Riemann Surfaces Associated to Subgroups of the Modular Group

Question

If $G$ is a subgroup of the modular group of index $\mu$, then the triangulation of $\Gamma\backslash\mathbb{H}^\ast$ (where $\mathbb{H}^\ast$ is the extended upper half plane) induces a triangulation on $G\backslash\mathbb{H}^\ast$. Here $\Gamma$ is the modular group $PSL_{2}(\mathbb{Z})$.
In the natural triangulation of $G\backslash\mathbb{H}^\ast$, $\sigma_0$ is the number of images of elliptic and parabolic points of $\Gamma$. It is convenient to write $\sigma_0 = \lambda_i + \lambda_\rho + \lambda_\infty$ where $\lambda_i\left(\lambda_\rho,\lambda_\infty\right)$ is the number of vertices equivalent to $i\left(\rho,\infty\right)$. Let $p_1,p_2,...,p_{\sigma_{0}}$ be the vertices of the triangulation.
For the vertex $p_k$, we determine the the number of 1-simplexes that meet at $p_k$ by distinguishing the following cases:
The Cases.
My question is: Can any one explain the reason behind these cases?
This has been taken from the book "Lectures on Modular Forms" by R.C. Gunning. The reason as to why these cases are true has not been provided in the book.
I have been spending a lot of time trying to figure out the reason behind this. I would be very grateful for absolutely any help that can be provided.

Answer 1

I think the point is that if you look at the action of $\Gamma = PSL_2(\mathbb Z)$ on $\mathbb H^*$, then it is free at almost all points: the stabilizer of $i$ in $G_{\mathbb R}=\mathrm{SL}_2(\mathbb R)$ is the circle $\mathrm{SO}_2(\mathbb R)$, and so as $G_{\mathbb R}$ acts transitively on the upper-half plane the stabilizer of any point in $\mathbb H$ is conjugate to $\mathrm{SO}_2(\mathbb R)$. In particular, the stabilizer of a point $z \in \mathbb H\subset \mathbb H^*$ is compact, hence as $\mathrm{SL}_2(\mathbb Z)$ is discrete, the stabilizers are finite and cyclic (since any finite subgroup of $S^1$ is cyclic). But it is easy to check that the only non-trivial finite-order elements of $\mathrm{PSL}_2(\mathbb Z)$ are ones that have order $2,4$ or $6$. The stabilizer of $i$ has order $2$ in $\Gamma$ while the stabilizer of $\rho$ has order $3$, and the stabilzer of $\infty$ is an infinite dihedral group.

The different cases are then just considering whether the stabilizer of the point $p_k$ lies in $G$ or not.