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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

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
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… 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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