Flocks of cones of higher degree

Flocks of cones of higher degree Peter Sziklai∗ Department of Computer Science Eötvös University, Budapest Pázmány P. s. 1/c, Budapest, H-1117, Hungary [email protected] Abstract: It is known that in PG(3, q), q > 19, a partial flock of a quadratic cone √ with q − ε planes, can be extended to a unique flock if ε < 14 q, and a similar and slightly stronger theorem holds for the case q even. In this paper we prove the analogue of this result for cones with base curve of higher degree. Keywords: flock, cone 1 Introduction A flock of a cone of PG(3, q) is a partition of the points of the cone different from the vertex into q disjoint plane sections. The case of the quadratic cone was investigated by several researchers. Associated with flocks of the quadratic cone are some line spreads of PG(3, q) and translation planes; q-clans and some elation generalised quadrangles of order (q 2 , q); hyperbolic fibrations; BLT-sets (when q is odd); and, when q is even, families of ovals in PG(2, q), called herds. Parts (i) and (ii) of the following theorem were proved in [10] in a short way, while part (iii) is due to Storme and Thas [11]. Theorem 1 Let q > 19. Assume that the planes Ei , i = 1, ..., q − ε intersect the quadratic cone C ⊂ PG(3, q) in disjoint irreducible conics. √ (i) If ε < 41 q then one can find additional ε planes (in a unique way), which extend the set {Ei } to a flock. (ii) If q = p is a prime and ε < 1 40 p + 1 then the partial flock can be extended to a unique flock. √ √ √ (iii) If q is even and ε ≤ q if q is a square and ε < 2 q if q is a nonsquare then the partial flock can be extended to a unique flock. In Section 2 of this paper we prove the following analogue of this result for cones with base curve of higher degree. The proof will be more complicated than in the quadratic case. ∗ Research is supported by OTKA F-043772, T-043758, T-049662, TÉT Hungarian-Spanish and Magyary grants. 1 √ Theorem 2 For 2 ≤ d ≤ 6 q consider the cone {(1, t, td , z) : t, z ∈ GF(q)} ∪ {(0, 0, 1, z) : z ∈ GF(q)} ∪ {(0, 0, 0, 1)} = C ⊂ PG(3, q) and let C ∗ = C \ {(0, 0, 0, 1)}. Assume that the planes √ Ei , i = 1, ..., q − ε, Ei 6∋ (0, 0, 0, 1), intersect C ∗ in pairwise disjoint curves. If ε < ⌊ d12 q⌋ then one can find additional ε planes (in a unique way), which extend the set {Ei } to a flock, (i.e. q planes partitioning C ∗ ). The proof (see below) starts like in the quadratic case. Using elementary symmetric polynomials we find an algebraic curve G(X, Y ), which “contains” the missing planes in some sense. The difficulties are (i) to show that G splits into ε factors, and (ii) to show that each of these factors corresponds to a missing plane. For (i) we use our new Lemma 3. For (ii) we have to show that most of the possible terms of such a factor do not occur, which needs a linear algebra argument on a determinant with entries being elementary symmetric polynomials; this matrix may be well-known but the author could not find a reference for it. To be self-contained we include that half of the proof which is common with the one in [10]. 2 Proof √ √ √ 1+d(d−1) q 1 + (d+1)q and n ≤ 1d q − d + 23 . Lemma 3 Choose three constants 1 ≤ d ≤ 6 q, α ≥ d+1 Let Cn be a curve of order n defined over GF (q), and denote by N the number of its points in P G(2, q). Suppose that Cn does not contain a component defined over GF (q) of degree ≤ d. Then N ≤ nqα. For curves without linear component a similar lemma can be found in Sziklai [9], which is a variant of a lemma by Szőnyi [12]. For curves without quadratic component see [10]. Proof: (i) Suppose first that Cn is absolutely irreducible. Then Weil’s theorem ([15], [6]) gives √ √ N ≤ q + 1 + (n − 1)(n − 2) q. We want q + 1 + (n − 1)(n − 2) q ≤ nqα, as it is quadratic in n √ it is enough to check it for n = d + 1 and n = d1 q − d + 23 , and it holds indeed. If Cn is not absolutely irreducible, then it can be written as Cn = D1 ∪ ... ∪ Ds , where Dj is P an absolutely irreducible component of order ij , so sj=1 ij = n. If Dj can not be defined over GF (q), then it has at most Nj ≤ (ij )2 ≤ ij qα points in P G(2, q) (see [6], Lemma 2.24). If Dj is defined over GF (q), then the Weil-bound implies again that Nj ≤ ij qα. Hence N= s X j=1 Nj ≤ s X ij qα = nqα. j=1 Proof of Theorem 2. Suppose that the plane Ei has the equation X4 = ai X1 + bi X2 + ci X3 , for i = 1, 2, ..., q − ε. Define fi (T ) = ai + bi T + ci T d , then Ei ∩ C = {(1, t, td , fi (t)) : t ∈ GF(q)} ∪ {(0, 0, 1, ci )}. Let σk (T ) = σk ({fi (T ) : i = 1, ..., q − ε}) denote the k-th elementary symmetric polynomial of the polynomials fi , then degT (σk ) ≤ dk. As for any fixed T = t ∈ GF(q) the values fi (t) are all distinct, we would like to find Xq − X Q , i (X − fi (t)) the roots of which are the missing values GF(q) \ {fi (t) : i = 1, ..., q − ε}. In order to do so, we define the elementary symmetric polynomials σj∗ (t) of the “missing elements” with the following formula: 2 Xq − X =   X q−ε − σ1 (t)X q−ε−1 + σ2 (t)X q−ε−2 − ... ± σq−ε (t)  X ε − σ1∗ (t)X ε−1 + σ2∗ (t)X ε−2 − ... ± σε∗ (t) ; from which σj∗ (t) can be calculated recursively from the σk (t)-s, as the coefficient of X q−j , j = ∗ (t)σ (t) − ... ± σ ∗ (t)σ 1, ..., q − 2 is 0 = −σj∗ (t) + σj−1 j−1 (t) ∓ σj (t); for example 1 1 σ1∗ (t) = −σ1 (t); σ2∗ (t) = σ1 (t)2 − σ2 (t); σ3∗ (t) = −σ1 (t)3 + 2σ1 (t)σ2 (t) − σ3 (t); ... (1) etc. Note that we do not need to use all the coefficients/equations, it is enough to do it for j = 1, ..., ε. Using the same formulae, obtained from the coefficients of X q−j , j = 1, ..., ε, one can define the polynomials σ1∗ (T ) = −σ1 (T ); σ2∗ (T ) = σ1 (T )2 − σ2 (T ); σ3∗ (T ) = −σ1 (T )3 + 2σ1 (T )σ2 (T )− σ3 (T ); ... (1∗ ) up to σε∗ . Note that degT (σj∗ ) ≤ dj. From the definition   X q−ε −σ1 (T )X q−ε−1 +σ2 (T )X q−ε−2 −...±σq−ε (T )  X ε −σ1∗ (T )X ε−1 +σ2∗ (T )X ε−2 −...±σε∗ (T ) is a polynomial, which is X q − X for any substitution T = t ∈ GF(q), so it is of the form X q − X + (T q − T )(...). Now define G(X, T ) = X ε − σ1∗ (T )X ε−1 + σ2∗ (T )X ε−2 − ... ± σε∗ (T ), (2) from the recursive formulae it is a polynomial in X and T , of total degree ≤ dε and X-degree ε. For any T = t ∈ GF(q) the polynomial G(X, t) has ε roots in GF(q) (i.e. the missing elements GF(q) \ {fi (t) : i = 1, ..., q − ε}), so the algebraic curve G(X, T ) has at least N ≥ εq distinct points in GF(q) × GF(q). Suppose that G has no component (defined over GF(q)) of degree ≤ d. √ √ 1+d(d−1) q 1 Let’s apply the Lemma with a suitable d+1 + (d+1)q ≤ α < 1d , n = deg G ≤ dε ≤ d1 q−d+ 23 , we have εq ≤ N ≤ dεqα < εq, which is false, so G = H1 G1 , where H1 is an irreducible factor over GF(q) of degree at most d. If degX H1 = dX ≥ 2 then degX G1 = ε − dX , which means that H1 has at most q + 1 + (dX − √ 1)(dX − 2) q and G1 has at most (ε − dX )q distinct points in GF(q) × GF(q) (at most ε − dX for each T = t ∈ GF(q)), so in total G has √ εq ≤ N ≤ (ε − dX + 1)q + 1 + (dX − 1)(dX − 2) q, √ a contradiction if 2 ≤ dX ≤ q + 1, so degX H1 = 1. One can suppose w.l.o.g. that both H1 and G1 , expanded by the powers of X, are of leading coefficient 1. So H1 is of the form H1 (X, T ) = X − fq−ε+1 (T ), where fq−ε+1 (T ) = aq−ε+1 + bq−ε+1 T + cq−ε+1 T d + δq−ε+1 (T ), where δq−ε+1 (T ) is an “error polynomial” with terms of degree between 2 and d − 1. At the end of the proof we will show that δq−ε+1 and other error polynomials are zero. 3 Now one can repeat everything for G1 , which has at least (ε − 1)q distinct points in GF(q) × GF(q) (as H1 has exactly q and H1 G1 has at least εq). The similar reasoning gives G1 = H2 G2 , where H2 (X, T ) = X − fq−ε+2 (T ) with fq−ε+2 (T ) = aq−ε+2 + bq−ε+2 T + cq−ε+2 T d + δq−ε+2 (T ). Going on we get fq−ε+3 , ..., fq (where for j = q−ε+1, ..., q we have fj (T ) = aj +bj T +cj T d +δj (T ), where δj (T ) contains terms of degree between 2 and (d − 1) only). Hence G(X, T ) = q Y (X − fi (T )). q−ε+1 For any t ∈ GF(q) the values f1 (t), ..., fq (t) are all distinct, this is obvious from   X q−ε − σ1 (t)X q−ε−1 + σ2 (t)X q−ε−2 − ... ± σq−ε(t)  (X − fq−ε+1 (t))...(X − fq (t)) = X q − X. For j = q − ε + 1, ..., q let the plane Ej be defined by X4 = aj X1 + bj X2 + cj X3 . We are going to prove that {Ej : j = 1, ..., q} is a flock. First we check the case “t = ∞”: we have to check whether the intersection points Ei ∩ C on the plane at infinity X1 = 0, i.e. the values c1 , ..., cq−ε ; cq−ε+1 , ..., cq are all distinct (for Γ we {z | Γ } | {z } Γ∗ know it). (Note that even if q planes partition the affine part of C ∗ then this might be false for the infinite part of C ∗ .) From (1∗ ), considering the leading coefficients in each defining equality, we have σ1 (Γ∗ ) = −σ1 (Γ); σ2 (Γ∗ ) = σ1 (Γ)2 − σ2 (Γ); σ3 (Γ∗ ) = −σ1 (Γ)3 + 2σ1 (Γ)σ2 (Γ) − σ3 (Γ); etc., so X q − X =    X q−ε−σ1 (Γ)X q−ε−1+σ2 (Γ)X q−ε−2−...±σq−ε (Γ) X ε−σ1 (Γ∗ )X ε−1+σ2 (Γ∗ )X ε−2−...±σq−ε (Γ∗ ) , which we wanted to prove. Now we want to get rid of the δj ’s, i.e. we are going to prove that δq−ε+1 , ..., δq = 0. Let s be the maximal T -exponent appearing in any of δq−ε+1 , ..., δq , so each δj (T ) = dj T s + ... (for j = q − ε + 1, ..., q; also 2 ≤ s ≤ d − 1 and there exists a dj 6= 0). In the equation G(X, T ) = X ε − σ1∗ (T )X ε−1 + σ2∗ (T )X ε−2 − ... ± σε∗ (T ) = ε Y (X − aq−ε+i − bq−ε+iT − cq−ε+i T d − δq−ε+i (T )), i=1 the coefficient of X ε−j T d(j−1)+s , j = 1, ..., ε is zero on the left hand side (i.e. the coefficient of T d(j−1)+s in σj∗ , it can be seen by induction from (1∗ ) for instance), and it is σj−1 (Γ∗ \ {cq−ε+1 })dq−ε+1 + σj−1 (Γ∗ \ {cq−ε+2 })dq−ε+2 + ... + σj−1 (Γ∗ \ {cq })dq on the right hand side. Hence we have a system of homogeneous linear equations for dq−ε+1 , ..., dq with the elementary symmetric determinant 1 σ1 \ {cq−ε+1 }) σ2 (Γ∗ \ {cq−ε+1 }) .. . (Γ∗ 1 σ1 \ {cq−ε+2 }) σ2 (Γ∗ \ {cq−ε+2 }) (Γ∗ ... ... ... 1 σ1 \ {cq }) σ2 (Γ∗ \ {cq }) (Γ∗ σε−1 (Γ∗ \ {cq−ε+1 }) σε−1 (Γ∗ \ {cq−ε+2 }) ... σε−1 (Γ∗ \ {cq }) 4 = Y (cq−ε+i − cq−ε+j ), 1≤i<j≤ε which is non-zero as the ci ’s are pairwise distinct. Hence the unique solution is dq−ε+1 , ..., dq = 0 and fj (T ) = aj + bj T + cj T d for each j = 1, ..., q. 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