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.



What is the Brauer group of the moduli space of (p.p.) abelian varieties?
**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](http://arxiv.org/pdf/1103.1855.pdf)
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 $\mathbb{Q}/\mathbb{Z}$. 


**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 non-identity element, there is a maximal ideal $\mathfrak{m}$ of $R$ such that the image after reduction to $R/\mathfrak{m}$ is also a non-identity 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 $(1/2)\mathbb{Z}/\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}$.


