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Let $G$ be a subgroup of $PSL(2,\mathbb{C})$, so that $G$ acts on $\mathbb{C}\cup\{\infty\}$ by linear fractional transformations. We say that $G$ acts properly discontinuously at a point $z\in \mathbb{C}\cup\{\infty\}$ if

  1. the stabilizer $G_z$ is finite, and
  2. there exists a neighborhood $U_z$ such that $g(U_z)=U_z$ for any $g$ in $G_z$ and $U_z\cap g(U_z)=\emptyset$ for any $g$ in $G\setminus G_z$.

Let $\Omega(G)$ be the subset of $\mathbb{C}\cup\{\infty\}$ at every point of which $G$ acts properly discontinuously. It is an open and $G$-invariant subset of $\mathbb{C}\cup\{\infty\}$. There is a theorem (see e.g. 1.5.2.5.1 Theorem from here) saying that $\Omega(G)/G$ can be endowed with a complex structure so that it becomes a Riemann surface provided that $\Omega(G)\ne \emptyset$ (and say $\Omega(G)$ is connected).

(The standard application is: $G=PSL(2,\mathbb{Z})$, $\Omega(G)=\mathbb{H}$ so that $\mathbb{H}/PSL(2,\mathbb{Z})$ is a Riemann surface.)

I was wondering whether this complex structure on $\Omega(G)/G$ is unique? If so, what fact does it follow from (or is it obvious)?

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    The uniqueness of the complex structure is first proved using covering space theory on the open subset of unramified points and the uniqueness at ramification points essentially comes from Riemann's theorem on removable singularities. – 2017-01-19
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    It depends on what you require from this structure: It is unique if you want the quotient map $\Omega\to \Omega/G$ to be holomorphic. (The latter is what appears in the link and what you did not copy in your question.) It is a good exercise in complex analysis to show that given a Riemann surface $X$ and a surjective map $f: X\to Y$, where $Y$ is a topological surface, there is at most one complex structure on $Y$ making $f$ holomorphic. – 2017-01-20
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    @MoisheCohen Thanks! Sorry do you know whether there exist complex structures on $\Omega(G)/G$ such that the quotient map is not holomorphic? Also, the theorem in the link alludes to the fact that it is important to consider the (unique) complex structure on $\Omega(G)/G$ that makes the quotient map holomorphic. Why is that so? That is, why is this structure of interest to us (and not the structures that do not make the map holomorphic)? – 2017-01-23
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    In most cases, the complex structure on the oriented surface $\Omega/G$ is not unique. The simplest example to consider is a cyclic loxodromic group. As for importance of the structure making the quotient map holomorphic: This structure remembers the conformal nature of the action of $G$ on $\Omega$. The other structures capture no such information, only the topology of the action. Considering them is still useful though in some situations, e.g. when studying the Teichmuller space of $G$ (i.e. the space of its quasiconformal deformations). – 2017-01-24

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