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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 $0$.

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    $\begingroup$ What is your definition of a fundamental domain? For most definitions, this stabilizer will be trivial. $\endgroup$ – Moishe Kohan Feb 20 at 12:00
  • $\begingroup$ @MoisheKohan I'll add the definition to the post. The definition I'm using is that 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. I'm confused by why the stabilizer would be trivial, when it isn't trivial for $\operatorname{SL}_2$ --- the condition that no two distinct points are $G(\mathbb{Z})$-equivalent doesn't rule out the existence of stabilizers. $\endgroup$ – Ashvin Swaminathan Feb 20 at 17:36
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    $\begingroup$ Note that the fundamental domain is not unique, not even up to $G({\bf Z})$ action; e.g. for your example of $G={\rm SL}_2$, instead of the iconic fundamental domain $|z| \geq 1, \left|{\rm Re}(z)\right| \leq \frac12$ you could use $|z| \geq 1, \frac12 \leq {\rm Re}(z) \leq 1$ (and uncountably many others, see also: Escher). Still the points with nontrivial stabilizer must be on the boundary, at least for this $G$. $\endgroup$ – Noam D. Elkies Feb 20 at 17:45
  • $\begingroup$ @NoamD.Elkies Does one say that a point in $G(\mathbb{R})/K$ has trivial stabilizer if the stabilizer is as small as possible (i.e., equal to the set of elements in $G(\mathbb{Z})$ that centralize $G(\mathbb{R})$? For other groups $G$, I guess it would be good to know whether the set of points with big stabilizer is always a measure-$0$ set. $\endgroup$ – Ashvin Swaminathan Feb 20 at 17:59
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    $\begingroup$ Sorry, I was thinking trivial in PSL_2, not SL_2. $\endgroup$ – Noam D. Elkies Feb 20 at 18:11
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Yes, the stabilizer is the intersection of the center of $G$ with $G(\mathbb Z)$.

The center in this setting is compact, hence lies in every maximal compact subgroup and stabilizes every point. So it suffices to show that only elements of the center stabilize points in the interior of $\Omega$. Let $g$ be a non-central element that stabilizes x. Choose an open neighborhood $U$ of $x$ contained in $\Omega$, whose translate under $g$ is also in $\Omega$, then for any $y$ in $U$, because $y$ and $gy$ are in $\Omega$, so by definition we have $y = gy$. So $y$ stabilizes all the points in a neighborhood of $X$, hence because $gy$ is a real-analytic function of $y$, stabilizes all points on $G(\mathbb R)/K$, thus lies in the center.

The "measure zero" question you answered in the comments also has a positive answer - the set of points with stabilizer larger than the center of $G(\mathbb Z)$ has measure zero. This set is the union over non-central elements $g$ of the fixed locus of $g$. Because there are countably many $g$, it suffices to show that the fixed locus of each $g$ has measure zero. The fixed locus of $g$ is a proper real-analytic subset and thus has measure zero, as desired.

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  • $\begingroup$ Thanks for this response! To fill in a few more details, your argument shows that the stabilizer of $x$ in $G(\mathbb{Z})$ is a discrete subgroup of the normal subgroup $\bigcap_{g \in G(\mathbb{R})} gKg^{-1}$. Since $G$ is semisimple, it is connected, and so the stabilizer of $x$ in $G(\mathbb{Z})$ must indeed be central. $\endgroup$ – Ashvin Swaminathan Feb 20 at 18:36

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