Genus zero Riemann surface is conformally equivalent to Riemann sphere

Statement

Any compact Riemann surface of genus zero (in other words, any compact Riemann surface that is topologically a sphere) is conformally equivalent to the [fact about::Riemann sphere]].

Equivalently, any simply connected compact Riemann surface is conformally equivalent to the Riemann sphere.

Related facts

• Riemann mapping theorem: This result has a similar feel to it: it says that any simply connected domain in $\mathbb{C}$ that is not the whole of $\mathbb{C}$ is conformally equivalent to the open unit disk.
• Every compact Riemann surface is a branched cover of the Riemann sphere

Facts used

1. Riemann-Roch theorem
2. Holomorphic on compact Riemann surface implies constant

Proof

Given: A compact Riemann surface $S$ of genus zero.

To prove: $S$ is conformally equivalent to the Riemann sphere.

Proof: Pick a point $p_0\in S$, and consider the divisor $D=p_0$. By the Riemann-Roch theorem, we have:

$\mathrm{d}\mathrm{i}\mathrm{m}L(-D)=\mathrm{d}\mathrm{e}\mathrm{g}(D)-g+1+\mathrm{d}\mathrm{i}\mathrm{m}L(D-K)$.

Ignoring $\mathrm{d}\mathrm{i}\mathrm{m}L(D-K)$, we get:

$\mathrm{d}\mathrm{i}\mathrm{m}L(-D)\ge 1-0+1=2$.

Thus, the space of meromorphic functions on $S$ with a simple pole at at most one point (namely, $p_0\in S$) and no higher order poles, is two-dimensional. Note that the space of holomorphic functions on $S$ includes only the constant functions, and is thus one-dimensional. Hence, there exists a nonconstant meromorphic function on $S$, having a simple pole at $p_0$ and no other poles.

In particular, this function gives an analytic mapping from $S$ to the Riemann sphere. $\mathrm{\infty }$ has only one inverse image under this mapping, so the mapping is a degree one analytic mapping, and hence a conformal isomorphism. So, $S$ is conformally equivalent to the Riemann sphere.