Combinatorial and Geometric Structures and Their Applications

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COMBINATORIAL AND GEOMETRIC STRUCTURES AND THEIR APPLICATIONS

Managing Editor Peter L. HAMMER, University of Waterloo, Ont., Canada Advisory Editors C. BERGE, UniversitC de Paris, France M.A. HARRISON, University of California,Berkeley, CA, U.S.A. V. KLEE, University of Washington, Seattle, WA, U.S.A. J.H. VAN ,LINT, California Institute of Technology, Pasadena, CA, U S A . G.-C. ROTA, Massachusetts Institute of Technology, Cambridge, MA, U S A .

NORTH-HOLLAND PUBLISHING COMPANY

- AMSTERDAM

NEW YORK

.

OXFORD

NORTH-HOLLAND MATHEMATICS STUDIES

63

Annals of Discrete Mathematics(14) General Editor: Peter L. Hammer Universityof Waterloo, Ont., Canada

Combinatorial and Geometric Structures and their Applications edited by

A. BARLOTTI Universitd di Bologna Bologna, Italy

1982 NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM

NEW YORK

OXFORD

Q

North-Holland Publishing Company, I982

All rights reserved. No part ofthis publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.

ISBN: 0 444 86384 2

Publishers NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM NEW YORK OXFORD Sole distributors for the U.S.A.and Canada ELSEVIER SCIENCE PUBLISHING COMPANY, INC. 52 VANDERBILT AVENUE NEW YORK, N.Y. 10017

Library of Congress Cataloging in Publication Data

Main e n t r y under t i t l e :

Combinatorial and geometric s t r u c t u r e s and t h e i r applications. 1. Combinatorial designs and configurations-Addresses, essays, l e c t u r e s . 2 . Combinatorial geometry--Addresses, essays, . l e c t u r e s . I. B a r l o t t i , A. (Adriano), 192311. S e r i e s . 5 11' .6 82-3515 QA166.25.C65 ISBN 0-444-86384-2 (U.S.) AACR2

.

PRINTED IN THE NETHERLANDS

PREFACE

A Symposium on "Combinatorial Structures and t h e i r Applications" was h e l d a t V i l l a Madruzzo (Cognola d i Trento), on October 20

- 25,

1980. The meeting was pro-

moted and sponsored by t h e "Centro I n t e r u n i v e r i t a r i o per l a Ricerca Matematica" (C.I.R.M.)

o f the " U n i v e r s i t a d i Trento" f o r the purpose o f o f f e r i n g an intense

week o f j o i n t s c i e n t i f i c a c t i v i t y t o some outstanding s c i e n t i s t s and a group of younger researchers. The e n t h u s i a s t i c co-operation o f a l l p a r t i c i p a n t s made t h e Conference a great success. Combinatorics i s an o l d branch o f mathematics. I n recent times the advent o f the e l e c t r o n i c age and t h e development o f computer technology has given great i m petus t o the study o f combinatorial techniques, both by p r o v i d i n g combinatorists w i t h a powerful new t o o l and a l s o by c r e a t i n g a new f i e l d of a p p l i c a t i o n f o r those techniques. As always happens i n the development o f science the a b i l i t y t o answer p r a c t i c a l questions i s g r e a t l y enhanced by the s c i e n t i s t making progress i n t h e study o f pure theory. This gives a c l e a r j u s t i f i c a t i o n f o r the tremendous a c t i v i t y and progress i n t h i s f i e l d . Combinatorics covers too broad a range o f subjects and so a f u r t h e r r e s t r i c t i o n had t o be made on t h e t o p i c s t o be considered. I t was decided t o confine the main t o p i c s t o F i n i t e Geometric Structures and i n p a r t i c u l a r t o Galois Geometries. Four l e c t u r e r s presented, i n a s e r i e s o f i n v i t e d addresses, the S t a t e o f the

A r t i n p a r t i c u l a r f i e l d s . Many i n t e r e s t i n g new r e s u l t s were a l s o given by a number o f c o n t r i b u t e d papers. A l a r g e p a r t o f the m a t e r i a l presented a t the Symposium appears i n d e t a i l i n t h i s volume. A few other papers r e l a t e d t o the t o p i c s considered have been adjoined. We are c o n f i d e n t t h a t many o f these papers w i l l form an i n v a l u a b l e basis f o r V

vi

Preface

f u r t h e r progress i n t h i s f i e l d .

I wish t o thank t h e d i r e c t o r s o f the C.I.R.M.,

Professor M. Miranda and

G. Zacher who organized the Symposium. Thanks must a l s o be extended t o the

G.N.S.A.G.A.

o f the C.N.R. which gave the support which made i t p o s s i b l e t o i n -

crease the number o f p a r t i c i p a n t s .

I am p a r t i c u l a r l y g r a t e f u l t o Professor Peter L. Hammer f o r encouraging me t o prepare t h i s volume and t o the referees f o r t h e i r i n v a l u a b l e assistance.

Adriano B a r l o t t i

CONTENTS

Preface

V

W. BENZ

On Finite Nonlinear Structures

1

G. TALLINI The Geometry on Grassmann Manifolds Representing Subspaces in a Galois Space

9

M. TALLINI SCAFATI

On k-sets o f kind (m, n) o f a Finite Projective o r Affine Space

39

J.A. THAS Combinatorics o f Finite Generalized Quadrangles: A Survey

57

L.M. ABATANGELO and G. RAGUSO On the n(2n t 1)-set of Class [O, 1, n, 2nl

77

A. BEUTELSPACHER Fi ni te Semi -symmetric Des i gns

83

A. BICHARA and G. KORCHMAROS Note on (q + 2)-sets in a Galois Plane o f Order q

117

A. BICHARA and F. MAZZOCCA On the Independence o f the Axioms Defining the Affine and Projective Grassmann Spaces

123

A. BICHARA and G. TALLINI On a Characterization of the Grassmann Manifold Representing the Planes in a Projective Space

129

M. BILIOTTI

On Inherited Groups of Derivable Translation Planes

151

P. BISCARINI and F. CONTI On (q t 2)-sets in 8 Non-Desarguesian Projective Plane of Order q

159

L. BORZACCHINI Subgraph Enumerating Polynomial and Ising Model

169

P.V. CECCHERINI and G. TALLINI Caps Related to Incidence Structures and to Linear Codes

175

vii

Contents

viii

M.J. DE RESMINI On Sets o f Type (my n) i n BIBD's w i t h A

2

2

183

G. FAINA and G. KORCHMhOS Desargues Configurations I n s c r i b e d i n an Oval

207

0. FERRI On Type ( ( q - 3)/2, A f f i n e Plane A2

211

(q

-

1)/2, q

-

1) k-sets i n an

Y9 L. GUIDOTTI and G. NICOLETTI

Minors i n Boolean Structures and Matroids

219

N.L. JOHNSON The T r a n s l a t i o n Planes o f Order 16 t h a t Admit SL(2, 4)

225

G. MENICHETTI Commutative D i v i s i o n Algebras

237

G. PELLEGRlNO and G. KORCHMAkO2 T r a n s l a t i o n Planes o f Order 11

249

C. SOMMA Generalized Quadrangles w i t h P a r a l l e l ism

265

G. TALLINI

On Line k-sets o f Type (0, n) w i t h Respect t o Pencils o f Lines i n PG(d, q)

283

Ann& of Discrete Mathemath 14 (1982) 1-8 CD North-Holland Publishing Company

ON FINITE NONLINEAR STRUCTURES

Walter Benz Mathemati sches Seminar der Universi tlt Hamburg 2000 Hamburg West Germany The inves i g a t i o n o f he geomtry o f plane sections o f a semi-quadric leads o the notion o f generalized c i r c l e geometries. I n the case o f the geometry of Mijb us for instance there i s exactly one c i r c l e through three d i s t i n c t points. This property does not c a r r y over t o the above mentioned general situation. What remains i s t h a t there e x i s t s a t l e a s t one j o i n i n g c i r c l e and t h a t a n b = b n c - c n a i s a consequence o f card(an b n c ) r 2 f o r d i s t i n c t c i r c l e s a,b,c. We are thus l e d t o the n o t i o n o f Tracks, namely t o sets a n b ( a + b c i r c l e s w i t h card(an b ) r 2 ) , which we have lntroduced i n [3],[4]. Almost nothing i s known i n a general s e t t i n g about the inner structure o f tracks, especially about t h e i r possible c a r d i n a l i t i e s i n the f i n i t e case. I n the f i r s t p a r t o f t h i s surveying a r t i c l e we l i k e t o stress some open problems i n t h i s context, concentrating ourselves on the geometries o f Mtibius and Laguerre. I n the second p a r t we are furthermore interested i n the theory o f Lorentz transformations o f the J l r n e f e l t world. This p a r t can be integrated i n the geometry o f Minkowski since Lorentz transformations are automorphisms o f t h i s geometry. Our maln goal here i s t o present correspondences o f basic theorems o f space-tlme geometry under the assumption t h a t the underlying f i e l d i s a Galois f i e l d . We also shall include remarks about connections between Lorentz transformations and the geomtry o f Laguerre.

5 1 Generalized Mllbius Planes. I n 1956,131, we have introduced the notion o f a Mobius plane as follows: Consider a s e t X o f points and a set Y o f non-empty subsets o f X o f c i r c l e s such t h a t the following conditions are s a t i s f i e d : (M I ) Through three d i s t i n c t points there i s a t l e a s t one c i r c l e . Given d i s t i n c t c i r c l e s a,b,c w i t h c a r d ( a f l b n c ) r 2 then a n b m a n c . (M I 1 ) Given a c i r c l e a and points A E ~ , B f a . Then there is exactly one d r c l e b3A.B w i t h a n b = { A ) . (M I11 ) There e x i s t four d i s t i n c t points, which are non-cocircular, 1.e. which are not on a common c i r c l e . Obviously, ( M I ) i s a consequence o f 1') Through three d i s t i n c t points there i s exactly one c i r c l e . (M I n the past we have c a l l e d structures (X,Y) s a t i s f y i n g ( M I ' ) , (M 11), (M 111) speclal M b i u s planes (Mobiusebenen i m engeren Sinn). I n the meantime i t has become more and more usual t o denote "special Mobius planes" by "Mobius planes". Therefore we l i k e t o say "generalized Mtibius planes" instead o f the o r i g i n a l Mobius planes. Generalized Mtibius planes occur q u i t e natural when studying the geometry of plane sections o f a semi-quadric 9 . The underlying antiautomorphism o f order a t most 2 i s the identy i f f ( M I ' ) i s satisfied. The following problem seems t o be s t i l l unsolved: A. Does there e x i s t a f i n i t e generalized Mobius plane, which i s n o t a Mtibius plane?

.

Consider a generalized Mdbius plane K. Given d l s t i n c t c i r c l e s a,b c a r d ( a n b ) > l Then a n b i s c a l l e d a track o f K (s.[31). 1

such t h a t

W.Benz

2

In [3] we have proved the following f a c t s : (1) card a 2 3 f o r every c i r c l e a (2) Through two d i s t i n c t points there is exactly one track. (3) Given a c i r c l e a and d i s t i n c t points P , Q € a Then a contains the track (PQ) through P,Q (4) A c i r c l e contains a t l e a s t three d i s t i n c t tracks. If x i s a track and A f x a point then there i s exactly one c l r c l e through A and x (5) Given c i r c l e s a,b then one and only one of the following cases holds true: a) a n b = 0 I b ) card(an b) = 1 , c ) a n b i s a track , d) a = b . (6) Given a c i r c l e a and a track x such t h a t c a r d ( a n x Then there i s exactly one c i r c l e b D x with card(an b{ (7) Given a point W of K. Define the tracks through W as points and t h e c i r c l e s through W as l i n e s . Then t h i s structure must be an affine plane, the socalled derivation of K i n W

.

.

.

.

.

.

Since every semi-quadric leads t o a generalized Mobius plane many propergeneral ized Mobius planes are known. If K is such anexample then i t turns o u t that card x = cardy for a l l tracks x,y of K Therefore we l i k e t o pose the following problem B. Let K be a (eventually f i n i t e ) generalized Mobius plane. Is then c a r d x = cardy true f o r a l l tracks x,y of K ? Consider a semi-quadric Q such t h a t the underlying generalized Mobius plane K i s miquelian. Then i t turns o u t t h a t Q i s a quadric and that thus K i s a Mobius plane. In [31 we have proved more: Given a miquelian generalized Mobius plane K such t h a t the following condition holds true (+) I f there exists a track x w i t h card x > 2 then card y > 2 f o r a l l tracks y of K . Then K must be a Mobius plane.

.

In case there e x i s t proper generalized Mobius planes, which a r e f i n i t e , or i n case t o study known r e s u l t s about f i n i t e Mobius planes by replacing ( M 1') by ( M I ) In this connection s. P. Dembowski [121, [14], and J.Kahn 1191.

A remains unanswered i t might be of i n t e r e s t

.

In 1161 G. Ewald has characterized generalized Mobius planes stemming from a semiquadric by means of a configuration theorem involving incidence and orthogonality. He has moreover given, [161 , a wide class of generalized Mobius planes by using his notion of weakly convex semi-surfaces.

5

2 Generalized Laguerre Planes.

Given a s e t X of spears and a s e t Y of subsets of X of cycles. The d i s t i n c t spears S , T € X are called p a r a l l e l , S//T, i f f there i s no cycle a such t h a t In case a,b a r e d i s t i n c t cycles w i t h a3S,T We also p u t S//S f o r a l l S E X c a r d ( a n b ) > l we c a l l a n b a track. Two tracks x,y a r e called p a r a l l e l , x//y, i f f there e x i s t s t o every spear S of respectively x,y a spear T of respectively y,x w i t h S//T The structure ( X , Y ) i s called a (now) generalized Laguerre plane (143) i f f the following conditions a r e s a t i s f i e d : (L I ) Through two non parallel spears S,T there i s exactly one track, (ST) Given non-cocircular spears S,A,B w i t h AHSffB then (SA)//(SB) (L 11) Given a cycle a and spears S E a , T t a , SHT, there e x i s t s exactly one cycle b w i t h b 3 S , T and a n b = {Sl, (L 111) Given a cycle a and a spear S g a there e x i s t s exactly one spear T E a w i t h T//S ( L IV) There e x i s t a cycle z and a spear V such t h a t V t z and card 2 2 3

.

.

.

. .

.

.

3

On finite nonlinear structures

A generalized Laguerre plane is a Laguerre plane iff card x = 2 for all tracks x.

Given a (commutative or non commutative) field F, char F * 2 , possessing an involutorial antiautomrphism t Consider the 3-dimensional affine space A3(F) over F and X := { (x,ySz)EFa lxxt t yyt = l }

.

.

If e is a plane of A3(F) with e n X * 0 and such that there is no line in e n X then we call e n X a cycle. This structure (X,Y) turns out to be a generalized Laguerre plane ([41)

.

Problem A is easily solved in the context of generalized Laguerre planes: Consider F=GF(9) and t : atbi+a-bi, a,bEGF(3), iz = - 1. In this case the above described generalized Laguerre plane consists of 216 spears, 729 cycles. There are 81 tracks through a given spear S. 45 of these tracks contain precisely 4 spears, the remaining 36 precisely 3 spears. We now like to present properties for generalized Laguerre planes, which to some extent are similar to those given for generalized Mobius planes in 5 1 . (1) Two distinct spears of a track or of a cycle are non parallel. (2) Given cycles a,b then a cannot be a proper subset of b (3) X is not a cycle. (4) The parallelity relation on X is an equivalence relatlon. The parallelity relation on the set of tracks is an equivalence relation. x//y implies card x = card y (5) card a 2 3 for all a € Y (6) S,Tca , S I T , implies (ST)ca ( 7 ) Given three pairwise non parallel spears P,Q,R Then there is exactly one cycle through P,Q,R if and only if (PQ)H(PR) (8) Let a,b be cycles. Then one and only one of the following conditions holds true: anb 0 , card(an b) = 1 , a n b is a track , d) a = b . Consider a spear W and let M(W) be the set of all tracks through W For xEM(W) define xW :=i YEM(W)I x / / Y ~ Let G(W) be the set of all classes xw , xEM(W) Then we get (9) a) M(W) * 0 , b card x w r 2 for all xEM(W) , c{ card G (W) 2 2 , d) Given x€M(W) and a cycle a 3 W . Then there exists y E x w with y c a Finally the following theorem holds true: Let K be a generalized Laguerre plane and let W be a spear of K Call the elements of M(W) points and the cycles through W and the elements of G(W) lines. If x is a point and g a line put “x on g” iff x c g for gEG(W) , x c g in case g is a cycle through W Then this is an affine plane, the socalled derivation D(K,W) of K in W

.

.

.

.

. .

.

.

.

.

.

.

.

C. If one knows the finite order of D(K,W) (and maybe other invariants of K) what are the possible cardinalities of the occuring tracks ? Quite similar to the construction,[l61, of G. Ewald of a general class ofgeneralized

W.Benz

4

Mobius planes i s the f o l l o w i n g c o n s t r u c t i o n concerning generalized Laguerre planes (r51): L e t E be an a f f i n e plane. A s e t C $ 0 o f p o i n t s i s c a l l e d a pseudo-oval i f f (t) For P E C there e x i s t s e x a c t l y one l i n e h through P w i t h h n C { P I Examples: a) Consider d i s t i n c t and p a r a l l e l l i n e s a,b o f E and p o i n t s U E a , V E b. Then (a u b) \ {U,V} i s a pseudo-oval b Consider two non p a r a l l e l l i n e s a,b o f E Then ( a u b ) \ ( a n b) i s a pseudo-oval. c { L e t E be the a f f i n e plane over a f i e l d F , char F 1 2 , possessing an i n v o l u t o r i a l antiautomorphism t Then ( x , y ) ~ F ' xxt + yyt = 1 i s a pseudo-oval. Consider now a 3-dimensional a f f i n e space A3 and l e t E be a plane o f A'. I f C i s a f i x e d pseudo-oval o f E and g a f i x e d l i n e o f A' w i t h c a r d ( g n E ) = 1, then d e f i n e t h e c y l i n d e r Z t o be the s e t o f a l l p o i n t s o f A' , which are on a l i n e h//g w i t h Cn h i 0 C a l l the l i n e s h c Z w i t h h//g generators. Then t h e followi n g theorem holds true: Define the p o i n t s o f Z t o be the spears and the sets e n Z t o be t h e cycles, where e i s an a r b i t r a r y plane o f A3 such t h a t e n Z $ 0 and such t h a t e cont a i n s no generator. Then t h i s i s a generalized Laguerre plane, which i s a Laguerre plane i f f there are no t h r e e d i s t i n c t and c o l l i n e a r p o i n t s i n C

.

.

{

. .

I

.

.

Remark 1 : Omitting c h a r a c t e r i s t i c 2, a semi-quadric, which leads t o a proper gener a l i z e d Mobius plane, can be c a n o n i c a l l y described as the s e t o f p o i n t s (x,,x,x,x,) o f a 3-dimensional p r o j e c t i v e space such t h a t XX ,: x,dx, t + XX,, t t X;X, = 0

.

-

Here t d notes the underlying i n v o l u t o r i a l a n t i a u t morphism. d i s an element w i t h d = d e , which i s assumed t o be n o t o f form kke As one e a s i l y can check there does n o t e x i s t such an element d i n the f i n i t e case. So the c o n s t r u c t i o n o f f i n i t e and proper generalized Laguerre planes does n o t c a r r y over t o t h e gener a l i z e d Mobius case.

.

.

Remark 2 : Consider t h e 3-dimensional p r o j e c t i v e space P3 over a f i e l d F, char F $ 2 , possessing an i n v o l u t o r i a l antiautomorphism t Then t h e geometry o f plane sections of t (x,y,z,w) c P3 xx +yyt zzt = wwt c e r t a i n l y leads t o a n o t i o n o f generalized Minkowskl planes. One should define a common basis f o r generalized c i r c l e planes as i t was proposed f o r c i r c l e geometries, s. H.-R. Halder, W. Heise 1181

{

-

I

1

.

5

3 Lorentz transformations.

F i r s t of a l l we l i k e t o describe t h e s i t u a t i o n over the f i e l d Given the Rn, n 2 2 Then

.

-

-

-

R o f reals.

:= (qq-Pi')'+ .*. + (qn-1 Pn-7 1' (4, Pnla i s c a l l e d the Lorentz-Minkowski distance of the p o i n t f l P(p, ,Pn) ,Q(q qqn) I n physics the n o t i o n PsQ f o r P,QER i s o f importancif P s Q o f the Rn stands f o r FJsO and P n S q The b i j e c t i o n s of the Rn i s c a l l e d a causal automorphism i n case P s Q i?f PssQs holds f o r a l l P,QERn The ma p i n g s : Rn+Rn i s c a l l e d a Lorentz transformation i f f = PsQs f o r a l l P,QERE I t turns o u t t h a t Lorentz transformations a r e b i j e c t i v e a f f i n e mappings o f t h e Rn

PV

.

. .

.

(.-.

. -

The two basic theorems of space-time geometry are the f o l l o w i n g : Theorem 1 (A.D. Alexandrov,[l]). L e t n be 2 3 and l e t s be a b l j e c t i o n of t h e

On finite nonlinear structures

-

Rn such t h a t fl = 0 i f f PsQs = 0 f o r a l l transformation up t o a d i l a t a t i o n .

P,QERn

5

. Then

s must be a Lorentz

Theorem 2 (A.D. Alexandrov, V.V. Ovchinnikova,[21). L e t n be 2 3 and l e t s be a causal automorphism o f t h e Rn Then s must be a Lorentz tranformation up t o a dilatation,

.

We have proved ([61,[71)

.

.

the following

Theorem 3 L e t n be 2 2 and l e t r < O be a f i x e d r e a l number. I n case n = 2 we a l s o a l l o w r > O Consider a mapping s : Rn+Rn such t h a t pV = r i m p l i e s = r f o r a l l P,QE Rn Then s must be a Lorentz transformation. J. Lester has proved t h a t Theorem 3 remains t r u e f o r n > 2 and r > O ( t o appear Arch. Math.),

.

I n [81 we have shown t h a t Theorem 1 i s equivalent t o t h e Fundamental Theorem o f the (n-1)-dimensional Laguerre geometry. By the way, Theorem 1,2 are n o t t r u e i n case n = 2 Theorem 1 has been generalized by J. Lester,[2Ol, t o t h e case o f a r b i t r a r y has p o i n t e d o u t connections m e t r i c vector spaces o f i n d e x 2 1 H. Schaeffer,[ZJ], between the Theorem o f June Lester and t h e geometry o f Laguerre. We now l i k e t o concentrate ourselves on the case n = 2 By changing the coordinate system of the Ra we can replace t h e Lorentz-Mlnkowski form x a - y a by x y as we w i l l do i n the sequel. We are thus working i n the sequel w i t h the distance

.

.

.

-

-

Pa) 6(P,Q) :s (q1 P1 )(q, f o r t h e p o i n t s P(P,,p,), Q(q,,q,) I n t h i s new s i t u a t i o n Ps Q simply reads as q, pl The f o l l o w i n g Theorem i s due t o F. Rado;

.

-

20

and qa

- pa 0 . 2

.

Theorem ([21]). Consider a commutative f i e l d F, char F92,3 L e t s be a b i j e c t i o n o f t h e plane Fa such t h a t 6(P,Q) = 1 holds t r u e i f f 6(Ps,Qs) = 1 f o r a l l P,QEFa Then s must be a c o l l i n e a t i o n o f Fa Mappings s i n t h i s Theorem of F. Rado a r e obviously o f form (x,y) + (x' ,y') such that x ' = a xt t b y ' = -1y t t c a or x ' = a yt t b , y'=;xttc where t E A u t F and a+O, b, c are i n F We thus can say t h a t s i s a Lorentz transformation up t o an automorphism t o f F I n the J a r n e f e l t world GF(p) , a b i g prime number, we a r e thus l e d t o Lorentz transformations. I t might be remar ed, t h a t Rado's Theorem i s n o t t r u e f o r F=GF(q) w i t h q E {3,4,8,9,161 (s.~lOl,III).

.

.

. .

!

I n 1103 we have posed t h e f o l l o w i n g Problem 1 : Given a comnutative f i e l d F Determine a l l mappings s of i t s e l f such t h a t P,QEF a 6(P,Q) = 1 i m p l i e s 6(PS,QS) = 1

.

As was pointed o u t i n [ l o 1 t h i s following Problem 2 : Given a f i x e d k 9 0 i t s e l f such t h a t v p,QEFa d(P,Q) (Here d(P,Q) stands f o r (ql

-

.

Fa i n t o

problem i s ( f o r char F + 2 ) e q u i v a l e n t w i t h t h e in F

. Determine a l l mappings

s s = k i m p l i e s d(P ,Q ) = k plIa (qa Pa)'.)

-

-

.

s o f Fa i n t o

W.Benz

6

L e t us c a l l F r e g u l a r i n case t h a t a l l the s o l u t i o n s o f Problem 1 ( f o r ( b i j e c t i v e ) c o l 1in e a t i ons

.

F) a r e

The f o l l o w i n g r e s u l t can be proved Theorem : F = GF(pn) i s r e g u l a r f o r

, a) p*2,3,5,7 b) p € ( 5 , 7 1 and n even c ) p = 7 a n odd and 31n d) p n = 7 GF(5) i s n o t regular.

.

, ,

In proving t h i s Theorem i n [10l(and i n a forthcoming paper o f mine) a r e s u l t of G. T a l l i n i , [251 , plays an important r m e . The cases F = GF(5), GF(7), GF(11) i n t h i s Theorem are due t o H.-J. Samaga,[221, who was applying a computer. Remark : In an e a r l i e r paper B. F a r r a h i , [ l 7 I a deals w i t h i n j e c t i v e s o l u t i o n s o f Problem 1 f o r GF(p) i n c l u d i n g a l s o other metrics. The i n f i n i t e dimensional case f o r F = R is included in E.M. Schr8ders[24I. T h i s l a s t Theorem corresponds t o our Theorem 3. We now l i k e t o f i n d a correspondence t o Theorem 1 i n case n = 2 and F = GF(pm). (Note t h a t n r 3 i s one o f t h e assumptions i n t h e Theorem of June Lester.) Three d i s t i n c t p o i n t s o f the Rn, n 2 2 , are p a i r w i s e o f Lorentz-Minkowski distance 0 i f f they are on a common l i g h t l i n e . Thus l i g h t l i n e s are mapped onto l i g h t l i n e s under b i j e c t i o n s , which preserve Lorentz-Minkowski distance 0 i n both d i r e c t i o n s . The l i g h t l i n e s i n our general s i t u a t i o n (observe n = 2) are t h e l i n e s p a r a l l e l t o the x-axis o r t o the y-axis. Thus preservance o f l i g h t l i n e s means almost nothing: If, f o r instance, f,g are b i j e c t i o n s o f F a then (x,y)-(f(x),g(y)) i s a biject i o n o f F2 preserving l i g h t l i n e s and thus distance 0 i n both d i r e c t i o n s . O f course Theorem 1 cannot be t r u e f o r n = 2 i n J a r n e f e l t ' s w o r l d GF(p), p a b i g prime number. But having i n t h e R-world two coordinate systems o f a l i n e moving against each other w i t h constant speed v ,one knows t h a t the f o u r world l i n e s of both o r i g i n s i n both l i n e - t i m e coordinate systems must be l i n e s . Taking t h i s i n t o account we a r r i v e a t t h e f o l l o w i n g mathmatical s i t u a t i o n (191) i n case F = GF(q), q=pm. Given two copies C,, Ca o f the a f f i n e plane over F = GFLq) 2 n d moreover two p e n c i l s o f l i n e s i n each copy, say L;sL,? i n C, and La+L, in C, , Consider a bijection s:X,+Xa , where X i denotes the s e t o f p0int.s o f C i Assume t h a t every l i n e o f L i i s mapped under s onto a l i n e o f L: , i = 1,2 Consider two d i s t i n c t l i n e s h,,ha and two d i s t i n c t l i n e s kq,ka o f C, n o t i n LluL;. By o f C, n o t i n L,'uL:

[ ~1 ]

. .

(similarly, A = a = we denote the cross r a t i o o f t h e p o i n t s o f i n t e r s e c t i o n of L: , L f , h,, h, w i t h Assume f i n a l l y , his = ki , i = 1,2 Then the f o l l o w i n g the i n f i n i t e l i n e o f C, holds t r u e Theorem ( [ 9 ] ) I f a cs a p r i m i t i v e (q-1)th r o o t o f u n i t y and i f t h e r e i s an automorphism t o f GF(q) such t h a t A = at , then s i s an a f f i n e mapping of C, onto C,

.

.

.

.

The assumption A = at i s o f course necessar since the cross r a t i o a has t o be transformed under an a f f i n e mapping i n A = ax , t € A u t GF(q) But a l s o t h e o t h e r assumptions o f t h i s Theorem a r e e s s e n t i a l l y necessary as we have shown i n [91, where t h e other occuring cases are discussed ( 5 . t h e two theorems below). In a r e a l

.

On finite nonlinear structures

7

s i t u a t i o n o f the R-world i t must be A = a corresponding t o t h e f a c t t h a t t = 1 i s the o n l y automorphism o f R , That a needs t o be a p r i m i t i v e (q-1)th r o o t o f u n i t y expresses t h e f a c t t h a t o n l y f o r c e r t a i n speeds v t h e connecting t r a n s f o r mation i s o f Lorentz type. For o t h e r speeds v complicated connecting transformat i o n s may occur according t o t h e

.

Theorem ([91) I f a i s n o t a p r i m i t i v e (q-1)th r o o t o f u n i t y , i f i n case a 1 the number q i s $ 4 and n o t a prime number, and i f furthermore t h e r e e x i s t s such t E A u t GF(q) such t h a t A = at , then t h e r e e x i s t s a b i j e c t i o n s : Xl-)Xa t h a t the f o l l o w i n g c o n d i t i o n s hold: (i ) s i s n o t an a f f i n e mapping , (ii) h: = ki , i = 1,2 , (iii)Every l i n e o f L? i s mapped under s onto a l i n e o f L': , i = 1,2

.

I n t h i s connection we f i n a l l y mention the Theorem ([91) I f A = a = 1 and i f q i s a prime number o r equal t o 4, then any (iii) o f t h e previous theorem b i j e c t i o n s : X 1 + X a having the p r o p e r t i e s (ii), must be an a f f i n e mapping o f C, onto Ca

.

.

REFERENCES: [ 11 Alexandrov, A.D., Seminar r e p o r t . Uspehi Mat.Nauk. 5(1950), no 3 (37), 187. [ 21 Alexandrov, A.D., Ovchinnikova, V.V., Notes o f the foundations o f r e l a t i v i t y theory. Vestnik Leningrad. Univ. 11, 95 (1953). [ 31 Benz, W., Zur Theorie der Mobiusebenen. I,11. Math.Ann. 134 (1958), 237-247, U. 149 (1963), 211-216. [ 41 Benz, W., FPhrten i n der Laguerregeometrie. Math.Ann. 150 (1963), 66-78. [ 51 Benz, W., Pseudo-Ovale und Laguerre-Ebenen. Abhdlgn.Math.Sem.Hamburg, 27 (1964), 80-84. [ 61 Benz, W., A Beckman Quarles Type Theorem f o r Plane Lorentz Transformations. Math.Z. 177 (1981), 101-106. [ 71 Benz, W., Eine Beckman-Quarles-Charakterisierung der Lorentztransformationen des Rn. Archiv Math. 34 (1980), 550-559. [ 81 Benz, W., Zuruckfiuhrung eines Satzes der Raum-Zeit-Geometrie auf den Fundament a l s a t z der Laguerregeometrie. Anzeiger (Usterr.Akad.Wiss.), Math.-Nat.Kl , Nr. 7, Jahrgang 1980, 117-121. [ 91 Benz, W., A Functional Equation i n F i n i t e Geometry. Abhdlgn.Math.Sem.Hamburg, 48 (1979), 231-240. [ l o ] Benz, W., On mappings preserving a s i n g l e Lorentz-Minkowskl-distance. I,II,III (I:Proc.Conf. i n memoriam Beniamino Segre, Rome 1981. 11,111: To appear J.of Geom. ) [lllDembowski, P., I n v e r s i v e planes o f even order. B u l l .Amer.Math.Soc. 69 (1963), 850-854. 1121 Dembowski, P., Mobiusebenen gerader Ordnung. Math.Ann. 157 (1964), 179-205. 1131 Dembowski, P., Automorphismen e n d l i c h e r Mobius-Ebenen. Math.Z. 87(1965) ,115-136. 1141 Dembowski, Pi, F i n i t e Geometries. Ergebn.d.Math. 44 (1968).Springer Verlag. [151 Ewald, G., Begrundung der Geometrie der ebenen S c h n i t t e e i n e r Semiquadrik. Arch.Math. 8 (1957), 203-208. 1161 Ewald, G., Ein SchlieBungssatz fur Inzidenz und O r t h o g o n a l i t a t i n Mobiusebenen. Math.Ann. 142 (1960), 1-21. 1171 Farrahi, B., 'On Isometries o f F i n i t e Euclidean Planes. Abhdlgn.Math.Sem.Hamburg 44 (1975), 3-11.

.

.

8

W.Benz

, Heise, W., Einfllhrung i n d i e Kombinatorik. Hanser-Verlag. Munchen-Wien, 1976. [I91 Kahn, J., I n v e r s i v e planes s a t i s f y i n g t h e bundle theorem. To appear Journ. Comb.Th. (A). [20] Lester, J., Cone preserving mappings f o r quadratic cones over a r b i t r a r y f i e l d s . Canad.J.Math. 29 (1977), 1247-1253. [211 Rado, F., On t h e c h a r a c t e r i z a t i o n o f plane a f f i n e isometries. Resultate d. Math. 3 (1980), 70-73. [22] Samaga, H.-J., Zur Kennzeichnung von Lorentztransformationen i n endlichen Ebenen. To appear J. o f Geom. [231 Schaeffer, H. , Automorphisms o f Laguerre Geometry and Cone Preserving mappings of M e t r i c Vector Spaces. Lecture Notes 792 (1980), 143-147. [241 Schrader, E.M., Zur Kennzeichnung d e r Lorentz-Transformationen. Aequationes Math. 19 (1979), 134-144. 1251 T a l l i n i , G., On a theorem by W. Benz c h a r a c t e r i z i n g plane Lorentz Transformations i n J a e r n e f e l t ' s World. To appear J. o f Geom. [18] Halder, H.-R.

Ann& of Discrete Mathematica 14 (1982) 9-38 0 North-HollandPublishing Company

THE GEOMETRY ON GRASSMANN MANIFOLDS REPRESENTING SUBSPACES I N A GALOIS SPACE Giuseppe T a l l i n i I s t i t u t o Matematico "G. Castelnuovo", U n i v e r s i t a d i Roma, I t a l y

1. ON THE GRASSMANN MANIFOLD Gr,d,q

=

ph,

L e t PG(r,q) be an r-dimensional ( r

> 3)

Galois space o f order q (q

p a prime, h a non-negative i n t e g e r ) . The Grassmann manifold repre-

senting the d - f l a t s (d-dimensional subspaces), 1

w i l l be denoted by Gr

-

.

r

-

2, i n PG(r,q)

-

Such a manifold as i t i s w e l l known i s an sq r+l 1 ,q) and has algebraic manifold, i n t e r s e c t i o n o f quadrics, i n PG( (d+l) 9

-

d

points, where

as =

S

1

j =o

qj

.

L e t Sd be a d - f l a t i n PG(r,q); i t s Grassmann coordinates w i l l be t h e r+l coordinates o f t h e p o i n t representing i t i n PG( (dtl) 1,q); therefore, a

-

p e n c i l o f d - f l a t s ( i . e . a l l t h e d - f l a t s through a ( d - 1 ) - f l a t contained i n t h e same ( d + l ) - f 1 a t ) w i l l be represented by a l i n e ; conversely, any l i n e on i s t h e image o f a pencil o f d - f l a t s . Thus, two d - f l a t s i n PG(r,q) r ,d ,q meeting i n a ( d - 1 ) - f l a t are represented by two p o i n t s on G such t h a t r ,d .q t h e l i n e through them i s completely contained i n Gr NOW, r e c a l l t h a t 9 *q i n PG(r,q) a c o l l e c t i o n o f d - f l a t s which p a i r w i s e meet i n a ( d - 1 ) - f l a t

G

.

consists o f e i t h e r d - f l a t s through t h e same ( d - 1 ) - f l a t o r d - f l a t s belonging t o the same ( d t l k f l a t ; next, t h e f o l l o w i n g d e f i n i t i o n s w i l l be made.

A (d,s)-star,

1

<s <

r

- d,

i n PG(r,q) i s the s e t o f a l l d - f l a t s

through a ( d - 1 ) - f l a t belonging t o a f i x e d ( d t s ) - f l a t . (Such a s e t i s a l s o c a l l e d an s-dimensional s t a r o f d - f l a t s through a ( d - 1 ) - f l a t ) .

9

G . Tallini

10

A (d,s)-dual-star, 1

d t 1 , i s the s e t of a l l d-flats belonging t o

4 s 4

a ( d t 1 ) - f l a t and passing t h r o u g h a fixed ( d - s ) - f l a t . (Such a s e t i s also called an s-dimensional s t a r of d-flats in a ( d + l ) - f l a t ) . Therefore, any s-dimensional subspace (s-space) on G

represents r,d,q either a ( d , s ) - s t a r or a (d,s)-dual-star in PG(r,q), and conversely.

For any a , 1 < s < r - d , c will denote the collection of s-spaces on S

G

each of which represents a ( d , s ) - s t a r i n PG(r,q). For any t , l < t < d + l ,

r,d,q will denote the collection of those t-spaces on G

each of which re-

C'

r ,d ,q presents a (d,t)-dual-star in PG(r,q). Obviously, C 1 = Ci i s the collection, t

.

R , of lines on G

Furthermore, G contains exactly two collections of r,dyq r ,d ,q maximal spaces, namely Cr-d and , an element T i n C r-d representing a l l the d-flats i n PG(r,q) through a fixed (d-7)-flat and an element T' in representing a l l d-flats in PG(r,q) belonging t o a fixed

d+l)-flat.

I t i s easy t o prove:

(1 *4)

Ti, T;E

(1.5)

T

(1 -6)

VUEC

E

(1.8)

S

t

VRER

T i # T;

T' E 1'

Cr-d,

VU'EC'

(1.7)

citl,

d t1

*3!TEZ

=+a!

*

r-d

=$

IT' n T ' I 1 2

4

1

e i t h e r T n T'

, P, or T n T '

E

R,

:UsT,

T' E C ' : U' C T' , dtl

=3!TEI:

r-d y

j!T1

E

zit,

: 11

5T

n TI.

i s the study (which the author started i n [ 2 1 ) of ryd,q point k-sets on G w i t h respect t o C ( s = l Y 2 , . . . , r - d ) and C ' ( t = l , 2 , . . . , r,d,q S t d t l ) , t h a t i s the study of k-sets of d - f l a t s with respect t o ( l i n e a r ) families The geometry on 0

of d-flats in PG(r,q), Such an investigation i s the subject of t h i s paper. Remark. On G

the two collections of maximal spaces a r e C (whose eler- 1 r , l ,q ments represent the s t a r s of lines i n P G ( r , q ) , such a s t a r consisting of a l l lines through a p o i n t ) and C; = P (whose elements represent the ruled planes

Grassmann manifolds representing subspaces in a Galois space i n PG(r,q),

11

a r u l e d plane being a plane considered as t h e s e t o f i t s l i n e s ) ;

thus , (1.3) becomes:

2. THE CHARACTERS OF A k-SET ON G

r ,d ,q

.

The number T ' o f Ts E C s ( 1 < s < r - d ) r ,d ,q m,d 8 ) w i l l be c a l l e d t h e index m c h a r a c t e r o f K

L e t K be a p o i n t k - s e t on G meeting K i n m p o i n t s ( 0

< rn <

w i t h r e s p e c t t o C s . L e t ml ,m2,..

<mi

G as; K i s s a i d t o be o f c&s

S

.,m R

be i n t e g e r s such t h a t 0

[mlYm2,...,m2ls

m # mlY...,mII;

if

K i s s a i d t o be a t y p e ( m l , m 2,...,m S m # ml ,m2,. .. ,m R and ' ~ ~ ~ ,0 d f#o r i = 1 ,2,. . . ,II.

w i t h respect t o C

S

)

II s

< m1 < m2 <

S

= 0

'I

Qd

if T

rn,d

= 0

.. . <

f o r any f o r any

Moreover, K has R c h a r a c t e r s

when e x a c t l y II o f i t s c h a r a c t e r s a r e non-zero.

may be s t u d i e d w i t h r e s p e c t t o Cs s t a r t i n g w i t h those k-sets on G rydyq having one c h a r a c t e r o n l y and t h e n i n c r e a s i n g t h e number o f non-zero c h a r a c t e r s . F o r a l l p a i r s o f i n t e g e r s r, d, w i t h r d

Let

T~

d 3 0, i t w i l l be s e t :

be t h e number o f p o i n t p a i r s i n K such t h a t t h e l i n e through them

(such a p a i r represents two d - f l a t s i n PG(r,q) meeting d o e s n ' t belong t o G r,d,q S and ' I ~ o f K satisfy the i n an h - f l a t , w i t h h < U-1); then t h e c h a r a c t e r s T m,d following equalities:

'r,d-l

'r-d,s

'

12

C. Tallini

(2.2)I

i s a s t r a i g h t f o r w a r d consequence o f the number o f (d,s)-stars i n

I = 'r,d-l 'r-d,s' Counting i n two ways the p a i r s (p,TS) s K, Ts E C s y and p E Ts ( t h e number o f TSIs i n Cs through a p o i n t p on

PG(r,q) being IC p

E

G

r,d,q (2.2)III

being ed

) e q u a l i t y (2,2)11 i s obtained. F i n a l l y , t o get ',-d-1 ,~-l i t ' s enough t o count i n two ways a l l p a i r s ((plyp2),Ts) where p1,p2

are d i s t i n c t points i n K, the l i n e p,p2 i s completely contained i n G

r,d,q' TS E C s y ply p2 E TS (the number o f Ts E C s through t h e l i n e p1p2 being

'r-d-2 ,s-2) ' I n a s i m i l a r way, t h e number

T,"~,

1

<

t

<

d t 1, 0

< n < 3ty

of TI

E

C'

t meeting K i n n p o i n t s w i l l be c a l l e d t h e index n character o f K w i t h respect

t o Xi. L e t nl ,n2,..

. ,n R be i n t e g e r s such t h a t

0

<

n1<

n2<

. . .< n R < et;

class [ nl ,n2, ...,nR] t i f T n' ,d = 0 f o r a l l n # if = 0 f o r any n # n1,n2,...,n and o f type (nlyn2 ,... ,n R t n ,d

w i l l be s a i d o f

I

T I

) I

T

It

n i ,d t o Ct.

# 0 for i

1,2,...,

When the c o l l e c t i o n Cr-d

T

'

n,d

~

(2.4)

'r-d 4

1

m= 1

12;

r-d 'm,d

-

R

and

s a t i s f y the f o l l o w i n g e q u a l i t i e s :

-

o f maximal spaces ( s = r d) i s considered,

(2.2) 's become:

1 m=O

nl ,n2,...,nRY

R. Therefore, K can be studied a l s o w i t h respect

As before, the characters

' 'n-d

K

'r,d-l

'

r-d m -rmYd= k * B S

m(m- 1) T m,d~

= - k ( ~k - 1)

-

2 -rd

.

Grassmann manifolds representing subspaces In a Galois space

And with respect to the collection

13

o f maximal spaces (t d t l),

( 2 . 3 ) ' s become:

1 n=O

,dtl n,d - 'r,d+l'

S S When d = 1 , writing T m,l = T m (1 < s < r - l), ( 2 . 2 ) ' s become:

T

'2

n,l

P n

= T

and

T, = T,

Considering the collections o f maximal spaces Cr-l and P = Xi, ( 2 . 4 ) ' s and ( 2 . 5 ) ' s become:

3r-1

1

m= 2

m(m-1)

m

=

k(k- 1 ) - 2 ~ ,

G. Tallini

14

92

I

1

n

n=l 3

T'

n

= k3r-2

,

2

1

n(n-1)

n=2

T'

n

.

= k(k- 1 ) - 2 ~

Next , we prove: Proposition

I.

On GrYdyq a k - s e t K o f t y p e ( m ) s y 1

.

Q

s G r-d- 1

i s either

Consequently, any k - s e t K on G such t h a t K # 0 , r,dyq r,d,q has always a t l e a s t two non-zero c h a r a c t e r s w i t h r e s p e c t t o Cs,

t h e empty s e t o r G Gr,dyq' 1GsGr-d-1. Proof.

K n Tr-d i s a t y p e (m) S s e t i n t h e p r o j e c t i v e space T 1 G s G r - d - 1; t h e r e f o r e ([ 21, prop. I ) , e i t h e r K n T r-d y r - d = 0, o r K n T I n t h e former case, m = 0 and K 0; i n t h e l a t t e r one, r - d - Tr-d m = as and K = G and t h e statement i s proved. r,dyq For any Tr-d

E Zrmd

.

I n a s i m i l a r way t h e n e x t p r o p o s i t i o n can be proved a k - s e t K o f t y p e (n);, 1 < t < d, i s e i t h e r t h e r ,d ,q . Consequently, any k - s e t K on Gr ,d ,q , such t h a t K # 0, empty s e t o r G r,dyq , has always a t l e a s t two non-zero c h a r a c t e r s w i t h r e s p e c t t o C t' ' Gr,d ,q l G t Q d . P r o p o s i t i o n 11.

3.

On G

k-SETS OF TYPE (ml ,m2)s,

1 G s G r

- d- 1

AFiD OF TYPE (nl , n 2 ) i y 1 Q t < d

ON Gr,d,q'

L e t K be a t y p e (ml,m2)s

s e t on G r,d,q G 8s, one o f t h e f o l l o w i n g must occur: (i1

m, = 0,

m2 = a s

,1<

s G r-d-1.

Since O<m,<m2

Grassmann manifolds representing subspaces in a Galois space (ii)

ml = 0,

(iii)

ml

(iv)

O<ml

For any Tr-d

> 0,

m i:8

2

s

m = * 2 s <m2
.

K n Tr-d

i n Cr-d,

i s a class [ m l ,m2Is

s e t i n the p r o j e c t i v e

1 G s C ( r - d ) - 1. When ( i ) holds, being K n Tr r-d' [Q, 4s]s, e i t h e r K n Tr-d = 0 o r K n Tr-d = Tr-d (no type space T

e x i s t s i n Tr-d);

15

thus, any Tr-d

i n Cr-d

-

o f class

( O Y

*s)s

set

e i t h e r contains no p o i n t i n K, o r

belong t o K. On the other hand, f o r any two Tr-d and tr-di n Cr-d, t h e r e e x i s t s 1 '2 1 o f elements i n Cr-d$ such t h a t T,.~ n T,-~ f; 0, a sequence T,-~, T,. d,..., i it1 h ,h-l),Tr-d 'r-d # 0 (indeed, i n PG(r,q), f o r any two Tr-dnTt-d=B ( i = 1 , 2 h o f ( d - 1 ) - f l a t s such ( d - 1 ) - f l a t s S and 5 , t h e r e e x i s t s a sequence S l y . . . , S 1 h t h a t dim(S n S ) = dim(Si n Sit') = dim(S n 5 ) = d - 2 ) . Therefore, i f a

TF-~

,...

Tr-d

E Cr-d

has an empty i n t e r s e c t i o n w i t h K ( o r Tr-d C K), t h e same i s t r u e

f o r a l l elements i n Xr-d,

so t h a t K =

0

K being o f type (0, 8 s ) s When ( i i ) holds, K n T r - d

-

hence, i f r d>3,

(or K = G ) which i s impossible, r ,d ,q

i s e i t h e r the empty s e t

or a s e t o f type (0,m2)s;

by [ 2 1 prop. X I V , KlqTr-d i s e i t h e r a p o i n t and q2 = 1 ,o r the

complement of a hyperplane i n Tr-d and m = qs. When m = 1 , K i s o f class [ 0,l ir-dY 2 2 r-d a non-empty c o l l e c t i o n F1 of # 0, t h a t i s K represents on G with T 1 ,d r,d,q d - f l a t s i n PG(r,q) which pairwise meet i n h - f l a t s w i t h h < d - 1. (For instance, S

When m 2 = q , d = 1, F i s a c o l l e c t i o n o f pairwise skew l i n e s i n PG(r,q)). 1 a c o l l e c t i o n F2 o f d - f l a t s i n PG(r,q) the complement o f K represents on G r,d,q such that, f o r any ( d - 1 ) - f l a t s S i n PG(r,q), e i t h e r a l l d - f l a t s through S

when

belong t o F 2 , o r the space j o i n i n g the d - f l a t s through S belonging t o F i s 2 an hyperplane. When d = 1, K i s t h e complement o f t h e image on G of a r y ,q l l i n e complex (see I41). I f r - d = 2, then s = 1 (as

< m2 < n T i t 1 # 0,

with 0 K

XIV), K n TAtl

1G s

<

r - d - 1); hence, K i s o f type (0,m2)1,

.

zit,

on G I f d 2 2 , choosing a Tdtl ' E such t h a t d+2,d,q i s o f type (0,m2)1 and, since d t 1 > 3 (by [ 2 1 , prop. K n TAtl

82,

i s e i t h e r a p o i n t and m2 = 1, o r the complement o f a hyperplane

o f the above Yq described c o l l e c t i o n F1 o f d - f l a t s i n PG(r,q); when m2=q, t h e complement o f K , i n T i t 1 and m2 = q.

Thus, when m2 = 1, K i s t h e image on Gd+2,d

being of type ( 1 , q t l ) , and t h e r e f o r e o f class [ l ,4r-d]r-d,

i s the image on

16

G. Tallini

of the above defined collection F2. G d t 2 ,d ,q If r - d 2 and d = 1 , K i s a type (O,m),

set, 0

< m < e l , on

G3

9 4, which i s investigated in [ 4 1 ; anyhow, when m = 1 , respectively m = q , K i s the

image of a collection of pairwise skew lines in PG(3,q), respectively the complement of the image of a l i n e complex. Case ( i i i ) i s the complement o f case ( i i ) . When ( i v ) holds, K n Tr-d (see prop. I). I f s

and, with

E

i s of type (ml ,m2)s i n Trmd, 1

< ( r - d) - 1

< ( r - d ) - 2 , then, by I71, prop. V , q must be an odd square

= +1,

moreover, K n T

r-d

mi

must be of type ( m i , m ~ ) l , with =

1t (q t

€45)-

Therefore, K i s of type ( m i

451/ Z

,mi)l; any subspace i n G r d , q meets

( I n ~ y m ~ )sl e t , hence (see [ Z I , XIV) K i s of class [ m l , m 2 I s r - d , where

and of class

s

t t [n,,n2]i

for all

t = 2,3,...,dtl,

K i n a type

for all s = 2 , 3 ,

...,

where

(3.4)

s s

more precisely, by prop. I and 11, K i s of type (ml,m2)s

for all s ' s ( s =

for a l l t ' s ( t = 2 , ...,d ) . 1 ) and of type ( n t , n t ) ' 1 2 t r-d the index my-d and mE-d characters of K with and T~ Denoting by

Z,...,r-d-

Grassmann manifolds representing subspaces in a Galois space

respect t o C

r-d

r-d

( K being of class [ ml

by ( 3 : 3 ) ' s ) , from (2.4)I and (2.4) I1 r-d

[T;-~

from which (as

T

1

~

-

= [ k 3d

m

-

k

I

0):

In a similar way, denoting by T'!~' characters of K w i t h respect t o

are given

ad] / (6) r-d

r-d 1 'r,d-l 1 / ( J 3 r - d

>- 0 ~ and T~r-d

, m2r-d

i t follows:

'r,d-l

(3.5)

r-d r-d ,m2 ]r-d where ml

17

Citly

and

T

Id''

the index n!tl

from ( 2 . 5 ) I and (2.5)

I1

dtl and n2

i t follows:

and thus

Finally, when ( i v ) holds and s = r - d - 1 , K n T in Tr-d = PG(r-d,q), with h recalling t h a t q = p , (3.9) Furthermore,

m2 =

0

I

< ml <

ml + p ,

1 K n Tred must

r-d i s of type (mlym2)r,d,l m2 < ar-d-l; therefore (see [ 7 1 , prop. 11),

with

0G I

h(r- d - 1).

be a solution of the equation ( c f . [ 71, n . 2):

, I

hence, the solutions of equation (3.10) must be integers. Thus K i s e i t h e r of type (M)r-d. where M is one solution of equation (3.10), or of type (MIyM2)r-dy where M1 and M2 a r e solutions of (3.10) and M1 < M2. When K i s of type (M)r-d (since

C. Tallini

18 T

~

m,d

=-0

~for

any m # M), from (2.4)I

(3.11)

k = M y

Also, by ( 2 . 2 ) I y (2.2)IIy T

r-d-1 m l ,d

T

r-d-1 m2,d

(3.12)

r-d-1 hence (as .cm i'

'

r,d-1 "d

-

-

'r,d-l

[ m 2 'r-d

- 'r'd-1

[ M 'r-d-1

o f equation (3.10), from (2.4)I

lTM,

r - d,d

~

M i ,d

'r-d-l]

-

/p

R

'

m 4 ]/pk ; 1 r-d

I'2

t 1

.

'r'd-1

-

- M1) 2 t h e d i s c r i m i n a n t

i t follows:

and (2.4)II

4 ' T

-

denoting by A = f.M2

K i s of type (M1,M2)r,d,

(3.14)

'

and (3.9), (3.11),

q*ml < M < q . m 2

When

i t follows:

>O) :

(3.13)

hence (as

and (2.4)II

k ed]/JZ;

>- 0~) :

These r e s u l t s w i l l be summed up i n t h e next p r o p o s i t i o n . Proposition 111.

L e t K be a type (mlym2)s s e t on G

Then o n l y the f o l l o w i n g cases may occur:

(A)

K i s e i t h e r the image on Gr,d

which p a i r w i s e meet i n h-flats,

r,d,q

, 1 < s < r-d-1.

of a c o l l e c t i o n F of d - f l a t s i n PG(r,q) 1 sq with h < d 2 ( K i s o f type (Oyl)l)y o r t h e

-

complement o f the image of F1 (K i s o f type (4, a,),).

19

Crassmann manifolds representing subspaces in a Galois space K i s e i t h e r the image on G

of a collection F 2 o f d - f l a t s in PG(r,q) r,d,q such t h a t , f o r any (d-1)-flat S in P G ( r , q ) , e i t h e r a l l d - f l a t s through S be(6)

long to F2, or the s e t theoretic union of d-flats in F2 through S i s a hyperor the complement of the image o f F2 ( K i s of

plane ( K i s of type ( 1 , type ( O , q ) l ) . (C)

When d = 1 , K i s a l i n e complex in P G ( r , q ) .

K i s a k-set on G

with m

>

2.

n < q - 1 , or

of type e i t h e r (O,n),, with

391 ,q

(mi

( D ) K i s a k-set on G r,d,q , w i t h q an odd square, of type and m i are given by ( 3 . 2 ) ' s and E = + 1 . Hence, K i s of type (mf,mz)s s

...,r-d-1,with ml every t = 2,3 ,...,d ,

S

2,3,

for

r-d r-d [ m l ,m

it1,

nl

with

with n?l

r-d

ml

S

and m2

with r-d

given by ( 3 . 3 ) ' s , and of type t t n l , n2 given by ( 3 . 4 ) ' ~of~ class

, m2

(m,41)1,

where

mi

f o r any

t t (nl,n2)i

d t l ,d+l by ( 3 . 3 ) ' s and o f c l a s s [ nl , d+,,

gi

given by ( 3 . 4 ) ' s ; furthermore, k s a t i s f i e s (3.6) and ( 3 . 8 ) .

K i s a k-set of type (mlYm2)r-d,ly 0 < ml < m2 < on Gr,dYq. Hence , R m 2 - m l = p ( 0 < II 6 h ( r - d - 1 ) ) and equation (3.10) must have integral solu(E)

tions; t h u s , K i s e i t h e r of type (M)r-d,

where M i s one solution of (3.10), or

where M1, !I2 a r e solutions of (3.10) and PIl

of type (M1,M2)r-d,

< !I2; in the

former case, k i s given by (3.11) and M s a t i s f i e s (3.13), i n the l a t t e r case,

k s a t i s f i e s (3.15).

since E

1 be assumed 2

< t < d (and so

d+1

1 < t 6 d , on G

> 3 ) . Under these hypothesis, by an arg-

ument similar t o the previous one (considering T;+l E

Cr-d)

.

When t = 1 , r,d.q Ei = R , K is of type ( n 1 ,n 2 )1 and prop. I1 holds. Therefore, i t may

Now, l e t K be a k-set, of type ( n l , n 2 ) C ,

E C'

d+l

the next proposition i s proved.

Proposition IV.

Let K be a k-set of type ( n l , n 2 ) C ,

2

instead of

Tr-d

< t < d , on G r,d,q ; then

either K i s one of the s e t s in prop. I 1 1 ( A ) , (B), ( C ) , ( D ) , o r ( E l ) K i s a k-set of type II

p

(0

(3.16)

(nl,n2);,

0 < n l < n2 < ad, on G

< R < h d ) and the equation x2 - x 1 + ( n l

t

n2 - 1 ) ad

so that n 2 - nl

r ,d ,q

I t n l n 2 adtl / ad-1

has integral solutions. Consequently, K i s of type e i t h e r (N)Atl,

=

o with N

20

G. Tallini

s o l u t i o n o f (3.16), o r (N1yN2);t1y N1

<

where N1 and N2 are s o l u t i o n s o f (3.16) and

N2. I n the former case,

(3.17)

k = N y r , d t l "r-d-l

'

and

i n the l a t t e r , k s a t i s f i e s the i n e q u a l i t i e s : (3.19)

4.

'1 'r,dtl

< < N2 'r,dtl

"r-d-l

"r-d-l

AND OF TYPE (1,n);

k-SETS OF TYPE

L e t K be a k-set o f type ( 1 ym)r-d-l,

1

ON GrYd

<mG

8r-d-1

Yq on G

r ,d ,q

.

I f m =

o f a col= a by prop. 111, r = d + 2 and ' K i s the image on Gdt2,d r-d-1' 4 l e c t i o n F2 o f d - f l a t s i n PG(dt2,q). Therefore, i t w i l l be aksumed t h a t 1 < m

< 8r-d-l.

Then, f o r a l l Tr-d's

sidered:

r - d = 2,

the s e t K n Tr-d i s o f t y p e (1 ym)r-d-l r-d ' i n the p r o j e c t i v e space Tr-d = PG(r-d,q). The f o l l o w i n g cases w i l l be con-

<

r - d = 3, and r - d > 3 .

i s a s e t o f type (l,m), i n the plane Tr-,=PG(2,q), r-d q; thus, by [ 51, prop. V I I I , q must be a square, m = J T t 1 and K n T r - d r - d = 2, K n T

When with m

in C

i s e i t h e r a Hermitian a r c o r a Baer subplane; therefore, e i t h e r IK n Tr-dl = q

Gtl,

or IKnTr-dI = q t G t l .

Furthermore, d = 1; indeed, i f d

> 2,

Hence, K i s o f class [ q t J i t l , q J ~ t l l r a Titl

E Cdtl

would be of type (1, q t l ) l i n Tit1 = P G ( d t l , q ) , s e t doesn't e x i s t

=

being chosen, t h e s e t K n T i t l with

d t l

3; b u t such a

(see [ 6 ] , prop. X ) . Consequently, K i s a k - s e t o f type

(1, q t l ) l and class [ q t G + l , q 6 t 1 l 2 on G3,1,q. i s a s e t o f type ( l y m ) 2 i n Tr-d = PG(3,q); thus When r - d = 3, K n T r-d (see [ 8 ] and [ 2 ] , prop. XXI), m q t l and KnTr-d i s e i t h e r a l i n e o r a ( q 2 t 1)-cap ( i . e . a quadric when q i s odd); therefore, e i t h e r I K n Tr-dl

=

= q t 1,

o r ( K n Tr-d( = q 2 + 1 . Consequently, K i s a k-set o f c l a s s [0,1,2,q+l]l,

of type

(l,qt1)2

When

r-d

> 3,

and o f class [ q t l , q 2 + 1 I 3 K

Tr-d

on G dt3,d,q' i s a s e t o f type (l,m)r-d-l i n Tr-d = PG(r-d,q);

hence (see [ 8 ] and [ 2 ] , prop. X X I ) , m = q t l

and

K n Tr-d

i s a l i n e , so

Grassmann manifolds representing subspaces in a Galois space IKn Tr-dl

= q + l . Therefore, K i s o f class [ O , l , q t l ] l y

21

type ( l y q + l ) r - d - l

and type (q+l),-d. From the Above argument and prop. I I I ( E ) , i t f o l l o w s : Proposition V .

L e t K be a k-set of type (lym)r-d-ly

1

<m

3r-d-1

on GrYdyq.

Then o n l y the f o l l o w i n g cases may occur:

(4.1)

(4.3)

K i s of type (lyal)l

on Gd+2,d,q ; thus i t i s t h e image, on

, o f a c o l l e c t i o n F 2 o f d - f l a t s i n PG(d t 2,q) Gdt2,d ,q prop. III(B)), i . e . K i s the dual o f a l i n e complex.

(see

( r - d 3 3) o f a c o l l e c t i o n F3 o f d - f l a t s r,d,q i n PG(r,q) such t h a t , f o r any (d- ) - f l a t S i n PG(r,q), the K i s t h e image on G

d - f l a t s through S belonging t o F3 form a pencil ( i . e . the space j o i n i n g them i s a ( d + l ) - f l a t ) . K (0,l ,q

t

l)l.

Furthermore

s o f types ( q t l)r-d and

,

k = %1 yr,d-l'4d For such a s e t t o e x i s t , 81 yr,d-l/8d

must be an i n t e g e r .

(For instance, when d = l , k = a r ; when d = 2 and r = 5, k = = (q3+1)a4; when d = 2

hand, when

and

r = 6 , k = ( q 3 t 1)a6; on t h e o t h e r

d = 2 and r = 7 , 8

y

/a

= 9

8

/a

, which

1 r,d-1 d 7 6 2 n o t an i n t e g e r ; thus, no such s e t e x i s t s on G ). 7,2,q (4.4)

is

K i s t h e image on Gd+3,d

o f a c o l l e c t i o n F4 c o n s i s t i n g o f Yq d - f l a t s i n PG(dt3,q) such t h a t , f o r any ( d - 1 ) - f l a t S i n PG(d+3,q),

the d - f l a t s o f the c o l l e c t i o n through i t form a

2

cone p r o j e c t i n g from S a (q t 1 ) - c a p belonging t o a 3 - f l a t skew w i t h S; when q i s odd, t h i s cone i s a second degree cone. 2 ( K i s o f types (q t 1)3 and (0,1,2)1). Furthermore,

2

Ydt3,d-1/ad = 'dt3 'dt2 'dtl / 3 2 ( 8 1 ' Thus, f o r such a s e t t o e x i s t , t h e rig,ht hand s i d e o f t h i s = ( q ")

22

G. Tallini

e q u a l i t y must be an i n t e g e r . (For instance, when d = 1, t h i s i s 2 2 impossible; i f d = 2 , then k = (q - q t l ) e4(q t l ) . ) o f a collection F consisting o f K i s the image on G 5 d+3 ,d d - f l a t s i n PG(dt3,q) such t h a t , f o r any ( d - 1 ) - f l a t S i n PG(d+3,q),

(4.5)

the d - f l a t s through i t , belonging t o F5, form e i t h e r a p e n c i l 2 o r a cone p r o j e c t i n g from S a (q t 1)-cap contained i n a 3 - f l a t (skew w i t h S ) . K i s o f types ( q t l , q 2 t 1 ) 3 and (0,1,2,qt1)1 furthermore: 'dt3

'd+2

8dt1/83 $2

'

<

;

2 'dt2 8dt1/82(81) ;

'dt3

The existence o f t h e above mentioned sets i s s t i l l an open question.

-'

The complement o f a type ( 1 8m)r-d-l

'r-d-1

set, 0

Proposition V I .

< m' <

8r-d-1.

set on G

i s a type ( m l , 8 r-d-1r,d,q Therefore, from prop. V i t f o l l o w s :

A type (m,q 8r-d-2)r-d-1

s e t on G

r ,d ,q the complement o f one o f the sets l i s t e d i n prop. V .

,0 G

m $ q 8r-d-2,

is

By a dual argument, the next r e s u l t can be proved.

Proposition V I I .

A type ( 1 ; n ) i set, 1 < n

< ed, on

o f t h e sets i n prop. V .

5.

TYPE (M)r-d AN0 TYPE (N);Itl

SETS ON G

G r,d,q

i s t h e dual o f one

r ,d ,q

w i t h one character o n l y w i t h respect t o c s , s G r - d - 1 rd,q ( d u a l l y , w i t h respect t o I;, t Q d) i s e i t h e r the empty s e t o r G (see r,d,q prop. I and 11). Generally, t h i s i s n o t t r u e when K i s a one character k-set A k-set K on G

w i t h respect t o

c r-d ( d u a l l y , w i t h respect t o ziti); f o r instance, t h e s e t

consistingof the l i n e s which e i t h e r a r e tangent o r belong t o a non-singular quadric Q i n PG(r,q),

r odd, i s o f type (er-2)r-1

(and, d u a l l y , t h e s e t con-

s i s t i n g o f the ( r - Z ) - f l a t s tangent t o Q i s o f type (8r-2);-1). section such sets w i l l be investigated.

In this

Grassmann manifolds representing subspaces in a Galois space

Let K be a type (R)r-d

now

=

Tr-d

m,d

o

for a l l m

+ MI,

k-set on G

r,d,q i t follows:

. From

( 2 . 4 ) I and ( 2 . 4 ) I I (since

k = M 'r,d-l /a d ''

(5.1)

therefore, for such a set t o exist, the right hand side of (5.1) must be an integer. By ( 2 . 4 ) I I I and ( 5 . 1 ) : 2

(5.2) thus, 2

-id

= k (k-adMtq

T~

ad,,);

i s a multiple of k .

When d = 1 , (5.1 ) becomes:

If r i s even, then

a1 and a r are coprime; therefore: (m an integer)

jI ~ = m1 s

(5.4)

I

m4r '

and from (5.2) i t follows:

If r i s odd, r = 2s t 1 , then a /9 = r 1

S

1 qZi

i =O

= a

ssq

2 and (5.3) becomes:

from which, taking into account ( 5 . 2 ) , (5.7)

Now, l e t K be a type f o l 1 ows :

(N)itls e t on G r,d,q . From

( 2 . 5 ) I and ( 2 . 5 ) I I i t

23

24

G. Tallini

k =

(5.8)

Yr,d+l/3r-d-l

therefore, f o r such a s e t t o e x i s t , the right hand side of (5.8) must be an integer. By ( 2 . 5 ) I I I and (5.8):

When d

1 , (5.8) becomes:

(5.10)

k = N 8

9

r r-1

/a

3

2 1

*

For r = 3,

2 k = N(q' + 1 ) ,

(5W3 and f o r r = 4,

2 k = N S4(q + 1 ) / a 2 .

(5W4

2 Since a and a a r e coprime, and the same i s true f o r a and (q + l ) , (5.10) 2 4 2 4 implies t h a t 32 must divide N ; thus, N = na2 ( n an i n t e g e r ) ; on the other h a n d , N N =

a2; therefore, e i t h e r n = 0 , or n = 1 , that i s e i t h e r N = 0 and K = 0, or a and K = G . Consequently, i n PG(r,q), r > 4, there e x i s t s no collec-

2 r , l ,q tion of l i n e s , F, such that any ruled plane contains exactly N lines belonging

to F, 0 < N < a2. (Indeed, i f PG(4,q) i s a given subspace in P G ( r , q ) , the collection F ' consisting of the l i n e s i n F belonging to PG(4,q) would be represented on G

491 s q i t was shown). Thus:

by a type ( N ) ;

s e t , 0 < N < S2, which i s impossible, as

, r 2 4 , no k-set having j u s t one character w i t h r y l,q respect t o E; = P e x i s t s , besides the empty s e t and Gr a 4 Proposition VIII. On G

.

By a dual argument:

, r 2 4 , no k-set having only one character w i t h ,r-2,q respect t o z 2 e x i s t s , besides the empty s e t and Gr ,r-2,q'

Proposition IX.

On Gr

Grassmann manifolds representing subspaces in a Galois space

Considering again a general value f o r d , l e t K be of type (M) type

and o f r-d Then ( 5 . 1 ) , (5.8), and (5.2), (5.9) must hold, so t h a t , i f K # 0 ,

(N)itl.

i . e . k # 0:

from which (eliminating M): (5.13)

thus, e i t h e r ar-d-2-ad-1 = 0 , 1.e. r

2d+l

and (5.13) i s an identity ( a s

) a n d , by (5.11), M = N , or ~ ~ - ~ - #~0 ,- i .8e . ~ - ~ '2d+l ,d-1 '2d+l ,d+l Therefore: r # 2 d t 1 , and, by (5.13), N = adtl, t h a t i s K = G r,dyq

.

Proposition X. either r

If K i s k-set of types (M)r-d

and

(N)i+lon Gr ,d ,q , then

2 d + l and M = N, or one of the following i s true: ( i ) K i s the

empty s e t , ( i i ) K = G

r,d,q'

NOW, l e t K be a type ( l ) r - d k-set on G

r ,d ,q

, that is

K is the image on

of a collection of d-flats in PG(r,q) such t h a t through any (d-1)-flat 'r,d,q in PG(r,q) there i s exactly one d - f l a t belonging t o the collection. Clearly, K i s of class [ O , l ] i t , .

If K i s of type ( l ) i t l , then, by prop. X, r = 2 d t l

If t h i s i s not the case, then K i s of type ld+l = and, by (2.5)I and ( 2 . 5 ) I I y T O,d Y ~ , -~ kar-d-l; + ~ therefore (as

and, by ( 5 . 1 ) , k = y2d+l,d-1/8d. (O,l);+l

0<

25

T1:I:

< Yr,d+l)

:

on the other hand, (5.14)

k =

'r ,d-1 "d

26

G. Tallini

thus ( r e c a l l i n g t h a t y r,d-1 /y r , d t l = ad t l ad/ a r - d 'r-d-1) 'd+l <: a r - d and so r > 2 d t l . W h e n d = l , (5.14) becomesk=ar/81, which i s a n i n t e g e r o n l y w h e n

r i s odd; t h e r e f o r e , when r i s even, no t y p e ( l ) r - l

s e t e x i s t s on G

when r i s odd, r = 2 s t 1 , f o r such a k-set,

*

r,l ,q'

S

(5.15)

k =

1

q 2 i = 8s,q2

i=O

(r

9

2stl);

consequently: Proposition X I .

two cases may occur: r ,d ,q hence r = 2 d + l , o r K o f t y p e (Oyl)itl and r > 2 d t 1 ;

For a t y p e ( l ) r - d k - s e t K on G

e i t h e r K i s o f type (l)itly

i n b o t h cases k i s g i v e n b y (5.14). When d

1, such s e t s d o n ' t e x i s t i f r i s

2 s t 1 , then k i s g i v e n by (5.15).

even: i f r i s odd, r

(An example o f such a

o f t h e l i n e s i n PG(2s t 1 ,q) j o i n i n g p a i r s o f r y ,q l 2 conjugate p o i n t s belonging t o two f i x e d skew s - f l a t s i n PG(2s t l,q ) which a r e s e t i s t h e image on G

conjugate i n a q u a d r a t i c extension o f GF(q)).

6.

k-SETS ON G r Y d y q HAVING TWO CHARACTERS WITH RESPECT TO THE MAXIMAL

~ AND - zdtl ~

COLLECTIONS c

L e t K be a k - s e t on G

r,dyq

. If K i s o f

c l a s s [M1y!12]r-d,

= k 2 - k [ 1 + a (M t M 2 - l ) tH1M2 ] yr,d-l d 1

'd

(6.2) = ( k 8d - M 1 Y r,d-1 )/(M2-M1)

From (6.2)

(since

then, by (2.4),

T

r-d M. ,d

.

0) i t follows:

1

and t h e e q u a l i t y e i t h e r a t l e f t o r a t r i g h t hand s i d e holds i f f K i s o f t y p e

21

Grassmann manifolds representing subspaces in a Galois space

either (Ml)r-d Or (M2)r-d' If K i s of class [ N l , N 2 1 ~ t l s

then, by (2.5),

(6.5)

From (6.5) i t follows: N1 'r,dtl

(6.6)

/%

r-d-1 < k < N 2 y r , d t l / 8 r-d-1

'

and the equality e i t h e r a t l e f t or a t right hand side holds i f f K i s of type either

(Nl)itlor (N2)it1.

If K i s of type (N)i+l and class [M1sM21r,dy

from (6.1), (5.8), and (5.9)

(taking i n t o account t h a t Y ~ , ~ - ~ / Yedtl ~ , 8d/%r-d ~ ~ ~ 4r-d-1) i t follows:

When r = 2d t 1 , (6.7) becomes: (6.8)

N2

- N(M1 +M2) +M1M2

and has the solution N

= 0

M, and N = M2.

, If N

M1 (since r = 2d+ 1 and

), by (5.8), k = M1 y r , d - l / % d ; therefore, i n (6.3) the

'2dtl ,dtl = '2dtl ,d-1 equality holds a t the l e f t hand side and K i s of type ( N ) r - d .

N = M 2 , by (5.8), k

M2~r,d-l/%d;

Similarly, i f

hence, in (6.3) the equality holds a t the

right hand side and K i s of type ( N ) r - d . When r # 2 d t 1 and K # 0, G r s d S q s i . e . 0 < N character s e t with respect t o Zr-d

< Bdtl,

K cannot be a one

(by prop. X ) ; thus, K i s of type (MlsM2)r-d.

Therefore, from (6.3) and (5.8) i t follows:

28

G . Tallini

An easy computation shows t h a t , denoting by f(N) the polynomial a t the l e f t hand side of ( 6 . 7 ) , the following equalities hold:

When M1 = 0 (under the assumption N > 0 ) , (6.7) gives (6.12)

(M2 < 'r-d) ;

N = 1 t ( M 2 - 1 ) 3d /8r-d-1

hence, from (6.9) i t follows

from which, i f r - d

> d t 1 , arWd < M2, and t h i s i s impossible; therefore,

r < 2 d t 1. If M2 = a

r-d y

beside the solution N =

edtl (see (6.11)), which i s im-

possible since N < ad+l, (6.7) has the solution

By (6.13) and (6.9),

r

adtl

8r-d-1

< ed8r-d,

i . e . qrSd

< qdt1, thus again

< 2 d t 1. IfO<M,<M2<3r-d

and

r > 2 d + l , i.e. d < r - d - 1 ,

then,from(6.10)

and (6.11), i t follows:

therefore, i f N i s any solution of (6.7), then e i t h e r N

N < M1

8dt,/3r-d

> M2

3dt1/4r-d

which contradicts (6.9). Thus, in any case, r

Proposition X I I .

If K i s a k-set on G

class [ M 1 , M 2 ] r - d ,

then e i t h e r r = 2 d t l

and K i s of type (M1,M2),-d,

r ,d ,q

,K

f

0, G r , d , q y

<

2d t 1 , and

o f type

and K i s of type (N),-d,

or

(N)itland

or r < 2 d t 1

where M1, M2, N s a t i s f y (6.7) and ( 6 . 9 ) ; also, i f

29

Grassmann manifolds representing subspaces in a Calois space e i t h e r M1 = 0 o r

M2

8

then N i s given by e i t h e r (6.12) o r (6.13).

r-d’

By a dual argument: If

Proposition X I I I . and class [ N1,N2]At1

, K # 0, GrYdyq, o f type (M)r-d r,d,q r = 2 d t 1 and K i s o f type (M)Atl, o r r > d

K i s a k-set on

then e i t h e r

and K i s o f type (N1,N2)At1,

moreover, i f e i t h e r N1

0

where

or

G

t

1

N1, N2 and M s a t i s f y :

H2 = 8

dtl ’

then e i t h e r

Next we prove: Proposition X I V .

.

When r i s even, no k-set o f types (M)r-l

Consequently, no k-set o f types (M)r-l r,l ,q w i t h r even. Gr,l,q on G

Proof. (6.18)

and (0,N);

exists

and (N, a2); e x i s t s on

Suppose, on the contrary, t h a t such a s e t K e x i s t s . Then, by prop. X I I I , M

1

+ (N-1)8r-2/31

(N

< a2)

-

Since r i s even (see (5.4)): (6.19)

also, a1 and ar,2 (6.20)

(m a p o s i t i v e i n t e g e r ) ;

M=ma,

are coprime, t h e r e f o r e ((N- 1)ar-2/a1

N = 1

t

n a1

being an i n t e g e r ) :

(n a positive integer).

G. Tallini

30

From (6.18),

t a k i n g i n t o account (6.19) and (6.20), i t f o l l o w s :

m a1 1 t n E 0 (mod

hence, (6.21)

1

t

n ar-2 ;

el),

that i s

n = - l t c a

(c a positive integer);

1 '

by (6.20) and (6.21): N = l - a Since N

1

2 t c a 2 = - q t c a 1 1 ' 2

< a2 (see (6.18)), - q t c a l < a2, i.e. When r i s odd, sets o f t y p e (M)r-l

Remark.

2 cel

< a,:

a contradiction.

and t y p e (0,N);

f o r special N's

do e x i s t ( c f . prop. X I ) . Proposition XV.

When r = 3s t 2

i s even, no k - s e t o f types (M)r-l

and

(N ,N ) ' , where N and a are coprime, does e x i s t on G ( f o r instance, 2 1 2 2 r,l ,q 2t 2 t h i s occurs when N2 = 1,2 ,q,qt l , q 2 t q - 1, q t q ) ; when r i s odd, such a s e t i s o f type (0,N2);. Proof. 8

1

When d = 1, (6.14) becomes: 8

M2 2

-

Ma2 [ a1 + (Nl

so t h a t N1N2 ar-l

8r-2

t

N 2 - 1)8r-2]

t

NlN2 8r-l

8r-2

= 0,

i s a m u l t i p l e o f a2. On the other hand, a2 and each o f

are coprime; therefore, N1 = ca2 ( c an i n t e g e r ) . N23 a r - ~ a 3 s + ~9 ar-2 = 83s from prop. X I V the From N1 < a2 i t f o l l o w s c = 0. Thus, K i s o f t y p e (0,N2);; statement follows. Taking the complements: Propisition XVI. (N1,N2);,

When r = 3 s t 2 i s even, no k-set o f types (M)r-l

and

; when r i s odd, such a where N1 and a2 are coprime, e x i s t s on G r,l ,q

Grassmann manifolds representing subspaces in a Galois space

s e t i s of type (N1y32);. Finally, assume K i s of types (M1,M2)r-d G r,d ,q

and

(N1%N2)it1

(therefore K # 0 ,

) . Under these assumptions, both (6.1) and (6.4) hold, so t h a t

Then, two cases may occur: either

or

which, when r = 2d t 1 , becomes:

Substituting (6.25) f o r k in (6.3) and (6.6) respectively, the following i n equalities are obtained:

and

If(6.23) holds, then

(6.29)

\

N1 N2 N1

t

N2

= M

M 8

1 2 dtl 1 + (I!, + M2

318

d

r-d

3

r-d-1

- 1)3d/8r-d-1

*

31

32

G. Tallini

and a l s o M1M2 = N1N28r-d8r-d-l / 8 d t l 8 d

1

(6.29')

M1 + M 2 = 1 + ( N 1 + N 2 - 1)8r-d-1/8d

;

hence, N1, N2 a r e solutions of equation (6.7) and M1, M2 a r e solutions of equation (6.14).

> 2 d t l and M, # 0 , M, # 8r-dy then, by (6.10) and (6.11), 18 f(M18dtl r-d KO and f(M28d+l/8rr-d)
If r

< 2 d t 1 , N1 # 0 and N2 #

8dtl,

then, denoting by g(M) the polynomial

a t the l e f t hand side i n (6.14), g(N1 8rr-d/8dt,) < 0 , g(N2 8r,d/8dtl) therefore (M1 and M2 being solutions of the equation g(M) = 0 ) :

A l s o , f o r a l l rls, from (6.29)'s i t follows:

(6.32)

M1 = 0 * N 1

(6.33)

M2

N2 = 1 t M2 -

0,

Br-d + N 2 = 8d+l 9

N1 = M1

and from ( 6 . 2 9 ' ) ' s : (6.323

N1 = 0

(6.33)

N2 = 8d+l

When r (6.34)

=

9

M1

0,

9

M2 =

M2 = 1 t ( N 2 - l)8r,d,l/8d,

8r-ds

M, = N 1 8r-d-1

2 d t 1 , from (6.29)'s i t follows:

M1 =

N1y

M2

N

2'

( r = 2d+l).

f8

d

'

< 0;

Grassmann manifolds representing subspaces in a Galois space

33

The next proposition sums u p these results. Proposition XVII.

.

Let K be a k-set of types (M1 ,M2)r-d

and (N1

,N2)it1on

Then e i t h e r (6.23) or (6.24) holds. In the l a t t e r case, k i s given by Gr,d,q (6.25), and (6.27) and (6.28)hold. In the former case, (6.32), (6.33), and ( 6 . 3 2 ' ) , (6.33') h o l d ; under the further assumptions M1

r > 2 d t 1 , then (6.30)'s hold; i f

if

> 0 and

M2

<8

r-d

'

r < Z d t l , then (6.31)'s hold; f i n a l l y ,

i f r = 2 d t 1 , then M1 = N1, M2 = N2. Note that by the same argument as in the second paragraph of s e c t . 3, the next r e s u l t i s proved.

No k-set of type e i t h e r (0,8r-d)r-d

Proposition XVIII. e x i s t on G

r,dyq.

k-SETS OF TYPE (0,m,8S)S, 1 G s

7.

or (0,8dt1);+1 does

< r - d - 1 , ON

G

r,d,q

F i r s t , the following proposition will be proved. Proposition XIX.

In PG(r,q) a k-set K of type (Oyn,as)s, 1

< s < r-1 e x i s t s

only i f s = 1 , and e i t h e r m = l o r m = q ; furthermore, K i s e i t h e r a t - f l a t o r i t s complement ( i n PG(r,q)), 1 Assume s

Proof. = inl,

0

< ml <

>1

G r-2.

and l e t S be an ( s - 1 ) - f l a t i n PG(r,q) such t h a t I S n K I =

8s-l. Any s - f l a t

belonging t o S. (Indeed, ml

through S meets K a t exactly m - ml points not

> 0 implies the intersection i s not empty,

implies i t cannot consist of eS points). Thus, ( m - m l ) 8 r - s

ml

(mar,s-

=

k)/(9r-s-

= k-ml,

m1<8

s-1 from which

1 ) so t h a t ml doesn't depend on the ( s - 1 ) - f l a t S. Hence,

K i s of type (O,ml , 8 s - l ) s - l . By induction on decreasing dimension, K i s o f type

(O,n,al

Il.

Let a be a plane joining a point i n K with a l i n e external t o K. The s e t C

in a so t h a t n = p ( c a non-negative integer).Similar1 l y , i f B i s a plane through a l i n e i n K and a point not in K , then B n K i s a

a n K

i s of type (0,n)

s e t of type ( n , a l ) l i n B y so that q t 1 - n

p

b

( b a non-negative integer).

G. Tallini

34

-

C

Thus, q + 1 p =

pb,

which i m p l i e s e i t h e r c = 0, i . e . n = 1, o r b = 0, i . e .

n = q. Therefore K i s o f type e i t h e r (Oyly81)l o r ( O y q y 8 1 ) 7 yand i n the l a t t e r

I f K i s o f type (Oyly81)ly then K i s

case i t s complement i s o f type (0,1,81)1. a t - f l a t and 1

6

r - 2, as both l i n e s contained i n K and l i n e s skew w i t h K

e x i s t . On the other hand, a t - f l a t can be o f type (0,m,8S)s o n l y i f s = 1 and m = 1. The statement f o l l o w s .

L e t K be a type (0,m,81)l i s a c l a s s [ O,m,sl

k-set on G

l1s e t i n Tr - d '

.

For any Tr-d E Z r - d ' K n Tr-d r,d,q Hence, i f r - d 3, then by prop. X I X , and

prop. I, X I V and XV i n [ 21, e i t h e r m = 1 o r m = q.

r - d = 2 and d

If

i n Tit1,

>

2, any Ti+,

and again (being d t 1

> 3)

E

~i~~ meets K i n a class [ 0,m,81

1; Set

e i t h e r m = 1 o r m = q (again by prop. X I X

and prop. I , X I V , XV i n [ 2 1 ) . I f r - d = 2 and

d = l , i.e.,

r = 3 and d = l , assuming 2 6 r n < q - l ,

any

in G 391 (O,m), o r (my$ ) l l l a t t e r one), m =

through an m-secant o f K meets K i n a s e t o f type e i t h e r

of t y p e (m,qtl)ly

then

plane

a

i n a. I n the former case (a s i m i l a r argument holds i n the c p , c a p o s i t i v e i n t e g e r . Then, any other plane a ' i n G 3,1 9 9 through an m-secant o f K meets K i n a type (O,m), s e t . (Indeed, i f K n a ' were q t 1 - m = pb, b a p o s i t i v e i n t e g e r , which would con-

t r a d i c t m = pc). L e t II be a l i n e belonging t o K ( s i n c e K i s o f type (0,m,81)l, 11 does e x i s t ) ; by the previous argument, both planes B~ and B

through II i n 2 n o t belonding t o bl; t h e r e

belong t o K. L e t P be any p o i n t on G G3,1 ,q 391 ,q e x i s t s a unique plane n on G through P meeting B1 i n a Tine t; since 391 ,q t 5 K, II belbngs t o K and, therefore, P E K; then K = G , a contradiction. 391 ,q Again, i t follows t h a t e i t h e r m = 1 o r m q. o r (0,q,81)l; Hence, i n any case, K i s o f type e i t h e r (flyly~l)l

i n the

l a t t e r case t h e complement o f K i s o f type (0,l y81)1. Thus,

, then e i t h e r m = 1 r4,q q. When m = 1, K i s the image o f a c o l l e c t i o n I o f d - f l a t s i n PG(r,q)

Proposition XX. or m

I f K i s any type (0,1n,8~)~k-set on G

such t h a t t h e d - f l a t s i n I through any ( d - 1 ) - f l a t S i n PG(r,q) - s t a r (-1 G t G r-d, t depending on S) and t h e r e a r e p e n c i l s

form a (d,t)Q

o f d-flats i n

PG(r,q) such t h a t one o f the f o l l o w i n g holds: ( i ) no d - f l a t i n Q belongs t o

I ; ( i i ) e x a c t l y one d - f l a t i n o belongs t o I ; ( i i i )

8 i s completely contained

i n I. When m = q, K i s t h e complement of t h e j u s t described s e t .

Grassmann manifolds representing subspaces in a Galois space

35

2 Q s < r-d-1. Any T r,dyqy r-d E z meets K in a class [O,m,sSIs set in T By prop. XIX, since s > 2 , r-d r-d' K n Tr-d i s either the empty s e t , or T , or a type ( O , m ) S s e t , or a type r-d (m,8s)s set in Tr-d. If K n Tr-d i s of type ( O , m ) , , then, by [ 21 prop. XIV NOW, l e t K be a type (O,m,sS)s k-set on G

(being r - d S

. If

3 ) , i t i s either a point and m = 1 , or the complement ofapoint

) then, by [ 2 1 , prop. X V , i t i s either s sy in T the complement of a point and m as - 1 , or a hyperplane T and r-d-1 r-d m Bs-1. Thus, either K i s o f type (0,1,9 s ) s and any T r-d 'r-d # Tr-d # Trmd) meets K a t a point, or K i s of type (0,s 8 ) and any Tr-d€ z s-1' s s r-d (0 # K n Tr-d # T r-d ) meets K in a hyperplane, or K i s of type ( 0 , ~ 1~, 9-s ) s S or of type (0,q y 8 s ) s so t h a t i t i s the complement of a set o f type either

and m = q

K n Tr-d

i s of type

(m,8

( O , ~ , S ~ or ) ~ ( 0 , 8 s ~ l y ~ s )I st .follows: A k-set o f type (0,m,8s)Sy 2 < s

< r - d - 1 , on

must r ,d ,q be of type either ( O y l y ~ l or ) l ( O , ~ , S ~ ) ~If. i t i s of type ( O y l , a , ) l y then Proposition XXI.

G

either m = 1 or m = 8 s-1' When m = 1 , K i s the image of a collection I, of d-flats in PG(r,q) such t h a t , for any (d-1)-flat S in PG(r,q), one of the following i s true: ( i ) no d-flat through S belongs t o l l ; ( i i ) exactly one d-flat through S belongs t o l l ; ( i i i ) a l l d-flats through S belong t o l l . When K i s the image of a collection l 2 of d-flats in PG(r,q) such t h a t , s-1 for any (d-1)-flat S in PG(r,q) one of the following holds: ( i ' ) no d-flat

m

9

through S belongs t o 12; ( i i ' ) a l l d-flats through S belong t o 12; ( i i i ' ) the d-flats through S belonging t o l 2 form a (d,r-d-1)-star. When K i s of type

(O,q,al)l

i t s complement i s one of the above described sets.

The dual statements of prop. XX and XXI are l e f t t o the Reader. Finally, we prove: Proposition XXII.

A type (0,m,8s)sy 2 < s

<

r - 2 , k-set K on G

i s the r , l ,q ) or the s e t of lines in

image of either a s t a r of lines ( i . e . K = T r-d 'r-d a fixed hyperplane in PG(r,q), or the complement of such a set. Dually, a r - 2, on G

i s the image of either the rYr-2,q collection of (r-Z)-flats belonging t o a fixed hyperplane, or the set of (r-2)k-set of type (0,m,as);,

2 G s

f l a t s t h r o u g h a given point, or the complement of such a s e t .

36

G. Tallini

(and m = 1) or 1 o r i t i s t h e complement o f

By prop. X X I , K i s the image o f e i t h e r a c o l l e c t i o n I

Proof.

a c o l l e c t i o n l 2 (and m = 8 such a s e t .

s-1

) o f l i n e s i n PG(r,q),

I f K i s the image o f a c o l l e c t i o n 11, then i n PG(r,q) t h e r e e x i s t s a s t a r

o f l i n e s through a p o i n t P a l l l i n e s o f which belong t o Il (otherwise, I1 would c o n s i s t o f pairwise skew l i n e s and K would be o f type ( 0 , l )

S

,a

contra-

diction). If a l i n e t i n I

e x i s t e d n o t through P, then through any P ' E t 1 there would be a t l e a s t two l i n e s belonging t o I1 (namely, t and PP') so t h a t

a l l l i n e s through P ' would belong t o Il.Consequently, through any o t h e r p o i n t

P" i n PG(r,q) t h e r e would be a t l e a s t two lines'belonging t o

(PP' and P I P ' ' )

so t h a t a l l the l i n e s through PI' would belong t o I1 and K = G d i c t i o n . Therefore, I1 consists o f a l l the l i n e s through P.

r y ,q l

,a

contra-

I f K i s the image o f a c o l l e c t i o n 12, then t h e r e e x i s t s a p o i n t i n PG(r,q)

through which t h e r e i s no l i n e belonging t o I2 (otherwise, K would be o f t y p e a c o n t r a d i c t i o n ) ; thus, f o r any p o i n t P i n PG(r,q) e i t h e r no l i n e

(8s-1,8s)s,

through P belongs t o I*, o r the space j o i n i n g a l l the l i n e s through P belonging to I

2

i s a hyperplane

Since I2 # P' E

T

exi s t s

0,

T,

which w i l l be c a l l e d the p o l a r hyperplane o f P.

there e x i s t s a p o i n t P such t h a t t h e ' l a t t e r holds. For any f i x e d

\ { P I , the l i n e PP' belongs t o 12; thus, t h e p o l a r hyperplane

T I

of P'

.

L e t a be a plane through PP' belonging t o

T

n

T'.

Since t h e two p e n c i l s

o f l i n e s through P and P ' i n a belong t o 12, a l l t h e l i n e s i n a belong t o l 2 ( r e c a l l t h a t K i s o f t y p e (0,1,81)1.

Hence, a l l l i n e s i n

TI

n

TI'

meeting P P '

belong t o 12. I t f o l l o w s t h a t the p o l a r hyperplane o f any p o i n t on PP' contains

TI

n

TI'.

Then two cases may occur: e i t h e r

TI

=

TI'

or n #

TI'.

the previous argument, the p o l a r hyperplane o f any p o i n t t h e r e e x i s t s a l i n e t i n I 2 n o t belonging t o

11

o f T meets PP' a t a p o i n t

TI

(r-2)-flats

#

2

TI

!on

=

TI',

then by

PP' i s

II.Assume

L e t T E t; the p o l a r hyperplane

whose p o l a r hyperplane, i . e .

l i n e PP, which i s impossible. Therefore, I If

TI.

If

TI,

must c o n t a i n t h e

consists o f a l l l i n e s i n

TI.

TI',

then t h e p o l a r hyperplane o f any p o i n t on PP' contains t h e

S =

TI

n

TI'

and d i s t i n c t p o i n t s have d i s t i n c t p o l a r hyperplanes. It

follows t h a t the correspondence between t h e p o i n t s on t h e l i n e P P ' and t h e hyperplanes through S, such t h a t t o any p o i n t on PP' t h e r e corresponds i t s p o l a r hyperplane, i s one-to-one and onto. Thus, through any p o i n t i n t h e space

37

Grassmann manifolds representing subspaces in a Galois space t h e r e would be some l i n e belonging t o 12, i . e . a l l p o i n t s would have a p o l a r hyperplane, which i s impossible. Therefore, the case n # n ' cannot occur and the statement i s proved. Let

P be the c o l l e c t i o n o f a l l non-empty subspaces i n PG(r,q). A mapping p : P E PG(r,q)

(7.1)

-+

rpE

P

,

such t h a t

w i l l be c a l l e d a generalized n u l l p o l a r i t y . A l i n e t i n PG(r,q) w i l l be c a l l e d an i s o t r o p i c l i n e w i t h respect t o p i f t h e r e e x i s t s a p o i n t P on t such t h a t t

5

xP

(and then the same i s t r u e

f o r a l l p o i n t s on t). o f t h e c o l l e c t i o n Z o f the i s o t r o p i c l i n e s Yq w i t h respect t o p i s a s e t o f class [0,1y81]1. Conversely, i f I i s a c o l l e c t i o n Clearly, the image on Gr

9

o f l i n e s i n PG'(r,q) which i s the inverse image o f a class [0,1y81]1 s e t on

Gr,l,q PG(r,q)

, then

a generalized n u l l p o l a r i t y i s obtained under which t o each P i n

t h e r e corresponds the set t h e o r e t i c union o f t h e l i n e s i n

I through P,

which t u r n s o u t t o be a subspace. Thus the i n v e s t i g a t i o n o f c l a s s [ 0,1,81 l1 k-sets

on G

r,l ,q

i s equivalent t o t h e study o f generalized n u l l p o l a r i t i e s .

REFERENCES [ 1]

B. Segre Lectures on modern geometry, Cremonese Ed. Roma, 1961.

[ 2 1 G. T a l l i n i , Problemi e r i s u l t a t i s u l l e geometrie d i Galois, Relaz. N. 30, 1st. Mat. Univ. Napoli (1973) 1-30.

[ 3 ] G. T a l l i n i , Graphic c h a r a c t e r i z a t i o n o f algebraic v a r i e t i e s i n a Galois

space, A t t i Convegno Teorie Combinatorie (Roma, 3-15 Settembre 1973) , tomo 11, Acc. Naz. L i n c e i 11976) 153-165

[41 G. T a l l i n i , I k-insiemi d i r e t t e d i uno spazio d i Galois s t u d i a t i r i s p e t t o

38

G. Tallini a i f a s c i d i r e t t e , Quad. Sem. Geom. Comb., n. 28, Univ. Roma (Settembre 1980)

.

[5]

M. T a l l i n i S c a f a t i , {k,n}-archi d i un piano g r a f i c o f i n i t o , con p a r t i c o l a r e riguardo a q u e l l i con due c a r a t t e r i , Note I , 11, Rend. Acc. Naz. L i n c e i (8) 3 (1966) 812-818, 1020-1025.

[ 6 1 M. T a l l i n i S c a f a t i , C a l o t t e d i t i p o (m,n) i n uno spazio d i Galois S r,q’ Rend. Acc. Naz. L i n c e i , (8) 53 (1973) 71-81.

( 7 1 M. T a l l i n i S c a f a t i , Sui k-insiemi d i uno spazio d i Galois S a due s o l i c a r a t t e r i n e l l a dimensione d, Rend. Acc. Naz. L i n c e i (8) 60r(?976) 782788.

[8 1 J.A. Thas, A combinatorial problem, Geometriae Dedicata

1 (1973)

236-240.

Ann& of Discrete Mathematics 14 (1982) 39-56 0 North-HollandPublishing Company

ON k-SETS OF KIND (m,n) OF A FINITE PROJECTIVE OR AFFINE SPACE M. T a l l i n i S c a f a t i I s t i t u t o Matematico "G. Castelnuovo", U n i v e r s i t a d i Roma, I t a l y

INTRODUCTION I n the f o l l o w i n g paper, f i r s t we e x p l a i n the r e s u l t s known up t o now about We w i l l h ( q = p , p prime),

the theory o f k-sets w i t h two characters i n a Galois space PG(r,q). then deal w i t h k-sets o f k i n d (m,n) i n an a f f i n e space A(r,q) determining some new r e s u l t s on t h i s subject.

More p r e c i s e l y , i n sect. 4 we t r e a t the case r = 2, g i v i n g f o r m,n,k a r i t h m e t i c a l l y necessary conditions f o r the existence o f such sets. 5 we consider the general case

r

>

3

the

I n sect.

and we prove t h a t , i f i n AG(r,q)a k - s e t

o f k i n d (m,n) which i s n o t a p o i n t o r i t s complement e x i s t s , then q must be an odd square; moreover m = (q-J:)/2,n

= (qt@T/2,k

= [ q r t (J5)'1/2.

1. THE k-SETS WITH TWO CHARACTERS OF A PROJECTIVE PLANE Let

'II

be a p r o j e c t i v e plane o f order q and K a k-set o f k i n d (m,n)

in

9 , w i t h 0 6 m < n 6 q t l (namely a s e t o f k p o i n t s such t h a t every l i n e meets

q i t i n m o r n p o i n t s ) . I f tmand tn denote the numbers o f m-secant and n-secant

l i n e s o f K, we have ( s e e [ l 8 ] , n. 1 ) :

(1.1)

I

2 tm+tn=q + q + l , mtn

t

ntn = k ( q t l ) ,

m ( m - l ) t m t n ( n - l ) t n = k(k-1). By e l i m i n a t i n g tmand tn from the above equations, we obtain: 39

M. Tallini Scafati

40

(1.2)

k

(1-3)

2

-k

2

[n+m+q(n+m-l)] tmn (q t q t l ) = 0

+l)n

-k

u = m

(q

u

k-n(qt1) n - m

n - m

.

’

i

n

-

\

v

m

v

n

= u m = u n

+

It f o l l o w s t h a t

a k-set K o f k i n d (m,n) e x i s t s , k must s a t i s f y q (1.3) and (1.4) hold, moreover n-m must d i v i d e q.

P r o p o s i t i o n I. I f i n n (1.2);

from (1.2) we have k = ( q t l ) ( n - l ) + l

I f m = O , t h a t i s i f K i s o f k i n d (O,n),

and (see p r o p o s i t i o n I)i t i s q :0 mod n. Examples o f such k-sets a r e any a f f i n e plane o f n

f o r n = q ; any p o i n t o f n f o r n = 1 (such two cases are 9’ q’ the only, i f q i s prime), t h e (q+Z)-arcs o f n , i f q i s even. We prove t h a t : q Proposition 11.

If K i s a k-set o f k i n d (0,n)

mod n), t h e s e t of 0-secant l i n e s o f K, i n k i n d (O,n’), Proof.

with n’=q/n.

q’

of n

q

(and then i t i s q :0

dual o f n

q’

form a k l s e t o f

A p e n c i l o f l i n e s w i t h center i n a p o i n t o f K consists o f n-secant

l i n e s o f K. A pencil o f l i n e s w i t h center i n a p o i n t outside K contains k/n = q+l

- q/n

n-secant l i n e s o f K and n ’ = q/n e x t e r i o r l i n e s o f K, so

t h e statement i s proved.

On k-sets o f kind (m, n) o f a finite projective or affine space

41

From p r o p o s i t i o n I 1 i t f o l l o w s t h a t from a ( q t 2 ) - a r c (q even) we o b t a i n a [ q(q-l)/ZI-set

o f k i n d (O,q/2)

(Cossu arcs, [ 51). Denniston 71 determined h a class o f s e t s o f k i n d (O,n), f o r each n d i v i s o r o f q = 2 i n n (see a l s o q [ 251, [ 261). T i l l now we d o n ' t know examples o f k-sets o f k i n d (0,n) ( w i t h h 1 < n < q) o f n , w i t h q odd. However f o r q = 3 (h > 1) i t has been proved q t h a t k-sets o f k i n d (0,3) and (O,q/3) do n o t e x i s t (see [ 5 1 , [ 2 4 1 ) . I f m = l , K i s o f k i n d (1, n). We can suppose n

< q (for n = q t l ,

K h coincides w i t h a l i n e ) . I n such a case, ifmoreover q = p , p prime, i t has

been proved([ 181 p r o p o s i t i o n VIII), t h a t q must be a square, n = / T t 1 and K i s e i t h e r a Baer subplane n

(and so k = q t & + l ) o r a h e r m i t i a n arc (namely

a (q J{+ 1)-set o f k i n d (1,

1)). We know h e r m i t i a n arcs i n non desarguesian

4-

fit

planes and hermitian arcs which a r e n o t hermitian curves i n desarguesian planes 131, [91. The k-sets o f k i n d (my q t l ) i n

go back again t o those o f k i n d (0,n) q which a r e t h e i r complements and s i m i l a r l y the k-sets o f k i n d (m,q) go back T

again t o those of k i n d (1,n). Therefore i t remains t o examine t h e k-sets o f k i n d (m,n) w i t h

From p r o p o s i t i o n I (see a l s o propositions I 1 and X o f [181), we have Proposition 111. s a t i s f y i n g (1.5), 1

< <

h-1

If i n n

a k-set o f k i n d (m,n) e x i s t s , w i t h q = p h , p prime q k must s a t i s f y (1.2), moreover i t i s n = m t p ' , w i t h

and so, i f q i s a prime, such sets do n o t e x i s t . A t l a s t we have:

, w i t h q n o t a prime i s q r a t h e r d i f f i c u l t . A n o t simple p r e l i m i n a r y question i s t o determine a l l t h e The construction o f k-sets o f k i n d (m,n) o f n

p a i r s o f i n t e g e r s m and n a r i t h m e t i c a l l y admissible, t h a t i s the p a i r s o f integers m,n f u l f i l l i n g (1.5) and (1.6),

such t h a t n-m=p',

with l
and such t h a t equation (1.2) has i n t e g e r roots. We have t h e f o l l o w i n g r e s u l t s (see [ l ] ) : i f i t i s

q

<8

such k-sets o f k i n d (m,n) do n o t e x i s t , i f q i s a

M. Tallini Scafati

42

square (q a 9), f o r any i n t e g e r m such t h a t 2 (ma

mtJq)

<mG

,

q- 41-:

the pair

i s a r i t h m e t i c a l l y admissible and then i t i s e i t h e r k = m ( 1 t q t J q )

o r k = (mtJY) (1 t q -

47).A

class o f examples o f such sets has been r e c e n t l y

constructed (see [6]). More p r e c i s e l y i f one considers i n

TI

two by two d i s j o i n t ( t h e i r existence i s proved i n [8], i f

9

m Baer subplanes

TI

q plane), t h e j o i n o f such subplanes i s a s e t o f k i n d (m,mtJT).

i s a Galois T h e i r comple-

ments g i v e other examples o f k-sets w i t h two characters.

2.

THE k-SETS WITH TWO CHARACTERS I N PG (r,q) ( r 3 3 )

L e t us examine the known r e s u l t s about the k-sets o f k i n d (man), w i t h 0

<m
o f a Galois space PG(r,q) w i t h r

> 3.

We proved ( i n [ 191, p r o p o s i t i o n V I I I ) t h a t : Proposition I V .

I n PG(r,q) ( r 3 3 ) a k-set o f k i n d (0,n) e x i s t s o n l y i f i t

i s e i t h e r n = l o r n = q and then i t i s e i t h e r K

{ P I o r K=PG(r,q)-PG(r-1,q).

S i m i l a r l y (considering the complements) a k-set K o f k i n d (m,qtl) e x i s t s o n l y i f i t i s e i t h e r m = l o r m = q and then i t i s e i t h e r K=PG(r-1,q)

-

o r K=PG(r,q)-

{PI.

Moreover we proved (see [ 23 I, p r o p o s i t i o n V) : t h a t Proposition V.

I n PG(r,q)

o f k i n d (m,q) f o r any m

( r 3 3 ) k-sets o f k i n d (1,n)

f o r any n G q and

1 do n o t e x i s t .

From p r o p o s i t i o m I V and V i t f o l l o w s t h a t the study o f k-sets o f k i n d (man) o f PG(r,q) i s reduced t o the study o f k-sets such t h a t :

About such sets we proved t h e f o l l o w i n g p r o p o s i t i o n (see [ 201, [ 21 I ) : Proposition V I .

L e t K be a k-set of k i n d (m,n) o f PG(r,q),

w i t h 2<m
On k-sets of kind (m,nl of a finite projective or affine space Then q must be an odd square. Moreover m,n,k

43

are given by:

We o u t l i n e the proof of such r e s u l t s . We begin by examinating the case r = 3 . Every plane of PG(3,q) meets K i n a k ' - s e t o f k i n d (m,n), f u l f i l l s (1.2).

We prove t h a t t h e r e are planes of PG(S,q),

where k '

t h a t meet K e i t h e r

i n a k - s e t o r i n a k2-set, where kl and k2 are the two r o o t s o f (1.2). If 1 we denote by K* the set o f planes t h a t meet K i n a kl-set, we prove t h a t i n the dual space PG*(3,q)

K* i s s t i l l a kn-set w i t h two characters o f k i n d

(u,v), where u and v can be expressed by means o f m,n,k.

Then w i t h n o t

simple arguments o f a r i t h m e t i c and geometric k i n d about K and K* we prove t h a t q must be an odd squareandm,n,k with r

>

are given by (2.2),

f o r r=3. I n PG(r,q),

4, as a k-set o f k i n d (man) meets every PG(3,q) i n a k ' - s e t again

o f k i n d (man), necessarily m and n are given by (2.2)1 and (2.2)2. Then we o b t a i n f o r k the value given by (2.2)3.

3.

PRELIMINARIES ON k-SETS OF K I N D (m,n) OF AN AFFINE SPACE A(r,q).

L e t K be a k-set o f k i n d (m,n) w i t h 0 G m A (r,q).

G

q o f an a f f i n e space

L e t t and t denote t h e numbers o f m-secant and n-secant l i n e s m n o f K i n A (r,q). We have (see [161):

where we p u t

(3.2)

I

m t +ntn = m

k8r-1yq

m(m-1)t + n ( n - l ) t n = k(k-11, m

a '

s

s,q

= q t q

s-1

t...tqtl

M. Tallini Scafati

44

(because r - 1 ,q l i n e s and the number of d i r e c t i o n s o f A(r,q) i s

and we remember t h a t the number o f l i n e s o f A(r,q) i s qr-’ every d i r e c t i o n contains qr-’

a

r - 1 ,q

) . By e l i m i n a t i n g tmand t

L e t be P E A(r,q)

-K

n

i n (3.1) we o b t a i n

and um ,

l i n e s o f K through P. I t i s un+um

m (3.4)

un

un the numbers o f m-secant and n-secant 3r-l,q

, nu

-

k ) / (n-,,,,,

= ( k - m 8r-l,q)

/ (n-m).

= ( n 8r-l,q

S i m i l a r l y , i f Q E K, l e t us denote by v

m

t

mu = k, whence we have

and vn the numbers o f m-secant

and n-secant l i n e s o f K through Q. It i s v n t v m = 8r-l,q = k - 1,

a

whence, from (3.4), we o b t a i n

=

,

wn be t h e numbers o f r-1 m-secant and n-secant l i n e s o f K through the d i r e c t i o n 6 . It i s wn+wm = q L e t then 6 be a d i r e c t i o n o f A(r,q)

nwn t mw = k , m

and l e t wm

, ( n - l ) v n t (m-l)v,

whence

w (3.6)

m

w

n

r-1

- k)

/ (n-m),

= ( k - m q r-1 )

/ (n-m).

=

(nq

It follows that:

Proposition V I I . (3.3) and (3.41,

I f i n A(r,q)

(3.5),

a k-set o f k i n d (m,n) e x i s t s , k must f u l f i l l

(3.6) hold.

,

On k-sets of kind (m, n ) of a finite projective or affine space

45

We prove (see a l s o [ 161) t h a t :

Let s be a n-secant l i n e o f K and P a p o i n t o f s n o t belonging t o K (such a p o i n t c e r t a i n l y e x i s t s , because i t i s n

< 4 ) . The l i n e s through P d i s t i n c t

- 1 have a t l e a s t m p o i n t s in common w i t h K, Yq so t h a t i t i s k > m (ar-l 1) t n = n tmq 3 Let t be a m-secant l i n e Yq r - 2 ,q o f K and Q a p o i n t o f t 17 K (such a p o i n t t) c e r t a i n l y e x i s t s , because i t i s from s , whose number i s 8r-1

.

-

m

>

0 ) . The l i n e s through Q d i s t i n c t from t, whose number i s 8

have a t most n-1 p o i n t s i n common w i t h K, so t h a t i t i s k t m = (n-1)s ar-*

4.

$9

tm.

- 1,

r - 1 ,q (n-1) ( 8

So we have (3.7).

r - 1 ,q

-1)

t

THE k-SETS OF K I N D (m,n) OF AN AFFINE PLANE A(2,q) ( q = ph, P PRIME) L e t Kbe a k-set o f k i n d (m,n) w i t h 0

A(2,q)

m

<

nG q

o f an a f f i n e plane

(desarguesian o r n o t ) o f order q, where we suppose q = ph , p prime

(though many arguments hold a l s o i f q i s n o t a prime power). I f m = O , namely i f K i s o f k i n d (O,n),

plane obtained by extending A(2,q).

In

TI

l e t us denote by

the projective 9 K i s s t i l l a k-set o f k i n d (0,n) IT

q (such t h a t t h e l i n e a t i n f i n i t y i s an e x t e r i o r l i n e ) , hence, by t h e r e s u l t s o f

sect. 1, we have k = ( q t l ) ( n - l ) + l and q E 0 mod n. Conversely, given a k - s e t K o f k i n d (0,n) i n a p r o j e c t i v e plane IT

i n every a f f i n e plane obtained from

IT

K i s s t i l l a k-set o f k i n d (0,n) q’ d e p r i v i n g i t o f an e x t e r i o r l i n e o f

q K. Therefore t h e study o f k-sets o f k i n d (0,n) o f A(2,q) i s reduced t o t h a t

the k-sets o f A(2,q) q

= 0 mod

o f A(2,q).

.

Considering the complements, we have t h a t q and o f k i n d (m,q) a r e such t h a t k = q 2 - (q+l)(q-m-1)-1

one already developped i n IT

(q-m) and t h e i r study i s reduced t o t h a t o f k-sets o f k i n d (0,n) Therefore i n the f o l l o w i n g we suppose:

By v i r t u e o f (3.7) we have

M. Tallini Scafati

46

.

n t m q G k 4 (n-1) q t m

(4.2)

From p r o p o s i t i o n V I I , i f we p u t t h e r e r = 2, we o b t a i n I f i n A(2,q)

Proposition VIII..

a k - s e t o f k i n d (m,n) w i t h m,n f u l f i l l i n g

(4.1) e x i s t s , k must s a t i s f y t h e equation 2 k -k [ 1 t ( q t l ) (ntm-1)] t m n q ( q t 1 )

(4.3)

0.

Moreover i t i s :

(4.4)

n(qt 1)- k n-m

-

(4.5)

v

m

= u m

= u

m'

m

u

'

-,

v

n n-m

w = u n n

n-m

-

k-m(qt1) n-m

-

n

= u t n

n

t

9 n-m '

-n.-mm

As, by v i r t u e o f ( 4 . 5 ) , q/(n-m) must be an i n t e g e r and b e i n g q = p

h

w i t h p prime, r e l y i n g on (4.2), we have:

R

(4.7)

1G R

n=mtp

Hence, b e i n g n/(n-m) n = s'p'

n/p

R

and m/(n-m)

m/p

(1

h-1.

i n t e g e r s , i t i s m = sp

R

( w i t h s,s' p o s i t i v e i n t e g e r s ) . From (4.7) we have t h a t s ' = s t l .

R

( r e l y i n g on (4.7) and being m = sp ) we have 2 R t h a t k/p must be an i n t e g e r , t h a t i s k = c p R , w i t h c p o s i t i v e i n t e g e r . Hence

Moreover by v i r t u e o f ( 4 . 4 ) we have:

(4.8)

R

m=sp

,n

= (st1)p'

,k

= cp

R

(s,c p o s i t i v e i n t e g e r s , 1 G II G h-1).

From (4.8) we have: (4.9)

R

P

m

G q-p',

1G R

Q

h-1.

On k-sets of kind (m,nl of a finite projective or affine space

47

Then a t any rate: (4.10)

P

<m
Q

q-p.

From (4.3) and (4.8) we i n f e r t h a t the integers s and c must s a t i s f y the f o l l o w i n g c o n d i t i o n : (4.11)

c 2 - c [ ( 2 s t l ) ( q t l ) - p h-&It s ( s t l ) q ( q t l ) = 0.

I t follows t h a t the d i s c r i m i n a n t A o f (4.11) must be an i n t e g e r square

(whence A

> 0),

A being as f o l l o w s :

Moreover, from (4.8) and (4.9) we have: (4.13)

l < s 6 p

h-& -2,

(4.14)

s(q t 1)

1

t

I

1G

c G s(qt1)

G

t

q-p

h-&

(1 G

.

The i n e q u a l i t y (4.13) can be improved ifi t i s &

(4.15)

16L

-

< h/2

3 [ph-&-1 - / q ( p

1

<

2

[ph-&-l + Jq(p

s G

h-1

2h-2&

< h/2. I n f a c t we prove t h a t : 2h-211

-q-l) / ( q t l ) ]

, or

-4-1) / ( q + I)]6 s 6 phia-2.

I f we p u t i n (4.12) x = 1 t 2s, we have: A/(qtl) = The parabola of

X

2

- 2ph-& X+q+p2h-211/ ( q + l )

2 0-

lR2 (x,y): 2

y = x -2p

h-&

x+q+p

2h-2&

/(q+1)

does n o t meet.the l i n e y = O ifand o n l y i f i t i s &

> h/2, w h i l e

i f it i s a
M. Tallini Scafati

48

i t meets t h e same l i n e i n two r e a l and d i s t i n c t points, namely:

h-P. x1 = p

- Jq(p2h-2e-

q-1) / (q t 1)

x2 = Ph-R t J q ( p 2 h - 2 L - q - l ) / ( q v l ) On the o t h e r hand by (4.13) i t i s

3

<

x G 2p

h-ll

-3.

y

. Hence, i f i t i s ll

h/2,

i n order t h a t A i s p o s i t i v e , i t i s necessary t h a t 3 G x 6 x whence, being x

1'

x2

or

<x

6 2ph-'-3

,

1+2s, we have (4.15)

Therefore we come t o the f o l l o w i n g conclusion: Proposition I X . (with q =

ph,

Let

K be a k-set o f k i n d (m,n) o f an a f f i n e plane A(2,q)

p prime), where m,n s a t i s f y (4.1). Then m,n,k

are given by (4.8)

1 6 R G h - 1 and s , c p o s i t i v e integers f u l f i l l i n g (4.11) and (4.13),

with

(4.15) and (4.14).

It f o l l o w s t h a t i n any case m and n must f u l f i l l (4.10).

So i n A(2,q)k-sets

o f k i n d (m,n),

with n

> q-p, or w i t h m < p and m # 0

do not e x i s t . We say t h a t a p a i r o f i n t e g e r s (m,n) i s admissible f o r t h e existence o f a k-set o f k i n d (m,n) o f A(2,q),

i f i t f u l f i l l s (4.1) and i f t h e conditions

a. and s, w i t h 1 G II 6 h-1 R R such t h a t i t i s m= sp and n = ( s t 1 ) p and equation

o f p r o p o s i t i o n I X hold, t h a t i s i f two integers 1

< s < ph-'-2,exist

(4.11) i n the unknown c admits i n t e g e r solutions, one o f which a t l e a s t s a t i s f i e s the i n e q u a l i t y (4.14). The problem o f determining t h e k-sets o f k i n d (m,n) i n A(2,q) p r e l i m i n a r l y o f the determination o f the admissible p a i r s (m,n) i n v e s t i g a t i n g the r e a l existence o f the r e l a t e d k-sets. I f q i s an odd square,

(4.16)

we p u t

s = ( 6 - 1 ) / 2,

consists

and then of

and

On k-sets o f kind (m, n ) o f a frnite projective or affrne space

49

the p a i r (4.17) i s admissible. I n f a c t now i t i s II = h/2 and s, II s a t i s f y (4.13), 1

being

< II < h-1, moreover equation (4.11) i n the unknown c, i f s i s given by

(4.16),

has the two i n t e g e r s o l u t i o n s c = (JT(qf1)) / 2, both s a t i s f y i n g

(4.14) (where we p u t II i n A(2,q)

h/2,

/ 2). The question a r i s e s wheather

s = (4:-1)

k-sets o f k i n d (m,n) w i t h q odd square r e a l l y e x i s t , r e l a t e d t o

the p a i r (m,n) given by (4.17), where

/ 2 1 4;= (q

(4.18) We v e r i f y t h a t f o r q

<

2

f

q ) / 2.

125 other admissible p a i r s d i f f e r e n t from those given

by (4.15) do n o t e x i s t . For q even, t h a t i s p = 2, i t i s

II # h - 1,

t h a t i s i n (4.8),

(4.9) and

< R < h-2. I n f a c t from (4.13), i f II = h - 1 and p = 2 we absurd 1 < s < 0. For q = 3 h ahd q > 9 i t i s II # h - 1, t h a t i s (4.9) and (4.13) i t i s 1 < II < h-2. I n f a c t from (4.13), i f and p = 3 , we o b t a i n 1 < s < 3-2, t h a t i s s = 1; i f s = 1 (4.12)

(4.13) i t i s 1 have t h e i n (4.8), II

h-1

-

becomes c2 3qc t 2q(q t 1) = 0, which must have i n t e g e r roots, whence A = q(q-8) must be a square and i t happens o n l y when q = 9.

5.

k-SETS

OF KIND(m,n) I N AN AFFINE SPACE AG(r,q), WITH r

3

<m

C n

I n AG(r,q)

l e t K be a k-set o f k i n d (m,n),

with 0

us consider t h e case m = O . We observe that, extending AG(r,q) o b t a i n a p r o j e c t i v e space PG(r,q), a t i n f i n i t y o f PG(r,q)

ifwe denote by PG(r-1.q)

(so t h a t AG(r,q) = PG(r,q)-PG(r-l,q)),

k-set K i s s t i l l o f k i n d (0,n). we have e i t h e r n = 1, namely

Then, as we suppose r K = {PI, or

> 3,

< q.

First let

i n order t o the hyperplane i n PG(r,q)

the

by p r o p o s i t i o n I V ,

n = q , namely K = PG(r,q)

- PG(r-l,q),

b u t i n t h i s case K should n o t have e x t e r i o r l i n e s and t h i s i s i m p o s s i b l e . So we have proved t h a t :

M. Tallini Scafati

50

Proposition X.

with r

I n AG(r,q),

except f o r n = 1 and then K t h a t i n AG(r,q)

> 3,

k-sets o f k i n d (0,n) do n o t e x i s t ,

{ P I . Considering the complements, i t ' f o l l o w s

k-sets o f k i n d (m,q) do n o t e x i s t , except f o r m=q-1

then i t i s K=AG(r,q)

and

- {PI.

Therefore i n the f o l l o w i n g we suppose:

Every plane meets K i n a k ' - s e t o f k i n d (m,n) and then, by v i r t u e o f proposition IX, i t i s :

m = sp

(5.2)

II

,

a.

(s t l ) p

n

,

k' = c'p

(1

< I < h-1)

w i t h s and c ' p o s i t i v e i n t e g e r s . Moreover k ' must be a s o l u t i o n o f (4.3). We observe t h a t any plane o f AG(r,q)

cannot meet K i n t h e same number o f

points, say k ' . I n f a c t , i f i t were so, f o r a f i x e d m-secant l i n e t o f K, r-2 i as the number o f planes through i t i s 8r-2 = C q and every plane through i=o t i s k'-secant o f K, we should have k = (k'-m)e t m . S i m i l a r l y , f o r a nr-2 -secant l i n e we have k = ( k ' n)er-2+n; e q u a l i z i n g we have t h e absurd con-

-

e r-2 (n-m)

dition

= n - m . I t f o l l o w s t h a t k ' , by varying o f t h e plane,

has a t l e a s t two d i s t i n c t values, s o l u t i o n s o f (4.3), i n t e g e r r o o t s of equation (4.3), planes

.

say klyk2

t h a t i s K i s k i n d (kly

(kl

< k2),

k2) w i t h respect t o

L e t 6 be a d i r e c t i o n o f AG(r,q). The improper s t a r o f l i n e s through 6 6 i s an a f f i n e space AG (r-1,q) whose "points" are t h e l i n e s o f t h e s t a r , the 6

" l i n e s " a r e the improper pencils o f l i n e s through 6, t h e "AG (h,q)l"s the improper s t a r s of l i n e s through 6 o f a space AG(ht1,q) 6

are

p a r a l l e l t o 6.

I n such an a f f i n e space AG ( r - l , q ) , l e t K6 be the k6-set, c o n s i s t i n g o f t h e 6 n-secant l i n e s o f K through 6. We purpose t o study the s e t K6 o f AG (r-1,q). As t h e number o f l i n e s o f AG(r,q) through 6 i s qr-'

and t h e number o f

n-secant l i n e s o f K i s k6, the number o f m-secant l i n e s o f K i s qr-'-k6, r-1 whence nk6 + m(q -k6) = k, so t h a t

On k-sets o f kind (m. n) of a finite projective or affine space

k6 =

(5.3)

'-'

k-mq-k n-m

51

mq r-1

P

II

6 L e t us examine how a " l i n e " o f AG (r-1,q)

meets K

Such a " l i n e " i s a 6' pencil o f l i n e s through 6 which l i e s on a plane a p a r a l l e l t o 6 o f AG(r,q). Such a plane a meets K i n a ki-set (4.3).

( i = 1,2) o f k i n d (m,n) where ki s a t i s f i e s

I t i s t h e r e f o r e the question o f determining the number w i o f n-

-secant l i n e s o f K p a r a l l e l t o 6 and belonging t o a. Such a number i s given by (4.6)2, where we r e l y on (4.4)2 and p u t t h e r e ki instead o f k, t h a t i s the requested number i s given by:

(5.4)

i II h-Il wn=ki/p -mp

.

1 2 Therefore we come t o the conclusion t h a t Kg i s a k6-set o f k i n d (wn,wn) 6 1 2 o f AG ( r - l , q ) , where k i s given by (5.3) and wn, wn are given by (5.4). 6 F i r s t l e t us prove t h a t :

(5.5) 1 I f i t were wn = 0, by v i r t u e o f (5.4), we should have kl =mq. are s o l u t i o n s o f (4.3),

i t i s klk2

= mnq(qtl),

As kl and k2

t h a t i s k 2 = n ( q + l ) ; moreover

k t k =q(n+m-l)+n+m = m q t n ( q t l ) , whence i t f o l l o w s q = m and t h i s i s absurd, 1 2 1 by v i r t u e o f (4.1). We o b t a i n t h e r e f o r e wn # 0. If i t were wf = q, by v i r t u e we should have k =q(pktm) = qn. As kl and k2 are 2 s o l u t i o n s o f ( 4 . 3 ) , i t i s klk2 = mnq(qtl), t h a t i s kl = m(q+l); moreover

o f (5.4) and (5.2),

q(ntm-l)+n+m = nqtm(q+l), whence i t f o l l o w s q = n and t h a t i s absurd, 2 by v i r t u e o f (4.1). We o b t a i n t h e r e f o r e wn # q. 1 2 6 I t f o l l o w s t h a t K6 i s a k6-set o f k i n d (wn,wn) o f AG (r-1,q) such t h a t : kl + k 2

(5.6)

1

< wn1 s: wn2 < q-1.

By previous arguments, a p o s i t i v e i n t e g e r u and an i n t e g e r A , w i t h 1

e x i s t , such t h a t :

< X < h-1,

M. Tallini Scafati

52

(5.7)

1

X

w 2 = ( 0 t l ) pX , n

wn = u p ,

1

G

X

G

h-1.

From (:5.4) and (5.7), we have:

(5.8)

k

1

= m q t up

At11

,

From (5.8) we have (as k 2 - k l

k

2

= mq

a,where A

+

o$tll

t +'I1.

i s the discriminant of ( 4 . 3 ) ) :

whence, by virtue o f (5.2), we have: (5.10)

( 2 s t l ) ( q + l )- p

h-11 2

1 - 4s(s+l)q(q+l) = = ( q t l ) [q-Zp h-& (1+2s)t (1t2s)2 I + p 2h-211 .

F?

I t follows t h a t : 2X

- P2h-29. ) / ( q + l ) = [ 4 - 2 P h-& ( 1 + 2 s ) + ( 1 + 2 s2) 1.

(5.11)

(P

We infer t h a t

(p2X-p2h-Z11) / ( p h + 1 ) must be an integer.

The three following cases may occur: (1)

2X > 2h

(11)

2X < 2h

(111)

2A = 2h

-

-

211, 211, 2 ~ .

In the f i r s t case, i f we p u t 2h

+

211- 2h =

A, we have t h a t

a is

a

positive integer, such t h a t : (5.12)

0

<

2h,

< h and 11 < h , we have X + whence A = 2A+211- 2h < 2h. I t i s : i n f a c t , being

X

11

.:2 h , so that 2X + 211 < 4h,

On k-sets o f kind (m, nl o f afinite projective or affine space

53

h As p2h-2e and p t 1 are coprime and t h e l e f t hand s i d e o f (5.13) must be an a h integer, (p-- 1 ) / ( p t 1 ) must be an i n t e g e r and t h e r e f o r e d > h. We have: = (P-a-h (Ph + l ) - $ - h - l ) / ( P

( $ - l ) / ( P + hl )

h t1)

Therefore ( $ - h t l ) / ( p

<

because, being A-h

h + I ) = P-a-h -(I a-h F+ l ) / ( P h t l ) .

must be an i n t e g e r . This f a c t c o n t r a d i c t s (5.12),

h, t h e numerator o f t h e above f r a c t i o n i s smaller than

t h e denominator. Therefore the case ( I ) does n o t occur. I n the case (11), i f we p u t 2h-2L-2X = b y we have t h a t b i s a p o s i t i v e i n t e g e r , such t h a t : (5.14)

O < b < 2h,

i n fact

(5.15)

b=2h-2L-2X

(P

2X

< -P

2h,

2h-29.

being 211+2X

>0

(see (5.2) and ( 5 . 7 ) ) . It i s :

) / ( p h + l ) = (P2X ( l - p b ) ) / ( p h + l ) .

h As the l e f t hand s i d e o f (5.15) must be an i n t e g e r and being p and p t 1 b h coprime, i t i s ( p - l ) / ( p t 1 ) an i n t e g e r , hence i t i s b > h. Then we have (2-1)

/ (ph+l) = ( P

b-h

h (P + I )

- Pb - h - 1 ) / (ph+l)

= pb-h

- (pb-h t 1) / pht l ) .

h Therefore ( p b - h t l ) / ( p t 1 ) must be an integer, b u t t h i s c o n t r a d i c t s (5.14 because, being b-h

< h,

the numerator o f t h e above f r a c t i o n i s smaller than

i t s denominator. Then a l s o the case (11) i s impossible. L e t us now examine the case (111). From (5.11), (5.16)

(1+2s)

2

-

2p

h-L

( 1 t 2 s ) t q = 0;

i f we p u t x = 1t2s, (5.16) becomes:

(5.17)

x

2

-

2ph-9.x+q = 0,

being 2X = 2h-2&, we have

M. Tallini Scafati

54

so the equation (1.17) i n the unknown x must have t h e i n t e g e r s o l u t i o n 1t2s. I t f o l l o w s t h a t i t s d i s c r i m i n a n t A must be an i n t e g e r square. We have:

(5.18) whence we i n f e r 2h

>h

(being A

> 0)

and moreover

h 2X-h A=p( p - 1 ) .

(5.19)

If2X = h, we have A = 0; from (111) we have L! = h/2 and equation (5.17) has the o n l y s o l u t i o n

x = ph'2

Jq,

so t h a t i t i s 1t2s =

6

and then (by

v i r t u e o f (5.2)):

s =

(5.20)

(61 ) / 2,

m = (q-~;i)/2,

n = (qtJ{)/2,

whence i t f o l l o w s t h a t q i s an odd square. I f 2X

>

h h, being A a p o s i t i v e square and being p and p2x-h- 1 coprime,

we have t h a t h i s even and p2X-h i s a p o s i t i v e square i n t e g e r , which we 2 denote by a

.

We p u t then h = 2hl.

It i s :

that is:

whence ph-hl

+a

p

- a- = 1 and then

2

a=O, w h i l e a # 0, because i t i s

2X > h. The o n l y possible s o l u t i o n f o r t h e existence o f a k-set o f k i n d (m,n) i n AG(r,q)

i s then m = ( q - J q ) / Z ,

n = (qtfi)/2

(see (5.20)).

I f we p u t

those values o f m and n i n (3.3) we o b t a i n f o r k t h e f o l l o w i n g values: (5.21)

k = (qr

(J:)')/2.

On k-sets of kind (m,nl o f a finite projective or affine space

55

We come t h e r e f o r e t o the conclusion t h a t t h e f o l l o w i n g p r o p o s i t i o n holds Proposition X I .

<

n

< q-1)

Ifi n AG(r,q)

(r

>

3) a k-set K o f k i n d (man) ( w i t h 1

< rn <

e x i s t s , then necessarily q must be an odd square, moreover m,n,k

are t h e f o l l o w i n g i n t e g e r s : (5.22)

n = (qtJ{)/2,

m = (q-G)/2,

k = (qr+(J{)r)/2.

REFERENCES S. A n t o n e l l i , S u g l i a r c h i d i t i p 0 (m,n) d i un piano g r a f i c o ne n. 23 I s t i t u t o Mat. Univ. Napoli, 1973, 1-13.

A. B a r l o t t i , Sui {k,nl-archi ( 3 ) 2 (1956) 553-556.

T

q'

Relazio-

d i un piano l i n e a r e f i n i t o , B o l l . U.M.I.,

F. Buekenhout, Existence o f u n i t a l s i n f i n i t e t r a n s l a t i o n planes o f order q square w i t h a kernel o f order q, Geometriae Dedicata, 5 (1976) 189-194.

P.V. Ceccherini, Su c e r t i {k,n)-archi d e d o t t i da curve piane e s u l l e ( r > 2), Rend. d i Mat., (2)6(1969) { k , n l - c a l o t t e d i t i p 0 (0,n) d i un S r,q 185-1 95. A. Cossu, Su alcune p r o p r i e t a dei I k , n l - a r c h i i n un piano p r o i e t t i v o sopra un corpo f i n i t o , Rend. d i Mat., (5)20(1961) 271-277. M. De F i n i s , Sui k-insiemi a due c a r a t t e r i i n un piano d i Galois, Semin a r i o d i Geom. Combinatorie, I s t i t u t o Mat. "G. Castelnuovol , Roma, quaderno n. 13, a p r i l e 1979, 1-11. R.H.F. Denniston, Some maximal arcs i n f i n i t e p r o j e c t i v e planes, J.Comb. Theory, (6) 3 (1969) 317-319. J.W.P. H i r s c h f e l d , C y c l i c P r o j e c t i v i t i e s i n PG (n,q), c e i , 17, Tomo I, Acc. Naz. L i n c e i (1976) 201-211.

A t t i Convegni L i n -

R. Metz, On a class o f U n i t a l s , Geometriae Dedicata, 8 (1979) 125-126. [ 101 B. Segre, Sui k-archi nei p i a n i f i n i t i d i c a r a t t e r i s t i c a due, Revue des

Math. pures e t appl.,

2 (1957) 289-300.

[ l l ]B. Segre, Le Geometrie d i Galois, Ann. Mat. pura e appl., 1-97.

(4) 48 (1959)

[ 121 B. Segre, Forme e geometrie hermitiane, con p a r t i c o l a r e riguardo a1

caso f i n i t o , Ann. Mat. pura e appl.,

( 4 ) 70 (1965) 1-201.

[ 131 6. Segre, I n t r o d u c t i o n t o Galois Geometries, Mem.Acc.Naz.

5 (1967) 133-236.

Lincei , (8)

56

M. Tallini Scafati

I 1 4 1 G. T a l l i n i , Problemi e r i s u l t a t i s u l l e Geometrie d i Galois, Relazione n. 30, I s t i t u t o Mat. Univ. Napoli, 1973, 1-30. [ 151 G. T a l l i n i , Graphic c h a r a c t e r i z a t i o n o f a l g e b r a i c v a r i e t i e s i n a Galois

space, A t t i Convegni L i n c e i , 17, Tomo 11, Acc. Naz. L i n c e i (1976) 153-165.

[ 161 G. T a l l i n i , Problemi d i imnersione nei sistemi d i Steiner, Seminario Geo-

m e t r i e Combinatorie, I s t i t u t o Mat. "G. Castelnuovo", Roma, quaderno n . 21 , novembre 1979.

171 M. T a l l i n i Scafati , Sui {k,n)-archi d i un piano g r a f i c o f i n i t o , Rend. Acc. Naz. L i n c e i , (8) 40 (1966) 1-6. [ 181 M. T a l l i n i S c a f a t i , Sui {k,n)-archi

d i un piano g r a f i c o f i n i t o con part i c o l a r e riguardo a q u e l l i con due c a r a t t e r i , Rend. Acc. Naz. L i n c e i , (8) 40 (1966) 812-818, 1020-1025.

, Caratterizzazione g r a f i c a d e l l e forme hermitiane d i d i Mat., 26 (1967) 273-303.

[ 191 M. T a l l i n i Scafati

un S

[ 201

, Rend.

r,q M. T a l l i n i S c a f a t i , C a l o t t e d i t i p o (m,n) i n uno spazio d i Galois SrYq, Rend. Acc. Naz. L i n c e i , (8) 53 (1973) 71-81.

[ 2 1 1 M. T a l l i n i S c a f a t i , The k-sets o f type (m,n) i n a Galois space SrYq, A t t i Convegni L i n c e i , 17, Tomo 11, Acc. Naz. L i n c e i (1976) 459-463. [ 221 M. T a l l i n i S c a f a t i , Sui k-insiemi d i uno spazio d i Galois a due s o l i

c a r a t t e r i n e l l a dimensione d, Rend. Acc. Naz 782-788.

[ 2 3 1 M. T a l l i n i S c a f a t i , I k-insiemi d i t i p o (m,n Rend. d i Mat. (14) 6 (1981).

L i n c e i , (8) 40 (1976)

d i uno spazio a f f i n e A

r,q'

Thas, Some r e s u l t s concerning { (.q. t l ).( .n - ); n)-arcs and { ( q t l ) ( n - l ) t t1;nl-arcs i n f i n i t e p r o j e c t i v e pianes o f order 9,J. Comb. Theory, 19 (1975) 228-232.

[ 241 J.A.

1251 J.A. Thas, Construction o f maximal arcs and p a r t i a l geometries, Geomet r i a e Dedicata, 3(1974) 61-64. Thas, Construction o f maximal arcs and dual ovals i n t r a n s l a t i o n planes , Europ. J Combi n a t o r i cs , (1 980) 1 , 189-1 92.

[ 261 J.A.

.

Thas, F. De Clerck, Some a p p l i c a t i o n s o f t h e fundamental characteri z a t i o n theorem o f R.C. Bose t o p a r t i a l geometries, Rend. Acc. Naz. L i n c e i , (8) 59 (1975) 86-90.

[ 271 J.A.

Ann& of Macnte Mathematics 14 (1982) 57-76 Q North-Hollandpublishing Company

COMBINATORICS OF FINITE GENERALIZED QUADRANGLES : A SURVEY Joseph A . Thas Seminar of Geometry and Combinatorics State University of Ghent Krijgslaan 2 7 1 B-9000 GENT (Belgium) Generalized quadrangles may be considered from two points of view, the group theoretical one and the combinatorial one. The present survey attempts to deal mainly with what might be called the combinatorial properties of the finite generalized quadrangles : first, given an abstract generalized quadrangle, what restriction on the parameters, etc., are there; second, what conditions of a combinatorial nature on an abstract generalized quadrangle force it to be isomorphic to one of the known examples. 1. INTRODUCTION

Generalized quadrangles were introduced in 1 9 5 9 by J. Tits in his During the last 10 years a lot of famous "triality paper" [ 5 3 1 interesting contributions to the theory of generalized quadrangles were written, and the importance of the subject is shown by the many applications to other domains, such as group theory, graph theory and designs.

.

Generalized quadrangles may be considered from two points of view, the group theoretical one and the combinatorial one. The present survey attempts to deal mainly with what might be called the combinatorial properties of the finite generalized quadrangles : first, given an abstract generalized quadrangle, what restriction on the parameters, etc., are there ; second, what conditions of a combinatorial nature on an abstract generalized quadrangle force it to be isomorphic to one of the known examples. Further, all quadrangles embedded in the space AG(n,q) are described. Finally, the survey contains some important characterizations of finite generalized quadrangles by properties of their automorphism group. I note that there are many links with the theory of the classical varieties in PG(n,q), the theory of projective and affine spaces, and the theory of the designs and the strongly-regular graphs. Since most of the proofs are very long and complicated, especially those related to characterization theorems, in my opinion it is better to delete all proofs and to refer to the literature. Finally, I do not claim that the survey is complete : it only contains a selection from the important combinatorial theorems concerning generalized quadrangles and may be considered as a short summary of the combinatorial part in the book about generalized quadrangles on which the author and S.E. Payne momently work, 57

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J.A. Thus

2 . AXIOMS AND DEFINITIONS

A (finite) generalized quadrangle (GQ) is an incidence structure S=(P,B,I) in which P and B are disjoint (non void) sets of objects called points and lines, respectively, and for which I is a symmetric point-line incidence relation satisfying the following axioms : (i) each point is incident with l+t lines (t>l) and two distinct points are incident with at most one line; (ii) each line is incident with l + s points ( s a l ) and two distinct lines are incident with at most one point; (iii) if x is a point and L is a line not incident with x, then there is a unique pair (y,H)EPXB for which x II! I y I L. It is easy to prove that I P I=v=(s+l) (st+l) and I B I=b=(t+l) (st+l) 1431

.

The integers s and t are the parameters of the GQ and S is said to have order (s,t); if s=t S is said to have order s. There is a point-line duality for GQ (of order (s,t)) which interchanges "point" and "line", interchanges s and t, in any definition or theorem. Normally, we assume without further notice that the dual of a given theorem or definition is also given. A grid (resp. dual grid) is an incidence structure S=(P,B,I) II i=O,...,s1 n j = O ,...,s,I,sl>O and s,>O (resp. with P={x.. I J II i=O ,..., tl A j=O ,...,t,), t,>O and t,>O), B={Lo B={L *Ls1 ' i J ILr iff i=r yt, I ) , x.. M, ,...,M, 1 (resp. P={xo ,..., xtl , yo I J (resp. L , Ix, iff i=r), and xi]IM, iff j=r (resp. L i jIy, iff j=r). iJ A grid (resp. dual grid) with parameters s1 , s 2 (resp. t, ,t2) is a GQ iff s I = s , (resp. tl=t,). Evidently the grids (resp. dual grids) with s1 =s, (resp. tl=t,) are the GQ for which t=l (resp. s=l).

,...

,...,

Let S be a GQ, a grid, or a dual grid. Given two points x,y of S, we write x-y and say that x and y are collinear, provided there is some line L for which x I L I y . And x+y means that x and y are not collinear. Dually, for L,MEB, we write L-M or L+M according as L and M are concurrent or non-concurrent, respectively. If x-y (resp. L-M) we also say that x (resp. L) is orthogonal or perpendicular to y (resp. M). For xEP put x1 ={YEP II y-XI, and note that x€xl. The trace of a pair of distinct points (x,y) is defined to be the set x1 "yl and is denoted tr(x,y) or {x,y)'. We have I{x,yI 1 I = s + l if x-y More generally, if ACP, A "perpendicular" and I {x,y}'l =t+l if x+y I XEA}. For xfy, the span of the pair (x,y) is is defined by A1 =n{xl 1 ~p(x,y)={x,y}~*={u~P11 UEZ1 VEX 1 "y 1 1 . If xjy, then {x,y)l' is also called the hyperbolic line defined by x and y. And the closure of the pair (x,y), xfy, is cl(x,y)={z€P I1 z1 :l{x,y}11 # $ }

.

.

A triad (of points) is a triple of pairwise non-collinear points. Given a triad T=(x,y,z), a center of T is just a point of T I . We say 1 T is acentric, centric, or unicentric according as IT I is zero,

Combinatorics of finite generalized quadrangles

59

positive, o r equal to 1. Isomorphisms(or collineations), anti-isomorphisms ( o r correlations), automorphisms, anti-automorphisms, involutions and polarities, o f generalized quadrangles, grids and dual grids are defined in the usual way. 3. THE CLASSICAL EXAMPLES Before giving other definitions and general properties of GQ, we describe the classical generalized quadrangles. The role of these quadrangles in the general theory may be compared with the role of the finite desarguesian planes PG(2,q) in the general theory of abstract finite projective planes. (a) Consider a non-singular quadric Q of projective index 1 1151 of the projective space PG(d,q), with d=3,4 o r 5 . Then it is easy to check that the points o f Q together with the lines of Q (which are the subspaces of maximal dimension on Q) form a GQ Q(d,q) with parameters s=q, t = l , v=(q+l) 2 , b=Z(q+l), when d=3; s=t=q, v=b=(q+l) (q3+1), when d=4; s=q, t=q2, v=(q+l) (q3+l), b=(q3+1) (q3+1), when d=5. Since Q(3,q) is a grid, its structure is trivial, Further, we recall that the quadric Q has the following canonical equation : xo x1 +x2x3 =0, when d=3 ; xi +xlx, +x3x4 =0, when d=4 ; f(xo ,xl) +x, x3 +x4 xg =0, where f is an irreducible binary quadratic form, when d=5. (b) Let H be a non-singular hermitian variety of the projective space PG(d,q2), d=3 or 4. Then it is easy to see that the points of H together with the lines on H form a GQ H(d,q2) with parameters 3 3 s=q’, t=q, v=(q’+l) (q +1), b=(q+l) (q +1), when d=3; 3 3 s-q” t=q , v=(q?+l)(q’+l), b=(q +l)(qs+l), when d=4. Recall that H has the canonical equation x;+l+x;

+1

+ . . . + x;+1=0.

(c) The points o f PG(S,q), together with the totally isotropic lines with respect to a symplectic polarity, form a GQ W(q) with parameters 2 s=t=q, v=b=(q+l) (9 +1). Recall that the lines of W(q) are the elements of a linear complex of lines of PG(3,q), and that a symplectic polarity of PG(3,q) has the following canonical bilinear form xo Y1 -XI Yo +x2 Y3 -x3 Y? * All these generalized quadrangles (all of which are associated with classical groups) are due to J. Tits 1 1 1 I

.

Theorem [ 51 ] (a) The GQ Q(4,q) is isomorphic to the dual of W(q). Moreover, Q(4,q) (or W(q)) is self-dual iff q is even.

60

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(b) The GQ Q(5,q)

is isomorphic to the dual of H(3,q2).

4. TWO FUNDAMENTAL THEOREMS Theorem 1431 is a GQ of order (s,t), then s+t divides st(s+l)(t+l). Tf,B,I) The following fundamental theorem is due to S.E. Payne [261. Theo rem Let X=Ixl ,..., x,,,}, ma2, and Y=Iyl,...,yn}, n22, be disjoint sets of pairwise non-collinear points of the GQ S of order (s,t), s>1, and let each point of X be collinear with each point of Y. Then If equality holds, then one of the following must (m-l)(n-l)<s? occur. (i) m=n=l+s, and each point of Z=P-(XUY) is collinear with precisely two points of XUY. 2 (ii) mfn. If m
.

Corollaries and remarks (a) By putting X=Ix, ,x, 1 and Y=Ix, ,x2 1' we find the celebrated 2 inequality t < s (if sfl), first discovered by D.G. Higman [ 141 If t=s2 then by (ii) any triad has exactly l+s centers. This was first discovered by R.C. Bose [ 6 ] . Conversely, if for fixed x and y, x+y, any triad (x,y,z) of a GQ S has a constant number of centers, then P.J. Cameron [ 9 I proves that t=s2. (b) Suppose that X=Ix, ,x2 and Y=Ix, ,x, 1' If 1 XI=p+l (and s > l ) there holds pt<s2. Moreover, if pt=s2 Hnd pet, then each point uiqcl(xl ,x2) is collinear with l+t/s=l+s/p points of Ix, ,x21 l . This inequality and interpretation of the equality were first discovered by J.A. Thas 1421. Moreover, using an argument analogous to that of P.J. Cameron in the last part of (a), he proves that if p
.

.

.

.

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5. REGULARITY, ANTIREGULARITY, SEMIREGULARITY AND PROPERTY (H) 11

If x-y, xfy, o r if x+y and I Ix,y1 I =t+l, we say the pair (x,yJ is regular, The point x is regular provided (x,y) is regular for all YEP, yfx. A point x is coregular provided each line incident with x is regular. The pair (x,y), x&y, is antiregular provided I z 1.JIx?y)1 I < 2 for all zEP-{x,y1, A point x is antiregular provided (x,y) i s antiregular f o r all F P with xcy. Further a point u is called semiregular provided that zEcl(x,y) whenever u is the unique center of the triad (x,y,z). And a point u has property (H) provided zEcl(x,y) implies xecl(y,z), with (x,y,z) a triad consisting of points in UL. Clearly, if a point is

Combinatorics of finite generalized quadrangles

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semiregular, then it has property (H). Theorem (a) Let x be a regular point of the GQ S=(P,B,I) of order (s,t). Then the incidence structure with point set xl-Ix1, with line set the set o f spans {y,z}ll , where y,z€xl-{x} and y+z, and with the natural incidence, is the dual of a net 1 1 1 1 of order s and degree t+l. If in particular s=t>l, there arises a dual affine plane of order s , and moreover the incidence structure xp with point set XI, with line set the set of spans {y,zlll, where y,zExl, and with the natural incidence, i s a projective plane of order s. (b) Let x be an antiregular point of the GQ S=(P,B,I) of order s, s f l , and let y#x be a point of x* with L being the line xy. An affine plane x(x,y) of order s may be constructed as follows. Points of n(x,y) are just the points of XI that are not on L. Lines are the point sets {x,z}ll-{x} , with x-z and y+z, and {x,uI1-{y}, with u-y and ufx. 1 Now let s 2 = t , sfl. Then for any triad(x,y,z) we have I {x,y,zI I = =s+l and I Ex,y,z}lLl <s+l. We say (x,y,z) is 3-regular provided I {x,y,z}lll =s+l. Finally, the point x is called 3-regular iff any triad containing x is 3-regular. Theorem Let S b e - a G Q . o f order (s,s'),,s#l, and suppcse that any triad contained in Ix,y} , x+y , is 3-regular, Then the incidence structure with point set {x,y}l, with line set the set of elements { z , z l , ~ " } ~with ~ , Z , Z ~ , Z " E { X , Y } ~ and , with the natural incidence, is an inversive plane of order s. A s an illustration of all these new concepts we have the following theorem on the classical examples. By 3 it is sufficient to consider Q(3,q), .Q(4,q)! Q(S,q), and H(4,q?). Of course the structure of Q(3,q) is trivial. Theorem (i) Properties of Q(4,q) : all lines are regular; all point; are regular iff q is even; all points are antiregular iff q is odd; all points and lines are semiregular and have property (H). (ii) Properties of Q ( 5 , q ) : all lines are regular; all points are 3-regular; all points and lines are semiregular and have property (HI * (iii) Properties of H(4,q2) : for any two non-collinear points x,y we haveI{x,y)lll=q+l; f o r any two non-concurrent lines L,M we have I I L , M I l l l = Z but (L,M) is not antiregular; all points are semiregular and have property (H); all lines have property (H) but no line is semiregular. Now there follow some very useful theorems about regularity, antiregularity, etc. Theorem 1 [ 511 (i) If l<s
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Theorem 2 Let S=(P,B,I) be a GQ of order (s,t), s>l and 0 1 . (i) If (x,y) is antiregular with s=t, then s is odd 151 I. (ii) If S has a regular point x and a regular pair (Lo ,Ll) of 1 non-concurrent lines for which x is incident with no line of ILo,L, 1 , then s=t is even [ 36 I. (iii) If x is coregular then the number of centers of any triad (x,y,z) has the same parity as l+t 131 I (iv) If each point is regular, then t+ll (~'-1)s~ 131 1.

.

Corollary Let S=(P,B,I) be a GQ of order (s,t), s>l and 0 1 . (a) If S has a regular point x and a regular line L, with XZL, then s=t is even 1361 (b) If s=t, if s is odd, and if S contains two regular points, then S is not self-dual 136 I. (c) If x is coregular and t is odd, then I Ix,yllll = 2 for all Y+X 131 1. (d) If x is coregular and s=t, then x is regular iff s is even 123 1, 131 1. (e) If x is coregular and s=t, then x is antiregular iff s is odd 131 ]

.

.

Theorem 3 1 5 1 ] Let S=(P,B,I) be a GQ of order (s,t), s>l and 0 1 . (i) If s=t and UEP is regular o r antiregular, then u is semiregular. (ii) If s>t, then uEP is regular iff u is semiregular. 6. OVOIDS AND SPREADS

An ovoid of the GQ S=(P,B,I) is a set 0 of points such that each line of S is incident with just one point of 0. A spread of the GQ S is a set R of lines such that each point of S is incident with exactly one line of R. It is trivial that GQ with s=l o r t=l have ovoids and spreads. A l s o it is easy to see that any ovoid o r spread has st+l elements. Some examples (a) Q(4,q) has ovoids (if QnPG(3,q) is an elliptic quadric, then it is an ovoid o f Q(4,q)); Q(4,q) has spreads iff q is even [ 36 I. (b) H(3,q-) has ovoids (an hermitian curve on H is an ovoid of H(3,qZ)) ; H(3,qz) has no spread 7 I . (c) H(4,q') has no ovoid 1491; we do not know whether o r not H (4,q2 ) has spreads

.

7. SUBQUADRANGLES

The GQ S'=(P';B',I') of order (s',t') i s called a subquadrangle of the GQ S=(P,B,I) of order (s,t) if P'CP, B'CB, and if I' is the restriction of f to (PIXB')V(B'XP'). If S ' f S , then we say that S' is a proper subquadrangle o f S. Now let S ' be a proper subquadrangle of S. Then for a line LEB precisely one of the following occurs : (i) LEB', i.e. L belongs to S ' ; (ii) LqB' and L is incident with a unique point x of P I , i.e. L is tangent to S ' at x ; (iii) LFB' and L is incident with no point of P', i.e. L is external to S ' . Dually,

Comblnatorics of finite generalized quadrangles

63

there are external points, tangent points and points of

S'.

The basic inequalities 1 2 2 1.1 40 1.. Let S ' be a proper subquadrangle of S, with notation as above. Then either s = s ' or s2s't'. If s = s ' then each external point is collinear with the l+st' points of an ovoid of S ' ; if s=s't', then each external point is collinear with exactly 1+s' points of S ' . The dual holds, similarly. Corollary I371 Let S'=(P',B',I') be a proper subquadrangle of S=(P,B,I) with S having order (s,t) and S ' having order (s,t'), i.e. s = s ' and t>t'. Then we have : (i) t>s; if t=s, then t'=l. (ii) If s > l , then t'<s; if t'=s>2, then t=s2 , (iii) If s = l , then ll and t'>l, then &kt'<s and s3' 2l, then t'=& (vi) Let S ' have a proper subquadrangle S" of order (s,t"), 9 1 . Then t"=l, t'=s and t=s?.

.

.

Automorphisms and subquadrangles 1241 The substructure S,=(P, ,Be ,Ie) of the fixed elements of an automorphism 0 of the GQ S=(P,B,I) must be given by at least one of the following : (i) Bg=$ and Po is a set of pairwise non-collinear points; (i)' P,=$ and Be is a set of pairwise non-concurrent lines; (ii) Po contains a point x such that x-y for every point YEP, and each line of Bo is incident with x ; (ii)' Be contains a line L such that L-M for every line MEBe, and each point of P, is incident with L ; (iii) S, is a grid; (iii)' S, is a dual grid; (iv) S, is a subquadrangle of order (s',t'), s ' > 2 and t'a2. Some examples of subquadrangles (a) Examples with s = s ' or t=t'. Q(4,q) has subquadrangles isomorphic to Q(3,q) (here s=t=s' and t'=l); Q(5,q) has subquadrangles isomorphic to Q(4,q) (here s2=t, s=t'=s') ; H(4,qZ). has subquadrangles isomorphic to H(3,q2) (here s = s ' , t=s3I2, t'=6). 2 (b) An example with s f s ' and tft'. Q(4,q ) has subquadrangles isomorphic to Q(4,q) (here s'=t', s=t=s't'). 8. ALL KNOWN NON-CLASSICAL GENERALIZED QUADRANGLES AND SOME OF THEIR PROPERTIES

[

(a) The generalized quadrangles T(0) of J. Tits Let d=2 (resp. d=3) and let 0 be an oval 1 1 1 ] (resp. an ovoid 1 1 I) of PG(d,q). Further, let PG(d,q)=H be embedded a s a hyperplane

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J.A. Thas

in PG(d+l,q)=P. Define points as (i) the points of P-H, (ii) the hyperplanes X of P for which I XnOI =1, and (iii) one new symbol (-). Lines are (a) the lines of P which are not contained in H and meet 0 (necessarily in a unique point), and (b) the points of 0. Incidence is defined as follows. A point of type (i) is incident only with Lines of type (a); here the incidence is that of P. A point of type (ii) is incident with all lines of type (a) contained in it and with the unique element of 0 in it. The point (m) is incident with no line of type (a) and all lines of type (b). It is an easy exercise to show that the incidence structure so defined is a GQ with parameters 2 s = t = q , v=b=(q+l)(q + l ) , when d=2; 2 3 2 3 s=q, t=q , v=(q+l)(q +l), b=(q +l)(q + l ) , when d=3. If d=2, the GQ is denoted by Tz(0); if d=3, the GQ is denoted by T3 ( 0 ) . If no confusion is possible, these quadrangles are also denoted by T (0) These examples are due to J. Tits [ l l ]

.

.

Isomorphisms (a) T2 (0) is isomorphic to Q(4,q) iff 0 is an irreducible conic (hence by the celebrated theorem of B. Segre [ 32 I, for q odd T2 (0) is always isomorphic to Q(4,q)); T2 (0) is isomorphic to W(q) iff q is even and 0 is a conic. We note that a complete classification of the GQ T2 (0),q even, depends on the classification of the ovals, which is probably hopeless. (b) T, (0) is isomorphic to Q(5,q) iff 0 is an elliptic quadric of PG(3,q) (hence by the celebrated theorem of A, Barlotti [ 31 , for q odd T3 (0) is always isomorphic to Q(5,q)). We note that a complete classification of the GQ T (0), q even, depends on the classification of the ovoids of PG(%,q), which is still open. Some combinatorial properties of T, (0) and T, (0) By the previous paragraph we may assume that q is even. (a) The point (-) is regular for T, (0) and 3-regular for T, (0); all lines of type (b) are regular for T(0). (b) T2 ( 0 ) has always ovoids (with the notation as above, if n is a plane o f PG(3,q) with * O O = @ , then (n-H)UI(m)) is an ovoid of

T2 ( 0 ) 1

(c) By (a9 T (0) has subquadrangles o f order ( q , l ) and of order (l,q), and T, (03 has subquadrangles of order (q,l). Moreover it is clear that T, (0) has subquadrangles of type T, ( 0 ’ ) .

(b) The generalized quadrangles of M . Hall, Jr., of S.E. Payne, and of R.W. Ahrens and G. Szekeres Let 0 be a complete oval [ 1 1 I, i.e. a (q+Z)-arc [ 1 1 I, of the projective plane PG(2,q), q=2”, and let PG(2?q)=H be embedded as a plane in PG(3,q)sP. Define points of the incidence structure as the points of P-H, and lines as the lines of P which are not contained in H and meet 0 (necessarily in a unique point). The incidence is that of P. rt is evident that the incidence structure so defined is a GQ with parameters SEq-1, t=q+l, v=q3 , b=q2 (q+Z). This example, first discovered by R.W, Ahrens and G. Szekeres [ l ] and independently by M. Hall, Jr. [ 13 I , is denoted by Tf ( 0 9 ,

Combinatorics o f finite generalized quadrangles

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The following example is due to R.W. Ahrens and G. Szekeres [ l ] . Let the elements of P be the points of the affine 3-space AG(3,q) over GF(q), with q odd. Elements of B are the following curves of AG(3,q) : (i) x=o, y=a, z=b, (ii) x=a, y=u, z=b, (iii) x=co'-bu+a, y=-2cu+b, z=u, where the parameter u ranges over GF(q) and where a,b,c are arbitrary elements of GF(q). The incidence I is the natural one. The structure so defined is a GQ with parameters s=q-1, t=q+l, v=qJ ; b=q' (q+2), and will be denoted by AS(q) Finally, we give an important construction due to S.E.Payne [ 211. Let x be a regular point of the GQ S=(P,B,I) of order s, s > l . Then P' is defined to be the set P-xl. The elements of B' are of two types : the elements of type (a) are the lines of B which are not incident with x ; the elements of type (b) are the hyperbolic lines {x,y}11 , with yfx. Now we define the incidence 1 ' . If YEP' and E B ' is of type (a), then yI'L iff yIL; if YEP' and LEB' is of type (b), then yI'L iff yEL. The incidence structure so defined is a GQ of order (s-l,s+l), and will be denoted by P(S,x).

.

Historical back round Generalized :uadraneles S=[P.B,I] of order ( q - l , q + l ) , q an arbitrary prime power, were first 'discovered by R . W . Ahrens-and G. Szekeres [l 1. In their paper they also note that the incidence structure (P*,B* ,I*), with P * = B * = B and LI*M, LEP* , MEB', iff L-M and L#M, is a symmetric 2-(q2 (q+2), q(q+l) ,q) design. These designs were new. Finally, they remark that for q=3 there arises a GQ with 27 points and 45 lines, whose dual can also be obtained as follows : lines o f the GQ are the 27 lines on a general cubic surface V in PG(3,C) [ 2 I, points of the GQ are the 45 tritangent planes or' V, and incidence is the natural one. Isomirphisms T e generalized quadrangles P(S,x) resp. P(S,L) of S.E. Payne arise from the regular points resp. lines of the generalized quadrangles of order q. By 8. (a) we can restrict ourselves to the GQ T,(O) of order q. For q odd, we have T2(0)nQ(4,q), in which case all lines are regular and all points are antiregular. Hence for q odd there arises only one GQ of S.E. Payne. The nicest model of that GQ is obtained : points o f the GQ are the points of by considering P(W(q),x> PG(3,q)-PG(2,ql, with PG(2,q) the polar plane o f x with respect to the symplectic polarity 8 defining W(q); lines of the GQ are the totally isotropic lines of 0 which do not contain x , and also all lines of PG(3,q) which contain x and are not contained in PG(2,q). Now we assume that q is even, Here the structure of T2(0) depends entirely of the nature o f the oval 0. The only fact we know in the general situation, is that the exceptional point (m) and all lines incident with it are regular, Without doubt a complete classification of the GQ P(S,x) and P(S,L), for q even, is hopeless. We emphasize that the GQ P(S,x) and P(S,L) are very important, since it can be proved [ 20 I that T; (O)aP(Tz ( O ' ) , ( - ) ) , with O'=O-{x 1 and xEO, and AS(q)'P(W(q) ,y). ,

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(c) The generalized quadrangles of W.M. Kantor A few years ago W.M. Kantor 1161 constructed new generalized quadrangles of order (q,q2), q a prime power with q=2 (mod 3), which arise from the generalized hexagons H(q) of order q associated with the groups G2(q). A generalized hexagon 1 1 1 1 of order q (21) is an incidence structure S=(P,B,I), with a symmetric incidence relation satisfying the following axioms : (i) each point (resp. 1ine)is incident with q+1 lines (resp. points) ; ' 3 4 5 (ii) I P I = I BI=l+q+q-+q +q +q ; (iii) 6 is the smallest positive integer k such that S has a circuit consisting of k points and k lines. As usual the distance of two elements cr,REPUB is denoted by A ( a , B ) o r x ( B , a ) . Hence we use the metric defined on the union of the sets of points and lines. The generalized hexagon of order 1 is the ordinary hexagon. Up to duality only one generalized hexagon of order q, q a prime power, is known. This generalized hexagon arises from the group G, (q), and was introduced by J. Tits in his celebrated paper on triality 1531. One of the dual choices of this generalized hexagon has a nice representation in PG(6,q) : its points are the points of a non-singular quadric Q, its lines are lines of Q (but not all the lines of Q), incidence is that of PG(6,q). The generalized hexagon with that representation will be denoted by H(q'). We now give the construction of the GQ of W . ? 4 . Kantor. Let L be a fixed line of H(q), q=Z(mod 3). Points of the GQ are the points of L (in H(q)) and the lines at distance 4 from L. Lines of the GQ are L, the lines of H(q) at distance 6 from L, and the points of H(q) at distance 3 from L Now we define the incidence: a point of L (in H(q)) is defined to be incident with L and with all the lines of the GQ at distance 2 (in H(q)) from it; a line at distance 4 from L is defined to be incident with the lines of the GQ at distance 1 o r 2 (in H(q)) from it. The incidence structure K(q) so defined is a GQ with parameters s=q, t=q' , v=(l+q) (1+~3), b=(l+q2)(l+q3J. The p r o o f of that result is group-theoretical [ 161, 1291, and at present there is no geometrical proof of it. For qf2, the quadrangles K(q) are isomorphic to no one of the other known GQ of order (q,q2? 1161, [ 2 9 I. Moreover it can be proved 1291 that for qP2 the line L i s the unique regular line of K(q).

.

9. GENERALTZED QUADRANGLES WITH SMALL PARAMETERS In this section we shall consider the GQ of order (s,t), with s4t and 6 4 . (a? It is trivial that f o r any t>l there is a unique example (evidently up to isomorphism). (b) By 4 either t=2 or t=4. In either case there is a unique quadrangle, The t=2 case Was probably first considered by C.T. Benson 141 , The t=4 case was handled independently at least four times, by J.J. Seidel [331, E. Shult [ 341, J.A. Thas 1 3 9 1 , S . Dixmier and F. Zara 1 1 2 1 (5) s=3. Here t=3,5,6 o r 9. S.E. Payne [ 251 , and independently S. D i x m z and F. Zara 1 1 2 1 have proved that f o r t=3 there is an example which is unique up to duality. For t=5 the uniqueness was proved by S . Dixmier and F . Zara [ 1 2 1 The same authors proved the non-existence of a GQ of order (3,6) 1 1 2 1. Finally, P.J. Cameron 1301, and independently S. Dixmier and F. Zara 1 1 2 1 , proved the uniqueness

c. e. .

.

Combinatorics of finite generalized quadrangles

61

of the GQ of order .(3,9). (d) Here t=4,6,8,11,12 o r 16. S.E. Payne 1 2 7 I, proved the uniqueness of the GQ of order (4,4). Examples are known for t=6,8 and 16. Nothing is known about the cases t=ll and 12.

e.

10. COMBINATORIAL CHARACTERIZATIONS OF KNOWN GENERALIZED QUADRANGLES Now we survey the most inportant combinatorial characterizations of the known generalized quadrangles. Special attention will be paid to the classical examples. (a) Characterizations of the classical generalized.quadrangles First we shall give the theorems characterizing one classical GQ, then these characterizing a few classical GQ, etc. Finally, two theorems characterizing all classical GQ are given. ( i ) (C.T. Benson [ 41). If s=t (>1) then S is isomorphic to W(s) iff all its points are regular. (ii) (J.A. Thas [ 441). If S is a GQ for which I I ~ , y I ~ ~ l S s + l > 2 for all x+y, then S is isomorphic to W(s). (iii) (F. Mazzocca [17], S.E. Payne and J.A. Thas 1301). Let S be a GQ of order s having an antiregular point x. If there is a point y-x, y#x, for which the associated affine plane n(x,y) (see 5) is desarguesian, then S is isomorphic to Q(4,s). (iv) (J.A. Thas [38]). If S has order s , s>l, and if S has an ovoid each triad of which is centric, then S is isomorphic to W(s) with s=2" (v) (S.E. Payne and J.A. Thas [ 301). If S has order s, s>l, and if S has a regular pair (L1,L2) of non-concurrent lines with the property that any triad of points lying on lines of {Ll,L2)l is centric, then S is isomorphic to W(s) with s=2I1. (vi) (S.E. Payne [ 2.81). Let S be a GQ o f order s , s > l . Suppose S has distinct points x,y, and distinct lines L,M with x I L I y I M , such that x,y,L,M are each regular and have desarguesian planes based at them (see 5). Then S is isomorphic to W(s) with s=21' , (vii) (G. Tallini [35]). Let S=(P,B,I) be a GQ of order (s,t). If L=(P,I{x,y~~~llx#yl, natural incidence), then a non-trivial subspace of L is a point set TfP which contains at least three points not belonging to a same span, and for which Ix,ylcT implies I x ,yllLCT

.

.

Theorem. A GQ S=(P,B,I) with s>t, s>l and 0 1 , is isomorphic to an H(3,q') iff the following conditions are satisfied (a) each point of S is regular; 11 11 1 1 (b) {x,y} UIx',y') Cnon-trivial subspace of L*Ix,yl UIx',y') C cnon-trivial subspace of L. (viii) (J.A. Thas 1391, [ 41 ] and [ 451). Let S be a GQ of order (s,s2), s>l. If s is odd and if S has at least one 3-regular point, then S is isomorphic to Q(5,s). If s is even and if S has at least one 3-regular point and all lines are regular, o r resp. if all points of S are 3-regular, then S is isomorphic to Q(5,s). (ix) (J.A. Thas [ 44 ] and 145 1 ) . If S is a GQ with parameters s,t ( 0 1 ) for which every centric triad of lines is contained in a proper subquadrangle S ' o f order (s,t'), then SzQ(5,s) (for s odd it is sufficient that S contains a point x such that every centric triad of lines, with a center incident with x , is contained in a proper subquadrangle S ' of order (s,t')). There easily follows that

68

J.A. Thas

SzQ(5,s) iff for each triad (u,v,w) with distinct centers x,y, the five points x,y,u,v,w, are contained in a proper subquadrangle S' of order (s,t') (for s odd it is sufficient that this holds for a fixed point x). ( x ) ( J . A . Thas 1451 and 1 4 8 1 ) . First we introduce the following axiom (D) : If L1 ,L1 ,MI ,M2 ,U1 ,U? , N 1 are distinct lines for which L,+M2 , L, -L2 , M, -M2 , L, 4 ,-44, , L, -U, -M2 , U, 4, , L,-N, -M1 and for which L,~L,, M,~M,, L,N, , u , ~ M ,, L ~ u ? ,u?ntq2, u , ~ u , , L , ~ N ,, N , ~ P I , are distinct, then there exists a line N, with L2-N,-M2 and N , - N , . Theorem Let 2<s1, 0 1 , sft,and s odd, is isomorphic to Q(5,s) iff (A), is satisfied for all lines L incident with some coregular point x. The G Q S with s>1, 0 1 , sft and all lines regular, is isomorphic to Q(5,s) iff (A), is satisfied for all lines L incident with some point x . (xii) ( J . A . Thas 142 1 ) . If s3=t? (>1) then S is isomorphic to H(4,s) iff I {x,yllll?&+l for all x+y. (xiii) ( J . A . Thas and S . E . Payne 1 511 ) , First consider again the structure L defined in (vii). Then L is a linear space in the sense of F. Buekenhout. A plane of L is a linear variety generated by three points not on a line of L , The following result is given with this notation in mind. Theorem 3 ' 1 Let S have order (s,t) with l < s Gt-. If each trace {x,y} , x+y, is a plane of L, then SzH(4,s). (xiv) ( J . A . Thas [ 44 I). Let S be a GQ of order (s,t) for which IIx,yl1I I>s2/t+l for all x+y. Then one of the following occurs : (a) s = l , (b) t=s2 , (c) S=W(s), (d) SzH(4,s). (xv) ( J . A . Thas [ 441, J . A . Thas and S . E . Payne 151 I). Let S be a GQ of order (s,t) in which each point has property (H). Then one of the following occurs : (a) each point is regular, (b) IIx,yllll = 2 for all x+y, (c) SzH(4,s). If S is a G Q of order (s,t) in which each point is semiregular , then we have one of the following : (a) s>t and each point is regular, (b) s=t and S=W(s) or each point of S is antiregular, (c) s, x+y, the number of centers of the triad (x,y,z), where z$cl(x,y), is a constant. Then one of the following occurs : (a) all points are regular, (b) s 2 = t , (c) SzH(4,s). (xvii) ( J . A . Thas 1441). Let S be a GQ of order (s,t), s > l , in

Combinatorics of finite generalized quadrangles

69

which the following condition holds : whenever (x,y,z) is a triad of points with x$Zcl(y,z), then Ixju{y,zll is contained in a proper subquadrangle of order (s,t‘). Then S’.W(s), Q(5,s) o r H(4,s). (xviii) (J.A. Thas 1441). Let S be a GQ of order (s,t) for which not every point is regular. Suppose that each set {xlu{y,z}l, where (x,y,z) is a centric triad with x$cl(y,z), is contained in a proper subquadrangle of order (s,t’). Then S’.Q(5,s), H(4,s), o r to Q(4,s) with s odd. (xix) (J.A. Thas 1481). The GQ S with s>l and_t>l, is a classical or a dual classical GQ iff it satisfies (A) o r (A) (notations are as in (xi)). (xx) (F. Buekenhout and C. Lefsvre [ 81). If a point set of PG(d,q), together with a line set of PG(d,q) form a GQ S of order (q,t), then S is a classical GQ. (b) Characterizations of the generalized quadrangles T, (0) (xxi) (J.A. Thas L451). If the GQ S o € order ( s , s 2 ) , 0 1 , possesses a 3-regular point x, then it is isomorphic to a T, (0). (xxii) (J.A. Thas [ 451). If S is a GQ of order (s,t), 0 1 , which contains a point x such that every centric triad of lines, with a center incident with x, is contained in a proper subquadrangle S ’ of order (s,t’), then S is isomorphic to a T3(0). There easily follows that S is isomorphic to a Tj(0) iff it contains a point x such that for each triad (u,v,w) with distinct centers x,y, the five points x,y,u,v,w, are contained in a proper subquadrangle S ’ of order (s,t’) (xxiii) (J.A. Thas 1451 and [ 481 ) . If K s < t and S contains a coregular point x such that (D) is satisfied whenever L, (resp. L2) contains x , then t=s> and S is isomorphic to a T,(O) (notations are as in (x)), (xxiv) (J.A. Thas [ 481 ) . Let S be a G(I with 9 1 , t>l and sft. Then S is isomorphic to a T, (0) iff (A), is satisfied f o r all lines L incident with some coregular point x (notations are as in (xi)).

-

1 1 . GENERALIZED QUADRANGLES IN FINITE AFFINE SPACES

By theorem lO.(xx) of F. Buekenhout and C. LefSvre (parts of which were independently proved by D. Olanda I1812 we know which GQ are ”embedded” in PG(n,q). It was that beautiful result which gave me the idea to attack the following problem : find all GQ whose lines are lines of an affine space AG(n,q), whose points are all the points of AG(n,q) on these lines, and whoseincidenceis the natural one. These GQ are said to be embedded in AG(n,q). The following theorems which are in J.A. Thas [ 4 6 1 and 1471 give the complete answer to the question. We note that the theorem on the embedding in AG(3,qg was independently proved by A. Bichara 151. Embedding in AG(Z,s+l). If the GQ S of order (s,t) is embedded in AG(Z,s+l), then S is a net of order s + l and degree 2. Embedding in AG(3,s+l). Suppose that the GQ S=(P,B,I) of order (s,t) is embedded in AG(3,s+l), and that P is not contained in a plane of AG(3,s+l). Then the following cases can occur (a) s = 1 , t = 2 (trivial case); ( b ) t=l and the elements of S are the affine points and affine lines of an hyperbolic quadric of PG(3,s+l), the projective completion

70

J.A. Thas

of AG(3,s+l), which i s tangent t o the plane a t i n f i n i t y of AG(3,s+l); (c) s=2, t = 2 (an embedding of the unique GQ with 15 points and 15 lines i n AG(3,3)); (d) S = q (0), i . e . P i s the point s e t of AG(3,s+l), and B i s the s e t of a l l lines of AG(3,s+l) whose points a t i n f i n i t y are the points of a complete oval 0 of the plane a t i n f i n i t y of AG(3,s+l); (e) S=P(W(s+l), x ) , i . e . P i s the point s e t of AG(3,s+l) and B=B,UB,, where B, i s the s e t of a l l a f f i n e t o t a l l y isotropic l i n e s with respect t o a symplectic polarity 0 of the projective completion PG(3,s+l) of AG(3,s+l) and where B, i s the class of parallel l i n e s defined by the pole x (the image with respect t o 6') of the plane a t i n f i n i t y of AG(3,s+l). Embedding i n AG(4,s+l). Suppose that the GQ S=(P,B,I) of order ( s , t ) is embedded i n AG(4,s+l), and that P is not contained i n a AG(3,s+l). Then the following cases can occur. (a) s = l , tE{2,3,4,5,6,71 ( t r i v i a l case); (b) s=t=2 (an embedding of the unique GQ with 1 5 points and 15 lines in AG(4,3)); (c) s=t=3 and w ( 4 , 3 ) ; (d) s=2, t = 4 (an embedding of the unique GQ with 27 points and 45 l i n e s i n AG(493)) Embedding i n AG(n,s+l), 1+4. Suppose that the GQ S=(P,B,I) of order ( s , t ) is embedded i n AG(n,s+l), +5, and t h a t P i s not contained i n an AG(n-l,s+l). Then the following cases can occur. (a) s=l and *I[ 11/21 , ,2"-,-11 ( t r i v i a l case) ; (b) s=2, t = 4 , n=5 (an embedding of the unique GQ with 27 points and 45 l i n e s i n AG(5,3)). Description of the f i v e sporadic cases 1 . s=t=2, n=3. Let w be a plane of AG(3,3), and l e t {Lo ,L, ,L2 } and I", ,?4y ,M, 1 be

. ..

two classes of p a r a l l e l l i n e s of u. Suppose that {q 1=h&nLi, Iyi}=MpJ,i and {zi 1=%"4 Further, let Y ,$ ,N, be three lines containing respectively xo ,yo ,zo

.

,

for which Nx$IMx ,Lo 1 , Ny${My ,Lo 1 , N,${M, ,Lo 1 , for which the planes NxM,, NyP4y , NzMz are p a r a l l e l , and for which ",LONx, LoNy,LoNz are d i s t i n c t . The points of N, are xo,x3 ,x4, the points of N are yo,y3 ,y4, and the points o f N, are z o , z 3 , z 4 , Y where notations are chosen i n such a way t h a t x3 ,y3 , z 3 , resp. x4 ,y4 , z 4 , are c o l l i near. The points of the generalized quadrangle a r e xo , ,x4, y o , . .,y4, zo ,. . . ,z4

...

and the l i n e s are Lo ,L, ,L, ,Mx ,My

.

,", ,N, ,NY ,NZ ,x3y4 ,x4y32x3~49x473 ,y3z4,y4z3*

2 . s=t=2, n=4. Let PG(3,3) be the hyperplane a t i n f i n i t y of AG(4,3), l e t ,w be a plane of PG(3,3), and l e t 1 be a point of PG(3,3)-(am. In u, we choose points

y,,,"02 ,mil ,y2,%,

,mz2

rnol

, i n such a way t h a t J$,,mil a r e collinear, t h a t m,, collinear, that m3, ,mI2 a r e collinear, and t h a t ,m2, ,ml3 are collinear. Let L be an affine l i n e containing 1, and l e t the a f f i n e points of L

rno2 ,m22 a r e

be denoted by po,p,,p3

,mo2

. The points

mol

of the generalized quadrangle are the a f f i n e

points of the lines porno, ,Porno2,Plmll , P l y , ,p,m,, ,P2%,' The l i n e s of the quadrang l e are the affine. lines of the hyperbolic quadric containing p o ~ 1 , p l ~ 1 , p 2 ~ 1 , the affine lines of the hyperbolic quadric containing porno, ,p,m,, ,p2$, , the affine l i n e s of the hyperbolic quadric containing p o ~ z , p , m l , ,p r y l , and the affine l i n e s of the hyperbolic quadric containing pornol ,p,m,, , p , y 2 .

Combinatorics of finite generalized quadrangles

71

3 . s = t = 3 , n = 4 . L e t PG(3,4) b e t h e h y p e r p l a n e a t i n f i n i t y o f A G ( 4 , 4 ) , l e t om be a p l a n e o f P G ( 3 , 4 ) , l e t H b e a h e r m i t i a n c u r v e (or a u n i t a l ) o f d m , and l e t 1 be a p o i n t o f PG(3,4)-(9,. I n w, t h e r e a r e e x a c t l y 4 t r i a n g l e s m o 1 m o z m o 3 , m 1 1 m 1 2 m 1 3 , m 2 1 m 2 2 m 2 3 , m 3 1 m 3 1 m 3 3 whose v e r t i c e s a r e e x t e r i o r p o i n t s o f H and whose s i d e s a r e s e c a n t s o f H . Any l i n e m o a m l b , a , b E I l , 2 , 3 ) , c o n t a i n s e x a c t l y one v e r t e x mzC of m 2 1 m 2 2 m z 3 , and one v e r t e x m3* o f m 3 1 m 3 2 m 3 3 (moamlb i s a t a n g e n t l i n e o f H ) . We remark t h a t t h e c r o s s - r a t i o ( m O a m l b m z c m 3 di)s i n d e p e n d e n t o f t h e c h o i c e o f a , b € I 1 , 2 , 3 1 . L e t L be a n a f f i n e l i n e t h r o u g h 1, and l e t p o , p l , p 2 , p , b e t h e a f f i n e p o i n t s o f L , where n o t a t i o n s a r e c h o s e n . points of the i n s u c h a way t h a t ( p o p l p 2 p 3 ) = ( m o a m , b m 2 c m 3 , )The g e n e r a l i z e d q u a d r a n g l e a r e t h e 4 0 a f f i n e p o i n t s o f t h e l i n e s pornO1, Po'"o2 ,Pom03 "I"'"l1 ~ P l m l z,Plrn13 , p z m ? l ~ P ~ ' "9pzm23 2 ~ ,P3'"31 ,P3m32 'P3m33 * The l i n e s o f t h e q u a d r a n g l e a r e t h e a f f i n e l i n e s o f t h e h y p e r b o l i c q u a d r i c c o n t a i n i n g pornoa, p l m l b , p r m z c , p 3 m 3 d , a , b = l , 2 , 3 . 4 . s = 2 , t = 4 , n = 4 . L e t PG(3,3) be t h e h y p e r p l a n e a t i n f i n i t y o f A G ( 4 , 3 ) , l e t 9,' b e a p l a n e o f PG(3,3) and l e t 1 b e a p o i n t o f PG(3,3)-w,. I n -9, we c h o o s e p o i n t s m,nx , n y , n Z ,n; ,nJ , n i ,n;,n;,n;' , i n s u c h a way t h a t m,nx , n y ,nz a r e c o l l i n e a r , t h a t m,ni , n i ,nz' a r e c o l l i n e a r , t h a t m,n;,n;,n;' a r e c o l l i n e a r and t h a t na , n i ,nbl, w i t h { a , b , c } = { x , y , z } , a r e c o l l i n e a r . Let L be a n a f f i n e l i n e t h r o u g h 1, and l e t x , y , z be t h e a f f i n e p o i n t s o f L . The p l a n e d e f i n e d by L and m i s d e n o t e d by w . The p o i n t s o f t h e g e n e r a l i z e d q u a d r a n g l e a r e t h e 2 7 a f f i n e p o i n t s o f t h e l i n e s xm, x n x , xn; ,xn;,ym,yny, y n J , y n " , z m , z ~ , z ~ , zThe ~ . 45 l i n e s o f t h e q u a d r a n g l e a r e t h e a f f i n e l i n e s Y o f w w i t h p o i n t s a t i n f i n i t y 1 and m , t h e a f f i n e l i n e s o f t h e h y p e r b o l i c q u a d r i c c o n t a i n i n g am, b n b , cnc ( r e s p . am, b n d , cn; 9 r e s p . am, b n t , cnd'), w i t h { a , b , c } = { x , y , z ) , and t h e a f f i n e l i n e s o f t h e h y p e r b o l i c q u a d r i c c o n t a i n i n g a n a , bnd , cn;' , w i t h { a , b , c } =

={x,y,z}

.

t = 4 , n = 5 . L e t PG(4,3) be t h e h y p e r p l a n e a t i n f i n i t y o f A G ( 5 , 3 ) , l e t H, b e a h y p e r p l a n e o f PG(4,3) and l e t 1 b e a p o i n t o f PG(4,3)-Hm. I n H, we c h o o s e p o i n t s mx ,my ,mZ ,nx , n y ,nz , n i , n i , n i , i n s u c h a way t h a t m x , m y , m z a r e c o l l i n e a r , t h a t m x , m y , m Z , n:,n;,nil nx , n y ,nz a r e i n a p l a n e u-, t h a t mx ,my ,mZ , n i , n i , n i a r e i n a p l a n e a r e i n a p l a n e c.C, t h a t ma , n b , n c UL, t h a t mx ,my ,mZ ,n;,n",n:' Y ( r e s p . ma ,nd ,nc , r e s p . ma , n r , n g ' ) , w i t h { a , b , c l = { x , y , z } , a r e c o l l i n e a r , and t h a t n a , n d , n d ' , w i t h { a , b , c } = { x , y , z } , a r e c o l l i n e a r . L e t L be a n a f f i n e l i n e t h r o u g h 1 , and l e t x , y , z b e t h e a f f i n e p o i n t s o f L . The p o i n t s o f t h e g e n e r a l i z e d q u a d r a n g l e a r e t h e 2 7 a f f i n e p o i n t s o f t h e l i n e s xmx ,ym, ,zmz ,xnx ,yny ,znz ,xn; , y n i ,z n i ,xn; ,yn;, zn;'. The 4 5 l i n e s o f t h e q u a d r a n g l e a r e t h e a f f i n e l i n e s o f t h e h y p e r b o l i c 5.

s=2,

72

J.A. Thas

quadric containing xmx , ymy , zmz, the affine lines of the hyperbolic quadric containing ama, bn, , cnc (resp. ama , bnd, cni , resp. ama, bn;, cn;'), with la,b,cI={x,y,zl , and the affine lines of the hyperbolic quadric containing ana, bnd , cni' , with {a,b,cl=Ix,y,zl. 12. MOUFANG CONDITIONS FOR GENERALIZED QUADRANGLES I do not want to close this survey without having mentioned Tits' famous theorem on Moufang GQ [541, [ 561. In fact, the theorem gives a characterization of GQ by properties of their automorphism group, but in my opinion the result is so fundamental and so beautiful that even a pure combinatorial survey has to contain it. First some definitions. Let S=(P,B,I) be a GQ of order (s,t). For a fixed point p define the following condition (M)p: F o r any two lines A,B of S incident with p, the group of collineations of S fixing A and B pointwise and p linewise is transitive on the lines (#A) incident with a given point x on A (x#p). Then S is said to satisfy condition (M) provided it satisfies (M) for all pointspeP. F o r a fixed line LEB, let P (fi)L be the condition that is the dual of ( M ) p , and let (8) be the dual of (M). If S satisfies both (M) and (fi), it is called a Moufang GQ Tits' theorem, A GQ S, with s>l and 0 1 , is a Moufang GQ iff it is a classical or a dual classical GQ. The proof uses deep results on algebras, and there is some interest in finding ways both to weaken the hypotheses and to avoid the heavy results on algebras. These two goals provided the motivation for the recent papers 1311 and[52] by S.E. Payne and J.A. Thas. They almost succeeded in proving Tits' theorem in an"e1ementary" combinatorial way, since the following was obtained : If S is a GQ of order (s,t) , l<s(t, satisfying both (M) and (k) , then one of the following holds : (i) SsH(4,s) and (s,t)=(qz,q3), (ii) either S o r its dual is isomorphic to W ( s ) and (s,t)=(q,q), (iii) t=s? with s a prime power. So there only remains to prove that in case (iii) we have SzQ(5,s). Moreover in case (iii) the authors proved a lot of strong properties about S, and they reduced the problem to the characterization of some easily described object in PG(n,q). As to the "local Moufang theorems", I propose the following sample results out of 1311. If s is prime and S of order (s,s2) satisfies (M)P f o r some point p, then SsQ(5,s). If S satisfies ($)L for every line L through some coregular point p, and if s>l and t>l, then s=qI1 , t=q11 , with q a prime power, and h'=h" o r h"a=h'(a+l) with a odd. If in particular s is prime, then S=Q(4,s) o r SEQ(5,s); if in particular every line is regular, then t=s' or SzQ(4,s).

.

It

Combinatorics of finite generalized quadrangles

73

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Baker H.F., Principles of geometry, Vol. I11 (Cambridge Univ. Press, 1 9 3 4 ) .

131

Barlotti A., Some topics in finite geometrical structures (Institute of Statistics Mimeo Series No. 4 3 9 , University of North Carolina at Chapel Hill, 1 9 6 5 ) .

1 4 1 Benson C.T., On the structure of generalized quadrangles, J. Algebra 1 5 ( 1 9 7 0 ) 4 4 3 - 4 5 4 . [ 51

Bichara A., Caratterizzazione dei sistemi rigati immersi in A 3 , q , Riv. Mat. Univ. Parma ( 4 ) 4 ( 1 9 7 8 ) 2 7 7 - 2 9 0 .

161

Bose R.C., Graphs and designs, in : Barlotti A. (ed.), Finite geometric structures and their applications (Cremonese, Roma, 1973).

1 7 1 Bruen A.A. and Thas J.A., Partial spreads, packings and hermitian manifolds in PG(3,q), Math. Z. 1 5 1 ( 1 9 7 6 ) 2 0 7 - 2 1 4 . 181

Buekenhout F . and LefSvre C., Generalized quadrangles in projective spaces, Arch. Math. 2 5 ( 1 9 7 4 ) 5 4 0 - 5 5 2 .

191

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1 Cameron P.J., Private communication, 1 9 7 4 .

111 1

Dembowski P., Finite geometries (Springer-Verlag, Berlin Heidelberg New York, 1 9 6 8 ) .

112 I

Dixmier S. and Zara F., Etude d'un quadrangle ggn6ralisg autour de deux de ses points non lies, 1 9 7 6 , preprint.

113 1

Hall M. J r . , Affine generalized quadrilaterals, in : Mirsky L. (ed.), Studies in Mathematics (Academic Press, New York, 1 9 7 1 ) .

ti4 1

Higman D.G., Partial geometries, generalized quadrangles and strongly regular graphs, in : Barlotti A. (ed.), Atti Convegno di Geometria e sue Applicazioni (University Perugia, 1 9 7 1 ) .

1

Hirschfeld J.W.P., Projective geometries over finite fields (Oxford Univ. Press, Oxford, 1 9 7 9 ) .

1 Kantor W . M . , Generalized quadrangles associated with G,(q), J. Comb. Th. (A) 2 9 ( 1 9 8 0 ) 2 1 2 - 2 1 9 . 117 1

Mazzocca F . , Sistemi grafici rigati di seconda specie,Relazione N . 2 8 , Istituto di Mat. dell'Univ. di Napoli, 2 2 pp.

118 1

Olanda D . , Sistemi rigati immersi in uno spazio proiettivo, 1st. Mat. Univ. Napoli, Rel. N. 2 6 ( 1 9 7 3 ) 1 - 2 1 .

14

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1191 Olanda D., Sistemi rigati immersi in uno spazio proiettivo, Rend. Accad. Naz. Lincei 62 (1977) 489-499. [

201

Payne S.E., The equivalence o f certain generalized quadrangles, J. Comb. Th. 10 (1971) 284-289.

[ 2 1 ] Payne S.E., Quadrangles of order (s-l,s+l), J. Algebra 22 (1972) 97-119. 1221

Payne S.E., A restriction on the parameters of a subquadrangle, Bull. Amer. Math. SOC. 79 (1973) 747-748.

1231 Payne S.E., Generalized quadrangles of even order, J. Algebra 31 (1974) 367-391. 241

Payne S.E., Skew translation generalized quadrangles, Congressus Numerantium 14 (1975) 485-504.

1251

Payne S.E., All generalized quadrangles of order 3 are known, J. Comb. Th. 18 (1975) 203-206.

[

[

261

Payne S.E., An inequality f o r generalized quadrangles, Proc. Amer. Math. SOC. 71 (1978) 147-152.

[ 2 7 ] Payne S.E., Generalized quadranglesof order 4, I and 11, J. Comb. Theory 22 (1977)267-279 and 280-288. [

281

Payne S . E . , Generalized quadrangles with symmetry PI, Simon Stevin 50 (1976-77) 209-245.

1291 Payne S.E., Generalized quadrangles as group coset geometries, Congressus Numerantium 29 (1980) 717-734. 1301 Payne S.E. and Thas J.A., Generalized quadrangles with symmetry, Simon Stevin 49 (1975-76) 3-32 and 81-103. 1311 Payne S.E. and Thas J.A., Moufang conditions for finite generalized quadrangles, in :Cameron P.J., Hirschfeld J.W.P. and Hughes D.R. (eds.),Finite geometries and designs (Cambridge Univ. Press, 1980). 1321 Segre B., Lectures on modern geometry (Cremonese, Roma, 1961). [

33 I

Seidel J.J., Strongly regular graphs with (-1,l ,0) adjacency matrix having eigenvalue 3, Lin. Algebra and Appl. 1 (1968) 28 1-298.

1341 Shult E., Characterizations of certain classes of graphs, J. Comb. Th. (B) 13 (19723 142-167. 1351 Tallini G., Strutture d'incidenza dodati di polarit2, Rendiconti del Seminario Matematico e Fisico di Milano 41 (1971) 1-41, 136 ] Thas J.A., 4-gonal configurations,in : Barlotti A. (ed), Finite geometric structures and their applications (Cremonese, Roma, 19731. 1371 Thas J . A . , 4-gonal subconfigurations of a given 4-gonal configuration, Rend. Accad. Naz. Lincei 53 (1972) 520-530.

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[381 Thas J.A., On 4-gonal configurations, Geometriae Dedicata 2 (1973) 317-326. 139 1 Thas J.A., On 4-gonal configurations with parameters r=q'+l and k=q+l, part I , Geometriae Dedicata 3 (1974) 365-375. 1401 Thas J.A., A remark concerning the restriction on the parameters o f a 4-gonal subconfiguration, Simon Stevin 48 (1974-75) 65-68. L41 1 Thas J.A., 4-gonal configurations with parameters r=q2 + 1 and k=q+l, part 11, Geometriae Dedicata 4 (1975) 51-59. 2

[42 I Thas J.A., On generalized quadrangles with parameterss=q t=q3, Geometriae Dedicata 5 (1976) 485-496.

and

1431 Thas J.A., Combinatorics of partial geometries and generalized quadrangles, in : Aigner M . (ed.), Higher Combinatorics (Reidel, Dordrecht-Holland, 1977). 1441 Thas J . A . , Combinatorial characterizations of the classical generalized quadrangles, Geometriae Dedicata 6 (1977) 339-351, [451 Thas J.A., Combinatorial characterizations o f generalized quadrangles with parameters s=q and t=ql, Geometriae Dedicata 7 (1978) 223-232. [46 I Thas J.A., Partial geometries in finite affine spaces, Math. Z. 158 (1978) 1-13 [471 Thas J.A., Geometries in finite projective and affine spaces, in : Bollobas B. (ed.), Surveys in Combinatorics (Cambridge Univ. Press, 1979). L48 I

Thas J.A., New combinatorial characterizations of generalized quadrangles, Europ. J. of Comb., to appear.

L491 Thas J.A., Ovoids and spreads of finite classical polar spaces, Geometriae Dedicata 10 (1981) 135-144. 150

I Thas J.A. and De Winne P., Generalized quadrangles in finite projective spaces, J. of Geometry 1 0 (1977) 126-137.

151 I

Thas J.A. and Payne S.E., Classical finite generalized quadrangles : a combinatorial study, Ars Combinatoria 2 (1976) 57-110.

1521

Thas J.A. and Payne S.E., Generalized quadrangles and the Higman-Sims technique, Europ. J. Comb. 2 (1981) 79-89.

153 1 Tits J., Sur la trialit6 et certains groupes qui s'en dgduisent, Publ. Math. T . H . E . S . , Paris 2 (1959) 16-60. 154 I

Tits J., Classification o f buildings o f spherical type and Moufang polygons : a survey, in : Segre B. (ed.), Atti Coll. Comb. Roma (Cremonese, Roma, 1977).

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1551

T i t s J . , B u i l d i n g s and B N - p a i r s o f s p h e r i c a l t y p e ( S p r i n g e r Verlag, Berlin,1974).

1561

T i t s J . , Q u a d r a n g l e s d e Moufang, t o a p p e a r ,

Annals of Discrete Mathematics 14 (1982) 77-82

0 North-Holland Publishing Company

- GRAZIA RFGUSO (-1 I s t i t u t o d i Geometria - Facolta d i Scienze Universita d i B a r i - Via N i c o l a i 2 - Bari L. M?RCA AEATANGEZO

- 1 T A L I A -

Given an n(2n+l)-set B of class b , i r n r 2 n I , nd3, of ~ ~ ( 2 ,,q=2', q) we show t h a t then e x i s t s a k-arc K, w i t h k=2n+l, in the dual plane of ~ G ( 2 , 2 ~ such ) that B consists of all the secants of K, and prove a necessary and sufficient condition f o r B i n order that K gives rise a closed arc (no. 3).

1. L e t M be an m-set (i.e. a set M of m points) of PG(2,q). We define ti (o
...

...,

,..

.

,...

..

Infiis paper we shall characterize the n(2n+l)-set (nb3) of c l a s s @,l,n,2n] of degree 2n, s t a r t i n g f r a n the following example:

...

,..

and

L e t K be a k-arc, w i t h k=q'+l , contained i n a proper subplane PG(2,q') of PG(2rq) , where q=2r. Then the number of secants of K , through a pint P (ePG(2,q)), is k-I o r it belongs to the set {0,1 ,(k-1)/23 according as PEK o r Pfl. Putting n=(k-1)/2 and b=n(2n+l), the set B of the secants of K i s a b-set of type (0,l ,n,2n) i n the dual plane of PG (2,q)

.

&t K be a (q+l)-arc (i.e. an oval) of EG(2,q). Putting n=(k-l)/2 and b--71(2n+l), the set B of the secants of K i s a b-set of type (O,n,2n) in the dual plane of FG(2,q). Our principal r e s u l t s

i n t h i s direction are contained i n the following theorem:

m m 1. I n ~ ( 2 , q ) q=2', , L e t B be an n(2n+l)-set 06 &h ~ , l , n , 2 ~ Then, . the se,t ofi (2n)-secants ofi B con6.titlLted a k-arc K i n f i t h e d u d pLane E ( 2 , q ) , wLth k=2n+l, buch that f i e nwnbeh 06 o e c m b 06 K thkough a p o i n t P ( P ) A 0, 1 o k n. Conuemelg, ouch a k-arc ofi PG(2,q) g i v u h,ihe i n t h e d u d pLane t o an n(2n+l)-set. 2. 1n p ~ ( 2 , q,) w.ith q=2rf the k-arcK, dedined by B, A a closed ahc, ptouided that no I-secants amhe fitom a point P(plB) 06 t h e i n t m e c t i o n 0 6 a (2nI-secant uLth an n-secant.

mm

(-)

Research p a r t i a l l y supported by G.N.S.A.G.A. 77

(CNR)

.

L.M. Abatangelo and G.Raguso

I8

Finally (no. 4) we discvss the cases n53. 2. L e t us make

S ~ I E propositions

t o p r w e the Theorem 1.

P r o p s i t i o n 2.1. F a t any point PEB, t h a t e x O t exacXLy ALvo (2n)-secants and 2n-1 n-secants which pano t h o u g h P. P r m f . For any p i n t PEB, l e t y and z be respectively the numbers of (2n)-secants and n-secants passing through P. S h p l e counting then yields the following diophantine equation: (2.1)

y(n-l)+ z(2n-1) = n(2n+l)-1 = (2n-1) (n+l).

From (2.1) we obtain y = 2n+3

-

+ (2-z)/(n-I).

22

As y must be an integer, it follaws that

- X(n-I)

z = 2

where X denotes an integer. Since y20, 220, we have then -1<X< 2/(n-I). Hmever n>3, so we have X-1 or X=O. Hence, the solutions of (2.1) are: (2.2)

y = o

z = n+l

for A=-I

(2.3)

y = 2n-1

2 = 2

f o r A= 0

This yields:

(2.4)

For any p i n t PEB, there are a t least two (2n)-secants passing through P.

Next we p r w e that case (2.2) cannot actually occur. In order to do this, we observe that, by (2.4) the (2n)-secants are not a l l concurrent. Suppose now that P ver i f i e s (2.2) and that s is a (2n)-secant, which does not pass through P. Then each of the 2n l i n e s PS, with SEsnB, is s t i l l a (2n)-secant. Therefore z)2n, that is n+l a2n. ?his i s impossible when 1-03.

From now on, l e t Of

Tn (resp. Tzn) be the set of a l l n-secants (resp. (2n)-secants) B. 'Ihen I Tnl=tn ,I Tznl"t2n The following Prop. 2.2 is an m i a t e conseguen-

.

ce of Prop. 2.1.

Praposition 2.2

tn = 4n2-1,

P r o p s i t i o n 2.3 The p o i n t

06

t 2 n = 2n+l.

intmectian

06

any ALvo (2n)-secants

fie4

on B.

Proof. It i s easy to show that, given any (2n)-secant s, every other (2n)-secant mets s in a p i n t of B. (Prop. 2.1 and Prop. 2 . 2 ) . Corollary 1. Thtee (2x1)-secants me n e w a concument.

79

On the n(2n+l)-setof class [0,1. n, 2n] W e are naw able to prove Theorem 1.

Putting k=2n+l, the set of all (2n)-secants of B is a k-set K (Prop. 2.2) in the dual plane E(2,q) of FG(2,q). Since, by hypothesis, B is an n(2n+l)-set of class @,I,n,24, K is a k-arc (Cor. 1) such that the number of secants of K, passing through a generic pint P(,&), belongs to the set {O,l, (k-l)/2}. Conversely, given a k-arc K, if the number of secants of K, passing through a generic point P($K) , belongs to the set {O,l, (k-1)/21 in the dual plane the secants of K form a b-set B (b=k(k-l)/2). Proposition 2.4. 7 6 +2n, any n(2n+l)-set B, n>3, 0 4 (O,l,n,2n).7 d q2n, B h 06 .type (O,n,2n).

d M 6

a d type

ptl,n,24

Proof. Let x be the number of I-secants issuing f r m any point P of B: (2.5)

x+y+z=q+l

where y and z are still the n h r s of n-secants and (2n)-secants respectively through P. From Prop. 2.1 : y = 2n-1

z = 2

and

,

fran (2.5): x = q-2n ,

Therefore x>O if and only if q>2n. Then: ti = (q-2n)(2n+l)n (2.6)

to = q2

+q+ 1

- tl

-

- tzn = q2 - q(2n2tn-I

+

2n (2n2-n-I) +I.

Since q is even, to is odd, and hence tofo. If 9;2n, for any pint P of B : x=O, therefore: t l = O and %=I. Proposition 2.5.

Fah

any n(2n+l)-set, n>3, 0 6 .type (O,l,n,2n)Lt 0 q22n2-n-1.

Proof. By (2.6):

-

to-1 = (q-2n)(q-2n’+n+1) tO>l

(q-2n)(q-2n2+n+1)> 0.

Since q>2n, this inplies q~n2-n-I. 3. we recall the definition of a closed arc fran ,]2[

4 [

1.

T be a projective plane w e r a field (2. L e t K be a k-arc and I a subset of satisfying the following properties:

Let

T,

L.M. Abatangelo and G.Raguso

80 (ii) (iii)

IS

for any s c a n t s of K: nII = k-2 for any tangent t of K w i t h # @ : I t n I l = k-I

tnI

It can be Shawn that f o r each point P of K there e x i s t s a unique tangent a t P (called the priviliged tangent a t P) , which mets I. Put R ={r n ( K U I ) :Ir n ( K U 1 ) l = k}, is called w i t h any r secant or tangent line of K. The gearetrical space ( K U I , R the a m b i e n t of K. According t o [47, we say t h a t the ambient of K is d o o e d i f :

(iv)

any tsm lines of R are concurrent a t a point of

KuI.

Note that the ambient of K is unique when k25. The following propositions prove our Theorem 2.

Proposition 3.1. On any n-secant, t h e h e d i n g e e (2n)-secant p a n a. .

A a ningle p a i n t

x(@)

, thlraugh

which a

Proof. This follows fran ITznl= 2n+l, and f m P r o p . 2.1. let L be the set of these points X. In the hypothesis of Theorem 2 , we have:

Proposition 3.2.

Proof. (a) let x be the n h r of n-secants through any point X. Since there e x i s t s a unique (2n)-secant through X:

Then, by Prop. (2.1 ) , there i s a unique point X on each (2n)-secant. Therefore :

ILI

= th = 2n+l

.

(b) We naw prove t h a t L is m t a i n e d in an external line. Given any three points XllX2,X3 of L, call S i the unique (2n)-secant through X i (i=1,2,3). Put: AI = s 2 n s 3

,

~2 = s l n s 3

I

A3 = s l n s 2

.

We may take AlA2A3 as the fundamental triangle of a hanogeneus coordinate system

i n FG(2,q). The equation of a l i n e r j , p a s s i n g through the point Aj, but d i s t i n c t , is of the form: fran the lines A,& ,

9%

(where j,k,m is a cyclic permutation of the integer 1,2,3). The e l m t Xj i s called the coordinate of r,. Given a (2n)-secant s, d i s t i n c t f m s1, s2, s3 let H . be the set of n-secants 7

81

On the n(2n+l)-set of class [0, 1. n, 2n] through A,

, i.e. H. 3 -- { r j = A ~ P( P E ( B U L ) n d ,

l e t A, denote the product of the coordinates of the lines r j of H j :

A - T

j -r.EH 1 1

Set

Xj

B1 = A2A3fls,

B2 = A3Aln s,

If b i denotes the coordinate of A i B i it is

AlA2A3 = blb2b3

(3.31

, by

B3 = AlA2fls

.

the extension of Ceva-Menelao

theoren,

7T

p E ~'1'2'3

Fran (3.21, (3.3) it follows Alh2h3 = -1

(3.4)

Next, l e t X be a generic point of L. "hen, f o r each i, w have H i = IAiXIXEL-{Xj,Xh}, I<J
L e t ai be the coordinate of the line A i X i

.

. By the previous discussion, we have

where

mus, by (3.4) ~ 1 ~ =2-1~ 3

Hence, X1,X2,X3 are three collinear points. This proves that L is contained in a line ro By the definition of L, ro must be an external line to B.

.

W e are now able to prove Theorem 2.

L.M. Abatangelo and G.Ragus0

82

~n the dual plane, I denotes the set consisting of a l l Then, by Prop. (2.21, 1 1 1 = 4n2 = (k-1)2, and by a s-le

to prove that (K,I) is a closed arc.

n-secants to B, p l u s ro . discussion it is possible

4. we conclude with the special cases of n = 1,2,3.

In t h e dual plane we define K as i n section 1 , i.e. as the set o f a l l (2n)-secants of B. I f n = 1 , then both B and K are 3-arcs. I f n = 2, thm K is either a 5-arc (not necessarily closed) or a set consisting of three collinear points. I f n = 3, then K is a 7-arc which cannot be closed.

REFERENCES

L.M. ABATANGELO - G. RAGUSO, Una cahattehizzazione g t ~ d i c ad e i h-ahcki di un piano di G d o d d'ohdine pahi che h i s i d t a n o e d d e h e ow& di un a o 0 3 p h no paophio, Rend. Mat (6) 2 (1981).

El

c2]

- -

L3 A

U. EiWttxcI M a t . , (6)

- G.

FAINA, Ahcki c h h i di un piano phoie.t.tiwo din-ito, Rend.

11. (1979).

A. OOSSU, Su d c u n e p h o p h i u i ? d e i ( k , n )-ahchi

un cohpo

6init0,

[4

1

[5

1 J.W.P. Oxford 1979.

c6

1 F.

Rend. Mat., (5) 20 (1961).

-

G. FAINA G. KDFC&S, ckiudi, €?end.M a t . , (6) mRscHFEIl),

&SZI,

hopha

i n t o h n a ad una c a n g W a d u g f i m c k i

Phojectiwe Geome,taien o v a FivLite F i e R h , Clarendon Press.

Inttroduzione &e

L7]

B . sM;RE, Le g e o m w e

18

B.

sax,

~~

2 (1981).

di un p h n o p o i e a X i u o

geomdhie &k.Lte, F e l t r i n e l l i , Milam 1978.

di G d o D ,

Ann. Mat. Pura Ilppl.,

~ e o t w l e b06 Modem Geommy, cremnese

48

(1959), 1-96.

~ a m a1961.

1 9 1 B. sM;RE, Int%odu&n t o G d o O GeornWed, A t t i Accad. Naz. d e i Lincei Mem. C1. Sci. F i s . Mat. Natur. (8) S (1967) , 137-236.

- EOJ

M. T W I N 1 SCAFATI, ( h , n ) - m c k i di un p h n o gtU&ico & k i A o con paa,ticoLahe higumdo a quek%. con due cahattehi, A t t i Accad. Naz. dei Lincei, Rend. (8) 53 (1972) 71-81.

-

111 - G. T W I N I , Ghadic c h m c t e h i z a t i o n od a dgeblraic w&&en dpace, A t t i dei convegni L i n c e i (17) Tarr, 11 Ram (1976).

1

in

a Gdoh

Annals of Discrete Mathematics 14 (1982)83-1 16

0 North-Holland Publishing Company

FINITE

SEMI-SYFIMETR I C

DESIGNS

A1 b r e c h t Beutelspacher Fachbereich Mathematik der U n i v e r s i t a t Mainz Saarstr. 21, D-6500 Mainz West Germany

Semi-symmetric designs g e n e r a l i z e Dembowski's semi-planes. under c e r t a i n c o n d i t i o n s a f i n i t e semiWe show t h a t symmetric design i s embeddable i n a symmetric design i n a n a t u r a l way.

-

-

CHAPTER 1. INTRODUCTION

5

1. M A I N RESULTS

Reinhold Baer was t h e f i r s t , who asked t h e question, whether any s e m i - a f f i n e plane i s " i n a n a t u r a l way" embeddable i n a p r o j e c t i v e plane. A s e m i - a f f i n e plane i s def i n e d as a n o n - t r i v i a l l i n e a r space S w i t h t h e f o l l o w i n g p r o p e r t y : I f p i s a p o i n t n o t i n c i d e n t w i t h t h e l i n e L, then t h e r e i s a t most one l i n e through p which does n o t i n t e r s e c t L. Dembowski 161 (see a l s o T o t t e n and de W i t t e [8]) proved 1962 t h a t any f i n i t e semia f f i n e plane S i s o f t h e form S = P-U, where P i s a ( p o s s i b l y degenerate) proj e c t i v e plane and U i s empty, c o n s i s t s o f a s i n g l e p o i n t , o f a l i n e L t o g e t h e r w i t h a l l p o i n t s on L, o r o f a l i n e L t o g e t h e r w i t h a l l p o i n t s o f L b u t one. I n t h e same paper Dembowski c o n s t r u c t e d i n f i n i t e s e m i - a f f i n e planes by a " f r e e process"; consequently, a s i m i l a r c h a r a c t e r i z a t i o n o f i n f i n i t e s e m i - a f f i n e planes i s hope1ess

.

I n t h e l a s t p a r t o f h i s book " F i n i t e Geometries" [7], Dembowski i n t r o d u c e d a much more general concept, namely t h e concept o f a semi-plane. A semi l a n e i s a f i n i t e incidence s t r u c t u r e S ( c o n s i s t i n g o f p o i n t s and l i n e s ) s a t d e f o l l o w i n g axioms : ( i ) Any two d i s t i n c t p o i n t s o f S a r e i n c i d e n t w i t h a t most one common l i n e ; any two d i s t i n c t l i n e s i n t e r s e c t i n a t most one p o i n t . C a l l two p o i n t s ( o r two l i n e s ) a r a l l e l c i d e n t w i t h a common l i n e ( o r , k i v vely).

f they c o i n c i d e o r a r e n o t i n d a p o i n t i n common, r e s p e c t i -

( i i ) Given a n o n - i n c i d e n t p o i n t - l i n e p a i r (p,L), then t h e r e e x i s t s a t most one l i n e p a r a l l e l t o L through p and a t most one p o i n t on L p a r a l l e l t o p. ( i i i ) There e x i s t a t l e a s t two d i s t i n c t l i n e s ; any l i n e i s i n c i d e n t w i t h a t l e a s t three points. Dembowski showed t h e f o l l o w i n g : I f S i s a % o n - e l l i p t i c " semi-plane, then S i s embeddable " i n a n a t u r a l wa loi n a p r o j e c t i v e plane. (For d e t a i l s , t h e reader i s r e f e r e d t o s e c t i o n 7.4 i n c f . i n p a r t i c u l a r 7.4.11 and 7.4.12.) By means o f a counterexample, Baker [l] showed t h a t " e l l i p t i c " semi-planes w i l l n o t behave so nicely.

fi];

83

A . Beutelspacher

84

The aim o f t h i s work i s t o generalize t h e n o t i o n o f a semi-plane i n such a way t h a t the r o l e o f the f i n i t e p r o j e c t i v e planes w i l l now be played by the symmetric designs. I n order t o do t h i s , we define: L e t ~ , v , x be non-negative i n t e g e r s . A semi-symmetric design o f typ ( p , ~ ) i s an incidence s t r u c t u r e S s a t i s f y i n g the f o l l o w i n g axioms: (1) Any two d i s t i n c t p o i n t s o f S are i n c i d e n t w i t h e x a c t l y A - 1 o r A blocks; any two d i s t i n c t blocks i n t e r s e c t i n e x a c t l y A-1 o r X p o i n t s . Two p o i n t s p, q ( o r two blocks B, C ) are c a l l e d p a r a l l e l , i f they COi n c i d e or a r e i n c i d e n t w i t h e x a c t l y A-1 common b l o m i n t e r s e c t i n e x a c t l y A-1 points, r e s p e c t i v e l y ) . I f p i s p a r a l l e l t o q ( o r B i s p a r a l l e l t o C ) , we w r i t e p I q (or, B 1 C, r e s p e c t i v e l y ) . ( 2 ) For any non-incident point-block p a i r (p,B) o f S there a r e 0 o r p blocks through p which a r e p a r a l l e l t o B y and 0 o r v p o i n t s on B which are p a r a l l e l t o p.

( 3 ) There e x i s t p o i n t s and l i n e s ; any p o i n t (or, any b l o c k ) i s i n c i d e n t w i t h a t l e a s t two blocks (or, two points, r e s p e c t i v e l y ) ; f o r any two points, t h e r e i s a block n o t i n c i d e n t w i t h them, and d u a l l y . Let S denote an incidence s t r u c t u r e s a t i s f y i n g (1). A point-block p a i r (p,B) o f S i s s a i d t o be o f t e (s,t), i f t h e r e are e x a c t l y s blocks through p which are p a r a l l e l t o B *exactly t p o i n t s on B which are p a r a l l e l t o p. Using t h i s d e f i n i t i o n , we can reformulate axiom ( 2 ) as f o l l o w s : ( 2 ) Any non-incident point-block p a i r o f ( 0 , ~ ) o~r (u,v).

S

has t y p e

(O,O), (u,O),

Our f i r s t Lemma and i t s C o r o l l a r y i n f o r m us, which p a i r s o f non-negative integers have a chance t o be the type o f a semi-symmetric design.

1.1 LEMMA. Denote by S a f i n i t e incidence s t r u c t u r e s a t i s f y i n g (1). I f ( s , t ) i s the type o f a non-incident point-block p a i r (p,B) o f S, then x d i v i d e s s - t .

\PI

Proof. Denote by

the number o f blocks through

points on B. Counting i n two ways t h e incidences we get

s(X-I) i.e. /PIX

-

p and by (q,C) w i t h

IBI

q I B

the number o f and p I C

t (Ipl-s)A = t(X-I) t (lBl-t)X,

s

= lBlX

-

t.4

class

We say t h a t t h e semi-symmetric design S i s i n ( I ) , i f t h e r e are i n t e g e r s p and v w i t h ( i ) p and v are m u l t i p l e s o f A; ( i i ) S i s o f type ( u , ~ ) . S i m i l a r l y , S i s s a i d t o be i n (11), i f t h e r e e x i s t integers p and v w i t h (i) d i v i d e s U-V, b u t divides n e i t h e r p nor V; ( i i ) S i s o f type ( u , ~ ) .

class

1.2 COROLLARY. Any f i n i t e semi-symmetric design

S

i s i n class ( I ) o r i n c l a s s (11).

Proof. Suppose t h a t S i s a semi-symmetric design o f type ( u , ~ ) such t h a t X M e s ,u, b u t X does n o t d i v i d e V . Then p-v i s no m u l t i p l e o f A, therefore, the preceeding Lemma implies i n view o f axiom ( 2 ) t h a t any non-incident point-block p a i r o f S i s o f type (0,O) o r (p,O). So, S i s o f type ( p , O ) and i n class ( I ) . S i m i l a r l y , i f X i s a d i v i s o r o f V . b u t does n o t d i v i d e p , S i s i n class ( I ) . F i n a l l y , suppose t h a t x d i v i d e s n e i t h e r v nor V . I f S i s n o t i n class (11),

85

Finite semi-symmetric designs then any non-incident point-block p a i r i s o f type class ( I ) . J

(0,O).

Therefore, S

is in

There are two classes o f f i n i t e semi-symmetric designs known.

+ + +

I. L e t St = (p ,8 ,I ) be a symmeiric 2-(v,k,A) design and denote by (p,B) a non-incident p o i n t - b l o i k p a i r o f S Then fyr any subset U +of Ip,Bl the i n c i dence s t r u c t u r e S = S -U w i t h p o i n t s e t p -U, block s e t B -U and the induced incidence r e l a t i o n i s a semi-symmetric design o f type ( x , ~ ) .

.

I n order t o introduce the second class, we need two d e f i n i t i o n s . I f D i s a block design and p, q are two d i s t i n c t p o i n t s o f D, t h e l i n e through p and q i s defined as the s e t o f a l l p o i n t s o f D which are i n c m t w i t h any block through p and q. It can e a s i l y be seen t h a t any block which does n o t c o n t a i n the l i n e L i n t e r s e c t s L i n a t most one p o i n t . We c a l l a l i n e L o f D p r o j e c t i v e , i f L has a p o i n t i n common w i t h any block o f D. 11. Denote by S+ a 3 y m e t r i c 2-(m 3+m 2tmt1,m 2tmt1,mtl) gesign and l e t L be a projective l i n e o f S Define the incidence s t r u c t u r e S SL as f o l l o w s :

.

St which are n o t i n c i d e n t w i t h L; the blocks o f St+L S+ which do n o t contain L; the incidence i s induced by the incidence o f S+. Then S++L i s a semi-symmetric design o f type (m,m) (A-1,x-I). the p o i n t s o f

S'cL

a r e those p o i n t s o f a r e those blocks o f

REMARK. I n any 3-dimensional p r o j e c t i v e space o f order m, any l i n e i s p r o j e c t i v e . (Inct, by the famous theorem o f Dembowski and Wagner ( c f . [7], 2.1.11), t h e f i n i t e p r o j e c t i v e spaces a r e characterized by the f a c t t h a t any l i n e i s project i v e . ) On t h e other hand, by the method i n d i c a t e d i n [7], 2.4.36, one can c o n s t r u c t many symmetric designs which are n o t p r o j e c t i v e spaces b u t contain a t l e a s t one projective line.

I n t h i s paper we s h a l l deal w i t h characterizations o f t h e above examples. I n order t o formulate our r e s u l t s , we need some d e f i n i t i o n s . L e t S be a semi-symmetric design. For a p o i n t p we denote by wpl the s e t o f a l l points p a r a l l e l t o p; f o r a block B the s e t IIBll i s defined analoguosly. We c a l l Hpll ( o r !Ell) t h e ( o i n t ) a r a l l e l class o f p (or, t h e (block) s s i s s a i d t o be tml,i f i t a r a l l e l class o f B y r e s p e c t i h . h L o n e element. The p o i n t p a r a l l e l i s m i s c a l l e d t r i v i a l , i f p o i n t p a r a l l e l class i s t r i v i a l . ( I n other words: The p o i n t p a r a l T e 7 i s m i s t r i v i a l i f any two d i s t i n c t p o i n t s a r e i n c i d e n t w i t h x common blocks.) Our f i r s t r e s u l t generalizes Dembowski ' s c h a r a c t e r i z a t i o n o f the f i n i t e semia f f i n e planes. THEOREM A. Denote by S a f i n i t e semi-symmetric design w i t h A t 2. I f the p o i n t parallelTsm o f S 1s t r i v i a l , then S i s 9 symmetric design qr t h e r e e x i s t s a and a p o i n t p o f S such t h a t S = S -{PI holds. symmetric design S

I n the examples I.and II.,and a l s o i n the case A = 1, the p a r a l l e l i s m i s t r a n s i t i v e , hence an equivalence r e l a t i o n ( c f . [7] , 7.4.2). However, i n semi-symm e t r i c designs w i t h 1 > 1 i t i s n o t a t a l l c l e a r , whether the p a r a l l e l i s m i s t r a n s i t i v e o r not. I n f a c t , i t w i l l be one o f the main steps i n t h e proofs o f t h i s paper, t o show t h a t under c e r t a i n assumptions the parallelism i s transitive.

-

-

We c a l l a p a r a l l e l class IIBII o f blocks t r a n s i t i v e , i f f o r any two blocks D with C 1 I B i t holds: D

i s parallel to

B

i f and o n l y i f

D

i s parallel t o

C.

C

and

A . Beutelspacher

86 I n other words: = IIBII.

IC1I

B

i s t r a n s i t i v e , i f f o r any block

C

parallel t o

B we have

T r a n s i t i v e p o i n t p a r a l l e l classes a r e defined s i m i l a r l y . Obviously, the f o l l o w i n g assertions hold: I f IIBII i s t r a n s i t i v e , then the r e s t r i c t i o n o f the p a r a l l e l i s m on (1Bll i s an equivalence r e l a t i o n . I f the p a r a l l e l i s m i s an equivalence r e l a t i o n , then any p a r a l l e l c l a s s i s transitive. Our second r e s u l t reads as f o l l o w s . THEOREM B. Denote by S a f i n i t e semi-symmetric design o f class ( I ) w i t h A > 1. f S contains 0 n o n - t r i v i a l t r a n s i t i v e p a r a l l e l class, then t h f r e e x i s t s a symm e t r i c design S , a nqn-incident point-block p a i r (p,B) o f S , and a subset U o f {p,BI w i t h S = S -U.

For semi-symmetric designs o f class (11) we could n o t prove a r e s u l t o f a s i m i l a r g e n e r a l i t y . This has two reasons: F i r s t l y , i f p < A , a p a r a l l e l class may n o t be "regular" o r " s i n g u l a r " (these terms w i l l be defined i n t h e n e x t section); on the other hand, we were n o t able t o show the existence o f a "complete" p a r a l l e l class. (A block p a r a l l e l class 7 i s s a i d t o be com l e t e , i f any p o i n t o f S i s i n c i dent w i t h a t l e a s t one block i n 7 ; dually&te p o i n t p a r a l l e l classes are defined. ) Now we can s t a t e our t h i r d r e s u l t . THEOREM C . Denote by S a f i n i t e semi-symmetric design i n class (11) having type r (a) If S contains a n o n - t r i v i a l t r a n s i t i v e b l o c k p a r a l l e l c l a s s and i f p 3 min{v,A-ll,

then p = u Q A-1. (b) Suppose u = A - 1 > 1 (or, I, A - 1 > l ) , suppose t h a t the p o i n t p a r a l l e l i s m i s t r a n s i t i v e and t h a t S contains a complete p a r a l l e l class. Then t h e r e e x i s t s a 3 2 2 symmetric 2-(m tqj t m t 1 , m tmt1,mtl) design St and a p r o j e c t i v e l i n e L o f St such t h a t S = S +L.

I n the next section we l i s t t h e basic f a c t s on semi-symmetric designs, which w i l l be used f r e q u e n t l y i n t h e sequel. I n chapter 2 t h e p r o o f o f Theorem A and i n chapter 3 the proof o f Theorem B w i l l be given. I n the f i r s t sections, we reduce the problem: I t i s shown t h a t t h e semi-symmetric designs i n question are o f type ( A , O ) , or (p,~), r e s p e c t i v e l y . I n the second p a r t s the proofs o f the respective Theorems w i l l be f i n i s h e d . S i m i l a r l y , i n t h e l a s t chapter, we prove Theorem C. I n [27 we have already proved Theorems A and B i n t h e special case u I, = A . However, i n order t o keep t h i s paper self-contained, we w i l l g i v e complete proofs o f a l l Theorems. We quote three papers which deal w i t h s i m i l a r problems. I n [3] we generalized Theorem A t o " s t r o n g l y resolvable" designs. Another g e n e r a l i z a t i o n o f t h e n o t i o n of a semi-plane i s due t o Bose and Shrikhande [43 and Cron and Mavron [5]: A A-plane i s a f i n i t e incidence s t r u c t u r e S s a t i s f y i n g the axioms ( I ) , ( 2 ) ' and a non-degeneracy axiom: ( 2 ) ' I f (p,B) denotes a non-incident point-block p a i r o f S, then t h e r e are A-I o r A blocks through p which are p a r a l l e l t o B and t h e r e are A - 1 o r A p o i n t s on B which a r e p a r a l l e l t o p. One sees immediately t h a t a semi-symmetric design i s a A-plane o n l y i f i t i s a semi-plane. I n our terminology, the authors prove: I f t h e p a r a l l e l i s m o f a x-plane

Finite semi-symmetric designs

with > 1 i s t r a n s i t i v e , then subdesign" i s removed.

S

81

i s a symmetric design, f r o m w h i c h a "Baer-

We f i n i s h t h i s i n t r o d u c t i o n by w a r n i n g t h e r e a d e r : O f t e n , t h e word bm-L-bqtntnWLic

duigiglz i s used w i t h a d i f f e r e n t meaning.

5

2. BASIC LEMMAS

Throughout t h i s paper, we s h a l l use t h e t e r m i n o l o g y o f Dembowski [7]. l a r , t h e f o l l o w i n g d e f i n i t i o n s w i l l be needed.

In particu-

Denote by S = (p,B,I) a f i n i t e i n c i d e n c e s t r u c t u r e and l e t p ,...,p be p o i n t s o f S. By ( p ,...,p ) we mean t h e s e t o f a l l b l o c k s o f S w h i c i a r e i k i d e n t w i t h each p o i n t i s d e f i n e d as t h e number o f b l o c k s I ply...,pmI ( i [? { l , . . . , m } ) ; i n c i d e n t w i t h ' e a c h pi :

b.

lP1'...YPmI = I (P1'...IPm)I. D u a l l y , f o r t h e b l o c k s B ,...,B o f S, (B l,...,B S i n c i d e n t w i t h each b l o i k Bi 'and I Bly...,Bn/ points.

) i s the set o f a l l points o f Benotes t h e number o f such

The number / P I o f b l o c k s t h r o u g h a p o i n t p i s c a l l e d i t s de r e e ; a p o i n t o f degree i i s a l s o c a l l e d an " i - p o i n t " . Analoguosly, t h e d e g Z h 7 a b l o c k and an i - b l o c k a r e d e f i n e d . 2 . 1 RESULT ( c f . Dembowski [7] , 1.1.2). Denote by S a f i n i t e i n c i d e n c e s t r u c t u r e w i t h v p o i n t s , a c o n s t a n t number r o f b l o c k s t h r o u g h each p o i n t , and a c o n s t a n t number k o f p o i n t s on each b l o c k . L e t h be t h e average number o f b l o c k s t h r o u g h two d i s t i n c t p o i n t s , i. e.:

where t h e summation r u n s o v e r a l l p a i r s Then

(p,q)

o f d i s t i n c t points.

(v-l)x = r(k-1).

-

-

denotes a f i n i t e semiFrom now on i f i t i s n o t stated otherwise S = (p,B,I) symmetric d e s i g n o f t y p e ( u , v ) w i t h x > 1. The number o f p o i n t s i n S i s v and b i s d e f i n e d as t h e number o f b l o c k s i n S. REMARK. I f S i s a-semi-symmetric d e s i g n (say o f t y p e ( p , ~ ) ) ,t h e n t h e dual i n c i d e n c e s t r u c t u r e S i s a g a i n a semi-symmetric d e s i g n ( o f t y p e ( v , ~ ) ) . I n p a r t i c u l a r , t h e p r i n c i p l e o f d u a l i t y h o l d s f o r t h e c l a s s o f f i n i t e semi-symmetric designs. The i n t e g e r s if

S

a

and

c

w i l l be d e f i n e d as f o l l o w s : I f S

i s i n c l a s s ( I ) , then

ah and v = cx; i s i n class (11), b u t n o t i n c l a s s ( I ) , then p =

U-v = ax.

The f o l l o w i n g P r o p o s i t i o n f o l l o w s i m m e d i a t e l y f r o m t h e p r o o f o f Lemma 1.1.1; t h i s a s s e r t i o n w i l l be used f r e q u e n t l y . 2.2 PROPOSITION. Denote b y (p,B) a n o n - i n c i d e n t p o i n t - b l o c k p a i r o f 7 a ) I f S i s i n class ( I ) , then the f o l l o w i n g assertions hold: If (p,B) i s o f t y p e (O,O), t h e n I p I = I B I ; if

(p,B)

i s o f type

(p,O),

then

I p l - a = IBI;

S.

A . Beutelspacher

88 if

(p,B)

i s o f type

(O,v),

then

i p I = lBl-c,

if

(p,B)

i s o f type

(P,v),

then

I p I - a = 161-c.

( b ) W-S i s i n c l a s s (11), then moreover, t h e f o l l o w i n g holds:

(p,B)

i s o f type

If

(p,B)

i s o f type

(O,O), then

I p I = 161;

if

(p,B)

i s o f type

( ~ , v ) , then

I p I = IB1ta.J

2.3 COROLLARY. L e t Define the integer 191

I CI

be i n c l a s s ( I ) and denote by B a b l o c k o f maximal degree. by IBI = n t c . Then f o r any p o i n t q o u t s i d e B i t holds

S n

of

C E

(p,v);

{ntc,ntcta,n,nta].

E

For any b l o c k

(0,O) o r o f type

S

t h e f o l l o w i n g holds:

{ntc,n ,nta,ntc-a,n-aYnt2c-a}.J

2.4 LEMMA. Denote by S a f i n i t e i n c i d e n c e s t r u c t u r e s a t i s f y i n g t h e axioms (1) -Then f o r any b l o c k B and any p o i n t p o f S i t holds:

IBI > A

and

I p I > A.

Proof. Assume t h a t t h e r e i s a b l o c k

B o f S w i t h I B / ,< A. Then by ( 3 ) A. If p and q a r e two d i s t i n c t p o i n t s o f B y t h e n - i n view o f ( 1 ) - any b l o c k o f S c o n t a i n s p o r q: a c o n t r a d i c t i o n t o ( 3 ) . D u a l l y , I p I > A fol1ows.J

2

6

IBI

6

2 . 5 COROLLARY. ( a ) For any two d i s t i n c t b l o c k s B and C o f S i t holds (6) (C). ( b ) L e t p and q be two d i s t i n c t p o i n t s o f S. Then t h e r e e x i s t s a b l o c k through p which does n o t c o n t a i n q.

q=

Proof. ( a ) If ( 6 ) t i o n t o 4.

i s a subset o f

(C)

I B I = lB,Cl

it follows

< A, a c o n t r a d i c -

( b ) i s t h e dual o f (a).,' L e t S denote an i n c i d e n c e s t r u c t u r e s a t i s f y i n g ( 1 ) . I f (p,B) i s an i n c i d e n t f o r t h e type o f (p,B). p o i n t - b l o c k p a i r of S, we w r i t e ( a ( p , B ) t l , B ( p , B ) t l ) This means: There a r e e x a c t l y a(p,B) blocks through p i n t e r s e c t i n g B i n A-1 p o i n t s and t h e r e a r e e x a c t l y B ( ~ , B ) p o i n t s q on B w i t h t h e p r o p e r t y t h a t p and q a r e j o i n e d by e x a c t l y A-1 blocks. 2.6 LEMMA. Denote by S a f i n i t e i n c i d e n c e s t r u c t u r e s a t i s f y i n g t h e axiom ( 1 ) . i s an i n c i d e n t p o i n t - b l o c k p a i r o f S having t y p e ( a + l , ~ t l ) , then ( l B l - l p l ) ( a - i ) = 6-a. Proof. Count t h e incidences (q,C)

with

p # q IB

( p l - i - ~ ) ( ~t- i~ )( x - 2 ) = ( l p p a ) ( A - q

and

t

p IC # B

i n two ways:

a(~-2).~

For t h e remainder of t h i s s e c t i o n we s h a l l deal w i t h p a r a l l e l classes i n f i n i t e semi-symmetric designs. F o r a b l o c k p a r a l l e l c l a s s n we denote by c i d e n t w i t h a t l e a s t one element o f n.

p(n)

t h e s e t o f p o i n t s which a r e i n -

From now on, l e t II = IIBII be a n o n - t r i v i a l b l o c k p a r a l l e l c l a s s o f S, where S i s a f i n i t e semi-symmetric design o f t y p e ( P , v ) . By M we denote t h e number o f blocks i n II.

Finite semi-symmetric designs

89

2.7 LEMMA. If p 2. x, t h e n t h e r e e x i s t s a t most one p o i n t p o f S such t h a t p i s i n c i d e n t w i t h each b l o c k o f n . P r o o f . Assume t h a t t h e r e were two such p o i n t s , say p and q. Then n i s a subs e t f (p,q) and t h e r e f o r e M < Ip,ql < A . Since n i s n o n - t r i v i a l , t h e r e e x i s t s a b l o c k C o f n w i t h C # B. By 2 . 5 ( a ) , t h e r e i s a p o i n t o f C o u t s i d e B; so, t h r o u g h t h a t p o i n t t h e r e a r e e x a c t l y p b l o c k s p a r a l l e l t o B. S i n c e no such b l o c k c o i n c i d e s w i t h B, we have M z u t l . Together we g e t u 4 A-1: a c o n t r a d i c t i o n . / We c a l l t h e p a r a l l e l c l a s s n singu!ar, if t h e r e e x i s t s e x a c t l y one p o i n t p i n c i d e n t w i t h any b l o c k o f n ; t h i s p o i n t p i s c a l l e d t h e s i n g u l a r p o i n t o f n. I f t h e r e e x i s t s no such p o i n t , n i s c a l l e d r e g u l a r . I n t h e n e x t Lemma, we c o n s i d e r s i n g u l a r p a r a l l e l c l a s s e s .

n

2.8 LEMMA. L e t

( t ):,<= X 2 , n o n - t r i v i a l , then

= IIB11

be a singular p a r a l l e l class w i t h the singular point

t h e n n = ( p o ) , p ( n ) = p , IIpdI v = A.

>

p 1B

P r o o f . ( a ) Through a p o i n t p w i t h =through po. So, p d A.

c (B);

i f , moreover, IIpoII

t h e r e a r e a t most

x

po. is

blocks which

( b ) Suppose now p = x > 2. Assume t h a t t h e r e e x i s t s a b l o c k , say C, t h r o u g h po which i s n o t p a r a l l e l t o B. Then t h r o u g h no p o i n t p w i t h p E ( C ) - ( B ) t h e r e i s a b l o c k o f n. (Foh: If p were a p o i n t o f p ( n ) . t h e n t h r o u g h p we would have t h e u = A b l o c k s o f n c_ ( p ) and t h e b l o c k C E (po); so p and po would have a t l e a s t x t l b l o c k s i R common.) Therefore, i f B ' i s an a r b i t r a r y b l o c k o f n - { B I , t h e n B ' and C i n t e r s e c t o n l y i n p o i n t s o f B: Since

(CSB') s ( B s B ' ) . C,B'I A-1 and I B , B ' I (C,B')

(B,B')

= A-1, we have

for all

I n p a r t c u l a r , f o r any b l o c k

B'

n-{BI.

E

B' # B p a r a l l e l t o

B i t holds

(B,B') c (B,C). Denote by q a p o i n t o f ( B ) - ( C ) . S i n c e A-1>, 2 , t h e r e e x i s t s a b l o c k A w i t h A # B t h r o u g h po and q . I n view o f t h e above argument, A i s n o t p a r a l l e l t o B. Since (A,B) # (B,C), we have IA,B,CI If

A

6

A-1.

plays the r S l e o f (B,B')&

(B,A)

C, one g e t s

for all

B'

E

n-IBI.

Hence (B,B') L, (AsBsC) and so (B,B')

= (A,B,C)

for all

B'

E

n-{BI.

T h e r e f o r e any b l o c k of n c o n t a i n s t h e A-1 a 2 p o i n t s o f (A,B,C): a c o n t r a d i c t i o n , s i n c e n i s supposed t o be s i n g u l a r . So, n = ( p ). T h i s i m p l i e s i m m e d i a t e l y p ( n ) = p . T h e r e f o r e t h r o u g h any p o i n t q o u t s i d e Bo t h e r e a r e e x a c t l y P = A b l o c k s o f n ; so, any such p o i n t q i s conn e c t e d w i t h po by a t l e a s t (hence e x a c t l y ) x b l o c k s . I n o t h e r words: Ilp II c, (B). SiRce u = A, S i s i n c l a s s ( I ) . T h e r e f o r e , i f IIp 11 i s n o n - t r i v i a l , we have v = cx >/ A. Denote by X a b l o c k , w h i c h does n o t p k s t h r o u g h p , c o n t a i n i n g a t l e a s t one p o i n t p a r a l l e l t o po. Then t h e p o i n t s on X p a r a l l e l t o po a r e

90

A . Beu telspacher

exactly the

A

points o f

(X,B).

This shows

v

= A.J

I n the f o l l o w i n g Lemmas, we s h a l l deal w i t h t r a n s i t i v e p a r a l l e l classes. 2.9 LEMMA. Let II be a t r a n s i t i v e b l o c k p a r a l l e l class w i t h e x a c t l y denote by X an a r b i t r a r y block o f n. (a) I f n i s regular, then

1x1 (11-1)

(b) I f

n

M blocks;

= (M-l)(A-1).

i s s i n g u l a r , then

( l X l - l ) ( p - l ) = (M-l)(A-Z). Proof. The t r a n s i t i v i t y o f n i m p l i e s i n p a r t i c u l a r t h a t through any p o i n t p d T f E r e n t from the s i n g u l a r p o i n t o f n (which e x i s t s o n l y i n case ( b ) ) t h e r e 0 o r !.I blocks o f n. (Namely: For any such p o i n t p t h e r e i s a block C E n w i t h p 1 C. Since n i s t r a n s i t i v e , we have n = IICll. Therefore, through p t h e r e are e x a c t l y 0 o r u blocks o f IICII n.) I n p a r t i c u l a r , any p o i n t o f X, which i s d i s t i n c t from the s i n g u l a r p o i n t , i s i n c i d e n t w i t h e x a c t l y p - 1 blocks o f n-IX}. Having t h i s i n mind, we can count t h e incidences (p,Y) w i t h p I X, Y E n-{X}, where p i s n o t the s i n g u l a r p o i n t o f n. This y i e l d s our a s s e r t i 0 n s . J

2.10 COROLLARY. I f p > 1, then any block o f a t r a n s i t i v e , s i n g u l a r o r r e g u l a r b l o c k p a r a l l e l class has t h e same degree.4 2.11 COROLLARY. Suppose A = 2 . I f p a r a l l e l class, then u = 1.4

S

contains a s i n g u l a r t r a n s i t i v e block

Let Y be a block w i t h Y 1 n. I f n i s r e g u l a r , we d e f i n e y = y ( Y ) t o be the number o f points on Y which are i n c i d e n t w i t h an element o f 1. I f n i s singul a r w i t h the s i n g u l a r p o i n t p , we d e f i n e y = y(Y) ( o r , y ' = y ' ( Y ) ) t o be the number o f p o i n t s on Y incideRt w i t h an element o f n, i f Y does n o t c o n t a i n po (or, i f Y i s i n c i d e n t w i t h po, r e s p e c t i v e l y ) .

2.12 LEMMA. Denote by n a t r a n s i t i v e p a r a l l e l c l a s s w i t h e x a c t l y M l e t X b e a block o f n. (a) I f n i s regular, then yp

(b) I f

n

and

MA

i s s i n g u l a r and i f yp

blocks;

= MA =

A a 3, then

-l) ( ( I X / - l ) ( pA-2

t

A-qA

'

( y t - l ) p = M(x-1) = ((lxl-l)(!.I-l) A - 2 A-2)( A - 1 ) . Proof. Count the incidences (p,C) w i t h p I Y, C E n , where ( i n case ( b ) ) -he s i n g u l a r p o i n t o f n. Using 2.9, the assertions fol1ow.J

p

is

2.13 COROLLARY. L e t n be a t r a n s i t i v e p a r a l l e l class and l e t X be a block o f n. I f p > A, then y , y t > 1x1. I n p a r t i c u l a r , the degree o f any block Y I II i s bigger than t h e degree o f X. Proof. By 2.7, n i s r e g u l a r o r s i n g u l a r . We assume y s 1x1 and y ' 6 1x1. This implies IXlu(A-1) a yu(A-1) or

(IXl(u-1) t A - l ) A ,

91

Finite semi-symmetric designs

IXll.I(A-2) a yIJ(A-2) = ( ( I X l - l ) ( I J - l )

t

A-2)Xy

or ( J X J - l ) p ( A - 2 ) >, (y0-1)u(X-2) = ( ( I X l - l ) ( u - l ) Using Since

t

A-2)(A-1).

a A, the f i r s t case y i e l d s the f o l l o w i n g c o n t r a d i c t i o n :

0 > -A(A-1) 5 IXI(lJ-A) >I 0. 1 < A ,< U, we g e t i n the second case from

A(2u-A) 2. A(u-Atl) 2. IXl(2u-A) the c o n t r a d i c t i o n 1x1 6 A. F i n a l l y , i n t h e t h i r d case we have 0 >I -(A-Z)(A-l) a ( I X l - l ) ( u - A + l ) > 0: a contradiction. J The following Theorem gives us a f i r s t c h a r a c t e r i z a t i o n o f semi-symmetric designs. Although the assumptions a r e q u i t e strong, the a s s e r t i o n w i l l be u s e f u l i n the next chapters.

(p,B,I) be a f i n i t e semi-symmetric design w i t h a t r a n s i t i v e 2.14 THEOREM. L e t S p a r a l l e l class n = [ell. I f n = 8, then S i s a symmetric 2-(v,k,A-l) design. Proof. Since 8 II,the p a r a l l e l class n i s r e g u l a r . This i m p l i e s t h a t through -point o f S there are e x a c t l y IJ blocks. Now 2.10 y i e l d s t h a t any block has the same degree, say k. Moreover, t h e dual incidence s t r u c t u r e S i s a 2-(b,u, A-1)design w i t h k "blocks" through any "point". By F i s h e r ' s i n e q u a l i t y we have

(-1 k >/p. Denote by (p,C) an i n c i d e n t point-block p a i r o f o f (p,C), we get using 2.6: = a t ( k - p ) ( A - l ) = IJ-1t ( k - p ) ( A - l ) . 1

S. I f

(at1,Btl)

i s the type

( 9 -

Now we d i s t i n g u i s h two cases. C u e 1. 6 = k-1. I n t h i s case, any two p o i n t s are p a r a l l e l , so S i s a 2-(v,k,x-l) design. Since any two blocks i n t e r s e c t i n e x a c t l y A - 1 points, S i s symmetric.

C u e 2. B < k-1. Using we get k-1 > B = 11-1 t (k-p)(A-l), or (k-p)(A-2) < 0. Since A >, 2 we have k < IJ:a c o n t r a d i c t i o n t o ( * ) . J ( 0 . )

2.15 COROLLARY. Denote by S a f i n i t e semi-symmetric design o f class ( I ) having u,v I f S contains a n o n - t r i v i a l t r a n s i t i v e block p a r a l l e l class n = type I l B l w i t h ) ] B l = ntc, then S i s a symmetric design. Proof. Since S i s i n class ( I ) and contains a n o n - t r i v i a l block p a r a l l e l class, w e v e IJ = aA b A . Then 2.7 y i e l d s t h a t II i s r e g u l a r o r s i n g u l a r . Because n t c i s the maximal block degree, i n view of 2.13, t h e r e i s no block o u t s i d e n. Now the a s s e r t i o n f o l l o w s from 2.14.4

.

92

A . Beutelspacher

CHAPTER 2 . SEMI-SYMMETRIC DESIGNS WITH TRIVIAL POINT PARALLELISM

I n t h i s c h a p t e r we s h a l l p r o v e t h e f o l l o w i n g > 1 and t r i v i g l THEOREM A. Denote b y S a f i n i t e s e m i - s y m n e t r i c d e s i g p w i t h and a p o i n t p o f S p o i n t p a r a l l e l i s m . Then+there i s a symmetric d e s i g n S with S = S o r S = S -{PI.

Under t h e assumptions o f t h i s Theorem, we can suppose t h a t S i s i n c l a s s (I); ~ a m d y : i f S happens t o be i n c l a s s ( 1 1 ) , t h e n any n o n - i n c i d e n t p o i n t - b l o c k p a i r i s o f t y p e (O,O), s o S i s o f t y p e (0,O) and so S i s i n c l a s s ( I ) as w e l l . Hence we have v = 0 and p = ax. I f t h e b l o c k p a r a l l e l i s m i s a l s o t r i v i a l , t h e n any n o n - i n c i d e n t p o i n t - b l o c k p a i r has t y p e (0,O); t h e r e f o r e any two elements o f S have t h e same degree, say k. It f o l l o w s t h a t S i s a 2-(v,k,x) design w i t h t h e p r o p e r t y t h a t any two d i s t i n c t b l o c k s i n t e r s e c t i n e x a c t l y A p o i n t s . Hence S i s symmetric and o u r Theorem i s proved. From now on, we suppose t h a t S c o n t a i n s a n o n - t r i v i a l b l o c k p a r a l l e l c l a s s . T h i s i m p l i e s p > 0, t h e r e f o r e a >, 1. I n t h e f i r s t s e c t i o n we show Theorem A w i l l be f i n i s h e d .

a = 1, w h i l e i n t h e second s e c t i o n t h e p r o o f o f

§ 1. PARAMETERS

a f i n i t e semi-symmetric d e s i g n o f t y p e ( ~ ~ 0w i)t h Denote b y S = (p,B,I) p = ax, a 3 1 and x > 1. We change t h e n o t a t i o n o f c h a p t e r 1 a l i t t l e b i t , b u t t h i s w i l l not lead t o misunderstandings.

1.1 PROPOSITION. There e x i s t s a p o s i t i v e i n t e g e r has degree n+a and any b l o c k o f S has degree

n n

such t h a t any p o i n t o f o r n+a.

P r o o f . Denote by n+a t h e maximal p o i n t degree o f S; l e t d e g r e e n+a. Then f o r any b l o c k B w i t h po 1 B i t h o l d s

IBI

E

po

S

be a p o i n t o f

{nta,nl.

T h e r e f o r e we have lql f o r any p o i n t

E

IBIi s n o t i n c i d e n t w i t h B, t h e n i s on B, 1.2.6 i m p l i e s

IPI (Namely: I f p case, where p since

In+a,nl q o f S. We remark t h a t f o r any p o i n t - b l o c k p a i r

(lpl-lBl)(x-1) B ( ~ , B ) = 0.) E

or

i t holds

I p I = IBl+a. I n t h e

= a >, 0,

Because any p o i n t has degree

IBI

IpI = IBI

(p,B)

n

or

n+a, f o r any b l o c k

B of

S

i t holds

In-a,n,n+a}.

F i r s t we show Any block 06

S

had dq'kee n

Oh

nta.

Fok: Assume t h a t t h e r e e x i s t s a b l o c k B

Bo has degree

n; i n p a r t i c u l a r , po

Any b l o c k B has a t most t h e degree an n - p o i n t p and we have n = I p /

of degree n-a. Then any p o i n t o u t s i d e i s o a p o i n t o f B.,

n. (Namely: O u t s i d e B (and B ) t h e r e i s I B l . ) If B does n o t pass t h r 8 u g h po, B

>,

93

Finite semi-symmetric designs

i s an n - b l o c k . Since t h r o u g h no Now we c o n s i d e r a b l o c k B which i s n o t i n c i d e n t w i t h p p o i n t o u t s i d e B t h e r e i s a b l o c k p a r a l l e l t o B y i t fol?ows

.

n-

p(llBII) C _ ( B o ) u ( B ) . Since through po t h e r e a r e p b l o c k s p a r a l l e l t o B, t h e p a r a l l e l c l a s s B i s non-trivial. L e t p be a p o i n t on B y which i s a l s o i n c i d e n t w i t h a b l o c k o f llBll-{B}. Using ( ~ p ~ - ~ B ~ ) ( =A -a(p,B) I) > 0, we g e t t h a t p has degree n t a ; t h e r e f o r e p i s a p o i n t of Bo. T h i s i m p l i e s : If B ' i s a b l o c k o f llBll-{B}, t h e n ( B ' ) i s contained i n (Bo): a c o n t r a d i c t i o n t o 1.2.5(a). Now we a r e a b l e t o p r o v e

Any p o i n t

06 S iA an

(ntal-point.

b. Since >, J B / f o r any n - b l o c k . We show t h a t u n d e r t h i s assump-

Ashme o n t h e canttray t h a t t h e r e e x i s t s an n - p o i n t b l o c k B of S, any b l o c k o f S t i o n , any p o i n t has degree n+a.

i s an

I n o r d e r t o do t h i s , l e t p be an a r b i t r a r y p o i n t o f S; w i t h o u t l o s s o f generality p # p Denote by B a b l o c k t h r o u g h p which does n o t c o n t a i n p The p a i r ,B) i s of t y p e ( p , O ) , so t h e r e i s a b l o c k 6' t h r o u g h po? which i s p a r a l l e l 90 B. We c l a i m t h a t any p o i n t p ' on B has degree n t a . ( I f p ' i s o f f B ' , t h e n t h e r e i s a t l e a s t one b l o c k t h r o u g h p ' which i s p a r a l l e l t o B ' (namely B ) ; so ( p ' , B ' ) is o f t y p e (v,O) and t h i s i m p l i e s I p ' ) = / B ' / t a = n t a . I f p ' i s on B ' , t h e n t h e r e e x i s t s a t l e a s t one b l o c k d i s t i n c t f r o m B t h r o u g h p ' , w h i c h i s p a r a l l e l t o B (namely B ' ) ; now 1.2.6 i m p l i e s

(B.

.

> 0, ( l p ' l - l B l ) ( ~ - 1 ) = .(p',B) t h e r e f o r e I p ' [ > J B I , i. e. Ip'I = n t a . ) I n p a r t i c u l a r , a n y - p o i n t p has degree n t a ; so, any p o i n t o f S i s an ( n t a ) p o i n t . Therefore, p does n o t e x i s t and hence any p o i n t o f S ' i s o f degree n t a .

Thus o u r P r o p o s i t i o n i s proved.4 REMARK. C l e a r l y , a b l o c k has a n o n - t r i v i a l p a r a l l e l c l a s s i f and o n l y i f i t s

degree i s n.

1.2 PROPOSITION. I f S i s a semi-symmetric d e s i g n o f t y p e ah > 0, t h e n a = 1.

p =

(p,O)

with

P r o o f . F i r s t we show: Thecre iA a beach 06 deghee n t a and a b l o c k 06 degtee n. Nmdy: I f any b l o c k has degree n+a, t h e n any n o n - i n c i d e n t p o i n t - b l o c k p a i r i s o f t y p e (O,O), so t h e p a r a l l e l i s m i s t r i v i a l : a c o n t r a d i c t i o n . Assume now t h a t any b l o c k o f S has degree n. Then by 1.2.6, t h r o u g h any p o i n t o f a b l o c k B t h e r e a r e e x a c t l y a ( x - l ) + l b l o c k s o f IIBII. I f M denotes t h e number o f b l o c k s p a r a l l e l t o B y we g e t

(M-l)(a-1) = I B J a ( x - 1 ) = n a ( h - l ) , hence M = n a t l . S i n c e t h r o u g h any p o i n t q w i t h ax b l o c k s o f (IBII, i t f o l l o w s (v-n)ax = ( M - l ) ( n - ( x - l ) ) therefore

= an(n-(x-I)),

vx = n ( n t 1 ) . On t h e o t h e r hand, (v-l)x = (nta)(n-I),

and so a-2 >, a-x = n ( a - 2 ) > a-2: a contradiction,

q 1B

there are exactly

p =

A. Beutelspacher

94

Now denote by B an n-block. For any p o i n t cr(p,B) = a(x-1). If B

p

on

B we have

has e x a c t l y M p a r a l l e l s , i t follows ( M - l ) ( A - l ) = IBltt(p,B) IBla(x-l),

or M-1 = an.

(1) Moreover,

(b-1

so (2) If

-

(b-l)x C

(3 1

(M-1))x

-

(M-1)

+

(M-l)(x-1)

= 1Bl(n+a-l),

n(n+a-1).

i s a block o f degree n+a, then t h e p a r a l l e l class o f ( b - l ) x = (n+a)(n+a-1).

C

i s t r i v i a l . So

The equations ( 2 ) and (3) imply together M-1 = a(n+a-1). Using (1) we see

5

a(a-1) = 0. Since

a > 0, t h i s i m p l i e s a = 1.4

2. SEMI-SYMMETRIC DESIGNS OF TYPE (X,O)

Together w i t h 2.1.2

,

the f o l 1owing Theorem proves our Theorem A .

2.1 THEOREM. Denote by S = (p,B,I) a f i n i t e semi-symmetric design o f type + ( X , O ) w i t h +A > 1. Then+there e x i s t s a symmetric design S and a p o i n t p o f S w i t h S = S o r S S -{PI. Proof. We suppose t h a t S i s n o t a symmetric design. Then, by 2.1.1, any p o i n t 57-S has degree n + l and f o r each block B o f S i t holds 181 = n + l i f and o n l y i f t h e p a r a l l e l class o f B i s t r i v i a l , IBI = n otherwise. This implies i n p a r t i c u l a r : I f B and B ' are two d i s t i n c t p a r a l l e l blocks, then I B I n = IB'I. Since S i s n o t a symmetric design, the proof o f 2.1.2 says t h a t t h e r e e x i s t s an n-block and a block o f degree n+l. I f B i s an n-block and M denotes the number o f p a r a l l e l s o f B y then t h e equation (1) of t h e preceeding s e c t i o n gives M n+l. (1) Equation ( 3 ) reads as f o l l o w s (b-l)x = 1ntl)n. (2 1 Denote by

(3) If E

(4)

e the number o f a l l n-blocks through a p o i n t o f ( v - l ) x = e(n-1) + (n+l-e)n = (n+l)n e. i s t h e number o f a l l n-blocks o f S, we get

-

EX = e ( n t 1 ) .

( F m : We count the incidences o f S: v ( n + l ) = (b-E)(n+l) + En = b ( n + l ) Using ( 3 ) and ( 2 ) i t f o l l o w s ((n+l)n e + x)(n+l) = ((nt1)n

-

which reduces t o equation (4).)

-

E.

+ x)(n+l)

-

EA,

S. Obviously,

Finite semi-symmetric designs

95

The f o l l o w i n g step i s c r u c i a l f o r our purposes.

7 6 B i~ a b l o c k luith a n o n - a 2 i v X p U & &A, t h e n IIBlI 0 t h e A & 06 a & n-bLochs 06 S. Namdy: Since any p o i n t has degree n t l , through any p o i n t o u t s i d e B t h e r e a r e e x a c t l y p = A blocks p a r a l l e l t o B. Thus (v-n ) x = (M- 1) (n-

(A1) ) .

Using ( l ) , we have vh = n ( n - x t l ) t nx = n ( n + l ) .

I n view o f ( 3 ) , t h i s implies e = A and by ( 4 ) we g e t E = n t l = 14. Since any block p a r a l l e l t o

B

i s an

n-block, our a s s e r t i o n i s proved.

I n other words: Two d i s t i n c t blocks are p a r a l l e l i f and o n l y i f they both have degree n. Denote by n the s e t o f a l l n-blocks o f S. flow we d e f i n e the incidence s t r u c t u r e pt := p " I n 1 , Bt := 8. The incidence r e l a t i o n p It B

n

:*

I+ B :++

St

+

+

t

( p ,B ,I ) as f o l l o w s :

I+i s defined i n the f o l l o w i n g way p I B B

E

n,

f o r a point if

B

p and a block

i s a block o f

B

of

S,

S.

By construction, any two d i s t i n c t blocks o f S+ i n t e r s e c $ n e x a c t l y x p o i n t s ; e = x implies t h a t througb any two d i s t i n c t p o i n t s o f S there are e x a c t l y A blocks. This means t h a t S i s a semi-symmetric design o f type (O,O), hence a symmetric design.

Since, by construction, S = S+-{n), our Theorem i s proved.4 We s t a t e the dual o f Theorem A e x p l i c i t e l y 2.2 THEOREM. Denote by S a f i n i t e semi-symmetric design w i t h x > 1 and t r i v i a l b l o c k p a r a l l e l i s m . Then S i s a symmetric design o r a symmetric design from which one block i s removed.4 The f o l l o w i n g a p p l i c a t i o n o f 2.1 w i l l be very u s e f u l i n the next chapter. 2.3 THEOREM. Denote by S = ( p , B , I ) a f i n i t e semi-symmetric design o f type i n class (I). Then S does n o t contain a s i n g u l a r t r a n s i t i v e p a r a l l e l class.

(p,v)

Proof. From now on,

we s h a l l use again the terminology o f chapter 1; i n p a r t i c u l a r , n+c i s the maximal block degree of S. We assume t h a t there i s a s i n g u l a r t r a n s i t i v e p a r a l l e l class n o f S. Without loss o f g e n e r a l i t y , we can suppose t h a t n i s a block p a r a l l e l class, say n = IlBll. Let po be the s i n g u l a r p o i n t o f n. By M we denote the number o f blocks i n n. Since !.I > 0 we have p = ax 5 A. I n view o f 1.2.8, t h i s implies p = x 2 2. Therefore, by 1.2.11, a > 2 and 1.2.8 y i e l d s n = (po).

Moreover, Ilp 11 i s t r i v i a l . (Foh: The t r a n s i t i v i t y o f n implies i n p a r t i c u l a r there are !.I = h blocks o f n = (p ); t h a t through'any p o i n t p w i t h p # p i n other words: po i s connected w i t h 8ny other p o i n t by e x a c t l y h blocks.7 L e t X be a block outside n. Then po 1 X and t h e p a i r (po,X) has type (0,O). Hence 1x1 ] p o l . f o r any block X I n.

A. Beutelspacher

96 By 1.2.13, i t follows

1x1

I B I . Since

>

M = Ip

0

s i n c e n+c i t holds

I

=

1x1

ri =

i s complete, we have

(p,)

ICI= n t c

-

Next we claim:

Therefore

C which i s p a r a l l e l t o

B

= ( ICl-l)(A-l),

(M-l)(A-2) = ( l C l - l ) ( u - l )

(5)

1x1.

n+c,

i s t h e maximal b l o c k degree. For any b l o c k

so

y(X) =

ntc- 1 X l '

(6) The p o i n t p U & m

S id non-,Oc.Lvid. O t h w L b e , by 2.1, t h e r e a r e two p o s s i b i l i t i e s . I f S i s a symmetric design, then S c o n t a i n s no n o n - t r i v i a l p a r a l l F l c l a s s a t a l l . I n t h e second case, S i s a punctured symmetric design: S = S -{PI. I n t h i s Situation, the only non-trivial Jhrough p. This p a r a l l e l c l a s s p a r a l l e l class o f S i s the set o f blocks o f S cannot be s i n g u l a r , s i n c e otherwise any b l o c k o f S through p would a l s o cont a i n t h e s i n g u l a r p o i n t po: a c o n t r a d i c t i o n . Denote by

p

06

a p o i n t w i t h a n o n - t r i v i a l p a r a l l e l class.

( 7 ) T h a e exAa2 a b l o c k B ' Rhhough, po duch t h a t B' COW a point pa..taUct Xo p, buR id not incident wcth p. Adbume t h a t t h e r e i s no such b l o c k B ' . For any p o i n t p ' p a r a l l e l t o p i t holds Ipo,p'I = A = Ipo,pI; t h e r e f o r e , by o u r assumption, f o r any p o i n t p ' p a r a l l e l t o p we have (PoYP') = (PoYP). Hence any o f t h e A 3 2 blocks through p and p would c o n t a i n a l l p o i n t s p a r a l l e l t o p. This i s a c o n t r a d i c t i o n , s h c e , as a n o n - t r i v i a l p a r a l l e l c l a s s , 11pB c o n t a i n s a t l e a s t x + l p o i n t s . From ( 7 ) we i n f e r i n p a r t i c u l a r n+c-1 = l ( B ' ) - { p o I l a v = C A . I n view o f I B I < n t c and 1.2.3, we g e t f o r any b l o c k

(8)

ICI

or

E

{n,n+a = ntl,n+c-a

=

n+c-l,n+2c-a

C

of

n:

n+2c-1},

I C I = n-a = n-1.

I n t h e f i r s t f o u r cases we have the following contradiction: c(h-I) a ntc-1

>,

ICI

>I

n, s i n c e

c

2

1. So, by ( 5 ) and (8) we g e t

cx.

So, we have o n l y t o c o n s i d e r t h e case I C I = n-1. F i r s t we c l a i m ( 9 ) Any point p d i 6 f i n c t dhom po has deghee n. Foh: Denote by C a b l o c k o f n which does n o t c o n t a i n p. Since through p t h e r e a r e u blocks o f ri, t h e p a i r (p,C) i s o f t y p e (u,O) o r (u,v). Therefore I p I = ICl+a = n - 1 + 1 = n, o r I p I = I C l t a - c = n-c. Assume t h a t p has degree n-c and consider a b l o c k X which c o n t a i n s n e i t h e r p n o r po. By 1.2.2, f o r t h e degree o f X i t h o l d s

1x1

{n-c,n-c-a,n,n-a}: a c o n t r a d i c t i o n t o 1x1 = n t c . We remark t h a t

E

(p,C)

i s of type

( ~ ~ 0 ) .

97

Finite semi-symmetric designs

F i n a l l y we show (10) The p o i n t p a h & U m

06

S

i~ . t t L i v i d .

Namdy: We prove t h a t t h e p a r a l l e l c l a s s o f any p o i n t p i s t r i v i a l . I n o r d e r t o do t h i s , we can suppose p # po. I f C i s a b l o c k through p n o t c o n t a i n i n g p, then t h e r e e x i s t s no p o i n t p a r a l l e l t o p on C. (Note'that (p,C) has type (u,O).) Denote by C a b l o c k through po and p. I n view o f 1.2.6 we g e t n ) ( x - 1 ) = 0; @(p,C) = a(p,C) t ( [ C l - I p l ) ( i - l ) = A-1t ( n - 1

-

hence p has no p a r a l l e l d i s t i n c t from p on C. Since any p o i n t i s on a b l o c k through po, a s s e r t i o n (10) i s proved. The c o n t r a d i c t i o n between ( 6 ) and (10) proves our The0rem.J CHAPTER 3. SEMI-SYMMETRIC DESIGNS OF CLASS ( I ) The aim o f t h i s chapter i s t o prove t h e f o l l o w i n g r e s u l t . THEOREM B. Denote by S (p,B,I) a f i n i t e semi-symmetric design o f c l a s s ( I ) w i t h A > 1. I f S c o n t a i n s a n o n - t r i v i a l t r a n s i t i v e p a r a l l e l c l a s s , then there, e x i s t s a symmetrJc design St and a n o n - i n c i d e n t p o i n t - b l o c k p a i r (p,B) o f S such t h a t S = S -U holds, where U i s a s u i t a b l e subset o f Ip,BI. Since S i s i n c l a s s ( I ) , we have u = ah and v = CX. We suppose always x > 1. Using Theorem A ( o r 2.2.2, r e s p e c t i v e l y ) , we can assume t h a t n e i t h e r t h e p o i n t p a r a l l e l i s m nor t h e b l o c k p a r a l l e l i s m i s t r i v i a l . I n p a r t i c u l a r , a and c a r e p o s i t i v e integers. L e t n denote a n o n - t r i v i a l t r a n s i t i v e p a r a l l e l c l a s s ; w i t h o u t l o s s o f g e n e r a l i t y , we suppose t h a t n i s a b l o c k p a r a l l e l c l a s s , say n = IIBII. I n view o f 1.2.8 and 2.2.3, n i s r e g u l a r . By M we denote t h e number o f b l o c k s i n n .

I n t h e f i r s t s e c t i o n we show t h a t t h e assumptions o f Theorem B i m p l y a = c. I n t h e second s e c t i o n f i r s t a = c = 1 i s shown; a f t e r t h i s , t h e p r o o f o f Theorem B i s no longer d i f f i c u l t .

5

1. NON-EXISTENCE THEOREMS

F i r s t , we s t a t e two Lemmas, which w i l l be used q u i t e a few times.

1.1 LEMMA. Suppose t h a t t h e r e e x i s t s a p o i n t Then a = 1.

p

in

S with

IpI = IBlta.

i s r e u l a r , t h e r e i s a b l o c k C o f n , which i s n o t i n c i d e n t = ICQta i m p l i e s t h a t (p,C) i s o f t y p e ( ~ ~ 0 )This . means: The p o i n t p i s i n p ( n ) ; furthermore: p has no p a r a l l e l on a b l o c k o f n n o t passing through p. On t h e o t h e r hand, c o n s i d e r a b l o c k C ' o f n through p. By 1.2.6 t h e r e a r e exactly

Proof. Since

with p. Now

n

/PI

B ( ~ , C ' ) = a(p,C') + ( l C ' I - I P I ) ( x - l ) = u-1 + ( l B I - I P I ) ( i - l ) = ax-1

points

p'

Now assume blocks o f

on

n

a

-

C'

a(x-1) = a-1 which a r e p a r a l l e l , b u t d i s t i n c t t o

1. Then t h e r e e x i s t s a p a r a l l e l p ' through p ' must a l s o c o n t a i n p:

>

ax = u ,< I p ' , p I

= A-1: a c o n t r a d i c t i 0 n . J

p p

p. i n p ( n ) . Then a l l t h e

98

A. Beutekpacher

1.2 LEMMA. Denote by X a block o f degree Suppose IpI = n or IpI = nta

and

/PI

IpJ = IBlta or Then \B\ < n

-

c(x-1)

5

161

and

ntc

and by

p a point outside

X.

a # c.

n-c.

Proof. The f i r s t assumption implies t h a t p has e x a c t l y v = C A p a r a l l e l s on T h e e c o n d assumption says t h a t no p o i n t o f Ilpll-{p) i s a p o i n t o f p ( n ) . I n view o f 1.2.13 we get together n+c = 1x1 5 y t v = y t cx > 161 t cX,

X.

i. e. (61 < n

-

c(X-l).J

By 1.2.3 we have the f o l l o w i n g p o s s i b i l i t i e s f o r 161

E

161:

Intc,n+a,n-a,n,ntc-a,nt2c-a}.

Using 1.2.15 we can exclude t h e p o s s i b i l i t y 161 = n t c . I n t h i s s e c t i o n we s h a l l prove: I f we suppose a # c, then a l s o the other f i v e cases cannot occur. We s h a l l see t h a t i t i s useful t o handle f i r s t t h e cases a # 2c and c # 2a. So, i n t h i s section, we suppose always a # c; moreover, f o r t h e next f i v e Propos i t i o n s , we suppose a d d i t i o n a l l y a # 2c and c # 2a.

ii (1)

we denote a block o f maximal degree '?t holds

(PI

E

n t c . Then f o r any p o i n t

{ntc,n+cta,n,nta}.

1.3 PROPOSITION. 161 # nta. Proof. Assume, B has degree

n+a.

Since n i s regular, f o r any p o i n t p o f S t h e r e i s a block containing p. Therefore, f o r the degree o f p we have

I pI

moreover, p

p outside

C

of

n

not

{nta,nt2a,nta-c,nt2a-c}; i s on a b l o c k o f n i f and only i f E

I p I E {n+2aynt2a-c1. So, by ( 1 ) and the general assumption on a and c, i t f o l l o w s I p I = n t a f o r any p o i n t p outside B This means t h a t through no p o i n t o u t s i d e B there i s a block of n. I n p a r t i c f l l a r , (6) i s a subset o f (Bo): a c o n t r a d i c t i g n t o 1.2.5.J

.

1.4 PROPOSITION. I B I # n-a. Proof. Assume 161 = n-a. Then f o r any p o i n t

I p I E {n-a,n,n-a-c,n-c}. By ( l ) , any p o i n t o u t s i d e Bo i s an 1.2 i t f o l l o w s n-1 = n-a = I B I < n-c, hence c < 1: a c o n t r a d i c t i o n . 4

p of

S

i t holds

n-point. Then 1.1 i m p l i e s

a = 1 and by

1.5 PROPOSITION. J B J # n. Proof. Again we assume t h a t our a s s e r t i o n i s f a l s e . Then f o r any p o i n t

p

of

S

99

Finite semi-symmetric designs we have and

p

I p I E tn,nta,n-c,nta-cl, i s on an element o f x i f and o n l y if I p I E {n+a,nta-c}.

By ( I ) , any p o i n t o u t s i d e B has degree n o r n t a . ( U t h w A e . any p o i n t o f f There i s a p o i n t o f degree Onta o u t s i d e 5., an n-point, which would i m p l y p ( n ) c _ (Bo) . ) Now 1.2 y i e l d s n = 151 < n-c.J

Bo

were

The p r o o f o f t h e n e x t P r o p o s i t i o n i s n o t so s h o r t . 1.6 PROPOSITION. I B I # ntc-a. Proof. We show t h a t t h e assumption 181 = n t c - a y i e l d s a c o n t r a d i c t i o n . Under t h i s a s s u m p t i o n , f o r any p o i n t p o f S we have [ p i E Intc-a,ntc,n-a,nl, and

p

i s i n p(n) IpI

E

i f and o n l y i f

Intc,nl.

By ( I ) , f o r any p o i n t

p

o u t s i d e an

(n+c)-block we know

1pI E In+c,nl. We w i l l g e t a c o n t r a d i c t i o n i n several steps. step 1 . a # 1. Asbwne a = 1. Then p = ah = A, hence or M = ntc. According t o 1.2.12, t h i s y i e l d s y = n+c, Since n t c i s t h e maximal b l o c k degree, any b l o c k o u t s i d e II has degree n+c; moreover, p(n) = p. We c l a i m t h a t t h e r e i s an n - p o i n t p. (16 not, any p o i n t would have degree n t c . Therefore, any n o n - i n c i d e n t p o i n t - b l o c k p a i r would have type (0,O) o r (p,O). I n p a r t i c u l a r , t h e p o i n t p a r a l l e l i s m were t r i v i a l : a c o n t r a d i c t i o n t o t h e general assumptions o f t h i s chapter.) This p o i n t p has e x a c t l y v p a r a l l e l s on any b l o c k o f x which i s n o t i n c i d e n t w i t h p . Now consider a b l o c k C o f II through p. Since a = 1, we g e t B(p,C) = .(p,C) + ( l C l - l p l ) ( ~ - 1 ) = p-1 t (c-a)(x-l) = C(A-1). L e t P be t h e p a r a l l e l c l a s s o f p and denote by m t h e number o f p o i n t s i n P . Counting t h e incidences ( q , C ) , where q i s p a r a l l e l t o p and C i s i n x, we get mh = mp = u ( c ( A - l ) t l ) + (M-p)v = h ( c ( h - l ) + l ) + (n+c-x)cx = c h ( x - 1 t n+c-h) + A, (2) in-1 = c ( n + c - I ) . I f X denotes an (n+c)-block, then f o r any o i n t x o f f X, t h e p a i r (x,X) i s o f type (0,O) o r (0,v). I n p a r t i c u l a r , IPXII i s t r i v i a l . Hence f o r any (n+c)b l o c k X through p i t holds

) 6 ( p y x ) = a(p,X) t ( ~ x ~ - ~ p ~ ) =( h0 - t~ C(h-1). Therefore, on any b l o c k through p t h e r e a r e e x a c t l y ~ ( h - 1 ) p o i n t s o f

a-tpl.

100

A. Beutelspacher

Hence

(m-l)(x-l) = nc(x-1), i. e. m-1 = nc. Together w i t h ( 2 ) we have c ( c - 1 ) = 0, hence c = 1 = a c o n t r a d i c t i n g a # c. The f i r s t s t e p i m p l i e s t o g e t h e r w i t h 1.1 t h a t t h e r e does n o t e x i s t an ( n t c ) - p o i n t i n S. I n p a r t i c u l a r , any p o i n t p o u t s i d e an ( n t c ) - b l o c k X has degree n. Therefore ( p y x ) i s o f t y p e (0,v) and t h i s i m p l i e s t h a t t h e p a r a l l e l c l a s s of X i s t r i v i a l . Using t h i s o b s e r v a t i o n we can prove S t e p 2. Any p o i n t which i s n o t i n p(n) has degree n-a. O t h m A e we would have a p o i n t q o f degree n t c - a . This p o i n t a p a r a l l e l i n s i d e p ( n ) . On t h e o t h e r hand, q i s on Bo and 8(qyBo) = a(q,Bo) + ( \ B o l - l q \ ) ( x - l ) = 0

t

q

cannot have

a(h-1).

Therefore n t c = lBol 8 y t 8(q,Bo)tl > ntc-a t a ( x - 1 ) t l : a contradiction. By A we denote t h e s e t o f p o i n t s o f S which a r e n o t i n p ( n ) .

S t e p 3. Any b l o c k Namely: The

X

outside

(n-a)-point

q

II c o n t a i n s any p o i n t q o f A . (n+c)-block

cannot be o u t s i d e t h e

X.

c 1. UZhthehwine A would c o n t a i n two d i s t i n c t p o i n t s , Say q and q ' . By t h e preceeding step, any b l o c k through q would a l s o c o n t a i n q : a c o n t r a d i c t i o n t o 1.2.5(b).

S t e p 4. ( A /

Step 5. A # p. Fob: I f A = 8 , then p ( n ) = p and any p o i n t i s an n - p o i n t . (The remark a f t e r Step 1 says t h a t t h e r e i s no (n+c)-point.) Since t h e r e a r e b l o c k s o f degree ntc-a (namely t h e b l o c k s i n n ) and s i n c e t h e r e e x i s t s a t l e a s t one (n+c)-block (namely Bo), i t f o l l o w s M-1

t

(ntc-a)(n-1) = (b-l)A = ( n t c ) ( n - l ) ,

hence M - 1 = a(n-1). On t h e o t h e r hand, we know (M-l)(x-1)

= (n+c-a) a x - 1 ) .

Together we g e t a(n-l)(A-l)

= (ntc-a (ah-l),

so 0 = (n-ax)(a-1)

c ( a x - I ) > 0, s i n c e through any p o i n t t h e r e a r e a t l e a s t t h e t

p =

ax

blocks o f

n.

S t e p 6. \ A \ # 1. Oththehwine, y = n t c - I . Then (A-1)yax = ( ~ - 1 ) y p = (A-1)Mx = ( A - 1

t

(ntc-a)(u-I))A,

hence ( x - l ) ( n + c - l ) a = A-1 + (n+c-a (ah-1) = A-I + (ntc-a)(ax-a) so

(x-l)a(a-l) = (a-l)(ntc-a) which i m p l i e s f i n a l l y

t

A-1,

t

(a-l)(n+c-a),

Finite semi-symmetric designs

101

(a-l)(ntc-ah) = -(A-1): a contradiction, since n+c + ah. Since Steps 5 and 6 contradict Step 4, our Proposition is proved./ 1.7 PROPOSITION. IBI # nt2c-a. Proof. We assume lBl = n+2c-a. Then for any point p it holds: IpI E ~nt2c-a,n+2c,ntc-a,ntc~; and p is incident with an element of n if and only if IpI E Int2c,ntcl. By (l), for any point p outside B it holds IpI = ntc. Therefore, for any is of type (0,O); in particular, the point p with p 1 Boy the pair (p:Bo) parallel class o f Bo is trivial. We claim: There is no point of degree nt2c in S. (O,thmLhe, by 1.1, w e would get a = 1. But nt2c-a IB[ < ntc implies c < a = 1: a contradiction.) In particular, any point inside p ( n ) has degree ntc. NOW, denote by p a point in p(n) on Bo. Since p is an (ntc)-point, w e have B(P,B,) = a(pyBo) t (lBol-l~l)(h-l) = 0. On the other hand, for any block C in II with p 1 C, the pair (p,C) is o f type ( p , v ) . So, there is a parallel p' to p outside Bo. But Ip'I = n+c = lBol implies that (p,Bo) is of type (0,O). This contradiction proves the Proposition.4

It remains t o show that the cases a

=

2c and c

=

2a cannot occur.

1.8 PROPOSITION. a # 2c.

Proof. We assume on the contrary a

2c. Then, in particular, a > 1. For any non-incident point-block pair (x,X) of S it holds: (x,X) is of type (0,O) if and only if 1x1 = 1x1; (x,X) is of type ( p , O ) if and only if l x l - 2 ~= 1x1; (x,X) is of type (0,v) if and only if 1x1 = 1x1-c; (x,X) is of type ( p , w ) if and only if 1xI-c = 1x1. For any point p not on Bo we have IpI E {ntc,nt3c,n,nt2cl. Therefore, for any block X of S it holds 1x1 E {ntc,n,n-c,n-Zcl. As usual, denote by B a block of II. We distinguish three cases. Case 1 . lBl = n. In this case, by 1.2.13, any block off II has degree ntc. Since ( B ( = n, any point of S has the degree Ip1 E {n,nt2c,n-c,ntc}, and p is on an element o f n if and only i f IpI E {nt2cYn+c). If X denotes an (n+c)-block, then for any point p with p 1 X w e have =

102

A. Beutelspacher

IpI

E

{ntc,nt3c,n,nt2c~,

hence IpI E {ntc,n,ntZcl. Since a # 1, 1.1 implies that there is no point of degree nt2c. In view of 1.2, outside an (ntc)-block, there is no n-point. Therefore any point off Bo has degree ntc; in particular, the parallel class of Bo is trivial. NOW, let us denote by q a point on B which is in p ( n ) . Then q has degree ntc; therefore (since the parallel clasg of Bo i s trivial), q has no parallel distinct from q on Bo. On the other hand, the parallel class of q is non-trivial; therefore there is a parallel q' to q outside B But this point q' has degree ntc; so, q cannot have a parallel on the (ntc)-b?ock Bo. This contradiction shows that the first case cannot occur. The second case is the most difficult one. Case 2. I B I = n-c. Now any point p has degree n-c,ntc,n-Zc, or n; moreover, p is inside p ( n ) if and only if IpI E {ntc,nl. For any point p outside an (ntc)-block X we have therefore IpI E {ntc,nl. In particular, the parallel class o f X is trivial and any point outside X is on an element of n. By 1.1, there does not exist an (ntc)-point. Assume that there is a point q of degree n-c. Then q is on B o y has no parallel in p ( n ) , but B(qyBo) a(qyBo) t (lBol-lql)(A-l) = 0 t 2c(x-1) parallels distinct from q on Bo. Consequently, ntc = l B o l % y + B(p,Bo)tl > n-c + 2c(~-l)+l: a contradiction. Thus, any point of S has degree n-2c or n. By A we denote the set of points which are incident with no element of n . We claim: Any block not in n contains any point of A. ( N a m ~ q :Any block X outside n has degree n or n+c; any point q in A has degree n-2c. So, q I X is impossible.) This implies 1A1 d 1; moreover: any block outside n has the same degree (namely ntc). Consider first the case A = 0. In this situation, any point is an n-point. Since the parallel class of Bo is trivial, we have (n+c)(n-1) = (b-1)x. Looking at the block B y we see on the other hand (n-c)(n-1) = (b-1)x - (M-l), so M-1 = 2c(n-1). Therefore, from (M-l)(A-l) = (n-c)(p-1) = (n-c)(2cx-l) we get 2c (n-1 ) (A- 1) = (n-c)(2cx- 1) ,

.

Finite semi-symmetric designs

103

hence

(2c-l)n = ~C'A - ~ C tA c. Furthermore, A = C implies y = ntc. Hence (~-.l)y2cx = (~-1)yu (A-1)Mx = (A -1t (n-c)(u-l))A, 2c(ntc)(x-l) = A-1 t (n-c)(2c~-l), or (2c-l)n = 4c2 A - A - 2c2 c t 1. Together we get 4c2 A A - 2c2 - c t 1 = 2c 2 A - 2CA t c, or 2c(A-l)(ctl) = A-1. Using A-1 # 0 we get the contradiction 2c(ctl) = 1. Therefore we have 1 ~ =1 1, i. e. y = ntc-1. This means (ntc-1)2c(x-l) = A-1 t (n-c)(2cx-l), (2cx-c)(2c-l) = (2c-l)n t A-1 > n(2c-1), hence n < ~ C A c. But through any point p outside A there are at least the u = ~ C Ablocks of n ; i. e. n = IpI a ~ C A . With this contradiction, the second case is finished. Case 3 . 1131 = n-2c. In this situation, for any point p of S we have / P I E {n-2c,n,n-3c,n-c~. Therefore, any point outside Bo is an n-point. But 1.1 says that there is no point of degree n. Thus our Proposition is proved.4

-

-

-

1.9 PROPOSITION. c # 2a. Proof. We show that the assumption c = 2a yields a contradiction. Under this assumption, for any non-incident point-block pair (x,X) of S it holds (x,X) is of type (0,O) i f and only if 1x1 = ( X I ; (x,X) is of type (p,O) if and only if 1x1-a = 1x1; (x,X) is of type (0,~) if and only if 1x1 = IXI-2a; (x,X) is of type ( u , v ) if and only if 1x1 = 1x1-a. For any block X o f degree ntc = n+2a (so, in particular, for Bo) and any point p outside X it holds (3) IpI E {nt2a,nt3a,n,nta}. Therefore, for any block Y of S we have I Y I E {nt2a,ntayn,n-a}. Thus our block B in n has degree nta, n, or n-a. By A we denote the set of all points of S outside p ( n ) . Case 1. I B I = nta. In this case, for any point p of S it holds J p JE Inta,nt2ayn-a,n~,

A. Beutelspacher

104

and

p

n

i s i n c i d e n t with an element o f

I p I E {ntZa,nl. By ( 3 ) , any p o i n t p o u t s i d e an

i f and o n l y i f

(nt2a)-block has degree

I p I E {n+a,nt2a,n}. I n view o f 1.2, such a p o i n t cannot have the degree Any block X which i s n o t i n il the (n+2a)-block X has degree p ( n ) . ) This implies 5 1.

We claim: A = fl. ( k h u m e on t h e c o W y we get (nt2a)(a-l) = (a-l)ah

-

Ibl =

p of

S

1. Then

A . (Any p o i n t

p

y =

p outside i s contained i n

n+Za-l. Using

(A-1) < (a-1)aX.

x

This implies immediately a # 1 (otherwise n+2a < ax. But through any p o i n t

nta.

contains a l l p o i n t s o f n o r nt2a; t h e r e f o r e

outside

A

= 1). Therefore

t h e r e are a t l e a s t

u elements o f

I[:

n+2a 3 I p I a v = ax.) The f a c t A = 0 y i e l d s t h a t any p o i n t o f S has degree n o r n+2a. Now we show t h a t any p o i n t i s an n-point. ( I f any p o i n t were o f degree n+2a, then t h e p o i n t p a r a l l e l i s m were t r i v i a l . Therefore we can assume t h a t t h e r e e x i s t p o i n t s o f degree n and p o i n t s o f degree n+2a. F i r s t o f a l l , 1.1 i m p l i e s a = 1. Using ( 4 ) we get y = nt2; i n view o f yu = MA we have M = nt2. Any ( n t 2 ) - p o i n t has t r i v i a l p a r a l l e l class, w h i l e any n-point p has e x a c t l y v p a r a l l e l s on any block n o t through p. I f C denotes a block o f n through p, we have 6(p,C) = a(p,C) + ( n + l n)(x-1) = 2 ( ~ - 1 ) .

-

I f we denote by m t h e number o f p a r a l l e l s t o

For any

p, i t f o l l o w s

u(2(X-1) + 1) t ( M - ~ ) v = mu, o r m = 2n+3. (nt2)-block X through p i t holds B(P,X) = a(P,X) + ( l X \ - l P l ) ( A - l ) = 2(X-1).

Thus i. e.

(m-l)(x-1) = lp12(x-1) = 2n(x-1), m = 2 n t l : a c o n t r a d i c t i o n . So, S contains o n l y p o i n t s o f degree

Since the p a r a l l e l class o f

Bo i s t r i v i a l , we have (n+Za)(n-l) = (b-1)X. On the other hand, (n+a)(n-1) = (b-l)X (M-1),

-

therefore

M-1 = a(n-1). Using

(M-l)(A-1)

= (n+a)(u-1)

we get t h e f o l l o w i n g c o n t r a d i c t i o n : (nta)(ax-1) = a ( n - l ) ( X - l ) = ( n - l ) ( a x - a ) < (n+a)(ax-l). Thus, t h e f i r s t case i s f i n i s h e d . Case 2. l B l = n. I n t h i s case, f o r any p o i n t p o f

S

i t holds

n.)

105

Finite semi-symmetric designs IpI

E

{n,nta,n-Za,n-al. p outside

So, by (3), f o r any p o i n t

Bo

we have

I p I E In,nta1. Then 1.2 implies immediately n = IBI

Casc 3.

IBI

<

n-c.

= n-a.

For any p o i n t IpI

p of

E

we would have

S

{n-a,nYn-3a,n-2a}.

So, any p o i n t outside

Bo

were an

n-point. By 1.2 we would have

n-a = I B I < n-c, therefore

2a = c < a: a c o n t r a d i c t i o n . 4

We summarize the r e s u l t s o f t h i s section. 1.10 THEOREM. Denote by S a f i n i t e semi-symmetric design o f class ( I ) having type ( p , v ) w i t h x > 1. Suppose t h a t n e i t h e r t h e p o i n t p a r a l l e l i s m nor the block p a r a l l e l i s m i s t r i v i a l and suppose moreover t h a t S i s n o t a symmetric design. I f S contains a n o n - t r i v i a l t r a n s i t i v e p a r a l l e l class, then p = v.4

5 2.

SEMI-SYMMETRIC DESIGNS OF TYPE

(u,p)

Denote by S = (p,B,I) a f i n i t e semi-symmetric design o f class ( I ) having type ( v , ~ ) with > 1, which i s n o t a symmetric design. Suppose t h a t n e i t h e r the p o i n t p a r a l l e l i s m nor t h e block p a r a l l e l i s m i s t r i v i a l . Moreover, we suppose t h a t S contains a n o n - t r i v i a l t r a n s i t i v e block p a r a l l e l class n = IIBII. The r e s u l t s o f t h e preceeding s e c t i o n imply under these assumptions ax = = w =' cx, hence a = c. By B we denote a block o f maximal degree nta, and by A t h e s e t o f a l l p o i n t s o f So which are n o t i n c i d e n t w i t h an element o f II. 2.1 THEOREM. I t holds

a = c = 1. Moreover, 161 = n and A = 0.

Proof. For a l l p o i n t s p outside an p o i n t s o f f Bn) - i t holds (1)

IpI

E

X

(so, i n particular, f o r the

{ntaynt2a,n}.

Therefore, f o r any block

IYI

(nta)-block

Y

of

S we have

E {nta,n,n-a). l B l # nta. Thus, we have t o d i s t i n g u i s h two cases. Case 1 . I B I = n. We have t o show: a = 1 and A = 0. I n t h i s case, f o r any p o i n t p o f S i t holds

By 1.2.15,

/ P I E {n,n+a,n-al, and any block outside n has degree (nta)-block we know

nta. By ( l ) , f o r any p o i n t

p o u t s i d e an

I p I E Inta,n}. We claim: Any p o i n t p o u t s i d e an (nta)-block X i s i n ~ ( I I ) . (O,i%mAe, p were an n-point and (p,C) were o f type (0,O) f o r any block p a r a l l e l t o B. Since ( p y x ) i s o f type (O,V), we get n t a = 1x1 >/ y + w > n t ax, a c o n t r a d i c t i o n . )

C

106

A . Beutelspacher

Consequently, any block outside n i s i n c i d e n t w i t h any p o i n t o f 1 A 1 Q 1. I n t h e case 1 ~ =1 1 (i.e. y = n t a - l ) , we get from

-

A; i n particular,

A-la;xli;A-l)

the equation ( n t a - l ) a ( x - l ) = A-1 t n(ax-1). Now

h

# 1 implies a # 1. Hence 2 n(a-1) = ( 1 - l ) ( a -a-1)

and t h e r e f o r e n u = ax blocks:

<

nta a IpI

(x-1)a. 3 11 =

This c o n t r a d i c t i o n shows

<

(x-l)a(a-l)

But t h e r e are p o i n t s

p

of

incident with a t least

S

ah. A =

0.

Now we c l a i m t h a t there e x i s t s a p o i n t o f degree n t a i n S. (&dune t h a t any p o i n t o f S i s an n-point. Then t h e p a r a l l e l class o f Bo i s t r i v i a l and so (nta)(n-1) = (b-1)x. Considering

B we get on the other hand

n(n-1) = ( b - l ) h

-

(M-1),

hence M-1 = a(n-1).

But, using (M-l)(x-1) = n(u-1) = n(ax-1) we get a(n-l)(A-I) = n(ax-i), which y i e l d s a c o n t r a d i c t i o n . ) But the existence o f an a = 1.

( n t a ) - p o i n t implies together w i t h 3.1.1 t h e a s s e r t i o n

Case 2. IBI = n-a. Our Theorem i s proved, i f we have shown t h a t t h i s case cannot occur. I n t h i s s i t u a t i o n , f o r any p o i n t p o f S i t holds IpI

E

{n-ayn,n-2al.

So, by ( l ) , any p o i n t o u t s i d e

Bo n-a = I B I < n-c = n-a:

has degree

n. I n view o f 3.1.2,

t h i s implies

a contradiction.4 By 2.1, we can suppose henceforth o f Theorem B.

a = 1. The n e x t Lemma i s c r u c i a l f o r t h e p r o o f

2.2 LEMMA. The o n l y n o n - t r i v i a l block p a r a l l e l c l a s s o f

S

is

n.

Proof. By 1.2.10 and 2.1, any block o f n 'has degree n. Therefore, by 1.2.13, -lock outside n i s ' a n (nt1)-block. I t remains t o show t h a t no two d i s t i n c t blocks o f degree n t l are p a r a l l e l . Assume t h a t t h e r e are two d i s t i n c t p a r a l l e l (nt1)-blocks, say X and Y. Then any p o i n t p on X which i s n o t i n c i d e n t w i t h Y has degree n t l . Such a p o i n t p has no p a r a l l e l d i s t i n c t from p i n p ( n ) , b u t v = h p a r a l l e l s on Y. Consequently, n t l = I Y I 3 y t v > n t x : a c o n t r a d i c t i o n . 4

107

Finite semi-symmetric designs

2.3 LEMMA. There e x i s t s a s e m i - s y m e t r i c design St and a p o i n t p o f St w i t h Moreover, t h e block p a r a l l e l i s m and t h e p a r a l l e l c l a s s %he property S = S -{PI. o f the p o i n t p are t r i v i a l . Proof. By 2.2, there i s a Ynique,no$-t$ivial p a r a l l e l class (p ,8 ,I ) as f o l l o w s thelncidence structure S

pt :=

d" I n ) ,

The incidence r e l a t i o n p It B n It B

n

of

S. We d e f i n e

t

8 := 8. It w i l l be defined i n the f o l l o w i n g way:

:*

p IB

:++

B

E

n

f o r p E p and B f o r a block B o f

E

8;

S.

By 2.1, A = b; so, through any p q i n t o f S there are e x a c t l y x bloqks o f n . I n other words: The p o i n t +n o f S i s j o i n e d t o any other p o i n t o f S by e x a c t l y A blocks. Therefore S i s a semi-symmetric design and the p a r a l l e l class o f the p o i n t n i s t r i v i a l . By construction, any two d i s t i n c t blocks i n t e r s e c t i n exactl y A p o i n t s ; hence t h e block p a r a l l e l i s m i s t r i v i a l .

Also, by construction, S = St-{nl.J Now the p r o o f o f Theorem B f o l l o w s mainly by quotations. By Theorem A ( o r 2.2.2, r e s p e c t i v e l y ) we can suppose t h a t n e i t h e r the p o i n t p a r a l l e l i s m nor the block p a r a l l e l i s m i s t r i v i a l . Moreover, by 1.2.14, n # 8. Under these assumptions, 3.1.10, 2.1 andt2.3 g i v e us t Q e existence o f a semi-symmetric design S and a point p o f S with S = S -{pl. Since, by 2.3, the block p a r p i l e l i s m o f St i s t r i v i p i , i n view,of 2+$.2, t h e r e e$istst$ symmetric design S and a block B o f S with, S = S -{B) o r S = S But i n t h e l a t t e r case, t h e p o i n t p a r a l l e l i s m o f S would have been t r i v i a l as w e l l . Therefore S S+-{pl Stt-{p,B}.

.

Because the p a r a l l e l class o f the p o i n t w i t h the b l o c k B.

p

in

St i s t r i v i a l , p i s n o t i n c i d e n t

Thus the proof o f Theorem B i s f i n i s h e d . 4 I n [2] we have constructed examples o f incidence s t r u c t u r e s showing t h a t t h e assumptions o f Theorem B cannot be weakened very much. CHAPTER 4. SEMI-SYMMETRIC DESIGNS OF CLASS (11)

I n t h i s l a s t chapter we s h a l l prove t h e f o l l o w i n g THEOREM C. Denote by S a f i n i t e semi-symnetric design o f class (11) having the t y p e v ) . (a) If S contains a n o n - t r i v i a l t r a n s i t i v e block p a r a l l e l c l a s s and i f u >, min{v,x-l), then u = v 6 A-1. (b) Suppose u = A - 1 > 1 (or, v = A - 1 > 1). I f the block parallelism3is2transi$ive and S contaigs a complete p a r a l l e l class, then+there e x i s t s p 2-(m t m tmt1,m t m t l , m t l ) design S and a p r o j e c t i v e l i n e L o f S w i t h S = S +L. REMARK. Note t h a t the conclusion of (a) holds i n p a r t i c u l a r , i f S contains a non1block p a r a l l e l class, which i s t r a n s i t i v e , o r contains a n o n - t r i v i a l t r a n s i t i v e p o i n t p a r a l l e l class.

be a f i n i t e semi-symmetric design o f type I n what follows, l e t S = (p,B,I) ( u , v ) i n class (11) which i s n o t a symmetric design. Since S i s n o t i n class ( I ) , A # 1. The i n t e g e r a i s defined by U-v = aA. Without l o s s o f g e n e r a l i t y , a i s

108

A . Beutelspacher

non-negative.

( I f not, consider t h e dual s t r u c t u r e o f

S.)

I n t h e f i r s t s e c t i o n we show: The e x i s t e n c e o f a n o n - t r i v i a l t r a n s i t i v e b l o c k p a r a l l e l c l a s s i m p l i e s p = u s A-1 o r u < 1-2.

5

1. PARAMETERS

I n o u r f i r s t Theorem, we do n o t assume t h a t t h e r e e x i s t s a t r a n s i t i v e p a r a l l e l class. 1.1 THEOREM. There e x i s t s a p o s i t i v e i n t e g e r n such t h a t any p o i n t o f degree n t a and any b l o c k o f S i s an n - b l o c k .

P r o o f . I f a = 0, t h e n any n o n - i n c i d e n t p o i n t - b l o c k p a i r has t y p e m; hence any two elements o f S have t h e same degree, say n. assertion i s true.

has

S

(0,O) o r Thus o u r

From now on, we suppose a b 1. Then any n o n - i n c i d e n t p o i n t - b l o c k p a i r S has t y p e (0,O) o r ( p , ~ ) ;moreover:

(p,B)

of

i s o f t y p e (0,O) i f and o n l y i f I p I = I B I ; (p,B) i s o f t y p e ( p , v ) i f and o n l y i f I p I = I B l t a . T h i s means: I f p i s n o t i n c i d e n t w i t h B y t h e n I p l - l B l E {O,a}; i n p a r t i c u l a r IPI 2 I B I . Denote by n t a t h e maximal p o i n t degree; l e t po be a p o i n t o f degree n t a . Then f o r any b l o c k C n o t t h r o u g h po we have (p,B)

ICI E i n t a , n } ; t h e r e f o r e , f o r t h e degree o f any p o i n t

p

/ P I E {nta,n}; f i n a l l y , t h i s i m p l i e s f o r any b l o c k

of

181

E

B

of

S

i t holds

S:

Inta,n,n-a}.

S t e p I . There i s no b l o c k o f degree

n-a

in

S.

k d m e t h a t t h e r e e x i s t s an

( n - a ) - b l o c k , say X. Then any p o i n t p o u t s i d e X has degree n, and f o r such a p o i n t p, t h e p a i r ( p a x ) i s o f t y p e ( p , ~ ) . Moreover, no b l o c k has degree n t a ( s i n c e o u t s i d e any ( n t a ) - b l o c k t h e r e i s an n-point). Using P = aX t v z A t u > A, we g e t t h a t any n o n - t r i v i a l b l o c k p a r a l l e l c l a s s i s r e g u l a r ( c f . 1.2.7 and 1.2.8). I n p a r t i c u l a r , t h e p a r a l l e l c l a s s o f X i s r e g u l a r . Therefore t h e r e e x i s t s a b l o c k X ' p a r a l l e l t o X n o t p a s s i n g t h r o u g h p The p a i r (po,X) has t y p e ( P , v ) . (Foh: There e x i s t s a p a r a l l e l t o X ' t b o u g h po (namely X); so (po,X') cannot have t y p e (O,O).) Hence

.

I X ' I = I p o l - a = n. On t h e o t h e r hand, any p o i n t q on (Throu h q t h e r e i s a p a r a l l e l t o I q l = q x ' I + a = n t a . ) NOW we c l a i m : Any p o i n t on X has degtree n t a

X not incident with X ' X ' ; so ( q , X ' ) has t y p e

i s an

(nta)-point.

(p,~):

.

h d m e t h a t t h e r e i s an n - p o i n t q on X: I f XI' i s a b l o c k p a r a l l e l t o n o t c o n t a i n i n g qo, t h e n . (qo,X") 9 s o f t y p e ( p , ~ ) ;t h e r e f o r e I X " I = Iq,(-a = n-a. I n p a r t i c u l a r , t h e r e i s no ( n t a ) - p o i n t o u t s i d e X I ' . T h i s means

or

(X)-(X,X') 5 (X,X"), (X) = (X,X') u (X,X").

X

but

109

Finite semi-symmetric designs

Consequently ,

n-a = 1x1 < I X , X ' I + I X , X " I (1) On the other hand, po i s a p o i n t o f 0 6 B(P,,X)

= a(p,,X)

Hence

-

-a+

and t h e r e f o r e

X

( l X l - l p o l ) ( x - l ) = a(p0,X)

t

(2) n+a-1 = [ p o l 1 3 a(po,X) Combining (1) and (2) we get 2a(x-l)

= 2(x-1).

>,

1 c n 6 2(x-1)

-

2a(x-1).

2a(x-l).

+

a,

so 2 ( a - 1 ) ( ~ - 2 ) L 1. Thus A 2 o r a = 1. I n the case x = 2, (1) implies [ X I s 2 A-1) = 2 = A: a c o n t r a d i c t i o n . Hence we can suppose A > 2 and a = 1. Denote by A ( o r A') the s e t o f p o i n t s o f degree n t a (or, n, r e s p e c t i v e l y ) on X. L e t n be the p a r a l l e l class o f X. Then the above arguments read as f o l l o w s I f X ' i s a block o f n which does n o t contain a l l p o i n t s o f A, then X ' i s an n-block and i s i n c i d e n t w i t h a l l n-points o f X. Conversely: I f X " i s a block o f Il which i s n o t i n c i d e n t w i t h a l l p o i n t s o f A', then X" i s an (n-a)-block and contains t h e r e f o r e a l l ( n t a ) - p o i n t s . This implies: Any block o f II contains A o r A'. As n i s r e g u l a r , A and A' cannot have a p o i n t i n common. Consequently, both, A and A', c o n t a i n a t l e a s t two points. (Clearly, < A-1. Using t

IAI,IA'~ IA'I = I A ~ A ' ~=

1x1

= n-1 I 2 ( x - l )

-

1

we get

~ A ~ , ~> Ax-2' I~ 1.) I f we denote by p and q ( o r by A' , r e s p e c t i v e l y ) , then

n c (PYq) u (P' 99' 1

p'

and

q')

two d i s t i n c t p o i n t s o f

A

(or,

Y

hence

In1

f

2(A-1)

t

1

I n view o f (2), t h i s i m p l i e s 2(X-1)

,<

.(po,X)

c I n [ - 1 < 2(A-l).

I n other words: Any block o f n passes through any p o i n t on X i s an (nta)-point.

po. This c o n t r a d i c t i o n shows t h a t

This implies t h a t any block Y w i t h Y # X has degree n. Denote by p a p o i n t outside X. Since ( p y x ) has type ( P , v ) , t h e r e e x i s t s a p a r a l l e l p ' t o p on X. Through p ' there i s a block Y which i s n o t i n c i d e n t w i t h p. Since t h e r e i s a p a r a l l e l t o p on Y (namely p ' ) , Y has degree n-a. With t h i s c o n t r a d i c t i o n , the f i r s t step i s done.

SXep 2 . There does n o t e x i s t a p o i n t o f degree n. kbume t h a t there i s an

.

n-point q I n t h i s s i t u a t i o n , any block X which i s not incident with q has degree n! therefore, f o r any such block, t h e p a i r (qo,X) i s o f type T0,O); i n p a r t i c u l a r , the p a r a l l e l class o f qo i s t r i v i a l . This implies t h a t t h e p a r a l l e l class o f any block B through q is trivial. ( k b w n e t h a t the p a r a l l e l class n o f B i s n o n - t r i v i a l . Since' n i s r e g u l a r , there e x i s t s a block B ' p a r a l l e l t o B which i s n o t i n c i d e n t w i t h qo. Then (qo,B') must be o f type ( p , ~ ) c o n t r a d i c t i n g our above observation.)

110

A . Beutelspacher

Therefore, any two b l o c k s B, B' through q have t h e same degree. (Namely: There e x i s t s a p o i n t p o u t s i d e B and B ' . Sinc@ (p,B) and (p,B') a r e o f type (O,O), i t f o l l o w s IBI = I p I = I B ' I . ) Hence any b l o c k through q has degree n t a . (There e x i s t s a b l o c k through qo which i s n o t i n c i d e n t w i t h o t h e ( n t a ) - p o i n t p .) This i m p l i e s t h a t any p o i n t p d i f f e r e n t from' q i s o f degree nta. L e t p be a p o i n t d i s t i n c t from q For any block B through q which does n o t c o n t a i n p, t h e p a i r (p,B) is'of t y p e (0,O). Denote by B' 'a b l o c k through p and qo. Since t h e p a r a l l e l c l a s s o f B' i s t r i v i a l , we g e t

.

B(p,B') = a(p,B') t ( ~ B ' ~ - ~ p ~ ) =( ~0.- 1 ) This means: The p o i n t p has no p a r a l l e l d i s t i n c t f r o m p on a b l o c k through qo. Since any p o i n t i s j o i n e d t o qo, p has a t r i v i a l p a r a l l e l c l a s s . Since t h e p a r a l l e l c l a s s o f qo i s t r i v i a l as w e l l , t h e p o i n t p a r a l l e l i s m o f S i s t r i v i a l . This c o n t r a d i c t i o n f i n i s h e s t h e second s t e p .

Step 3. There does n o t e x i s t a b l o c k o f degree n t a .

.

kdume on t h e cowYuvLy t h a t t h e r e i s an ( n t a ) - b l o c k B For any p o i n t p o f f B , t h e p a i r (p,B ) i s o f type' (0,O); i n p a r t i c u l a r , no p o i n t o f B i s p a r a l l e ? t o a p o i n t o f f OB Moreover, t h e p a r a l l e l c l a s s o f Bo i s trivial.'Therefore, f o r any p o i n t q on0 Bo i t holds

.

B(q,Bo) = a(q,Bo) hence f o r any p o i n t q on

( ~ B o ~ - ~ q ~ ) ( A -=l 0) t ( n t a Boy t h e p a r a l l e l c l a s s o f q

t

-

( n t a ) ) ( x - 1 ) = 0; i s trivial.

But: Since t h e p a r a l l e l i s m i s n o n - t r i v i a l , t h e r e e x i s t s an n-block, say B. By t h e above remarks, a p o i n t q on B which i s n o t i n c i d e n t w i t h B has no p a r a l l e l on B: a c o n t r a d i c t i o n t o t h e f a e t t h a t (q,B) i s o f t y p e ( p , ~ ) . Thus t h e p r o o f o f 1.1 i s f i n i s h e d . 4

1.2 COROLLARY. I f a # 0, then any n o n - i n c i d e n t p o i n t - b l o c k p a i r o f S i s o f

type

(P,v).J

Now we consider t r a n s i t i v e p a r a l l e l classes. Since we suppose p a v throughout, p o i n t p a r a l l e l classes and b l o c k p a r a l l e l classes p l a y a d i f f e r e n t r S l e ; p o i n t para1 1e l classes a r e r a t h e r unpleasant.

1.3 LEMMA. Suppose a > 0. I f t h e r e e x i s t s a t r a n s i t i v e r e g u l a r p o i n t p a r a l l e l c l a s s IT, then v 6 A-2. Proof. Denote by m t h e number o f p o i n t s i n

(3)

71.

Using 1.1 and 1.2 i t f o l l o w s

(m-l)(A-l) = ( n t a ) ( v - l ) ,

and mA = ( n t a ) v ,

hence

(4)

m-1 = nta-A.

Together w i t h ( 3 ) we g e t Thus u < A . I t remains t o show t h a t t h e assumption d i c t i o n . Under t h i s assumption, ( 5 ) reads

1 , e. n t a = ~ ( ~ - 1 i. Now (4) i m p l i e s 2 m = (A-1)

n

A*

-A-

u =

A-1 leads t o a c o n t r a -

a.

.

Since any non-incident p o i n t - b l o c k p a i r i s o f t y p e

(u,u), we g e t

111

Finite semi-symmetric designs

b(x-1) so

=

bv

=

m(nta) = (A-1)2 X(A-l),

.

(6) b = X(X-1) 2 From bn = v(nta) we get v = (x-l)n = ( ~ - l ) ( h 2-x-a). If M denotes the number of blocks in a block parallel class, it follows (b-l)x = M-1 + n(n+a-1) (7) and (8) (v-n)p = (M-l)(n-x+l). From (7) we infer - 2h + a(x 2-x-l). M-1 = (b-l)x n(n+a-1) = Using this equation, (8) yields (x-Z)(x 2 -x-a)(ax+x-1) = (v-n)p (M-l)(n-x+l) 2 2 2 = (x -2x+a(x -~-l))((x-l) -a). Therefore ( ~ - 2 ) x ( x - 1 ) ~ a(x-2)x 2 (A-1) - a 2 x(x-2) a(x-2)(x-1) = x(x-~)(x-l)~ ax(x-2) + a(x 2 -x-l)(x-l)' - a2(x2-x-1).

-

-

-

-

Since a # 0, we conclude a(x-1) = a(x 2-A-1 2 2 t ( A -A-l)(A-l)

.

-

x(x-2)) = ( A - 2 ) x2 (A-1) + (x-2)(x-l)

-

So, 1-1 would divide x(~-2), i. e. A - 1 = 1. But in view of (6), A b = 2: a contradiction.4

x(x-2) =

2 implies

By the following Theorem, part (a) of Theorem C is proved. 1.4 THEOREM. Suppose that there exists a non-trivial transitive parallel class n of s. (a) If n is a block parallel class, then a = 0, i. e. p = v. (b) If n is a point parallel class, then a = 0 or v 6 A-2. (c) If a = 0, then u = v s A-1. Proof. (a) If we assume a a 1, then we have p = ax + v > A ; hence n is regular. N o w i m i l a r arguments as in the first part of the proof of 1.3 show p < A : a contradiction. (b) In view of 1.3 we can suppose that n is not regular. Then 1.2.7 and 1.2.8 imply v 5 x. Since S is in class (11), it follows v s A-1. Suppose a 5 1. We show that the assumption v = A-1 yields a contradiction. In order to do this, we distinguish two cases. Case 1. n is singular. Then there is a "singular block" B such that any point of n is incident with Bo. Clearly, the following equationOholds: (91 mx = (n+a)v = (nta)(x-1), where m denotes the number of points in n. First, let us consider the possibility n = (Bo). Then m = n; hence (9) yields n+a ax = U-v < p: a contradiction. Thus n is a proper subset of (Elo). Therefore the following equation is true: (10) m(x-1) = (nta-l)v = (nta-l)(x-l),

A . Beutelspacher

112

so

m = n+a-1. Together w i t h ( 9 ) we g e t (n+a-l)A = ( n + a ) ( x - 1 ) . Since n+a and n t a - 1 a r e r e l a t i v e l y prime, n+a n+a s A: a c o n t r a d i c t i o n .

has t o d i v i d e

Cane 2 . II i s n e i t h e r r e g u l a r n o r s i n g u l a r . I n t h i s s i t u a t i o n , t h e r e e x i s t two d i s t i n c t b l o c k s , say n 5 (B,C). It f o l l o w s A-1 = v Q

1x1

IB,CI

\<

\<

B and

A; i n particular, C, w i t h

A.

Since any b l o c k o f S i n t e r s e c t s n i n a t l e a s t u = A-1 p o i n t s , t h e r e a r e two p o i n t s such t h a t any b l o c k i s i n c i d e n t w i t h one o f t h e s e p o i n t s : a c o n t r a d i c t i o n . ( c ) Since a = 0, we can suppose w i t h o u t l o s s o f g e n e r a l i t y t h a t II i s a b l o c k p a r a l l e l c l a s s . Denote b y B a b l o c k o f n . L e t us assume IJ a A. Since S i s i n c l a s s ( 1 1 ) , we have u > A; i n p a r t i c u l a r , n i s r e g u l a r . I f we denote by M t h e number o f b l o c k s i n n , we g e t ( M - l ) ( ~ - l )= n(u-1) and MA = y p ;

hence M-1 = n This i m p l i e s

(n-y)p.

-

-

A(A-1) (n-YIu(A-1) < A n n 1-1 < A. T h i s c o n t r a d i c t i o n shows = A

T h i s means

-x-

-

A( A-1). n p si

A-1.4

1.5 COROLLARY. L e t S be a f i n i t e semi-symmetric d e s i g n o f c l a s s ( 1 1 ) w i t h !.I-v = aA, where a i s an i n t e g e r . I f S c o n t a i n s a n o n - t r i v i a l t r a n s i t i v e p a r a l l e l c l a s s and if m i n I u , u l 5 A-1, t h e n any element o f S has t h e same degree.4 1.6 COROLLARY. Suppose then

Inl-1

n

a = 0. I f

-A-

i s a transitive regular parallel class o f

(n-y)u.J

1.7 LEMMA. Suppose t h a t S c o n t a i n s a n o n - t r i v i a l t r a n s i t i v e p a r a l l e l c l a s s Then t h e f o l l o w i n g a s s e r t i o n s a r e e q u i v a l e n t :

A-1; ( b ) v = A-1; ( c ) S i s o f t y p e ( ~ - l , ~ - l ) . P r o o f . It i s s u f f i c i e n t t o show t h a t ( a ) and ( b ) a r e e q u i v a l e n t . v ( b ) " : I f we assume a a 1, t h e n A-1 = p = aA t u > A, (a)

p =

a c o n t r a d i c t i o n . Consequently, v = p = A-1. " ( b ) + ( a ) " : I f n i s a b l o c k p a r a l l e l c l a s s , t h e n 1 . 4 ( a ) y i e l d s a = 0. I n t h e case where n i s a p o i n t p a r a l l e l c l a s s , i n view o f 1 . 4 ( b ) , t h e f a c t v = 1-1 i m p l i e s a = 0, i. e. u = v = A-1.4

n

S,

113

Finite semi-symmetric designs

5

2 . THE CASE

A-1

p =

I n what follows, we denote by S = (p,B,I) a f i n i t e semi-symmetric design w i t h A - 1 containing a n o n - t r i v i a l t r a n s i t i v e block p a r a l l e l class n w i t h e x a c t l y M blocks; by B we denote a block i n n. By 4.1.7, S has type ( A - 1 , ~ - 1 ) . I n view o f 4.1.5, any element o f S has the same degree, say n. p

2.1 LEMMA. I f A > 2 , then n i s r e g u l a r . Proof. (a) Assume t h a t n i s s i n g u l a r w i t h the s i n g u l a r p o i n t

po. Then

(M-l)(x-2) = ( n - l ) ( p - 1 ) = (n-l)(A-Z), hence M = n. This implies n = (p ) . Using i s p a r a l l e l t o po. Therefore, for'any block i t holds

1-1 = v = a contradiction.

1x1

p =

X

A-1, we see t h a t any p o i n t o f S which i s n o t i n c i d e n t w i t h po

= n:

(b) Assume t h a t t h e r e a r e a t l e a s t two p o i n t s i n c i d e n t w i t h any element o f n . Denote by 6 t h e number o f those p o i n t s which a r e on any element o f n . Obviously, M 8 A . By v we denote the number o f p o i n t s which a r e i n c i d e n t w i t h some element o f n. Clearly, ( i - 6 ) p = M(n-6), so

i(x-1) = I f M s A-1, then

i p =

Mn

-

(W-U)~.

-

i(~-l) = Mn (M-p)6 6 Mn s (A-l)n, i. e. i < n: a c o n t r a d i c t i o n , since n i s supposed t o be n o n - t r i v i a l . Therefore M = A ; hence i ( x - 1 ) = An 6 = ( x - l ) n t n-6, o r n-6 = ( A - I ) ( i - n ) . On the o t h e r hand, we have ( M - l ) ( ~ - l - 6 ) = (n-6)(p-l) (n-s)(A-2). Therefore (n-s)(A-2) = ( M - l ) ( ~ - l - 6 ) = ( ~ - 1 ) ( ~ - 1 - 6 ) , (~-1)(i-n)(A-2) hence (i-n)(A-2) = A-1-6 < A-2: a c o n t r a d i c t i o n . 4

-

From now on, we suppose always

h >

2.

2.2 LEMMA. There e x i s t s a p o s i t i v e i n t e g e r

z with

n = ( h - l ) ( z ( ~ - l ) t 1). Proof. By 2.1, the p a r a l l e l class n i s r e g u l a r . Therefore ( M - l ) ( ~ - l ) = n(p-1) = n(x-2). Thus A - 1 has t o d i v i d e n; denote by y t h e p o s i t i v e i n t e g e r w i t h n = y(A-1). We get M-1 = y ( ~ - 2 ) . Using MA = y p = y(A-1) (y-1). we see t h a t A-1 i s a l s o a d i v i s o r o f M = y(x-2) t 1 = y ( x - 1 ) S o , there e x i s t s a non-negative i n t e g e r z w i t h y-1 Z ( A - 1 ) . It f o l l o w s n = ( ~ - 1 ) y= ( A - l ) ( z ( h - 1 ) t 1 ) . Since n > A-1, z i s p o s i t i v e . 4

-

114

A. Beutelspacher

2.3 COROLLARY. The f o l l o w i n g equations hold: M = ( A - l ) ( ~ ( x - 2 )+ l ) ,

+ l), n-y = 2-1, vx = (1-1) 2 (Z(A-1) t 1 ) 2

y = h(Z(A-2)

-

(A-l)(Z-l).

Proof. By ( M - l ) ( x - l ) = n(x-2) and M A = y ( h - 1 ) , t h e f i r s t and the second equat i o n f o l l o w by 2.2. The second equation implies together w i t h 2.2 the t h i r d . The M-1 t n ( n - l ) . J l a s t equation f o l l o w s i n view o f (v-1)h 2.4 COROLLARY. The i n t e g e r

Praof. By sor o f

2.3,

A

z(z-1)

i s a multiple o f + 1)2

(A-l)((A-l)(z(A-l)

divides

(A-l)(Z(h-l) t 1)2 This implies the assertion.4

-

-

x. (2-1));

hence

A

i s a divi-

(2-1).

2.5 PROPOSITION. The f o l l o w i n g assertions are equivalent:

n

i s complete, (b) y = n, ( c ) z = 1, (d) n = A ( A - 1 ) . Proof. The f i r s t two assertions are equivalent, by d e f i n i t i o n . The equivalence o f m n d ( c ) f o l l o w s from 2.3; 2.2 gives us t h e l a s t equivalence.4 (a)

2.6 PROPOSITION. We have

z = 1 or

z a A.

Proof. Suppose z > 1. Then II i s n o t complete, by 2.5; denote by p a p o i n t outside p(n). Since (p,C) i s o f type (0,O) f o r any block C o f II,no p o i n t p a r a l l e l t o p i s i n s i d e p ( n ) . Because the p a r a l l e l class o f p i s n o n - t r i v i a l (doh: e i t h e r any o r no p a r a l l e l class i s t r i v i a l ) , t h e r e i s a p o i n t p ' p a r a l l e l t o p w i t h P ' # P. Denote by X a block through p ' which i s n o t i n c i d e n t w i t h p. Then (p,X) i s o f type (A-1,x-1). Therefore p has A-1 p a r a l l e l s on X, no one o f which i s a p o i n t o f ~ ( I I ) .Thus 2 - 1 = n-y A-1.4 2.7 PROPOSITION.If S contains a t r a n s i t i v e complete p a r a l l e l class, then t h e r e exists a positive integer m with n = m2t m , v = m3t m2 , x = rntl. Proof. Using m := x-1, the a s s e r t i o n f o l l o w from 2.1, 2.5, 2.2 and 2.3.4 The f o l l o w i n g Theorem i s important. 2.8 THEOREM. Ifthe block p a r a l l e l i s m o f parallelism i s transitive.

S

i s t r a n s i t i v e , then a l s o t h e p o i n t

Proof. Denote by p an a r b i t r a r y p o i n t o f S. We show t h a t any two d i s t i n c t p o i n t s ofpll are j o i n e d by e x a c t l y A-1 blocks. Denote by m t h e number o f p a r a l l e l s t o p. For any p o i n t p a r a l l e l class IT i t holds

-

I n l - 1 = (v-1)x n(n-1); d u a l l y , f o r any block p a r a l l e l class n(n-1). I n l - 1 = (b-1)x

-

n

we have

Since any two elements of S have t h e same degree, we know v = b. Thus any two p a r a l l e l classes have the same c a r d i n a l i t y . I n p a r t i c u l a r , f o r any block p a r a l l e l class II we have 1111 = m. Since any block p a r a l l e l class i s t r a n s i t i v e , ws get

Finite semi-symmetric designs

115

On any block not through p there are 0 or exactly A-1 points parallel to p. Denote by C a block through p. By the assumptions of our Theorem, the parallel class of C is transitive; hence a(p,t) = u-1 = A-2. Therefore B(P,C) = a(p,C) + (lCl-lpl)(A-1) = a(plC) = A-2. In other words: Any block of S contains 0 or exactly A-1 points parallel to p. Consider now the incidence structure 7 with point set Ilpll, whose blocks are those blocks of S which are incident with at least one point of Ilpll. Then I has exactly v' = m points, point degree -r' n and block degree k' = A-1. In view of 1 . 2 . 1 , for the average number A of blocks through two distinct points of 1 it holds 1 - r W W ; l ) - "!-^;2'* By ( l ) , this implies 1 = A-1. But through any two distinct points of llpll there are at least A-1 blocks; hence any two points of p are incident with exactly A-I common blocks. In other words: The parallel class of llpll is transitive. Since this holds for any point p, the point parallelism is transitive.4 By the following Theorem, the proof of Theorem C will be finished. 2.9 THEOREM. Denote by S a finite semi-symnetric design of type ( A - l , ~ - l ) with transitive block parallelism. I f S contains a complete parallel class, then there exist2 a symmetric p-(m 3+m 2+m+l,m 2tm+l,wl) design S+ and a projective line L o f S with S = S +L. Proof. By 2.8, also the point parallelism of S is transitive. Therefore, any parallel class is regular; in view of 2 . 5 , any parallel class is complete. By 2.7 and 2.3 it follows 2 3 2 2 n m +m, v = m +m , A = m t l , 1111 = m for a positive integer m. The number e of block (or, point) parallel classes is b = wl. e = Inl + + + We define the incidence structure St (p ,8 , I ) as follows: p+ := pu{l~B1l I B E 81, 8' := 8~{llpll I p E p l . The incidence relation I+ will be defined in the following way: p I+ B :* p I B for p E p and B E 8; IIBII I+ c :o c 11 B for B,C c 8; p I+ IlqII :o p II q for p,q E p; 11B11 I+ llpll : for all B E 8 and all p E p. Obviously, S+ is a semi-symmetric design of type (mtl,m+l) = (x,A) with trivial parallelism, hence a (symmetric) 3 2 2 2-(m +m +mtl,m +mtl,m+l) design. Moreover, the set L {IIBH I B E 81 is a line o f S+ intersecting each block; hence L is projective. Since, by construction, S = St+L, everything is shown.4

116

A . Beutelspacher

REFERENCES

1. Baker, R.D., An E l l i p t i c Semiplane, J. Combinat. Theory (A) 25 (1978) 193-195. 2. Beutelspacher, A. , A G e n e r a l i z a t i o n o f Dembowski 's Theorem on Semi-planes, Geom. Ded. 10 (1981) 59-72. 3. Beutelspacher, A., Embeddings i n S t r o n g l y Resolvable Designs, Arch. Math. 34 (1980) 171- 180. 4. Bose, R.C. and Shrikhande, S.S., Baer Subdesigns o f Symmetric Balanced Incomp l e t e Block Designs, i n : Ikeda, S. and o t h e r s (eds.), S t a t i s t i c s (Tsusho, Tokyo, 1976).

Essays i n P r o b a b i l i t y and

5. Cron, J.C. and Mavron, V.C., Generalised Semiplanes and C e r t a i n D i v i s i b l e P a r t i a l Designs, Math. Z. 161 (1978) 265-275. 6. Dembowski, P . , S e m i a f f i n e Ebenen, Arch. Math. 13 (1962) 120-131. 7. Dembowski, P., F i n i t e Geometries (Springer, B e r l i n Heidelberg - New York, 1968).

-

8. Totten, J. and de W i t t e , P . , On a Paschian C o n d i t i o n f o r L i n e a r Spaces, Math. Z. 137 (1974) 173-183.

Annals of Discrete Mathematics 14 (1982) 117-122

0 North-Holland Publishing Company

NOTE ON (q+Z)-SETS I N A GALOIS PLANE OF ORDER q Alessandro B i c h a r a and GBbor Korchmaros I s t i t u t o Natematico "G. Castelnuovo", U n i v e r s i t a d i Roma, I t a l y Department o f Mathematics, T e c h n i c a l U n i v e r s i t y o f Budapest, Hungary

and L e t 0 be a ( q + Z ) - s e t i n a p r o j e c t i v e p l a n e PG(2,q) l e t 0 5 R be t h e s e t o f t h e p o i n t s o f R w i t h t h e p r o p e r t y t h a t every l i n e containing a p o i n t o f 0 i n t e r s e c t s R i n two p o i n t s . I n t h i s paper we p r o v e t h a t : I f 101 > 2, t h e n q i s even; i f 101 > q/?, t h e n 0 = a ; f o r each q even, t h e r e e x i s t s an R such t h a t 101 = q/2.

L e t R be a s e t o f q+2 p o i n t s i n t h e p r o j e c t i v e p l a n e PG(2,q) o v e r a r f i n i t e G a l o i s f i e l d GF(q), q=p and p i s prime.

A l i n e o f PG(2,q)

(or chord),

..., p o i n t s

i s c a l l e d external, unisecant ( o r tangent), bisecant

..,, u-secant, ... , t o R depending on whether i t has 0,l ,2,. . .,u, i n common w i t h R.

F o r each P E R we d e f i n e cp(P) as t h e number o f chords o f R t h r o u g h P. I t i s c l e a r t h a t cp(P)

< qtl.

I f cp(P) = q t l t h e n e v e r y l i n e t h r o u g h P i s a

c h o r d o f R. Now d e f i n e 0 as t h e s e t o f t h o s e p o i n t s P € 0 f o r w h i c h cp(P) = qtl.

I f 0 = R t h e n R c o n t a i n s no t h r e e c o l l i n e a r p o i n t s ( i . e . R i s a ( q t 2 ) - a r c ) , hence q i s even ( c f . [ 51, [ 111, [ 1 2 1 ) . F o r any q even, a ( q t Z ) - a r c i s given, f o r example, b y t h e p o i n t s o f an i r r e d u c i b l e c o n i c p l u s i t s nucleus

.

I n t h i s Note we s h a l l o b t a i n some r e s u l t s on t h e c a r d i n a l i t y 101 o f Q: Theorem 1.

I f (01 > 2

Theorem 3.

F o r each q even, t h e r e e x i s t s an R such t h a t

t h e n q i s even.

117

= q/2.

A . Bichara and G. Korchmaros

118

F o r numerous o t h e r a s p e c t s o f ( q t 2 ) - s e t s see B i b l i o g r a p h y . We n o t e an immediate consequence o f o u r r e s u l t s . F ( x ) o v e r GF(q) i n t h e i n d e t e r m i n a t e x, d e f i n e @(F )

Given a polynomal

as t h e s e t o f t h o s e elements s o f GP(q) f o r which t h e polynomal F ( x ) - ( m ( x - s ) t tF(s))

has e x a c t l y two r o o t s i n GF(q) f o r e v e r y m E GF(q)

Xm = (l,O,O),

Ym = (0,1,0)

-

10). I l r i t e

and

n ( F ) i s a ( q t 2 ) - s e t i n PG(2,q) and

(p(Y,)

= q+l. Then we have

Corollary. I f I o ( F ) I > 2 t h e n q i s even.

>

Corollary 2.

If lo(F)I

C o r o l l a r y 3.

F o r each q even, t h e r e e x i s t s a polynomal F ( x ) o v e r GF(q) such

t h a t l o ( F ) I = (q/2)

-

q/2

then o(F)

GF(q).

1.

P r o o f o f Theorem 1. The b a s i c i d e a we use i s due t o B. Segre ( c f . 1 5 1 , 1111, 1121). L e t a ( a ) be t h e p r o d u c t o f t h e non-zero elements o f GF(q). Then ( c f . ibid.) (11

a ( a ) = -1. Given a ( q + 2 ) - s e t ~2 w i t h

of

> 2,

l e t A,,

A2, A3

be a y t h r e e p o i n t

o; t a k e A1A2A3 as fundamental t r i a n g l e o f a homogenous c o o r d i n a t e system

i n PG(2,q). The e q u a t i o n o f a l i n e p a s s i n g t h r o u g h t h e v e r t e x A . o f t h e fundamental J t r i a n g l e b u t d i s t i n c t f r o m t h e s i d e s AjAk and A . A can be expressed i n t h e from

J m

(where j k m i s a c y c l i c p e r m u t a t i o n o f t h e i n t e g e r s 1,2,3).

Note on ( q Q h e t s in a Galois plane of order q I f the p o i n t P: (al,a2,a3)

of

then t h e equations o f the l i n e s x2 =

n i s d i s t i n c t from the fundamental points, r 2 = A2P, r3 = A P

rl = Alp,

x3 =

X1X3,

119

,

X2X1

are i n t u r n

3

x1 = X3x2

where

Xi

# 0

(i

1,2,3).

rl, r2, r3, we have

As P l i e s on each o f t h e l i n e s

For each o f the q-1 p o i n t s o f 0 , d i s t i n c t from A1,A2 o b t a i n 3 elements X1,

X2 and X

product o f the q-1 products X

3

and A

3'

X X = 1. L e t S 1 2 3 C l e a r l y A . ( i = 1,2,3)

such t h a t X

we w i l l denote t h e

X = 1. w i l l take 1 1 2 3 each value o n l y and once only. Thus i n S each non-zero element o f GF(q) occurs p r e c i s e l y three times, i.e. S

But, from ( 2 )

=

3

(.(a))

.

we have S = 1 and from ( 1 )

n(a)

-1, consequently 1 = -1,

t h e r e f o r e q i s even. Proof o f Theorem 2.

Assume t h a t Theorem 2 i s f a l s e .

l e a s t t h r e e c o l l i n e a r p o i n t s A, B, C.

Then 5 2 - 0 admits

at

We consider the s e t

Clearly, E i s an arc and f o r each P E 1 the number o f tangents o f E a t P i s q

-

101. According t o Theorem 1

,q

i s even, therefore, by a theorem o f B.

Segre ( c f . ill]n. 181) C can be associated with an algebraic envelope Tt o f classe t = q - 101 but

, such

t h a t the I E / t d i s t i n c t tangents o f E belong t o Tt,

rt contains no chord o f E .

A . Bichara and G. Korchmdros

120

Now j o i n C t o t h e p o i n t s Q E Q. when Q, # Q.,

CQ, # CQ,

From t h e d e f i n i t i o n o f Q we have t h a t

Thus t h e l i n e s CQ w i t h

E Q a r e tangents t o

t h e i r number i s 101. S i n c e t = q - 101 and by h y p o t h e s i s

C and

>q/2, i t f o l l o w s

t h a t t h e r e e x i s t more t h a n t t a n g e n t s t o rt t h r o u g h C. Ile have t h e n t h a t t h e p e n c i l w i t h c e n t r e C belongs t o theorem

rt

rt. I n view o f above-mentioned S e g r e ' s

c o n t a i n s no c h o r d o f Z. Thus, A, B y C c a n n o t l i e on t h e same l i n e .

P r o o f o f Theorem 3.

The ( q t Z ) - s e t w i t h

I@[=

q/2 t o be d i s c u s s e d h e r e i s due,

e s s e n t i a l l y , t o 11. T a l l i n i S c a f a t i [ 161. Given any GF(q) w i t h q even, d e f i n e C1 and C 2 b y

C, = {b E GF(q) C 2 = {b E GF(q)

l 2

x txtb=O

has two r o o t s i n G F ( q ) l ,

2

x txtb= 0

has no r o o t s i n G F ( q ) l .

Then (see [ 1 6 1 ) : (i) (ii)

IC,I = q/2, if b ,b E C1 t h e n a l s o

ICIJ

1

2

bl

t

b, E C1.

Thus t h e s e t

i s a ( q t Z ) - s e t o f PG(2,q) and a s i m p l e c a l c u l a t i o n , which we o m i t , shows t h a t from ( i i ) i t f o l l o w s t h a t Q = I(b,b

ACKNOYJLEDGEMENT.

T h i s r e s e a r c h was p a r t i a l l y s u p p o r t e d b y GNSAGA o f CNR and

was done w h i l e t h e second a u t h o r was a t t h e U n i v e r s i t y o f Rome s u p p o r t e d by a g r a n t w i t h i n t h e I t a l i a n Hungarian c u l t u r a l exchange program.

Note o n (q+d)-setsin a Galois plane of order q

121

REFERENCES [

1 1 A. B a r l o t t i , Some topics i n f i n i t e geometrical s t r u c t u r e s , Chapel H i l l , N.C. Mimeo s e r i e s n . 439, (1965)

[

21 R.J. Bumcrot, F i n i t e hyperbolic spaces, A t t i del Convegno d i Geom. Comb. e sue Appl., Perugia (1971) 113-130.

t 3 1 M. Hall, Ovals i n the Desarguesian Plane of Order 16, Annali di Mat. Pura ed Appl., 102 (1975) 159-176. [ 41

J.W.P. Hirschfeld, Ovals i n Desarguesian Planes of Even Order, Annali di Mat. Pura ed Appl., 102 (1975) 79-89.

51 F. Karteszi, Introduction t o f i n i t e geometries, Akademiai Kiadb, (Budap e s t ) , 1976. 61 G. Korchmaros, S u l l e o v a l i d i t r a s l a z i o n e i n u n piano di Galois d ' o r d i n e pari , Acc. Naz. dei XL, 3 (1977/78) 55-65. [

71 G. Korchmaros, Gruppi di collineazioni t r a n s i t i v i sui punti di un'ovale ( ( q t 2 ) - a r c o ) di S con (1 p a r i , Atti del Sem. :!at. F i s . dell'Univ. Modena, 27 (1978) 861906.

[

81 H. LUneburg, Uber projektive Ebenen, in denen jede Fahne von e i n e r nichtt r i v i a l e n Elation i n v a r i a n t gelassen wird, Abh. Math. Sem. Hamburg, 3 (1965) 37-76.

[ 91

B. Segre, Le geometrie di Galois, Annali di Mat. Pura ed Appl. 48 (1959) 1-96.

101 6. Segre, Ovali e curve u nei piani di Galois di c a r a t t e r i s t i c a due, Rend. Acc. Naz. Lincei, 2 (1962) 785-790. [ 11 I

B. Segre, Lectures on Modern Geometry, Cremonese (Roma) 1961.

8

[

121 B. Segre, Introduction t o Galois geometries, Mem. Acad. Naz. Lincei, ( 1967) 135-236,

[

131 B. Segre-U. Bartocci, Ovali ed a l t r e curve nei piani d i Galois d i c a r a t t e r i s t i c a due, Acta Arithm., (1971) 423-449.

[

141 B. Segre-G. Korchmaros, Una proprieta degli insiemi d i punti d i un piano d i Galois c a r a t t e r i z z a n t e q u e l l i formati dal l e singole r e t t e e s t e r n e ad una conica, Rend. Accad. Naz. Lincei, 62 (1977) 613-618.

[151 G. T a l l i n i , Sui q-archi di un piano l i n e a r e f i n i t o di c a r a t t e r i s t i c a p = 2 , Rend. Accad. Naz. Lincei, 23 (1957) 242-245. [161 M. T a l l i n i - S c a f a t i , Archi completi in u n S , con q p a r i , Rend. Accad. 2 4 Naz. Lincei, 37 (1964) 48-51.

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Annals of Discrete Mathematics 14 (1982) 123-128 Q North-Holland Publishing Company

ON THE INDEPENDENCE OF THE AXIOMS DEFINING THE AFFINE AND

PROJECTIVE GRASSMANN SPACES

A. Bichara and F. Mazzocca I s t i t u t o Matematico "G. Castelnuovo", U n i v e r s i t a d i Roma, I t a l y I s t i t u t o Matematico, Facolta d i Ingegneria, U n i v e r s i t a d i Napoli , I t a l y

We prove t h a t i n the p a r t i a l l y planar spaces the axioms d e f i n i n g the a f f i n e and p r o j e c t i v e Grassmann spaces are independent.

1.

I n [ 21 and [ 31 ( c f . a l s o [ 1 ] and [ 71) the a f f i n e and p r o j e c t i v e Grassmann

spaces are defined and i n v e s t i g a t e d and i t i s proved t h a t such spaces having f i n i t e index are isomorphic t o t h e Grassmann spaces r e l a t e d t o an a f f i n e and p r o j e c t i v e space, respectively. Since t h e axioms d e f i n i n g t h e above mentioned spaces a r e complex, we p o i n t o u t the problem o f the independence o f them. I n t h i s paper we posit i v e l y solve such a problem. We now r e c a l l some d e f i n i t i o n s and the main r e s u l t s holding i n such spaces. The Reader i s a l s o r e f e r r e d t o [ 11, [ 21, t 31, [ 41, [ 5 1 and 71. A p a r t i a l l i n e space i s a p a i r (S, R), where S i s a non-empty s e t and R i s a covering o f S whose elements have order a t l e a s t two and such t h a t any two e l e ments i n S belong t o a t most one element i n R. Call t h e elements i n S p o i n t s and

lines.

Any two d i s t i n c t p o i n t s are c o l l i n e a r i f they belong t o the elements i n R a same l i n e . A subspace o f (S, R) i s a subset T o f S whose p o i n t s a r e pairwise c o l l i n e a r and such t h a t t h e l i n e through any two p o i n t s i n T i s contained i n T. C a l l (S, R) a proper p a r t i a l l i n e space i f t h e r e e x i s t two n o n - c o l l i n e a r points; otherwise, c a l l

(S, R) a l i n e space. Furthermore,

(S, R)

i s connected i f

any two d i s t i c t p o i n t s belong t o a polygonal. C a l l any proper p a r t i a l l i n e space such t h a t any three p a i r w i s e c o l l i n e a r p o i n t s are contained i n a subspace a p a r t i a l l y planar space. I~OW, l e t (G, F) be a p a r t i a l l y planar space such t h a t t h e f a m i l y M o f the maximal subspaces i n (G,

F ) s p l i t s i n t o t h r e e d i s j o i n t classes S1, S2, T and,

s e t t i n g S = S1 u S2, the f o l l o w i n g axioms hold: (i)

SES,TET*

either

S n T = 0

123

or

SnTEF;

124

A . Bichara and F. Mazzocca

(VEBLEN AXIOM I N S ) i f S1 and S2 are d i s t i n c t elements i n S, e i t h e r meet(iii) i n g a t a p o i n t o r both belonging t o S2, then any two d i s t i n c t elements i n S, both meeting S1 and S2 a t d i s t i n c t points, e i t h e r i n t e r s e c t o r both belong t o S2; (iv)

(G,F)

i s connected;

(v)

the f a m i l y S2 e i t h e r i s t h e empty s e t o r i s a convering o f

G;

C a l l r = (G, F, S1, S2, T) an a f f i n e Grassmann space (AGS) i f S2 # otherwise, i f S2 = 0, c a l l r = (G, F, S,T) a p r o j e c t i v e space (PGS). An a f f i n e ( o r p r o j e c t i v e ) Grasmman space

r

0;

has f i n i t e index h i f there e x i s t s

a saturated f i n i t e chain C o f subspaces o f s i z e h+l, w i t h a l i n e as minimal and a subspace T E T as maximal. Ift h e r e e x i s t s such a chain C, then the s i z e o f every chain, s i m i l a r t o the previous one, i s h t l ; so, such a d e f i n i t i o n makes sense. L e t A be a f i n i t e ( o r i n f i n i t e ) a f f i n e space o f dimension a t l e a s t t h r e e and, l < h < dim A 1, l e t Gh(A) and Fh(A) be the f a m i l y o f the h-dimenVh E N,

-

sional subspaces i n A and the f a m i l y o f t h e (proper o r improper) p e n c i l s o f h-dimensional subspaces i n IA, r e s p e c t i v e l y . The p a i r ( G h (A), F h(A)) turns o u t t o be a p a r t i a l l y planar space and the f a m i l y o f i t s maximal subspaces can be p a r t i h h t i o n e d i n t o the classes S (A), S (A) and T h(A); namely

S 1 E Sh1(A ) <=> <=>

S

1 i s the f a m i l y of a l l t h e h-dimensional subspaces i n IA

which c o n t a i n a given (h-1)-dimensional subspace; S

2

i s a complete f a m i l y o f p a i r w i s e p a r a l l e l h-dimensional

subspaces i n A; T E T ~ ( A )<=>

T i s the f a m i l y o f a l l h-dimensional subspaces i n A con-

t a i n e d i n a given subspace o f dimension (h+l).

h - t h Grassmann space r e l a t e d t o A; i t i s n o t hard t o show t h a t r h (A) t u r n s o u t t o be an AGS o f f i n i t e index h. Analogously we d e f i n e the h - t h Gras_smann space

I

r e l a t e d t o a p r o j e c t i v e space P , which turns o u t t o be a PGS o f f i n i t e index h. We remark t h a t i f P i s a p r o j e c t i v e space o f f i n i t e dimension on a comnutative f i e l d , then there e x i s t s an isomorphism between t h e space r h (P ) and the Grassmann manifold representing the h-dimensional subspaces i n P , The above mentioned examples a r e the only examples o f Grassmann spaces o f

The independence of the axioms defining Grassmann spaces finite

index, since we cen summarize t h e r e s u l t s i n [ 2 ]

125

and [ 3 ] i n the f o l l o w i n g

way, as was mentioned above: Theorem. I f r i s any a f f i n e ( p r o j e c t i v e ) Grassmann space o f f i n i t e index h, then there e x i s t s an a f f i n e space A (a p r o j e c t i v e space P ) , such t h a t r and r h(A) ( r and r h( P ) ) a r e isomorphic. I n t h i s section we prove t h a t the axioms ( i ) - ( v i ) i n s e c t i o n n.1 are inde2. pendent. To t h i s end, f o r each o f them we s h a l l g i v e an example o f space s a t i s i f y i n g a l l the above mentioned axioms, w i t h t h e exception o f the selected one. I Example I: L e t (G, F, SD

SeD T ) be any AGS and f e i F a l i n e . Moreover, l e t R

have a l i n e space s t r u c t u r e ( c f . 1 ) on t h e p o i n t s e t o f f. Set F'

(F- if} )

U

R;

the p a i r (G, F ' ) s t i l l t u r n s o u t t o be a p a r t i a l l y planar plane, whose f a m i l y o f a l l maximal subspaces s p l i t s i n three d i s j o i n t classes S1, S 2 , T. I n (G, F', S1, S2, T) a l l t h e axioms o f t h e AGS hold w i t h the exception o f ( i ) .

Example 11. L e t (G, F, S1, S 2 , T ) be any a f f i n e Grassmann space and p any o b j e c t n o t belonging t o G. Take any maximal subspace S E s2 and s e t

F' = F The p a i r (GI, F '

U

R

S;=

IS2

-

{S} ) U { S U { p }

}.

turns o u t t o be a p a r t i a l l y planar space and t h e f a m i l y o f i t s

maximal subspaces s p l i t s i n the d i s j o i n t classes

S1,

Sh and T. I n (G', F ' ,

S1, Sh, T) the axiom ( i i ) does n o t hold, since each l i n e {p, q) , qE S, i s contained i n the unique maximal subspace S U { p l Furthermore i t i s n o t hard t o show t h a t a l l the other axioms i n AGS hold.

.

Example 111. L e t Abe any a f f i n e space o f order a t l e a s t three, t h a t i s each l i n e i n A must contain a t l e a s t t h r e e p o i n t s . More, l e t rh (A) = (G, F, S1, S2, T ) b e t h e h-th Grassmann space r e l a t e d t o A and p any p o i n t i n G. Set

A, Bichara and F. Mazzocca

126

the p a i r ( G I , F ' ) turns o u t t o be a p a r t i a l l y planar space and the f a m i l y o f i t s maximal subspaces i s t h e s e t t h e o r e t i c union o f t h e d i s j o i n t sets S i , Si, T I . Moreover, i t i s n o t hard t o prove t h a t (GI, F', S i , Sh, T ' ) s a t i s f i e s a l l t h e axioms o f an AGS, w i t h t h e exception o f ( i i i ) . Example I V .

L e t (GI, F, S1, S2, T) and (GI,

G n G ' = 0. It i s easy t o see the p a i r (GuG',

nar space; more, (G

U

GI,

F

U

F',

S1

U

Si,

F',

S i , S i , T ' ) be any two AGS's and

F u F ' ) i s already a p a r t i a l l y p l a S2 U Sh, T U T I ) s a t i s f i e s a l l

the axioms o f an AGS, w i t h t h e exception o f ( i v ) . Example V.

L e t Pn be any p r o j e c t i v e space o f dimension n and ( R ( P n ), F ( P n ) )

the p a r t i a l l y

planar space, whose p o i n t s and l i n e s a r e t h e l i n e s and the p e n c i l s

o f l i n e s i n Pn , respectively. L e t Pk be any proper subspace o f Pn o f dimension k, k f n-1; take the s e t R (Pn- Pk) o f the l i n e s i n Pn n o t belonging t o Pk a n d t h e geometric s t r u c t u r e

F(Pn- Pk) which i s induced by ( R (P,),

F (P,))

on R (Pn- Pk).

Set

pn

-

S1 = the f a m i l y o f a l l l i n e s i n Pn

-

Pk through a p o i n t b e l o n g i n g t o

= the f a m i l y o f a l l l i n e s i n Pn

-

Pk through an ( i d e a l ) p o i n t

pk,

S2 belonging t o Pk, T

= the f a m i l y o f t h e r u l e d planes i n Pn

a l l the axioms i n AGS hold i n ( R ( P n exception o f ( v ) .

-

Pk), F ( P n

-

-

Pk;

P k ) , S1, S2, T), w i t h

the

Example V I . L e t P be any p r o j e c t i v e space o f dimension three, G t h e l i n e s e t o f P and F the s e t o f t h e p e n c i l s o f l i n e s i n P . L e t IT be any plane i n P ; c a l l S1 the f a m i l y o f a l l l i n e s i n P through any p o i n t i n P n, S2 t h e f a m i l y o f a l l l i n e s i n P through any p o i n t i n IT and T t h e f a m i l y o f a l l r u l e d planes, The p a i r

-

(G, F ) t u r n s o u t t o be a p a r t i a l l y planar space, whose maximal subspaces s p l i t i n three d i s j o i n t classes S1, S2, T. Furthermore, i n (G, F , S1, S2, T) the axioms

( i ) - ( v ) hold, b u t ( v i ) does n o t hold. The above mentioned examples prove t h e independence of t h e axioms d e f i n i n g the a f f i n e Grassmann spaces. Taking as a s t a r t i n g p o i n t any PGS, i t i s p o s s i b l e t o repeat t h e constructions

The fndependence of the axioms defining Grassmann spaces

127

in examples I , 11, 111, IV, w i t h the exception of taking S i n S1 in example 11. In such a way, one can prove t h a t the axioms of the projective Grassmann spaces are always independent. Acknowledgement.

This research was p a r t i a l l y supported by Gnsaga of CNR.

REFERENCES A. Bichara, F. Mazzocca, On a characterization of the Grassmann spaces representing the lines in an affine space, Simon Stevin, to appear, A. Bichara, F. Mazzocca, On a characterization of the Grassmann spaces associated w i t h an a f f i n e space, Proc. of Inter, Conf. on Combinatorial Geometries and t h e i r Appl , Roma, Giugno 1981, t o appear. A. Bichara, .On a characterization of the Grassmann space representing the hdimensional subspaces i n a projective space, Proc. of Inter, Conf. on Combinatorial Geometries and t h e i r A p p l , , Roma, Giugno 1981, to appear. P. M. Lo Re, D. Olanda, Grassmann Spaces, Journal of Geometry, to appear. F. Mazzocca, D. Olanda, A graphic characterization of the lines of an affine space, Proc. of Inter, Conf. on Combinatorial Geometries and t h e i r Appl., Roma, Giugno 1981, t o appear. B. Segre, Lectures on modern geometry, Cremonese Ed., 1961. G. T a l l i n i , On a characterization of the Grassmann manifold representing the lines i n a projective spaces, Proc. of the second I s l e Thorus Conf. 1980,London Math. SOC., Lectures Notes, s e r i e s 49, Cambridge University Press, 1981. G. T a l l i n i , Spazi parziali di r e t t e , spazi polari. Geometrie subimmerse, Quaderno n.14 del Sem. Geom. Comb., 1st. Mat. Univ. Roma.

.

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Annals of Discrete Mathematics 14 (1982) 129-150 0 North-Holland Publishing Company

ON A CHARACTERIZATION OF THE GRASSHANN MANIFOLD REPRESENTIRG THE PLANES I N A PROJECTIVE SPACE

Alessandro Bichara and Giuseppe Tal l i n i I s t i t u t o Matematico "G. Castelnuovo", U n i v e r s i t i i d i Roma, I t a l y 1.

INTRODUCTION The Grassmann manifold representing the planes i n a p r o j e c t i v e space w i l l

be characterized as a p a r t i a l l i n e space (G,

F ) whose maximal subspaces s a t i s f y

some s u i t a b l e conditions (see s e c t i o n 2 ) . The p r o o f uses the f o l l o w i n g : from

F ) i t i s possible t o construct - i n a natural way - another p a r t i a l l i n e space, say (S, R), which turns o u t t o be isomorphic t o the Grassmann manifold (G,

representing the l i n e s i n a p r o j e c t i v e space [ 3 1 . The r e s u l t s t h a t w i l l be proved here can be generalized t o t h e Grassmann manifold representing the d - f l a t s i n a p r o j e c t i v e space. The p r o o f o f such a general theorem, which goes by i n d u c t i o n ( s t a r t i n g from d = 2 , since the step from d

1 t o d = 2 i s q u i t e special), i s the subject o f another paper by the

authors.

2.

PRELIMINARIES L e t G be a non empty set, whose elements w i l l be c a l l e d p o i n t s , and

F a

(proper) non empty c o l l e c t i o n o f (proper) subsets o f G, which w i l l be c a l l e d

l i n e s . The p a i r (G,

F ) i s s a i d t o be a p a r t i a l l i n e space (PLS) i f the f o l l o w -

ing hold ( [ Z I ) : ( i ) Any two d i s t i n c t p o i n t s i n G belong t o a t most one l i n e i n F. ( i i ) Any l i n e i n F contains a t l e a s t two p o i n t s o f G. ( i i i ) F i s a covering o f G. Two d i s t i n c t p o i n t s p and p ' i n G w i l l be s a i d t o be c o l l i n e a r i f they belong 129

A . Bichara and G. Tallini

130

t o a l i n e , denoted by (p,p'),

i n F; by p

- p'

i t i s meant t h a t p and p '

(p# p ' ) are c o l l i n e a r ; otherwise, p and p ' are non-collinear: p f p ' . (G, F)

w i l l be c a l l e d a proper p a r t i a l l i n e space (PPLS) i f t h e r e

e x i s t two non-collinear p o i n t s i n G. A p o i n t subset H o f a PPLS (G, F) i s c a l l e d a subspace i f any two o f i t s p o i n t s a r e c o l l i n e a r and the l i n e j o i n i n g them i s completely contained i n H; a subspace H o f (G, F) i s a maximal subspace i f i t i s n o t p r o p e r l y contained i n any subspace o f (G, F ) . A PLS (G, F) w i l l be c a l l e d i r r e d u c i b l e i f t h e f o l l o w i n g holds: Any l i n e i n F c o n t a i n s a t l e a s t t h r e e p o i n t s o f G.

(ii')

F i n a l l y , a PPLS (G, F) i s s a i d t o be connected i f For any two d i s t i n c t p o i n t s p, p ' i n G t h e r e e x i s t s a polygonal

(iv)

path j o i n i n g them. L e t P(r,K)

(r

> 4)

be an r-dimensional p r o j e c t i v e space over a f i e l d K.

L e t G be t h e c o l l e c t i o n o f a l l t h e planes i n P(r,K) and F c o n s i s t s o f p e n c i l s o f planes i n P(r,K)

(a p e n c i l o f planes being the s e t o f a l l t h e planes

through a l i n e contained i n a 3 - f l a t P(3,K) i n P(r,K)),

Then (G,

F) i s a

proper i r r e d u c i b l e PLS which i s isomorphic t o Grassmann manifold G

rY2,K (As i t i s w e l l known, Gr,2,K i s an i r r e rtl ducible algebraic (3(r-2))-dimensional manifold i n an(( ) 1)-dimensional representing the planes i n P(r,K).

-

p r o j e c t i v e space; moreover, t h i s manifold i s an i n t e r s e c t i o n o f quadrics.) Any n o n - t r i v i a l ( i . e . p r o p e r l y containing some element o f F) subspace in

(G, F ) i s e i t h e r

( a ) the c o l l e c t i o n o f a l l the planes through a l i n e L belonging t o an h - f l a t (through L), h 3 4 (such a c o l l e c t i o n w i l l a l s o be c a l l e d an L - s t a r ) or (b) the c o l l e c t i o n o f a l l planes belonging t o a 3 - f l a t ; o r ( c ) the c o l l e c t i o n o f a l l the planes through a p o i n t i n a 3 - f l a t ;

such

a s e t o f planes w i l l be c a l l e d a s t a r o f planes. Therefore, a maximal subspace i n (G, F) i s e i t h e r an il-star, and the c o l l e c t i o n o f such maximal subspaces w i l l be denoted by S, o r the s e t o f a l l planes i n a 3 - f l a t i n P(r,K),

and the c o l l e c t i o n o f these maximal sub-

spaces w i l l be denoted by T. The PPLS (G, F) i s connected and s a t i s f i e s : Ale

If t h r e e p o i n t s i n G are pairwise c o l l i n e a r , then t h e r e e x i s t s a

The Grassmann manifold representing the planes in a projective space

131

subspace i n (G, F) through them.

No l i n e i n F i s a maximal subspace. Furthermore, t h e r e e x i s t two A2. c o l l e c t i o n s , say S and T, o f maximal subspaces i n (G, F ) and any maximal subspace belongs e i t h e r t o S o r t o T. Moreover: (1)

SES,TET*eitherSnT=fl,

(11)

VfEF*3!SES93!

(111)

I f S,

SO, S"

element i n S, d i s t i n c t from

E

TET:

or

SnTEF.

f c S , f C -T .

S p a i r w i s e meet a t d i s t i n c t p o i n t s , then any

S" and meeting both S and S' a t d i s t i n c t p o i n t s ,

meet S" too.

T

E

There e x i s t three subspaces f, n,

A3. T and

T

covers f and i s covered by

Axioms A1,

Remark 1.

7

i n (G, F) such t h a t f

E

F,

T.

A2, A3 and t h e connectedness hypothesis, besides being

s a t i s f i e d by Grassmann manifold representing t h e planes i n P(r,K),

are a l s o

s a t i s f i e d by the generalized Grassmann manifold representing the planes i n an e i t h e r i r r e d u c i b l e (and p o s s i b l y Pascalian) o r r e d u c i b l e p r o j e c t i v e space P even o f i n f i n i t e dimension. Remark 2.

The s e t t h e o r e t i c union o f Grassmann manifolds representing t h e

planes i n skew p r o j e c t i v e spaces i s a non-connected PLS s a t i s f y i n g axioms A1, A2, and A3. I n order t o characterize the (generalized) Grassmann manifold representi n g the planes i n one o r more p r o j e c t i v e spaces (see Remark 2 ) , PPLS's (G, F) s a t i s f y i n g A1, A2, and A

w i 11 be proved:

3

w i l l be studied. Namely, the f o l l o w i n g r e s u l t s

I f (G, F) i s a connected PPLS s a t i s f y i n g axioms A1, A2, and A3,

Theorem 1.

then t h e r e e x i s t a p r o j e c t i v e space (1, L ) and a mapping F from the c o l l e c t i o n %

P o f a l l planes i n (1, L ) i n t o G such t h a t : (i)

F i s one-to-one and onto;

( i i ) F consists o f e x a c t l y t h e images under F o f the p e n c i l s o f planes in

?; ( i i i ) S consists o f e x a c t l y the images under F o f L - s t a r s i n (1, L),

L E L; and ( i v ) T consists o f e x a c t l y t h e images under F o f the c o l l e c t i o n s o f planes

132 in

A . Bichara and G. Tallini

?,

each o f them being formed by a l l t h e planes i n a 3 - f l a t i n (G, F ) . Thus, (C,

L ) i s i r r e d u c i b l e i f f (G, F ) i s i r r e d u c i b l e . F i n a l l y ,

if

(I, L ) i s f i n i t e l y generated and Pascalian, then (G, F ) i s isomorphic t o Grassmann m a n i f o l d r e p r e s e n t i n g t h e planes i n (C, Theorem 2 .

L).

L e t (G, F ) be a PPLS s a t i s f y i n g axioms A

A2, and A3. Then each 1' connected component o f (G, F ) i s a PPLS s a t i s f y i n g axioms A1 and A2. I f f o r each connected component o f

(G, F ) A3 holds, then (G,

F) i s the set

t h e o r e t i c union o f g e n e r a l i z e d Grassmann m a n i f o l d s r e p r e s e n t i n g t h e planes i n p a i r w i s e skew p r o j e c t i v e spaces.

3. SOME PROPERTIES OF A CONNECTED PPLS Let

(G, F )

be a connected PPLS s a t i s f y i n g axioms A1, A 2 , and A

sect. 2 . Then: P r o p o s i t i o n I. The c o l l e c t i o n s S and T a r e skew. L e t M and

3

in

PI' ( w i t h M # M I )

be two maximal subspaces having two d i s t i n c t common p o i n t s p and p ' ; then

M and M ' belong t o d i f f e r e n t c o l l e c t i o n s o f maximal subspaces. Thus, two d i s t i n c t maximal subspaces belonging t o t h e same c o l l e c t i o n have a t most one common p o i n t . Proof.

I f a maximal subspace M" were contained i n S n T, then

M" n M"

MI';

a c o n t r a d i c t i o n t o A21, as MI' p r o p e r l y c o n t a i n s a l i n e . Since p and p ' belong t o M y they a r e c o l l i n e a r ; l e t f be t h e l i n e j o i n i n g them; o b v i o u s l y , f

5M

and f c

M' , t h e two maximal subspaces belong

t o d i f f e r e n t c o l l e c t i o n s ' a n d t h e statement i s proved. P r o p o s i t i o n 11.

L e t S and S ' be two d i s t i n c t maximal subspaces i n S, having

a common p o i n t p. I f T i s a maximal subspace i n T , meeting-S and S ' a t t h e l i n e s f and f ' , r e s p e c t i v e l y , then f and f ' meet a t p, so T passes through p. Proof.

f and f ' are d i s t i n c t l i n e s (otherwise, two d i s t i n c t maximal subspaces

133

The Grassrnann manifold representing the planes in a projective space S and S' i n S would meet a t t h e l i n e Since any l i n e i n

f = f ' , which i s impossible by prop. I ) .

F has a t l e a s t two points, t h e r e e x i s t p o i n t s p '

E f and

p" E f ' such t h a t p, p ' and p" are p a i f w i s e d i s t i n c t . These p o i n t s are p a i r wise c o l l i n e a r (indeed, p, p '

E

s,

p, p"

€

s',

and p ' , p"

T);

€

thus, by A1,

there e x i s t s a subspace H through them and N i s contained i n a maximal subspace M. I f 11 belonged t o

M

s,

then M = S (by prop. I, M n

2 Hence M

s i m i l a r l y , M = S ' ( s i n c e t1 n S '

= S);

t o S, then S = S' a c o n t r a d i c t i o n .

{p,p"l).

s ' 3 {p,p')

implies

Therefore, i f I1 belonged

$ S and, by A*,

H E T . Since !1

belongs t o T and contains the d i s t i n c t p o i n t s p ' and p" i n T E T , by prop. I,M = T. Moreover, p E M and 14 = T imply t o S; thus,

{p)

5T n S

f

and

p E T. The p o i n t p i n T belongs

p E f; s i m i l a r l y , p E f ' ; therefore, t h e

d i s t i n c t l i n e s f and f ' meet a t t h e p o i n t p and t h e statement i s proved. Proposition 111. Any T i n T i s a p r o j e c t i v e space. Proof. I t i s enough t o prove t h a t i n the l i n e space T Veblen-Nedderburn axiom holds : L e t fl and f

be two l i n e s i n T meeting a t the p o i n t p3; i f f3 and f4 2 are d i s t i n c t l i n e s i n T, each o f them meeting both fl and f2 a t p o i n t s d i s t i n c t from p3, then f3 and f4 meet a t a p o i n t . = f2 n f4; then

Through

..,4)

t h e r e i s e x a c t l y one maximal subspace S E S (see i A211). Such maximal subspaces a r e p a i r w i s e d i s t i n c t ( i f i # j and Si = sj, fi ( i = 1,.

then the maximal subspace Si would share w i t h T E T the p o i n t s i n fi which i s impossible by A21). Now, fi c Si,

i = 1,2

and {p

l 3

= fl

U

f

j'

n f2 imply

I p3 } -C 1S n S2; since S1 # S2, by prop. I, {p31 = S1 n S2. By the same argument, {pl} = S2 n Sg and {pz} = S1 n S3; moreover, S meets S1 and

4 S2 a t t h e p o i n t s q1 and q2, respectively. The t h r e e maximal subspaces S1, S2, and S pairwise meet a t d i s t i n c t p o i n t s (see (3.1)) and S4 meets S1 and S

2

3 a t d i s t i n c t points; therefore, by A2111, S4 meets S3 a t a p o i n t

134

A . Blchara and C. Tallini

q: {q} = S3 n S4. Since T meets l y , and S3 n

S4

S3 and S4 a t the l i n e s f3 and f 4' respective-

= {q}, by prop. 11, f 3 and f 4 meet a t the p o i n t q and the

statement i s proved. The p r o j e c t i v e space which a r e members o f T , c o n t a i n p r o j e c t i v e planes forming a c o l l e c t i o n 11 o f subsets o f G. Clearly, any element i n 11 i s a subspace i n (G,

F ) which i s contained i n a maximal subspace belonging t o

Proposition I V .

T.

L e t T and T ' be two elements i n T through a p o i n t p i n G.

, and S3 a r e three p a i r w i s e d i s t i n c t elements i n S through p 2 f { E F. I f t h e l i n e s fi, and such t h a t T n Si = fi E F , and T ' n S i i = 1,2,3, belong t o t h e same plane a i n T, then a l s o the l i n e s f; belong

Assume S1, S

t o a unique plane i n T I .

Proof.

I t i s enough the prove

L e t a' be the plane i n T' through f ' and f;. 1 t h a t f ' belongs t o a ' .

3

L e t f and f ' be two l i n e s n o t through p, the former i n a , t h e l a t t e r i n a ' . Since a and a' a r e p r o j e c t i v e planes, f meets fl,

meets

fi and f;.

Set {qi}

= f n fi,

i = 1,2,3,

I t i s easy t o check t h a t the f i v e p o i n t s qi,

and

'

f2 and f3, and f '

{q!} = f ' n f!, j = 1,2.

J

J

are pairwise d i s t i n c t ;

qj moreover, i f S and S' are the maximal subspaces i n

9 through

f and f ' ,

, S 2 , S 3 are p a i r w i s e d i s t i n c t 1 and {qi) = S n S i , Iq!) S' n Sj ' Since S1, S2, and S3 pairwise meet a t J d i s t i n c t p o i n t s and S' ( # S) meets S, and S2 a t d i s t i n c t p o i n t s , S and S' r e s p e c t i v e l y , then the f i v e subspaces S, S', S

have a common p o i n t q, which i s obviously d i s t i n c t from q1 and q ' Thus, 1' t h e three maximal subspaces S, S1 and S' pairwise meet a t d i s t i n c t p o i n t s

3 ( # S') meets S and S1 a t t h e d i s t i n c t p o i n t s q3 and p ( r e s p e c t i v e l y ) . Therefore, S3 and S' have a common p o i n t q ' . Since Sg n T ' = f i , S' n T ' = f '

and S

and I q ' ) = S3 n S', by prop. 11, f ' and f ' meet a t 4. Hence, f ' and f i a r e 3 coplanar. The plane through them contains f ' and t h e p o i n t p on f i ; thus, i t i s a'

and f i belongs t o a ' .

The Grassmann manifold representing the planes in a pmjective space

F)

4. THE PARTIAL LINE SPACE (S, R ) ASSOCIATED WITH (G, Take p E G, a E

n,

with p

135

o f S, c o n s i s t i n g P ,a o f those max'imal subspaces i n S meeting a a t l i n e s i n F through p, i . e .

r = {SES: P ,a i s uniquely defined. Proposition V. I f a , a'

(4.1

E

n

E

a; then the subset r

and S n a E F 1

S 3 p

and p

E

,

a , p ' E a ' , then

1

Proof. Since a i s a p r o j e c t i v e plane,

through t h e p o i n t p i n a there are a t

l e a s t two d i s t i n c t l i n e s fl and f 2 o f

F.

The maximal subspace S1 and S2 i n

r and now i t w i l l be shown P ,a t h a t S1 # S2. I f S1 = S2, then t h i s member o f S would share w i t h a maximal

S through fl and f 2 ( r e s p e c t i v e l y ) belong t o

subspace T i n T a s e t I containing the d i s t i n c t l i n e s

fl and f2, which

and S1 # S2; (4.1) f o l l o w s . i s impossible by A21. Therefore, S1, S2 E r P ,a L e t S1 and S2 be two d i s t i n c t elements i n r n r S1 and S2 p,a p',a'. meet a a t l i n e s i n F through p and a' a t l i n e s i n F through p'; thus p and p ' belong t o

S, n S2. By prop. I, p = p ' .

Mhen a

a ' , (4.2) i s obviously true. Assume a

#

a ' . L e t T and T ' be

the maximal subspaces i n T through a and a ' respectively; then (otherwise, S1 would meet T a t two l i n e s i n

T # T'

F through p, one belonging t o

a, t h e other t o a ' , which c o n t r a d i c t s A21) and T n T ' =

{PI.

Furthermore,

by prop. I V , any element Sg i n r , d i s t i n c t from S1 and S2, meets T' a t P ,a being coplanar w i t h S1 n T ' and S2 n T ' belongs t o a l i n e i n F, which

-

r

P,a"

hence, rp,a

5

-

rpl,al. By the same argument,

(4.2) follows.

r p',a'

C r

-

p,a

and

{r : a E ll , p ~ a o} f subsets o f S i s defined; P ,a since i t i s n o t a proper c o l l e c t i o n (see prop. V ) , l e t R be the proper Thus, the c o l l e c t i o n

c o l l e c t i o n associated w i t h it.

136

A . Bichara and G. Tallini

The pair (S, R ) is a PLS. Moreover, two d i s t i n c t elements

Proposition VI.

in S a r e collinear i n (S, R ) i f f they have a common point i n G. Proof. Let -

S and S ' be two d i s t i n c t elements in S. If S n S '

{p}, then l e t

f be a line in S through p; through f there i s a maximal subspace T E T meeting

s'

a t one point a t l e a s t ; thus, i t meets

s'

a t a line f '

F. In the

E

projective space T , the d i s t i n c t l i n e s f and f ' a r e joined by a plane therefore, S and S ' belong to r

.

CI E

II ;

P 9Q If S n S ' = 0, no element in R through S and S' exists.

Let f be a l i n e in S E S. Through f there i s a maximal subspace T E T . If p E f , l e t f ' be a l i n e in T through p , d i s t i n c t from f . Through f ' there i s a maximal subspace S'

E

S , which i s obviously d i s t i n c t from S and meets

S a t p ; thus, through S there i s an element of R (joining S and S'). Hence, R i s a cover of S. Moreover, f o r any r E R ,

I rl > 2

two elements i n R have a t most one common p o i n t

(s, R )

(see (4.1) and any (by (4.2 ); i t follows t h a t

i s a PLS.

Let p be a point in G . The collection S

of a l l elements n S through P p i s a subspace of (S, R ) . (Indeed, any two d i s t i n c t elements in S are P

collinear i n (S, R )

and the l i n e through them i s completely contained in S ) . P Let p be a point i n G and T an element i n T through p.

Proposition VII.

s in

(S, R ) i s isomorphic t o the s t a r F consisting P P ,T of the lines in F through p and belonging to the projective space T. Thus,

Then the subspace

s i s a projective space and i s of f i n i t e dimension h i f f T i s of f i n i t e P dimension

h

t

1.

Proof. -

Any element in S meets T a t a l i n e i n F (see A21). P P,T Let cp be the mapping defined by

q:SES+SnTEF Clearly,

cp

P,T'

i s one-to-one and onto (see A211). Moreover,

cp

-1

maps pencils

onto lines, i n R , belonging t o S (Indeed, any pencil i n F conP ,T P' P,T s i s t s of a l l lines in T through p , b e l o n g i n g t o a plane n; such a pencil in F

i s the image under

cp

of the l i n e r

).

P Now, i t will be proved t h a t cp maps lines in S onto pencils i n F P P,T' ,Q

137

The Grassmann manifold representing the planes in a projective space be a l i n e i n S and S and S ' two d i s t i n c t p o i n t s on it; they p,a' P meet T a t ( d i s t i n c t ) l i n e s f and f ' , which are j o i n e d by a plane a through Let r

= r (see prop. V); furthermore, v(rp,a) = v ( r p ) Psa' p,a ,,I i s the p e n c i l c o n s i s t i n g of the l i n e s i n F through p and belonging t o a . P,T It f o l l o w s t h a t cp i s an isomorphism between S and the s t a r o f l i n e s through P p i n t h e p r o j e c t i v e space T.

p i n T; then, r

From prop. Proposition

VII,

VIII.

i t f o l l o w s immediately:

L e t T and T ' be any two d i s t i n c t elements i n T through

a p o i n t p i n G. Then T i s o f f i n i t e dimension h t 1 i f f TI i s o f f i n i t e dimension

h t 1.

Next, we prove: Proposition

IX. Any T i n

T i s a 3-dimensional p r o j e c t i v e space. Furthermore,

the c o l l e c t i o n S o f a l l elements i n S through a p o i n t p i n G i s a p r o j e c t P i v e plane, which i s a subspace o f (S, R ) .

Proof. I f T Assume T #

i s the space

r, and

7

l e t q1

i n A3,

then T i s a 3-dimensional p r o j e c t i v e space.

-

be a p o i n t i n T

and q

2 (G, F ) i s connected, t h e r e e x i s t both a f i n i t e subset

p o i n t s i n G and a f i n i t e subset P1 = 41 3 Pn = q2 and l i n e fi

fi

3

{fl,

{pi,

..., f n - l 1

i = 1, 2,

pitll,

a p o i n t i n T. Ip

l,..., p n l

o f lines i n

. . ., n - 1.

have t h e common p o i n t p1 = 9,;

since

T

of

F such t h a t Through any

there i s e x a c t l y one T . i n T (see A211). The subspaces 1

Since

and T1

has f i n i t e dimension equal t o three,

V I I I , T1 i s a 3-dimensional p r o j e c t i v e space. Since TinTitl 2 [pitll ( i = 1, ...,n - 2 ) and Tn-l n T2 {pnl, a l l Tiis ( i = 1, n-1) and T

by prop.

...,

are 3-dimensional p r o j e c t i v e spaces. F i n a l l y , t o prove t h a t S i s a p r o j e c t i v e plane, i t i s enough t o r e c a l l P t h a t t h e r e e x i s t s an element T i n T through p (through p t h e r e i s a t l e a s t one l i n e f i n F which i s contained i n a maximal subspace i n T); T i s a 3-dimensional p r o j e c t i v e space and from prop. V I I the statement follows.

138 5.

A . Bichara and G. TaNini

THE SUBSPACES OF (S, R ) F i r s t we prove: Given t h r e e p a i r w i s e c o l l i n e a r p o i n t s i n (S,R), Sly

P r o p o s i t i o n X.

S2, and

S3, t hro ugh t h e same p o i n t p i n G (when c o n sidered as subspaces o f ( F , G ) ) ,

t h e r e e x i s t s a p r o j e c t i v e p l a n e i n (S, R ) through them. Proof. -

By prop. I X , S

and S3.

P

i s a p r o j e c t i v e p l a ne i n (S, R), t hrough Sly

S2,

Given t h r e e independent p a i r w i s e c o l l i n e a r p o i n t s i n (S,

Proposition X I .

n o t t hro ugh t h e same p o i n t i n G (when c o n s i dered as subspaces o f

R),

(G, F ) ) ,

t h e r e e x i s t s , i n (S, R ) , a p r o j e c t i v e p l a n e t hrough them.

P roof .

Under t h e assumptions, t h e t h r e e p o i n t s (which o b v i o u s l y e x i s t )

a r e p a i r w i s e d i s t i n c t . (Indeed,

p1 = p2 would i m p l y p1 E S1

and p1 E S2,

= p ) . Thus, t h e t h r e e l i n e s (pi,p.) i # j, i,j = 1,2,3, J 2 3 i n F are p a i r w i s e d i s t i n c t (otherwise, S , and S 3 would c o n t a i n t h e l i n e 1 ' s2 (pl,p2) = (pl,p3), which c o n t r a d i c t s A 1 1 ) . Consequently, t h e r e e x i s t s a 2 subspace i n (G, F ) c o n t a i n i n g ply p2, and p3 (see A1); t h i s subspace i s a

and t h en

p1 = p

member a o f subspace

T

n.

( S i n c e i t meets S1 a t a l i n e , i t i s cont ained i n a maximal

E T and i s t h e p r o j e c t i v e p l a n e spanned i n

Consider t h e f o l l o w i n g subset o f

:

ab t u r n s o u t t o be a p r o j e c t i v e p l a n e . I n f a c t , i f S i

elements i n lines

~ l h ,t h e n

fl and f

T b y pl, p2, and p3).

and S;

a r e two d i s t i n c t

t h e y meet t h e p r o j e c t i v e p l a n e a a t two ( d i s t i n c t )

o f F and f, and f 2 meet a t a p o i n t p i n a ; thus, S i and S; 2 belong t o t h e l i n e r i n R. L e t r and r , p # p ' , be two Pya Pya P' ,a

The Grassmann manifold representing the planes in a projective space ( d i s t i n c t ) l i n e s i n a s . Since t h e l i n e (p,p')

in

F on

CY

139

e x i s t s , the maximal

subspace S E S through i t i s the o n l y element belonging t o

r P,"

The statement f o l l o w s .

nr PI,"'

As a c o r o l l a r y t o prop. X and X I : Proposition X I I .

Any subspace o f (S, R ) i s a p r o j e c t i v e space.

6. THE COLLECTION

P

OF MAXIMAL SUBSPACES I N (S, R)

L e t p be a p o i n t i n G and S

Proposition X I I I .

(1

F ) i s a PPLS, such a subspace e x i s t s ) . p and S n S 0) i s e i t h e r a l i n e i n R

(E G) and n o t through p (since (G,

Then, the set

{S

L

E S: S 3

an element o f S through q

9 o r t h e empty set. By prop. I X , the elements i n S through p form a p r o j e c t i v e

R); thus, t h e r e i s always some element i n S through p which i s skew w i t h S * consequently, (S, R) i s a proper p a r t i a l space (see V I ) .

plane i n (S,

(1'

Proof.

Assume t h a t , through p, t h e r e are two d i s t i n c t elements o f

S' and S", both meeting S a t a p o i n t : q

S' n S = q ' , S" n S

q q ' # 9". (Indeed, q ' = q" i m p l i e s t h a t through t h e l i n e (p,q') there are two d i s t i n c t elements of

q

= 9".

say

S,

Then,

= (p,q")

S, a c o n t r a d i c t i o n ) . Since the three

p o i n t s p, q ' , and q" a r e pairwise c o l l i n e a r , there e x i s t s a subspace

a

in

n

through them. The l i n e r i n (S, R) consists o f those elements i n S P ,a through p, meeting S a t a p o i n t on (q',q"); moreover, any p o i n t on (q',q'') q i s j o i n e d t o p by a l i n e belonging t o some element i n r PP' Now, assume S"' i s an element i n S through p, n e i t h e r skew w i t h S q' nor belonging t o r ; then, S"' meets S a t a p o i n t q"' n o t on (q',q"). P ,a q Since the p o i n t s q', q"', and p a r e pairwise c a l l i n e a r , they are contained i n a plane a'

of

n,

and q"' E (q',q"), (p,Ci').

which i s d i s t i n c t from CY (otherwise, a n S a c o n t r a d i c t i o n ) ; these two planes i n

IT

= a' n S q 9 meet a t the l i n e

L e t T and 1' be the maximal subspace i n T through a and a ' ,

respectively. Then,

T

= T' (by prop.

contains the l i n e (p,q')

I , t a k i n g i n t o account t h a t T n T '

= a n a'). The three p o i n t s q ' , q", and q"' belong

140

A . Bichara and G. Tallini

t o T nS thus, the p r o j e c t i v e plane through them i n T i s completely contained q' i n S , which i s impossible (see A21). q F i n a l l y , i t must be proved t h a t i f an element S i n S through p and i n P c i d e n t w i t h S e x i s t s , then there e x i s t s a l i n e r , any element o f which q P ,a i s an element o f S through p and i n c i d e n t w i t h S Set S n S 2 {q'}; q P qthrough the l i n e (p,q') t h e r e i s a maximal subspace T o f T, meeting S a t a q l i n e through q ' , a l l whose p o i n t s are c o l l i n e a r w i t h p ; and t h i s l i n e and

.

p span a plane belonging t o Prbposition X I V .

Let S

P

n.

The statement f o l l o w s ,

be the s e t o f a l l elements i n S through a p o i n t p

i n G. I f S belongs t o S and doesn't pass through p, then the subset

(S'

'L

S

i n (S, R ) means S' i s i n c i d e n t w i t h S i n (G, F ) ) o f S

empty o r a l i n e i n R . Furthermore, S P

Proof.

P i s a maximal subspace i n

i s either (S, R ) .

The f i r s t p a r t o f the statement f o l l o w s from prop. X I I I . Again by

i s a subspace i n (S, R ) , containing some element which i s P non-collinear w i t h S. Therefore, no subspace containing both S and S e x i s t s P i n (S, R ) and S i s a maximal subspace.

prop. X I I I , S

P

Thus, the c o l l e c t i o n P = {S : p E GI o f maximal subspace i n (S, R ) , P any o f them being a p r o j e c t i v e plane (see prop. I X ) , arises. Proposition XV.

The c o l l e c t i o n

P = ISp : p E GI i s proper.

Furthermore,

any two d i s t i n c t elements i n P share a t most one p o i n t o f S.

Proof.

L e t p and q be any two d i s t i n c t p o i n t s i n G. The maximal subspace

and S i n (S, R ) share a l l the elements i n S through both p and q. I f p P q and q are non-collinear i n (G, F), then there i s no element o f S through

S

them. I f p and q are j o i n e d by a l i n e f

E

F, then ( i n ( G , F ) )

there exists

e x a c t l y one maximal subspace belonging t o S which passes through f and so through p and q. I n both cases,

S n S G 1, and the statement follows. I P 91

141

The Grassmann manifold representing the planes in a projective space

7. THE COLLECTION E OF MAXIFlAL SUBSPACES I N (S, l?) F i r s t we prove: Proposition X V I .

L e t S and S ' be two d i s t i n c t elements i n S meeting a t a

p o i n t p i n G. I f S1 and S2 are d i s t i n c t elements i n S both meeting S and S ' a t d i s t i n c t p o i n t s i n G, then S1 and S E S belonging t o the l i n e

3

s2

are c o l l i n e a r i n (S, R ) and any element

(S1, S2) i n R e i t h e r meets both S and S ' a t

d i s t i n c t p o i n t s of G o r belongs t o the l i n e ( S ,

Proof.

S') i n R.

From A I 1 1 i t follows t h a t S1 and S2 meet: I q l = S1 n S2, and thus

2

they a r e c o l l i n e a r i n (S, R ) (see prop. V I ) . Since n e i t h e r S1 nor S p, then q # p. I f q E S, any element

s

contains

2 # S on the l i n e (S1,S2) i n l?, passing

through q, meets S a t t h a t p o i n t . I f q 4 S, t h e s e t o f a l l elements i n S through q and i n c i d e n t w i t h S i s e i t h e r a l i n e i n R o r t h e empty s e t (see prop. XIV); since both S, and S2 pass through q and are i n c i d e n t w i t h S, any element on the l i n e (S1, S 2 ) i n R i s i n c i d e n t w i t h S. By the same argument, any element on (S1, S2) i s proved t o be i n c i d e n t w i t h S ' ; thus, i t i s o n l y

S2) passing through p, then i t belongs t o the l i n e (S, S ' ) . Under these assumptions, S3 contains both p and { S t c E S: S" 3 p q, and, thus, meets S1 a t q. Hence, S3 E L ' , where L ' and S" n S1 # 01; a l s o S and S ' belong t o L ' (by hypothesis). By prop. X I V , S3, S and S' belong t o a l i n e i n R ; so, S3 E (S,S') and the statement i s t o be proved t h a t i f S3 i s an element on (Sly

proved. L e t S and S ' be two d i s t i n c t elements i n S, which are c o l l i n e a r i n (S, R ) ; then S and

S' share a p o i n t p i n G. Denote by o(S,S') t h e s e t con-

s i s t i n g o f a l l elements i n S t h a t e i t h e r belong t o the l i n e (S,S')

i n R,

o r meet both S and S ' a t p o i n t s i n G d i s t i n c t from p. Proposition X V I I .

I f S and S ' a r e d i s t i n c t elements i n S which are c o l l i n e a r

i n (S, R ) , then the s e t

o(S, S ' ) i s a subspace i n (S, R ) and i t properly

contains the l i n e (S, S') i n R. Proof

.

Let T be a member o f T through the p o i n t p, where { p l = S n S';

5y

142

A. Bichara and G. Tallini

A21, T meets S and S' a t the lines f and f ' in F , respectively. I f q

E

f,

q ' E f ' , q , q ' # p , then q # q ' and q and q ' belong to a l i n e f " i n T. Through

f " there i s an element S" E S meeting S and S' a t q and q ' , respectively. Thus, S" belongs t o a(S,S') \ (S,S'); therefore, the l i n e (S,S') i s properly contained i n G(S,S'). To prove t h a t a(S,S')

points S l y S2

E

o(S,S')

i s a subspace in (S, R ) ( i . e . any two d i s t i n c t

are collinear in (S, R ) and the l i n e joining them

i s completely contained in o(S,Sl)), three cases will be considered. ( i ) If S1 and S2 both belong t o the l i n e (S,S'), then there i s nothing

t o prove. ( i i ) If

f1 and

S2 meet S and

S' a t d i s t i n c t points, then the statement

follows from prop. XVI. ( i i i ) If S1 meets both S and S' a t d i s t i n c t points and S2 belongs t o the line (S,S'), then

-

by an argument simular t o t h a t in prop. XVI

- it

is

easy to prove t h a t S and S are collinear and any element on ( S l y S2) belongs 1 2 to o(S,S'). Proposition XVIII.

Let S and S' be two d i s t i n c t elements i n S which are

collinear i n (S, R ) , i . e . they meet a t a point p i n G. Then there e x i s t exactly two maximal subspaces o f (S, R ) t h r o u g h S and

s';

the f i r s t one i s S ( i . e . P

i t consists o f a l l elements i n S through p ) and belongs t o P; the second one i s a(S,S'). Proof.

These two subspaces share exactly the elements on the l i n e ( S , S l ) .

Let H be a subspace of (S, R ) containing both S and S' and so the

l i n e (S, S l ) in R. Then: e i t h e r H consists of elements in S a l l of them through p and H i s a subspace of S , o r there exists some S" E S contained i n H and P not passing through p. Since H i s a subspace and S", S, and S' belong t o H,

S" i s collinear w i t h b o t h S and S ' , i . e . meets S and S' a t d i s t i n c t points. By prop. XIV, the elements in S which are collinear with S" a r e exactly the

P

elements on the line ( S , S l ) i n R; hence, H n S = (S, S'). Therefore, any P element i n H (S, S'), being collinear w i t h both S and S', meets S and S' a t points i n G d i s t i n c t from p (and from each other). Thus H C o(S, S') and the statement follows. Proposition XIX.

Let S and S' be two d i s t i n c t elements i n S, which are col-

143

The Grassmann manifold representing the planes in a projective space l i n e a r i n (S, R ) . IfS1 and S 2 are d i s t i n c t elements i n u(S,S'), then they meet and u(S1, S 2 ) = u ( S , S'). Proof. Since S, and S2 belong t o the subspace and thus have a common p o i n t

(I E

u ( S , S'), they a r e c o l l i n e a r

G. By prop. X V I I I , the subspace

u(S, S ' ) ,

containing S1 and S2, i s contained i n e i t h e r the maximal subspace S , o r the q maximal subspace u(S1, S2). Since the elements i n u ( S , S l ) d o n ' t a l l pass through a same p o i n t ,

u(S, S ' )

5 u(S1,

S 2 ) . E q u a l i t y f o l l o w s from

a(S, S ' )

being a maximal subspace i n (S, R ) .

8.

FURTHER PROPERTIES OF

P AND C

I n the PLS (S, R ) two c o l l e c t i o n o f maximal subspaces have been defined (see prop. X I V and X V I I I ) :

P = IS : p P

E

GI, a proper c o l l e c t i o n ;

X = {u(S, S') : S, S' E S , S # S', S n S' # 01. Proposition XX.

For any r i n R , t h e r e e x i s t a unique S

u i n E such t h a t r C S and r 5 u.

P

i n P and a unique

P

Proof.

Since any two d i s t i n c t elements on r a r e i n c i d e n t , t h e statement

f o l l o w s from prop. X V I I I and X I X . Proposition X X I . or a l i n e i n R.

If S E P

P and u

E

X, then S n u i s e i t h e r t h e empty s e t

Proof, -

The Set S n u consists o f a l l elements i n S through p which are P c o l l i n e a r w i t h any element i n u. Take S E S n u and l e t S' be any other P element i n a; then (see prop. X I X ) u = u(S, S') and S E S I f {p) = S n S ' ,

P'

then the statement f o l l o w s from prop. X V I I I . I f p 4 S ' , then the s e t o f a l l elements i n S

- by prop.

XIV

-

i n c i d e n t w i t h S ' i s a l i n e r i n R , containing P S. L e t S" be a p o i n t on r, d i s t i n c t from S. Since S" meets S and S ' a t d i s t i n c t

144

A . Bichara and G. Tallini

p o i n t s , S " E u ( S , S'); t h e r e f o r e ,

( b y prop. XIX) u(S, S ' ) = u ( S , S " ) and t h e

p r e v i o u s argument proves t h e statement. Any two d i s t i n c t elements i n

Proposition X X I I .

c s h a r e a t most one p o i n t o f

S. I f u , u ' and u" a r e t h r e e p a i r w i s e d i s t i n c t elements i n l u ' n u''1 = 1, t h e n Iu n u"I

=

1. Consequently, i f {uj : j

i s a f i n i t e sequence of elements i n

1 t h e n Iu n u o ntl Proof.

>

c

such t h a t Iui n uitlI

and Iu n u ' l = 1,

E

{O,l,...,ntl}}

2 1 (i=0,

...,n ) ,

1.

From prop. X V I I I i t f o l l o w s t h a t any two d i s t i n c t elements i n Z share

a t most one p o i n t i n S. l l r i t e I S ' ) = u n u ' and IS") = u ' n u". I f S ' = S", t h e n t h e s t a t e m e n t i s o b v i o u s l y t r u e . I f S' # S " , t h e n S' and S " , b e l o n g i n g t o t h e subspace u ' , a r e c o l l i n e a r i n (S, R ) and l e t r be t h e l i n e j o i n i n g i n P t h r o u g h r, which P meets u ' a t t h e l i n e r ( s e e prop. XXI) and u a t a l i n e s' t h r o u g h S ' and u" them. By p r o p . XX, t h e r e e x i s t s a maximal subspace S

a t a l i n e s" t h r o u g h S". The l i n e s s' and s" i n R a r e d i s t i n c t ( o t h e r w i s e , b y prop. XX, u = u " , w h i c h i s i m p o s s i b l e ) and b e l o n g t o S , a p r o j e c t i v e p l a n e P ( b y prop. I X ) . I t f o l l o w s t h a t t h e r e e x i s t s e x a c t l y one element S i n S bel o n g i n g t o s' n so'; c l e a r l y , S belongs t o u ( a s S

u" (as S

E

s" and s" 5

0");

an element i n u Proof.

i

Let u

1

s' and s' C u ) and t o

thus, IS} C u n u". S i n c e u # u'', b y t h e f i r s t

p a r t o f t h e statement, IS1 = u n u"; t h e r e f o r e , Proposition X X I I I .

€

ICJ n u"1

= 1.

and u2 be two d i s t i n c t elements i n C .

( i = 1, 2 ) , t h e n I S 1 n S21

>

I f Si

1 i m p l i e s (ul n u,l I

>

is

1.

I f S1 = S2, t h e s t a t e m e n t i s o b v i o u s l y t r u e . Assume S1 # S2; t h e n

S1 n S2 c o n s i s t s o f a p o i n t p i n G. L e t fl be a l i n e o f

F in

S1 t h r o u g h p

t h e r e i s e x a c t l y one T E T which 1 meets S2 a t a l i n e f 2 o f F. T i s a 3-dimensional p r o j e c t i v e space ( s e e p r o p . (and such a l i n e does e x i s t ) ; t h r o u g h f

I X ) ; thus, t h e r e e x i s t s a l i n e f i n T t h r o u g h p d i s t i n c t f r o m fl and f2. Through f t h e r e i s an element S i n S , w h i c h i s c l e a r l y d i s t i n c t f r o m S1 and S2. Therefore, t h e maximal subspace

floreover, ul n u(S1, S)

3

u(S

S ) and i' I S 1 l , u(S1,S) n u(S,S2)

From p r o p . X X I I t h e s t a t e m e n t f o l l o w s .

u(S, S 2 ) i n Z e x i s t .

2

I S } , and u(S,S2) n u2 ?IS2}.

145

The Grassmann manifold representing the planes in a projective space Proposition X X I V .

and u ' i n X share e x a c t l y one

Two d i s t i n c t elements u

p o i n t i n S.

Proof,

By prop. X X I I , i t i s enough t o prove t h a t there e x i s t s a c o l l e c t i o n

,...

,...

: j E t0 , n t l I l o f elements i n Z such t h a t lul nu. I 1 (i=O ,n) j 1t 1 and uo = u, untl = u'. Take S E u and S ' E u'. Assume S # S' (otherwise t h e

{u

q2 a p o i n t i n S ' , w i t h

statement i n t r i v i a l ) and l e t q1 be a p o i n t i n S,

F ) i s connected, t h e r e e x i s t a f i n i t e subset o f G, {p1,...,pnt21, and a f i n i t e subset o f F, Ifl ,. . ,fntl I, such t h a t q1 = p1 , phtll (h = 1, ...,n t l ) . Through each l i n e fh i n F q2 = pnt2 and fh 1 {p,, q1 # q2.

Since (G,

.

there i s a maximal subspace Sh i n S (see A211). The c o l l e c t i o n {S1,...,S

ntl e i t h e r are equal o r have

consists o f elements i n S such t h a t Sk and S ktl' the maximal subspace uk = u(Sk,Sktl) t i o n {ak : k = l,...,nI,

>

lat n

1

(t

>

ktl i t may be assumed Sk # Sktl.

H.1.o.g.

e x a c t l y one common p o i n t p

at n u

1,

3 tS I ttl

ttl

l,...,n-1).

(k

Writing u

...,n)

..., n-1);

u and u

=

> 1. Since

Then,

i n I: e x i s t . I n the c o l l e c -

( t = 1, 0

1

ntl

hence,

= u', l e t ' s prove

and S n S 2 E u 1 1 1and q2 E S', {qll, by prop. X I I I , Iuo n ulI > 1. Since q2 E Sntl 'ntl 'n and S' E u', lan n u') > 1 and t h e statement i s proved.

t h a t luo n ull

9.

1 and lan n untlI

S E uo, S

THE PROOFS OF THEOREH 1 AN0 2 By the previous sections, t h e space (G,

F ) i s associated v i t h t h e PPLS

(S, R) (see prop. X I I I ) s a t i s f y i n g :

Given t h r e e pairwise c o l l i n e a r points, t h e r e e x i s t s a subspace conA;) t a i n i n g them (see prop. X and XI).

No l i n e i s a maximal subspace; moreover, t h e r e e x i s t two c o l l e c t i o n s , say c and P, o f maximal subspaces i n (S, R) such t h a t an:' maximal subspace belongs e i t h e r t o C o r t o P (see prop. X I V , X V I I I and XX) and A')

(I) (11) (1111

a,

U'

E

c,

u # u'

u E Z, n E P

v

r E R

=s>

* d!

u

*

l o n o ' \ = 1 (see prop. X X I V ) ;

either E

C,

u n n = 0 or

j! n

E

P :

r

5

u n n E R (see prop. X X I ) ;

u, r

5n

(see prop. X X ) .

Therefore (see [ 31 ), ts being the s e t o f a l l elements i n C through S E S

Its( > 2 ; furthermore, I: =

I t s : S E S I i s a proper c o l l e c t i o n and the p a i r

,

A . Bichara and C. Tallini

146

i s a p r o j e c t i v e space. The mapping p: L

(C,L)

+

S, defined by

p ( l l s ) = S, i s

one-to-one and onto and maps p e n c i l s o f l i n e s i n L onto elements o f R and i t s -1 inverse mapping p maps l i n e s i n R onto p e n c i l s o f l i n e s i n L. F i n a l l y , P maps r u l e d planes onto elements o f P and p-’ maps elements i n P onto r u l e d planes. be an element i n et be the c o l l e c t i o n o f a l l planes i n L). L e t -P andLLTI the s e t o f a l l l i n e s i n then p(LTI) i s an element S i n P. Consider

: P’+

F

the mapping

(C,

TI

IT;

P

G, defined by

I t i s easy t o check t h a t :

Proposition XXV.

F i s one-to-one and onto.

Next, L e t L be a l i n e i n L. The image under F o f t h e s e t o f a l l

Proposition XXVI. planes i n

7 through L

i s the element S = p ( L ) i n S. The image under F - l o f an

element S i n S i s the s e t o f a l l planes i n

7 through

the l i n e

L

p-l(S) i n

L. Therefore, S i s the c o l l e c t i o n o f the images under F o f t h e sets o f a l l

planes i n (C,

Proof. L c

TI

If Q

L ) through any l i n e i n L.

7 and

IT E

$? E

p(LTI) = S

L T I ep ( L )

B

then P’ p(LTI) Q S

- by E

S

P

- F(n)

(9.1) Q

p

E

= p.

Moreover,

S, and the statement follows.

Two d i s t i n c t elements p1 and p2 i n G are c o l l i n e a r i n

Proposition X X V I I .

(G, F ) i f f they are t h e images under F o f two planes i n (C,

L)

which meet

a t a l i n e o f L. Proof.

Set ri

F

-1

(p.), i .1

1,2.

By prop. X X V I , F(ni) E S , i = 1,2;

t h e subspace S o f (G,

I f rl n n

2

i s a l i n e L i n L, s e t S = ~ ( l ) .

hence, the p o i n t s p1 and p2, belonging t o

F), a r e c o l l i n e a r i n (G, F).

Conversely, i f p1 and p2 a r e j o i n e d by the l i n e f i n

F , l e t S be the

The Grassmann manifold representing the planes in a projective space

maximal subspace i n S through f.

Obviously, pl,

147

p, E S; hence n 1 and n 2

meet a t the l i n e p - l ( S ) i n L and the statement follows. Proposition X X V I I I .

If

7

7 which

i s the s e t o f a l l t h e planes i n

a 3-dimensional subspace i n (C, L ) , then the image under F o f

7

belong t o

belongs t o T .

Furthermore, the inverse image under F o f an element i n T i s t h e s e t o f a l l the planes i n a 3 - f l a t o f

Proof.

First it w i l

be proved t h a t i f T

F),

E

T, then F-’(T) c o n s i s t s o f a c o l -

contained i n the same 3 - f l a t i n (1, L ) . Since T i s a

l e c t i o n o f planes a1 subspace of (G,

(c, L ) .

t s p o i n t s a r e pairwise c o l l i n e a r ; hence, F-’(T)

o f planes p a i r w i s e meeting a t l i n e s i n L (see prop. X X V I I ) .

consists

Thus, F-’(T) con-

s i s t s o f e i t h e r planes i n a 3 - f l a t o f (I, L ) , o r planes through a l i n e . L e t ’ s prove t h e l a t t e r case i s impossible. L e t

pl,

-

p,

and p be t h r e e independent 3 by A2 T c a n ’ t be a l i n e ) . The

-

p o i n t s i n T (such p o i n t s do e x i s t because -1 -1 planes F (p,) = n1 and F (p,) = n, meet a t a l i n e L i n L. F-l(p3) meets

IT^

The plane r3=

and IT, a t l i n e s d i s t i n c t from L and from each other. (Indeed,

i f n l s IT,, and n3 passed through L , then t h e t h r e e p o i n t s pl, would belong t o the element

p ( L ) = S i n S, which would share w i t h T t h e p o i n t s

on the l i n e s j o i n i n g a l l t h e p a i r s o f p o i n t s pi, this i s

impossible (see A,I).)

the 3 - f l a t j o i n i n g nl and n,;

p2, and p3

pj,

i # j, i,j

1 , 2 , 3 , and

Thus, F - l ( T ) consists o f a l l t h e planes i n t h i s space w i l l be denoted by ( n l ,

IT,). Let n

i = 1,2,3. Then n n n = L i is i is = 1,2,3, are t h r e e l i n e s i n L, two of them a t l e a s t being d i s t i n c t (as nl, IT, be a plane i n (nl, n,),

d i s t i n c t from n

and n3 d o n ’ t form a p e n c i l ) . Assume Ll # 1, (a s i m i l a r argument holds i n t h e other cases).

The t h r e e p o i n t s p = F(n), p1 and p2 a r e p a i r w i s e c o l l i n e a r

(as IT, nl and n, p a i r w i s e meet a t l i n e s i n L). subspace i n (G,

F) containing p, pl, and p.,

Thus, by A1, t h e r e e x i s t s a

L e t H be a maximal subspace i n

(G, F ) through these points. H cannot belong t o S, because n = F-’(p), -1 -1 and n, = F (p,) d o n ’ t pass through t h e same l i n e (see prop. n1 = F (p,)

XXVI).

Hence, H i s an element i n T through p1 and p, and so H = T (T being

the o n l y maximal subspace i n T through the l i n e (pl,

p,))

and p E T.

Thus, the image under F o f any plane i n IT^, IT,) i s an element i n T, -1 and conversely, i.e. F (T) i s t h e s e t o f a l l planes i n (nl, n,).

148

A . Bichara and G. Tallini

planes i n

By a s i m i l a r argument, the image under F o f t h e set o f a l l t h e

a 3-flat o f

(c, L) i s proved t o be an element T i n T, and t h e p r o o f i s complete.

Proposition X X I X .

Three planes

IT^, IT2 , and IT3 i n

(C,

L ) form a p e n c i l ( i . e .

they belong t o the same 3 - f l a t and have a common l i n e ) i f f t h e i r images under p2, and p3’ are t h r e e p o i n t s i n G which belong t o the same l i n e i n F .

F, pl,

IT^, and IT3 form a p e n c i l , then they pass through a l i n e .t i n L and belong t o t h e 3 - f l a t (v1, IT^) spanned by v and IT Set S = P ( L ) and 1 2‘ Proof.

If v1 ,

l e t T be t h e image under F o f the s e t o f a l l the planes i n X X V I and X X V I I I , {p1,p2,p31

5

S, {p1,p2,p3)

e i t h e r the empty s e t o r a l i n e i n

5

T.

IT^).

(n1,

Since (by A21)

By prop.

T nS is

F , t h e p o i n t s ply p2, and p3 belong t o a

line. Conservely, i f

ply

p2 and p3 belong t o t h e l i n e f i n

T E T be the two maximal subspaces i n (G,

IT^,

n2, and

IT^

pass through the l i n e p-’(S)

a l l the planes i n a 3 - f l a t i n

(c, L).

F, l e t S

E

S and

F ) through f. The t h r e e planes and belong t o t h e s e t F-’(T) o f

Therefore,

IT,,

n2 and

and the statement i s proved.

IT

3

form a p e n c i l

From the r e s u l t s i n t h i s section theorem 1 f o l l o w s . F i n a l l y , i f (G, F ) i s a non-connected PPLS s a t i s f y i n g A1 and A2 and G P i s the connected component o f p (p E G) ( i . e . the s e t o f a l l t h e p o i n t s t h a t

F), then

can be reached from p by a polygonal path c o n s i s t i n g o f l i n e s i n

F containing a p o i n t i n G i s completely contained i n G and any P P subspace i n (G, F ) which i s n o t skew w i t h G i s completely contained i n G P’ P Hence, denoting by F t h e s e t o f a l l t h e l i n e s i n F which are contained i n G P P’ the p a i r ( G F ) i s a connected PPLS s a t i s f y i n g axioms A1 and A,. Theorem 2

any l i n e i n

follows.

P’

P

ACKNOWLEDGEHENT.

L

This research was p a r t i a l l y supported by GNSAGA o f CNR.

The Grassmann manifold representing the planes in a projective space

149

REFERENCES

[ 11 [

B. Segre, Lectures on modern geometry, CremoneseEd. Roma (1961)

21 G. T a l l i n i , Spazi p a r z i a l i d i r e t t e , spazi p o l a r i , Geometrie subimmerse, Quaderni Sem. Geom. Combinatorie, 1 s t . Mat. Univ. Roma, n. 14 (gennaio 1979).

[3]

G. T a l l i n i , On a c h a r a c t e r i z a t i o n o f t h e Grassmann m a n i f o l d r e p r e s e n t i n g t h e l i n e s i n a p r o j e c t i v e space, i n : P.J. Cameron, J.!4.P. H i r s h f e l d , D.R. Hughes (eds.) F i n i t e Geometries and designs. London l l a t h . SOC. Lect. Notes S e r i e s n. 49. Cambridge U n i v e r s i t y Press (1981) 354-358.

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Annals of Discrete Mathematics 14 (1982) 151-158 0 North-HollandPublishing Company

ON INHERITED GROUPS OF DERIVABLE TRANSLATION PLANES Mauro Biliotti Universita di Lecce

Summary. We examine some questions about the structure of inherited groups of derivable translation planes in connection with a recent classification [ 13 of derivable nets. In a forthcoming paper [l]

G.

Lunardon and the author introduce

a classification of derivable nets of finite affine translation planes. This classification is based on the number of the Baer subplanes through a common point which are contained in a derivable net

of a translation plane TI and which are left invariant by the dilation group of II fixing that point. The knowledge of this number seems to be very useful to gain information about the collineation group of II leaving invariant the derivable net, the so called "inherited group", which acts as a collineation group also in the derived plane

of II. There is indeed a result of Foulser [3]

asserting that

if a Baer subplane contained in a derivable net of II is not left invariant by every dilation of II fixing a point in it, then the collineation group of II which fixes pointwise this Baer subplane has order at most two. It is the aim of this paper to produce some results which can be obtained by the above arguments. In the first section we consider the subgroup of the inherited group generated by the affine elations of

and we show that the

structure of this group depends on the type of the derivable net of

II and that it may be very constrained. In the other section we get some results about the subgroups of the inherited group fixing two incident lines in the derivable net of II. 151

M. Biliotti

152

0. Let I'i be a finite translation plane of order q

2

,

k q = p , p a

prime number, and let N be a derivable net of II. Denote by

fi the af-

fine translation plane obtained by replacing N by its unique replacement net

i.Let

0 be a fixed point of II (and of

f i ) . We shall deno-

te by B(N) the set of the Baer subplanes of II which contain 0 and are contained in N. Moreover let C(II,N) be the collineation group of II fixing 0 and N and let D ( n ) be the dilation group of II fixing 0. A similar symbolism is given for the plane

y.

Below we recall

some known results. RESULT 0.1 [l].

The number of the Baer subplanes

-are left invariant b~

D(n)

is

0, 2

of B(N)

which

q+l.

The net N is said to be of type I, 11.0 or 11.2, according to the number of D(n)-invariant Baer subplanes of B(N) is q+l, 0 or 2.

The collineation group of

RESULT 0.2 [3].

a -Baer subplane of B(N) which is not -

fl fixing pointwise

D ( I I ) -invariant, has order ,< 2.

1. Denote by E(F,i) the subgroup of C(fi,i) generated by the affine elations of

F.

THEOREM 1.1. Let E(n,i) # . If N is of type I, then one of

the following holds: a) E ( n , i ) is an elementary abelian p-yroup; t tlk ; b) E(Eri) SL(2rp 1 r c) p = 2 d) p = 3

and and

k 5 dihedral group of order 2s where slp k l ;

E(if,i) &s

SL(2r5).

E(fi,i)

If N is of type 11.0, then e) p = 2

and

where hlk

and

is

g dihedral group. Moreover ( D ( I I ) k 2(ph+l) llE(f,fi) 1 1 2(p +1).

E(n,i)

If N is of f) p = 2

and

where hlk

11.2,

E(F,i)

and

g dihedral group. Moreover I D ( I I )

h 2(p +1) llE(Zri) 1 1 2(pk-1).

i.Therefore

E(fi,i)

= p

I

= p

then

Proof. An affine elation of axis in

1

2h

2h

fi leaving R invariant, has its

fixes every Baer subplane of B(i). From

this it follows that E(5,K) is faithful on every Baer subplane of B(i)

. Since the Baer

subplanes of B(i) are desarguesian (see

131),

On inherited groups of derivable translation planes

then E ( E , N )

153

k is (isomorphic to) a subgroup of SL(2,p ) generated by

elements of order p. If N is of type I, then no restriction can be made on the structure of E ( i , i ) except that given by Dickson's theorem (see [53

, Hauptsatz

8.27)

. Thus we

have a) , b) , c) and d)

Let N be of type 11.0. Since an affine elation of C(:,i) its axis in

if

.

has

then it acts on II as a Baer collineation fixing

pointwise a Baer subplane C contained in N . But, by Result 0.2, this Baer collineation is involutorial since, by assumption, C is not

E

D(n)-invariant. Thus affine elations of are involutorial, Ti has k even order and E(E,i) 6 SL(2,2 1 . Let S be a Sylow 2-subgroup of E('i,i).

S is an elementary abelian group. Suppose there exists an

element u L S which fixes pointwise a Baer subplane C of

fi.

Then

there are only two possibilities, namely C is also a Baer subplane

n.

of II, or C is a line of

But we have seen that u fixes every Baer

subplane of B(i) and hence every line through 0 of II. So C cannot be a line of II. Likewise, if C is a Baer subplane of contained in N and hence it is a line of

if

n,

then it is

contrary to our assum-

ptions. Therefore every element of S is an affine elation. Furthermore, since S is an elementary abelian group, all the elations of S have the same axis. If I S 1 > 2, this contradicts Result 0.2 and so I S 1 = 2. But I E ( n , i )

I

> 2 since

i

does not contain C(n,\)-inva-

riant lines and thus, by Dickson's theorem, E(i,i) is a dihedral k group of order 2s, where s12 +1. Let u be an affine elation of E ( Y , \ )

and let C be the Baer sub-

plane of N fixed by u. Denote by Q a quasifield which coordinatizes

II. We may choose our quadrangle of reference inside C (and with 0 as origin) so that Q is a right two-dimensional vector space over its subfield K which coordinatizes C (see r3-J). Since N is of type 11.0, then the kernel N(Q) of Q is not contained in K (see [l]). Furthermore

0

induces on Q an involutorial automorphism which fixes

K pointwise and leaves N(Q) invariant. Thus N(Q) is a quadratic extension of N(Q)

n

K. By [I]

,

Teorema 2, the stabilizer of each Baer

subplane of B(N) in D(II) is the subgroup D1 of D(II) whose elements are the dilations of II induced by elements of N(Q)* fl K. So, if

'

154

M. Biliotti

h ID(n)I = 22h-l, then ID I = 2h-l and D(II) = D1xD2 with ID2 I = 2 +l. 1 Since D(n)' = D(II) I we then have also that :D = D2. Moreover every element of D2 does not fixe any line of

ithrough

0 and so

-E(II,N)

0

acts

f.p.f. on D2. Thus E2 = D 2 ~ < a >is a subgroup of and h 2(2 +1) IE(ii,i) I . Let E(i,i) = DX. Since E(t,i) acts faithful on

I

ak contained in i,then there are only two possibilities, namely D does not fixe any line of ithrough 0 and ID1 ak+lr or D fixes exactly two lines of i through 0 and a desarguesian plane of order

1

I

( DI Zk-l. But N is of type 11.0 and hence D2 6 D does not fixe any

line of

i through

0. Thus the first possibility occurs and

IE(nlN)I 12(2k+1). The proof of f) is similar. REMARK 1.2. Suppose that E(5,i) = DX(a>

is a dihedral group and

that N is of type 11.0 or 11.2. The question arises if every element of D is a dilation of II. In all the examples which are known to the author, this is the case. An affirmative answer was obtained for derived semifield planes (see [ 2 1 , Theorem 2.3)

.

REMARK 1.3. Derivable nets of type 11.0 or 11.2 yield several restrictions on the existence of affine homologies in the derived plane. If N is of type 11.0, then an affine homology of C ( i f , i ) with axis in

is necessarily involutorial. If N is of type 11.2, then

an affine homology of C(i,i) either has its centre and its cocentre in the points at infinity of the lines of

icorresponding to the

D(ll)-invariant Baer subplanes of B(N), or it is involutorial and its centre and cocentre do not lie in

i.

2. The type of a derivable net N allows us to gain information about the structure of the subgroups of C(II,N) fixing two distinct lines of N through 0. We call such a group an "N-autotopism group". THEOREM 2.1.

Let A

be an N-autotopism group of

plane Il of even order, with derivable or 11.2, then a Sylow 2-subgroup -

of

=N.If

translation

N is of type 11.0

A admits a normal 2-complement.

Proof. Let S be a Sylow 2-subgroup of

A.

We shall prove that S

On inherited groups of derivable translation planes is in the center of its normalizer in

A,

by applying Burnside's theorem (see [5],

155

so that the thesis follows Hauptsatz 2.6). We start

to observe that the number of the Baer subplanes of B(N) which are k not D(ll)-invariant, is 2 +1 or pk-l. Since these Baer subplanes are permuted by

then S fixes at least one Baer subplane C of B(N)

S,

which is not D (TI) -invariant. If

S

acts faithful on C, then it is contained (up to isomor-

phism) in a Sylow 2-subgroup of the autotopism group of the desarguesian plane C and hence it is a cyclic group. Then it is well known (see If

S

[s],

S

0

S

is in the centre of its normalizer.

of S which fixes C pointwise, has order two. There-

0

is in the center of

lows that either So

S

does not act faithful on C, then, by Results 0.1 and 0.2,

the subgroup fore

Satz 2.7) that

S

and S/So is cyclic. From this it fol-

is a cyclic group, or

S

S

is the direct product of

by a cyclic group. We have to examine the second case. Let

S o = , T~ =

I. Since

T

fixes C pointwise, then C is the only Baer

subplane of B(N) which is left invariant by

T

also fixes C . By Results 0.1 and 0.2, S

NA(S)

and hence by 0

S.

So

is the subgroup of

NA(S) which fixes C pointwise, so that T is in the center of N ( S ) . A

But S contains exactly two other involutions, besides

T,

and hence

all involutions in S must be centralized by every element of

-

NA(S)

S. A s S is abelian, then all involutions of S are in the

center of

NA(S).

Thus, by applying a result of Huppert (see [ 4 ] ,

Hilfssatz 1.5), we conclude that S is contained in the center of NA(S). This completes the proof.

REMARK 2.2.

normal complement

A

N

of a Sylow 2-subgroup of

-

A

leaves invariant at least two (desarguesian) Baer subplanes i 1' C2 k of B ( i ) , so that we have N/NO < rL(2,2 1, where No is the subgroup of N fixing

i1

of ll with axis

(or

il

i 2) pointwise. But

No

is a group of homologies

and centre in the point at infinity of

f2, fur-

thermore No acts as a group of homologies with the same centre and axis also on every Baer subplane of B ( N ) and therefore it is a cyclic group. So the structure of culty.

N

could be determined without diffi-

M. Biliotti

156

We may still observe that results of Theorem 2.1 do not extend to N-autotopism groups with N of type I. Assume indeed that II is a Hall plane and that N is the derivable net of il whose replacement gives rise to a desarguesian plane. Then it is well known that N is of type I and that an N-autotopism group of II contains SL(2,q). As an immediate result of the classification of derivable nets, we have also the following proposition. PROPOSITION 2.2.

II &

translation plane with derivable

net N and let :be its derived plane. line -of

N through 0, then each

If C(II,N) dilation of ?i is

fixes exactly

one

also a dilation 9 1 1 .

Proof. It suffices to prove that the derivable net

-

of f is of

type I. N cannot be of type 11.0, since, in this case, there is a dilation of

ii

is

which does not fixe any line of N through 0. If

of type 11.2, then the line fixed by C(II,N) must be a D(f)-invariant Baer subplane of B(i), But then C(II,N) fixes also the line of N through 0 corresponding to the other D ( ? ) -invariant Baer subplane of B(i) , contrary to our assumptions. Thus, by Result 0.1,

is of

type I. COROLLARY 2.3.

The

dilations

of

derived semifield plane are

dilations also in the corresponding semifield plane. Corollary 3.2 was proved also by Johnson and Rahilly in [S].

REFERENCES Biliotti, M. and Lunardon, G., Insiemi di derivazione e sottopiani di Baer in un piano di traslazione, Atti ACC. Naz. Lincei, C1. Sci. Fis. Mat. Nat. (8) 69 (1980) (to appear). Biliotti, M. and Menichetti, G., Derived semifield planes with affine elations, J. Geometry (to appear). Foulser, D.A., Subplanes of partial spreads in translation planes, Bull. London Math. SOC. 4 (1972) 32-38. Huppert, B., Subnormale Untergruppen und p-Sylowgruppen, Acta Sci. Math. Szeged 22 (1961) 46-61.

On inherited groups of derivable translation planes

157

151 Huppert, B. , Endliche Gruppen I (Springer-Verlag, Berlin Heidelberg New York, 1967).

[6] Johnson, N.L. and Rahilly, A . , On elations of derived semifield planes, Proc. London Math. SOC. (3) 35 (1977) 76-88.

This Page Intentionally Left Blank

Annals of Discrete Mathematics 14 (1982) 159-168 Q North-Holland Publishing Company

ON (q+z)-SETS

I N A NON-DESARGUESIAN PROJECTIVE PLANE OF ORDER q

-

P. B I S C A R I N I F . CONTI Dipartimento d i Matematica

U n i v e r s i t a d i Perugia,

("1

Italy

$ 1 . 6. Segre [71 and G. T a l l i n i [9] discovered t h e by now c l a s s i c r e s u l t t h a t i n a Desarguesian p r o j e c t i v e plane o f o r d e r q, PG(2,q), i s n o t a complete q-arc. A. B a r l o t t i [l] l a t e r proved t h a t t h i s r e s u l t cannot be extended t o a nonDesarguesian p r o j e c t i v e p l a n e e x h i b i t i n g a complete o r d e r 9. R.H.F.

9-arc i n a Hughes plane o f

Denniston [3], w i t h a computer, determined a l l t h e complete 9-arcs

i n t h e known non-Desarguesian planes o f o r d e r nine. G. M e n i c h e t t i [6] has more r e c e n t l y proved t h a t i n every p r o j e c t i v e plane o f o r d e r q=2 k ,k,4, t h e r e a r e complete q-arcs. It i s n o t knownif complete q-arcs i n every non-Desarguesian plane

Hall o f odd

order q e x i s t o r not. I n t h i s paper we study a problem r e l a t e d t o complete q-arcs. L e t n be a p r o j e c t i v e plane o f o r d e r q; l e t a s e t , n , o f q t 2 p o i n t s o f

n

l e t a l i n e r o f n where r meets n i n e x a c t l y i p o i n t s be c a l l e d an i - s e c a n t . I n t h e f o l l o w i n g , a 1-secant o r 2-secant l i n e

be c a l l e d a ( q t 2 ) - s e t ; (i=0,19,.,)

i s c a l l e d a tangent o r a secant r e s p e c t i v e l y . Given a ( q t 2 ) - s e t , n i n n we denote by $(P) t h e number o f secants pass i n g through a p o i n t P E n ; o b v i o u s l y $ (P) i q t l . q t l a n u c L u and we denote by 0 We c a l l a p o i n t PE 0 such t h a t $ ( P ) the set o f nuclei o f 62. Recently A . Bichara and G. Korchmaros o b t a i n some p r o p e r t i e s o f c a r d i n a l l t y of a s e t o f n u c l e i o f a ( q t 2 ) - s e t i n a Desarguesian p r o j e c t i v e plane PG(2,q)

q = ph , p prime. Among o t h e r t h i n g s , t h e y f i n d t h e f o l l o w i n g r e s u l t : Theorem

(")

-

I f 101 '>2 then q i s even.

Research p a r t i a l l y supported by G.N.S.A.G.A. 159

(CNR).

P. Biscarini and F. Conti

160

I n t h i s paper we prove t h a t t h i s theorem i s n o t t r u e i n a non-Desarguesian p r o j e c t i v e plane. 3 a method o f c o n s t r u c t i o n f o r ( q t 2 ) - s e t s With t h i s aim we introduce i n w i t h a t l e a s t three n u c l e i beginning from a complete q-arc. I n 5 4 we determine a l l 11-sets w i t h more than two n u c l e i i n the t h r e e known non-Desarguesian planes o f order 9 obtained from t h i s construction. As regards the construction, i t i s o f i n t e r e s t t o study t h e s e t o f p o i n t s

which l i e on hioh

tangents o f a complete q-arc, and which we c a l l extha ex&

(qt1)/2

points. I n $ 5 and §6 we s h a l l prove the f o l l o w i n g r e s u l t s :

-

THEOREM 1 L e t N be a s e t o f e x t r a e x t e r i o r p o i n t s o f a complete q-arc,w, i n a p r o j e c t i v e plane, n , o f order q. I f q 2 1 1 , IN1 = n 2 5 then N i s an n-arc and secants o f N are tangents o r e x t e r i o r lines to w , THEOREM 2

-

I f a complete q-arc, w , has a t l e a s t t h r e e e x t r a e x t e r i o r p o i n t s

such t h a t the l i n e s j o i n i n g these p o i n t s are e x t e r i o r l i n e s o f r i s e t o a (qtZ)-set w i t h a t l e a s t t h r e e n u c l e i .

42.

0,

then w

gives

We r e c a l l some w e l l known p r o p e r t i e s o f a complete q-arc. L e t l’i be a p r o j e c t i v e non-Desarguesian plane o f odd order q and l e t w be a

complete q-arc i n 11

.

-

For any p o i n t P E 11 w we denote by t ( P ) t h e number o f tangents o f w t h a t contain P; c l e a r l y t h e number o f secants t o w passing through P i s (q-t(P))/Z. Define ei t o be t h e number o f p o i n t s o f II which l i e on e x a c t l y i tangents to w

.

-

Since w i s a complete q-arc , w i t h q odd, any p o i n t P ~ l l w l i e s on a t l e a s t one secant and on a t l e a s t one tangent, o r r a t h e r an odd number o f tangents; we then have

1 t ( P ) 14-2

and ei = 0

f o r i = 0,

i >q-2,

i = 2k.

Every p o i n t P E w l i e s on e x a c t l y two tangents and q-1 secants. We s h a l l c a l l conjugate the l i n e s t h e r e a r e tangents i n same p o i n t o f w

.

§ 3 . L e t P be a p o i n t o f w and pl, p2 the conjugate tangents. I f Aepl, A # P and B e p 2 , B # P, then the s e t tl = w u { A , B } i s a ( q t 2 ) - s e t and P i s a nucleus.

L e t P,Q be d i s t i n c t p o i n t s o f w , denote by pl, i n P and by ql, q2 the conjugate tangents i n Q. I t i s e a s i l y proved t h a t i f A = p i n q j and B = p j n q i

p2 t h e conjugate tangents i,j=1,2

, then

the set

0 = w u {A,B}

i s a (qtZ)-set which has the p o i n t s P,Q as n u c l e i . I n a general way, i f E i s a subset o f p o i n t s o f w , such t h a t P,Q # Eand f o r every p o i n t S E I: we have A E s1 and B E s2 , where s1 ,s2 denote t h e conjug& t e tangents i n S,

then n = w u { A,B)

i s a ( q t Z ) - s e t and @ ={ P,Q,Sh

; v ShEE

}.

161

On (q*2)-setsin a non-Desarguesian projective plane

5 4 . Now we apply t h e procedure o u t l i n e d i n 5 3 t o complete 9-arcs i n known nonDesarguesian planes o f o r d e r nine. R.H.F. Denniston [3] determined a l l classes o f complete 9-arcs i n these planes; t o be p r e c i s e he proved t h a t i n t h e Hughes plane complete 9-arcs f a l l i n t o t h r e e t r a n s i t i v i t y classes under t h e c o l l i n e a t i o n group, i n t h e t r a n s l a t i o n plane and l i k e w i s e i n i t s dual t h e r e i s a s i n g l e t r a n s i t i v i t y c l a s s . F i r s t we consider t h e Hughes plane, II, o f o r d e r 9. Using t h e symbols and terminol o g y o f Denniston, i n II

t h e t h r e e classes o f complete 9-arcs can be repl-esented

by t h e f o l l o w i n g :

w5 =

~6 = w7 =

{ B,F,Q3,Q6,N3,o3,W4,R5,V7 { {F Y N ~ Y N ~ Y U ~ Y

} } U ~ Y} ~ ~ Y ~ ~ , S ~ ’ W ~

For any one o f these complete 9-arcs t h e r e a r e s i x e x t r a e x t e r i o r p o i n t s and an e x t r a e x t e r i o r p o i n t

i e s on f i v e tangents and two secants o f t h a t g i v e n

arc. Now we consider

w5.

I n t a b l e I we show t h e tangents through t h e e x t r a e x t e r i o r p o i n t s

P7,Y5 ,T4,

T ~ Y YL V ~ . Table

I

From an examination o f t a b l e I i t f o l l o w s e a s i l y t h a t t h e tangents b4, p a7, i5 i n 6, F,Q3,Q6 r e s p e c t i v e l y l i e on P7, and t h a t t h e i r conjugate tangents l i e on Y5.

From 5 3 , then, n = %u{P7,Y5} i s a 11-set and 0 = { B,F,Q3,Q6} the set o f i t s nuclei. Moreover t a b l e I shows t h a t 5 = w 5 u { T4,T6} i s a 11-set w i t h 0 = { N3,03,W4,R5} being t h e s e t o f n u c l e i .

is

We n o t e t h a t n and E a r e t h e o n l y 11-sets o f II having a t l e a s t t h r e e n u c l e i obtained from w5, s i n c e i t i s easy t o see t h a t no o t h e r p a i r s o f p o i n t s

,

P. Biscarini and F. Conti

162

e x i s t i n II which are i n t e r s e c t i o n s o f t r i p l e s o f conjugate tangents.

We examine now t h e 9-arc w6= { B,H,P3,P,,R,,03,Q5,U6,N3} and i n t a b l e I1 such t h a t there e x i s t a t l e a s t t h r e e we show t h e p a i r s of p o i n t s of n w6 conjugate tangents pass through them.

-

Table

I1

Looking a t t a b l e 11, i t can be i n f e r r e d t h a t 11-set w i t h

n"

=

w6u

@ ' = {B,H,P3,P7} { S4J3 } , n " ' =

n u c l e i , given t h a t we have We note t h a t n ' , n"

n'

being the s e t o f i t s w6 u

w 6 u {S8,T7} nuclei

{ T3,W5}

i s an

and t h a t

are 11-sets w i t h e x a c t l y t h r e e @ " ' = { H,P7,03} respectively. @'I= { B,P3,R,} and and n " ' are the o n l y 11-sets o f n containing w6

and having a t l e a s t t h r e e n u c l e i . S i m i l a r l y , we consider the 9-arc Table

w

=

I11

{ F,N3,N4,U4,U,,0,,06,S4,W7)

On (q+2)-setsin a non-Desarguesian projective plane

163

The 9-arc w7 can be completed i n o n l y one way i n an 11-set w i t h f o u r n u c l e i , and i n f o u r ways i n 11-sets w i t h t h r e e n u c l e i . I n t h i s manner we can o b t a i n t h e 11-set nl having t h e p o i n t s N3,N4,U4,U5 n u c l e i by adding t h e e x t r a e x t e r i o r p o i n t s V4,V7.

n5= w,

as

Moreover n2 = w7U{PgyH} , n3 = w7u{p6,A } , n4 = w 7 u { R ~ , w ~ }and a r e t h e I l - S e t S w i t h e x a c t l y t h r e e n u c l e i o b t a i n e d from w 7 ‘J { R3,S6}

and we have @ 2 = { F,N4.,W7} , Q3 = { F,U5,S4} , @4 = { N ~ Y O ~ Y , U ~@5 } ={N4,03,U4} r e s p e c t i v e l y as i s e a s i l y proved by l o o k i n g a t t a b l e 111. Now we consider t h e t r a n s l a t i o n plane

v and i t s dual

A

, which

can be c o o 1

d i n a t i z e d by a n e a r - f i e l d o f o r d e r n i n e . The elements o f t h e n e a r - f i e l d may be denoted as i n [3] by d i g i t s from 0 t o 8, homogeneous p o i n t - c o o r d i nates i n V ( 1 ine-coordi nates i n A ) being enclosed i n round brackets and coordinates o f t h e dual k i n d i n square brackets. See [3] f o r t h e t a b l e s o f a d d i t i o n and m u l t i p l i c a t i o n . I n v t h e r e e x i s t s o n l y one c l a s s o f t r a n s i t i v i t y o f complete 9-arcs which can be represented by t h e f o l l o w i n g (see [3] ) : w8

=

1

(001 ) ( 1 31 ) (231 ) (101 ) (01 1 ) (751 ) (251 ) (41 1 ) ( 1 10)

I n t a b l e I V we have t h e s i x e x t r a e x t e r i o r p o i n t s o f

w8

1

and t h e tangents

passing through them. Table

IV

From t a b l e I V we can deduce t h a t w8 may be completed i n two d i s t i n c t 11-sets w i t h f o u r n u c l e i : El = w8 u {(761),(461)} and Z 2 u,U{(481),(771)} w i t h O1 = { (001),(231),(751),(411)} and = {(131),(101),(011),(251) Anologously i n t h e dual plane A , t h e o n l y c l a s s o f complete 9-arcs can be represented by w9 - I [ l O O ] 11011 [1101 [1121 [162I El681 [171] [1771 [147]1

.

-

I n t a b l e V we show t h e e x t r a e x t e r i o r p o i n t s and t h e tangents f o r them.

P. Biscarini and F. Conti

164

Table V

From t a b l e V we can e a s i l y i n f e r t h a t we have two 1 1 - s e t s w i t h f o u r n u c l e i E2 = W9 u { [124],[128] } , and o b t a i n e d by W9 ; El = 09u { [0171,[013]} and { [100],[101] ,[162],[168]} r e s p e c t i v e l y . a1 = { [1101,[112],[171] ,[1771 } , F i n a l l y , we n o t e t h a t

O2 =

El,

E2

and

-

,

El

-

C2 a r e t h e o n l y 1 1 - s e t s w i t h

l e a s t t h r e e n u c l e i c o n t a i n i n g a complete 9 - a r c o f

v

and

at

A respectively.

§ 5 . We want t o s t u d y t h e c o n f i g u r a t i o n o f e x t r a e x t e r i o r p o i n t s o f a complete q-arc, w , i n a p r o j e c t i v e non-Desarguesian p l a n e o f odd o r d e r q. In

02

we r e c a l l e d t h a t f o r e v e r y p o i n t P E n-w

t h e number o f t a n g e n t s i n P, we have q

-

t(P)

points

LEMMA 1

-

Proof.

The number o f secants o f w

.

q

-

2

, implying

t(P) L 1 <

, where

t(P)

that

- 1' 2 c a r r i e s a t most two e x t r a e x t e r i o r

Every e x t e r i o r l i n e r o f is

o b t a i n e d by c o u n t i n g t h e number o f secants o f

.

q(q-1)/2 T h i s number can be p a s s i n g t h r o u g h ev--" p o i n t o f

w

r and a d d i n g up t h e s e numbers. T h i s sum can a l s o be w r i t t e n as f o l l

(2) where

T

denotes

:

T f i + p 3 ! . . L 4

P

2

denotes t h e number of e x t r a e x t e r i o r p o i n t s c o n t a i n e d i n t h e e x t e r i o r li-

ne r and t h e sum i s t a k e n o v e r a l l p o i n t s P E such ~ that From ( 2 ) and ( 1 ) we have

t(P) # (qt1)/2

.

On ( q"2het s in a non-Desarguesian projective plane

165

LEMMA 2

- Every s e c a n t r o f

Proof.

Using t h e same procedure as i n LEMMA 1, we c o u n t t h e number o f

secants o f

c a r r i e s a t most one e x t r a e x t e r i o r p o i n t .

w

d i s t i n c t f r o m r i n two ways.

W,

Then we have

where

't

denotes a g a i n t h e number o f e x t r a e x t e r i o r p o i n t s o f r and t h e sum must

be extended o v e r a l l p o i n t s

PEW

-

r

such t h a t

t(P) # (qt1)/2

.

From ( 4 ) and ( l ) , f o l l o w s t h a t

hence

and c l e a r l y

< 2 LEMMA 3

-

I f t h e number o f e x t r a e x t e r i o r p o i n t s o f w i s g r e a t e r t h a n f o u r

e v e r y tangent, r, t o

and q-11,

w

P r o o f , L e t r be a t a n g e n t t o

c o n t a i n s a t most two e x t r a e x t e r i o r p o i n t s . w

, and R

= r n w

.

I n t h i s case we may a g a i n use t h e method o f LEMMA 1, c o u n t i n g t h e numbt, ,f secants t o

where

'I

w

n o t p a s s i n g t h r o u g h R i n two ways.

We have

denotes t h e number o f e x t r a e x t e r i o r p o i n t s o f r and t h e sum i s extended

o v e r a l l p o i n t s P o f r d i s t i n c t f r o m R and such t h a t

t(P)

# (qt1)/2

.

Using ( 1 ) one t h e n d e r i v e s f r o m ( 6 ) t h a t

and hence

We suppose, f o r t h e sake o f argument, t h a t t h e l i n e r has e x a c t l y f o u r e x t r a e x t e r i o r p o i n t s . Then t h e t a n g e n t s , d i s t i n c t f r o m r and p a s s i n g t h r o u g h

P. Biscarini and F. Conti

166

a t t h e p o i n t s d i s t i n c t from R. Since, these f o u r p o i n t s , a r e a l l tangents t o by hypothesis, t h e number o f e x t r a e x t e r i o r p o i n t s i s a t l e a s t f i v e , t h e r e e x i s t s one p o i n t t h a t does n o t belong t o r and f o r t h i s p o i n t we have a t most f i v e tangents. But f o r every e x t r a e x t e r i o r p o i n t we have (qt1)/2 5 5

q 59

and t h a t i m p l i e s

(q+1)/2 tangents, hence

c o n t r a r y t o t h e hypothesis t h a t

q 2 11

.

We now prove t h a t t h e r e cannot be t h r e e e x t r a e x t e r i o r p o i n t s c o l l i n e a r t o a tangent

r

of

w

.

T h i s again i s proved by t h e r e d u c t i o ad absurdum method.

r

We suppose t h a t t h e tangent

to

contains exactly three extra e x t e r i o r

points. The number o f tangents d i s t i n c t from r through these t h r e e e x t r a e x t e r i o r points i s from

3 ( ( q t 1 ) / 2 -1) = 3 ( q t 1 ) / 2 ,

r i s (q-1)/2 and

they i n t e r s e c t r p a i r w i s e i n t h e same p o i n t . T h i s i m p l i -

es t h a t we have a t most ( q - 1 ) / 4 t h a t does n o t belong t o Since we have

them t h e number o f o t h e r tangents d i s t i n c t

3 tangents through every e x t r a e x t e r i o r p o i n t

t

r ,

( q t 1 ) / 2 tangents through every e x t r a e x t e r i o r p o i n t ,

it

follows t h a t q + l < g - 1 , 3 2 4 hence

q 5 9, c o n t r a r y t o t h e hypothesis t h a t

11.

q

Therefore LEMMA 3 i s proved. Now we a r e a b l e t o prove THEOREM 1 because i t i s an immediate consequence

.

o f LEMMAS 1, 2 , 3

16.

Now we can prove THEOREM Let

a p :

P

w + w

.

n

-

w ;

R

if

(PR) n w

= {RJ

Q

if

(PR) n w

= {R,Q}

(PR) i s t h e l i n e j o i n i n g P and R. We have I R E w : ap(R) = R 1 = t ( P )

p o i n t , we have

IRE

Note t h a t i f (i)

we d e f i n e t h e f o l l o w i n g mapping

such t h a t

ap (R) where

2

be an a r b i t r a r y p o i n t of

w

: ap(R) = R

ap(R)

(PS) n w

=

aS(R)

I

, then

= (qt1)/2 = R

,

P # S

.

i f P i s an e x t r a e x t e r i o r

, we

have e i t h e r

= {R}

or (ii)

(PR) and(SR) a r e t h e conjugate tangents a t R and t h e s e t w u { P,S } i s a ( q t Z ) - s e t w i t h nucleus R

Now we consider t h e hypothesis o f THEOREM 2 , denoting by

.

N1 ,

N2,

N3

167

On fq+2hsetsin a non-Desarguesian projective plane t h e e x t r a e x t e r i o r p o i n t s of to

t h e l i n e s (N.N.)

W,

1 J

i,j=1,2,3

being e x t e r i o r l i n e s

w.

I f R i s a f i x e d p o i n t f o r two d i s t i n c t mappings

aNi

, aNj

i ,j=1,2,3

then

t h e conjugate tangents a t R must pass one through Ni and t h e o t h e r through N j Since

P i s an e x t r a e x t e r i o r p o i n t t h e mapping

p o i n t s then

w

fixed

has ( q t 1 ) / 2

aNi, aNj having a t l e a s t one common f i x e d p o i n t .

p o i n t s , t h e mappings Having f i x e d

ap

.

N1, N2, i f

a N and a N have a t l e a s t t h r e e common f i x e d 1 2 i s a (qt2)-set w i t h a t l e a s t three nuclei.

u { N1 ,N2}

On t h e o t h e r hand ift h e common f i x e d p o i n t s o f two, then s i n c e ( q t 1 ) / 2 2 5 points, e i t h e r w i t h

aN

Thus THEOREM 2

, the

or with

1 i s proved

mapping aN 2

a

.

N3

a N and a N a r e a t most 1 2 has a t l e a s t t h r e e common f i x e d

REFERENCES B a r l o t t i , A. , Un'osservazione i n t o r n o a un teorema d i B. Segre s u i q - a r c h i , Matematiche, 21 (1966) 23-29. Bichara, A. and Korchmaros, G. order q

. ( t o appear).

, Note

on ( q t 2 ) - s e t s i n a G a l o i s plane

of

Denniston, R.H.F., On a r c s i n p r o j e c t i v e plane o f o r d e r 9, Manuscripta Math.,

4 (1971) 61-89.

A c l a s s o f non-Desarguesian p r o j e c t i v e plane, Canadian J. Math. , 9 (1957) 378-388 Hughes, D.R.,

.

On a r c s i n a f i n i t e p r o j e c t i v e plane, Canadian J. 19 (1967) 376-393

M a r t i n , G.E.,

.

M e n i c h e t t i , G. , q-archi c o m p l e t i n e i p i a n i d i H a l l d i o r d i n e Acc. Naz. Lincei,56 (1974) 518-525

.

Segre, B . , Curve r a z i o n a l i normali e Mat. pura e appl., Segre,

Math.,

Zk

, Rend.

k - a r c h i n e g l i spazi f i n i t i ,

Ann.

39 (1955) 357-379.

B., Lectures on modern geometry, Cremonese Roma (1961).

T a l l i n i , G., Sui q - a r c h i d i un piano l i n e a r e f i n i t o d i c a r a t t e r i s t i c a p = 2 , Rend. Acc. Naz. L i n c e i , 23 (1957) 242-245.

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Annals of Discrete Mathematics 14 (1982) 169-174 0 North-HollandPublishing Company

SUBGRAPH ENUMERATING POLYNOMIAL AND ISING MODEL Luigi Bor zacchini lstituto d i Geometria Universitf di Bari Bari, Italy. In this paper it is shown a theorem concerning the calculus of the subgraph enumerating polynomial ( SEP ) of a graph. Then such a theorem is utilized for a combinatorial approach to Ising model to enumerate the closed graphs ( shared by their edge-number ) on a finite two-dimensional square lattice. To any non-oriented multigraph G = ( V, X 1, where V is the s e t of vertices and X the set of edges, we can associate a "subgraph enumerating polynomial" ( SEP ) so defined:

where C

is the number of subgraphs ( in the terminology of BERGE (1) )

i, j of G with i vertices and j edges.

In a previous paper ( BORZACCHINI ( 2 ) we have proved useful1 theorems concerning the SEP and i t s calculus. In the first section of this paper we will demonstrate a theorem generalizing those results, in the second section we will set some questions of Ising model in terms of SEP and in the third section we will applicate the theorem of first section to this model. 1. W e can also write (1) a s follows:

where the sum is taken over all subsets of V and where XA is the set of the edges of G with both incident vertices in A. Let be T any subset of V and R any subset of ( T' is the complementary set of T in V ): to any ( x, y ) 6 X such that x cs R and y E T, let's associate a loop ( y, y); let's denote with L the set of such loops. Let be the graph

T,R

= ( T, XT u LT,R ) and let denote P

GT, R demonstrate the following theorem:

T, R

THEOREM 1. For any T GV:

PG(u,v) = RST

.IR1

V

169

(u, v) the SEP of GT,R, we can

1st 'T,

R(u, v,

L.Borzacchini

170

Proof. Let be T any subset of V, any A C V can be partitioned in two subsets: R = A n T and L = A n T . Hence:

where XRL is the set of edges of G with a vertex in R and a vertex in L.

observing that the second sum is P

(u,v) we get the thesis. T, R With this theorem the calculus of the SEP of any graph can be reduced to is a SEPs of graphs with a lesser number of vertices because every G T, R graph with I T I vertices. 2. Ising model is a central tool of statistical mechanics. Let's denote with ( seen a s a graph ) a two-dimensional square lattice with m vertices L

m,n on the vertical side and n vertices on the orizontal side; combinatorial approach to this model ( see THOMPSON (3) ) can be seen a s the enumeration having of the partial graphs ( in the terminology of BERGE (1)) of L m, n even degree in every vertex: such partial graphs will be called afterwards "closed graphs". Our aim is the calculus of the number of closed graphs shared by their edge-number, i. e. the calculus of the enumerating polynomial:

where s is the number of closed graphs in L having r edges, by m,n, m, n the SEP of Lm -l , n -

If G is planar and F

G

the set of faces of G and let's denote with D(G) the

"dual graph" of G, i. e. D(G) = ( W , Y 1, where there is a (1-1)-correspondance d: FG-+W, and for any edge of G between two faces A and B we set in Y the edge ( d(A) , d(B) ). We can now show the following lemma:

LEMMA. Any closed graph is ( a s to the faces ) bichromatic. Proof. Any face of a closed graph G has an even number of edges ( because

is the union of square-meshes of L 1. Hence the degree of the corresponm, n ding vertex in D(G) is even. D(G) is then bichromatic ( as to the vertices ) and hence G is bichromatic ( a s to the faces ).

Let's now denote the faces of a closed graph coloured differently from the infinite face as the "inner faces" of the closed graph. If we denote with FOG

171

Subgraph enumerating polynomial and Ising model

the set of finite faces of G and with Wo = d(FoG) and Yo with the same rule of Y applied only to the finite faces, we can define the graph Do(G) = ( Wo

,

Yo).Now we can show the following theorem: THEOREM 2. There exists a (1-1)-correspondance f between the set of closed graphs in L and the set of subgraphs in Lm-l,n-l and, if H is a closed m, n with r edges and f(H) the corresponding subgraph in Lm-l,n-l graph of L m,n r = 41 - 2 j with i vertices and j edges, then: Proof. Let's remember that the finite faces of L

m, n

a r e its square-meshes.

Then. DO(Lm,n) = Lm-1,n-l If H is a closed graph of L m , n , H can be seen a s the set M(H) of square-

I

I

whose union is H. Let be AH = x vertex of Lm-l,n-l meshes of L m, n B € M(H) and d(B) = x Let be LH the subgraph of Lm-l, n-l relative to

1.

AH: the correspondances H+A H 4 L H a r e obviously bijective and we get the first part of the theorem. If LH has i vertices and j edges, H has i squa-

re-meshes in its inner faces and then equally coloured in Lm,n; an edge have a (v,w) in LH means that the two square-meshes f(v) and f(w) in L m, n common edge on their boundary. such an edge, beeing f(v) and f(w) equally coloured, is not an edge of the closed graph. Then every vertex in LH becomes 4 edges of its corresponding square-mesh, but every edge in LH means the deletion of an edge between two square-meshes and hence of two edges from the total number of edges obtained multiplying 4 times i. Hence we obtain the second part of the theorem. From this theorem we have the following result: COROLLARY 2 . 1 . For any Lmrn, if P

m, n

(u, v) denotes the SEP of L m, n

41-2 j=r

The calculus of the SEP of a two-dimensional square lattice is then sufficent to calculate S (w). This approach to Ising model, in comparison with other m, n approachs, shows some advantages: i) it enumerates closed graphs shared by their edge-number; ii) it is true for finite lattices without hypotizing periodic boundary-conditions, i.e. without wrapping the lattice on a cylinder.

172

L. Bonacchini

(u,v) is yet very hard. For a sake of simplicity let's m, n set constant one dimension of the lattice ( say m ) and let's denote one of the sides of lenght m a s "development side". Let's now order the vertices of this , where for every j: a. e {0,1], side ( say top-down 1. Let then be L n, a l , . . , am J adjoining a loop on the i-th a graph with mxn vertices obtained from L m, n vertex of the development side if and only if ai = 1. 3. The calculus of P

.

If Pn

, from theorem 1 applied (u,v) is the SEP of L n, %, . . , am , taking a s the set of vertices of the development side, am

.

, %, .. . , am

to any Ln

,%, . . .

we can obtain: P

n, %,

. . . , am(u, v)

+. . . -..7 , ., b

-

U l

( bl,

1 5 1

ulRl V pT, R(', v, RS T +bm % b l + . . , + a b +b b +...+bm-lbm m m 1 2 =

V

bm)

'n-1, bl,

.. . ,bm(u, v)

(2)

and the sum is over all m-uples (bl, If we set: h = 1

and if

+

. . . ,bm), where

zm a. 2'-1 J

j=l

k = 1

for every j: b. t {O, 13. J

+F: j=l

b.J 2

j-1

[C], denotes the i-th component of the vector C, we can write (2):

(3)

where: Ah,k

b1 +. . . +bm =

V

b b +. 1 2

,

. +bm-lbm+%bl+. . . +ambm

With a vectorial notation (3) becomes: Pn(u,v) = A Pn-,(u,v)

where A is the matrix with entries Ah, k, and hence: P,(u,

V) =

An

where Po is a constant vector with 2

m

components [ p o l

= 1, for every i.

Subgraph enumerating polynomial and king model

and then in vectorial notation: Q(x,u,v) = ( I - X A )

-1

173

Po

Setting

we can write: Sm(w, X) = where B h, k

= W

4(b +. 1

[C

. . +bm-l)

I

-

-1

xB1

Po]

- 2(blb2 +. . . +bm-2bm-l+ybl+. . . +am-lbm-l

)

REFERENCES

(1) Berge C . , Graphes et Hypergraphes ( m o d , Paris, 1970 ) (2) Borzacchini L. , Subgraph enumerating polynomial and reconstruction conjecture, Rend. Acc. S. F. M. Napoli, Ser. IV, Vol. XLIII ( 1976 )

(3) Thompson C. J. , Mathematical Statistical Mechanics ( Princeton University Press, Princeton N. J., 1979 )

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Annala of Discrete Mathematics 14 (1982) 175-182 0 North-Holland Publishing Company

CAPS RELATED TO I N C I D E N C C STRUCTURES AND TO LINEAR CODES

P.V.

C e c c h e r i n i and G. T a l l i n i

I s t i t u t o Ftatematico "G.

Castelnuovo", U n i v e r s i t a d i Roma, I t a l y

A s e t K o f p o i n t s o f PG(r,q) i s c a l l e d a cap o f k i n d s o f PG(r,q) i f s t 2 i s t h e minimum number o f dependent p o i n t s o f K and K spans PG(r,q). kle d i s c u s s some p a r t i c u l a r cases i n which caps can be used f o r s t u d y i n g p r o p e r t i e s o f i n c i d e n c e s t r u c t u r e s and o f l i n e a r codes.

1. CAPS RELATED TO INCIDENCE STRUCTURES Let I

(IF',

B y I) be a f i n i t e i n c i d e n c e s t r u c t u r e , with

I f P E P , B EIB, p u t (P) =

A subset

{B

EIB

I P I B}

,

(B)

{P EIP

IP

I B}

.

Pc P (resp. B c IB) i s s a i d t o b e o f even t y p e i f

V B E IB, L e t us o r d e r

I ( B ) n PI

IP and IB: IP

i s even ( r e s p . V P E = (Pl,

P2,

..., P,,),

IP, I ( P ) n BI i s even).

ID

=

(a1,

B2,

..., Bb);

and l e t M be t h e v x b i n c i d e n c e m a t r i x o f I o v e r GF(2): M = (mij),

If

B

E

m

ij

= 1 i f Pi I B j'

m

ij

= O if PiJB

j '

IB, l e t us denote t h e c o r r e s p o n d i n g column o f M ( v e c t o r o f 175

ii

) and

P. V: Ceccherini and G. Tallini

176

by t h e same simbol B. S i m i l a r l y b w i l l a l s o denote t h e c o r r e s p o n d i n g row o f M ( v e c t o r o f Z,) and t h e

t h e c o r r e s p o n d i n g p o i n t o f PG(v-1,2) P (E IP)

c o r r e s p o n d i n g p o i n t o f PG(b-1,2).

It i s easy t o p r o v e t h e f o l l o w i n g ,

A s e t P o f columns ( r o w s ) o f M i s dependent i f f t h e s e t P o f

Lemma 1.1.

b l o c k s ( p o i n t s ) c o n t a i n s a non-empty subset o f even t y p e . F o r any f i n i t e i n c i d e n c e s t r u c t u r e I = (

Theorem 1.2.

IP, IB, I ) w i t h i n -

c i d e n c e m a t r i x M y we have (2)

max { / P I IP

5 IP,

P c o n t a i n s no non-empty s u b s e t o f even t y p e )

=

= max { I B I

18 :-By B c o n t a i n s no non-empty s u b s e t o f even t y p e }

=

= rank H = d t 1, say.

Actually respect t o

d t1

<

v,b.

lle say t h a t 7 =

(IP, ID, I )

i s o f even t y p e w i t h

IP ( r e s p . t o IB) if t h e r e e x i s t s a non-empty s u b s e t P

5IP

(resp.

B C I B ) o f even type, i . e . i f d + l < v ( r e s p . i f d t l < b ) . I f 7 i s symmetric (i.e.

i f v = b ) , t h e n i t i s o f even t y p e w i t h r e s p e c t t o

B. I f v < b ( r e s p . i f b even t y p e w i t h r e s p e c t t o IB ( r e s p . t o IP).

even t y p e w i t h r e s p e c t t o If I = (

IP

iff it i s o f

< v), then I i s o f

IP, IB, I ) i s of even t y p e w i t h r e s p e c t t o IB, we can d e f i n e

(3) s t 2 = min { I B I

1 P z B 5 B, B o f even t y p e ) .

I f I i s o f even t y p e w i t h r e s p e c t t o

I n o t h e r words, by Lemma 1.1,

IP, we can d e f i n e

s t 2 ( r e s p . t + 2 ) i s t h e n t h e minimum number

o f columns ( r e s p . rows ) o f M which a r e l i n e a r l y dependent.

A s e t K o f k p o i n t s o f a PG(d,q) = S i s a k-cap o f k i n d s o f S, S

denoted k d 99

, if

s t 2 i s t h e minimum number o f dependent p o i n t s o f K and

K spans S. The l a t t e r c o n d i t i o n i m p l i e s k 2 d t 1. We can summarize t h i s by

the following.

Caps related to incidence structures and to linear codes

L e t I = ( IP, IB, I ) be o f even t y p e w i t h r e s p e c t t o

Theorem 1.3. to

177

IB

(resp.

P) and n o t t r i v i a l , i.e. such t h a t :

Then t h e columns (resp. rows) o f any i n c i d e n c e m a t r i x 11 o f I g i v e r i s e t o a cap in

bd,2

(resp. v t d,2

in

PG(b-l,?)),

a r e t h e numbers, depending o n l y on I and n o t on

where v,b,d,s,t by ( 1 )

PG(v-l,?)

- (4).

M y given

The above bS (resp. vt ) w i l l be c a l l e d t h e block-cap (resp. t h e d 92 d,2 p o i n t - c a p ) r e l a t e d t o I. Remark 1.4.

s, t > 0 by ( 5 ) . Yoreover s > 0 i f f B,B'

We have

B # B ' implies (B)# (B'),

P

and t h e incidence w i t h

i.e. E;

PI

and

iffblocks may be i d e n t i f i e d w i t h subsets of

i n t h i s case we say t h a t

i n c i d e n c e o r t h a t 1 i s a geometric space. F i n a l l y P #

EIB

i m p l i e s (P) # ( P I ) , i.e.

t

I

>0

i s a geometrical iff

P, P ' EIP and

i f f I i s , as we w i l l say, a "separated"

i n c i d e n c e s t r u c t u r e . Note a l s o t h a t i f s , t

> 0,

i . e . i f 1 i s a separated

(IB, I ( P ) l p E I P )

geometric space, then t h e "dual" s t r u c t u r e 19s

i s also a

separated geometric space. Theorem 1.5. L e t t h e i n c i d e n c e s t r u c t u r e I =

(IP, IB, I ) s a t i s f y t h e f o l l o w -

i n g condition: (Ph YP

) Any h p o i n t s ( h

2) a r e i n c i d e n t w i t h a t most one b l o c k and every

block i s incident w i t h a t l e a s t p points. Then every s e t B

[ (p-l)/(h-1)]

+

2

SIB

o f even type c o n t a i n s e i t h e r zero o r a t l e a s t

blocks, so t h a t any [ ( p - l ) / ( h - l ) ]

i n c i d e n c e m a t r i x M o f 1 a r e independent. Proof. L e t B SIB -

be o f even t y p e w i t h

t 1

columns o f any

P. V, Ceccherini and G. Tallini

178

0 and l e t

181 = b ' G [ ( p - l ) / ( h - l ) l

B

..., Bbl).

{B1, B2,

we know t h a t

t 1,

i.e. with p-(h-l)(b1-2)

Using I ( B 1 ) n (Bi)l

G h-1,

B1 i s incident w i t h a t l e a s t p-(h-l)(bl-2) Bi ( i = 2,..,,b1-1)

are n o t incident w i t h

>

h;

i =2,...,b'-lY

2 h p o i n t s which

and which a r e consequentely i n -

c i d e n t w i t h Bb, because 8 i s o f even type; t h i s c o n t r a d i c t s t h e c o n d i t i o n (phyp)' By d u a l i t y we have t h e f o l l o w i n g .

Theorem 1.6.

Let the incidence structure I = (

IP, IB,

I)s a t i s f y t h e f o l l o w -

i n g condition:

(Bk

YO

)

Any k b l o c k s ( k

> 2)

are i n c i d e n t w i t h a t most one p o i n t and

any p o i n t i s i n c i d e n t w i t h a t l e a s t u blocks.

Then every s e t P [ (u-l)/(k-l)]

matrix

t

SIP

o f even t y p e c o n t a i n s e i t h e r zero o r a t l e a s t

2 p o i n t s , so t h a t any [ ( u - l ) / ( k - l ) ] t 1 rows o f any i n c i d e n c e

M o f I a r e independent.

C o r o l l a r y 1.7.

Let the incidence structure

7 =

(IP, IB, I )

s a t i s f y con-

) (resp. c o n d i t i o n (B

) ) . I f I i s o f even t y p e w i t h r e s p e c t k 90 t o IB (resp. t o IP), then t h e r e l a t e d block-cap bS (resp. p o i n t - c a p dy2 vt ) has parameters s a t i s f y i n g d,2

dition

(Ph

YP

d 3 S 2

[ (~-l)/(h-l)I

(resp. d

t 3 [ (u-l)/(k-1)l).

2. EXAMPLES: CAPS RELATED TO LINE SPACES L e t us i l l u s t r a t e t h e above r e s u l t s by two examples. Theorem 2.1. where

L e t I = (S,C

U

R ) be t h e geometric space ( c f . Remark 1.4),

S i s t h e s e t o f t h e p o i n t s , C i s t h e s e t o f a l l non-singular conics

and R i s t h e s e t o f a l l t h e l i n e s of t h e Galois p l a n e PG(2,q). Then s a t i s f i e s (P5

Yq 1

I

), so t h a t any [ q / 4 1 t 1 columns o f t h e i n c i d e n c e m a t r i x

179

Caps related to incidence structures and to linear codes o f 1 are independent.

A f i n i t e p a r t i a l l i n e space (S, R) i s a f i n i t e geometric space s a t i s f y ) a r d such t h a t the s e t R o f blocks ( c a l l e d l i n e s ) i s a

i n g c o n d i t i o n (P 2 3 2

cover o f S. Theorem 2.2.

L e t 7 = ( S , R) be a f i n i t e p a r t i a l l i n e space such t h a t I R I = k R i n R. Then 1 s a t i s f i e s (P

f o r every

incidence m a t r i x o f 1 are independent.

2,k

) , so t h a t any

k

columns o f any

By d u a l i t y : Theorem 2.3.

Let 1

(S, R)

f o r every P i n S. Then

I (P)l

be a f i n i t e p a r t i a l l i n e space such t h a t

=r

I s a t i s f i e s (B ), so t h a t any r rows o f any i n 2,r

cidence m a t r i x o f 7 are independent. Theorem 2.4.

L e t now 1 = (S, R)

be a f i n i t e l i n e space such t h a t

IS1

= v,

IRI = b, I R J = k f o r every R i n R. ( I n other words: 1 i s a Steiner system S(E,k,v)

and b = v ( v - l ) / ( k ( k - 1 ) ) ) .

Denote by

r the number

(v-l)/(k-1)

of

l i n e s through a p o i n t ; then we have: (i)

E 5 S o f even type:

For any s e t r z k r 0

(ii)

(mod2)

E = 0 or E = S ;

9

r z 0, I: z 1 (mod 2)

3

E = 0.

r I 0, k

9

t h e v rows o f the incidence

i 1

(mod 2)

m a t r i x M o f 1 are independent;

r

k :0

vt i n PG(b-1,2) d,2 r e l a t e d t o 1 i s a (dtZ)-simplex i n a d-dimensional space I

(mod 2)

the point-cap

( i .e. t=d=v-2).

Proof.

( i ) L e t r z 0 (mod 2). I f 0 # E # S, then t h e r e e x i s t p o i n t s

and Q E

E. Because E i s o f even type and r :0 (nod 2 ) , we f i n d t h a t I E l i s

P

E

S

\

both an even and an odd number when we count \El by t h e l i n e s through P and a l s o by the l i n e s through Q. I n conclusion, E =

0 or E

= S,

t h e second

case being obviously excluded i f k i s odd. ( i i ) Ifr z 0, k z 1 (mod 2), then E = 0, so t h a t the v rows o f M are

E

180

P.V. Ceccherini and G. Tallini

independent by Lemma 1.1.

If r

5

k E 0

(mod 2 ) , then 7 i s of even type with

respect t o points (by ( i ) ) , and we can consider the point-cap

The

vt

e q u a l i t y d t 1 = v - 1 follows from

d,2' ( i ) and from ( 2 ) , assuming P S \

t h e complementary set of a point

E S;

from ( i ) and from ( 4 ) , assuming

P

furthermore,

tt2

{PI,

v follows

P = S.

3 . CAPS AND LINEAR CODES

rtl be t h e ( r t 1)-dimensional vector space over the Galois Let V = ( F ) q f i e l d F of order q . A l i n e a r code C o f V i s c a l l e d a p e r f e c t e-error-correct9 ing code i f

a& where

S(a,e)

V

and

d = min Cd(x,y)

I x,yEC,

d(x,y) i s the number of d i f f e r e n t components of x and y ,

S(a,e) = Ix

E

V

I

x#yl>2et1,

and

d(a,x) G e l .

I t i s well known ( c f . [ 2 1 , [ 6 1 ) t h a t t h e only p e r f e c t

codes a r e

t h e following: (i)

perfect single-error-correcting codes (e.g. Hamming codes);

(ii)

t r i v i a l p e r f e c t codes in case r t 1 = e , and q = 2, r t 1 = 2e t 1 ;

(iii)

the Golay ternary code: e = 2, q = 3 , r t l = 1 1 ; and the Golay binary code:

e=3, q=2, r t l

23.

I t follows t h a t : Theorem 3.1.

The only p e r f e c t 2-error-correcting l i n e a r codes ( w i t h minimum

5 ) a r e the following: t h e r e p e t i t i o n binary code 5 C1 = {(O,O,O,O,O), ( l , l , l , l , l ) l of (F2) and the Golay ternary code C2. Our purpose i s t o give a simple proof of Theorem 3.1, involving connec-

distance d

t i o n s between codes and caps. For d e t a i l s see [ 1 I , where actual Theorem 3.1 was obtained as Corollary 3.7; we take now the opportunity t o note t h a t t h e condition "not containing b a s i s vectors" i n t h a t c o r o l l a r y is obviously a consequence of t h e o t h e r hypothesis.

181

Caps related to incidence structures and to linear codes be as above and S = PG(r,q) be t h e r e l a t e d p r o j e c t i v e L e t V = (F )rtl 9 x ( ~ ) a r e (column) v e c t o r s space o f dimension r and o r d e r q. I f x ( ’ ) ,

...,

p a i r w i s e independent and spanning V, l e t us denote by t h e same symbol K t h e f o l l o w i n g : t h a t o r d e r e d k - s e t o f V, t h e c o r r e s p o n d i n g o r d e r e d k - s e t o f S ,

.

Then r t 1 < k, r a n k K = r + 1 m a t r i x o v e r Fq k and t h e l i n e a r code C = C(K) o f ( F ) d e f i n e d by Kx = 0 has dimension 9 k - ( r t l ) and does n o t c o n t a i n any b a s i s v e c t o r . Conversely, i f r t 1 G k and and t h e c o r r e s p o n d i n g ( r t 1 ) x k

C i s any ( k , k - ( r t 1 ) ) - c o d e

n o t c o n t a i n i n g any b a s i s v e c t o r , t h e n we may g i v e

a C a r t e s i a n r e p r e s e n t a t i o n f o r C o f t h e t y p e K x = 0, where (rt1)xk

matrix with

rank

K = rtl

and w i t h columns

K i s a suitable ( 1 x ) x(k)

,...,

which a r e non-zero v e c t o r s . As b e f o r e , l e t us denote by t h e same symbol K = K(C)

t h e above m a t r i x and t h e c o r r e s p o n d i n g o r d e r e d k - s e t s o f V and o f

PG(r,q).

Obviously

C(K(C)) = C .

Moreover K i s an ( o r d e r e d ) k-cap o f k i n d

denoted k S , i f and o n l y i f t h e code C has d i s t a n c e d = s t 2 , rYq so t h a t C i s an [ (stl)/2]-error-~orrecting code.

s i n PG(r,q),

F o r example, codes C1 and C2 i n t h e s t a t e m e n t o f Theorem 3.1

a r e C1 =

3

i n PG(3,2) c o n s i s t i n g o f t h e f o u r K1 i s t h e 5’ 392 i n PG(4,3) c o n s i s t i n g fundamental p o i n t s and t h e u n i t p o i n t ; K2 i s t h e 113 4Y3 o f t h e f i v e fundamental p o i n t s , t h e u n i t p o i n t and t h e f i v e p o i n t s ( O , l y - l y - l y l ) y

= C(K1), C2 = C(K2), where

( 1 ,O,-l

,1 , - l ) ,

( 1 ,1,0,-1

,-I),

( 1 ,-I ,-I,O,l),

( 1 ,-1,l ,-1 ,O): see T a l l i n i

[

51,

5 3.

L e t us now come back t o t h e g e n e r a l case. I f C i s an e - e r r o r - c o r r e c t i n g code i n

( F )k, then 9 k-dim C Y

i=O where e q u a l i t y occurs i f an o n l y i f C i s a p e r f e c t e - e r r o r - c o r r e c t i n g code. T h i s i m p l i e s t h a t t h e number k o f p o i n t s o f a k-cap K = k S s a t i s f i e s rYq

where e q u a l i t y occurs i f and o n l y if C(K) i s a p e r f e c t [ ( s t 1 ) / 2 1 - e r r o r c o r r e c t i n g code. cap

K = k3 rYq

F o r s = 3 , t h i s i m p l i e s t h a t t h e number o f p o i n t s o f

satisfies:

a

182

P.V. Ceccherini and G. Tallini

where e q u a l i t y occurs i f and o n l y i f the code [ 5 1 and M i g l i o r i [ 31 have proved t h a t :

tained i n a cap o f k i n d 2"

C(K) i s p e r f e c t . But T a l l i n i

"k = p(3,r,q)"

i f f " K i s n o t con-

"K i s one o f the caps K1 o r K2I1.

iff

Hence Theorem 3.1 completely follows. F i r s t o f a l l the code C i n the d = 5 ; moreover, since C i s p e r f e c t , and then

C = C(K1) = C

ACKNOWLEDGEMENT.

1

or

S

and s = 3 because kr,q k = p(3,rYq), so t h a t K = I$ o r K = K

statement must be o f t h e type C = C(K)

with

C = C(K2) = C

K =

2'

2'

This research was p a r t i a l l y supported by GNSAGA o f CNR.

REFERENCES 111 P.V. Ceccherini, G. T a l l i n i , Codes, caps and l i n e a r spaces, i n P.J. Cameron, J.W.P. Hirschfeld, D.R. Hughes (eds.), F i n i t e geometries and designs, London Math. SOC. Lect. Notes Series no. 49, Cambridge, U n i v e r s i t y Press (1981) 72-80. [ 2 ] J.H.van L i n t , On the nonexistence o f p e r f e c t 2- and 3-Hamming-errorc o r r e c t i n g codes over GF(q), I n f . Contr., (1970) , 396-401.

16

[3]

G. M i g l i o r i , C a l o t t e d i specie s i n uno spazio r-dimensionale d i Galois, Rend. Accad. Naz. L i n c e i , 60 (1 976) 789-792.

B. Segre, Le geometrie d i Galois, Ann. Mat. Pura Appl. 48 (1959) 1-97. [ 51 G. T a l l i n i , On caps o f k i n d s i n a Galois r-dimensional space, [ 41

Acta A r i t h . ,

[6]

7 (1961)

19-28.

A. Tietavainen, On the nonexistence o f unknown p e r f e c t codes over f i n i t e f i e l d s , S I A M J. Appl. Math.,g (1973) 88-96.

Annals of Discrete Mathematics 14 (1982)183-206

0 North-Holland Publishing Company

ON SETS OF TYPE (m,n)

I N BIBD'S WITH A

>

2

M a r i a l u i s a J . de Resmini I s t i t u t o Matematico "G. Castelnuovo", U n i v e r s i t S . d i Roma, I t a l y

Sets o f t y p e (m,n) i n BIBD's w i t h A > 2 a r e s t u d i e d , necessary c o n d i t i o n s f o r t h e i r e x i s t e n c e a r e g i v e n and some c l a s s e s o f BIBD's c o n t a i n i n g such s e t s a r e c o n s t r u c t ed. Furthermore, t y p e (1,n) s e t s i n symmetric BIBD's a r e c h a r a c t e r i z e d b y p r o v i n g t h a t such a s e t i s e i t h e r a Baer subdesign o r a H e r m i t i a n subset. F i n a l l y , some i n f i n i t e c las s e s o f t y p e (m,n) s e t s , m > 2, a r e c o n s t r u c t e d i n SBIBD's (A > 2 ) .

1.

INTRODUCTION A BIBD (Balanced I n c o m p l e t e U l o c k Design) B(v,k,A)

i s a p a i r (S, B ) ,

where S i s a v- s e t , whose elements a r e c a l l e d p o i n t s , and B i s a c o l l e c t i o n of b k-subsets of S

-

blocks

-

such t h a t any two p o i n t s o f S a r e cont ained

i n e x a c t l y A b l o c k s . When A = 1, a B(v,k,l)

i s a S t e i n e r System S(2,k,v).

The number b o f b l o c k s and t h e number r o f b l o c k s t hrough a p o i n t a r e given by (1.1)

b =

AV(V- 1 ) k(k 1)

-

r =

Y

-

A(V 1 ) k - 1

Y

whence t h e necessary c o n d i t i o n s f o r t h e e x i s t e n c e o f B(v,k,A) (1.2)

AV(V-1)

0

A(v- 1) i 0

(k(k- l)),

( k - 1).

These c o n d i t i o n s a r e a l s o s u f f i c i e n t f o r k = 3,4,5 ex c ept io n o f t h e n o n - e x i s t i n g B(15,5,2)),

for k

e x c e p t i o n o f t h e n o n - e x i s t i n g 8(21,6,2))

and f o r

A

i 0,6,7,12,18,24,30,35,36

(42)

follow:

and any A ( w i t h t h e 6 and A > 1

k

= 7

(with the

and

and A > 30 n o t d i v i s i b l e by e i t h e r 2

183

M.J.de Resmini

184

o r 3 1101. I n I41 and [ 51 sets o f type (m,n) by any block e i t h e r i n m o r i n

-

n points

i . e . subsets o f the p o i n t s e t met

-

i n S t e i n e r systems S(2,k,v)'s

were considered and some necessary and s u f f i c i e n t conditions f o r existence were proved. (Some i n f i n i t e classes o f S(2,k,v)'s

their

containing

type (m,n) sets were constructed.)

>

Here sets o f type (m,n) i n DIBD's w i t h A

2 w i l l be studied, necessary

existence conditions w i l l be given and some classes o f BIBD's containing sets o f type (man) w i l l be constructed. The special case o f symmetric BIBD's (SBIBD's), i . e . b = v, r = k,

w i l l a l s o be considered. Namely, type (1,n)

sets w i l l be characterized

extending t o SBIBD's w i t h A 2 2 a w e l l known theorem 1111 i n p r o j e c t i v e planes. Furthermore, some i n f i n i t e classes o f type (m,n) sets, m 2 2, w i l l 2 some r e s u l t s proved i n [ 31

be constructed extending t o SBIBD's w i t h A f o r X = 1.

Repeated blocks w i l l be allowed, b u t s o l u t i o n s n o t containing them

w i l l be found i n most cases.

2. SETS OF TYPE (man) Given a B(v,k,A)

on a v-set S, an h-subset ti o f S i s defined t o be an

,...,mS] ,

h-set o f class [mo,ml meets H e i t h e r i n m

0)

o r i n ml,

0G m <m <

...,

0 1 or i n m

S

... < mS

k,

i f any block

points. (A block meeting H

i n m. p o i n t s w i l l be c a l l e d an in.-secant). I f t m i s t h e number o f mj-secants, J J j then i t i s easy t o v e r i f y [ 6 ] t h a t the f o l l o w i n g e q u a l i t i e s must hold: k

1

m .=O

J

(2.1

1

tmj = b

185

On sets of type (m,n) in BIBD's with A > 2 t m . # 0 f o r a l l m . ' s i n J , then an h-set o f J J m ] , m E J, i s c a l l e d o f type (mo, m ) and t h e m . ' s are i t s s j S J

If J c {0,1,2,..,,k} class [ m

0

and

,...,

characters.

...,

0 G m

Here h-sets o f type (m,n),

<

n 4 k , (i.e.

having two c h a r a c t e r s )

-

f o r s h o r t , (h;m,n)-sets

-

w i l l be considered; f o r them equations (2.1) become:

mt

t

ntn

=

r h

m(m-l)tm + n ( n - 1 ) t n = Ah(h- 1 )

.

I

For these equations t o be c o n s i s t e n t , h must be an i n t e g r a l r o o t o f t h e equation :

-

A h2

(2.3)

h(r(ntm-1)

t

A)tbmn

.

0

Therefore, i t s d i s c r i m i n a n t A

(2.4)

= (r(ntm-l)+A)*

-

4Xbmn

must be a non-negative square. Let

u,

and

um be t h e numbers o f n-secants and m-secants through a p o i n t

i n s \ H and vn and v

t h e numbers o f n-secants and m-secants through a p o i n t i n H, m then t h e f o l l o w i n g e q u a l i t i e s must h o l d : I

I

I

J I

(2.5)

'mtUnZr

I

mum

+ nun

(2.6) =

v m t v n = r

')(m-1 )vm c

t

(n-1 )vn

X (h-1 )

Thus

m

I

hence ,

n

-

Ah rm n - m

m

r - A

n - m Y

.

M.J. de Resmini

186

i s a necessary c o n d i t i o n f o r a type (m,n) set t o e x i s t and n - m r

proper d i v i s o r o f

- A.

must be a

When m = 0, equation (2.3) has the s o l u t i o n :

that i s h = (

(2.9)

n - 1 ) ( v - 1 ) + 1 = (n k - 1

- l)v t k- n k - 1

Thus (2.7) and (2.9) provide t h e f o l l o w i n g necessary c o n d i t i o n s f o r an (h;O,n)-set

t o e x i s t i n a B(v,k,X):

(2.10)

n ( (r-A)

(2.11) (When

(v

x

-

= 1, (2.11)

1) (n

-

1) i 0 ( k

,

-

1)

.

i s always s a t i s f i e d . )

I f H i s an (h;m,n)-set

i n a B(v,k,X)

on

S, then S \ H i s a ( v - h;k

- n,k

-m)-

-set H I , which w i l l a l s o be c a l l e d t h e 2-characters complement o f H ( t o p o i n t o u t t h a t the s e t t h e o r e t i c complement o f H i s considered as a type (ml,nl)set). Remarks. ( i ) For any BIBD an (h;O,l)-set

i s a p o i n t and i t s 2-characters

complement i s a ( v - 1; k - 1 , k ) - s e t . ( i i ) For any B I B D an (h;O,k)-set h =

*I>+

x

cannot e x i s t ; indeed,

1, n = k + h = v,

so t h a t i t c a n ' t be o f type (0,k).

3,SETS

OF TYPE (0,2) I N B(v,3,X)'s

When the block s i z e o f a BID0 i s k = 3, o n l y type (0,2)

sets can e x i s t ;

On sets of type (m,n) in BIBD's with

2

187

f o r such a set, by (2.8), h =- v t l 2

(3.1)

and the necessary conditions (2.10) and (2.11) become:

,

2 1 + 3 -

(3.2)

(3.3)

v

1 (2)

.

Proposition 3.4. The f o l l o w i n g are necessary conditions f o r a B(v,3,X) v t 1 ; 0,2)-set t o e x i s t : e i t h e r ing a

contain-

( 7

(i)

X

=0

(6)

and

v E 1

(ii)

X

!0

(2)

and

v :1 o r 3 (6)

z 0 ( 3 ) and v

( i i i ) ,A (iv)

v

3

3

or

=3

7 (12),

Proof. I t ' s enough t o consider

(4),

or

(4),

( v f 7 o r 3 (12)),

or

or

any 1. together the necessary conditions (1.2),

(3.2)

t2 must be an integer.

and (3.3) keeping i n mind t h a t

Some conditions i n prop. 3.4 w i l l be proved t o be a l s o s u f f i c i e n t , b u t f o r the construction the f o l l o w i n g r e s u l t i s needed. Proposition 3.5.

v+ 1 (2; 0,2)-set

L e t H be a

v t 1

i n a B(v,3,X)

on S. Then, t a k i n g

2, X) c o n s i s t i n g o f h copies o f as blocks i t s 2-secantsY H i s a D ( T ; v+ 1 v- 1 , 3, X) whose S(2,2, Furthermore, HI = S \ H i s a subdesign B(2 blocks are the 0-secants o f H.

7).

Proof. Since

H i s of type (0,2),

the

= H I are completely contained i n

o f a B(-

v- 1

, 3, X)

and

uo

X blocks through any two p o i n t s i n S \ H

HI and

to

i s equal t o the number o f blocks

i s equal t o the number o f blocks through one

o f i t s points. Since H contains a l l p a i r s from a repeated

x

times, i t obviously consists o f

X

-set, each o f them v+ 1 copies o f S(2,2, 2)

-

Proposition 3.6. For any p a i r ( h , v ) s a t i s f y i n g e i t h e r ( i i i ) o r ( i v ) i n prop. 3.4 there e x i s t s a B(v,3,X) v- 1 a subdesign B ( 7 , 3 , X ) .

containing a (-

""2

,0,2)-set

whose complement i s

=

M.J. de Resmini

188

P r o o f . L e t S be a v - s e t such t h a t e i t h e r ( i i i ) o r ( i v ) i n prop. 3.4 h o l d s . S i n c e v t 1 v t1 can be coni s even, on a -subset H o f S a resolvable S ( 2 , 2 , 2 ) 2 s t r u c t e d and i t s b l o c k s can be p a r t i t i o n e d i n t o ( v - 1 ) / 2 p a r a l l e l classes,

-

-

each c o n t a i n i n g ( v t 1 ) / 4 b l o c k s . Moreover, c o n d i t i o n s (1.2) b e i n g s a t i s f i e d , v- 1 be t h e on S \ H can be c o n s t r u c t e d . L e t B1 ,...,B a B( 7 , 3 , X ) (v-1)/2 V t 1 p a r a l l e l c l a s s e s o f t h e r e s o l v a b l e S ( 2 , 2 , 2 ) ; w r i t i n g them down X t i m e s i n any f i x e d o r d e r and adding t o a l l b l o c k s i n a p a r a l l e l c l a s s t h e same v- 1 1 , t h e r e q u e s t e d B(v,3,A) i s c o n s t r u c t e d . element o f S \ H = {1,2,...¶ 2 A c y c l i c p e r m u t a t i o n on t h e elements i n S \ H i s used. The c o n s t r u c t i o n i s

-

sketched be1 ow:

2

............................. L

v-1 B (--Xt2) 1 2 (The symbol

v-1 B (--Xt3).. 2 2

v-1 %-l 2

v-1

At-t 2 - (2-

B . j means a l l t h e b l o c k s o b t a i n e d adding

j

1

the p a r a l l e l class v- 1 When X G -j-

t o a l l blocks i n

.)

Bi

,

1).

t h i s c o n s t r u c t i o n p r o v i d e s a B(vY3,A) w i t h o u t r e p e a t e d t h i s doesn't

b l o c k s ( p o s s i b l y , t h e r e a r e r e p e a t e d b l o c k s i n t h e subdesign v- 5 ). happen when X

7

To g i v e a c o n s t r u c t i o n o f a B(v,3,X)

containing a type

0,2) s e t when

e i t h e r ( i ) o r ( i i ) i n prop. (3.4) h o l d s , t h e f o l l o w i n g d e f i n t i o n ( t h a t seems new) w i l l be made.

A BIBD B(v,k,X) be p a r t i t i o n e d i n t o

w i l l be c a l l e d a p s e u d o r e s o l v a b l e BIBD r / X c l a s s e s , each c o n t a i n i n g

any p o i n t appears e x a c t l y

A

i f i t s b b l o c k s can

Av/k b l o c k s and such t h a t

t i m e s i n each c l a s s .

Thus, necessary c o n d i t i o n s f o r a p s e u d o r e s o l v a b l e B(v,k,A) PRB(v,k,X)) (3.7)

t o e x i s t are: v - 1 - 0

(k-1)

(Note t h a t ( 3 . 7 ) ' s i m p l y ( 1 . 2 ) ' s

and

.)

X v - 0 (k)

.

( f o r short

189

On sets of type (m,n) in BIBD S with h>-2 When

X = 1, a pseudoresolvable BIBD i s a r e s o l v a b l e BIBD.

Suppose t h a t e i t h e r ( i ) o r ( i i ) i n prop. 3.4 holds. I f a v t l pseudoresolvable B ( ~ , 2 , A ) , H, e x i s t s , then a B(v,3,X) e x i s t s c o n t a i n i n g v- 1 H as a type (0,2) s e t , t h e complement o f H b e i n g a subdesign B ( 2 , 3 , X ) . P r o p o s i t i o n 3.8.

Proof. Conditions -

(3.7) a r e s a t i s f i e d and ( v + 1 ) / 2 i s always odd. On a v t1 -subset H o f a v - s e t S a P R B ( ~ , Z , X ) being constructed, i t s b l o c k s a r e

p a r t i t i o n e d i n t o ( v - 1 ) / 2 classes each c o n t a i n i n g ATv t l

(c)2

b l o c k s and a l l

X times. To a l l b l o c k s i n a c l a s s t h e same

p o i n t s , any o f them appearing

p o i n t o f S \ H i s added; thus, through any p o i n t i n S \ H t h e r e a r e v t 1 1= u2 blocks. To t h e blocks so formed t h e b l o c k s o f a 8 ( 9 , 3 , X ) on 4 S \ H a r e added and t h e r e q u i r e d design i s completed. Since any c l a s s c o n t a i n s each p o i n t

X times, t h e r e a r e e x a c t l y X b l o c k s through any p a i r o f p o i n t s

x€H,yES\

H. V t l

a construction w i l l

As t o t h e e x i s t e n c e o f pseudoresolvable B(?,Z,X),

now be given f o r

X

= 2.

Recall t h a t t h e complete graph

K2n+l

has n Hamiltonian c i r c u i t s no two

7

( ( v t 1 ) / 2 odd); i t has 3 (v+l)/2 e d g e - d i s j o i n t Hamiltonian c i r c u i t s , c o n t a i n i n g ( v t 1 ) / 2 edges each. Taking t w i c e

o f them s h a r i n g an edge. Consider

K

a l l such c i r c u i t s ( t h e edges being p a i r s , 1.e. b l o c k s o f s i z e 2) t h e (v-1)/2 v+ 1 classes o f a pseudoresolvable B ( ~ , 2 , 2 ) a r e obtained. Using such design as a t y p e (0,2) s e t , a B(v,3,2)

can be c o n s t r u c t e d (as i t was shown i n prop.3.8)

and i t has no repeated b l o c k s .

vtl i s odd, a PRB(--,Z,X) can s t i l l be 2 constructed w i t h t h e h e l p o f t h e ( v - 1 ) / 4 p a i r w i s e e d g e - d i s j o i n t Hamiltonian When X > 2, X :0 ( 2 ) , and ( v t 1 ) / 2

A

” b l o c k s and c i r c u i t s o f K(vtj as f o l l o w s . Each c l a s s c o n t a i n s -22 4 v+ 1 2i s t h e l e n g t h o f a Hamiltonian c i r c u i t when X = 2. The r e q u i r e d classes 4 can be formed u s i n g X/2 c i r c u i t s f o r each o f them and r e p e a t i n g these classes twice. With such a PRBIBD t h e c o n s t r u c t i o n i n prop. 3.8 p r o v i d e s a B(v,3,X) w i t h repeated b l o c k s

(X

> 2

even).

Remark. Beside t h e c o n s t r u c t i o n s shown i n prop. 3.6 and 3.8,

t h e r e i s t h e con-

s t r u c t i o n by t r i a l and e r r o r by which i t i s p o s s i b l e t o g e t BIBD’s w i t h o u t repeated b l o c k s (provided t h a t X

G

( v - 5)/2),

c o n t a i n i n g a t y p e (0,2)

s e t and

M.J.de Resmini

190

a subdesign. Examples were c o n s t r u c t e d f o r (v,A) = (9,6), (1392)a (19a2)a (1394)a (9921, (21,2),

(13,6),

(7,319 (11a3)a (15,3),

(17,6),

(15,5),

(7,2),

(1995)

and a r e a v a i l a b l e by t h e a u t h o r (see a l s o [ 6 1 ) . O f course, BIBD's w i t h o u t r e peated b l o c k s a r e n o t isomorphic w i t h those h a v i n g repeated blocks; thus t h e r e a r e several examples o f non-isomorphic s o l u t i o n s .

4.SETS OF TYPE (m,n).IN B(vY4,A)'s F o r B(vY4,X)'s s e t s o f t y p e (0,3),

(1,3)

and (0,2) w i l l be considered

( t h e o t h e r p o s s i b l e types being t h e complements). Type (0,3). From (2.8) i t f o l l o w s

(4.1

1

h =

2v 1 3 t

Therefore, w i t h t h e h e l p o f necessary e x i s t e n c e c o n d i t i o n s (1.2) (2.11),

(2.2),

and (2.5) ( t 3

, (2.10) ,

u3 must be i n t e g e r s ) , i t i s easy t o prove:

and

P r o p o s i t i o n 4.2. containing a

Each of t h e f o l l o w i n g i s a necessary c o n d i t i o n f o r a B(v,4,h) 2vt 1 (-;0,3)-set t o exist: 3

(i)

v z 4 o r 13 (36), any A;

(ii)

v :4 o r 13 (18), A :0 (2);

(iii)

v :4 (9), A :0 (4);

(iv)

v :1 o r 10 ( 1 8 ) ,

(v)

v E 1 o r 28 (36), h E 0 (3);

(vi1

v :16 o r 25 (36),

(vii)

v

(viii)

v :1 o r 4 (12),

5

1 (3),

x

=0

A

x

(6);

:0 (3);

0 (12);

x

5

0 (3).

Some of t h e c o n d i t i o n s i n prop. 4.2 a r e a l s o s u f f i c i e n t ; b e f o r e p r o v i n g it, t h e f o l l o w i n g r e s u l t i s s t a t e d ( t h e p r o o f i s q u i t e obvious, see prop. 3.5).

H be a t y p e (0,3) s e t i n a B(v,4,A)

on S. Then H, t a k i n g 2vt 1 as b l o c k s i t s 3-secantsY i s a B(-,3,X); furthermore, S \ H i s a subdesign 3 v- 1 B(2,4,A) having as b l o c k s t h e 0-secants o f H. P r o p o s i t i o n 4.3.

Let

191

On sets of type (m,nl in BIBD's with 1 2 2

Now a construction o f B(vY4,A)'s containing a type (0,3) s e t w i l l be shown. If v

Proposition 4.4.

and

X s a t i s f y e i t h e r o f t h e conditions ( i ) , ( i i ) ,

2v+ 1 then a B(v,4,X), U , containing a (-;0,3)-set 3 v- 1 A ) , e x i s t s . Moreover, i f X whose complement i s a subdesign B(-,4, 3 ( v - 10)/18, U doesn't contain repeated blocks. o r ( i i i ) i n prop. 4.2,

G

(v-7)

-

x

Proof. L e t S be a v-set, v s a t i s f y i n g the hypothesis; then 2 v + ' -1 3 (6) so 3 2vt 1 v-1 t h a t a resolvable S ( 2 , 3 , 3 ) e x i s t s . Furthermore, a B(-,~,A) exists. 2v+l on H and L e t H be a 2v t 1 -subset o f S. Construct a resolvable S(2,3,-j-) 3 v-1 a B(-,4,X) on S \ H. The blocks o f the former can be p a r t i t i o n e d i n t o 3 B , each o f them c o n t a i n i n g ( 2 v t 1 ) / 9 (v 1)/3 p a r a l l e l classes B1, ( v - 1 )/3 blocks. W r i t e down these p a r a l l e l classes, i n any f i x e d order, h times. Add-

-

....

-

i n g t o a l l blocks i n the same class the same element i n S \ H f o r each par

-

t i t i o n , c y c l i c a l l y permuting the elements o f S \ H when passing from one par t i t i o n t o the next one,

t3 blocks o f

B(v,4,x)

are constructed; the remaining

ones are the blocks o f t h e subdesign. The construction i s sketched below, : taking S \ H = {1,2,..,,

v- 1 B1 3

B2 1

v-1 B (--Xt2) 1 3

v-1 B (--A 2 3

............ B ( v - l ) / 3 v-1 (3-

............................

(The sums are modulo

(v-1)/3

+3)

.... B( ~ - 1 ) / 3(--v-3 1

and t h e n o t a t i o n B. j means t h a t 1

v-1 3

At-t

1)

j E

S1 H

i s added t o a l l blocks i n class Bi .) v- 1 When X 6 - there are no repeated blocks i n the above t a b l e (since a l l Bi

blocks i n each pair

x E H, y

E

S

are d i s t i n c t ) and i t i s s t r a i g h t f o r w a r d t h a t through any \ H

t h e r e are X blocks and

u3 = X ( 2 v t 1)/9.

(There may

be repeated blocks i n the subdesign .) The o t h e r cases i n prop. 4.2 w i l l now be considered. Proposition 4.5.

I f v and X s a t i s f y e i t h e r ( i v ) o r ( v ) i n prop. 4.2 and i f a

pseudoresolvable B ( v , 3 , A )

-

e x i s t s , then a B(v,4,X)

can be constructed

M.J. de Resmini

192 2v+ 1 which c o n t a i n s a (-;0,3)-set 3

v- 1 and a subdesign B ( 3 , 4 , h ) .

v- 1 s a t i s f y e i t h e r ( i v ) o r ( v ) , a B(--,4, X) e x i s t s ; moreover 3 2 v t l v-1 PRB(my3yh) e x i s t s ( ( 3 . 7 ) ' s h o l d ) , t h e n i t s -5-33 3 3 2,v+ 1' blocks, can be p a r t i t i o n e d i n t o ( v 1 ) / 3 classes each c o n t a i n i n g 3 3v-1 blocks. Then adding t o a l l b l o c k s i n each c l a s s t h e same p o i n t o f B ( 7 , 4, A),

Proof. Since v and h 2v+ 1 z 1 (6). I f a

t3 b l o c k s

-

-

a r e obtained which, t o g e t h e r w i t h t h e blocks o f t h e subdesign, f o r m

t h e b l o c k s o f a B(vY4,X) c o n t a i n i n g a t y p e (0,3) s e t . As t o t h e e x i s t e n c e o f PRB(v',3,X),

necessary c o n d i t i o n s (3.7) become

either 1 or 2

(4.6

X

(4.7

X z 0

E

(3)

(3)

and

v' z 3

(6)

and

v' E 1

(2).

When ( 4 . 6 ) ' s hold, s i n c e a RB(v',3,1)

,

or

e x i s t s , a PRB(v',3,X)

s t r u c t e d i n t h e f o l l o w i n g way. The b l o c k s o f RB(v',3,1) into

(v'

-

1)/2

p a r a l l e l classes, each c o n t a i n i n g

t u r n X d i s t i n c t classes among these, v' c o n t a i n i n g v ' / 3 b l o c k s . I f X
can be p a r t i t i o n e d

v ' / 3 blocks. Taking i n

( v ' - 1 ) / 2 new classes a r e formed each

- 1,

blocks (two classes share a t l e a s t v ' / 3 as a t y p e (0,3) s e t i n a

can be con-

a l l these classes c o n t a i n d i s t i n c t b l o c k s ) ; thus w i t h such a PRB (v',3,X)

t h e l a t t e r has no repeated b l o c k s ( w i t h t h e

exception o f p o s s i b l e repeated b l o c k s i n t h e subdesign). When (4.7)'s h o l d and X = 3 a PRB(v',3,3)

w i t h repeated b l o c k s can be

c o n s t r u c t e d t a k i n g a l l i t s classes i d e n t i c a l w i t h t h e f o l l o w i n g : 1 2 3, 2 3 4, 3 4 5,

4 5 6,

5 6 7,

..., v ' - 2

When such design i s taken as a type (0,3)

v'-1 v',

s e t i n B(v,4,3),

v'-1 v ' 1,v'l

2

.

t h e t3 blocks ob-

t a i n e d completing these repeated b l o c k s by elements o f t h e complement a r e a l l d i s t i nct. When X :0 PRB(v',3,X)

and A # 3, t h e number o f b l o c k s i n each c l a s s o f t h e

(3) V'

i s X-

=

3 3 times t h e previous one.

v'

so t h a t each c l a s s can be formed r e p e a t i n g X /3

Analogously t o prop. 4.5 t h e n e x t one

i s proved.

193

On sets of type (m,n) in BIBD's with A22 I f v and A s a t i s f y e i t h e r ( v i ) , o r ( v i i ) , o r ( v i i i ) i n prop. 2v t 1 e x i s t s , t hen a 4.2, t h e l a t t e r w i t h v f 4 ( 3 6 ) , and i f a PRB(-,3,A) 3 v- 1 B(vY4,X) e x i s t s which c o n t a i n s a t y p e (0,3) s e t and a subdesign D ( 3 , 4 , X ) .

P r o p o s i t i o n 4.8, ,

Remark. By t r i a l and e r r o r B ( v Y 4 , X ) ' s w i t h o u t repeat ed b l o c k s c o n t a i n i n g a v- 1 2vt 1 subdesign can be c o n s t r u c t e d and o f t e n t hey ( ~ ; 0 , 3 ) - s e t and a (-,4,A) 3 a r e n o t is omo rp hi c w i t h t h o s e o b t a i n e d by p r o p . 4.4, 4.5, and 4.13. I f i s w o r t h

'' s a t i s f i e s 3

n o t i c i n g t h a t when

h = 2v

f o r a 8(hY3,X) c o n t a i n i n g a t y p e (0,2) t o c o n s t r u c t a B(v,4,h)

t h e n ecessary and s u f f i c i e n t c o n d i t i o n s s e t t o e x i s t , i t i s sometimes p o s s i b l e

c o n t a i n i n g a t y p e (0,3) s e t which i n t u r n c o n t a i n s a

set [61.

t y p e (0,2) Type (1,3).

F o r an ( h ; l , 3 ) - s e t

t o e x i s t i n a B(v,4,A),

h must be a s o l u t i o n o f

(2.3) which now w r i t e s Ah2

(4.9)

-

h(3r+A)t3b = 0

,

and i t s d i s c r i m i n a n t must be a non-negative square. Since (4.10)

A

= X

2

V

y

v must be a square; then h =

(4.11

v

r

2

4;

It i s n o t d i f f i c u l t t o prove [ 6 1 :

P r o p o s i t i o n 4.12.

If an ( h ; l , 3 ) - s e t

b l o c k s i t s 3-secants S(2,3,h)

-

i s a B(v,3,X)

H e x i s t s i n a B(v,4,X),

( i1

XV(V-1) F 0

(ii)

X ( v - 1 ) :0

(12), (3)

y

-

t a k i n g as

fi,

Fo r a B ( v , ~ , x ) c o n t a i n i n g an ( h ; l , 3 ) - s e t

f o l l o w i n g must h o l d :

H

c o n s i s t i n g o f X copies o f a r e s o l v a b l e

and t h e same i s t r u e f o r t h e complement

P r o p o s i t i o n 4.13.

t hen

t o exist, a l l the

194

M.J. de Resmini (iii)

v

a square,

(iv)

h :3

(v)

r - X : O

(6)

and

-

v

h E 3

(6),

and

(2).

From prop. 4.13 i t f o l l o w s v = 3 6 ( 2 ~ t 1 ) ~ and X E

P r o p o s i t i o n 4.14.

a r e necessary c o n d i t i o n s f o r a B(v,4,X)

0

(3)

c o n t a i n i n g a t y p e (1,3) s e t t o e x i s t .

C o n d i t i o n s i n prop. 4.14 a r e a l s o s u f f i c i e n t (when a p r o p e r s e t s w i l l show.

as t h e main r e s u l t on t y p e (1,3)

For a l l v = 3 6 ( 2 w t 1 )

P r o p o s i t i o n 4.15.

2

,

w

0,1,2r.,y V t

there exists

Jv

w i t h X = 3 ( 2 w t l),which c o n t a i n s a (-;1,3)-set 2 v - JT a c t e r s complement, a (-;l ,%)-set. 2 B(v,4,X),

Proof.

Since

v 4; :3 2 f

(6),

there e x i s t resolvable

t h e s e t formed b y X copies o f a r e s o l v a b l e S(2,3,

a

and i t s 2-char-

v f JK )'s. 2

S(2,3,

H be t h e s e t c o n s i s t i n g o f h copies o f a r e s o l v a b l e S(2,3, and

X i s chosen),

72w

2 72w t 7 8 w t 2 1 ) 2

t

66w

t

15) =

Let

z.

S

The blocks o f S can be p a r t i t i o n e d i n t o 36wL t 3 9 w t 10 p a r a l l e l classes, each 2 c o n t a i n i n g 24w t 2 6 v t 7 blocks; t h e b l o c k s o f ' s can be p a r t i t i o n e d i n t o 2 2 36w t 33w t 7 p a r a l l e l classes, each c o n t a i n i n g 24w t 22w t 5 blocks. Consider

X copies o f S H

(G)

('s),

each of them p a r t i t i o n e d i n t o i t s p a r a l l e l classes and c a l l

t h i s s e t . A same element belonging t o t h e complement w i l l be added t o

a l l b l o c k s i n a same p a r a l l e l class; t h e r e f o r e Nt =

N-

Y p a r a l l e l classes of

S

I iI Y p a r a l l e l classes o f

and

's

IS1

must be i n t e g e r s . Since h = 3(2w Nt = 3w N, and N-

t

2

t

and

1), N

-

= 3w t 1.

a r e t h e numbers o f times t h e elements o f t h e complement must be r e -

On sets of type Im, n) in BIBD's with X r 2

195

peated. Through any p o i n t n o t i n H t h e r e are

u 3

(2w

2 1 ) ( 3 6 ~ t 45w

t

t

14)

3-secants o f H and U

24w

2

3

t

26w

t

Nt

7

.

(Indeed, any p a r a l l e l c l a s s o f S c o n t a i n s e x a c t l y

24w2 t 26w

+ 7 b l o c k s and

t o a l l these blocks t h e same element i s added,) S i m i l a r l y * U

24w Since

2

t

3 22w

+

Nt and N-

-,

= N

5

2 u = ( 2 w t 1 ) (36w t 2 7 w t 5 ) . 3

%

where

a r e coprime w i t h t h e numbers o f b l o c k s i n t h e p a r a l l e l classes,

t h e r e a r e no repeated b l o c k s i n Il(vY4,A). (A simple computation shows t h a t a l l parameters a r e c o r r e c t . For more d e t a i l s , see [ 61.) Remark. By prop. 4.15,

a S t e i n e r system S(2,4,v)

cannot c o n t a i n a type (1,3)

set. Thus a previous r e s u l t i s improved 141, [51. Type (0,2).

From (1,2),

(2.10),

and (2.11) i t f o l l o w s

P r o p o s i t i o n 4.16.

v t2 c o n t a i n i n g a (-;0,2)2 t o e x i s t are either

Necessary c o n d i t i o n s f o r a B(v,4,X) 2v 2 s e t (and i t s 2-characters complement, a (-;2,4)-set) 2

-

(1)

X

i

1

(2)

and

v 54

(12),

(ii)

X

f

0

(2)

and

v

(3)

5

1

or'

.

Whether these c o n d i t i o n s a r e a l s o s u f f i c i e n t i s s t i l l t o be proved. Anyhow, t h e f o l l o w i n g examples can be quoted: S(2,4,16)

c o n t a i n i n g a (6;0,2)-set

(i.e.

an o v a l i n AG(2,4)).

Then t h e r e p l i c a t i o n 6(16,4,X),

formed t a k i n g X copies o f S(2,4,16),

s o l u t i o n for any X.

I

S(2,4,28)

c o n t a i n i n g a (10/0,2)-set

is a

[ 7 1 and again i t ' s p o s s i b l e t o take

a r e p l i c a t i o n BIBD. B(16,4,3)

c o n t a i n i n g a (6;0,2)-set

contained i n a type (0,3)

s e t [ 61.

M.J. de Resmini

196

Moreover, B(7,4,2) 8(13,4,4)

, B(10,4,2),

, B(7,4,4),

6(13,4,2)

c o n t a i n i n g a type (0,2)

6(10,4,4)

, and

s e t have been c o n s t r u c t e d ( b y t r i a l and e r r o r )

[ 61.

5.SETS OF TYPE (m,n) I N B ( v Y 5 , X ) ' s Sets o f type (0,4) w i l l be considered f i r s t ; f o r such a s e t (2.9) becomes

moreover: 3vt 1 L e t H be a ( ~ ; 0 , 4 ) - s e t i n a B(v,5,X).

P r o p o s i t i o n 5.2.

Then, t a k i n g as

b l o c k s i t s 4-secantsY H i s a B(3v+1,4,X) and i t s complement i s a subBIBD 4 v- 1 B(---,5,A), whose b l o c k s a r e t h e 0-secants o f H. 4 Proof. See prop. -

3.5 and 4.3.

With t h e h e l p o f (1.2), P r o p o s i t i o n 5.3. v- 1 i s a B(4,5,X),

and (2.11) i t i s easy t o prove

(2.10),

For a B(v,5,X)

c o n t a i n i n g a t y p e (0,4)

set, whose complement

t o e x i s t , one o f t h e f o l l o w i n g c o n d i t i o n s must h o l d :

,

( i1

v z 5 or 21

(80)

(ii)

v E 5

and X E 0

(iii)

v :1 o r 5

(iv)

v

(v)

v :5

(40)

(vi1

v z 5

(8),

(16)

(20)

any A; (5);

and X : 0 ( 4 ) ;

z 1 ( 4 ) , v > 21, and X and

X

E 0

v 2 21,

E 0

(20);

(2); and

X z 0 (10).

The necessary e x i s t e n c e c o n d i t i o n s i n prop. 5.3 a r e a l s o s u f f i c i e n t ; when e i t h e r ( i ) o r ( i i ) h o l d s a B(v,5,X)

w i t h o u t repeated b l o c k s w i l l be c o n s t r u c t e d

( t h e r e may be repeated b l o c k s i n t h e complement if v i s t o o s m a l l ) ;

when e i t h e r

(iii), o r ( i v ) , o r ( v ) , o r ( v i ) h o l d s t h e r e w i l l g e n e r a l l y be repeated blocks. P r o p o s i t i o n 5.4.

I f e i t h e r (i.) o r ( i i ) i n prop. 5.3 holds, then a B(v,5,X)

= B

197

On sets of type (m,nl in BIBD's with A22

can be c o n s t r u c t e d which c o n t a i n s a (=;0,4)-set H and a subdesign 4 v- 1 v- 1 B(-,5,A). moreover i f A 4 - and ( i ) holds, t h e r e a r e no repeated blocks 4 4 i n B; if (ii) holds, t h e o n l y p o s s i b l e repeated b l o c k s i n B a r e those o f t h e subdesign. Proof.

Since

3v+l 7 - 4

(12),

there e x i s t s a resolvable

S(2,4,

3vt ' ) . 4

Taking

i s c o n s t r u c t e d as

as H A copies o f t h i s S t e i n e r system, t h e r e q u i r e d B(v,5,A) i n prop. 3.6 and 4.4.

I f e i t h e r (iii) o r ( i v ) i n prop. 5.3 holds and i f a pseudo-

P r o p o s i t i o n 5.5.

B ( t Y 4 , A ) e x i s t s , then a B(v,5,A) e x i s t s which c o n t a i n s 4 v-1 B ( y y 4 , A ) as a s e t o f type (0,4) and whose complement i s a subdesign B ( 4 , 5 , A ) . resolvable

Proof.

See prop. 3.8 and 4.5.

As t o t h e e x i s t e n c e o f t h e PRB(v',4,~)'s i n prop. 5.5, when A = 4 such a BIBD can be c o n s t r u c t e d so t h a t t h e B ( v , ~ , A ) i n which i t i s contained as a type (0,4) s e t has no repeated blocks, w i t h t h e e x c e p t i o n o f p o s s i b l e repeated blocks i n t h e subdesign. Namely, take a l l classes o f blocks i n B ( v ' ,4,4)

equal

t o the following: 1 2 3 4,

2 3 4 5,

3 4

5 6,

...,

v'-3 v'-2 v'-1 v',

v ' - 2 v ' - 1 v ' 1, (When A

0

(4)

or

A

=0

(20),

v'-1 v' 1 2,

v' 1 2 3.

A # 4, t h i s c l a s s can s t i l l be used, b u t

a (vY5,A) B I B D w i t h repeated b l o c k s i s obtained, as i n s e c t i o n s 3 and 4 .) P r o p o s i t i o n 5.6.

I f e i t h e r ( v ) o r ( v i ) i n prop. 5.3 h o l d s and i f a pseudore-

solvable

B ( E Y 4 , A ) e x i s t s , then a B(v,5,A) c o n t a i n i n g i t as a t y p e (0,4) 4 v- 1 s e t can be c o n s t r u c t e d and i t s complement i s a subdesign B ( 4 , 5 , A ) . When

A

= 2, a pseudoresolvable

B(v',4,2)

as r e q u i r e d i n prop. 5.6 can

be c o n s t r u c t e d t a k i n g a l l classes equal t o t h e f o l l o w i n g one: 1 2 3 4,

3 4 5 6, 5 6 7 8,

Taking such B(v',4,2)

..., v ' - 3

as a type (0,4)

v ' - 2 v ' - 1 v ' , v'-1 set, a B(v,5,2)

V'

1 2.

i s c o n s t r u c t e d as

M.J.de Resmini

198

i n sections 3 and 4 and i t has no.repeated blocks, w i t h the exception o f possible repeated blocks i n the subdesign. (When e i t h e r

h f 0

(2)

o r X:O

(lo),

can s t i l l be used, b u t t h e r e w i l l be repeated

X # 2, t h e above B(v',4,2)

blocks i n B(v95,X), as already shown. ) For type (0,3) s e t s i t i s easy t o v e r i f y : P r o p o s i t i o n 5.7.

I f H i s a type (0,3) s e t i n a B(v95,X), then, t a k i n g as blocks v+l i t s 3-secants, H i s a B(-,3,X). 2

Furthermore, from (1.2), P r o p o s i t i o n 5.8.

(2.10) and (2.11)

i t follows:

For a B(v95,X) containing a type (0,3)

s e t t o e x i s t , one o f

the f o l l o w i n g necessary conditions must hold: ( i1

v :5 or 41

(60),

(ii)

v

1 or 5

(20),

(iii)

v :1 o r 5

(iv)

v

=5

(10)

(12)

and x = O

(v)

v :1

(4),

v 2 13,

and A E 0

(15);

(vi 1

V

:1

(2),

v 2 11,

and X :0

(30).

=

any A ; and A

=0

and A f 0

(3);

(6);

(5);

No general c o n s t r u c t i o n o f B(v,5,X)'s

v+ 1 c o n t a i n i n g ( ~ ; 0 , 3 ) - s e t s has been

found y e t , b u t examples can be w r i t t e n by t r i a l and e r r o r . S i m i l a r r e s u l t s h o l d f o r type (0,2) sets i n B(vS5,h)'s. Since now h =

v+3 7,

such a s e t consists o f a l l p a i r s from an h-set, each o f them repeated X times. Furthermore, P r o p o s i t i o n 5.9.

E i t h e r o f the f o l l o w i n g i s a necessary c o n d i t i o n f o r a v+ 3 B(v95,X) containing a (-;O,2)-set to.exist: 4

=5

(i 1

v

(ii)

v

(iii)

v E 5

(8)

and

(iv)

v :1

(4)

and A :0

or

21

(40),

any X ;

1 or

5

(20)

and Xz 0

X

0

(2);

(5); (lo),

( v 2 9).

On sets of type (m,n) in BIBD 's with A 2 2

199

When ( i v ) i n prop. 5.9 holds, a special example can be constructed [ 6 1 : B(9,5,10)

H whose complement i s a B(6,5,4)

containing a (3;0,2)-set

a

having as

blocks the 0-secants o f H. The necessary conditions f o r t h e complement o f a 3v- 3 when v 4 u t 1, A = 10 and 2 < u G 30 type (0,2) s e t t o be a B(-,5,X') 4 are s a t i s f i e d a l s o by t h e p a i r s (X',u) = (5,3), ( 6 , 7 ) . I n B ( v , ~ , ; \ ) ' s also type (1,4) and (1,3)

can be considered, b u t they seem

n o t t o e x i s t . ( l l i t h the help o f r e s u l t s i n section 2, i t ' s easy t o f i n d necessary existence conditions, b u t cone o f them has been so f a r proved t o be also suff i c i e n t [61.)

6.SETS OF TYPE ( O a k - 1) I N B(v,k,X)'s The s i z e h of a type (0,k h =

-

1) s e t i n a B(v,k,X)

v(k-2)tl k- 1

is

,

and i t i s s t r a i g h t f o r w a r d t o prove t h e f o l l o w i n g r e s u l t . Proposition 6.2.

I f a B(v,k,A)

on S contains a type ( O a k - 1) s e t H, then,

t a k i n g as blocks i t s ( k - 1 ) - s e c a n t s , H i s a B(h,k- 1,X) and i t s complement i s a subdesign B ( v - h,k,X)

(whose blocks are the 0-secants o f H).

Since (1.2) and (2.10) must h o l d and h i n (6.1) must be an i n t e g e r , i t ' s

a matter o f computations t o f i n d necessary existence conditions; namely: P r o p o s i t i o n 6.3.

For a B(v,k,h)

containing a type ( O a k - 1 ) s e t t o e x i s t , one

o f the f o l l o w i n g must hold:

(41)

2 v E k o r k ( k - 1 ) t 1 ( k ( k - 1 ) ) , any 2 v E k ( ( k - 1) ) and X f 0 (k);

(iii)

v :1

(iv)

v :1

(11

Remark.

or k ( k ( k - 1 ) ) (k-1)

and

and

X ;

X :0 ( k - 1 ) ;

X z 0 (k(k- 1)).

When prime f a c t o r s o f both k and k - 1 are considered, other necessary

M.J. de Resmini

200

conditions may a r i s e besides those i n prop. 6.3. Necessary conditions i n prop. 6.3 are a l s o s u f f i c i e n t provided t h a t s u i t a b l e smaller BIBD's do e x i s t . By an argument s i m i l a r t o t h e one used i n sections 3, 4,and 5, i t i s easy

t o prove t h e general r e s u l t s s t a t e d below. Proposition 6.4.

I f e i t h e r ( i ) o r ( i i ) i n prop. 6.3 holds and i f a resolvable

S(2,k-1 ,h) and a B(v-h,k,A)exist,

then a B(v,k,A)

whose complement i s a subdesign B ( v - h,k,A),

containing an (h; 0,k-1)-set,

can be constructed. Furthermore,

i f A G v - h, the o n l y repeated blocks are the possible repeated blocks i n

B ( v - h,k,X).

Proof.

The type (0,k- 1) s e t consists o f X copies o f a resolvable S(2,k- 1,h).

The blocks o f each copy are p a r t i t i o n e d i n t o p a r a l l e l classes and t h e same p o i n t from the complement i s added t o a l l blocks i n a same class. By a c y c l i c permutation on the elements o f t h e complement repeated blocks a r e avoided, provided t h a t X G v - h. (Cf. prop. 4.4 and 161.) Proposition 6.5. and a B(v-h,k,X)

I f e i t h e r ( i i i ) o r ( i v ) i n prop. 6.3 holds and i f a PRB(h,k-1,A)

e x i s t , then a B(v,k,A)

as a type ( 0 , k - 1 ) Proof.

can be constructed, having the former

s e t and t h e l a t t e r as subdesign.

See prop. 3.8,

4.5 and [ 6 1 .

As t o the existence o f PRB(h,k- l , x ) ' s ,

when

X = k - 1 and h = k ( k - 2 ) w + k - 1 ,

such a BIB0 can be constructed t a k i n g a l l i t s ( h - 1 ) / ( k - 2) classes o f blocks equal t o : 1 2 3

... k-1,

2 3 4

... k-1

... k ( k - 2)Wt k - 2

k ( k .2 ) ~ k(k-Z)w+l

... k-2,

3 4 5

... k-1

k ktl,

............

k ( k - 2)Wt k - 1,

... k ( k - 2 ) w t k - 1

k(k-Z)W+k- 1 1 2

k, 1,

.................

and when such PRBIBD i s taken as a type (0,k- 1) s e t i n a B(v,k,k-

1), v =

= k ( k - l ) w + k , the l a t t e r w i l l have no repeated blocks besides t h e possible

repeated blocks i n t h e subdesign.

A s i m i l a r c o n s t r u c t i o n can be done when v

!1

( k ( k - 1 ) ) and

X

= k - 1.

20 1

On sets of type (m, n) in BIBDf with A22

7. ON SETS OF TYPE (m,n) I N SYMMETRIC BIBD's F i r s t l y r e c a l l t h a t (v,k,A) b = v, and t h e r e f o r e r

and k

-A

symmetric BIBD (SBIBD) i s a design i n which

k, and any two blocks meet i n A p o i n t s . Hence

BIBD the order i s r - A).

i s the order o f the design ( f o r a (v,k,k)

The we1 1 known Bruck-Ryser-Chowla theorem provides necessary conditions f o r a SBIBD t o e x i s t .

Assums and van L i n t [ 1 ] already studied sets o f class [ 0,l , 2 ] i n

Remark.

- c a l l e d ovals

SBIBD's. Such sets

by the above mentioned authors

(0,2) when the order i s even and o f type (0,1,2)

- are o f

type

when i t i s odd. Their approach

i s q u i t e d i f f e r e n t from the one here, b u t some r e s u l t s (included here f o r sake o f completeness) are obviously t h e same. Anyhow, sets o f type (0,2) w i l l be considered as l e a s t as possible. For a type (0,n) with m

<X

set, by (2.10),

n

<

k-

A; hence, no s e t o f type (m,k)

e x i s t s i n a SBIBD. Moreover:

Proposition 7.2.

A type (0,k- 1) s e t i n a (v,k,A)

SBIBD i s the complement o f

a block.

Proof,

Straightforward.

O f course, when the order o f a SBIBD i s a prime, o n l y t r i v i a l sets o f type (0,n) e x i s t ( i . e . n

1 and

n

k

- A).

To generalize t h e idea o f oval, t h e f o l l o w i n g d e f i n i t i o n i s made.

A set o f type (0, X t l ) i n a (v,k,A)

SBIBD w i l l be c a l l e d a A-oval;

by

(2.8) i t s s i z e i s h = k t 1. (Obviously, 1-ovals are t h e usual ovals i n even order p r o j e c t i v e planes and those considered i n [ 1 ] when the order i s even.) Therefore, i f a (v,k,A)SBIBD = ( k t l , k(k+l), A At1 (0, A t 1 ) s e t i n $.

(v',b',k',X')

S contains a A-oval, t

then a B I B D w i t h parameters

1,A) e x i s t s which i s embeddable as a type

M.J.de Resmtni.

202

Furthermore , simp1e computations show: P r o p o s i t i o n 7.3.

I f a (v,k,2)

b i p l a n e contains a 2-oval (i.e.

a type (0,3) s e t ) ,

then v = (3wt2) (3wt1) 2 Proposition 7.4.

1,

t

k = 3w

t

2,

and

h = 3w

t

3

.

I f a (41 t 3, 2A t 1, A ) SBIBD e x i s t s , then any o f i t s r e s i d u a l

designs i s a A-oval and two A-ovals ( i n the same design) need n o t t o be i s o morphic (see [ 12 I). Now some r e s u l t s on type (1,n) sets i n SBIBD’s w i l l be proved, keeping i n mind t h a t such a s e t , t a k i n g as blocks i t s n-secants i s a B(h,n,A)

(otherwise,

i t c o u l d n ’ t be o f type (1,n)).

L e t U be a (v,k,A)

SBIBD; Dose and Shrikhande 121 defined a (v”,k*,A)

sub-

design U)+ o f U t o be a Baer subdesign i f i t i s a symmetric B I B D and kzc’

(7.5)

= 1

-A

Jk

t

(see also [8]). With the help o f (2.2) and (2.3), i s a (vJc;l,

kk)-set i n

i t ’ s easy t o check t h a t a Baer subdesign

D [61. The converse i s a l s o true, as t h e next p r o p o s i t i o n

w i l l show. Proposition 7.6.

L e t H be a type (1,n) s e t i n a SB(v,k,A).

-

If

H i s a SBIBD

( t h e blocks being i t s n-secants), then n = 1 t J k - X and H i s a Daer subdesign. Furthermore, there e x i s t type (1,l

-

Jk

t

- A)

sets which are n o t Baer subdesigns;

such a s e t w i l l be c a l l e d a Hermitian set. Proof.

(7.7)

When the BIBD i s symmetric and m = 1,-(2.3) Ah2

-

h (kn

and (7.1) holds. I f H i s an (h,n,A)

(7.8)

h

t

A) t v n = O

SBIBD, then

n(n-l)+

x

becomes:

203

On sets of type (m,n) in BIBD's with X 22 Moreover, (2.7) being now

k

(7.9)

I

( n - 1)

-

(n

(k-A),

.

l)t t X

S u b s t i t u t i n g i n (7.7) f o r h, k, v the values given by (7.8),

(7.9) and (7.1)

respectively, the f o l l o w i n g equation i s obtained: 2 2 n ( 1 - t ) t n ( t t t - h -1)

(7.10)

-

t

2

0,

At-tth

t

whose d i s c r i m i n a n t i s (7.11)

A

2 2 ( t - t t h -1)

.

Therefore 2 -(t tt-A-1) f (tZ-t+h-l) 2(1 t )

-

n =

n =t-X i s meaningless when X > 1 t-1 the i n t e r e s t i n g s o l u t i o n i s

The s o l u t i o n

(7.12)

and t r i v i a l when A = 1;

n = l t t ; t

since

n = 1

k - A , from n- 1 k- X n-1

t-

(7.12) i t f o l l o w s

, i.e.

(7.13)

- 2n - ( k - A - 1) = 0, t J k - h .

n2 n = l

whence

Thus, H i s a Baer subdesign and

-

(7.14)

h = ' + Jh k - h

Besides the r o o t (7.14), (7.15)

h ' 1-

Jk-x

x

JE+ I = - 4-+

x

k -

x

equation (7.7) has t h e r o o t (k

-

1)

t

1

*

M.J. de Resrnini

204

(which d o e u correspond t o the .complement o f ti). Jk- A A ( 7 ( k - 1 ) t 1;1, 1 t J k ) - s e t i n a (v,k,A)

-

SBIBD, k

X a

square, w i l l be c a l l e d a Hermitian subset because i t i s a Hermitian a r c when

[lll.

X = l

2 and a square order a w e l l known

Remark, Prop. 7.5 extends t o SBIBD's w i t h X theorem i n p r o j e c t i v e planes, proved by

M.

T a l l i n i S c a f a t i i n [ 11 1.

With t h e help o f Baer subdesigns, Hermitian subsets and maximal type (0,n) sets i n (v,k,X) type (m,n),

m

k

SBIBD's,

> 2,

-

X a square, i t i s p o s s i b l e t o c o n s t r u c t sets o f

extending t o SBIBD's ( w i t h X

> 2)

some r e s u l t s by !I.

de

F i n i s [ 3 1 , as the next propositions w i l l show. Proposition 7.16.

a square) having no common p o i n t ( i . e . -set (where h

Proof.

[ 2 , k* + 1, 2k*l, that

tEki= 0.

Remark.

skew). Then H

i s given by (7.14) and

I f i s easy t o check t h a t i f

k* t 1 )

(2h;2,

-

U

i s a (2h;2,

H'

k * t 1)-

k* by (7.5)).

(h;l,k")

s a t i s f y equation ( 2 . 3 ) , then

s a t i s f y the same equation. Since

H U HI

i s a s e t o f class

i t i s enough t o prove t h a t i t i s o f type ( 2 , k*

Solving equations(2.1)

t

l),i.e.

the r e s u l t i s obtained.

I f h i s the s i z e o f a Baer subdesign, then 2 X , so t h a t k = a t x,

-

= Jk

SBIBD ( k - X

L e t H and H ' be two Baer subdesignsof a (v,k,X)

hlv,

and w r i t i n g a =

Therefore, a SBIBD whose order i s a square may be p a r t i t i o n e d i n t o Baer subdesigns ( t h i s i s t r u e and w e l l known when 1 = 1). The example a t the end o f t h i s s e c t i o n shows such a p a r t i t i o n . C o r o l l a r y 7.17. (v,k,h)

If H1,

SBIBD, k - X

s = 2,3,...,

..., Hs

are

s

a square, then HIU

a 2 - a t X - 1.

p a i r w i s e skew Baer subdesigns o f a ...U

H

S

i s a type (s, s t a ) set,

205

On sets of type (m, n) in BIBD 's with X 2 2

Proposition 7.18. L e t H and H I be two skew Hermitian subsets i n a (v,k,A) w i t h k - h a square. Then the set t h e o r e t i c union o f t h e i r p o i n t s ,

.-

(i 2; t Jk-A, Proof.

2(1 t 4k-X))-set,

-

- A)) J k - A)];

2(1 t

= 2h' and h '

-

I f (h';l,l

2(1 t J k

where

t J k - A ) s a t i s f y equation (2.3),

.,

i,i s

SBIBD an

i s given by (7.15).

-

then (2h'; 2 + J k - A,

s a t i s f y the same equation. H i s a s e t o f class [ 2, 2 t J k - A

,

s o l v i n g the simultaneous equations (2.1),

i s found, so t h a t

the solution t 2 = 0 ( 2 + J k - A , 2(1 - J k - A ) ) and t h e statement

i s o f type

i s proved. I f t h e r e e x i s t s pairwise skew Hermitian subsets i n a (v,k,X)

C o r o l l a r y 7.19.

- A a square,

SBIBD, k (sh';(s

- 1 ) + kS,

then the s e t t h e o r e t i c union o f t h e i r p o i n t s i s an 2 G s < k/kfe (where h ' i s given by (7.15) and kfg

skft)-set,

by (7.5)). Proposition 7.20. SBIBD ( k

-A

L e t H and H ' be two skew type

a square). Then

i

-

(0, J k - A ) s e t s i n a (v,k,A)

- A,

H u H' i s a s e t o f type (Jk

- A),

2 Jk

having s i z e h=2h=2 Proof. triple

J k - x (k- Jk-A) A

-

s a t i s f y equation (2.3), the same i s t r u e f o r t h e - (K; J k - A , 2 J k - A ) . H i s a s e t o f class 0, J k - A , 2 J k - A ] ,

Since (h;O, J k - A )

-

[

b u t s o l v i n g the simultaneous equations (2.1)

so t h a t

-

(Jk-A

i s o f type I f H1,

C o r o l l a r y 7.21.

; (s

(sa

-

,

..., HS,

parwise skew sets o f type (0,a)

-

2Jk-A) 2

the s o l u t i o n to = 0

and the statement i s proved.

<s
i n a (v,k,A)

t

A/a

SBIDD, then H1

3

4

1 3 6

7 814

3

5 8

1 2 6

91015

1 4

5 14 15 16

1 5 7

91112

4 6

7 91316

5

U

...

U

H

S

i s an

SBIBD ( a biplane) whose blocks a r e [ 9 1 :

611

4

-

(a = J k - A ) , are s

l ) a , sa)-set.

Example. Consider the (16,6,2) 2 3

i s found,

7 10 12 15 91315

M.J.de Resmini

206

5 6

1 4 8 10 11 13

8 10 12 16

1

2

3 12 13 16

2

2'

4

8 91214

3 9 10 11 14 16

2

5 7 10 13 14

6 11 12 13 14 15

7 8 11 15 16

A p a r t i t i o n i n t o Baer subdesigns i s provided by t h e f o u r Baer subdesigns: {3,5,9,11},

~1,2,10,13~, {6,7,8,161,

{4,12,14,151.

Any two o f them form an

(8;2,4)-set. Two skew Hermitian subsets are: 11,2,4,6,8,161 union i s a (12;4,6)-set, Both {4,5,7,8}

whose complement, {5,10,11 ,151,

and {1,6,11,161

union i s an (8;2,4)-set

and {3,7,9,12,13,141;

their

i s a (4;0,2)-set.

a r e type (0,2) sets; t h e i r s e t t h e o r e t i c

(which cannot be s p l i t i n t o two Baer subdesigns). REFERENCES

[ 11

E.F. Assmus jr. and J.H. can L i n t , Ovals i n P r o j e c t i v e Designs, J. Comb. Th. A 21 (1979) 307-324.

[2]

R.C. Bose and S.S.Shrikhande, Baer subdesigns o f symmetric balanced i n complete block designs, i n S.Ikeda e t a l . eds. "Essay i n P r o b a b i l i t y and S t a t i s t i c s " , Shinka Tsusho Co. Tokyo, 1976, 1-16.

[3]

M. de F i n i s , On k-sets o f type (m,n) i n p r o j e c t i v e planes o f square order, i n : P.J. Cameron, J.W.P. H i r s h f e l d , D.R. Hughes (eds.), F i n i t e geometries and designs, London Math. SOC. Lect. Note Series n. 49. Cambridge Univers i t y Press (1981) 98-103.

[4]

M.J. de Resmini, Sui k-insiemi a due c a r a t t e r i nei sistemi d i S t e i n e r S(Z,!L,v), Quad. Sem. Geom. Com. n. 25 (Marzo 1980), 1st.Mat.Univ. Roma.

[ 51

M.J. de Resmini, On k-sets of t y p e (man) i n S t e i n e r systems, i n : P.J.Cameron, J.W.P. H i r s h f e l d , D.R.Hughes (eds.), F i n i t e geometries and designs, London Math.Soc.Lect. Note Series n.49. Cambridge U n i v e r s i t y Press (1981) 104-113.

[ 6 ] M.J. de Resmini, Sugli insiemi a due c a r a t t e r i nei BIBD con X > 1, Quad. Sem.Geom.Comb. n. 29 (Settembre 1980), 1st. Mat. Univ. Roma.

[7]

M.J. de Resmini, There are a t l e a s t t h r e e non-isomorphic S(2,4,28)Is, appear i n Journal o f Geometry.

to

Haemers and M. Shrikhande, Some remarks on subdesigns o f synnnetric designs, J. S t a t i s t . Planning and Inference 3(1979)361-366.

[ 81 W.

[9]

M. H a l l Jr., Group Theory and Block Designs, i n Proc. I n t . Conf. Theory o f Groups, Canberra 1965, Gordon & Breach, 1967, 115-144.

[ 101 H. Hanani, Balanced Incomplete Block Designs and Related Designs, Oiscr.

Math.

fi (1975)

255-369.

[ 111 M. T a l l i n i S c a f a t i , {k,n)-archi

d i un piano g r a f i c o f i n i t o , con p a r t i c o l a r e riguardo a q u e l l i con due c a r a t t e r i , Rend.Acc.Naz.Lincei (8) (1966) 812818, 1020-1025.

[ 1 2 ] N.M. Singhi, (19,9,4) Hadamard Designs and t h e i r Residual Designs, J.Comb. Th. A 16 (1974) 241-252.

Annals of Discrete Mathematics 14 (1982) 207-210 Q North-Holland publishing Company

DESAW;UES CONFIGURATIONS INSCRIBED IN AN OVZXG

Giorgio Faina Department of Mathemtics, University of Perugia 06100 Perugia, Italy &br

KorchnAros

Department of Mathematics, University of Budapest 1111 Budapest, Hungary

Letu C be an oval - (q+l)-arc- in the desarguesian projective plane PG(2,q), q=p and p is a prime. By a d-line we shall mean each line of PG(2,q) which is the axis of a Desargues configuration of triangles A (i=1,2) w i t h vertices Ai,Bi,Ci on C. We shall denote by D the set of 411 d-lines (with respect to C). It is w l l knawn that if C is a conic then D consists of (i)

all tangents to C for q even,

(ii) all secants and all external lines to C for q odd. To canplement this fact, we shall prove in t h i s paper the follcwing THD.3FG3.l I.- If C i s not a conic, then

D consists o f a l l l i n e s of

PG(2,q).

First of all we note that, for q=2,4 and, by Segre's theorem [ 71 , also for every q cdd, the only ovals in pG(2,q) are conics. The proof of the Theorem consists in showing that i f

q

(>?)

C is a conic.

i s even and there e x i s t s a line

r not belonging to D, then

-

We consider the affine plane obtained by deleting the line r and all points incident with it. Let a s y h 1 such that y!pG(2,q). case I: Icnrl=o. Following Liineburg [6] , it can be constructed an inversive plane I(C,r) of order q as follows:

a point

of I(C,r) is either an affine point or

m,

a c i r c l e of I(C,r) is either an affine line plus m, or the images of C under a collineation of PG(2,q) fixing r pointwise. It can be sham that I (C,r) is miquelian. For a proof of this result, see [2] , or [3]. on the other hand, it is easy to show that I(C,r) is miquelian if and only if C is a conic. Therefore, in this case C must be a conic. Case 2:

ICnrl=2. With a slight modification of the L i i n e h r g mthod mentio201

G.Faina and G.Korchmriros

208

ned before one can construct a Minkmski plane M(C,r) of order q. To see this, it is convenient to introduce the concept of horizontal and v e r t i c a l lines. By a horizontal (resp. v e r t i c a l ) line we shall mean each Let Cnr={X,,Y,I. affine line with *roger point Xm (resp. Y-). We shall call each other affine line an oblique line. For every horizontal line h (resp. vertical line v), we define Ihl = IQIQEhIU{hIr IvI = IQ~QEV}U{V). We shall demte by H (resp. V) the set of all horizontal (resp. vertical) lines. Finally, We put H+ = HuI-1

l b w we construct M(C,r)

point of M(C,r) a vertical line,

A

is either an affine point, or

a horizontal generator a v e r t i c o l generator

. -, or a horizontal line, or

+,

of M(C,r) is either a set of type Ihl or V of M(C,r) is either a set of type

+,

IvI or H

a block of M(C,r) is either an oblique line plus m , or a (q-1)-arc C' plus a horizontal line h and a vertical line v, where C' is the bmge of C under a allineation of FG(2,q) fixing r pointwise, and h (resp. v) is the tangent of the oval C1uWm,YmI in X, (resp. Y,). theorem (see [ 4 ] ) states that, for every p w e r q of 2, there exists (up to ismrphisns) only one Minkawski plane of order q. Thus M(C,r) is isarprphic to the geometry G(Q) of the plane sections of a non degenerate ruled quadric Q in PG(3,q). On the other hard., it is easy to shod that M(C,r) is isanorphic to G(Q) if and only if C is a conic. Therefore, also in this case, C mst be a conic.

A wellknown

Case 3: lcnr \=I. we shall prove that t h i s case cannot actually occur. The proof will be by contradiction. Let T be the unique point of C on r and N the nucleus of C. For any point P of C-{TI, let Cp={C-{PIIu{NI. Cp is an oval of PG(2,q) and r is a secant of Cp. Clearly r is not contained in the set D (associated to $1. Then it follows, by case 2, that Cp m s t be a conic. Evidently CpjCpl and Icpn%, I=q-l, when P+PI. But, if q>7, I$n$, Iy-1 implies that cP=cpl, contradiction. This finishes also the proof of the Theorem.

man "heorem I follows imnediately: THEOREM 11.-

The only ovals o f PG(2,q),q even, admitting a Pascal l i n e which i s not a tangent are conics.

209

Desargues configurations inscribed in an oval (This Theorem 11 was f i r s t proved i n

.-

[5] )

.

Proof Given an oval C of PG(2,q) , suppose that C admits a Pascal l i n e r, i n the sense of Buekentmut [I]. AS q is even, i f r is not a tangent to C, then r be a d-line w i t h respect to C. Thus, by Theorem I, C n u t be a conic. REFERENCES:

[I] Buekenhout, F., E t u d e intrins-edes

ovalegRend. Mat. 25 (1966) 363-393.

[2] Faina, G. and Korclm&os, G., Su una classe d i ovali d i un piano desarguesiano d i ordine pari che danno lucqo ad un piano inversivo, A t t i Sem. Mat. Fis.

Univ. M e n a ( t o appear)

.

Alcuni r i s u l t a t i n e l l a geometria f i n i t a , in: Barlotti, A. (ed), Prcceedings of the Trent0 Conference (North-Hollanl, Amsterdam, 1981 )

[3] Glynn, D.G.,

.

, Survey on Sharply k+ansitive Sets of Perrrutations and Minkmki-m-Structures, A t t i Sem. Mat. Fis. Univ. Wena 27 (1978)

[4] Heise, W. and Cuattrccchi P. 51 -57.

[5] Korchrtkos, G.,

Sulle ovali d i traslazione nei piani d i Galois d'ordine pari, 5 vol. I11 (1977/78) 55-65.

Rend. Accad. Naz. d e i XL

tjb, projektive men, i n denen jede Fahne von einer n i c h t t r i vialen Elation invariant gelassen wird, Abh. Hamburg 29 (1965) 37-76.

[6] Luneburg, H.,

[7] Segre, B., Ovals i n a f i n i t e projective plane, Canad. J. Math. 7 (1955) 41 4-41 6.

This Page Intentionally Left Blank

Annals of Discrete Mathematics 14 (1982) 211-218 Q North-Holland Publishing Company

k-SETS I N AN AFFINE PLANE Agq

ON TYPE ((9-3)/2,(9-1)/2,q-l)

Osvaldo F e r r i I s t i t u t o Matematico

-

Universita d i L'Aquila

- L'Aquila,

Italy

For any type ((q-3)/2,(q-1)/2,q-l), k-set i n an a f f i n e plane A (q>3,odd), w i t h k < q(q-l)/2+q/5, i t i s proved t h a t 2 ,q k=q(q-1)/2. Furthermore such a s e t c o n s i s t s o f t h e external p o i n t s o f a parabola. I n t h i s paper k-sets K i n a f i n i t e a f f i n e plane A

( o f odd order q) 2 ,q having three characters w i t h respect t o the l i n e s (see [6l)-namely o f type 1.

((q-3)/2,(q-1)/2,q-1)1-are if k

< q(q-l)/2+q/5

considered, and the f o l l o w i n g r e s u l t i s proved:

andq >3, then

k=q(q-1)/2 and K i s the s e t o f a l l

p o i n t s external t o a parabola. (Ifhen A2

i s n o t a Pascalian plane, any q-arc sq belonging t o a ( q t 1 ) - a r c tangent t o t h e l i n e a t i n f i n i t y i n the p r o j e c t i v e

w i l l be c a l l e d a parabola). Moreover, some special 2,q values o f q w i l l be d e a l t with. plane associated w i t h A

2.

L e t K be a type ((q-3)/2,

(q-1)/2,q-1)1

k-set i n an a f f i n e plane, o f

order q ,A2 ,q ,q odd ; then : Proposition I. When q=3, K i s a conic.

Proof.

When q=3, K i s a k-set o f type (0,1,2)l;

On the other hand, k

>

t h a t e i t h e r k = 2, o r

therefore, k

= 4.

2, because t h e r e e x i s t 2-secants o f K; i t f o l l o w s k = 3, o r

k = 4 and

K

i s a hyperbola, a parabola,

an e l l i p s e , respectively. Now l e t us prove: Proposition 11.

, o f type ((q-3)/2, 99 k > [ q ( 4 - 1 ) / 2 1 -1.

I f K i s a k-set i n A2

w i t h respect t o t h e ' l i n e s , then

21 1

( q - 1)/2, q - l ) l

212

0.Fern

Proof. Consider a ( q - 1)-secant o f K ( e x i s t i n g under t h e hypothesis on t h e t y p e o f K). It contains a p o i n t Q i n A

n o t belonging t o K. The o t h e r l i n e s through 2 4 Q a r e a t l e a s t [ ( q - 3 ) / 2 I - s e c a n t o f K, t h e r e f o r e k q - 1 t q ( q - 3 ) / 2 = =

and t h e statement i s proved.

[q(q-1)/2]-1,

By P r o p o s i t i o n 11, k = [ q(q-1)/2] - l t n ,

(2.1) P r o p o s i t i o n 111. I n A2 with k

< q(q-1)/2

Proof. L e t -

t

(q

’9

> 3) i f

n

>

0 an i n t e g e r .

K i s a t y p e ((q-3)/2,(q-1)/2,q-1)1

k-set,

q / 5 , then k = q(q-1)/2.

t

t t denote t h e c h a r a c t e r s o f t h e k - s e t K w i t h r e s p e c t t o t h e 1’ 2’ 3 l i n e s ( t h a t i s t h e numbers o f l i n e s meeting K i n (q-3)/2,(q-1)/2,q-l points

r e s p e c t i v e l y ) , by (2.1) ( t a k i n g i n t o account (18) i n [ 6 1 n. 1, w r i t t e n f o r A

2 4

):

From (2.2) we have:

t

,

3

= q +

4(n-1) ( n t q t l ) 2 9 -1

From t h e t h i r d e q u a t i o n (2.3) i t f o l l o w s t h a t : (2.4)

m=

4(n-1) ( n t q t 1) q2

-

1

213

On type ((q-3112, (q-ljJ2, q-1) k-sets must be an i n t e g e r . n n l ( -)= m 2 Remark t h a t , i f m i s an i n t e g e r , then 2 q+l = m q+l2 2(n-1) are i n t e g e r s too, t h a t i s , i f 2(n-1)(nt2) q- 1 then b o t h tl and t2 a r e i n t e g e r s .

fi -

-

When n = O , from (2.4) i t f o l l o w s only i f e i t h e r q = 3 o r

q

m=-

-

q-1 Since we assume q

5.

2(n-1) and t i s integer, 3

and t h i s r a t i o i s an i n t e g e r

>

3, we c o n s i d e r

q = 5 only.

w i t h tl = 14, t = 12, t = 4 . L e t 6 be 1 2 3 a p a r a l l e l c l a s s i n t h e plane and l e t wl, w2, w3 be t h e numbers o f l i n e s

Thus we have a 9-set o f t y p e (1,2,4)

belonging t o 6 and meeting K i n (q-3)/2,

(q

- 1)/2,

q

-1

points respectively;

obviously:

(2.5)

w1+w2+w

,

= 5

3

w t 2 w +4w = 9 1 2 3

.

S u b t r a c t i n g t h e f i r s t equation from t h e second, w2+3w

(2.6)

= 4 ;

3

= 4/3 ( a c o n t r a d i c t i o n ) and w2 > 2 i m p l i e s w3 G 2/3, 3 the o n l y p o s s i b l e s o l u t i o n i s w2= 1; thus, by (2.6), w3= 1. Therefore, i n any

since

w2 = 0 i m p l i e s w

p a r a l l e l class, t h e r e i s e x a c t l y one (q o f p a r a l l e l classes i s s i x , t does n o t e x i s t . Hence n

> 1.

3

- 1 ) = 4-secant

o f K; s i n c e t h e number

6, b u t t = 4 ; i t f o l l o w s t h a t such a 9-set 3

When n = 1 , we have m=O, k = q ( q - l ) / 2 ,

t,=q(q-1)/2,

t = q . This case w i l l be considered i n t h e n e x t s e c t i o n . 3 Assume n 3 2; then

and an easy computation shows t h a t

(2.8)

1 n > - q t l . 5

From (2.1) t h e statement f o l l o w s . From (2.4) we o b t a i n :

t2=q(q+1)/2,

214

0.Ferri

n =

(2.9)

-q

t

J(q

t

212 t m(q2

2

- 1) ,

w i t h m > 1;

2 thus, A = ( q t 2 ) t m ( q 2 - 1) must be a square. When m = 1, 2 A = 2q t 4 q

(2.10)

t

3 ;

when m=2, 2

(2.11)

A = 3q t 4 q t 2 .

When m = 3, A = ( 2 q t l ) 2 y so t h a t , by (2.9),

,

n = (q+1)/2

(2.12)

and t h i s case i s a r i t h m e t i c a l l y possible; f o r any q and n = ( q t 1 ) / 2 , 2 k = ( q 2 - 1)/2, t =q-1, t = q -q-2, t = q t 3 . I t f o l l o w s : 1 2 3 Proposition I V . n= 1 or

n

>

I f 2q

2

4q t 3 and 3q

t

(qt1)/2; thus, i f n

<

2

t

4q t 2 are n o t squares, then e i t h e r

(qt1)/2 ( t h a t i s

k

< q(qt1)/2t(q-1)/2),

then n = 1 and the o n l y a r i t h m e t i c a l l y p o s s i b l e value i s k=q(q-1)/2. Proposition V.

h L e t q = p (p a prime, h a p o s i t i v e i n t e g e r ) be t h e order o f n

i f 2 and 3 are n o t squares i n GF(p), then t h e integers 2 q L t 4 q t 3 ( - 3 modp) A2,q' 2 and 3q t 4q t 2 (E 2 mod p) are n o t squares and e i t h e r n = 1 o r n 3 (q t 1)/2; *

consequently, i f n

<

( q t 1)/2, then n

possible value i s k = q(q

- 1)/2,

The f i r s t values o f p, namely p

1 and again t h e o n l y a r i t h m e n t i c a l l y

< 29

were examined and the f o l l o w i n g

r e s u l t was proved. Proposition V I . i n (2.1).

Proof.

If q= p

h

and

It follows that, i f k

then e i t h e r n = 1 o r n 2 ( q t 1 ) / 2

p = 3,5,19,29,

< q(q-

1)/2

t

(q-l)/2,,then

k=q(q-1)/2.

2 F i r s t consider p = 3 . Assume m = 1 and 2q t 4 q t 3 a square b2, t h a t i s

On type ((q-3)/2, (q-ljJ2, q-1) k-sets

2 2q t 4 q t 3 = 3 (2q.3h-1t4*3h-1+1) 3 ( 2 q 0 3 ~ - ’t 4-3h-1

215

= b2, and 3 d i v i d e s b, hence b = 3 c . Thus

1) = 9c2, t h a t i s t 4q t 1 = 3c2, a contradiction. 2 2 Assume m= 2 and 3q t 4q t 2 = d , a square. Consequently 2 E d (mod 3), a t

2

c o n t r a d i c t i o n ( 2 n o t being a square i n GF ( 3 ) ) . The statement i s proved f o r t h e numbers 2 and 3 are n o t squares i n GF(p) and t h e

p = 3. Llhen p = 5,19,29,

statement f o l l o w s from V. Proposition V I I .

If q = p

h

and

p=7,17,

then e i t h e r n = l o r n

> q/3.

Proof.

Both when p = 7 and p = 1 7 , 3 i s n o t a square i n GF(7) and i n GF(17), 2 b u t 2 i s a square. It f o l l o w s t h a t 2q t 4q t 3 c a n ’ t be a i n t e g r a l square;

thus, i n (2.9),

m

> 2, t h a t i s n > q/3, as an easy computation shows.

As f o r the prime numbers p 6 29, d i f f e r e n t from the ones examined above, t h a t i s p = l l , p.13,

p = 2 3 , both 2 and 3 are squares i n GF(p) (so t h a t

Proposition V does n o t hold) and-indeed- the f o l l o w i n g a r i t h m e t i c a l l y p o s s i b l e cases are obtained:

3.

p=q=ll,

n=3,

k=57,

tl=36,

t2=84,

t =12; 3

p=q=23,

n=9,

k=261,

t =83, 1

t =444, 2

t3=25,

We w i l l now deal w i t h the case n = l . From (2.3),

and, from (2.1),

k = q ( q - 1)/2.

Fix a point Q

4

K, and denote by ulsu2,u3

Q meeting K i n ( 9 - 3)/2, (q there i s a t most one ( q

- 1)/2, q - 1 points,

- 1)-secant:

l i n e through Q being a t l e a s t a [ (q = ( 9 - l ) ( q t 1)/2

Clearly,

(3.2)

for n = l :

> q(q-

t h e numbers o f l i n e s through r e s p e c t i v e l y . Through Q

assume t h e r e a r e two, then any other

- 3)/2]-secant,

k

>

2(q-1)

t

(q-l)(q-3)/2 =

1)/2, a c o n t r a d i c t i o n . Thus, e i t h e r u3 = 0 o r u = 1. 3

I

u,

t

u2 t u3 = q t l

ul(q-3)/2+ u2(q-1)/2+ u 3 ( q - l ) = q(q-1)/2,

216

0.Fern

from which, when u

(3.3)

u3 = 1

1

= q-1,

u2 = 1,

= (q-1)/2,

u

u

= 1.

3

u3 = 0,

When

u

(3.4)

1

2

u

(q+3)/2,

3

= 0.

\ K 2Yq holds; thus t h e complement o f K c o n s i s t

L e t P, r e s p e c t i v e l y I , be t h e s e t c o n s i s t i n g o f t h e p o i n t s f o r which (3.3),

r e s p e c t i v e l y (3.4),

Q

E

A

two d i s j o i n t sets, P and I.

1

L e t P be a p o i n t o f K and denote by vl,v2,v3 through P meeting K i n ( 4 - 3)/2,

t h e numbers o f l i n e s

( q - 1)/2, q - 1 p o i n t s , r e s p e c t i v e l y .

Obviously:

(3.5)

vq-; 1

v2 + v 3 = q t l

(2 - 1)

v1 t (2q-1 1 ) v 2 t (q-2)v

From t h e second e q u a t i o n (3.5),

3

-!&ILL- 2 1.

t a k i n g i n t o account t h e f i r s t one, we

obtain:

(2.61 If v 3 > 4 ,

( 4 - l ) v 3 - 2Vl = q t 1. then 2v, > 4 ( q - l ) - ( q t l )

c o n t r a d i c t i o n . Therefore, 0 v1 = - ( q t 1 ) / 2 , v

3

<

v3

= 3q-5,

hence 2 ( v l t v 3 )

< 3 (v3 an i n t e g e r ) . I f v3

= 0 then

t h a t i s absurd; i f v3 = 1, then v1 = -1, impossible. I f

= 2 , then:

I f v = 3 , then: 3

(3.8)

Z3(qtl), a

v1 = q - 2 ,

v2 = 0,

v

3

= 3.

Let 6 be a p a r a l l e l class i n the plane, and denote by w o f l i n e s i n 6 and meeting K i n (q-3)/2, Obviously:

I

(3.9)

w + w 1

2

+w3=q

Wl(q- 3)/2

t

w2(q-1)/2+w3(q-l) = q(q-1)/2.

From the second equation (3.9), 2w2

(3.10) Assume w3

> 2;

if w3

w1 = 0, 1,

t a k i n g i n t o account the f i r s t one:

+ (q+l)w3

= 2q.

from (3.10) i t f o l l o w s 2 q = 2 w 2 + ( q + l ) w 3

impossible. Thus, e i t h e r w (3.11)

i n (q-l)/Z,

w the numbers 1' 2"3 i n q-1 p o i n t s , respectively.

3

0, o r w3 = 1.

w2

> 2w2t2(q+1)>2(qtl),

I f vl3=O, then:

w3 = 0;

q,

then:

By (3.11) and (3.12) no two (q-1)-secants o f K are p a r a l l e l ; hence the (q-1)-

-secants ( t h e i r number being q, see (3.1)) determine q d i s t i n c t p a r a l l e l classes. But t h e p a r a l l e l classes i n the plane are q t 1; thus t h e r e e x i s t s e x a c t l y one p a r a l l e l c l a s s 6" no l i n e i n which i s a (q-1)-secant,

that i s

there e x i s t s e x a c t l y one p o i n t a t i n f i n i t y i n the p r o j e c t i v e plane n obtained from A2

q'

adding i t s l i n e a t i n f i n i t y , f o r which (3.11) holds,while

,q f o r any o t h e r p o i n t a t i n f i n i t y (3.12)

i s s a t i s f i e d . Consequently, no p o i n t

i n K can e x i s t s f o r which (3.8) holds; indeed, through every p o i n t P i n K t h e r e i s a t l e a s t one [ ( q - l ) / 2 ] - s e c a n t o f K, namely the l i n e through P belonging t o the p a r a l l e l class @( ( r e c a l l (3.11)). Hence, f o r t h e p o i n t s P i n K o n l y (3.7) i s s a t i s f i e d . NOW, consider, i n n

the s e t K ' obtained q' adding t o K the q p o i n t s a t i n f i n i t y d i s t i n c t from @; thus, k ' = I K ' I = = q(q-1)/2

t

q = q(qt1)/2. Any l i n e i n A2

so K ' , a t (q-1)/2 points, by (3.11);

belonging t o 6* meets K, and ,q the l i n e a t i n f i n i t y , Em, meets K ' a t

0.Ferri

218

q p o i n t s . It f o l l o w s t h a t

in

TI

-

9' C i n IT

K' i s a [ q ( q t l ) / Z ] - s e t o f t y p e ( ( q - l ) / 2 , ( q t l ) / 2 , q ) l

t h us (see [ 3 ] ) , i t i s t h e s e t o f a l l p o i n t s e x t e r n a l t o a ( q + l ) - a r c

.

Therefore, K i s t h e s e t o f a l l p o i n t s e x t e r n a l t o t h e parabola P i n q c o n s i s t i n g of t h e q p o i n t s i n C \ { a n C}, and t h e f o l l o w i n g has

A2 ,q , been proved:

Proposition V I I I .

00

I n A2

Yq k - s e t K, w i t h k < q ( q - 1 ) / 2

,q

odd, f o r a t y p e ((q-3)/ 2,

+ q/5,

it i s

k=q(q-1)/2;

(q-l)/Z,

q-l)l

f u r t h e r m o r e (when q > 3),

K i s t h e s e t of a l l p o i n t s e x t e r n a l t o a p a rabola, (namely, a q - a r c which,

i n t h e associated p r o j e c t i v e plane the l i n e a t i n f i n i t y ) .

ACKNOWLEDGMENT.

1~

q'

completes i n t o a ( q + l ) - a r c t angent t o

T h i s r e s e a r c h was p a r t i a l l y support ed b y GNSAGA o f CNR.

REFERENCES [ l ] 0. F e r r i , Su una c a r a t t e r i z z a z i o n e g r a f i c a d e l l a s u p e r f i c i e d i Veronese d i un S5 , Rend.Acc.Naz. L i n c e i ( 8 ) 61 f a s c . 6 (1976) 303-310. ,q [ 2 1 0. F e r r i , Le c a l o t t e a due c a r a t t e r i r i s p e t t o a i p i a n i i n uno spazio d i Galois S , R i v . Un. Parma ( 4 ) 6 (1980) 3 99 [ 3 1 0. F e r r i , Una c a r a t t e r i z z a z i o n e g r a f i c a d e l l ' i n s i e m e d e i p u n t i e s t e r n i ad una o v a l e i n un p i a n o TI ( q d i s p a r i ) , Rend. Mat. Univ. Roma ( 1 ) 14, s e r i e V I (1981). q [41

B. Segre, L e c t u r e s on modern geometry. Cremonese, Roma (1961).

1 5 1 G. T a l l i n i , Problemi e r i s u l t a t i s u l l e qeomet rie d i G a l o i s . Rel. n. 30, 1 s t . Mat. Univ. N a p o l i (1974). [ 6 1 G. T a l l i n i , Graphic c h a r a c t e r i z a t i o n o f a l g e b r a i c v a r i e t e s i n a Ga o i s space, i n Conv. I n t e r . Geom. Combinat. (Roma 3-15 s e t t . 1973) Tomo 11, Acc. Naz. L i n c e i (1976), 153-165. [ 7 1 G. T a l l i n i , Spazi c o m b i n a t o r i e s i s t e m i d i S t e i n e r , R i v . Mat. Univ 5 (1979), 221-248. Parma ( 4 ) -

Annals of Discrete Mathematics 14 (1982) 219-224

0 North-HollandPublishing Company

MINORS I N BOOLEAN STRUCTURES AND MATROIDS L. GUIDOTTI and G. NICOLETTI I s t i t u t o d i Geometria U n i v e r s i t a d i Bologna

1 INTRODUCTION Forbidden minors are a useful device f r e q u e n t l y used i n Combinatorics and o t h e r branches o f Mathematics. This technique consists o f d e s c r i b i n g a c l a s s o f s t r u c tures as those which do n o t admit any given substructures, u s u a l l y named forbidden

.

minors I n t h i s way, the problem o f recognizing i f a p a r t i c u l a r s t r u c t u r e belongs t o a given class i s reduced t o checking whether i t contains some o f the excluded minors, o r not. I n t h i s paper, we introduce f i n i t e matroids as a class o f h e r e d i t a r y o r co-here! i t a r y systems, subjected t o some forbidden minors-condi t i o n s . Thus, we provide some new axiomatizations o f f i n i t e matroids, which seem t o be u s e f u l i n g i v i n g a more systematic approach t o the various cryptomorphic axiomatizations o f matroids. Hereditary systems and matroids a r e here defined i n terms o f subsets o f a f i n i t e boolean algebra: our language may cause some d i f f i c u l t i e s a t t h e beginning, b u t many conceptual s i m p l i f i c a t i o n s are made possible by i t . I n p a r t i c u l a r , i t can be e a s i l y seen t h a t the whole theory o f d u a l i t y i n matroid theory i s nothing b u t an instance o f t h e boolean d u a l i t y . 2

PRELIMINARIES AND NOTATION

I n t h i s paper, we w i l l deal w i t h f i n i t e boolean algebras. By t h e term " d u a l i t y " we w i l l r e f e r t o the boolean d u a l i t y . I f B denotes a f i n i t e boolean algebra, we w i l l denote by the same symbol B i t s underlying set. The l e a s t and the greatest element o f B w i l l be denoted by OB and lB, respectively, o r by 0 and 1 when no misunderstanding can a r i s e . We r e c a l l t h a t the h e i g h t o f an element x . o f a f i n i t e boolean algebra B, whose a t o m a r e ai,

i s defined as f o l l o w s : hg(x):=card{ai:

The height o f

ai(x).

B i s defined as h( B) := hB( 1 ) 219

220

L. Guidotti and G. Nicoletti

For any given x,ycB,

w i t h xsy, we w i l l s e t

Ix,Y(

:={zcB:xrz
Then, Ix,yJ i s a subalgebra o f B, w i t h respect t o the p a r t i a l order defined i n B. A descending s e t o f a f i n i t e boolean algebra B i s a ( p o s s i b l y empty) subset A o f

B such t h a t , f o r every x,ycB,

i f XLYEB, then xcA. Dually, an ascending s e t o f B i s

i f X ~ Y E B , then xcA.

a ( p o s s i b l y empty) subset A o f B such t h a t , f o r every x,ycB, For every subset A o f B, we s e t max(A):={xcA: x i s maximal i n A}, min(A):={xcA: x i s minimal i n A}, V (A) :={xcB:

t h e r e e x i s t s ycmax( A) ,xkyl,

A(A) :={xcB: t h e r e e x i s t s ycmin(A) ,xsy},

r( A) :={xEB:

XCAI.

A s t r u c t u r e , o r system, over a f i n i t e boolean algebra B i s a p a i r S:=(B,A),

r e A i s a subsev o f B. Two given systems Si:=(Bi,Ai),

i=1,2, w i l l k said

whe

to

be

isomorphic whenever t h e r e e x i s t s a boolean isomorphism f:B1-+B2 such t h a t f(A1)=A2. I f S1 and S2 are isomorphic, we w i l l s e t S1=S2.

dual

o f a s t r u c t u r e S:=(B,A) i s the s t r u c t u r e S":=(B",A), where 6" denotes The the dual o f the boolean algebra B. We have, obviously, Soo=S.I f S:=(B,A), we s e t

V(S) := (B ,V(A) , A(S) :=(B,A(A)) 9 r(S):=(B,r(A)). I t i s e a s i l y v e r i f i e d t h a t f o r any system S t h e f o l l o w i n g i d e n t i t i e s hold:

v(so)=(v(s))o, A(SO)=(A(S)1O ,

r(s")=(r(s))".

Now letS:=(B,A)

be a system over the boolean algebra B: a subsystem o r substru

c t u r e o r minor o f S w i l l IfSi:=(Bi,Ai),

be any system S(x,y) w i t h x,ycB,

xsy, and

-

S(X,Y) :=( I X,Y I ,An1 x,yl 1. i=1,2, a r e two given s t r u c t u r e s , we w i l l s e t S1<-S2 whenever

t h e r e e x i s t x,ycB2, xsy, such t h a t Sl~S2(xsy). Given a class C o f s t r u c t u r e s and a ( f i n i t e ) f a m i l y o f s t r u c t u r e s F, we w i l l say t h a t F i s a f a m i l y of forbidden ( o r excluded) minors f o r C i f , f o r every s t r u c t u r e T we have: T i s i n C i f and only i f s f o r every S b , S / T . I t i s worthwile t o remark t h a t d u a l i t y and t h e operator

r

preserve minors, and

consequently forbidden minors:

Proposition 1 Let S i : = ( B . , A . ) , i=1,2 be two systems: then S j sS2 if and only if 2 2 Slo5Szo and SIs S2 if and only if r(Sl)
22 1

Minors in Boolean structures and matroids 3

FORBIDDEN MINORS FOR HEREDITARY SYSTEMS

A h e r e d i t a r y system ( o r independence system) i s a system S=(B,A) where A i s a descending f a m i l y o f B; co-hereditary systems a r e defined d u a l l y . The systems (B,0),

(B,{OBI),

(B,B)

a r e h e r e d i t a r y s t r u c t u r e s which w i l l be c a l l e d

the degenerate, t r i v i a l and d i s c r e t e h e r e d i t a r y systems, r e s p e c t i v e l y . Dually, (B,0), (B,{lBI), (B,B) a r e co-hereditary systems, which w i l l be c a l l e d the

erate, trivial

a

and d i s c r e t e co-heredi t a r y sistems, r e s p e c t i v e l y .

We w i l l denote by P and Po the t r i v i a l co-hereditary s t r u c t u r e and the t r i v i a l h e r e d i t a r y s t r u c t u r e over a boolean algebra o f h e i g h t 1, r e s p e c t i v e l y .

P

P

PO

Figure 1

P i s the unique system over a boolean algebra o f h e i g h t 1 which i s P? Furthermore, P i s t h e forbidden minor f o r t h e c l a s s o f h e r e d i t a r y systems, and d u a l l y f o r P: as s t a t e d i n t h e f o l l o w i n g

We remark t h a t

n o t a h e r e d i t a r y system, and d u a l l y f o r

proposition:

Proposition 2 Let S:=(B,A) be a system over a f i n i t e boolean aZgebra B: then S is a hereditary system if and onZy if [email protected], S is a co-hereditary system if and only if Po+$. 4

FORBIDDEN MINORS FOR MATROIDS As i s w e l l known, f i n i t e matroids can be defined i n various cryptomorphic ways.

Here, we a r e i n t e r e s t e d i n regarding them as h e r e d i t a r y ( o r co-hereditary) s t r u c tures, subjected t o some a d d i t i o n a l conditions. More p r e c i s e l y , an independent matroid ( i - m a t r o i d f o r s h o r t ) i s a h e r e d i t a r y system M=(B,I), where I s a t i s f i e s the f o l l o w i n g "augmentation axiom" : f o r every x,ycI, hB(x)hB(y), t h e r e e x i s t s ZES such t h a t x>z and xvy>zVy. The elements o f S w i l l be s a i d t o be the spanning elements o f M.The degenerate coh e r e d i t a r y systems wi.11 be c a l l e d the degenerate s-matroids.

L. Guidotti and G. Nicoletti

222

A dependence matroid ( o r d-matroid ) i s a co-hereditary system M=(B,D), where D s a t i s f i e s t h e f o l l o w i n g axiom: f o r every dl,d2ED, i f dlAd2 C D then t h e r e e x i s t s d3ED such t h a t dgdlVd2 and he( d3)=hB(dlvd2)-l. The elements o f D w i l l be c a l l e d t h e dependent elements o f M.The d i s c r e t e co-hered it a r y systems w i 11 be regarded as degenerate d-matroids. A non-spanning matroid ( o r n-matroid ) i s a h e r e d i t a r y system M=(B,N) where N s a t i s f i e s t h e f o l l o w i n g axiom: f o r every nl,n2EN,'

i f nlVn2LN,

then t h e r e e x i s t s n3EN such t h a t n p l A n 2

hB(n3)=hB(nlAn2)t1. The elements o f N w i l l be c a l l e d t h e non-spanning elements o f

M.

and

The d i s c r e t e he

r e d i t a r y systems w i 11 be s a i d t o be degenerate s-matroids. The l i n k s between t h e classes o f s t r u c t u r e s now defined a r e described by t h e f o l 1owi ng proposi ti ons :

Proposition 3 Let S:=IB,A) be a system over the f i n i t e boolean algebra B.Then:

i) ii) iii) iv) v) vi) Proof

S is a d-matroid S is an i-matroid S is an a-matroid S is an n-matroid S is an s-matroid S is an a-matroid See [4],

[6]

.

i f and only i f rlS) is an i-matroid;

if and only i f V($)

is an s-matroid;

i f and only i f r(S) is an n-matroid;

i f and only if r(S) is an s-matroid;

if and only i f ii(S) is an i-matroid; i f and only i f r l S ) is an d-matroid.

Proposition 4 Let S:=(B,AI

be a system over a f i n i t e boolean algebra B. Then:

il S is a d-matroid if and only if So is an n-matroid; ii) S is an i-matroid i f and only i f So is an s-matroid. Proof I t f o l l o w s from d u a l i t y between axioms d e f i n i n g d-matroids and n-matroids, i-matroids and s-matroids, r e s p e c t i v e l y . Now,let M1 be a d-matroid, and s e t M2:=r(M1), M3:=v(M2), M4:=r(M3); then we w i l l r e f e r t o Mi, i=l,2,3,4 as t h e cryptomorphic representation o f t h e same Et r o i d M. Then M:, i=1,2,3,4 a r e cryptomorphic representations o f a second matroid o f M. M'which w i l l be c a l l e d the I t i s easy t o check t h a t a l l t h e h e r e d i t a r y s t r u c t u r e s over boolean algebras o f h e i g h t 1 o r 2 are i-matroids, and dually, a l l the co-hereditary s t r u c t u r e s 2 ver boolean algebras o f h e i g h t 1 o r 2 a r e s-matroids. Now l e t B be a boolean algebra o f h e i g h t 3 whose atoms a r e ai, i=1,2,3. The system Q:=(B,A) defined by t h e f a m i l y A=IOB,al ,a2,a3,alva31 i s a hereditary s t r u g under boolean isomorphisms - the unique h e r e d i t a r y system ture; moreover, Q i s

dual

-

223

Minors in Boolean structures and matroids over B which i s n o t an i - m a t r o i d . Dually, t h e s t r u c t u r e o_"=v(O_) i s phisms

-

-

under isomor

t h e unique c o h e r e d i t a r y system which i s n o t a s-matroid.

2

p" Figure 2

Furthermore, we have: Theorem 1 Let S : = ( B , A ) be a hereditary system over the f i n i t e algebra B; then S

i s an i-matroid if and only i f Q@. Dually, l e t S:=(B,AI be a oo-hereditary system over B; then S i s a s-matroid i f a n d only i f p"$S. P r o o f L e t t h e h e r e d i t a r y system S be an i - m a t r o i d , and l e t aEB, biEB, bi>a, hB(bi)=hB(a)+l, i=l,2,3, and s e t c1=b2vb3, c2=blvb3, c3=blvb2, d=clvc2=c2vc3= =c1vc3 * L e t us suppose b,cA and cleA: then,by t h e augmentation axiom, t h e r e e x i s t s ZEA such t h a t blx and Z A Y = ( V ~ V X ) A Ay=(v1Ay) > x ~ y ,showing t h a t t h e augmentation p r o p e r t y holds. By t h e previous theorem, we can say t h a t

{P,2>i s

f o r t h e c l a s s o f i - m a t r o i d s , and d u a l l y , {P",o,") f o r t h e c l a s s o f s-matroids.

a f a m i l y o f f o r b i d d e n minors

i s a f a m i l y o f f o r b i d d e n minors

Ro=r(v),

L e t us now consider t h e systems R=r(Q) and p i c t u r e d i n F i g u r e 3. I t i s easy t o see t h a t R i s t h e unique c o - h e r e d i t a r y system o v e r a boolean algebra o f h e i g h t 3 which i s n o t a d-matroid, and d u a l l y Ro i s t h e unique h e r e d i t a r y s y z

L. Guidotti and C.Nicoletti

224

tem over a boolean algebra o f h e i g h t 3 which i s n o t an n-matroid.

R

R0

Figure 3 Applying Proposition 1 we get immediately:

Let S:=(A,BI be a co-hereditary system over the f i n i t e algebra B; Theorem 2 then S is a d-matroid if and only i f &@. Dually, l e t S:=(A,BI be a hereditary 8yc tem, then S is an n-matroid if and only if RqS. As a consequence o f the l a s t theorem, {Po,R} and {P,Ro} are f a m i l i e s o f f o r b i d den minors f o r the classes o f d-matroids and n-matroids, r e s p e c t i v e l y . REFERENCES

[l]Brylawski ,T.H. ,Kelly,D.G. and Lucas,T.D., Vatroids and combinatorial Ceomg t r i e s , Lecture Note Series, U n i v e r s i t y o f North Carolina, Chapel Hi11.(1974). [2] Crap0,H.H.

and Rota,G.C.,

t o r i a l Geometries, I1.I.T.

On the foundations o f Combinatorial Theory: Combin2 Press Cambridge (1970).

[3] Las Vergnas,M. ,Sur l a d u a l i t e en t h e o r i e des matroides,C.R.Acad.Sci.(Paris)

13

(1970) 804-806. [4] Pezzoli ,L.,

Sistemi d i indipendenza modulari, B.U.M.I.

( 5 ) 18-8 (1981)167-180.

151 Tutte,W.I.,

I n t r o d u c t i o n t o t h e theory o f matroids (American Elsevier, New

York, 1970). [6] Welsh,D.J.A.,

F a t r o i d theory (Academic Press, London, 1976).

Annals of Discrete Mathematics 14 (1982) 225-236 Q North-Holland Publishing Company

THE TRANSLATION PLANES OF ORDER 1 6 THAT ADMIT S L ( 2 , 4 ) N . L . Johnson'

Department o f Mat hemat i c s The U n i v e r s i t y o f Iowa Iowa C i t y , Iowa 52242 U.S.A. 1. I N T R O D U C T I O N

Our main r e s u l t is: Theorem. A t r a n s l a t i o n p l a n e o f o r d e r 1 6 a d m i t s S L ( 2 , 4 ) as a c o l l i n e a t i o n group of t h e t r a n s l a t i o n complement i f and o n l y i f n is D e s a r g u e s i a n , Hall o r Dempwolff. Foulser-Johnson-Ostrom

[ 2 ] have shown t h a t i f a t r a n s l a t i o n p l a n e

n

of o r d e r 22r a d m i t s SL(2,2') where t h e i n v o l u t i o n s are e l a t i o n s t h e n n I s D e s a r g u e s i a n . Thus, i f t h e Sylow 2-subgroups f i x Baer s u b p l a n e s rri p o i n t w i s e and t h e ni are a l l i n t h e same d e r i v a b l e n e t t h e n n is Hall. By Johnson [ 5 ] , t h e ni must b e l o n g t o t h e same n e t p r o v i d e d 2' # 4 . I n any c a s e t h e ni must s h a r e a t l e a s t two components. Dempwolff .[1] h a s c o n s t r u c t e d a t r a n s l a t i o n p l a n e o f o r d e r 1 6 admitt i n g S L ( 2 , 4 ) where t h e Sylow 2-subgroups f i x Baer s u b p l a n e s ni p o i n t w i s e and t h e rri s h a r e p r e c i s e l y two components. Johnson [ 4 ] h a s shown i f a t r a n s l a t i o n p l a n e n o f o r d e r 22r a d m i t s S L ( 2 , 2 r ) where t h e Sylow 2-subgroups f i x Baer s u b l i n e s ( a s e t o f 2' p o i n t s on a l i n e ) p o i n t w i s e t h e n n is O t t - S c h a e f f e r and r I s odd. A c t i n g on a t r a n s l a t i o n p l a n e o f o r d e r 1 6 , t h e f i x e d p o i n t s p a c e of 2-subgroup o f S L ( 2 , 4 ) is e i t h e r (1) a l i n e , ( 2 ) a Baer s u b p l a n e , ( 3 ) a Baer s u b l i n e , o r ( 4 ) a s u b p l a n e o f o r d e r 2.

a Sylow

By t h e r e s u l t s mentioned we need o n l y t o c o n s i d e r c a s e ( 4 ) o r c a s e ( 2 ) where t h e Baer s u b p l a n e s do n o t s h a r e a l l of t h e i r components. 2. THE FIXED POINT SPACES ARE SUBPLANES OF ORDER 2 Throughout, n w i l l d e n o t e a t r a n s l a t i o n p l a n e of o r d e r 1 6 which a d m i t s a c o l l i n e a t i o n group & i s o m o r p h i c t o S L ( 2 , 4 ) . ( 2 . 1 ) L e 3 . Let a t r a n s l a t i o n p l a n e rr o f o r d e r 1 6 admit a c o l l i n e a t i o n group & i s o m o r p h i c t o S L ( 2 , 4 ) (in i t s t r a n s l a t i o n complem e n t ) . The s e t o f p o i n t s f i x e d by e a c h Sylow 2-subgroup i s one o f the following:

(1) a l i n e , ( 2 ) a Baer subDlane 'This

.

r e s e a r c h was p a r t i a l l y s u p p o r t e d by a g r a n t from NSF. 225

226

(3)

(4)

N.L. Johnson

a Baer s u b l i n e ( a s e t o f a s u b p l a n e of o r d e r 2 .

4-points

of a l i n e ) ,

P r o o f . Assume n o t (l), ( 2 ) , o r ( 3 ) . L e t Q = {01,u2,a1u2,1] denote a Sylow 2-subgroup o f &. O b v i o u s l y , e a c h i n v o l u t i o n ai fixes a Baer s u b p l a n e n p o i n t w i s e . al f i x e s nu and e i t h e r I n d u c e s an ai e l a t i o n o r Baer i n v o l u t i o n on n C l e a r l y ghen, rul n nU2 is a u2 * s u b p l a n e o f o r d e r 2 t h a t is f i x e d p o i n t w i s e by a l , a 2 and t h u s ala2. Thus, we see f o r p l a n e s of o r d e r 1 6 we a r e r e d u c e d t o t h e c a s e where t h e Sylow 2-subgroups f i x s u b p l a n e s o f o r d e r 2 p o i n t w i s e o r where two s u b p l a n e s do n o t s h a r e a l l components. We s h a l l need t h e f o l l o w i n g f u n d a m e n t a l theorem o f G a l o i s .

-( 2 . 2 )

Theorem ( s e e Huppert [ 3 ] , 8.28, p . 2 1 4 ) . L e t m > 1 be t h e d e g r e e o f r e p r e s e n t a t i o n of PSL( 2 , p f ) as a t r a n s i t i v e p e r m u t a t i o n .

m

Then

2

p f +1

except i n t h e following cases:

m = 2, 3, m = 3, 5, m = 5, 7, m = 7, 9, m = 6, 11, m = 11.

(1) pf = 2 ,

(2)

pf =

(3) (4) (5)

P = Pf = Pf =

(6)

p

f

f

=

L e t n be a t r a n s l a t i o n p l a n e o f o r d e r 16 a d m i t t i n g a c o l l i n e a t i o n group 3 i s o m o r p h i c t o S L ( 2 , 4 ) . We may assume t h a t e a c h Sylow 2subgroup o f 1 f i x e s p o i n t w i s e a s u b p l a n e o f o r d e r 2 . S i n c e S L ( 2 , 4 ) is i s o m o r p h i c t o P S L ( 2 , 5 ) , ( 2 . 2 ) t e l l s u s t h a t any s e t o f f e w e r t h a n f i v e e l e m e n t s of a s e t permuted by PSL(2,5) must be p o i n t w i s e f i x e d . S i n c e t h e r e a r e o n l y s e v e n t e e n p o i n t s on t h e l i n e a t i n f i n i t y , we may complete t h e p r o o f of o u r main r e s u l t by c o m b i n a t o r i a l arguments. An a l t e r n a t e approach would be t o compute t h e p o s s i b l e ' G F ( 2 ) S L ( 2 , 4 ) modules and u s e r e p r e s e n t a t i o n t h e o r y . T h i s l i n e o f s t u d y i s b e i n g 2 considered f o r SL(2,q) i n t r a n s l a t i o n p l a n e s o f o r d e r q by F o u l s e r , Johnson, and Ostrom b u t even w i t h r e p r e s e n t a t i o n t h e o r y , 1.1 = 1 6 becomes a s p e c i a l c a s e . Here, i n o r d e r t o s i m p l i f y o u r a r g u m e n t s , i t is more c o n v e n i e n t t o t a k e a more c o m b i n a t o r i a l app r o a c h e v e n though some c a l c u l a t i o n s are i n e v i t a b l e .

(2.3)

Lemmn. Under t h e above h y p o t h e s e s , t h e l e n g t h s o f t h e o r b i t s of t h e group I on t h e l i n e a t i n f i n i t y 1, o f TI s a t i s f y one of the following:

(1) 1 , 5 , 5 , 6 ; ( 2 ) 1,6,10; ( 3 ) 1,1,5,10;

(4)

1,1,15.

P r o o f . L e t Q be a Sylow 2-subgroup. Q o r d e r 2 p o i n t w i s e . L e t Q = {al,a2,ula2,1}.

f i x e s a s u b p l a n e no o f Each i n v o l u t i o n ai

The translation planes of order 16 that admit SL(2,41

227

f i x e s a Baer s u b p l a n e pointwise. C l e a r l y , nui n nu$ = no nui f o r i #j. ai must permute n u j - no and c l e a r l y c a n n o t i x any o f these points.

So,

Q

has o r b i t s of l e n g t h s

1,1,1,2,2,2,4,4

on a,.

The n o r m a l i z e r of Q h a s a c y c l i c group 2. o f o r d e r 3 t h a t i s r e g u l a r on Q - (1). Thus, QG h a s an o r b i t o f l e n g t h 6 on Am. The two Q - o r b i t s o f l e n g t h 4 c a n n o t be f i x e d by 8 . Thus, we have e i t h e r t h e p o s s i b i l i t i e s l i s t e d r p o s s i b l y ( 4 ) 5 , 5 , 6 or ( 6 ) 5 , 1 2 ( s i n c e I&\ = 3 . 4 . 5 ) . That is, i f one o f t h e Q - o r b i t s of l e n g t h 4 is moved i n t o t h e o r b i t o f l e n g t h 6 we c o u l d c o n c e i v a b l y have c a s e s ( 2 ) , (3), ( 4 ) , o r ( 6 ) s i n c e w e would have a n o r b i t of l e n g t h 1 0 , 1 2 , o r 1 5 . Case ( 5 ) . 5 , 6 , 6 . L e t t h e o r b i t s be r e p r e s e n t e d by Q (Sylow 2-subgroup) f i x e s e x a c t l y one p o i n t o f

So

r5,1'i,rz. r5 and two

I?;

( s e e a b o v e ) . There are f i v e Sylow 2-subgroups s o no p a i r is g e n e r a t e d by any p a i r can f i x a common p o i n t on 1, ( s i n c e L e t P1,P2 be f i x e d by Q on ri. The o f Sylow 2 - s u b g r o u p s ) . s t a b i l i z e r o f P1 h a s o r d e r 1 0 . Every e l e m e n t 8 o f o r d e r 5 is e r e g u l a r on t h e Sylow 2-subgroups. Thus, Q and Q # Q f i x P1. of

But t h e n

(Q,Q

e)

= &

must f i x

P1.

Case ( 6 ) . 5 , 1 2 . Let t h e o r b i t s b e r e p r e s e n t e d by r5,r,,. Every Sylow 2-subgroup f i x e s two p o i n t s P1,P2 o f r12. Again t h e s t a b i l i z e r of a p o i n t has order t h i s c a n n o t be t h e c a s e . So we see t h a t ( 2 . 4 ) Lemma.

&

5

and t h e above argument shows t h a t

a l w a y s f i x e s a p o i n t on t h e l i n e a t i n f i n i t y .

Case ( 4 ) o r ( 2 ) c a n n o t happen.

P r o o f . Assume ( 4 ) . A Sylow 2-subgroup Q f i x e s e x a c t l y one p o i n t of l e n g t h of t h e o r b i t o f l e n g t h 1 5 . Thus, t h e r e must b e a &-orbit 5 i n t h i s case. Assume ( 2 ) . A Sylow 2-subgroup f i x e s e x a c t l y two p o i n t s o f t h e o r b i t of l e n g t h 1 0 a n d c l e a r l y no two Sylow 2-subgroups can f i x a common p o i n t i n t h i s o r b i t . The t e n p o i n t s are t h e n p a r t i t i o n e d i n t o f i v e p a i r s o f p o i n t s which a r e permuted by &. The n o r m a l i z e r of a Sylow 2-subgroup is t h e s t a b i l i z e r o f a p o i n t and t h u s a p a i r . B u t , t h e n o r m a l i z e r must f i x t h e p a i r p o i n t w i s e . L e t Q be a Sylow 2-subgroup which f i x e s t h e p o i n t s P and R of t h e o r b i t o f l e n g t h 1 0 . L e t y map P t o R . Then Q y f i x e s R and Q a l s o f i x e s R. T h i s i s a c o n t r a d i c t i o n u n l e s s Q y = Q. B u t , s i n c e t h e normali z e r must f i x P g& R we have a c o n t r a d i c t i o n .

( 2 . 5 ) Lemma. L e t d be a & - f i x e d component. o r b i t l e n g t h s on d are 1 , 1 , 5 , 1 0 . L e t Agad, o r b i t s of l e n g t h 5 and 1 0 on d.

Then t h e ( p r o j e c t i v e ) AloaP denote t h e

P r o o f . Each Sylow 2-subgroup f i x e s e x a c t l y one a f f i n e p o i n t # 8 on P. If two Sylow 2-subgroups f i x e d t h e same p o i n t P t h e n S L ( 2 , 4 ) would f i x P . S i n c e a n e l e m e n t o f o r d e r 5 i s a 2-primi4 t i v e d i v i s o r of 2 -1, t h e e l e m e n t s of o r d e r 5 must t h e n f i x 8p

N.L.Johnson

228

p o i n t w i s e . But t h e n t h e i n v o l u t i o n s would be e l a t i o n s . So t h e r e is a n o r b i t of l e n g t h 5 on d . A s above, t h e n o r m a l i z e r o f a Sylow 2-group h a s an o r b i t o f l e n g t h 6 and o f l e n g t h 4 on ie. Thus, s i n c e t h e o r b i t of l e n g t h 4 cannot be f i x e d by ( 2 . 2 ) , ( 2 . 5 ) is proved. & a c t s i r r e d u c i b l y on a 4-dimensional v e c t o r s p a c e o v e r GF(2) and has o r b i t s o f l e n g t h s 5 and 1 0 . Assume w e have c a s e ( 1 ) o r ( 3 ) of ( 2 . 4 ) .

So,

( 2 . 6 ) Lemma. such t h a t i f then

proof. ((7)

We may choose a basis f o r a &-invariant component d r h a s o r d e r 3 and (5 o f o r d e r 2 n o r m a l i z e s (7)

7 must f i x e x a c t l y two p o i n t s o f t h e o r b i t o f l e n g t h n o r m a l i z e s two Sylow two s u b g r o u p s ) .

5

By ( 2 . 5 ) , 7 must f i x a s u b p l a n e p o i n t w i s e s o must f i x a Baer subp l a n e nr p o i n t w i s e . Thus, 7 f i x e s e x a c t l y one p o i n t o f t h e o r b i t o f l e n g t h 1 0 on d .

T h i s makes (n7 n 4 a 2-dimensional s u b s p a c e . L e t n,, n ie b e {[O,O,l,Ol,[O,O,O,ll,[O,O,l,l]~}. I t ' s e a s y t o check o u t t h a t on

Since

7

7-orbit

.

is c o m p l e t e l y r e d u c i b l e on d by Maschke's theorem l e t where W is r - i n v a r i a n t . C l e a r l y , W - (oj is a

W B (n7 ll d ) = ie

The n o r m a l i z e r 72g((r)) p e r m u t e s t h e permutes t h e 7 - i n v a r i a n t s u b s p a c e s .

r - o r b i t e s of

r

and a l s o

Note t h a t s i n c e t h e r e a r e e x a c t l y 1 0 g r o u p s o f o r d e r 3 t h e above two n o n z e r o v e c t o r s i n t h e o r b i t h5,ie o f s t a t e s t h a t t h e sum o f o f l e n g t h 1 0 . Thus, t h e 7 - o r b i t l e n g t h 5 is i n t h e o r b i t Alo,ie is n o t a s u b s p a c e . o E nb( ( 7 ) ) and la1 = 2. We a s s e r t t h a t e i t h e r a l l of t h e a r e s u b s p a c e s o r e x a c t l y one is. (5 fixes a ? - o r b i t s i n A1o,d 7 - o r b i t , s a y {A,B,C]. If A B 7 , C t h e n A t B 7 ,B t C 5 C t A . W must b e one o f t h e t h r e e ? - o r b i t s i n Aloe I f W # [ A , B , C ] and {A,B,C} is n o t a s u b s p a c e t h e n {A+B,B+C,C+A] is a ? - o r b i t t h a t s i n c e ( A t B ) t B t C = C t A and t h e sum o f any p a i r cannot be i n A 5 ,m is n o t i n A 5 , & . But i f o f i x e s A and maps of v e c t o r s o f A 5 ,d B ++ C t h e n A t B A A t C and B t C BtC. I n o t h e r words, r-subspace. {AtB,BtC,CtA] U 0 must be a o - f i x e d Let

-

The tmnslatton planes of order 16 that admtt SL(2, 4) Thus, we may assume =

[E] and

[i g]

u =

ID1

where

[El

=

= 2.

both The 7 - o r b i t both 0.

In

(as i n ( 2 . 6 ) ) f i x e s

u

and must i n t e r c h a n g e

W

i] = [: :] s i n c e = [ y i], [i !], o r [A

u R =

nT

and

UT = T 2 u

:I.

then

=

and

R

x4

= 1.

A5.

so

X2 =

and and

6.

are n o t

X4

0,ll.

is n o n s i n g u l a r s o

s

[ y k].

=

u

If

[' ' ,"i]

fixes

CX,,X2,X3,X4]

so that

= [X,,X2,X3,X4]

[X1,X1tX2,X4,X3]

X2 = 1 = X

and

[O,O

X1 X3

s o h a s t h e form

x = 0

Clearly,

must f i x e x a c t l y one p o i n t of

[i :]

n

and

[O,O,l,O]

Ft[y

R

is n o t a s u b s p a c e s o

A5, y = o

Then

{[O,l,O,O],[l,O,O,O],[l,l,O,O],~].

is

W

229

fixes

X1

= 0

and t h e

[O,l,l,l]

O2 1 0

other points i n

h5

are

and

[l,O,l,l]

[l,l,l,l]. Since t h e

t h r e e i n v o l u t i o n s i n a & ( ( ~ ) )a r e g i v e n by t a k i n g t h e d i f f e r e n t values f o r R, ( 2 . 7 ) is p r o v e d .

-(--2 . 8 )

Case (3),

Lemma.

1,1,5,10

of ( 2 . 3 ) cannot happen.

P r o o f . Choose x = 0 , y = 0 as & f i x e d components. Choose b a s e s as i n p r e v i o u s lemma and assume y = x is a 7 - f i x e d component i n t h e o r b i t o f l e n g t h 5. By ( 2 . 7 ) , we may c h o o s e u n o r m a l i z i n g T to fix

[l,l,l,l]

[! i].

R = of length

u 5

on

y = 0.

u =

So

[if] on

f i x e s p r e c i s e l y t h i s p o i n t on s o t h e Sylow

fix this point.

2-subgroup

a

Thus, i f

E Q

Iy

= 0

y = 0

y = 0

[zi it]

containing

Q

is

where

i n the orbit

u then

must a l s o

u

fixes

[ l , l , l , l ] and i n t e r c h a n g e s [l,l,O,O] and [ O , O 1,1] on y = 0 (nu n Y = o is ~ ~ , ~ 1 , 1 , 0 , 0 1 , c 0 , 0 , 1 , i 1 , ~ ~ , ~ , ~ , ~ 1 ~ .

Since

u

commutes w i t h

u

it follows t h a t

CiR

= RCi

and t h e r e f o r e

N.L. Johnson

230

(1)

C1,1lCl

t C1,llC3

(2)

C1,1ICl

t [0,01C3 =

(3)

[O,OlC,

w e must have

L-l,ll, CO,Ol, and + [ 1 , 1 1 C 3 = [1,11, C1 = 8

=

[:

or

C3 = I

and

or

R.

-

Squaring

0,

w e have C12 t C C = I . S i n c e C12 = 8, C 2 C 3 = I s o C 2 = C?j and 2 3 And, C C t C 4 C 3 = 8 s o t h a t C1 = C,, since a l l thus C2 = C3.

3 1

Ci

the

commute.

ly = 0

So

c2 ,

is

a

If

does n o t f i x

[c2 C J y = x (we c a n make t h i s a s s u m p t i o n ) t h e n on x = 0 , u must f i x one o f [l,l,O,O],[O,O,l,l],[l,l,l,l] and i n v e r t t h e o t h e r p a i r (see p r o o f t o ( 2 . 8 ) s i n c e y = x is f i x e d b y 7). But 'si f i x e s [1,1,1,1] on Y = 0 s o c a n n o t f i x [1,1,1,1] on X = 0.

fixes

If on

X = 8, t h e n if

a

IX = 8 i s

and = %

so

Itd2d3 = I,

IX

Thus,

C1,lIdq =

[1,1I,

or

I2

a=

and

d 3 18.

is e i t h e r

d3

-

07

= 0

"1

d 3 dl or

where

T h i s e l e m e n t must have o r d e r

Squaring

C1,ll.

Also,

[dl

R

IX

= 0,

[ l , l , l , l ] on

[l,l,O,O]

we have

= C1,11,

or

[s] 0

so

is n o n s i n g u l a r and is

Case I .

a?.

CO,Ol,

[l,lld2 = dl,d4,d2

dl

[l,lld3 = Cl,lldl

So

[::

and i n t e r c h a n g e s

[O,O,l,l]

we obtain

[t t].

And

dld2+d2dq = 8

or

[dk

= I,

d2td d

SO

1 2 3

dl = d q .

where e a c h

d3,d2,

I.

C12 = 8, :C

2,3,5

= dl

= d:

s i n c e it I s i n

= I.

Form

SL(2,4).

The translation planes of order 16 that admit SL(2, 4 )

then

= I

(d,A)2

= d3dl. so

4

lor1 = 3

then

47

Iy

if

dl = R.

and

-ur 1 y

= 0

So = 0 =

dl = R.

so that

( ( r ~ )IX ~ = 0

then

has order So

d d +d d A = 8 3 1 1 3

Thus, # 2.

If dl = R

which implies

is

d Ad A = d3dlA 2A 3 1 d3dl = 8. This cannot be, Thus,

I

,

[d3RA2y

dl = I

23 1

:].

=

R =

[ y i]

[: i].

must have

or

[:::

where

has even order so that

[: i].

[I1 1[]1

CIA =

[ y :]

1 1

Choosing 0 1] 1 =

A

=

[: :],

C1 =

[ y t]

[t

: C

= 8.

:I.

=

1 0 [l 1[]0

=

[: :][i

-01

0 1

O]+Ll

:I

=

[:

O][l

:I.

1 1

C1 = 8

(:TI3

where

I

IA

=

I

Y = 0 = 2,

C2 = I So,

C2 = I

or

R.

or

But,

C, = I,

:I [: :] [: ![I: :] [: :I* [: :I+[: :I [h :I

If

-0

Elementary

[: :].

either way we obtain a contradiction. That is, if

[: :I+[: [: :] +[: t] = [i i] # 12.

If

we must have

(ClA)2 =

t (C AC tC C ) C A = I2 1 2 2 1 2

(C1AC2+C2C1)C2A =

IX

and

calculations show that this leads to a contradiction. Since has (1,l)-entry ((CIA) (C,A)+A(ClAC2+C2Cl)C2A.

A

(%I2

Thus,

=

But

C2 = R,

0 1 0 1 11 [I 0111 11 =

So we must have

But, analogy provides that we have a similar contradiction since we only transpose the lower matrix. This proves (2.8). (2.9) Lemma.

Case 1 of (2.3) cannot happen.

Case 1,5,5,6. (Recall the orbit of length 6 is an orbit under the normalizer of a Sylow 2-subgroup). Let y = x be the &fixed component. Let T fix y = x x = 0, y = 0 of one of the orbits of length 5. Then u E 7l,( ( 7 ) ) of order 2 maps x = 0 -4 y = 0.

232

So

N.L. Johnson

[el,

a =

B

.'.

B2 = 8.

Ar

B =

4x4

a

matrix over

and

]B :

IB411 2

GF(2).

since

may b e chosen

7

BIA = A 2 B1.

Let

y = xM and y = xR be f i x e d by 7 i n t h e second 5 - o r b i t . Then y = xM A y = xfl. a must f i x p r e c i s e l y one p o i n t on y = x b y 7 S i n c e 1,1,5,10 on y = x . lBll 1 = 2 . T here are t h r e e i n v o l u t i o n s i n nb((7)) s o we may t a k e B1 a r b i t r a r i l y as a n e l e m e n t of order 2.

.'.

Since

m3 = m A

= mlA,

and

fixes

T

3

m4

"1

3 by

are nonsingular.

g l=' [ 8

m4

where

Now l e t

a1,a2,u3

u1 =

Thus,

ml,ml

a contradiction since

m3 = 8.

previous c a l c u l a t i o n t h a t

M

E (A).

M1 E ( A ) . If

ml

Thus and

-

ml

M =

R&((T)).

I A

7

-1 a17 =

[mk m,]:

tg is n o n s i n g u l a r .

be the three involutions i n

ml

a r e e q u a l we have

Then

r

So

8 8 1 A

0

8 1

Let

233

The translation planes of order 16 that admit SL(2, 4)

It f o l l o w s t h a t

A 2BIA = B1A2

r -1ur =

and

01

"l["lA

"1.

[*I.

S i n c e a l l i n v o l u t i o n s i n l?g.(r) i n v e r t y = xM i [ml "1["1 rlAi so, have 0 B4 8 iiil B4 m4

B4

"3.

A2jtk

M = [ml

m4

= 1.

jq

=

T h a t is,

[El8

Ak = ( A 2 ' ) - l

m4O] .

But,

Mtfl =

[:

= A'.

we must

m 1B 1A iml = BIAi

i =1,2,3. We know' ml = A', ml = Ak f o r some j s o t h a t , i n p a r t i c u l a r , mlBlml mlBl = A j B l = B,A2j

Thus,

y = xfl

and

So

m4:~4]

for

and k . And = BIA 2 j A k

-

ml = ml.

B1 *

So

must be nonsingu-

l a r i n o r d e r t h a t we have a s p r e a d .

We have t h u s e l i m i n a t e d a l l p o s s i b l e c a s e s o f ( 2 . 3 ) .

3 . THE DEMPWOLFF PLANE OF ORDER 1 6 We assume t h a t t h e Sylow p o i n t w i s e b u t ni and

2-subgroups Qi f i x Baer s u b p l a n e s ni i f j , do n o t s h a r e a l l t h e i r compoa

n e n t s . I t is e a s y t o s e e no and n l s h a r e t components. S i n c e any p a i r o f Sylow 2-subgroups g e n e r a t e s S L ( 2 , 4 ) , i t f o l l o w s t h a t ni,nj, i f j , s h a r e e x a c t l y t h e same t components. Thus, t h e r e must be ( 5 - t ) ( 4 t 1 ) t t components s o 1 7 2 ( 5 - t ) 5 t t . Thus, t 2 2. I f t = 3 t h e n t h e group would l e a v e a s e t o f 4 p o i n t s on i n v a r i a n t and t h e n must f i x e a c h o f t h e s e p o i n t s . If t = 4 t h e r e must be a n o r b i t o f l e n n t h 8 on 4 - . Thus, t = 2. By lemma 141,

pA#yJ

( 5 ) , we may choose c o o r d i n a t e s s o t h a t

Q, =

t r i c e s isomorphlc t o

where

GF(4)

and

Q, =

rqm]

A

is i n a f i e l d

f

is a c o n s t a n t i n

K

a

of

K

2x2

so

ma-

N. L.Johnson

234

[! i]

A E

.

,8

And, the normalizer of

(PA) =]+A[

c

CA

:] = [: :] if and only if ba] for a = 0 o r 1. Also, p A

=, a!,[

is

so that

= A-lC

:[I:

A-l

[i

Note ' 0

where

Qo,Q,

ICl

a = d

C = 8.

or

= 2

and

b+a = C.

fixes each component

of both no and nl. The common components of no and nl are X = 8, y = 8 and y = x is a component of n o . If C = 8, no and nl share 2 3 components which implies they share all components so that TT is H a l l by Foulser-Johnson-Ostrom. Since A-lCA = CA 2 , we

C

may assume Since

[*

[ y :],

=

Qo

and

pA

'y! =[ x

[?;[I:

!]. and

and

Q1

y =

]'-A

=

It follows dim3 = 0

[ At

or

Im31 = 2.

So

so

-

m2 = C(f- 1mltmq).

1

- -

m1,m4 E (A)

or

m4

So the components of

nl

have the

It is straightforward that

Aty

[:: ":I["

= 0,

C(f-lAitG(Ai)) where G(Ai) = 8 or is in (A). CA Y G(Ai) No element of Q1 can fix a component of no-X = 0, Y = 0. Thus, must map the set of three such components onto the remaining Q1 twelve components not on nl. Thus, the set of nine images (Ql-(1)) (no-(X=O,Y=O)) = set (Qo-(l)) (n,-(X=O,Y=O) ) Let form

Dj

y = [x

and

no,

n 0'

' O ] , i =1,2,3 F(A ) , A F(A1) = 8 o r has order 2. Let the components have the form

be fixed by

m3 m4

3

no

be a component of

[i

m2 = 8, ml = m4 € (A)

F(1) = 8

= Cm

[m3 A[],.

and

the components of where

I 8 I] =

ml m2

m4]

rectly that

-ml

fix the components of

ml m2

3 m [ ] ' A

"'1

y ='"[x m3 m4

Let

= C(f-'A'+G(A')).

The images of

y = x

[$ ,;ij)]

.

under

Q0-W

235

The translation planes of order 16 that admit SL(2, 4)

y = xM

have t h e form

y = x[

images o f

where

M =

a

A I OL] u n d e r F(A ) , A

.:s,l[;:I*

:][:j

Q l - ( l )have t h e form

The y = xN

e q u a t i n g t h e (1,l) and (1,2) e n t r i e s we h a v e : t D Ai

A’

(1)

J

= A’ t A k f C A a t A k f F ( A a )

and D

(2)

Case a .

Let

J

= Aatk

tAkfCAatk

tCAkfF(Aa)Ak tAkfAa.

L = 0 = k.

Then from (l), A j t So AJ t D Ai = I t f C . I f j # 0 t h e n A’ + I DjAi + f C j? ( A ) . Hence j = 0 and D Ai = fC

J

D Ai = I t f C ( F ( 1 ) = O ) . j = A2j. But = DOAi = C ( f - l + G ( I ) ) A i

.

= I tfC t f . Then f C ( A 2 i t I ) = From (2), Do = I t f C t f . So fCA-i I t f E ( A ) o r 8 b u t f C ( A 2 i t I ) p ( A ) u n l e s s i t i s 8. T h e r e f o r e I = f and i = 0 . So G(1) = 0 and f = I .

k = 1, k = 0 . From (l), A’ + D Ai = A t f C A + f F ( A ) = J A t C A t F ( A ) . So j = 1 s i n c e C A t F ( A ) L (A). Thus, DIAi = D1 = A t C A +CF(A) + A = C A +CF(A). From (21, C(AtB(A))Ai = CA t F ( A ) . So, D1 = (CA+F(A))A2‘ = C A +CF(A) and h e n c e CA(A2’+I) =F(A)(A2itC). S i n c e A2‘ t C i s a l w a y s s i n g u l a r , F(A) = 8 and i = 0 . Thus, G(A) = 8.

Case b .

Case c .

2

L = 2, k = 0 .

From (l), A’ t D Ai = A 2 t C A 2 t F ( A ) j 2 i From (21, D Ai = C A 2 tF(A2) = C(A2 t G ( A ) ) A Then j = 2. 2 2 2 SO D2 = A 2 t C A 2 t C F ( A 2 ) + A 2 = CA t C F ( A ) , D 2 = C ( A 2 + G A 2 ) ) = CF(A2).

Hence

G(A2)

2 2 G ( A ) = F(A ) = 8.

.

= F(A2).

The images of for

ponents of

no

and

rrl

Since

F(A2) f ( A )

y = x[t:j

j , k =0,1,2.

so CG(A 2 )

we have

under

Qo-(l) are

These images p l u s t h e com-

[k]

completely f i l l o u t t h e plane.

T h i s p l a n e i s t h e u n i q u e p l a n e (which t h e n must b e t h e Dempwolff plane) s a t i s f y i n g our assumptions.

is a collineation

N.L. Johnson

236

o f n which commutes with & so admits GL(2,4). So, we note that the three planes that admit SL(2,4) also admit GL(2,4). Note that our arguments also give an alternate construction of the Dempwolff plane.

REFERENCES [l] Dempwalff, U., Einige Translationsebenen der Ordnung 16 und Ihre Kollineationen (to appear). [2] Foulser, D.A,, Johnson, N.L., and Ostrom, T . G . , Characterization o f the Desarguesian planes of order q2 by SL(2,q) (submitted).

[31 Huppert, B., Endliche Gruppen I (Springer-Verlag, Berlin-Heidelberg-New York, 1967).

[4] Johnson, N.L., The translation planes of Ott-Schaeffer (to appear).

[5] Johnson, N.L., Addendum to "The geometry of SL(2,q) lation planes of even order" (submitted).

in trans-

Annals of Discrete Mathematics 14 (1982) 237-248 0 North-HollandPublishing Company

COMMUTATIVE D I V I S I O N ALGEBRAS Giampaolo Menichetti ( - ) I s t i t u t o d i Geometria U n i v e r s i t y o f Bologna Italy I n t h i s paper we g i v e a geometrical method f o r the c o n s t r u c t i o n o f commutative f i n i t e dimensional d i v i s i o n algebras over a f i e l d

K w i t h char K # 2. L e t V be a vector space over a f i e l d K, char K # 2, w i t h dim V = n < w . We denote b y & = (V,f) the algebra over K defined by a given b i l i n e a r map f : V 2 + V. For every

X,YE

d = (V,f)

we p u t L,(y)

=

f ( x , y ) and R,(y)

a l e f t ( r i g h t ) zero d i v i s o r i f f d e t L, = 0 ( d e t R,

= f(y,x).

Then x # 0 i s

= 0).

We remark t h a t an algebra d w i t h no zero d i v i s o r s i s a pre-semifield i n t h e sense o f Knuth [71. I f d a l s o possesses the u n i t y element, then we have a d i v i s i o n algebra o r a s e m i f i e l d . Using a s u i t a b l e isotophism over a given algebra d w i t h no zero d i v i s o r s , we can d e r i v e a d i v i s i o n algebra 9

-f ( x , y )

= f(Ril(x),

Lil(y)),

. For example,

i f d = (V,f),

then 9 = (V,F),

i s a d i v i s i o n algebra f o r every a ~

u n i t y element i s e = f(a,a).

da # ,0. The

For t h i s reason we may consider o n l y algebras

without zero d i v i s o r s . Let U = =

z

n- 1

i=O

{ uo,

xiui,

u1 y

,..., n- 1

2

i=O

be a f i x e d basis i n V. I f x , y ~ d= (V,f),

yjuj,

then

where fk(x,y)

=

n- 1

-8

1

,J=o

cijkxiyj,

( - ) Research p a r t i a l l y supported by G.N.S.A.G.A. 237

k = 0,1,

(CNR)

... ml,

238

G. Menichetti

and cijk€

K a r e the m u l t i p l i c a t i o n constants o f d w i t h respect t o t h e basis

U.

Moreover t h e matrices associated w i t h Lx and Rx i n t h e basis U a r e

n- 1 = (rki),

xF !

respecti v e l y

2

rki =

.

k,i = 0,1,

cijkxj,

j=O

L e t u be a permutation o f t h e i n d i c e s i,j,k du = (V,f,) where

...,n-1,

and l e t d = (V,f).

We p u t

f U b j'Uj) For example, i f

5

=

(5 { k),

then dB i s the anti-isomorphic algebra o f d .

RESULT 1 (see Bruck 161 ) . 16 d 0 an dgebha ulith no zeho d i w 0 o h 6 , ,then d.

had no zeho div.&o~hb doh each pmnuhaXon

u

06

t h e indiceb i,j,k,

I n t h e f o l l o w i n g we suppose t h a t d i s commutative, i . e .

REMARK. d ' and d" ahe anti-.&omohpkic dgebhad A commutative algebra d = ( V , f )

g:v+v, If x =

(1) where

(2)

n- 1

2

i=O

xiui,

.

determines the quadratic map

g(x) = f(x,x).

then n- 1 g(')

=

2

k=O

gk(x)uk n-1

gk(x) = fk(XsX) =

- 2 CijkXiXj,

1 ,j=O

k = O,l,...,rl-l.

Conversely, every quadratic map g determines a b i l i n e a r map f by 1 f(X,Y) = z[g(x+Y)-g(X)-g(Y)l*

A map g a l s o gives a quadratic tronsformation $I i n the p r o j e c t i v e space IP(V) = IPn-'(K) o f the n-tuples X E (xo, x1 x,,-~). Namely

,...,

239

Commutative division algebras

n- 1 where dgk/dxj = 2 2cijkxi. Hence i-0 L = Itx= 2J(X). -X

So the n a t u r a l p r o j e c t i o n 7: V - + I P ( V ) , ~ ( x =) X, maps a zero d i v i s o r x o f d t o a p o i n t X belonging t o t h e hypersurfacedt'defined by the equation det\l(X) = 0. I n p a r t i c u l a r we have t h e f o l l o w i n g

divL6oh6 i d 6 the jacobiun 06 the quadhutLC & c ~ n ~ ~ o h m a t i o n ~L6~ (non-zeho ~) at each p o i n t 06 IP (V). zeho

ye denote by Y t h e l i n e a r system n-1 zkgk(x) 0. Zk' K , (4) k=O generated by t h e quadrics Qk: gk(X) 0.

2

I t i s easy t o v e r i f y t h e f o l l o w i n g c l a s s i c a l

RESULT (see B e r t i n i [ 4 ] ) . The h y p m w r d a c e Jt' w i t h equation detJ(X)

= 0 L6

the

LOCUS06 the p o i n t s X E IP ( V ) whobe p o k b hypehplanea ulith h e a p e c t t o the quadhics Qk, k = 0,1,. ,n-1 , me U n M y d e p e n d e n t .

..

From t h i s we have t h a t Jt' contains the base p o i n t s o f the l i n e a r system Y a n d the s i n g u l a r p o i n t s o f i t s degenerate quadrics.

As usual we c a l l detJ(X) t h e j a c o b i a n

06

The m a t r i x o f the quadric ( 4 ) i s M(2) = (aij), i,j = 0,1,

-

9.

...,n-1,

where

=

and Z

z

( ~ ~ , z ~ , . . . , z ~ - ~Hence ). M(2) coincides w i t h the m a t r i x I-;,

n- 1

= Jziui,

=o

1

follows t h a t

o f t h e l i n e a r map

Li:

yHfall(z,y)

i n d " . From t h i s and Result 1 i t

C.Menichetti

240

me non d e g w e h a t e ( i . e . t h e dischiminant 06 (4) i~ ZULO

4 6

Z~=Z~=...=Z~-~=O),

REMARK. The eaSt c o n d i t i o n h p f i e s ththat t h e QUUCLJLLCSQk:gk(X)=O,k=O,l,.

. . ,n-1,

me finineahey independent.

Propositions 2 and 3 imply t h a t The j a c o b h n aU 0 6 t h e QU&U

PROPOSITION 4.

id6

i n IP(V)

a

06

QU&&C

t m n b 6 0 h ~ a t i o n( 3 ) 0 non-zeho

betonging t o t h e f i n e a h bydtem (4) me non-de-

g enehate.

This p r o p o s i t i o n i s t h e geometric equivalent o f Result 1 w h e n d i s commut a t i ve. We can consider ( 4 ) as a p r o j e c t i v e space I P ( 9 ) isomorphic t o t h e space o f the n-tuples 2 = (zo,zl, z n- 1) . Hence t h e equation d e t bl(Z)=O defines a hypersurfacep’of IP ( 9 ) .

...,

The n x n symmetric m a t r i x S(A) -

= (aij),

adi

= a..EK,

depends on the m-tuple A = (aO0,ao1,.

.. ,a n-,,n-l),

J1

i,j = 0,1,

...,n-1,

m = n(nt1)/2. We denote b y Y

the hypersurface o f IPm-’(K) defined by t h e equation d e t S(A) = 0.

every quadric o f IPn-’(K)

n- 1

2

..

aijxixj = OHA (aO0,ao1 ,. ,an-l ,n-1 ) i,j=O i s mapped t o a p o i n t o f IPm-l(K). I n p a r t i c u l a r a

By means of t h e f u n c t i o n

q~ :

f

p o i n t A E Y i s the image o f a degenerate quadric. I P ( 9 ) i s the subspace o f IPm-’(K) generated by the n p o i n t s ,Cn-l,n-l , k ) , k = 0,1, ,n-1. HenceH”is t h e i n t e r s e c t i o n o f Y

...

(COok,COlk,

...

w i t h I P ( 9 ) . So we have t h a t PROPOSITION 5 .

hab no IP(Y)C

ZULO

A commLt&ue

diu0oh6

d g e b & a d = (V,f),

f(x,y) =

n-1

2

k=O

fk(x,y)uk,

i d 6 t h e i n t m e o t i o n f l o 6 Y U h i-12 c o M u p o n d i n g bubbpace

I P ~ - ~ ( K ) ha^ no h a t i o n a t points oueh K.

Some technical r e s u l t s are useful f o r applying t h e previous propositions t o the study o f commutative algebras w i t h no zero d i v i s o r s . Indeed many problems are easier when working i n a s u i t a b l e algebraic extension o f K. Here we consider the case t h a t K possesses a Galois c y c l i c extension K* o f degree n 2 2 . Q a n d every f i n i t e f i e l d are examples o f such f i e l d s . Some explanatory examples are considered below.

Commutatfve division algebras

241

REMARK. K * & a c o m m W v e n-dimenhiovml div&ion aLgebha (V,fo), fo(x,y) = xy, o v a K.

I f u i s a p r i m i t i v e element o f K * over K, then Uo basis o f V which we s h a l l consider instead o f U.

.

.

{l,u

.....un-'

}

is a

L e t G be the Galois group o f K * over K. Thus G = < z > a n d IGI = n. Moreover i z ( U ) E K * , i 0,1, n-1, and

....

z'(u) = u + i

= 0.

Hence the Vandermonde m a t r i x

...

U2

un-1

...............................

\ I

i s non-singular. n- 1 L e t x = 2 xjuJ be an element o f V and j=O n- 1

(5)

j=O

.

....

....

be i t s image under zi, i € { O , l , n - l } . The equations zi(x) 0, i=O,l, n-1, d e f i n e n hyperplanes ni o f IPn-'(KU), l i n e a r l y independent and w i t h no r a t i o nal p o i n t s i n K. We denote by Pi, i e { O , l n-1}, t h e v e r t i c e s o f t h e

...

.....

hypertetrahedron T = {ai : i = O,l, ,n-1} which i s the i n t e r s e c t i o n o f t h e n-1 hyperplanes nk, k = 0,1, n-1, k f i.

....

.

The group G acts t r a n s i t i v e l y on t h e planes and on the v e r t i c e s o f T. Hence n-1 (Pi, i = 0,1, n-1) a r e conjugates w i t h we say t h a t ni, i 0,1, respect t o K.

We remark t h a t the columns o f

(wo,wl

g-'

g i v e the coordinates o f Pi.

1 ........................

u-1

and Po

....

....

I

.... ,w,,-~).

From ( 5 ) we have x = .j

n- 1

2

i=o

zi(wj)zi(x),

j

.

O,I

.....n-I.

Namely

242

G. Menichetti

Hence

n- 1

g(x) = f(x,x)

where bhk=

n-1

2

i,j=O

n-1

h,k=O

i,j=O

2

g(x) =

(6)

n-1

'-'

or

2 x.x.f(ui,uj) i,j=O 1 J

=

h bhkc

k

h , k=O

f(ui,uJ)zh(wi)zk(w.)

E K*.

J

i d i We observe t h a t f ( u ,u ) = f(uJ,u ) , i,j = O,l, n-1. Indeed bhk bkh, k,h = 0,1,..., n- 1 i j k h bkh' i,J=of ( u vu )'G ( w i ) z ( w j ) n- 1

...,n-1,

implies t h a t

2

i,j=O n-1

h f ( u ,u ) z ( w j ) Z ( w i ) = bhk* i,J=O Analogously we prove t h a t bhk' bkh, h,k = O,l, JI-~, i m p l i e s t h a t f ( u i,u j ) = =

j

i

f ( u ,u ), i,j = 0,1,

...

...,n-1.

REMARK. In ( 6 ) ,the map g

given

c ~ 6a

mapping

06

K * i n t o LtheRd.

I n t h e f o l l o w i n g Y I d e n o t e s t h e l i n e a r system

which correspond t o t h e extension K * o f K. Analogously we d e f i n e X * , Y * , PROPOSITION 6. The qu&cb n-1 Q i: gi( X ) = 2 gk(X)zi (u k ) = 0, i = 0,1,

(7)

k=O

...,n-1,

genetlate Y x .

mood.

1is

non-singular. Hence every quadratic form gk(X) i s a l i n e a r

combination o f the gi ( X )

.

Using (6) we can w r i t e the equation o f Qi i n t h e form gi(x)

=

2

n-1 zi(bhk)zhti(x)zk+i(X) h, k=O

Furthermore

(8)

X;

h

= z (x)

,h

= 0,1,

...,n-1,

= 0.

etc.

243

Commutative division algebras

.

are p r o j e c t i v e coordinates i n IP n-l (K*) Hence n- 1 g”(X1) = ri(bhk)x;+ix;+i, h ,K=O

2

...,

i = 0,1,

... ,n-l,

(xb,xi, x;-~) and x,!, = x i , etc.) are t h e equations o f (XI system whose fundamental p o i n t s are Pi , i = 0,1,.. ,n-1.

.

Q’i n a coordinate

EXAMPLE 1.

x

We suppose g(x)

2

. Then f(x,y)

= xy and hence

I

= K*.

The quadrics

i i 2i Q : g ( X ) = z ( x ) = 0, h = O,l,...,n-l, which degenerate i n t o two coincident hyperplanes o f T, generate Y * . We observe t h a t P

js

j E {O,l,

...,n-1},

i s conjugate w i t h a l l the p o i n t s

..

n. i n t h e (degenerate) p o l a r i t y defined by every quadric Qi , i = 0,1,. ,n-1. J H e n c e p i s t h e l i n e a r system o f a l l t h e quadrics o f IPn-’(K) whose T i s a s e l f -

PE

p o l a r hypertetrahedron ( f i g . 1 explains the case n = 3 ) .

fig.1

.

From (1) we can deduce t h e equations o f t h e quadratic transformation $ For 2 2 2 2 example, i f n = 3, then x = (xo+ xlu + x2u ) = go(x) + gl(x)u + g2(x)u i m p l i e s go(x) = x i t coc2x2 2 + 2cox1x2 gl(x)

= (co+ c1c2)x22 + 2xox1 t 2ClX1X2

,92(x) = x1

+

(Cl+

c;,x;

+

2xox2t 2C2X1X2,

244

G. Menichetti

where u3= c2u2+ clu t c Y'has

0

, c 1 . K.~

no base points.

..,n-l},

i€ { 0,1,.

deg(H*) = n and every p o i n t PEni,

i s singular f o r a t least

the quadric Qi. Hence **degenerates i n t o the hyperplanes rci, This r e s u l t can be proved a l s o as f o l l o w s .

n-1

(9)

ZZ~X;' = 0,

z;E K*,

x r 2 = g"(x),

i

i=O

where (10)

. . ,n-1 .

t h e coordinates ( 8 ) . Then t h e l i n e a r s y s t e m Y * i s defined

We use i n IP"'(K*) bY

i = 0,1,.

-J '

= (dg"/dxA)

hence = 2'xix;.

detJ'

O,I

,..., n-I.

= Zdiag(xb, x i

,...,x,!,-~),

..x,!,-~.

From

J * = (dgk/agIi)J' we deduce t h e equation of**.

(dx,',/dxj)

The m a t r i x o f the quadric ( 9 ) i s

M'(Z') 2'

=

(zi, z i ,..., z,!,-~),

= diag(z6, z i

,...,z,!,-~),

and ( Z ' ) = zbzi.

det!'

. .z,',-~.

(7) and (10) imply

S u b s t i t u t i n g i n (4) we have n-1 n-1 ( ZZ.ZJ(Wi))X!2 J j=o i=o

.

2

Then 2'

j

Hence

=

= 0.

n-1 ZZiZj(Wi).

i-0

degenerates i n t o t h e n hyperplanes

n-1

.

a.: ZzizJ(wi) J

i=o

= 0, j = O,l,

...,

n-1, conjugate w i t h respect t o K.

...,

..,

The images o f Qi, i = 0,1, n-1, i n IPm-l(K*) are n p o i n t s Si, i = 0,1,. n-1, l i n e a r l y independent and conjugate w i t h respect t o K. These a r e s i n g u l a r p o i n t s o f Y * w i t h m u l t i p l i c i t y n-1. #"''* i s the i n t e r s e c t i o n o f Y * w i t h the l i n e a r

Commutative division algebras

. ,n-1.

i = O,l,..

v a r i e t y generated by Si,

245

EXAMPLE 2. Suppose t h a t i n (6), b = blO= 01 f(x,y) = X Z ( Y ) + YZ(X).

1 and b..= 0 otherwise. Then g ( x ) = 2xr(x) and 1J

The l i n e a r system Y* i s generated by t h e quadrics Q’: gi(x) = r i(x)zitl(x), i = O,I (11)

,..., n-I.

Each o f these degenerate i n t o two d i s t i n c t hyperplanes ni and nitl o f T.

i = O,l,

Every p o i n t P€ni,

...,n-1,

belongs a l s o t o # * .

Indeed: i f P i s com-

mon t o a t l e a s t two d i s t i n c t hyperplanes, then i t i s a s i n g u l a r p o i n t f o r some quadric (11); iJP

l i e s o n l y on hyperplane ni, then i t s p o l a r hyperplanes w i t h

respect t o two d d s t i n c t quadrics (11) coincide w i t h n i .

i = 0,1,

...,n-1.

We deduce t h a t ni€M*,

Two cases a r e possible:

1. deg(H*) n and henceM*coincides w i t h t h e hypertetrahedron T; 2. the equation o f M * i s an i d e n t i t y i n IP”’(K*). O b v i o u s l y d = (V,f) has no zero d i v i s o r s o n l y i n t h e f i r s t case. L e t us consider the p o i n t I

(1, l,...,l)eIPn-’(K*).

The p o l a r hyperplanes o f

I w i t h respect t o quadrics (11) are

Pi : zi(x) I n the p r o j e c t i v e coordinates

pi:

xi

t zi+’(x)

= 0, i = O,I

(a),

t

0, i = 0,1,

,...,n-I.

...,n-1.

These hyperplanes are l i n e a r l y independent i f f t h e n x n m a t r i x i d e n t i t y m a t r i x and

-I =

0 1 0

I

t

E,

where

... 0

i s non-singular. I t i s easy t o v e r i f y t h a t det(l

+ E)

0, n even,

2, n odd.

Hence the commutative algebra d =(V,f), w i t h dimKV = odd and f(x,y) = xz(y) yz(x), has no zero d i v i s o r s . The corresponding d i v i s i o n algebra 9 = (V,P),

+

246

G.Menichetti

f(x,y) = f(Rj'(x), L;'(x)), where R1(x) = Ll(x) t w i s t e d f i e l d (see A l b e r t i l ] , [ 2 ] , 1 3 ] ) .

= x t z(x), i s a comnutative

We suppose n i s odd. The quadrics (11) i n t e r s e c t each o t h e r i n the (n-3)/2-dimensional

linear

variety

Ao: x

=

z 2 ( x ) = z 4 ( x ) =...=

n-1

and i n i t s conjugates w i t h respect t o K, A1, A 2 , . .

(x) = 0

., An,1.

I n o t h e r words 9 ' is

...,

the l i n e a r system o f a l l the quadrics o f IP"'(K) through Ai, i = 0,1, n-1. For example, i f n = 3, then p i s t h e n e t o f a l l t h e conics passing through t h r e e fixed p o i n t s Ai, i = 0,1,2, l i n e a r l y independent and conjugate w i t h respect t o K (fig.2).

fig.2 The equations o f t h e quadratic transformation

.

n- 1 x.x.u.z(uJ) i, j = O 1 J 1

2

=

c$

come from xz(x) =

n-1 z g h ( x ) u h , as i n Example 1, h=O

.W* degenerates i n t o n hyperplanes which are conjugate w i t h respect t o K. Indeed t h e m a t r i x o f t h e quadric

is

M'(Z')

-

= diag(z6,

Moreover, f o r n odd, d e t ( E

t

t

zi

) = 2.

,...,~ ; - ~ ) (t grt ) .

Commutative division algebras

241

In this case H"' is the intersection of Y * w i t h the (n-1)-dimensional linear variety through n singular points of Y * (with multiplicity n-2), independent and conjugate with respect to K. We remark that the previous propositions allow us to deduce some geometric propositions from results concerning the commutative algebras. For example, it is well-known (cfr.[5] and 181) that a finite-dimensional real algebra with no zero divisors must have dimension n 1,2,4 or 8. From this and Proposition 2 we deduce that the jacobian of all quadratic transformation in a projective space IP"'(IR), n # 1,2,4 or 8, is equal to zero at some points. REFERENCES

[ l ] Albert,A.A. , On nonassociative division algebras, Trans. Amer. Math. SOC. 72 (1952) 296-309.

, Finite

[21

noncomnutative division algebras, Proc. Amer. Math. SOC.

9 (1958) 928-932.

, Generalized

(31

twisted fields, Pacific J. Math. 1 1 (1961) 1-8.

[4) Bertini ,E., Introduzione alla geometria proiettiva degli iperspazi (Principato. Messi na , 1923). [5] Bott,R. and Milnor,J., On the parallelizability of the spheres, Bull. Amer. Math. SOC. 64 (1958) 87-89. [6] Bruck,R.H. , Some results in the theory of linear non-associative algebras, Trans. Amer. Math. SOC. 56 (1944) 141-199.

[7] Knuth,D.E., 182-217.

Finite semifields and projective planes, J. Algebra 2 (1965)

[8]Milnor,J., Some consequences of a theorem of Bott, Ann. of Math. 68 (1958)

444-449.

This Page Intentionally Left Blank

Annals of Discrete Mathematics 14 (1982) 249-264 0 North-Holland Publish@ Company

TRANSLATION

PLANES

OF

ORDER

Giuseppe PELLEGRINO

Gabor KORCHMAROS

University of P e r u g i a , I t a l y

An

2

11

University of Budapest,Hungary

i n t e r e s t i n g problem concerning f i n i t e t r a n s l a t i o n p l a n e s is t h e following:

To find spreads G,from some regular spread F of PG13,ql,by replacing a subset F 1

of F with a partial spread G' d i s j o i n t from F ' . This problem has been exhaustively discussed only i n t h e case where G i s subre= gular,i.e.

G is o b t a i n a b l e from F by r e v e r s i n g a s e t o f d i s j o i n t r e g u l i of F.

In r2-1 -- A.A.Bruen pointed o u t another remarkable case.He supposes t h a t F ' admits a set R of r e g u l i with t h e following p r o p e r t i e s : q is odd;

R consists of (q+l)(q+3)/4 lines;

any two reguli of R have two lines i n common;

(1)

any three reguli o f R have no line i n common; every l i n e of F' belongs to a regulus of R. Using t h e w e l l known incidence p r e s e r v i n g

isomorphism between t h e miquelian i n =

versive plane I P ( q ) with i t s p o i n t s and c i r c l e s , a n d a r e g u l a r spread F of PG(3,q) with i t s l i n e s and r e g u l i (see [21 , t h . 4 . 3 and n.3 of t h i s paper),Bruen states:

Starting from any partial spread F',contained i n a regular spread F of pC(3,q) and satisfSing the properties (l),one can construct on an e l l i p t i c quadric a fa= w i l y 8' o f conics

ties:

- called

i n 121 _ - a chain of circles

- with

the following proper=

0' contains exactly (q+l) (q+3)/4 points; (2)

any two c i r c l e s of 0' have exactly two points i n c o m n ; any three c i r c l e s of 8' have no point i n common.

Conversely,starting from a chain 8' of circles,one can construct a partial spread F' which belongs t o a regular spread F and admits a s e t R of reguli which sat;=

sfies

(1).

Concerning p a r t i a l spreads G' covering t h e same p o i n t s as F',we do n o t have e x i = stence or

no e x i s t e n c e theorems.However Bruen

249

12.1 proves

t h a t i f such a G' e x i s t s

250

G.Pellegrino and G. Korchmaros

then a s e t o f (q+1)/2 l i n e s i n t h e r e v e r s e of each Now,suppose t h a t we have a chain

@ 'of

r e g u l u s of R belongs t o G'.

c i r c l e s i w e can c o n s t r u c t t h e corresponding

F ' and t r y t o compose a p a r t i a l spread G ' so t h a t G = ( F - F ' l U G '

i s a non r e g u l a r

spread. Furthermore t h e g e n e r a l problem of c o n s t r u c t i n g chains o f c i r c l e s i n PG(3,q) i s \

s t i l l open.We know examples only i n spaces o f small order.The c a s e s q=3,q=5 are

t r e a t e d i n [2] - - and 131 .In [2] Bruen a l s o g i v e s a n example f o r q=7 and shows,for

states t h e following 2 p r o p e r t i e s f o r t h e r e s u l t i n g t r a n s l a t i o n p l a n e s II o f o r d e r q : a l l t h e s e c a s e s , t h e e x i s t e n c e of t h e r e l a t i v e spread G ' . H e

when q=3, II

i s an Andre plane;

when q>3, II

i s n o t an Andre p l a n e and i s n o t isomorphic to a subregular

plane. I n 151 G.Korchmaros g i v e s a f i r s t example o f a chain of c i r c l e s f o r q = l l .

I n t h i s paper w e study t h e p a r t i a l spread a r i s i n g from Korchmaros example and show t h e e x i s t e n c e of t h e r e l a t i v e p a r t i a l spread G'.We plane TI a s s o c i a t e d to t h e spread ( F - F ' l U

show t h a t t h e t r a n s l a t i o n

G' i s d e r i v a b l e and,by derivation,we

f i n d f o u r planes TI (i=1,2,3,4).Even each II i s derivab1e;in t h i s way w e o b t a i n i i two n o t isomorphic p l a n e s II and II .Next t h e f u l l c o l l i n e a t i o n group of each 12 34 o f these planes i s determined.Finally,we p o i n t o u t t h a t t h e p l a n e s n,TI ,11 1 3' 5 2 ' 1134 are pairwise not isomorphic.

W e give a l i s t of n o t a t i o n s t h a t w i l l use throughout t h e paper. I f A,B a r e s e t s , t h e n A-B denotes those elements o f A non i n B.The c a r d i n a l i t y o f a s e t X i s denoted by

Ia {b I b

1x1

. L e t GF(q) be a Galois f i e l d o f odd orderiwe p u t

x

= {a

i s a square element o f GF(q)l

v

=

i s a non-square element o f GF(q)}.

< u , v , . . > denotes t h e group generated by u,v..Generally

we use homogeneous coordi=

n a t e s x , y , z , t e t c . t o denote p o i n t s i n PG(3,q).Sometimes t h e l i n e j o i n i n g t h e point (x,y,z,t) t o the point (x',y:z',t')

i s denoted by r ( x , y , z , t ) , (x:y:z:t')].

1.- L e t GF(11) = ~0,1,2,3,4,5,6,7,8,9,10~. Consider i n P G ( 3 , l l ) t h e e l l i p t i c q u a d r i c with equation 2 2 (1.1) Q : x + y = zt i and t h e s e t I o f t h e following p o i n t s

25 1

Translation planes of order 1 l 2

I : I P =(1,O,O,O),P 1 =(6,0,10,1),P 2 =(7,0,2,1),P 3=(10,0,7,1), P =(8,0,8,1),P =(2,0,6,1),Pm=(0,1,0,0)}. 4 5 (i= ,0,1,.,5) be the polar plane of the point P

Let II

i

C

i

1

with respect to Q,and let

the conic n n Q. i

According to

151

,the conics C

i

form

a chain C of circles.In the following table

we give the coordinates of the points of each C

1

(by this table the reader can

check that C forms a chain) Table 1

We give a l s o the equations of the conics C

1

Table 2 2 x +lOzt=O ,y=O

cm: c

*

;

co:

y2+10zt=0 ,x=o

(x+4t)2+y 2 =3t2,z=3x+9t;

2 ’

c .

;

c1

2

2

2

: (x+5t) +y =4t ,z=x+y;

(xtt)2ty2=5t2,z=9x+4t

i

3 ’

2 2 C4 : (~+3t)~+y’=t’ , z=5x+3t ; C5 : (x+9t) +y =9t2 ,z=4x+5t , Now we consider the four points =(0,3,1,1) ; S =(0,8,1,1) , S =(8,3,10,1) , S =(8,8,10,1) 1 2 3 4 and denote I! the conic Q n o (i=1,2,3,4),where U i s the polar Plane O f S 1 i i 1 with respect to Q.The points belonging to the conic D are given in the follo= S

1

wing table

G. Pellegrino and G.Korchmiiros

252

I n t h i s s e c t i o n w e s h a l l determine t h e c o l l i n e a t i o n group G of P G ( 3 , l l ) ha=

2.-

ving t h e following p r o p e r t i e s

c

( i ) : G maps Q o n t o i t s e l f and l e a v e s

invariant.

I t is c l e a r t h a t ( i )is e q u i v a l e n t t o

( i i ) :G maps Q onto i t s e l f and l e a v e s I i n v a r i a n t . As Pi

( i = O , l , . . , 5 ) belongs t o t h e p l a n e x

( i i )w e have

2.1.

G fixes P ,

n

Moreover,as

and t h e plane x

2.2.

G leaves C ,

w i t h equation

y=O , b u t P ,

TI. ,from

Y

.

Y is t h e p o l a r plane of ,P

Y 2.1. it follows,

Y

with r e s p e c t t o Q and , C

=n nQ,from Y

invariant.

Denote by G ' t h e r e s t r i c t i o n map o f G onto x . I n o r d e r t o determine G ' , w e consi= Y d e r t h e following subsets o f I : I ' = { P ,P ,P ,P ,P 1 , I = {I;UPo}. y 1 2 3 4 5 y By (1.2),we have t h a t I ' belongs t o t h e conic r w i t h t h e following equation Y (2.1) r : x2+3zt=o,Y=o.

r

and a r e tangent a t O ( O , O , O , l ) and Y(O,l,O,O);hen= Y ce t h e p o l a r l i n e s o f Po with respect t o r and C , are t h e same.The equation of

The conics

and C ,

this l i n e is z = 0 .

l i e on x

Translation planes of order 112

253

By 2 . 2 . any c o l l i n e a t i o n o f G ' l e a v e s Cm invariant.Furthermore w e have:

-

2.3. Any c o l l i n e a t i o n J, of G',which leaves

Proof. L e t

J,(C,)

r

invariant,also fixes P 0

.

{cmn r l ={O,Yl,we have e i t h e r $ ( 0 ) = O f $ ( y ) = yo r

C m f $ ( r ) =r.As

( r e s p e c t i v e l y t ) t h e common t a n g e n t o f cm Y and r a t 0 (resp.Y);we a l s o have e i t h e r J , ( t ) = t , $ ( t ) = t o r $ ( t ) = t , J , ( t Y ) = 0 0 Y Y O Y =t . A s P = t n t ,it follows t h a t J, f i x e s P $(O)=Y ,J,(Y)=O.Now w e denote by t

O

O

Y

0

.

Next w e s h a l l prove 2.4. Any c o l l i n e a t i o n J, o f G ' l e a v e s BOOf.

r

invariant.

By way o f c o n t r a d i c t i o n , l e t us suppose t h a t t h e r e is i n

J, such t h a t

$(r) #

a collineation

r.Then w e have

(2.2)

J,(Pn) = P

;

u (n)

$ ( P o ) # Po

where u ( n ) is a permutation o f t h e i n t e g e r s Of1,..,5.We $(P5)=Po,$(P )=P 0

GI

c o n s i d e r t h e c a s e where

(a similar process must be c a r r i e d o u t f o r t h e o t h e r c a s e s

1

too) .By 2 . 2 . ,

.

J,(Cm) = C ,

(2.3)

Now w e study t h e a c t i o n of J, on r.By ( 2 . 2 ) t h e conic

$(r)

must s a t i s f y t h e f o l l o =

wing conditions: a)

$ ( r ) contains

P

;

l ,i.e. 2 3 4 5 Presently it is J, (0)# = {P , p , p , p

$ ( r ) c o n t a i n s t h e s e t { $ ( P ) f $ ( )~,$ 1 2 l$(r) n r l = 4 ; c ) J , ( P ~ )is a p o i n t of 0 J, (0)# Y,otherwise we have I$ ( r ) r l r b)

which is a contradiction.Similar1y $(Y) # Y

$(Y) # 0. 2

Then according t o (2.3),we p u t $(O) = A = (a,O,a , I ) where aft? # O.L e t t (resp. t A

B

,

( Y ) = B = (B,O,B

2

be t h e tangent t o Cm a t A (resp.B) and p u t

,I)

,

C =

= t n t .Then from (2.3) we have again $ ( P ) = C = P1' ( s e e c ) ) . A B 0

Therefore

$ ( r ) belongs

t o t h e p e n c i l o f conics determined by C ,

and by t h e p o l a r

l i n e c o f C t o r e s p e c t to Cm .Now C = t n t = (a+Bl0,aB,2) ;moreover A B 2 (2.4) a8 = 10(a+B) # 0 s i n c e C must belong t o

r.

So t h e equation o f t h e p e n c i l c o n t a i n i n g L(a+B)x + 10(z+aBt)] i n t h i s pencil

W e obtain Put

+

$ ( r ) is

2 1OX(x +10zt) = 0 ;

$ ( r ) is

J,(r)

determined by t h e condition a ) . 2 2 : (z+aBt) + 9 ( a + B ) ( z + a B t ) x+ (a+B) z t

=

0

h # 0 by (2.4I.The a f f i n e equation o f $ ( r ) : (z+10h2)2 + 2hx(z+lOh2 ) + h 2 z = 0.

a+8 = h ,where

(2.5)

2

. $ ( r ) becomes

G.Pellegrino and G. Korchmbros

254

$ ( r )m u s t s a t i s f y

n l'l,we have t o determine t h e b) .In o r d e r t o c a l c u l a t e 4 3 2 2 -1 r o o t s of t h e equation 90 +60 +30 +20+1 ' = 3(0+9)2.(30 +30+1)=0 ( O = x h ) ,

Now

which has i n ~ ~ ( 1 o1n )l y the double r o o t 13=2.~hus I@(r) n rl<4;and t h i s i s a con= tradiction. From 2.1.

and 2 . 3 . it follows

2.5. The c o l l i n e a t i o n s o f G' f i x P

;the c o l l i n e a t i o n s of G f i x P

0

I t i s e a s y t o check t h a t any c o l l i n e a t i o n of

n

Y

which f i x e s P

0

0

and Pm.

and l e a v e s Cm i n =

v a r i a n t is e i t h e r x ' = x , z ' = h z , t ' = h

-1

t , o r

x ' = x , z ' = k t , t ' = k

-1

z

where h,k # 0 a r e elements o f GF(11). Such c o l l i n e a t i o n s w i l l l e a v e I ' i n v a r i a n t i f and only i f Y Then,from 2.1. it follows

x,

h(#O)f

k

v

E

.

2.5. G c o n s i s t s o f t h e following c o l l i n e a t i o n s x' = x (2.6)

x' = x

, y'

= uy

, y'

= uy

, 2'

= hz

, z'

= kt

, t' , t'

= h = k

-1 -1

2

( u =1 ; h(#O)E

t

2

z

( u =1 ; k

E

x

)

v).

To e s t a b l i s h t h e a b s t r a c t s t r u c t u r e of G,we consider t h e following c o l l i n e a t i o n s

of P G ( 3 , l l )

h

:

x' = x

, y'

= y

,

z' = hz

I

k

:

x' = x

, y'

= y

,

z' = k t

,

j

:

x' = x , y ' = 1Oy

(2.7)

, z'

= z

t ' = h-'t -1 t' = k z

, t'

= t

.

(

h(#O)f

(

k

x)

E V)

I t i s easy t o v e r i f y t h a t

H ={h Ih(#O)E K = { k Ik

E

XI

i s a c y c l i c group o f o r d e r 5;

v } i s a s e t containing 5 i n v o l u t i o n s ;

H U K i s a d i e d r a l group of o r d e r 10;

j commiltes with b o t h h and k. From t h i s it follows 2.6.

i s o f o r d e r 20.

MOreover,since every element of

2.7.

G =

i s o f t h e type (2.6),we conclude

.

Now we s h a l l determine t h e subgroup U

( i = 1 , 2 , 3 , 4 ) o f G which f i x e s D

i

I t i s easy t o v e r i f y t h a t

u

1

=

u = IlOj,l} 2

;

u

3

=

u

4

= (13

i'

255

Translationplanes of order 112 where,according to (2.71,

lOj

: x'

1

= x

, y'

= 1Oy

:x"x,y'=y

, z'

= 10t

, z ' = z

, t'

= 1Oz

. t ' = t .

Moreover j exchanges D

w i t h D and D with D 1 2 3 4' W e summarize t h e r e s u l t s o f t h e p r e s e n t s e c t i o n with t h e following

Theorem 1 : The coltineation group G of PG13,ll) which maps Q onto i t s e l f and leaves C invariant i s the direct product o f the group {I, j l with the diedral

group {h,k

I h(#OE

x,k

E

v } o f order 10.

The subgroup ui (i=1,2,3,4) of { l , l O j ) when i = 1 , 2

G

which leaves D invariant i s :

; { l }when

There is a collineation i n

G

i = 3,4.

i

exchanging D with D and D 1

2

oith D

3

4

e s h a l l use t h e B-ruck's i n c i d e n c e , a s d i s c u s s e d in ([2]n.4).The 3.- W

.

following

is a summary.Suppose q be odd.Denote by s a non-square element o f GF(q) and l e t 2 2 2 = {a+ibl a , b c G F ( q ) , i = s} GF(q) c GF(q ),where GF(q 2 For m y element o f GF(q ) d e f i n e , a s usua1,the norm n ( a + i b ) p u t t i n g n ( a + i b ) = 2 = a2+b .Then n ( a + i b ) E GF(q) f o r a l l a+ib.We d e f i n e a norm set f o r a l l p E GF(q) :

.

I

N ( p ) = {a+ib n(a+ib)=p}.

-

The miquelian i n v e r s i v e p l a n e I P ( q ) is defined as follows.The p o i n t s are t h e ele= 2 2 ments of GF(q ) and a symbol c.GF(q ).The circles are o f two types:each circle o f type 1 h a s t h e equation (a)

{ a + i b +h(c+id)}Um

where h ( c + i d ) denotes t h e s e t { h ( c + i d )I A

E

GF(q)}

;

each c i r c l e o f type 2 has equation

(0)

.

a+ib + N(P)

L e t us consider t h e map Sa which a s s o c i a t e s t h e l i n e o f PG(3,q)

(3.1)

to t h e p o i n t = o f I P ( q )

(3.2)

,(0,1,0,0)]

[(l,O,O,O) ;

and t h e l i n e

[ ( a , s b , O , i ) , (b,a,i,O]

to t h e p o i n t a + i b o f I P ( q ) .

By Bruck's theorem (see [2],th.4.3),the

l i n e s (3.1) and (3.2) form a r e g u l a r

spread F o f PG(3,q)j Sa is an incidence p r e s e r v i n g isomorphism between I P ( q ) with

i t s p o i n t s and c i r c l e s and F with i t s l i n e s and r e g u l i . Each s u b s e t Ra of t h e form

G.Pellegrino and G. Korchmdros

256

R : [ ( (a+Xc),s (b+Xd),O, 1) ,( (b+Xd),(a+Xc)I 1,0)] u [ (1,O,O,O) ,(0,l,O,O)] a (A E GF(q)) is a regulus of F .In fact n maps the circle (a) of type 1 to R (3.3)

.

Also each subset R

of the form B R :~~((a+c),s(b+d),O,l),(b+d),(a+c),1,0~] I c+id E N(p)) (3.4) 8 is a regulus of F .In fact Sl maps the circle (6) of type 2 to R B ’ Following 12,th.4.41 ,the opposite regulus R ’ (resp. R‘ of R B is given by (3.5) (3.6)

R’ :[((a+Xc),O,s(b+Xd),l),((b+Xd),l,(a+Xc),0~] R’

*

B ’

(resp. R

B

U ~~O,OfOfl~,~lfOIO,O~]

(q-1)/2 {[(a+c,bs+bs (q+1)’210, 1) I (b+d,a+cs lllo)] Ic+id

E

N(P))

It is well known that the Geometry of the plane sections of an elliptic quadric is isomorphic to inversive plane IP(q).This isomrphism,with the notations introo

duced in n.l,can be obtained,as usua1,by the stereographic projection of Q from the centre (0,0,1,0)onto the plane n

with equation z = 0.

We apply these results to our case,where we assume s = 10. First of al1,we give the table of N(p) ( p E GF(11) Table 5 N(O) = {O}; N( 1) = { 1 ,10 ,iI 101,3+51,3+6i ,8+5iI8+6i,5+3il5+8i,6+3i,6+8il

;

N(2) = { l + i , l + ~ ~ ~ , ~ ~ + ~ , ~ ~ + ~ ~ ~ l ~ + ~ ~ f ~ + 8 ~ , ~ + ~ ~ N(3) = {5,6,5i,6i,4+3i,4+8i,7+3il7+8i,3+4i,3+7i,8+4i,8+7i~ ; N(4) = {2,9,2i,9i,1+5i,1+6i,10+5il10+6i,5+if5+10il6+if6+10i~ ; N(5) = { 4 , 7 , 4 ~ , 7 ~ , 1 + 2 ~ , 1 + 9 ~ , 1 0 + 2 ~ , 1 0 + 9 ~ l 2 + ~ f 2 + 1 0 ~;l 9 ~ ~ l 9 ~ 1 0 ~ ~ N(6) = {5+5i,5+6i,6+5i,6+6ill+2ifl+9iflO+2iflO+9il2+il2+lOil9+i,9+lOi~ ; N(7) = {3+3i,3+8i,8+3i,8+8i,5+21,5+9if6+2i,6+9i,2+5i,2+6i,9+5i,9+6i~ ; “8)

= ~~+~~,2+9i,9+2i,9+9i,4+5il4+6i,7+5i,7+6il5+4i,5+7i,6+4i,6+7i~

N(9) = {3,8,3i ,ail2+4i,2+7il9+4i,9+7i,4+2i,4+9iI7+2i,7+9iI N( 10)= C4+41,4+7il7+4i,7+71,1+3i ,1+8i,10+3i ,10+81,3+i ,3+10i,8+1,8+1011 The chain of circles C ,defined in n.1,gives rise to the following chain C’ of IP(q) (3.5) does not hold.In this case one can determine R’ in a U ~~l,O,O,O),(O,l,O,O)] the following way.The lines of Ra are [~X,0,0,1~,~0~,1,0)] ( 4 ) For a=b=d=O,c=l

and belong to the quadric with equation xz=yt;since this quadric contains R‘ a

Translation planes of order ] I 2

251

n

By (3.3) and (3.4) we are able to calculate the reguli corresponding in circles of the chain

c

.Thus

F'' is the union of the lines which form these regu=

li.Similarly,from (3.5) and (3.6) we obtain the opposite reguli R' of R i i' Table 6 R~ + = n t ,y=nz I n=0,1,..,10)~{t~O,z=o} =Ix=nz ,y=nt I n=0,1,..,10)u{zEo,t=o}

R

;

;

..,101U

R1 ={x+lOny+ (2+6n)z+(5+9n)t=O ,nx+y+(5+9n)z+ (9+5n)t=O I n=O, 1 , U{lOy+6z+9t=O R

2

I

x+9z+St=O)

={x+lOny+(6+7n)z+(4+5n)t=O

3

;

,nx+y+(4+5n)z+(5+4n) t=OIn=O,1, ..,iO)u

U { lOy+7z+5t=O ,x+Sz+4t=01 R

;

={x+l~ny+(7+10n) z+ ( 1+4n)t = ~,nx+y+ ( 1+4n)z+ (4+n)t = In ~ = I ~1 I

u { 10y+10z+4t=n ,x+.l;..+t=OI

;

I

..,10 I u

..

R~ =Ix+l~ny+( 10+8n)z+ (3+n)t = ~,nx+y+ (3+n)z+ ( 1+3n)t = ~ n=O ,I, ,Io

u

, x+z+3t=O) ={x+i~ny+( 8 + ~ nz+ ) (9+3n)t=O ,nx+y+ (9+3n)z+( 3+9n)t = I ~n=0 ,I,..,10 )U u{lOy+8~+3t=O

R

5

U{ 10y+2~+3t-O

R; ={x=ny

,

.

,x+3z+9t=O1 ;

t=nzl n=O,i,. ,101

u {y=o ,

R' ={x=ny ,z=lontl n=0,1, ..;IOIU

R' ={x+lOny+8nz+3t=O 1 R I ={x+lOny+Znz+gt=O 2 R' =(x+lOny+6nz+St=O 3 R' ={x+i0ny+7nz+3y=0 4 R' ={x+lOny+lOnz+t=O 5

t4 the

Z=O}

;

I y=o , t=o)

;

.,101 U {lOy+8z=O ,x+7t=0) ,nx+y+10z+10nt=oI n=O, 1 ,..,IO)U 1oy+2z=o ,x+lOt=O1 ,nx+y+8z+8nt=oln=O,1,. .,10) U { y+5z=O ,x+8t=0) ,nx+y+7z+7nt=01n=Of1,.

.,

u{ .,lo] U

,nx+y+7z+~nt=0ln=0,I,. 101 ,nx+y+6z+6nt=0

I n=O, 1,.

lOy+7z=O ,X+2t=O) lOy+lOz=O x+6t=01

Let us consider the half reguli given in the following table

; ; ;

;

.

G. Pellegrino and G. Korchmciros

258

Table 7

4

R'

= {x=ny ,t=nz In=0,2,5,6,9)U(y=O

R'

= {x=ny ,z=lOt

.j R; R; R;

4

4

R' 4 R'

,z=O)

;

I n=1,3,4,7,8,10} I

= {x+l0ny+~nz+3t=0 ,nx+y+7z+7nt=0

= (x+l0ny+~nz+3t=0 ,nx+y+lOz+lont=o

I

= {x+lOny+6nz+5t=O ,nx+y+8z+8nt=O = (x+10ny+7nz+3t=0 ,nx+y+7z+~nt=0

I

n=lf3,4,7,8,10}

I

n=1,3,4,7,8,10}

n=1,3,4,7,8,10}

; ;

n=1,3,4,7,8,10}

;

.

= {x+lOny+lOnz+t=O ,nx+y+6z+6nt=0 1 n=1,3,4,7,8,10} 5 It can be verified,by some calculations which are long but easy,that these half

reguli form

a partial spreadG' which covers the same points as

Similarly the conic

(3.8)

F'

.

D ,defined in n.l,gives rise to the circle

i

-

D' * 3i+N(8) ; D' 8i+N(8) ; ' 1 ' 2 ' D' * 8+3i + N(8) ; D' : 8+81 + "8) 3' 4

From (3.3,4,5,6) we obtain the regulus T: and its opposite regulus T',given by I i the following Table 8

T = 1 T

2

(Zn+lO)x+(5n+9)y+z+nt=O

,(3n+lO)x+(4n+3)y+(10nt

ln=0,1,..,10} u IZx+Sy+t=o , ~ x + ~ ~ + I o z + I o ~;= o }

= {(5n+9)x+(Zn+lO)y+nz+t=O ,(4n+3)~+(3n+lO)y+(lOnt

/n=0,1,..,Io~uI ~ x + ~ Y + z = o, ~ X + ~ ~ + I O Z + I O ~; = O ) T = 3

I (n+10 x+ (n+1

y+ 1Oz+2n t=O

,(n+1

x+ ( ion+ 1 y+gnz+lo t=O

I

In=o,l,..,l~Iu {x+y+Zt=O ,x+loy+9z=o1 ; T = { (n+l)x+ (n+10)y+Znz+lot=O ,(n+10)x+ ( 10n+l)y+z+2nt=o 4 Jn=O,i...,10} u {X+Y+ZZ=O ,x+10y+2t=0} ; Ti = T' = 2

( 3n+9)x+ (4n+61 y+lOnz+ ( 10n+10)t=O

T' = 4

,(n+10)x+ (8n+9)y+ ( l+n)z+10t=O I

ln=O,1, ..,lo} u { ~ X + ~ ~ + I O ~ + ,x+8y+z+lot=O) IO~=O ; (4n+6)x+ ( 3n+9 y+ ( 10n+ 10)z+lOnt=O ln=0,1,..,10}

T' = 3

I

u

, ( 3n+2 ) x+ ( l+lOn)y+nz+ ( 10n+10 ) t=O I

( ~ X + ~ ~ + I O Z + I ,O ~~x=+O~ o ~ + z + I o ~ =; . o }

I (n+10)x+ (n+l)y+lOz+Znt=O ,(n+l)x+ ( 1+10n)y+gnz+lot=O I

In=O, I,..,101

u

Ix+y+Zt=O ,x+lOy+9z=d} ;

I (n+l)x+(n+1O)y+1Onz+10t=0 ,(l+lon)x+(n+l)y+2z+9nt=ol Jn=O,1,.

.,101 u {x+y+loz=o , lox+y+gt=O) .

259

Translation planes of order 112 We conclude this section by pointing out

Theorem 2. The f o l l o w i n g line subset are not regular spreads of P G ( 3 , l l ) IF-F'IU G'

i )

Ti)) U (G'U T') (i=1,2,3,4) i (F-(F'U TiU Ti+l)) U (G'U TI1 U T !1+1

ii)

(F-(F'U

iii)

(

i=I or i = 3 ) .

4.- We proceed to give the collineation group W of PG(3,ll) which maps

F onto

itself and leaves

F' invariant.We recall that F' is the union of the reguli R

(i-,O,l,..,S).We

put

(4.1) h ' : x'=x (4.2)

k':

x'=lOkz+4kt

(4.3) j': x'=y (4.4) m':

, y'=y , z'=hz

, y'=x

x'=mx+lOy

Let r be any line of (4.5)

,t'=ht

(h(#OlE

x)

, y'=7kz+lOkt , z'=lOx+7y , , z'=t , t'=z

, y'=x+my , z'=mz+t

i

t'=4x+lOy

,t'=lOz+mt

(m

E

(k

E

v)

GF 11))

F .Firstly we suppose that r is given by (3.1) .Then

h'(r) = j'(r) = r

.

Next suppose that r is given by (3.2),then (4.6)

-1 -1 h'(r) = ~ a h - l , l O b - l , O f l (bh ) f ,ah ,1,0)]

(4.7)

j'(r) = [(a,b,O,i), (10b,a,l,O)]

k'(r) is the line passing through the points (4k,10kf10+4b,4a+b) and ( 10k,7kf7a+10b, 10a+4b) .Therefore

[~0,0,0,1~,~0,0,1,0~]when (4.8) k'(r):

r =[Cl,O,O,O),CO,l,O,O)]

,

[(1,0,0,0), (0,1,0,0)] when r =[(0,0,0,1),(0,0,1,0)] , 2 2 -1 2 2 -1 2. 2 -1 2 2 -1 [(ak(a +b ,10b(a +b ) ,O,l),(bk(a +b ) ,&(a +b 1 ,1,0)] when

r =[(a,lOb,O,l),(b,a,l,O)]

(a,b#O).

By (4.4) m'(r) is the line passing through the points (ma+b,a+lOb,l,m) and (mb+lOa,b+ma,m,lO).Therefore (4.9) (4.6)..(4.9)

{I',m'l

m

E

m'(r) = r

show that h',j',k',m', GF(ll)}

. maps

F into itself.It is easy to prove that

is a cyclic group of order 12.By (4.5) and (4.9),this group

fixes each line of F. As the group of all the collineations of PG(3,q) fixing each line of a spread of

PG(3,q) as order at w s t q+l,it follows that {l',m'Im

E

GF(11)) is the full col=

G.Pellegrino and G. Korchmdros

260

l i n e a t i o n group of P G ( 3 , l l ) f i x i n g each l i n e o f F.By Bruck's r e s u l t (see n.3)

a

(1,0,0,0),(0,1,0,0)

:

(a,lOb,O,l),(b,a,l,O)

* *

(0, O , l , O ) 2 2 (a,b,a t b , I )

is an incidence map p r e s e r v i n g isomorphism between F w i t h i t s l i n e s and r e g u l i

and Q with i t s p o i n t s and circles.

I',m'l

NOW,^^ {

m

E

G F ( I I ) } is the f u l l c o l l i n e a t i o n group of ~ ~ ( 3 ~ f1i x1i n)g

each l i n e o f F ,we have

W/{l',m'

I

m

GF(ll)}=G ,

E

where G i s the full c o l l i n e a t i o n group of P ~ ( 3 , l l )which maps Q o n t o i t s e l f and l e a v e s C invariant.By theorem 1, G = .Now i t i s easy t o v e r i f y t h a t < h',jr,k'7

2:

,

Furthermore any c o l l i n e a t i o n g ' ( E < h ' , j ' , k ' > ) acts o n F i n the same way as the

corresponding g ( e < h , j , k > ) acts on Q;i.e. $2

:

g'(f)

W/Il',m'lm

It follows t h a t Since m'

* g(P)

i f and o n l y if E

m

E

.

~ ~ ( 1 1 *) )< h ' , j ' , k ' >

commutes with h ' , k ' , j '

following groups: { l ' , m ' l

f

:

*

P

(f

E

F,P

E

Q).

, w e have t h a t W is t h e d i r e c t product of t h e

GF(11)) with < h ' , k ' , j ' >

.From Theorem 1 w e have

also t h a t < h r , k ' , j l > is t h e d i r e c t product of { l ' , j ' l w i t h {h,'k'lh(#O)E X,k A s o u r n e x t step we s h a l l determine t h e subgroup Vi

invariant.As i n

ves T

i F as U

i

on Qlwhere U

i

n t h e regulus

=v is

1

2

Similarly

corresponds t o t h e c o n i c D ,V a c t s on i i is t h e subgroup o f G l e a v i n g D i n v a r i a n t (see n . 1 ) . T

i

i

the d i r e c t product o f t h e group

v 3=V 4

vl

( i = 1 , 2 , 3 , 4 ) of W which l e a =

From t h e above p r o p e r t i e s o f W and by theorem 1,we have t h a t V re V

E

is the group

{l',m'lm

E

{l',lO'j'}

,

z'=4x+lOy

,

= V

2

with { l ' , m ' l m

;furthermo= E

~ ~ ( 1 1 ) )

GF(11)) where,with e a r l i e r n o t a t i o n , l '

i s t h e i d e n t i c a l c o l l i n e a t i o n and

1 0 ' j ' : ~ ' = 4 Z + t, y ' = z + I t

1

.

tm=10x+7y

j exchanges D with D , a s w e l l as D with D , i t follows #at j' exchanges 1 2 3 4 T with T and T with T 2 3 1 4' W e can summarize t h e above d i s c u s s i o n by t h e following Since

Theorem 3. The c o l l i n e a t i o n group W of PG(3,ll). which mps F onto i t s e l f and lea=

ves F ' invariant i s t h e d i r e c t p r o d u c t of the following groups { l ' , j ' I ; {l',rn'l

m

E

GF(11)); {h',k'l

h(#O)

E

Xrk

E Y

1

The f i r s t kjo groups are c y c l i c of order 2 and 12 respective1y;the t h i r d group

Translation planes of order 112

261

i s diedral of order 10. The subgroup V ( i = l , 2 , 3 , 4 /

of W which leaves T invariant i s given:

i

i

by the d i r e c t product o f the groups { l ' , l O ' j ' l bythegroup { l ' , m ' I m

and { l ' , m ' I m

E

eF(11)) when i=1,2;

~ ~ ( 1 1 when ) ) i=3,4.

E

There i s i n W a cotlineation exchanging T with T and T with T 2

1

4

3

'

5.-We shall establish the collineation group Z of PG(3,ll) leaving the spread

IFF') U G' invariant.By Bruen's result (see 121 ,th3.5) ,Z is the subgroup of W which maps G' onto itself.It is easy to check that h ' ( G ' ) Moreover m'

A

leaves

= G'.

invariant if and only if m belongs to the following set

i;'

.

={0,2,5,6,9)

= G' and j'(G')

Now we prove that k' does not leave R'=(ROb,R',R',R',R',R',R'} invariant;from 0 1 2 3 4 5 this it will follow that k'(G') # G'. In order to do it,firstly we consider the collineation 10' of k'.mw [(6,9,0,1), (9,6,1,0)] in 1(7,6,0,1), (6,5,1,0)] belong to R';so

is an h in

x

#

lO'(R')

{h',j'I

h(#O)

E

XI

and

;and the latter line does not

we have k'=lO'h' since for every k in v there

R'.Next

such that k=lOh.Then,as h'fR')=R',it

By theorem 1 ,
10' maps

Ih(#O)E X,m

{l',m'Im

E

E

A}>

follows k'(R')

# R'.

is the direct product of the groups

A.).Therefore we obtain

Theorem 4 . The collineation group z o f P G ( 3 , l l ) leaving the spread IF-F')u G' invariant i s the d i r e c t product o f the following c y c l i c groups {1',,7"1 ; C l ' , m ' l m

E

A},

Ih'lh(Z0)

E

XI ,

of order 2,6 and 5 respectively. Let L (i=1,2,3,4) be the subgroup of Z which leaves the spread fF-(F'u Ti)) u i U(G'UT') invariant.As this spread is obtainable from IF-F') UG' by reversing i

T by its opposite regulus T ' , L i

i

i

is the subgroup of V

i

which leaves T

i

invariant.

Then,from theorem 4 it follows

of z which leaves the spread fF-(F'u T )) U

Theorem 5. The subgroup Li(i=l,2,3,4) U(G'UT:)

i

invariant i s :

the d i r e c t product of the groups { l O ' j ' , l ' l

and { l ' , m ' l m

the group { l ' , m ' ] m

E

Finally,let L

(i=1 or i=3) be the subgroup of L

i,i+l

(F-(F'UTiUTi+l))U

A}

E

8)

when i=1,2;

,when i=3,4.

(G'UT;UT'i+l)

invariant.

i

which leaves the spread

262

G. Pellegrino and G. Korchmbros

As t h i s spread

is o b t a i n a b l e from ( F - ( F ' U

i t s o p p o s i t e regulus T' L i+l' i , i + I riant.From t h i s w e have

Theorem 6. Let L

i'i+l

T ) ) U ( G ' u T') by r e v e r s i n g T i i i + l by is t h e subgroup o f L which l e a v e s T inva=

i

li=l or i = 3 / be t h e subgroup of

( F - t F ' u T.u T ~ + ~ )U)tG'

u T; u T'

) i+l

invariant.Then

L

i+l

i

L

which Zeaves t h e s p r e a d

i,i+1

= L

i'

6.- From any spread F o f PG(3,q) w e o b t a i n a t r a n s l a t i o n p l a n e 11 o f order q

2

i n the following way (see [ l ] ,e.g.) .Let S=PG(4,q) and p u t C=PG(3,q) .The p o i n t s of 11 a r e those of s-2

; t h e l i n e s o f Il are t h e p l a n e s o f S which meet C i n a li=

ne of F and do n o t belong t o C . I f w e consider S-C a s a v e c t o r space of dimen= s i o n 4 over GF(q),the p o i n t s o f 11 are v e c t o r s of S-C and t h e l i n e s o f 11 through t h e zero v e c t o r 0 (sometimes c a l l e d components) are c e r t a i n 2-dimensional sub= spaces;the o t h e r l i n e s are t r a n s l a t e s of t h e components.When q is prime,in this r e p r e s e n t a t i o n t h e c o l l i n e a t i o n s f i x i n g 0 a r e non s i n g u l a r l i n e a r transforma= t i o n s on t h e v e c t o r space.This is a s t r i c t l y a f f i n e r e p r e s e n t a t i o n .

Every c o l l i n e a t i o n of 11 is a product o f a t r a n s l a t i o n w i t h a c o l l i n e a t i o n f i x i n g t h e p o i n t O.The s t a b i l i z e r o f 0 i n the c o l l i n e a t i o n group of II w i l l c a l l e d t r a n = s l a t i o n complement o f 11. We w i l l study the t r a n s l a t i o n planes a s s o c i a t e d to t h e spreads,given i n theorem 3 , i n terms o f i t s t r a n s l a t i o n complement. F i r s t o f a l l we d i s c u s s t h e plane 11 a r i s i n g from t h e spread I F - F ' I U G ' . The t r a n s l a t i o n complement A

In

of II c o n t a i n s t h e following c o l l i n e a t i o n s

h j

:

t ' = 5 , r l a = ~

, T ' =

T

:

c a = q , ~ a = 5, G ' = T

,'I.=

6

I!

:

5 ' = mE+lOn, rlc= t+mn

facthrJi,m a c t

6' = me

+T

,

T * = 1oc+mT

on E i n t h e same way as h ' , j r I r n ' and by t h . 4 t h e l a t t e r c o l l i =

neations leave ( F - F ' ) u G' i n v a r i a n t . I t is easy t o v e r i f y the following f a c t s :

{h,dlh(#)

lm

E

E A}

XI

2

{h',j'lh(#O)

E

XI

is not a group f o r S-I

d e r 60 which contains t h e group D

, but

<{zlm

E

A}> is a c y c l i c group o f or=

o f the d i l a t a t i o n s whose c e n t r e is 0.

Since,by theorem 4 , t h e f u l l c o l l i n e a t i o n group Z o f C which l e a v e s f F - F ' I U G' i n v a r i a n t is t h e d i r e c t product of ~ l ' , m ' ~ mE A } with {h',j'Ih(#O)

E

xIrwe have

263

Translation planes of order 112 t h e following

Theorem 7. The translation complement of the translation plane ll ,arising from the spread

(F-F') U G '

i s the d i r e c t product of a diedral group of order 10 with

a cyclic group of order 60 containing the d i l a t a t i o n group. Next we d i s c u s s t h e t r a n s l a t i o n plane ll. (i=1,2,3,4) a r i s i n g from t h e spread ,

1.

t h e t r a n s l a t i o n complement o f II i 0 i' We f i r s t l y p o i n t o u t t h a t II 2 II and ll 2 114 ,according t o t h . 3 and ([2]th.3.1). 1 2 3 As (F-(F'U T )) U (G' U T') is o b t a i n a b l e from (F-F') U G' by r e p l a c i n g T with i i i i t s opposite regulus T;,lli is o b t a i n a b l e from II by d e r i v a t i o n . I t is well known (F -(Flu Ti)) U (G'U T ' ) .Denote by

( s e e [4] plement A

)

A'

t h a t i n t h i s case A i is the i n h e r i t e d group o f t h e t r a n s l a t i o n com= of ll . I n particularOAi

is a subgroup o f A

.According to theorems

4,5 and ? , w e have

Theorem 8. The translation complement of n (F-IF'u Ti)) U (C'U T;)

i

(i=1,2,3,4)

arising from the spread

i s given by the d i r e c t product of the group (1,Xl with

a cyclic group of order 60 containing the d i l a t a t i o n group. F i n a l l y we d i s c u s s t h e t r a n s l a t i o n p l a n e ll (i=1 o r i=3) a r i s i n g from t h e i,i+l i,i+l ).Denote by A )) u (G'U T ' U T' the t r a n s l a t i o n comple spread (F-(F'U T U T i i+l i i+l ment of ll .Since (F-(F'lJ T . U T )) U (G'U T ' U T ' ) is o b t a i n a b l e from 1 i+l i i+l i,i+l with i t s opposite regulus T' (F-(F' U T )) U (G'U TI) by r e p l a c i n g T I T[i+l i i i+l is o b t a i n a b l e from II by derivation.As abOve,we t h e n have t h a t A i r i + t + : s a i 0 group o f Ai .By theorems 6 and 0 it follows

Theorem 9. The translation complement of ll spread

(F -(Flu T

i

u Ti+l)) U

i,i+l

(G'u T;UT;+~)

(i=l or i=31 arising from the

i s the same as i n theorem 8 .

From o u r d i s c u s s i o n we have t h a t n,II 1l II3 lll 1211134 a r e n o t p a i r w i s e isomorphic.

G.Pellegrino and G.Korchmaros

264

BIBLIOGRAPHY

1 1 1 A.BARL-I

,Representation

and c o m t m c t i o n of projective planes and other

geometric structures from projective spaces, Jahresber.Deutsch.Math.Veren.,77 (1975),243-38. 121 A.A.BRUEN,InVePSiVe geometq and some new translation planes I,&om. Dedic.,7, (1977),81-98 131 A.A.BRUEN-J.A.THAS,

~Zock,chainaand configurationa in finite geometries,

Atti Accad.Naz.Lincei,Rendiconti,(8)

59 (1975),744-748.

[a] P .DEMBOWSKI ,Finite Geometries ,Springer Verlag,Berlin-Heidelberg-New York 1968. 151 G.KORCHMAROS,ExampZe of chain of circles Ser.A,31 (1981),98-100.

in PGf3,&),q=7,ll. ,J.Com.Theory

Annals of Discrete Mathematics 14 (1982) 265-282 Q North-Holland Publishing Company

GENERALIZED QUADRANGLES WITH PARALLELISM C l e l i a Somma I s t i t u t o Matematico "G. Castelnuovo", U n i v e r s i t a d i Roma, I t a l y

1.

INTRODUCTION I n t h i s paper we again discuss generalized quadrangles ([lo1

, [111,

[121, [141) and define, f o r the f i r s t time, generalized quadrangles (S, R) w i t h p a r a l l e l i s m , t h a t i s generalized quadrangles such t h a t i n R t h e r e i s an equivalence r e l a t i o n s a t i s f y i n g the Euclidean axiom and such t h a t every trapezium i n (S, R) i s a parallelogram. Some examples show t h a t such a c l a s s o f generalized quadrangles i s nonempty; therefore, study r e l a t e d t o t h i s class i s o f some i n t e r e s t . I n the f o l l o w i n g section n. 3 we show some p r o p e r t i e s which h o l d f o r such a class; i n f a c t , we prove t h a t there e x i s t s a r e l a t i o n between the parameters p and n o f (S, R) ( [ l o ] , 111 1, [121) and thus the c a r d i n a l i t y o f S and the c a r d i n a l i t y o f R are determined by j u s t the parameter p

(cf.

p r o p o s i t i o n V I I I ) . Moreover, we show t h a t no generalized quadrangle w i t h p a r a l l e l i s m i s o f the f i r s t k i n d (see p r o p o s i t i o n I X and 1101 and [ l l l ) . I n s e c t i o n n. 4, t a k i n g some p r o p e r t i e s , which h o l d f o r some generalized quadrangles w i t h p a r a l l e l i s m (see s e c t i o n n. 2) as a s t a r t i n g p o i n t , we introduce two graphic conditions ( c f . (N) and ( N ' ) )

and we show t h e i r

equivalence; moreover, t h e generalized quadrangle (S, R) w i t h p a r a l l e l i s m , i n which one o f them holds, i s o f the second k i n d (see p r o p o s i t i o n X I I ) . Then we show, under t h i s hypothesis, (S, R) i s 1-embeddable i n an a f f i n e space AG(3,p)

o f dimension three and order p; thus, (S, R)

i s isomorphic

t o one o f the " c l a s s i c " generalized quadrangle 1-embedded i n such a space. Indeed, t h e r e e x i s t s an isomorphism between (S, R) and the generalized quadrangle W ( p ) , 1-embedded i n AG(3,p) lines W

and r e l a t e d t o a l i n e a r complex o f

o f the p r o j e c t i v e space PG(3,p) i n which AG(3,p) i s embedded.

I n t h i s way we characterize the generalized quadrangle W(p); moreover, 265

266

C. Somma

using r e s u l t s proved i n 1 3 1 and [ 4 1 , we characterize the generalized quadrangle h h 1-embedded i n AG(3, 2 ) , 2 = P and h > 1, and r e l a t e d t o a ( p + Z)-arc i n t h e h plane a t i n f i n i t y 71 o f AG(3, 2 ) which contains a non degenerate conic.

2.

DEFINITIONS AND EXAMPLES L e t S be a f i n i t e and non-empty s e t o f p o i n t s and R a proper and non-empty

f a m i l y o f proper subsetsof S, c a l l e d l i n e s . Two d i s t i n c t p o i n t s P and Q i n S are c a l l e d c o l l i n e a r , i n which case we w i 1 w r i t e P

%

Q, i f t h e r e e x i s t s a l i n e i n R

hem n o t c o l l i n e a r and w r i t e P $ Q. Further-

containing them; otherwise, we c a l l

more, l e t F denote the s e t o f l i n e s o f R through a p o i n t P o f s. We say the P p a i r (S, R) a non-degenerate generalized quadrangle i f S = U r and the rER

f o l 1owing axioms hold :

(2.2)

[Vr€R,VP€S-r

s € R/ P€ s ,

]*[3!

I r n s l =11;

I n a generalized quadrangle ( [ 101, [ 11 I ) every l i n e has the same P c a r d i n a l i t y p a 3 and every p e n c i l F PY and moreover

(2.6)

IRI = n [ ( p

The integers p and

-

1) ( n

-

E

S, has the same c a r d i n a l i t y n

3 3

1) t 1 1 .

n are c a l l e d parameters o f (S, R).

A degenerate block t ( P ) w i t h pole P i s the s e t o f p o i n t s i n S which are e i t h e r c o l l i n e a r o r c o i n c i d e n t w i t h the p o l e P;

261

Generalized quadrangles with parallelism

i n t e r s e c t i o n of two degenerate blocks w i t h n o t c o l l i n e a r poles and i t i s c l e a r t h a t t h e p o i n t s on a t r a c e are pairwise n o t c o l l i n e a r ([ 101, [ 11 I ) . We remark t h a t i n the l i t e r a t u r e t h e term t r a c e r e f e r s a l s o t o the i n t e r s e c t i o n o f two degenerate blocks w i t h c o l l i n e a r poles. I n the c l a s s i f i c a t i o n o f T a l l i n i

([ 101, [ 11 I ) , (S, R ) i s a generalized quadrangle o f the f i r s t k i n d i f t h e r e

-

i s e x a c t l y one t r a c e throubh two d i s t i n c t n o t c o l l i n e a r p o i n t s P and Q, o f

t h e second k i n d i f any t r a c e l i e s on e x a c t l y two d i s t i n c t degenerate blocks and o f t h e t h i r d k i n d otherwise. An isomorphism between two generalized quadrangles (S, R) and ( S l , R ' ) i s any one-one mapping f : S the l i n e s i n R onto t h e l i n e s i n R ' and such t h a t

f-l : S '

+

+

S' which maps

S maps t h e l i n e s

i n R ' onto t h e l i n e s i n R. We c a l l a generalized quadrangle ( S , R) d-embedded i n a f i n i t e a f f i n e space AG(r,q) o f dimension r and order q i f R i s a proper and n o t empty f a m i l y o f subspaces o f AG(r,q)

o f dimension d, d

> 1. Moreover,

(S, R ) i s c a l l e d d-embeddable i n a f i n i t e a f f i n e space AG(r,q) i f t h e r e i s an isomorphism between (S, R) and a generalized quadrangle d-embedded i n AG(r,q). NOW, l e t (S, R) be a generalized quadrangle whose parameters are P and n; d e f i n e i n R an equivalence r e l a t i o n //,

(2.7)

[V

P

c a l l e d p a r a l l e l i s m , such t h a t

E S, V r E R, P @ r ] -[3!

s E R, P E s , s / / r l

(Euclidean axiom)

The axiom (2.8) i s equivalent t o the f o l l o w i n g

(PI

i n (S, R) any trapezium i s a parallelogram.

We c a l l a generalized quadrangle w i t h p a r a l l e l ism any generalized quadrangle whose parameters are p and n s a t i s f y i n g e i t h e r (2. 7) and (2.8) o r (2.7) and

(P); the equivalence classes, o r p a r a l l e l i s m classes, a r e c a l l e d d i r e c t i o n s

C Somma

268 too.

The study o f generalized quadrangles w i t h p a r a l l e l i s m i s j u s t i f i e d a s l examples do e x i s t . Now, we r e c a l l a f i r s t construction. L e t AG(3d,q) space of dimension 3d,d>l,

be an a f f i n e Galois

and even order q,q=Zh and h a 2 , l e t PG(3d,q) be

the p r o j e c t i v e space i n which AG(3d, q) i s embedded and IT the hyperplane a t d ,P be n = ( q t 2 ) p r o j e c t i v e subspaces i n f i n i t y o f AG(3d, 4). I n IT l e t P1 n o f dimension ( d - 1) such t h a t t h e space j o i n i n g any t h r e e of them i s t h e

,.. .

iperplane n. L e t R ' be t h e family, which members a r e the subspaces o f PG(3d,q) d i = 1, ...,(q + 2 ) , and n o t l y i n g on o f dimension d containing a subspace P i' 'TI. Now, l e t R be t h e f a m i l y o f a f f i n e subspaces which a r e i n t e r s e c t i o n o f AG(3d, q) w i t h each element i n R ' ; i n [14] i t i s proved t h a t t h e p a i r ( S , R ) , U r = AG(3d, q), i s a generalized quadrangle, c l e a r l y d-embedded i n E R AG(3d, 4 ) .

S =

(S, R ) i s w i t h p a r a l l e l i s m : i f two elements

subspace Pi a t i n f i n i t y , we w r i t e x

E

x and y

i n R have t h e same

Y and i t i s c l e a r t h a t t h e r e l a t i o n E

i s an equivalence r e l a t i o n . Moreover, E s a t i s f i e s (2.7) and (2.8).

(2.7):

let

x be any element i n R and l e t A be a p o i n t i n S-x. It i s c l e a r t h a t t h e r e e x i s t s one, and o n l y one, element y i n R through A such t h a t x

E

y.(2.8):

let

x and y be two elements i n R through a same p o i n t A. Take a p o i n t B o f S belonging t o x; there e x i s t s one, and o n l y one, element y ' i n R through B and such t h a t y '

y. The space

c1

j o i n i n g x and y i s o f dimension 2d; y ' l i e s on

a too. Now, l e t A ' be any one p o i n t on y ' ; through A ' t h e r e i s one, and o n l y one element x ' i n R which i n t e r s e c t s y i n a unique p o i n t B'. Then, i t i s c l e a r that IxrIx'(

= 0 ( c f . ( 2 . 2 ) ) and, s i n c e x ' l i e s on a, i t a r i s e s t h a t

X'E

x.

Clearly, tne q u a d r i l a t e r a l ABA'B' i s a parallelogram, hence (P) holds. When d = 1, we o b t a i n the f a m i l y $(q) containing generalized quadrangles, say Q(q), which are 1-embedded i n a

threedimensional a f f i n e space AG(3, q)

o f even order q and r e l a t e d t o a ( q t Z ) - a r c Q i n t h e plane n a t i n f i n i t y o f AG(3, 4) ( [ l l , [ 3 l a [ 4 l a [151). Another fami l y o f generalized quadrangles w i t h para1 1e l ism i s constructed i n the f o l l o w i n g way. L e t AG(3,

(1)

be an a f f i n e Galois space of dimension

t h r e e and order q, q 3 3, and l e t W be a l i n e a r complex o f l i n e s ( [ 91) i n t h e p r o j e c t i v e space PG(3, q) i n which AG(3, q) i s embedded. Take t h e plane a t

269

Generalized quadrangles with parallelism

i n f i n i t y ,n

.,P

of AG(3, 4); the l i n e s o f W l y i n g on ,n

are l i n e s through a p o i n t

Now, l e t R ' be the f a m i l y c o n s i s t i n g o f the l i n e s i n W which are n o t con-

tained i n ,n

and o f t h e l i n e s i n PG(3, q) n o t l y i n g on

plane i n the p o i n t P;,

T,

and meeting such a

i f we i n t e r s e c t each element i n R ' w i t h AG(3, q),then

we o b t a i n a new f a m i l y R. The p a i r ( S , R ) = W(q), S = U r = AG(3,q), i s a *R generalized quadrangle which i s 1-embedded i n AG(3, q) ([ 1 I, 31, 141, [ 151). It i s shown i n [ 31 and [ 41 t h a t t h e r e e x i s t s an isomorphism between the

generalized quadrangles W(q) and Q ( q ) , i f the order q i s even and i f t h e ( q t 2)-arc Q contains a n o t degenerate conic o f n ([ 91). Thus, by such an isomorphism, W(q) i s a generalized quadrangle w i t h p a r a l l e l i s m , i f q i s even. However, we are able t o show W(q) i s s t r a i g h t w i t h p a r a l l e l i s m f o r any order q. L e t E be the f o l l o w i n g r e l a t i o n i n R: x

e i t h e r X,

E Y

= ,Y

or

X,(

P,)

, ,Y

a r e d i s t i n c t and

c o l l i n e a r p o i n t s i n ,n

i f x and y are elements i n R, ,X and

and Y,

are the p o i n t s a t i n f i n i t y r e s p e c t i v e l y

,P i s the p o i n t o f nb, which above we defined. It i s r o u t i n e matter t o show t h a t the r e l a t i o n E i s an equivalence r e -

l a t i o n ; furthermore, (2.7) and (2.8) hold. L e t x be a l i n e o f R whose p o i n t o f i n f i n i t y i s P, and l e t A be a p o i n t i n

S n o t l y i n g on x;

then t h e r e e x i s t s

one, and only one, l i n e x ' through A whose. p o i n t a t i n f i n i t y i s c l e a r l y , i t i s the o n l y one l i n e i n R such t h a t the p o i n t ,X

x

E

XI.

a t i n f i n i t y o f the l i n e x i s d i s t i n c t from P,

belonging t o S-r,

take the planes through the l i n e X(,

),P

,P and,

If, a t t h e contrary, and A i s a p o i n t d i s t i n c t from

~ .,l

L e t a be t h e o n l y one o f these planes which contains the p o i n t A . I f x l i e s on a , then t h e l i n e x ' through A, whose p o i n t a t i n f i n i t y i s X,,

i s a line

i n R, c l e a r l y t h e r e belongs t o a and i t i s the o n l y one l i n e i n R such t h a t x

E

XI.

A t l a s t , i f x does n o t l i e on a , then t h e r e e x i s t s on such a plane

an o n l y one l i n e y i n R through A, whose p o i n t a t i n f i n i t y i s , ,Y

pole o f a

i n t h e symplectic p o l a r i t y ([ 9 ] ) , which defines the l i n e a r complex o f l i n e s W. I t i s c l e a r t h a t t h e p o i n t Ym belongs t o and

,P

hence x

E

y

T,

and i t i s c o l l i n e a r w i t h ,X

and (2.7) holds.

Next, we show (2.8), t h a t i s ( P ) , holds. L e t x and x ' be two d i s t i n c t

270

C Somma

lines i n

R such t h a t x

E x ' . I f ,P

i s the p o i n t a t i n f i n i t y o f both x and x ' ,

R intersecting x '

so through any p o i n t Y on x t h e r e i s an o n l y one l i n e y i n

i n a p o i n t A. O f course, the l i n e y l i e s on the plane a j o i n i n g the l i n e s x and x ' ; moreover, the p o i n t a t i n f i n i t y ,Y

o f y i s d i s t i n c t from P,

and i t

i s the p o l e o f the plane a . L e t Y ' be a p o i n t on x d i s t i n c t from Y ; the l i n e y' i n

R through Y ' and i n t e r s e c t i n g x ' i n a p o i n t A ' , does n o t l i e on plane

c l e a r l y i t has Y- as i t s p o i n t a t i n f i n i t y and then i t i s such t h a t y

EY'.

Hence, t h e trapezium Y Y ' A ' A i s a parallelogram. I f the l i n e s x and x ' a same p o i n t a t i n f i n i t y , ,X

,X

# P,,

a,

have

then the plane a j o i n i n g them has ,X

as i t s pole. Moreover, a l l the l i n e s through any p o i n t on x i n t e r s e c t i n g x ' there belong t o a and t h e i r p o i n t a t i n f i n i t y i s ;P,

thus, such l i n e s a r e

pairwise p a r a l l e l . Then, a l s o i n t h i s case, each trapezium having two opposite sides on t h e l i n e s x and x ' i s a parallelogram. Now, we o n l y must consider t h e f o l l o w i n g case. L e t the p o i n t a t i n f i n i t y ,X

and Xb, o f the two l i n e s x

and x ' be d i s t i n c t , n o t coincident but, o f course, c o l l i n e a r w i t h , P, is x

E

x'.

as i t

Clearly, we remark i n t h i s case x and x ' are two skew a f f i n e l i n e s .

As everybody knows, t h e l i n e s o f a l i n e a r complex which i n t e r s e c t two skew l i n e s o f the l i n e a r complex c o n s t i t u t e a regulus o f a quadric. I t f o l l o w s t h a t the l i n e s o f W i n t e r s e c t i n g both x and x ' c o n s t i t u t e a regulus o f a quadric

Q; such regulus contains the l i n e X(,

X.);

Then t h e quadric Q meets

in a

T ,

conic, which s p l i t s i n the l i n e (Xm, XA) and i n a l i n e r. The l i n e X(,

) ;X

belongs t o the above defined regulus, hence a l l t h e l i n e s i n t h i s regulus i n t e r s e c t t h e l i n e r, t h a t i s they p a i r w i s e are i n r e l a t i o n

E.

Then, t h e r e

f o l l o w s each trapezium having opposite sides on x and x ' i s a parallelogram. Furthermore, we remark the generalized quadrangle W ( q ) i s o f t h e second kind, t h a t i s any t r a c e i n W ( q ) l i e s on e x a c t l y two d i s t i n c t degenerate blocks ([ 101, [ 11 I ) . L e t A be any p o i n t i n AG(3,q) and l e t a be t h e a f f i n e p o l a r plane o f A; the degenerate block t ( A ) contains a l l the p o i n t s on a i n AG(3, q) and the a f f i n e p o i n t s on t h e l i n e j o i n i n g A w i t h . P,

Now, take any

p o i n t A ' , n o t c o l l i n e a r w i t h A; l e t a' be the a f f i n e p o l a r plane o f A ' . i s clear t h a t the a f f i n e l i n e complex o f l i n e s W and P,

It

r = a n a' does n o t belong t o t h e l i n e a r

i s n o t t h e p o i n t a t i n f i n i t y o f r; thus, r

does n o t

belong t o W(q). Therefore, the t r a c e 1 = t ( A ) n t ( A ' ) contains the l i n e r and two p o i n t s B and B', n o t l y i n g on r, which r e s p e c t i v e l y are t h e i n t e r s e c t i o n s

27 1

Generalized quadrangles with parallelism

o f a and a ' w i t h the l i n e s through A ' and through A, having P, as t h e i r p o i n t a t i n f i n i t y . I f t h e r e e x i s t s a p o i n t A", plane a " , p o l a r plane o f A", p o i n t between

such t h a t 1 C t ( A " ) , since t h e a f f i n e

contains the l i n e r and a t most an o n l y one

B and B', then necessarily e i t h e r

either A"=A', or A"=A

a"

a ' , o r a" = a.

Thus,

and W(q) i s o f the second kind, c l e a r l y .

Moreover, we remark what f o l l o w s . The l i n e r above seen l i e s on e x a c t l y q + l planes, and o n l y one plane o f these, say l i n e s , whose p o i n t a t i n f i n i t y i s . P,

T,

contains a p e n c i l o f p a r a l l e l

Take any one o f the o t h e r q planes, say

l e t Ai be t h e pole o f ai. The p o i n t s Ai, i = l,...,q, l i e on an a f f i n e i' l i n e s, which c l e a r l y does n o t belong t o W(q); furthermore, i f X and X ' a r e

c1 *

any two p o i n t s belonging t o s, then i t i s easy t o show t h a t t h e t r a c e t ( X ) n t ( X ' ) contains the l i n e r. Then, the generalized quadrangle W(q) s a t i s f i e s such properties; t h i s remark suggested the study o f generalized quadrangles w i t h p a r a l l e l i s m , which s a t i s f y a s i m i l a r graphic condition. This

m i nutel y

.

study i s i n f u r t h e r s e c t i o n n. 4,

Moreover, we remark t h a t the generalized quadrangle Q(q) , 1-embedded h i n a f i n i t e a f f i n e space AG(3, q), q = 2 , h > 1 , and r e l a t e d t o a ( q + 2 ) - a r c

o f the plane IT,

IT

a t i n f i n i t y o f AG(3, q) containing a n o t degenerate conic o f

t h e r e i s o f the second k i n d and i t s a t i s f i e s a s i m i l a r property. The study

o f such a generalized quadrangle Q(q), and t h e p r o o f o f t h e above mentioned properties, c o n s t i t u t e s the s e c t i o n n. 2 i n [ 3 1 and [41, t o which the Reader i s referred. Therefore, since i n [ 11 and [ 151 t h e r e i s the c h a r a c t e r i z a t i o n o f generali z e d quadrangles 1-embedded i n a f i n i t e a f f i n e space AG(3, q ) and i n 131 and [ 41 t h e r e i s the c l a s s i f i c a t i o n o f such generalized quadrangles, then, by

r e s u l t s proved i n 111, [3], [41, [151 and by what precedes, every generalized quadrangle 1-embedded i n AG(3, q), q

3.

>

3, i s w i t h p a r a l l e l i s m .

STUDY OF PROPERTIES, WHICH A GENERALIZED QUADRANGLE WITH PARALLELISM SATISFIES

Next, we study t h e f i r s t properties, which (S, R) s a t i s f i e s , and i n t h i s way we mean a generalized quadrangle w i t h p a r a l l e l i s m , now and i n what follows.

C Somma

212

We remark t h a t i n (S, R ) any trapezium i s a parallelogram. F i r s t o f a l l , i t i s clear: Proposition

I.

Any two p a r a l l e l l i n e s are d i s j o i n t .

Let N be the number o f l i n e s i n a p a r a l l e l i s m class; i t i s c l e a r ( c f . ( 2 . 5 ) ) t h a t pN = IS1 = p [ ( p - l ) ( n - 1 ) t l l . Therefore:

Proposition 11. P a r a l l e l i s m classes have the same c a r d i n a l i t y N= ( p - l ) ( n - 1 ) t 1. By p r o p o s i t i o n I 1 and by axiom (2.6), since t h e p a r a l l e l i s m classes par-

t i t i o n R, so c l e a r l y i t f o l l o w s : Proposition 111.

There are e x a c t l y

n = I R l / N p a r a l l e l i s m classes.

NOW, l e t B be any p o i n t i n S and r a l i n e i n R n o t through

R; so, t h e r e

i s an o n l y one l i n e s i n R through B i n t e r s e c t i n g r ( c f . ( 2 . 2 ) ) .

Say

IT

the

s e t t h e o r e t i c union o f the p l i n e s through the p p o i n t s on r, which are p a r a l l e l t o s. Next, i n t h i s way we prove: Proposition IV-.

The s e t t h e o r e t i c union IT' o f t h e p l i n e s through t h e

points on s , which are p a r a l l e l t o r, coincides w i t h IT; furthermore,

IIT~

p

= p

2

and o n l y the l i n e s , which are i n the two above seen p e n c i l s o f p a r a l l e l l i n e s , are the l i n e s o f Proof.

R

belonging t o IT.

L e t s ' be a l i n e i n

(cf. proposition

I), so

IT

p a r a l l e l t o s and A = r n s .

Since

s n s' =0

the p o i n t B does n o t belong t o s ' ; thus, t h e r e e x i s t s

an o n l y one l i n e r ( B ) through B, which i n t e r s e c t s s ' i n a p o i n t B ' . Say A = r n s ' ; by axiom

(P), the trapezium A B B ' A ' i s a parallelogram, hence r ( B )

i s p a r a l l e l t o r. Moreover, the l i n e r ( B ) i n t e r s e c t s a l l the o t h e r l i n e s si l e t Ai be the i n t e r s e c t i o n p o i n t o f r and si,

t h a t i s Ai

i n IT.Indeed,

= r n si;

more, l e t

ri(B) be the o n l y one l i n e i n R through 8 ' i n t e r s e c t i n g si and Bi = si n r i ( B ' ) . Take t h e q u a d r i l a t e r a l AiA'B'Bi;

t h e sides AiBi

and A ' B ' belong t o p a r a l l e l

Generalized quadrangles with parallelism

lines, hence A.A'B'B. i s a trapezium and, by axiom ( P ) , 1

Thus, r(B)

1

273 s a parallelogram.

r . ( B ' ) and a l l the lines i n n' belong to n , t h a t i s n' 1

5 n.

In a similar way, exchanging the roles of n and n ' , we show n c n ' , hence n' = n .

I t is routine matter to show the l a s t section of the statement.

-

Call net the above defined s e t n. In a similar way, as in proof of proposition IV and by proposition IVY we can a s s e r t : If a net n contains two points on a l i n e , then the l i n e

Proposition V . belongs t o n. Proposition IV.

Let n and

71'

be two nets and l e t r and s be two d i s t i n c t

lines i n R; i f there i s ( r u s ) c ( n n

then n = n ' .

TI'),

Proof. Suppose Ir n s I = 1 ; then the lines r and s define a net, which necessa r i l y i s both

71

and

T'.

On the other hand, i f r n s = 0, then r and s belong t o a same paralleF ism class, because there are a t l e a s t two nets, which contain ( r u s ) . Take any point A on r; there exists an only one l i n e through A intersecting s in a point 8. As i t contains both A and B y which a r e points belonging t o n and t o n', such a l i n e l i e s on n and on n', by proposition V . T h e n , the statement i s proved. Next, we prove: Proposition VII.

Let r and s be two incident l i n e s , which define a net n;

there e x i s t s a one-one mapping between n and any parallelism class containi n g neither r nor s. Furthermore, any l i n e , which i s not parallel e i t h e r t o

r or t o s , there intersects n, necessarily in

only one p o i n t .

Proof. Let 6i and 6 . be the parallelism classes t o which belong r and s respectJ

ively; c l e a r l y , the index i i s d i s t i n c t from the index j . Now, l e t Ah be a parallelism class, h $ I i , j } . Let f : A

E

n

+

rh(A) E A h

be the mapping

which maps any point A in n onto the l i n e rh(A) belonging t o 6h and containing A. By propositions IV and V , is s t r a i g h t follows rh(A) meets n in j u s t the

214

C. Somma

p o i n t A, hence f i s an i n j e c t i o n . Take i n

any

TI

l i n e x and a p o i n t B on x;

there are n l i n e s through

B y which a r e pairwise d i s t i n c t . Exactly two o f these n l i n e s l i e on

T I , and

one o f them i s the l i n e x; a l l the o t h e r l i n e s have i n common t h e p o i n t B only. We remark t ( B ) means the degenerate block o f pole 8. Now, l e t F'(B1

-

be the s e t ( t ( B )

x)

U

( p - l ) ( n - 1 ) t 1. When B runs

B; i t a r i s e s I F ' ( B ) I

on r, we h a v e p l ( F ' ( B ) ( d i s t i n c t p o i n t s ; moreover, IS1 3 p t F ' ( B ) I . By ( 2 . 5 ) , S = p l F ' ( B ) I and thus, each l i n e , which i s p a r a l l e l n e i t h e r t o r nor t o s ,

intersects

TI,

necessarily i n an o n l y one point, by p r o p o s i t i o n V . Hence, f

i s a one-one mapping. From propositions 11, I V Y V I I s t r a i g h t i t f o l l o w s : Proposition V I I I .

Every p a r a l l e l i s m class contains N = ( p - 1) ( n - 1 ) t 1 = p 3 2 3 and I R I p t2 p p t 2 , (SI = p l i n e s ; therefore, n

2

.

Furthermore, we show: Proposition I X .

Any generalized quadrangle (S,

o f the f i r s t k i n d ([ 101, [ 1 l l ) ; i . e .

R) w i t h p a r a l l e l i s m never i s

through any two d i s t i n c t and n o t c o l l i n e a r

p o i n t s t h e r e are a t l e a s t two d i s t i n c t traces. Proof. n

Since i n a generalized quadrangle o f the f i r s t k i n d i t ever a r i s e s

p ([ 101, [ 11 I ) , so, t a k i n g i n t o account p r o p o s i t i o n

V I I I , t h e statement

i s proved.

4. EMBEDDING I N A FINITE AFFINE SPACE AG(3,

p).

L e t (S, R) be a generalized quadrangle w i t h p a r a l l e l i s m , which parameters a r e p and

n. I n the preceding section we showed i n (S, R) t h e r e i s an

equivalence r e l a t i o n , which p a r t i t i o n s the l i n e s i n R i n t o n = p t 2

classes

o r d i r e c t i o n s (see propositions I 1 and V I I I ) . Now, take care one o f these p a r a l l e l i s m classes, say 6; l e t A and A ' be any two d i s t i n c t and n o t c o l l i n e a r p o i n t s and l e t

1 be t h e trace, which i s the s e t t h e o r e t i c i n t e r s e c t i o n o f

215

Generalized quadrangles with parallelism

the two degenerate blocks t ( A ) and t ( A ' ) o f pole A and A ' , r e s p e c t i v e l y . By axiom ( 2 . 2 ) , there e x i s t s a p o i n t B i n t ( A ) l y i n g on the l i n e through A ' belonging t o 6 and a p o i n t B ' i n t ( A ' ) l y i n g on the l i n e through A belonging t o 6; o f course, B # B ' , since 6 i s an equivalence class and A # A ' . So, t h e p o i n t s B and B ' belong t o the t r a c e 1, c l e a r l y ; the a l l o t h e r p

p o i n t s on 1

c o n s t i t u t e a subset o f 1, c a l l 6 - a f f i n e t r a c e . C a l l 6 - a f f i n e block t ' ( A ) o f pole A the s e t t h e o r e t i c union o f a l l t h e l i n e s through A n o t belonging t o 6. Then, i t i s c l e a r : Proposition X.

The s e t t h e o r e t i c i n t e r s e c t i o n o f two 6 - a f f i n e blocks o f n o t

collinear points i s a 6-affine trace. I n what follows, (A, 6 ) w i l l mean the l i n e i n R through A belonging t h e p a r a l l e l i s m class 6. Now, suppose t h e r e e x i s t s i n ( S ,

R) a p a r a l l e l i s m class N such t h a t t h e

f o l l o w i n g graphic c o n d i t i o n holds, r e l a t e d t o N:

(N)

t h e r e i s an o n l y one N - a f f i n e t r a c e through two n o t c o l l i n e a r p o i n t s . NOW, we show t h a t the c o n d i t i o n

(N) r e l a t e d t o a f i x e d d i r e c t i o n N i s

equivalent t o the f o l l o w i n g :

( N ' ) t h e poles o f the N - a f f i n e blocks through a N - a f f i n e t r a c e a(1) a r e e x a c t l y the p o i n t s on a N - a f f i n e t r a c e a ' ( 1 ' ) . F i r s t we show: Proposition X I .

I f ( N ) holds i n (S, R ) r e l a t e d t o a d i r e c t i o n N, then (N')

holds too. Proof.

L e t a(1)

be a N - a f f i n e t r a c e and l e t A and B be any two n o t c o l -

l i n e a r p o i n t s such t h a t t ' ( A ) n t ' ( B ) = a ( 1 ) . Since A $ B y so by ( N ) there exists

o n l y one N - a f f i n e t r a c e through A and B . O f course, t h e l i n e s (A,X)

and (B,X) through any p o i n t X on a(1) l i e on the N - a f f i n e block t ' ( X ) ; so the p o i n t s A and B belong t o t ' ( X ) .

Take any o t h e r p o i n t Y on a ( l ) , Y # X;

216

C. Somma

i t i s r o u t i n e matter t o show i n a s i m i l a r way A and B belong t o the N - a f f i n e

block t ' ( Y )

Of

pole Y. Thus, the p o i n t s A and B belong t o the N - a f f i n e t r a c e

t ' ( X ) n t ' ( Y ) , which i s , by (N), a ' ( 1 ' ) precisely; so a ' ( 1 ' )

C

t ' ( X ) . Exchang-

i n g t h e r o l e s of a(1) and a ' ( l ' ) , we can assert t h a t each o f t h e p N - a f f i n e blocks, t h e pole o f which i s a p o i n t on a ' ( l ' ) , t h e r e contains a(1). Since two n o t c o l l i n e a r points X ' and Y ' on a(1) belong t o e x a c t l y .p + 2 degenerate blocks ( [ l o ] , [ l l l ) , since one o f them contains t h e l i n e through X ' o f N and another one the l i n e through Y ' o f N, so, by (N), t h e statement i s proved. Next we show: Proposition X I I .

The generalized quadrangle (S,

R) w i t h p a r a l l e l i s m , such

t h a t ( N ' ) holds, i s o f the second kind, t h a t i s any t r a c e l i e s on e x a c t l y two d i s t i n c t degenerate blocks (t 101, [ 11 I ) . Proof.

L e t A and A ' be any two d i s t i n c t and n o t c o l l i n e a r p o i n t s i n S. It i s

c l e a r t h a t , by above d e f i n i t i o n s , t h e t r a c e 1 = t ( A ) n t ( A ' ) consists o f a N - a f f i n e t r a c e a(1) and o f two d i s t i n c t p o i n t s B and B 1 , l y i n g on the l i n e s through A and A ' belonging t o N, r e s p e c t i v e l y . BY c o n d i t i o n ( N ' ) , the poles of the N-affine blocks through a(1) are e x a c t l y t h e p o i n t s on a N - a f f i n e t r a c e a ' ( l ' ) , t o which both A and A ' belong. I f t h e r e e x i s t s a p o i n t A" i n S such t h a t 1

C

t ( A " ) , then, by c o n d i t i o n ( N ' ) ( c f . p r o p o s i t i o n X, t o o ) , A"

belongs t o the N - a f f i n e t r a c e a ' ( 1 ' ) through A and A ' . The l i n e (A", N) through A " and belonging t o N i n t e r s e c t s a ' ( 1 ' ) i n A"; "(A",

N). On the other hand, i f 1

C

therefore, A " = a'(1')CI

t ( A " ) , so one a t l e a s t o f t h e p o i n t s B and

B' belongs t o the l i n e (A", N). I f B E (A", N), then (A", N) i s t h e l i n e (8, N) through B belonging t o N, since the l i n e s i n N p a r t i t i o n S; moreover,

A" = a ' ( 1 ' ) n ( B y N ) . Furthermore, the p o i n t A l i e s on (6, N); since A $ A " and both points belong t o a l ( l I ) , then A = A". = A",

I n a s i m i l a r way we show A ' =

i f B E (A", N). Then t h e t r a c e 1 l i e s on t h e o n l y two degenerate blocks

t ( A ) and t ( A ' ) ; hence (S, R ) i s a generalized quadrangle o f t h e second kind. NOW, take any N - a f f i n e t r a c e a(1); there e x i s t s

o n l y one l i n e belong-

i n g t o N through each p o i n t l y i n g on a ( l ) , o f course. I n t h i s w a y we o b t a i n p

l i n e s , c l e a r l y p a i h i s e d i s j o i n t ; we c a l l

N-net

t h e s e t t h e o r e t i c union o f

Generalized quadrangles with parallelism such l i n e s . I t i s r o u t i n e matter t o show t h a t Proposition X I I I .

211

a N-net i s a net. It i s c l e a r :

There e x i s t s an o n l y one N-net through a N - a f f i n e t r a c e .

Next, we show: Proposition X I V .

Let ( S ,

R) be a generalized quadrangle w i t h p a r a l l e l i s m , i n

which ( N ' ) holds r e l a t e d t o a f i x e d d i r e c t i o n N; the N - a f f i n e blocks through a N - a f f i n e t r a c e a(1) pairwise i n t e r s e c t i n t o a(1) and the s e t t h e o r e t i c union o f such N - a f f i n e blocks and the N-net through a(1) i s the whole s e t S .

Proof.

O f course, any two N - a f f i n e blocks through a(1) o n l y i n t e r s e c t i n t o

the p o i n t s belonging t o a(1) ( c f . p r o p o s i t i o n X, t o o ) . L e t IT be t h e N-net through a(1); i t i s c l e a r t h a t t o each N - a f f i n e block through a(1) there be-

2

long e x a c t l y P -P p o i n t s n o t l y i n g on IT.By c o n d i t i o n ( N ' ) , t o the s e t t h e o r e t i c union o f t h e N - a f f i n e blocks through a(1) there belong e x a c t l y

2

p ( p - p ) p o i n t s n o t on IT. Thus, t h e statement i s proved.

O f course, from (N) there f o l l o w s :

Proposition XV.

Any two N - a f f i n e traces i n t e r s e c t i n g i n t o two d i s t i n t poi t s

l i e on t h e same N-net. Next, we prove: Proposition X V I .

If ( S ,

R) i s a generalized quadrangle w i t h p a r a l l e l i s m , i n

which (N') holds, then (N) holds. Thus, t a k i n g i n t o account p r o p o s i t i o n X I , t h e r e f o l l o w s ( N ) i s equivalent t o ( N ' ) . Proof.

Take any two n o t c o l l i n e a r p o i n t s A and A ' , and l e t B and B ' any two

d i s t i n c t p o i n t s belonging t o the N - a f f i n e t r a c e

a ( 1 ) = t ' ( A ) n t ' ( A ' ) and

c l e a r l y n o t c o l l i n e a r . It i s easy t o see t h a t t h e N - a f f i n e t r a c e a ' ( 1 ' ) = -t'(B)n

t ' ( B ' ) contains both A and A ' . Since any two n o t c o l l i n e a r p o i n t s

l i e on e x a c t l y p t 2 degenerate blocks ([ 101, [lll),by (N'), a l l the poles

C Somma

278

o f the N - a f f i n e blocks through a'(1') l i e on a(1) and such N - a f f i n e blocks pairwise i n t e r s e c t i n a ' ( 1 ' ) ( c f . p r o p o s i t i o n XIV); through A and A ' t h e r e i s necessarily only. the N - a f f i n e t r a c e a ' ( 1 I ) . L e t L denote the t r a c e s e t o f a generalized quadrangle (S, R ) w i t h p a r a l l e l i s m ¶ i n which (N) holds; since (S, R ) i s o f t h e second k i n d ( c f . p r o p o s i t i o n X I I ) , f o r any t r a c e 1 E L t h e r e e x i s t o n l y two n o t c o l l i n e a r p o i n t s X and Y such t h a t t ( X ) n t ( Y ) = 1.

The l i n e s belonging t o N through X and Y i n t e r s e c t

1 i n two points; thus, we can consider

o n l y one N - a f f i n e t r a c e a(1). L e t

V 1 E L, and Ro = R u R ' .

R ' denote the f a m i l y o f " t h e N - a f f i n e traces a ( l ) ,

Since ( S , R ) i s a generalized quadrangle, and thus a p a r t i a l geometry, from '3 ( N ) , t a k i n g i n t o account IS1 = p ( c f . p r o p o s i t i o n V I I I ) , t h e r e follows: Proposition X V I I . Let of

n

n

(S,

RO)

3 i s a Z-Steiner system S(~,P,P) ( [ 7 1 , t81, [ 1 3 1 ) .

denote t h e f a m i l y o f

N-nets and N - a f f i n e blocks; c a l l the members

"planes" and t h e members o f R" " l i n e s " . Next, we show:

Proposition X V I I I . Every "plane" more, i f

IT

IT

i n II contains e x a c t l y p

2

points; further-

contains two d i s t i n c t p o i n t s A and B y then the " l i n e " i n Ro through

them l i e s on such a "plane". Proof.

Since each n e t contains P

forward computation shows

IIT~

2

points (cf. proposition I V ) , a straight-

= p 2 , f o r every "plane" IT i n II.

Next, we prove the second statement. L e t c o l l i n e a r p o i n t s belonging t o

IT, then

IT

be a N-net. I f A and B are

t h e l i n e ( A , B) i n R through them be-

longs t o R" and, by p r o p o s i t i o n V , t h e r e l i e s on

IT;

furthermore, i t i s c l e a r

t h a t t h e r e i s n o t any N - a f f i n e t r a c e through A and B. I f A and B are n o t c o l l i n e a r p o i n t s , then the N - a f f i n e t r a c e a(1) through them l i e s on a N-net IT', t o which both A and B belong. By d e f i n i t i o n , one of the two p e n c i l s o f p a r a l l e l l i n e s i n R l y i n g on n' belongs t o the p a r a l l e l i s m class N. On the other hand, since

71

belongs t o II, so i t contains a p e n c i l o f p a r a l l e l l i n e s belong-

i n g t o N; as A and B belong t o n, thus l i e s on IT.

IT

=

IT'

and the N - a f f i n e t r a c e a(1)

279

Generalized quadrangles with parallelism Let

IT

be a N - a f f i n e block t l ( A ' ) o f pole A ' . I f A and

B are c o l l i n e a r

p o i n t s and A ' coincides w i t h one o f them, then the statement i s already proved.

B A ' does n o t belong t o (A, B),

I f t h i s i s n o t the case, then we show the l i n e (A, 9) i n R through A and

contains necessarily the p o i n t A ' . Indeed, i f then t h e r e e x i s t two d i s t i n c t l i n e s i n

R, and p r e c i s e l y ( A ' , A ) and ( A ' , 9)

(these l i n e s t h e r e e x i s t since A ' i s the pole o f a N - a f f i n e block and thus c o l l i n e a r w i t h each p o i n t on i t ) through A ' and i n t e r s e c t i n g ( A , 9); so we have a c o n t r a d i c t i o n ( c f . (2.2)).

I f A and

B a r e n o t c o l l i n e a r points, then

l e t a(1) be the N - a f f i n e t r a c e through them; l e t t l ( B ' ) one o f t h e p N - a f f i n e blocks through a(1). I f

9' = A ' , then the statement i s proved. On the o t h e r

hand, i f 6 ' # A', then the N - a f f i n e blocks t ' ( B ' ) and t l ( A ) i n t e r s e c t i n a N - a f f i n e trace, which necessarily coincides w i t h a ( l ) , since i t i s through A and 9. Hence, a ( 1 ) c t l ( A l ) = n; as a(1) i s i n R o y so the whole statement i s shown. Proposition X I X .

The "planes" i n TI through a " l i n e " r i n

sect i n r and t h e i r t h e o r e t i c union i s the whole s e t

R" pairwise i n t e r -

s.

Proof. I f r i s a N - a f f i n e trace, then i t l i e s on an o n l y one N-net, o f course ( c f . p r o p o s i t i o n X I I I ) and on e x a c t l y P N - a f f i n e blocks, which poles belongs t o a N - a f f i n e t r a c e r ' ( c f . ( N ' ) and p r o p o s i t i o n X I ) . It i s c l e a r t h a t two

such N - a f f i n e blocks i n t e r s e c t i n r ( c f . p r o p o s i t i o n XIV); moreover,

each N - a f f i n e block o f them and

IT

o n l y i n t e r s e c t i n r. A s such a N - a f f i n e

block contains ( p 2 - p ) p o i n t s n o t belonging t o IT, so t h e s e t t h e o r e t i c union 2 2 p p o i n t s n o t on n . The s e t o f the p N - a f f i n e blocks contains p ( p p) = p

-

t h e o r e t i c union o f

IT

-

and the p N - a f f i n e blocks, t h a t i s t h e s e t t h e o r e t i c

union o f a l l the "planes" through r, contains p

J

points, thus i t i s the whole

s e t S. NOW, l e t r be a l i n e i n R belonging t o the p a r a l l e l i s m class N. The o t h e r p t

1 p a r a l l e l i s m classes w i t h r g i v e r i s e t o e x a c t l y p t 1 nets, pairwise i n -

t e r s e c t i n g i n r. By propositions I V and X V I I I , i t i s easy t o see ever a p e n c i l o f l i n e s belonging t o N l i e s on any such a net; thus, a l l these nets are N-nets. O f course, r does n o t belong t o any N - a f f i n e block; thus, a l l the 2 "planes" i n il through r o n l y are such p t 1 nets. Each o f them contains p d i s t i n c t p o i n t s n o t l y i n g on r, hence the s e t t h e o r e t i c union o f a l l

the

-p

C. Somma

280

"planes" through r c o n s i s t s o f ( p t 1) ( p z - P) + P = P Next, l e t r be a l i n e i n N-net

IT

3

points.

R n o t belonging t o N. There e x i s t s an o n l y one

through r, t o which r and the d i r e c t i o n N g i v e r i s e , since through

each p o i n t on r t h e r e i s an o n l y one l i n e belonging t o N. It i s c l e a r t h a t TI i s a "plane" i n

n.

and i t does n o t e x i s t any other n e t through r belonging t o

NOW, l e t t ' ( A ) a N - a f f i n e block through r; o f course, r C ( t ' ( A ) n IT). I t

i s c l e a r t h a t a l l the l i n e s i n R l y i n g on

IT

and i n t e r s e c t i n g r t h e r e belong

t o t h e d i r e c t i o n N; moreover, by d e f i n i t i o n , any one o f them i s n o t contained i n t ' ( A ) . Thus, ( t ' ( A ) n IT)

r meets

IT

-

r = 0; therefore, each N - a f f i n e block through

i n the p o i n t s on r only. Since the c a r d i n a l i t y o f r i s p and each

N-affine block through r has the pole on r, then t h e r e e x i s t e x a c t l y p N - a f f i n e blocks through r, pairwise d i s t i n c t ; o f course, two o f them.intersect i n r only. Therefore, through r t h e r e are IT and p N - a f f i n e blocks only, t h a t i s ( p + 1) "planes", which pairwise i n t e r s e c t i n r. Furthermore, each o f them

contains p 2

-p

p o i n t s n o t l y i n g on r and t h e i r s e t t h e o r e t i c union consists

of (p2-p)(ptl)+p = p

3

d i s t i n c t p o i n t s . Thus, t h e whole statement i s proved.

Now, l e t C be the f o l l o w i n g f a m i l y o f subsets o f S: C the "line" i n

E

C i f , an o n l y i f ,

Ro through two d i s t i n c t p o i n t s o f C t h e r e l i e s on C.

By p r o p o s i t i o n X V I I I , t o C t h e r e belong a l l t h e members i n II; f u r t h e r more, i t i s c l e a r t h a t t h e empty s e t in

R o t h e r e belong t o

C.

0,

the singletons o f S and t h e " l i n e s 1 1

Then, i t i s easy t o see C i s a f l a t - s e t i n S; thus,

the p a i r (S, C ) i s a f i n i t e f l a t - s p a c e (161, 1131). Taking i n t o account propositions X V I I I and X I X , i t

i s n o t hard t o show i n

(S, C ) t h e f o l l o w i n g axiom holds ([61, [131):

[YX, Y if 2, Z

5 S,

GS, C E c /

X$

c,

XE

-

Y u C ] -r

YE

xu

c,

means the i n t e r s e c t i o n o f a l l the members i n C containing the

subset Z. Hence: Proposition XX. C i s h combinatorial geometry on S ([61, [131). Then, we say (S, R") i s a combinatorial S t e i n e r system ([131). Since t h e l i n e s i n

R are members i n

C o f dimension one, then the general-

i z e d quadrangle (S, R) w i t h p a r a l l e l i s m , i n which (N) holds, t h e r e i s l-embedd-

Generalized quadrangles with parallelism

28 1

ed i n the combinatorial geometry (S, C ) (2); now, we show: Proposition X X I . C i s the geometry i n a three-dimensional a f f i n e space AG(3,p) o f order p ; moreover, the generalized quadrangle (S, R) 1-embedded i n (S, p ) i s r e l a t e d t o a l i n e a r complex o f l i n e s W i n t h e p r o j e c t i v e space i n which AG(3,p)

i s embedded.

Proof.

Taking i n t o account p r o p o s i t i o n I V , section n. 6, i n [ 131 (see [ 21 and

1151, too) and p r o p o s i t i o n XX, since t h e r e e x i s t l i n e s i n R d i s j o i n t and belonging t o a same "plane" i n ll, t h e r e e x i s t proper subspaces o f dimension d > 1 3 the members o f ll -, since IS1 = P and (S, R O ) i s a combinatorial S t e i n e r

-

system, so i t f o l l o w s t h a t C i s t h e geometry i n an a f f i n e space o f dimension three and order p . Since the N - a f f i n e blocks a r e members o f dimension two i n C, c o n t a i n i n g p t l

l i n e s o f R through a s i n g l e point, so, t a k i n g i n t o account t h e r e s u l t s

proved i n [ 11 and [ 151, the statement i s proved. We a t l a s t remark, t a k i n g i n t o account t h e r e s u l t s proved i n 131 and f41, h t h a t i f p = 2 , h > 1, the generalized quadrangle (S, R) i s r e l a t e d t o a ( p t 2 ) - a r c o f t h e plane a t i n f i n i t y

IT

o f AG(3, p ) , which contains a n o t de-

generate conic o f IT.

ACKNOWLEDGEMENT.

This research was p a r t i a l l y supported by GNSAGA o f CNR.

REFERENCES

[ l l A. Bichara, Caratterizzazione dei sistemi r i g a t i imnersi i n A3,q, Riv. Mat. Univ. Parma (4)

4

(1978) 277-290.

I 2 1 A. Bichara, Sistemi r i g a t i immersi i n uno spazio combinatorio, Sem. Geom. Comb. Univ. Roma, 6 (1978). I 3 1 A. Bichara, F. Mazzocca, C. Somna, C l a s s i f i c a z i o n e dei sistemi r i g a t i imnersi i n AG(3,q), Sem. Geom. Comb. Univ. Roma, (1978).

2

[ 4 1 A. Bichara, F. Mazzocca, C. Somna, On the c l a s s i 6 i c a t i o n o f generali z e d quadrangles i n a f i n i t e a f f i n e space AG(3,2 ), B o l l . U.M.I. (5) 17-8 (1980) 298-307.

282

C Somma

[ 5 1 F. Buekenhout, Une caracterisation des espaces affins basee sur la notion 111 (1969) 367-371. de droite, Math. 2 . [ 6 1 H.H. Crapo and G.-C. Cambri dge , 1970. [

Rota, Combinatorial geometries, The M.I.T. Press,

71 M.J. de Resmini, Sistemi d i Steiner e questioni collegate, Sem. Geom. Comb. Univ. Roma, 12 (1978), Vol. I , 11, 111, IV.

[ 8 ] M. Hall, J r . , Combinatorial theory, Waltham, 1967. 191 6 . Segre, Lectures on modern geometry, Cremonese, Roma, 1961.

[ l o ] G. T a l l i n i , Ruled graphic systems, Atti Conv. Geom. Comb. Perugia, (1971) 385-393. [

11 I G. Tallini , Strutture d’incidenza dotate di polaritti, Rend. Sem. Mat. F i s . Milano, 41 (1971) 75-113.

[ 1 2 1 G. T a l l i n i , Problemi e Risultati s u l l e geometrie di Galois, Rel. n . 30, 1st. Mat. U n i v . Napoli, 1973. [131 G. T a l l i n i , Spazi combinatori e sistemi d i Steiner, Riv. Mat. Univ. Parma ( 4 ) 5 (1979) 221-248. [

141 J.A. Thas, On 4-gonal configurations, Geometriae Dedicata, 31 7-326.

2 (1973)

“15! J.A. Thas, Partial geometries i n f i n i t e a f f i n e spaces, Math. 2 . (1978) 1-13.

158

Annals of Discrete Mathematics 14 (1982) 283-292 0 North-HollandPublishing Company

ON LINE k-SETS OF TYPE (0,n) WITH RESPECT TO PENCILS OF LINES I N PG(d,q)

Giuseppe T a l l i n i I s t i t u t o Matematico "G. Castelnuovo", U n i v e r s i t a d i Roma, I t a l y

1.

INTRODUCTION

be the Grassmann manifold representing the l i n e s i n a Galois 4 The i n v e s t i g a t i o n space PG(d,q) and El t h e c o l l e c t i o n o f a l l l i n e s on G d4 w i t h respect t o c1 i s equivalent t o the i n v e s t i g a o f p o i n t k-sets on G dYq t i o n o f l i n e k-sets i n PG(d,q) w i t h respect t o p e n c i l s o f l i n e s . L e t Gd

.

; the index s character o f K w i t h respect L e t K be a p o i n t k-set on G 1 d,q meeting K a t s p o i n t s . K i s t o Z w i l l be the number 'I o f l i n e s on G S d3 9 1 saidtobeofclass[ml,m2 mtl,,0<ml<m2< < m t 6 q t l (the 1 i f T~ # 0 f o r any s # m!s being i n t e g e r s ) w i t h respect t o l i n e s on G 1 d,rl K w i l l be c a l l e d a s e t o f Qp(ml, m2, mt)l with # mly m2, mt;

,...,

...

...,

...,

respect t o l i n e s on Gd T~~ # 0, i E {l,2,

if

T~

..., t).

S

= 0 for all s

# m,,

m2,

..., mt

and

F i n a l l y , K w i l l be a t character s e t i f exact-

l y t o f i t s characters are d i f f e r e n t from zero.

can be studied i n terms o f t h e i r i n c r e a s i n g numbers sq of non-zero characters and i n [61 t h e f o l l o w i n g r e s u l t s were proved. Point k-sets on Gd

Proposition I. A one-character k-set i s e i t h e r t h e empty s e t o r G dYq Proposition 11.

No type (O,qtl)l

Proposition 111.

Any type (O,q)l

Gd,q

k-set e x i s t s . k-set i s the complement o f a p o i n t set on

representing a l i n e complex i n PG(d,q).

Proposition I V .

Any type ( l , q t l ) l

k-set consists o f t h e p o i n t s on Gd

presenting a l i n e complex i n PG(d,q). 283

4

re-

C. TaNini

284

Proposition V .

A type (0,1)1 k-set represents a s e t o f pairwise skew l i n e s

i n PG(d,q) and any type ( q , q t l ) l

set i s i t s complement.

-

1, p o i n t k-sets on Gd 2
contains e i t h e r zero or n l i n e s i n the s e t . Consequently, t a k i n g the complements, type (m,qtl)ly

2.

2G m

< q - 1 , k-sets on

Gd

3q

are investigated.

ON LINE k-SETS OF TYPE (0,n) WITH RESPECT TO PENCILS OF LINES I N PG(d,q) For short, a l i n e k-set o f type (0,n) w i t h respect t o p e n c i l s o f l i n e s

w i l l be c a l l e d a (0,n)-p-type

l i n e k-set.

F i r s t , the f o l l o w i n g p r o p o s i t i o n w i l l be proved. Proposition V I .

L e t K be a proper l i n e k-set i n PG(d,q) such t h a t

the s e t t h e o r e t i c union o f the l i n e s i n K

-

-

H being

the f o l l o w i n g hold:

(2.1)

The span o f H, [ H I , i s PG(d,q).

(2.2)

For any p o i n t P i n H and any l i n e r i n K, r $ P, t h e r e are e x a c t l y n l i n e s i n K through P meeting r, n i s a f i x e d i n t e g e r and 2 4 n

Under these assumptions,

< qt

where

1.

K i s a (0,n)-p-type set, i . e . any p e n c i l o f

l i n e s i n PG(d,q) contains e i t h e r 0 o r n l i n e s i n K. Proof.

I n PG(d,q),

take any p e n c i l F c o n s i s t i n g o f t h e l i n e s through a p o i n t

P and belonging t o a plane a . I f P $? H, then there i s no l i n e i n K through P and F n K = 0. Assume .P belongs t o H; one o f the f o l l o w i n g must be t r u e :

a) No l i n e i n K belongs t o a ; hence, F contains no l i n e i n K. b ) I n a t h e r e i s a l i n e r belonging t o

K and P

$ r; i !

by (2.2),

t h e r e are

e x a c t l y n l i n e s through P meeting r. c ) a contains a l i n e r E K through P and a p o i n t P ’ i n H \ r. By ( 2 . 2 ) ,

On line k-sets of type (0, nl 2,

since n

0:

285

c o n t a i n s a l i n e r ' i n K t h r o u g h P ' and n o t t h r o u g h P; t h u s ,

t h r o u g h P t h e r e a r e n l i n e s i n K i n c i d e n t w i t h r ' ( r b e i n g one o f them). d)

(Y

c o n t a i n s a l i n e r E K t h r o u g h P and

a! 0

H = r.

Case d ) w i l l be proved i m p o s s i b l e and ( r e c a l l i n g prop. I , K b e i n g p r o p e r ) t h e statement w i l l f o l l o w . A p l a n e t h r o u g h r E K w i l l be c a l l e d a t a n g e n t p l a n e o f K i f i t meets H e x a c t l y a t t h e p o i n t s on r, a s e c a n t p l a n e o f K i f i t c o n t a i n s a t l e a s t one point i n H \ r . Let B

and B2 be two d i s t i n c t secant p l a n e s o f K t h r o u g h r . Any p l a n e 1 t h r o u g h r b e l o n g i n g t o t h e 3 - f l a t j o i n i n g B1 and B2 w i l l now be proved t o b e a secant p l a n e o f K. S i n c e B1 i s a secant p l a n e o f K, t h e r e i s a p o i n t P1

\ r and t h r o u g h P t h e r e a r e n l i n e s i n K i n c i d e n t w i t h r and, t h u s , 1 1 b e l o n g i n g t o B1; l e t s be one o f them. S i m i l a r l y , t h e r e e x i s t s a p o i n t P 2 i n B n H \ r and t h r o u g h i t t h e r e a r e n l i n e s i n K m e e t i n g s and one o f them 2 i s skew w i t h r. T h e r e f o r e , any p l a n e i n S t h r o u g h r a t l e a s t - l e t i t be t in H n

-

meets t a t a p o i n t i n H \ r ; t h u s , any p l a n e i n S t h r o u g h r i s a secant p l a n e of K. Consequently, t h e s e t t h e o r e t i c u n i o n o f a l l secant p l a n e s o f K, and [ H I C PG(d',q);

t h r o u g h r, i s a subspace PG(d',q) PG(d,q),

by ( 2 . 1 ) ,

PG(d',q)

=

i . e . any p l a n e t h r o u g h r i s secant t o K and t h e s t a t e m e n t i s

proved.

A s i m i l a r argument proves: Proposition V I I .

L e t K be a l i n e k - s e t i n PG(d,q) and H t h e s e t t h e o r e t i c

u n i o n o f t h e l i n e s i n K. Assume t h e f o l l o w i n g h o l d :

(.2.4)

Through any p o i n t P i n H and any l i n e r i n K, r e i t h e r ml,

o r m2,

..., o r mi

l i n e s t h r o u g h P i n c i d e n t w i t h r,

where t h e m l s a r e f i x e d i n t e g e r s and 2

J

Then K i s o f c l a s s [ 0 ,ml ,in2,.

9 P, t h e r e a r e

< ml<

..,m ?J I w i t h

...< m i ( q t l .

respect t o pencils

o f lines. Consider a p o i n t k - s e t K on Gd

Yq

,of

t y p e (O,n),

w i t h r e s p e c t t o C1

286

G. Tallini

2 < n G q - 1 . L e t Cd-,

be the c o l l e c t i o n o f (d-1)-dimensional subspaces on

any o f which represents a (d-1)-star o f l i n e s i n PG(d,q) ( i . e . a l l the Gd,q The s e t K ' = Td-l n K l i n e s through a p o i n t i n PG(d,q)) and l e t Td-l E i s a p o i n t s e t i n Td-ly no k-set o f type (n)

1

o f class [O,nIl

w i t h respect t o l i n e s i n Td-l.

w i t h respect t o l i n e s ,

1

< n < q,

Since

exists i n a project-

i v e space, one o f the f o l l o w i n g must be t r u e :

(i)K ' i s o f type ( 0 ) 1 w i t h respect t o l i n k s i n Td-l; ( i i ) K ' i s o f type (O,n),

hence, K' =

w i t h respect t o l i n e s i n Td-l,

0.

so t h a t d - 1 =

= 2 (see [ 5 1 , prop. XIV). It f o l 1ows : Proposition V I I I . I n PG(d,q), d > 4, no (0,n)-p-type

l i n e k-set e x i s t s (2

I n PG(d,q),

d 2 4, no l i n e k-set K e x i s t s such t h a t the

set t h e o r e t i c union H o f the l i n e s i n K s a t i s f i e s (2.1) and (2.2), 2

<

nG

< q - 1. A c o l l e c t i o n K o f l i n e s i n PG(d,q) w i l l be c a l l e d a p a r t i a l geometry embedded i n PG(d,q) i f the s e t t h e o r e t i c union, H, o f t h e l i n e s i n K satisfies :

(2.5)

Through any p o i n t i n H there are N l i n e s belonging t o K.

(2.6)

Given any l i n e r i n K, through any p o i n t P i n H \ r t h e r e are n l i n e s meeting r.

The i n t e g e r n i s c a l l e d the parameter o f the p a r t i a l geometry. When n = 1, p a r t i a l geometries i n PG(d,q) are generalized quadrangles ( r u l e d systems) embedded i n PG(d,q).

They were completely characterized by

D. Olanda i n 131,

On line k-sets of type (0,nl [ 4 1 and by F. Buekenhout and C. Lefgvre i n [ 1 I .

287

Therefore, assume n

>

2.

By prop. VI, a partial geometry K i s a (0,n)-p-type l i n e k-set i n P G ( d , q ) . Thus, such a geometry doesn't e x i s t when n = q + 1 (see prop. 11); when n = q ,

K i s the complement, in P G ( ' ) ( d , q )

( t h e s e t o f a l l lines in P G ( d , q ) ) , of a

line complex (see prop. 111). I t follows (recalling prop. VIII): Proposition X . Let K be a partial geometry, of parameter n 3 2 , embedded in PG(d,q). If n

PG(d,q). If

> q , then n d

= q and

K i s the complement o f a l i n e complex in

q , i . e . no partial geometry e x i s t s i n P G ( d , q ) ,

4, then n

withda4and 2 < n < q - 1 . Thus, the investigation of partial geometries in PG(d,q) turns out t o be the investigation of partial geometries in PG(J,q), with 2 4 n

< q - 1 , as

i t will be seen in the next section.

3.

ON (0,n)-p-TYPE LINE k-SETS IN PG(3,q) ( 2

< n < q - 1)

Let K be a (0,n)-p-type l i n e k-set i n PG(3,q)(q = p h , p a prime, h a representing the l i n e s in K form a 3Y q I t will be assumed 2 G n w i t h respect t o lines on G3 4

non-negative i n t e g e r ) ; the points on G k-set o f type ( O , n ) ,

< q - 1. Let

T

be a plane on G3

following i s true:

4

.

and consider the s e t K '

K n n ; one of the

a ) K ' i s of type ( 0 ) l i thus, K' = 0. b ) K ' i s of type ( O , n ) l ;

e

(3.1)

h > 2 , n = p ,

hence: l
and K ' = ( q t l ) ( n - l ) + l = N .

Consequently , Proposition XI.

No (0,n)-p-type l i n e k-set e x i s t s i n PG(3,q), q a prime.

, q = ph, p a prime, h a 2 , o f type 4 ( O , n ) , with respect t o Z1, i s of class [ O , N I 2 , N = ( q t 1 ) ( n - l ) + l , w i t h

Proposition XII. A point k-set on G3

288

G. TaNini

e , 1 < 1 < h - 1.

respect t o planes on G and n = p 3Yq L e t K be a (0,n)-p-type

l i n e k-set i n PG(3,q),

q = ph, p a prime, h

>

2;

il

p and any s t a r and any r u l e d plane share w i t h K e i t h e r

by prop. X I I , n

zero o r N = ( q t l ) ( n - 1 ) t l l i n e s . 2 2 Denote by ‘c0 and -rN the numbers o f s t a r s sharing w i t h K zero and N l i n e s , 2 respectively; the i n t e g e r s r 2 and T~ w i l l be c a l l e d the characters o f K w i t h 0

respect t o ( l i n e ) stars; denoting by H the s e t t h e o r e t i c union o f the l i n e s i n K, the f o l l o w i n g equations hold:

(3.2) (3.3)

T

T

*

0

O

2

= q3tq t q t l t T2

N

-

IHI,

=q3+q2tqtl

.

Thus ,

Furthermore, 2

(3.5)

N [HI

N

(3.6)

N(N-1)

IHI

T

T~

= k(qt1);

N ( N - 1 ) I-; = k ( k - 1 ) - 2 ~,

being the number o f skew l i n e p a i r s i n K. From (3.5), N = ( q t l ) (n-1)

i t f o l lows

Therefore, k being an i n t e g e r , by (3.7),

t

1,

recalling that

On line k-sets of type (0, n)

289

Proposition XIII. The size of the set theoretic union, H , o f the lines in K i s a multiple of q t 1 , i.e.

IHI

(3.8)

= m (qtl)

,

m

an integer.

Furthermore, by (3.5) and prop. X I I I , (3.9)

k=mN.

P and r PN be the numbers of planes in PG(3,q) containing zero and P N lines in K, respectively; :T and T~ are called the characters of K with respect t o ruled planes. Clearly, Let

T~

(3.10)

Tp

o

Proposition XIV.

P 2 ‘NrTN

T2 0 ’

‘

The set theoretic union, H, o f the lines in K i s a point

set of class [ m , q + m 1 2 with respect t o planes i n PG(3,q); that i s , any plane contains either m or q t m points in H.

Proof.

Let a be a plane in PG(3,q); by prop. XII, i t shares with K either

zero or N lines. Assume a contains no line in K (thus, the plane f ( a ) representing a on O3 ,q

i s external t o K ) . Clearly, any line in K meets a a t a point; moreover,

the s t a r of lines through P (3.9), la n H I (3.11)

E

H shares N lines with K; therefore, recalling

N = k = m N; in other words: i f a contains no line in K , then la n H I = m.

Assume a contains N lines i n K; in the dual plane a* these lines form a type (O,n), N-set. Thus, the number o f points i n a n H belonging t o lines In N K contained in a i s i ( q + 1 ) and this i s the number of n-secants of a type

N-set; moreover, through any other p o i n t i n H n a there are N lines in K, no one of them belongs t o a . Therefore, (O,N)l

N (N-n);(qtl)tN(lanHI

N -;(qtl))

= k-N =mN-N,

G. Tallini

290

from which

Nl'crn H I

N2

N(m-1)

N2

n ( q + 1 ) - - (n q t l ) t N ( q t l ) ;

t-

hence ,

N = N(m+q).

( a n HI

Thus , i f a contains lines in K, then [ a n H I = q t m .

(3.12)

From (3.11) and (3.12) the statement follows. Let a be a plane containing lines in K; the number of points in H n a

through which there i s no line in K belonging t o a i s a

(3.13) thus, being ] H I

(3.14)

= la

n HI

- -nN( q + l )

- -Nn( q + l ) ;

< ( q t 1 ) ( q 2 t 1 ) (and recalling (3.8)):

i n N

m =-(q+1)

Since [ H I

= q t m

-

a

q-

0

.

PG(3,q) and through any point in H there are N lines in K, the

previous argument shows t h a t (3.15)

a = O-=>[ Vr E K, VP E H \ r

3 3!

n lines in K through P

incident with r ] o K i s a partial geometry. Assume a

0 and l e t s be any line in PG(lI(3,q) not contained in H such

t h a t s n H # 0; then, there exists some line r in K meeting s. Let plane joining r and s. The lines belonging both t o K and t o

a

form the dual

of a type (O,n), N-set; moreover, since a = 0, through any point in there are exactly n lines in K on

a.

Therefore, s meets

a n

be the

OL

a

nH

H a t N/n points.

Hence, any line n o t in H and meeting H in a non-empty s e t , meets H a t N/n

On line k-sets of type (0. n) points; thus, H i s o f class [O,N/n,q+l]l.

29 1

On the other hand, H cannot be a

one character s e t w i t h respect t o l i n e s , otherwise (being H # 0) H = PG(3,q) 2 q +l; thus, by (3.14)2y q 2 + 1 = ( q + 1 ) N/n q, i . e . and, by (3.8), m

-

n = q + l (as N = ( q + 1 ) ( n - l ) + l ) ,

a c o n t r a d i c t i o n (being n < q - 1 ) .

some l i n e external t o H must e x i s t , otherwise and i t s complement o f type (0,q

+ 1 - N/n)l , so

Moreover,

H would be o f t y p e (N/n, q + l ) l t h a t (see [ 51, prop. X I V ) e i t h e r

q + l - N / n = 1, o r q + l - N / n = q, i . e . e i t h e r n = q, o r n = 1, a c o n t r a d i c t i o n , as 2

< n
L e t B be a plane containing a l i n e external t o H and a p o i n t

1.

-

i n H. The s e t B n H i s o f type (O.N/n)l i n B; consequently, N/n = pc = q t 1 h-9. q/n = q + 1 p ( c p o s i t i v e i n t e g e r ) , a c o n t r a d i c t i o n , since 1 < II < h - 1.

-

Therefore ,

-

a>O.

(3.16) BY (3. 5) and (3.16), Proposition

g.

No p a r t i a l geometry embedded i n PG(3,q) e x i s t s when 2

<

n

<

< q - 1. The previous r e s u l t s a r e summed up i n the n e x t proposition. Proposition X V I . PG(3,q)

(q =

ph,

L e t K be any (0,n)-p-type, p a prime, h an i n t e g e r

a.

>

2

< n < q - 1, l i n e k-set i n

1). The f o l l o w i n g must hold:

(i 1

h.2,

(ii)

Both any r u l e d plane and any s t a r o f l i n e s share w i t h K e i t h e r

zero o r N = ( q + 1 )

(iii)

n = p ,

(n- 1 ) + 1

1<11
l i n e s and these l i n e s form a t y p e (O,n), N-set. 2 P P Denoting by T and T~ ( T and ~ T ~ ) , r e s p e c t i v e l y , t h e numbers 2

0

o f s t a r s o f l i n e s ( r u l e d planes) sharing w i t h K zero and and by H the s e t t h e o r e t i c union o f t h e l i n e s i n K, (3.17)

IHI

(3.18)

-N( q + 1 ) n

(3.19)

= m(q+l),

k=mN,

- q<m
N

l i n e s , respectively,

292

m = q2 + 1

(3.20)

Furthermore, e i t h e r or

-

C. Tallini

H = PG(3,q).

H i s a s e t o f type (m,q t m ) 2 w i t h respect t o planes,

H = PG(3,q). Since n e i t h e r a t y p e (0,3)1 nor a type (0,3’1-1)1

h s e t e x i s t s i n PG(2,3 )

(see [ 21, [ 7 ] ) , from prop. X V I i t f o l l o w s : Proposition X V I I .

Neither a (0,3)-p-type h 3 (h 2 2 ) . e x i s t s i n PG(3,q) when q

nor a (0,3

h-1

)-p-type l i n e k-set

REFERENCES [ 1 I F. Buekenhout, C. Lefevre, Generalized Quadrangles i n P r o j e c t i v e Spaces,

Arch. Math. XXV, 5 (1974) 540-552.

[ 2 1 A. Cossu, Su alcune p r o p r i e t a d e i (k,n)-archi i n un piano p r o i e t t i v o sopra un corpo f i n i t o , Rend. Mat. ( 5 ) , 20 (1961) 271-277. 131

D. Olanda, Sistemi r i g a t i imnersi i n uno spazio p r o i e t t i v o , Relaz. n.26,

141

D. Olanda, Sistemi r i g a t i immersi i n uno spazio p r o i e t t i v o , Rend. Acc.

1st. Mat. Univ. Napoli, (1973) 1-20.

-

Naz. Lincei, (8) 62 (1977) 489-499.

[ 5 1 G. T a l l i n i , Problemi e r i s u l t a t i s u l l e geometrie d i Galois, Relaz. n.30, 1st. Mat. Univ. Napoli, (1973) 1-30. 161 G. T a l l i n i , I k-insiemi d i r e t t e d i PG(d,q) s t u d i a t i r i s p e t t o a i f a s c i d i r e t t e , Quad. Sem. Geom. Comb. n. 28, Parte I,Giugno 1980. 1st; Mat. Univ. Roma, 1-17. [ 7 1 J.A.Thas, Some r e s u l t s concerning { ( q + l ) ( n - 1);n)-arcs and { ( q + l ) ( n - 1) + 1;nl-arcs i n f i n i t e p r o j e c t i v e planes o f order q, J . Comb. Th. (A) 19 (1975) 228-232.

ANNALS OF DISCRETE MATHEMATICS VOl. 1:

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