We give a geometric proof of the upper bound of q(2n+1) + 1 on the size of partial spreads in the polar space H(4n + 1, q(2)). This bound is tight and has already been proved in an algebraic way. Our alternative proof also yields a characterization of the partial spreads of maximum size in H(4n + 1, q(2))

Vanhove, Frédéric

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Ghent University Academic Bibliography

Advances in Mathematics of Communications doi:10.3934/amc.2011.5.157Volume 5, No. 2, 2011, 157–160A GEOMETRIC PROOF OF THE UPPER BOUNDON THE SIZE OF PARTIAL SPREADS IN H(4n+ 1, q2)Fre´de´ric VanhoveDepartment of Mathematics, Ghent UniversityKrijgslaan 281, S22B-9000 Ghent, Belgium(Communicated by Ivan Landjev)Abstract. We give a geometric proof of the upper bound of q2n+1+1 on thesize of partial spreads in the polar space H(4n+1, q2). This bound is tight andhas already been proved in an algebraic way. Our alternative proof also yieldsa characterization of the partial spreads of maximum size in H(4n+ 1, q2).1. IntroductionA classical finite polar space is an incidence structure, consisting of the totallyisotropic subspaces of a projective space with respect to a non-degenerate sesquilin-ear form or a non-degenerate quadratic form. All dimensions will be assumed to beprojective from now on, and we will also refer to m-dimensional subspaces as simplym-spaces. In particular, the 0- and 1-dimensional subspaces of such a polar spaceare known as its points and lines, respectively. The generators are its subspacesof maximal dimension. A partial spread of a classical finite polar space is a set ofgenerators with no two incident with a common point. If a partial spread actuallypartitions the point set of the polar space, it is said to be a spread.The Hermitian variety H(n, q2) is a particular type of classical finite polar space,consisting of the subspaces in PG(n, q2), the points of which all have homogeneouscoordinates (x0, . . . , xn) satisfying the equation xq+10 + . . .+xq+1n = 0. In this polarspace, the generators are (n−1)/2-dimensional, if n is odd, or (n−2)/2-dimensional,if n is even, and the number of points is given by |H(n, q2)| = (qn+1 + (−1)n)(qn −(−1)n)/(q2−1). We refer to [4] for proofs and much more information on Hermitianvarieties and polar spaces in general.Thas [6] proved that in H(2n + 1, q2) spreads, or thus partial spreads of sizeq2n+1 + 1, cannot exist, which has made the question on the size of a partial spreadin such a polar space, an intriguing question. Improved upper bounds on the sizeof partial spreads in H(2n+ 1, q2) were proved in [2].On the other hand, partial spreads of size qn+1 + 1 in H(2n + 1, q2) were con-structed for all n ≥ 1 in [1], by use of a symplectic polarity of the projective spacePG(2n + 1, q2), commuting with the associated Hermitian polarity. In the Baersubgeometry of points on which these two polarities coincide, a (regular) spread ofthe induced symplectic polar space W (2n + 1, q) can always be found, and these2000 Mathematics Subject Classification: 05B25, 51E23.Key words and phrases: Hermitian varieties, partial spreads.This research is supported by the Research Foundation Flanders-Belgium (FWO-Vlaanderen).157 c©2011 AIMS-SDU158 Fre´de´ric Vanhoveqn+1 + 1 generators extend to pairwise disjoint generators of H(2n + 1, q2). Max-imality of partial spreads of H(2n + 1, q2) constructed in this way was also shownfor n = 1, 2 in [1] and for all even n in [5].In [3], De Beule and Metsch proved that the maximum size of a partial spread inH(5, q2) is q3 + 1, and they also obtained additional information on partial spreadsmeeting that tight bound. In particular, they found that every generator ofH(5, q2),not meeting any element of such a partial spread S in a line or more, meets exactlyq2 − q + 1 elements of S in a point.Using techniques from algebraic graph theory, we recently proved in [7] that thesize of a partial spread in H(4n+ 1, q2) is at most q2n+1 + 1, and this bound is thustight as well. It turns out that a geometric property of partial spreads of maximumsize in H(5, q2) can be generalized, and in fact paves the way for a new, completelygeometric proof of the upper bound in H(4n+ 1, q2).2. ToolsWe first state a lemma by Thas [6].Lemma 2.1. Let pi1, pi2 and pi be three mutually disjoint generators in H(2n+1, q2).The set of points on pi1, that are on a (necessarily unique) line of H(2n + 1, q2)meeting both pi and pi2, form a non-singular Hermitian variety in pi1.Corollary 2.2. Let pi1, pi2 and pi be three mutually disjoint generators in H(2n +1, q2). The number of generators meeting pi in an (n − 1)-space, and meeting bothpi1 and pi2 in a point is |H(n, q2)| = (qn+1+(−1)n)(qn−(−1)n)q2−1 .Proof. We let ⊥ denote the Hermitian polarity of PG(2n + 1, q2), associated withthe polar space. It is obvious that every generator meeting pi in an (n − 1)-space,can meet pi1 and pi2 in at most one point. On the other hand, through any pointp1 ∈ pi1, there is a unique generator 〈p1, p⊥1 ∩ pi〉 meeting pi in an (n − 1)-space.Hence we have to determine the number of points p1 ∈ pi1 such that the generator〈p1, p⊥1 ∩ pi〉 also meets pi2 in a point.First suppose that a point p1 ∈ pi1 is such that the generator 〈p1, p⊥1 ∩ pi〉 meetspi2 in a point p2. In that case, the line p1p2 is a line of H(2n+ 1, q2), meeting pi aswell, as p⊥1 ∩ pi is a hyperplane of 〈p1, p⊥1 ∩ pi〉. Conversely, suppose a point p1 ∈ pi1is on a line of H(2n+ 1, q2), meeting pi in p and pi2 in p2. In that case, both p1 andp are in the generator 〈p1, p⊥1 ∩pi〉, and hence so is the entire line p1p, including thepoint p2. The desired result thus follows from Lemma 2.1.3. The proofTheorem 3.1. The size of a partial spread S in H(4n + 1, q2), n ≥ 1, is at mostq2n+1+1. If |S| > 1 and pi ∈ S, then every generator meeting pi in a (2n−1)-space,will meet the same number of other elements of S in just a point, if and only if|S| = q2n+1 + 1. In that case, that number must be q2n.Proof. Let S be a partial spread of size at least 2 in H(4n+ 1, q2). Consider a fixedelement pi ∈ S. Let {Ni|i ∈ I} be the set of generators meeting pi in a (2n−1)-space.As the number of (2n − 1)-spaces in a generator equals (q4n+2 − 1)/(q2 − 1), andthe number of generators through any (2n − 1)-space in H(4n + 1, q2) is given byq + 1, the cardinality of I is q4n+2−1q2−1 q.Advances in Mathematics of Communications Volume 5, No. 2 (2011), 157–160On the largest partial spreads in H(4n+ 1, q2) 159Note that any generator Ni and any generator in S\{pi}, are either disjoint ormeet in a point. For every Ni, i ∈ I, let ti denote the number of generators inS\{pi}, meeting Ni in a point. We now count the number of pairs (Ni, pi′), with pi′an element of S\{pi} meeting Ni in a point, in two ways. As through every point p′on an element pi′ of S\{pi}, there is a unique generator meeting pi in a (2n−1)-space,we obtain:(1)∑i∈Iti = (|S| − 1)q4n+2 − 1q2 − 1 .Now we count the number of ordered triples (Ni, pi1, pi2), with pi1 and pi2 two distinctelements of S\{pi}, both meeting Ni in a point. We know from Corollary 2.2 thatfor every two distinct elements of S\{pi}, there will be exactly |H(2n, q2)| generatorsNi, meeting both of them in a point. Hence we obtain:(2)∑i∈Iti(ti − 1) = (|S| − 1)(|S| − 2)(q2n+1 + 1)(q2n − 1)q2 − 1 .Combining (1) and (2), we find:(3)∑i∈It2i = (|S| − 1)q2n+1 + 1q2 − 1((q2n+1 − 1) + (|S| − 2)(q2n − 1)).As (∑i∈I ti)2 ≤ (∑i∈I t2i )|I|, with equality if and only if all ti are equal, this implies:(|S|−1)2(q4n+2 − 1q2 − 1)2≤ (|S|−1)q2n+1 + 1q2 − 1((q2n+1−1)+(|S|−2)(q2n−1))q4n+2 − 1q2 − 1 q,with equality if and only if all ti are equal. Since we assumed that |S| > 1, we cancancel factors on both sides to obtain:(|S| − 1)(q2n+1 − 1) ≤((q2n+1 − 1) + (|S| − 2)(q2n − 1))q,implying that |S| ≤ q2n+1 + 1, with equality if and only if all ti are equal. In thatcase, their constant value must equal (∑i∈I ti)/|I| = q2n.4. RemarkThis technique fails when applied to partial spreads in H(4n + 3, q2), where ityields a negative lower bound on the size instead.AcknowledgementsThe author is grateful to Jan De Beule for his helpful suggestions while preparingthis manuscript.References[1] A. Aguglia, A. Cossidente and G. L. Ebert, Complete spans on Hermitian varieties, Des.Codes Cryptogr., 29 (2003), 7–15.[2] J. De Beule, A. Klein, K. Metsch and L. Storme, Partial ovoids and partial spreads in Her-mitian polar spaces, Des. Codes Cryptogr., 47 (2008), 21–34.[3] J. De Beule and K. Metsch, The maximum size of a partial spread in H(5, q2) is q3 + 1, J.Combin. Theory Ser. A, 114 (2007), 761–768.[4] J. W. P. Hirschfeld and J. A. Thas, “General Galois Geometries,” The Clarendon Press,Oxford University Press, New York, 1991.[5] D. Luyckx, On maximal partial spreads of H(2n+1, q2), Discrete Math., 308 (2008), 375–379.Advances in Mathematics of Communications Volume 5, No. 2 (2011), 157–160160 Fre´de´ric Vanhove[6] J. A. Thas, Old and new results on spreads and ovoids of finite classical polar spaces, Ann.Discrete Math., 52 (1992), 529–544.[7] F. Vanhove, The maximum size of a partial spread in H(4n + 1, q2) is q2n+1 + 1, Electron.J. Combin., 16 (2009), 1–6.Received March 2010; revised August 2010.E-mail address: fvanhove@cage.ugent.beAdvances in Mathematics of Communications Volume 5, No. 2 (2011), 157–160

A geometric proof of the upper bound on the size of partial spreads in H(4n+1, q²)

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A geometric proof of the upper bound on the size of partial spreads in H(4n+1, q²)

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