A Characterization of Translation Hyperovals

View metadata, citation and similar papers at core.ac.uk brought to you by CORE

provided by Elsevier - Publisher Connector

Europ. J. Combinatorics (1997) 18, 893±899

ANTONIO MASCHIETTI

In this paper, a characterization of translation hyperovals in any translation plane of even order and one of regular hyperovals in Desarguesian planes of even order are given. These characterizations make use of the concept of a strongly regular secant line. c 1997 Academic Press Limited

1. INTRODUCTION

An oval in a projective plane is a subset of points satisfying the following two properties: O (i) any three points of are never collinear; (ii) has exactly one tangentO line at each of its points. O If the plane is ®nite and has order q, then an oval consists of q 1 points. The problem of classifying ovals is still open, even in the case+ of Desarguesian planes. In the ®nite case, if PG(2, q) is the Desarguesian projective plane over the Galois ®eld GF(q) and q is odd, then every oval is the point set of a conic of PG(2, q). This is the famous Segre's theorem [12]. This theorem does not hold when q is even; in this case, there are many different classes of ovals which are not conics. For an account on ovals the reader is referred to [5, 8]. Comparable to Segre's theorem is a theorem due to Buekenout [1, 2]: if Pascal's theorem holds for an oval in a (®nite or in®nite) projective plane, then the plane is Pappian and is a conic. O O This theorem is very interesting, since a graphic property of an oval allows one to determine the type of the plane. Many generalizations exist of Buekenhout's theorem; an account of these may be found in [3, 8]. If is an oval in any projective plane, a line ` is called a Pascal line of if all the hexagonsO inscribed in have the following property: if two diagonal points lie onO`, then the third one is also on `.O Buekenhout [2] raised the problem of determining the set of the Pascal lines of an oval in a ®nite projective plane. In PG(2, q), q even, the Pascal lines of an oval form one of the following sets (see [4, 7]): (1) the empty set; (2) one tangent, and the oval is a translation oval; (3) all lines, and the oval is a conic. In this paper we study a similar problem. We recall the concept (introduced in [10]) of a regular triple with respect to a hyperoval (a set of q 2 points, any three of which are not collinear) in a projective plane of even order q and that+ of a strongly regular secant line. Then Theorem 2 below, which holds in any translation plane, and its corollaries, stated in a Desarguesian plane, yield: In PG(2, q), q even, the strongly regular lines of a hyperoval form one of the following sets:

(10) the empty set; (20) one secant, and is a translation hyperoval; (30) all the secant through a point of , and is a regular hyperoval, that is a conic completed by its nucleus, the meet of all the tangent lines to the conic.

0195-6698/97/080893 + 07 $25.00/0 ej960137 c 1997 Academic Press Limited

894 A. Maschietti

2. HYPEROVALS WITH A STRONGLY REGULAR SECANT LINE

Let be a hyperoval of a projective plane of even order q, 5q . We will denote the point set of 5q by the same symbol 5q ; furthermore, if P and Q are distinct points of 5q , the symbol PQ will denote the line on P and Q.

DEFINITION 1. Let be a hyperoval of 5q . An unordered triple X, Y, Z of distinct points { } of 5q is said to be -regular if for every point P of 5q at least one of the lines PX, PY, PZ is\-secant.

For every point P of 5q ,weset \

E(P) Q 5q Q P, and the line PQ is -exterior ={ ∈ \ | 6= } S(P) Q 5q Q P, and the line PQ is -secant P . ={ ∈ \ | 6= }∪{ } Then the triple X, Y, Z is -regular iff E(X) E(Y ) E(Z) . { } ∩ ∩ =∅

LEMMA 1. Let X, Y, Z be an -regular triple and P any point of 5q .IfPX is -exterior and PY is -secant,{ then }PZ is -exterior.

PROOF. The proof is straightforward. 2

REMARK 1. Because of the complete symmetry of the de®nition of a regular triple, the conclusion of the lemma does not change if we permute the points X, Y and Z. It follows that if the lines PX and PY are both -secant, then the line PZ is -secant.

It is easy to check that the incidence structure the points of which are the points of 5q not on and the blocks of which are the subsets S(P), for every P 5q , is a Hadamard 2-design (see [9]). ∈ \ We collect together some known results about regular triples, the proofs of which can be found in [9] and [10].

RESULT 1. Let be a hyperoval of 5q and () the associated Hadamard design: H (1) X, Y, Z is a regular triple if and only if X, Y, Z is a line of (). { } { } H (2) If X, Y, Z and X, Y, Z 0 are both regular, then Z Z 0. (3) For{ every X} and {Y there is} Z such that X, Y, Z is regular= iff q 4. In this case () is isomorphic to the 2-design of points{ and (hyper)planes} of PG(=3, 2). H (4) If X, Y, Z is regular and q > 4, then X, Y and Z are on the same secant line of . (5) The{ triple }X, Y, Z is regular iff q 0 (mod 4) and every line neither on X, Y nor Z meets E(X{) E(Y )} (and also E(Y ) ≡ E(Z) and E(X) E(Z))inq/4 points, while the lines through∩X, Y or Z meet this set∩ in either 0 or q/∩2 points.

From now on, we suppose that q 8, because of (3) of Result 1. ≥ DEFINITION 2. A secant line s of is said to be strongly regular if for any two distinct points X and Y on s (s ) there is a point Z s (s ) such that the triple X, Y, Z is -regular. \ ∩ ∈ \ ∩ { }

Examples of hyperovals with a strongly regular secant line are the translation hyperovals of a (not necessarily Desarguesian) projective plane of even order [10].

REMARK 2. Since a regular hyperoval of PG(2, 2d ) is a translation hyperoval with respect to any line on its nucleus, it follows that any line on the nucleus is strongly regular. A Characterization of Translation Hyperovals 895

From now on, we suppose that ` is a strongly regular -secant line. We set ` ∞ ∩ ∞ = O, N and ∗ O, N . For every line ` ` on N, we de®ne the following incidence { } ` = \{ } 6= ∞ structure, ` (): D ∞ pointsÐthe points of `∗ ` O, N ; ∞ = ∞ \{ } blocksÐthe points of `∗ ` (` ), denoted by [P], [Q] and so on; incidenceÐX is incident= with\ [P]∩ if the line PX is -secant.

Following [6], we will denote by n(2) the 2-design of points and hyperplanes of the projective geometry of dimension n overP the ®eld GF(2).

` PROPOSITION 1. ` () is a 2 (q 1, q/2 1, q/4 1) symmetric design, and is iso- D ∞ − − − − morphic to d 1(2), for some integer d 3. In particular, the order q of 5q is a power of two, q 2dP. − ≥ = PROOF. See Theorem 3.2 in [11]. 2

Let P be any point of 5q ( ` ). We denote by SP the set of points X `∗ such that \ ∪ ∞ ` ∈ ∞ the line PX is -secant. Then SP is the point set of the block [P]of ` () (here ` is the D ∞ line PN). We remark that SP q/2 1 and that if U and V are distinct points on the line | |= − ` and U, V ` , then SU SV q/4 1. 6∈ ∩ | ∩ |= − For every point P 5q , we set E P E(P) ` and for every point Q 5q , ∈ \ = ∩ ∞ ∈ \ Q P, let rPQ E E and sPQ S S . Since E q/2, it follows that 6= =| P ∩ Q| =| P ∩ Q| | P |= q q sPQ rPQ 1 + 2 − = 2 − ; whence sPQ rPQ 1. = − LEMMA 2. Let P be a point of 5q ( ` ) and ` a line on P, different from the lines PO and PN. Then, letting Q vary in `\ (`∪ `∞ ), Q P, Q ,if` is an exterior line, \ ∩ ∞ 6= 6∈ q2 rPQ , (1) Q = 4 X 2 2 q (q 2) rPQ + , (2) Q = 16 X while, if ` is a secant line, q(q 2) rPQ − , (3) Q = 4 X 3 2 q rPQ . (4) Q = 16 X PROOF. Let ` be an exterior line. Equality (1) is obtained counting in two ways the pairs (Z, Q) E P [` (` ` )], such that Q P and the line QZ is exterior. In fact, if ∈ × \ ∩ ∞ 6= ` ` Z, then Z E P and ∩ ∞ = ¯ ¯ ∈ q 1, if Z Z (Z, ) q − = ¯ | · |= 1, otherwise, ( 2 − ( , Q) rPQ. | · |= Hence equality (1) follows. 896 A. Maschietti

To obtain (2), we count the (ordered) triples (Z, Z 0, Q) E P E P [` (` ` )], such ∈ × × \ ∩ ∞ that Z Z 0, and Q P and the lines QZ and QZ0 are both exterior. Then the number (Z, Z ,6=) equals either6= q/2 1, if Z Z or Z Z,orq/4 1, since Z, Z belong to a 0 ¯ 0 ¯ 0 regular| triple· | and Z, Z , Z is− never regular,= as the= lines PZ, PZ− , P Z are all three exterior. 0 ¯ 0 ¯ Furthermore, { } ( , , Q) rPQ(rPQ 1). | · · |= − Thus q q q q q2 rPQ(rPQ 1) 1 (q 2) 1 1 (q 2) (q 2). Q − = 2 − − + 4 − 2 2 − − − = 16 − X Using (1), equality (2) follows. Now let ` be a secant line and Z ` ` . Equality (3) is obtained proceeding as for (1). ¯ = ∩ ∞ To obtain (4), we count the (ordered) triples (Z, Z 0, Q) E P E P [` (` ` )], such that Z Z , Q P and the lines QZ, QZ are both exterior.∈ Then× × \ ∩ ∞ 6= 0 6= 0 q 1, if Z, Z , Z is regular, 2 0 ¯ (Z, Z 0, ) q − { } | · |= 1, otherwise,  4 − ( , , Q) rPQ(rPQ 1). | · · |= − Hence equality (4) follows. 2

THEOREM 1. Let ` be a strongly regular secant line of . Then there are q 1 other ∞ − hyperovals 1,...,q 1 such that: − (1) i ` O, N , for every i 1,...,q 1; ∩ = ∩ ∞ ={ } = − (2) P, Q i iff S S ; ∈ P = Q (3) ∗,1∗,...,q∗ 1, where i∗ i O, N , is a partition of 5q ` . − = \{ } \ ∞ ` PROOF. We know that the 2-design ` () is isomorphic to d 1(2), for every line ` ` D ∞ ` P − m 6= ∞ on N.Ifm ` is another line on N, then ` () is isomorphic to ` (). Moreover, these 6= D ∞ D ∞ two designs have the same lines, which are the -regular triples. Then for every point P `∗ there is exactly one point Q m such that S S. It follows that the equivalence relation∈ ∈ ∗ P = Q on 5q ( ` ) \ ∪ ∞ P Q if S S ∼ P = Q has q 1 equivalence classes, each of which has q points. We− prove that each equivalence class is a q-arc. It suf®ces to show that for every point P 5q ( ` ) on each line through P there is at most one point Q P such that S∈ S\. This∪ is∞ clear when the line is either the line PO or PN. Let t be6= a line on P, P = Q different from the lines PO and PN. Let Q P be a point on t such that S S. Then ¯ 6= P = Q if t is an -exterior line, equalities (1) and (2) of Lemma 2 become ¯

2 q 2 q rPQ (q 2) , r (q 2) , = − 4 PQ = − 16 Q Q Q Q X6= ¯ X6= ¯ whence rPQ q/4 and sPQ q/4 1. The same result is obtained when t is an -secant = = − line. Therefore we may deduce that on every line on P 5q ( ` ), distinct from the lines PO and PN, there is exactly one point Q P such that∈ S \ S∩, while∞ on the lines PO and 6= P = Q PN for any two distinct points Q and R we have S S q/4 1. Thus each equivalence | Q ∩ R |= − class is a q-arc. If i∗ is an equivalence class, then the (q 2)-set i i∗ O, N is a hyperoval. Properties (1) and (3) follow. + = ∪{ } 2 A Characterization of Translation Hyperovals 897

DEFINITION 3. The set () ,1,...,q 1 is called the hyperoval bundle de®ned by . P ={ − }

LEMMA 3. Let ` be a strongly regular secant line of the hyperoval , () the hyperoval bundle and a hyperoval∞ such that ` . Then the followingP propositions are ¯ ¯ equivalent: ∩ ∞ = ∩ ∞

(a) (); ¯ ∈ P (b) for every X `∗ ` ( ` ), if a line on X (different from ` ) is both -secant (resp., -exterior)∈ ∞ = and∞\-secant,∩ ∞ then every -secant (resp., -exterior)∞ line on X is an ¯ -secant line; ¯ (c) and have the same regular triples. ¯

PROOF. (a) (b) Let i () and `X ` any line on X.If`X is both an ⇒ ¯ = ∈ P 6= ∞ -secant and i -secant line, then X S for every P , because of Theorem 1, part (2). ∈ P ∈ i∗ Therefore i∗ S(X). Similarly, if `X is -exterior and i -secant, then i∗ E(X). (b) (a) It⊂ suf®ces to prove that if P, Q , then S S. By way of⊂ contradiction, ⇒ ∈ ¯ ∗ P = Q let X S, but X S. Then the line QX is -exterior and -secant. By hypothesis, ∈ P 6∈ Q ¯ E(X), but the line PX is -secant, a contradiction. ¯ ∗ ⊂ (a) (c) Let i (). We prove that if X, Y, Z is an -regular triple, then ⇒ ¯ = ∈ P { } X, Y, Z is also an i -regular triple. { } Let P be any point of 5q ` and suppose both the lines PX and PY i -exterior. We \ ∞ prove that the line PZ is i -secant. If both PX and PY are -secant, then, since (a) is equivalent to (b), from (b) it follows E(X) and E(Y ). Therefore E(X) E(Y ).As X, Y, Z is -regular, so i∗ ⊂ i∗ ⊂ i∗ ⊂ ∩ { } E(X) E(Y ) S(Z); whence i∗ S(Z), which implies that the line PZ is i∗-secant. The∩ other cases,⊂ that is when both⊂ PX and PY are exterior or PX is -secant and PY is -exterior, give the same result. ` ` (c) (b) For every line ` ` on N, the projective geometries ` () and ` () ⇒ 6= ∞ D D ¯ determined by and , respectively, are isomorphic. As and have∞ the same regular∞ ¯ ¯ triples, which are the lines of ` () and ` (), for every block [P]of ` () there is ` ` ¯ ` ` D ∞ D∞ D ∞ exactly one block [P]of () such that S S ¯ ; moreover, P iff P . Thus an ¯ D` ¯ P = P 6∈ ¯ ¯ 6∈ isomorphism σ between ` ∞() and ` () is the¯ following: σ is the identity on the point ` ` ¯ D∞ D ∞ set and σ([P]) [P]ifS S ¯ . = ¯ P = P We remark that the block set of¯ ` () contains the block [A], where A ` , while D` ¯ ¯ = ∩ ¯ ∗ the block set of ` () contains the∞ block [A], where A ` .IfA A, then the two ` ¯ ∗ ¯ projective geometriesD ∞ have the same block sets. In this case= σ ∩is an automorphism,= which is the identity, since σ ®xes every point. This means that SP SP¯ for every point P N on `. Let () , ,..., and () , ,...,= be the hyperoval6= bundles 1 q 1 ¯ ¯ ¯ 1 ¯ q 1 de®nedP by ={and respectively− } and `Pa line={ on N different from− } ` . Denoting by the ¯ ¯ i hyperoval of () passing through the point A ` , let X be any∞ point of ` such that ¯ ∗ ∗ the line ` PAX is -secant. Then the line ` =is ∩-secant. The hyperovals and∞ have X ¯ X ¯ i the same regular= triples and the projective geometries ` () and ` ( ) have the same ` ` ¯ i D ∞ D ∞ ¯ i block sets. Therefore, because of the previous remark, SP SP , for every point P on `. This means that the line PX is -secant iff it is -secant. Since= the line ` is both -secant, ¯ i X ¯ then every -secant line on X is also -secant. Thus we have proved that for every point P ¯ i ¯ on ` N the line PX is -secant iff PX is -secant. ¯ Similarly,\{ } we can prove that if ` is an -exterior and -secant line on X, then every X ¯ -exterior line on X is -secant. 2 ¯ 898 A. Maschietti

3. THE CHARACTERIZATION THEOREM

d THEOREM 2. Let 5q be a translation plane of even order q 2 , d 3, with translation line ` . A hyperoval is a translation hyperoval iff the line ` =is a strongly≥ regular -secant line. ∞ ∞

PROOF. In [10] it has been proved that the axis ` of a translation hyperoval is a strongly regular secant. Conversely, assume that ` is a strongly∞ regular secant of the hyperoval . We shall prove that is a translation hyperoval∞ with axis ` . We denote the translation group ∞ of 5q by T . We set O, N ` . For every elation g{ with}= center∞O∩ (resp., N) and axis ` , the hyperoval g has the same ∞ regular triples of . Since the plane 5q admits all the elations with center O and axis ` , then g ∞ by Lemma 3 the hyperoval bundle de®ned by is O () g TO , where TO is the P ={ | ∈ } subgroup of all the elations with center O and axis ` . Similarly, since 5q has all the elations ∞ h with center N and axis ` , the hyperoval bundle is also N () h TN , where TN ∞ P ={ | ∈ } is the subgroup of all the elations with center N and axis ` . Thus O () N (). Hence g ∞h P gh = P for every g TO there is just one h TN such that ; that is, . We have thus a group of elations∈ with axis ` which∈ stabilizes and= is regular on the points= of O, N . Therefore is a translation hyperoval.∞ \{ 2}

COROLLARY 1. A hyperoval of PG(2, 2d ), d 3, is regular iff has two distinct strongly regular secant lines. The nucleus of is the meet≥ of the two secant lines.

PROOF. This follows from Theorem 2, since a hyperoval of PG(2, 2d ) which is a translation hyperoval with two distinct axes is a regular hyperoval, the nucleus of which is the meet of the two axes. Furthermore, as already noticed (see Remark 2 in Section 2), for a regular hyperoval every secant line on the nucleus is strongly regular. 2

COROLLARY 2. Let be a regular hyperoval with nucleus N of PG(2, 2d ), where d 3. Then only the lines through N are strongly regular. ≥

PROOF.Ifs is strongly regular line not on N, then sg is again a strongly regular secant line, for every g PSL(2, 2d ). Therefore all the secant lines of are strongly regular. Hence is a complete∈ conic of nucleus X, for every X . This happens only if q 2orq 4 (see, e.g., [5]). ∈ = = 2

REMARK 3. It is known (see, e.g., [5]) that a translation hyperoval in the Desarguesian plane PG(2, 2d ) has the canonical form

n (1, t, t2 ) t GF(2d ) (0, 1, 0), (0, 0, 1) , 1 n d 2,(n, d) 1. ={ | ∈ }∪{ } ≤ ≤ − = The axis of is the line ` with equation x0 0. One easily checks that the triple X (0, 1,αk), Y (0, 1,βk)∞, Z (0, 1, αβ/(α =β)), with k 2n 1, α β, α 0, β =0, is a regular triple.= Moreover,= the hyperoval bundle+ de®ned by= contains− the6= hyperovals6= 6= 2n d d λ (1, t λ, t ) t GF(2 ) (0, 1, 0), (0, 0, 1) ,λGF(2 ). ={ + | ∈ }∪{ } ∈

REFERENCES

1. ÐÐF. Buekenhout, Plans projectifs aÁ ovoides pascaliens, Arch. Math., 17 (1966), 89±93. 2. ÐÐF. Buekenhout, Etude intrinsequeÁ des ovales, Rend. Math. Applics., 25 (1966), 333±393. 3. ÐÐG. Faina, Geometric and algebraic theory of B-ovals: results and problems, in: Giornate di Geometrie Combinatorie, G. Faina and G. Tallini (eds), UniversitaÁ degli Studi di Perugia, 1993, pp. 175±212. 4. ÐÐG. Faina and G. Korchmaros, Desarguesian con®gurations inscribed in an oval, Ann. Disc. Math.,14 (1982), 207±210. A Characterization of Translation Hyperovals 899

5. ÐÐJ. W. P. Hirschfeld, Projective Geometries over Finite Fields, Clarendon Press, Oxford, 1979. 6. ÐÐD. R. Hughes and F. C. Piper, Design Theory, Cambridge University Press, Cambridge, 1985. 7. ÐÐG. Korchmaros, Sulle ovali di traslazione in un piano di Galois di ordine pari, Atti Accad. Naz. XL Rend., 5±III (1977), 55±65. 8. ÐÐG. Korchmaros, Old and new results on ovals of ®nite projective planes, in: Surveys in Combinatorics 1991, A. D. Keedwell (ed.), London Mathematical Society Lecture Note Series, 166 (1991), 41±72. 9. ÐÐA. Maschietti, Hyperovals and Hadamard designs, J. Geom., 44 (1992), 107±116. 10. ÐÐA. Maschietti, Regular triples with respect to a hyperoval, Ars Combin., 39 (1995), 75±88. 11. ÐÐA. Maschietti, On Hadamard designs associated with a hyperoval, J. Geom., 53 (1995), 122±130. 12. ÐÐB. Segre, Ovals in a ®nite projective plane, Can. J. Math., 7 (1955), 414±416.

Received 19 July 1995 and accepted 13 January 1997

A. MASCHIETTI Dipartimento di Matematica `G. Castelnuovo', UniversitÁadegli Studi `La Sapienza', p.le A. Moro, 2, Roma, Italy E-mail: [email protected]