Ovals in Desarguesian planes

This paper surveys the known ovals in Desarguesian of even order, making use of the connection between ovals and hyperovals. First the known hyperovals are and the inequivalent m of small order arc found. The ovals contained in each of the known are determined and presented in a uniform way. Computer for new hyperovals reported. 1. OVALS AND HYPEROVALS Let PG(2, q) be the 'DC," "<"""'0" -:)"t, projective plane over the field G F( q) of order q, where q is a power of a prime p. An oval of PG(2, q) is (1 set of q + 1 points, no three of which are collinear. The points of a non-degenerate conic in PG(2, q) form an oval. When q is odd, the converse is true, so that every (q + 1 )-arc is the set of points of a non-degenerate conic ([12; 5, 8.2.4]). vVhen q is even, examples of non-conic ovals are known, and a complete classification of ovals has not yet been effected. A line of PG(2, q) meets an oval in either 2 points, 1 point or 0 points, in which case it is called a secant, a tangent or an external line ,iVhen q is even, the set of tangents to an oval all pass through a common point. This point can be adjoined to the oval to give a set of q + 2 points, no three collinear. Such a set is called a hyperoval and the unique point which is adjoined to an oval to obtain a hyperoval is called the mLcle1Ls of the oval. An a,ccount of ovals and hyperovals appears in Hirschfeld [5]. Australasian Journal of Combinatorics 1(1990)t pp.149-159 Given a hyperova.l, an oval can be obtained by deleting one of the points of the hyperoval. This deleted point is the nucleus of the resulting oval. There are up to q + 2 ovals which can be obtained from a hyperoval in this way, but we only distinguish those which are distinct under the action of the automorphism group pr L(3, q) of PG(2, q). Sets of points which are images of one another under elements of q) are called eq'ui'valeni. 1.1 Theorem Let H be a hyperoval in PG(2, q), q even, and let G be the stabiliseI' of H in pr L( 3, q). The ovals obtained by the P and Q of Hare if and only if P and Q lie in the same orbit of G on H. Proof: Let 0 1 and O 2 be ovals such that First suppose that and for some element u E q). Since u maps lines to and P and Q are the intersections of the tangents of 0 1 and O 2 respectively, u(P) = Q. Thus so u E G and thus P and Q are in the same orbit of G on n. Conversely, suppose that P and Q are in the same orbit of G 011 H. Then there exists an element u E G such that u(P) = Q. Since u fixes H, u(Ol) = O 2 and the result follows. Thus to study ovals when q is even, it is useful to first find hyperovals, then determine the possible ovals by finding the stabiliseI' of each hyperoval. For the rest of the paper we suppose that q is evell, ~o that q 2h for some integer h. 2. THE KNOWN HYPEROVALS OF PG(2,q), q = 211 A polynomial with coefficients in G F( q) \v11ic11 induces a permutat.ion on the 150 There a useful canonical elements of G F( q) is called permutation 'fII11.'11'11. ,fj 'ff I. '/ll.I. form for hyperoval in terms of IJ,,"A. LU'UUVClH'-'" poly'nomlals, follows. 2.1 Theorem [5, written A o in q) where q is even can be D(f) {(I, t,f(t»): t E GF(q)} U {(O, 1,0), 0,1)} where f a IJ ... ·Je.Lu'u.u",U."·H of and f(1) 1. for each E