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Let X be a projective variety (say, irreducible) and E a vector bundle on X or rank r. Is it true that there exists a codimension 2 closed subset Z in X such that restriction of E(n) (for n large enough) to U = X - Z has a trivial sub-bundle of rank (r-1)? Is this written somewhere? What happens when X varies in a flat family over a base S?

UPD: corrected the question - an ample twist of E is supposed to have a trivial sub-bundle.

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Stated like this it can't be true. Indeed if $X$ is normal then by Hartogs lemma each embedding $O_{X-Z} \to E_{|X-Z}$ (i.e. a section of $E_{|X-Z}$) should extend to $O_X \to E$. But a priori $E$ can be without global sections. For example $E = O(-1) \oplus O(-1)$ on $P^2$. Maybe you want to have a subbundle which is trivial up to a twist?

UPD. If a twist is allowed then the answer is yes. Indeed, after an appropriate twist we can assume that $E$ is generated by a vector space $V$ of global sections, that is a map $V\otimes O_X \to E$ is surjective. Let $E' = Ker(V\otimes O_X \to E)$. It is a vector bundle on $X$. Consider the Grassmannian $Gr(r-1,V)$ of vector subspaces of rank $r-1$. Let $U$ denote the tautological subbundle. On the product $X\times Gr(r-1,V)$ we have a composition $$ U\boxtimes O \to V\otimes O\boxtimes O \to O\boxtimes E. $$ Let $Y$ be its discriminant locus (the scheme of points where the rank is $\le r-2$). The fiber of $Y$ over a point $x \in X$ consists of all $(r-1)$-dimensional subspaces which intersect with the subspace $E'_x \subset V$ of codimension $r$. A simple parameter count shows that it has codimension 2 in the Grassmannian. Since this is true for all $x \in X$ we conclude that $codim Y = 2$. Hence for generic point $u \in Gr(r-1,V)$ the fiber $Y_u \subset X$ also has codimension 2. But since the corresponding map $O^{r-1} \to E$ is an embedding out of $Y_u$, we just take $Z = Y_u$ and restrict the above map to $X - Z$.

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Could even take $O(-1)\oplus O(-1)$ on ${\mathbb P}^1$. –  t3suji Nov 24 '10 at 4:59
    
Sorry - of course to a twist! I want the restriction of E(n) for any n large enough, to contain a trivial subbundle –  Vladimir Baranovsky Nov 25 '10 at 3:25
    
Thank you! I couldn't see why Y had codimension exactly two (and not less), but looking at fibers over X does the job. –  Vladimir Baranovsky Nov 29 '10 at 16:03

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