Let $F,G$ be quasicoherent sheaves of $\mathcal{O}_X$modules on a scheme $X$.

What is exactly the difference between the derived functors $\mathcal{Ext}^i(F,G)$ and $Ext^i_X(F,G)$?

By definition $Ext^i_X(F,G)$ arises from the the global section functor $U \to \Gamma(U, F^{\vee} \otimes G)$ while $\mathcal{Ext}^i(F,G)$ from the Hom functor $U \to \mathcal{Hom}_U(F|_U,G|_U)$.

Both are calculated using injective resolutions $I_{\bullet}$ of $F$ or $G$.

Futhermore the functors are essentially the same since

(*) $$\Gamma(U, F^{\vee} \otimes G)= \Gamma(U, \mathcal{Hom}(F, \mathcal{O}_X) \otimes G)= \mathcal{Hom}(\mathcal{O}_X, \mathcal{Hom}(F, \mathcal{O}_X)\otimes G)= \mathcal{Hom}_X(\mathcal{F}, \mathcal{G})$$

So why generally $\mathcal{Ext}^i(F,G)$ and $Ext^i_X(F,G)$ should be distinguished?

Till now - according to line (*) - they arise from the same functor, or am I wrong?

  • 1
    $\begingroup$ Just as an example, take $X=\mathbb{P}^1_k$, $k$ a filed. Then $\mathcal{E}xt^1(\mathcal{O},\mathcal{O}(-2))=0$ while $Ext^1(\mathcal{O},\mathcal{O}(-2))=k$. In general there is only a spectral sequence connecting the two. $\endgroup$ – Mohan Oct 8 '18 at 11:34
  • $\begingroup$ @Mohan: Yes. But where is then the error in my reasonings? Is the line $\(*\) $ in generally wrong? Since if (*) would be true then $\mathcal{Ext}^i(F,G)$ and $Ext^i_X(F,G)$ would be derived from the same (up to natural isomorphism) functor, right? Your conterexample shows indeed that that can't be true. $\endgroup$ – KarlPeter Oct 8 '18 at 11:54
  • $\begingroup$ The middle equality in (*) is wrong. It is not $\mathcal{H}om(\mathcal{O}_X,-)$, but $\mathrm{Hom}(\mathcal{O}_X,-)$. $\endgroup$ – Mohan Oct 8 '18 at 12:29
  • $\begingroup$ @Mohan: Ah yes, of course. $\mathcal{H}om(F,G)$ was by consruction a sheaf while $Hom(F,G)$ is just the set of morphism between the two sheaves. I guess that's the cruical point. $\endgroup$ – KarlPeter Oct 8 '18 at 17:18

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