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How to compute the dimension of the linear system of quadric hypersurfaces that pass through a given projective variety $S$?

A concrete one: $S$ is a $5$-dimensional projective variety in $\mathbb{P}^9$ which is the transverse intersection of $4$ quadric hypersurfaces.

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By "pass through" do you mean "contains" or "intersects"? I usually only see that phrasing when talking about individual points. –  Charles Siegel Feb 4 '10 at 15:02
    
Charles -- probably this means "contains" because otherwise we don't get a linear system in general. –  algori Feb 4 '10 at 18:31
    
Here is an answer assuming "passes through" means "contains": to each projective variety there corresponds a homogeneous ideal (it is formed by all polynomials that are zero at each point of the subvariety) and the dimension of the linear system of quadrics containing the subvariety is the degree 2 part of the ideal. In the above example (complete intersection of 4 quadrics) we get 4. –  algori Feb 4 '10 at 18:38
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1 Answer

Your question is equivalent to the computation of $H^0(\mathcal{I}_S(2))$.

In the example you give, $S$ is a complete intersection of $4$ quadrics and so the resolution of its ideal sheaf $\mathcal{I}_S$ is given by the Koszul complex (I write $\mathcal{O}$ instead of $\mathcal{O}_{\mathbb{P}^9}$):

$0 \to \mathcal{O}(-8) \to \mathcal{O}(-6)^{\oplus 4} \to \mathcal{O}(-4)^{\oplus 6} \to \mathcal{O}(-2)^{\oplus 4} \to \mathcal{I}_S \to 0$.

Tensoring with $\mathcal{O}(2)$ we obtain:

$0 \to \mathcal{O}(-6) \to \mathcal{O}(-4)^{\oplus 4} \to \mathcal{O}(-2)^{\oplus 6} \to \mathcal{O}^{\oplus 4} \to \mathcal{I}_S(2) \to 0$.

Splitting this exact sequence into short exact ones it is immediate to check that

$H^0(\mathcal{I}_S(2))=H^0(\mathcal{O}^{\oplus 4})=4$,

as Algori states in his comment.

Therefore the linear system of quadrics passing through $S$ has dimension $4-1=3$.

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