# Why is this a Riemann surface?

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.

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 $$|z| in $$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).

• Thanks for your answer! But my question is actually about the case when the action is not free, could you please describe it? Nov 13, 2020 at 14:51
• I've added an explanation. Nov 13, 2020 at 19:54
• I'm still confused. f and h are both maps on actual numbers, instead of on their corresponding points in $D/ \Gamma$, right? Because otherwise they can't multiply g. But if f,g,h are all maps on actual numbers, how does this overlap map ''pass through'' the quotient space? I think since the group does not act freely, we can't find a local bijection between the space and its quotient. Another question is that, we can locally choose a branch of h only if z is away from zero. What if z=0? Nov 14, 2020 at 7:18
• Your comment seems to have three different questions. I'll try to answer them all. "Passing through" the quotient space is not something overlap maps do. Instead, what overlap maps do is to "pass between" the parameter sets for two charts in the quotient space. Nov 14, 2020 at 13:38
• You're right, we can't find a local bijection between the space and its quotient. What my answer shows is a local bijection between a local quotient of the space and a **local neighborhood" of the quotient. Nov 14, 2020 at 13:39

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 $$u\to \prod_{\gamma \in Stab(z)} (\gamma(u)-\gamma(z))$$.

• I still don't understand what happens at z (which has non-trivial stabilizers), can you eplain it in details? thanks! Nov 13, 2020 at 14:59
• For a fixed $t\in \Bbb{H}$ and $r$ small, is $\{ \frac{at+b}{ct+d}\in Aut(\Bbb{H}), |t-\frac{at+b}{ct+d}|\le r\}$ compact ? Use that $\Im( \frac{at+b}{ct+d})= \frac{\Im(t)}{|ct+d|^2}$, thus $|ct+d|\approx 1$ and $|at+b|\approx |t|$. The compactness of this set and the discreteness of $\Gamma$ implies that for $u,v$ close enough to $z$ then $\Gamma u=\Gamma v$ iff $v=\gamma(u)$ for some $\gamma\in Stab(z)$, thus $\Gamma u\to \prod_{\gamma \in Stab(z)} (\gamma(u)-\gamma(z))$ is a chart. Nov 13, 2020 at 16:59
• @ZichenGao Anything still unclear ? $f(u)= \prod_{\gamma \in Stab(z)} (\gamma(u)-\gamma(z))$ has a zero of order $n=|Stab(z)|$ at $z$ so (a branch of) $f(u)^{1/n}$ is biholomorphic near $z$, ie. there are $n$ values of $v$ such that $f(u)=f(v)$, thus all of the form $v=\gamma(u)$ with $\gamma\in Stab(z)$. Nov 14, 2020 at 11:53