proof of the uniformization theorem

Our proof relies on the well-known Newlander-Niremberg theorem which implies, in particular, that any Riemmanian metric on an oriented $2$ -dimensional real manifold defines a unique analytic structure.

We will merely use the fact that $H^1(X,\mathbb R)=0$ . If $X$ is compact, then $X$ is a complex curve of genus $0$ , so $X\simeq\mathbb P^1$ . On the other hand, the elementary Riemann mapping theorem says that an open set $\Omega\subset\mathbb C$ with $H^1(\Omega,\mathbb R)=0$ is either equal to $\mathbb C$ or biholomorphic to the unit disk. Thus, all we have to show is that a non compact Riemann surface $X$ with $H^1(X,\mathbb R)=0$ can be embedded in the complex plane $\mathbb C$ .

Let $\Omega_\nu$ be an exhausting sequence of relatively compact connected open sets with smooth boundary in $X$ . We may assume that $X\setminus\Omega_\nu$ has no relatively compact connected components, otherwise we fill the holes of $\Omega_\nu$ by taking the union with all such components. We let $Y_\nu$ be the double of the manifold with boundary $(\overline\Omega_\nu,\partial\Omega_\nu)$ , i.e. the union of two copies of $\overline\Omega_\nu$ with opposite orientations and the boundaries identified. Then $Y_\nu$ is a compact oriented surface without boundary.

Fact: we have $H^1(Y_\nu,\mathbb R)=0$ . We postpone the proof of this fact to the end of the present paragraph and we continue with the proof of the uniformization theorem.

Extend the almost complex structure of $\overline\Omega_\nu$ in an arbitrary way to $Y_\nu$ , e.g. by an extension of a Riemmanian metric. Then $Y_\nu$ becomes a compact Riemann surface of genus $0$ , thus $Y_\nu\simeq\mathbb P^1$ and we obtain in particular a holomorphic embedding $\Phi_\nu\colon\Omega_\nu\to\mathbb C$ . Fix a point $a\in\Omega_0$ and a non zero linear form $\xi^*\in T_aX$ . We can take the composition of $\Phi_\nu$ with an affine linear map $\mathbb C\to\mathbb C$ so that $\Phi_\nu(a)=0$ and $d\Phi_\nu(a)=\xi^*$ . By the well-known properties of injective holomorphic maps, $(\Phi_\nu)$ is then uniformly bounded on every small disk centered at $a$ , thus also on every compact subset of $X$ by a connectedness argument. Hence $(\Phi_\nu)$ has a subsequence converging towards an injective holomorphic map $\Phi\colon X\to\mathbb C$ .

Proof of the "fact": Let us first compute the cohomology with compact support $H^1_c(\Omega_\nu,\mathbb R)$ . Let $u$ be a closed $1$ -form with compact support in $\Omega_\nu$ . By Poincaré duality $H^1_c(X,\mathbb R)=0$ , so $u=df$ for some "test" function $f\in\mathcal D(X)$ . As $df=0$ on a neighborhood of $X\setminus\Omega_\nu$ and as all connected components of this set are non compact, $f$ must be equal to the constant zero near $X\setminus\Omega_\nu$ . Hence $u=df$ is the zero class in $H^1_c(\Omega_\nu,\mathbb R)$ and we get $H^1_c(\Omega_\nu,\mathbb R)=H^1(\Omega_\nu,\mathbb R)=0$ . The exact sequence of the pair $(\overline\Omega_\nu,\partial\Omega_nu)$ yelds $$ \mathbb R=H^0(\overline\Omega_\nu,\mathbb R)\to H^0(\partial\Omega_\nu,\mathbb R)\to H^1(\overline\Omega_\nu,\partial\Omega_\nu;\mathbb R)\simeq H^1_c(\Omega_\nu,\mathbb R)=0, $$ thus $H^0(\partial\Omega_\nu,\mathbb R)=\mathbb R$ . Finally, the Mayer-Vietoris sequence applied to small neighborhoods of the two copies of $\overline\Omega_\nu$ in $Y_\nu$ gives an exact sequence $$ H^0(\overline\Omega_\nu,\mathbb R)^{\oplus 2}\to H^0(\partial\Omega_\nu,\mathbb R)\to H^1(Y_\nu,\mathbb R)\to H^1(\overline\Omega_\nu,\mathbb R)^{\oplus 2}=0 $$ where the first map is onto. Hence $H^1(Y_\nu,\mathbb R)=0$ .

Bibliography

J.-P. Demailly, Complex Analytic and Algebraic Geometry.