Ovoid (polar space)

In mathematics, an ovoid O of a (finite) polar space of rank r is a set of points, such that every subspace of rank r-1 intersects O in exactly one point.{{citation

| last = Moorhouse | first = G. Eric

| editor1-last = Klin | editor1-first = Mikhail

| editor2-last = Jones | editor2-first = Gareth A.

| editor3-last = Jurišić | editor3-first = Aleksandar

| editor4-last = Muzychuk | editor4-first = Mikhail

| editor5-last = Ponomarenko | editor5-first = Ilia

| contribution = Approaching some problems in finite geometry through algebraic geometry

| doi = 10.1007/978-3-642-01960-9_11

| location = Berlin

| mr = 2605578

| pages = 285–296

| publisher = Springer

| title = Algorithmic Algebraic Combinatorics and Gröbner Bases: Proceedings of the Workshop D1 "Gröbner Bases in Cryptography, Coding Theory and Algebraic Combinatorics" held in Linz, May 1–6, 2006

| url = https://books.google.com/books?id=sstt1cj7Nv8C&pg=PA285

| year = 2009| isbn = 978-3-642-01959-3

| citeseerx = 10.1.1.487.1198

}}.

Cases

=Symplectic polar space=

An ovoid of W_{2 n-1}(q) (a symplectic polar space of rank n) would contain q^n+1 points.

However it only has an ovoid if and only n=2 and q is even. In that case, when the polar space is embedded into PG(3,q) the classical way, it is also an ovoid in the projective geometry sense.

=Hermitian polar space=

Ovoids of H(2n,q^2)(n\geq 2) and H(2n+1,q^2)(n\geq 1) would contain q^{2n+1}+1 points.

=Hyperbolic quadrics=

An ovoid of a hyperbolic quadric Q^{+}(2n-1,q)(n\geq 2)would contain q^{n-1}+1 points.

=Parabolic quadrics=

An ovoid of a parabolic quadric Q(2 n,q)(n\geq 2) would contain q^n+1 points. For n=2, it is easy to see to obtain an ovoid by cutting the parabolic quadric with a hyperplane, such that the intersection is an elliptic quadric. The intersection is an ovoid.

If q is even, Q(2n,q) is isomorphic (as polar space) with W_{2 n-1}(q), and thus due to the above, it has no ovoid for n\geq 3.

=Elliptic quadrics=

An ovoid of an elliptic quadric Q^{-}(2n+1,q)(n\geq 2)would contain q^{n}+1 points.

See also

References