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A theorem of Auslander and Goldman states that for a regular integral scheme $X$ the inclusion of the generic point $\mathrm{Spec}\ K \to X$ induces an injective map $Br(X) \to Br(\mathrm{Spec}\ K)$.

Let $U \subset X$ be a dense open subset of a complex analytic space. In which situations (i.e. under which conditions on $U$ and $X$) is it known that the map $Br(X) \to Br(U)$ is injective?

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Certainly that map is not usually injective. By the Grauert-Oka principle, the Brauer group of a Stein manifold equals the topological Brauer group of that manifold. So, for instance, injectivity fails for the inclusion of $\mathbb{C}^2$ into $\mathbb{CP}^2$. – Jason Starr Apr 16 '12 at 21:08
Dear Jason, what is the statement of Grauert-Oka principle that you refer to? I can see why a cohomological analytic Brauer group would be equal to cohom. topological Brauer group, but how do I prove the equality for "true" Brauer groups? – Dima Sustretov Apr 16 '12 at 22:27
@Dima: Here is a reference. MR0098198 (20 #4660) Grauert, Hans Holomorphe Funktionen mit Werten in komplexen Lieschen Gruppen. (German) Math. Ann. 133 1957 450–472. 32.00 Every element of the "true" Brauer group has associated (in more than one way) a principal $\mathbf{PGL}_n$-bundle which has a section if and only if the element is zero. Grauert's theorem proves that this bundle has an analytic section if and only if it has a continuous section. – Jason Starr Apr 17 '12 at 12:03
up vote 2 down vote accepted

This got a bit long for a comment, so here's an answer. The answer is it is almost never injective. In general, for compact $X$, $Br(X)$ is infinite (at least if $H^2(X,O_X)\neq 0$), while $Br(U)$ is finite if $U$ is Stein. I'm not sure how Jason's comment above gives a counterexample, as both $\mathbb{C}^2$ and $\mathbb{CP}^2$ have trivial analytic Brauer group. But, you could do something similar. Let $X$ be an K3 surface, and let $U\subset X$ be a Stein submanifold. Then, $U$ has the homotopy type of a $2$-dimensional CW-complex, so by the Oka principle, the analytic Brauer group of $U$ is zero. But, the analytic Brauer group of $X$ is pretty big (it's a copy of several $\mathbb{Q}/\mathbb{Z}$. For more details about this sort of thing, see Section 2 of Schroer's 2005 paper on topological methods for analytic Brauer groups.

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@Benjamin, thank you for you answer. To clarify: why does U have a homotopy of a 2-dimensional complex? – Dima Sustretov Jun 29 '12 at 0:02
This is closely related to the Lefschetz hyperplane theorem, as proven by Andreotti and Frankel: Specifically, Theorem 1 of their paper says that an $n$-dimensional stein manifold $X$ has no integral cohomology in degree higher than $n$ and that $\Hoh_n(X,\mathbb{Z})$ is torsion-free. In particular, this holds for smooth affine complex $n$-folds, from which they deduce the Lefschetz hyperplane theorem. It's a 5-page paper worth looking at. – Benjamin Antieau Jun 29 '12 at 1:01
Sorry, that should read $H_n(X,\mathbb{Z})$. – Benjamin Antieau Jun 29 '12 at 1:03

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