# Fundamental domains of infinite-index subgroups of $SL(2,\mathbb{Z})$

While discussing modular forms associated to different subgroups $\Gamma$ of $SL(2,\mathbb{Z})$, there appeared to be a heuristic relationship between the index $[SL(2,\mathbb{Z}) \colon \Gamma]$ and the area of the Riemann surface $\Gamma/\mathbb{H}$, where $\mathbb{H}$ denote the hyperbolic plane. Specifically, it appears that when the hyperbolic area of the fundamental domain of $\Gamma$ is infinite, then $\Gamma$ has infinite index in $SL(2,\mathbb{Z})$.

Are any results that describe such a relationship? And if so, can we say anything for more general subgroups of $SL(2,\mathbb{R})$? (That is, for subgroups not necessarily contained in $SL(2,\mathbb{Z})$.) Specifically, I'm wondering about subgroups $G(\lambda) < SL(2,\mathbb{R})$ generated by the isometries $z \mapsto z +\lambda$ and $z \mapsto -1/z$, for $\lambda \in \mathbb{R}$.

Edit: @hunter has answered in the case where $\Gamma$ is a subgroup of $SL(2,\mathbb{Z})$. However, it is still not clear how this may generalize to subgroups of $SL(2,\mathbb{R})$.

In general given discrete subgroups $\Gamma_2 < \Gamma_1 < PSL(2,\mathbb{R}) = \text{Isom}(\mathbb{H}^2)$, there is an equation $$\text{Area}(\mathbb{H}^2 / \Gamma_2) = [\Gamma_1:\Gamma_2] \cdot \text{Area}(\mathbb{H}^2 / \Gamma_1)$$ So if $\text{Area}(\mathbb{H}^2 / \Gamma_1)$ is finite, as is the case when $\Gamma_1 = PSL(2,\mathbb{Z})$, then $\text{Area}(\mathbb{H}^2 / \Gamma_2)$ is finite if and only if $\Gamma_2$ has finite index in $\Gamma_1$. On the other hand if $\text{Area}(\mathbb{H}^2 / \Gamma_1)$ is infinite then $\text{Area}(\mathbb{H}^2 / \Gamma_2)$ is also infinite.

The reason for this equation is that the inclusion $\Gamma_2 < \Gamma_1$ induces an orbifold covering map (a branched covering map) from $\mathbb{H}^2 / \Gamma_2$ to $\mathbb{H}^2 / \Gamma_1$ of degree equal to the index $[\Gamma_1:\Gamma_2]$.

• I'd add this is a general fact about finite-index Riemannian covers and does not rely on any hyperbolic geometry (beyond the mention of $PSL(2,\mathbb{R})$). – Neal Oct 10 '14 at 1:26

$\def\Z{\mathbb{Z}}$ $\def\union{\cup}$

If $\Gamma$ is of index $n$ in $SL_2(\Z)$ (where we allow $n = \infty$) and $D$ denotes the fundamental domain for $SL_2(\Z)$, then a fundamental domain for $\Gamma$ is given by $$\gamma_1 D \union \gamma_2 D \union \ldots \union \gamma_n D$$ where the $\gamma_i$ are a choice of coset representatives. As the action of $SL_2(\Z)$ preserves the volume form on the upper half plane, we see that the volume is just $n$ times the volume of $D$ (in particular the volume is infinite if and only if the index is infinite, as you guessed).

(For the purpose of this answer, because we don't want to think about the boundaries too hard, "fundamental domain" means "whatever fundamental domain means plus or minus sets of measure zero.")

• as for the general case, I'm not sure what generalization we could ask for. One thing that would be interesting is if there were an algorithm, that, given a discrete subgroup of $SL_2(\R)$, determined whether or not it had cofinite volume. I don't know the answer to that question. – hunter Oct 9 '14 at 14:42
• Thanks for the answer - the case I have in mind is the fundamental domain of $G(\lambda)$ for some $\lambda \in \mathbb{R}$, where $G(\lambda)$ is the subgroup of $SL(2,\mathbb{R})$ generated by $\begin{pmatrix} 1 & \lambda \\ 0 & 1\end{pmatrix}$ and $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Any thoughts in this case? (I realize some statement in full generality is perhaps too much to ask for.) – msteve Oct 9 '14 at 14:50