then we already know that G has order s 3 — s, so suppose s f t (so that t = s2 by Theorem 10.5.3). Set C = {L0, LI, . . . , Ls} and suppose Gj is the group of symmetries about Li for all feasible i. Suppose p is a point of S not in Q, and consider the following s + 1 lines JVj := proj p L;. If A is the G-orbit which contains p, then every point of Nt not in fl is also a point of A. Now consider an arbitrary point q ^ p on N0 which is not on a line of C. Then again every point of proj^Li, i =£ 0, not in Q, is a point of A. Hence we have the following inequality: |A| > l + (s + l ) ( s - l ) + ( s - l ) 2 s , (10.1) from which it follows that |A| > s3 - s2 + s. Now fix a line U of £-"-. Every line of S which meets this line and which contains a point of A is completely contained in A U Q,. Also, G acts transitively on the points of U. Suppose k is the number of lines through a (= every) point of U that are completely contained in A U SI (as point sets). If we count in two ways the number of point-line pairs (u, M) for which u G A, M ~ U and ulM, then it follows that k(s + l)s

= |A| > s(s2 - s + 1)

3 ~+ n —— = s - 2 + — - , s+1 s+1 and hence, since k e N, we have that k > s - 1. Thus |A| > s 3 - s and so also \G\. M

, ^ ^ k >

S2 S 1

In the rest of this section, we can suppose that s > 3; if s = 2, then S = Q(5,2) by Section 1.8. In that case, there are s + 1 = 3 subGQs of order 2, isomorphic to Q(4,2) and mutually intersecting in the base-grid. The basegroup stabilizes each of these subGQs (as it is generated by symmetries), and acts freely on the points of S not in f2; whence G = SL 2 (2). Similar properties hold for the case s = 3.

10.6. SPGQs of Order {s,s2)

211

For y/i = s > 3, the strategy of handling the situation is essentially the same as that followed in Section 10.4. The aim again is to prove that G is a perfect central extension of G/N, when (G/N, C) is not sharply 2-transitive. The latter case has to be handled separately (in an entirely different way than in the case t < s2). Then, one eventually comes to the following result, using Lemma 10.6.1. Proposition 10.6.2 Suppose that S is an SPGQ of order (s, s2), s > 3, with base-span C = {U, V}1-1, base-group G and base-grid T = (fi, £ U Cx, I'). Furthermore, let N be the kernel of the action of G on C. Then G/N is isomorphic to PSL 2 (s). Also, G acts freely on S \ fi. Proof. See Section 7.10 of [206], in particular the Lemmas 7.10.2 — 7.10.7. • We know that G acts freely on the points of <S \ fl. Assume that G has order (s + l)s(s - 1). Let A be an arbitrary G-orbit in S \ f2, and fix a line W of £-"-. By the fact that G acts freely on the points of S \ tt, that \G\ = (s + l)s(s — 1) and that G acts transitively on the points incident with W, we have that any point on W is incident with exactly s - 1 lines of S which are completely contained in A except for the point on W which is in Q., and that every point of A is incident with a line which meets W (recall that G is generated by groups of symmetries). Now define the following incidence structure S' = S'(T) = (V, B', I'). • LINES. The elements of B' are the lines of S' and they are essentially of two types: (1) the lines of

C^uL^;

(2) the lines of <S which contain a point of A and a point of Q,. • POINTS . The elements of V are the points of the incidence structure and they are just the points of f2 U A. • INCIDENCE. Incidence I' is the induced incidence. Then S' is an SPGQ of order s. By Theorem 10.5.1, S' is isomorphic to the GQQ(4,s).

212

Chapter 10. Configurations of Translation Points

Theorem 10.6.3 Suppose S is an SPGQ with base-span C and of order (s, s2), s ^ 1. Assume that G has size s3 — s. Then there exist s + 1 subquadrangles of order s which are all isomorphic to Q(4, s), so that they mutually intersect in the base-grid F. Proof. From each G-orbit in S \ fl arises a subGQ of order s which is isomorphic to Q(4, s), and there are exactly s + 1 such distinct G-orbits. •

The Main Structure Theorem for SPGQs of Order (s, s2) We are ready to prove the main result for SPGQs of order (s,s2) [206, Chapter 7]. Theorem 10.6.4 Suppose S is an SPGQ of order (s, s2), s ^ 1. Then S contains s + 1 subquadrangles, all isomorphic to the classical GQ Q(4, s), which mutually intersect in the base-grid T. Also, the base-group G acts freely onS\Q, \G\ = s3-sandG^ SL2(s). Proof. The fact that G acts freely on S \ Q, implies that G is a perfect group, and also that G/N ^ PSL 2 (s) if ./V is the kernel of the action of G on £; see [206]. Moreover, ./V is contained in the center of G (recall the proof of Lemma 10.4.2). Thus, G is a perfect central extension of PSL 2 (s), and this leads to the fact that G = SL2(s) and that \G\ = s3 - s; see again [206]. The result follows from Theorem 10.6.3. •

10.7 Translation Generalized Quadrangles with a Line of Translation Points Although it was conjectured in 1983 by Payne that every SPGQ of order (s, s 2 ), s > 1, is isomorphic to Q(5, s), see PROBLEM 26 of [115], such a similar result as in the case t < s2 cannot hold. For, let /C be the quadratic cone with equation X0Xi = X2 of PG(3, q), q odd. Then the q planes irt with equation tX0 - mfXx + X3 = 0, where t £ GF(q), m a given nonsquare in GF(q) and a a given field automorphism of GF(g), define a Kantor-Knuth semifield flock T of K; see

10.7. Generalized Quadrangles witii a Line of Translation Points

213

Section 4.5. In [116], Payne noted that the dual Kantor-Knuth flock generalized quadrangles are span-symmetric (this observation is rather hidden in [116]), and this infinite class of generalized quadrangles contains nonclassical examples; T is not a linear flock (and then the GQ is nonclassical) if and only if a is not the identity. Moreover, every nonclassical dual Kantor-Knuth flock GQ contains a line L for which every line which meets L is an axis of symmetry (see [206]). Hence there is some line each point of which is a translation point. In order to initialize a theory for thick span-symmetric generalized quadrangles of order (s, s2) with s ^ 1, and to answer several important questions in a broader context, K. Thas studied the following problem in [202, 205, 206]. PROBLEM.

Classify all thick generalized quadrangles with distinct transla-

tion points. In the papers [202], [205] and [206, Chapter 7], this program was carried out completely. Before coming to a synthesis of the main result, cf. Theorem 10.7.2 below, we need the next theorem. Theorem 10.7.1 (Fong and Seitz [59, 60]; Kantor [85]; K. Thas [199]) 7/<S is a GQ of order (s, s2), s > 1, each line of which is an axis of symmetry, then S is isomorphic to Q(5, s). Proof (Taken from K. Thas [199]). Let U and V be any two nonconcurrent lines of S. Then S is an SPGQ with base-span {U, V}-11- and by Theorem 10.6.4, {U, V} is contained in s + 1 subGQs of order s. Now 5.3.5 of [128] implies the theorem. • Theorem 10.7.2 Suppose S is a generalized quadrangle of order (s,t), s ^ l ^ t , with two distinct collinear translation points. Then we have the following possibilities: (i) s = t, sis a prime power and S = Q(4, s); (ii) t — s2, sis even, sis a prime power and S = Q(5, s); (iii) t = s2, s = qn with q odd, where GF(q) is the kernel of the TGQ S = <S(°°) with (oo) an arbitrary translation point of S, q > 4n 2 -8n+2 and S is the point-line dual of a flock GQ S^) where T is a KantorKnuth flock;

214

Chapter 10. Configurations of Translation Points

(iv) t = s2, s = qn with q odd, where GF(q) is the kernel of the TGQ S = <S(°°) with (oo) an arbitrary translation point ofS, q < 4 n 2 - 8 n + 2 and S is the translation dual of the point-line dual of a flock GQ S(F) for some flock T. If a thick GQ S has two noncollinear translation points, then S is isomorphic to one of Q(4, s), Q(5,s). Conversely, if S = T(O) is a good thick TGQ of order (s, s 2 ) in odd characteristic, then S has a line of translation points. Proof. Let S be as assumed. If S contains two noncollinear translation points, then by Theorems 10.5.1, 10.5.3 and 10.7.1 S is isomorphic to one ofQ(4,s), Q(5,s). Suppose that S contains two collinear distinct translation points u and v. Then for s = t, the statement follows immediately from Theorem 10.5.1. If s ^ t, then t = s2 by Theorem 10.5.3. By transitivity, every point of L -.— uv is a translation point, and hence every line of L1- is an axis of symmetry. If we fix some base-span C for which L e £-*-, then by Theorem 10.6.4 there are s + 1 classical subGQs of order s which mutually intersect precisely in the base-grid r = (f2, £ U £-"-, I') (with the usual notation). Fix an arbitrary translation point (oo)lL, and consider the TGQ S(°°) = T(O) with base-point (oo). Then by Theorem 7.5.7, O is good at its element TX which corresponds to L. If s is even, the result follows by Theorem 7.8.3. Now let s be odd. Then Theorem 5.1.2 implies that the translation dual T(0*) of T(O) satisfies Property (G) at the flag ((oo)',L'), where (oo)' corresponds to (oo), and V to L. If we now apply Theorem 4.10.6, then it follows that T(0*) is the point-line dual of a flock generalized quadrangle S^J7). The theorem now follows from Theorem 5.1.14. Conversely, suppose that S = T ( 0 ) is a good thick TGQ of order (s, s 2 ), s odd, with TT e O a good element. Let x be any point of T ( 0 ) incident with •K. Then by Theorem 6.2.1 and Theorem 6.3.5, x is coregular. By Corollary 6.6.9, we conclude that a; is a translation point. The theorem is proved.

•

Corollary 10.7.3 A GQ of order (s,t), s =£ 1 ^ t and s even, has two distinct collinear translation points if and only if S is isomorphic to Q(4, s) or Q(5,s). •

10.7. Generalized Quadrangles with a Line of Translation Points

215

Corollary 10.7.4 Suppose S = T(£>) is a TGQ of order (qn, qm), where q is odd ifn = m. Suppose (oo) is the base-point of the TGQ. Let S* = T{0*) be its translation dual with base-point (oo)'. Then we have one of the following possibilities: (i) |Aut(5)| = |Aut(«S*)|; (ii) |Aut(<S)| = (qn + l)|Aut(<S*)| with n ^ TO; (iii) |Aut(5*)| = (qn + l)|Aut(5)| with

n^m.

Proof. The Aut(5)-orbit of (oo), with S = {V, B, I), is either V, a line, or {(oo)}. If n = TO with q odd and if the Aut(5)-orbit of (oo) is not {(oo)}, then by 1.3.6 (iv) of [128] each line of S is regular, so 5 ^ Q(4,g n ) by Theorem 2.2.2, hence S = S*. If every point of S is a translation point, then S is classical, S = S*, and so Aut(<S) = Aut(<S*). In all other cases we have one of (i), (ii), (iii) by Theorem 3.11.1. • The following corollary, which shows that all possibilities in Corollary 10.7.4 occur, was deduced in K. Thas [205] in the context of TGQs arising from flocks. Corollary 10.7.5 (1) Suppose S = T(O) is a TGQ of order (s, s 2 ), s even, and let (oo) be the base-point. Let S* = T(0*) be the translation dual with base-point (oo)'. Then |Aut(<S)| ^ |Aut(5*)|. (2) Suppose S = T(0) is a TGQ of order (s, s2), s odd, which arises from a flock but which is not the point-line dual of a Kantor-Knuth semifield flock GQ. Let (oo) be the base-point of T(0). Let S* = T(0*) be the translation dual ofT(0) with base-point (oo)'. Then |Aut(<S*)| = (s+l)|Aut(5)|. Proof. Suppose we are in Case (1). Then by Theorem 10.7.2, both <S and S* either have precisely one translation point, or all their points are as such. The result follows from Theorem 3.11.1. If we are in Case (2), then S* is the translation dual of the point-line dual of a flock GQ — hence S* is good, and so by Theorem 10.7.2 it has a line of translation points. By Theorem 10.7.2 and Theorem 6.4.4, <S has only one translation point, and the result follows from Theorem 3.11.1. •

Chapter 10. Configurations of Translation Points

216

10.8 On the Classification of Translation Generalized Quadrangles Combining the obtained results, we arrive at the following theorem. Theorem 10.8.1 Let S = T(0) be a TGQ of order (qn,qm) with translation point (oo). Then we have the following possibilities. (i) n = m, and either S = Q(4, qn), or (a) S has one translation point, q is even, and Aut(<S) = Aut(<S)(oo); (b) S has one translation point, q is odd, and if u and v are arbitrary points ofT(0) and T(0*), not collinear with (oo) and (oo)' (where (oo)' is the translation point of S* = T(0*)), respectively, then Aut(<S)„ = [Aut(<S)(oo)]„ 3* Aut(5*)„ = [Aut(S*)(oo)»]„. Also, |Aut(5)| = |Aut(5*)|. (ii) n < m, and then we have one of the following cases. (a)

S^Q(5,qn).

(b) S and S* are not classical, have two distinct collinear translation points, and then q is odd, m = 2n, and S = S* = S{F)D, where T is a Kantor-Knuth semifield flock. (c) S and S* are not classical, S is not the point-line dual of a KantorKnuth semifield flock GQ, has a line of translation points, and then m = 2n, q is odd, and S is the translation dual of the point-line dual of a flock GQ. In this case we have that |Aut(<S)| = (qn + l)|Aut(5*)|. Moreover, we have that [Aut(5) ( o o ) ] u = Aut(S*)„ = [Aut(S*) (oo) ,]„, where (oo) is an arbitrary translation point on the line of translation points [oo] of S, where (oo)' is the translation point of S*, and where u and v are arbitrary points o/T(£>) and T(0*), not collinear with (oo) and (oo)', respectively. (d) S andS* are not classical, S* is not the point-line dual of a KantorKnuth semifieldflock GQ, has a line of translation points, and then

10.8. On the Classification of Translation Generalized Quadrangles

217

m = 2n, qis odd, and S* is the translation dual of the point-line dual of a flock GQ. Similarly as in Case (ii) (c), we have that |Aut(5*)| = (qn + l)|Aut(5)|. Also, [Aut(S*) (oo) ,]„ = Aut(5) u = [Aut(5) ( o o ) ] u , where (oo)' is an arbitrary translation point on the line of translation points [oo]' of S*, where (oo) is the translation point of S, and where u and v are arbitrary points ofT(0) and T(0*), not collinear with (oo) and (oo)', respectively. (e) S and S* both have precisely one translation point, and then Aut(S) = Aut(<S)(oo) and Aut(<S*) = Aut(<S*)(ooy, where the notation is obvious. Furthermore, |Aut(5)| = |Aut(<S*)|, and Aut(S)u^Aut(S*)v, where u and v are arbitrary points of T(O) and T(0*), collinear with (oo) and (oo)', respectively.

not

If H — Aut(<S)(00) is the stabilizer of (oo) in Aut(«S), then we have that H = T xi Hx, with T the translation group of <S(°°) and x any point not collinear with (oo). Proof. First suppose that n = vn. Then by Theorem 10.7.2, S has one translation point if S is not classical. Part (i) then follows from Theorem 3.11.1. Next suppose that n < m, and that S is not isomorphic to Q(5,
218

Chapter 10. Configurations of Translation Points

In the next chapter, we will consider GQs which are TGQs with respect to every point without any finiteness assumption. We will develop our ideas in the more general context of Moufang GQs.

Chapter 11

Moufang Quadrangles with a Translation Point In this chapter, we study generalized quadrangles which are translation with respect to every point. In fact, these are a special case of the class of so-called Moufang quadrangles. We will formulate and prove a lot of equivalent, but seemingly much weaker, conditions than the Moufang Condition. From time to time, and when appropriate, we apply these results to translation Moufang GQs. We also show a basic correspondence between general Moufang GQs and translation Moufang GQs, namely, that every Moufang GQ contains a full or ideal thick translation Moufang GQ. All these results will be shown without any finiteness assumption. This guarantees that all arguments be purely geometric, or combinatorial group theoretic. In the finite case, other proofs exist (but not necessarily shorter ones!) making use of deeper group theoretic theorems such as the classification of Moufang sets and of irreducible split BN-pairs of rank 2 (which are not available in the infinite case).

11.1 Notation Two nonintersecting lines L, M will sometimes be called opposite; likewise two noncollinear points x, y will sometimes be called opposite. For every point x and line L, we denote the set of elements incident with x and L by [x\ and [L], respectively. With this notation, the mapping pL>M • [L] —» [M] : x H-> projMx is a bijection for every line M opposite L. We shall use this map below, so we treat pL,M as standard notation. 219

220

Chapter 11. Moufang Quadrangles with a Translation Point

Let <S be a finite GQ of order (s,t), s,t > 2. A subquadrangle of order (s,t'), for some t' e N, will be called a full subGQ; a subquadrangle of order (s',t), for some s' e N, will be called an ideal subGQ. In general, a full subquadrangle <S' of a (not necessarily finite) generalized quadrangle S is a subquadrangle with the property that every point of S incident with some line of S' belongs to S'. Dually, one defines in general an ideal subquadrangle. Now let <S be a thick GQ of order (s, t), let S' be a full subGQ of order (s, t'), t' < t, and let S" be a full subGQ of S' of order (s, t"), t" < t'. Then t = s2, t' = s and t" = 1; see Lemma 1.1.1. In particular, S" does not admit a proper full subGQ. We start with some general lemmas on subGQs and coUineation groups, some of which are trivial in the finite case. We gather these in the next section. In Section 11.3 we introduce a lot of conditions on the coUineation group of a GQ S and its action on S. All these conditions assume a transitive coUineation group with a so-called Tits system, and therefore we study this situation in Section 11.4. In the subsequent sections, we prove the promised equivalences of conditions we introduced, and show that every Moufang GQ contains a subGQ which is a TGQ.

11.2 Some General Elementary Lemmas The first lemma is a characterization of GQs, grids and dual grids. Lemma 11.2.1 Let S = (V,B,I) be an incidence structure satisfying the following properties. (i) Each element is incident with at least 2 elements. (ii) There exist a point x and a line L not incident with x. (iii) For each x e V and each L e B, with x not incident with L, there exist a unique point y and a unique line M for which xlMIylL. Then S is either a generalized quadrangle, or a grid, or a dual grid. Proof. Suppose, by way of contradiction, that there are two points x\, x2 both incident with two distinct lines Li, L2. Then every line incident with x\ must be incident with x2, as otherwise we contradict (iii). Dually, every

11.2. Some General Elementary Lemmas

221

point on any line incident with both xi,x2 must be incident with every line through both xi, x2. Conditions (i) and (ii) imply that there exists a line L not incident with any of the points on L\. Then the line through x\ meeting L must also be incident with x2, but it contains a point — namely, the point on L — not incident with L\, a contradiction. Suppose now that S is not a grid, nor a dual grid. Left to show is that there are a constant number of points on a line and, dually, a constant number of lines through a point. If all lines are incident with exactly two points, then it is easy to see that we obtain a dual grid; likewise, if all points are incident with exactly two lines, then we have a grid. So we may suppose that there is a line L with three points, and a point x incident with three lines. Condition (iii) readily implies that opposite lines carry the same number of points, and opposite points are incident with the same number of lines. We claim that every line M ~ L, M ^ L, is opposite some line K ^ L. Indeed, let x\,X2, X3 be three points incident with L, and let Mtlxi be arbitrary but distinct from L, i = 1,2,3. We show that there is a line opposite both L and Mi. Let yjlMj, j = 2,3, yj ^ Xj. It is clear that x cannot be collinear with all four of x2, £3,2/2,2/3- But since x2 and x3 are opposite 2/3 and y2, respectively, at least one of y2,2/3 is incident with at least three lines. One of these three lines does not meet L nor Mi. The claim is proved. Now Mi is incident with the same number of points as L, and the lemma is proved.

We refer to Condition (iii) of the previous lemma as the Main Axiom of GQs. For thick GQs, we have the following alternative definition. Lemma 11.2.2 Let S = (V,B,I) be an incidence structure satisfying the following properties. (i') Some point is incident with at least three lines, and some line is incident with at least three points. (ii') There exists an ordinary quadrangle in S, i.e., there are four points xi,x2,X3,Xi G V with x\ ~ x2 ~ £3 ~ £4 ~ x\, where xi 7^ £3 and X2 ^ £4.

(iii) For each x e V and each L e B, with x not incident with L, there exist a unique point y and a unique line M for which xlMlylL. Then S is a thick generalized quadrangle.

222

Chapter 11. Moufang Quadrangles with a Translation Point

Proof. By (iii), the lines x\x-i and £3x4, and the lines £2X3 and x4xi in (ii') do not meet. Let L be any line of S. If L is incident with one of the points Xi, i e {1,2,3,4}, say with x\, then we can apply (iii) to the pair {x3,L} and obtain a second point on L. If L is not incident with one of xi,...,X4, then we can apply (iii) to L and each of xi,..., 24. It is easy to see that this gives rise to at least two distinct points on L. Similarly, one shows that each point of S is incident with at least two lines. Hence we have proved (i) of the previous lemma. As (ii) of that lemma is an immediate consequence of (ii'), we conclude that S is either a thick GQ, or a grid or dual grid. But the latter two do not satisfy (i')The lemma is proved.

•

The next lemma is evidently true in the finite case, comparing orders. Proposition 11.2.3 Let S = (V, B, I) be a thick GQ and let S' be a thick subGQ of S. If S' contains at least one line L such that every point inS on L is also a point of S', then S' is a full subGQ. Also, if S' is both full and ideal in S, then it coincides with S. Proof. If M is a line in S' opposite L, then the above map PL,M is a bijection in both S and S'; hence all points of M in S belong to <S'. The first result now follows by repeatedly applying this observation and noting that the graph (B, rf) is connected (it even has diameter 2). For the second assertion, let z be an arbitrary point of S, and let L be a line of S'. If zlL, then z belongs to S' since S' is a full subGQ. Suppose now z IL. Since S' is full in S, the point projL2; belongs to S'. Since S' is ideal in S, the line proj z Llproj L ^ belongs to S'. Since S' is full in S, the point z must now belong to S'. Similarly all lines of S belong to S'. The proposition is proved.

•

The connection between coUineations and subGQs will become apparent from the following proposition. Proposition 11.2.4 Let ipbea collineation of the GQ <S. Let V (B') be the set of invariant points (lines) of S under the action of ip. If V contains two opposite points of S, and B' contains two opposite lines of S, then V' and B' are the point and line set, respectively, of either a subGQ S' of S, or a grid or dual grid.

11.2. Some General Elementary Lemmas

223

Proof. Let I' be the restriction of I to the sets V and B'. According to Lemma 11.2.1, it suffices to show that S' = (V',B',I') is an incidence structure in which every element is incident with at least two elements, in which there is a point of V not incident with some line of B', and in which the Main Axiom of GQs holds. The latter is clear since, if a point x is fixed, and a line L not through x is fixed, then proj^L and projLa; are uniquely determined by x and L and therefore fixed. We now show that every line of <S' is incident with at least two points of S'. Let L be any line of S'. By assumption, there are two opposite points xi, x 2 in <S'. Hence the points p r o j ^ , i = 1,2, belong to S'. If proj L xi ^ proj L x 2 , then we are done. Suppose now that proj L xi = projLa:2. By assumption there are also two opposite lines in <S'. Consequently not all elements of B' contain the point proj L a;i. Let M e B' be such that it is not incident with proj L a;i. If M meets L in another point, then we are done, so we assume that M is opposite LinS'. Then the line M cannot be concurrent with both proj^L and proj^L, since a triangle would appear. We may assume without loss of generality that M is not concurrent with proj L. Since x\ and proj^rri are collinear, the Main Axiom of GQs implies that y\ := proj M ;n and 2/2 := P r oj M (P r o JL : r i) a r e different points, belonging to V. Hence also proj L yi and proj L xi = proj L (proj M (proj L a;i)) are two distinct points on L belonging to S'. Hence every line of S' is incident with at least two points of <S' and dually, every point of S' is on at least two lines of S'. Now it is also clear that S' contains a point x and a line L which are not incident. • Here is an immediate useful corollary. Corollary 11.2.5 Let ipbe a collineation in a thick GQ S fixing an ordinary quadrangle, fixing all points on a certain line of that quadrangle, and fixing all lines through a certain point (vertex) of that quadrangle. Then ip is the identity. Proof. By the previous proposition, ip fixes pointwise a subGQ which is by assumption thick. By Proposition 11.2.3, this subGQ coincides with S. Hence ip fixes everything and is the identity. • We end this section with two general combinatorial results. The first one is folklore, the second one is an observation of Cuypers, see [21].

224

Chapter 11. Moufang Quadrangles with a Translation Point

Lemma 11.2.6 Let S = (V, B, I) be a thick GQ and let x e V be arbitrary. Then the graph (T, E) with vertices the points ofS opposite x, and adjacency given by collinearity is connected. Proof. Let y, z be two points opposite x. If some point in {y, z}-1 is not coUinear with x, then y and z belong to the same connected component of T. If all points of {y, z}1- belong to Xs-, then replacing y by any point y' coUinear with y and opposite x yields a pair {y1', z} for which {y1', z}1- is not contained in x1-. Now y is adjacent to y', which is in the same connected component as z. • Lemma 11.2.7 Let S = (V, B, I) be a thick GQ not of order (2,2). Let {x, L} be a flag ofS, with x e V and L e B. Define the following graph (T, E). The vertex set T is the union of the set of points of S opposite x and the set of lines ofS opposite L. The edge set E is the set offlags ofS whose elements are both contained in T. Then the graph (F, E) is connected. Proof. It is clearly enough to show that every two vertices of (r, E) that correspond with points of <S can be connected by a path in (F, E). So let y, z be two points of S opposite x. Since the order of <S is unequal (2,2), we may assume, without loss of generality, that every point of S is incident with at least four lines. Since there is exactly one line through each point opposite x which is not opposite L, we see that the valency of y in (r, E) is at least 3. Similarly for z. We may obviously assume that y and z cannot be joined by a path of length 2 or 4 in (r, E). This means that y and z are not incident with a line opposite L, and that the intersection point (if it exists) of any line through y opposite L with any line through z opposite L is coUinear with x. We now assume first that projLy = yro)Lz, but proj^L ^ projzL. Since there is one line through each of y, z not opposite L, we see that there are at least three points xi, X2, ^3 not on L coUinear with all of x, y, z. Now let y' be a point coUinear with y, y' ^ y, such that x3Iyy' and such that y' is opposite x (hence belonging to T). Clearly, y and y' can be connected inside (r, E) (by a path of length 2). If y' is coUinear with z, then y',z,xz are the vertices of a triangle, contradiction. So y' is not coUinear with z, and at least one of the lines Li := projyZXj, i = 1,2, say L\, is not concurrent with L (since they cannot be equal). Also, the intersection point z' of Li and zx\ is opposite x. We now have the path (z,zz',z',Li,y',yy',y) of length 6 connecting z with y.

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225

If, still under the assumption that pro]Ly = projLz, we have proj^L = projzL, and so yz is a line, then we apply the previous paragraph to a point y* opposite x, collinear with projLy, and not on yz. We obtain that y and y* are connected in (T,E), and y* and z are connected in (T,E), hence y and z are connected by a path in (r, E). We now assume that projLy ^ projLz. Then the line M := projz(projJ/L) is opposite L and the point projMy is opposite x. Hence we have a path from z to projM,z inside (F, E). But by the foregoing, we also have a path from projMy to y, hence from z to y. The lemma is proved. • If S is a GQ of order (2,2), then it is easy to show that (T, E) of the previous lemma has exactly two connected components, each being an ordinary quadrangle in S.

11.3 The Moufang Property and Analogues We introduce some more notation. Let <S = (P,B, I) be a GQ. An apartment is an ordinary quadrangle in S. It consists of four points and four lines. A root is a set of five "consecutive" elements of an apartment. Hence a root contains either two points xi,x2 and three lines Li,L2,L3, with L1IX1IL2IX2IL3, or dually, it contains three points and two lines. It will be useful to distinguish between these dual notions. Therefore, we will call the former a root, and the latter (dually) a dual root. A root without its extremal lines will be called a panel, more precisely, the interior (panel) of the root. Similar but dual definitions for dual panels. So a panel is a set of two distinct collinear points, together with the joining line. Note that the Main Axiom of GQs implies that every pair of flags (a flag being an incident point-line pair) of <S is contained in at least one apartment of S. We are now ready to define the Moufang Property for generalized quadrangles. Let S be a generalized quadrangle with full collineation group G = Aut(S). Let 7r be a panel. Then we say that TT has the Moufang Property (or, equivalently, n is a Moufang panel) if for some root a with interior IT, the group G^l of collineations (called root elations and, dually, dual root elations) fixing every element incident with any element of n acts transitively on the set

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Chapter 11. Moufang Quadrangles with a Translation Point

of apartments containing a. Let n = {x, y, L}, with x, y e V and L £ B. Let M be any line through x distinct from L. Suppose IT is a Moufang panel, and let a be a root with interior ix such that GH acts transitively on the set of apartments containing a. Let u, v! be two points on M distinct from x. Suppose the unique line of a incident with y and unequal L is N. Then N is opposite M and we may consider the distinct points u\ = proj^u and u[ = pro)Nu'. Using the Main Axiom of GQs, it is quite easy to see that MI and u[ determine unique apartments £ and £', respectively, containing a. Hence there is a collineation 9 e G^l mapping S to £', and hence mapping u\ to u[. But since M is fixed under 9 by assumption, 9 maps u to w', and hence, for every line M' through y distinct from L, 9 maps the (unique) apartment containing TT, U and M' to the (unique) apartment containing TY, v! and M'. We have shown that the definition of Moufang panel is independent of the root involved in that definition. We can even say a little more. Suppose that 9 e G^', with TT as above. Suppose that 9 leaves an apartment through n invariant. Then by Corollary 11.2.5, 9 is the identity. Hence, in general, G ^ acts freely on the set of apartments through a, with a a root with interior ir, and it acts sharply transitively on that set precisely when TT is a Moufang panel. If every panel and every dual panel of the GQ S has the Moufang Property, then we say that S has the Moufang Property, or that S is a Moufang GQ. If every panel is a Moufang panel, or if every dual panel is a Moufang dual panel, then we say that <S is a half Moufang GQ. If S is a Moufang GQ, then the group generated by all root elations and dual root elations is called the little projective group ofS. Note that a symmetry about a line L is automatically a root elation, and it will sometimes be called an axial root elation (with axis L). Dually we define central root elations (with center some point x). It is now clear that every line of a GQ in which every point is a translation point, is an axis of symmetry, and hence the GQ is half Moufang. Let {x, L} be a flag of the GQ S. Let Mix and ylL such that y is not incident with M. As above, it is easy to see with the aid of Corollary 11.2.5, that the group G^X'L^ of collineations fixing all lines through x and fixing all points on L acts freely on the set of apartments containing {x,y,L, M}. We call the flag {x, L} a Moufang flag if the group G^x^ acts transitively (and hence sharply transitively) on the set of apartments containing {x, y, L, M}. As before, one shows easily that this definition is independent of the chosen line M through x, M =£ L, and of the chosen point y on L, y ^ x.

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227

We see that the definition of a Moufang flag is a self-dual one, hence there is no need to introduce something like a "dual Moufang flag". If every flag of the GQ <S is a Moufang flag, then we call S 3-Moufang, where the number 3 refers to the length of the sequence (y, L, x, M) as a path in the incidence graph of S (which is the graph (V U B, I)). The 3-Moufang Condition is a completely self-dual one, and so one might think that a notion of "half 3-Moufang" cannot be defined. But actually, one can take a kind of greatest common divisor of the 3-Moufang Condition and the half Moufang Condition to obtain the following definition of a half 3-Moufang generalized quadrangle. Let {x, L} be a flag of the GQ S. Let zlMlx and KlylL such that z ^ x, K 7^ L and y is not incident with M. As before, the group df acts freely on the set of apartments containing {x, y, z,L,M}, and the group GK acts freely on the set of apartments containing {x, y, K, L, M). When we have transitivity for each z, then in the first case, we say that the flag {x,L}is half 3-Moufang at x, while in the second case we talk about half 3Moufang at L. The GQ S is called half 3-Moufang if either every flag {x, L} of S is half 3-Moufang at x, or every flag {x, L} of <S is half 3-Moufang at L. Now let x be any point of the GQ S. Recall that if the group G ^ of collineations fixing all lines through x acts transitively on the set of points of S opposite x (here, this is not necessarily a regular action!), then we say that x is a center of transitivity. Dually, we define an axis of transitivity. If all points are centers of transitivity, and all lines are axes of transitivity, then we say that S is a 2-Moufang GQ. If either all points are centers of transitivity, or all lines are axes of transitivity, then we call S a half 2-Moufang GQ. Let {x,L} again be a flag of the GQ S, with x a point and L a line. Another flag {y, M} is called opposite {x, L} if the point y is opposite x and the line M is opposite L. If the group GX,L of all collineations of S fixing the flag {x, L} acts transitively on the set of flags opposite {x, L}, then we say that {x, L} is a transitive flag. If all flags are transitive, then we say that S is a 1-Moufang GQ, or, equivalently, that <S satisfies the Tits Condition, or that <S is a Tits GQ. The last two names are motivated by their group theoretic counterparts, the Tits systems, see below. If S satisfies the Tits Condition, then the corresponding coUineation group G acts transitively on the set of ordered pairs of opposite flags of S. For convenience, we shall sometimes also refer to the (half) 4-Moufang Condition for a (half) Moufang GQ.

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Chapter 11. Moufang Quadrangles with a Translation Point

Let S = (V, B, I) be a Tits GQ with full collineation group G. Let {x, L} be any flag in S, with x e V and L e B. Set B := GXjL. Let £ be an apartment containing {x, L}. Define iV := Gs, the (setwise) stabilizer of £, and set H = Br\N. Then iJ is precisely the group of collineations fixing X elementwise. Hence H < N. Now, if there exists a nilpotent normal subgroup U of B such that f/i? = B, then we say that S has the FongSeitz Property. A Fong-Seitz GQ is a Tits GQ with this additional group theoretic property. From group-theoretic point of view this is a natural condition. Note that, a priori, nothing guarantees uniqueness of U, but one can actually show that it is unique (see De Medts, Haot, Tent and Van Maldeghem [48]); in the finite case it is not so difficult to see that U is the unique Sylow p-subgroup of B, when G contains the little projective group of the corresponding Moufang GQ — as we will later prove, every Fong-Seitz GQ is a Moufang GQ. Hence B is always a split extension of the nilpotent group U, and B = U:H. All Moufang GQs have been classified by Tits and Weiss in the monograph Moufang Polygons [229]. Technically, the classification was completed in 1997. In the finite case, the classification of Moufang GQs follows from the classification of Fong-Seitz GQs by Fong and Seitz [59, 60]. They show that the Moufang Condition implies the Fong-Seitz Property, and then they use entirely group theoretic methods to classify the Fong-Seitz GQs (in fact, they classify all Fong-Seitz generalized polygons, but the part of quadrangles is the largest by far). The proof is highly nontransparent from a geometric point of view, and it is approximately 100 pages long. Corollary 11.6.7 and Theorem 11.7.4 reduce in a completely geometric and elementary combinatorial group theoretic way the Fong-Seitz Condition to the Moufang Condition (including the infinite case!), and hence combined with the classification theorem of Tits and Weiss (mentioned above), this classifies all Fong-Seitz GQs. In the finite case, one can also appeal to the alternative geometric classification of Moufang quadrangles by Payne and Thas [128, Chapter 9], Kantor [85] and K. Thas [199], see also Thas [188]. All this replaces the bulk of the papers of Fong and Seitz [59, 60] by elementary methods, and generalizes it at the same time to the infinite case. In this chapter, we essentially prove that all the conditions introduced above, except for the Tits Condition and the half 2-Moufang Condition, are equivalent. In particular, this will imply that a GQ in which every point is a translation point is a Moufang GQ. Also, we will show that every Moufang GQ has a thick subGQ which is a TGQ. For all this, we will not need finiteness, so we prove things in the most general setting.

11.4. Tits Generalized Quadrangles and Tits Systems

229

Concerning the Tits Condition, it can be proved, using the classification of finite simple groups, that also this condition is, in the finite case, equivalent with the Moufang Condition; see Buekenhout and Van Maldeghem [33]. In the infinite case, there are a lot of counter examples, the most striking ones being without doubt those constructed by Tent [150]. But we will not use the classification of finite simple groups in this book, and so we will not prove the result of Buekenhout and Van Maldeghem mentioned above. We refer to work of K. Thas [209, 210, 211] on the classification of finite Tits GQs without the classification of finite simple groups. Concerning the half 2-Moufang Condition, it can be shown that it is also equivalent to the Moufang Condition, at least in the finite case. This has been done by K. Thas and Van Maldeghem [215], but the proof of that result is beyond the scope of this book.

11.4 Tits Generalized Quadrangles and Tits Systems As we saw in Chapter 3, a 4-gonal family is the group theoretic equivalent of the notion of an EGQ. Likewise, we will axiomatize group theoretically the notion of a Tits GQ. This gives rise to groups with a BN-pair, also called Tits systems. This notion was introduced by Tits [218, 220] for spherical buildings, and we specialize it here to the class of GQs. At this stage, it is convenient to introduce a generalized distance between flags of S as follows. Let F = {x, L} and F' = {x', L'} with x, x' e V and L,L' e B be two flags of S. Then 5(F,F') = z £ {-3,-2,-1,0,1,2,3,4}, where \z\ = \{d(x,x') + d{L,L')) and if \z\ ^ 4, then z has the same sign as d(x', x) + d(L', x) - d(x',L) - d(L', L) (distances are measured in the incidence graph of <S and the sign of 0 is "+")• Note that 8(F, F') = -5(F', F) if 5(F, F') is even but not equal to 4; otherwise 8{F, F') = S(F', F). In the next proposition we will be dealing with double cosets. It is convenient to think of these as unions of ordinary cosets. For instance, the double coset HgK for subgroups H, K of some group G and g e G consists of the union of all left cosets hgK for h ranging over H, or of all right cosets Hgk for k ranging over K. Note that, in contrast with ordinary cosets, double cosets need not contain an equal number of elements. For instance, \HgH\ > \HH\ whenever g does not normalize H. We will also consider a quotient group G/N as consisting of cosets of TV in G. With this convention, it makes sense to multiply elements of the

Chapter 11. Moufang Quadrangles with a Translation Point

230

quotient group with elements or subgroups of G; it is just the ordinary multiplication of subsets in a group. Proposition 11.4.1 Let S = (V, B, I) be a thick Tits generalized quadrangle and let G be a collineation group of S such that G acts transitively on the set of ordered pairs of opposite flags of S. Let F = {x,L} be any flag in S, with x e V and L e B, and let E be an apartment containing F. Define B := GxM N := G E and H = B n N as above. Then N n (Gx\ GL) and N n (GL \ Gx) are nonempty. Choose arbitrarily sx and sL, respectively, in these sets. Then (G, B, N) and SX,SL satisfy the following properties. (BN1) G=

(B,N).

(BN2) H < N andW := N/H = (sxH, sLH) is isomorphic to Dih 8 , the dihedral group of order 8. (BN3) BsBwB

C BwB U BswB, for allw EN and s e {sx, sL}.

(BN4)

sBs^B,fors€{sx,sL}.

Also, B does not contain any nontrivial normal subgroup ofG and H coincides with

P| w~lBw.

Proof. Let F' = {x1, L'}, x' e V, V e B, be the flag in E that is opposite F. Also, let M ^ L and M' / L' be the "second" lines in E incident with x and x', respectively. By the Tits Property there is some element sx e Gx mapping the pair ({x, L}, {x', L'}) to the pair ({x, M}, {x\ M'}). Clearly sx stabilizes E and belongs toJVfl (Gx \ GL). Similarly we find sL e N n (GL \ Gx). We may treat the elements sx and SL as arbitrary elements in N n (Gx \ GL) and sL e N n (GL \ Gx), respectively. Remark that the Tits Property is equivalent with G being transitive on the set of flags of S, together with requiring that B acts transitively on the set of apartments containing F. We now show that N acts transitively on the flags of E. By flag transitivity, there is a collineation g mapping any flag F' of E onto F, and by our previous remark, there is a collineation g' e B mapping E 9 back to E. Then gg' e N and maps F' onto F. We now show (BN1). Let g e Gbe arbitrary. Let E' be an apartment of S containing the flags F and F9. By the Tits Property there is some gx e B

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231

mapping E' to E. Since N acts transitively on the flags of E, we can find w£N with {F991 ) w = F. Hence ggxw € B and so G = BNB. We turn to (BN2). The action of N on E is clearly isomorphic to a subgroup of the natural action of Dih 8 on a square. By our remark above, this action in fact coincides with that natural action. Since Dih 8 acts sharply transitively on the "flags" of the corresponding square, we see that H = B n N is the elementwise stabilizer of E, and hence a normal subgroup of the setwise stabilizer N. Moreover, H is the kernel of the permutation group (N, E) (S seen as a set of flags) and as such N/H ^ Dih 8 . Returning to our proof above of (BN1), one sees that, if we define wz e N/H as the coset of H in N whose elements map F onto the flag of E at generalized distance z, then an arbitrary element g e G that maps F onto a flag at generalized distance z from F belongs to BwzB (which defines a unique double coset of B since H < B). Conversely, it is easy to see that every element of BwzB maps F to a flag F' at generalized distance S(F, F') — z from F. Hence G is the disjoint union of the double cosets BwzB, for z £ {-3, - 2 , - 1 , 0 , 1 , 2 , 3 , 4 } . Now let F' be a flag with 6(F, F') = i. If i < 0, then one easily sees that S(F, F'SL) = 1- i. If i > 0, then either 5{F,F'SL) = 1- i, or S(F, F'SL) = i. Now, one checks that wi-i = W%SL = WiW\. This shows BWBSLB C BWB U BWSLB, for all w e W. Taking inverses, (BN3) for s = SL follows. Similarly, (BN3) is also true for s = sx (and sxH = w-\). Regarding (BN4), we may choose a g e B which does not fix the unique point incident with L contained in E and distinct from x. This is possible by the thickness assumption. For this g, the image of F under SLgsL clearly differs from F (and lies at generalized distance 1 from F). Hence sLBsL ^ B. Similarly sxBsx ^ B. If K < G is contained in B, then K = K9 < B n B3, for every g e G. Hence K stabilizes every flag F9, for g ranging over G, and so K fixes every flag (by flag transitivity of G) and must therefore be trivial. Since w~1Bw is the stabilizer of the flag Fw, and Fw ranges over all flags of E as w ranges over N, every element of f]weN w~1Bw belongs to N and hence to H. Conversely, every element of H fixes every flag of E and hence belongs to w~1Bw, for all w e N. The proposition is proved.

•

The group W is called the Weyl group of (G,B,N), and the elements sx and SL representatives of the standard generators ofW.

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We now would like to prove a converse of the previous proposition. To that aim, we define the notion of a group with a BN-pair of Type B 2 , also called a Tits system of Type B 2 . Since we will not encounter any other types of such groups, we will shortly talk about groups with a BN-pair and Tits systems. Let G be a group, and let B and N be two subgroups of G. Set H = B n N. Then (G, B, N) is called a Tits system, or (B, N) is called a BN-pair in G, or G is called a group with BN-pair (B, N), if there exist elements sx and SL of G, with {s2, s |} C if, such that (BN1) up to (BN4) of Proposition 11.4.1 hold. The group W in (BN2) will be called the Weyl group and the cosets sxH and s^H the standard generators ofW. If S is a Tits GQ, and (G,B,N) is as in Theorem 11.4.1, then we call (G, B, N) a natural Tits system associated with S. If (G, P, AT) is a Tits system with Weyl group W and if s^ and sL are corresponding representatives of the standard generators of W, then we define the following incidence structure SG,B,N = (P, &, I)- Define Px = (B, BSx) andP L = {B,BSL). • POINTS. The elements of V are the right cosets of Px. • LINES. The elements of B are the right cosets of PL. • INCIDENCE. For g,h e G, the point Pxg is incident with the line PLh if Pxg fl P^h ^ 0 (and in this case we may choose g = h). The group G acts (on the right) as a collineation group on SG,B,N, and it is flag transitive since every flag can be written as {Pxg, Ptg} (and is the image under g of the "standard" flag {PXl PL}). If we denote the point Px by x, then the point Pxg can be written as x9. Similarly, we write PL as E, and every line can be written as L9, for some g e G. Before continuing, we prove a lemma, which will later lead to the Bruhat Decomposition. First a definition. Let W be the Weyl group of some Tits system, and let sx,sL be representatives of the standard generators. Then the minimum number of factors needed to write an element w e W as a product of sxH and SLH is called the length of w, and denoted £(w). Since W is a dihedral group of order 8, we have £(w) e {0,1,2,3,4}, for an arbitrary element w oiW.

11.4. Tits Generalized Quadrangles and Tits Systems

233

Lemma 11.4.2 If (G,B,N) is a Tits system with Weyl group W, then for w, w' G W, one has BwB = Bw'B if and only ifw = w'. Proof. Clearly, if w = w', then BwB = Bw'B. Now assume that w ^ w' and assume that BwB = Bw'B. First suppose that w = s is a standard generator of W. Note that B C BsBsB, since sBsB contains sHsH = H < B. Then (BN3) tells us that B C BsBsB = BsBw'B C Bsw'B U Bw'B. Hence either sw' C B or w' C B, implying that either w' = s or w' is the identity. The former contradicts our hypothesis, the latter implies BsB = B, implying s is the identity, again a contradiction. So we may assume that 1 < £(w) < £(w'). Let w = sw", with s a standard generator, and £{w") = £(w) - 1. Then Bsw"B = Bw'B implies the existence of b, b' G B with sw" = bw'b', hence w" = s~1bw'b' G BsBw'B C Bsw'BuBw'B. So either Bw"B = Bsw'B, or Bw"B = Bw'B. An easy induction on the length of w implies now that either w" = sw', contradicting w = sw" as s is an involution, or w" = w', contradicting £{w') > £(w) > £(w"). The lemma is proved.

•

Now let s e {sx,sL} and note that (BN3) implies BsBsB c BsBUB. Since (BN4) says that BsBsB ^ B, we deduce that at least one element w of BsB belongs to BsBsB, and hence multiplying w at the left and at the right with B, we obtain BsB C BsBsB. We now see that BsBsB = BsBUB. It follows that BSXB\JB = BsxBsxB = Px and BsLBuB = BsLBsLB = PL. We now claim that G is generated by Px and PL. Indeed, by the previous paragraph, sxH c Px and sLH c PL. Also, (BN2) implies that N is generated by sxH and sLH, and so N < (PX,PL)- Hence, since B < Px n PL, we obtain (B, N) < (Px, PL} and (BN1) completes the proof of our claim. Next we claim that Px n PL = B. Indeed, since sx £ B, the double cosets BsxB and B are distinct, and likewise BsLB and B are distinct. Also, Lemma 11.4.2 implies that BsxB ^ Bs^B. The claim is now clear. As a consequence, we may identify the flags of SatB,N with the right cosets of B. Denote by £ the subgeometry induced by the set of flags {B, Bsx,

BSL,BSXSL,

BSLSX,

BSXSLSX,

BSLSXSL,BS

SL}-

Just as we showed BsxB ^ BSLB above using (BN2) and (BN3), the same conditions also guarantee that those eight flags are different. Notice that

Chapter 11. Moufang Quadrangles with a Translation Point

234

(using the equality Dih 8 )

BSXSLSXSL

=

which follows from W =

BSLSXSLSX,

B
QPxsLnPL, C PxsL

BSLSXSL BsxsLsxsL BSXSLSX

DPLsxsL,

Q PXSLSXSL

n

C PXSLSXSL

n PL

PLSXSL,

C PX ^x)

BsLsx Bsx

C Px

nPL CPXDPLSX.

Hence E induces a quadrangle in SG,B,N, if we can show that the point x is not collinear with X3LSXSL and dually, that the line L is not concurrent with t 8 ' 8 1 8 ' (and then it follows that xSL is not collinear with xSLSx by applying SL to both points and remarking that BSLSXSLSL C BSLSXH = BHSLSX = Sx BSLSX, and likewise for the lines i and LS:cSL). Now PXSLSXSL

= BSXBSLSXSL

U

C BsxsLsxsLB

U

BSLSXSL BSLSXSLB,

which follows from multiplying the right hand side of the first line at the right with B and then applying (BN3) with s = sx and w = sLsxsL. Any line P^g incident with Px shares a coset of B with Px. The cosets of B in Pj; are of the form B and Ps^fe, with b e P . Hence we may put # = s x 6 (leaving the more trivial case of g e P to the reader). Hence the coset P L # is equal to Bsxb U BsLBsxb, and hence belongs to the union of the double cosets BsxB\JBsLsxB (indeed, (BN3) implies that BsLBsxB C BsLsxBU BsxB; taking inverses of the left hand side, applying (BN3) again, and then again taking inverses, we deduce BSLBSXB C BSLSXB U BSLB. SO BsLBsxB c BsLsxB by taking intersections, using Lemma 11.4.2). This union should now contain a right coset of P that is also contained in one of the double cosets BSXSLSXSLB or BSLSXSLB. This implies, however, that one of these double cosets coincides with one of P s ^ P or BSLSXB, contradicting Lemma 11.4.2. The dual argument is similar, and so we conclude that E is a quadrangle. It is easy to see that sx and sL preserve E and act as generating reflections of the dihedral action of N on E.

11.4. Tits Generalized Quadrangles and Tits Systems

235

Now we claim that G = BNB. Since G = (B, N), it suffices to show that every element bnb'n'b", with b, b', b" G B and n, n' e N, belongs to BNB. Noticing that every element of N can be written as a finite product of sx and SL times an element of H < B, the claim easily follows from (BN3) using induction on the length £(n) of n. Together with Lemma 11.4.2, this gives rise to the Bruhat Decomposition, which states that

G = (j BwB. wGW 9

Consider any flag F . This flag corresponds to the coset Bg. By the foregoing paragraphs we can write g = b'nb, with b,b' G B and n G N. Hence Bg = Bnb and so b^1 fixes F and maps F9 into E. It now follows rather easily that the geometry SG,B,N does not contain any digon {xi,x2,Li,L2}, with xi,x2 G V and Li,L2 G B with XilLj, i,j G {1,2}. Indeed, by flag transitivity, we may assume x\ = xaxAL\ = L. By the previous paragraph, we may also assume that the flag {x2,L2} is contained in E, a contradiction. Now suppose that SQ,B,N contains a triangle xilL3Ix2lLi!x3lL2Ixi, with the x% points and the Lt lines, i = 1,2,3. We may again assume x\ = x and L2 = L, and by the previous paragraph, we may assume that the flag {x2,Li} belongs to E. Since E is a quadrangle, we must have x2 = xSLSxSh. But then L 3 is incident with both x and xSLSxSL, contradicting the fact that these points are not even collinear. Now let x' be any point not incident with some line L'. We may assume x' — x and L' G E. It follows that x' is incident with at least two lines (the lines through x in E), that L' is incident with at least two points, and that there is a flag {y, M} with x'lMlylL'. The flag {y,M} is unique by the previous paragraph. Hence we have shown that SG,B,N is a GQ (note that by flag transitivity, it cannot be a grid or a dual grid without order). The group G acts as a collineation group, by multiplication at the right, on SQ,B,N- Since every element of the kernel K of that action must in particular fix B, we see that K is a normal subgroup of G contained in B. Conversely, if K' < G, and K' < B, then, since the stabilizer of the flag Bg, g G G, clearly coincides with B9, the group K' stabilizes every flag of SG,B,N, and hence belongs to the kernel K. So K is the biggest normal subgroup of G contained in B, and as such equal to K = p | Bg. geG

236

Chapter 11. Moufang Quadrangles with a Translation Point

The group G/K acts faithfully on SG,B,N and the stabilizer of the flag F is B/K. Now we determine the stabilizer of E. Since N stabilizes E and acts transitively on the flags of E, it suffices to determine the elementwise stabilizer S of E, and then NS is the global stabilizer of E. So suppose some b e B stabilizes E. Since E is uniquely determined by the opposite flags F and FSxSLSxSL, this is equivalent with b e B fixing FSxSLS*SL. Hence S = B n B S i S i S i S l . Note that then b automatically belongs to Bw, for every w &W. Hence we can also write S=

p|

BW.

w€W

We have proved the following theorem. Theorem 11.4.3 Let (G, B, N) be a Tits system with Weyl group W. Then the geometry SG,B,N defined above is a generalized quadrangle satisfying the Tits Condition. Setting K= f]B9 geG

andS=

f]

Bw,

w€W

G/K acts naturally and faithfully by right translation on SG,B,N- Also, B is the stabilizer of a unique flag F and NS is the stabilizer of a unique apartment containing F, and the triple (G/K, B/K, NS/K) is a natural Tits system associated with SG,B,N• In the literature, the Tits system (G,B,N) is called saturated precisely when N = NS, with S as above. Replacing N by NS, every Tits system is "equivalent" to a saturated one. It is now clear that the notion of a Fong-Seitz quadrangle is equivalent to the notion of a saturated Tits system (G, B, N) with the property that there exists a normal nilpotent subgroup U of B with B = UH. If we do not insist on (G,B,N) being saturated, then we must require that B = US, with S as above. In any case, we call such Tits systems split.

11.5 Properties of Moufang Quadrangles In this section, we will show that the Moufang Property implies all properties that we introduced in Section 11.3.

11.5. Properties of Moufang Quadrangles

237

From now on until the end of this chapter we always assume that the GQs we are dealing with are thick, except if explicitly mentioned otherwise. First we prove some easy assertions. They belong to folklore, but an explicit reference is Van Maldeghem [232], Chapter 6. Proposition 11.5.1

(i) Every Moufang GQ is a 3-Moufang GQ.

(ii) Every half Moufang GQ is a half 3-Moufang GQ. (iii) Every 3-Moufang GQ is a 2-Moufang GQ. (iv) Every half 3-Moufang GQ is a half 2-Moufang GQ. (v) Every half 2-Moufang GQ is a Tits GQ. (vi) Every k-Moufang GQ is always a half k-Moufang GQ, k = 2,3,4. Proof. Let <S = (V, B, I) be a GQ with full collineation group G. Let n := {xi,Li,X2} andV := {L0,xi,Li} be a panel and a dual panel, respectively, with x\,x2 G V, L0,Li e B. First suppose that ir and n' are Moufang. Then the groups G ^ and G^ ' generate the group G^1'1,1' which acts transitively on the set of apartments containing -K U TT'. Hence the flag {xi,Li} is a Moufang flag. This shows

(0. Next, suppose that w is Moufang. Then the flag {xi,Li} is clearly half 3-Moufang at L1. This shows (ii). Suppose now that the flags {xi,Li} and {x2,Li} are Moufang. Then the group generated by G^XuLl^ and G^X2'Ll^ clearly acts transitively on the set of apartments containing xi,Li,x2. This shows (iii). We now show (iv). More exactly, we assume that for every line Llx\, the flag {x\,L} is half 3-Moufang at L, and we show that the group G^Xl^ acts transitively on the set of points opposite x\. If 2/1,2/2 are two distinct collinear points opposite x\, then the line M joining t/i and y2 meets some line L' incident with x\. The element of G^Xl'L'^ mapping ?/i to y2 evidently belongs to G^Xl\ and (iv) follows from Lemma 11.2.6. Suppose now that S is half 2-Moufang, and that all points are centers of transitivity. Since the stabilizer of a dual panel {L0, x\, L{\ acts transitively on the panels {x2,L2,x3}, with LiIx2lL2Ix3, x\ ^ x2 ^ 23 and L\ ^ L2, using the group G^Xl\ we see that G acts transitively on the set of flags of

238

Chapter 11. Moufang Quadrangles with a Translation Point

S, and that the stabilizer of the flag {xi,L0} flags opposite it. This proves (v). Finally, (vi) is trivial.

acts transitively on the set of •

The Moufang Property is inherited by every subGQ. In the finite case, this means that classical and dual classical GQs only have classical and dual classical subGQs. Also this is folklore, but we present a proof. Proposition 11.5.2 Any thick subquadrangle S' = (V',B', I) of a Moufang quadrangle S = {V, B, I) is a Moufang quadrangle. Proof. Let a be a (dual) root in S' with interior (dual) panel -K, and let E and E' be two apartments in S1 containing a. Let u be the (dual) root elation with respect to a which maps E onto S'. Then the intersection <S" of S' with its image S'u under u clearly satisfies the assumptions of Lemma 11.2.2, and hence is a thick subquadrangle in both S' and S'u. Moreover, looking at the elements incident with one of the elements of -K, we see that S" is full and ideal in both S' and S'u. Lemma 11.2.3 implies that S" coincides with both S' and S'u, hence S' = S'u and u is a (dual) root elation in S'. It is now straightforward to conclude that S' is a Moufang GQ. • We will now show that a Moufang GQ is a Fong-Seitz GQ, see also (33.4) of Tits and Weiss [229]. Theorem 11.5.3 Every thick Moufang generalized quadrangle is a Fong-Seitz generalized quadrangle. Proof. By Proposition 11.5.1, we already know that a Moufang GQ is a Tits GQ. So let S be a Moufang GQ and take a flag {x,L}, then we only have to construct a nilpotent normal subgroup U of the stabilizer B of {x, L} such that UH = B, with H the intersection of B with the stabilizer N of some apartment E of S containing x and L. The construction is quite easy, and also the fact that UH = B will be easily proved. But the proof that U is nilpotent will need some arguments. Fix an apartment E containing {x,L}. We now give new names to the elements of E. We put E = {x\, x%,..., x$}, where we read the subscripts modulo 8, with Xilxi+i, for alii e Z mod 8, and where x2 = x and x3 = L. Hence, for convenience, we abandon here the notational convention

11.5. Properties of Moufang Quadrangles

239

of capital letters for lines and small letters for points. Instead, we have even index for points and odd index for lines. The group of (dual) root elations related to the (dual) panel {XJ_I, x i ; xi+i} will be denoted C/j. We putU = {U1,U2,U3,U4). First we show that U = U\U2U3U^. Let Uj 6 Ui and ui+\ e C/i+1 be arbitrary. Since Ui fixes all elements incident with XJ_I, and u i + 1 fixes Xi-i, we see that [uj,u i+ i] fixes all elements incident with x;_i. Similarly, [ui,ui+i] fixes all elements incident with xl+2- Both u; and u i + i fix all elements incident with either Xi or XJ+I, hence also [UJ,UJ+I] does. We conclude that [ui,ui+\] fixes S and an ideal and full subGQ, hence [ui,ui+1] = id = [ui+i,Ui\. Now let m-i G I7i_i and u i + i G Z7i+1. As above we see that [m-i,Ui+i] fixes all elements incident with x;_i, with Xj and with x i + i . Hence [ui-i,ui+i\ G Ui. Similarly, [Ui+U L/j_i] < Ui. Finally, let u\ G U\ and u 4 G t/4. Then u := [111,144] fixes, as before, all elements incident with both x 2 and x 3 . Put x 5 = x%. Since x 4 is fixed by u, x'5 is a line through x 4 . By the Moufang Condition, there exists a dual root elation u2 G C/2 mapping x 5 to x 5 . Likewise, there exists a root elation u3 G C/3 mapping xg to x§. The collineation ra^Mj1 now fixes all elements incident with x 2 or x3, and fixes xg and x 5 , hence E, and so has to be the identity by Corollary 11.2.5. It follows that u = u2u3 = u3u2 e U2U3 = U3U2. Similarly, [C/4,t/i] < U2U3. Now note that b~1a = a[a, 6]6 _1 , for arbitrary group elements. Hence, if ui,v!i G Ui, for % G {1,2,3,4} arbitrary, then, putting \u\vv!1,u^\ = v!2v!3, we obtain (u\u2u3Ui)~

(u^u^u^u^) = U3 u^[

U'IU'Z'U'IU^ 1

=

U2 u2u'3u'4,

1

,

1

(11.1)

1

Ui u[[Ui u'1,U3\u 2U2 U2U2 U3 1

>

X [u2 U 2,U4,]u':iU^1u'i,

(11-2)

which clearly belongs to U\U2U3UA- Hence UXU2U3U\ is a subgroup of U. On the other hand, since Ui < U\U2U3UA, for i = 1,2,3,4, we also have U < UXU2U3U±. Hence U = UXU2U3U^. Note that, since t/j+i and Ui centralize each other, UiUi+i is a group, for alii G Z mod 8. Also, the group Ui-i normalizes UiUi+i, so Ui-\UlUi+\ is a group, too, for all i G Z mod 8. We further claim that, if we write an element u of U as u = u\u2u3u±, where Ui G Ui, i = 1,2,3,4, then this is unique. Indeed, suppose also u =

240

Chapter 11. Moufang Quadrangles with a Translation Point

u'^u'zu'i, with v!i &Ui,i = 1,2,3,4. First we show that, if 0,10,20,30,4 — id, with at e Ui, i = 1,2,3,4, then at = id for all? e {1,2,3,4}. Indeed, all of ai,a2,a3 fix x\, and a.4 fixes zi only if 04 = id. Hence 04 = id. Similarly, a3 = id by considering the action on x8. Similarly a2 = ai = id. Now from Equation (11.2), we deduce that id = u~lu = (m ... U4)"1u[ ... u'A implies that u^ui = id, hence m = u[. This implies [uj" 1 ^,^] = [M^1 14,1x4] = id, and so u2' = u3 = id, and now it follows rather quickly that u2 = u2, u3 = u'3 and u4 = u'A. Hence the claim is proved. So we know exactly what an element of U looks like. To prove nilpotency, we want to show in particular that the Ui are nilpotent. That is our next task. We will show that each Ui is nilpotent of class at most 2, and that one of U\ and U2 is abelian. If both U\ and U2 are abelian, then there is nothing to prove, since by the Tits Property the groups U2i are conjugate amongst themselves, and also the groups U2i+i are mutually conjugate, for all i e Z mod 4. So we may assume that for instance U\ is not abelian. We first claim that every element of Vi := [U\, U\] fixes every line meeting x\. Indeed, let x be any point on x\ (distinct from x8 and from x2). Let vi = [ui, u[] be an element of V\. Let u3 e U3 be such that Then w"3 fixes all lines through x and through x2, and fixes all points o n i i . Moreover, it has the same action on the set of points incident with x3 as u\. Hence [u\3, u[]v^x fixes x3 pointwise. But it belongs to U\, hence vi = [ui3,u[], and the latter fixes all lines through x. The claim is proved. Next we show that [Vi, U\] is trivial. Let m e U\ and v\ e V\ be arbitrary. Consider any nontrivial element W4 in C/4, then the conjugate v,™4 has the same action on the set of points incident with x3 as u±. Hence [ui,ui] has the same action on that point set as [w"4, vi\. But the latter fixes all lines concurrent with x\ (since v\ does, and w"4 fixes x{), and fixes all points on a;"4, which is not equal to xi. This implies that [w"4, vi] is the identity, and hence so is [ui, vi]. We have shown that Ui is nilpotent of class 2. Since Vi fixes xj- elementwise and U3 stabilizes xj- globally, the commutator [Vi, U3] fixes Xi elementwise and hence is trivial, since it is also a subgroup of U2. Hence, by conjugating, [V2i+i, U2i-i] = [V2i+1, U2i+3] = id, C/4X

(11.3)

for all integers i modulo 4. Let u\ e U* and M4 e be arbitrary. Then [ui, u4] = u^u"4. The latter expression shows that, if we fix u4, then, since u\ fixes xi pointwise, the action of [1/1,1*4] on the points of x\ is determined by M"4. But C/"4 acts transitively on the points of x\ distinct from x2, hence we see that, writing [111,114] = 1*21/3, with m € Ui} i = 2,3, we may choose,

11.5. Properties ofMoufang Quadrangles

241

for each 114, an arbitrary u3, and then some (unique) u\ £ XJ\ and u2 e U2 exist for which [ui, U4] = u2u3. Since u3 and u"4 have the same action on the points of x\, we see that u3 = u\iU, for the unique u e UgUi that maps Zg4 to X4. Since now Vi and V3 are conjugate (as U\ and t/3 are), we deduce that u3 e V3 if and only if ui e Vi. Dually, we may fix ux, choose u2 and get unique u3 and m with [^1,114] = u2u3. We now apply this as follows. We choose v\ e Vi nontrivial, but fixed. We let u2 and u'2 be arbitrary but nontrivial in U2. Our aim is to show that [«2,u2] = id. We consider the unique u3 e U3 and 114 e t/4 for which [v\, -u4] = u2u3. We now first claim that [u2u3,u2] = id- Indeed, [u2u3,u'2] = [[vi,U4],u'2], and this is equal to [u4,vi]u'2~1v^1u^1viU4U2. First notice that, by a simple computation and the fact that U4 centralizes U3, we have [u2, u^1] = [u'2, u^ 1 ]" 4 = [u'2, Ui]~l. Secondly, remember that also Vi centralizes U3. Hence we have V\U\%)!2 = v!2V\1i2 =

U4U2U^ U4

u'2Vi[u2,U~ll]U4

= M2[u2,U4"1]l'iU4

(11-4)

and also Vi 1u4~1u'2 = Vi lu'2[u'2,Uiju^1 = u'2v^lu^l[u'2,U4\.

(11-5)

So one calculates [U2U3,U2] = [[vi,U4],U2] = [u4,Vi]u'2

Vi1U±1ViU4U2 1

1

(11-6) 1

= [u4,vi]u'2~ v^ uJ u'2[u2,u^ ]viU4 1

1

1

= [u4,Ui]u 2 u'2V^ U^ [u'2,U4][u'2,U^ }viU4 = [u4,Vi]v^1U^1ViU4

= [w4,fi]bi,w4] = id,

(H-7) (11-8) (11.9)

(11.10)

where we use Equation (11.4) to go from Equation (11.6) to Equation (11.7), and Equation (11.5) to go from Equation (11.7) to Equation (11.8). Our claim is proved. But now id = [u2u3,u'2] = [u2,u2}U3[u3,u2} = [u2,u'2]Ua, implying that u2 and u'2 commute. Hence U2 is abelian if U\ is not. Hence from now on, we may assume, without loss of generality, that U2 is abelian, regardless of C/i being abelian or not. Note that the Tits Property

242

Chapter 11. Moufang Quadrangles with a Translation Point

implies that C/2i is abelian, for all i. We now show that [U2, U4] centralizes C/i (and also f/5 by the same token). Indeed, let m G Ui be arbitrary, i = 1,2,4, then we have (using the fact that u2 centralizes [C74) U\] < U3U2 because U2 is abelian) [[u 2 ,ti4],Wl] = [u21U^1U2Ui,U1] l

=

[uj1,^]"2"4^,?/!]

U4

= [u^ ,ui] [u4,ui] =id. Hence by a shift of the indices, we obtain that [U2%, U2l+2] centralizes both U2i+3 and U2l-i. We will denote the centralizer of l ^ - i and U2i+3 in U2i+i by W2i+i, for all i. Note that we have just shown that [U2il U2l+2] < W2i+i, and that Equation (11.3) implies that V2i+i < W2i+\. Now, we claim that Wi
= U1U2U3[UIU2U3,UA\

= UiU2U3[Ui,U4]U2U3[u2,U4]U3

G UlU2U3,

since [ M ! , ^ ] " 2 " 3 G (U2U3)U2U3 = U2U3 and [u2,uA]U3 e U33 = U3, for all Ui e Ut,i = 1,2,3,4. Similarly U2U3UA < U. Hence also U2U3 < U. We also remark that, defining V\ as the trivial group [U\, U\] in case that U\ is abelian, we always have that [Vi, U3] is trivial. Also, since [Vi, f/4] < U2V3 if Ui is nonabelian, as shown above, and since this remains true if U\ is abelian (and V\ is trivial in this case), we see that Vj"4 < ViU2V3 (and the latter is indeed a group), for all U4&U4. Now we take u := u\u2u3Ui and u' := u[u2u3u'4, with Ui,u't E Ui, i = 1,2,3,4, arbitrary, and we calculate [u, u'}. Note that, in general, [ab,cd] = [a,d}b[a,c\db[b,d}[b,c}d.

(11.11)

Assuming, as above, that U2 and C/4 are abelian, and taking into account that u2 and ?4 centralize each other and that they centralize Ui, and hence that [uiU2,u[u'2] = [ui,ui], and similarly [u3U4,u3u'4] = [143,1*3], we calculate: [«,«'] = [uiU 2) U , 3U 4 ] U8U *[«lU 2) uWy uiuiu3U4 [«3«4,«[ l «4][«3«4,«W2] uiui , = [ M l U 2 , U 3 U 4 ] M 3 , i 4 [ W l , U i] u 3«4«3«4 [ u 3 ; U y [ w 3 U 4 j U /^ ] «>^ a L 1 2 )

II.5. Properties ofMoufang Quadrangles

243

We now examine each factor of Equation (11.12). We will often use the following fact. Since for m eUi, and UJ G Uj, one has Uj[uj,Ui] = u™\ UpCUjpjM].

(11.13)

Firstly, [U1U2,U'3U4} = [ M i ^ y " 2 ^ ! , ^ ] ^ " 2 ^ , ^ ] ^ , ^ ] ^ , and so this belongs to (U2U3)U2.U^U2.U3 = U2U3.U2U3.U3 = U2U3. Hence [uxU2,v!zu'AU3Ul e U2U%3Ui. The latter equals U^3UiU^Ui = 4 U% U3, which is in U2[U2,U4}U3 = U2U3 by (11.13) above. Secondly, [u1,u'1]u3u*UsU4 = [ui,u'i]^ U 3 " 4 , s i n c e [Vi,U3] is trivial. We further deduce that [ui,u'1]"4«3«4 e (V1U2V3)U3U4 < ViU2U3 (again using (11.13)). Thirdly, [u 3 ,ti 3 ] G V3. Finally, [u3u4,u'1u2)u3u4 G U2U3, similarly to the first factor. Consequently, [u,u'\ G ViU2U3. We now examine [U, ViU2U3\. With the above notation, this means that we must evaluate [uiU2U3Ui,v'1u2u3\, with v[ G Vi arbitrary. We redefine u' as v'^u^. We can do that evaluation by putting u[ = v[ and u'A = id in Equation (11.12). We obtain, keeping into account that u2 centralizes U\ and U3, hence [uiu2,u'3\ — [ui,u'3], and that v[ centralizes U\, hence [u\, v[] = id, [«,«'] = [u1,u'3\UaU*[u3,u'3][u3U4,v'1y''2]u'3-

(11-14)

We examine each factor of Equation (11.14). Firstly, [ui,u'3]U3Ui G U2W3. Secondly, [1*3,1*3] G V3 < W3. Finally, applying Equation (11.11), noting [u3,«i] = [1*3,1*2] = id, and remarking that [u4, u2] commutes with u3 since both u4 and i*2 do, we compute (also using (11.13)), [u3U4, v'-^ul^ = [w4, W'2][M4, v^]"3 G W3.U2V3 = U2W3. We have shown that for the second central derivative of U holds [U, U][2] < [U, U2U3] < U2W3. So, in order to have an estimate for the third central derivative [U, U][3], we consider an arbitrary w'3 G W3 and redefine v! = u'2w3. We can now substitute id for v[ and w'3 for u3 in Equation (11.14) and calculate: [u,u'} =

[uiy3]»™[u3,w3][u3u4,v!i]<,

= [u3,w3}[u3u4,u'2}w3.

(11.15)

Both factors of the right hand side of Equation (11.15) are contained in W3. So [U,U][3] < W3. But W3 by definition centralizes U\, and additionally centralizes U2 and U4. Hence we see that, putting v! = w'3, and hence substituting id for u'2 in Equation (11.15), [u,u'] = [u3,w'3],

(11.16)

244

Chapter 11. Moufang Quadrangles with a Translation Point

and so [U, J7][4] < V3. Choosing now v3 G V3 arbitrarily, we see that v3 centralizes Ui,U2,U3 and t/4, hence v3 e Z{U) and so [U, U][5] is trivial. This shows that U is nilpotent (and of class at most 5). We now show that UH = B. Take an arbitrary element b of B. Put x'6 — x\ and x'7 = x7. We claim that there is an element of U mapping {x 6 ,x 7 } onto {x'6,x'7}. Indeed, let u\ G U\ be the unique root elation mapping £4 onto x'4 := proj xg- Dually, let w4 be the unique dual root elation of U4 mapping x\ onto x\ := proj x^. Then U1U4 maps {x4,xi} onto {x 4 ,xi}. Let u3 G U3 be such that it maps XglM4Ixi onto x'8 := proj^,, x2. Finally, let u2 G U2 be such that u^ (which fixes pointwise) maps X5lU4U3Ix4 to x'5 := proj^^xl,, or equivalently, x^lU4U3 to x'7. Then we see that u := uxu4u3u^ G U maps {x 6 ,x 7 } onto {x'6,x'7}. Our claim is proved. But now bvr1 G H, so b G HU, hence HU = B, or, taking inverses, UH = B. In order to show that U
•

We will now show that every Moufang GQ contains a full or ideal subTGQ. Theorem 11.5.5 Every Moufang generalized quadrangle contains a thick full or ideal translation subquadrangle.

11.5. Properties ofMoufang Quadrangles

245

Proof. We use the same notation and conventions as in the proof of Theorem 11.5.3 above. In particular, we assume that U2 is abelian. Suppose that for some i e Z mod 8, an element ut of U* commutes with all elements of Ui+2, and let u e Ui+2 be arbitrary. All elements incident with xf_x are fixed by both Ui[ui, u\ (since the latter equals uf, and m fixes all elements incident with Xj_i) and [ui, u] (since it is the identity). Hence Ui fixes all elements incident with xf_x. Since u was arbitrary and Ui+2 acts transitively on the elements incident with x, distinct from xi+i, it follows that Ui fixes xf- elementwise. Hence Ui is a symmetry about xt. Hence, if W3 is trivial, then, since [U2, U±] < W3, all elements of U2 commute with all elements of C/4, and the previous paragraph implies that U2 consists entirely of symmetries about x 2 . By transitivity, every point is a center of symmetry, and every line is a translation line. Hence S is a TGQ. Now suppose that W3 is not trivial. Then by definition every element of W3 commutes with all elements of U\ and so, every element of W3 is a symmetry about x3. We now redefine W2i+\ as the subgroup of U21+1 containing all symmetries about x2i+1, for all i e Z. Our assumption implies that none of W2i+i is trivial, and Lemma 11.5.4 implies that they are all conjugate to one another. We also remark that [V2, V4] < W3. We define the following point set in x2 : n=

({x2,x6}±)W3U({x2,x6}±)w\

Now let A be the set of lines of S incident with at least one point of ft, and let II be the set of points of S incident with at least two lines of A. Then we now show that S* = (II, A, I) is a subGQ of <S. We do that in a number of steps. 1 Wl STEP 1. If x'0 e {a^,^}" \ {^0,^4}, then we claim that x'0 = x'0Ws. Indeed, by the Moufang Property, there exists a unique collineation u £ Q[x2x'0,x'0,x'0xe] mapping x0 to x 4 . Since u fixes all points on the line x2x'0, we deduce that x'0Wl C x'0W3. Using u~l, the claim follows. As a consequence, the groups W\ and W3 preserve Q and hence act on <S* as collineation groups. STEP

2. We claim that both U0 and U4 stabilize tt and hence act as collineation groups on S*. Indeed, this follows readily from (x^3)Ui = (XQ4)WS, for all U4, € Ui, and the fact that XQ4 G {X 2 ,X 6 }- L . Similarly for U0.

STEP

3. Every line of S through any element of n belongs to A. Indeed,

Chapter 11. Moufang Quadrangles with a Translation Point

246

this is clear for the points of n. Now let x be the intersection of two elements L, M of A. We may assume that L,M are not incident with x2, and that they are concurrent with the distinct lines L', M', respectively, incident with x2. Using the action of U0 and [74, we can map {Lf, M'} onto {xi,x3}. Subsequently, we can combine this with the actions of W\ and W3 to eventually map the intersection of L and L' onto x4, and the intersection of M and M' onto x0. Hence we may assume that x belongs to {x^x^}1-. Let Y be any line through x, and let z be the projection of x2 onto Y. It is necessary and sufficient to prove that z belongs to Q,. Clearly we may assume that z ^ {XQ, X±}. Let u2 £ U2 map XQ onto x and let UA e U4 map x\ onto a^z. Set z' := proja,2^x6. Then Z

/[u4,«2] _ ,U~1U~1U4U2 — Z

_ U~1UiU2 — Xn

_ u4u2 — XQ

_ Ju1 — /6

_ _ — £.

Since [C/4, t/2] < W3, the proof of STEP 3 is complete. 4. The incidence structure S* is a GQ. Indeed, let x be any point of S*, and L any line of 5*, not incident with x. By STEP 3, the line proj^L belongs to <S*, and if projLx does not belong to fi U {x2}, then by definition of LT, it belongs to II since it is incident with two elements of A. Applying Lemma 11.2.2, the result follows. Clearly <S* is ideal, and a;3 is an axis of symmetry. Since S* is also a Moufang GQ by Proposition 11.5.2, every line is an axis of symmetry and S* is a Moufang TGQ. The theorem is proved. • STEP

We end with another result on Moufang TGQs. Theorem 11.5.6 All root groups of a Moufang TGQ are abelian. Proof. Let S be a Moufang GQ with a translation point. Since the translation group of a TGQ is uniquely determined by the symmetries about the lines through the translation point, the group U\U2Uz is the translation group with respect to the translation point x2, using the notation of Theorem 11.5.3. Since the translation group is abelian, both U\ and U2, as subgroups of U1U2U3, are abelian. Hence all root groups are abelian. •

11.6. Half3-Moufang Quadrangles

247

11.6 Half 3-Moufang Quadrangles In this section, our principal goal is to show that a half 3-Moufang GQ is a Moufang GQ. This result is due to Haot and Van Maldeghem [68]. So let S = (V, B, I) be a thick generalized quadrangle with automorphism group G, satisfying the half 3-Moufang Condition. More exactly, we assume that for all dual roots {y0,2/1,2/2,2/3,2/4}, with 2/0I2/11 • • • l2/4> a n < i w r t n 2/o, 2/2,2/4 G V, the group G^ 2 ' acts transitively on the apartments containing 2/0, ••• ,2/4Our first aim is to show that S is half Moufang — more exactly, that all panels are Moufang. As in the previous section, we fix some apartment £ := {x0,xi,...

, x 7 } , a; 0 lxil • • -Ia^Ixo,

where x0 e V, and where we read the subscripts modulo 8. We prove some lemmas, under the assumptions just stated. Lemma 11.6.1 All sequences (x,y,y',z), with xlyly'lz, x e B, x ^ y', and y ^ z, form a single orbit under G. In particular, all groups Gz'v are conjugate. Proof. For any given (x, y) G B xV, with xly, it suffices to prove that the group GXiV acts transitively on the set of flags {y1, z), with yly'lz, y ^ z and 2/ 7^ x. So let, with obvious notation, {2/1,21} and {y'2,z2} be two such pairs. First suppose that y[ ^ y'2. Then we choose arbitrarily some element z' in {z1,z2}i- \ {y}- Set xx := proj^z' and x2 := proj z ,x. There is a collineation u e Gy'X2' mapping z\ to z2, and hence y[ to y'2. If 2/i = y'2, then we consider any pair {y'3, z3}, with yly'3lz3, y[ ^ y3 ^ x, and 23 ^ 2/, apply twice the previous paragraph and get the result. • Lemma 11.6.2 If in S the span {x,y}±A- of some opposite points x,y contains at least 3 elements, then all dual panels of S are Moufang. Proof. We may assume without loss of generality that {x, y} = {x2,x6}. Let x'6 e {x2,x6}J-L, with x2 ^ x'6 ^ x§. Let x'5 denote the line incident nxes x with x 4 and x'6. As Gxt i 2, x&}, the span {x2,x6}±± has to be stabilized as a set, but as the lines through x4 are fixed as well, this implies that the span is fixed pointwise, and hence in particular x'6 is fixed. Consider

248

Chapter 11. Moufang Quadrangles with a Translation Point

an arbitrary element g e G ^ 3 ^ and choose an element h € G^i ma P~ ping x2 to X6 {h exists by the half 3-Moufang assumption on the dual root {x0,x0xf;,Xg,X5,x4}). The commutator [g,h] clearly belongs to G%*<xs<xe\ and hence is trivial. Consequently g = gh e G^X3'Xi'x^. • Let Q denote the set of lines incident with XQ. Lemma 11.6.3 Let x range over the set of points incident with x\, x ^ XQ, and let y range over the set of points not on x\ but collinear with x. If the action of Gy'x' on Q. is independent of x and y, then all panels of S are Moufang. Proof. It suffices to show that there is an element g e G^XO'XI'X2^ mapping x6 to an arbitrary point z ^ x0 on x 7 . Let us start with an arbitrary nontrivial collineation u e Gf^ 'X2'. Then there is a unique point z' on x£ collinear with z. Hence, if we denote by x'2 the unique point on xi collinear with z', then the collineation u' e G^,1'*2' mapping x" to x-j maps Xg to z. The composition uu' fixes all points on x\ and — by assumption — it also fixes all lines incident with xo, since the action of u on f2 must be the inverse of the action of v! on Q.. Moreover, uu' maps x6 to z. Also, the action of uu' on the set of lines through x2 is the same as the action of u' on that set (since u fixes every line through X2). Interchanging now the roles of x0 and X2, we see that the collineation u" e Gz mapping x%' back to x 3 has an action on the lines through X2 that is inverse to the action of uu' on that set. This implies that uu'u" e G^x°'Xl'X2K Since uu'u" maps x& to z, the assertion follows. • So, with the notation of the previous lemma, we will first fix x, vary y, and prove independence of the appropriate action; afterwards, we vary x and make particular choices for the points y. Lemma 11.6.4 Let y be any point not on xi collinear with x2. Then the action ofGy'X2' on His independent ofy. Proof. First we note that we may assume y to be incident with x3. Indeed, this follows immediately from the fact that the group GLX60,X acts transitively on the lines through x2 distinct from x\, and so any group Gz1'*^, with z any point not on x\ but collinear with x2, can thus be seen as a

11.6. Half3-Moufang Quadrangles

conjugate of some Gy'X2, lines through x0.

249

with ylx3, under a collineation which fixes all

Now, if the action of Giy1'*2' on tt were not independent of the choice of y, with y incident with x3, then we may assume that for some ylx3, y ^ x2, the action of the group Gx := GJf1,a:21 on fl differs from the action of the on group G*2 := Gxt ^- Since this statement does not involve x4, x5, XQ and XT, we may rename x^ as y and hence assume Gi = Gl*1^2'. Note that both Gi and G2 act faithfully on fl. Suppose first that there is an element u e Gx U G2 such that u commutes with every element of G\ U G2. We claim that Gi and G2 must have the same action on Q. Indeed, if not, then there is a collineation gi e G\ such that its action on $7 is not induced by any element of G2. Let g2 € G2 be such that g2 maps xf back to x7. Noting that (x%)9lS2 = {xf 92)u — x%, we see that gxg2 € G^, 2 ^. If x% were not contained in {x2,x6}-L-L, then gig2 would fix at least three points on some line through x2, implying that 9192 would fix an ideal subGQ, by Lemmas 11.2.2, 11.2.3 and Proposition 11.2.4. This contradicts the fact that gig2 does not fix all lines through xo- Hence we have a point span of at least three elements. Lemma 11.6.2 now implies that S is half Moufang with respect to dual roots, i.e., Gi = G2. Hence we may assume that the centralizer of G\ U G2 in G\ U G2 is trivial. Note that G\ and G2 normalize each other. We claim that G\ cannot have a commutative action on fl Indeed, if G\ were commutative, then also G2 would be commutative (as by Lemma 11.6.1 the groups G\ and G2 are conjugate, and they both act faithfully on Q). If only the identity in Gi has the same action on fi as some element of G2, then G\ and G2 centralize each other, contradicting the assumption that the centralizer of G\ U G2 in Gi U G2 is trivial (alternatively, by Proposition 9.1.2(iii) two abelian groups acting regularly on a set Q. and centralizing each other must have the same action on f2). Hence there is some nontrivial element c\ in G\ having the same action on Q as an element c2 in G2. Now c\ centralizes G\ since G\ is commutative. Since c\ and c2 have the same action on Q, this implies that [c2, Gi] acts trivially on fi. Since [c2, G\] < G\C\ G2, this implies [c2, G{\ = {id}. Consequently both c\, c2 centralize Gi U G2, again a contradiction with our assumptions. The claim is proved. Next we claim that only the identity in Gi has the same action on fl as some element of G2. Indeed, suppose by way of contradiction that there is a u\ G Gf inducing the same action on 17 as some w2 e G2. Since u\ cannot lie in the center of G\ UG2, we may suppose there is a g £ G\ UG2 such that

250

Chapter 11. Moufang Quadrangles with a Translation Point

the commutator [ui, g] ^ id (and this is equivalent to the assumption that the action on tt of that commutator be nontrivial). Suppose g e G2 — the case g e G\ is similar, if one interchanges the roles of x0 and a;4, noting that the action of Gi and G2 on tt is permutation equivalent with their action on the set of lines through X4. Consider an arbitrary h e Gx°'x , then gh induces the same action on fi as g. It is clear that all the commutators [wi,g], [u2,g] and [112,gh] induce the same action on £1, and each of them fixes all points of x3. This easily implies [ui,g] = [u2,ff] = [u2,gh] ='• u (use Corollary 11.2.5). Since [u2,gh] fixes the line x% pointwise and since h is arbitrary, we see that u, which is not trivial, fixes all points collinear with X2. So, the image of xe under u must lie in the span of x2 and x6 which forces the generalized quadrangle to be half Moufang with respect to dual roots by Lemma 11.6.2. But then again G\ = G2, a contradiction. Hence the regular actions of G\ and G2 on fi normalize each other and share only the identity. This easily implies that they centralize each other, and the actions on fl are opposite, see Proposition 9.1.2(iii) and (iv). We conclude that, for arbitrary ylx3, y ^ x2, the action of Gy'X2' on O either is the same as the action of G2 on fi, or it is opposite. Suppose both really occur. So for some ylx3, y ^ x2, the action of Gi = Gy'X2' on tt is opposite the action of G2 on fl, and for some 2:1x3, z ^ x2, the action of G3 := Gz1 on fi is the same as the action of G2 on £1 Since Gi n G2 is trivial, no nontrivial element of G2 can fix all points on x\. This implies that G2 n G3 is trivial. But G2 and G3 normalize each other, hence they centralize each other. This means that the action of G3 on O — which is the same as the action of G2 on Q — centralizes the action of G2 on SI, hence this action is commutative! This contradicts a previous claim. We conclude that all actions of Gy'X2' on il, ylx3, y ^ x2, either are the same as the action of G2 on £1, or are opposite. In particular, the action is independent of y. The lemma is proved. • Lemma 11.6.5 If x'2 is an arbitrary point on x\, x'2 ^ x0, and x'4 is the unique point on x$ collinear with x2, then the action of Gi^1 on£l coincides with the action ofG,1'*2' on Q. Proof. Let U2 be the permutation group acting on tt given by the action of G^l • Define U2 as the permutation group on Q, given by the action of £rFi.z2J_ we assume that U2 ^ V2 and seek a contradiction.

11.6. Half3-Moufang Quadrangles

251

Let UQ be the permutation group acting on Q defined by G^X4'X7\ and note that by Lemma 11.6.4, this action is the same as the one induced by GH6,X7^ on n. Now notice that G[x4'X2] is normal in GXo,X2tXi, that G[x46'Xr] is normal in GXo,XetXl, and that G[^'xJi is conjugate to c i ? ' * 7 ' in the group H generated by both (indeed, if g is nontrivial in G^!'X7\ then there is a unique element h e (GX4'X2')9 mapping x 3 to x5 and hence x 2 to xe and xi to :r7; then (G[£'x'])h = G I ? ' * 7 1 ) . Consequently, U2 and f/6 are conjugate in K := (t/ 2 , f^), U2 < KX1 and £/6 < KXl (although K is defined as a permutation group acting (only) on S7, it is clear what we mean with KXl and KXl). By Lemma 9.1.3, (K, Q,) defines a Moufang set with root groups U2 and C/6. Likewise, (if', Q), with X ' = {U^ U6), defines a Moufang set with root groups U'2 a n d ^6- We now want to apply Theorem 9.1.5 to obtain a contradiction. So we show that U2 and V^ normalize each other. Choose arbitrary nontrivial u2 e U2 and u'2 e t/^

an

d l e t them be in-

duced by the collineations g e GX4'X2' and g' e G*,1'*2', respectively. Then g9' belongs to G[xx;X2\ which has the same action on Q as G[X4'X2] by Lemma 11.6.4. Hence u22 e U2 and t/^ normalizes U2. Similarly, U2 normalizes U2. Now Lemma 9.1.5 leads to a contradiction.

•

Lemmas 11.6.3, 11.6.4 and 11.6.5 complete the proof of the following theorem. Theorem 11.6.6 If every flag of a GQ is half 3-Moufang at its point, then all panels are Moufang. • We have the following easy corollary, see also Tent [151] and Haot and Van Maldeghem [67]. Corollary 11.6.7 Every halfMoufang GQ is a Moufang GQ. Hence every half 3-Moufang GQ is a Moufang GQ. Proof. Assume all dual panels are Moufang. Then obviously all flags are half 3-Moufang at their points. According to Theorem 11.6.6, every panel is Moufang. • There is an interesting consequence for TGQs.

252

Chapter 11. Moufang Quadrangles with a Translation Point

Corollary 11.6.8 If every point of a GQ is a translation point, then the GQ is Moufang. Equivalently, if every line is an axis of symmetry, then the GQ is Moufang. •

11.7 2-Moufang Quadrangles and Fong-Seitz Quadrangles We apply the results of the previous section to the 2-Moufang quadrangles and Fong-Seitz quadrangles. We will show that the 2-Moufang Condition implies the 3-Moufang Condition, and that any Fong-Seitz quadrangle is half Moufang. Hence in both cases, the quadrangle is a Moufang GQ. The next result was first observed by Van Maldeghem [231]. Lemma 11;7.1 If the thick GQ S = (V, B, I) has a center x of transitivity incident with an axis L of transitivity, then the group of whorls about both x and L acts transitively on x' \ x^, for each x'lL, x' ^ x. Proof. Given the flag {x, L}, x € V, L e B, xlL, and given two points y, z opposite x with proj L y = proj L z, we must show that there exists a collineation of S fixing all lines through x, fixing all points on L, and mapping y to z. We prove this first in a special case, namely, when there are lines M, N opposite L, and a point w opposite x with ylMlwlNlz. By the 2-Moufang Condition, there exists a collineation ux fixing all lines through x and mapping y to w. Also, there exists a collineation uL fixing all points on L and mapping M to N. Note that UL fixes w, as proj w L is fixed by UL (otherwise a triangle arises). Also, UL maps y onto z. Hence [u" 1 , ur] maps y to z and belongs to G^X'L\ As the unique GQ of order (2,2) is Moufang, we may assume that the order of S is not (2,2). Let (T, E) be the graph with set of vertices the points of S opposite x and the lines of S opposite L, where adjacency is just incidence. Suppose that S is not a 3-Moufang GQ, and let y,y' be two points at minimal distance, say In, in (r, E) such that proj L y = proj L y' and such that there does not exist a collineation of G^^l mapping y to y' (this distance is well-defined by the connectivity of (r, E), see Lemma 11.2.7). Hence In > 4. Let (y = y0,Yi,y2,Y3,... ,y2n = y') be a minimal path in (T,E) connecting y and y'. Then z := proj y (proj L y) and y satisfy the assumptions of our special case above, and so there exists an u e G^X'L^ mapping y to z. Now z and y' are at distance at most 2n - 2, and hence

11.7. 2-Moufang Quadrangles and Fong-Seitz Quadrangles

253

there exists an u' e G^x^ mapping z to y'. Consequently uu' e G^X'L^ maps y to y', a contradiction. The proof of Lemma 11.7.1 is complete.

•

There are two interesting corollaries. Theorem 11.7.2 Every thick 2-Moufang generalized quadrangle S = (V, B, I) is a 3-Moufang generalized quadrangle. • Corollary 11.7.3 If in a GQ every point is an elation point and every line is an elation line, then the GQ is Moufang. • Now we turn to Fong-Seitz GQs. The next theorem was proved by Tent and Van Maldeghem [153]. Theorem 11.7.4 Every thick Fong-Seitz generalized quadrangle S = (V, B, I) is a half Moufang generalized quadrangle. Proof. We use the same notation as before. In particular, we have an apartment E consisting of the elements x±, x2, • • •, x$. Let B be the stabilizer in G = Aut(S) of the flag {xltxg}, and let TV = G s . The fact that S is a Fong-Seitz GQ means that we have a nilpotent subgroup U < B such that UH = B, for H = B n N. Note that H is the stabilizer in B of the flag {rr4, x5}. Furthermore, we know that B acts transitively on the set of flags opposite {xi, x8} (since S is in particular a Tits GQ). Let F be an arbitrary flag opposite {x\, xs}. Let b e B map F to {X4, x5}. Let u e U and h e Hbe such that uh = b. Then Fu = Fbh~1 = {x4,x5}h~1 = {x4,x5}. Hence U acts transitively on the set of flags opposite {x\,x%}. We claim that UH' = B, for every conjugate H' of H in B. Indeed, let H' = H9, g € B. For arbitrary b e B, there is a unique element u e U mapping {x4, x5}9b to {XA, x5}9. Hence {b~lu)9 fixes {x4, xs} and so belongs to l H, implying that b~ u e H', or b e uH'. The claim follows. Consequently we may vary the apartment E through {x\,x%} in our arguments. Since U is nilpotent, it has a nontrivial center Z(U). We claim that every element of Z(U) belongs to G^Xl'XsL Indeed, suppose some element v e Z(U) does not fix the point x2 (which can be considered as an arbitrary point on x i ) . Then there is a unique line x'5 through X4 meeting the line xl. By the transitivity property of U, there exists anu eU mapping {x 4 , x'5 } to {x 4 ,x 5 }. Since v e Z(U), [u,v] = id, so x2 = x[2'v^] = {xv2)uv^, hence

254

Chapter 11. Moufang Quadrangles with a Translation Point

x\ is fixed under u. Similarly, x%, is fixed under u. Since also x 4 is fixed under u, this implies that x'5 is fixed under u, so u = id and v = id, a contradiction. We conclude that Z(U) fixes x\ pointwise. Together with the dual, the claim follows. We now show the following assertion. This is the only place where we use the full strength of the hypothesis that U is nilpotent. 1. The group G^ 1 ' 2 ^ acts transitively either through x2 distinct from x\, or on the set of points on Also, G[Xl) acts transitively on the set of lines through and G^ 8 ' acts transitively on the set of points incident from x$. CLAIM

on the set of lines x7 distinct from xs. x2 distinct from x\, with x7 but distinct

Let U be nilpotent of class n and let i > 0 be maximal with respect to the property that [U, t/][i_1] does not fix all elements incident with x\ and those incident with xs. Since [U, £/][n-i] < Z(U), and all elements of the center fix all elements incident with x\ and those incident with xs, we have i < n. Without loss of generality, we may assume that v e [U, f ][j-i] does not fix all elements incident with xs. If v fixes x7, then consider u eU mapping x7 onto some element x'7 not fixed by v. Then clearly, [v, u _ 1 ] would not fix x7. But is has to because [w,tt_1] e [U, U}^ and i was chosen to be maximal. Varying x7 (as is allowed), we conclude that v does not fix any line through xs, except for x\, of course. Choose x'G arbitrary on x7 with x'6 ^ xs- We may rechoose x2 such that x^ is collinear with x\, and put x'5 = proj^./ x3. Let x'4 be the unique point on x'5 collinear with xg. Clearly the flags {x^,x5} and {x'^x'5} are opposite {a; 8 ,a;i}. Hence there exists an u e U mapping {xi,x5} onto {x'4,x'5}. As x7 is concurrent with both x5 and x'5, we see that u fixes x7, and hence maps x6 to x'6. Since the commutator [u, w -1 ] belongs to [U, U]^, it fixes all lines through x&. This is impossible if u did not fix x7. Hence u fixes x7. Since xvQ is collinear with both a;4 and x'A, it must also be fixed under u. Hence x[f%M = xv6u~ly~lu = x'6. But [y-^u] 6 [?/,£%], implying that it belongs to G^ 8 ^ 1 '. So the first part of the claim is proved (since x'6 was arbitrary). The second part is now proved similarly (considering the dual, and choosing i now to be maximal with respect to the property that [U, f][i-i] does not fix all elements incident with x\). CLAIM 1 is proved.

11.7. 2-Moufang Quadrangles and Fong-Seitz Quadrangles

255

By CLAIM 1, we may assume, up to duality, that G^XuX8^ acts transitively on the set of lines through x2 distinct from x\. Of course, by transitivity of B, the same is true for any point x'2 on x\, x2 ^ xs. The same claim implies that, interchanging the roles of x\ and x7, the group G^ 8 ' acts transitively on the points of x\ distinct from x%. Hence G^ 8 ' acts transitively on the set of lines concurrent with x\ and not incident with x 8 . We now show that the group G^Xr'Xs'Xl^ is nontrivial. As this will be an important step, we call it CLAIM 2. 2. The group G^XT'Xa'Xl^ is nontrivial — in other words, there is some nontrivial dual root elation. CLAIM

Indeed, first we prove that this is true if Z{U) contains some nontrivial root elation a. Since a e G^ 1 ^ 8 ', we necessarily have that a fixes all lines through some point (distinct from x%) o n i i ; without loss of generality we may take this point to be x2. Considering an arbitrary u e U, it follows easily from [a, u] — id that a fixes all lines through x^. By the transitivity of U, we see that a is an axial elation with axis x\. Let h be some nontrivial element of G^ 3 '* 4 ' not fixing x\ (this exists by CLAIM 1 above, interchanging the roles of x% and x 4 and of x\ and x3). Then, again by CLAIM 1, there is some u e G^ 1 ' with x\u = x3. We now see that both [a,h] and ahu belong to G^ 2 ^ 3 ', and that they have the same action on the points incident with x\ (this action is the one of ah, because u acts trivially on the points of xi). The latter implies that v := [a, h]~lahu G G^ 1 ^ 2 ^ 3 !. There remains to show that v is nontrivial. We first show that the image of xs under a cannot be equal to the image of x5 under ah. Indeed, suppose by way of contradiction that 5 . Then h fixes x$. Since it also fixes x5, it stabilizes {xs,x%}. But a is axial, hence { a ^ , ^ } - 1 = {x5,xi}-L, and h maps {x^^xi}1- to {2:5, x^}. The latter cannot be equal to {x$,xi}^. We conclude that x" =£ x% . Since ahu fixes all lines through a;4, we deduce that v indeed does not act trivially and CLAIM 2 is proved in this case. So we may assume that Z(U) does not contain nontrivial root elations. We remark that, in particular, this implies that Z(U) does not fix any line of S through x2 except for x\. Next we show CLAIM 2 in the case that there is a nontrivial root elation. So let a now be a nontrivial element of G^Xs'Xl-x^. Let w be any element of G fixing x\ and mapping xs to x2; we denote U* := Uw. By our previous remark, there exists v e Z(U*) with x'7 := x^ ^ x 7 . By CLAIM 1, there exists u £ G^ with x\ = x7. Without loss of generality, we may also assume that x2 = x6. Put x'6 := Xg.

256

Chapter 11. Moufang Quadrangles with a Translation Point

Now consider the product of three root elations (3 := aa~uauv and set x'8 = x2~u • We verify easily that (3 e G^x'-"Xs'XlK Moreover, if (3 were trivial, then v would belong to G^x's'XuX2\ contradicting our hypothesis that Z(U*) does not contain any nontrivial root elation. Thus we may now assume that G^8'*1' does not contain any nontrivial root elation. This means that G^X8'Xl^ must act faithfully on the set of elements incident with ir7 and on the set of elements incident with x2. We claim that G^8'*1' = Z(U). Indeed, let U* be as above, and let a and (3 be two arbitrary nontrivial elements of G^7'*8' and Z(U*) < G^Xl'X2\ respectively, with the only restriction that a does not fix x2. Clearly, [a,/3] e G^s.zi] our hypotheses imply the existence of some u e G^7' mapping Xj (which is different from x7 since otherwise (3 would be a root elation inside Z{U*)) to x\. It is easy to see that [a, (3} and a?u induce the same action on the set of elements incident with x-j. But since they both belong to G^Xb'Xl\ and since the latter acts faithfully on the elements incident with XT, we conclude that [a,/3] = ofiu. On the other hand, our arguments do not really use the fact that (3 e Z(U*), except for the fact that (3 e G^Xl'x^ and that (3 does not fix xy. Also, we only used Part 2 of CLAIM 1, which is independent of the choice of the duality class of S made after the proof of CLAIM 1. Hence our assumption that a does not fix x2 permits to interchange the roles of a and (3 in the above argument (interchanging the roles of X7,x% with those of x2,x\, respectively) and so [a,j3] is conjugate to both a and (3. Hence a e G^X7'X^ \ GX2 and f3 e Z(U*) are mutually conjugate. Consequently, a is conjugate to some element of Z(U). Since the latter is defined independently of x2, we see that every element of G^ 7 ^ 8 ' is conjugate to some element of Z{U) (taking into account that G^ 7 ^ 8 ' acts faithfully on the set of points incident with x\ and hence every nontrivial element of it maps sOme point of x2 onto a different one). It follows that each nontrivial element w of G^8^1! is conjugate to some element w' of Z(U), clearly under an element of B (since the flag {x%, x\} is the unique flag of S with the property that w and w' fix all elements incident with at least one of the elements of the flag). Since Z(U) is normal in B, we conclude that G^X8,Xl^ = Z(U), showing our claim. It now follows that Z(U) acts transitively on the set of lines incident with x2, distinct from x\, and thus every element in U fixing some line through x2 distinct from x\ must fix every line through x2. In particular, the stabilizer of x-i in U, which acts transitively on the set of points collinear with xs and opposite x2, belongs to G^2'. Since CLAIM 1 implies that G^2' also acts transitively on the set of points incident with x±, but different from x2,

11.7. 2-Moufang Quadrangles and Fong-Seitz Quadrangles

257

we conclude that G^ 2 ' acts transitively on the set of points opposite x2. Since Z(U) acts faithfully and nontrivially on both the set of elements incident with x2 and the set of those incident with x 7 , we may assume, by changing E if necessary, that some element z e Z(U) does neither fix x 4 nor xe- Then the line x'5 := x\ is opposite x 5 and the point x'6 := projx, x 4 is opposite x2. Let g e G'E21 map x 6 to x'6. Then g fixes x 4 (since g fixes x3 and proj^xe = proj^Xg). Set x\ := p r o j E | z | = proj x | x 6 = x\. If X49 = x 4 , then proj^xg = proj^xg, implying the existence of the triangle {x 6 , proj^Xg, x | } . Hence g does not fix x | and consequently x!f'" ' = xf UffU = x 4 s " ^ x 4 . So [g,u~l] is a nontrivial root elation, contradicting the previous part of the proof of CLAIM 2. CLAIM 2 is proved.

We now finish the proof of Theorem 11.7.4. First we show that there are fixed elements x'2lx\, x'2 ^ x 8 , and x' 7 Ix 8 , x'7 ^ xi, such that for each nontrivial h e G^Xs'Xl\ there exists a collineation a e G ^ S ^ I , ^ ] i n ( j u c m g the same action as h on the set of points incident with x'7. We choose some nontrivial element j3 in Ql^,x6,x7] ^[S exists by CLAIM 2), we put x'7 = xf , and we let u e G^ 1 ' be such that x^u = X7. Define Xjlx'i as x2 = XQ. It is easily seen that h' :— [h,(3]u 1/3 and h induce the same action on the points of x 7 . There are two possibilities. Firstly, if h fixes x 6 , then [h, /?] is clearly a nontrivial root elation related to the panel { x 6 , x 7 , x 8 } , hence we may put a — h'. Secondly, suppose that h does not fix x 6 . By CLAIM 1 and our assumptions preceding CLAIM 2, there is

some v e G^ 6 ! mapping x£ to x\. Clearly, [h, /?] and {3~hv induce the same action on the set of lines through x 6 , hence [h,0\(3hv e G^ 6 '. Since fi~hv belongs to G^-Xl\ it immediately follows that 7 = [h, (3}(3hv is a root elation with respect to the panel {x 6 , x7, x 8 } inducing the same action on the set of points incident with x\ as [h,/3], and hence also as hPu. Consequently we may now put a = j u & . Now let g e G^ 8 ^ 1 ' be arbitrary, and let a e G^Xa,XuX^ induce the same action on the points of x'7 as g and guaranteed to exist by the previous arguments. Then 0 = gar1 e G ^ ' ^ 8 ^ 1 ' and has the same action on the set of lines incident with x2 as g. Since by CLAIM 1 and the assumptions made preceding CLAIM 2, G^ 8 ^ 1 ' acts transitively on the set of lines through x2 distinct from xi, we conclude that S is half Moufang and the theorem is proved. •

Chapter 11. Moufang Quadrangles with a Translation Point

258

As a corollary, we obtain a characterization of groups acting on Moufang quadrangles and containing the little projective group. Recall the definition of a split Tits system on page 236. Corollary 11.7.5 Let (G,B,N) be a split Tits system. Denoting K := p| G B9, then the GQ SG,B,N is Moufang, G/K acts faithfully on it and contains the little projective group. Proof. It is easy to see that (G/K, B/K, N/K) is also split. Moreover, if U < B is nilpotent with B = UH, H = B n N, then B = UHS, with S = f]neN Bn. Hence the corresponding saturated group with BN-pair is also split. The assertion now follows from Theorem 11.4.3 and Theorem 11.7.4. • Remark 11.7.6 In the finite case, Fong and Seitz [59, 60] show in a completely group theoretical way that a split Tits system is related to one of the classical quadrangles (but they talk about their groups rather than about the geometries). The way the proof works is as follows. They observe that a split Tits system contains two kinds of Moufang sets, and then they use the classification of finite Moufang sets to get control over the Tits system as a group. The observation just mentioned is the group theoretical analogue of the next proposition. Proposition 11.7.7 Let S = (V,B,I) notation as before, the pairs M2 :=

be a Moufang GQ. Using the same

{{x0,xA}L,{G^x^)xe{xo^},)

and are Moufang sets. Proof. We prove the assertion for M2- Since, for every x e {x0, x^}-1, every element of Q1^OX,X,XX4] n x e s ^ Q ^ XQ a n c j x^ t h e g r o U p S Q[X0X,X,XX4] a c t on {x0, Xi}-1. Moreover, the group G^XoX'x'xx^ fixes x and acts sharply transitively on {a; 0 ,x 4 }- L \ {x}. Finally, the groups G^oX'x'XX4l and G^XoV'y'yx^ are conjugate, for all x,y e {x0, XA}1-, and G^XOX'X'XX4^ is normal in GX0}X^X4.

11.8. Conclusion

259

For the finite Moufang GQs, the Moufang sets arising are always related to PSL2{q), except for H(4, q2), where M(H(2, q2)) turns up for M2. Indeed, in all cases but this one, the root groups are abelian and act on a projective line in the appropriate ambient projective spaces of the natural embeddings of the classical GQs. However, in H(4,g 2 ), the perp of a pair of noncoUinear points is a Hermitian unital, and hence geometrically isomorphic to H(2, q2). Our assertion above is now clear.

11.8

Conclusion

From the previous sections we conclude: Theorem 11.8.1 Let S be a thick generalized quadrangle. Then S is a Moufang GQ if and only if it is a Fong-Seitz GQ if and only if it is a halfMoufang GQ if and only if it is a 3-Moufang GQ if and only if it is a half 3-Moufang GQ if and only if it is a 2-Moufang GQ. • A large class of examples of Moufang quadrangles is given by the so-called classical and dual classical quadrangles. These are the geometries related to classical groups of Witt index 2. In the finite case, all Moufang quadrangles are classical or dual classical, and they are precisely the classical and dual classical examples W(q), Q(4, q), Q(5, q), H(3, q2), H(4, q2) and H(4, q2)D of Section 1.4. Since the first four classes are TGQs or dual TGQs with respect to every point or line, it follows from Corollary 11.6.8 that these are Moufang. For H(4,g 2 ), one needs to identify the root elations and dual root elations with certain collineations of the surrounding ambient projective space PG(4, q2). We leave this to the reader. An explicit proof can be found in Chapter 4 of Van Maldeghem [232]. In fact, in the finite case, the group U1U2U3U4 of the previous sections is exactly the unique Sylow p-subgroup in the stabilizer B of a flag in the quadrangle, in the collineation group induced by the linear automorphisms of the projective space, where q is a power of the prime p.

Chapter 12

Translation Ovoids in Translation Quadrangles

In general, ovoids in EGQs and TGQs do not have special properties inherited from the elation or translation group of the GQ they are living in. Even in the case where the ovoid contains the elation or the translation point, this does not seem to put additional structure on the ovoid. For instance, it is known that it is hopeless to try to classify the ovoids of H(3, q2) (which is an EGQ with respect to every point), because there are too many (wild) examples. Although the situation is much better in Q(4, q), there is yet no hope to see a general pattern for the ovoids in this TGQ. However, when an ovoid arises from two particular general constructions (namely, from polarities, or subtended in a subGQ), then the fact that the ambient quadrangle is an EGQ or TGQ does put additional structure on the ovoid. In particular we will show that in these cases the ovoid is an elation or translation ovoid. Applied to the classical GQs, this gives rise to the classical and semiclassical ovoids (the latter being denned as the ovoids arising from polarities in classical GQs; in the finite case these are the the Suzuki-Tits ovoids which we will construct in Section 12.3 below), and the corresponding Moufang sets naturally arise. It is worthwhile to note that in this way we can geometrically describe three of the four infinite classes of finite proper Moufang sets. Among these three, we will take a closer look at the ones related to the Suzuki groups. In this chapter, we assume the GQs to be thick, unless explicitly mentioned otherwise. 261

262

Chapter 12. Translation Ovoids in Translation Quadrangles

12.1 Ovoids, Elation or Translation with respect to a Flag or a Point In this section we define elation ovoids and translation ovoids, with respect to flags and points in arbitrary GQs. Our definition will generalize the definition of translation ovoid with respect to a point given in Section 7.2. Let <S be a GQ and O an ovoid of <S. Let x e O be arbitrary, and let L be a line incident with x. Let G be the full automorphism group of S and let H be a subgroup of Go- Then we say that O is elation with respect to the flag {x,L} and with respect to H if there is a normal subgroup E < HL, called elation group ofO, acting sharply transitively on O \ {x}. We usually assume that H is given and omit the phrase "with respect to H". If O is elation with respect to every flag containing the point x £ O, such that all corresponding groups E coincide, and such that E < Hx, then we say that O is elation with respect to the point x (and with respect to H). In this case E is a group of elations of S about x. For, if some element e € E fixed a point z opposite x, then e would fix all the points of O collinear with z. If in the previous paragraph, the group E is abelian, then we say that O is translation with respect to x (and H), and we call E a translation group ofO. Now suppose S = Q(4, q) and let O be a translation ovoid with respect to a point x and with respect to the stabilizer of O inside the little projective group of S, in the above sense, and denote by T the corresponding translation group. Let L be any line incident with x and let y and y1 be two points of O with projLy = projLy'. Note first that T is a p-group, for the unique prime dividing q. Hence there exists an element 6 e T of order pn mapping y to y'. Since T is a subgroup of the little projective group of S, it follows easily that 9 fixes L pointwise. Hence O is a translation ovoid with respect to x in the sense of Section 7.2. The converse is easy. Hence the definitions of the present chapter generalize those of Section 7.2. We also have the following immediate result. Lemma 12.1.1 Suppose an ovoid O of the generalized quadrangle S is elation with respect to some group H and with respect to two flags {x,L} and {x', L'}, with x ^ x', and with corresponding groups E and E'. Suppose also that (E, E')x fixes L and (E, E')x> fixes V. Then we can choose E' such that (O, E, E') is a Moufang triple.

12.1. Ovoids, Elation or Translation with respect to a Flag or a Point

263

Proof. The group (E, E') acts doubly transitively on O, hence we may assume that E' is conjugate to E (the condition that (E, E')xi fixes L' implies that the conjugate E9, with g <s {E, E') mapping x to x', also fixes L'). Now the condition that (E, E')x (which is a subgroup of Hx) fixes L combined with E < Hx implies that E < (E,E')L = {E,E')X. Similarly E' < (E, E')L, = (E, E')x, and the result follows. • In the finite case, being elation with respect to two different flags containing different points is a rather strong restrictive property, in view of the classification of Moufang sets. If an ovoid is translation with respect to two different points, then either the full stabilizer of the ovoid contains a sharply 2-transitive group, or it contains PSL 2 (g) in its natural action. This follows from Theorem 9.2.1. Example 12.1.2 The above definitions can best be illustrated with the smallest example. Indeed, let O be an ovoid in W(2). We may use the description of W(2) with as point set the set of pairs of {1,2,3,4,5,6}, and with as line set the set of partitions of {1,2,3,4,5,6} into three pairs (with natural incidence relation). We can take for O the set of pairs containing the element 6, and we can put x = {5,6}. Then Klein's fourgroup {id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} acts sharply transitively on O \ {x} and fixes all lines through x. It is obviously a normal subgroup of the stabilizer of a; and O in the full automorphism group of W(2). Now let L be an arbitrary line incident with x. We may take L = {{1,3}, {2,4}, {5,6}}. The group generated by the 4-cycle ( 1 2 3 4) acts sharply transitively on O \ {x}, and it is a normal subgroup of the stabilizer H of the flag {x, L} in the full collineation group of W(2) stabilizing O. Indeed, one checks easily that H has order 8 (as there are 45 flags in W(2) and 2 ovoids through x) and is generated by (1 2)(3 4) and (12 3 4). So we see that O is elation with respect to every flag containing a point of O, but there are very different elation groups, and there are elation groups with respect to a flag that are not elation groups with respect to any point. This hybrid behavior can be explained by the fact that an ovoid of W(2) is at the same time a classical and a semiclassical ovoid. Indeed, in the standard embedding of W(2) in PG(4,2), every ovoid spans a hyperplane, and every ovoid also arises from a polarity.

264

Chapter 12. Translation Ovoids in Translation Quadrangles

12.2 Self-Polar Elation Generalized Quadrangles Let <S = (V, B, I) be a (finite) EGQ with elation point x and elation group G. In this section, we investigate the case where S is self-polar. It follows that <S has order s, for some s > 2. Let p be a polarity of S. An absolute point for p is a point a which is incident with its image ap. Similarly one defines an absolute line for p. It is wellknown that the set of absolute points of p is an ovoid of <S, see 1.8.2 in Payne and Thas [128], Indeed, let L be any line of S. If L is absolute, then Lp is an absolute point incident with L. If L is not absolute, then projLLp is an absolute point with image pro]LPL under p. Also, if a and b were two collinear absolute points, then (ab)p is the intersection point c of ap and bp. Since there are no triangles in S, we may assume without loss of generality that ap = ab, But then c = b, contradicting bp ^ ab (by the bijectivity of p). So let O be the ovoid of absolute points of p. We claim that we may assume that x e O. Indeed, suppose x £ O. Since xp is an elation line, all points in (proj^pz)-1 not on xp are elation points, and this easily implies that also projxPx is an elation point. But we have remarked in the previous paragraph that this is an absolute point for p. So we may rename this point by x and the claim is proved. Since xp is an elation line incident with x, we first study this situation in the next proposition. Proposition 12.2.1 Let S be a GQ of order s with an elation point x and associated elation group G and an elation line L incident with x and associated elation group H. Let a be an arbitrary point opposite x and R an arbitrary line opposite L but incident with a. Put b := proj^a; and S = projaL. Then we have (i) The group K := G D H has order s2. (ii) The group (G, H) is equal to GRKbKsHa = GRKHa = GRH = GHa and has order s4. It acts sharply transitively on the set offlags opposite {x,L}. (iii) The group G is the only "full" elation group about x and the group H is the only "full" elation group about L. Consequently, G is normal in Aut(cS)a; and H b normal in Aut(S)L-

12.2. Self-Polar Elation Generalized Quadrangles

265

Proof. The proposition is certainly true for s = 2, as there are unique elation groups in this case, and they arise from the root groups when we view the GQ as a Moufang GQ. So we may assume that s > 2. Since i is a center of transitivity and L is an axis of transitivity, Lemma 11.7.1 tells us now that the group J < Aut(<S) of whorls about both x and L has order s2 and acts sharply transitively on the set of flags {y, M}, with y ~ projLa, ylL and M ~ proj^i?. We now show that J < K. By symmetry, it suffices to show that J < G. Let p be any prime dividing s and let g e Gs have order pn, for some positive integer n. Let u be the unique element of J mapping a to a3. Since every nontrivial orbit of g on the set of s — 1 points on L distinct from x and projLa has size a multiple of p, and s - 1 is not divisible by p, there must be at least one trivial orbit, say {z}. The coUineation gu~x fixes the quadrangle containing the elements a, R, x, L, all lines through x, and the additional point z on L, hence by Proposition 11.2.4 it fixes a thick subGQ of order (s',s). Lemma 1.1.1 implies s' = s and hence g = u. Since the Sylow p-subgroups of Gs, for p ranging over all primes dividing s, generate Gs, we have shown that Gs < J, and clearly Gs = Js- Hence Js < G. The very same argument shows that Gb,PmJLa < J, implying Jb < G. Since Jb n Js is trivial, we see that | J b J s | = \Jb\ • \ Js\ = s2, hence JbJs = J and J
266

Chapter 12. Translation Ovoids in Translation Quadrangles

elation group and hence acts freely on the s 3 points of S opposite x). But s* also divides \V\ {c}\ = s - 1; this implies s* = 1 and the claim follows in this case. Assume now that not all orbits of e in V \ {c} have the same size. Then some multiple en of e fixes at least two, but not all elements of V. By composing with a suitable element of K (namely the element bringing R&n back to R), we see that we obtain a nontrivial collineation fixing the quadrangle {c, x, 6, proj^c}, all lines through x and at least three points on L. This collineation is, by Proposition 11.2.4, the identity, a contradiction. The claim is proved. We now compose e with a suitable element of K to obtain a collineation, which we denote again by e, fixing R and acting fixed point freely on the set V. Note that e might now not be an elation, but this is not important for the rest of the proof. Since we still have e ^ G, the action of e on V does not coincide with the action of any element of GR. If W is the group of all whorls about x, then we consider WR and we note that the foregoing implies that WR acts transitively but not sharply transitively on V. Also, if some element w e WR fixes at least two points of V, then a thick subGQ of order (s', s) is fixed pointwise, and as before this implies w = id. So WR acts faithfully as a Frobenius group on V. The Frobenius kernel necessarily coincides with GR, contradicting the fact that also e belongs to it. Hence e must already belong to GR. Together with the dual arguments, this proves (iii). • We now go back to our polarity p and the ovoid of absolute points O. We denote the spread of absolute lines by T. It will also be useful to denote the set of flags consisting of an absolute point and an absolute line by T. Hence Op = T, Tp = O and Tp = T. We may assume x e O, as we showed above, and we put L = xp. From the previous proposition follows that the elation groups G and H with respect to x and L, respectively, are uniquely determined, and hence Gp = H and Hp = G. Also, denoting K = GnH,we have Kp = K. Proposition 12.2.2 If S is an elation generalized quadrangle with elation point x and if p is a polarity with absolute point x and corresponding set of absolute points O, then O is an elation ovoid with respect to the flag {x, xp} and with respect to the stabilizer of O in the full collineation group ofS. Proof. With the previous notation, we consider an arbitrary element u of the group D := (G, H) mapping an element F € T to another element Fu e T. We prove that u stabilizes T. As p interchanges G and

12.3. Suzuki-Tits Moufang Sets

267

H by conjugation, it normalizes D. We claim that u and p commute. Indeed, FPUP = (FU)P = Fu, and hence by Proposition 12.2.1(ii), u coincides with up, proving our claim. So for any flag F' e T, we now have [F'U)P = (FIP)U = F'u, which means that F'u e f , So u stabilizes T. By Proposition 12.2.1(h), u acts freely on T \ {{x, L}}, and E := Dp acts sharply transitively o n 0 \ {x}. Since D < Aut(<S){^L}, we also have E
12.3 Suzuki-Tits Moufang Sets 12.3.1 Polarities in the Symplectic Quadrangle W(q) We specialize the situation of Proposition 12.2.2 to the case of W(q). Since for q odd, all lines of W(g) are translation lines, but no point is a translation point, W(q) cannot be self-polar in this case. For q even, W(q) is selfdual (see Theorem 1.4.1(0); this can be seen by projecting Q(4, q) from its nucleus n in PG(4, q) onto a 3-dimensional subspace not containing n. We now determine precisely when W(g) is self-polar. To do this, we need to represent W(q) with intrinsic coordinates. There is a general theory of coordinatization of an arbitrary generalized quadrangle, and the next proposition is a result of this. But we will not need to introduce the general theory here.

Chapter 12. Translation Ovoids in Translation Quadrangles

268

Proposition 12.3.1 The symplectic quadrangle W(q), for any power q of the prime 2, is isomorphic to the following geometry {V, C, I). Let oo be, as usual, a symbol, not belonging to GF(g). • The POINT SET is equal to

P = {(oo)}U{(a) || {{k,b) || b,keGF(q)}U{(a,£,a')

aeGF(q)}U || a, a',£ G GF(q)},

• the LINE SET is

£ = { [ o o ] } u { [ f c ] || * e G F ( « ) } U {[a,£] || a,£eGF(q)}U{[k,b,k'}

\\ b, k, k' € GF(g)},

and • INCIDENCE is given by

(a, I, a') I[a,£]l (a) I [oo] I (oo) I [k] I (fc, b) I [k, b, k'\, for all a, a', b, k, k', £ £ GF(g), and {a,£,a')l[k,b,k'}

<^^ ^

6 + a' — aA; , L / _ 2, , , ^ + fc' = a yfc

and no other incidences occur. Proof. To describe W(q), q even, we consider PG(3,g) and the alternating form X0Yi + XXY0 + X2Y3 + X3Y2 with respect to some given basis. We define the following bijection j3 from the points of P G ( 3 , q) to V. Let (x0,x1,x2, x3) be the coordinates of an arbitrary point of P G ( 3 , q). If xi =£ o, then we define . (x0,x1,x2,x3)p

.0 =

(x3 x0 x2x3 x2\ —, — + —— — . \Xi

X\

X\

Now suppose that x\ = 0 . If x3 ^ 0, then {x0,0,x2,x3)p

=

—, — x3 x3

X\ J

12.3. Suzuki-Tits Moufang Sets

269

If x\ = x3 = 0 and x 2 ^ 0, then x x2

( 1 0 , 0 , ^ 0 ) " = ^0

Finally, (1,0,0,0)" = (°°) We also define (3 from the set of lines of W(q) to C as follows. Note that a line of P G (3, q) with Pliicker coordinates (p 0 i, P02, P03, P12, P13, P23) is a line of W(g) if and only if p 0 i = P23 and p%x = P02P13 + Po3Pi2- Hence the lines ofW(g) are in bijective correspondence with the 4-tuples [po2,Pi3,Po3,Pi2], up to a nonzero (multiplicative) constant. So let [po2,Pi3,Po3,Pi2] be the 4-tuple of Pliicker coordinates of an arbitrary line ofW(g). If P13 ^ 0, then we define Ell (E91

[P02,Pl3,P03,Pl2]/3

P03Pl2\

P13 ' VP13

P13

1/2

/

P03 P13

Now we suppose piz = 0. If P12 / 0, then bo2,0,p 0 3,Pl2] /3

1/2

P03 Pl2

P02 Pl2

If P13 = P12 = 0, and P03 7^ 0, then [P02,0,PO3,0] / 3 =

P02 i?03

Finally [1,0,0,0]" = [00]. Some elementary observations show that (a,e,a')l[a,e]l(a)l[oo}l{oo)I[k]l{k,b)l[k,b,k'}, for aRa,a',b,k,k',e

£ GF(q).

Let k,b,k' be arbitrary in GF(q). Then the line [k,b,k'] of W(g) has as inverse image under /? the line with Pliicker coordinates P02 = b2 + kk', p13 = 1, p 0 3 = fc' and P12 = fc (and then poi = P23 = b). One verifies easily that this line contains the points of P G ( 3 , q) with coordinates (b, 0, k, 1) and (fc', 1,6,0). Now, the inverse image under /3 of the point (a, ^, a') e "P is clearly the point of P G ( 3 , q) with coordinates {(. + oa', 1, a', a). This point

270

Chapter 12. Translation Ovoids in Translation Quadrangles

belongs to the line generated by (b, 0, k,l) and (fe', 1,6,0) if and only if ab + k! = £ + aa! and ak + b = a'. Substituting in the former the value of a' obtained by the latter, the assertion follows and the proposition is proved. • We can now determine when precisely W(q) is self-polar. This is due to Tits [219], but we provide a different proof. Recall that the corresponding ovoids are semiclassical ovoids. Proposition 12.3.2 W(q) is self-polar if and only if q = 22e+l, for some nonnegative integer e. Proof. If q is odd, then W(q) is not self-dual. So we may suppose that q is even. Suppose first that W(q) is self-polar and let p be a polarity. We use the description of W(g) given in Proposition 12.3.1. By conjugating with suitable elements of the full collineation group of W(g), one verifies that we may assume that (oo^ = [oo], that (0,0,0) p = [0,0,0], and that (l)p = [1]. It follows that (a)p = [k] induces a permutation a f-> k := a9 of GF(gr) with 0e = 0 and l6 = 1. Also, we have (0,0) p = [0,0]. Clearly the chain (a) ~ ( a , 0 , 0 ) l [ l , a , a 2 ] l ( l , a ) - (0, 0, a) ~ (0,a)l[0] must be mapped onto [a9] - [a0,O,Q]l(l,a9,a29)l[l,a9]

~ [0,0,a0} ~ [0,a fl ]l(0).

This implies that [0,0,^] p = (0,0,£ 8 ~ ) and hence [a,£], which is concurrent with [0,0,^], is mapped onto (a9,£9 ). Since (a,£,a') is the unique point incident with [a, £} en collinear with (0,0, a'), the line (a, £, a')p is the unique line incident with [a,£]p = (a9,£9 ) and concurrent with the line (0,0,a') p = [0,0, a' 9 ]. Consequently (a,£,a'Y

[a9,£9~\a'e],

=

and dually (ke~\be,k'e~1),

[k,b,k']P = for all a, a\b,k,k',£e

GF(q).

Expressing that (a,£, a')l[k, b, k'] if and only if (a,£, a')pl[k, 6, k']p, we obtain the identities {ak + b)9 + b9 ^a9{k9^)2, 2

9 1

91

{a k + £) ~ +£ '

e e

=a k ~\

(12.1) (12.2)

12.3. Suzuki-Tits Moufang Sets

271

Putting k — 1 in Equation (12.1), we deduce that 8 is additive. Putting a = l and b — 0 in Equation (12.1) we see that [k9'1)2 = ke. Equation (12.1) now reduces to (ak)s = aek9, implying that 6 is a field automorphism satisfying (ke)e = k2. In this case also Equation (12.2) is satisfied. Conversely, if we have a field automorphism 9 satisfying (ke)e = k2, for all k e GF(g), then we may define a duality of W(g) as the action of p on V and £ above and verify easily that this is a polarity. We conclude that W(g) is self-polar if and only if GF(q) admits an automorphism 8 with (ke)e = k2, for all k e GF(g). Putting 8 = 2 e + 1 , for some nonnegative integer e, we see that this is equivalent to requiring that 2q is a perfect square (and in this case q = 2 2e+1 ). The proposition is proved. • Remark 12.3.3 The above proof is, without much change, also valid in the infinite case, in which case the conclusion is that the field must be perfect and admit an automorphism whose square is the Frobenius map. Such automorphisms have been called Tits automorphisms in [232]. Remark 12.3.4 The proof of the above theorem can be shortened considerably if one uses the general result of Payne (see 1.8.2 of Payne and Thas [128]) stating that a finite GQ of order s admits a polarity only if 2s is a perfect square.

12.3.2 The Suzuki-Tits Ovoids Since W(g) is elation with respect to all of its points and lines, we can apply Proposition 12.2.3 in the case q = 2 2 e + 1 and obtain a Moufang set. We now show that this Moufang set is nothing else than the Suzuki-Tits Moufang set defined in Subsection 9.2.3. We first give an explicit description of the set O of absolute points of the polarity p above of W(q) in terms of coordinates in PG(3, q). So let 8 : GF(g) -> GF(q) : x ^ x2"+1 be the Tits automorphism of GF(q), and let p be the associated polarity of W(q), given by the action on V as follows: (oo)' = [oo], (a)" = [ae], {k,bf = [ke" ,b%

(a,£,a')" = [ae ,ee~\a'e],

272

Chapter 12. Translation Ovoids in Translation Quadrangles

for all a,a',b,k,£ e GF(q). By the calculations in the proof of Proposition 12.3.2, p defined in such a way induces indeed a polarity, and the action of p on the lines is given by [oo]" = (oo),

[kr = [ke~\

M ' ^ a V

1

) .

(ke~\be,k'e~1),

[k,b,k']P =

for all a, b, k, k',£ e GF(q). Since (oo) is an absolute point, all other absolute points are not collinear with (oo), and hence they are labelled with three coordinates. One now computes that the point (a, £, a') is incident with its image [a9,£9~\ a'6} if and only if £ = a'e + a2+e'. Hence £> = {(oo)}U{(a,a 2 + 9 + a ' V )

II

a,a'eGF(q)}.

Using the formulae in the proof of Proposition 12.3.1, we can translate this into projective coordinates of P G ( 3 , q) and obtain a set X, with X = {(l,0,0,0)}u{{a2+9+aa'+a,9,l,a',a) || a,d G GF(g)}. This is exactly the set X of Subsection 9.2.3, used to define the Suzuki-Tits Moufang sets. In order to completely see the equivalence between the Moufang set arising from p and the Suzuki-Tits Moufang set of Subsection 9.2.3, we have to prove that the respective root groups coincide on X. We show this for the root group Uoo related to (oo). One can easily check that the following transformation uXtX> of V U £, defined as (oo) i-> (oo), [oo] i-> [oo], (a) h^ (a + x), [k}^{k

+ xe],

(k,b) ^ (k + xe,b + kx + x' +

x1+e),

[a, I] ^ [a + x,£ + a V + x'6 + x2+6}, (a,£, a') i-> (a + x, £ + a V + x'S + x2+e, a! + x' + ax9), [k, b, k'] i-> [jfc + x9, b + kx + x' + x1+9,k'

+ x'e + kx2},

for all a, a', b, k, k', £ e GF(q), is a collineation of W(q). It preserves the ovoid O (in fact, it even commutes with the polarity p) and it maps the point (a,a2+e + a'6, a1) of O onto to point (a + x, (a + x)2+e +

12.4. Subtended Elation Ovoids

273

(a'+x'+axe)e, a'+x'+axe) oiO. In Subsection 9.2.3 we have written these points as (a, a') and (a+x, a'+x'+axe), respectively, and hence we see that uXiX> has exactly the same action as (x,^)^ of Subsection 9.2.3. Together with a similar calculation for Uo, which will coincide with the root group related to (0,0,0) e V (corresponding to the point (0,1,0,0) of PG(3, q)), we conclude that the finite Suzuki-Tits Moufang sets are precisely the Moufang sets arising from polarities of the finite symplectic quadrangles. Remark 12.3.5 The calculations as performed above are also valid in the infinite case, and without much change also in the case when the field K is not perfect and the quadrangle is defined over a subspace L of the vector space K over the field K0. For more information about this situation we refer to Theorem 7.3.2 of [232] and [234]. Remark 12.3.6 Above we mentioned that uXtXi preserves the ovoid and even commutes with the polarity. Obviously, when an automorphism commutes with p, it stabilizes O (since it stabilizes the set of fixed flags of p). Reciprocally, every automorphism of W(g) stabilizing O commutes with p, as shown by Tits [219]. The latter result was generalized to the imperfect case by Van Maldeghem, see 7.6.10 of [232].

12.4 Subtended Elation Ovoids In this section, we give ourselves a thick elation generalized quadrangle S of order (s, t) and a proper full subquadrangle S* of order (s, £*) containing the elation point p. In the examples, three cases will occur: (s,t,t*) = (q,q,l), (s,t,t*) = (q,q2,q) and (s,t,t*) = (q2,q3,q)- But the next results are also valid for infinite s,t,t*. Take an arbitrary point x in S collinear with p, and not contained in S*. Let Ox be the ovoid of S* subtended by x and let H be the stabilizer of x and S* in the full automorphism group of S. Also, let E be the group of elations about p (in the given elation group of S^) fixing the point x. Lemma 12.4.1 With the above notation, Ox is an elation ovoid of S* with respect to p and E. Proof. Let xi, x2 G Ox be arbitrary but distinct from p. Let 8 e E be the elation of S mapping xi to x2- Then 9 fixes x, and S* n S*9 is an ideal (lines of <S* through p) and full (points of S* collinear with p) subquad-

274

Chapter 12. Translation Ovoids in Translation Quadrangles

rangle (contains x2) of both S* and S* . Proposition 11.2.3 implies that S*9 = S*. • In many situations, the elation group of an EGQ has the property that, whenever an elation fixes a point y ^ p collinear with the elation point p, then it fixes all points on the line py. This is in particular the case for all elation groups of Moufang quadrangles arising naturally from the (dual) root elations. Hence the following lemma does not describe an exceptional situation. Lemma 12.4.2 With the above notation, and under the assumption that E fixes all points on the line px, the ovoid Ox of S* is an elation ovoid with respect to p and H. Proof. Let h £ Hp and 8 e E be arbitrary. Choose z e Ox distinct from p. Let 9' e E be such that it maps zh~l9h to z. Then h~l9h6' fixes all lines of S through p, all points on px, and the point z, and so is the identity by Corollary 11.2.5. Consequently Hp normalizes E. • As a corollary we obtain the next proposition. Proposition 12.4.3 quadrangle. Suppose stabilizing S*, fixing points on px and p'x,

With the above notation, suppose that S is a Moufang p ^ p' e O and let E and E' be the collineation groups all lines through p and p', respectively, and fixing all respectively. Then ( 0 , E, E') is a Moufang triple.

Proof. By Lemma 12.4.2 all that remains to be shown is that E and E' are conjugate in H. But this follows immediately from the 2-transitive action of (E, E') < H on O and the uniqueness of both E and E' with respect to the sets of their fixed elements. • We now investigate when exactly the previous proposition generates translation Moufang sets (i.e. E and E' are abelian). Of course, if S* is rather small (for instance, not thick), then this situation is more likely to occur. We will present some examples below reinforcing the impression that it will be very difficult to state a general sufficient condition for E to be abelian. However, if one restricts to the "maximal" situation, i.e., the situation in which S* is as large as theoretically possible, then there is a clear connection between the commutativity of E and properties of the standard elation group of S with elation point p.

12.4. Subtended Elation Ovoids

275

So, what do we mean by "as large as theoretically possible"? Notice first that the larger O is, the larger S* is. Clearly, with the above notation, the size of O is equal to the number of lines through x that meet <S*. So we say that <S* is an extreme full subquadrangle if every line of S is incident with at least one point of S*. Proposition 12.4A We use the notation of Proposition 12A.3, and assume that S* is extreme full in the finite Moufang quadrangle S. Then the following are equivalent. (i) The groups E and E' are abelian. (ii) The Moufang triple (O, E, E') gives rise to a translation Moufang set. (iii) All root groups ofS are abelian. (iv) All points of S are translation points of S. (v) All points ofS* are translation points ofS*. In each case, the group U0 of dual root elations corresponding to the dual panel (L,p,px), with L any line through p different from px, is isomorphic to the group E. Proof. Note first that E has the same faithful action on the lines of S through x as C/0. Consequently these two groups are isomorphic as permutation groups, and hence also as abstract groups. We now show that (i) implies (ii), that (ii) implies (iii), that (iii) implies (iv), that (iv) implies (v) and that (v) implies (i). By definition, (ii) and (i) are equivalent. Now assume (ii) (and hence also (i) holds). Let U\ be the group of root elations corresponding to the panel (p, px, x). Since there are only two conjugacy classes of root groups in S (conjugate in the little projective group of <S), and since UQ, U\ belong to different classes, it suffices, in view of the isomorphism E = U0, to show that U\ is abelian. Suppose by way of contradiction that U\ is not abelian. By Theorem 11.5.5 we know that there is a thick full or ideal translation subquadrangle S** (with a translation line) of S. In view of the fact that U\ is nonabelian, it cannot belong to a translation group in any translation subquadrangle and so <S** is ideal.

276

Chapter 12. Translation Ovoids in Translation Quadrangles

Hence, if S has order (s,t), S* has order (s,f) and S** has order (s**,t), then by Lemma 1.1.1 we have (since S* is not necessarily thick) t* < s t > s**, a contradiction. Now assume (iii). The finite Moufang quadrangles with abelian root groups have either regular lines or regular points (this is easy to check given that fact that S is a classical quadrangle). Since S has a full subquadrangle, Lemma 1.1.1 implies that s < t. If s < t, then no point can be regular, hence all lines are regular and all points of <S are translation points. If s = t, then S* is a grid and hence all lines are regular. The result follows. The implications (iv) => (v) => (i) are obvious.

•

Remark 12.4.5 The foregoing proposition remains valid in the infinite case, when the quadrangle arises from algebraic or classical groups. Indeed, in the former case, the root groups have a certain dimension (as algebraic groups). Just like in the previous proof, the assumption that the root group U\ is nonabelian leads to a subquadrangle S**. Let the dimensions of the root groups of S, S* and <S** be given by (s, t), (s, £*) and (s**,t), respectively, where (s, t) = (dim(t/i), dim(t/o)), and the others similarly. Then, by [233], we know that s** < t < s, and also t = s + t* > s, a contradiction. Also (without going into details here), these quadrangles are related to ordinary root systems of Type B 2 , which implies that certain root groups commute. This, in turn, implies that either points or lines are regular (for details, see Chapter 5 of [232]). If the lines are regular then the assertions follow. If the points are regular, then the existence of a nonthick ideal subquadrangle implies s < t (by [233] again). The existence of the full GQ S* implies s > t, so s = t and s* = 0. This now implies, as in the above proof of Proposition 12.4.4, that all lines of S are regular, and hence that all points of S are translation points. If the GQ <S arises from a classical group that is not an algebraic group, then one has to perform a direct calculation, which we shall not do here. Finally, if the GQ S arises from a group of mixed Type F 4 (see [104]), then all root groups are abelian, but the GQ does not have any regular point or line. This is the only situation in which counter examples to Proposition 12.4.4 can exist in the infinite case (but we do not know whether they really occur). Remark 12.4.6 With each point of a Moufang quadrangle corresponds a Moufang set in a canonical way. For, let y be a point of the Moufang quadrangle S and choose an arbitrary point y' opposite y. For every line Lly, we consider the group UL of all (dual) root elations with respect to the

12.4. Subtended Elation Ovoids

277

dual panel (L, z, V), where y'lL'lzlL. If X is the set of lines incident with y, then (X, {UL)LIV) is a Moufang set (and clearly, the isomorphism type of this Moufang set is independent of the choice of the points y and y'). If S' is a subquadrangle, then, putting x = y, the Moufang set found in Proposition 12.4.3 can be viewed as a sub Moufang set of this one, and in case the subquadrangle is extreme full they are even isomorphic. In the latter case, this shows that the isomorphism class of the Moufang set found in Proposition 12.4.3 is independent of the chosen point x. So Proposition 12.4.3 implies that every full subquadrangle of a Moufang quadrangle S gives rise to a Moufang set in this subquadrangle. We will call this Moufang set induced by S. We now apply the situation of Propositions 12.4.3 and 12.4.4 to three concrete examples. Example 12.4.7 Let <S be isomorphic to Q(4, q) and let <S* be a subquadrangle of order (q, 1) (a grid). Then S induces a translation Moufang set in S*, the little projective group of which is isomorphic to PSL 2 (g). This follows immediately from Remark 12.4.6. If we view S* as a ruled nondegenerate quadric Q in PG(3,g), then the corresponding subtended ovoid O is a conic contained in Q. Example 12.4.8 Let S be isomorphic to Q(5,gr) and let S* be a subquadrangle of order (q,q) (isomorphic to Q(4, g)). Then <S induces a translation Moufang set in S*, the little projective group of which is isomorphic to PSL2(c/2). This also follows immediately from Remark 12.4.6. If we view <S* as a nonsingular parabolic quadric Q in PG(4, q), then the corresponding subtended ovoid O is an elliptic quadric contained in some 3-dimensional subspace of PG(4,g) as the intersection of that subspace with Q. Example 12.4.9 Let S be isomorphic to H(4, q2) and let S* be a subquadrangle of order (q2, q) (isomorphic to H(3, q2)). Then <S induces a Moufang set in S*, the little projective group of which is isomorphic to PSU3(g). This also follows immediately from Remark 12.4.6. If we view S* as a nonsingular Hermitian variety H in PG(3, q2), then the corresponding subtended ovoid O is a Hermitian curve in some plane of PG(3, q2) (the intersection of that plane with H). In conclusion to Sections 12.3 and 12.4, we can say that three of the four infinite classes of finite proper Moufang sets arise from ovoids in appropriate Moufang quadrangles. The remaining class (related to the Ree groups)

278

Chapter 12. Translation Ovoids in Translation Quadrangles

arises from ovoids in finite generalized hexagons, but a description of how this happens is beyond the scope of this book. See, however, Chapter 7 of [232] for some details.

Chapter 13

Translation Generalized Quadrangles in Projective Space We now consider and aim to classify all embeddings of TGQs in finite projective spaces £ with the property that the translation group is induced by automorphisms of £ stabilizing the TGQ. Hence, in this chapter, we will have to deal with explicit coUineations of (Pappian) projective spaces, and we will use the following terminology and notation. After introducing homogeneous coordinates for the points in the projective space PG(d, K), d > 1 and K a (commutative) field, in the usual way, every point is represented by a nonzero (d + l)-tuple (x0, x\,..., xd) (determined up to a nonzero scalar multiple) that we also shall consider as a 1 x (d + 1)-matrix. The Fundamental Theorem of Projective Geometry then states that every collineation ip : (x0, x\,,.., xd) i-> (x'0, x[,..., x'd) can be written as (X'Q,^,...,^)

=

(x0,xi,...,xd)eM,

where M is a nonsingular (d + 1) x (d + l)-matrix with entries in the field K and 9 is some field automorphism, which we call the companion field automorphism of ip. In general, we say that ip is a semilinear collineation. If 9 is the trivial automorphism, then we say that the collineation ip is linear; if moreover det M is a (d+ l)st power in K (and note that M is determined up to a scalar multiple), then we say that ip is special linear. The set of special linear coUineations forms a normal subgroup of the group of all coUineations, called the little projective group of PG(d,K). We will use without further notice the well-known property that, if a linear collineation fixes at least three points on a line, then it fixes all points on that line. 279

280

Chapter 13. Translation Generalized Quadrangles in Projective Space

13.1 Generalities about Lax Embeddings In this chapter we give ourselves a GQ S = (V,B,I) laxly embedded in some projective space E := PG(d, K), for some field K. "Laxly embedded" means that we can identify the point set V with a generating set of points of E, and the line set B with a certain set of lines of E in such a way that, if a point x eV and a line L e B are incident in S, then they are also incident in E. Note that we do not require the converse of this property (but it will nevertheless follow, see below). We neither require that lines in B which do not meet in S have no intersection point in E, and this can actually happen. To illustrate this, we present the following well-known example. Example 13.1.1 Let H be a hyperoval in PG(2,4), i.e., a set of six points no three of which are collinear. The 15 points not in H, together with the 15 lines meeting H in exactly two points, constitute a lax embedding of the symplectic quadrangle W(2) of order 2. We will call this example the hyperoval embedding of W(2). Observe that every line external to H contains the five points of an ovoid of W(2), and since there are six such lines, every ovoid arises in this way. These six lines define a dual hyperoval H* that we call dual to TC. Also, the full group of collineations and correlations (these are the isomorphisms of PG(2,4) to its dual) stabilizing H U H* is a group of order 2 • 6!, the latter being exactly the order of the full group of automorphisms and anti-automorphisms of W(2) (an anti-automorphism being an isomorphism from W(2) to its dual). Hence every automorphism and every anti-automorphism of W(2) is induced by a collineation or correlation of PG(2,4) stabilizing H or mapping it to its dual, respectively. Note that the little projective group of W(2) is induced by the little projective group of PG(2,4), i.e., by special linear collineations. We say that S is polarized at some point x € V if the set of lines of <S through x does not span E. If S is polarized at every point in V, then we briefly say that S is polarized. Also, if for the line L e B, each point of E on L belongs to S, then we say that L is a full line of S. If all lines of S are full, then we briefly say that S is full in E, or is fully embedded (in E). There exist GQs which are not fully embedded, but with full lines. Example 13.1.2 Let K be an infinite field with a countable number of elements, let d = 4, consider five points pi,P2,P3,P4,P5 in E generating E, and let Li be the line generated by pt-2 and pt+2, considering the subscripts

13.1. Generalities about Lax Embeddings

281

modulo five. We now define a free 4-gonal closure of this pentagon S0 as follows. We number the points of the line L4 using the nonzero natural numbers, assigning 1 to pi and 2 to p2. Furthermore, we select a nonempty finite subset F of points of the line L5 neither containing p2 nor p3. Suppose we arrive in Step n of the construction at some finite set of points and lines in £, forming a connected geometry Sn without triangles, such that (CO) there is a natural number k such that precisely the first k points of L4 are contained in <Sn, (CI) no point is contained in F, (C2) incidence in Sn is inherited from E, (C3) for every pair of points a, b of Sn not incident with L4, the planes (a, Li) and (b, L4) coincide (and note that these planes are always well-defined since we assume that incidence in Sn is induced by the incidence relation in £) if and only if the line ab of E also belongs to Sn and so does the intersection point ab n L4. We execute Step n + 1 in the following manner. Whenever a point x and a line L of Sn are at distance 5 in the incidence graph of Sn, then we will choose an appropriate "new" point y on L (so y belongs to E, but not to <Sn) and declare y and xy to be elements of <Sn+i. It is obvious that we never introduce triangles, and that the infinite union of all our geometries is a generalized quadrangle. We must now show that we can make our choices such that Sn+i satisfies (CO), (CI), (C2) and (C3). Therefore, let \1> be the set of points of Sn, let k be the number of elements of * incident with L4, and let $ be the set of planes of E containing L4 and at least one point of <S„ not incident with L4. Since there are only finitely many points and lines in Sn, there are only finitely many pairs (x, L) as above to consider. We number them and treat them in the given order. Each time we treated one pair, we check the validity of the conditions (CO), (CI), (C2) and (C3). Hence, we may assume that these conditions hold for all points and lines so far constructed, and we consider the new current value of k, add the new points to * and the new planes, if any, containing L4 and a newly constructed point not on L4, to the set $. Suppose first that L ^ L4 and x is not incident with L4. If L is concurrent with L4, then the point x is not contained in the plane (L,Lt) by (C3). Also, whichever choice for y we make, (CO), with the same k, and (C3) will

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always still hold. Hence we only have to avoid that the line xy contains some point of * different from x, and we must also make sure that y <£ F. But these conditions only rule out a finite number of possibilities for y; as there are infinitely many points on L, we can make a legible choice for y. Note that we do not require that xy does not meet a line of Sn (or another "already defined line" of <Sra+i), and we indicate below why we cannot build in this extra condition, which we do not need for our purposes here anyway. If L is not concurrent with L4, then, since L is incident — in our extension of Sn sofar obtained — with at least 2 points by our construction, Condition (C3) assures that (L,L4) is 3-dimensional. Hence there are only a finite number of points of E on L that belong to some member of $. Also, the plane (x,L) meets L 4 in at most one point, it contains only a finite number of elements of * , and we must avoid that these points are on xy. But clearly there is a choice for y meeting these conditions, and then (CO) and (CI) hold trivially, and (C2) and (C3) by construction. Suppose L = LA,. Then the plane (x,L4) does not contain any element of * besides x and the points of L4, by (C3). We now choose for y the (k + l)st point of LA. Then (CO) is satisfied, and trivially also (CI), (C2) and (C3) hold. Note that we have no control over the fact that xy meets some line of Sn, hence we do not necessarily obtain a lax embedding with the extra property that two lines of the quadrangle meet if and only if they meet in the projective space. Suppose finally xlL4. Then, similarly as above, L does not meet L 4 and so only finitely many points of E on L belong to some member of <J>. If we avoid those points for y, then (C3) holds. The other conditions hold trivially. Now the union of all Sn over N is a (thick) generalized quadrangle S laxly embedded in E. Moreover, the line L4 is full because of Condition (CO) and the fact that S has infinitely many points on every line (this is easy to see and left for the reader to prove). But the line L5 is not full because the points of F belong to L5 in E, but they do not belong to S. Lemma 13.1.3 Let the GQ S be laxly embedded in E. Then the points of S incident in E with some fixed line L of S are precisely the points of S incident with L in S. Proof. Suppose a; is a point of S not incident with the line L of <S. Assume by way of contradiction that x and L are incident in E. In S, there is a unique line M meeting L and incident with x. Clearly, in E, this line has to

13.1. Generalities about Lax Embeddings

coincide with L, a contradiction.

283

•

A celebrated result of Buekenhout and Lefevre [32] classifies all fully embedded generalized quadrangles in finite projective spaces. We mention it without proof. Theorem 13.1.4 Every fully embedded generalized quadrangle in a finite projective space is a classical GQ in its standard embedding as described in Section 1.4. • Dienst [51, 52] classifies all fully embedded generalized quadrangles in arbitrary projective space (and only classical GQs turn up, i.e., GQs associated to classical groups; in any case all examples are Moufang). A crucial step in the above classifications is the fact that every full embedding is automatically polarized. We record a proof of this fact here, not using finite dimensionality. Hence, in the following lemma, we emphasize that the dimension d of E can be any cardinal number. Lemma 13.1.5 If S is fully embedded in E, then the embedding is polarized. Proof. Let, by way of contradiction, p be a point of <S at which the embedding is not polarized. Hence every line px of E, with x a point of S opposite p, is contained in a finite dimensional space T,x spanned by some lines of S through p. We may select x and E x is such a way that the dimension of Hx, say TO, is minimal. Clearly, TO > 1 as otherwise x is on some line L\ of S incident with p and hence not opposite p. So there are m lines Li, I / 2 , . . . , Lm of <S through p generating E^. Denote by T,'x the space generated by L2,1/3, •.., Lm. Note that TO > 1 implies that Y,'x is well-defined and has dimension at least 1. The line M of S through x meeting Li is not contained in E'x (by minimality of m), but it is contained in T,x. Hence M meets H'x in a unique point x', which is a point of S belonging to a space E'x of dimensionTO- 1 spanned by some lines of S through p. This contradicts the minimality of TO. Hence such m cannot exist (sinceTO^ 1), and consequently every point opposite p must lie outside the space spanned by the lines through p. Hence the embedding is polarized at p after all. • Polarization of an embedding at every point is a useful property. For instance, in 3-dimensional space, it immediately implies that all points are regular.

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Proposition 13.1.6 Let the not necessarily finite GQ S be laxly embedded in P G ( 3 , K), for some field K. If the embedding is polarized at every point, then every point is regular. Proof. If S is a grid, then the result is trivial. So we may assume that there are at least three lines incident with every point. For any point x of S, we denote by £x the plane of PG(3,K) spanned by x1- (and this is indeed a plane by Lemma 13.1.3). If £x = £y for two distinct points x, y, then the plane £x would induce a subquadrangle which is clearly full and ideal in S, hence coincides with S, a contradiction. Now let {x, y} be an opposite pair of points of S. The intersection of £„ and £w, for arbitrary u,w e {x, y}x, u ^ w, is precisely the line xy of P G ( 3 , K). Hence {x, y}11 C xy. Conversely, let z be a point of <S on the line xy of P G ( 3 , K). Since £w ^ £2 ^ £„, and since S is not a grid, there is a line L of S through z not contained in £„ and not in £w either. But then proj u L is the intersection of £u and (u,L), and hence passes through z. Similarly for proj^L. This means that z e {u^w}-1. Since u,w were arbitrary, we conclude that z 6 {x, y}-1"1". • A natural question to ask is whether the property of being polarized is already enough to allow a classification. Thas and Van Maldeghem [194], in the finite case, and Steinbach and Van Maldeghem [146, 147], in the general case, answer affirmatively this question and the latter classify all polarized embedded quadrangles in arbitrary projective space. They are all Moufang GQs and in each case, the little projective group of the GQ is induced by the projective group of the projective space. In the finite case, a polarized embedding is almost always a full embedding in some subspace of E, obtained by restricting the (possibly infinite) field K to a finite subfield. There is only one counter example to this phenomenon, and it is given by an exceptional embedding of W(2) in PG(4, K). We describe it later. Here, we want to prove a kind of converse to the classification result of Steinbach and Van Maldeghem, namely, roughly speaking, we aim to show that polarization of an embedding follows from an assumption on the collineation group of the GQ in question. In particular, if the finite GQ is Moufang and all (dual) root elations are induced by the little projective group of the projective space E, then the embedding is polarized, except in some small well-understood cases. But first we give some examples of nonpolarized embeddings.

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285

Example 13.1.7 Let S := W(g) be embedded in E := PG(3,g) (in the standard way). Let p be a point of <S, and let it be the plane of E containing all lines of S through p (the plane it exists since the embedding is polarized a t p ; it is the image of p under the symplectic polarity corresponding to S). Then we define the quadrangle S' as follows. • The POINT SET is the set of points of E \ -K.

• The LINES are all lines of S not through p, plus all lines of E through p, but not in TT. It is easy to see that <S' is isomorphic to the Ahrens-Szekeres quadrangle AS(q) of order (q - l,q + 1), see 3.2.6 of Payne and Thas [128]. The definition given here entails a lax nonpolarized embedding, and the lines of S' incident with p in E are examples of lines not meeting in the quadrangle, but intersecting in the projective space. Example 13.1.8 A second class of (nonpolarized) examples is given by the T2 (0), with O a hyperoval in some plane it of the 3-dimensional space P G ( 3 , q), with q even. Here, every line meets another line in P G ( 3 , q) that is not concurrent in the quadrangle. The two previous examples show that, even if one assumes that every line of the embedded GQ misses only one point of the projective space, the embedding does not have to be polarized. Before proceeding, we remark that all the finite examples we have given so far (including the natural embeddings of the classical quadrangles as quadrics and Hermitian varieties in finite projective spaces) have an ambient projective space — that is, the space generated by the points of the quadrangle — of dimension at most 5. This is not a coincidence and can be proved quite easily. Proposition 13.1.9 If a GQ S of order (s,t), t > I, is laxly embedded in PG(d, q), then d < 5. Proof. Suppose the GQ S of order (s, t), i > 1, is laxly embedded in ~PG(d,q), with d > 5. Consider two opposite lines L,M of S. The intersection of S with the projective space (L, M) is a full subquadrangle S' of order (s, t'), with t' < t. Considering a line N oi S concurrent with L but not belonging to S', we intersect <S with the 4-space (L, M, N) and obtain

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a subquadrangle S" of order (s, t") with t' < t" < t. Considering a line K of <S concurrent with L but not belonging to S", we intersect S with the 5-space (K, L, M, N) and obtain a subquadrangle S'" of order (s, t"') with t' < t" < t'" < t. Now Lemma 1.1.1 implies t' = 1, *" = s and *'" = s2 > t, hence t = t'" and 5 ' " = S. The assertion is now clear. • Other examples of nonpolarized embeddings are given by first extending the ground field of the projective space in which a classical GQ is naturally embedded, and then projecting from a suitable point onto some hyperplane (and one may repeat this procedure a number of times). A remarkable result of Thas and Van Maldeghem [195] now shows that every lax embedding of the classical quadrangles Q(4,s), Q(5, s), H(3,s) and H(4,s) in PG(d,q), d > 3, gcd(s, q) ^ 1, arises from the standard embedding as described above. Since we will need this result for Q(4, s) and d e {3,4}, and for Q(5, s) and d = 5, we sketch a proof of it now. Lemma 13.1.10 Every lax embedding of Q(4, s) and Q(5,s) in PG(4, q) and PG(5,q), respectively, with gcd(s, q) ^ 1, arises from the standard embedding by extending the field GF(s) over which the projective space is defined. Every lax embedding of Q(4, s) in P G ( 3 , q), with gcd(s, q) ^ L arises from the standard embedding in PG(4, s) by extending the field over which the projective space is defined and then projecting from a well-chosen point onto a hyperplane. Sketch of Proof. First consider S = Q(4, s), laxly embedded in PG(4, q), with gcd(s, q) ^ 1. Let L be a line of S and let n be a plane of PG(4, q) skew to L. We project the set of lines of S concurrent with L from L onto •K and obtain an embedding of the dual affine plane A corresponding with the regular line L into IT. The condition gcd(s,q) =£ 1 implies, by Limbos [99], that the "parallel points" of this dual affine plane are collinear in -K, i.e., the lines of S through a fixed point x of <S on L are contained in a 3-space. This is exactly polarization at x. Moreover, [99] also implies that, under our assumptions, A extends uniquely to a projective subplane V of w. Hence GF(s) is a subfield of GF(q), and the points of any line M of S opposite L form a projective subline over GF(s) of M. Now it is quite easy to show that every grid of S of order (s, 1) generates a 3-dimensional subspace (if such a grid were contained in a plane, then by considering a subquadrangle in an appropriate 3-dimensional space we obtain a contradiction to Lemma 1.1.1), and is fully embedded in some 3-dimensional subspace over

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287

GF(s) (and this subspace is uniquely defined by four points of the grid lying in a quadrangle and a fifth point of the grid not collinear with any of the four previous points). We fix one such subspace PG(3, s) and note that any line M opposite L, but not belonging to the grid in PG(3, s), together with PG(3,s) generate (over GF(g)) the whole space PG(4, q) as otherwise the space over GF(q) generated by PG(3, s) would already contain S — this follows from Lemma 1.1.1. Now we add the points of S incident with M to obtain a unique subspace PG(4, s) of PG(4, q), containing PG(3, s) and the points of S on M. It is then easily seen that all the points of the grids of S containing M and sharing two (intersecting) lines with PG(3, s) are contained in PG(4, s). Repeating this argument for M substituted by other lines of S not in PG(3, s), but whose points (of <S) have already been proven to be in PG(4, s), we eventually obtain that all points of S are contained in PG(4, s), i.e., the embedding is full in a subspace over GF(s), and hence it is isomorphic to the appropriate natural embedding therein. Now consider S = Q(5,s) laxly embedded in PG(5,g). By the previous paragraph, every subquadrangle <S' = Q(4, s) of S is fully embedded in the standard way in some subspace PG(4, s) (one uses again Lemma 1.1.1 to show that such a subGQ is never contained in a 3-dimensional subspace). For a point x of S outside S', the subtended ovoid Ox of <S' is an elliptic quadric in some 3-dimensional subspace PG(3, s) of PG(4, s). Hence the lines of S through x are contained in the 4-dimensional space spanned by Ox and x. Hence the embedding is polarized at x, and so at every point. The fact that the embedding is full in some subspace PG(5, s) over GF(s) is proven in a similar way as above for Q(4, s). Finally, let S ^ Q(4,s) be laxly embedded in PG(3,g), with gcd(s, q) ^ 1. We will be very sketchy here and leave some nontrivial details to the reader. We can again show that the points of S on any line L of PG(5, q) containing two collinear points of S form a projective subline of L over the field GF(s). Next we construct a PG(3, s) containing all points of <S of a grid Q of order (s, 1) (here one needs to prove first that there exists a grid Q which generates PG(3, q)). We then embed PG(3, q) into a PG(4,q) as a hyperplane, choose a point c in PG(4,g) outside PG(3,g), and consider a line L of <S not contained in the grid Q. The line L has some point x in common with Q, and we choose a line L' in the plane (c, L) through x, and project the points of S on L from x onto L'. We call this "lifting". We can now lift uniquely all points of every grid Q1 of S containing L and sharing two intersecting lines L\, L2 with Q, where the lift of a point y is the unique point in the intersection of (c, y) and (Li,L2,L'). Doing

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this a number of times, substituting L with appropriate other already lifted lines, the lemma eventually follows. • For finite lax embeddings, dimension five plays a special role, as in this dimension all lax embeddings can be classified. Since we will use this result of Thas and Van Maldeghem [195], we sketch a proof of it here. Lemma 13.1.11 If the GQ S of order (s,t) is laxly embedded in PG(5,g), then S = Q(5,s). If gcd(s,q) = 1, then s = 2 and the embedding is as described in Example 13.3.4 below. Otherwise the embedding arises from the standard embedding by extending the field GF(s). Sketch of Proof. If gcd(s, q) > 1, then, in view of the previous lemma, we only have to show that S = Q(5, s). Let L, M be two opposite lines of S. Then the 3-space (L,M) induces a subGQ S' of order (s,t') of S. Choose a line N in S meeting S' in a unique point. Then the subspace (L, M, N) induces a subGQ <S" of order (s,t") of S containing S'. By Lemma 1.1.1, this implies t = s2, t" = s and t' = 1. Moreover, the abundant number of ways to choose S" can be used to conclude, by 5.3.5 of Payne and Thas [128], that S ^ Q ( 5 , s). If gcd(s,q) = 1, then the previous argument still holds to conclude that S = Q(5, s). But now we cannot appeal to Lemma 13.1.10 anymore. Nevertheless, the assumption that allowed us to use the results of Limbos [99] in the proof of Lemma 13.1.10 can be replaced by the assumption s > 3. The case s = 2 is now handled by directly assigning coordinates to all points, in a systematic way, along the lines of the proofs of Theorems 13.2.2 and 13.4.1. The case s = 3 cannot occur since in this case any subquadrangle S' isomorphic to Q(4,3) is embedded as in Example 13.3.3 below (this requires some calculations with coordinates), and from that description one deduces that any classical ovoid of S' spans the whole 4-dimensional space H spanned by S'. But there are at least two subquadrangles isomorphic to Q(4, s) containing the points of the same ovoid, and this implies readily that S would be contained in the hyperplane H, a contradiction. •

13.2 Planar Translation-Homogeneous Embeddings of Translation Generalized Quadrangles The extreme case of a nonpolarized embedding occurs in projective planes (and we speak of planar embeddings: no planar embedding of any GQ can

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ever be polarized in any point). All classification results in the literature concerning embeddings of GQs in projective spaces exclude the planar case, basically because there are no subspaces to put structure on the embedded GQ. However, we can put some structure on the embedding by hypothesizing that the GQ is a TGQ and that the translation group is induced by coUineations of the projective plane. That is exactly what we are going to do, and we will do this in arbitrary dimensions. If the dimension is equal to 2, then we will always assume that the projective plane is Desarguesian. First we give precise definitions. Let S = (P, B, I) be a translation generalized quadrangle embedded in PG(d, EC), for some field K and some cardinal number d, and let x be the corresponding translation point. If every translation with respect to x is induced by a collineation of PG(d,K), then we say that the embedding is translation-homogeneous (with respect to x). If all these coUineations of PG(d, K) are induced by linear transformations or special linear transformations of the underlying vector space, then we say that the translationhomogeneous embedding is linear or special, respectively. If K is an infinite field, then it may have an essentially arbitrary Galois group, which may contain, in principle, the whole translation group. This is a serious obstruction for our methods, and hence we will only consider the finite case when dealing with translation-homogeneous embeddings in general. Likewise, in the linear case, if K is a nonabelian skewfield, then we have no control over the group of inner automorphisms (which induces a linear group), and so we discard this case. In the present section, we will deal with planar translation-homogeneous embeddings of TGQs in finite Desarguesian projective planes. Such embeddings can never be linear since every nontrivial symmetry about a line L would be a collineation with an axis L having s + 1 centers (the points of the quadrangle on the line L, viewed as a line of the projective plane). We first present some examples. They are all related to W(2).

Example 13.2.1 Let q be a power of 2, and let £±,£2 be two distinct elements of GF(q2) such that £x + £\ = 1, lx ^ £\ and £x£\ = £2£\. Put k = £1 + £2. Identify the point set of W(2) with the pairs of {1,2,3,4,5,6} and the lines with the partitions of {1,2,3,4,5,6} into pairs. Then the fol-

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lowing identification defines a lax embedding of W(2) in PG(2, q2). {1,2} ~ (0,1,0) {3,4} <-(1,1,0) {4,5} <-(1,1,1) {2,3}^(fc,«i,l) {2,4} ^(k,£\,l)

{5,6} ~ (1,0,0) {3,6} <-(1,0,1) {1,3} ~ (*«,£?, 1) {1,5} ^(k",£q2,l) {2,5}^(fc,£2,l)

{4,6} <-(0,0,1) {3,5} ~ (0,1,1) {l,4}~(fcVi,l) {l,6}<-(fc«,^ + l,l) {2,6}^(fc,£2 + l,l)

In order to check that the above really defines a lax embedding, one only has to verify that approximately half of the lines of W(2) are induced by lines of PG(2, q2) taking into account that the coUineation (a, 6, c) t-y {aq,bq,cq) of PG(2,g 2 ), a,b,ce GF(q2), stabilizes the above set of points. We leave the easy detailed computations to the reader. Let us consider the point {1,2} as a translation point of W(2). The corresponding translation group is generated by the symmetries (1 2)(3 4)(5 6), (1 2)(3 5)(4 6) and (1 2)(3 6)(4 5). Hence it is also generated by the automorphisms (1 2), (3 4)(5 6) and (3 5)(4 6). The latter three are induced by the semilinear collineations of PG(2, q2) with companion field automorphism I H J ; ' and matrices

as one can easily check. Hence we have a translation-homogeneous embedding with respect to {1,2} of W(2) in PG(2, q2), for an arbitrary even prime power g / 2 (and it is easy to generalize this embedding to infinite projective planes over fields in even characteristic admitting an involutory automorphism). We call this embedding the generalized hyperoval embedding with parameters (£i ,£2). We say that a translation-homogeneous embedding of a TGQ <S in PG(d, K) is faithful if the identity is the only coUineation of the projective space that fixes all points of the TGQ. In the finite case, this is equivalent with saying that the embedding is not contained in a PG(d, K'), with K' a proper subfield of K (after remarking that thickness and connectivity of S imply that the point set of S is not contained in the union of a nontrivial subspace and a complement). In general, it is equivalent with saying that the embedding is not contained in a PG(d, K'), with K' a subfield of K, that is fixed pointwise under a nontrivial field automorphism of K. We now have the following characterization result.

13.2. Planar Translation-Homogeneous Embeddings

291

Theorem 13.2.2 Every faithful translation-homogeneous embedding of a translation generalized quadrangle in a finite Desarguesian projective plane is isomorphic to a generalized hyperoval embedding of W(2) with some parameters (£1,^2). The faithful embedding is translation-homogeneous with respect to two different translation points if and only if it is isomorphic to the hyperoval embedding of W(2) in PG(2,4). Proof. Let <S = {V,B, I) be a TGQ of order (s,t) with translation point x, embedded in the plane PG(2, q). Since we assume that the embedding is faithful, V is not contained in any subplane over any subfield GF(g') of GF(g). This is equivalent with saying that no nontrivial collineation of PG(2, q) fixes T pointwise (since we are dealing with finite fields). A direct consequence of this is that the companion field automorphism of any collineation stabilizing V is uniquely determined by the action on V and so the map that assigns the companion field automorphism to a semilinear collineation of PG(2,g) that induces an element of the translation group T of <S(X) is a group morphism T from T to the automorphism group of GF(q). The translation group T is an elementary abelian p-group for some prime p. So T has exponent p, and s and t are powers of p. Noting that Aut(GF(q)) is a cyclic group and hence has at most one subgroup of exponent p, we see that the mapping r : T —> Aut(GF(q)) assigning to any translation of S^ the companion field automorphism (see above) of the corresponding collineation of PG(2, q) has a kernel of size at least s2t/p. Hence the intersection U of T (the latter viewed as collineation group of PG(2,g)) with PGL 3 (g) is nontrivial and has size at least s2t/p. As each element of U fixes at least three lines of P G ( 3 , q) through x, each such element is a central collineation of PG(2, q) with center x, and hence it is also an axial collineation with some axis A. It is well-known (see e.g. Hughes and Piper [75]) that x is incident with A in PG(2, q) if and only if the central collineation has order p', where p' is the unique prime dividing q. In this case, the collineation is called an elation of PG(2, q). We now claim that all elements of U are elations of PG(2, q), i.e., p divides q. By the previous, this is equivalent with saying that U contains at least one nontrivial elation ofPG(2,g). Indeed, since the order of U exceeds s, the group U cannot have a free action on the s points, different from x, of an arbitrary line L of S through x. Let 9 G U be a nontrivial collineation fixing some point y on L, with y ^ x. Since the order of 9 is p, it must fix a third point of S on L. Then 9

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fixes all points of PG(2, q) on L and hence is an elation of PG(2, q) (since it also fixes all lines through x). The claim is proved. Let L be any line of S through x. The group S of symmetries about L does not contain nontrivial elements of U, hence T is injective on S and so S is a cyclic group of order s. This easily implies s = p. Hence t G {p,p2}We claim that t = p. For, suppose by way of contradiction that t = p2. Let Li and L2 be different lines of S through x and let 81 and 92 be nontrivial symmetries about L\ and L2, respectively, chosen such that they have the same image under T. Set 9 = 9i92l. Then 8 maps a point u, with a / i , to a point u' ^ u. Also, 0 fixes every element of {x, u, u'}-1, and the latter contains p + 1 elements {UQ, UI,...,UP}. Since 6 has order p and belongs to U, it fixes all points on the lines xuiy i = 0 , 1 , . . . ,p, of PG(2, q), clearly a contradiction. The claim follows. Since s = t = p, Corollary 3.5.6 implies that S = Q(4,p). We now claim that p = 2. Suppose by way of contradiction that p is odd. Then there exists a collineation a e T fixing every point of <S on the lines Li and L2 (with the above notation), and fixing every line of S through x. Clearly a £ U and ar belongs to the unique subgroup of order p of Aut(GF(g)). Now we may rechoose 9\ € S\, with Si the group of symmetries about L\, so that Q\ = ar. The composition aO^1 belongs to U, and hence is an elation in PG(2, q) with axis Li and center x. Now let a be any point not collinear with x, and let b be its image under a6^1. Since a and b are not collinear, since a and b are collinear with a common point y on L\, and since p is odd, there is a second point y' ^ j/ in {x, a, 6}- 1 . But y' is fixed under cr^f 1 , a contradiction. The claim is proved. Hence S is isomorphic to W(2). We now give explicit coordinates to the points of the embedded W(2). We assume that the point set of W(2) is abstractly given by the pairs of {1,2,3,4,5,6} and the line set by the set of partitions of {1,2,3,4,5,6} in pairs. Without loss of generality we may assign the following coordinates (for some a, b, b',ki,k2,£i,£2 e GF(q)): {1,2}-» (0,1,0), {3,4}-» (1,1,0), { 5 , 6 } ^ {3,5}-> (0,1,1), { 4 , 6 } - (0,0,1), { 3 , 6 } {4,5} - (a,6',l), {2,3} - (*i,£i,l), {2,5} -

(1,0,0), (a,6,1), (k2,£2,l).

Denote the image of any element k e GF(q) under the involutory automorphism of GF(g) by k. We may assume without loss of generality that x = {1,2} = (0,1,0). Set y = {5,6} = (1,0,0). The stabilizer in T of y has order 4. If the two elements 0i, 92 of that stabilizer not fixing the

13.2. Planar Translation-Homogeneous Embeddings

293

point {4,6} both have nontrivial companion field automorphism, then the composition 9 is a linear collineation and fixes the base triangle. Since q is even, and the order of 9 is equal to 2, this leads to a contradiction. Hence we may assume that 9\ is linear. But it fixes three lines through x, so it is central with center x; it fixes three points on the line xy, so the axis is xy. This easily implies that the image of (a, b, 1) under 9\ is (a, 6 + 1,1). Hence b' = b + 1. Moreover, interchanging the roles of the first and third coordinates, we also see that ^ = ^ + 1, implying a = 1. We can also take the image of (fci,4, 1) under 6\. Note first that 9\ does not fix the lines through y, as it would otherwise have two different centers. So 9\ corresponds to the permutation (3 4)(5 6) of {1,2,3,4,5,6}. Thus we obtain {2,4} = {2,3}Sl = (h,eu l)*1 = (huh + 1,1). Likewise {2,6} = (fc2,4 + M ) . Now the (semilinear) collineation 9 corresponding with the permutation (1 2) fixes the points (1,0,0), (1,1,0), (0,1,0), (0,0,1) and (0,1,1), hence has identity matrix. So it follows that {1,3} = {2,3}e = (k1:£ul)e

=

(ki~h,l).

Likewise {1,4} = (k1,£1 + 1,1),

{l,5} = (fc 2 ,4,l),

{l,6} = (fc2,£2 + l,l).

Now we express that collinearity in the quadrangle implies collinearity in the projective plane. Expressing that the points {5,6}, {2,3}, {1,4} are collinear, we see that 4 + 4 + 1 = 0. Doing the same for {3,4}, {2,5}, {1,6}, we obtainfc2+fc2+ 4 + 4 + 1 = 0. One checks easily that the points on an arbitrary line of W(2) incident with {4,6} or {3,5}, but not with both, are collinear in PG(2,
294

Chapter 13. Translation Generalized Quadrangles in Projective Space

In particular, if follows that b = b. Let 03 be the (semilinear) collineation corresponding to the permutation (1 2)(3 5)(4 6). Then, since (0,1,0), (0,0,1) and (0,1,1) are fixed, and since (1,0,0) and (1,1,0) are permuted, 83 maps (xi,x2,1) onto (xi,x\ + X2,1). Applying this to the points {2,3} and {1,5}, we see that fci = k2 and h + £2 + k2 = 0. Hence b = 0, and we can put k = k\ — k2 = £\ + £2. Noting that from k{£2 = k2£i follows that £1£i = £2£2, and that £i = £2 implies k = k and hence {1,4} = {2,3}, we conclude that we have a generalized hyperoval embedding of W(2). We now note that in the generalized hyperoval embedding above, the points {1,2}, {1,3}, {1,4}, {1,5}, {1,6} are incident with one line in PG(2, q), and similarly for the points {1,2}, {2,3}, {2,4}, {2,5}, {2,6}. We may conclude that whenever a planar embedding of W(2) is translationhomogeneous with respect to some point x, then every ovoid of W(2) containing x is contained in a line of PG(2, q). If the faithful embedding is translation-homogeneous with respect to two distinct collinear translation points, then we may assume without loss of generality that, for the previous embedding, in addition to {1,2}, also {5,6} is one of these two points. By the previous paragraph, the point {1,6} is incident with the line of PG(2, q) through (1,0,0) and (0,0,1). This easily implies that £2 = 1. But then l\l\ = 1 and so li satisfies the quadratic equation X2 + X + 1 = 0, which implies that £x, and hence also k, are contained in GF(4). By faithfulness, q = 4, and it is easy to see that we now have the hyperoval embedding. Finally, if the embedding is translation-homogeneous with respect to two noncollinear points, then it is translation-homogeneous with respect to every (translation) point and the result follows from the previous paragraph. The theorem is proved. • Remark 13.2.3 The second part of the previous theorem has a more geometric proof using a result of Temmermans (unpublished) that characterizes the hyperoval embedding of W(2) as the unique (up to field extension) planar embedding of W(2) with the property that the points of each ovoid of W(2) lie on a line of the plane.

13.3. Exceptional Non-Planar Translation-Homogeneous Embeddings

295

13.3 Exceptional Non-Planar TranslationHomogeneous Embeddings In this section we classify the translation-homogeneous embeddings of some small quadrangles. These will give rise to exceptional embeddings. The reason to treat them separately is because our general classification technique will fail for these small parameters. It turns out, however, that precisely in these cases, exceptional embeddings exist. We start with the case s = t = 2, i.e., the symplectic quadrangle W(2). We first describe some embeddings. Example 13.3.1 Some Embeddings of W(2) in P G ( 3 , q). Suppose that the prime power q is an odd perfect square and let £, £ ^ 0, be an element of GF(q) that is mapped to — £ under the unique involutory field automorphism. As before, we denote that automorphism by k \—> k. We define the following embedding of W(2) in P G ( 3 , q). {1,2}-(0,0,0,1) {3,4}-(1,0,0,1) {4,5} - ( 0 , 1 , 0 , 0 ) {2,3}-(-1,-1,-1, {2,4}-(-1,1,1,* +

{5,6}-(1,0,0,0) {3,6}-(0,1,0,1) {1,3}-(1,1,1,£+1) 1) { l , 5 } - ( - l , - l , M - l ) { 2 , 5 } - (1,1,-1,^+1)

{4,6}-(0,0,1,0) {3,5}-(0,0,1,1) {1,4}-(1,-1,-M-1) {1,6}- (-1,1,-1,^-1) { 2 , 6 } - (1,-1, M + l ) One checks that the linear collineations with matrices (1 0 0 0\ /-l 0 0-1 0-1 0-1 and 0 0-1-1 0 0

\o o o 1/

0-l\ 0-1 1 0

V o o o

1/

respectively, together with the unique involutory semilinear collineation having identity matrix, generate a group of order eight acting on W(2) as the translation group with respect to the translation point {1,2}.

296

Chapter 13. Translation Generalized Quadrangles in Projective Space

The same group acts similarly on the following embedding of W(2). {1,2}-(0,0,0,1) {3,4}-(1,0,0,1) {4,5}-(0,1,0,0) { 2 , 3 } - (2£- 1 , - 1 , - 1 , * {2,4}-(2£-l,l,l,£+l)

{5,6}-(1,0,0,0) {3,6}-(0,1,0,1) { 1 , 3 } - ( 2 £ + l , 1,1,^+1) {l,5}-(2£-l,-l,l,£-l) { 2 , 5 } - (2^+1,1,-1,^+1)

{4,6} - ( 0 , 0 , 1 , 0 ) {3,5}-(0,0,1,1) { 1 , 4 } - (2* + l, - 1 , - 1 , ^ - 1 ) {l,6}-(2£-l,l,-l,£-l) { 2 , 6 } - (2^+1,-1,1,^+1) If we call two embeddings equivalent whenever there exists a collineation of the ambient projective space taking the points and lines of the first embedded GQ to the points and lines, respectively, of the second embedded GQ, then the two embeddings defined above are not equivalent. Indeed, in the first embedding every nontrivial axial root elation with axis through {1,2} has nontrivial companion field automorphism, while in the latter example this is true for only one nontrivial axial root elation, namely the one with axis {{1,2}, {3,4}, {5,6}}. Example 13.3.2 A Class of Non-Polarized Translation-Homogeneous Embeddings. Consider the standard embedding of the generalized quadrangle Q(4, q) in PG(4, q), q even, and extend the groundfield GF(q) of the projective space to a field GF(g'), keeping the quadrangle unchanged. Select a point x of Q(4, q) and let n be the nucleus of Q(4, q) in PG(4, q). Select a point y of PG(4,g') on the line xn such that y does not belong to PG(4,q), except if y = n, and project Q(4, q) from y onto a hyperplane H of PG(4, q') not incident with y to obtain an embedding of Q(4, with the corresponding linear projective group of PG(4, q'). Then T fixes the points x and n, and since every element of T is linear and has order 2, the group T fixes y and so it also "projects" from y. Hence the projected embedding is translationhomogeneous. It is polarized at every point z that has the following property: the hyperplane, tangent to the quadric Q(4, q), at the pre-image z' of z under the projection, contains y.

13.3. Exceptional Non-Planar Translation-Homogeneous Embeddings

297

In case y ^ n, this is equivalent to saying that this hyperplane contains x; hence, if y ^ n, then the embedding is polarized exactly at the projections of the points of a;-1. If y = n, then the embedding is polarized at every point and it is the standard embedding of W(q) in a subspace PG(3, q) of H. Example 13.3.3 Small Quadrangles in 4-Dimensional Projective Space. The quadrangles W(2) and Q(4,3) admit embeddings in PG(4, q) for various q. These embeddings are translation-homogeneous with respect to every (translation) point. We start with W(2). The Universal Embeddings of W(2). Let K be an arbitrary field of characteristic different from 3 and identify the basis vectors of a 6-dimensional vector space V over K with the six ovoids of W(2). The coordinate 6-tuple associated with an arbitrary point x of W(2) has 1 in every position of an ovoid in which it is not contained, and —2 in the positions of ovoids it belongs to. Since every point belongs to exactly two ovoids, the points are contained in the hyperplane with equation 5Zi=i xi = 0> a n d s m c e every line of W(2) is incident with exactly one point of each ovoid, the sum of the coordinates of the three points incident with an arbitrary line of W(2) equals zero. In other words, collinear points belong to the same vector plane. Consequently we obtain a lax embedding of W(2) in a hyperplane H of PG(V). More concrete, and using the model for W(2) of pairs and partitions into pairs of the set {1,2,3,4,5,6}, we may identify the ovoids of W(2) with the numbers 1,2,..., 6 (and the ovoid i contains the points {i,j},j =/= i), and then the number i is identified with the basis vector ei. The point {i,j} now is identified with the vector J2t=i ak£k, with a, = a,j = —2 and a^ = 1 forfc i {i,j}. The transposition (12) induces a symmetry (with center) in W(2), and it is easy to check that this symmetry is also induced by the linear transformation of V with matrix /

1 3 2 3 1 3 1 3 1 3 1

2 3 1 3 1 3 1 3 1 3 1

1 3 1 3 2 3 1 3 1 3 1

1 3 1 3 1 3 2 3 1 3 1

1 3 1 3 1 3 1 3 2 3 1

-I

1 3 1 3 1 3 1 3

298

Chapter 13. Translation Generalized Quadrangles in Projective Space

Since this linear map fixes all vectors in the 4-dimensional space with equations f Xi + X2 + X3 + X4 + X5 + X6 = 0, \ X3 + X4+X5 + X6 = 0, it induces a special linear coUineation in H. We can do this for every transposition, and thus prove that the embedding is a special translationhomogeneous embedding with respect to every (translation) point. This description is also valid in characteristic 2. In this case, the coordinates of all the points of W(2) satisfy the quadratic equation X^=2 X)}=i XiXj = 0 which represents a nonsingular parabolic quadric in H isomorphic to Q(4,2) in the 4-dimensional subspace H' of H over the subfield GF(2) ofK. If K has characteristic 3, then we consider PG(6, K) and associate with the point {i, j} of W(2) the point with coordinates all 1, except in the positions i and j , where we write — 1. This defines a lax embedding of W(2) in the 4-dimensional projective space obtained by projecting the hyperplane H" with equation X\ + X2 H 1- X7 = 0 from the point c := (1,1,1,1,1,1,0) onto an arbitrary 4-dimensional space in H" not containing c. In fact, it is desirable to leave that 4-space unspecified so that it is easier to see the automorphisms (the symmetric group acting on the first six coordinates in a natural way). As in the previous case, one can show that the embedding is a special translation-homogeneous one. In all cases, the embeddings described here are polarized. This is obvious since there are only three lines through each point in W(2), and these cannot generate a 4-dimensional space. These embeddings are called universal because every lax embedding of W(2) in PG(3, K) arises from a projection of the universal one from a wellchosen point in PG(4, K) after embedding PG(3, K) in PG(4, K). This is proved in [195] and we do not repeat the proof here (it goes along the lines of the proof of Theorem 13.4.1 below). We remark that the hyperoval embedding of W(2) is not a projection of the universal lax embedding. The Universal Embeddings o/Q(4,3). We label the base vectors of a six-dimensional vector space Ve over the field GF(3) with the ordered pairs {(01), (02), (03), (12), (13), (23)} of the set {0,1,2,3} and write coordinates in lexicographic order. In the projective space PG(Ve) we consider the Klein quadric with equation X01X23 -

13.3. Exceptional Non-Planar Translation-Homogeneous Embeddings

299

^02-^13 + -^03^12 = 0. In PG(3,3), we consider the symplectic bilinear form f3 given in ordinary coordinates by X0 Xi 2/o 2/i

+

x2 x3

XQ

2/2 2/3

2/0 2/3

X3

and let the image of the line of PG(3,3) incident with the points (rr0, x\, x2, x3) and (y0,2/1:2/2,2/3) under the Klein correspondence be given by the line in PG(V 6 ) with coordinates Xij = Xiyj — Xji/i. The totally isotropic lines with respect to (3 are mapped under this correspondence onto the set of points of the quadric Q(4,3) given by the equations -^02^13 -^03

XQ\X\2

+ -^0)1^23 + -^12-^23,

X0l + X2Z-

An arbitrary line — which we can label with a homogeneous 4-tuple (a 0 , ai, a2, a3) that represents a point of PG(3,3) — of Q(4,3) is obtained by intersecting this quadric with the plane spanned by the points {ai,a2, 03,0,0,0), (ao,0, 0 , - 0 2 , - a 3 , 0 ) , (0,a 0 ,0, ai, 0, - a 3 ) , (0,0,ao,0,ai,a 2 ). We now write down all the points of Q(4,3) in coordinates. But we substitute sometimes —1 by the letter k, and consequently we sometimes write 2 as 1 - k, 0 as 2 - k or as 1 + k and 1 as A;2. The reason to do so will become clear below. (0,1,0,0,0,0) (1,0,1,0,0,0) (1,1,1,0,0,0) (l,fc, 1,0,0,0) (0,0,0,1,0,0) (0,1,0,1,0,0) (0,^,0,1,0,0) (0,0,1,0,0,1)

(0,1,1,0,0,1) (0,0,0,0,1,0) (1,0,1,0,1,0) (k,0,k, 0,1,0) (0,0,0,1,1,0) (1,1,1,1,1,0) {k,k,k, 1,1,0) (0,0,0, A:, 1,0)

{0,0,k,0,l,k) (l,l + fc,2,l,2-fc,l) (l,A:,2,l,2-fc,l) (1,1,2,1,2-^,1) (k,k,l + k,k,l,l) (0,k,l,k,l,l) (l,fe,2,fc,l,l) (1-fc, l,2-k,l-k, 1-k,

(l,fc,l,fc,l,0) (k,k2,k,k,l,0) (0,fe,l,0,0,l) {k,k,l + k,0,1,1) (0,0,1,0,1,1) (1,1,2,0,1,1) (l,k,l + k,0,l,k) (k,k2,2k,0,l,k)

(1,1,2,1 — fc,l —fe,1) (0,1,1,1-fc,l-fc,l) (1 —fc,1,2 — fc,l, 1 —fc,1) (l,l + fc,2,l,l-/c,l) (0, A, 1 , 1 , 1 - A , 1) {k,k,l + k,1,1,1) (0,1,1,1,1,1) fc,2,l,l,l) 1) ( l , l +

300

Chapter 13. Translation Generalized Quadrangles in Projective Space

Now let K be a field admitting an element r satisfying r 2 - r + 1 = 0. Then it is a tedious exercise to check that, with k = r, the above 40 points define a lax embedding of Q(4,3) in the hyperplane H of PG(5, K) with equation -^03 = XQI + -^23-

One can now check that the little projective group, which contains the translation group with respect to any (translation) point, of this lax embedded GQ S is induced by the projective special linear group PSL5(K.) of H. In any case, no other collineation than those of the little projective group are induced by PGL 5 (K) unless char(K) = 3. For K ^ GF(g), with q not divisible by 3, the group PrL 5 (g) induces the full automorphism group of S if and only if q is a perfect square and ^fq = - 1 mod 3. These are tedious exercises, and expressions with matrices can be found in [195], to which we refer for more details. Hence the embedding is a special translation-homogeneous embedding with respect to any point. This embedding is called universal because every embedding of Q(4,3) in a 3-dimensional projective space over IK is a projection of this one, for all fields K, see [195] (where the proofs are only carried out for finite fields, but they hold without much change for arbitrary fields). Example 13.3.4 The Universal Embeddings of Q(5,2) in PG(5, K), with K Any Field. Let <S = Q(5,2) and let <S' be a subquadrangle of order 2. We embed <S' as in Example 13.3.3. Recall that S can be constructed as follows from W(2). Add to the point set of W(2) two copies of the set of ovoids — since we can identify each ovoid with a number i e {1,2,3,4,5,6}, we may write these twelve additional points as i and i', with i e {1,2,3,4,5,6}. The thirty extra lines are formed by the triples {i, {i, j}, j'}, for i, j e {1,2,3,4,5,6}, i ^ j . In other words, the ovoid i, the ovoid j ' and the unique point in the intersection of these ovoids form a line. So it suffices to attach coordinates to these 12 additional points in such a way that the points of each of the additional thirty lines are collinear. This can be done in the following way. First let the characteristic of K be different from 3. Assign to the point i the 6-tuple with 0 in every position, except for position i, where we put 1. Assign to the point i' the 6-tuple with 1 in every position, except in position i, where we put - 2 . This defines a lax embedding of Q(5,2) in PG(5, K). If the characteristic of K is equal to 2, then the embedding is full in the subspace PG(5,2) over the subneld GF(2), and the points of S are precisely those points of PG(5,2) whose coordinates satisfy the quadratic equation £ ? = 2 E ^ i xixj = ° a b o v e -

13.4. Non-Planarly Embedded Small Translation Generalized Quadrangles

301

If the characteristic of IK is equal to 3, then give the point i in PG(6,K) coordinates (0,..., 0,1,0,..., 0), where the 1 is in the position i, and the point i' gets the coordinates (0,..., 0,1,0,..., 0,1), where the first 1 is in the position i. We now still project from the point c = (1,1,1,1,1,1,0), but do not restrict to points of H". We obtain a lax embedding in the 5-dimensional projective space over K obtained from PG(6, K) by considering the subspaces through c. Note that no element of this embedding lies in the hyperplane with equation Xx-\ \- XQ = 0. This means we always have an embedding in an affine space, and, in particular, for |K| = 3, the embedding is full in that affine space. For every field K, the above lax embedding is translation-homogeneous and linear, even special. This follows easily from the following two observations. (i) All collineations of S' extend uniquely to linear collineations of PG(5,IK) preserving S (these are just the permutations of the coordinates) . (ii) By the uniqueness of this embedding, i.e., the embedding is uniquely determined by S' and one additional line (the two points not in S' of which may be chosen freely; we will prove this below in Theorem 13.4.4), the embedding has an automorphism group acting transitively on the subGQs of order 2. (iii) The little projective group of <S is a simple group, and if the embedding were not special, then the intersection with the special linear group of PG(5, K) would be a nontrivial normal subgroup of the little projective group of S. Again, this embedding is universal in the sense that every embedding of S in PG(d,K), d € {3,4}, can be considered as projection of this universal one (and for d = 5 every embedding is equivalent to the universal one).

13.4 Non-Planarly Embedded Small Translation Generalized Quadrangles We start with classifying the translation-homogeneous embeddings of W(2).

302

Chapter 13. Translation Generalized Quadrangles in Projective Space

Theorem 13.4.1 If W(2) is faithfully and translation-homogeneously embedded in PG(3,g) (with respect to a single translation point), then either q is an odd perfect square and the embedding is as described in Example 13.3.1, or q is even and the embedding is as described in Example 13.3.2. Proof. Let W(2) be embedded in P G ( 3 , q). We assume that the embedding is translation-homogeneous with respect to the point x = {1,2}. We first treat the case where the embedding is not polarized at x. Then, without loss of generality, we may assign coordinates as follows: {1,2} - ( 0 , 0 , 0 , 1 ) {5,6} - ( 1 , 0 , 0 , 0 ) {4,6} - ( 0 , 0 , 1 , 0 ) {3,4} - ( 1 , 0 , 0 , 1 ) {3,6} - ( 0 , 1 , 0 , 1 ) {3,5} - ( 0 , 0 , 1 , 1 ) {4,5}-(0,l,0,0){l,3}-(a,6,C,d), for some a,b,c,d G GF(q). Note that none of a,b,c,d is zero since otherwise the point {1,3} is contained in a plane it together with two lines through x. Applying the translation group with respect to x, this implies that all points opposite x are contained in n, and hence, since the intersection of W(2) with 7r is a subquadrangle of W(2), this would mean that the embedding is planar, a contradiction. Taking the lines through {1,3} into account, there must be elements a',b',c' G GF(g) such that {2,4} - (a',b,c,d)\{2,5}

- (a,6,c',d)|{2,6} -

(a,b',c,d).

The line {{3,5}, {2,6}, {1,4}} forces {1,4} to have coordinates (a, b', c + k,d+ k), for some k G GF(q). The line {{5,6}, {1,4}, {2,3}} implies {2,3} = (a", b',c + k,d + k), a" e GF(q). Considering the lines through {2,3}, we then have {1,5} = (a", 6', c", d + k) and {1,6} = (a", b"', c + k, d+ k), for some b", c" G GF(q). Expressing that the respective points of the lines through the points {3,5}, {3,6} and {3,4} are collinear in P G ( 3 , q), an elementary calculation reveals that

{

a' = a" = a + fc, b' = b" + k = b + k, d = c" + k = c+k.

Note that the unique nontrivial central root elation 9 with center {1,2} has nontrivial companion field automorphism. This implies that either the axial

13.4. Non-Planarly Embedded Small Translation Generalized Quadrangles

303

root elation 9' with axis {{1,2}, {3,4}, {5,6}} has trivial companion field automorphism or the composition 99' does. Since the matrix of that linear coUineation is determined by the image of the points in or1, this matrix is equal to (I 0 0 0\ 0-1 0-1 0 0-1-1

\o o o \j Since {1,3} e ' = {2,4} and {1,3} ee ' = {1,4}, we either have (a, - 6 , - c , d b — c) = (—a — k, —b, —c, —d), or (a, —b, —c,d — b — c) = (a,b + k,c+k,d-\-k). Since clearly k ^ 0, both cases imply that q is odd. Also, q must be a square. We denote the unique involutory field automorphism as usual by £ i-> I. I: (a, - 6 , - c , d - b - c) = (a, & +fc,c +fc,d + k). Then 6 = c = -fc/2 and the point (a, - 6 , - c , d — & - c), which is just equal to {1,3} e ', coincides with (a + k, b, c, d), which is {2,4}. One checks that this implies that CASE

a

a+k

I

and

a+k d

d+k

Similarly, the linear coUineation 9" with matrix (

•

-1

o

V

i/

maps the point {1,3} either to {1,5} or to {2,5}, and then 69" maps {1,3} to either {2,5} or {1,5}, respectively. So we have again two cases. First suppose that {1,3} e " = {1,5}. Then 2a = -k. Putting l=-\- -2d/k, we obtain as embedding the first example of Example 13.3.1. Next suppose that {1,3} e " = {2,5}. Then d = -fc/2 Putting we obtain the second example of Example 13.3.1.

2a+k Ik >

CASE II: (a, -b, —c,d — b — c) = (—a — k, - 6 , —c, —d). In this case, the nontrivial axial root elation 9' with axis {{1,2}, {3,4}, {5,6}} has trivial companion field automorphism. Then, if 6" is an axial root elation, the composition 06'0" is an axial root elation with nontrivial companion field automorphism, and this is equivalent to the second example of Example 13.3.1.

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But, if 9" is not an axial root elation, then 99" is, and we again obtain an embedding equivalent to the second example of Example 13.3.1. Now suppose that the embedding is polarized at x. Let L be any line of W(2) incident with x. The pointwise stabilizer of L in the translation group T has order 4, and hence there is some element 9 of the translation group T induced by a linear coUineation of P G ( 3 , q), fixing all points of W(2) — and consequently also of P G ( 3 , q) — on L, and not fixing x1- pointwise. In the plane of P G ( 3 , q) spanned by x1- the coUineation 6 is induced by an axial coUineation with axis L. the center must necessarily be x, as 9 fixes at least three lines of that plane through x (namely, the three lines of W(2) through x). So q is even and without loss of generality we may assign coordinates as follows: {1,2}-(1,0,0,0) {4,6} - ( x , 1,1,0) {3,6} - (1,1,0,0) {4,5}-(0,1,0,0)

{5,6}-(0,0,1,0) {3,4} - ( 1 , 0 , 1 , 0 ) {3,5} - (x + 1,1,1,0) {1,3}-(0,0,0,1),

and we may put {2,3} = (a, b, c, 1), for some a,b,c e GF(q). But now the coordinates of all other points are uniquely determined, in fact in two different ways. For instance, {1,5} is the unique point on the line ( ( x , l , l , 0 ) , ( a , 6 , c , l ) ) and in the plane ((1,0,1,0), (0,1,0,0), (0,0,0,1)) (this plane contains the point {3,4} and the line {{1,3}, {2,6}, {4,5}} and hence also the line {{1,5}, {2,6}, {3,4}}), which implies that it has coordinates (a+cx, a+b+c+bx, a+cx, x+1); also it is the unique point on the line ( ( x , l , l , 0 ) , ( a , 6 , c , l ) ) and in the plane ((1,1,0,0), (0,0,1,0), (0,0,0,1)), which implies that it has coordinates (a + bx,a + bx,a + b + c + cx,x + 1). Hence we already deduce b = c. Considering similarly the point {1,6}, it follows that b = c = 0 and we obtain the following coordinates. {1,3} {2,3} {1,5} {2,5}

- (0,0,0,1) - (a, 0,0,1) — ( a , a , a , x + l) — (ax,a,a,x + 1)

{2,4} - (0,0, a, x + 1) {1,4} - (o(x + 1), 0, a, x + 1) {2,6} — (0,a,0,x + l) {1,6} — (a(x + l ) , a , 0 , x + l)

Applying the recoordinatization (X 0 , Xi, X 2 , X 3 ) H-> (X 0 , Xi, X 2 , aX 3 ), we may assume that a = 1. Now we embed P G ( 3 , q) in P G ( 4 , q) and see that S is the projection from the point (x, 1,1,0,1) onto PG(3,
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equation X 4 = 0) of the following embedding of W(2). {1,2} - ( 1 , 0 , 0 , 0 , 0 ) {3,4}-(1,0,1,0,0) {4,5}-(0,1,0,0,0) {2,3}-(l,l,l,x+l,l) {2,4} - ( 0 , 0 , 1 , 3 + 1 , 0 )

{5,6}-(0,0,1,0,0) {3,6}-(1,1,0,0,0) {1,3} - ( 0 , 0 , 0 , 1 , 0 ) { 1 , 5 } - ( 1 , 1 , l,a:+l,0) {2,5}-(0,0,0,3+1,1)

{4,6}-(0,0,0,0,1) {3,5}-(1,0,0,0,1) { 1 , 4 } - ( 1 , 1 , 0 , 1 + 1,1) { 1 , 6 } - ( 1 , 0 , 1 , ^ + 1,1) { 2 , 6 } - ( 0 , 1 , 0 , 3 + 1,0)

Applying the recoordinatization (X0,Xi,X2,X3,X4) H-> (X0,Xx,X2,X3/ (3 + 1 ) , X4), we now easily see that all these points belong the the intersection of the nondegenerate quadric with equation X0X3 + XXX2 + X1X4 + X2Xi = 0 with the projective 5-space over the field GF(2) obtained by restricting the coordinates to GF(2). The proof of the theorem is now complete. • By Theorem 1.4(ii) of [195], every embedding of W(2) in P G ( 4 , q) is isomorphic to the universal one, and hence independent of being translationhomogeneous. So we may state the following result, and we do so for general fields K. Theorem 13.4.2 Every lax embedding of W(2) in PG(4,K) is isomorphic to the universal embedding of W(2) in PG(4,K) and hence polarized at every point and translation-homogeneous with respect to every point (and the embedding is linear). Proof. The proof is almost identical to the proof of Theorem 13.4.1, except that we can now choose freely the coordinates of all points of 2+ in some 3-dimensional subspace, and additionally two points opposite x outside that 3-space. The coordinates of all other points are now determined and uniqueness follows similarly as in the proof of Theorem 13.4.1, except that we do not have to identify the embedding with a projection. We leave the details to the reader. They can be found in [195]. • Now we come to Q(4,3) in PG(4, q). Also here, one can classify without the hypothesis of being translation-homogeneous and the result is that every lax embedding is translation-homogeneous with respect to every point. Theorem 13.4.3 Every lax embedding o/Q(4,3) in PG(4, q), with q not necessarily divisible by 3, is isomorphic to the embedding described in Example 13.3.3. In particular, q = 0 or 1 mod 3.

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Proof. The proof of this lemma is a straightforward calculation well-chosen but essentially arbitrary coordinates to the points, expressing that the points collinear in Q(4,3) are also collinear in We will not perform this exercise here, but refer the reader to more details.

assigning and then PG(4, q). [195] for

The fact that q must be congruent to 0 or 1 modulo 3 is because there must exist an element r of the field satisfying r2 — r + 1 = 0, implying r 3 = - 1 , with r ^ - 1 if the characteristic is different from 3. • Finally we treat Q(5,2) in P G ( 5 , q). As in the previous case, one can classify without the hypothesis of being translation-homogeneous and the result is that every lax embedding is translation-homogeneous with respect to every point. Since we already dealt with the case q even in Lemma 13.1.10, we could concentrate on the case q odd, but we can in fact do the general case. Theorem 13.4.4 Every lax embedding of Q(5,2) in P G ( 5 , K), with K any field, is isomorphic to the embedding described in Example 13.3.4. Proof. Let <S = Q(5, 2) and let S' be a subquadrangle of order 2. We treat the case where the characteristic of the underlying field K is different from 3 (otherwise the arguments are completely similar, but the calculations differ). Without loss of generality, we may embed S' as in Example 13.3.3. Also, we may choose the coordinates (1,0,0,0,0,0) for the point 1 and (1, - 2 , 1 , 1 , 1 , 1 ) for the point 2'. For i G {4,5,6}, the points 2', i', {3, i} and {3, 2} are coplanar (since they lie on two lines through 3) — note that we know the .exact coordinates of three points of that plane n, the points 1, i' and {1, if are collinear, and we know the coordinates of two points of this line K. It follows that we can calculate the coordinates of the intersection i' of 7r and K. By interchanging the roles of 3 and 4 we obtain the coordinates of 3'. Now, for i e {1,4,5,6}, the coordinates of the point i are calculated by considering the intersection of the lines through 2', {2, i} and 3', {3, i}, respectively. Interchanging the roles of 2 and 4 and of 3 and 5, the coordinates of 2,3 follow. Finally, we can now also easily calculate the coordinates of 1'. The proof of the theorem is complete. •

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13.5 Non-Planarly Embedded Translation Generalized Quadrangles We now treat the general case of a TGQ of order (s,t) translation-homogeneously embedded in PG(d,g), d > 3, that is, d e {3,4,5} (cf. Proposition 13.1.9). By Lemma 13.1.11, we may assume d < 5. Also, the previous section allows us to assume that t > 2 for d = 3 and t > 3 for d = 4. In other words, we may assume t > d, with d G {3,4}. Lemma 13.5.1 If the TGQ <S of order (s, t) is translation-homogeneously embedded in PG(d, q), d e {3,4}, t > d, with respect to the translation point x, then the embedding is polarized at x. Proof. Without loss of generality, we may assume that the embedding is faithful. Suppose by way of contradiction that the embedding is not polarized at x. Let T be the corresponding translation group (an elementary abelian p-group, for some prime p), and let T be the map that assigns to every semilinear collineation the companion field automorphism. Since T is not cyclic, V cannot be one-to-one if restricted to T, hence there is some nontrivial linear collineation 6 in T. We first claim that p divides q. Since t>d,we can choose d lines Lx, L 2 , . . . , La through x in S that generate PG(d, q). Let L be a line of S through x different from Lu for all z e { 1 , 2 , . . . , d}. Let I c { 1 , 2 , . . . , d} be (settheoretically) minimal with respect to the property that L is contained in the space H generated by {Li || i e I}. Let y ^ x be a point of S on L. The stabilizer Ty has order st and is not cyclic. As above, this means that there exists a nontrivial linear collineation 9' in Ty. But since p divides s, 9' fixes at least three points on L, and hence fixes all points of PG(d, q) on L. Now, the condition on L and / implies easily that 6' fixes H and induces a central collineation in ff with center x. But since all points on L are fixed, the axis contains L and hence also x. So, if 9' is not the identity in H, then it is an elation in the projective space H and the claim follows. Now assume that 0' is the identity on H. Then 9' is central in PG(d, q). Also, it is certainly not a symmetry about xy. Suppose that s > p. Then, since Aut(GF(g)) contains at most p—1 elements of order p, there are two different symmetries about xy with same companion field automorphism. Taking the "difference" of these, this implies the existence of a "linear" symmetry about xy. Hence s = p. Now Corollary 3.5.6 implies that S is classical and isomorphic to either Q(4,p), p > 2, or Q(5,p). In the first case, no collineation of <S

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fixes all the points of three lines incident with x (by the antiregularity of the point x), in the second case we argue as follows. If at least two lines incident with x are not contained in H, then the center of 6' is x, since the center must lie on every such line. But in that case, the claim is proved again. Hence all lines through x but one are contained in H, and 0' fixes all points on these lines. Again, there is no such coUineation in Q(5,p) but the identity. Hence the claim is proved. So, in the previous paragraph, we found a linear coUineation 6' e Ty of order p, and clearly it cannot fix every point of or1. So there is a point z ~ x not fixed under 6'. Since p divides q, and since 0' fixes all lines of <S through x, the coUineation 8' is central in PQ(d,q) with center x. Let M be a line of <S incident with z but not with x. Then M9' is contained in the plane n = (x, M) and the latter induces a subquadrangle S' of order (s, £'), if > 1, entirely contained in II. Since T fixes the lines of S through x in n, and these lines span n, every element of T maps u, with u / i a point of S' in II, to a point of II. By the transitivity of T on points opposite x, we see that this implies that S = S', a contradiction. The lemma is proved. • We continue with a general lemma for TGQs of order s for even s. Lemma 13.5.2 Let S^ be a translation generalized quadrangle of order s, with s even. Let L\, L2, L3 be three arbitrary lines incident with x, and let y be an arbitrary point opposite x. Abo, let z be any point on L\ distinct from x and distinct from projLiy. Then there exists a subquadrangle of order 2 containing x, y, z and Li,L2, L3. Proof. Let M2 = pro)yL2 and let z* = projM22;. Let s = 2™ and let 0 be a pseudo-oval in PG(3n — 1,2) corresponding with S. Let 7TJ G O correspond with Lt, i = l,2,3. Let ir be the nucleus of O (see Theorem 3.9.1). Embed PG(3n-1,2) as a hyperplane in PG(3n, 2) and let y' and z' be the points of PG(3n, 2) \ PG(3n -1,2) corresponding with y and z*, respectively. Then y'z' meets ir2 in a unique point a2- The 2n-space generated by a2,iri,TT3 meets it in a point a (this follows immediately from the definitions of a pseudo-oval and the nucleus). The line aa2 meets the space (7Ti,7r3) in a point a', and there is a unique line oia 3 , with ai e iti and a3 e 7r3, containing a'. Hence we have found a unique plane £ := (ai,a 2 ,a 3 ,a) meeting all of 7ri, 7r2,7r3,7? in a single point, and containing the given point a2. Now the projective space spanned by f and y' defines the subquadrangle of order 2 with the required properties. •

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It is easy to see that the previous lemma is equivalent to the following corollary. Corollary 13.5.3 Let S^ be a translation generalized quadrangle of order s, with s even. Then every ordinary pentagon through x is contained in a subquadrangle of order 2. • We now come to the first main result of this section. Theorem 13.5.4 Let S^ be a translation generalized quadrangle of order (s,t), t > 2, laxly but translation-homogeneously embedded in PG(3,g), with respect to x. Then S = W(s), with s even, and the embedding is as described in Example 13.3.2 above. Proof. Again, we may assume that the embedding is faithful. By Lemma 13.5.1, we know that the embedding is polarized at x. Let p be the unique prime dividing s. We claim that p divides q as well. Let y ~ x, z ~ x, y ^ x and z 2, or S = Q(5,p). In both cases S does not admit central root elations. Hence there is a linear collineation 9 e Ty not fixing z. It is easily seen that the restriction of 9 to H is a nontrivial elation with center x and axis xy. The claim follows. Suppose now, by way of contradiction, that the embedding is not polarized at y. Then applying T, we see that the embedding is not polarized at any point of xy, excluding x. An arbitrary linear symmetry about xy then has s centers on xy, implying it is necessarily the identity, a contradiction. Hence no nontrivial symmetry about xy is linear. If s > p, then, since GF(q) contains at mostp-1 automorphisms of order p, there are two such symmetries with the same companion field automorphism, implying that there is such a linear symmetry. Hence s = p. If t = p2, then the same argument shows that we again have linear central coUineations with center x, contradicting the fact that in this case S = Q(5,p). If t — p, and hence S = Q(4,p), then as in the proof of Theorem 13.2.2, there is a linear collineation fixing all points on two lines through x and not fixing any other point collinear with x, a contradiction. Hence the embedding is polarized at every point

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collinear with x. For such a point u, we denote by £u the plane containing Consider again y, z as above. The planes £y and £,z meet in a line L, which contains {y, z}±. Let u be an arbitrary point of {y, z}1- \ {x}, and let v be arbitrary in {u, x}1- \ {y, z}. Since £v contains u and x, it contains L and hence v1- must necessarily contain {y, z}^. Consequently v G {y, z}±± and we conclude that {x, u} is a regular pair of points. Varying y, z, we see that a; is a regular point. Hence, by Theorems 2.1.3 and 2.1.5, s = t and s is even. If we can show that <S = W(s), then knowing that the embedding is polarized at every point collinear with x, we see that, using Lemma 13.1.10 and the construction of Example 13.3.2, the embedding is necessarily isomorphic to the latter. Hence we show S = W(s). To achieve this, it suffices to prove that all lines are regular. Since all lines through x are already regular by the fact that S is a TGQ, we only have to show that two nonconcurrent lines L,M of S not incident with x and not meeting the same line of S through x form a regular pair of lines. So let L and M be two opposite lines of S not incident with x and not meeting the same line of <S incident with x. The lines L and M are concurrent with some lines L' and M' of <S, respectively, which are incident with x. Let y and z be the respective intersections (so LlylL'lxlM'lzlM). We first claim that the point set of S on L (and also on M) is projectively equivalent with the point set of a projective line over the subfield GF(s) of GF(q) (and hence that the former is indeed a subfield of the latter!). Note that the lines L and M' are not contained in a plane of P G ( 3 , q), since otherwise, by the action of T, all points of S opposite x would be contained in that plane, and so the embedding would be planar. So L and M' are contained in a nonsingular hyperbolic quadric (by the regularity of M') containing all the lines o f < S i n { L , M ' } - L u { I / , M'}±±, and we see that the point sets of L and M' are projectively equivalent. So, to prove our claim, it suffices to show that the points of S incident with M' form a projective line over GF(s). In PG(3,g), we may coordinatize the line M' in such a way that x is assigned the inhomogeneous coordinate (oo), and two arbitrary but fixed points ZQ ^ x and z\ ^ z0, z\ ^ x, of S on M' get the coordinates (0) and (1), respectively. Choose a fixed point y0 on V, with y0 =£ x. Let K'lx, M' ^ K' ^ L', be a variable line through x, and let ulK', a / i , be the point on K' that belongs to {y0, z0}±±. Let K be any line

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incident with u but different from K'. The hyperbolic quadrics containing {M',K}^ U {M',K}^ and {L^K}1- U {L',K}X± induce a projectivity pKi from the set of points S(M') of <S on M' to the set of points of S on L', fixing a; and mapping ZQ onto y0- Since x is fixed, this projectivity is a projection from some point c of PG(3,q) on the line y0z0 of PG(3,g). Since t > 2, and since Aut(GF(q)) contains at most one involution, there exists a linear symmetry 9 with axis K'. Obviously, 9 defines in the projective plane £x a central collineation with axis K'. Since 9 fixes L',M',u and if, it maps ZQ to y$. It follows that c is the intersection of the lines y0z0 and y%z\. Since 6* is an involution of S, and can also be viewed as an involution of P G ( 3 , q) by faithfulness, the point c is a fixpoint of 9, and hence is contained in the axis, which is K'. Considering two different lines K[ and K2 playing the role of K', we see that p := PK[P^ is a nontrivial projectivity preserving S(M'), and fixing x and z0 (the projectivity is indeed nontrivial since the centers c\ and c2, respectively — playing the role of c above — are distinct!). In coordinates, if p maps (1) to (a), then it maps an arbitrary point (6) of M' to (ab). Since there are s — 2 choices for K2 ^ K[, given K[, and since no two choices define the same projectivity = PK'IPK\> K3IX> M> ^ K3 ^ L'> i m P l i e s t h a t PK^PK] is trivial, hence K'2 = K'3), we obtain, adding the identity, a set of projectivities fixing (oo) and (0) and acting sharply transitively on S(M') \ {x,z0}. Since the projectivity group of M' acts sharply 3-transitively, we deduce that this set of projectivities is actually a group. Varying z0 on M' in S, we obtain a group of projectivities acting sharply 2-transitively on S(M') \ {x}. The Frobenius kernel of this action is necessarily a subgroup of the additive group of GF(g) (identified via the inhomogeneous coordinates on M'). So we see that the inhomogeneous coordinates of the points of S(M') \ {x} form an additive subgroup of GF(q), and removing (0), they form a multiplicative subgroup of GF(g) \ {0}. We conclude that they form a subfield GF(s) of GF(<7). Our claim is proved. (PK'IPKI

Now let L0 ~ M0 ~ N0 be three arbitrary lines of S with L0 7^ N0 and such that x is not incident with either. We also assume that L0 and N0 do not meet the same line incident with x. We will show that, considering a^proj^Lo and projj.iVo, Corollary 13.5.3 implies that {L0,M0,N0) is PG(3,g). For, otherwise there is a subGQ S* of order 2 containing x, LQ, M 0 , N0 and embedded in a plane, say TT. Considering a point u0 of S* opposite x, we see that x1- and UQ are contained in -K, and so S is contained in n, a contradiction. Hence the points of <S on the lines L0,M0, N0 define a unique projective subspace P G ( 3 , s) of P G ( 3 , q). We claim that, for ev-

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ery point wolL0, the points of S on the line projWo7V0 belong to PG(3, s). Indeed, it is clear that |{projxL0,proj;cMo,proja;./Vo}| G {2,3}, and either wo ~ x or wo / x. Hence there are four cases to consider; we will consider the generic case and leave the others to the reader (in fact they will follow immediately from the arguments of the generic case, permuting the notation). Hence we assume that L'0 := projxXo, MQ := projxM0 and NQ := projxN0 are all distinct, and that x w0. Then the quadrangle <S* above is uniquely defined. Put y0 -.= projLox, z0 •= projM()x and u0 := proj^x. Inside S*, there is a grid containing L0,M0,N0; two other lines of that grid are proj^ j/o and projLotto, and, viewed as lines of PG(3,q), they also belong to PG(3, s). Now the point ZQ belongs to PG(3, s) and hence so does the unique transversal P0 through z0 meeting both proj^ y0 and projL UQ, and its intersection points with the latter two. Since a projective subline over GF(s) is determined by three points, we see that all the points of the grid in S belong to PG(3, s). Now, considering the subGQ of order 2 containing Lo,vro)WQN0, N0 and x and repeating this argument, the claim follows. The same claim can be proved for the cases where L0 and N0 do meet the same line incident with x, or when one of the lines L0, M0, N0 is incident with x, and the point w0 is on a line through x or is x, or not. Indeed, this is easy since every line through x is regular, and every grid of S through such a line is easily shown to be contained in a subspace over GF(s), noting that each such grid spans PG(3, q). Now we finally prove that the pair {L,M}, with L and M two opposite lines of S not incident with x and not meeting the same line through x, is regular. Let N be any line of <S meeting both L and M. Then there is a unique projective subspace, which we may again denote by PG(3, s), over GF(s), containing the points of S on the lines L,M,N. Now we build the subquadrangle So generated by the points of <S on L, M, N. This subquadrangle can be constructed as follows. Each step consists of looking at already constructed points and lines at distance 5 in the incidence graph of the thus far constructed subgeometry, and adding the unique line of S passing through the point and concurrent with the line (and not belonging to the so far constructed subgeometry). It is easy to see that this construction indeed yields a subGQ. But by our above claims, So is entirely contained, and hence fully embedded, in PG(3, s). Now there are two possibilities (since S has order s). (1) For some choice of L and M, the order of So is s, and then S = So

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is fully embedded in PG(3, s), which implies that the embedding is polarized by Lemma 13.1.5. So every point is regular by Proposition 13.1.6, S = W(s), and we are done. (2) For all pairs {L, M}, the subGQ <S0 has order (s, 1), which means that all lines are regular, and we are done again.

• Our second main result treats the 4-dimensional case. Theorem 13.5.5 Let the TGQ «S(x) = S of order (s,t) be translation-homogeneously embedded in PG(4, q) with respect to x, and suppose t > 3. Then S = Q(4, s) and the embedding arises from the natural embedding in PG(4, s) by field extension. Proof. We may assume that the embedding is faithful. By Lemma 13.5.1, we know that the embedding is polarized at x. We first claim that no three lines of S through x are coplanar. Indeed, suppose by way of contradiction that the three lines Li,L2,L3 through x are coplanar, say, they belong to the plane ir. Choose any point y opposite x, not contained in IT. Then H := (n,y) intersects S in a translation subquadrangle S' = (V',B',I), translation-homogeneously embedded in the 3-dimensional projective space H, because every translation of S^ mapping y to a point of S in H, and viewed as a collineation of PG(4, q), clearly preserves H. Hence S' = W(s) by Theorem 13.5.4 and t = s2 by Lemma 1.1.1. We also know from Theorem 13.5.4 that the embedding of <S' in H is polarized at each point z ~ x. Consider such a point z ^ x. Let irz be the plane containing z1- n V, and choose any line Llx in S, with z IL. Then H' := (irz,L) intersects S again in a translation-homogeneously embedded subquadrangle (with respect to x) S" of order s, which is, by Lemma 13.5.1, polarized at x. Hence the plane (xz, L) contains exactly s +1 lines of S. Continuing like this, we see that, for every pair of lines L, M through x, the plane (L, M) intersects S in s +1 lines through x. In this way we obtain a linear space with s2 + 1 points (the lines through x) and every block containing s + 1 points (consequently every point is contained in s blocks). Counting the number of incident point-block pairs in two ways, and putting b equal to the total number of blocks, we arrive at the contradiction (s + 1)6 = s(s2 + 1). Our first claim is proved. Our first claim implies that every linear element 6 of the translation group T fixes all lines through x of PG(4, q) which belong to the 3-space £x generated by the lines of <S through x.

314

Chapter 13. Translation Generalized Quadrangles in Projective Space

Let p be the unique prime dividing s. We claim that p divides q as well. Let y~x, z~x, yj^x and z ^ y. Let T be the translation group with respect to x. Suppose first that all linear collineations of Ty fix z. Hence every such linear collineation fixes the plane -K' = (x, y, z) pointwise. Then, as before in the proof of Lemma 13.5.1, s = p and so <S = Q(4,p), p > 2, or S = Q(5,p). In both cases S does not admit central root elations. Hence there is a linear collineation 9 e Ty>z not acting trivially on £x. Then it induces an elation of £x with center x and axis ir', and the claim follows. If, on the other hand, Ty contains a linear collineation that does not fix z, then this collineation is again central in £x with center x, and the axis contains xy, hence the claim again follows. We now claim that t < s2. For, suppose byway of contradiction that t = s2, and let 9 be a nontrivial linear collineation in Ty p). Let u be a point opposite x and collinear with both y and z. Then {x, u, u6}1- contains s + 1 elements (see Section 2.5) and 9 fixes all points on every line xx', with x' G {x, u, it6}-1. This now easily implies that 9 fixes £x pointwise, a contradiction, since this would mean that \{x, u, ue}-L\ = s2 + 1. Our claim is proved. Next we claim that t G {s, sp}. Suppose by way of contradiction that sp < t < s2. By Theorem 3.10.1(ii), p > 2. Since both s and t are powers of the same prime p, we have t > p2s. Let y and z be as in the previous paragraph, and let M be an arbitrary line of S with xlM and y IM Iz. Let F : Tj,^ —> Aut(GF(g)) be the group morphism that assigns to every element of Ty ps. Hence S cannot act fixpoint freely on the set of points of S on M, without x. Hence there is at least one nontrivial linear element 8 G TVtZ fixing some point of S on M distinct from x. Since the lines xy, xz and M span £x, and since the order p of 8 divides q, this implies that 8 fixes all points of £x, and so 6 has some center c in PG(4, q). So 8 is a symmetry with center x, while by the dual of Theorem 3.3.2 a TGQ of order (s, t) can only have symmetries about a point if st(t + 1) = 0 mod s + t, and this, together with Theorem 3.8.1(ii), implies s = t. Next we claim that all lines of <S are regular. Indeed, consider two arbitrary opposite lines L, M of S and consider the subGQ induced by (L, M), which has order (s,t"), for some positive integer t". Assume first that t" > 1. Then t" > s and t > st", implying t2 > s3. Clearly s ^ t, so t = ps and, as

23.5. Non-Planarly Embedded Translation Generalized Quadrangles

315

t ^ s2, we have p2 = s and p3 = t, contradicting Theorem 3.8.1(ii). Hence t" = 1 and the pair {L, M} is regular. Our claim is proved. Corollary 3.8.3 now implies that S = Q(4, s), since we already proved t^s2. Lemma 13.1.10 concludes the proof of the theorem. • Finally, in finite 5-dimensional space, all embeddings are translation-homogeneous, as follows from combining Lemmas 13.1.10 and 13.1.11. Theorem 13.5.6 If the generalized quadrangle S of order (s,t) is laxly embedded in PG(5,q), then S = Q(5,s). If gcd(s,q) = 1, then s = 2 and the embedding is as described in Example 13.3.4. If gcd(s,q) ^ 1, then the embedding arises from the standard embedding in PG(5, s) by including the latter in PG(5,q). • We now have the following corollary. Corollary 13.5.7 If the translation Moufang generalized quadrangle S of order (s, t) is embedded in PG(d, q), d > 2, and all symmetries are induced by collineations ofPG(d, q), then either the embedding is full in a subspace over GF(s) (in which case this full embedding is the appropriate natural one), or one of the following holds. (i) d = 2, S = W(2) and the embedding is the hyperoval embedding in some subplane PG(2,4) (and q = 4 corresponds with the only faithful case). (ii) d — 4, S = W(2) and the embedding is the universal embedding in PG(4, q), as described in Example 13.3.3. It is automatically special. (iii) d = 4, S = Q(4,3) and the embedding is the universal embedding in PG(4, q), as described in Example 13.3.3. It is automatically special. (iv) d = 5, S = Q(5,2) and the embedding is the universal one in PG(5, q), with q odd, as described in Example 13.3.4. It is automatically special.

Appendix A

Open Problems The problems mentioned here are mainly concerned with translation generalized quadrangles, but several other interesting open problems will also be stated.

A.1 Chapter 1 Problem A.1.1 Are Q(4,g) and W(q) the only GQs of order q, q odd and q> 5? Problem A. 1.2 Is there just one GQ of order (q - 1, q + I), for any odd q, q>5? Problem A. 1.3 Is H(4, q2) the only GQ of order (q2, q3), q > 2? Problem A. 1.4 7s there a unique GQ of order (4, t), t e {6,8,16}? Problem A.1.5 Does there exist a GQ of order (4, t), t e {11,12}? Problem A. 1.6 Does there exist a GQ of order 6? Problem A.1.7 Classify all ovoids of Q(4, q). All known ovoids of Q(4, q) are listed in Section 7.4. Problem A.1.8 Classify all spreads of Q(5, q). 317

318

Appendix A. Open Problems

Many spreads of Q(5, q) are known, see e.g. Payne and Thas [128], Thas [170], and Thas and Payne [190]. Problem A.1.9 Does H(4, q2), q > 2, have a spread? By Brouwer [20], H(4,4) has no spread.

A. 2 Chapter 2 Problem A.2.1 Does there exist a GQ of order (s,t), with s < t < s2, for which all lines are regular? Problem A.2.2 Is every GQ of order (s,t), 1 < s < t, with all lines regular, isomorphic to Q(5, s)? Thas and Van Maldeghem [195] proved that all lines of a GQ S of order (s, s + 2), s > 2, are regular if and only if s = 2 and S = Q(5,2). Problem A.2.3 Does there exist a GQ of order s, s / 1, with regular point x, for which the corresponding projective plane nx is not Desarguesian? Problem A.2.4 Is every GQ of order s, s odd and s > 1, for which each point is antiregular, isomorphic to Q(4, s)? Problem A.2.5 Give a purely geometrical proof of the theorem of Bagchi, Brouwer and Wilbrink [6] on antiregular points in a GQ of order s, s > 1 and odd; see Theorem 2.4.6.

A.3 Chapter 3 Problem A.3.1 Does there exist a TGQ of order {qa,qa+1),

q odd and a > 1?

Problem A.3.2 Is every TGQ of order a, q •£ I, isomorphic to a T 2 (C) of Tits, that is, is any generalized oval 0(n, n, q) regular? Problem A.3.3 Is 0(n, n, q), q odd, always isomorphic to its translation dual 0*{n,n,q)?

A3.

Chapter 3

319

Problem A.3.4 What can be said about 0(n, n, q) ifq is even and the (n — 1)spread T, respectively %, defined in Section 3.9, is regular? In the odd case it follows that 0(n, n, q) is classical, see Theorem 3.9.6. Problem A.3.5 Is every TGQ of order (q,q2), g ^ 1, with all lines regular isomorphic to Q(5, q)? Problem A.3.6 Does there exist a TGQ of order (q, q2), with q even, not isomorphic to a T 3 ( 0 ) of Tits, that is, is any generalized ovoid 0(n, 2n, q) with q even, regular? Problem A.3.7 Is 0(n, 2n, q), with q even, always isomorphic to its translation dual? Problem A.3.8 A weak generalized ovoid is a set of q2n + 1 (n - 1)dimensional subspaces of P G ( 4 n - l,q), every three of which generate a (3n - l)-dimensional space. Is every weak generalized ovoid a generalized ovoid? Problem A.3.9 Classify all thick TGQs of order {q2,q% where GF(g) is the kernel of the TGQ. In [97], the authors classify all TGQs of order (4,16) (they are all isomorphic to Q (5,4)). Problem A.3.10 Classify all EGQs with an axis of symmetry incident with the elation point. Problem A.3.11 Let («S(oo),G) be a thick EGQ. Is G always a p-group for some prime? By work of Frohardt [62], the next problem is equivalent. Problem A.3.12 Let (S^°°\G) be a thick EGQ. Is G always nilpotent? Problem A.3.13 Classify EGQs 6S ( o o ) , G) of order (s,p), with p a prime. EGQs of order (p, t) with p a prime were classified by Bloemen, Thas and Van Maldeghem; see [13].

320

Appendix A. Open Problems

A.4 Chapter 4 Problem A.4.1 Does there exist a semifieldflockof the quadratic cone K, of PG(3, q), q odd and q ^ 3h, which is not a Kantor-Knuth flock? Problem A.4.2 Does there exist a GQ of order (q2, q), q =fi 1, which is neither the point-line dual of a TGQ nor of a flock GQ? Problem A.4.3 Classify all herds ofhyperovals ofPG(2,q), q even, arising from flocks of the quadratic cone in PG(3, q). Interesting results on this topic can be found in O' Keefe and Penttila [91]. Problem A.4.4 Is any GQ of order {q,q2), q even, which satisfies Property (G) at a pair {x, y}, with x and y distinct collinear points, isomorphic either to a T 3 (C) of Tits or to the point-line dual of a flock GQ? In the odd case the GQ is always the point-line dual of a flock GQ, see Theorem 4.10.7.

A. 5 Chapter 5 Problem A.5.1 Is one of 0(n, 2n, q) and O* (n, 2n, q) always good? Problem A.5.2 7s 0(n, 2n, q), with q even, always good? Problem A.5.3 Is 0(n,2n,q), with 0(n,2n,q) good and q even, either the point-line dual of a flock GQ and hence classical by Theorem 4.5.1, or isomorphic to a T 3 (0') of Tits? Problem A.5.4 IfO(n, 2n, q), q odd, is not a Kantor-Knuth egg, is necessarily q = 3h? Problem A.5.5 If a good 0(n, 2n, q), q odd, is not a Kantor-Knuth egg, is necessarily q = 3h? This is equivalent to Problem A.4.1, by Corollary 5.1.4 and Section 4.7. Problem A.5.6 Is an egg 0(n, 2n, q), q even, with at least two good elements regular?

A.6. Chapter 6

321

A. 6 Chapter 6 Problem A.6.1 Let x be a regular point of the GQ S of order (s, t), s >t > 1, and let N® be the corresponding dual net. Under which conditions do particular automorphisms of'A/jf extend to automorphisms of S?

A. 7 Chapter 7 Problem A.7.1 If O is an ovoid of Q(4,g), q odd, which is not a KantorKnuth ovoid, is necessarily q = 3h? Problem A.7.2 Let S be a GQ of order (s2, ss), having a subGQ S' isomorphic to H(3, s2). Assume that in S' each ovoid Ox subtended by a point x of S\S' is a Hermitian curve. Prove that S = H(4, s2). Note that Theorem 7.5.4 is the corresponding theorem for quadrics in even characteristic; see Remark 7.5.5 when the characteristic is odd. Problem A.7.3 Let Sbea TGQ of order (s, s2), s even, with O = 0(n, 2n, q), and assume that S' is a subGQ of order s containing the point (oo). 7s <S necessarily isomorphic to a T3(0') of Tits? Problem A.7.4 Prove Proposition 7.9.2 without using Theorem 3.3.1. Problem A.7.5 Let S = T(0), with O = 0(n, 2n, q) and q even, be a TGQ such that O is good at its element ir and assume that S' is a subGQ of order s = qn ofS which contains the point (oo). By Theorem 7.7.10, S' = T(0'), with O' = 0'(n, n, q) a pseudo-oval on O. Does O' necessarily contain ir? Problem A.7.6 Work under the same hypotheses as in the previous question. Is O necessarily regular, that is,isS = T 3 (C) with O some ovoid o/PG(3, s)? Problem A.7.7 Classify all GQs of order (s, s2) with a thick subGQ S' of order s fixed pointwise by an involution. If s is even, then S' = Q(4, s), and if s is odd, then each point of the subGQ is antiregular, see [182].

322

Appendix A. Open Problems

A.8 Chapter 8 Problem A.8.1 Is any translation generalized oval regular? Problem A.8.2 Prove, without using the classification of finite simple groups, that Theorem 8.5.1 also holds for q odd. Problem A.8.3 Generalize Theorem 8.5.1 by means of other group actions. Problem A.8.4 Classify generalized ovals admitting a nontrivial involution. Problem A.8.5 Classify generalized ovoids admitting a nontrivial involution. Problem A.8.6 Suppose O is a generalized oval in P G ( 3 n -l,q), let ix € O, and let r be the tangent space of O at -K. Suppose 9 is the set of involutions ofPG~L3n(q) that fix r pointwise and stabilize O. How big should | 6 | be at least to conclude that O is a translation generalized oval? Problem A.8.7 Same question for ovals.

A.9 Chapter 9 Problem A.9.1 Are the root groups of any proper Moufang set necessarily nilpotent? Problem A.9.2 Is there an absolute constant N e N such that, if the root groups of a proper Moufang set are nilpotent, then the nilpotency class is at most N? Problem A.9.3 Can one classify all proper translation Moufang sets? Problem A.9.4 Are the root groups uniquely and unambiguously determined by the permutation group defined by the (proper) Moufang set?

A. 10 Chapter 10 Problem A.10.1 Classify all span-symmetric generalized quadrangles of order (s, s2), s ^ 1.

All.

Chapter 11

323

Problem A.10.2 Classify all span-symmetric generalized quadrangles of order (s, s2), s =£ I, with a regular line not contained in the base-grid. One may want to consult Chapter 7 of the monograph [206] for the known results on both problems. Problem A. 10.3 Classify EGQs with a line of elation points. When the elation group is abelian (that is, the EGQ is a TGQ), this problem has been solved by K. Thas — see Chapter 10.

A. 11

Chapter 11

Problem A. 11.1 Prove, without the classification offinitesimple groups, that a finite GQ admitting a group acting transitively on pairs of opposite points and on pairs of opposite lines, is classical or dual classical. A major step towards the solution of this problem is contained in the papers [209, 210, 211], where the author classifies the finite Tits GQs. Problem A. 11.2 Prove, without finiteness condition, that half 2-Moufang GQs are necessarily Moufang (or provide a counter example). Problem A. 11.3 Is every GQ all points and lines of which are regular, necessarily a Moufang GQ?

A. 12

Chapter 12

Problem A.12.1 If a not necessarily finite EGQ is self-polar (or self-dual), then are the elation groups unique? For the finite case, the answer is given in Proposition 12.2.1. Problem A.12.2 Is every polarity in an arbitrary or finite GQ uniquely determined by the set of absolute points? Problem A. 12.3 Do there exist a GQ S, a proper full subGQ S', and two points of S not in S' subtending nonisomorphic ovoids of S' ?

324

Appendix A. Open Problems

A. 13 Chapter 13 Problem A. 13.1 Can one classify all linear translation-homogeneously embedded TGQs S^ in Pappian projective spaces? Problem A.13.2 Can one classify all lax embeddings of W(2) in Pappian or Desarguesian projective planes? Problem A. 13.3 Can one classify translation-homogeneously embedded TGQs S^L\ with L a line, in finite projective spaces? Problem A. 13.4 Can one classify all (linear) "elation-homogeneously" embedded EGQs in finite projective spaces? Problem A. 13.5 Can one classify all (lax) embeddings of the symplectic quadrangle W(s) in the ^-dimensional projective space PG(3, q), q > s? Problem A. 13.6 Can one classify all lax embeddings ofGQs infinite projective planes?

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New

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79

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Index (B,N), 232 (G,B,N), 230 (G,X),28 (G,F),38 (X, (Ux)xex), 181 (r,E),224 (oo), 39 (J, J*), 31

[£], 219 [<7,/i],27 [n - l]-oval, 39 [n - 1]-ovoid, 39 N , 219 3 D 4 (K), xv r = ( f i , £ u £ - L , l ' ) , 200 K, 37 K 3 , 192 PG(1,K), 188 PG(d,?),4 P G w ( n + m - l , g ) , 39 PG(i)(n-l,g),39 PSL 2 (p"), 187

(«S(P),G), 29

^ = S , 87 A-1, 2 D - T , 88 G',28 G°PP, 184 G-1, 203 Gx,27 H xi T, 59 H(x,L,p), 99 L~M,2 L(i), 89 M3(JFsr), xvii W(T), 89 JVG(G,-), 201 T*,89 Ux, 181 Z(G), 28 [0]-oval, 39 [0]-ovoid, 39 [A,B},27 [G, G](„), 28 [G, G][„j, 28

PSU3(F,CT),

190

PSU 3 (p"/ 3 ), 187 PSU 3 (?), 190 R(3 n '), 187 R(K,6>), 194 R(?), 194 Sym(X), 184 Sz(2™'), 187 Sz(K,L,6>), 191 Sz(g), 6, 191 o, 72 S(F, F'), 229 Sti), 97 &, 121 £(w), 232 <8, 201 A ( 0 ) , 176 337

Index

338

AS(g), 285 H(d,q2),5 H«,15 Q(d,g),4 T(O), 39 T(C*), 50, 55 T(n,m,q), 39 T*(0), 50, 55 T*(n,m,g), 55 T2(0),7 T2(0), 7 T3(0), 7 W(«), 5 id, 27 C ~ C, 88 Co, 65 Ci,65 C„, 70 C2n, 70 J7, 25 .F(C), 88 f o , 148 0(C), 87 Hxy, 82 J , 24 J(O), 25 _LX,

5 ' ( r ) , 211 S(G, J), 31, 72 5(.F), 73 5(£) = ( P ) B ) I ) , 2 0 <S
= {V,B,

I),

232

?

<S<8 =(7 gj,B<8,I«8), 202 <->iy

£J

i d , 146, 149 K2, 146, 149 ^(ai,a2).

M(H(2,q2)), 190 M2, 258 A4 3 , 258 JV, 14 A/"*, 15 0 , 5 , 39 0(6, c), 122 0(n,m,q), 39 O*, 50, 55 0*(n,m,q), 55 0 2 , 146 0 s r ( 2 n ' ) , 187 Or, 148 P(<S,z), 207 P W , 149 VL, 150 S = (7>,B,I),1

8 5

£,200 £(£>), 19 £(5) = (V, B' = B[ U B'2,I'), 20 £ = (P,iBiUB2,I)J19 7W(PG(1,K)), 188 M(R(K,6>)), 193, 194 .M(Sz(K,L,6>)), 190 7v4(H(2,F,<7)), 190

=

\'xyi

&xy-> *-xy)j

TX, 149 TP, 149 71173 W H ) 197 W^(p"/3), 187 UR, 196 Wfl(3"'), 187 1,27 i, 300 11,63 1,63 gcd, 286 tr, 122 n(x,y), 18

oJ-

Index

339

TTi, 39

pro]Lx, 1 proj^L, 1 PL,M>

2 1 9

n, 39 /*(A,A'), 123 gh,27 gtW, 121 St* (A), 122 SL,231

s*, 231 zlL, 1 zli, 2 x ~y, 2 x-1, 2 (BN1), 230 (BN2), 230 (BN3), 230 (BN4), 230 (MSI), 181 (MS2), 181 absolute line, 264 point, 264 additive g™-clan, 74 ambient projective space, 285 anti-automorphism, 280 antiregular, 16 apartment, 225 k-arc, 5 automorphism, 4 group, 4 axial root elation, 226 Axiom of Veblen, 15 axis, 226 of6>, 169 of a translation generalized oval, 170

of a translation generalized ovoid, 170 of a transvection, 139 of symmetry, 33 of transitivity, 227 base-grid, 200 base-group of a span-symmetric generalized quadrangle, 200 of a translation generalized quadrangle, 34 of an elation generalized quadrangle, 29 base-line of an elation generalized quadrangle, 29 base-point of a translation generalized quadrangle, 34 of an elation generalized quadrangle, 29 base-span, 200 BLT-set, 77 Blueprint, 217 BN-pair, 229 in G, 232 of Type B 2 , 232 Bruhat decomposition, 232, 235 building, xx k-cap, 5 carrier, 61 center, 2, 226 of6>, 169 of a group, 28 of an elation, 29 of an elation generalized quadrangle, 29 of symmetry, 33

Index

340

of transitivity, 163, 227 nth central derivative, 28 central root elation, 226 Chevalley group, xv type group, 188 circle set, 19 <7-clan, 73 classical egg, 44 generalized quadrangle, 4, 259, 283 inversive plane, 25 Laguerre plane, 20 ovoid, 145 pseudo-oval, 43 collineation, 4 group, 4 commutator, 27 subgroup, 27 companion field automorphism, 279 Condition (Kl), 31 Condition (K2), 31 conic plane, 117 conjugate, 27 correlation, 280 cover, 150 covering map, 150 degree, 14 derivation, 69 derived flock, 69 plane, 19, 25, 104 dual BLT-set, 77 classical generalized quadrangle, 259 (t + 1) x (t + l)-grid, 4 hyperoval, 280

panel, 225 root, 225 elation, 225 egg, 39 egglike, 25 EGQ, 29 elation about a line, 29 about a point, 29 generalized quadrangle, 29 group, 29 of an elation ovoid, 262 ofPG(2,g),291 ovoid with respect to a flag, 262 with respect to a point, 262 point, 29 elementary pseudo-oval, 42 pseudo-ovoid, 42 equivalent g-clans, 88 conies, 63 embeddings, 296 exceptional flock, 63 extreme full subquadrangle, 275 faithful translation-homogeneous embedding, 290 flag, 225 flock, 25, 61, 64, 65 generalized quadrangle, 73 m-fold cover, 150 Fong-Seitz Condition, 228 generalized polygon, 228 generalized quadrangle, 228 Property, 228

Index

action, 183 4-gonal closure, 281 Frobenius group, 28 kernel, 28 Frobenius' Theorem, 28 full automorphism group, 4 collineation group, 4 embedding, 280 line (of an embedding), 280 projective group of a Moufang set, 182 subquadrangle, 220 Fundamental Theorem of g-Clan Geometry, 87 Ganley flock, 72 generalized quadrangle, 76 generalized distance, 229 hexagon, xv hyperoval embedding with parameters (£i,e2), 290 linear representation, 46 octagon, xv oval, 39 cone, 58 ovoid, 39 cone, 58 pointed conic, 167 polygon, xv quadrangle, 1 4-gonal basis, 201 family of type (s, t), 31 good, 101 at a line, 101 egg, 101

341

element, 101 line, 101 translation generalized quadrangle, 101 grid, 4 (si + 1) x (s2 + l)-grid, 4 grid-symmetric generalized quadrangle, 209 group of Lie-type, xv of mixed type, xv with a BN-pair, 229 half 2-Moufang Condition, 228 generalized quadrangle, 227 3-Moufang generalized quadrangle, 227 Moufang generalized quadrangle, 226 herd of (k + 2)-arcs, 86 of hyperovals, 86 Hermitian Moufang set, 190 unital, 187, 197 hyperoval, 5 embedding, 280 ideal subquadrangle, 220 improper Moufang set, 182 incidence relation, 1 induced Moufang set, 277 inequalities of Higman, 2 interior panel, 225 internal plane, 19, 25, 104 inversive plane, 24 Kantor-Knuth

342

flock, 69 generalized quadrangle, 76 ovoid, 146 Kantor-Ree-Tits ovoid, 146 kernel, 5, 37 ofV|, 118 of a generalized oval, 49 of a translation generalized quadrangle, 37 Knuth semifield plane, 146 Laguerre plane, 19 lax embedding, 280 length, 232 Lenz-Barlotti Class III.2, 205 classification, 205 line, 1 linear, 25 collineation, 279 embedding, 289 flock, 61, 62, 64 little projective group, 279 of a Moufang quadrangle, 226 of a Moufang set, 182 Liineburg plane, 146 Main Axiom, 221 Maschke's Theorem, 206 Miquelian, 25 mixed building of Type F 4 , xxii Mobius plane, 24 Moufang flag, 226 generalized quadrangle, 219, 226 panel, 225 projective plane, xxi set, 181 triple, 185

Index

1-Moufang generalized gle, 227 2-Moufang generalized gle, 227 3-Moufang generalized gle, 227 4-Moufang generalized gle, 227

quadranquadranquadranquadran-

natural Tits system, 232 net, 14 T-net, 139 nilpotent group, 28 nonstandard example, 188 nth normal derivative, 28 normalized g-clan, 87 nucleus, 5, 49, 172 ofVf, 118 opposite flags, 227 points and lines, 219 order, 2 of a Laguerre plane, 19 of a net, 14 of an inversive plane, 25 ovals, 5 ovoid, 3, 5, 6, 62, 145 ovoide a translation, 172 panel, 225 parallel class, 139 parameters, 2 partial flock, 86 spread, 200 Payne derivative, 208 Penttila-Williams ovoid, 146 Penttila-Williams-Bader-LunardonPinneri

Index

flock, 72 generalized quadrangle, 11 perfect group, 28 permutation group, 27 perp, 2 planar embedding, 288 point, 1 point-line duality, 2 pointed generalized conic, 167 polarity, 264 polarized embedding, 280 at a point, 280 projection, 1 projective group of a Moufang set, 182 translation generalized oval, 173 proper Moufang set, 182 Property (*), 172 Property (G), 81 at a flag, 81 at a line, 81 at a pair, 81 pseudo-conic, 43 pseudo-oval, 39 pseudo-ovoid, 39, 43 pseudo-quadric, 43 quadratic Galois extension, 189 quadric Veronesean, 116 Question (1), 92 Question (2'), 93 Question (2), 92 Question (3'), 96 Question (3), 96 Question (4), 98 rank 1 group, 185 Ree group, 194 Ree-Tits

343

Moufang set, 193, 194 ovoid, 146 unital, 187, 197 regular pair, 11 point, 11 pseudo-oval, 42 pseudo-ovoid, 42 3-regular point, 22 triad, 22 representative, 231 Roman generalized quadrangle, 11 root, 225 elation in a Moufang set, 181 of a generalized quadrangle, 225 group in a Moufang set, 181 saturated Tits system, 236 semiclassical ovoid, 261 semifield flock, 71 semilinear collineation, 279 semisimple algebraic group, xv skew translation generalized quadrangle, 96 soluble group, 28 solvable group, 28 span, 2 span-symmetric generalized quadrangle, 200 special embedding, 289 linear collineation, 279 SPGQ, 200 spherical building, xx split BN-pair of rank 1, 187

344

of rank 2, 219 Tits system, 236 spread, 3 of symmetry, 207 standard elation, 96 generators, 232 Steinberg representation, xxi STGQ, 96 sub Moufang set, 181 subGQ, 2 subquadrangle, 2 subtended ovoid, 146 Suzuki group, 6, 191 Suzuki-Tits Moufang set, 190, 191 ovoid, 6, 187 Sylow p-subgroup, 28 symmetry about a line, 33 about a point, 33 tangent circle, 19, 25 line, 5 plane, 6 plane of V|, 117 space of a 4-gonal family, 31 of a generalized oval, 39 of a generalized ovoid, 39 of a translation ovoid, 172 TGQ, 34 Thas flock, 63 Thas-Payne ovoid, 147 thick, 4 thin, 4 Tits automorphism, 271 building, xv

Index

Condition, 227, 228 endomorphism, 190, 192 generalized quadrangle, 227 ovoid, 6, 145 system, 227, 229, 232 trace, 2 2-transitive generalized oval, 177 (x, L)-transitive projective plane, 205 translation, 34 dual, 50, 55 generalized oval, 169, 170 ovoid, 169, 170 quadrangle, 34 group, 34 of a translation ovoid, 262 line, 33 Moufang set, 181 oval, 170 ovoid, 146, 170, 172 with respect to a flag, 146 with respect to a point, 147 point, 33 of a projective plane, 139 translation-homogeneous embedding, 289 transvection, 139 unital, 196 universal central extension, 205 embedding, 298, 300, 301 ofQ(4,3),298 ofW(2), 297 Veronese variety, 116 Veronesean quadric, 116 vertex of a generalized oval cone, 58

Index

of a generalized ovoid cone, 58 weak generalized ovoid, 319 Weyl group, 231, 232 whorl about a line, 29 about a point, 29

Series in Pure

ics-Vol. 26

Translation Generalized Quadrangles The three basic works on generalized quadrangles are (co-)authored by the writers of this book: "Finite Generalized Quadrangles" (1984) by S E Payne and ) A Thas, "Generalized Polygons" (1998) by H Van Maldeghem, and "Symmetry in Finite Generalized Quadrangles" (2004) by K Thas. Central in the present work are translation generalized quadrangles, but also many of the most important results in the general theory of generalized quadrangles that appeared since 1984 are treated. Translation generalized quadrangles play a key role in the theory of generalized quadrangles, comparable to the role of translation planes in the theory of projective and affine planes. The notion of translation generalized quadrangle is a local analogue of the more global "Moufang Condition", a topic of great interest, also due to the classification of all Moufang polygons, which was completed by ) Tits and R M Weiss in 1997. Attention is thus paid to recent results in that direction, in many cases not restricted by finiteness assumptions. Translation Generalized Quadrangles is essentially self-contained, as the reader is only expected to be familiar with some basic facts contained in the monograph "Finite Generalized Quadrangles" by S E Payne and I A Thas. The authors however repeat the most frequently used results from that source. Also, some proofs w h i c h are either too long or too technical are left out, or just sketched. Further, there is a list with 71 open problems.

/orld Scientific