# Algebraic de Rham cohomology vs. analytic de Rham cohomology

Let $X$ be a nice variety over $\mathbb{C}$, where nice probably means smooth and proper.

I want to know: How can we show that the hypercohomology of the algebraic de Rham complex agrees with the hypercohomology of the analytic de Rham complex (equivalently the cohomology of the constant sheaf $\mathbb{C}$ in the analytic topology)? Does this follow immediately from GAGA? If not, how do you prove it?

I think that this does not follow immediately from GAGA because, while the sheaves $\Omega_X^i$ are coherent, the de Rham $d$ is not a map of coherent sheaves (it is not multiplicative). Am I correct in my thinking?

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## 4 Answers

I don't think you can get this directly from GAGA. The reference that I know for this result is Grothendieck, On the de Rham cohomology of algebraic varieties. It is short, beautiful, and in English.

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Thanks! Hooray for English. – Kevin H. Lin Mar 12 '10 at 3:47
As many people are pointing out below, you can get this from GAGA for proper varieties, and certainly Grothendieck's proof uses GAGA among other tools (see footnote 6). – David Speyer Mar 12 '10 at 4:02

If $X$ is smooth and proper, GAGA does in fact suffice (despite the observation that $d$ is not $\mathcal{O}_X$-linear: One obtains a comparison map of hypercohomology spectral sequences; it is an isomorphism on the $E_2$ page by GAGA, and thus on the $E_\infty$ page.

It is to prove the general case (i.e., $X$ smooth but not necessarily proper) that one needs to do additional work.

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Some care is needed in the application of GAGA even in the proper case since the lack of $\mathcal{O}$-linearity for the maps in the deRham complex implies that one needs to be attentive to how a comparison morphism is actually defined and why it is an isomorphism. None of this requires anything deep, but for a rigorous argument one cannot just blindly "plug it into" some general nonsense about analytification of complexes of sheaves of modules. – BCnrd Mar 12 '10 at 6:02
That's a great point --- the existence of the comparison morphism isn't formal! This is made clear by noting that there's no formal construction where one "plugs in" the algebraic deRham complex and gets out the analytic deRham complex; at some point you need to know how to differentiate holomorphic functions not coming from algebra. So yes, to define the comparison map we need to verify that the level-wise "analytification" maps are compatible with the differentials (i.e., that you can differentiate an algebraic form by treating it as a holomorphic form). – Anatoly Preygel Mar 12 '10 at 7:30
Brian -- the key observation is that whenever we have a ringed space $(X,O)$ and a sheaf $F$ of $O$-modules, then when computing the cohomology it doens't matter whether we consider $F$ as a sheaf of $O$-modules or just as a sheaf of abelian groups. So the analytic to algebraic comparison morphism exists for general reasons: just consider the algebraic and analytic de Rham complexes as complexes of sheaves of abelian groups on different sites, analytic and Zariski. It is a quasi-isomorphism for any $\Omega^i$ by GAGA, hence a quasi-isomorphism. – algori Mar 12 '10 at 13:42
Algori: I'm afraid what you write is not precisely correct. Firstly, you have to pull back the algebraic complex from the Zariski topology to the analytic one to get a comparison map for general reasons.' Next, the maps of sheaves is definitely not a quasi-isomorphism. If you'd like a quasi-isomorphism, the easiest thing is perhaps to build up a map between two Cech complexes in the obvious manner, using affine open coverings (it works for the cohomology of the analytic sheaves because these are Stein manifolds). It's rather important to remember that GAGA theorems are all global. – Minhyong Kim Mar 12 '10 at 15:53
Algori, this might help: Let $f: X_{an} \to X_{zar}$ be the natural map of locally ringed sites, $f^{-1}$ the pullback on sheaves of groups, and $f^*$ the pullback on sheaves of modules. Then, GAGA is relating the cohomology of $\mathcal{F}$ on $X_{zar}$ with the cohomology of $f^* \mathcal{F}$ (not $f^{-1} \mathcal{F}$) on $X_{an}$. Then $f^{-1} \Omega_{zar}^*$ has a differential, but we need to provide the analytic deRham differential and a map of complexes $f^{-1} \Omega_{zar}^* \to \Omega_{an}^*$ that is identified with the natural map $f^{-1} \to f^*$` levelwise. – Anatoly Preygel Mar 13 '10 at 0:50

Various people have answered the question, and also brought up some of the subtleties in applying GAGA. So I won't rehash all that. So let me just suggest the additional reference:

Deligne, Équations différentielles à points singuliers réguliers

especially Chapter II, section 6. These issues are dealt with carefully in a more general setting of de Rham cohomology with coefficients in a regular integrable connection. The result is no doubt true for a regular holonomic D-module, and it would be nice if someone wrote this down carefully. But perhaps I'm straying too far from the original topic.

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This does follow from GAGA via the spectral sequences associated to the dumb filtrations on the algebraic and analytic de Rham complexes of sheaves, see p. 96 of tome 29 of PMIHES in a paper of Grothendieck (1966).

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algori, can you fix your link? I'd do it for you, but I can't figure out what you mean to link to. – David Speyer Mar 12 '10 at 4:03
It's fixed. – Harry Gindi Mar 12 '10 at 4:08
David, fpqc -- I am trying to give a link to Grothendieck's paper on algebraic de Rham cohomology, but for some reason there is some rubbish before http, which I'm unable to delete. – algori Mar 12 '10 at 4:09
I fixed it already. – Harry Gindi Mar 12 '10 at 4:09
FWIW (my understanding of) the underlying reason for this logic is that arXiv sometimes generates pdfs on the fly and dumps them in a temporary directory and offers you a link to that. So linking "directly to the pdf" equals linking to a temp file that in a few months' time will be gone. Who knows if Numdam does the same---but who cares, because whether they do or they don't, linking to the journal page will surely always work. – Kevin Buzzard Mar 12 '10 at 7:25