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Let $X$ be a regular projective (complex) surface and $S$ be a finite set of closed points on $X$. Denote by $j:X\backslash S \to X$ the open immersion. Assume further that $H^1(\mathcal{O}_X)=0$. Is $H^0(R^1j_*\mathcal{O}_{X \backslash S})=0$?

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    $\begingroup$ Is there any reason to believe something like this should be true? I don't see why there should be any natural map between the two and it's false for $S$ empty... $\endgroup$
    – dhy
    Commented Feb 28, 2016 at 2:13
  • $\begingroup$ @dhy I have edited the question. $\endgroup$
    – Ron
    Commented Feb 28, 2016 at 2:29
  • $\begingroup$ $j_*\mathcal{O}_{X\smallsetminus S}$ is isomorphic to $\mathcal{O}_X$. $\endgroup$
    – abx
    Commented Feb 28, 2016 at 3:56
  • $\begingroup$ @abx I am a bit confused. I know this fact. I do not understand, whether this proves or disproves the question. Could you please elaborate a bit more? $\endgroup$
    – Ron
    Commented Feb 28, 2016 at 5:37
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    $\begingroup$ $R^1j_*\mathcal{O}_{X\smallsetminus S}$ is a sheaf supported on $S$, with fiber at $s\in S$ the group $H^2_{\{s\} }(\mathcal{O}_X)$, which is nonzero. See SGA2, §I and II. $\endgroup$
    – abx
    Commented Feb 28, 2016 at 7:46

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Since $S$ is finite, $X$ being projective is a red herring, so is anything about the (global) cohomology of $\mathscr O_X$, and (to some extent) it being a surface, or regular.

If $X$ is $S_2$, and $\dim X\geq 2$, then the sheaf version of Exercise III.2.3 in [Hartshorne] (or the exercise itself noting that by $S$ being finite one may start with restricting to affine schemes) implies that there is a long exact sequence $$ 0\to \mathscr H_S^0(\mathscr O_X) \to \mathscr O_X \to j_*\mathscr O_{X\setminus S} \to \mathscr H_S^1(\mathscr O_X) \to \\ \to \mathscr H_{\emptyset}^1(\mathscr O_X) \to R^1j_*\mathscr O_{X\setminus S} \to \mathscr H_S^2(\mathscr O_X) \to \mathscr H_{\emptyset}^2(\mathscr O_X) \to \dots $$ Since $X$ is $S_2$, $\mathscr H_S^0(\mathscr O_X) =\mathscr H_S^1(\mathscr O_X) =0$ and $\mathscr H_{\emptyset}^i(\mathscr O_X) =0$ for all $i>0$, so this falls apart to the isomorphisms $$ j_*\mathscr O_{X\setminus S}\simeq \mathscr O_X $$ and $$ R^ij_*\mathscr O_{X\setminus S} \to \mathscr H_S^{i+1}(\mathscr O_X) $$ for all $i>0$. (In particular $R^ij_*\mathscr O_{X\setminus S}$ is supported on $S$). Since $S$ is finite, $R^ij_*\mathscr O_{X\setminus S}$ is the sheaf associated ot its own global sections, in particular $R^ij_*\mathscr O_{X\setminus S}\neq 0$ if and only if $H^0(X,R^ij_*\mathscr O_{X\setminus S})\neq 0$.

By Grothendieck vanishing $H_x^{d}(\mathscr O_X)\neq 0$ for $d=\dim \mathscr O_{X,x}$ for any $x\in X$, which shows that if $\dim X=d$ near $S$ (e.g., $X$ is equidimensional, or at least that the dimension of the union of irreducible components of $X$ containing $S$ is $d$), then $H^0(X,R^{d-1}j_*\mathscr O_{X\setminus S})\neq 0$.

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  • $\begingroup$ Thank you very much for the very detailed and helpful answer. I also wanted to ask, does this also mean $H^1(\mathcal{O}_{X\backslash S}) \not-0$? $\endgroup$
    – Ron
    Commented Mar 1, 2016 at 17:13
  • $\begingroup$ Yes. It's the same as $H^0$ of $R^1$. You should be able to extract this from the above proof. $\endgroup$ Commented Mar 1, 2016 at 17:43

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