4
$\begingroup$

I have a really quick question. I am interested in $G=SL_2(\mathbb{Z}[1/p_1,...,1/p_n])$, where $p_1$,..., $p_n$ are prime numbers. Since $G$ is a subgroup of $SL_2(\mathbb{R})$, it acts in the hyperbolic plane $\mathbb{H}$. My question is: are all the isotropy groups of this action finite?

Thank you

$\endgroup$
  • $\begingroup$ The answer is no, in general. Why do you want to know? $\endgroup$ – Venkataramana Feb 23 '16 at 0:03
  • $\begingroup$ Actually I would like to verify that those groups satisfy the following two properties: (M) every finite subgroup is contained in a maximal finite subgroup and (NM) If M is a finite maximal subgroup then N(M) its normalizer it is equal to M. $\endgroup$ – user88026 Feb 23 '16 at 1:02
6
$\begingroup$

This answer addresses the question posed in the comments.

It follows from Bass-Serre theory that every finite subgroup of $G$ is conjugate into $SL_2(\mathbb{Z})$ (see Section II.1.4 of Serre's book Trees). So maximal finite subgroups will be order $4$ or $6$, generated by an element of trace $0$ or $1$, and thus property (M) holds. More specifically, to every $p_i$ is associated an action on a tree. A finite subgroup of $SL_2(\mathbb{Q})$ has to fix a vertex of this tree, and by Bass-Serre theory is conjugate into $SL_2(\mathbb{Z}_{(p_i)})$. Applying this to each $i$, one sees that a finite group is conjugate into $SL_2(\mathbb{Z})$, for which the finite subgroups are well-known to be cyclic.

The second property (NM) is false in general. Consider the matrix $\left[\begin{array}{cc}a & b \\-b & a\end{array}\right]$, $a^2+b^2=1$, then it normalizes the maximal finite subgroup generated by $\left[\begin{array}{cc}0 & 1 \\-1 & 0\end{array}\right]$. Such matrices are plentiful coming from Pythagorean triples.

On the other hand, for certain denominators $p_i$, I think property (NM) will hold. If $-1$ and $-3$ are not quadratic residues $(\mod p_i)$ for all $i$, then I think that the stabilizers of maximal finite subgroups will be trivial. Equivalently, $p_i\equiv -1 (\mod 12)$ for all $i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.