Let $D$ be a bounded hermitian symmetric domain with automorphism group $G(\mathbb R)$. In the example I have in mind, $D$ is Siegel upper half-space of degree $g$ and $G(\mathbb R) = \mathrm{Sp}(2g,\mathbb R)$.


Let $O$ be an order in a totally real number field. Example: $O=\mathbb{Z}[\sqrt{d}]$ with $d\in \mathbb Z$ positive.


Let $\Gamma$ be a torsionfree finite index normal subgroup of $G(O)$. Fix an embedding $G(O)\subset G(\mathbb R)$. Note that $\Gamma$ acts on $D$. 


Are all isotropy groups of $\Gamma$ mutually isomorphic?


The answer is positive if $O = \mathbb Z$, as then the isotropy groups of the action are compact and discrete (hence finite), and thus trivial. 


In general, I suspect that the isotropy groups will be free $\mathbb Z$-modules of finite rank. I'm asking whether the rank can be vary with the points of $D$.



Actions of torsionfree discrete subgroups on hermitian symmetric domains
This is not true. The isotropies need not even be Abelian.


Consider the Hermitian form $h(z\_1,z\_2)=\mid z\_1\mid ^2+\sqrt{d} \mid z\_2 \mid ^2$ as a Hermitian form with respect to the quadratic extension $E/K$, where $K=\mathbb{Q}(\sqrt{d})$ and $E=K(\sqrt{-1})$. Then $SU(h)$ is a semi-simple algebraic group over $K$ and $SU(h)(K\otimes \_{\mathbb{Q}}\mathbb{R})=SU(2)\times SU(1,1)$; thus the first factor is compact, and corresponds to the embedding induced by $\sqrt{d}\mapsto \sqrt{d}$ from $K$ into $\mathbb{R}$. The group $SU(h)$ embeds into $Sp\_4(\Omega) $ since it preserves the imaginary part $\Omega$ of the Hermitian form $h$. 


Hence $SU(h)(O\_K)$ embeds in $Sp\_4(O\_K)$. The image of $SU(h)(O\_K)$ in $Sp\_4(\mathbb{R})$ is inside the compact group $SU(2)$ and hence fixes a point $p$ in the Siegel upper half space. Since $SU(h)(O\_K)$ is a lattice in $SU(1,1)$ it follows that $SU(h)(O\_K)$ is (highly) non-abelian.


If we take some other point $q$ in the Siegel upper half space, the isotropy of $SU(2)$ can be strictly smaller and the isotropy can even be abelian. 


