A special case of the uniformization theorem

Question

I am interested in a proof of the following fact :

Suppose that $X$ is a Riemann surface homeomorphic to the Riemann sphere. Then $X$ is conformally equivalent to the Riemann sphere.

Of course, this follows from the uniformization theorem which states that every simply connected Riemann surface is conformally equivalent to either the open unit disk, the complex plane or the Riemann sphere.

However, I was wondering if it was possible to prove this without using the full classification given by the uniformization theorem.

Any relevant reference is welcome.

Thank you and best regards, Malik

Answer 1

An alternative is to simply use Riemann-Roch, which does not depend on the Uniformization Theorem.

R-R says that $\ell(D)-\ell(K-D) = \mathrm{deg}(D)-g+1$, where $\ell(D)$ is the dimension of the space of meromorphic functions $f$ on your surface $C$ such that $(f)+D\ge0$.

If $C$ is homeomorphic to the sphere, then, for topological reasons, $g=0$, so when $D=(p)$ for any $p\in C$, the formula simplifies to $\ell(D)-\ell(K-D) = 2$. In particular, $\ell(D)\ge2$, which implies that there is at least one nonconstant meromorphic function $f$ that has a simple pole at $p$ (and no other poles). This then defines a holomorphic mapping $f:C\to\mathbb{CP}^1$ that is $1$-to-$1$ (because it only assumes the value $\infty$ once), and this gives the conformal equivalence.

Answer 2

There are such proofs. See, for example Goluzin, Geometric theory of functions (Appendix). He uses the following fact. Let $h$ be an analytic diffeomorphism of the circle onto itself. Then there is a Jordan analytic curve $\gamma$ such that a conformal map $f_1$ of the unit disc onto the inside of $\gamma$ and a conformal map $f_2$ of the exterior of the unit disc onto the outside of $\gamma$ are related by $f_1(z)=f_2\circ h(z)$ for $z$ on the unit circle.

This fact has a very simple proof, using the Riemann mapping theorem finitely many times. (See Goluzin). It immediately implies that a Riemann surface homeomorphic to the sphere is conformally equivalent to the sphere.

By the way, this proof is the adaptation of Lavrentiev's proof of the existing of homeomorphic solution of the Beltrami equation, which also implies the uniformization for the sphere. In the case when Beltrami coefficient is analytic, this is usually credited to Gauss. But Gauss has only one sentence on that:-)

Answer 3

Here is a nice proof that appears in a paper of Mazzeo and Taylor. Consider the distributional derivative $\delta'$ (supported at a point $p \in X$) of the Dirac delta distribution. Since $\langle \delta', 1\rangle = 0$, on a compact manifold we can solve $\Delta u = \delta'$. Since $X \setminus p$ is simply connected (this is where the topology figures in) we can form a meromorphic function in the usual way with $u$ as its real part. Observe that this meromorphic function will have one simple pole at $p$. This already defines a holomorphic map $f : X \mapsto \hat{\mathbb{C}}$, which also is a diffeomorphism as $f$ has degree 1.

Somehow the application of the Riemann-Roch theorem (which is probably harder to prove than the Uniformization theorem) seems a bit of a overkill (a matter of personal opinion I am sure).