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Let $F,G$ be coherent sheaves on a smooth Noetherian scheme $X$ of dimension $n$. I can see that $Ext^k(F,G) = 0$ for $k > 2n$. Is this sharp? Or can it perhaps be improved to get vanishing for $k>n$?

The argument I have in mind is to use the spectral sequence $H^i(\mathcal{Ext}^j(F,G)) \Rightarrow Ext^{i+j}(F,G)$. Since $\mathcal{Ext}^j(F,G)$ can be computed locally on affine charts using a projective resolution, we have $\mathcal{Ext}^j(F,G) = 0$ for $j > n$ by smoothness. Moreover $\mathcal{Ext}^j(F,G)$ is coherent, so by the Grothendieck vanishing theorem we have $H^i(\mathcal{Ext}^j(F,G)) = 0$ for $i > n$. So the $E_2$ page of the spectral sequence is already zero outside the box $(i,j) \in [0,n] \times [0,n]$, and in particular for $i+j > 2n$.

But perhaps something more subtle happens?

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  • $\begingroup$ A trivial remark : if $F$ is locally free (i.e a vector bundle) then this $Ext^k(F,G) \cong H^k(X,F^* \otimes G)$ which vanishes for $k > n$. $\endgroup$ – Nicolas Hemelsoet Jan 29 '18 at 17:55
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    $\begingroup$ If $X$ is projective over a field, you can use Serre duality to prove the vanishing for $k > n$. $\endgroup$ – Sasha Jan 29 '18 at 20:41

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