The Riemann Surface $G/H$

Question

Let $G$ be a subgroup of $SL(2,Z)$ that is of finite index and $H$ be the upper half-plane.

How is the quotient topological space $G/H$ defined (understood)?

Answer 1

As a topological space $G\backslash\Bbb{H}$ is just the space whose points are subsets of $\Bbb{H}$ of the form $\{ gz,g\in G\}$ and whose open sets are the image of $G$-invariant open subsets of $\Bbb{H}$.

It becomes a Riemann surface with its (pre)sheaf of holomorphic functions: the functions on some open of $G\backslash\Bbb{H}$ that come from an holomorphic function on the corresponding open subset of $\Bbb{H}$.

$gs = s$ for some $g\ne \pm I\in SL_2(\Bbb{Z})$ implies that $s=\gamma(i)$ or $s=\gamma(e^{2\pi /3})$ for some $\gamma\in SL_2(\Bbb{Z})$. Thus for a small enough disk $D\ni s\not \in SL_2(\Bbb{Z})i,SL_2(\Bbb{Z})e^{2i\pi /3}$ any holomorphic function $f$ on $D\subset \Bbb{H}$ gives an holomorphic function $F(Gz)=f(z)$ on $GD\subset G\backslash\Bbb{H}$.

When $f$ has a simple zero at $s$ then it is a chart for the Riemann surface $G\backslash\Bbb{H}$ around $Gs$.

The case $s=i$: When $S=\pmatrix{0&1\\-1&0}\in G$ then $f$ must be of the form $h(z)h(-1/z)$ with $h$ holomorphic for $F(Gz)$ to be well-defined around $Gi$, and it is a chart when $h$ has a simple zero at $i$.

The case $s=e^{2i\pi/3}$: When $ST\in G$ then $f$ must be of the form $h(z)h(STz)h((ST)^2 z)$ to be well-defined at $e^{2i\pi/3}$ and again it is a chart when $h$ has a simple zero at $e^{2i\pi /3}$.