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Let $X$ be a smooth projective surface over a field $k$. Is there a way to compute $H^1(X - \{x\},\mathcal{O}_{X-\{x\}})$ in terms of similar invariants for $X$? Actually I'd like to remove even more points, finitely many points.

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The general reference for what follows is SGA 2, §1 to 4. Let $S\subset X$ be a finite subset, and $U:=X\smallsetminus S$. There is an exact sequence $$H^1_S(X,\mathcal{O}_X)\rightarrow H^1(X,\mathcal{O}_X)\rightarrow H^1(U,\mathcal{O}_U)\rightarrow H^2_S(X,\mathcal{O}_X)\rightarrow H^2(X,\mathcal{O}_X)$$ Now $H^i_S(X,\mathcal{O}_X)=\bigoplus_{x\in S}H^i_{\{x\} }(X,\mathcal{O}_X)$. This is zero for $i<2=\mathrm{depth}(\mathcal{O}_x)$, and $H^2_{\{x\} }(X,\mathcal{O}_X)$ is a dualizing module over $\mathcal{O}_x$, hence infinite-dimensional over $k$. Thus $H^1(U,\mathcal{O}_U)$ is infinite-dimensional over $k$, and can be described as an extension of the kernel of the homomorphism $\bigoplus_{x\in S}H^2_{\{x\} }(X,\mathcal{O}_X)\rightarrow H^2(X,\mathcal{O}_X)\ $ by $\ H^1(X,\mathcal{O}_X)$.

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  • $\begingroup$ You mean $H^1(U,O_U)/H^1(X,O_X)$ is the kernel of this homomorphism. $\endgroup$
    – Sasha
    Jun 24, 2016 at 12:43
  • $\begingroup$ Dear abx, what you write still looks incorrect: I think that $H^1(U,\mathcal{O}_U)$ is an extension of the vector spaces you mention and not a quotient. $\endgroup$ Jun 25, 2016 at 11:41
  • $\begingroup$ Oops, right again! Many thanks, I edit. $\endgroup$
    – abx
    Jun 25, 2016 at 14:47

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