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Let $\mathfrak{F}$ be a sheaf of abelian groups on a smooth scheme $X$. Suppose for some $i>1$, there exists a surjective morphism $H^i(\mathfrak{F}^{\otimes i-1}) \to H^i(\mathfrak{F}^{\otimes i})$. Is it true that this means $h^i(\mathfrak{F}^{\otimes i-1})=h^i(\mathfrak{F}^{\otimes i})=0$?

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closed as unclear what you're asking by Fernando Muro, Daniel Moskovich, David Roberts, Carlo Beenakker, Andrés Caicedo Jun 30 '13 at 18:17

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I'm confused. Why would that be true? Tensor over what sheaf of rings (I suppose over $\mathbb{Z}$)? Are the $i$ in $H^i$ and $F^{\otimes i}$ supposed to be the same $i$? What is $h^i$ if $F$ is not a sheaf of vector spaces over some field? Why would $h^i$ be finite? In a common situation where $X$ is projective over a field $k$ and $F$ is a coherent sheaf of $O_X$-modules, the assertion "there exists a surjective morphism" means just that $h^i(F^{\otimes (i-1)}) > h^i(F^{\otimes i}). Furthermore, there exist obvious counterexamples. –  Piotr Achinger Jun 23 '13 at 22:17
In addition to what Piotr said, in what topology is $\mathcal F$ a sheaf? If it's Zariski, then it's very unnatural to ask for $\mathcal F$ to be a sheaf of abelian groups; for the cohomology groups to be "geometrically meaningful" it should be (quasi-)coherent. In the étale topology the situation is slightly better but one should still in general demand that $\mathcal F$ is torsion and make assumptions about the characteristic. My advice would be to learn more about cohomology in general and cohomology of algebraic varieties in particular until you know what kind of question you want to ask. –  Dan Petersen Jun 24 '13 at 6:07

1 Answer 1

No. Take $X$ to be an Abelian variety of dimension $g$, and $\mathfrak{F} = \mathcal{O}_X$. Then $h^i(\mathcal{O}_X) = \binom{g}{i}$.

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Since $\mathfrak F$ was a sheaf of abelian groups, probably the tensor product means over $\mathbb Z$. But the same idea gives a counterexample: take $\mathfrak F = \underline{\mathbb{Z}}$ and again $\mathfrak F^{\otimes i} = \mathfrak F$ for all $i$. –  Dan Petersen Jun 24 '13 at 6:09

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