Are the only Riemann surfaces which are quotients of $\mathbb{C}$the cylinder and the toruses? Why?

Question

Consider the Riemann surfaces $\mathbb{C}^\times=\mathbb{C}\setminus\{0\}$ and $\mathbb{C}/\Lambda$, where $\Lambda$ is a lattice in the complex plane (i.e. a discrete additive subgroup of $\mathbb{C}$). Now, we can show that $\mathbb{C}^\times$ is biholomorphic to the quotient $\mathbb{C}/\mathbb{Z}$, where $\mathbb{Z}$ is considered as an additive subgroup, so both of these are quotients. I've heard somebody say, pretty long ago, that these are the only Riemann surfaces that arise as quotients of the complex plane. Is this true? If so, why?

Answer 1

By the uniformisation theorem, a Riemann surface is a quotient of a simply connected Riemann surface $S$ by a discrete subgroup consisting of biholomorphic maps on $S$, consisting, apart from the identity, of maps without fixed points. Note that $S$ is biholomorphic to $\Bbb C$, the Riemann sphere, or the upper half-plane.

The biholomorphic maps from $\Bbb C$ to itself are $z\mapsto az+b$ with $a$, $b\in \Bbb C$, $a\ne0$. The ones without fixed points are those with $a=1$, the translations. The only discrete subgroups of fixed-point-free biholomorphic maps are (i) the trivial group, (ii) cyclic generated by a translation, (iii) consisting of translations forming a rank $2$ lattice.

The quotients of $\Bbb C$ by these groups are (i) $\Bbb C$, (ii) $\Bbb C^\times$ and (iii) complex tori.