On the Smallest Non-Trivial Tight Sets in Hermitian Polar Spaces

Jan De Beule,Klaus Metsch

doi:10.37236/6461

Abstract

We show that an $x$-tight set of the Hermitian polar spaces $\mathrm{H}(4,q^2)$ and $\mathrm{H}(6,q^2)$ respectively, is the union of $x$ disjoint generators of the polar space provided that $x$ is small compared to $q$. For $\mathrm{H}(4,q^2)$ we need the bound $x<q+1$ and we can show that this bound is sharp.

Highlights

Let q be a prime power and let GF(q) be the finite field of order q
We show that an x-tight set of the Hermitian polar spaces H(4, q2) and H(6, q2) respectively, is the union of x disjoint generators of the polar space provided that x is small compared to q
The elements of the finite classical polar space P associated with f are the totally singular or totally isotropic subspaces of PG(n, q) with relation to f, according to whether f is a quadratic or sesquilinear form

Summary

Introduction

Let q be a prime power and let GF(q) be the finite field of order q. From the definition of tight sets, we see that there are q2 + x choices for Q ∈ P ⊥ ∩ T One of these points is P but P is not contained in any of the pairs we are counting. If T contains a line of H(4, q2), T \ is an (x − 1)-tight set of H(4, q2) In this case the induction hypothesis implies that T is a union of x lines. We have to show that T is the union of x lines This follows from Lemma 2.4 provided that we can show that |P ⊥ ∩ Q⊥ ∩ T | x for any two non-perpendicular points P and Q of T. Since T is a tight set with parameter x < q + 1, it follows that all q3 + 1 points of the Hermitian curve belong to T. It remains to show that T contains a plane. seaxtisiAstfssysiaunmgpoe|hino⊥tn∩PthTe∈|c>oTntxtrh(aqar2ty+ltihe1sa)t.oTnTmhdeooreresefontrohetaLnceom√ntm2aqianl3ian.6epslsahhnoebw. esTinthgheanstebcxyanLtetqmo−mHa√(632,.q4q,2+)th2ae,nrdae contradiction to the assumed upper bound on x

A new tight set

Findings

Open problems