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What is the Brauer group of the moduli space of principally polarized abelian varieties of a given dimension? I am primarily interested in the "open" moduli space, i.e. not a compactification. The question has several levels of generality depending on how general is the ground field (ring even?) but I know nothing, so already the knowing the analytic Brauer group over the complex numbers would be very interesting for me.

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For an analytic space $X$, by the analytic Brauer group of $X$ do you mean the group $ H^2(X,\mathcal{O}_X^{\times})$, its torsion subgroup, or the group given by equivalence classes of Azumaya algebras on $X$? –  ulrich Sep 27 '12 at 6:33
The first. But I will be happy with an answer to any of these. –  Felipe Voloch Sep 27 '12 at 7:41

1 Answer 1

up vote 6 down vote accepted

I am posting as an answer instead of a comment, even though I think this might be wrong, because I could not format the weblinks properly in comments.

Since the (orbifold) moduli space is a quotient of the Siegel upper half space by the (orbifold) fundamental group $\textbf{Sp}_{2g}(\mathbb{Z})$, it seems to me that the analytic Brauer group of the moduli space should be $\text{Hom}(H_2,\mathbb{Q}/\mathbb{Z})$, where $H_2 = H_2(\textbf{Sp}_{2g}(\mathbb{Z}))$ is the kernel of the universal central extension of $\textbf{Sp}_{2g}(\mathbb{Z})$. According to the proof of Proposition 2.3 of Finite quotients of symplectic groups vs mapping class groups by Funar and Pitsch, it is well-known that $H_2$ equals $\mathbb{Z}$ for all $g\geq 3$. So this implies that the analtyic Brauer group is $\mathbb{Q}/\mathbb{Z}$.

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I assumed that the equality $Br(X)=Hom(H_2(\pi_1(X)),\mathbb{Q}/\mathbb{Z})$ would be some standard fact I could look up, but I've failed to find it. Do you have a reference? Thanks. –  Felipe Voloch Sep 29 '12 at 22:39

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