Say I have two locally free sheaves $F,G$ on projective variety $X$. I know the cohomology groups $H^i(X,F)$ and $H^i(X,G)$. Is this enough to give me information about $H^i(X,F\otimes G)$? In particular, if $H^i(X,F)=0$, what conditions on $G$ guarantee that also $H^i(X,F\otimes G)=0$?
Cohomology of a tensor product of sheaves
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algebraic-geometry
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1Have you seen http://mathoverflow.net/questions/34673/kunneth-formula-for-sheaf-cohomology-of-varieties? – 2010-11-10
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4Just for reference you'll need more than both vanishing to get that the tensor product vanishes. Take for instance $X=\mathbb{P}_k^n$, $F=\mathcal{O}(-n)$, and $G=\mathcal{O}(-1)$, then $H^n(X, F)=0$, $H^n(X, G)=0$ but $H^n(X, F\otimes G)=k$. – 2010-11-10
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You need to make some positivity assumptions on $E$ and $F$, as your conclusion is just not true in general. The only case I know of is Le Poitier's vanishing theorem, which says that if $E \otimes F \otimes \omega_X^{-1}$ is ample on a smooth projective variety $X$, then $H^i(X,E \otimes F)=0$ for $i \geq rs$, where $rk(E)=r$ and $rk(F)=s$. This is satisfied for instance if $\omega_X^{-1}$ is nef, and $E$ and $F$ are both ample on $X$, but even here you need $rs$ to be small relative to the dimension of $X$ if you want to say something meaningful.