Non-compact Riemann surfaces with zero Euler characteristics

Question

The question arises because of a statement in John H.Hubbard's book "Teichmuller theory and Applications to Geometry, Topology and Dynamics, vol 1" Page 130, Chapter 4.4.It says that:

The only Riemann surfaces with $\chi=0$ are cylinders or compact surfaces of genus 1.

The main trouble is the only here, especially the non-compact case. For compact case, we can use classification of closed (compact without boundary) surfaces to verify. But what about the case compact with boundary?

For non-compact case, all I know is the following strong version of uniformization theorem (see Theorem 1.8.8 in John's book):

1. The Riemann sphere $P^1$ is the universal covering only of itself
2. The plane C is the universal covering of itself, or of punctured plane, or of all compact Riemann surfaces homeomorphic to a torus
3. All other Riemann surfaces have universal covering space biholo/analytically isomorphic to D

By this theorem, all I want to show is that Case 3 never happens, I want to use Riemann-Hurwitz at first, but soonly I realized this may not be a finite covering map, just considering the case 2, you will know why Riemann Hurwitz can't use here.

So what should I do? Any help is welcome.

Answer 1

There are in fact a few cases of Riemann surfaces with Euler characteristic zero having $D$ as their universal cover: a punctured open disc; and an annulus in $C$, i.e. $$A = \{z \mid r_1 < z < r_2\} \quad r_1 > 0 $$ These, together with the ones you listed, are in fact all possible examples of Riemann surfaces of Euler characteristic zero. And notice that from a purely topological perspective, these surfaces are all homeomorphic to each other.

If you want a full proof, you're going to need some topology. The first step is to restrict attention to a noncompact Riemann surface $S$ such that each end of $S$ has a neighborhood holomorphically equivalent to one of the two examples I just listed: a punctured disc or an annulus. The reason this restriction is needed is because if $S$ has any other kind of end then its first homology has infinite rank and therefore the Euler characteristic is not even defined.

Once that fact about ends has been proved, one uses the classification of surfaces to prove that $S$ is homeomorphic to a closed surface of some genus $g \ge 0$ with some finite set of points removed; and then one proves that the Euler characteristic equals zero if and only if $g=0$ and two points were removed, which happens if and only if $S$ is homeomorphic to an annulus.

Finally, one applies the uniformiztion theorem to prove that $S$ is holomorphically equivalent to one of the examples already described.