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 $D$.


1 Answer 1


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.

  • $\begingroup$ Thank you for your answer. This is actually very surprising (to me). I have only one question: is it possible at all for $\Gamma$ to act with only abelian stabilizers (if $K$ is real quadratic)? $\endgroup$
    – John
    Oct 19, 2016 at 7:29
  • 1
    $\begingroup$ @John. It does not seem possible for a finite index subgroup $\Gamma $ of $Sp_{2g}(O)$ to have only Abelian stabilizers, since such a $\Gamma $ will always contain a finite index subgroup of $SU(h)(O_K)$ for the $h$ as above. $\endgroup$ Oct 19, 2016 at 8:19

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