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Let $O$ be the ring of $S$-integers in a real quadratic number field. Let $G$ be an $S$-arithmetic subgroup of $SL_2(O)$ whose intersection with $SL_2(\mathbb Z)$ is not of finite index in $SL_2(\mathbb Z)$.

Assume the action of $G$ is properly discontinuous for simplicity.

Can the fundamental domain of the action of $G$ on the complex upper half plane $\mathbb H$ have finite volume?

What if we assume $G$ to be of finite index in $SL_2(O)$?

What if we remove the hypothesis that $G\cap SL_2(\mathbb Z)$ be of infinite index in $SL_2(\mathbb Z)$?

The reason I ask is simply because I wish to understand whether the "quotient" of $\mathbb H$ by $G$ could behave "well" in certain situations.

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    $\begingroup$ Out of interest, can you give an example of an $S$-arithmetic subgroup whose intersection with $SL_2(Z)$ does not have finite index in $SL_2(Z)$? $\endgroup$
    – znt
    Commented Dec 8, 2016 at 18:44
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    $\begingroup$ The first question is possibly to understand when the action of $G$ on the hyperbolic plane is proper? if you're interested in the non-proper case (e.g. $G=\mathrm{SL}_2(\mathbf{Z}[\sqrt{2}])$), possibly it would be useful to specify what is meant by "fundamental domain". $\endgroup$
    – YCor
    Commented Dec 8, 2016 at 19:16
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    $\begingroup$ This question seems confused. If $G$ is $S$-arithmetic for a nonempty set $S$ then by definition it will be commensurable with $SL_2(O_S)$ which is far larger than $SL_2(O)$, so it cannot be a subgroup of $SL_2(O)$. If we take $S = \varnothing$ then $G$ is automatically commensurable with SL_2(O) and it follows immediately that $G \cap SL_2(Z)$ is commensurable with $SL_2(O) \cap SL_2(Z) = SL_2(Z)$. $\endgroup$
    – David Loeffler
    Commented Dec 9, 2016 at 8:04
  • $\begingroup$ @znt That's a good question. I don't have a specific example unfortunately. $\endgroup$
    – Ciro
    Commented Dec 9, 2016 at 10:03
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    $\begingroup$ @YCor Yes, that's a good point. Can we say something if we add the condition that $G$ acts properly discontinuously on $\mathbb H$? What if we also assume $G$ torsion-free? $\endgroup$
    – Ciro
    Commented Dec 9, 2016 at 10:04

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No arithmetic subgroup of $SL_2(O_S)$ will ever act properly on the upper half-plane. These groups are simply too big, and the orbits will not be discrete subgroups.

To see this explicitly, consider the orbit of $z \in \mathbb{H}$ under the subgroup of elements of $G$ of the form $\begin{pmatrix} 1 & x \\ 0 & 1\end{pmatrix}$, which is a finite-index subgroup $L$ of the additive group of $O_S$. Then the orbit of $z$ is just a coset of the image of $L$ in $\mathbb{R}$, which is a dense subgroup of $\mathbb{R}$.

If you want proper actions you have to make $G$ act on $\mathbb{H} \times \mathbb{H}$ (if $S = \varnothing$), as @MatthiasWendt points out; or the product of this with some non-archimedean spaces too if your set $S$ is nonempty

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