Timeline for Algebraic de Rham cohomology vs. analytic de Rham cohomology
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Mar 14, 2016 at 20:01 | vote | accept | Kevin H. Lin | ||
| Mar 14, 2016 at 20:01 | |||||
| Mar 13, 2010 at 0:50 | comment | added | Anatoly Preygel |
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.
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| Mar 12, 2010 at 18:25 | comment | added | algori | Brian -- I'm unable to give a precise meaning to your remark that " the general nonsense of analytification of sheaves of modules doesn't literally include the d-maps in the deRham complexes," | |
| Mar 12, 2010 at 18:23 | comment | added | algori | Minhoyng, I was talking precisely of the pullback from analytic to Zariski sites. This induces an isomorphism $H^*(X^{Zar},\Omega^i)\to H^*(X^{an},Omega^i)$ (hence my use of the term "quasi-isomorphism", which may have been misleading); of course, $\Omega^i$ on the right and on the left stands for sheaves wrt different topologies. So the (hyper)cohomology map $H^*(X^{Zar},\Omega^*)\to H^*(Z^{an},\Omega^*)$ is an iso by comparing the spectral sequences. Notice that this comparison map is induced by adjunction and the map of complexes of sheaves of abelian groups on $X^{an}$. | |
| Mar 12, 2010 at 16:29 | comment | added | BCnrd | Algori, I hope Minhyong's comments help to clarify to you what the point of my comment was (which seems to have been missed in your comment). Let me try again: the general nonsense of analytification of sheaves of modules doesn't literally include the d-maps in the deRham complexes, due to their non-linearity, so one has to make (admittedly easy, but still necessary!) arguments specific to that situation to set up the relation between the two complexes before putting it into the spectral sequence for underlying complexes of abelian sheaves). | |
| Mar 12, 2010 at 15:53 | comment | added | Minhyong Kim | 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. | |
| Mar 12, 2010 at 13:42 | comment | added | algori | 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. | |
| Mar 12, 2010 at 7:30 | comment | added | Anatoly Preygel | 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). | |
| Mar 12, 2010 at 6:02 | comment | added | BCnrd | 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. | |
| Mar 12, 2010 at 3:48 | history | answered | Anatoly Preygel | CC BY-SA 2.5 |