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Let $E$ be a rank $r$ bundle on a quasiprojective scheme. There is a natural map

$$ \Lambda^r H^0(E) \to H^0(\Lambda^r E) $$ It does not have to be surjective as can be seen from the case when $E$ is a direct sum of line bundles. However, is it true that $E$ can always be twisted by an ample line bundle so that this map becomes surjective?

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1 Answer 1

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Assuming $H^0(X,\mathcal O_X)$ is a field $K$, yes. We will prove a slightly more general fact: For all pairs of coherent sheaves $A$, $B$, for some ample $L$, $H^0(X,A \otimes L) \otimes H^0(X,B \otimes L) \to H^0(X,A \otimes B \otimes L^2)$ is surjective.

This is more general because it is sufficient to show $\otimes^r H^0(X,E\otimes L) \to H^0( X,\otimes^r (E \otimes L) )$ is surjective, but this can be done by applying the $A\otimes B$ version $r-1$ times.

The idea of the proof is that we twist to make $A$ and $B$ generated by global sections. We then show that the obstruction to surjectivity lies in a cohomology group, and we twist again to make that cohomology group vanish.

The proof is as follows: By twisting by an ample line bundle, we may first assume $A$ and $B$ are generated by global sections. So $H^0(X,A)\otimes \mathcal O_X \to A$ is surjective and $H^0(X,B) \otimes \mathcal O_X \to B$ is surjective. So

\[C= \left(H^0(X,A) \otimes H^0(X,B)\right)\otimes\mathcal O_X = \left( H^0(X,A)\otimes\mathcal O_X \right) \otimes \left( H^0(X,B) \otimes \mathcal O_X \right) \to A \otimes B \]

is surjective because the tensor of two surjective maps of coherent sheaves is surjective. Let $D$ be its kernel. Twist $A$ and $B$ by a line bundle $L$ such that $D \otimes L^2$ has trivial cohomology. Also choose $L$ sufficiently large that $H^0(X,L)^{\otimes 2} \to H^0(X,L^{\otimes 2})$ is surjective, which is possible because the section ring of ample divisors is finitely generated. Then $H^0(X,C \otimes L^2) \to H^0(X,A \otimes B \otimes L^2)$ is surjective, because $H^1(X,D\otimes L^2)=0$.

But $C$ is a constant sheaf, so $H^0(X,C\otimes L^2) = H^0(X,A) \otimes H^0(X,B) \otimes H^0(X,L^2)$. Composing with the surjective map $H^0(X,L)^{\otimes 2} \to H^0(X,L^{\otimes 2})$, we obtain a surjective map:

\[ H^0(X,A) \otimes H^0(X,L ) \otimes H^0(X, B) \otimes H^0(X,L) \to H^0(X,A\otimes L \otimes B \otimes L)\]

This clearly factors through $H^0(X,A\otimes L) \otimes H^0(X,B \otimes L)$, which makes the desired map surjective as well.

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Very nice, thank you! –  Vladimir Baranovsky Apr 12 '13 at 15:38

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