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Let $\mathcal{E}$ and $\mathcal{F}$ be two coherent sheaves on a polarized projective variety $(X,\mathcal{O}_X(1))$.

Denote by $E=\Gamma_*(\mathcal{E})=\oplus_{k\in\mathbb{Z}}\mathcal{E}(k)$, $F=\Gamma_*(\mathcal{F})$, and $R= \Gamma_\ast(\mathcal{O}_X)$.

Is it true that $$\oplus_{k\in\mathbb{Z}}Ext^i_{\mathcal{O}_X}(\mathcal{E}, \mathcal{F}(k))=Ext^i_R(E, F)?$$

If the equality does not hold in general, when and to what extent it will hold?

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I assume you mean to ask whether: $$\oplus_{k\in\mathbb{Z}}Ext^i_{\mathcal{O}_X}(\mathcal{E}, \mathcal{F}(k))=Ext^i_R(E, F)?$$ For this question, the answer is no, set $\mathcal{E} = \mathcal{O}_X$. Choose $\mathcal{F}$ such that $H^i(X, \mathcal{F}) \neq 0$ for some $i > 0$. Then obviously $Ext^i_R(R, F) = 0$. However $Ext^i_{\mathcal{O}_X}(\mathcal{E}, \mathcal{F}(0)) = H^i(X, \mathcal{F}) \neq 0$.

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  • $\begingroup$ What if we assume that $H^i(X, \mathcal{F})=0$ for all $i>0$? $\endgroup$
    – Fei YE
    Commented May 30, 2012 at 6:00
  • $\begingroup$ This example will work as long as some $H^i(X, \mathcal{F}(k)) \neq 0$. $\endgroup$
    – Karl Schwede
    Commented May 30, 2012 at 10:34
  • $\begingroup$ Sorry. What I really mean is that $H^i(X, \mathcal{F}{k})=0$ for all $k$. Thank you for your answer. $\endgroup$
    – Fei YE
    Commented May 30, 2012 at 10:50
  • $\begingroup$ If $i = \dim \mathcal{F}$, then $H^i(X, \mathcal{F}(k))$ generally will not be zero for any $k \ll 0$. $$\text{ }$$ Furthermore, if $\dim \mathcal{F} = 0$ (ie, a skyscraper sheaf), then the module $F$ is never finitely generated (unless it is zero). The reason is we have non-zero infinitely negative degrees. $\endgroup$
    – Karl Schwede
    Commented May 30, 2012 at 22:26
  • $\begingroup$ What you can do is the following. Choose a locally free resolution of $\mathcal{E}$ made up of $\oplus \mathcal{O}_X(m)$. Now try to figure out if the $\Gamma_*$ of this resolution is still exact (at least in the low degrees you care about). If it happens to still be exact, then you are are more or less fine... $\endgroup$
    – Karl Schwede
    Commented May 30, 2012 at 22:28

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