Exact sequences of $\mathcal{O}_X$-modules and splitting

Question

Let $X$ be a scheme (a curve over a field for example). Exactness of a sequence $\mathcal{F}\to \mathcal{G}\to \mathcal{H}$ of $\mathcal{O}_X$-modules can be checked on stalks : it is exact if and only if $\mathcal{F}_x\to \mathcal{G}_x\to \mathcal{H}_x$ is exact for every $x\in X$.

If $0\to \mathcal{F}\to\mathcal{G}\to\mathcal{H}\to 0$ is split exact, then so are each of the $0\to \mathcal{F}_x\to \mathcal{G}_x\to \mathcal{H}_x\to 0$, but the converse isn't necessarily true I think.

My question is this : are there useful ``local'' or other criteria (or maybe an obstruction theory of some sort) for checking split exactness of sequences of $\mathcal{O}_X$-modules?

Answer 1

I don't think local is way to go here, because being split on the level of the whole sheaf tautologically requires the existence of a global section to the surjection.

However that's exactly what $Ext^1 (\mathcal{H}, \mathcal{F})$ is for (your obstruction theory if you wish). Split exact sequence corresponds to $0$ element in the Ext group. You can use full force of homological algebra to check it. In particular, since you were interested in the possibility of the local answer, often you can utilize the local-to-global spectral sequence: $$H^q(X, \mathcal{Ext}^p (\mathcal{H}, \mathcal{F})) \implies Ext^{p+q}(\mathcal{H},\mathcal{F}).$$

Answer 2

Just to parrot off the @Bananeen, I think the answer is "Yes, if $X$ is affine". Here's a proof along the lines of the last answer.

$H^i(Hom_{O_X}(H,F))=0$, $i>0$ because the higher cohomology of any coherent sheaf is 0 (see here(Hartshorne III.3.5)) for an easy proof. Use this in the local to global spectral sequence as follows.

The relevant part of the exact sequence of low order terms is

$H^1( \mathcal{Ext^0(H,F)}) \to Ext^1(\mathcal{H,F}) \to H^0(\mathcal{Ext^1(H,F)}) \to H^2(\mathcal{Ext^0(H,F)})$

By the above, the first and last terms are zero so

$Ext^1(\mathcal{H,F}) \cong H^0(\mathcal{Ext^1(H,F)})$

To understand what this means, lets recall what the global sections of the sheafification of a presheaf are. They are essentially local sections of a presheaf that are compatible. Usually all you have is a map from global sections of the presheaf to global sections of the sheafification and one cannot know if its injective or surjective.

Now, consider the presheaf $U \to Ext^1(H(U),F(U))$. Its sheafification is $\mathcal{Ext^1(H,F)}$. The global sections of the presheaf make up the LHS of the above iso. The global sections of the sheafification make up the RHS of the iso. Hence sheafification commutes with global sections in our case!

Our extension maps to zero in $Ext^1_{\mathcal{O(U)}}(\mathcal{H}(U),\mathcal{F}(U))$ for all sufficiently small $U$ by hypothesis. Hence it maps to the 0 global section of $H^0(\mathcal{Ext^1(H,F)})$. And therefore it maps to $0$ in $Ext^1(H,F)$. So the extension splits.

(Also if anyone knows a simpler way of showing this fact, I'd appreciate a comment :) )