## How does one view the De Rham spectral sequence as a Grothendieck spectral sequence?

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I was rereading basic results on de Rham cohomology, and this led me inevitably to the fact that $H^q(X,\Omega^p)$ converges to $H^*(X)$ for any smooth proper variety (over any field). How does one view this spectral sequence "maturely" as a Grothendieck spectral sequence?

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A naive comment: find the resolution, and this will give you the derived functor. – David Roberts Jun 27 2011 at 3:00
As far as I remember. the de Rham compleX is the hyper-resolution of the constant sheaf in the infinitesimal topology; see webcache.googleusercontent.com/… – Mikhail Bondarko Jun 27 2011 at 5:13

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In addition to what Mariano and Torsten have said, the Hodge-de Rham spectral sequence starts at $E_1$ and the Grothendieck spectral sequence starts at $E_2$. Hence it's unlikely that the former would be a special case of the latter. – Dan Petersen Jun 27 2011 at 6:18
Dan's point is excellent, I do not know of any nice interpretation of the $E_2$-term of the Hodge-de Rham spectral sequence which would come out of a composed functor interpretation (hence acting as an argument against such an interpretation). – Torsten Ekedahl Jun 27 2011 at 6:44

## 2 Answers

If by "Grothendieck spectral sequence" you mean the spectral sequence associated to the composite of functors (fulfilling the Grothendieck condition) then I am skeptical as to whether this is possible. Also I do not see that there would be any particular point in being able to view it in that light (unless the functors involved would be interesting in themselves). As you seem to be looking for some general principle that would give the dRss I would like to a mention one such which seem to give most spectral sequences in use very quickly. Usually one constructs the dRss by taking a Cartan-Eilenberg resolution of the de Rham complex and then consider the spectral sequence associated to a double complex. However, one can instead use the Massey exact couple construction; once one has an exact couple one automatically gets a spectral sequence. One systematic way of constructing such exact couples is to start with a triangulated category $T$, a sequence of morphisms $\cdots\to X_{i-1}\to X_i\to X_{i+1}\to\cdots$ and a homological functor $H$ on $T$. This gives a spectral sequence starting with $H^i(Y_j)$ and going towards $\varinjlim_iH^\ast(Y_i)$. (Convergence is not assured but is OK for instance of $Y_i$ is $0$ for small $i$ and equal if $i$ is large.) Starting with the naive truncations of de Rham complex gives the dRss (and starting with canonical truncations in the algebraic case gives the conjugate spectral sequence).

This setup works very generally, for instance it is how Adams first constructed the Adams spectral sequence.

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You probably want one (the second one...) of the two hypercohomology spectral sequences which compute $\mathbb H^\bullet(X,\Omega^\bullet)$, the hypercohomology of the de Rham complex. A reference for this is Weibel's book.

I doubt you can view this as a Grothendieck spectral sequence, but it has sufficiently much hyper in it to be considered mature, I guess.

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In fact "hyper" is so mature that it seems to be no longer used... – Torsten Ekedahl Jun 27 2011 at 6:46

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Kids these days... What can you expect?! :) – Mariano Suárez-Alvarez Jun 27 2011 at 7:05
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