6
$\begingroup$

If $\mathcal F^\bullet$ is a chain complex of sheaves, to compute the hypercohomology, you take the cohomology of an injective resolution of $\mathcal F^\bullet$, i.e., a chain complex of injectives $\mathcal I^\bullet$ which is quasi-isomorphic to $\mathcal F^\bullet$.

It seems to be a well-known fact that for the algebraic de Rham complex $\Omega^\bullet_{X/S}$ you can compute this hypercohomolgy in practice by taking the total cohomology of the Cech/de Rham double complex $(\check{C}^i(X,\Omega^j), d, \check{d})$.

Does anyone know of a reference for this (somewhere in EGA?), or is there an easy proof ?

$\endgroup$
5
$\begingroup$

This has nothing to do with the details of the problem; it is a generalization of Leray's theorem to hypercohomology. If $\mathcal{F}_{\bullet}$ is a complex of sheaves, and $U_i$ is an open cover of $X$ such that $H^q(U_{i_1} \cap \cdots \cap U_{i_r}, \mathcal{F}_j)=0$ for $q \geq 1$ and all $(i_1, \ldots, i_r)$ and $j$, then the hypercohomology of $\mathcal{F}_{\bullet}$ is computed by the cohomology of the Cech-$\mathcal{F}$ double complex.

I don't have my books on hand to find you a reference, but it isn't hard.

$\endgroup$
  • 2
    $\begingroup$ (Presumably, the double complex in the question is constructed starting from such a good covering and not an arbitrary one) $\endgroup$ – Mariano Suárez-Álvarez Aug 15 '12 at 4:49
  • 2
    $\begingroup$ EGA $0_{\rm{III}}$ 12.4.7. $\endgroup$ – user22479 Aug 15 '12 at 5:56
  • 3
    $\begingroup$ For the algebraic De Rham complex, you take the cover to consist of affines, in which case all the intersections will be as well (for a separated scheme). After this, you just set up an obvious sheafy version of the double complex, which then gives you an acyclic resolution. You can prove that an acyclic resolution computes the same cohomology as an injective resolution the same way you do it for a single sheaf as you might find, for example, in Lang's algebra book. $\endgroup$ – Minhyong Kim Aug 15 '12 at 6:01

Your Answer

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

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