*. 5icci (Ed.)
7eoria dei numeri Lectures given at the Centro Internazionale Matematico Estivo (C.I.M.E.), held in Varenna (Como), Italy, $ugust ,
C.I.M.E. Foundation c/o Dipartimento di Matematica “U. Dini” Viale Morgagni n. 67/a 50134 Firenze Italy
[email protected]
ISBN 978-3-642-10891-4 e-ISBN: 978-3-642-10892-1 DOI:10.1007/978-3-642-10892-1 Springer Heidelberg Dordrecht London New York
©Springer-Verlag Berlin Heidelberg 2011 st Reprint of the 1 ed. C.I.M.E., Florence, 1955 With kind permission of C.I.M.E.
Printed on acid-free paper
Springer.com
CENTRO INTERNATIONALE MATEMATICO ESTIVO (C.I.M.E)
2° Ciclo - Varenna, Villa Monastero – 16-25 Agosto, 1955
TEORIA DEI NUMERI
H. Davenport:
Problemes d’empilement et de decouvrement ......................
1
L. J. Mordell:
Equazioni diofantee .............................................................. 45
P. Erdös:
Some problems on the distribution of prime numbers ......... 79
G. Ricci:
Sul reticolo dei punti aventi per coordinate i numeri primi ....................................................................... 89
C. A. Rogers:
The geometry of numbers ..................................................... 97
H.D A V E H P 0 R T
PROBLEMES DIEMPILEMENT
ET
DE DECOUVREMENT
ROMA-Istituto Matematico dell'Universita,1955-ROMA
1
H.Davenport
PROBLEMES D'EM'PILEMENTET DE DECOUVREMENT
Le sujet que je me propose de traiter dans ces conferencea of fre beaucoup de problemes simples et interessants, qui pour l a plupart attendent toujours une solution.
Jusqu'a present
on n'a pas trouve beaucoup de methodes applicables a ces mes, et
a mon
probl~
avis, un ricne terrain attend encore la decouv erte
d 'idees nouvelles.
1.
Definitions.
Les concepts d'empilement Ie plus compact et de recouvrelQCnt les moins compact sont tres generaux, et nos defini t ions seront done d'une portee plus large que nous n'est necess aire plus tard. Le sujet se rapporte
a 1 'espace
numerique reel
a.
n dimc))wions
et nous nous servirons de la notation suivante. N:ous desi gncronS' un point quclconque de l'espace par
et nous definissons
A~ = ~ +I
( A un nombre
(Ax1 ,···, AXn)
= (X 1+Y1'···'Xn+Yn)·
Pour chaque ensemble E de points, nOlus noterons ble de tous les points rci ~ designe
ULn
A~,
~
+
AE l' ensem-
ou ~ est un point quelconque de E.
nmmbre reel
j
soit positif, soit negatif.
chuque point E de lies pace, notons points
re el),
E+E
Pour
llensemble de tous les
E, ou 2! est un point quelconque de E.
Soit S un ensemble borne de points, mesurable au sens de 3
H.Davenport
- 2 -
Lebesgue et
a mesure m(S)
> O.
Suit C Ie cube
1 2
Soient
~1'.'"
des points tels que les ensembles
~k
S + ~1'.'"
S +~
soient tous disjoints et soient tous contenus dans Ie cube ou
A est
A c~
un grand nombre positif,
Nous disons que ces ensembles sont empiles dans le cubeAC.
La mesure totale des ensembles est k m(S), et Is mesure (c'est-a.:-
A. C
dire Ie volume) de
I
ainsi k m(S) 'Am.
A'rI.
est
La densi te de 1 I empilement est
Si ko designe la plus grande valeur possi blede
k, nous definissons la densite de l'empilement Ie plus compact de S dans
A C comme
suit:
(1)
<S( 5; A0) ~
On a 0
1•
On demontre facilement que la limite
existe, et nous appelons l~us
compact pour S, ou Ie coefficient dlempilement de S.
Gonstatons dlabord deux proprietes simples de ~ (5; AC), con5idere comme fonction de diminuer
a mesure
que
i\. .
Primo, Ie nombre ko ne peut pas
A augmen-te,
que
4
et SiA 1 .( A2
il slensuit.
H.:!Davepport
- 3 -
Secundo, on a
pour tout entier positif m.
Oar on peut regarder le cube mAC
comme une somme de mn cubes, dont chacun est congru a
A0;
et si
on repete dans chacun de ces cubes l'empilement Ie plus compact pour Ie cube).. 0, on obtient pour mAO un empilement possib1 8 la meme densi te qu' auparavant. Ces deux proprietes montrent
S( 5 ; AC)
est une fonction presque monotone de
s tence de la limite s'ensuit facilement.
I~emarquons
que nous aurions obtenu la
A , et
a
q'J.0
I' exi-
On a, bien entendu,
m~me
valeur pour
~~ 5 )
si nous avions suppose Ie cube 0 ferme au lieu d'ouvert. La definition de
S~ ( 5 )
que nous venons de dOll..'1er est un
peu speciale, en tant qu"'elle emplo:fJe un systeme particulier de coordorillees et une espece particuliere de corps (c'est-a-dire eli cube). Ces limitations se laissent facilement lever.
Premie~ement,
il est evident qu1une translation du syateme de coordonaees n'a 8ucun effete Deuxiemement, on peut remplacer Ie cube C par tout ensemble J qui est quarrable,
mesurable au sens de Jordan.
c'est-a-dire~
Dans ce cas-la, nous definissons SCS i que ~ (j ,\ C) mais nous remp1a@ons
.•
>\
A J)
de 1a
par
A"'m(J). Il es t
m~me
fa gon
f acile de demonttrer que
Car, puisque Jest mesurable au sens de Jordan, il est contenu 5
H.Davenport
- 4 -
dans la reunion d1un eftJemble fini de cubes non-empietants
C1 "" ,C r , et contient lui-m~me un ensemble fini de cubes non-empietants c~, ... ,CS' tels que et
et
pui ssent se faire arbitrairement petits. Avec une notation evidente, nous avons
at
. terme 0 ( (\ \ n-1) est introdui t afin de tenir compte du nom'l)rc ou, Ie
+~,~ qui sont contenus dans }l J maQs non entierement contenus dans un de AC 1 , ••• , .~ Cr , En faisant tendre A des corps
vers
CX)
S
nous obtenons
~( C/1 \ . ~
0,' -t('0') (
rm
(C 15 )
J)
ffll(C) + ... +
W\
(Ct)
tm (3)
ce qui donne (4).
La relation (4) montre en pa:cticulier que jnvariant affine de S, c'est·,· 3.-dire que· $"(S1)::
X
S(S)
b~S2)
est si S...I
S(:
transforme en 32 au moyen d'une trasformation lineaire de l'e s pac(: a determinant non-nul, Car, dans une telle trasformation, 1:.1 mesure de J et :La mesure de D se multiplli:ant tous les deux pcr 6
H.Davenport
- 5 -
le mame nombre,
a savoir
le determinant de la transformation.
Nous passons maintenant
a la
definition dtun autre nombra:
la densite du recouvrement Ie moins compact pour S, ou Ie coeffi-
c~ont
de
r~couvrement
de
S,
que nous designerons par
~ (S). s
nnte toujours un ensemble do points Vlorne et mesurable, mais nors allons supposer maintenant que S ai t au moins un point interic)ul', afi11 d'assurer qib.e la densite soit toujours finie. Soient des points tels que la reunion des ensembles
3,1" , ,,3,1
S + 3,1 , ••• , S + 3,1
contienno tout 10 cube vrent
A. C.
AC.
Nous disons que ces ensembles reCO l1-
Soi t 10 la moindre valeur possible de 1.
s ons la densite du recouvrement Ie moins compact de
Nous definis-
AC par S com-
me suit:
On demontre dans difficulte que
~
cxistc, Ul1
L'analoguc de (4) ticmt, d'ou i l suit que
~ (S)
nst allssi
invariant affine de S.
On a
~(S) ~ -1
(6 )
Les definitions de
S~S)
at de
~
~
(S) peuvent se formuler
[,utroment. Designons par .£1 '].2" •• un ensemble infini de points, tol que les cnsGmbles S + .l?1' S + .l?2""
soient tous clisjoin-Il;r,
::?r()llons la donsi to superioure de la reunion de ces ensembles, e-t11otons
~~ (S)
la borno superioure de cettG dGnsite supericure,pri-
se pour tout ensemblE) :2.1 '.£2" •• permis. On demontre aisemont qJ,le
7
H.Davenport
- 6 -
cGd;lJe dGfini tion equi vaut
m~me
panr
~(s).
a la
ae
dofini tion donnee plus tard. E-t
2. Empilements et rocouvrements reguliers. On appelle resea~'ensemble de points qu'on obtient de l'ensemble de tous les pmints a. coordonnees entieres au moyen d 'un.e transformation lineaire de l' espacG.
Ainsi les points ! du r'€liieaU
s '0 btiennent de n form.e s lineaires
en uonnant aux variables, u 1 , ••• un toutes :J.es valeurs entikre,s . Les coefficients cl .. sont des nombres reels arbi traires, a conlJ eli tion que det
a n-1
dimensions] Dans Ie langage des matrices, les pOints! d'un reseau
s 'obtiennent de
ota A des i gne une matrice reelle non-singulHre et ou !:! est Ie point (vecteur) general a. coordonnees entieres. Une substitution £
= M y., ou
},1
ments entiers
est une matrice unimodulaire (c.-a.-d. matrice d'de-
a determinant
±
1)~
transforme l'ensemble de tous
lc::s pOints entiers :ll en l'ensemble de tous les pOints entiers 11 s'ensuit que Ie reseau reste Ie
m~me
~.
si on remplace la matri ce
A par Ia matrice AM. On.
donne Ie nom do determinant d 'un roseau
a.
la valeur absolue 4e
determinant de A. Nous designons Ie determinant d 'un reseau;\ 8
H.l)avenpOl't
- 7-
d(
1\).
IJe determinant d'un reseau est susceptible d'une t~imple
interpretation geometrique: i l est Ie reciproque de Ie densite du roseau. Cette densite se definit de fagon naturelle
COIDnIG
Ie
quotient limitant du nombre de points d'un reseau dans un grand corps
di~.se
par Ie volume du corps. Or, un corps de volume V
dans l'espace des points.! correspond V/d ( £
1\ )
a un
corps de volume
dans l' espace des points :9:, et comme la densi te des poi1"lts
a coordonnees
entieres est egale
a 1.
il s·ensuit que la densi-
to du resoau /\, dans l' espace des points .! est egale
a 1/d ( A ).
Soit maintenant S un ensemblG donne (borne et mesure,blo avec
m(S)
> 0)
et supposons qu'un reseauA ait la propriete que
tous les ensembles S + j2, ou p parcourt les points de disjoints. Alors S, (1)
a
A
A , f3oi~J1t
donne un empilement regulier pour I' ensemble
densite m(S)/d( /\ ). Nous definissons
s(S) =~«tt ci(A)5) 'm (
et nous appelons
~ (S)
1\
Ia densi te de l' empilement regulier le
plus compact pour S, ou Ie coefficient d'empilement regulier de S. Nous definissons de fagon analogue la densite du recouvrcmont regulier Ie moins compact pour S, ou Ie coefficient de ment reguliar de S, que nous notons
!i}
recouv.~
(S). Il est evident que
puisqne les empilements reguliers forment un sousensemble de t ow, les empilements, et de
3.
m~me
quant aux recouvrements.
Les corps convexes.
Dorenavant, nous allons no us limiter pour la plupart au cas oD. l' ensemble S est un corps convexe sym~trique. Un corps conve~e
9
H.Davenport
- 8 -
s;r.metr1que K est un ensemble de pOints, ouvert 1 ) et borne, tel
t (2 - ~)
que
est un point de K quand 2 et S sont des pointu
En particulier
-K. K. Il est bien connu que tout corps
est quarrable, c'est-a-dire
pos~ede
ae
conve~
un volume au sens clasSique
de Jordan. Nous allons donc employer la notation V(K) au l i eu m(K).
Le probleme de trouver
S
ae
(K) pour un corps convexe f;;ymetrlque
eqtdvaut au probleme de' trouver Ie determinant critique de 1f, voila un des problemes les plus importants de la geometri8 des noro.bres. Car la condition que les corps K + :£ n'empietent pa& (p. ()-Gant un pOint quelconquc de 1\ 10 r eseau
1\
.:l - :£1 et
ainsi Ie point
~
-:.£2
~,
et K + :£2 ont un point commun £,
se trouvent tous les deux dalile K, et
~(P1-P2) se trouve dans K. D'autre part, s'il
0Aiste un point S de .& +
a 1a condition que
n'a.it pa.s d'autre point que l'origine 0 dans le
corps 2K. En affet, 81 K + ~J.lo:L'S
) equiv aut
A autre
que 0 dans 2K, elora les corps ~ es
t
.9. contiennent tous les deux le point
dn.ns Ia definition de
1\ de maniere
a.
~
(K) dans (1) du
A ) minimal,
rendre d(
/\ n I ai t d I autre point que 0 dans 2K.
Sl et empietent. Done,
S2, i l nous faut choisie
toujoura a condition que Le minimum de d ( 1\ ) sous
cette condition est, po.r det1ni tion, 1e determinant critique du corps 2:K, designe par
I:!
(2K). Done
(1)
Le simple fait que iln
S (K)
,
1 nous donne le theoreme c1assiql4e
l.1inkowski, fondamental pou.r la geometrie des nombres:
D. (K) ~
2-n V(K).
---------------------1) Pour les valeurs de
~ U (K) etc.
dans K sa frontiere ou non.
i l n' importe rien si on il'.cllH,
Les formulations deviennent un pea
plus simple s1 l'on inclue la frontiere quand on traite lEls qlle .. stions de recouvrement mais non quand on traite les questions
a eJupilement. I
10
H.Davenpo:c-l;
- 9 -
Dans los raisonnements relatifs aux corps convexes,U -est [OlXvent utile de s e servir de la norme (fonction de distanc e) C!. ' U~1
corps .
lious a.efinissons la norme F(!) d lun corps convexe symetr::' que It CO Jil.ffiE: suit. Si ! est un po int queleonque autre que 0, i l exist/? Lm nCllmbrc posi tif et bien de f ini .LU
tel que
fl' ontiere de K, et nous posons F(!) :::
d.efinition nous me ttons 1<' (0)=0 . pour lesquels F (x) f )?cmtHre).
<1
f
f!
se trouve svr
-1. Pour compl ete r l a
K consiste alors des points ::::
(ou F ( x ), 1 si I 'on comprend dans K s a
On demont re fa cilement que F(x) possede les :proprie~
pour .! 10 ; (ii ) Ji' (! + ;[) (iH)]i' (),2S)
F(!) + F (;~L) ;
/ AI F (!)
pour tout nombre reel
A
Re ci proquement, si :B' <'~) est une fontJtion possedant ce s troi.!' 1'l'opriStes(2: l'inegalite
sy~
F(!)< 1 d6finit un corps conve xe
metrique . Pour;un corps convexe s;ymetrique donne K, et pour l1o M e ,
1\
(2 )
lL'1
resec1\.\
,nous definis sons Ie minimum M de F(.!) pmur 1\ par
/'1 _
tmArrt-
F(
12 t 1\ r-* 0
t'
.L\utrement dit, M est le plus grand nombre tel que le reseau
ii
n ' Hit aucun point sauf 0 dans Ie corps K. Done
(2)
On :leut delffuire de ce trois proprHtes que ~,(!) est un8 fonttin\. CO'lt ·In;1 8.
11
H. Davenport
- 10 -
Suivant (1),
5(K):::
,~
1\
Nous def inissons aussi Ie minimum inhomogfme Mr de F(2E) pour
(5)
1\ en posant
M r
~
1
F ( 12
'Y'VWYI-
f- f 1\
- j) )
OlI Ie maximum est pris pour tous les points
~
de 11 espace.
Aub~e
ment dit, Mr 8st Ie plus petit nombre avec cette propriet e :
pour
chaque point ~ de l' espace il existe un point E de /\ t el que
- %) K. Done ]!' (2.
: : . Mr.
Le r eseau
ir "
fonrni t un recouvrement pour
(6 )
Ceci est simplement une autre f ormulation de la definition de
~(K),
et elle reste valable si K est convexe ou non;
~
cet
egard el18 differe de ( 4). 4. peux i negalites de Rogers. C. A.Rogers(1) et ablit les qui r elient
~(K) a
T HEO~
1.
S(K)
et
inegalite s~enerales sui~t es,
srt
(K)
a ()(K).
Pour t out corps K, convexe et symetrique,
11 s ' ensui t que 1) J .London Uath . Soc . 25 (1 950), 327-331 . 12
~
a
H.Davenport
- 11 -
La demonstration de (1) est fort simple et intuitive. Nous retournons
a la
definitions de
S· (K)
donnee dans Ie
~
1. Suppo-
so ns que :£1' ••• d!.k soient des points de I' espace tels que les corps
soient tous disjoints et soient tous contenus dans un gram1 cul
Ac.
Supposons que k p~enne sa valeur maximale. Si E est un
po int quelconque, et si le corps K + E est contenu dans
AC,
alors 10~ corps K + E empiete necessairement uri des corps indiC{y.ts plus haut. I!G
Gorps K + E est contenu dans
Ao designe
C, oiL
A C si
E est dans Ie cube (,\ -
Ail)
une constante convenablement choisie, par eXf>Pkt-
')le Ie cliametre de K. Si
K+R empiete K + ."ei , E est conte"1U
d~lrS
2K + :£ .• Ainsi les corps ~
2 K + :£1' ••• ' 2K + Ek l' ccouvrent
tout Ie cube
(A- Ao)c. II s'ensuit que
Ceci donne Ie resultat enonce, en faisant
A tendre
vers l'iI1fi.ni.
La demonstration de (2) est un peu plus difficile. Nous :rn'el1on s un reseau compact pour K.
1\
qui fournisse un empilement regulier Ie plus
Nous laissons
a c5te
la demonstration
qu't~1
tel reseau existe, car son existence n' est aucunement esse:1ti elJ.e au raisonnement do 1:r
du
~ 3,
Soit
F(~)
la norme de K.
Par la definiti on
il existe un point z .de l'espace tel que M I
13
H. Davenport
- 12 -
pour tout E de /\ P OT
• Considerons maintenant le point
1a definition de lil I ; i l exi s te un point Ep de
(4 )
F (.f- o - 3Z) -<. _......
Ains i F(}Eo
-~)~lmI'
conv exite de F que
1\
3~.
ToujO"l.tl'cS
tel que
IVl 1
et i l s'ensuit de la propriete do
3
? ui s que tout reseau est symetrique par rapport
a 0,
nous pouvons
ch'mger E en -E, et nous obtenons ainsi
No us considerons mainten ant un reseau nouveau, qui cou,iste. de t ous les points E de 1\ et de tous les points E ± :C'cmar quons que
*
3-
E.o' n ouo
Eo n I est pas un point de!\ , car si Eo =3E., leG
in( 9s1 ites (3) et (4) seraient en contradiction. II est cl air (lue 1e s ysteme de points E, E ± } Eo constitue en effet un
1\ ,
et que
t
cl (1\) ~
at (/\) ,
SoH 1.1 le minimum de F pour!\ , selon
GOrt 8 que i (E) ~ M pour tout E
F(E,) pour tout ].,10 de " .
Le reseau
1\ fut
~
res~~'V..
min
10
)~ definition du ~ 3 9 ik • Alors, par (5),
de,l\
(M'3 MI )
Ainsi 1e minimum !II, de F pour!\ satisiAi'Xe
choisi afin de donner un empilement rEgu.-
l i tH' 10 pl us compact pour K, et ainsi I
14
H.Davenpor1;
- 13 -
sU:lvant
(4) du ~ 3. De plus
$(~ ) ~ y. (\<) M 1~ j
'Y\.
ct{/\,)
En comparant cos relatiotis, on oyti ont
Ceci ii:lplique I' ine gali t e
01', sel on (
6 ) du
~ 3,
'h,
~ (k) ~ V(I<-) M1 . d. (/\)
co q,\li d6montre ( 2 ).
5. Uno minoratio_Il..'p'our Nous passons maintenant
.L
2)\-1
iLH~ gali te
a la
S(K).
demonstration de l'inegaiite
I
Que no us d eduisons du theoreme do !.hnkowski-Hlawka dt>
la ge ometrio des nombres. La plus simple formulation de co theor eme est la suivant e . THEOJlB1IE_.?~
Si S est un ensemble de points, bornDt
Q.ll.Y_~~b_.qu.L possMe un volume V(S) au sens de Jorda&~~_s:i,. V(S)<1
!J-__ '2'iste _un
r es e au de de termi nant
gui n I a pas d I au ~£.£....Jl..oiJl~t;
dans S sauf .JLout -etro O. Admottons ce th eo reme pour Ie moment. Soit K un corps COl1vexo symetrique avec V(K) q~ e lc onquo
< 2.
Coupons K on deux par un plan
passant par O. Alors, 10 th eoremo affirme qu'il
o::i8te un r es eau
a determinant
1 s ans aucun point sauf 0 danS
15
H.Davenport
- 14 -
.6 (K)
ce qui implique
~ 1. Au moyen de raisonnements sur
l'homogeneite et la continuite, cette proposition peut se fo rmuler dans des termes generaux ainsi:
a (1) ci-dessus. Le rusultat 1 )But encore s'ameliorer ) jusqu'a une faeteur constante, mais ~ 3, ceei equivaut
Ti' apres (1) du
rion n' est connu d 'un ordre de grandeUr essentie11ement l1cilJ_eur, Nous passons maintemmt
a 1a
demonstration du theoreme d e
Ui l1kovrski-Hlawka sous la forme enoneee ci-dessus.
II est, bien
cntendu , susceptible de generalisation et modification.
Nous
l a demonstration trouv8e par Rogers, dans la formulation dr; Cassels 2 ). La demonstration se base sur la consideratio'n d'lIn ,:UiV011S
a determinant
sys teme de reseaux
1.
de ([ ' une moyenne, qu'au moins un de -;;6 vonlue.
On demontre, C0S
gr~ce
a I ' ,)tu-
reseaux possede 1a
prop~i!
Nous considerons Ie systeme de reseaux donne pRr les
l:o:_'mes lineaires x = ;\(u 1+ 5.tu ), ... , x .. = A (u 1+ f 1l1.)5 \ 1-n 1"1' n n- I n....) h- n un' lei n-1 sont des parametres qui val'ien-t xn= 1\
I\.
entr'e 0 et 1, et
51" .. , 5
est un autre parametre qui
V8rs l'infini. Soi t N(
~" )1
... ,
f/"--1 ) 1e nombre
a la
fin tendre
de points du reseau danS
L' cmsernble S, sans compter I' origine si 1 'origine se trouve dans
S. 3i ;\ est suffisalllillent grand, il n' y aura pas de tel pointbns S quand un=O. Ainsi, si c c~racte ristique
e (x1 , •.• ,xn)
designe la fO llction
de S, nous avons
=- ~ , ' '2- () (A (u. ( +~ ~h\. {., ~*O -r. ,A(u..",_,+Sn._,w'h.)J A.1_}\.t.Vn.)
N (5,)
"-) ~
h _ \)
I
J. . ,
j
1
1) voir H.DAVENPORT et C.A.ROGERS,l)uke Math.J. 14 (1947),3 67-374.
2) !2'oc. Cambri~e Philos Soc. 49 (1953), 165-166.
16
H. Davenpor~,
- 15 -
pour toute fonction qui evanouit en dehors d1un intervall e fi ni. Apres avoir applique ce principe n-1 fOis, nous obtenons
I
~ ,,~. f' J![A'1,J'''' !l~ '7"
=:
I-I{
L . A
\A.
:;0
n.
_
j'\l() .
".
Je ~ Xl
(
~OO
.)0
J'
.
I\"-~,- ) oL,\"
~ 'Yt ~ I
I
\\- h.. ) ) A !L'I1,
I
A1-n~
~ f-
0'0
I
- f.b
i~Yl
I
. -1
n:
0, la somme d ' i nt egrale
en n-1 dimensions tend vers une integrale en n obtient
dln-'
ol~ ."
I
A~co, et quand donc
Qnand
"'-I )
dimension~,et
on
tJQ
-J (t('1,) " ,(1,,) d't], '''&\'1", -(PO
[1e fait que untO dans 1a somme ne change rien ClerEiere integrale est V(S)
<
1. Donc si
a 1a
A est
limi teJ
La
suffisamment
zrand , il existe au moins un systeme de valeurs de
~ 1"'"
g
n-1
,Jour le quel N( C1
est-El-dire
f , ... ,
)i
~.
)
n-
N( ~I , ••• , ~ n-1 )=0. 17
1)
(1,
Il existe ainsi un resDau dt.!.
- IG -
H.Davenport
sysmeme qui n'a aucun point dans S, sauf
peut-~tre
O.
Ce qui
demontre le theoreme. 11 serait
a propos
de citer un memoire de Siege1 3 ), d3ns
lesuel il definit une mesure dans l'espace de
~
les
rese~ux
et ou i l demontre le theoreme de Minkowski-Hlawka en prenant une moyenne etendue sur la totalite de cet espace. Dans un memoire sous presse, C.A.Rogers a reussi les moyennes de diverse s puissances de N(
a
eval~e~
~I"'" Sn-1)' II
obtient ainsi des resultats meilleurs, quoique 1 'ordre es s en tiel de 1a grandeur reste le
qu'auparavant.
m~me
6.~!ltj..Qre~
de BJj,chJeldt sur
l'empilement de spheres. Il es t evident que l'espace
an
~~K)=
S(K)=1
si K est un cube dans
dimensions. II existe aussi d'autres polyedres
(polyMres qui reniplissent 1 'espace) qui ont la Si K est une sphete (une boule) dans l'espace ('i(
es t evident que 0 (Ei:) det erminer
~(K)j
< 1,
an
pro prietl).
dimensi ons~
il
et clest un probHme interess ant tie
qui dans ce
cas-~' .
que de n. Nous noterons ce nombre obtenu sur ce sujet est dfi
m~me
~
est un nombre qui ne
~epend
'1\.' Le resul tat princi pal
a Blichfeldt 2 )
THEOR:;:;1,1E 3: (1)
3)h-rmals of Math. (2),46 (1945), 340-347. 1)Pour un resume de ce qui est connu a ce sujet, voir Keller, Enzyklop. Math.Wiss. I 2, Heft II, Teil III (1954), ~ 5. 2) Math. Annalen 101 (1929), 605-608. Il putlia le resultat en 1914, mais avec uno limitation aux empilements reguliers, qu 'il a levee dans le memoire Ulterieur.
18
_ 17 _
H.Davenport
Considerons un ernpilement de spheres ayant la densi te voisim~
S , que
1, empilemelttt
nous pouvons supposeI' arbi trairement
'~ d e 'S • L' idee principale de 1a demonstration, c I D"t 'Y\..
que, si l'on elargit
{2,
a rayon
C(~s
spheres de sorte qu'elles aient
ra~Toll
au lieu de 1, alors, quoique les nouvelles spheres empiet
('lIes ne peuvent empieter
de beaucoup.
On peut mesurer leur
empi9tement de diverses fagons. Par exemple, il n'est pas difficile ue demontrer qu'aucun point de rieur
a
l'~space
plus de n+1 des spheres 8largies.
systeme de spheres amene directement
qui nlest pas tout
a rayon V2 a l'inegalite
a fait
ne peut
~tre
inte-
Ainsi la densite uu
ne depasse pas n+1. Ce fait
nous
aussi bonne gue (1) mais I' est presqu.e.
Hous n ' al1ons pas poursuivre cette maniere d'aborder le problchne. :;' nis nous allons exposer ia d(~illonstration de (1) donnee pal' Blichfeldt. Blichfeld t remplace chague 'llhere par une sphere concentrique, distance c I !}st que
l'
a rayon {2,
remplie de matiere
matE~ri elle.
a densi te
a
2_1'2
du centre. Le point essentiel de sa demonstratic:'l,
1:.§.. densi t_~. -1.()_iE11e
c_~_E?ll.!_..?lo~au
de_.1.~~~_tiere
a tout
poini...9:.~_1_!
c;spa-
plus. _
Afin de demontrer cette assertion, considerons un point C:l1.uJ.conque de l! espace et prenons ce point
(sans nuire
a Jl:a
s; eneralite) comme 1'origine O. i30ient ll.1""'~ tous les centres de spheres qui se trouvcnt ~ un.e distance
O1'l
2 de O. Alors la densite totale de matiC;re
\ ll. \ designe In di stance d' un point ll. de O. 19
a
0
...
,~-
0E: qu' il nou8 faut dono demontl"er, c' est que rY'Y'l
'\
L.
(2 )
~ .:: 1 La demonstration se basera , bien entendu, sur le fait que
l.E.i
-
.E. j
I
~ 2
si ilj, ce qui exprime le fait que les spheres
originel1es de rayon 1 n'empictent pas. On a
G0
qui demontre (2). Nous avons demontre l'enonce ci-dessus que 1a densite
tot o.le de 1a matiere
a un
point que1conque de l'espace ne depas-
s e pas 2. Or, Ie systeme de spheres materielles s'obtient du systene originel de spheres, qui avm.tt densi te
S , en rempla~ant
chaque sphere de volume V(K) par une sphere materielle de masse 1 disons, M. La densi te moyenne de la matiere est donc Il s' ensui t que
AS V(k)
n
M S
V(k) ~
2
ne nO U8 rest e qu' a calculer M. Si nous partageons 1 a sphere
materiel1e en de minces concnes, 1imitees par deux spheres COBcentl'iques, nous obtenons 20
H.Davenport
- 19 -
'En substi tuant dans (3) nous avons (1), si nous rap pllons que
S est
arbi trairement voisin de
D~ •
Quelqaas am61iorations Hgeres de (1) furent trouvees par Blichfeldt
lui-m~me
ne changent rien
et plus recemment par Rankin 3 ), mais elles
a l'ordre
de grandeur du resultat.
Les inc Gu -
l i t(SS de Minkowski-Hlawka et de Blichfeldt montrent que
1 2 'YH
11 semble probable que que n -1
00,
(S:
)1/"/1. tend
vel' une limi te
mais jusqu'ici on n'a pas reussi
a Ie
a mes1il.re
demontrer4),
3) Annals of I.lath. (2),48(1947),1062-1081.
4) Un raisOlmemertt de Chubanty [comptes Rendus, 235 (1952),
529-53 2 ]
qui menerai t au resul tat
malheureusement inexact.
21
(S~
)I/~
t,
est
H.Davenport
- 20 -
7. Une majoration pour
~: le
No tons J_'espace
an
0'I'V •
coe fficient de recouvrement d 'une !IPhere dans
dimensions, et ~" "vVle coefficient de recouvrement
re gulicr. Il s 'onsuit des ineealites de Rogers, ajoutees de Blichfeldt, que
~ ~
3 l\ ., 5Y1-
'l<
c:y7f .: :~ V vv ~
<
2'YV J~ <:
(Jt
+
E)
VI.
< ( 2,
a
celle
, , 23 ) 'rt )
(Vi. ... £ )).. <
llour n gr and . La premiere ine gali te est susceptible d I ameliora-
t i on, si l'on adapte Ie raisonnement de Rogers afin de tenir eOl1lptc
du fait, qu'il slagit ici d'une sphere. Uno methode nouvolle pour trouver une maJ'oration de
ilonn80 par Davenport 1 ) en 1952.
~ tnt .y[,
.....
Il demontra que, pour n suffi-
et G.L.Watson a recemment ameliore cette inegalite jusqu'a
~ <. (1.11)n, en so seJ71rant de la m~me methode avec dos details
" "1'1., diff~5ren ts.
Cos resul ta-t;s mon-trent, qu' il est possi bio de recott-
'lI'Ur, par des
spher~s
an
dimcnsiom d 'une fagon bien plus ejiiCa,...
en qu'il n'est possible de los empiler. La demonstration de (1) est un peu compliqu8G et exige de.s eal culatio:1s laborieuscs. Je ne donnerai done que la demonstratiol'l1 'un resultat plus faible 2 ), puisqu'elle illumine Ie princ ipe 1)
R~nd.
Circolo Llatematico di PaleJllno (2), 1(1952), 92-107.
2) Davenport, pomptes Rendu8, 23, (1951), 511-573.
22
- 21 -
H~DavenpGrt
g:£neral.
THEORElI'IE 4.
I
o·
(2)
'~J (~ )~ \ Yv
/\NVYI
'-
6
... 1· 1j3
densite
~
'
!\
Nous r appelons d'abord que tout reseau rocouvrement de l'espace par des spheres
..
a rayon
engendre
Ult
MI , et quo la
de ce recouvrement est donnee par '\'i,
V.z:. VMr
0)
J0) ou Vest Ie volume d 'Ulle sl'Jhere
a rayon
Nous voulol1s choisil'
1.
un roseau pour lequel MI est aussi petit que possible par r apport a (d( (\ ) )f/)'v. Nous utilisons, une contraction bien simple: no us definisDOll S
Ie reseau "n
).
(}V' I ';' I,
't
A
+;;
l.l-~"
par les formes lineaires a)
I "
')
(,,'
"- ~ -
t .:::
\'("y, - 1
+
tv ) (.I~
00, N est un entier positif. Evidemment d (!\
?<-)\...='
N
(,(.1'\,
)==~.
Par definiti on,
0 11
! est un point de /\ et
z un point
quelconque de l' ~space
a
n c1.i;olensions. Or, ( 1.1, I
+~Y\,-l N
2
j
)+ .. ,+ftA, ~
22
+~"-_z.. )-t.-(J±I' I-Z) ,.,-1 tot I{ 11. 1'1.-1'
3) L' a bsence de symetrie dans cette definition n'est qu ' apparGllte; si nous multiplions par N nous obtenons Ie reseau de tous les points entiers qui satisfont
a x
n
(mod N). 23
H.Davenport
- 22 -
H01J.8
vou10 ns choisir des entiers u 1 ' •.• ,un qui rendent cette
somme minima.le. I.rettons un_v(mod N), ou 0 "v
~ N.
To ut en
gardant v fixe pour le present, nous choisis8ons u ' ... ,un afin l de rendre minimum l' expression ci-dessus. Si ~, desi gne, pour tout nombre reel entre
~
D'oll (Mr ) ~
l
,1a valeur absolue de la differenc e
et l' entier le plus proche de
11' - ?, 11 ~
oL'-
II
\I~
S'
nous obtenons
-Z, W+ ". + IW - z", 11
2
2~ 7 ~'" (II ~ -z, II \ ... + 1\ ~ -l~ \r), -
-
desi gne tout point de I' espace et v=O, 1 t •• • ,N-1. Le choi:X: de v s'effeotue maintenant par la cons id eration
d ' clile valeur moyenne.
'~~ f(1f) ~ ~
Pour toute f enction f(v), on a ~I
L f(v-) "V::. 0
Par consequent,
oV.
La f onction
'f «($ ) ainsi
def'i nie est evidemment paire at 24
H.D8ovenport
-23 -
*.£
periodique, 80dmettant 180 periode Elle est facile a calculer pour 0 ~
5
1 ~
2N'
si nous prenons
)J
impair, afin de simplifier. Si N=2N 1+1, on a
1W=iJ {~\t.~-sY+k (;1- sY\ ~ ~ (tl S~+ ~ [ V- = f ~ + /t} ~ N'" . . ). 12 .
NI
2__
4)
,
,Y
A/4
\T: I
~ =2~.
Cetto Gxpression atteint son maximum quand
D'ou il
slensuit que
(l + ~) 6 N~ .
I l!A~!'t\. .
En substituant dans (3), nous obtenons
V designe ici le volume d1une sphere
a rayon
1 dans l'espa-
ce 8. n dimensions, et on sai t que sa valeur est
V::.
(r(t ))""
)
en prenant N grand. 25
H.Davenport
- 24 -
Ce qui demontre Ie theoreme. Les resultats plus precis, tels que (1), s'obtiennent
an
en contruisant un reseau
dimensions au moyen d'un reseaD
a
k dimensions (k== 4, par exemple). La demonstration donnee ci-des-" sus n'est que Ie cas k==1 de la demonstration plus generale. Nous remarqumns que, dans
qr
cxacte de do
~a
demonstration ci-dessus, la valeur
~) n'avait pas d'importance. Cette fonction
S est ossontiollement une constante si on prend N grand,
la constante etant la valeur moyenno de
II t u 2
pour une v~
riabl0 reolle t, co qui est 112 • Dans la demonstration plus generale, cotto constanto devient la valeur moyenne du carre de la distance d 'un point du point Ie plus proche d 'un reseau pa:r'ti-
ak
culier
dimensions.
Lo calcul de cette valour moyenno pour
un reseau general semllle offrir des diflficul tes, et les resv_ltats cites' plus hauts furont obtonus par une serie d'epreuves avec a"GS
reseaux speciaux. Dne consequence de la demonstration plus generale nous
laisse affirmer que
oxiste.
II ost possible que la valeur de cette limite (qUi
est, bi en entendu, une constante absolue), est 1. II est interossant do comparer la demonstration du theoremo 4 avoc la demonstration de la minoration de remo de l1inkowski-Hlawka.
b:
')1..
dans Ie theo-
Tous les deux theoremes sont des
th80remes sur l'existence; l'un affirme l'existence d'un reseau d'ompiloment passablement bon (4), l'autre l'existence d'un 4) Lo theoreme de lUinkowski-Hlawka est, bien entendu, d 'un caract ere plus general que notre theoreme 4; nous 10 considerons ici seuloment eJ.'un point de vue special. 26
- 25 -
H.Davenport
reseau de r ecouvr8ment passablcment bon.
L'oxistcnce d 'un reseau
d 'empilcment se demontra indirectement au moyen d'un raisonnomcnt sur los moyennes; en contraste, la construction d'un reSefl.1A de recouvroment fut explicite.
Neanmoins, un raisonnement sur
le s moyennes fut employe afin de montrer que ce re s eau four nit en verite un bon recouvrement. 3. Une minoration pour ~"" • I1 est cl air, que le coefficient
lier d'une sphere
an
~~de
recouvrement
r~gu
di mensions est nec essairement plus gr and
que 1, mais il ne paraft pas facile de demontrer beaucoup pl us pour n gr and .
Le seul result at connu dans cette direction est
et Davenport 1), et i l se formule comme suit.
du a Bambah
THEORElI1:E 5.
On a
(1 ) 0 1l
f:.~l' ~ 0 ~ n
,....; 00. demonstration r epose sur 1a consideration d'un systeme
L8
de polyedres qui pav ent l'espace, systeme qui est associe reseau
A
a examiner de
~ous
co sys teme de polyedres.
Nous noterons
!\ •
1)
l ' ens emble
D(~),
et on a
n J.~ondnn Math.So~~
qQ~
Les pOints qui se trouvent plus
gres de 0 que d 'un point particulier .!lIO de
( ;:: )
n
1es points de l 'espace qui se trouvent plus pres de 0
(te tout autre point de es paco
a t out
dans l' esp ace S, n dimensions. Voronoi 2 ) fut 1e pr81nier
D (q) 27 (1952), 224-229.
2) J ournal fUr [,1 ath. 133 (1907), 97-178.
27
1\
forment un deHli-
_ 2(1 _
H. Davenport
est done convexG, et symetrique autour de O. Puis que tout
a une
point de l'esface se t rouve ou autre de
1\ , n
est borne.
10 demi-espace D(30) contient
distance bornee d'un point
Si 30 est suffisamment loin de 0,
n , et ainsi l' intersection dans
18
formule (2) est effeetivement une intersection d'un nombre fini
n
do demi-espaces.
(l
est done un poly-eare.
Il est clair que, si .l2. est un point de
1\ ,
Ie
polyMr<~
+.l2. eomprend les points de l'espace qui sont plus pres de Q
qu e de tout autre point de done tout
l'esp~ce
Les polyMres
1\
sans empieter.
n +:£ remplis sont
II s'nnsuit d'une considera-
tion dedensite que
v(n) Lo polyedre
f\
~ Gt(!\).
determine, de fag on tres simple, la plus
grande sphere pour laquelle
1\
fournisse un empilement: c I est
la plus grande sphere qu'on puisse inserire dgrs (\ B desi8ne cette sphere, les spheres B+.l2. (B
Car, si
£/\ ) n1empietent
pas;
d1autre part, toute sphero plus grande que B contiendrait un point
a l'interieur
point do
A que
nlempietent.
de (l , qui serait done plus pres dlun autre
de 0, ee qui est impossible si les spheres
On voit de fagon analogue que la plus petite sphi?-
re pour laquelle
1\ fonrnisse un recouvrement est simplement
~_~l~ite sphere qulon puisse circonscrire
a [)
nous demontrons maintenant (suivant Vororubi) que Ie de faees de ne depa_sse pas 2(2 n _1 ).
n
Nous remarquons d'abord que ehaque face correspond point du reseau, et que la face eorrespondante
a un
no~
~ t~~
point 30 de
I~eonsiste des points qui sont equidistants de 0 et de 30 et qui sont plus loins de tout pmint de
1\
1\
autre que 0 et 30. Or,
est synetri que par r apport au point ~3o' et la face est clol'l.e
aussi
s~etrique
par rapport
a '1230•
Chaque face a ainsi un cen-
tre de symetrie, qui se trouve au pied de la perpendiculaire tire.e 28
_ 27.
H.Davenport
de 0 a la face. SoiEmt
:J::
!i1"'"
respondent awe faces de
± SN les points distincts de
1\
qui cor-
n . Choisissons une base pour 1\ , et
r9partissons tous les points de 1\ entre 2n classes, selon les parit6s des entiers u 1 '.o. ,un qui definissent Ie point par rapport a la base. Les points 0, S1""'.9.:n appartiennent done tous
a des
classes distincts.
Car si 0 et !l1 appartenaient
m@me classe, Ie point ~ !l , serait un point de
1\
n
impossible puisqu'il se trouve sur une face de
a 18. m~me
:12 appartenai.cnt un point de 1\
classe, Ie point
,ce qui est Si S1 et
~ (9.(!l2) serait
, ce qui est impossible puisque '29.1 et
les centres de deux faces de
fl
a 18.
tS2 sont
(et q11 ±q2)' et Ie point
n.
iC9.1+g,2) est ainsi un point interieur de 11 s'ensuit que n ::H1 ~ an, at Ie nombre des faces est 2N ( 2 (2 -1). Hous arrivons maintenant
a la
demonstration du theoreme
5. Sait S la plus petite sphere qui contienne du recouvrement fourai par Ie reseau
v(s) =-
ct0) .?:,YanLa.'\Ll2..1us 2 (2
La densite
est alors
V(Sl. V(n)
II suffit done de demontrer que si n
1\
fl .
~faces I
r1
est un polyedro
inscrit da.'1s une sphere San
dimensions. et si Ie plbed de 18. perpendicul8.ire de 0 a
fa2~~_e
trouve
a l'interieur
vV((S)(1) >.i3 - f~
cony_~~~, c~l!£.
de Is. face, alors 3)
3) 10 resultat reste valable sans cotto condition, mais dans cc cas-la, la demonstration es-l; plus difficile.
29
- 20 -
Nons pouvons supposer que
5
H.Davenport
a r ayon 1, de sorte que
r(-t)~
r('ti,h.) Nous noterons AV ( V =1, •.• ,N) l'aire (c'est-a.-dire Ie volu~ me
a n-1
dimensions) de la face generale de
a la
ce perpendiculaire de 0 Decomposlmt
n
n
,et h V la di stan-
face.
en pyramides, nous obtenons
(4) Nous remarquons, pour y revenir plus tard, que, puisque la face generale est contenue dans une sphere
a rayon VI -
avons
~ 2.v
, nons
1-\ < )/
all J
n-
1 desi gne Ie volume d 'une sphere a rayon 1 dans n-1 dimen-
sions.
Si on prolonge la pyramide a sommet 0 qui a pour base la
face generale, jusqu'a ce qu'elle recontre la surface de la re, elle coupe de la sphere une "pyr8.mide spherique i ' S tv
(6 )
J~~ ~
l
sphe-
• On a
5)1
~aJ
N$ tre M.che principale, c'est de trouver une minoration de V(Sv ) en terJIles de Ay et h
V
•
Nous considerons une face particuliere de Ie suffi.xe
Y
pour Ie present.
perpendi culaire de 0 un :~oint
a la
a l'interieur
et
omettol\.~
Designons par 0' Ie pied de ta
face; comme nous avons deja vu, 0 ' est
de la face.
L'element d'aire (c'est-a.-d.ir e
de volume a n-1 dimensions) de la face,
30
a distance
r de 0',
H.Davenport
s' exprimc par r n - 2 dr dcJ , ou soli~e
a n-2
dimensions,
(L w
a sommet
designe l' element d' an£:l(~
O. Done
(7)
ou 1 'integration exterieure s'etend sur l'espace
a n-2
dimen-
s ions des vecteurs tires de 0' dans 1a face, et ou R designe 1a distance de 0' jusqula la frontiere de 1a face, prise Ie lOll(; c1'un tel vecteur.
L' element d' aire r n - 2 dr d west la base d 'un element 1e volume h r n - 2dr dwde la pyramide qui correspond a la face.
t
L'element correspondant de volume de la pyramide spherique s 'o btiont par une dilatation, et puisque Ie rayon de 1a est 1, tandis que la distance de 0
0.. }- +- ILl.
a l'element
spher~
de 1a face est
(8)
B:il substituant ceci en (8) et en comparant Ie result at avec (7), nous obtenons
31
H.Davenport
- 30 -
01.1,
en reintrodui sant Ie suffix e
Y
Voici une mino:i1ation de V(S v) en t ermes de Ay et h y
ce que neue
avons cherche. 11 ne reste quIa deduire (}) de (4),(5),(6),(9). Definis·son s Cy par (10)
AJ.ors , puisqtj.e hy
<:
1 , I' ine gali te (5) implique C
y < 1,
et en
out re -(
-ev~.
...
y
Cdy
)
L 'emploi de ceci dans (9) nous donne
(' C at
(11)
V(s J
~ ?~ )C(I_VC;(;_t.fJj"'ll
Le fonction
CP
·
(x); definie par
4(Jo) '" (1_ (I _~, h~ )""'.
J
est convexe pour 0 <. x ..( 1, parce qu.'elle s'exprime comme
~
L.
~+'Y'rt~ QM- X
i'vl-=o avoc
Q'W\. > 0, et possede ains i une s econd deri vee positive
Selon un t heoreme bien connu sur les valeurs moyennes avons (12)
tv
.2.-L cr(C y ) ) /Y'}J.::.I
-- --- --------------
'
th 'f
(C)
(4), naUS
J
4) Theoreme de Jensen; voir Hardy, lji t tlewood et Polya, Ine_
- 31
au
N
-
(13)
N
I
c',i C -- ,~I N
::
)1._1
H.Davenport
~
-
IY J~_I
1- ~}lAv.::r. )1:,1
%V(fl) N
J,.. -,
(6),(11),{12),(13), on a
D' Bpr~s
J~) ~ /It,
}
( 14)
f~ J([)(_~(t....:;...v...;.cU"':""'_--="7":-~ CJ(~ ~ t '''/2.;:.C
y
D))
I
I
_
~, tv ~ (I~ ~'(I~(.5) )"-1>. 1
_ V('n) (.. -
C,tt
Jo(~_Cb(1-tS))'rt/2.
Nous pouvon s supposeI' que V(
n )>t
(3) tient. D@nc, d'apres (13),
C>l
J , car autrement n
'Yt. Jb, N J 11-1
:.2.
Puis que
on a J
n/J n- 1
1 (2 j(' /n)7, d'oD.
tV
.
pour n grand. Puisque qu.e
CJ > 1
4
~
,
V(n)
(~
\
.c-
.J..
4
]IT
<. 2(2 n -1) et
S==2/(n-1),
pour n g:and, de sorte que
>J 1 o
a '1les ure Que n
I
1'1/\":; .> -N
C
~
(~t
(L-t0-t"J)'Yl"l
00,
on a
(-1 _ t ~)). 'ttl 2. --) 33
i l s'ensuit
H.Davenport
- 32 -
l a c.onvergonoe etant uniforme dans tout intervalle to ou 0
Ainsi. ~
.
J" > ({ - E "-)
ou
v(n) t "'..; 0 pour
n
-1
00.
(
Jo
I
t - 4" cU "
t
Ce qui demontre Ie
t.} 1,
(~- E "- ) , tlH~oreme.
1e probleme de trouver une minoration de CJue 1 est plus difficile.
~
~:
plus grand.\",
II nous faut maintenant consid erer
W1
rocouvl'ement non-regulier. La construction du systeme de pol yMres reBce applic able, mais 11 est plus difficile d'obtenir une majo r gtion du nombre de faces (nombre qui ne sera plus le t ous les polyedres).
Eogers 5) ont reussi
m~me
pour
En depit de cette difficulte. Erdos et
a demontrer
que
> ~ 00. Leur demonstration utilise l LYl resul tat 6 general de Rogers ) sur Ie maximum d tune integrale etendue sur ~m
Ol l
ttl""' 0 quand n
ensemble convexe de points.
o
J.London
I,~ath.Soc.
28 (1953), 287-297.
6) J.Londcm Math.Soc. 28 (1953), 293-297, 410-416.
34
- 33 -
~.
Le eas n
H.Davenport
=2
.
Considerons maintenant Is. question de l'empilement le plus compact de cercles dans le plan (regulier ou non).
La conjuctu.re
qui se presente, c'est que cet empilement a lieu; quand les ccntl't>r forment un "reseau equilateral", c'est-a.-dire un reseau dont la cellule fond amenta1e consiste en deux triangles equilateraux. Thue 1) fut le premier
a demontror co fait en 1892. Depuis ce
temps-la, on a donne plusiens demons§rations, mais meme aujourd 'hui il ne semble y avoir aucune demonstration vrainwn"t simple et intuitive. De meme, on conjecturerait que le recouvrement Ie l!lDins co;npact par des eereles sur Ie plan a lieu quand les centre:s fo'(mrmt un res eau equilateral.
Kershnor 2 ) fut 1c premier
a demoni;l.'er
ce f :J,it en 1939. Los donsites d'empilement ey de recouvremont pour un
a calculer.
~uilatoral sont faciles
r6sea~
Si le cote d 'ub: tri angle
equilateral est egal ~ 1., 1e det~rminant du reseau est.L ·V j • Le rayo~
du plus grand corcle qui se laisse empiler est
1 22' et
la
dens ite dlempilement est done
lrr
Ji-
1'3
-::-V .l.
if - -LVi
L0 r ayon du plus petit phm est
3- \j 1
1-
..:... II
..L- :.
-!- V~
-------~-------
=
I
V1
::
0)
3 06~
.'
.
cercle qui nous permette ie reeouvrer le et 1a densite de recouvrement est done
-Yrl211"
1) Pour la reference, voir L. Fejes Toth, Lagerungen in der Ebe~e ~uf_9:.~J~.ii£..l_und
2) Ji.rlwrican J.of
im Raum iS pringer 1953). math. 61 (1939),665-671.
35
H.Davenport
- 3+ -
Nous enongons les resultats cites plus haut sous forme de theoreme, et nous esquissons une demonstration. Cette demonstra-tion a le m.erite que ilies
m~mes
principes generaux s'appliquent
aux deux problemes. Theoreme 6.
IT
-Vii.
(1)
(2)
2.17 -.
V2l
Considerons un systeue de cercles
a E1 ,E2 , ••• ,
a rayon
1, aux centres
qui forment un empilement dans le plan. Nous
pouvons associer
a
chaque point Ei le polyg~ne convexe
n
consiste en tous les points du plan qui sont plus proches
(~i) qui
a
quIa aucun autre centre du systeme •
"0.
.....1
Euisque l' aire de chaque cercle est
rr , il suffi t, pour demon-
trer (1), de montrer que l'amre moyenne de
11
(Ei) est ~ VI.Z
Cette moyenne se forme de la fagon habituelle en consider8.nt les points qui se trouvont dans un grand carre at £8 prenant 11..118
limite. De fagon o.n.alogue, si
avons un systeme de cercles ft
110US
rayon 1, aux centres a E1 ,E2" •• , qui recouvrent le plan, il suffit pour demontrer (2), de montrer que l'aire moyenne de
t V2J.
n(Ei )
est $ :Dans ce cas-la, le po1ygone
cl~
de
r~on I
f\ (E·) J..
arec centre
a EJ...•
11
a deux
l.
Dans ce cas-ci, 1e po1ygone
est inscrit dans Ie circle
Nous avons recours
(E.)' est circonscri t au cer-
a rayon
1 avec centre
a ~"
~
propositions, 1 'une geometrictCl e,
I' autre analytique. IJa proposition geometrique, c' est aue de tous les polygones a un nombre donne de cotes qui sont cincon-
.
serits a
lL'1
corcle donne, le
polyg~ne
36
regulier a la plus petite
H.Davenport
- 35 -
a un
aire, et que de tous les polygones
nombre donne de cotes
qui sont inscrits dans un cercle donne, Ie polygone regulier a la plus grande aire.
Ainsi, si (1 (£i) a n i cotes, on a dans le premier cas Aire
n
>-I" C(n.), J.
(,l?.) l.
ot dans Ie second cas Aire
fl
(P.)
<
-:t....
l(n.), l.
au C(n i ) designe l'aire d'un polygone regulier scrit a un cercle
a rayon
a ni
1, at ou r(n.) designe l'aire d'un pol.
l ygo ne re gulieT an. cotes inscri t dans un cercle J.
Or,
C(rn~)
1 ('I'':'
)
tr1.
:::.
-
-
a rayon
1.
1T
to\M '"
1\1'v
1 1't1' v
c~tes circon-
I).{M
2
}
llI. nt'
La proposition analytique dont noue avons besoin est J.'il'1egalite de J ensen, que nous avons deja recontre dans Ie ~ 8. C(;tte inegali t e affirme que , si f U(x)
~
0 pour x ) X, aUrs
pour taus nombres x 1 , ... ,xr tous) X, et que si f'(x),(O, I'inegnlite dans le sens inverse est valable. Si C(x)=x tan
2.1T 1 Cf/() x. =~
X"
)t
,
on a
)<..
•
11
(.(n'
11
I).1M
X
>0
et si r (x)= ..!- vsin 2IT ,~oo a
1·eX. , =
.LA
~
-II
J
Il s' ensu.it que 37
- -
H.Davenport
.~.-:
~v
(~i)
[1
Aire moyenne de
.>.
-
dans Ie premier cas, et
n(~i)
Aire moyenne de
C(n)
/"
L.
I (n)
clans Ie second cas, ou n designe 180 valour moyenne de n 1..• Los polygones CE'i) remplissent tout Ie plan sans empieter.
n
De co seul fait on peut deduire que
n 4 6.
La deduction se
base
sur la relation d'Eulor 3
+ F
+ 2
0
::=
entre les nombres de sommets, de faces et de cotes d'un COl1voxe dans l'espace
c1 ' •.• ,OF
polyed~e
a trois dimensions. 3i on designe par
les nombres de c$tes des faces differentes, on a 3 F ~ 0 1 +••• + CF
=2
1,... ,as 1es
3i n deaigne par C
C.
nombros de c$tes qui so roncontTe:nt
aux sommcts differents, on a
3S
~ q
o+
2
D'ou 0
s
+••• + C
< .?3
"
.~
0 + F
~
2 0
,
3 F - 6.
Lo nombre moyen de dIMs par face s'exprime par /
i"
~-I t
F
CF
::L.
-
2.C ...< G F
8i maintemmt nous pcrmettons au
-
po~dre
-F
12-
<6
de degenerer dans un re-
couvroment du plan par des polygonos convexos, au moyen de rendre uno des faces du polyedre arbitraircment grand, nous obtcnons le resultats eno nce ci-dessus pour Ie nombre moyen des c$tes de polygones.
Afin de completer la demonstration, il foudrait
examiner en plus de detail Ie proces de limite qui entre 38
dans
H.Davenpo rt
- 37 -
l 'emploi du terme "moyen". Nous pas sons ce point, qui ne pr8sente d ' ail1eurs aucune difficulte serieuse. Le sujet d'empilemen:b. et de recouvrement dans Ie plan a ete etudie avec beaucoup de sueces.
Premierement, quand aux
carcl es , on a deIDontre les analogues des inegalites (1) et (2) pour les empilements at recouvrements contenus dans certaines domaincs, par exemple dans un hexagone convexe 3 ). Deuxiememon:t, on a etendu les r esult ats au cas plus general dans lequel les ~arc l cs
sont remplac es par une region K convexe et symetri que .
1e coeffici ent d'empilement s'exprime par plus petit hexagone circonscri t mont slexprime par
~~~~)'
a K.
~~~~
, ou H est 1e
Le coefficient de r ecouvre·-
ou H' est Ie plus grande hexagone
inscrit dans K.
Ces resultats sont due
ete etendus
au cas ou les notations, aussi bien que les
m~me
a Fejes Toth. lIs ont
translations, sont permises. lTeanmoins, quand aux re couvrements, il y a une limitation : i l f8ut supposer dans l a demonstration qulaucuns deux des regions convexes ne se croisent. Il cst probable que Ie resul tat reste vrai sans cette limitation. Pour (1':1
lLn
expose des travaux sur ces sujets, voir Ie Chapitre 3
livre de Fejes TO'th.
3) Segre et lViahler, American Math. r'J onthly, 51 (1944), 261-270.
39
H.Davenport
10. D'autres resultats pOp! n petit. Les problemes de determiner
st'
et':Jj ne sont pas encore
r ssolus, et ils semblent presenter de grandes difficultes. Les
b) et ;0-)
problemes de determiner
(probHmes de reseau) sont
plus tractables, car on peut les abordee au moyen des methodes do 18 theorie des nombres. Les valeurs de SM-sont connues pour
8,
aux travaux de Korkine et Zolotareff et do 1 o Blichfold G ) sur les minima des formes quadratiques. Cos savants
11
~
gr~ce
onto determine la majoration exacte ~It- pour le minimum
['J0yI'
les valeurs entieres, non toutes nullesr, des variables] d'une forme quadrati que positive
an
variables, de determinant 1.
Or , une tolle fonne peut s'exprimer comme
Ot'l
2
+ .. ,
-t 'Je. 2.
'YV
ou x 1 ' ••• ,x~ sont des formes lineaires a determinant 1. Les points (x 1 "" ,xn ) qui resultont de valeurs entieres des variables independantes forment un reseau
a determinant
1, et dans
la notation empioyee auparavant, on a
Ie maximum etant pris sur tous les reseaux de determinant 1.
s"'" ~e s
-= Jl1.
valeurs connues ,d e
¥1Vv'
D~c
yl1..( 2.
t""
I
ct les valeurs resul tantes de n sou'/:
Pour les references, voir Koksma, Diophantiehhe Approximationen (Springer, 1936), Ch. 2.
40
H.Davenport
- 39 -
les suivantee: n
2
21/3
2 III.
4
8'/5
(¥ )'16
6
S:
7
~LI }'/r
8
'i
11 n'y a
Jr.. :. 0 I 906<1 · 213 rr
-
:; 0
f4 0
,.
3Vi I Jr'iT -.. 0 6I~". I I
5
=
"'
\1'"~
3
in
~ =- -i
t",,"12-
=
auc~
~
It; 2-
-~ !L-
4S\t'%
lL' 10;
-
OI46S" .
I.
~
031-4 .. I
::.
JT4 :.
o .z ~) . . I
02)'2 .. . }
3g~ co!maissancc directe sur 1a question 8i
quand n ) 2, mais
a mon
avis i1 eat peu probable qUG
cotte egalite tiGnno pour n grand. La valeur exacte de
~3 ~t
determine par Bambach2 ) en
19~3:
Sa demonstration est difficile, Gt une demonstration plus simple est
a souhaiter. 11. Prob1emes divers.
En conclusion, je voudrais passer en revue quelquesuns des problemas non resolus du sujet. Je nG discuterai pas les problemes
a deux
ou
a trois
dimensions, parce qu'il
y en
a un expose
excellent dans le liyre de Fejes Toth. A mon avis, lee problemes qui offrent le plus grand interet
it:,
sont les suivantes: quels sont les vrais ordres de grandeur de
b}\. $"Alfr ) 13 ~ ) J
41
H.Davenport
- 40 -
A cet egard, il serait d'inter@t d'avoir
quand n est grand ?
Une demonstration, soit par des exemples particuliers, soit indirectement, que
S;> 61<; ,
ou ;;:.(
En second lieu on ales problemes d'estimer; aussi precisement que possible. les quatres coefficients
correspondants d'un
corps convexe quelconque dans l'espaoe a n dimensions. L'inegalite (:K) , 3n- 1 , consequence de l'inegalite 11(K) ~ 3n - 1 ,S (K) de Rogers, est certainement loin de la verite, et
11
elle doit
~tre
susceptible d'une grande amelioration.
L'empilement et Ie recouvrement par des corps non-convexes ont re9u peu d'attention, sauf pour Ie l'espace par des corps congrus a ete etudie
a cause
a
pro~leme
de recouvrer
Ix 1 ,. • .,xn I ~1,
probleme qui
du r81e qu'il joue dans Ie probleme
non-resolu de Minkowski sur un produit de formes lineaires non-homogenes. II y a aussi quelques modifications de ces problemas, qui meritont
etudies. On peut definir, pour tout
peut-~tre d'~tre
ontier positif n, la densite d'empilement r-fois Ie plus compac t, 0-(;
la densite de rocouvroment r-fois Ie moins compact.
premiere,
®n
Pour 18.
exige que tout point de l'espace se trouVD dans an
plus r des corps, et pour la seconde,
o~rlge
que chaque point
<18
l'espace se trouve dans au moins r des corps' On peut designer les densites ainsi definies par
IJes seuls resul tats Il
G.
etudie
d~nt
j
'aie connaissance sont dus
l'empilement double (r:::2) de spheres et il a
;E) J. London r:lath.Soc. 28 (1952), 297-304.
42
1)
a L. Few •
H. Davenport
- 41 '"
demontre que \' {l)-X ()tv
~
inegalite qui correspond
a celIe
de Blichfeldt. II a aussi
0.emontre que
>2. (13 ). 'Yl-/2
.
Chabauty2) a introduit Ie concept d'empilements semi-reguliers, qui Bont intermediaires entre les empilements reguliers et les empilements generaux.
Cependant, comme
~ous
ignorons
a present la vraie relation entre ces deux genres d'empilement il est difficile de juger l'utilite de ce concept. 2) Cpllogue d' algebre et theorie des nombres (Paris, 1950),27.
o ________________
43
0
to UA Z I 0 N I
It 0
ill
D I 0 FAN TEE
a-IfJtituto IIla"tematico dell'Universita,1955-R
45
0 ill
a
... 1 -
EQUAZIONI
L. J. Eorcl(~ll
DIO.F'AN1'EE
1.
Siano x 1 ,x 2 ' .•. , xn' diciamo (x), n variabili e sia f(x 1 ,x2 ,··· ,xn) ' diciamo f(x), un polinomio in queste vqriabili a coefficienti ruzionali . Possiamo, s enza dilfiinuire 1a general it a , supporre ahe i coefficienti siano interi, poiahb si occuperemo di equazioni f(x)=O. Una qU8stioni evidentc PROBL.G;.':"~
e il
- Trovare tutte Ie soluzioni di f(x)=O
1) i n numeri razionali 2) i n inte;ri,
c
Supponiamo dapprima che f(x) sia un polinomio omogeneo: Ie
due qucs tioni (1) (2) sono evidentemente 10. stessa. Abbiamo adesso una soluzione banale (x)=(O). Inoltre dobbiamo considerare soltanto Ie soluzioni in-tere per Ie quali x1 ,x2 "" ,xn non hanna divisori com1l.."Yli ccc etto 1, aioe (x 1 ,x2 , ... ,x)=1. Supponiamo adesso che f(x) sia un polinomio non omogeneo. Ponendo
otteniamo
illl '
equazione omogenea
Abbiamo adesso evide ntemcnte una
corrispondenz ~l
1-1 tl'a i
valori razi onali di (x) e i valori interi di (y) dove y n+ 1/0 e (Y1'Y2'···'Yn+1)=1. II problema generalc e coincidente con l'intera teoria dei numeri e richiede tutte Ie risorse della matematica. Esso suggerisce altre questioni. Possiamo trovare condizioni necessaria ragionevolmente semplici per 10. risolubilita di f(x)=O? Bono (juGs te condizioni suf47
I,. J. Mordell
- 2 -
ficienti? Possiamo trovare qualche soluzione e dCt queste possiamo dedurre altre
tutte Ie altre
0
0
'xua infinita di altre?
TEOREMA. Soluzioni razionali 0 intere di f(x)=O possono esistere solo se f(x)=O pub essere sOddisfatta da valori rea11 di (x) •
La dimostrazione TEOREIvIA.
e evidente.
Una condizione necessaria per l'esistenza di
soluzioni intere di f(x)=O
e che
la congruenza
f(x) ~ 0 (mod ~,O abbia soluzioni per tutti gli interi La
dimos'~1'azione
m.
e evic1ente.
Le pro prieta elementari delle congruenze mostrano che POSsiamo considerare sol t D.nto M=po( dove p prende i valori primi ed
d.=1,2, ••• Una condizione necessaria per l'esistenza di solu-
TEORlli~A.
zioni intere (x)l(o) dell'equazione omogenea f(x)=O
e che
la
congruenza f(x)
~ 0
(mod. M)
abbia soluzioni per cui
(x 1 ,x2 , ••• ,xn )=1 per tutti gli 11 e, in particolare, quando Iij::::pCl. , abbia soluzioni diverse da quelle
per cui x 1 ;; 0 (mo d p), ••• , x -
= 0 (mo d p).
n -
Infatti in una equazione omogenea, possiamo supporre
(x1 ,x2 , ••• ,xn )=
=1 e questa (x) deve soddisfare f(xhO (mod M). Quindi, anche, se la congruenza f(x):::O (mod pc( ) ,perol maggiore di un certo
cX(J ,1'i-
chiGde che
=•.• :-: Xn-~ 0 (mod p), 1 = - x 2al101'8 l'equazionc omogenea f(x) =0 ha soltanto la soluzione x
banale (x)=O.
48
L.J.t1ordell
- 3 -
2. E9VAZIONI PRIVE Dr SOLUZIONI I NTERE. Adesso consideriamo il PROBLEW.lA - Trovare , mediante considerazioni sulle congruenze; equazioni senza soluzioni intere nale (x)=O.
0
con la sola soluzione ba-
2
L'equazione x 1 = a + b x 2 2 Que sta richiede la risulubilita di x 1 =a (mod p) per ogni r
2
numero primo p dispari divisore di b, e quella di x :;a(mod 2 ) se 2r e 1a pili alta potenza di 2 che divide b. La teoria elementare delle congruenze mostra che queste condizioni so no sufficienti se (a,b)=1. Abbiamo i seguenti risultati noti quando b
e una
potenza di
2.
La congruenza x~ _ a (mod 2r)
e risulubile
e sempre
risolubile per r=1;
solo se a .. 0,1 (mod 4) per r=2; ed
se a ;0,1,4 (mod 8) per r=3 e quindi
e risolubl1e
e risolubile
solo
anche per r
maggiore di 3. Per un numero primo dispari p, supponiamo per semplicita (a,b)=1j allora la congruenza x1=a (mod p) e riso l ubile soltanto q81ando a e un residuo quadratico di p e si puo scrivore (a/b) =1; 2
e la congruenza non
e risolubile
quando a non
e residuo
quadrati-
co di p, scritto (a/b)=-1. Questi risult~ti significano che ogni divisore primo p di 2 x1-a per x 1 intero deve dividere a oppure deve essere rappresentato da uno dei numeri finiti della progressionc aritmctica Esempio: se un numero primo dispari p divide X~+1 allor~ p e 1~11a forma 4 n + 1. Infatti se x21 ;-1 (mod p), allora xP1-=(-1)~(mod p), p,:,1 oppure 1 (-1)~ percio p ~ 1 (mod 4).
=
49
L.J.Mordell
- 4 -
Evidentemente un'equazione f(x 1 ,x2 , •• "xn )= x
e impoBsibile.
2
- a
in interi, se f(x) non ha divisori primi del tipo
ricordato. Esempio: Se (a,p) '" 1 e (a./p)= -1, e se
a '):!
.".2
.A1 -
""2 '"
allora llt1 :: 0 (mod p), x 2 ;; 0 (mod p) e quindi x3 aO (mod p).
Poiche se x 2 =0 (mod p) a110ra x 1 =0 (mod p). Se x2i (mod p) possiamo porre x 1 =. x x 2 (mod p) .. allora 2 2 (x "a) x 2 =0 (mod p) 2 2 2 e questo e impoBsibile. Quindi l'equazione x 1-ax2=p x} puo avere
°
solo la soluzione banale illroviamo adesso
x1;x2~x3=O,
alc~~e
quando (a/p)=-1.
equazioni prive di soluzioni intere
o con 1a soluzione banale (x)=O, considerando congruenze con qualche M speciale. Esempio con M=2 r Esempio M=4. 2
2 2 x1 + x2
= 4x 3
+ 3
2
Adesso x 1 _ 0,1; x2 £ 0,1 e quindi Esempio:
x2 + x 2 0,1,2 e non 3 2 2 1 2 2x 1+x 2=(4x 3+3) x 3•
Possiamo supporre (x 1 ,x2 ,x 3)=1. Allora x3 e evidentemente non dispari per quanto precede. Se x) e pari allora x1=x2zO (mod 2). Esempio M = 8. 2
Qui x 1 ;:0,1,4 e quindi 2 2 x 1+2x 2
~
0,1,2,3,4,6 e non 5 50
0
7
- 5 -
L. J . i ,ordell
2 2 Esempio: x, - 2x 2 = 8 x 3 + 3,8 222 Esempio:x1 + x 2 + x3 = 8 x 4 +
non 7. Esempio: Questa equazione non ha soluzioni intere, eccetto x,=x 2=x 3=O, se a,b,c, hanno 10 stesno segno. aosl pure non vi sonG soluzioni se a,b,c sono tutte dispari e
a~b;c
(mod 4); oppure se a/2,b,c sono
tutte dispari ed e b+c ;4 (mod 8) oppure b+c :. a (mod 8).
x;+x~+x~;o (mod 4) richiede X':.X 2=X 3=0 imod 2). Per 1a seconda parte usiamo aX~+b~~+CX~;O La prima parte segue perche
(mod 8). Evidentemente x 1 ,x 2 ,x 3 non sono tutti dispari poiche a+b+c
i 0 (mod 8), ne due di essi possono essere
p~ri. Quindi sola-
mente uno 8 pari. Questo non puo essere x, poiche b+ctO (mod 8), ne puo essere x 2 poiche eX 3 e dispari, e non puo essere x3 poiche bX 2 e dispari. Esempio: x~+x~+x~ x,=x2=x3=0.
=2
x 1 x 2 x 3 ha soltanto la soluzione intera
Poiche x"x 2 ,x3 non possono essere tutti dispari, ne
possono esserlo due di essi e neppure uno solo. Quindi x"x 2 ,x 3 sono tutti pari. Poniamo x,=2X"
222 etc.; al.lora X,+X2,X3=4 X'X2X3; come sopra
x,X2 X 3 sono tutti pari. Quindi x 1 ,x2 ,x 3 sono divisibili per una potenza arbitrariamente elevata di 2, e quindi debbono essere tutti nulli. Esempi con M=3 r •
2
2
Es. M==3. Oa+' )x,+(Eb+')x 2",,3x3 dove a. b sonG interi e (x 3,3)=1, por asempio x~=' e
,2 ! = x 1 + x 2 ::: 0.
COSl
Poiche x~ =0,1, questo richiede x,=O, x 2:.O ehe e evidentemente impossibile Es. M=9, Adesso x, \O,~
e quindi x 13+x 2 3+ x/=. 0;:!:1, !2,!3 e non +4. 51
L.J.Ho!"dell
~ 6 -
Esempio con M=7, x 3 =7x 3:2,±3 impossibile Es. x 3 +x3 :: 7x3±3 31 :a ) 3 Adessa x 1 =O,±1, e quindi x~+x2=o,±1,±2 e non ±3 Es.
f(x)=(7a+1)x~+(7b+2)X~
+
(7c+4)X~ + (7d+1)X,x2x 3=O ha l'unica
soluzione (x)=O. Qui X~+2X~+4x~+x1x2x3=O e noi dimostriamo che x 1=X 2-=X 3-=0 • Se
~3-0, Xf+2X~=':O
e questo riclliede x 1=x 2;O.
Se x3tO, possiamo porre x 1=xx3'x 2=yx 3 coslcche risulta x 3+2y3+4 + xy
::0
Evidentemente xtO, yiO e quindi x 3:±1, y 3=±1, Questo conduce a quattro casi ~mpossibili, poiche ( 1 ) x 3;1,y 3",,1 da xy =0 3 3 (29 x3",,-1, y3=-1 da xy :-1 mantre x y =.:1
-
3 3 3 0) 1(3=1; y:::-1 da xy =.:-3, cio~~ x y3;1 (4) x 3",,-1, y 3=1 da xy :: 2, ciae x 3'y 3 ;;, 1 •
Esemp1 nei Cl)lali M e una potenza di un numerm primo di una forma speciale Ii=)? 2
Es. x1+1=pxt
se
PE3 (mod 4) e impossibile poiche
un numero positivo a :3 (mod
4) ha s empre
un
("'1!p);:-~.
Poiche
divisore primo p;)
(mod 4), l'equazione 2
x 1 + 1 :: aX 2
e pertanto Es.
impossibile.
x~+x~=px2 e impossibile quando p e un intero primo :3 (mod 4)
a meno che non sia x 1=x3;0 (mod p) Es. X~-2=PX2 se p=.:±3 (mod 8) poiche (2!p)=-1
X~+2::px2
se
p=5,7 (mod 8 ) poi che (2!p)=-1. 52
- 7 -
L. J .Mordell
2
T.r
~
2
2
2
2
Es. x1+x2-P(x3+x4)=0 dove p~3 (mod 4) Qui e x~+x~= ~ (mod p) e ~os~ x 1=p X 1 , x=pX 2 • A110ra P(x~+x4) =0 (mod p ) e pertanto x 3;x4=0 (mod p) Quindi 1e soluzioni dell'equazione omogenea avrebbero sempre un fattore comune. Es. rfJ=9 j x~ +2X~ +4X~ = 9X~. Qui x3+2X~+4X~ ;0 (mod 9)
X~=0,±1'
Poiche
la sola soluzione di questa
e
x 1=0 (mod 3), x 2=0 (mod 3), x3~0 (mod 3) e quindi x 4;0 (mod 3) . xf+2X~=7
Es. 111=7 2 ,
(X~+2X~)
Qui xi+2X~=O (mod 7) ogni soluzione della quale
e x 1;x2=o
Qullndi
7(x~+2x~)~0
e c081
x 3=x 4=O
(mod 7) (mod 72 )
(mod 7)
M=p",3 (mod 4) / =x 3 + 7 Qui xi-o (mod 2) poiche $.l1ora X;; 1
(mod 8)
2
poiche allora y::2 (mod 4)
x~-1(mod 4)
e pertanto
l;7
(mod 4). Ma e y2 + 1 = (x+2) (x 2-2x+4)
Adesso
i
-2x+4 )
e cos~ x-l'2
0
>0
e x+2 =3 (inca 4) e qudmd;4
!1ofl
puo dividere y2+ 1• Es. dove k
= x3
/
+k 2 3 a - b e a;-1 (mod 4), b::O (mod 2)
=
0
b non possiede
divisori primi =3 (mod 4). Poiche adosso y2 + b 2 = x3 + a 3
e cos~ y
2
=
3 x - 1 (mod 4).
Evidentemente x
io
(mod 2), xa-1 (mo d 4) 53
8 COSl
x.1 (mod 4).
ai
- 8 -
< x 2-ax
Ma 0
M =p
+ a 2 ;:;3 (mod 4) e
COS1.
non puo dividere
l+b 2
r
r.1 =p 3• Esempio di Eulero x 31 + pX 23 + p 23 x3 = o. Qui x 3+px 3 + p2 x 3 =0 (mod p3) e 1
2
a-
x 1 :0 (mod p), e aLlora
COS1.
successivamente x 2;O (mod p) e ~3=0 (mod p).
3. SOLUZ!ONI
RAZIONALI
Queste possono essere talvolta determinate senza processo ar:ttmetico 22
x + Y =: 1 2 2 y = 1 - x
Es. Scriviamo
allora ogni soluzione (x,y) con ylO e quindi
xl1,
conduce ad un
unico insieme (a,b), (-a,-b) tale cne (a,b)=1 e a =1-x b Y
e dove ab=O. Quindi mediante divisione 8
i e
=
COSl.
x=
~
y
a 2_b 2 a~+b2
2ab y- 2 2 a +b •
,
La soluzione esc1usa y=O, x= '-f 1 pUG essera inclusa in questa poiche x=1, y=O si ottiene Can a= ~1j b=O~ x=:1, y=O "
"
"
1\= 0 , b=
11.
Scrivendo b=at, dove t e un parametro razionale che prende valore infinito quando a=O, tutte 1e soluzioni razionali sono date da 1_t 2 x-- 1+t 2 ' Es.
2
x +xy + y
Poniamo y=1-t x, dove t 2
2
2t y=-1+t2
=1
e un
parame-tro razionale. A110ra
x- + x(1-tx)+(1-tx) 54
2
= 1
L. ,T. i!lo rcl ell
- 9 -
da cui x + 1 - tx - 2t + t 2x ::: 0 x:::
2t - 1 2
t -t+1
Es.
ecc.
x 3 + y3 ::: x y y::: tx, X(t 3+1)::: t
ecce
Esempio: x 3.y3+z3::: 1•
2
Sia r una radice cubica complessa delltunita, cioe r +r+1=O. Allora (x+y) (x+yr) (x+yr2) ::: (1-z) (1-zr) (1_zr 2 ). Pertanto una corrispondenza 1-1 fra ogni soluzione razionale (x,y,z) e Ie coppie razionali dei parametri (a,b)
e data
da
')
x+Vl"'2 x+yr iIo a+br ~-2 ::: a+br • 1-zr ' 1-zr Evidentemente, dati x,y,z, noi abbiamo un unico insieme (a,b), e, dato (a,b) noi abbiamo (x+y) (a+br) (a+br 2 ) ::: 1-z. e queste tre equazioni lineari in x,y,z determinano x,y,z in funzione di a e b. Esempio: L'equazione x 3+y3+ z3=1 ha la soluzione parametrica
9t4, yo;:. 3t-9t4, z:::1-9t 3 Esempio: L'equazione x3+y3+z 3=2 ha la sola soluzione ,rarametrica x:::
x::: 1+6 ;l3 t y=1-6t 3 , z:;-6t 2 • Esempio: Una soluzione di x4+y4+4z4:::1 in numeri razionali
-s ,
da
dove t
4 x::: t t 4+2
e un
2:1;3
y=t 4+2
z:;;..l.t.. t 4+2
parametro.
Vi sono
~articolari
soluzioni per Ie quali
2
Allora
,
x +2yz ::: 1 y4+4z4:::1_x4:::1_ (1-2yz)2 55
e data
1. J. lliorde11
;. 10 -
2
quindi
2 2
4 yz
(y +2z)
=
y2+2z2
=
2~
Poniamo y=t z, a110ra (t 4+ 2 ) z=2t. Poiche (y2_2z2)2~ 4yz (1-2yz) = 4~zx 2 2
2
t 4 _2
x- y -2z =
2t t 4_2
z
0
= Esempio:
X4+y4+Z4
t 4+2
=2
=2
x4+y4+(x+y)4
Poniamo z=x+y, al10ra
222
2( x +xy+y ) 2
x +xy+y Esempio:
y +kz
2
Ponendo
n
= x ,
= 1
e da~a
Una soluzione parametrica 2
2
=2
ponendo y=1-tx.
k 1'azionale
y + z'{:k
= (a+bV-k)n
y - z -k
= (a-b
_k)n
a 2+kb 2
dove a,b sono
param~t1'i
razionali, otteniamo soluzioni razionali
di questa equazione.
4. SOLUZIONI
INTE_~
Equazioni delle quali si trovano facilmente soluzioni inte1'e. o~
0
bsemp~o:
2
x + y
2
= z2•
Basta considerare Boluzioni nelle qua1i a1cuna coppia di numeri x,y,z, hanno un fat tore comune. Poiche se (x,y)=k a1101'a x=kI, y=kY e z=kZ dove X2+Y'=Z2
La soluzione
e
e data da 2 2 :x:=a -b , +- \± y=2ab, ±
(X,Y)=1.
complet~
56
2
z=a +b
2
L. J .Illordell
... 11 -
dove a,b ~ 0,
a
e una b J 0.
± z==a 2+b 2 ,
+ y==a 2-b 2 ,
.. _ x'--2 a b ,
e
- ,_ 1"
qualunque coppia di interi di parita diverse, e (a,b)
= 1.
Le soluzioni con x) 0, y
~
0, z
*° sono ottenute omettendo
tutti i doppi segni , e imponendo la nuova condizione a
i
b
# O.
Evidentemente x,y non sono ambedue dispari altrimenti allors
2
x =y
2
='
(mod 4) e z
2
~2
(mod 4); quindi
~y
sono di parita diversa
e pertanto z e dispari. Consideriamo Ie soluzioni con y pari. Scriviamo
y2 == .2 _
i
== {z-x) (z+x).
Al1ora, se d=(z-x,z+x), d necessariamente divide (2x,2z) e quindi z,x sono dispari, d=2. . d·1 z+x=_+2 a 2 , QUln dere a
~
y2__~J a 2b2 dove nOlo poss;amo pren~
+2b 2 ,
z-~._
q O.
0, b
Quindi x=± (a 2_0 2 ), z=±(a 2 +b 2 ), y=±2ab,+2ab 2
2
2
x +y = 2 z ,
Es.
(x,y,z)=1 Evidentemente x,y sana ambedue dispari. Poniamo x= X+Y,
y=X-Y
AHora X,Y sono interi e sono di p!U\Ltidiverso.. e (X,Y)==1; quindi X2+y2 == z2 2 2 e X == a - II , Y = 2ab, 2
2
x = a -b +2ab, x = 2ao+a 2-b 2 ,
TEOREW~.
2
2
y=a -b -2ab, y==2ab- (a 2_b 2 )
L'equazione x 4+y4: z2 non ha soluzioni intere eccetto cbe
nel caso xy=O. E' sufficiente considerare Ie soluzioni con
x~
0, y
~
0,
zOe (x,y}=1. E' chiaro che x,y ~on possono entrambi essere dispari poiche allora e2 a 2 (mod. 4) e cosl baste considerare solo x ;: 1 (mod 2);
~
57
0 (mod 2).
- 12 -
222 2 Quindi, x =a -b , Y =2ab, Poiche
2
2
2
x +b =a,
2 2 z=a +b , dove (a, b)=1, a
b;O (mod 2), a;1 (mod 2).
I
Anche (x,b)=1 perche se p mo
x=p
dove (p,q)=1,
L.J.Mordell
2
-
p~O,
(x,b), p
2
2pq,
q , b= q ~O
y2
I
a= p
~
b J. 0 •• F
a, pi (x,y). Cosl noi abbia2
+ g
2
e p,q hanna parita diversa. Ora
= 2ab = 4pq
(p2 + y2).
"rO~Cl~ . 11>. p,q,p.q 2 2 sono pr~mi . oR 1 oro a copple, .? era p=r , q=s 2 , p 2+q 2~ t 2 , 1' O, s~O, t ~O, cioe
Q
Anche
1'4 + 54 = t 2 x = 1'4 - 13 4 , Y = 2rst, z
= a2
+ b2 =
x y == 0 ,
Se
13 4
1'8 + 6 1'4 l'
+ s8.
st (r-s)=O;
e cosl s01uzioni con xy==O conducono a soluzioni can rs=O poiche r=~ porta a t 2=2r 4• Quindi ogni s01uzione x,y,z can xylO porta
a una so1uzione r,s,t con rslO e z=
t
< \rZ
pOiche
(r 4 + s4)2 + 4r 484 ') (r4+ s4)2 = t4.
Questo processo pUG essere continuato e condurra a una nuova sOluzions r 1 *s1,t 1 con 1',8110, t1
e noto
t~=O;
d1
come "metoda delia
descente in,ginie". Es.
y2
= X4 * 1
ha solo Ie so1uzioni razionali X=O, Y= ±1 perche se X = ~ con (x,y)=1, a1101'a Y = ~2 e coai z2=x 4+y4 etc.
Y
y
5. RICERCA DEI PUNTI RAZIONALI SU UNA CURVA: CASO DELLE QQm.9HE PROBLEMA- Data una soluzione 1'azionali di f(x,n 2 ••• n 4 )=O, dedurre altre soluzioni 1'azionali. Innanzi tutto l'equazione f(B 1 ,n 2 , ••. ,n4 )=0 rappresenta una curva. Noi la scriviamo nella forma f(x,y)=O e riguardiamo una 58
- 13-
eoluzione razionale (n,y) di
L.J.Mordell
f(x, y)=O
come un punta razionale
P su1la curva f(x,y)=O. Se noi scriviamo f(x 1 ,x2 , ••• ,n 4 )=o in forma omo genea f(x,y,z)=O bast era considerare le soluzioni intere (n,y,z)l(o,o,O) di f(x,y,z)=O e aneora parlare di punti razionalm su1la curva f(x,y,z)=O. TEOREMA - Siano date due ourve f(x,y)=O g(x,y)=O di gradi r,s. Se r s - 1 delle intersezioni sono punti razionali, allora la rimanente
~ntersezione
e ancora
un punto r azionale.
Eliminando y allora x e determinato da una equazione di grado rs a coefficienti r a zionali. Poiche r s-1 delle sue radici sono note, la rimanente grado e cosi
e determinata
e ugualmente
da una equazione di primo
razionale.
Analoghi risultati valgono per Ie intersezioni di tre superficie. Es. Sia f(x,y)=O una conica con un pun to razionale noto P,(x,y)=
=
(Xo'Yo)' Noi possiamo prendere per g(x,y)=O la retta y-YO=t(~TJO'
passante per S dove t
e Un
parametro razionale. La retta incontra
la conica in un altro punto Q e cosi Q e Un punto razionale le cui coordinate sbno espresse pe~ mezzo del parahletro ogni punto razionale Q su f(x,y)=O
e dato
t.
E' chiaro c~e
in questo modo e cle
una corrispondenza 1-1 tra i punti Q e i valori razionali di t. Anche P corriaponde al valore di t per il qUale la retta
e tangente
in P e anche Ie intersezioni della retta x=x o distinte da P corrispondono a t=oo, ma q}l.ando x=xo e una t angente, allora P corrisponde a t=oo. Dunque x ;:
y =
A1t
2
+ B1t+C 1
At2 + B t + C
A2t
2
+ B2t + C
a
At2 + Bt + C
dove gli A1 ,B. ,C . scno costanti razionali da la soluzione generale ~
1
di f(x,y) =0. 59
L.J.Mol'del1
- 14 -
6.
PUNT I BAZIOYALI SU UNA CUBICA Sia f(x,y)=O una cubica irriducibile. La curva puc avere un punta doppio e cosi
e di
come pUG non averlo e salora
gener~ero,
e
di genere uno. TEOREMA- Tutti i punti razionali della cubica di genere zero poasono essere espressi mediante un parametro razionale. Vi pUG essere un solo punta doppio, perche se P'P 1 fossero punti doppi la congiungente ~ P 1 dmcontrerebbe la curva in quat ... tro punti e coal apparterrebbe alIa cubica che risulterebbe rlducibile. Poiche P
e determinato
'"U.
d):!.
dalle eqaazioni
~ ~:o::f "O~
e un punto razionale (~oYo)' Allora ogni altro Q e determinato parametricamente da1la equazione
P
puntQ
~azionale
y - Yo ~ t(x - ~o)
pOiche il punta doppio coUte. per due delle intersezioni della retta con Es.
1s cubice.. ax3 + by3 = cxy,
l'origine
y = tx
Poniamo
x
e tIll
punto doppio.
allora
_...£:L
- at 3+b
L'origine corrisponde tanto a t=O come a t=oo TEOREMA- Da un noto insieme di punti razional:i. P"P2,,,,Pn di una. cubica, supponiamo di e~:here une, possono in generale ottenersi altri punti. La tangente in P1 incontra la cubica in tre punti due de! quali sonG in Pl ' e coal il te:czo punta P11 e razionale e di verso da P1 se P1 non
e un
punto di Hesso. 60
Ugualmente la tangente in 1>11'
1.J. Uordell
- 15 -
puo in geherale condUDre a un nuovo punto razionale P1,1,1
etc.
Anche la retta P1P2 interseca la curva in un terzo punto :razionale P12 in generale diverso da P1 e P2' Questa procodimE:.nto di tangenti e secanti pub esaere appli ... cate a tutti i punti di S ad anche a tutti i punti trovati in questc modo.
TEOR&MA DELLA BASE FINITA (Mordell). Tutti i ptlnti razionali di una cubica di genera uno possono essere dedotti fa un numero finito di punti col procedimento delle tangenti e delle secanti. 2 3 Es. Y = x + k Data una aoluzione razionale (p,q),
e
(p,q)
q~O.
La tangente in P il punto
2
y-q=
e cosi
.llL 2q
x3 + k =
(x-p) 2
[q+ ~P q (x-p)]
2 •
Poiche due radici sono p,p, 1a somma delle radici da per x 1a terza redice 4
2
4
Y:lq~ (iL - 3p) I 2 q 2 Queste s~ilo 1e coordinate di Q, i1 "taA~enzia1ell di P. x=
~
- 2p e a110ra
Alternativamente. Posto x = P + X, a110ra
y2 Posto
2
+ 3 p
2
X
:<
+ 3 P X2 + X-'. 2
Y=
aUora
Es.
=q
~ 4q2
y3
q +
lJL 2q
N
etc.
3p+ X
= a 3x 3
+ bx
2
+
e1.: + d.
Posto b
Es.
Y = ax ~2 '0 etc. Il IItangenziale" del punta P (~, ax 3 + by3 + cz3 + dxyz 61
1 ,~ ) su
=0
- 16 -
Q,
Sorivendo
L. J • Mordell
-c {o3 ') ), y= ,. (g ~ 3-a::,&'), z= '~ (af'-b~3).
t .~ x=) (b I't)
x=x/~ , etc. J basta provare ilrisultato quando
a=b=e=1. Per prima cosa, Q appartiene alIa tangente in P cioe x (3
~1 +d ~ ~
) +. ;. =0
pOiche
Inoltre Q appartiene alIa cubica poiche
2: ~ ~ ('1 \- S~Y+ at s~ ~ TT (~~ - ~ ~) ;:
n(,>- S3)(~\ ~3.r S3+~ ~~~)~
O.
TEOEEMA- Se a,b,c sono interi diversi da zero e a due a due prim! fra loro, non divisibili per il quadrato di un numero primo, e al piu uno fra a,b,e e :1, e se d
e un
intero, allora la curva
ax 3 + by3 + cz 3 + dxyz
= 0
possid4e un'infinita di punti razionali
0
non ne possiede aleuno.
Supponiamo (x,y,z)=1. Allora x,y,z sono a due a due primi fra loro. Infatti se p
e un
numerd primo e p\(x,y), allora p2/e~3. cioe
plz contro l'ipotesi. Inoltre x,y,z~Ot i~fatti se z=O, ax 3+by 3=O ha l'unica soluzione x=y=O poiChe a,b sono liberi da quadrati e
abh1. Supponiamo ora chc sulla curva vi sia 8010 un numero finito ~1
di punti razionali, indichiamo con lx,y,z) quello per cui
Ixyzl ha i1 massimo valoree Allora per il tangenziale X,Y,Z
IXYZ I ~ Ixyz \ • Ora ~ X ::,c (b ~3_
b Z_
CZ 3)
.I
S' Y~ 'a' (C'l~- &l1.3)J
z( a, x,3_ b ~)) 62
L.J.Mordell
- 17 -
dove
~
e un intero tale che
Ora
Se
(X, Y, Z)= 1.
..
primo can xyz. Perche se p fosse un nuroero primo e
p/~ , ply. ,allora p e primo con yz, e poiche pic, p/b contr~ llipotesi.
p/cyz3, p/bzy3,
Quindi
l XYZ lilt ~ z,
Poiche xyzf.O,
\ ~i
J
cioe C'JJ 3_ ct')('3 fl. "-
b by3-cz3=O,±
Quindi
's 61?
1.:0. i.
S ; cz 3-ax3=O,± S;ax3_by3=O,. S'
dove i segni sono indipeniinti I'uno dall'altro. Noi possiamo escludere 10
mi fra
lor~ e
Anche ±
° poiche a,b,o sono a due a due Pri
bcf.±1, etc.
~
membri delle tre
puc essere esclud,perche la somma dei primi q~uazioni
e zero.
Ee!inizione; Due curve s ana dette equivalenti se possono trasformal 8i l'una nell'altra con trasformazione birezionale a coefficient1 raziohali, Se Ie curve sana f(x,y)=O, F(X,Y)=O. abbiamo x :: a+X,Y) , X =
A(x,y),
y =
b(X,Y)
Y = B(x,y)
dove a,b,A,B sono funzioni razionali dei loro argomenti a coefficienti razionali. TEOREMA - Una
cu~ica
con un punta razionale
cubica sotto la forma normale di Weierstrass
63
e equivalente
a una
L.J.I:1orde11
- 18 -
Sia P il punta razionale e Q il tangenziale di P; possiamo 8upporre che 11 punta Q sia nell'origine. La retta y=tx per Q incontra la cubica in aLtri due punti determinati da
x2 + 2Mx + N
L
=0
dove L, M, N sono polinomi in t di grado 3,2,1 rispettivamente ed a coefficienti razional i . Ne segue ( Lx + M)2 ; M2 - LN. 2
II polinomio di quarto grado in t M - L N=O ha una radice razionale t=to poiche P e razionale. Ponendo t=to+1/T' (Lx+M)2=M 2 -
14 f(T),
dove f(T) e un polinomio di terzo grado in T. . M~diante una sost1tuzione T = C X + D e cos~ c~ ~ 3 2 3 f(T) assume la forma (4X -g2X-g3)' Allora se Y =41 -g2X-g).
- LM =
Lk + M versa.
;i : ,
Y
(eX + D)2
•
B' chiaro che da X.Y otteniamo
XtY
e viC$-
7. PUNT! RAZIONALI SU UNA QUARTICA Consideriamo soltanto il caso che i1 genere della curva sia
e che sia data nella forma y2 = ax 4 + bx 3 + ox 2 + dx + e
Una forma piu conveniente sarebbe
z2= ax 4 + bx 3y + cx 22 y + dxy 3 + ey 4• Sappiamo dalla teoria delle quartiche che vi sono dne invarianti
1
g2 = ae - -4 bd + 3 a g).
£
£
4 c
4
6
c
£
6
( K)2,
£. 6
£ 4
e
4
64
IJ.J.Mordell
.., 19 -
TEOREliA
Ques-ta quartica , quando sia nota un suo plmto razionale,
e equivalente
alIa eubiea normale y2 = 4X3 - g2X -g3 •
Cia signifiea ehe possiamo supporre ehe e alha un quadrato esatto,
0,
;J.
scrivendo~,
X
to esatto.
1
- in luogo di y,x, ehe a sia un quadrax
Se a=O, il risul tato
e immediato,
Se ariO, ponendo
y Va
i n luogo di y, possiamo supporre a=1. Quindi, mediante una trasfoE. mazione affine in x, possiamo supperre b=O, Scriviamo a110ra la quartica 2
Y
ora, essendoei il Runto razionale x=Q), y=m , Ora
g
2
2
= e + 3c ,
o
-c
-0
d
d
e
2. .5 -ce-ci..,..c:
" d"l d 2= 4c 3-g2c-g3' qUln Allora la eubiea normale h8. il punta (X,Y)=(e,d) e possiamo
applicare il proe'2dimcnto della tangente e clella secante (eecetto quando a=O),
e data
La corrispondena fra la quartiea e 10. cubiea 2
Y = -X + 2X + c,
2x== ~ X-e
Sostituendo al posta di y nella qnartiea, abbiamo x 4+2x 2 (2X+e)+(2X+e)2=x 4- 6cx 2+4dx+e ossia
x2(X-C)+dx==X2+cX+ 2x(X-c)=-d
2 c ~e
± [d 2+4(X 3-c;)X) 65
+ (e 2-o)(x-c)]
~
da
I" J .llordcll
- 20 -
3 1 ==-d ±(.U -g2X- g )"2" 1 • Y == (4X 3-g 2X-g a)2
Di qui, se
possiamo prendere 2x(X-e)=Y-d. Esempio:
y2*ax 4+bx 3+ex 2+dx+e
Sia data 1a emrva'
e sia (p,qf'O) un punto razionalo sulla eurva. Ponendo X= p+X, possiamo supporre p=O, e ehe e sia un quadra2
to esatto, e=f f'O. 2
Portiamo dove
y=Ax +Bx + f 2fA+B 2=C,
2fB=d, 2
Allora A x+ 2AB = ax + b dB. in generale un nuovo valore per x. Se a
e un
quadrato csatto, a== 0{
:J,
, possiamo porre
2
y= o(x +C x + D dove e
2ot.C=l!
d. x
101
j
+ e = 2
0(
D + C 1::::. ~
Dx+C 2
Esempio/ Si supponga di conoseere due punti razionali (p,q),(p',ql);
'x. :. /'LX i ~I X+1 A110ra 1 ~equazione prende 1,,\ forma ~ y~
t 4 = ~X
+ bX
3
+ eX
e si possono apuliearc ambeduc i
2
+
rrH,-~oJi
di cui sopra.
Esempio: 2 2 2 2 Y (a 1x +b1~+c1)+y(a2x +b 2x+e 2 )+a 3x +b 3X+c 3=0
Riso1vasi rispotto ad y. Al10ra
2y(a1x2+b1x+e1)+a2x2+b2x+~2*vrR dove
66
L. J. Nlordell
- 21 -
Quindi 1a curva
e equivalente
alIa curva
y2::Ax4+Bx3+Cx2+Dx+E sia (x"y,) un punto noto. Ponendo x=x 1 ' l'equazione quadratica in y ha una soluzione Y" e cosl un'altra ra-
A~ernativamente,
dice raziona1e
e data
da 2
. ~2~'±£2~'±£2-Y +y = - ( 2 1 2 a,x,+b,x,+c, e c061 abbiarno un nuovo punta (x"y 2 ). Pongasi Y=Y 2 nell'equazione; allora l'equazione quadratica in x ha 01 t:ee alla radioe x" 2
x 2+x,=-
una radice razionale x 2 data da
b'Y 2 + b 2y 2+b 3 2 3,J'2 + a2~2+a3
Quindi abbiamo in generale una nuova soluzione (x 2 'Y2)' Ana10gamente troviamo i punti (x 2 'Y3)' (x 3 'Y3)'(x 3 ,y 4 ) etc. Esempio. La quartica 2
2
2
2
Y =a,x +b 1X1 +c 1 , z =a 2x +b 2x+c 2 con i l punta razionale (x"y"z,).
2 2 Poasiamo prendere x,=O, poi sostituire c"c 2 con 1'["c 2 • Nella prima equazione, poniamo y=tx+c, e c06i 2
a,lll:+b,=t X+2tc 1 b,-2tc, X=--t 2 _a , 2 2 2 2 1 2 .2 2 2 e C061 z (t -a 1 ) =s =a 2 (b 1 -2tII1 ) +b 2 (b,-2tc,)(t -a 1 )+(t -a,)
°2 •
Questo polinomio di quarto grado in t diviene un quadrato quando tJ,-2tc,=0, cioe per t=b 1 / 2c ' a meno che c,=O. (Similmonte, si eccettua il 6a60 che c 2=0). So c,=c 2=0, poniamo y= Y/X,z~Z/X ,x='/X' e
2
Z =a +b X 2 2 67
2
-
e
II.J.Mordell
22 ..
COSl
T,a soluzione eli questa equaziore e data da un ben noto teorema delle Teoria dei numeri. Alternai;ivamente 2 2 2 2 c 2y -c 1 z = x Prendiamo
C2~+ C1
(jZ::,
2'~1 (""x.tq)C~
i - C-i ~ =
q
X.
y=c 1+px. 2c"2-
e cosl)brevemente Aibl ora ( c +rx ) 2 1
~2 b a 1 .... + 1 x + c 12 dB. in generale un val.ore razionale per x, e altrettanto per y e z • 22 23456 . Esemp~o. y = a + bx + ex + dx + CX + fx + gx-,
=
Abbiamo qui una soluzione x=O, y=a, ma non possiamo trovat' e alcun'altra soluzione. Non si ha vantaggio ponendo y = a + px + qx2 + rx 3 • 2
Infatti, se poniamp 2ap=b,
p +2a2= c,
2pq+2ar=d che de-
terminano p,q,r, allora abbiamo un'equazione quadratica in x. Congettura. In generale l'equazione non ha se non un
numer~
finito
di soluzioni razionali.
8. PUNTI RAZIONALI
su gcmm
SEI\i:fLICI SUPERFICIE
cumCRE.
Esempio;
3
z::;x
2
2
-ay
Soluzioni parziali sono date da x+y \}a=(p+q Es. w3 = ne
Va) 3
TT (x+ye +z e
2)
cubic~Cfo/
una soluzione
,
e ora
x- Yay= (p-q Va) 3 , dove
e , If ' 't' sono
dat a ponendb
68
2
z=p -aq
2
radici della equazio-
- 23 ..
It. J. Mord~ll
~ eT Z 9\ (11. + (i e+ .~ e~) 3 't of ~ lP t '.4'1= (f· + ~ If -t" ~!f2. )~ 'X. + I ¥'+ 2tr'~ (ttt q'\f + 'ttt}'r~ t
')C
EsemE.Q.:
(x+y+z)3 - dxyz =
Supponiamo omogenee x,y,z
Allora (x+y+z)3=dyz(x+z) in coo~dinate
dyz2=m
e una
;n,
cubic a di genere zero con punta doppio in
(-1, 0 ,1).
Poniamo
x+z= PY , a110ra (P+1)3 y3= dpy2z quindi
l!
A110ra, essendo dyz 2=m si ha ~ Una soluzione
=~-
paro.metri~~ r~~l~>l~~le e ora
data ponendo P=dmt 3•
Per esempio, ponendo
x
= y
y= Z + X,
+ Z
z= X + y,
d=24,
m~8n
al10ro.
(X+Y+Z)3 - 3(Y+Z) (Z+X) (X+y) = n e cosi abbiamo una soluzione dell'equazione di X3 + y3 + z3 = n Esempio: 2
z -ry Siano
2
= x
3
e , tf' If!
+ px + q
R~ley
t(
1-0.
Ie radill[i dell' equaz ione cubica
X3 + px + q = 0 Poniamo
)H~ Vt "TT [~+ ;.cjJ\(1 H " e}/~J ~'fl'f
2dove,
~ Vi : ;.
~ ,~
,)A.
n
Q,If, 'Y
L~ t 4~. e~(Y) +v-e)J2"
~ 1/ Bono
.
lHJ.r&fJe
tri raziono.li.
69
J
!..3.!I1ordell
- 24 -
Allora, y,z Bono numeri razionali e
L'equazione originaria
e soddisfatta
se
x-a =(~ t~e~y'- ~ (~+1re)l
e due analoghe equazioni in 't' ' dato che X \. Sviluppando la prima delle tre e ponendo
e4= _ pe L _ e uguagliando i coefficienti di
q
g
eO,
e,
tv ')( +~ 2llT (x-e) . el~/l'
gl; si ha.
x=e-IZ.'I"I2. )1 2
= ~)A,
o=
+ l rz. 'lr'Y\
",u. ~ _p. u..:2. _ Iz.
ilJ'1
Ie quali
e cosi abbiamo una soluzione che di pende da due parametri.
TEOREMA GENERALE. Sia 113. superficie S data da
f(x,y,z)=O in coordinate non
omogenee oppure da g(x,y,z, w)=O in coordinate omogenee. Supponiamo che f e g siano irriducibili e che dipendano da tre paran.etl"i essenziali, per escludere il caso cho la superficie sia un cono o un prisma.
TEOREMA La superficie cubica ha infinitipunti razionali, quando ne abbia uno. Proviamo dapprima cne se ha un punto razionale, allora ha infiniti, oppure pUG essere ri z
2
= f (x,y) 70
tta nella forma
0
ne
I,. J. Mordell
- 25 -
e un
eove f
polinomio cubico.
Sia (Xo'Yo,Zo,W o ) una soluzione razionale di g(X,Y,Z,W)=O. Poniamo
X=Xoz+X,
Allora 2
1z +Ivlz+N = 0 dove, 1,M,N sono polinomi omogenei rispettivamente di primo, secondo e terzo grado in x,y,w. Qui
1,M non sono ambcdue i denticamente mlllli, poiche in tal
casola superficie sarebbe un cono Se 1 ca z=-N!M.
un cilindro.
0
e identic amente nullo abbiamo una soluzione parametriSe 1 non e identicamente nullo, prendiamo x,y,w tali
che 1=0, e allora abbiamo una soluzione parametriea , eccetto ebe se M e divisibile per 1; in t ale caso scriviamo = 4M 1
2
_ N
Can una conveniente sostituzione, poniamo w in luogo di L, e z in luogo di z+ 2M
cosl abbiamo
L
(~)2 w
= f( !, X)
w w
1'equazione z dove f(xtY)
e un
2
= f(x,y)= ax
3
+ •••••• ,
polinomio eubico dipendente essenzialmente dalle
sue due variabilit ha lila soluzione razionale
parametrica~
Possiamo supporre, previa sostituzione lineare in x,y che a;iO, ed anche a=1. Allora, scriv endox-py-q in luogo di x, possiamo supporre che l'equa zi one as suma Ia forma z
!
II secondo membra 8 f'un zi oue di una sola varia bile, per
esempio x +
Ay
(
A..
costante ) s e
2
sol t anto se
a = b = d = e = f = 0
71
L.J.Mordell
- 26 -
caso che viene escluso. Poniamo x=y
2
+ 2ty dove t e un parametro; allora
z2 = y6+6ty5+(12t2+a)y4+(8t3+2at+b+d)y3+(2bt+c+e)y2 + + (2ct+f)y + g. Poniamo dove
z = y3 + Ay2 + By + C
2C+2A~8t3+2at+b+d
2A ... 6t,
ciOe
1t 2 J3=~
A = at,
2
3 c= .;: ,b.,;. .+d:: ;. -. .; a; .,:t:--.. ;ot_ 2
Sostituendo nell'espressione che da z2, abbiamo 2 P Y + Qy ... R § 2
dove
p=l1: 4
=0
2
+ ~ (b+3d)t + C + e - a4 2
.it:.5 + 2at 3.12 (b+d)t 2 + 21 (a 2+4 b )t+f - 21a (b+d).
Q= 2
Allora
232
R = g- 4(b+d-at -t) 1
= -
41 t 6
+ ••.•••
Osserviamo che
zPt -Q ::.~t~ _ ~ (~t ~d,) t2.-+(.2.e-o})t + ~ (~tol)-t espressione che non e identicamente nulla, eccettuato il caso che a=o, e=o, #=0, b+9d=O. Risolvende l'equazione in y abbiamo, eccetto che per un numero finite di valori di t, due valeri di y finiti e distint1, poiche ne P,
ne
Q2_4PR=3t10+ ..••.• =
, sono identicamente nul-
I i in t.
Quindi
;2iP~ ~ ~
-
Q ± VX
p.lX.::.~P3+4P-~ 2- 2 1. t· 72
::.
L.J.Mordell
- 27 -
z
e
==
y3 +
Brevemente X= cl ± 0/\ 'VA , y= (3 ± ~ ~ , z= (f ± ,("V; . Questi due punti definiscono una retta, ed ogni suo punto e dato nella forma
dove
e e un parametro. Questa retta interseca la superficie in un terzo punto, evi-
dentemente raziona1e poiche l'equazione cubic a in
e la
determina il punta d'intersezioue ha il fattore ~L_ Questo punto non
e all'infinito
quale
Ll
eccetto che per un numero
finito di valori di t; e infatti, a1trimenti aarebbe
oasia (2Pt_Q)3+ap2(2Pt_Q)+d 3p 3=0 identicamente. Dai coefficienti di t 12 abbiamo d=O. Se 2Pt-Q=0 identicamente, allora a=O, e =0, f=O, b+qd=O, cioe b=d=O, che 8e
e proprio 2
i1 caso
es~luso.
2
(2Pt-Q) +aP =0 identicamente, il coeffieiente di t
8
da
a=O e quihdi 2Pt-Q=0 identieamente, ece .ecc. TEORE11A:
La equazione f(x,y,z)=O dove f(x,y,z)
e un
polinomio del
terzo grado in x,y,z, simmetrico in x,y ha una infinita di soluzioni razionali
(aammmendo ehe l'equazione riguardi tre variabili
essenZiali] • Nella forma omogenea con x=X!W etc., la equazione d&.venta g(x,y,z,w)
= O.
I termini indipendenti da z,w assumono la forma
a (X 3 + y3) + b [X2 y + X y2) 73
- 28 -
L. J • Mor-dell
Quindi una soluzione particolare (1 , -1
,!B, 0)
e i vi dentemente
et c.
9. SOLUZIONI INTERE DI ALCUNE EQUAZIONI
y 2+1=x 3 , ha la sola soluzione intera w=1, y=O, 8criviamo
Es.
(y+i) (y-i) = x 3 Dobbiamo ora studiare Ie pro prieta ar:ttmetiche dei numer;i. complessi. Si chiami z=a+ib un intero complesso dove a e b sono nwneri interi. Gli unici divisori dell'unita sonG: ir,
r=o,1,2,3, cioe ±1,
±i.
Noi chiamiamo z un numero primo se z ha come divisori solo r i ed anche .( i r) x. Dove E=O,1,2,3. Poiche 2=i(1_i)2, il numero due non a meno di un fattore unitario. Anche 1-i 2
E' noto che quando p
p=c +d ed ora
2
e un
e primo, ma un quadrato e un nwnero primo.
"primo" della forma 4n+1 • allers
dove c e d sono interi razionali, quindi p=(c+id)(c-id) ~±id
I numeri primi della
sono numerm primi complessi.
forma 4n+3 restano "primi" nella teoria dei numeri complessi. La legge fondamentale dell'aritmetica
va~e
per questi "comple!
si interi" ed ogni nwnero compl esso intero PUQ decomporsi in modo Iillico in fattori "primi". non considerando i fattori unita =i r ed altresl l'ordine dei fatteri primi. Esempio:
r
Se XY=Z , ed X ed Y non hanno fatwri comuni, allora
X-: .
~
0(
.A.tL-oI
A ) Y:;:. (., It
It
tz.
B) ~ ::
dove A,B sono interi complessi.
74
I.A )7\.1. AB J<¥)=~~'~'~I
- 29 -
L.J.Mordell
Ritorniamo a11a nostra equazione: y2+1=x 3 , evidentemente y e pari e 1a x
e dispari
e quindi y+i ed y-i non hanno fattori
comuni, perche essi dividerebber@ y+i=(y-i)=2i. Di conaeguenza y+i=i r (a+ib)3 dove a e b sono razionali interi. Anche y_i=i- r (a_ib)3, poiche i=(_i)3 la i r puo esaere assorbita in :(-a+ib)3 e in£1ne 1=3a 2b_b 3=b(oa;2_b2). Quindi a=o,b=-1, ossie x=1, y=O. Esempio: La equazione y2+2=x 3
~a una sola soluzione intera x=3,
y=±5. Dobbiamo ora studiare numeri algebrici della forma z=a+b -2. Le loro proprieta sono analoghe a quelle di z=a+ib. Qui y
e
dispari e la x e pari. Scrivendo (y+ y:2 )(y- V-2 )=x 3 , noi abbiamo ora y+
v-::2
=
=(a+b ~)~ dove a e b sono razionali interi. Quind! 1e bOa 2-2b 2 ) ossia b=1, a=±1 anche x=a 2 + 2b 2
;t f. 0' ( Mc'~
==
3 ecc.
;L.) ) 't f?> (M.<.x:f it) J CllM W?k: X: f (>\W d 4)
~1+" ::'(i<. _t)(x +,)( 1
+1) I (MJ.~ ?<~x:+.{ .= ~ (V\W~) 2
e quindi non puo dividere y +4. Esempio: y2; 5 = x 3 II precedente metodo non puo essere applicato a questa equazione, perche ia legge fondamenta1e dell'aritmetica non vale per numeri algebrici della forma z=a+b~. Esempio (2+V-5) (2- 'f:§)=l; si noti che 2± 2• della forma (m::td
·\f=5)
~
non sono
Esempio da tre differenti modi, essenzialmente diversi, di decomposizione in fattori. 75
L.J.Mordell
.- 30 -
Qui i1 nmnero tre non puo decomporsi nella forma 3;o(a+bf:5) (C+d'f="S), e sso non puo considerarsi "primo \I, perche (4+ (4-
v:5)
V-s)
e di visi bile per 3 senza che alcun fattore 10 sia.
10.
TEOREW..ADELLA Bi\._~E_ FINlY-
2 4 + bxy 3+ 2 z;ox ex y 2 + dxy 3 + ey 4 Sia
& una r adice dell' equazione
Noi dobbiamo ora studiare Ie proprieta aritmetiche dei nmneri algebrici della forma A+
B\t
+
c 9\
D
e?> ,
dove A,B, ...
sono nmneri razionali. La legge fondamentale del1'aritmetica non vale per tall numeri algebrici. Questo causa gravi difficolta. Si scriva l'equazione data come 3egue: 2 .. 3 S I z =(x- ey)(x + ~b+ e) Si puo provare che e
2
xy+ ..... )
dove p,q,r,s appartengono a un insieme finito di nmneri razionali e doveP,Q,R,S sana clei razionali interi che deebono essere determinati. Poiahe () soddisfa una equazione de
quarto ordine, sv-ilup-
pando i l secondo membro ed esprimendo 6 6 , (}5, ()I(in termini di 1 n3 flO, (}1, (j2, e~ ~ J e quindi raffrontando i termini in 8 ,~in enmambi i membri, noi trovia mo due equazioni inP,Q,R,S. Risolvendo queste equazioni troviamo che P,Q,R,S possono essere espressimediante tre parametri x 1 ,y j ,z1' dove
2
4
3
2
2
3
4
z1 = x 1 + bX1Y1+cx1Y1+dx1Y1+eY1
76
L.J.iI!ordell
- 31 -
Si ha anche
Inoltre si vede che continuando con questo procedimento si arriva a soluzioni di grandezza limitataj ne segue che esse sono in numero finito. Puo provarsi che questa procedimento dimento (metodo) della
tar~gente
e equivalente
al prOQe-
s ecan te per Ie equazioni cu'biche.
TEOREMA: L'equazione z2
p2+
x+
y+Ax3+Bx2y+Cxy2+Dy3
dove i coefficienti sono interi eO. (
)=1 ha una infinita
di soluzioni. Con unG conveniente sostttuzione lineare, possiamo porre A=1,
}Iv
=0,
Se p=o, poniamo x=o, z 2=Dy 3 etc. Se
p~O,
2
poniamo x=4p X, y=2p
Poniamo 2
z= p+ 2pX - 2p X Allora z
Poniamo
2
0
=p~+4p
22324
y = t
X-8p X
+4~
X
Xe
X= 2. Up t 3 + etc.
77
• Allora
- .32 -
L.3.MordeU
B I BL I 0 GR AF I A CAR.'VIICHAEL - Diophantine Analysis, London , Chapman and Hall, 1915.
"
- Analyse Indeterminee - Paris, Les Presses Universitaires, 1929.
DICKSON
- Histoty of the the~~ Numbers, vol.2 -Washington, Carnegie Institution, 1920.
MORDELL
- A Chapter in the theory of NUmbers-Cambridge, The University Press, 1947.
MORDELL
- Integer S21utions of cubic Equations in three variables, "Rendiconti di L1atematica e delle sue applicazioni" Serie V, vol.XIV, Roma, 1955. - L'analyse indeterminee de degre superieur, Gauthier-Villars et Cie 1929.
P~ris,
B.SEGRE
- On the rational solutions .our homogeneous cubic Eguations in fO}l,I'---Y§.r:Lables i Mathematicae Notae,
SKOLEM
- Diophantische Gleichungen, Berlin,Springer 1938.
_0 XI.
78
SOME PIlOBL3Il1S ON THE DISTR.IBUTION OF PRIME NillilBEllS
ROl1A-Isti tuto Mat ematico· dell' Uni vcrsi ta, 1955-RmflA
79
P~Erdos
- 1 -
SOME PHO BLETdS ON THE DISTRIBU),ION OF PHIlliE NUMBER~
In this lecture I ,vill talk anout some recent; questions on the distribution of primes. It is not claimed that the problems I will discuss are necessarily imporkmt ones. I will mainly speak about pro blems which have occupied me a great deal in the last few years. Denote by
If (X)
the num.ber of princs not exc() oding x. The
prime number thuoremltates tha t7r(X) / Xho gX 7 1 a s x.......+ (I). It K give s a ILUcn better approximation is well known that
L {19
to
Tr (X)
'I;o;i.
than x/log.k. The s t atoment tJJa t for X.
L )(
r lr(x)-
(1)
~1
> X (c:.
)
-.i-- \ < X. ~/1. +~ ~K
is oqui valen t t o t he "iemann hypoth e sis. In our lectures we will not (ieal with problems like (1) but will rather co n sid er pro blems of distribution of primos i n th e small
(G. g.
probl ems on
consewutive prime numbers). An old and very di f ficult pro blem on prime nmubo r s requires
to find to overy int eg er n a pr ime number p > n, or i n other words to bo abl o to write down arbi traril;y large Tlrim e numbors. The largest primo number which is known is 22281_ 1, this number Was "Qr
It is an unsolved problem whether
·~here
are irrfinltely
m~ny primes of the form 2P-1. A few years a go Mills 1 ) proved. that ·t;he re exists a constant
rc3J
c ) 1 so that for al l .i :nt eg ers n , it seems i mpossible to daten.in e every
JZ:
>1
C
is a prime. Unfortunately
expl ici t ely. Ver ;r likely for
there are infini tcly many inte r;ers n for which [ cJ
is composite. Put d =p 11
n+
1-P • It immedi at 81;,' fo llows from the prime n
81
- 2 -
P.Erd~s
number theorem that limd flog n ~ 1.
Backlund
2)
proved that the 3 above limit is ~ 2 and Brauer and Zeitz ) proved that it is~ 4. westzynthius 4 ) proved that iiin d flog n == PJ. I proved 5) using n
n
Brunts method that for infinite4r many
J?I.
and then 6 ) proved
the same resu_lt much simpler without the
use of Bmn' s method.
Finally Itan.k:in 7 ) proved ·that for infini te-
ly many n (2 )
It seems very hard to improve (2). In the opposite direction Ingham8 ) proved that for n > n o d > n 5/ 8• Cramor9 ) conjecn
tured that
Using the Riemann hypothesis Craiher9 ) proved that
t"'..
~K2 ~
(€o:a 'lncyr.
C n
I conjectured 10) that
l
(3 ) w .
I
ol~ >c~illts
tliK~\ eorem,
th . ''J.S
'I'V
J:
-<
C1'L
(e,a ~ )
1.
K=1
an immediate conse quence of the prime number ( 3 ) l- f . 1 e1' conJec . t ure10 ) ' t rue'lS b os t POSSl. bl C:. A Slmp
which I also can not prove is the following: Let n be any integer, donote by a 1 ~a2"" a
Cf{~ .
L
(CL~ - (LC_i)2. < C
i.. 1Yl..
~ =.2
Another conjecture which sooms very deep is that d flog n has n
82
P,Erdes
- 3 -
a continuous distribution function 1. e, that for evary c
~
0 the
density f(c) of integers n for which dn/log n < c exists and is a continuous function of c, for which f(O)=O, f(oo )=1, Here is anothor problem which has the same rolation to tho pr evious one as (4) has to (3): Put n k=2.3,5 .. ,Pk,a"a 2 , .. ·a Cf1 (nk) arEvihe intGgers, < nk relatively prime to n k • Denote by A(c,nk ) the number of a's satisfying
Then lim A(c,nk )/ (((nk ) exists and is a continuous function of c. k=Tt follows from t ho primo numb.;r theor om that ~d/log n
~ 1,
and 1 10 ) prov od that lim dn/lo g n <: 1. It is of course to
be expected that lim d flog N=O, i n fact a well known ancient -- n conjocture st at 'J s that d ==2 for infinitely many n. 1 11 ) further n
provod that
I could hot prove that
Turan and 1 12 ) pro,v ca. that dn+ 1 > (1+ t. )d n and dn +1 <: (,- t. )d n have both more than c 1n solutions in integ~rs m ~ n. One would of course oonjecture that xim
dn+ ,/dn =00 and that lim - dn +,/dn =0 and that the donsi ty of integers n fo I' which d 1 > d is 1/2, but those
n+
n
con~ecturos seem very difficult to prove. Renyi and 1'3)
provod that tho numbor·of solution of d 1=d, m n is less than 3/2 m+ m c.n/(log n} , the right orgor of magnit ude for the numb
n+
1=d
n
has infinitely lIl2by solutions. All the 83
- 4-
P,Erdt'Js
results mon·tionod horu ar c obtain0d by Brun's method. In our paper with Turan 12 ) we roma rk that we are unable to prove that the inequalities d
2> d 1> d have infinitely many n+ · n+ n solutions. In fe,ct we arc even unable to prove that at le [',st one
of tho ino qualities d u + 2 > d n + 1.'>d n , d n+2·< d n+ 1< d n ho.v o ly m~ny solutions.
~nfinito-
It scems c ortc.in tho..t dl/log n is everywhere dense in the int erval (0,00). We prove tho follwwing
~2.~~ 4) £f.J2.osi ti ve
.[3e~-"f .1i_1&i.1!.
'rhe
form
a
set
j:!!f:.'l.SUl~.
Remark ,
Despite the fact that the set of limit points of
d Ilog n has }:losi tive measu r e, I can no·t; decide about mny given n
number whether it actually belongs to our set, thus in particular I do not know if 1 is a limit point of the number's d Ilog n. n
First of all it follows from the prime number theorem that there are at least ::::/4.10g x values of
lit
for which
of d Ilog m m (or what is the same thing of d Ilog x) of the m satisfying (5) m have positive measure, (A point z is in S iiff there exi s ts an We shall show that the set of limit points S
infini te se quence x k ~
Q)
and mk
~ 00
~~ X~ < ,\0r",. <.I< 11: .)d <;.f"",·X x\<.....,2")\, Zf",. .';\1. ""1 /: I ${] i~tlt,d ·1n-"" j''1 (. . I:( Ad" StA ci-ecvlA:J ~ (0,2) , 7
~
I~
If
OlD'
ri'lwol'em
IS
can be cov'JrwI for ev(;ry
f a lse then hy the Heinu-Borel theorem S
E
by a finite Eumber of intervals the
sum of the l cn rjjh of whi ch is l ess than k==1,2, ..• ,N,
'11< fillY
'
1-
l-~"
f ,
Let (a k , bIc ),
(b . -a, ),( t be the intervals which cov or S. Let 1
.1
Clearly for all suffici ul1tly large x tho dm which
84
- 5 _.
satisfy
f02
some
l{
1~
P!Erdaa
N
~
(6 ) Now VIe estimate the nUlllber of integers m(p
/6/,
satisfy
A we11 kno·lm rcsul t of
SChllir~lIni:~.r.nI5)
of solu:H ons of d =-1 , p :< 111 In"'"
c.{ .___x._
n
( &~ )( )2 !~ It
11:
(1
<- ::)
m-
which
states that the nu!nbcr
is Insi:l t;llb'l
+ ~ (tv)
.
·
Thus the mi.",1",,! of 30::LCltions i rc
7:l
of p
<.:
Ci.
III
ill '
s ati sfie ~;
/6/; is
less than
(7)
since it is well known that
1:.e pien
(I
+~)
~ C (e - CL )~_ + (~ (T
E /7/
contradicts
/5/.
)
thus our Theorem
is proved ., Before finis hing my lecture I mention a few problems on additive number theory. ROlllEl.:t) off k
form a +p
~
16 )
proved that the (trmsi ty
i nteger , is positive .
Let
a1,q 'l~".
L\k \
£t k +-I
.
'-
integers of the
This re sult is surpri.sing
since th 8 number of solutions of ak+p .~ x Gene:ca.l1sing this reflnl t I proved
0][.
17)
is lE)sS than Pt
x.
c
tho following r8Sul t
be:i\n infinite seqllence of integers satisfying
,then thE: ne cessary and sufficient condHi ,) n that the
densi tv , of p+a .
1~
< C l<
tl. k
is 'oosi ti.ve is thRi;
n (1 ~
'p-Ia",
+ ~)
r-
fOT 8.
<~ , 85
certain c and all k
- 6 -
P,Erd~s
Kalmar 18 ) raised the problem whether for every a> 1 the density of integers of the form
:- k p+ La]
is positive. At present I caJ.not
answer this question. Following a qqestion of TUI'BR I prov,1d 17) that i f f(n) denote the number of solutions of n=p+2 k then. lim f(n)=m, in fact for infinitely Illany n
f(n)
> c,loglog
n. One would guess that
f(n)/log n -""?0r nut I can not even prove that there do not exist infini tely many intege rs n so that for all 2k< n, n_2 k is a prime, it S8ems th",t n=105 is tho largest such Let sfying a k
3.
<
' k
< c,
3') <-
<' ..
inte~cr.
be an infini tc sequence of integers sati-
Demote again by f (n) the munber of so lutions of
It seems likely that Iiill f(n)=oo. 191 18) Van del' Corput . a~d I proved that there exist infin.i te-
n=p+ak ,
ly many odd integers n not of tho form 2k +p, in fact r18) proved that
t~ero
e~.ists
an arithmetic progression consisting only of even k
numbers no term of which is of the form 2 +p. I also wan tad to pro va that for evory r there exists an ari thmutic progression no term of whieh is of tho form p+q
I'
where q
r
has not mora than r prime
factors , Thls result would easily follow i f I could prove the following conjecturo on congruences which sooms
intor~sting
in
itself: Ta every constant c thero exists a systom of congruences
so that -e"VJ!:'ry intoger satisfios at laast ono of tho congruences
(8),
'r he simI)lcst such system
5(mod
6), 7(mod 12).
i~'
O(mod 2), O(mod 3), 1 (mod 4),
Doan Swi ft constructed such a systom
Vii th
0=6, nut tho general
question sooms vory cl:ifficult. tl provo d th a t th ore o~s . t s a num b orr so · . k, 20 'roceny ) Llnlll k1 that m'ory intcg\)r is or[ tho for~ P1+P2+2 + •.• +2 kr • It seoms vory
86
p • .srd~s
- 7 likel, that for every tho form
.
p~2
I'
there arc infinitely many intogers not of
k
+ .. •• +2 r.
Ono final problom: Is it truo that to every c 1 ,c 2 and sufficiently largo x ·thoro oxist mor e) than ~1· log x consocutive primos
~
x so that tho lIiifferonc o betwoon any two is groater than
c,}
If 01 canb r] chosen sufficiontly small this is well known21:
F 0 0 T NOT E S 1) Bu..ll.Amol'. l.l£.th.Soc. 53,604 (1947). 2) Cormlwnt a ti one s in honorom. Ernst Leonardi lindellH, Helsinki 1929. 3) SHz. Berliner Math. (}es.29, 1930, 116-125. Soo also H.Zoits, :nemental'e Butrachtungon Uber oine zahlenteoretischo Behauptung von Legendre. 4) Comm " Ph~TS. -Math •• Helsingiol's) 5,(25/1931. 5) Quart0rly Jonrnal of Math. Oxford Series, 1935, 124-128. 6) Schrin;on des Math. Seminars und des Institutes f11r angewandt o I:Iath. Bel' Univ.Berlin .1, 1938, 35-55.
7) London Math.Soc. Journal,
11,
1938, 242-247.
8) Quarterly- .Journal of Math. Oxford Series 9) Acta Arithmc-:;:i.ca,
~,
.§., 1937, 255-266.
1937, 23-45;
10) Duko Math.~ournal §., 1940, 438-441. 11) Publ.tiath •
.1,
1949, 33-37.
12) BulLAmor .Math.Soc. 54, 19 48 , 371-378. ibid 541 1943 , 88 5-889. 13) Simon Stevin
\Hs.
800
further P.Erd~s
Natm:zrk. Tijdschr. 2'(, 1950, 115-125.
14) Professor Ricci informed me in tho discussion follwwing my locture~ that this r esult is an Jeasy consequence of some previous r os ults of this. (Hivi sta di l'Iiat.Univ.iarma, 2. (1~54) 3-54) . 15) E.1andau, die GoldbachschG Verm1J.tung und del' Schnirelmannsche Satz , Gottingor Machrichtcm, 1930, 285-276. In fact Schnurcl-' mann provos that the numb er rod!: solutions of P1- P2= ·t , P1 < x
87
- 8 -
P.Erd6s
16) 11ath.Ann.alon, 109 (1934) 6G8-678. 17) Summa Brasilionsis Mathcm(::.ticao, II (1950) 113-123. 18) Oral ComnlUnitJation. 19) Van dar Corput, Simon Stovin, 27 (1950) 99-105,(in Dutch). P.Erdtls (in I-Iungarnan) lUath.Lapok III (1952) 122-128. 20) Trudy Math. lnst.Stoklov, 38, 1951, 152-t69. 21) P.irdl'Js, Acta Szogod, 15 (1949), 57-63. Soo also M. Cugiani Annali di Matcmatica SariG IV, 38 (1955), 309-320.
88
GIOVANNI
C='!S~~~==~~'!ft_~:II
RICCI
Iie.'="et.
e:cr
__ _ _ _ _ _
SUL RET ICOLQ DEI PUNTI AVENTI PER OOORDINATE I NUMERI PRIM!
....,.------
ROM! - Istituto Matematico dell'Universith, 1956
89
G. Ricci
SUL RETICOLO DEI PUNTI AVENTI PER COORDINATE I NUi:.1EBI PRIMI
1.
Oonsideriamo la successione crescente dei numeri interi
naturali primi P1' P2' P3~~" che, insieme a quell a dei loro o)posti, si ordina in una successione bilatera (p . -p.)
••• , P.i' P.i+1' ••• , P-2, P- 1,
.-3.
p"
=
:I.
P2 , ••• , Pi - 1' Pi' •••
Riferito i1 piano a un sistema cartesiano ortogonale O(u,v), ai consideri 11 sistema dei punti le cui coordinate (u,v) sono ambedue
intfJr~
primi, presi col segno posi tivo
0
negativo:
e, in
p~rticolare,
B.l.
denotiamo con B. gli speciali punti
= A (i,i
3.
+ 1)
i: (
p., P.+,). 3.
l.
II sistema dei punti A (i,j) costituisce un reticolato irrego1are, 1a cui irregolarita ~ribuzione
e connessa
con quella della di-
dei numeri primi.
Poniamo il seguente problema: "Assegnato un campo C sul piano 0 (u,v) quale relazione sussiste fra 1'area e 1a posizione di C col numero dei nodi A (i,j) contenuti in C?". Nel caso in cui 6i consideri l'insieme di
~
i punti
coordinate intere, il problema analogo a questo e classico nella Geometria dei Numeri ( 1 ); rna per i1 retioo lato irregola8
re che abbiamo presQ in considerazione i1 problema si presenta molto difficile.
(1) Una espOSlzlone riassuntiva sugli studi e sui risultati riguardanti questa problema classico si trova in G. RICC~,Fi gure, reticoli e computo di nodi, Rendiconti Sem. mat. e fis. Milano, ~, pp.165-205 (1948). Qui s1 trovano anche indicaz10ni bibliografiche. 91
- 2 -
G. Ricci 2.
OSSERVAZIONI PRELDUNAll.I.
11 sistema dei punti (nodi) A(i,j)
a ciascuno degli assi u bisettrici v
= u,
= 0,
= -u.
v
1a distribuzione dei nodi
1
(~)= [
r' ~
1%
..
=0
rispetto
e rispetto a ciascuna delle
Su ognuno di questi assi di simmetria
e regolata
(f ~
) ~ t.i. 0
Dal :t'eorema di G. HOHEISEL.
)( (~t ~ (t )
v
e simmetrico
-11(~ ) tV
+0
dal IIPrimZClhlsatz ".
[~t"(- ,e,t8~)J '" t
9
--->-.(ff<~<~) ~~
!
S iv~ ~ ( iJ < 8 < -1 ) I
e da altri analoghi. Alcune proposizioni classiche sui numeri primi hanno una ovvia interpretazione su questo sistema di
nodi("reticolo~);
per es.empio: 1) I nodi sulla retta v = u+2 corrispondono alle coppie (Pi'
P1+1 ), con Pi +1= Pi +2, che sono le ooppie di"numeri primi gelIlelli".
v) 0, u+v = 2n corrispondono alle "pa.rtizioni di Goldbach" p + p' = 2n. 3) Un elassieo teorema di Viggs Brun ci dice che, au ogni parallela alla bisettrice degli assi, il numero dei nodi aventi 2) I nodi sul eegmento u)
0,
I
r
aecissa u ~ ~ non supera C ~ ~~ ,rna d' al tronde non e noto se di tali nodi ne esistano in numero finito 0 infini to. Questo teorema ci dice che le bisettrici v =
Z u,
come
assi di simmetria sono privilegiate e che Ie lore parallele sono estremamente pili povere di nodi in confronto di esse.
3.
NORl~LIZZAZIONE
DEL PIANO.
Quando si voglia confrontare l'area di un campo con i1 numero dei nodi A (i,j) in esso contenuti conviene Itnormalizza92
- 3 -
G. Ricci
re"
i1 piano mediante una. trasformazione che sia in acco:rdo con
)
l'andamento di 1{(~ ~ Poniamo x = x(u), y
{tAl J
.
'( =• ~Iqlt~)
I
f~ + = y(v) e precisamente J'" clfJ, 't" fnt~M+t) Cl() .
0
otr.:Nella . :I4ii~I"IH) . ~" ~(I ..i+eJ rasformazione (u,v)--t (x,y) I
ti Bcorrono ciascuno in
i due assi coordina-
se
stesso, con una contrazione intorno
all'origine: poniamo A(i,j)==' (p.,P.) -+ l.J
r~
dove
Ill.:::
P(i;j)~
(x . ,y.) l.J
oJ~
te.~tl ~
It')
L'elemento d'area nel piano (x,y) risulta dxdy =
~
1+
'0 (u,v)
~
du • dv logu" log v
dove
e 'Y (u,v) = (J u logu
+
1 v 210g v) •
8i consideri per ogni punta P(i,j) la minima delle distan-
ze dai punti rimanenti;
e estremamente
probabile che tale di-
stanza abbia come minima limite 10 zero al crescere indefinitamente di i+j, rna questo fatto non e ancora dimostrato. In ogni modo la distribuzione di P(i,j) risulta normalizzata rispetto alle lunghezze su ogni parallela agli assi e rispetto alle aree. 4. CONFRONTO FRA LIAREA DEL CAlI1PQ E IL NUIv1ERO DEI NODI P(i,j) CONTENUTI.
Sia C un qualunque campo del piano normalizzato (x,y); 93
- 4 G. Ricci denot1amo con S(O)la sua area e con N(C) il numero dei nodi P(i,j) contenuti in C;
denotiamo inoltre con N (C) il numero o dei nodi particolari Q. P(i,i+1) contenuti in C. II nostro ~
problema consiste nel confrontare N(C) con 8(C). Lo scopo di questa espoaizione
e di
rispondere a questo problema per tipi
particolari jel campo C, interpretando geometricamente alcune
propo~izioni
stabilite in
t4~
nostra memoria (1).
Consideriamo come campo C ilseguente parallelogrammo (aperto) T = T(a, b;
r ,A. )
x+)A- <. yt x+
(a.( x(b;
orientato perallelamente alla bisettrice y
A)
=x
e all'asse del-
le y.
POiche Bulla retta y = x si trovano tutti i nodi P(i;1)
e evidente
che in base al Primzahlsatz
-A , A)
per ogni campo T(O,b; b = S(T'
lion
e noto
£ ) quando
se questo valga per
£.<m
o"A
~ t
risulta
b~bo( [.. ).
~ >" t.
Veniamo alle proposizioni annunciate. Tear. I.
N(T)< 8S(T). (1+
f1 (1» per a
0< S ~ 1 fissi e b...., + 00 Tear. II. Se I' : : 0, A 15/16,
= (1-
S)b; °~< A
I
che
N(T) ~ No (T)
Teor. III.
>
a ::: (1- b)b esiste
J
c>o tale
c· SeT).
Se No(T)
('\J
o ~ f < ~ , °< S ~
S(T)/(A
-1) per a = (1- 5)b;
1 fissi e b ~ + 00 allora sono verificate necessariamente ambedue Ie condizioni:
)1..{ 15/16, (1) G.RICCI, Sull'andamento della differenza d1 numeri prlml consecutivi, Rivista di Hatem. Univ. Parma,2, pp.3-54 (1954); a questa memoria rimandiamo per la bibliografia (p.54). 94
- 5 -
G. Ricci
~~o.~ . J:V.
set = 0, A)
= (1-&
per a
1
(0<.3 {1),
}h,
>t
N(1J!) ) No (T)
~~ eb
\6 f
allora. ..
~bha.stanza ~e e
S(T).
In particolere de que atc ul t1m.o teoreJll8 s1 r1cava Tear. V.
Se ,., = 0, N(T) ~
per a
= (1-
f)
A= 1 aUora
N (T) 0
b (0
>
e
O· 555 • S(T)
<&~
1) e b abbastanza grande.
Le proposizioni pr~cedenti ci conaentono di individuare famiglie di campi C per Ie quali, a1 orescere indefinit~ente del campo Bussistono relazioni del tipo ~1 See)
<
N(e)
<
02 S(C) (0 1 , c 2 costanti> 0) cioa
per Ie quali i1 numero N(e) dei nodi in esso contenuti ha l'ordine eli grandezza dell 'area S(C).
95
THE G E 0 MET R Y
0 F
N U MB E R S •
ROhlA-Isti tuto Uatematico dell ' Universita,1955-R011A
97
C.A.Rogers
- 1 -
THE Introduction:
GEOMETRY
~HEOREMS
OF
OF
NUIABERJ?
BLI~LDT
AND
M11~I.
The Geometry of Numbers is concerned wHh a study of the relationships between point lattices and sets of points. In this course of
lectur~s
I want to give an account of some of the basic
general results of the subject. I vall confine my attention to the study of what may be called homogeneous problems, and will say nothing about non-homogeneous
problems; partly because
eight lectures are not too many to devote to the homogeneous problems, and partly because i t is the theorems
0 11
the homogeneous
problems which form the basis of the subject. The results on non-homogeneous problems tend to be rather special or rather simple. We shall begin our study of lattices by investigating some of tho propertios of a special ihattice, the lattice of all points :!!=(u 1 ,u2 , •.• ,un ) in n-dimensional Euclidean space with integral coordinates u 1 ,u2 , ••• ,un • Later we shall see that, in a very precise sense, this l attice is a typical lattice. We first pro+\
ve a result due to Blichfoldt ; ~~Ki~M
1. (Blichfeldt). Let S be any bounded set of points
1£=(x1ix2, ••• ,x) in n-dimensional sllace. Suppose that
5
has a
volume 1(S) ') 1. Then th \}re are two points 1£,X in S, auch that the difference point 1£ - X::: (x 1 - y l' • d
,
xn ... Yn)
has integral coordinates. Here the term vo lume has boon left vague on purpose; the theorem and its proof remain valid for any reasonable definition 9.f_~9.!~~1._~!.s;!._f~.r
tho Jordan content or
thc~
LGb csguG measure.
+) H.F.BLICHFELDT, Trans. Amer.li1ath. Soc. , 40 (1914), 227-256.
99
C.R.Rogers
- 2 -
Proof. J!'or arty p(i);int .§!: = (a 1 , ... ,an) with integral coordinates, let
Let
5 (.§!:)
S(.§!:)-~
denote the set of points ~ of S with
a,
x 1 < a 1 + 1, ••• ,an' xn <: an +1. donote tho translated set of points ~-~ with! in ~
S(a). The, using V to donote th o volume of a set, it is clear that
Further, each set
is cont\1inod in tho cube C of points
S(§:)-~
I with
< 1, ••• ,0
O~Y1
<1.
~Yn
As S is b@unded, all but a finite number of tho sots be ompty. If all these sets 1
S(~)-~
~
=
2 V(S(~) ~
i So at least two of tho sots common point, s ay
~.
-x
=
)
~)
-
V(S(~~))
will
wero disjoint, we would have
Ls (~) -~}
V( C) :;:t V( U
=-=
S(~)-.§!:
= v(s)
s(§)-~,
>1/
<say S(.:£)-.:£, S(£.)-,2., have a
Then the points
-z
+ b
,
belong to S, and tho difference point ~ - I =.:£ - £.
= (b 1-
c,; ••• ,bn- c n ) has integral coordinat es, as r oquired. We now want to use this theorem due to Blichfeldt to p1:'ove MinJwwski's first theorem. A set
But wo first need two definitions.
Kof points in n-dimensional
specr; is said to be conveN,
i f it has the property that, if ~ and I aro any two pmint s of
then every point
~
~ = =
of the form
~ + (1-
(A. X1+(1- \
A) x. )y 1 ' ... , 100
AXn+ (1-
~) yn) ,
K ,
C.A.Hogcrs
- 3 -
where
~
0
A~
joining .! and
1 ( i. c . every point
yJ
~ of the line segment
lios in K. A set S is said to be symmetrical or
symmetrical in tho oriGin.Q. '" (0,0, ... ,0), i f i t has tho property that, i f ! is
point of S, then the point
an~'
-~"'(-X1'-X2"" ,-xn ) is also in S. With those definitions wc can stat 0 and prONe Minkowski tIs first theorem.
THEonE)i 2. b OlUld ud convex
(l.1inkowski ' .
. . :l symme-Gr~c&..uL
IS
se't
fLcst theorem). ., h
W~-CLVO
If K is a
1 umo gruu"cr tllan 211 , there .j.
is a point of K, other t hat the; o:;:'i gin, with integral coordinates. Proof. Consider tl18 S()t of all po ints of the fo rm 1 2!
1 1 , •.• ,2 Xn) = (1Z'X 1 '2X2
'
whure x 1.ies in K. This get is denoted by ·}K. It is cloar that the volume of ~K i~ ("~)n timns that of K. is groater than 1.
~hus,
So the volume of !K
by Blichfoldt's thoorem, thore are
!:;-x.
points !;;,,1.. in
1l-K,
coordinates.
By the definition of' tK, the !Jo ints 2!,21.. lie in
K.
such that the differenco point
By tho symmoi;ry of Kt tho poillts 2!; -2K. lie in
K~
has integral By the
convon ty of X, tho point t(2!) + i (-21..) = ! - 1.. liGS in K.
But lIIhis point has integral coordinate,s, and it is
not MO origin, as x :.,md y aro dist:iinct. This provo£' the theorem. This s imple theorem of Minkov:ski is the fOll.l1dation stone on which tho Geometry of Numbors i s bUlillt. It can be restat ed in various ways; there is 8.llother way of stating the theorem which is ;nore convellir-mt when one wishes to apply the theorem.
THEOREM 3. If K is
8.
bounded convex symmetrical set with
n
volume 2 , there is a point i n or on the boun dary 0:::' K, other that the ori gin, with integral coordinates.
101
C.A. Ro gers
Proof. Since, for each positive i nte ger r, tbe set (1 +
1) r
(1 )
K
is a bo-unded convex s ymmetric al set, with volume
(1+~) n
> 2n
2n
it follows, by Theorem 2, that this set contains a point,
~r
say, other than the or igin with integral coordinates. Por each r, the point -r u lies in the bounded set 2K. But there are only a fini t o munber of points with inte gral co ordinat os in the set
2K. So ther e must be a point v wi th int egr al coordinates in 2K, and such that
1. :: l:!r for an inf i nite sequenc e of value s of r.
Then, 1. i s different
from t he origin, and belongs t o the set (1), for some arbi trar'i.ly large value, s of r. This implies t hat the point (2)
v
lies in K, f or some arbitrarily l ar ge values of r. But as r tends to i nfinity, thi s point (2) converge s to t ho point 1.. Hen ce 1. is a limit of point s of K, and so lio s i n Ie or on the boundarw of K.
Thus 1. satisfi es the requir ei.lonts of t he theorem. Follo wing rH nkol'1ski, we 2.pply_his t heorem to provo a theorem
on simultaneous ,Diophant ine approximation , i.e. on the approximation of a number of given ittational by r a tional approximations with t he same denominator. THEOREM'
4.
being irrational.
Let <-(p at}" •• • , ol." be any real numbers, Then there are an infinite humber of sets of
integers u 1 ' 1.12 " • • , un' v with v
)'f\ Pro~
1, sat i sf ying the conditions
~
I t
1)J1\.
I"
,
-.!)\< tL -
1101.. M.. .. V
1
-
We obtain the i nfinite number of solutions by an
inductive process. If we have no solutions, we start the inductive pr ocess by taking t =
i.
Otherw:£se, i f we have a finite 102
C.A.R5gers
- 5 -
number of solutions, we take
t
t ==
min Iv ()(C u 1
\
,
where the minimum is taken ovelO the finite number of solutions. Then, t is positive, since takes the value O.
eXt is
irrational, while v never
Further,. since any solutions has
t.
it is clear that t ~
Iv 0(.1 -u11~1,
Now consider the region K in (n+1)-dimensional space consisting of the points with coordinates (x 1 ,x2 , ••• 'Xn'y) satisfying -
!
t,
b(1 Y
~
o(2 Y \
!Xn I
~nYI ....< ~
y
t,
-1:;,
t
-n
•
It is easy to SGe that K is a parallelepiped with the origin as
centre, with volumG 2 n+ 1
Thus K is a closed bounded convex
symmetrical set with volume 2n+1. So by Theorem 3, there will be point, say (u l ,u 2 , ... ,un ,v), in K, ot her them 2., with integral coordinates. By changing all tho signs, i f' l1ocess8.ry we can
~
ensure that v ~ O.
lu.-
Then we have v ~ 0 , and
c;(.vl
1.
:I.
I
iv I ~
c::.
"'-
(i=1
t
,2, ... ,n),
tn.
If v were equal to zero, it would follow that
\u.1 :I.
I
so that u 1 ' u
A
Co
,
•••
,
(i=1,2, •.• ,n),
1
" t --
"2
u , v woul d all be n
Z G1I10,
of those i ntegers. Thus
v
1:-
1
and
/' -1/11 t ~v •
Konsequontly
103
contrary to our choice
C.li.Rogers
- 6 -
0(.
t -
v
....<
1
t
and u 1 ,u 2 , ••• ,un' v satisfy our requirements. \ v (l(1-U11
Further, since
~ t,
it follows f rom our choice of t , that tho solutions is not one that we alruady had.
Honce; it is clear tha t the inductive
process will lead to an infinit y of solutions of the inequalities. While it Flay bo shown tha-t Theorem 4 will still romain valid if the constant
in tho numerators of tho right-hand sides
of the inequalities is r epl aced by a slightly smaller numbor+), it is not possible to prove a theorem of this type with the exp one~t
Perron
(n+1 )/n r epl aco d by anythihg larger. This was ShOwn by who proved that there is, fo r each n, a positive constant
c n and al gebraic numbers 0( l ' quali ties
0/... 2" •• ,
Lattices: TIE_BASIS
'<
n such that the ine-
TJ~.9Ki:'1iI
SO far we hav e only been concerned. with the lattice
pOints with integral coordinates.
I\of
The fol lowing definition de-
fines the ge!1er8.1 n-dimensional la·t;tice in ter-JD.s of this special lattice. Definition. A lattice ~~_:l2~!~~::!.1._!~~£~_£~~
1\
in n-dimensional space is a set
be obtained b;y applyi!1g a non-singular 11-
+) See P.MULL:SNDER, Proc.Nederl.Akad.v.Wet.,52 (1949) ,50-60 .
t)
O.PERRON, Math. Annalen, 83 (1921), 7'7 - 84.
104
C. A.Rogers
- '7 -
near
transform~~tioa
to the set of po i nts with integral coordina-
tes. This definition has the advantage that the lattice of points with integral coordinates is particularly simple and has many obvious proper-ties, further it is clear that many of these properties are left
by a lillear trasformation. For example,
invari ~n t
the distance between any two po iats with int egral coordinates is at least 1.
r
1. =
r
Now, although a non-singular linear trasformation
t rasforming the general ])oin-t; .!= (x 1 , •.. ,x) i nto the point
2S given by
does not preserve the distance
botwe en two points
theBo \-,ill bo constant s c, C with 0
~,.?S;
such that
It follows that the dist ance between any two points y,y' of the lattice
r' 1\0 is
at least c.
Thus uny lr. ttice i8 a discrete set of
points. Another trivi al result is that, if of /\ 0 ' then
-~
and
~
- 1. are
.~
and y.. are any points
a1 so po i n ts of /\0'
It follows
immediately that the same result holds for any lattice It should be noted that the same lattice
represented i n the form transformations
r
{I
for an infini ty
~f
always be
different linear
However we can prove that tlllie ab solute value
of the determinant of cular choice of
r /\
1\ may
depends only on
J\
r " Let d( r ) denote th.i_s 105
and not on the partiabsolute value of
C.l..Ro gers
... I) -
the determinant of
=
r
2
A0'
where
r
r . Suppos e that we have
/\ ""
r'1 A °
r 2 are non-s ingular linear
1 and
=
transfor-
mations . Then, using t he usual notation
r -11 r 2 l eave s
So the transforma tion (0;:
(\ 0 invari ant . Thus
the trasforms
tV ~{ ; 8 h
j
•
,
•
I
e~ \\
of the points
belong to
~1
(1,0, •.. , 0 ),
-2
e
(0,1, ••• ,0),
e -n
(0, 0 , ... ,1),
Ao'
the matri x
B
( a)
and so have inte:'.~ral coordinates. This means that has i ntegral elemelltfl .
sitive int eg er. But, since
i
Henc e d (
..
Gf /\0= /\0 '
0 )
must be a po-
i t follows , in the
same way , tha t d( f) -1 ) is also a pos it i '.' ':; int eger . Now we have
This implies t h at
~hus
and the value of the d etenninant d ( tice
1\
r ) dep ends
onl y on the lat-
and not on tts represen t a tion in the form
106
1\= r 0 0 ,
So
r:; . A. Hogere
- 9 -
we are lead to ma.ke the followi 118 definition. Eefinition. The determinant d( f\. ) of a. lattice:
4(
ned to be
r
1\
r ), where
1\
is defi-
\-. is any linear transformation such
that = ;\0' We shall repeatedly need to make use of tho concept of a basis
A. • A
for a lattice ~
=
set of n linearly independent Iloints
(i) (x 1 '
(i)
X2
(i) , ... , xn)
will be called a basis of the lat-Qicc
(i=1 ,2, ••. ,n)
1\ , if f\
consists of the
points + ... +
+ where
,un take all possible sats of integral values. It
u1'~2""
is cloar, for example, that the points (3) form a basis for the lattico Aoof all points with integral coordinates. With any set of n linearly independent points !1,f2 , •.. ,fn we can associate t'he corresponding non-singular lineal_' traEformation \' given by (i)
x_
~ij
(i,j
~
=
1,2, ... ,n).
Then the sot of all points of the form.
+ un-n x
where u 1 ,u" ... ,u take all po ssible sots of integral values, is ;: n precisoly the set ,-' SO tho points x 1 ,x 2 " .• ,x form a basis for
1\ '
A(.)
--
-
-n
if, and only i f, the corresponding matrix
formod from their coordinates, tr8.nsforms the lattico lattico
A .
f\o
r
into the
We denot e: the absolute value of the doterminant
of the coordinates of points x 1 ,xr" ••• ,x by the symbol -c. --n d (!1 '!2 ' •.• , 2Sn ) ; it folloxs immediately that, if 2S 1 '!2"'" a lattice
1\ ,
~il
then d (x - 1 ' -x 2 " • ., -xn )
107
d(
1\ ).
form a basis for
- 10 -
C.A.Ro gers
When we are studying a particular latt ice" i t i s often convenient to work with some s pecially chosen basi s .
The follo ..
wing theorem gives one method of choosing a basis, having a specially convenient relationship to any given set of n linearly independent points of the lattice. THEOREM 5. (The Bas.isTheorem). Let 2:1 '!2"" '!n be n linearly inde pendent points of a lattice A • It i s possible to find a basis X1 ,X2, ••.
,Xn
of/\ of the f orm (i
1, 2 , ••. , n )
where
o~
c( \
5 ~ J i/~
1=-' L
Proof. We not e t hat the theo r em f0 2111S the caSE:: k=n of the following proposition, of
A
Proposition Pk , of the form
It is po ssi bl r3t o find. points X1 ,:1.2'"
• ,Xk
\.,
o'
\1 X- ~
... =
\{' :: ' \ ,..}.
-
~ ~
..>: 1.
(; 12
, , •.• , k ) ;
where
\is
1
() < d\. ~i I J~' such that avers' poi nt ! of ," which
f or real
L
CLm
to 8xpre s f3ed in the form
f..- l ' A 2 , ... , 'A" can also be fJ xpr ossed i n t he form
! = u1X1 + u 2 for integers u 1 ,u 2 '. · . ,Uk'
~2
+ , .. + utXk
108
O.A. Rogers
... 11 -
)ie prove Proposition Pk for k=n by induction on k. First consider the case when k=1. Consider the set L1 of points ~ of
1\
which can be expressed in the form
A 1~1
~ =
where 0<>--1 ~ 1. The point x 1=1.x 1 is an example of a point of L1 • Since!\ is discreet there are only a finite number of such pOints. Take
iL1 = ~ 11~1
1\ 1
to be the point TJ 1for v:i1ich say ~ 1 1 •
t akes its smallest possible value,
Then Y1 is a point of
1\ and 0 <0(11 ~ 1. We have to
prove that, if
~ is a point of
1\
==
,\
1~1
then ~
==
u 1 iL1 Suppose that X=
for 80me integer u 1·• Cho08e u 1 to be the inte ger such that
o ~ \1-u1~11 «X
A1-X1
is a point of /\ •
1 1·
Then the point
is a point of
e - u1iL1 1\ of the
from the choise of
ct..
== (
"1 - u 1 ol11) .e1
form we are constcle:C'ing. It follows
11 that we must have
so that as required. Now suppose that 1
n and that the Proposition Pk-1
holds. We deduce that Proposition I'k also holds. :t3y:2roposition P k - 1 we can find points Y1.,x'2'''',x'k-1 of the appropriate fom such that every point ~ of
1\ which cun be expressEd in the fom 109
- 12 -
C.A, Rogers
can also be expressed in the form
for integers u"u 2"",1\_1' Consider the set Lk of points! of
1\ which can be
expressed in the form
where (5)
The point !k
=
O'!1 + •.• + O'!k_1 + 1.~
i8 an example of such a point.
Since the set of points (4) where
;.., , ••• , "k satisfy (5) is a iIlounded set, there will only be a finite number of points in Lk • Let
be the point of Lk for which '" k assumes its smallest possi ble value
at
kk' To establish Proposition Pk it suffices to consider any point! of 1\ which can be expressed in the form (4) and to show that ! can also be expressed in the form
We can certainly choose integers u k ,v 1 ,.,. ,vk_ 1 such that Of Ai<. - \.A..K'::.(.I(K <<:.I~~,<
110
C.A. Ro gers
.. "3 "."
Then the point ~
(~1
- v1
~1
~c(k1
-
- v2
~2
- ••• - v k_1
-v 1 ).!1-+-···+(
~ k-1-~ + (
~
~-1
-
~ ~k
C(k(k-n-v k-1).!k-1
k-ukC:<
kk)~
would be a pOint of Lk with too small a value for~k' if we had
Hence
and
~
can be expressed in th& form )(:. ,u ,.1.J', +1 .
-
J
-
>.
where (i==1, ••• , lc-1).
But by Proposition Pk-1 the point
can be expressed in the form
for integers u 1 ' ... ,uk_1 • This gives the appropriate representation for~. Thus the Proposition :?k fiolds and the theorem follows by induction. Corollary. Let of a lattice of
A such
1\ ,
It
~1'~2""'~
be n line arly independent points
:i.s posr;;ible t o find a basis ;i1 ,;i2'" ., ~
that 111
- 14 -
C. A.Rogeils
(i=1 ,2, ••• ,n),
for some integers ~ ij' Proof. Take X1 ,X2 "",Xn to be a basis provided by the theorem. Solving the vector equations for -1 x 'Y2""'x It: -n we obtain
for some real numbers ~ ij'
Since X1 , •••
it follows that the coefficients
13UCC:CSSlVE Having studied
lB lJ ..
'Xn is
a basis for
are integers.
l:1INIMA
s~e
of the elementary properties of lattices
we now investigute some of the relationship between lattices and sets of points.
We are in a positicn to state and prove the gene-
ral form of Minkowski's fil'st theorem. THEIlliREM 6. (Minkowski's First Theorem). Let
1\ be
metrical set.
~ny
lattice and let K be any bounded convex sym-
SUI'Pose that the volume V( K) of K satisfies (6 )
Then there is a point of K.
!
other than ~ of
A
in oroon the boundary
If strict inequality holMs in (6 ) , then there is such a
point 2S in K.
r is
Proof. The lattice (\ can be expressed in the form
a non-singular linear trasf ormation. Then the linear tran§.
formation oj. set
~ I~where
r
r -1.transforms •
K into the bounded convex symmetrical
K, and transforms the boundary of K into that of
The volume of
r "Kis -t
112
r -K. -I
C.A. Rogers
- 15 -
so that
v (r-1K)
I-
2n ,
with strict inequality, if (6) holds with strict inequa.lity. Hence, it follows from Theorems 2 and 3, that there will be a point ~ other than .2. of (\0 • which is in or on the boundary of 1 K, and which is in r-~'K i f strict inequalit~lOlds in (6).
r-
is now clear that the corrisponding point,!
It
=r ~ of ,,=
1-' Ao
satisfies our requirements. It is convenient at this stage to introduce a number of definitions. A lattice set
A will be said to be admissible for
S, if there is no point 2S, different from .2,', of
set S will be said to be of the finite
t~.
1\
(!
in S, A
if there is at least
one lattice which is asmissible for S, and will be said to be of the
infini~.£...J.y~,
cri tical
if
deteJ'mi~.~.
there is no lattice admissible for S.
The
of a set S of the finite type is defined
to be the lower bound of the ·determinants d( .~ ) of the lattices
1\ admissible for S, and is denoted by
6
(8).
The critical
determinant A (8) of a set S of the infinite type id assigned the conventiona+ value + co • While these definitions have been stated for arbitrary sets S, we shall be mainly interested i n fairly simple sets. In part&cular we will be concerned with star bodies and convex star bodlies.
By a star body I will mean a non-empty open set with
the property that, if ,! is any point or boundary point of S, then every point of the formA,! where -1 <.. ~ -< 1 is also a point of S. An open convex set will be called a convex body.
A convex star b£
iy is any set which is both a star body and a conveK body, i.e. a
sy~netric
convex body.
In terms of these definitions Minkowski's first theorem takes the following form. THEOREM '7, V(K), then
If
K is bounded convex star body with volume
A (K) -
" 2-n V(K). /1 113
(7)
C.A. Rogers
- 16 -
Froof. If /\ is a lattice admissible for K, it follows from Theorem 6 that d(
1\ ) 'r
6
So the lower bound
2-n V(K).
(K) of such numbers d(
A)
satisfies (7).
We will also need 'the concGpt of the successive minima of a set
S for a lattice A Definition.
If 1
~
of a set S for a lattice of the numbers
r
k
! n the k th minimum
A, is
A.
fk
if
)iA <. ,Pi' the only point of
I\. )
S contains at least k
H is clear, from -this definition, that if
only point of the lattice"
f'Ak (S,
defined to be the lower bound
such that the set
linearly independent points of
l" k=
r <. r
l.. then the
which can be in S will be
r-
-1 (\
.Q..
Thus,
which can be in S is 2"
so that
or
ConsequeAtly we have (8)
Combining this with Theorem (7) we obtain the result that
r
't'l.. 1
()
\jlZ
~2
'\,\
(9 )
for any convex star body K. These inequalities (8) and (9) are of little intrinsic interest, the first being an immediate consequence tions
Q[[
the defini-
and the second being a trivial cOlls equence of Minkowski's
first theorem; but they point the way teal some beautiful results 114
C.A. Ro gers
- 17 -
which have far reaching consequences in the Geometry of Numbers. The main known results, arranged in increasing order of generali-
r
ty and decreasing order of precision, are as follows. THEOREM A. Let I~ l' J\~ 2"'" n be the successive minima for a lattice /\ of a set K, which is either (a) a sphere with centre £ in n-dimensional space, or (b) a convex star domain in the plane, or (c) a convex star body in 3-dimensional spacE Then (10 )
TREORE!.l B. (Minkowski I s second Theorem). Let t-"1
r 2"
••
,r-
be the successive minima of a convex star body K for a iattice
1\ • Then (11 )
THEOR:cM C.
Let
I~ 2' " .,
Pn be the
sl1c cessive mini-
ma for a set S. Then ( 12)
These
~heorems
have been assigned letters as they are
Besults , which I do not intend to prove. The results in cases (a) and (b) of Theorem A are due to IvIinkowski} that in case (c)
is due to A.C.WOOdS.+)
+) Not yet published.
115
n
- 18 -
C.A. Rogers
Theorem B is due to Minkowski, but simpler proofs have been given by Davenport+) a.nd bI Estermann+). Th.eorem C was found independently by Chaubauty ) and by myself++), but owes much to earlier work of Jarnik-tr+). I t is worth noticing that the constant on the right hand sides of these inequalities (10),(11),(12) are all best POSSible. l
)
It has been conjectured that the inequality (10) holds for any convex star body K. If we adopt the terminology of Polya's 'Plausible Reasonmng' , we can say that, the fact that the truth of the inequality (10), together with the y,now inequality (K) , V(K) implies the known result (11), renders thres conjecture more
plausible. However Wood's proof of this conjecture, when n=3, is by a method which would become hopelessly complicated and involved, aven when n=4.
I myself am inclined to believe that
the conjecture is false; but I should admit that, despite quite a lot of effort, I have not been able to find a
cour~er
example.
Perhaps my failure is connected with the fact that I was trying to conntruct such a counter example in 3-dimeasional space. As far as the applications, which we wish to make in these lectures, are concerned there is nothing to choose between the tlureo theorems. Each would serve equally well for our purpose. Furiibef the precise values of the constants on the right ....harid sides are not important. So I have made the arbitrary decision to prove the following weaker form of Theorem. C, which c'Jmbino S !~~_~~~~~~~~~~_;~~~ch so often go together) of generality and
+) H.JJA";;ENPOHT, Quart.J.oi Math. 10 (1939), 116-121. Jour-n~DdOn Math. Soc. 21 (1946), 179-182. ComJ2i.~..BGnd..u._~, 228 (1949), 796-797.
t) T.ESTERMANN, $) C.CHAUBAUTY,
++) C.A.ROGERS, Froe. K. Ned. Akad. v.Wet. 52 (1949):;t 256-263. ~)
I)
V.JARNIK
,Vestnik Kralovske Ceske Spolecnosti Nauk (Fraha t 1941) See K.I.lAF!LER. !'roc.K;Ne{L.Akad.v.Wet.,52 (1949),633-642. 116
- 19 -
C.A.Rogers
simplici ty. THEOREM 8. Let ,f-J! tll!"" fn. be the successive minima for a lattice/\of a set S, with 0 < 4(s) + IX). Then
<
/t t
2.. ' .
f
rt
6 (s) ~ 2 'tl- 1 01 ( A)
(13 )
Proo!. We choose integers m1 ,m 2 , ••• ,mn of the form
mk
=
(k
= 1, •• .,n),
where a[k is chosen to be the integer such than (k=1 , ••• ,n).
( 14)
Then
J
We choose a number
with
Now we may eX]?ect that there will be 60me points of the lattice
1\ in the
set
For each k, with 1 ~k~ n, let Ilk be the set consisting of the point e and all points of in t he set o ~S. Now, i f .! is any point of
/\
1\
in ~ mle S then the point
' 1ct.....J'.+i-:3..\<..,
X
is a point of
1\
in
V~+1
::.. IYnK-H..
W\~
~
S. So we have (15)
Let d(k) be the number of linearly independent points in Lk , 117
C.A.Rogers
- 20 -
TheJi.t as
V1Yl.~
k. It follows from the inclusion relation
(15) that it is possible to choose linearly independen.t points
~1 '!2 t· •• ,
!n
of /\ , such that for each k, with 1 -{ k (n, and
with d(k)"> 0, the points
!d(k)
~1'!2"'"
are linearly independent points of L k ,
Then by Theorem 5 it is
possi ble to choose a basis 11 ,I2 ""'iLn of / \
such that for
each d with 1 ' d ~ n the points
are linearly dependent on the points
il1,I2 , ... , Then, for each k \1'1 th 1
'i k
~
Lk will be linearly dependent
Ia
nand d (k) /' 0, the points of the points
o~
1 1 ,I2 , ... • Id(k) which are inclusGd among the points
11' 1 2 , ••• , I k -1 We now study the lattice
1\
I
genGrated by the paints
we have
So
, ~ ,. 2.
I
118
C.A.Ro gers
- 21 -
::.
(16 )
Consider any pOint! other than ill of
1\'.
Then
x= for some integers u 1 ,u2 ' ••• , un which are not all zero. So for some k with 1 ' k .... <. n we have
Uk ~ 0, ~+1
= ~+2 =
...
u =0 • n
Now the point
V~ is a point of
1\
.!.
"1J~
ink
m1 ' X1 + ••• + uk Xk
which is not linearly dependent on the points
So this point
is not in the set that
1\
J~
S. Sonse quently ! is not in S. This shows
is admissi ble for S. HDllce
60 using (14) and (16) we have
Slince
V may
be t ake s as
c lOS G 'lS
wo pleas e to fithiS proves (13).
119
-
THE SPACE
OF
C.A.Rogers
22-
LATTICES :
Mahler's compactness theorem. Now we return to the study of lattic (:;s , but insteas of studying the properties of a given lattice we study properties of the space of lattices.
Although some of the ideas involged go back
to Minkowski, th e first person to make aformal study of the space of lattices and to introduce the appropriate topology was mahler
+)
•
Following Mahler we will introduco a topology into the space of lattices by introducing the concept of the conv ergence of a sequC3llce of Ja:tl;ices
I\. l' 1\ 2"
.• to a lattice /\ •
1\
llJefini tion. The sequence of lattices 1\ l' 2"" said to converge t o the l a ttice 1\ ,if there are bases
for the lattices / \ ,
;\1' /\2"" as r
is
such that for k=1,2, ••• ,n, ~
+
( 18)
CD.
To justify this definition it is necessary to prove that, if " l' +)
A 2"
.. , converge both to
f\
and to ,,:, then
1\ = /\'
; the
K.li1AHLER, :froc. Royal Soc. A, 187 (1946), 151-1 87 and Proc. K.
Ned • .Akad. v,Wet.. 49 (1946), 331-343, 444-454, 524-532,624-631.
120
C.A.Rogers
- 23 -
proof is not difficult and will be omitted, Having introduced the concept of convergence we are able to introduce tho ideas of closed set. open set, compact set·)
~~
oontinunua function in the usual way. These definitions lead immediately to the follow:tng almost trivial. result. THEOREM 9. The functions d ( /\ ) is a continuous function. Proof. Suppose thay we have a lattice lattices
1\ l' 1\
converging to /\
2""
1\
and a sequence of
• We have to prove that
But there will be bases of the form (17) for the lattices
1\1'
where the condition (18) is satisfied. So, as we havf:l
"2"'"
r - i +00 ,
d(
~e
1\,
A r )=d
(x(r)
(r)
- 1 ' ~2
x(r»
' • • ., -n
required. We shall also need the concept of a
~ounded
set of lattices.
Definition. A set L of lattices is said to be bounded, if there are constant c 1 (a) i f
(b) if
~
1\
>0
and
°2 /
O. such that:
is any point other than £. of
any
lattice of L, then
\x I ')Ci is any le."tticG of L 7 then
0\(1\) ~
(2
.)It is convenient to work with tho old-fashioned definition of
compactness expro.s sed in terms of the convergonce of some subsequonce of an infinite sequence of points of the set rather than in terms of the coverings of tho sot by opcm sets. 121
-24-
C.A.Rogers
With these definitions Mahler's compactness takes the following form. TmROErnITd 10. Any closed bOUllded set of lattices is compact. P~~of.
Let L be any closed boundes sot of lattices.
Let c 1 ,c 2 be the corresponding positive cohstants provided by the definition of bOUl1dedness. ~ake S to be the sphere of all pointe
as with
<
I~ 1 C i By theorem 7 we have
so that certainly
£'1 (S) :> Now, if
1\
0
is any lattice of L, it follows from Theorem 8 that (19 )
But hy the choise of 0 1 and 02 we have (20)
Hence =:
say.
Co:,
So there are necessarily n linearly independeht points
~1'!2""'2£u of!\ in the set c 3 S , i.e. with
But by
Theore~
5 we can choose points 122
~1'~2""'~
forming a
C. A. Rogers
,.. 25 -
basis of
A
ana such tha t l.
\F where
0
./
"
'\ I ~, ::1., L- ~
-S d v~
(i
~S
1,2, •.. ,n),
(i, j =,2, ... ,n).
z.... i
So there is a basiS ,;'[,1 ',;'[,2" ..
'Xn
of
A such
that
1\
where c 4=nc 2 c 3 is independent of the choice of in L. We note a1so that the inequalities (19) and (20) imply that d( 1\ ) ~c5' where c 5 is a constant. Now suppose that
1\ l ' 1\
2""
is any sequence of lattices
of L. Then there vvill be a corresp!i:mding sequence of bases (r=1,2, ... ), such that (r=1,2, .•• ).
l' .
For a suit a bl o subsequence r 1 ,r 2 , ... of -the po sitive integers we
have
for some po ints ,;'[,1'X2 , ••• ,
Xn'
Clearly
as r = r h
123
C.A.Rogers
- 2.6 -
for all r, it follows that d (x'1
,x'2 ' •.•
'Xn)
'~ c 5 /
o.
So x'1 ,x'2" • "Xu are linearly independent and form tho basis
1\ .
of a lattice
I\~
-)
1\
1\'7.. belongs
Since
Thus by the defini tion of convergence we have
to Land L is closed it follows 'that
1\
is in
L. This proves that L is compact. Before
VIC
apply t his r esult to obtain a theorem of interest
in th e Geometry of Numbers it is convenient to prove the following lemma. LEMI,':A. L8t S be any star body of tho finite typo. Then tho
set A of all lat0ic GS
1\
which are ac1.Jllissi ble for S, and which
satisfy is closed and botUlded. f'roof. Lot
!\ l '
/\ 27'"
convorging to a lattice
J\ .
be a sequ enc e of lattices of A We have to provo that /\ necessarily
lies in A. I n the first placo
for all r. So, by Theorem 9,
So. if
1\
d( /\
)~
21.\(S).
is not in A, then thoro will be a pOint! of
other t han .2. in S. Take bases of th0 form (17) for the 1B.tticcei
f\ '
j\ l'
/\ 2'" •• so that thoy satisfy (18). Then, for some
int ogors u 1 , u 2 ' •.. ,un not all zero we have +
Sinco S is open, pre:vi
x
(r)
(r)
+ .....2 ~2
124
+ ••• + u
x.
n --n
C.A.Rogers
wiihl bo a point of choice of
1\"1.
"'l
othGr than Q. in S. Thiu is contrary iIllb the
to bc a lattice of' A. H8nce
1\ is
nccossarily
admis s ible for S and so belongs to A. This proves that A is closed. To prove that A is bounded wo have
can choose a posi ti VG munb"r
.! with I .! I 3 6: (S)
~1'
Il1c~rcly
to note that we
so small that tho set of points
~ ~1 is contained in S, and that we can take c 2 '"
< +00.
COl'lI1iHary.
set A is comp a ct.
Th~;
Proo±:. '['he rosul t follows by Thuorclll 10. This rGsult leads us to on8 of Mabler t s main results. If S is. any star body of the finito type,
Theore~_J~
th8re is at least one lattice
A
which is admissible for Sand
which has detGrminant
A. ) = tJ.
d(
(S).
!iemark. Such a lattico admissible for S wi1t;h d( /\ )= /\ (S) is called a critical lattice of S. Proof. Let A be the set of all lattices. with d(/\ ) -'
2
6
1\
admissible; for S
(S).
As 6(S) is dofined to be tho infimwa of thl) detorminants of tho latticos
1\
admissiblD for S it is clour tll8.t A iB not empty.
-
Indeed thore will bo a soqucnco of la-G t ic()s /\ l ' admis si ble 1'or S such that d( <Xl.
~, ... all
,::
1\ )-)/\ (S) r
as r --1
J\
--
So apart from a fini to >11.lil'oor all those la ttid (JS will
belong to A. S:LnCl) A is compact, by the corollary to the lomma, it follows that there will be a subsequenc e of thGse lattices converging to some lattice /\ in A. Since d( function by Thcoro)ll. 9 we will have d( Thus
1\
1\ )
limd(A) r
satisfies our requirements.
125
1\ ) is
/\
( S) •
a continuous
C.A.Rogers
- 28 -
9ri tical
~.§Ittice8
".2.!..s_onvex star bodi es..
Having proved that every star body of the fi nite type possesses at least one critical lattice, we proceed to investigate such lattices. However we shall co nfine our attention to the case of a convex star body. Our main result i s t he foll owing theorem due to Swinnerton-Dyer+). THEOlUM 12,
Let K be a bounded convex st ar body, and let
be a critical lattice of K. Then at least '~n (n+1) pairs of points
:t
.2E of
A lie
Proof. Let
A
on '!iha bound a ry of K. = 1/\ 0 wbere
;\0 is
the lattice of points
witb integral coordi:qates. Tbenl\o is a critical l att ice of the bounded convex star domain pairs of
1\0
pairs of /\
1<0
=
the number of point
on tbe boundary of KO is equal to tbe number of point on the boundary of K. Thus it suffices to prove the
A = /\
theorem in the spacial case when Suppose then that /\ g~al
r -1 K fmd O.
is tbe lattice of points with inte-
coordinates. Wc co nsider a lattic(; t\
/\::1/\ I
where the matrix
r =(
....,
of the transformation
r
is of
the form
or
\ + ij'
where i f i=j
otherwise +) H. P. F. SWINNERTON-DTIR, Proc. Cambridge Phil. Soc. 49 (1953) f
161-162,
Our proof follows his closely.
126
C. A.Rogers
- 2\l -
and
I 'S
1 c(,~J Let
± ! 1"
1
(i,j=1,2, ... , n) .
(21)
± ~ be the point pairs of /\ on t he bounda-
'"
'±!N
r y of K, and let ±! ' 1 " " given by
r'
x'
- r
be t he corr espondi ng l;oints of /\' r = 1,2, ... ,N.
x -1"
We want to pl ace r estri ctions on t he numbers 01.- . . which
lJ
will ensure that these po i nt s -xl, x' 2" .• , ..... ~'. are no'\; i n K. By I l~ the the ory of co nvex set s we c an choose t ac - planes to K containing the points !1 '!2""
The equati ol1u of th e s e tac-planes
, ~.
can be written i n the fo rm
r
= 1,2, ... ,N,
us ing the obviuo s sc alar product. notatio n . Then the condi tiohs that the po int s that
~
!i '!2'" " !N
r
x -r
should lie
=' -3 r -Xr
OJ:
t h ese t ao - pl ane s a re
r=1,2, ... , N.
These co ncli tions will be stiltis fi ec\ provi dod
i. e. provided
n
n
r..
(1')
r.1 . . a. ,.(.. 1 J -1
(r) X. = 0, -J
r=1 ,2, • .• , N.
i=1 j=1 f\ ' Consi der t he d eterminant of t he l a t tice 1 \ • This i s
11 -t
d(/\')
~
~ 01.,11
k ct 4 l . '
E ~'2
1. + £d.- 'LL 1 f':
'
127
t
~iY\..
~ ~2n
(22)
C.A,Rogers
for some numbers C1 ,C 2 , ... ,Cn independent of (. • ',V e have M
f:2. C{ j i
C1 ":·
1=1 C 1.
=-
,~
L, i ,.J' =1
i .( ..,t 1'1
- r.:j~i r: 0/. .':J J
iq Provided we can choose
0/. .. so that ~J
CX .• :=('~ •• J.J J1.
i,j '" 1 ,2, ••• ,no
(23)
while not all c) .. are zero, it will follow that efuther C1 is not lJ zero, or C1 is zero and n ., 11 2 oL •. r -_..i. " (l("' _. -2. - 2 -1 it j :: 1 IJ
X:
ti ·
.i <J
In e i t .her case we shall have d ( small numbers
f.
(\' '; "'-.. -"_
of the appropriat.3 sign.
~:.-o:c
all sufficiently
Provided N < ir;l(n+1) the
equations (22) and (23) together constitute a system of less than tn(n+1) + tn(n-1)
'"
n
2
homogeneuus linear IIquazions for the n 2 variablescX ..• Thus, provided N
<
1.J
tn(n+1) we can choose fixed values for 0( ij not
all zero satisfying the conditions (21), changing the signs of all the
DI. ..
that
lJ
n
C
j
-
2: c[':£o 1:0.1. .1 J.
128
(22) and (23); and by
if necossary we can ensure
a.A.Rogers
- J:l -
We suppose~hat N < ~-n(n+1) and study the corresponding lattice;\
for 'these fixed values of
values of £
lJ
whicp are sufficiently small to ensure
t< Since d(
cI.. .. ani for positive
l\ )
th~t
<
i
d(!\. )
= 1 and /\
is critical for
A (K)
1
>
d(
6 (K),
So, by the definition of
K we have
1\ ) • there is, necessarily a point
I
X'=- X'
of!\ other than.2. in K. Since K is bouIlded, as
t
c
tends
to zero through a suitably chosen sequence of positive values the point
Xi
will converge to a point,
Now each ele~ent of the matrix element of the unit matrix I as €.. inverse of of
r
r'
Xo
say, in the closure of K.
r
tends to t he corresponding
t ends to zero. Since the
is the adjoint of J~ divided by the determinant
, i t follows that each elGment of the matrix
to the correspond"ing element of the UJli t matrix as
r
-1 tends
E.
tends to
t
zero. Consequently tho I,oint
converges to the point
Xo
as E.
tends to zero throuGh the spe-
cially chosen sequence. But this point -j
r "1
I
1
~
is a point, other than .2., of the lattice
1\
So Xo is a point,
other than £., of / \ and .. 1
r '(
I
-~
for all suffici ently small numbers
f
of the sequence. As
a critical lattice of K, and as the point
Y.o
of
A.
1\
is
liGS in the
closure of K, it must belong to the boundo.ry of K, and must be of the form + ~ r
129
e.A. Rogers
- 32 -
for some r with 1
'::::r~
==
N. So we have
r
(+x) I • --r == +x ~ l'
Now by our choice of the matrix (r:X .. ) this point lies in
lJ
a tac-plane to K. But, as K is an open set, no point of K can lie in a tac-plane to K.
This is contrary to our choice of
y! • (.
Consequently we must have N ~-tn(n+1) and at least tn(n+1) pairs of pOints of
1\
lie on the boundary of K.
There is a complementary result due to Minkowski showing that a lattice, which is aclmissi ble for a convex star body, cannot ~ve
too
m~~y
pOints on the
~oundary
of the body. We state the
sesult without proof.
THEOREM D.
A
Let K be a bounded convex star body, and let
be a lattice admissible for K. Let N be -the number of pairs
of points ±x of /\ on tho b01mdary of K. Then N
~ t
{3n -1 J
and, if K is stric-tly convex
We remark that it is possible to show by ellw.rr,ples that the numbers tn(n+1), t {~n.. 1} and 2n _1 occUl1 ring in these thoorems are all bes-t possible. Howevor I am inclined to thinlc that the last word has not been
sa~d
on this subjGct.
I suspect that it
possible to specify a value of N, for oach integer n, with N very much less than 2n _1 for largo values of n, with the sho~dbe
property: - i f
r
is a critical lattice of a convex star body K,
it is possible to choose a set of a mOijt N pairs of points +x of ~ on the boundary of K such that, given any system of neigh-
bourhoods for those points, there will bo a neighbourhoods of}\ 1
in tho space of lattices, such that every lattice!\ in the
neighbourhoods ofil with dU).' ) .c:. d( A) has a point in at least one of the choson neighbourhoods of -the chosen points ±!.
130
C.A.Ro gers
- 33 -
Thore is anoth er goneral r esult due to lilinkowski which gives us information concerning the poi nts of an admissible lattice of a convex sta.r body whic h can lie on the boundary of the body.
U. Suppose that Il is a lattice admissible for a
TfIEOREM
convex star body K and that boundary of K.
!1'~2""'~
are points of on the
~hen
(24)
with strict inequality when Proof.. I f xi' x 0 -
-
Co
, •••
]Ii
is strictly: convex.
,x are linearly dependent, then -n
and there is nothing to prove. So we may suppose that ~1'~2"'" x are linearly independent. Then t here is a non-singular linear
~a.nsformation 9
transforming the points
~1 '~2'" • ,~
into the
points x 1 ,x2 , •.• ,x ; the matrix of (0 will be that of the coordina-xl tes of the points ~1'~2" . . ,~. Thus =:
We study the set KO
=:l
d(x ,x , • . • ,x ) • _j- 1 - 2 -n
e
K. This s et is clearly a cOnvex star
body vii th the point s ti 1 ; ~2 H •• , ~ on its boundary. So KO contains the set P o ~ al l p~ints of the f orm
~ == A1 1~1 + A2 ~2
+ •••
+,\ it ~
where
This is the set of all points (x 1 ,x 2 , •.• ,xn ) with
I
')C
II + I y. 201 to., \ X Y11 <. -1
it is called a generalized octahedron. The volume of P is clearly 2n times the volume of the re gi 0n defined by 131
C.A.Rogers
- 34 -
The region defined by this system of inequalities has volume 1/n! So the volume of P is 2n /nl Thus the volume of KO is at leas~ 2n /nl Further the volume of Ko will exceed 2n/nl, i f Ko is strictly convex. Buttho VOlunlOS of K and KO arc connected by the equation Hence Y(K)=d(2S1'!2""'!n)V(KO~ --, ./ / d (!1'!2 '
2n/n t,
.. • , 2S.n )
(25)
with strict' inequality when K is strictly convex. Since
1\
is
admissi ble for K; it follows from Minkowski's first theorem that
(26 ) The required resu.lt (24) follows from (25) and (26). Further
we
see that (24) holdw with strict inequality when K is strictly convex. CONVEX STAR DJMAINS:
Convex star bodiell. in
3-dimen~.i0nal
sJ2..?Lc9..
We will now discuss tho significance of our general results in tho special case when n=2 and then proceed to make
SOIOO
re-
marks about the ccrresponding results in 3-dimensional space. When we are working in
8,
plane the term 'body I seems inappropria-
te, so we shall replace the term 'body' oJ the term 'domain', Suppose that
1\
is a critical l attice for a convex star domain K.
Then, by Theorem 12, there are at least three pairs of points of on the boundary of
K.
let
~1'
±!2' 132
±!a
he three such pairs.
C. A.Ro gers
- 35 -
By theorem 13 the determinant d(~1'~2) of any two points
of
A
on the boundary of K has one of the values 0, d (
1\ ),
2d (
A).
Consider first the possibility that there are two points
!\
!1'~2
of
on the boundary of K wi th d(!1'!2)
We maJ
SUPPO SE:
=
2d(
1\ ).
whlllhout loss of 'geEerali tJ that
A is
the lattice
of points with integral coord,i.na t es. Let t l:e coordina tes of ~1;!2
be (a,b) and (c,a). Then d (!1 '!2)
Iad
==
- bc
I
= 2d (
1\)
2,
==
so that qd
Now, if a was
e~en
- bc
==
1:
2.
and c was odd , this equation would imply that
b was even, so t hat (ta,ib) would be a point of in K.
Aother than Q
Similarly, if c was even and a was oMd, (lc,ta) would be
a point of /\ other than
0
i n K.
Thus a and c are both even or
both odd. Similarly band d are both even or both odd. Thus, in any case, the points t!1 + t!2; t!1- t!2 are points of /\ • As K is convex these points must be boundary points of K. It follows without difficulty that the boundary of K consists of the parallelogram·with vertices at the points ±!1' 1: !2 and cp:r).tains the eight points
of
1\ . Let us exclude from this consJ.deration this special case
when K is a parallelogram and the yertices and the mid"points of the sides of
1\ are points of the critical lattice J\
the determinant of any pair of points !1'!2 of
J\
. Then
on the boundary
of K has the value 0, if !1 '!2 are linearly dependent, and the value d (
1\ ),
if !1' !2 are lL:,early inde pe ndent. So i f ±!1 ' 133
Fig . i
134
- 36 -
~ ~2'
±
~3
C. A. Rogers
are distinct pairs of point s of
1\ on the
boundary of
K, we have d(~'~3) = d(~3'~1)=d(~1'!2) = d(
In particular the points
~1'!2
generality we may suppose that
generate
A is
I' ).
A . Without loss of
the lattice of points with
integral coordinates and that .!1'.!2 have coordinates (1,0) and (0,1). Let !3 have coordinates (a ,b). Then as d(!1
'.!3)=dC~2'.!3)
= d(
1\)
1,
we have b
1
=
0
a
a
a
=
±
:1
1•
-1) were points of /\ on the boundary 0 f K,
If both (1,1) and (1 ,
we would have two po:iints of minant 2 '" 2d( /\ ) , ration.
± 1,
b
0
=
=
~
A on
the boundary of K with deter-
possibility we are excluding from conside-
So just one of the point pairs ±(1,1), ±(1, -1) will lie on the bound ary of K. Changing .!2 into -.!2' if necessary, we may suppose that the pOints ±(1,1) lie on the boundary of K. It is clear from the considerations of the last two paragraphs that i f
A is any critical lattice for K then it is poe-
sible, by applying a suitable linear trasfolmation K and
J\ ,
t~
ensure that is the lattice of points with integral coordinatee, while the paints ±(1,0), ±(0,1), lie on the boundary of K.
!(1,1)
(27)
From a different pOint of view, this
means that by applying a linear -t;rasformation of determinant d(
1\ ) to
the points (27) we can obtain a set of six points of/\ 135
-
C.A.R
37 -
The points (27) form a hexagon of area 3.
0),\ the boundary of K.
This exagon is symmetrical dm 2. and its sides are
~~rallel
to and
half as long as its diagonals. We call such a hexagon a vector hexagon.
It follows that, if
there are six points of
1\ is
A on the
any critical lattice of K, then
bound~ry
of K forming a vector
hexagon of area 3d( /\). In this case
6 (K )
(28)
= d(l\) :::} VCR)
where V(h) is -t;he area of vector hexagon with tl1s vertices on the bouadary of K. It so happens that, if H is IDly vector hexagon with its vertices oh the boundary of K, then the lattice!\ generated by the vertices of H is admissible for K. Let
~1
and
~2
be any two
vertices of H which are neither opposite nor adjacent.
Then we
can apply a linear traasformation transforming .!, and .!2 inte the points (1,0), (0,1). remaining
By the vector property of the hexagon the
vertices will be either -(1,0), -(0,1), (1,1), (-1,-1)
or -(1,0), -(0,1), (1;-1), (-1,1).
We can reduce the second case to tho first by cha.nging -!2'
~2
into
So we can suppose that the points ~(1,0),
±(0,1), ±(1,1)
lie on the boundary of K. It followS that there are no points of
K on the f01lowing segments (see Fig.1), since the pOints given in brackets In each cas e lle on the boundary of: (1 ,0) to (2,1)
(0,-1 ) and (1,0);
(2,1) to (1,1)
(0,1)
(1,1) to (1 ,2)
(1 ,0)
(1,2) to (0,1)
(-1,0) and (0,1 h
(0,1) to (-1 ,1 )
(1 , 1 )
(-1,1)to (-1, 0 )
136
\U1d (1,1); and (1,1); and (0,1);
(- 1 ,-1 ) and (-1,0).
-
38 -
~.A.RogGrs
Similarly no points of K lie on the segments obtained from these by symmetry in £. ConsGquently K is contained in the star domain bounded by the segments joining tho points (1,0), (2,1), (1,1), (1,2), (0,1), (-1,1) ( -1
7
°).
(--2
,-n,
(-1 ,-1), (-1;"2 ), (0, -1) , ( 1 , -1 ) t
(1 ,0).
It is now easy to vorify that the lattice /\ generated by the vertices of R is tho lattice of points with integral coordinates and is admissible for K. l"urther, tho determinant of d(
1\.) =
1\
satisfies
} VCR)
where VCR) is tho ar o') of the hexagon. Thus, if R is any vector hexagon with its verticoB on the boundary of K, then
~
(K)
~
d(l\)
=}
vOl)
(29 )
It is now cl(,ar that tho foJ.lowing thoorGm follows from a comparison Iilf tho r 0sults (28 ) and (29). THEOREM 14. If
1< is
6 (K)
==
a convex star domain,
Oot
min} VCR)
where the minimum is takGn over all vector hexagon R With their vertices on the boundary of K. Coroll~.
If K is a circl e of radius r, then 2 r •
Proof. Tho only v ector hexagon inscribGd in K are regular hexagons with arGa :2 r •
______!!?:~~._~.~.._~~I)thu'
formul a , due t o RGinhardt +), for the
"'") E. REINHARDT. !tb.!J:, . J:[a~~.J?9m. Rambtg:g., 10 (1934), 216-230, see
also K.Mahler, Pro.S-.X,Nod. ~Lkad. v. Wet. 50 (1947), 692-703.
137
C.A.Rogors
critical doterminant of a convex star domain K. This is
L1(K)
=
t
min V (H)
where V(H) donotes the area of a symmotrical hexagon H circumscribed about H, and whero the minimum is taken ovor all such hoxagons. It is clear that tho situation is much morc complicatod in 3-dimensional spaco. Minkowski +) gavo a detailed discussion of the critical lattices of a 3-dimonsi anal convox star body; his results could bo summed up in the form of a rathor
complicatG~
formula
analogous to (30) wh0re ono has to maximiz0 tho volumcs of threo special
sorts of polyedra inscrib0d in tho convex body and then
make a choice betwe"m the mawima found in this way. Minkowski IS method is so complicated that it has only been used by Minkowski++) to find tho critical detorminant of a sphere and of an octahedron, and by Whitworth+++) to find tho critical determinants of a double cone and of a cube with a pair of its corners cut off. Minkowski's results ha~e beon extended, in part, to 4-dimonsional ++++) space by Wolff t but his results are (probably necessarily) so involved that he can only apply them to the case of a four dil.monsional sphere with difficulty. Thore is no known analoguo of Reinhardt's formula (31) even in 3-dimensional space; although tho first step towards such a formula has beon taken by Davenport
t
and van dor corput+ +) Any geuc,ral rosults abou·c the critical lattices of convex star bodics in 3;dilllonsions or of special types of such bodies
+)Nachr~G8s~Wiss.G8ttin~n 1904, 311-355.
++) Loc.cU •
• ++) J.V.WHITWORTH,Proc.London.hlath.Soc. (2),53 (195~),422-442, and hnali di Mat. (4) 27 (1948), 29-37. ++++) K.H.VTOLFF, Monaths.liIath. 58 (1954),
38-5~.
+:j:+) J.G.van del' CORPUT and H.DAVBNPORT, Proc. K.Ned.Akad.v.Wet. 49 (1946), 701-707. 138
C.A.Rogers
- 40
(e.g, bodies of revolution) would be of considerable interest.
CRITICAL DET:;:;B.MINANTS 011 STAR BODIES: Arithmetic minima
alBebraic forms
o~
So far wo have been concerned with goneral theoretical results and with wpecial results for convex star bodies. I now want to discuss results for some special star domains and star bodies. The bodies with which we shall be concerned will be regions which correspond to certain classes of algebraic forms. By an algebraic form we simply mean a homogeneous polynomial in a number of variables u 1 ,u2 ' ••• ,un' By a class of algebraic forms we mean the set of all. polynomials, which can be 0 btained from a given polynomial (32)
by substituting for u 1 ,u2 "" ,un linear combinations,
of u1tU2 •••• ,un the determinant of the coefficients having the value 1. The arithmetic minimum M(f) of a form f(u 1 ,u2 , •.• ,un ) is defined to be the lower bound of the values taken by f(u 1 ,U2 ,···,Un ) for integral values of u 1 ,u2 ' ••• ,un not all zero. If a form f(u 1 ,u2 , ... ,un ) has degree h then it is easy to show that the upper bound M of the ari"l;hmetic minima of the forms of the same class as f(u 1 ,u2 , •.• un ) is of the set S of points
co~~ected,
(x1'x~, ... t:.
with critical determinant
,xn ), defined by the inequality
139
-
by the equation
M
41
C.A.Rogere
= {~(S)~
-h/n •
We omit the proof, which is straightforward. Since our results are more interesting from the arithmetic rather than from the geometrical point of view, we state our results in terms of the arithmetic minima of tho forms of a claas rather than in torms of the critical determinant of the corrcapcinding star body.
Our fitst result forms part of a much more complicated theorem due to Markoff+). THEOREM 1 5.
If
f( u,v )
= au2
+ 2buv + cv
r)
is an indefini t~dratic form with d at'
2
==
b'" ... ac ) 0, then there
e integers u,v bo h !/lero such that If(U,V)
I < 2V(d78)
unless f(u,v) is equivalent to some multiple mf u
2
or of u
2
-uv-v
2
-
2 2v .
In these cases there are integral values of the variables such that
Remark.
Two forms are said to be equivament if one can be
transformed into the other by an integral unimodular substitution. Proof. Let m be the lower
bo~~d
of the values taken by
+) A.MARKOFF. Math. Annalen 15 (1879), 381-406 and 17 (1880)
379-399. See J.W.S.CASSZLS,
Annal?~f
140
Math., 50 (1949),676-685.
-
k(u,v) that m
I
for integers u,v not both ~ero. We may clearly suppose
> o.
If we had
C.A.Ro gers
42
Write
J)
£= d/m
2•
2, then we would have
So we may suppose that () {, 2. The n we may choose [. so small that
(1 +
J )(1- E )2, 1.
(1 +
~ )(1-
Having made this choice of
if 0
[)2)9/4, i f
E. ,
<S .f 5/4
5/4<S~2.
will be pos s ible to choose
it
integers uO,vO' not both zero, so that m 1 -
£
We may suppose that uo,v o have no common f actor; suoh a COmmDn factor could be removed without disturbing ·the inequa:}.ity (33). The, it is possible to choose int eg ers wo ,to so that uoto - vOwO Write
F(u,v)
= Au2
=
1.
+ 2Buv + Cv 2
where
B= ±
~ auOwO
+ b(UotO + vow\;t) + cVoto}'
J
2 -,
2
C '" + \ awo + 2bwoto + ct a the sign being chosen so that
I
A :;: f(uo,vo)l Then, using the product rule for multiplic ation of matrices, j
=
so that 141
B2 _ AC
o Vo
B
A
U
C
B
Wo
(Uoto
.
- w v )
o
a.
a . A.Rogers
.:1-3
t
0
b
a
c
b
2
2
( b - ae) = b
0
2
t
v
o
- Be
Further
o ."
-
A
< _._ID__ 1 -f.
'
while, for a:.l i nte gcrs u ,v, v/ e hav e
I f(Uou+v"v,w_u+t r,v)! ~ unless Wri te
U
=
U
I
'T
=
U
\ .1
ill
O.
e =B/A Clnd .fA =ID/ A;
Then we [lave
1-t
~b vi 2
,
d .:...
(U+ :" A ,-' A2.
1 =-1 A u.2+- 2B A
provided n ,IT
3.l' (~
l.\V
Y2
I '
+{ ·~An'-·2 -
Sl ... A C }
•
A
V
2.1
not 'bo t h zero .
III pa:L'tic1:l1ar
for all integ er8 u. NOH , ths "Cut Cl f' n·'l.IDb',n'E' x for which
is :.\n interval 11 of
le~6th
142
0
,.. 44
-
while the set 12 of numbers x for which
i- b1i is empty,
if
&-<
~-JA
~'.n intol'Va1
1, lmd is
of longth
PZ =' 2Jbrt~J~ '
c·
o '/ 1, Let n be the number of l)oints of th .J form x =- u +
e
(u i1}tegra1)
in th,.: intc:ciolof th'2 interval 11' 'j.'hcln c1 nrly
J\~ PI.-1~-i+2({-E)li-;8By the choice of £ we have n n
> 1, > 2,
i f 0 ",
5 :!
i f 5/4<:
6>
5/4 ~ 2,
But, by (34) , any such point. ,.. in trw :be in 1 2 ,
Thus, wo must have
i1~t(;mor
of 11 must
b~) 1 and
p2~n-1,
E~
If O..c.
5/~-,
this. illlj,llios th,lt
e ~i
2 Is-=-:; ~. so that
b
2.
is neC6I1SE1:::-1.1y e qual to
inequali t ,y implies that
rz--"
2Vb-1. ?
5/4..
If 5/4 <.
S~
2, the
e2 ~I::
so th£~t [; is necessarily 8CjuFI to 2. I n the case when
e-
D =5/4,
have /" =1, £ =(1 and that
8
==h -
it
i t is oas~' t o vurify that we must
e must
differ from
we have tho idGUti ty
143
i
by an intogor. If
- 45
e
F(u-hv,v)=±A { (u-hv+ =±A
f
(u-iv)
C.A. Rogers
_.
2
-
v)
2
-bf 2
452 v
C'.
v
2}
}
=±A { u 2 _ uv _ v 2 } •
2 2 So f(u,v) is oquivalent to a multipl o of the form u -ny-v • In tho c ase when ~ =2, :i. -t follows i n the s nmc way that}A = 1 ,
f
=0,
e
and that
I
must have: an int up'al v alu", sr,y h . Thon we
have J!' (u - hv, v)"'::A {. (u·-hv+
? 2} e v) 2 - ¢/-v
2 ')) { U -2v ~J.
=±A
So, in this case, f(u, v ) is equivalent to a multipL, of tho form u 2_2v 2 • This completes thu p:coof. This I'csul t
fo rillS tho
theorem of Morkoff
fiI'st;
dl'l' a much morc complicates
~Ja rt;
may be stated in ·the followi ng way.
~lhich
J'HEQ1~M'p.!... If
f (u,v)
au:
2.
+ 21Juv + cv
2
is an indGfinit6 qu adratic forr,; 'with d=1}'-3C> 0, than thoro are integers u,v not; both zero sucL that
If(u,v)1
.~
2 {( dll-;)". . \ 1,
unlo ss f (11, v) is oqui V::11011t to some mul tiplc of one c1'
J.
certain
infini t\; soquence of forms
2 2 u (,-uv-v u
2
- 2v
2
Ho l'O t ho sGlluonc e of forms can 'be E: pecifi ed exactly in terms of tho solu·tiona of trw Diophan-tino <)quc:tion
oc
2
(: 2 + Y + z
but tho result s arG too
= 3x
y z
compl i c ~ tec1
In a numbor of other
ca&1e~,
for mo t o oxplain.
wl)cn we study th0 ari thmotic
minima of a class of Cllge:braic forms it is )oss i ble to obtain
144
- 46
C. \ .
simila r results. Supp ose we study t h e: can be obt ai!l.cd f rom
co.
CL1SS
HOg0I'S
of al:'. f orms which
fixnd fo rm
f (u 1 ,1.1,) , (_
•••
,u ) n
by substituting fo.l' u 1 ' u~~ , ••• , un' linoar combimttio m
,
of
1.1 1 ,u 2
..
, ••• ,un , where the det ormimmt of t h o co efficients hus the
value 1. I n some cases, there will be wuch substitutions, which give rise to a form g (u 1 ' u 2 , ••• ,un ) ,
whose co cf f icicmts ar .:: i n r ation.,l r a-t io S 5 ::mrJ whicll is s uch thht 1\1 (f)
fo r all forms f of -the
~
M ( g)
In
cl ~, ss .
SOllie
of thuse cases the re sult
will bo ii s ola t od I; Ll the:: s ens e -tha t the s e will b'.) a nu.rnbo r
such tha t M( f ):f Lt , c..
for all forms f of tho el iss , which are not oqui'f alent under on int cgr c.l l1..'1imodula r substi tutic n t o :!::g. Examples of al ge braic f a TIns , wher o ,thoro is SLLch a f orm g with coefficionts i n rationa l I? tios , but wnur o t he r:crmlt is not isol a tod, a re providod by tho cases i:: 2 2 f(x 1 ,.· .,xn ) x 1 + x 2 +... + x n ' all v 2,lues of n(but g is only known ex:)licitoly for n ::: 8 ); and by tho cubic forms
145
C.A.Ro gers
- 47 -
(b ordell)
x 1 x 2 (3[1 + x Z )
2
2
x 1 (x1 + x2
(Mordell)
studi ed b,Y Her!.:it€ a n d lat. er iYlvestigated thorOl!ghly by Mord ell +). ~i;xallp l es
of a;L g ebra ic
with coef:i:' i:::ients in
fcl' ~~, s ,
r<~ t i o l1al
where there i s Gu_ch a form g
r a ti os , 2nd where the res ult is iso-
lateci i n t t e abcv tl smls 8 , a rc providod by the fo rms
,
x,, x 2 2 2 x1 + x? 2
x1 + 2
-:i2 2
x1 + x2
x 2_
.J
y
(Morkoff)
J
2 + x)
-
(Markoff)
2
- -)
.J
-
2
(Oppen he im)
:\;
2
x4, (Dav cmpo r t) (To rhheim)
An otil 'J r example vlhich lils iso lat e d in a wo ::d . e } ' SGn s e 'by
th(~
iEl
++)
plY Jvided
form (Dav enpo rt)
General llctho d s f or di fJ cu ::l sing so me of t hese i s olation re sults have be en d ev 8lo p 8d 1)~, Ma}1~er a nd Davenport ane!- Ro gers +++ ) '-I r b :JI S· amd CC',.:Cl'icd f \U""1e 'J' e a ss e ]_s anU. Wlnner t on- DY8r +-H+ ) • l:-1owev er th e
aillmo unt of ·);,.)llll' ali ty inv olved i s
E£:['V 8 1'1 y
limited
8Ed
it would
bo lDost i nt eres t i ng, i f Clny reaL l y gewn"dl results could bo ob-
t aine d. FoT' eX£lT1pl c , i t qpp0 8r S to be certain ~~~£_. ~~_ ~~?~~~ion.
r() s nlt ; onl y i f the
VOhUii€
t ~ lat
thet';)
C 3 11
00
of the corrcs pond ihg
+ )'llomr ::LL"'For a bried ac c ount of Uo hi s tdlr;y of th l r e sults discu! sed i n thi s sec ti on
SG e
H.DAV£)I;JPORT,P.!~Q.
2.£.J:1_l'!..tl1C:lllati..£.i DPs , 1950 , VoL 1, 166-174. ++ ) RJ c(m tly announcod i ll t h e Bu~~\'~_ E@tll . §.Q2.•
+++ ) ROGEES- Phil. Trap s. Royal SO C" A. , 242. (1 95)0),311-344. ++++ ) 'phil. Trans, Hoyal So~. ,A , 248 (1655) , 73- 96 .
146
-
C.A. Rogers
~8
st9.:1' body, d c fillO :J. by
If (x 1 ,x2 , ••• xn )! is infini -te, but
1,
-!:,
t.his has not been proveQ,
THE NUMBER OF LAOlT'ICB POINTS HT A STAR :BODY:
,! .he Minkowski-Hlawka Theorem It is
t :::,:L lrial
a
COl1S \) (jlJ_ene i;
of the definItion of the cri ti-
cal determ.inant of a set S t'0n't, if S is sylnflllt.rical in
~
a.l'J.d
d(l\ )<: £1 (S), -1;(])Gn th.ere is a pair of poin'~s
:!:! of
in S. A few years ago I
I~
made tho f ollowing conjecture. Con:i"' 9t~:t.
Let
ill
'be a poai tive
s t:t which is 3:J'lllJ!K, tric'.'tl in 0, and le-t ill
Then thcrl' <'I r e;
III
d,(.J\
inte~,
A.
bf;
8.
let S be a stur
18,-ttice wHh
)'<.tl (S)
distinct pairs ±!'1 ' •• ' ,:r.~m of points of
I wus only able to !)rove+ ) this conjec-cu:l.'e, 'Nhen c~rtain
s pecial forlllS. kl'oic ,'3so:L' }U8.wku tells
of his Her'x Schmidt has r:roYcc1
large i!ltug:eI' s
If\,
p~_
in S.
is of
that a studGnt
con;; ect-u.re 1'01' all sul'fic:Le ntly
~;hi s
In th,e spoc iS!l eLse whel1 .m. i s a p:d u\G number
I ,\'2.5 aiJl,,, 'C0 protJe the foll o·,v:ing·
1-'J!1!.9B.f.M'
~e
Ii)
,f\..
Let p be a pI'ime,
81it;htl~ s'~:c'olltnr
resul. t.
S be a. otar ,Jet, which is
l {~t
S:lTlllliletrical in _,?-, and let / \ be a lattice with ]I
d(
1\- )<.
Then either (i) then; if.! -':,f -""1",.
",PO " l.' "i-'-n • VoJ
L\ '"r.imiti r
" :!::!1' +) --~1"'" L",
disti nct llfdr s +a'l' +:&2"" -
lier,:; of'
/I.
-
it
- '-
~ (;:;) .
+"",., -""'~1
"'-~ '"
n,l. " ,'
ve l'n
noi.nt a ""::"1
U,
l.'
(>+' J...
A
'-'llC;' U ~ L_
or' (,'"' \--" ) +" "'.1,)e"".1. ';,
t}l',,-I... 1I
-~" C>~"
'0+1 __
,+~ 1 of Tjl'imitille noitl'GU ot'1\. i n S. -"':\1;+" "
point ~ of a lattice j\ is c(lllecl
~ it io n ot po::w ibL! to e X::::)j:-\~~<J a
147
}_1\
CJ
pl'imi'tive point
lh~' fonu
.. 49
Th(1 proof oZ
'~his
C.A.Rogers
-
:i_s n ot v,u:y :l:j-fficult; it depends
r(~sult
on showing that if the conclusion wero falsE:: , there would be a I
sub-lat t ice 1\.
Vii tIl C:let . ; ~L'r!t1nant
1\
of
d (!l.
I
)
= po. (/\ ) ~.6
and with no point other thnn of
L\
in S,
0
(5 )
con:;r ~ry
to the l!.efini tion
(s).
It is not [[i:dicult to
Mi nkowski-IUawka theorem, which
TEE.9A.€.H .. G._ cal L1 £"
Let
S
·1;his l'e ,1111 '[; -to prove the
lJ.3l':
b? iJt:1tcd as follows.
I.laJ'-
bc a bOlmded stnr sot, which is symmetri.-
uncl which has Jordan lllcas ure V(3). Then
Y( 5)
~ ($):5 2 -S (n) where
This tlwoY'cm vms Zirst oco ·,.ll1ci atGd by qui te lik ."l y tlLt he
)1,,0.
a valid proof
1'01'
Mi.l.lk ows:·~i,
and it is
i -:;, a1 tIlO'i).Zh he never
publis:'lcd a proof. Blichfeldt i'·, onn of his sho:rt Eu"(; eu stated thClt he lliqd proved a
slightl~.·
proof was never publishe d. . Hl a wka+) and is
pn?ci '::. 0 rElsul t
CiOJ.... e
proof a number of othe r :tll'coi's have 1 ++) • d· U0 to 'Casses
+;-~~~~~;,~~,
SL ~c :.: b6t11
Hluwka
:l)l'.~)lished
his
publ.i.shec;l 8o m8 simpler
:Pl'oDaH;y the: s ill'.plost preo :!:' is that
nJ ' . proo'1'" .L l.l .S
lS
• • glven III
eli
].l g.htJ S .. ' .y
' " f8rent cui
Eath. Zeit, 49 ('i944 ), 285-31 :2 .
++) J. '11:. S. CASSELS, Pro~Q.!:tp.l.brid i:; e_...;?h:iJ-.211..!J3..2'£., 165-Hi6.
'hut again the
The 1i 1.'21; publishl';d prooi' is due to
r a ther complic:.l"Gcd .
cU1d somG mo r e complicated.
1
49 (1953),
This proof is oi;;,i1n1' tta: unpubli s hed f.,roofs found
independently by H.W0yl and C.A.Ho g e.L'G .
148
C. ;1.• Rogers
- 50 -
form i n ? ro ie ssor Davenport' s le cture n o';c s. To see that Theorem G is a cons equenc e of I" .~ ;t
suffici cmt to argue in thu foll owing way.
ThG o~c m
l\. €;
F it is
bo the lattice
t.f>.oof points whose co ordinates are int e gl'al multiples of
E •
for any fixed positiv~ of primi tiv;.; point s of
1\ ~
Let m(
c )
f.
be the number of pairs
i n S. Let p=p( E ) bo t h e. largest pri-
me numb er s atisfying
Then , since the 1'ntio cf cons l;cuti vu p:cimes tends to 1,
A
0
S is bounded",
can choos e It so
VIC
1<1:;: [; 0
that S
~;;
contained
in tho cube
I X~ 1<. R ) . Nov! ~ p(
E:
small
p(
then
f; p~
. '.1
).-. m as !.....,.. O. f!o f rO\' i ch;d i. is cn:C f i c i<:mtly e ) /" H, and, i f ~ is ,my point oth0r t han 2. of Are
is not in the cub e and so i s not i n S.
Tl~us ,
by Theorom
F, we have
met) '7p(E) provi ded
E. is s uff:!' clently
Sl:id l.
HenCl
But since S is a syliim Etl'i cal Jo rdan me asuI': lbl e star' se t, it i s an elemontary exorcise in tho us e of t h e MlJbius function fA- (n) to prov o that
~ ~;--O
f
(I
)')1
~~((~)
(f)::: C
Thus
[\ (s)
v(s) 2 ~ en)
149
- 51 -
C,A.Rogers
as re qui r ed. 1\. nUlllbcl' of rD.oa.ificationf3 and imprOVDJll\.mts of the Min..l{owaki-
H1 2.wk a theorom :.ir e knol"m. I w:LIJ. end by quoting a :cesul. t I have obtained r ccentl/ ). Tho o:r'cm.l!.
Let S be a sy:Jilletrical s ot in n-di menaiomJ. spa-
C0 with a fini tu Lubes gu8 measur e V(S ). Them ; provided n is suffici cntly large,
tf
'3 V(s) vn~·
This is an improvement on 'r hea rem G for larg e values of n, since
~.
For a breif account of nu arly all tho r usul.ts discus-
sed above, togothQr with many other i nt el'esting res1..11 ts, see: -E. Hlawka, Grundbegriffe der Geometrio c1(;r iahlen. JabisboI1icht
d. Dout. rIlath-Verein. 57 (1954), 37-55, Soo also the recent book "G eometric dor Zahlen"
b ~T
Kel ler .
+)Submittcd for ))ubJ.ication in th e goc..!...L91lClon l:1at '1.SoC.
150
