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For a projective variety $X$, Serre's vanishing theorem says that $H^i(X, \mathcal{F}(n))=0$ for any coherent sheave, $i>1$ and sufficiently large $n$. I am wondering, is there a similar type of vanishing theorem on quasi-projective varieties, namely, let $Y$ be a quasi-projective variety, what can we say about the vanishing of $H^i(Y, \mathcal{F}(n))$ under the same setting as projective case.

Or is there a similar type of theorem for local cohomology, say, when is $H^i_{pt}(X, \mathcal{F}(n))$ vanishing?

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Hi, what's F(n) in this context? – Lars Apr 28 '10 at 19:27
By $F(n)$, I mean $F\otimes \mathcal{O}_X(n)$. – Fei YE Apr 28 '10 at 20:06
Thanks, sorry, I was confused. – Lars Apr 28 '10 at 20:07
up vote 8 down vote accepted

Even for quasi-affine variety you don't have a vanishing theorem except Grothendieck's vanishing theorem on a noetherian topological space of finite dimension.
Consider, for example, affine plane without point $\mathbf{A}^2\backslash 0.$ Then the structure sheaf is ample but the first cohomology $H^1(\mathbf{A}^2\backslash 0, \mathcal{O})$ is not trivial and even it is infinite dimensional.

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Thank you very much for the reply. But why the structure sheaf $\mathcal{O}_{\mathbf{A}^2\setminus \{0\}}$ is ample. What is the definition of ampleness in this setting? – Fei YE Apr 28 '10 at 20:09
If you embed $A^2\setminus\{0\}$ into $P^2$, the restriction of $O(1)$ will be isomorphic to $O$, so it is ample. – Sasha Apr 28 '10 at 20:14
See a definition in Hartshorn book page 153. – user2464 Apr 28 '10 at 20:20

Another quick couple comments although I certainly agree with ploughshare.

If $X$ is projective over an affine variety (but itself is only quasi-projective) then serre vanishing still holds.

With regards to the local cohomology question $H^i_{pt}(X, F)$ essentially never vanishes for all $i$, no matter what $F$ you pick (as long as it's non-zero). A theorem of Grothendieck (see for example Bruns and Herzog, ``Cohen-Macaulay Rings'', Theorem 3.5.7) says that if $M$ is an $R$-module of a Noetherian local ring $(R, m)$ then $H^d_{m}(M) \neq 0$ where $d$ is the dimension of $M$.

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Another comment on the local cohomology part of the question: Besides Karl's perfect answer one might also point out that the twisting does nothing to the cohomology group. The twisting line bundle is trivial in a neighbourhood of the point, so locally near the point $\mathcal F(n)\simeq \mathcal F$.

On the other hand, your question is exactly about local cohomology: If $Y$ is quasi-projective and its projective closure is $X$, then the groups $H^i(X,\mathcal F(n))$ and $H^i(Y,\mathcal F(n))$ are part of a long exact cohomology sequence with the third vertex being local cohomology supported on $X\setminus Y$.

Finally, ploughshare's example is actually just an explicit manifestation of Karl's comment via the (equivalent of the) above sequence: $\mathbb A^2$ is affine, so the cohomology of $\mathbb A^2\setminus {0}$ is the same as the local cohomology at the point $0$. There is a shift of indexes, so the non-vanishing of $H^1(\mathbb A^2\setminus 0, \mathcal O)$ is equivalent to the non-vanishing of $H^2_0(\mathbb A^2, \mathcal O_0)$ which follows from the theorem of Grothendieck Karl mentioned.

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