the hermitian variety h(5 4) has no ovoid

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The Hermitian variety H (5, 4) has no ovoid Jan De Beule Ghent University joint work with: Klaus Metsch Gießen University The Hermitian variety H(5, 4) has no ovoid – p.1/8

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The Hermitian variety H(5, 4) has no ovoid

Jan De Beule

Ghent University

joint work with: Klaus Metsch

Gießen University

The Hermitian variety H(5, 4) has no ovoid – p.1/8

Introduction

The Hermitian variety H(d, q2) is the set of points ofPG(d, q2) satisfying the equation

Xq+1

0 + Xq+1

1 + . . . Xq+1

d = 0

When d = 2n + 1, 2n respectively, H(d, q2) contains points,lines, . . . , n-dimensional subspaces of PG(d, q2),(n − 1)-dimensional subspaces of PG(d, q2) respectively.

The Hermitian variety H(d, q2) is a example of a so-called

classical polar space. The subspaces of maximal dimen-

sion are also called generators.

The Hermitian variety H(5, 4) has no ovoid – p.2/8

Ovoids

An ovoid of a Hermitian variety H(d, q2) is a set O of pointsof H(d, q2) such that every generator meets O in exactlyone point.

The Hermitian variety H(5, 4) has no ovoid – p.3/8

Ovoids

An ovoid of a Hermitian variety H(d, q2) is a set O of pointsof H(d, q2) such that every generator meets O in exactlyone point.

If H(d − 2, q2) has no ovoids, then H(d, q2) has no ovoids.

The Hermitian variety H(5, 4) has no ovoid – p.3/8

Known results

J. A. Thas: the Hermitian variety H(2n, q2), n > 1, has noovoids.

The Hermitian variety H(5, 4) has no ovoid – p.4/8

Known results

J. A. Thas: the Hermitian variety H(2n, q2), n > 1, has noovoids.

G.E. Moorhouse: H(2n + 1, q2), q = ph, p prime, h > 1 hasno ovoids if

p2n+1 >

(

2n + p

2n + 1

)2

(

2n + p − 1

2n + 1

)2

Ovoids of H(5, q2) are not excluded.

The Hermitian variety H(5, 4) has no ovoid – p.4/8

Known results

J. A. Thas: the Hermitian variety H(2n, q2), n > 1, has noovoids.

G.E. Moorhouse: H(2n + 1, q2), q = ph, p prime, h > 1 hasno ovoids if

p2n+1 >

(

2n + p

2n + 1

)2

(

2n + p − 1

2n + 1

)2

Ovoids of H(5, q2) are not excluded.

A. Klein: H(2n + 1, q2) has no ovoids if n > q3.

The Hermitian variety H(5, 4) has no ovoid – p.4/8

Ovoids of H(3, 4)

Suppose that O is an ovoid of H(3, 4). There exists a planeπ, π ∩ H(3, 4) = H(2, 4), such that either

π ∩ H(3, 4) = H(2, 4) = O, or

O = (H(2, 4) \ L) ∪ (L⊥ ∩ H(3, 4)), L a line of π,L ∩ H(3, 4) = H(1, 4).

The Hermitian variety H(5, 4) has no ovoid – p.5/8

Ovoids of H(3, 4)

Suppose that O is an ovoid of H(3, 4). There exists a planeπ, π ∩ H(3, 4) = H(2, 4), such that either

π ∩ H(3, 4) = H(2, 4) = O, or

O = (H(2, 4) \ L) ∪ (L⊥ ∩ H(3, 4)), L a line of π,L ∩ H(3, 4) = H(1, 4).

Every partial ovoid of H(3, 4) containing 8 points can beextended to an ovoid of H(3, 4).

The Hermitian variety H(5, 4) has no ovoid – p.5/8

Ovoids of H(3, 4)

Suppose that O is an ovoid of H(3, 4). There exists a planeπ, π ∩ H(3, 4) = H(2, 4), such that either

π ∩ H(3, 4) = H(2, 4) = O, or

O = (H(2, 4) \ L) ∪ (L⊥ ∩ H(3, 4)), L a line of π,L ∩ H(3, 4) = H(1, 4).

Every partial ovoid of H(3, 4) containing 8 points can beextended to an ovoid of H(3, 4).Every plane π meets O in 0, 1, 2, 3 or 6 points.

The Hermitian variety H(5, 4) has no ovoid – p.5/8

Ovoids of H(3, 4)

Suppose that O is an ovoid of H(3, 4). There exists a planeπ, π ∩ H(3, 4) = H(2, 4), such that either

π ∩ H(3, 4) = H(2, 4) = O, or

O = (H(2, 4) \ L) ∪ (L⊥ ∩ H(3, 4)), L a line of π,L ∩ H(3, 4) = H(1, 4).

Every partial ovoid of H(3, 4) containing 8 points can beextended to an ovoid of H(3, 4).Every plane π meets O in 0, 1, 2, 3 or 6 points.

If π is a plane, |π ∩ O| = 3, then the points of π ∩ O are

collinear.

The Hermitian variety H(5, 4) has no ovoid – p.5/8

Ovoids of H(5, 4)

Suppose that O is an ovoid of H(5, 4). Let p be a point ofH(5, 4) \ O. Then |p⊥ ∩O| = 9. If π is a plane in p⊥,π ∩ H(5, 4) = H(2, 4), then |〈p, π〉| ∈ {0, 1, 2, 3, 6, 9}

The Hermitian variety H(5, 4) has no ovoid – p.6/8

Ovoids of H(5, 4)

Suppose that O is an ovoid of H(5, 4). Let p be a point ofH(5, 4) \ O. Then |p⊥ ∩O| = 9. If π is a plane in p⊥,π ∩ H(5, 4) = H(2, 4), then |〈p, π〉| ∈ {0, 1, 2, 3, 6, 9}Suppose that O is an ovoid of H(5, q2). Consider a plane π

that meets the variety in H(2, q2) and put m := |π ∩O|.Suppose furthermore that 1 ≤ m < q3 + 1. Let A, resp. B,be the set consisting of all points x ∈ O \ π such that 〈π, x〉meets H(5, q2) in a cone sH(2, q2), resp. an H(3, q2).

We have |A| = (q2 − 1)(q2 − 1 + m) and|B| = q2(q3 − q2 + 2 − m).

If q = 2 and x is a point of (π ∩ H(5, 4)) \ O, then|x⊥ ∩ B| ∈ {0, 3, 6, 7, 8, 9}.

The Hermitian variety H(5, 4) has no ovoid – p.6/8

The last steps

Suppose that O is an ovoid of H(5, 4). Then |π ∩O| ≤ 3 forevery plane π, π ∩ H(5, 4) = H(2, 4) and |α ∩ O| < 6 forevery 3-dimensional space α, α ∩ H(5, 4) = H(3, 4).

The Hermitian variety H(5, 4) has no ovoid – p.7/8

The last steps

Suppose that O is an ovoid of H(5, 4). Then |π ∩O| ≤ 3 forevery plane π, π ∩ H(5, 4) = H(2, 4) and |α ∩ O| < 6 forevery 3-dimensional space α, α ∩ H(5, 4) = H(3, 4).

The Hermitian variety H(5, 4) has no ovoids

The Hermitian variety H(5, 4) has no ovoid – p.7/8

References[1] A. Klein. Partial Ovoids in Classical Finite Polar Spaces. Des. Codes Cryptogr.,

31:221–226, 2004.

[2] G. Eric Moorhouse. Some p-ranks related to Hermitian varieties. J. Statist. Plann.Inference, 56(2):229–241, 1996. Special issue on orthogonal arrays and affinedesigns, Part II.

[3] J. A. Thas. Ovoids and spreads of finite classical polar spaces. Geom. Dedicata,10(1-4):135–143, 1981.

[4] J. A. Thas. Ovoids, spreads and m-systems of finite classical polar spaces. InSurveys in combinatorics, 2001 (Sussex), volume 288 of London Math. Soc.Lecture Note Ser., pages 241–267. Cambridge Univ. Press, Cambridge, 2001.

The Hermitian variety H(5, 4) has no ovoid – p.8/8