Algebraic versus Analytic Brauer Group - MathOverflow

Algebraic versus Analytic Brauer Group

2010-01-19T23:30:11Z

Let $X$ be a smooth projective algebraic variety over $\mathbb{C}$. Then I think that someone (Serre?) showed that the Cohomological Etale Brauer Group agrees with the torsion part of the Analytic Brauer Group $H^{2}(X,\mathcal{O}^{\times})$. This latter group is calculated in the classical (metric) topology on the associated complex manifold with the sheaf of nowhere vanishing holomorphic functions.

However there can easily be non-torsion elements in $H^{2}(X,\mathcal{O}^{\times})$: for instance consider the image in $H^{3}(X,\mathbb{Z}) \cap (H^{(2,1)}(X) \oplus H^{(1,2)}(X))$.

Could there be a topology more refined than etale but defined algebraically which can see these non-torsion classes? Notice that one can also ask the question for any $H^{i}(X,\mathcal{O}^{\times})$. For $i=0,1$ the Zariski and etale work fine.

Why do things break down for $i>1$?

Answer by James Borger for Algebraic versus Analytic Brauer Group

2010-01-20T23:32:35Z

I'd be surprised if such a topology were in the literature. (I'm no expert on the Brauer group, but once I thought a little about it.) So, it's unlikely you'll get a yes answer to your question. To give a no answer you'd of course have to turn it into a precise, mathematical yes/no question. It would probably be interesting if you could.

Answer by Damian Rössler for Algebraic versus Analytic Brauer Group

2011-09-19T08:47:27Z

I think the article by B. Toen "Derived Azumaya algebras and generators for twisted derived categories", arXiv:1002.2599, gives a pointer to a possible answer to your question.

EDIT I have weakened the assertion...