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.

1$\begingroup$ 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 $X$? $\endgroup$ – ulrich Sep 27 '12 at 6:33

$\begingroup$ The first. But I will be happy with an answer to any of these. $\endgroup$ – Felipe Voloch Sep 27 '12 at 7:41
Edit. The main idea below is incorrect. I explain the mistake below the original post.
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 wellknown that $H_2$ equals $\mathbb{Z}$ for all $g\geq 3$. So this implies that the analtyic Brauer group is $\mathbb{Q}/\mathbb{Z}$.
Correction. Let $X$ be a smooth, complex orbifold with universal cover $f:\widetilde{X}\to X$ and with fundamental group $\Gamma$. Denote the universal central extension of $\Gamma$ as follows, $$1\to H \to \widetilde{\Gamma} \to \Gamma \to 1.$$ For every homomorphism $\widetilde{\rho}:\widetilde{\Gamma} \to \textbf{GL}_n(\mathbb{C})$ mapping $H$ into the center of $\textbf{GL}_n(\mathbb{C})$, say $$\rho_H:H\to \mathbb{C}^\times,$$ there is an associated group homomorphism, $$\rho: \Gamma \to \textbf{PGL}_n(\mathbb{C}),$$ and vice versa. There is also an associated $\textbf{PGL}_n(\mathbb{C})$torsor over $X$ defined as the quotient of the diagonal action of $\Gamma$ on $\textbf{PGL}_n(\mathbb{C})\times \widetilde{X}$. The class of this $\textbf{PGL}_n(\mathbb{C})$torsor only depends on $\rho_H$. In fact, there is a modification of $\widetilde{\rho}$ (not changing $\rho$) so that the image of $\rho_H$ is in the torsion subgroup $(\mathbb{C}^\times)_{\text{tor}} \cong \mathbb{Q}/\mathbb{Z}$. The Brauer class of $\rho$ comes from the corresponding element in $\text{Hom}(H,\mathbb{Q}/\mathbb{Z})$.
Unfortunately, the kernel of $\rho_H$ contains the kernel of the map of $\widetilde{\Gamma}$ to its profinite completion. Indeed, there is a finitely generated $\mathbb{Z}$subalgebra $R$ of $\mathbb{C}$ that contains all the matrix coefficients of $\text{Image}(\widetilde{\rho}).$ The ring $R$ is a Jacobson ring. Thus, for every element of $H$ that is mapped to a nonidentity element, there is a maximal ideal $\mathfrak{m}$ of $R$ such that the image after reduction to $R/\mathfrak{m}$ is also a nonidentity element. Since $\textbf{GL}_n(R/\mathfrak{m})$ is a finite group, it follows that the element of $H$ is not in the kernel of the profinite completion. Although the group $\Gamma = \textbf{Sp}_{2g}(\mathbb{Z})$ is residually finite, the group $\widetilde{\Gamma}$ is not, and the kernel of the map to the profinite quotient is precisely $2H$ as a subgroup of $H$ (writte in additive notation). Thus, although $\text{Hom}(H,\mathbb{Q}/\mathbb{Z})$ is $\mathbb{Q}/\mathbb{Z}$, the part of the group that arises from Brauer elements is $(1/2)\mathbb{Z}/\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}$.

$\begingroup$ 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. $\endgroup$ – Felipe Voloch Sep 29 '12 at 22:39

2$\begingroup$ This is old, but just I came back to it. There is a mistake in my answer, as I explained above. The only part of $\mathbb{Q}/\mathbb{Z}$ that comes from Brauer classes is $(1/2)\mathbb{Z}/\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}$. $\endgroup$ – Jason Starr Aug 4 '16 at 23:44