1
$\begingroup$

Let $X$ be a smooth algebraic variety of dimension $n$ over a field $k$. Let $\Omega_X$ be the sheaf of differentials (over $k$). Then we may consider the deRham complex $$\Omega_X^\bullet= \mathcal{O}_X\to \Omega_X\to \wedge^2\Omega_X\to \cdots $$

The deRham cohomology is then the hypercohomology of $\Omega_X^\bullet$.

What I am confused about is how exactly one is allowed to take this hypercohomology. Sheaf cohomology on schemes is computed for $\mathcal{O}_X$ modules, but the arrows in the complex $\Omega_X^\bullet$ are not $\mathcal{O}_X$ linear (only $k$ linear). How do you take the hypercohomology then?

Thanks for any help or direction.

$\endgroup$
  • 1
    $\begingroup$ Hypercohomology of complexes of sheaves of abelian groups. $\endgroup$ – Johannes Huisman Jun 29 '16 at 13:40
  • $\begingroup$ Do you mean compute as complexes of sheaves of abelian groups in the Zariski topology? Does this give the correct answer. Cohomology of quasi-coherent sheaves is computed as cohomology of $\mathcal{O}_X$ modules. $\endgroup$ – user45150 Jun 29 '16 at 15:40
  • 2
    $\begingroup$ Yes, in the Zariski topology. You can compute cohomology of $\mathcal O$-modules either way, it gives isomorphic cohomology groups. $\endgroup$ – Johannes Huisman Jun 29 '16 at 16:41
  • $\begingroup$ A reference for the last comment is Hartshorne Proposition III.2.6. $\endgroup$ – Nefertiti Jul 8 '16 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.