# A special case of the uniformization theorem

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

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.

• I did not know that one could use Riemann-Roch to prove this. Thank you! – Malik Younsi Jul 8 '14 at 22:50

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:-)

• Thank you very much for the reference. However, I don't see how the conformal welding fact that you mention immediately implies that a Riemann surface homeomorphic to the sphere is conformally equivalent. The opposite is clearly true though. Could you add some details? – Malik Younsi Jul 4 '14 at 18:24
• Look in the reference I gave. – Alexandre Eremenko Jul 4 '14 at 21:02
• Yes, I did. The proof that the fact that you mention implies that a Riemann surface homeomorphic to the sphere is conformally equivalent to the sphere is 5 pages long, which is why I thought I was missing something when you wrote immediately implies – Malik Younsi Jul 5 '14 at 0:09
• He proves the general case. If you restrict his proof to the case of the sphere, it will be shorter than 5 pages. – Alexandre Eremenko Jul 10 '14 at 6:55
• Incidentally, Lavrentiev's arguments are a variation on the earlier conformal welding arguments by Caratheodory, who used them in order to prove the uniformization theorem. I suspect that Caratheodory is the original source of this approach. – Misha Jul 10 '14 at 7:39

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).

• This is a nice proof using PDEs! I am not familiar with the fact that "Since $\langle \delta', 1\rangle = 0$, on a compact manifold we can solve $\Delta u = \delta'$" Do you know a reference for this? – Malik Younsi Jul 23 '14 at 18:26
• @MalikYounsi Yes, you can actually prove that for $\Delta u = f$ to be solvable, $f$ must be orthogonal to the constants. Hint: Necessity is obvious by integration by parts. For sufficiency, see for example, Michael Taylor's PDE Volume I, pg 362. You can work the proof yourself. – MBM Jul 24 '14 at 1:12