Why is this a Riemann surface?

Question

I am reading chapter 3 of Donaldson's Riemann Surfaces. In the end of this chapter he tries to show that any discrete subgroup $\Gamma$ of Aut(D) can make D/$\Gamma$ into a Riemann surface. I understand that every point in D/$\Gamma$ has a neighborhood homeomorphic to a ball, but what is the overlap map between two charts? I can't see why it is holomorphic.

Answer 1

Roughly speaking, each chart $U \subset D/\Gamma$ is parameterized by an open subset $V$ of $D$ itself.

For two charts $U_1,U_2 \subset D/\Gamma$ parameterized by open subsets $V_1,V_2$ of $D$, the overlap map is the restriction of an element of $\Gamma$.

Since $\Gamma$ is a subgroup of $\text{Aut}(D)$, each element of $\Gamma$ is a holomorphic map, and so its restriction to any open subset is holomorphic.

Caveat: I wrote "roughly speaking", because this answer is so far valid only in the special case that $\Gamma$ is acting freely on $D$. The general case can also be described, but it's complicated by the fact that for a point $z \in D$ which is stabilized by a nontrivial cyclic subgroup of $\Gamma$, the chart around the point of $D/\Gamma$ corresponding to $z$ is not actually parameterized by an open subset of $D$ itself. If you want to know about that case I could describe it too, but I suspect that your question might be mostly about the case of a free action.

Edit: Here's a few more rough details about the case of an action which is not free.

For a point $P \in D$ at which the stabilizer subgroup $\text{stab}(P)$ is a cyclic group of order $n\ge 2$, and letting $p \in D/\Gamma$ be the corresponding point in the quotient, the coordinate chart around $p$ parameterized by first translating $P$ to $0$ by some element of $\text{Aut}(D)$, then taking a small open ball $|z|<r$ in $D$ around $0$, and then taking $w = z^n$. The chart around $p$ is then parameterized by the open ball $|w| < r^n$.

So the overlap maps are a bit more complicated. The most complicated case is one chart centered on a point $p$ with a $w=z^m$ coordinate ($m \ge 2$), and another overlapping chart centered on a different point $p'$ with a $w'=z^n$ coordinate ($n \ge 2$). In this case the overlap map would look like this: $$w' = f \circ g \circ h(w) $$ where $h$ is a branch of $\sqrt[m]{z}$, and $g \in \Gamma$, and $f(z)=z^n$. That's a composition of three holomorphic maps, and so the result is holomorphic (I have shoved under the rug the elements of $\text{Aut}(D)$ where we carry out the translation from $P$ to the origin and from $P'$ to the origin; so it's better to think of $g$ as the composition of the inverse of one translation, followed by an element of $\Gamma$, followed by the other translation).

Answer 2

With $Stab(z)$ the subgroup of $\Gamma$ leaving $z$ fixed, which is finite because discrete in some group isomorphic to $PSO_2(\Bbb{R})$, for $u$ close enough to $z$ then $\Gamma u\to \prod_{\gamma \in Stab(z)} (\gamma(u)-\gamma(z))$ looks like a chart from a neighborhood of $\Gamma z$ to a neighborhood of $0$. Thus away from the $z$ with non-trivial $Stab(z)$ we have that $s\to \Gamma s$ is a chart and the transitions functions are the obvious ones, at the remaining $z$ the transitions functions involve some algebraic functions obtained by inverting locally the rational function $u\to \prod_{\gamma \in Stab(z)} (\gamma(u)-\gamma(z))$.