Let $G$ be a semisimple group (the cases of primary interest to me are where $G$ is a special linear group or a special orthogonal group), let $K$ be a maximal compact subgroup of $G(\mathbb{R})$, and let $\Omega$ be a fundamental domain for the left-action of $G(\mathbb{Z})$ on $G(\mathbb{R})/K$.

(**Edit** -- here is the definition of fundamental domain I'm using: A fundamental domain for $G(\mathbb{Z})$ acting on $G(\mathbb{R}/K)$ is a subset $\Omega \subset G(\mathbb{R})/K$ such that $\Omega$ is the closure of an open subset, $\Omega$ intersects every orbit, and no two distinct points in the interior of $\Omega$ are $G(\mathbb{Z})$-equivalent.)

**Question**: For $P \in \Omega$, let $S_P \subset G(\mathbb{Z})$ be the stabilizer of $P$. If $P$ is restricted to lie in the interior of $\Omega$, is $\#S_P$ a constant? If so, is it possible to compute this constant for, say, $G = \operatorname{SL}_n$ or $G = \operatorname{SO}(p,q)$?

**Some Initial Thoughts**:

- The answer is yes when $G = \operatorname{SL}_2$. In this case, it is shown in Serre's
*Cours d'arithmétique*that the stabilizer of any $P$ in the interior of $\Omega$ is given by $\{\pm 1\}$, which is incidentally the center of $\operatorname{SL}_2(\mathbb{Z})$ (perhaps this continues to hold for $\operatorname{SL}_n(\mathbb{Z})$ for $n > 2$?) - The stabilizer in $G(\mathbb{R})$ of $P$ is a conjugate of the subgroup $K$. Thus, $\#S_P$ is the (necessarily finite) number of integral points in this conjugate subgroup, but I'm not sure how that varies with $P$.
- In his paper
*Ensembles fondamenteaux pour les groupes arithmétiques*, Borel constructs a finite union $U$ of Siegel sets such that the set of elements $\gamma \in G(\mathbb{Z})$ for which $U \cap \gamma \cdot U \neq \varnothing$ is finite. I guess this means that $\#S_P$ is bounded independent of $P$, but I'm not sure if the argument in Borel's paper (or in his previous paper together with Harish-Chandra entitled*Arithmetic Subgroups of Algebraic Groups*) allows one to effectively compute $\#S_P$.

**Edit -- Partial Progress**: Please find below some partial progress toward answering the above question; please let me know if the following is incorrect or can be improved!

We say that a point $P \in G(\mathbb{R})/K$ has big stabilizer if $S_P$ contains an element of $G(\mathbb{Z})$ that does not centralize $G(\mathbb{R})$. Let $P \in G(\mathbb{R})/K$; as remarked above, the stabilizer in $G(\mathbb{R})$ of $P$ is given by $PKP^{-1}$, and so the stabilizer of $P$ in $G(\mathbb{Z})$ is the set of integral points in $PKP^{-1}$. Let $\Sigma$ be the finite set of elements in $G(\mathbb{Z})$ that stabilize $P$ but that do not centralize $G(\mathbb{R})$, and let $U'$ be an open neighborhood of $1 \in G(\mathbb{R})$. For each $s \in \Sigma$, observe that the set $V_s := \{g \in G(\mathbb{R}) : gs = sg\}$ is a hyperplane in $G(\mathbb{R})$ of positive codimension; let $U := \bigcap_{s \in \Sigma} (U' - V_s)$.

Since $\Sigma$ is finite and $K$ is compact, by choosing $U'$ to be sufficiently tiny, we can arrange that for any $h \in U$, the point $hP$ does not have big stabilizer. Since $U$ is obtained by deleting finitely many hyperplanes from an open neighborhood of $1 \in G(\mathbb{R})$, it follows that a "generic perturbation" of $P$ results in a point without big stabilizer. Since $P$ is arbitrary among points with big stabilizer, we deduce that the closure of the set of points in $G(\mathbb{R})/K$ that have big stabilizer has empty interior, and so it has measure $0$.