0
$\begingroup$

Hi guys, I'm studying Cech cohomology of sheaves and I've the following doubt: If you have a coherent sheaf $\mathcal{F}$ in a compact complex variety $X$ then the cohomology groups are finite dimensional vector spaces (that's ok).

But it's true that we can reduce ("contract" or something) the analysis of the whole space to the support of the sheaf (to study there the cohomology)?

For example, suppose that $\mbox{Supp}(\mathcal{F})=\{p\}$ is a single point. What we can say about the cohomology of $X$ and the cohomology of $\{p\}$? It's true that the stalk $\mathcal{F}_p$ is a finite dimensional vector space?

Thank you very much!

$\endgroup$
4
  • $\begingroup$ It is. $H^0(X,\mathcal F)=\mathcal F_p$ is finite dimensional, $H^i(X,\mathcal F)=0$ if $i>0$. $\endgroup$ Feb 24, 2013 at 16:35
  • $\begingroup$ And how I can proof that $H^0(X,\mathcal{F})=\mathcal{F}_p$? Thanks for answer! $\endgroup$
    – Peter
    Feb 24, 2013 at 16:39
  • $\begingroup$ I don't think this is so far from the definition. $\endgroup$
    – S. Carnahan
    Feb 24, 2013 at 17:46
  • $\begingroup$ For the proof that one can restrict to the support of the sheaf from the point of view of Cech cohomology : one can "sheafify" the construction of the Cech complex and it is clear that the sheaves in this complex are supported on the support of F. $\endgroup$
    – user25309
    Feb 24, 2013 at 18:22

1 Answer 1

2
$\begingroup$

Let $P\in X$ be the only point such that $\mathcal{F}_P\ne 0$ and consider the immersion $i:\{P\} \rightarrow X$ and consider the sheaf $\mathcal{S}: \{P\} \mapsto \mathcal{F}_P$; consider now the morphism of sheaves $\phi: \mathcal{F} \rightarrow i_* \mathcal{S}$ such that $\phi(X)$ is the canonical map to the stalk and for any $U\subsetneq X$ open $\phi(U)=0$. Then $\phi$ induces isomorphisms on the stalks, hence $\phi$ is an isomorphism of sheaves and hence we are done.

$\endgroup$
1
  • $\begingroup$ \phi(X) is the canonical map to the stalk, sorry. $\endgroup$ Jul 7, 2019 at 21:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.