# Construction of Riemann surfaces as quotients

There is a theorem in Chapter 4 Section 5 of Schlag’s complex analysis text. I went through it a month ago, but only half understood it, and it is my hope that passing through it again, this time with writeup, will finally shed light, after having studied in detail some typical examples of such Riemann surfaces, especially tori, the conformal equivalence classes of which can be represented by the fundamental region of the modular group, which arise from quotienting out by lattices on the complex plane, as well as Fuchsian groups.

In the text, the theorem is stated as follows.

Theorem 4.12.  Let $\Omega \subset \mathbb{C}_{\infty}$ and $G < \mathrm{Aut}(\mathbb{C}_{\infty})$ with the property that

• $g(\Omega) \subset \Omega$ for all $g \in G$,
• for all $g \in G, g \neq \mathrm{id}$, all fixed points of $g$ in $\mathbb{C}_{\infty}$ lie outside of $\Omega$,
• for all $K \subset \Omega$ compact, the cardinality of $\{g \in G | g(K) \cap K \neq \phi\}$ is finite.

Under these assumptions, the natural projection $\pi : \Omega \to \Omega / G$ is a covering map which turns $\Omega/G$ canonically onto a Riemann surface.

The properties essentially say that the we have a Fuchsian group $G$ acting on $\Omega \subset \mathbb{C}_{\infty}$ without fixed points, excepting the identity. To show that quotient space is a Riemann surface, we need to construct charts. For this, notice that without fixed points, there is for all $z \in \Omega$, a small pre-compact open neighborhood of $z$ denoted by $K_z \subset \Omega$, so that

$g(\overline{K_z} \cap \overline{K_z}) = \emptyset \qquad \forall g \in G, g \neq \mathrm{id}$.

So, in $K_z$ no two elements are twice represented, which mean the projection $\pi : K_z \to K_z$ is the identity, and therefore we can use the $K_z$s as charts. The $g$s as Mobius transformations are open maps which take the $K_z$s to open sets. In other words, $\pi^{-1}(K_z) = \bigcup_{g \in G} g^{-1}(K_z)$ with pairwise disjoint open sets $g^{-1}(K_z)$. From this, the $K_z$s are open sets in the quotient topology. In this scheme, the $g$s are the transition maps.

Finally, we verify that this topology is Hausdorff. Suppose $\pi(z_1) \neq \pi(z_2)$ and define for all $n \geq 1$,

$A_n = \left\{z \in \Omega | |z-z_1| < \frac{r}{n}\right\} \subset \Omega$

$B_n = \left\{z \in \Omega | |z-z_2| < \frac{r}{n}\right\} \subset \Omega$

where $r > 0$ is sufficiently small. Define $K = \overline{A_1} \cup \overline{B_1}$ and suppose that $\pi(A_n) \cap \pi(B_n) \neq \emptyset$ for all $n \geq 1$. Then for some $a_n \in A_n$ and $g_n \in G$ we have

$g_n(a_n) \in B_n \qquad \forall n \geq 1$.

Since $g_n(K) \cap K$ has finite cardinality, there are only finitely many possibilities for $g_n$ and one of them therefore occurs infinitely often. Pass to the limit $n \to \infty$ and we have $g(z_1) = z_2$ or $\pi(z_1) = \pi(z_2)$, a contradiction.