OVOIDS OF PG(3, q), q EVEN, WITH A CONIC SECTION

Transcription

1 OVOIDS OF PG(3, q), q EVEN, WITH A CONIC SECTION MATTHEW R. BROWN ABSTRACT It is shown that if a plane of PG(3, q), q even, meets an ovoid in a conic, then the ovoid must be an elliptic quadric. This is proved by using the generalized quadrangles T () ( a conic), W(q) and the isomorphism between them to show that every secant plane section of the ovoid must be a conic. The result then follows from a well-known theorem of Barlotti. 1. Introduction and definitions Throughout this paper we will assume that q is even. We will be considering ovoids of PG(3, q) which contain a conic as a plane section by using different representations of the classical generalized quadrangle of order q. An oal of PG(2, q) is a set of q1 points of PG(2, q), with no three collinear. Let be a line of PG(2, q); then is incident with zero, one or two points of and is accordingly called an external line, a tangent or a secant to. An elementary count shows that there is a unique tangent to incident with a given point of. If is tangent to and incident with the point P, then we say that is the tangent to at P. Ifq is even, then the tangents to are coincident in a fixed point, the nucleus of (see [9, Lemma 8.6]). The classical example of an oval in PG(2, q) is the set of points satisfying an irreducible quadratic equation, called a conic (more precisely this is a non-degenerate conic). A hyperoal of PG(2, q) is a set of q2 points, with no three collinear. An oval of PG(2, q) together with its nucleus is a hyperoval of PG(2, q). An ooid Ω of PG(3, q) is a set of q1 points of PG(3, q), with no three collinear. From this point we will assume that q 2, so that an ovoid is a maximal-sized set of points, with no three collinear. Let π be a plane of PG(3, q); then π meets Ω in a single point or in an oval of π and is called accordingly a tangent plane or a secant plane. The intersection of Ω and a secant plane of Ω is called a secant plane section of Ω. There is a unique tangent plane to Ω containing a given point P Ω. This plane is the tangent plane to Ω at P. (See [1, 2, 18] for the above.) If q is odd, then ovoids of PG(3, q) have been classified as the non-degenerate elliptic quadrics of PG(3, q) (see [1, 18]). For q even, q 2h, the two known isomorphism classes of ovoids are the non-degenerate elliptic quadrics, which exist for all h 1, and the Tits ovoids (see [10, Chapter 16]) which exist for h odd, h 3. Most results characterizing ovoids of PG(3, q) have been in terms of the secant plane sections. Barlotti [2] proved that if every secant plane section of an ovoid is a conic, then the ovoid must be an elliptic quadric. In [22] Segre strengthened this result (for q 8) to say that if at least (qq2q)2 secant plane sections of an ovoid are conics, then the ovoid must be an elliptic quadric. The motivation for further results of this type characterizing ovoids is the connection of ovoids with inversive planes Received 19 November 1998; revised 15 July Mathematics Subject Classification 51E20, 51E12 (primary), 51E21 (secondary). J. London Math. Soc. (2) 62 (2000) London Mathematical Society 2000.

2 570 MATTHEW R. BROWN (see [5]). In particular these results concern bundles, pencils and flocks of ovoids. A bundle (respectively, pencil) of an ovoid Ω is a set of q1 (respectively q) distinct secant plane sections of Ω whose corresponding set of secant planes intersect pairwise in a fixed line meeting Ω in two points (respectively one point). A flock of Ω is a set of q1 secant plane sections of Ω partitioning all but two points of Ω. Prohaska and Walker [21] proved that if every element of a bundle of an ovoid Ω is a conic, then Ω is an elliptic quadric, while Glynn [7] showed that if every element of a pencil is a conic, then the ovoid is an elliptic quadric. The corresponding result for flocks was proved by Brown, O Keefe and Penttila [3]. In this paper we prove that if an ovoid of PG(3, q) has a secant plane section that is a conic, then it must be an elliptic quadric. The proof of this result is independent of the bundle, pencil and flock results, relying on only the result of Barlotti [2, 5.2.7] which states that if every secant plane section of an ovoid is a conic, then the ovoid is an elliptic quadric. This result answers in the affirmative a conjecture made by Glynn in [7]. All of the above results consider cases where some selection of secant plane section(s) consists entirely of conics. The work of Penttila and Praeger [20] and O Keefe and Penttila [15] shows that if an ovoid has a pencil of translation ovals then it is either an elliptic quadric or a Tits ovoid. Further, in [16] O Keefe and Penttila show that if an ovoid has each secant plane section an oval which is contained in a translation hyperoval, then the ovoid is necessarily either an elliptic quadric or a Tits ovoid. The classification of ovoids of PG(3, q), q even, has been completed for the cases q 32 (q 4in[2], q 8in[6], q 16 in [13, 14] and q 32 in [17]). In the cases q 4 and q 16 there are only the elliptic quadric ovoids while in the cases q 8 and q 32 the only ovoids are the elliptic quadrics and the Tits ovoids. Note that the classification of ovoids for q 32 implies that for q 32 an ovoid of PG(3, q) containing a conic must be an elliptic quadric. For an excellent introductory survey of results on ovoids of PG(3, q), see [12]. A (finite) generalized quadrangle (see [19] for a comprehensive introduction) is an incidence structure (,, I) in which and are disjoint (non-empty) sets of objects called points and lines, respectively, and for which I ()() isa symmetric point line incidence relation satisfying the following axioms: (i) Each point is incident with 1t lines (t 1) and two distinct points are incident with at most one line. (ii) Each line is incident with 1s points (s 1) and two distinct lines are incident with at most one point. (iii) If X is a point and is a line not incident with X, then there is a unique pair (Y, m) for which X I m I Y I. The integers s and t are the parameters of the generalized quadrangle and is said to have order (s, t). If s t, then is said to have order s. If has order (s, t), then it follows that (s1) (st1) and (t1) (st1) [19, 1.2.1]. If (,, I) is a generalized quadrangle of order (s, t) then the incidence structure (,, I) is a generalized quadrangle of order (t, s) called the dual of. The classical generalized quadrangles of order q (q a prime power) are Q(4, q), which arises as the points and lines of the non-singular (parabolic) quadric in PG(4, q), and W(q), which is defined as the singular points and lines of a symplectic polarity of PG(3, q). From [19, 3.2.1] we have Q(4, q) W(q) and Q(4, q) W(q) if and only if q is even.

3 OVOIDS OF PG(3, q), q EVEN, WITH A CONIC SECTION 571 Another class of generalized quadrangles of order q are those constructed by Tits (first appearing in [5], see [19]). Let be an oval of PG(2, q) and let PG(2, q) beembedded as a hyperplane in PG(3, q). The points are (i) the points of PG(3, q)pg(2, q), (ii) the planes of PG(3, q) which meet PG(2, q) in a single point of, and (iii) a symbol (). The lines are (a) the lines of PG(3, q) which meet PG(2, q) in a single point of, and (b) the points of. Incidence is as follows: a point of type (i) is incident only with the lines of type (a) which contain it, a point of type (ii) is incident with all lines of type (a) contained in it and with the unique line of type (b) on it, and a point of type (iii) is incident with no line of type (a) and with all lines of type (b). The generalized quadrangle is denoted by T (). From [19, 3.2.2] we have T () Q(4, q) if and only if is a conic and T () W(q) if and only if q is even and is a conic. An ooid Ω of a generalized quadrangle of order (s, t) is a set of points such that each line of is incident with precisely one point of Ω. It follows that Ω has st1 points. If Ω is an ovoid of PG(3, q), then Ω defines a polarity of PG(3, q). If π is a plane of PG(3, q) which is tangent to Ω at the point P, then the polarity interchanges π with P, and if π meets Ω in the oval with nucleus N, then the polarity interchanges π with N. This polarity is necessarily a symplectic polarity of PG(3, q) and so the lines tangent to Ω form a linear complex of PG(3, q). Also, Ω is an ovoid of the generalized quadrangle W(q) constructed from the symplectic polarity that it defines. Conversely, if Ω is an ovoid of W(q), then by [23] Ω is also an ovoid of PG(3, q). 2. The isomorphism from T () to W(q) The following isomorphism from T () tow(q) is constructed by composing the inverse of the isomorphism from Q(4, q) tot () in the proof of [19, 3.2.2] with the isomorphism from Q(4, q) tow(q), for q even, in the proof of [19, 3.2.1]. We then apply an automorphism of W(q) to put the isomorphism from T () tow(q) into a preferred canonical form. We introduce explicit coordinates here for use later. First let q be even and let π be the plane of Σ PG(3, q) defined by the equation x 0. Let be the conic in π defined by the equations x x x x 0, that is, (1, t, t/,0):t GF(q)(0, 1, 0, 0), with nucleus (0, 0, 1, 0). We construct T () from Σ, π and as in Section 1. Now embed Σ in PG(4, q) as the hyperplane defined by the equation x 0. Let be the non-singular (parabolic) quadric of PG(4, q) defined by the form x x x x x 0. Thus the singular points and lines of form a generalized quadrangle Q(4, q). Note that. If P is a point of, then let P denote the tangent space to at P, and so we have (0, 0, 0, 0, 1), that is, (0, 0, 0, 0, 1) is collinear, in Q(4, q), with every point of. Thus we may now define an isomorphism from Q(4, q) tot () which essentially acts by projecting from (0, 0, 0, 0, 1) onto Σ. Let φ be the inverse of this isomorphism. We now explicitly determine the action of φ on the points of T (). The affine points of T () (that is, the points of Σπ) have the form (x, x, x,1,0)forx, x, x GF(q) and φ :(x, x, x,1,0) (x, x, x,1,x x x ). If π X is a point of T () which is a plane of Σ meeting π in a line tangent to, then φ maps π X to the point X of (0, 0, 0, 0, 1) such that π X X Σ. Using hyperplane

4 572 MATTHEW R. BROWN coordinates in PG(4, q), let π X [t,1,0,s,0][0, 0, 0, 0, 1] be a plane of Σ meeting π in a line tangent to at (1, t, t/, 0, 0). Since [t,1,0,s,0][0, 0, 0, 0, 1] (1, t, t/,0,s) [0, 0, 0, 0, 1] we have Similarly φ :[t,1,0,s,0][0, 0, 0, 0, 1] (1, t, t/,0,s). φ :[1, 0, 0, s,0][0, 0, 0, 0, 1] (0, 1, 0, 0, s). The point () oft () is mapped to the point (0, 0, 0, 0, 1) of Q(4, q) byφ. Now we consider an isomorphism φ from Q(4, q) tow(q). The point (0, 0, 1, 0, 0) is the nucleus of. Let H be the hyperplane of PG(4, q) with coordinates [0, 0, 1, 0, 0], and so (0, 0, 1, 0, 0) H. Projecting from (0, 0, 1, 0, 0) onto H the points (respectively lines) of Q(4, q) are mapped onto the points (respectively lines) of a generalized quadrangle isomorphic to W(q), defined by the form x y x y x y x y 0. This is the map φ. Thus the action of φ on the points of Q(4, q) is φ :(x, x, x,1,x x x) (x, x,0,1,x x x), for x, x, x GF(q), :(1, t, t/,0,s) (1, t,0,0,s), for s, t GF(q), :(0, 1, 0, 0, s) (0, 1, 0, 0, s), for s GF(q), :(0, 0, 0, 0, 1) (0, 0, 0, 0, 1). The map φ φ is an isomorphism from T () tow(q) that has the following action on the points of T (): φ φ :(x, x, x,1,0) (x, x,0,1,x x x), for x, x, x GF(q), :[t,1,0,s,0][0, 0, 0, 0, 1] (1, t,0,0,s), for s, t GF(q), :[1, 0, 0, s,0][0, 0, 0, 0, 1] (0, 1, 0, 0, s), for s GF(q), :() (0, 0, 0, 0, 1). We now apply the automorphism of W(q) that swaps the x and the x coordinates to put the isomorphism from T () to W(q) into the preferred canonical form presented in the following lemma. LEMMA 2.1. Let π be the plane of PG(3, q), qeen, defined by the equation x 0. Let be the conic in π defined by the equations x x x x 0, that is, (1, t, t/,0):t GF(q)(0, 1, 0, 0) with nucleus N (0, 0, 1, 0). Construct T () from Σ, π and. Let W(q) be the generalized quadrangle defined as the singular points and lines of the symplectic polarity of PG(3, q) with form x y x y x y x y 0. Then there exists an isomorphism φ from T () to W(q) that acts on the points of T () by (x, x, x,1) (x, x, x x x, 1), for x, x, x GF(q), [t,1,0,s] (1, t, s, 0), for s, t GF(q). [1, 0, 0, s] (0, 1, s, 0), for s GF(q), () (0, 0, 1, 0).

5 OVOIDS OF PG(3, q), q EVEN, WITH A CONIC SECTION Ooids of PG(3, q) containing a conic 3.1. Projectie ooids of T () Let π be a plane of Σ PG(3, q) and let be an oval in π. Let T () be the generalized quadrangle constructed from Σ, π and and let Ω be an ovoid of T (). If () Ω then of the q1 points of Ω, qq are affine points (that is, points of Σπ) and q1 points are planes of Σ meeting π in distinct lines tangent to. If the qq affine points of Ω have the property that no three are collinear in Σ, then we say that Ω is a projectie ovoid of T (). The reason for this term will become clear from the following lemma. LEMMA 3.1. Let Ω be a projectieooid of the generalized quadrangle T () and let be the qq affine points of Ω. Then is an ooid of PG(3, q). Proof. We check that has the property that no three of its points are collinear in PG(3, q). Let P, Q, R be three distinct points of. IfP, Q, R or P, Q, R, then we have the property that they are not collinear in PG(3, q). Hence, without loss of generality, let P, Q and R. Since P, Q is a line of π and R π it follows that P, Q, R are not collinear in PG(3, q). Now (again without loss of generality) let P and Q, R. Since any line of PG(3, q) incident with P and not contained in π is a line of T (), it follows that P, Q, R are not collinear in PG(3, q). Thus is an ovoid of PG(3, q). We also have the following converse. LEMMA 3.2. Let Ω be an ooid of PG(3, q), and let the oal be the intersection of the plane π with Ω. Then (Ω)π P :π P is the tangent plane to Ω at a point P is an ooid of the generalized quadrangle T () constructed from, π and PG(3, q). Proof. This is essentially the reverse of Lemma 3.1. Now we restrict our attention to the case of an ovoid Ω of PG(3, q) that contains a conic section. By Lemma 3.2, Ω gives rise to an ovoid Ω of the generalized quadrangle T (). By using the isomorphism from T ()tow(q) we may map Ω onto an ovoid Ω() of W(q). It follows from [23] that Ω() is also an ovoid of PG(3, q). We will now start with Ω in a canonical form and apply the explicit form of the isomorphism generated in Section 2 to construct the ovoid Ω() of PG(3, q). Note that while we shall eventually prove that Ω() and Ω are elliptic quadrics it is not immediate at this point that Q() and Ω are projectively equivalent in PG(3, q). Let Ω be an ovoid of PG(3, q) meeting the plane π in a conic. We may assume that the symplectic polarity of PG(3, q) defined by Ω has form x y x y x y x y 0 (by applying the Klein correspondence to [11, Theorem ]). We may also assume that the plane π has equation x 0 and that the conic has equation x x x x 0, that is, (1, t, t/,0):tgf(q)(0, 1, 0, 0) and has nucleus N (0, 0, 1, 0) (see [3; 11, Theorem ]). By Lemma 3.2, Ω gives rise to an ovoid Ω of T () where Ω (Ω)[t,1,0,t/]:tGF(q)[1, 0, 0, 0].

6 574 MATTHEW R. BROWN Using the isomorphism φ in Lemma 2.1, let φ(ω ) Ω(). Thus the symplectic polarity associated with Ω() has the form x y x y x y x y 0. Also [t,1,0,t/] (1, t, t/, 0) for t GF(q) and [1, 0, 0, 0] (0, 1, 0, 0), that is, Ω(). Now since Ω() and Ω() defines the same symplectic polarity as Ω it follows that we may repeat the above to generate another ovoid of PG(3, q) from Ω(). In fact we may repeat this process ad infinitum. However for our purposes here one iteration of the process will be sufficient. We will now use φ and the fact that φ(ω ) Ω() is an ovoid of PG(3, q) to generate conditions on some of the secant plane sections of Ω. It should be noted that in [7] Glynn investigated the properties of the restriction of φ to AG(3, q) (without explicitly using the generalized quadrangle isomorphism) and also observed that φ(ω ) is an ovoid of PG(3, q) containing Restrictions on a secant plane section of the ooid Ω of PG(3, q) In this section we will derive algebraic conditions on some secant plane sections of an ovoid Ω of PG(3, q) containing a conic. To do this we use the o-polynomial representation of hyperovals of PG(2, q). By the fundamental theorem of projective geometry any hyperoval of PG(2, q) is isomorphic to a hyperoval containing the fundamental quadrangle (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1). Consequently we may write in the form ( f(t), t,1):t GF(q)(1, 0, 0), (0, 1, 0), where f is a permutation polynomial over GF(q). Note that from [9, Section 1.3] the natural map from the polynomials over GF(q) with degree less than q to functions from GF(q)toGF(q) is a bijection. Thus, as with f above, we will abuse notation and use the same symbol to represent both a function and the unique polynomial of degree less than q that generates the function. Any polynomial with degree less than q that arises from a hyperoval as above will be called an o-polynomial (following Cherowitzo [4]). From [9, Theorem 8.22] we have the following theorem. THEOREM 3.3. If q 2, a permutation polynomial f with f(0) 0 and f(1) 1 is an o-polynomial if and only if for each element s GF(q), the function f s (x) [ f(xs)f(s)]x, with f s (0) 0, is a permutation polynomial oer GF(q). In Section 4 we will state a result by Glynn which gives an alternative description of o-polynomials and is more convenient for calculations that follow. Returning to our considerations in Section 3.1, let Ω be the ovoid of PG(3, q) as in Section 3.1 and let π a be the plane defined by the equation x ax, for a 0. Note that in the case where a 0 the plane π is the tangent plane to the ovoid Ω of PG(3, q) at the point (0, 1, 0, 0) and hence a point of the ovoid Ω of T (). We will now abuse notation and let φ, which is an isomorphism from T () to W(q), also represent the map on AG(3, q) PG(3, q)π that it induces. Thus φ:(x, x, x,1) (x, x, x x x, 1).

7 OVOIDS OF PG(3, q), q EVEN, WITH A CONIC SECTION 575 Now φ fixes the set of points of any plane of AG(3, q) whose projective completion contains N, the nucleus of. In particular φ fixes the set of points of the affine part of the plane π a (that is, the affine plane π a AG(3, q)) and the restriction of φ to this affine part of π a is φ πa :(a, x, x,1) (a, x, x ax, 1). Now let a Ωπ a ; then a is an oval of π a containing (0, 1, 0, 0) and with nucleus (0, 1, a, 0) (since (0, 1, a, 0) is the image of π a under the polarity defined by Ω). For convenience we will represent the plane π a in the coordinates x, x, x by omitting the x coordinate. Thus we write a (F(t), t,1):t GF(q)(1, 0, 0), for some function F. Let BaF(0) and AaF(1)1aF(0). Since the line spanned by (F(1), 1, 1) and (F(0), 0, 1) does not contain (1, a, 0), the nucleus of a, it follows that A0. Consider ( f(t), t,1):t GF(q)(1, 0, 0), where f(t) af(t)tb. (1) A This is an oval projectively equivalent to a and on the fundamental quadrangle; so f is an o-polynomial. Now φ πa :(F(t), t,1) (F(t), taf(t), 1) and so () (F(t), taf(t), 1):t GF(q)(1, 0, 0) a (F(t), taf(t), 1):t GF(q)(1, 0, 0) is the oval that is the intersection of π a with the ovoid Ω(). It follows that () has a nucleus (1, a, 0). By similar reasoning to that used for the oval a,if g(t) taf( t)b, (2) A then (t, g(t), 1):t GF(q)(1, 0, 0) is projectively equivalent to () and g is an a o-polynomial. Combining (1) and (2) and switching to the indeterminate x, we have g(x) A(xx)f(x), where A 0. So far we have only considered planes of PG(3, q) meeting π in a line tangent to at the point (0, 1, 0, 0). We now extend our considerations to any plane meeting π in a line tangent to. Let ψ t be the collineation of PG(3, q) that acts on points of PG(3, q) by ψ t :(x, x, x, x ) (tx x t/x,(t1) x tx t/(t1) x, t/(t1) x t/x x tx, x ). Then ψ t fixes, commutes with the symplectic polarity defined by Ω and interchanges the point (0, 1, 0, 0) with the point (1, t, t/, 0). Thus if we let Ω t ψ t (Ω), then we may apply the discussion of this section to the ovoid Ω t. Consequently we have the following lemma.

8 576 MATTHEW R. BROWN LEMMA 3.4. Let Ω be an ooid of PG(3, q) and π be a plane of PG(3, q) such that πω is a conic. If π is any plane of PG(3, q) such that ππ is a line tangent to but not tangent to Ω, then the oal πω is projectiely equialent to an oal ( f(t), t,1):tgf(q)(1, 0, 0) with nucleus (0, 1, 0), where f is an o-polynomial satisfying the equation g(x) A(xx)f(x), (3) for some o-polynomial g and A GF(q)0. In Section 4 we will show that if (3) is satisfied, then g(x) x and f(x) x. Now consider o-polynomials f and g satisfying (3). Since f and g have no constant term and no odd degree terms we may write f(x) f xf x f q xq, g(x) g xg x g q xq. From this it follows that f A 0 and similarly that g q/ A 0, that is, the coefficient of x in g(x) is non-zero. Further, since g has no odd degree terms it follows from (3) that f f f q 0, that is, f has no terms of degree 2n where n is odd. Since f has no term with degree greater than q2 it follows that g q/+ g q/+ g q 0, that is, g has no term of degree greater than q2. Thus f(x) f xf x f q xq f x (q )/ f xi, (4) i i= where f 0, and g(x) g xg xg x g q/ xq/ q/ g xi, (5) i i= where g q/ The classification of ooids of PG(3, q), qeen, containing a conic We now work towards proving that any ovoid of PG(3, q) containing a conic is an elliptic quadric. The bulk of this work is in extending Section 3.2 to prove the following result. THEOREM 4.1. Let Ω be an ooid of PG(3, q), qeen, and π be a plane of PG(3, q) such that πω is a conic. Then eery secant plane section of Ω that intersects in exactly one point is a conic. This result allows us to prove the classification. THEOREM 4.2. Let Ω be an ooid of PG(3, q), qeen, and π be a plane of PG(3, q) such that πω is a conic. Then Ω is an elliptic quadric.

9 OVOIDS OF PG(3, q), q EVEN, WITH A CONIC SECTION 577 Proof. Let πω be the conic and let be a secant plane section of Ω not meeting in exactly one point (and so meets in exactly zero or two points). Let the plane containing be π and the nucleus of be N. Let P be a point of ππ not contained in Ω. The plane P, N, N (where N is the nucleus of ) meets Ω in an oval such that 1. Theorem 4.1 implies that is a conic and consequently that is a conic. Hence by [2, 5.2.7] Ω is an elliptic quadric. To prove Theorem 4.1 we now show that if an o-polynomial f satisfies (3) for some g and A, then, in fact, f(x) x, g(x) x and A 1. By Lemma 3.4 this proves Theorem 4.1 and hence, by the above, we have Theorem 4.2. In Section 3.2 we saw that by using (3) and the fact that an o-polynomial has no odd degree terms we were able to restrict the o-polynomials f and g to the forms given in (4) and (5), respectively. The result that an o-polynomial has no odd degree terms is generalized by a result due to Glynn, which we will use repeatedly in the proof that f(x) x. First we define a partial ordering on the set of integers n where 0 n q1 and q 2h. If b h b 2i i and c h c 2i i i= i= (where each b i and each c i is either 0 or 1) then b c if and only if b i c i for all i. THEOREM 4.3 [8]. A polynomial f of degree at most q2 satisfying f(0) 0 and f(1) 1 is an o-polynomial if and only if the coefficient of xc in f(x)b (mod xqx) is zero for all pairs of integers (b, c) satisfying 1 b c q1, b q1 and b c. Note that b 1 gives the result that an o-polynomial has no odd degree terms. Now we shall show that in general f(x) x. Let q 2h, h 3 and let f be an o-polynomial over GF(q). We say that f is 2k-sparse if 1 k (h1)21 and f(x) f xf k+ x k+ f k+ x k+ f h k+ k+ x h k+ k+ h k f x f i i= k+ xi k+. (6) Let g be another o-polynomial; then we say that g is 2k-spare if 1 k (h1)21 and g(x) g h x h g k x k g k x k g h k k x h k k LEMMA 4.4. then f(x) x. h k g h x h g i i= k xi k. (7) Iffisano-polynomial oer GF(2h), h 3 and f is 2 (h )/ +-sparse, Proof. If k (h1)21, then 2h k1 is12 when h is odd and 0 when h is even. Thus from (6) we have f(x) f x x. Note that it follows from (4) that f is 2-sparse and from (5) that g is 2-spare. We will show that (3) and Theorem 4.3 allow us to proceed inductively to the case when f is 2 (h )/ +-sparse and g is 2 (h )/ +-spare, from which it follows, by Lemma 4.4, that f(x) x.

10 578 MATTHEW R. BROWN LEMMA 4.5. Let f and g be o-polynomials oer GF(2h), h 3, satisfying (3). If 1 k (h1)2, fis2k-sparse and g is 2k-spare, then f is 2k+-sparse and g is 2k+-spare. Proof. First we shall apply Theorem 4.3 to f. Let b 2k+1 and c 2k+1α2k+ where α 0,, (2h k 1)2k2h k 1 (these values of α ensure that c q1). We consider the expansion of f(x)b before reduction modulo xqx. Note that this expansion has the property that deg f(x)b (2k+1) (2h k+2k+) 2h+2k+2k+2h k+. (8) Now we consider which xd xc (mod xqx) have degree small enough to appear in f(x)b, that is, d deg f(x)b. By (8) we have d 2h+2k+2k+2h k+. Nowd cγ(2h1), for γ 0 and from our restriction on d we have 2k+1γ(2h1)α2k+ 2h+2k+2k+2h k+ γ 12 h+2k+2h k+ 2h1 since α 0 γ 4 52k+2 h k+. 2h1 Since k 1 it follows that γ 4, that is, γ 0, 1, 2 or 3. Now if γ is even then d 2k+1α2k+γ(2h1) 2k+γ2hα2k+(1γ) is odd, but since f(x) has no odd degree terms neither does f(x)b. Hence it follows that if γ is even, then xd does not appear in f(x)b. This leaves us with the possibilities that γ 1orγ3. Now suppose that γ 3. Then d cγ(2h1) 2k+1α2k+32h3 4 (mod 2k+) 2k+4(mod2k+). Now recall that h k f(x) f x f i i= k+ xi k+ and so if f(x)b contains a term of degree d c3(2h1), then since c3(2h1) 2k+4 (mod 2k+) it follows that this term of degree d must have the form h k ( f x)t ( f i i= k+ xi k+ )t i, where t 2k2 and t h k t i= i b. Since b (2k2)12k we now have the following upper bound on d: d 2k+4(12k)(2h k+2k+) 432h2k+α2k+ c3(2h1) d, a contradiction. Thus we have γ 3 and so we are left to consider the case γ 1. That is, we look for terms of degree 2k+1α2k+(2h1) in f(x)b.

11 Now OVOIDS OF PG(3, q), q EVEN, WITH A CONIC SECTION 579 2k+1α2k+2h1 2k+2α2k+2h 2k+2 (mod 2k+), and so if T(x) is a term of degree c2h1 off(x)b, then f k (x) k T(x). Since 2k+1 b (1 2k )2k and every term of f(x)b is of the form k ( f xu ui i) i, i= it follows that T(x) ( f x) k S(x) k, where S(x) f u k+ xu k+. Thus (u2k+)2kα2k+2h and hence u α2h k. Since k (h1)2 it follows that h2k1 0 and so u is an integer as required. Thus T(x) ( f x) k ( f α k+ + h k x α k+ + h k ) k is the unique term of degree c2h1 in the expansion of f(x)b before reduction mod xqx. By Theorem 4.3 we have f k f k α k+ + h k 0 and since f 0 it follows that f α k+ + h k 0. Recall that α 0,,2h k 1, and so we have f h k f k+ + h k f k+ + h k f h k+ k+ 0. (9) As a consequence of (3) we also have g h k g k + h k g k + h k g h k k 0. (10) Now we turn our attention to g and apply Theorem 4.3 with b 2k+1 and c 2k+1α2k+, α 0, 1,,2h k 1. We consider the expansion of g(x)b before reduction mod xqx. Note that g(x)b has degree deg g(x)b q 2 b 2 h (2k+1) 2h+k2h. (11) Now we consider which xd xc (mod xqx) have degree small enough to appear in g(x)b, that is, d deg g(x)b. By (11) we have d 2h+k2h.Nowdcγ(2h1), for γ 0, and so we have 2k+1α2k+γ(2h1) 2h+k2h γ 2 h+k2h 2k+1 2h1 γ 2k 2 k2h 2k+1. 2h1 since α 0 Thus γ 2k1. Now cγ(2h1) 2k+1α2k+γ(2h1) (γ1) (mod 2k).

12 580 MATTHEW R. BROWN Since every term of g(x) has degree congruent to 0 mod 2k so does every term of g(x)b (before reduction mod xqx). Thus for xd to appear in the expansion of g(x)b we require that (γ1) 0 mod 2k, that is, γ µ2k1 for some integer µ. We also have 0 γ 2k1 and so we conclude that γ 2k1. Thus our task now is to search for terms of degree d 2k+1α2k+(2k1) (2h1) in g(x)b. We now show that most of the terms of g(x)b have degree less than d. Consider terms of g(x)b of the form (g q/ xq/) k+ (g i k xi k ), (12) for 1 i 2h k1. From (10) we know that g i k 0 for i 2h k and so the largest possible degree of a term of the form (12) is 2h (2k+3)(2h k 1) 2k+ 2h+k2h2k(2h k2h k 2) 2h+k2h since k 1 2h+k2h2k+2k (α1) 2k+2h+k2h2k d. Thus any term of g(x)b of the form in (12) has degree less than d. The only terms of g(x)b that have degree greater than all terms of the form in (12) are (g q/ xq/) k+ and the terms (g q/ xq/) k+ (g i k xi k ) for 1 i 2h k 1. First we consider (g q/ xq/) k+ g k+ (x q/ h ) k+. Since the degree of this term is 0 mod 2h we also require that d (α1) 2k+2h+k2h2k 0(mod2h ). From this it follows that we must have 2h (α1) 2k+2k. However, α 2h k 1, so (α1) 2k+2k 2h k 2k+2k 2h k2k 2h for k 1, from which it follows that 2h (α1) 2k+2k and we have a contradiction. We are now left to consider the case (g q/ xq/) k+ (g i k xi k ) g k+ g q/ i k x h+k h +i k. Equating the degree of this term with d we have 2h+k2hi2k (α1) 2k+ 2h+k2h2k, which implies that i 2α1. Thus the coefficient of xd in g(x)b and hence of xc when g(x)b is reduced modulo xqx is g k+ g q/ α k+ + k. By using Theorem 4.3 and the fact that g q/ 0 we have g α k+ + k 0 for α 0,,2h k 1, that is, g k g k+ + k g k+ + k g k+ + k g h k k 0. (13) From (3) and (13) we have f k+ f k+ + k+ f k+ + k+ f k+ + k+ f h k+ k+ 0. (14) By (9), deg f 2h k2k+, and by (14) there are no terms of degree α2k+2k+ for α 0,, 2h k 1 (note that this is not all of the possible values of α), that is, f k+ f k+ + k+ f k+ + k+ f k+ + k+ f h k k+ 0.

13 Thus f must have the form OVOIDS OF PG(3, q), q EVEN, WITH A CONIC SECTION 581 f(x) f xf k+ x k+ f k+ x k+ f h k k+ x h k k+, where f 0, that is, f is 2k+-sparse. For g, by (10) the degree of g(x)g q/ xq/ is at most 2h k 2k and by (13) there are no terms of degree α2k+2k for α 0,,2h k 1. Thus g is 2k+-spare. THEOREM 4.6. If f(x) q/ f xi and g(x) q/ g xi are o-polynomials i= i i= i oer GF(2h), h 3 with f 0, g q/ 0 and satisfying then f(x) x and g(x) x. A(xx)f(x) g(x), A 0, Proof. Since f and g satisfy A(xx)f(x) g(x) it follows that f 0, g q/ 0, f is 2-sparse and g is 2-spare. Thus using Lemma 4.5 inductively it follows that f is 2 (h )/ 1-sparse and g is 2 (h )/ 1-spare. From Lemma 4.4 it then follows that f(x) x, and consequently that g(x) x. Now, finally, we restate the classification of ovoids of PG(3, q) containing a conic. THEOREM 4.2. Let Ω be an ooid of PG(3, q), qeen, and π be a plane of PG(3, q) such that πω is a conic. Then Ω is an elliptic quadric. Acknowledgements. The author would like to thank Stan Payne and Christine O Keefe for their suggested revisions to the paper. The work was supported by the Australian Research Council. References 1. A. BARLOTTI, Un estensione del teorema di Segre-Kustaanheimo, Boll. Un. Mat. Ital. 10 (1955) A. BARLOTTI, Some topics in finite geometrical structures, Institute of Statistics Mimeo Series 439 (University of North Carolina, North Carolina, 1965). 3. M. R. BROWN, C. M. O KEEFE and T. PENTTILA, Triads, flocks of conics and Q (5, q), Des. Codes Cryptogr. 18 (1999) W. CHEROWITZO, Hyperovals in Desarguesian planes of even order, Ann. Discrete Math. 37 (1988) P. DEMBOWSKI, Finite geometries (Springer, Berlin, 1968). 6. G. FELLEGARA, Gli ovaloidi di uno spazio tridimensionale de Galois di ordine 8, Atti Accad. Naz. Lincei Rend. 32 (1962) D. G. GLYNN, The Hering classification for inversive planes of even order, Simon Stein 58 (1984) D. G. GLYNN, A condition for the existence of ovals in PG(2, q), q even, Geom. Dedicata 32 (1989) J. W. P. HIRSCHFELD, Projectie geometries oer finite fields (Oxford University Press, 1998). 10. J. W. P. HIRSCHFELD, Finite projectie spaces of three dimensions (Oxford University Press, 1985). 11. J. W. P. HIRSCHFELD and J. A. THAS, General Galois geometries (Clarendon Press, Oxford, 1991). 12. C. M. O KEEFE, Ovoids in PG(3, q): a survey, Discrete Math. 151 (1996) C. M. O KEEFE and T. PENTTILA, Ovoids of PG(3, 16) are elliptic quadrics, J. Geometry 38 (1990) C. M. O KEEFE and T. PENTTILA, Ovoids of PG(3, 16) are elliptic quadrics II, J. Geometry 44 (1992) C. M. O KEEFE and T. PENTTILA, Ovoids with a pencil of translation ovals, Geom. Dedicata 62 (1996) C. M. O KEEFE and T. PENTTILA, Ovals in translation hyperovals and ovoids, European J. Combin. 18 (1997)

14 582 OVOIDS OF PG(3, q), q EVEN, WITH A CONIC SECTION 17. C. M. O KEEFE,T.PENTTILA and G. F. ROYLE, Classification of ovoids in PG(3, 32), J. Geometry 50 (1994) G. PANELLA, Caratterizzazione delle quadriche di uno spazio (tridimensionale) lineare sopra un corpo finito, Boll. Un. Mat. Ital. 10 (1955) S. E. PAYNE and J. A. THAS, Finite generalized quadrangles (Pitman, London, 1984). 20. T. PENTTILA and C. E. PRAEGER, Ovoids and translation ovals, J. London Math. Soc. (2) 56 (1997) O. PROHASKA and M. WALKER, A note on the Hering type of inversive planes of even order, Arch. Math. 28 (1977) B. SEGRE, On complete caps and ovaloids in three-dimensional Galois spaces of characteristic two, Acta Arith. 5 (1959) J. A. THAS, Ovoidal translation planes, Arch. Math. 23 (1972) Department of Pure Mathematics and Computer Algebra Ghent Uniersity Gent B9000 Belgium mbrowncage.rug.ac.be