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Let $X$ be a compact complex manifold. By definition, $Pic(X)={\rm H^1}(X,\mathcal{O}^\times)$. We know a lot about this group. What is known about the groups ${\rm H^n}(X,\mathcal{O}^\times)$ for $n\ge 2$?

A bit more specialized question. It is well known that for a nonsingular projective complex variety $X$ the natural map $${\rm H^1}(X,\mathcal{O}^\times)\to{\rm H^1}(X,\mathcal{M}^\times)$$ is trivial. What is known about the kernel of the same map for $n=2$ or $n=3$? (Here $\mathcal{M}^\times$ is the sheaf of nonzero meromorphic functions, and the topology is the strong one).

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$H^2(X,\mathcal{O}^{\times})$ is often called the (cohomological) Brauer group. There is a vast literature on it. –  Daniel Loughran May 11 '11 at 8:55
    
Thank you. But I would like to have some examples at hand. And, what about $n=3$? –  Alex Gavrilov May 11 '11 at 9:06
    
Indeen, after little googling I found some papers about this. Though, in the most of them the cohomology is etale. –  Alex Gavrilov May 11 '11 at 9:38
    
And, apparently, all they care about is the torsion. –  Alex Gavrilov May 11 '11 at 10:22
    
@Alex I am no expert, but I believe if you are looking at compact complex manifolds as you originally stated then $H^i(X,\mathcal{O}^{\times}) = H^i(X,\mathbb{G}_m)$, since the complex topology is (morally) as good as the étale topology. However with arbitrary varieties it is better to work with $H_{ét}^i(X,\mathbb{G}_m)$ than $H^i(X,\mathcal{O}^{\times})$. –  Daniel Loughran May 11 '11 at 10:57

2 Answers 2

First of all, it probably depends on how you define $H^1(X, \mathcal{O}^\times)$. I don't see any reason why derived functor cohomology should agree here with Cech cohomology.

I think that $H^i(X, \mathcal{O}^\times)$ is a functor of order $i+1$ in the sense of Mumford "Abelian Varieties" (2.6, Remark preceding the proof of the theorem of the cube), at least for complex projective varieties. That is, there is a higher analogue of the theorem of the cube for $H^i(X, \mathcal{O}^\times)$. For this, we look at the exponential sequence as in the aforementioned Remark.

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To clarify the question: $H^n$ there is the Cech cohomology group (in the strong topology). Thank you for the idea, but if there exists an analog of the theorem of the cube, I would prefer to read about it. To get it for myself may be not so easy. –  Alex Gavrilov May 11 '11 at 9:32
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By a theorem of Godemont, on a Hausdorff paracompact space, Cech and derived cohomology always coincide. Assuming that "strong topology" is the analytic topology, that means you don't have to worry about this issue. See mathoverflow.net/questions/19312/… for further discussion. –  David Speyer May 25 '11 at 18:42
    
Yes, I meant the analytic topology, of course. Actually, I was pretty sure that these cohomology coincide, though did not know where to look for this. Thanks. But, what I am looking for are not some functorial properties but nontrivial results PUBLISHED somewhere. To read it! –  Alex Gavrilov May 26 '11 at 8:36

Here is a reference: Grothendieck's three exposés in Dix Exposés sur la Cohomologie des Schémas (and the references therein). One can find there e.g. computation of $H^i_{ét}({\rm Spec}\text{ } \mathcal{O}_K, \mathbb{G}_m)$ for spectra of rings of integers in number fields.

MR0244269 (39 #5586a) Grothendieck, Alexander Le groupe de Brauer. I. Algèbres d'Azumaya et interprétations diverses. (French) 1968 Dix Exposés sur la Cohomologie des Schémas pp. 46–66 North-Holland, Amsterdam; Masson, Paris, 14.55

MR0244270 (39 #5586b) Grothendieck, Alexander Le groupe de Brauer. II. Théorie cohomologique. (French) 1968 Dix Exposés sur la Cohomologie des Schémas pp. 67–87 North-Holland, Amsterdam; Masson, Paris, 14.55

MR0244271 (39 #5586c) Grothendieck, Alexander Le groupe de Brauer. III. Exemples et compléments. (French) 1968 Dix Exposés sur la Cohomologie des Schémas pp. 88–188 North-Holland, Amsterdam; Masson, Paris (Reviewer: J. S. Milne), 14.55

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