Caps on classical varieties and their projections Jürgen Bierbrauer Department of Mathematical Sciences Michigan Technological University Houghton, Michigan 49931 (USA) Antonio Cossidente Dipartimento di Matematica Università della Basilicata via N. Sauro, 85 85100 Potenza (ITALY) Yves Edel Mathematisches Institut der Universität Im Neuenheimer Feld 288 69120 Heidelberg (GERMANY) 1 Proposed Running Head: Caps on classical varieties Corresponding Author: Antonio COSSIDENTE Dipartimento di Matematica Università della Basilicata via N. Sauro, 85 85100 POTENZA (ITALY) cossidente@unibas.it 2 Abstract A family of caps constructed by Ebert, Metsch and T. Szönyi [8] results from projecting a Veronesian or a Grassmannian to a suitable lower-dimensional space. We improve on this construction by projecting to a space of much smaller dimension. More precisely we partition P G(3r − 1, q) into a (2r − 1)−space, an (r − 1)−space and q r − 1 cyclic caps, each of size (q 2r − 1)(q − 1). We also decide when one of our caps can be extended by a point from the (2r − 1)−space or the (r − 1)−space. The proof of the results uses several ingredients, most notably hyperelliptic curves. 3 1 Introduction Let P G(N, q) be the projective space of dimension N over the finite field GF (q). A k–cap K in P G(N, q) is a set of k points, no three of which are collinear [14], and a k–cap is complete if it is maximal with respect to inclusion. The maximum value of k for which there exists a k–cap in P G(N, q) is denoted by m2 (N, q). The number m2 (N, q) is only known, for arbitrary q, when N ∈ {2, 3}; namely, m2 (2, q) = q + 1 if q is odd, m2 (2, q) = q + 2 if q is even, and m2 (3, q) = q 2 +1, q > 2. A cap of size m2 (3, q) in P G(3, q) is called an ovoid. The only known infinite classes of ovoids are the elliptic quadrics and, if q is an odd power of 2, the Suzuki–Tits ovoids [18]. Aside of these general results the precise value of m2 (N, q) is known only in the following cases: m2 (N, 2) = 2N , m2 (4, 3) = 20, m2 (5, 3) = 56, and m2 (4, 4) = 41 [9]. Finding the exact value for m2 (N, q), N ≥ 4, and constructing an m2 (N, q)–cap seems to be a very hard problem, [13]. A natural asymptotic problem is the determination of µ(q) = lim supN →∞ logq (m2 (N, q)) . N As a cap cannot be larger than its ambient space, clearly µ(q) ≤ 1. It is an interesting open problem to decide if µ(q) < 1. Segre’s recursive construction [16] yields µ(q) ≥ 2/3. No better lower bound seems to be known for general q. An exception is the ternary case. It follows from [2] that µ(3) > log3 (2.21). An interesting method to construct caps is to study intersections of quadrics or of hermitian varieties, or in general of classical varieties such as Veronese varieties and Segre varieties [6],[5]. In this paper we construct a class of cyclic caps in P G(3r − 1, q), r ≥ 2, (i.e. caps admitting a regular action of a cyclic group of automorphisms) obtained by projecting certain caps contained in the Grassmannian of lines of P G(2r − 1, q). These caps have size (q 2r − 1)/(q − 1) and it turns out that they are unions of Veronese varieties. This explicitly constructed family of caps reaches the best known general lower bound µ(q) ≥ 2/3. These caps are therefore just as large as the largest known caps, in an asymptotical sense. We also study extensions of these caps. 4 2 Singer cycles and their lifting Let V be an (n + 1)–dimensional vector space over the Galois field IFq , and let S be a Singer cycle of GL(V ). It is known [15], that S is conjugate in n GL(n + 1, q n+1 ) to the diagonal matrix D = diag(α, αq , . . . , αq ), where α is a primitive element of IFqn+1 . Denote byˆ: GL(V ) → P GL(V ) the canonical epimorphism. The image of hSi underˆis a subgroup hŜi ⊂ P GL(V ) of order (q n+1 − 1)/(q − 1). Let N = n(n + 3)/2. Denote by V the Veronese variety of P G(N, q), namely the Veronesean of all quadrics of P G(n, q), see [14]. It follows from [14] that V is a (q n + q n−1 + · · · + q + 1)–cap. Moreover, since Aut(V) is isomorphic to P GL(n + 1, q), each linear collineation of P G(n, q) can be lifted to a collineation of P G(N, q) leaving V invariant, see [5]. Consider case n = 3. The lifting S̄ of a Singer cycle S of GL(4, q) to a collineation of P GL(10, q) fixes the Veronese variety V of P G(9, q). D, say α2 conjugates over The collineation S̄ is conjugate to the block diagonal matrix diag(S 2 , S q+1 , T ), where T is a Singer cycle of GL(2, q). From a geometric point of view, this means that S̄ induces a collineation ψ of P G(9, q) which fixes (setwise) two 3–dimensional projective subspaces Σ1 and Σ2 and a line, say ℓ. The order of ψ is q 3 + q 2 + q + 1 and its action outside the fixed subspaces is semiregular. In particular, S̄ fixes a 5–dimensional subspace Σ3 , inducing a collineation e whose matrix representation is E =diag (S q+1 , T ). It is easy to see that E is exactly the exterior square Λ(2) (S) of S, and it fixes (by definition) the Grassmannian G1,3 of lines of P G(3, q). The collineation e has been studied in [10], [4]. It has order q 3 +q 2 +q +1, fixes a 3–dimensional subspace, the subspace Σ2 , inducing a collineation of order q 2 + 1, whose orbits are elliptic ovoids [7], and a line (the line ℓ). Each of the remaining orbits of e on the point set of Σ3 is a cap of size q 3 +q 2 +q+1. These are the caps studied by David Glynn [10]. Each such Glynn cap is the complete intersection of G1,3 with the quadratic cone of vertex the line ℓ and basis an elliptic quadric in Σ2 . In particular, a Glynn cap can be projected onto an elliptic ovoid of Σ2 from the line ℓ. In [5], it has been proved that a Glynn cap is the projection of a Veronese variety of P G(9, q). Hence, projecting a Veronese variety V of P G(9, q) from a suitable 3subspace of P G(9, q) disjoint from V we get a Glynn cap, and the Glynn cap can be further projected onto an elliptic ovoid. In [8] the authors, using the geometry of lines of P G(n, q) for arbitrary n, generalize Glynn’s construction to any Grassmannian of lines of P G(n, q), obtaining caps of size q n + · · · + q + 1. We refer to these caps as EMS–caps. 5 It was shown in [3] that each EMS–cap can in fact be obtained projecting a suitable Veronese variety. Unfortunately EMS–caps are very small with respect to their ambient space. We pursue the program sketched in the introduction and give an equivariant construction by projecting EMS–caps to a subspace of small dimension. The general setting is described in the following section. The proof is in Section 4. 3 The general setting Let S be a Singer cycle of GL(2r, q), r ≥ 2. Its canonical form in GL(2r, q 2r ) 2r−1 is D = diag(α, α2 , . . . , αq ), where α is a primitive element of IFq2r . Consider the second exterior square Λ(2) (D) of D. Then Λ(2) (D) is a diagot nal matrix whose entries are αq +1 , 1 ≤ t ≤ r and its conjugates over IFq . r Note that αq +1 is an element of IFqr . It follows that Λ(2) (S) has the rat−1 tional canonical form Λ(2) (S) = diag(S q+1 , . . . , S q +1 , T ), where T is a Singer cycle of GL(r, q). Then Λ(2) (S) induces a linear collineation φ of P G(2r2 − r − 1, q) of order (q 2r − 1)/(q − 1), fixing the Grassmannian of lines, G1,2r−1 of P G(2r − 1, q). Also it fixes r − 1 subspaces Σ1 , . . . , Σr−1 , each of dimension 2r − 1, and an r − 1-dimensional subspace π. Outside the fixed linear spaces (and subspaces generated by them) hφi acts semiregularly. This means that the orbits of hφi not in the fixed linear subspaces all have length (q 2r − 1)/(q − 1), and in particular those contained in G1,2r−1 are ((q 2r − 1)/(q − 1))–caps [8] (apart from one shorter orbit if r − 1 ≡ 2 mod 3). We note that hφi also fixes r − 1 subspaces of dimension 3r − 1. We improve upon the EMS–construction by showing that projection to the subspace, where the collineation group hφi has matrix representation diag r−1 (S q +1 , T ), produces caps. The details are given in the next section. 4 Construction of a family of cyclic caps LetLF = IFq2r and L = IFqr , where r ≥ 2. Consider the direct sum V = F L, an 3r−dimensional vector space over IFq . We write the elements of the corresponding P G(3r −1, q) with homogeneous coordinates (a : b), where a ∈ F, b ∈ L. Let α be a primitive element of F. The action of the Singer 6 cycle determined by α lifts to V as follows: g : (a, b) 7→ (aαq r−1 +1 , bαq r +1 ). This induces an action on P G(3r − 1, q) in the canonical way. The Singer cycle has order (q 2r − 1)/(q − 1) both in its action on P G(2r − 1, q) and in the lifted action on P G(3r − 1, q). The orbit containing (a : b), where ab 6= 0, will be denoted by O(a : b). Let N, T : F −→ L be norm and trace, respectively. The projective subspaces spanned by F and L are denoted P G(F ), P G(L). The subgroup of order u of the multiplicative group of F is denoted Z(u). The group Z(q r +1) consists of the elements of norm N = 1. Let Q = q r and U = F ∗ q+1 = Z((q 2r−1 /(q + 1)). Theorem 1. For every 0 6= a ∈ F, 0 6= b ∈ L, the orbit O(a : b) containing (a : b) ∈ P G(3r − 1, q) under the lifted action of the Singer cycle has full length (q 2r − 1)/(q − 1) and is a cap. Proof. Assume O(a : b) is not a cap. There must be three collinear points (a : b), (axq r−1 +1 : bxq r +1 ), (ay q r−1 +1 : by q r +1 ) in O(a : b). As these points are different, we have x, y ∈ / IFq and x/y ∈ / IFq . There are coefficients λi ∈ IFq (by force all nonzero), such that aλ1 + aλ2 xq r−1 +1 bλ1 + bλ2 N (x) + aλ3 y q r−1 +1 + bλ3 N (y) = 0 = 0 Obviously we may as well assume a = b = 1. Choosing λ3 = −1 and reordering we obtain yq r−1 +1 N (y) = λ2 x q r−1 +1 = λ2 N (x) 7 + λ1 + λ1 We compute N (y q N (y q r−1 +1 r−1 +1 ) in two ways. Using the second equation we obtain ) = N (y)q = λ22 N (x)q r−1 +1 r−1 +1 = (λ2 N (x)q r−1 + λ1 λ2 (N (x)q r−1 + λ1 )(λ2 N (x) + λ1 ) = + N (x)) + λ21 . The first equation yields N (y q r−1 +1 ) = N (λ2 xq r−1 +1 = λ22 N (x)q + λ1 ) = (λ2 xq r−1 +1 + λ1 λ2 (xq r (q r−1 +1) r (q r−1 +1) + λ1 )(λ2 xq + xq r−1 +1 r−1 +1 + λ1 ) = ) + λ21 . Comparing these expressions and eliminating the obvious common terms we obtain xq r−1 (q r +1) + xq r +1 = xq r (q r−1 +1) + xq r−1 +1 . Collect all terms on one side, eliminate the common factor xq r−1 +1 . Fortu- nately the polynomial factors: 0 = xq 2r−1 +q r −q r−1 −1 − xq 2r−1 −1 − xq r −q r−1 + 1 = (xq r −q r−1 − 1)(xq 2r−1 −1 − 1). If the first factor vanishes, then xq−1 = 1, hence x ∈ IFq , contradiction. Assume the second factor vanishes. As gcd(q 2r − 1, q 2r−1 − 1) = gcd(q 2r−1 − 1, q − 1) = q − 1 we obtain the same contradiction. In the following sections we determine when a cap O(a : b) can be extended by points from P G(L) or P G(F ). 8 5 Extensions by points from the (r −1)−space P G(L). q 2r − 1 − caps in q−1 P G(3r − 1, q) as constructed in Theorem 1. Let Q = (0 : d) ∈ P G(L). Then Theorem 2. Let r ≥ 2 and O(a : b) one of the cyclic O(a : b) ∪ {Q} is a cap if and only if r is odd and q is a power of 2. Proof. Assume P1 = (ay q r−1 +1 : by q r +1 ) and P2 = (a(xy)q r−1 +1 : b(xy)q r +1 ) are collinear with Q. Let λ1 , λ2 , λ3 ∈ IFq be the coefficients of a linear combination. Here y, x are arbitrary nonzero elements of F, but we want P1 6= P2 , equivalently x ∈ / IFq . The first coordinate shows λ1 + λ2 xq choose λ2 = −1. It follows λ1 = xq r−1 +1 r−1 +1 = 0. We can ∈ IFq . Let g = gcd(q 2r − 1, q r−1 + 1). Observe that xg ∈ IFq . As q 2r − 1 = (q r−1 + 1)(q r+1 − q 2 ) + (q 2 − 1) we have gcd(q 2r − 1, q r−1 + 1) = gcd(q r−1 + 1, q 2 − 1). Let r be odd. We have q r−1 + 1 = (q 2 − 1)(q r−3 + q r−5 + · · · + 1) + 2, showing gcd(q r−1 + 1, q 2 − 1) = gcd(q 2 − 1, 2). Let r be even. We have q r−1 + 1 = (q 2 − 1)(q r−3 + q r−5 + · · · + q) + (q + 1), hence gcd(q r−1 + 1, q 2 − 1) = q + 1 in this case. This shows the following: Lemma 1.    q+1 if r is even   g = gcd(q 2r − 1, q r−1 + 1) = 1 if r odd, q even     2 if r and q odd. 9 Let r be odd and q even. We have g = 1, hence x ∈ IFq and P1 = P2 , contradiction. This shows that each Q ∈ P G(L) yields an extension cap. Let r and q both be odd. We have x2 ∈ IFq . It follows that x ∈ IFq2 . As 2 xq = x it follows λ1 = xq r−1 +1 = x2 and xq r +1 = xq+1 . The second coordinate section shows (bx2 − bxq+1 )N (y) + λ= 0, or b λ3 = N (y) (xq+1 − x2 ) ∈ IFq . d Choose x in the cyclic group of order 2(q − 1), but outside IFq . Then x2 = λ1 ∈ IFq , xq+1 − x2 6= 0. As y is arbitrary in F, we can choose y such that λ3 is indeed in IFq . This shows that no cap extension is obtained in this case. Let r be even. We have xq+1 ∈ IFq , equivalently x ∈ IFq2 . It follows λ1 = xq+1 and N (x) = x2 . The second coordinate-section yields b λ3 = N (y) (x2 − xq+1 ) ∈ IFq . d Choose x ∈ IFq2 \ IFq . Then x2 − xq+1 6= 0 and we can choose y ∈ F such that λ3 ∈ IFq . Once again no extension cap is obtained. 6 Extensions by points from the (2r−1)−space P G(F ). q 2r − 1 − caps in q−1 P G(3r − 1, q) as constructed in Theorem 1. Extension caps O(a : b) ∪ {Q}, Theorem 3. Let r ≥ 2 and O(a : b) one of the cyclic where Q = (c : 0) ∈ P G(F ), do exist if and only if r = 2 and q a power of 2. The remainder of this section is dedicated to a proof of Theorem 3. Choose P1 , P2 and coefficients λ1 , λ2 , λ3 ∈ IFq as in the proof of Theorem 2. 10 Without restriction λ2 = −1. The second coordinate section yields λ1 = N (x) = xq r +1 ∈ IFq . This shows that x ∈ Z((q r + 1)(q − 1)). The first coordinate section shows a r−1 r−1 r λ3 = y q +1 (xq +1 − xq +1 ) ∈ IFq . c r−1 r Observe that xq +1 − xq +1 6= 0 as x ∈ / IFq . We distinguish several cases, which are treated in order of increasing difficulty. Let r be odd and q a power of 2. As g = 1 in this case (see Lemma 1) r−1 we have that y q +1 varies over F when y does. We can choose y such that λ3 ∈ IFq . It follows that no extension cap is obtained. r−1 Let r and q both be odd. We have g = 2. This means that y q +1 varies over the squares in F. Observe that the elements of IFq are squares in F. We have to show that when x ∈ Z((q r + 1)(q − 1)) \ IFq , among the values r−1 r xq +1 − xq +1 there are squares as well as non-squares. r−1 r Let λ = N (x) ∈ IFq and y = xq +1 . Then y q = xq +q = λxq−1 , hence 2r 2r−1 y = y q = λ(xq−1 )q . It follows f (x) = y − λ = λ(xq−1 − 1)q 2r−1 . 2r−1 Let χ be the quadratic character of F. Then χ(f (x)) = χ(xq−1 − 1)q )= q−1 q−1 r χ(x − 1). Here x varies over Z(q + 1) \ {1}. This yields an equivalent formulation for our claim that no extension caps are obtained in the case under consideration. We have to show that v − 1 represents both squares and non-squares, where v varies over Z(q r + 1) \ {1}. As it is clear that squares are represented (choose v = −1), it suffices to show that there is some v ∈ Z(q r + 1) \ {1} such that v − 1 is a non-square in F. Let w = v − 1 6= 0. The statement to be proved is equivalent to the following: there is some nonsquare w ∈ F such that N (w + 1) = 1. As N (w + 1) = N (w) + T (w) + 1 we can write equivalently N (w) = −T (w) or T (w/N (w)) = −1. As N (w) is a square in F, our statement is equivalent to the following: • There is a nonsquare w ∈ F such that T (w) = −1. The elements w of trace T (w) = −1 are those of the form w = −1/2 + y, where y = 0 or y Q−1 = −1. We have N (w) = 1/4 − y 2 = 1/4 − z, where 11 either z = 0 or z a non-square in L. Observe that w is a square in F if and only if N (w) is a square in L. Assume the statement we wish to prove is not true and fix a non-square c ∈ L. Then the equation 1/4 − ca2 = b2 has a solution b for every a ∈ L. Clearly these are way too many solutions for this quadratic equation, which, as we know, has Q + 1 projective solutions. This concludes the proof in the case that r is odd. r−1 Let r be even. As g = q + 1 we have that y q +1 varies over the subgroup r−1 r U. Observe IFq∗ ⊂ U. We have to prove that xq +1 − xq +1 contains elements of each coset of U as x varies over Z((q r +1)(q −1))\IFq . The same arguments as in the case when r and q are odd show that we can reformulate our claim as follows. • Let r be even and either r > 2 or q odd. For every coset γU there is some z ∈ Z(q r + 1) \ {1} such that z − 1 ∈ γU. Consider first the case that q is odd. The choice z = −1 shows that U itself is represented. Observe that Z consists of the elements of norm N (z) = 1. Let v ∈ Z and w = v − 1. Our claim is that every coset γU contains an element w such that N (w + 1) = 1. As before this is equivalent with −1 = T (w/N (w)) = T (1/w). In other words, we have to show that every coset γU contains an element of trace −1. The elements x 6= 0 of trace T (x) = 0 are those satisfying xQ−1 = −1. As T (−1/2) = −1 the element x ∈ F with T (x) = −1 are precisely those of the form x = − 21 + y, where y = 0 or y Q−1 = −1. We want to show that every coset of U contains such an element. The subgroup U consists precisely of the elements u ∈ F such that N (u) is a (q + 1) − st power (this is clear: as Z(Q + 1) is the kernel of N, the image N (U ) has order |U |/|Z(Q + 1)| = (Q − 1)/(q + 1)). We have N (− 21 + y) = 41 − y 2 with y as above. Our claim is that every coset of the group N (U ) of index q + 1 in L contains one of our elements 14 − y 2 . Here y varies over y = 0 and the elements satisfying y Q−1 = −1. It follows that z = y 2 varies precisely over z = 0 and the elements in L satisfying z (Q−1)/2 = −1, in other words the non-squares in L = IFQ . Assume our claim is not true. Then there exists an element c ∈ L such that with cxq+1 = 41 − z we have that z never is a non-square or 0, in other words 1 y 2 = − cxq+1 4 12 has solutions y ∈ L, y 6= 0 for every x ∈ L∗ = IFQ∗ . This is a hyperelliptic curve of genus (q − 1)/2 (see [17], page 197). We observe that c cannot be a (q + 1) − st power for if it were we would be able to choose x 6= 0 such that y = 0 is the only solution of the equation, which we assume not to be the case. What can we say about the number N of rational points if our assumption is satisfied? There is one point (0 : 1 : 0) at infinity (and the curve has a singularity there). If y = 0, then by assumption we have no solution. It follows that every x 6= 0 yields precisely two values for y. Together with two solutions when x = 0 this yields a number og rational places N ≥ 1 + 2 + 2(Q − 1) = 2Q + 1, contradicting the Hasse-Weil bound. Our last case is: r even, q a power of 2. Recall that we have to show the following when r > 2 : for every 0 6= γ ∈ F there is some z 6= 1 such that z − 1 ∈ γU and N (z) = 1. Using w = z − 1(= z + 1) instead, an equivalent expression is N (w + 1) = 1. A by now familiar argument shows that this is equivalent to T (w/N (w)) = 1. This yields the following equivalent condition: • For every 0 6= γ ∈ F there is w ∈ γU such that T (w) = 1. Fix w0 such that T (w0 ) = 1. The general element of trace T = 1 has the form w = w0 + a, where a ∈ L. Write c = N (γ). We have to show that there is some a ∈ L such that N (w0 + a) = cxq+1 , where x ∈ L. We have N (w0 + a) = (w0 + a)(w0Q + a) = N (w0 ) + aT (w0 ) + a2 = N (w0 ) + a + a2 . Let tr : L −→ IF2 be the absolute trace. As a varies over L, the element a + a2 varies over the elements of trace tr = 0. We have reached the following equivalent reformulation: • Fix w0 ∈ F such that T (w0 ) = 1. For every 0 6= c ∈ L there is x ∈ L such that tr(cxq+1 + N (w0 )) = 0. As T (w0 ) = 1 we have w0Q = w0 +1, hence N (w0 ) = w0 (w0 +1) = w0 +w02 . Apply the definition of the trace. This yields tr(N (w0 )) = 1. Our claim simplifies as follows: • For every 0 6= c ∈ L there is x ∈ L such that tr(cxq+1 ) = 1. Assume this is not the case for some c ∈ L. Consider the trace-form defined on L by tr, where we view L as a vector space over IF2 . With respect to this non-degenerate bilinear form we have Lq+1 ⊂ c⊥ . It follows that 13 Lq+1 is contained in a proper subgroup of the additive group of L. As the linear combinations of (q + 1) − st powers are closed under addition and multiplication, this means that Lq+1 is contained in a proper subfield of L. Clearly, this cannot happen unless Q = q 2 , in which case Lq+1 = IFq . We see that for r > 2 such extension caps cannot exist, whereas in case r = 2 they positively exist. Remark 1. Theorem 3 is also confirmed by a result in [4] (see Proposition 9), where it is proved that if q is even, a cap O(a : b) in P G(5, q) can always be extended by the points of an elliptic ovoid in P G(F ), where now F is a 3–dimensional projective space. Remark 2. We have not been able to decide the completeness of our caps. Computer experiments suggest the conjecture that O(a : b) can never be extended by a point (c : d), where c 6= 0, d 6= 0. 7 Some geometric links Proposition 1. Each cap constructed above is a projection of an EMS–cap. Proof. We give the proof in case r = 3. It is easy to see that the proof generalizes to arbitrary r. The collineation group hφi fixes two 5–dimensional subspaces K and F and a plane L, inducing collineations whose matrix representations are S q+1 , Sq 2 +1 and a Singer cycle T , respectively. Take an EMS-cap E inside the Grassmannian G1,5 in P G(14, q). We note that E spans P G(14, q). Consider the cone C with vertex K (hence its generators are 6–dimensional subspaces of P G(14, q)) generated by K and a point of E), and project E on the space Σ = hF, Li. Since K is φ–invariant as well as E, the cone C is φ-invariant. 14 The Grassmann dimensional formula shows that each generator of C meets Σ in exactly one point. Call X the set of all such points. Clearly X is φ–invariant, and in particular it must be a hφi–orbit. We have three possibilities. The set X cannot be the plane L as E would have to be contained in an 8–dimensinal subspace, contradicting the fact that E spans P G(14, q). Likewise, X is not contained in L as X is not contained in an 11-dimensional subspace. This shows that X must be one of our caps. Let O be one of the caps constructed in Theorem 1. Consider the subgroup hCi of hAi of order q 2 + q + 1. Its canonical form in GL(9, q 6 ) is a 2 diagonal matrix whose entries are η q +1 and all its conjugates over IFq , again 2 η q +1 and all its conjugates over IFq , and η 2 and all its conjugates over IFq , 3 where η = αq +1 ∈ IFq3 . In geometrical terms, hCi fixes the 5-space F inducing a regular 2–spread S and the plane L permuting its points in a single orbit (indeed gcd (2, q 2 + q + 1) = 1). Also hCi fixes each 5–subspace generated by π and a plane of S. The collineation induced by C on each of these 5– subspaces is exactly the lifting of a Singer cycle of P GL(3, q) to a collineation of P G(5, q) fixing a Veronese surface [1]. Each of these 5-subspaces is either disjoint from O or meets O in at least a Veronese surface. We have proved the following: Proposition 2. Each cap constructed in Theorem 1 is union of q 3 + 1 Veronese surfaces. Remark 3. Note that the previous proposition extend to higher dimensions. In the general case we have Veronese varieties instead of Veronese surfaces. The canonical form of C in GL(9, q 3 ) coincides with the canonical form of the Kronecker product of a Singer cycle of GL(3, q) by itself. We conclude with the following: Corollary 1. Each of the Veronese surfaces partitioning O is contained in a Segre variety S2,2 of Σ. 15 8 The link to cyclic codes Consider the cap O(1 : 1). Write the points of this cap as columns of a matrix q 2r − 1 G with 3r rows. There are n = columns. The presence of the Singer q−1 cycle (regular on the columns) shows that this generates the dual of a cyclic q−ary code, more precisely: the cyclic code is the dual of the trace code of the F −ary code generated by the rows of G. This cyclic code has parameters [ q 2r − 1 q 2r − 1 , − 3r, 4]q . q−1 q−1 In the language of the theory of cyclic codes it has as defining set the cyclotomic cosets containing q r−1 + 1 (of length 2r) and q r + 1 (of length r). Apparently the standard bounds for cyclic codes do not suffice to show that the minimum distance is indeed 4. 9 Cap partitions Denote by κ(P G(n, q)) the minimum number of caps into which P G(n, q) can be partitioned. Theorem 1 shows that the points of P G(3r − 1, q) can be partitioned into • the points of a subspace P G(2r − 1, q), • the points of a subspace P G(r − 1, q), and • q r − 1 caps of size (q 2r − 1)/(q − 1) each. This yields the following recursive bound: κ(P G(3r − 1, q)) ≤ q r − 1 + κ(P G(2r − 1, q)). Recent results on the ternary case are in [11]. Cap partitions will be studied in a subsequent paper. 16 References [1] R.D. Baker, A. Bonisoli, A. Cossidente and G.L. Ebert, Mixed partitions of P G(5, q), Discrete Mathematics, to appear. [2] A.R. 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