# Riemann Surfaces formed as quotients of $\mathbb{C}$

I was trying to solve a problem given in Simon Donalson's book on Riemann Surfaces.

The problem asks to classify the Riemann Surfaces formed as quotients of $\mathbb{C}$.

If $\Gamma$ is the (discrete) group by which we're forming the quotient, I was able to prove that $\Gamma \subset \{z \mapsto az + b: |a| = 1\}$ (otherwise the quotient won't be Hausdorff).

Since what I need to prove is that $\mathbb{C}/\Gamma$ is isomorphic to $\mathbb{C}/\mathbb{Z}$ or $\mathbb{C}/\Lambda$ for some lattice $\Lambda$, I was trying to show that $\Gamma \subset \{z \mapsto z+b\}$

This attempt was unsuccessful. In fact, if $\Gamma_m = \{z \mapsto ze^{2i\pi \frac{k}{m}}:k \in \mathbb{Z}\}$ then I was able to equip $\mathbb{C}/\Gamma_m$ with a conformal structure, which turned out to be equivalent to $\mathbb{C}$ (via the map $z \mapsto z^m$).

Similarly, I was able to define charts on $\mathbb{C}/\Gamma$ even if some points had non-trivial stabilisers (via the technique Donaldson uses to define charts when the group action isn't free).

If only one point has a non-trivial stabiliser, the case more or less becomes the same as $\Gamma= \Gamma_m$ and so the quotient is $\mathbb{C}$.

So, the case I'm stuck is when there are multiple points (possibly infinitely many) with non-trivial stabilisers.

Thanks

EDIT:

As Andrew has pointed out in the comments, it is likely that the group action was supposed to be free, in which case the conclusion is easy to see.

In case someone does have any interesting results when the group action isn't free, feel free to add an answer about it (no pun intended).

• Since the complex plane itself (the quotient by a cyclic group acting by rotations) is not in your list of "allowable quotients", is it possible Donaldson is speaking only of quotients that are covering maps, i.e., quotients by fixed-point free actions...? – Andrew D. Hwang Jul 9 '17 at 14:48
• @Andrew Hmm... Considering that he invokes this result only when the universal cover is $\mathbb{C}$, this is probable. I interpreted the question as to describe the quotients of $\mathbb{C}$ which are not isomorphic to $\mathbb{C}$. Thanks, this was helpful. – Deeparaj Bhat Jul 10 '17 at 6:05
• What happens when $\Lambda = \mathbb{Z}+i\mathbb{Z}$ or $\mathbb{Z}+\omega\mathbb{Z}$ together with $z \mapsto e^{i \pi /2}z$ or $z \mapsto e^{2i \pi /3}z$ ? – reuns Jul 10 '17 at 7:13
• @user1952009 Wouldn't that (the first one) be the Riemann Sphere? I think the quotient would be compact and simply connected and hence must be the Riemann Sphere. – Deeparaj Bhat Jul 10 '17 at 10:21
• If $a \neq 1$, the affine map $z \mapsto az + b$ has a fixed point, so the quotient is not a covering map. ;) – Andrew D. Hwang Jul 10 '17 at 10:41