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_zs as charts. The gs as Mobius transformations are open maps which take the K_zs 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_zs are open sets in the quotient topology. In this scheme, the gs 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.

 

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