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 $H^1(X,ℝ)=0$. 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 $\mathrm{\Omega }\subset ℂ$ with $H^1(\mathrm{\Omega },ℝ)=0$ is either equal to $ℂ$ or biholomorphic to the unit disk. Thus, all we have to show is that a non compact Riemann surface $X$ with $H^1(X,ℝ)=0$ can be embedded in the complex plane $ℂ$.

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

Fact: we have $H^1(Y_{\nu },ℝ)=0$. 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 ${\overline{\mathrm{\Omega }}}_{\nu }$ 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 ${\mathrm{\Phi }}_{\nu }:{\mathrm{\Omega }}_{\nu }\to ℂ$. Fix a point $a\in {\mathrm{\Omega }}_0$ and a non zero linear form ${\xi }^{*}\in T_{a}X$. We can take the composition of ${\mathrm{\Phi }}_{\nu }$ with an affine linear map $ℂ\to ℂ$ so that ${\mathrm{\Phi }}_{\nu }(a)=0$ and $d{\mathrm{\Phi }}_{\nu }(a)={\xi }^{*}$. By the well-known properties of injective holomorphic maps, $({\mathrm{\Phi }}_{\nu })$ is then uniformly bounded on every small disk centered at $a$, thus also on every compact subset of $X$ by a connectedness argument. Hence $({\mathrm{\Phi }}_{\nu })$ has a subsequence converging towards an injective holomorphic map $\mathrm{\Phi }:X\to ℂ$.

Proof of the ”fact”: Let us first compute the cohomology with compact support $H_c^1({\mathrm{\Omega }}_{\nu },ℝ)$. Let $u$ be a closed $1$-form with compact support in ${\mathrm{\Omega }}_{\nu }$. By Poincaré duality $H_c^1(X,ℝ)=0$, so $u=df$ for some ”test” function $f\in 𝒟(X)$. As $df=0$ on a neighborhood of $X\setminus {\mathrm{\Omega }}_{\nu }$ and as all connected components of this set are non compact, $f$ must be equal to the constant zero near $X\setminus {\mathrm{\Omega }}_{\nu }$. Hence $u=df$ is the zero class in $H_c^1({\mathrm{\Omega }}_{\nu },ℝ)$ and we get $H_c^1({\mathrm{\Omega }}_{\nu },ℝ)=H^1({\mathrm{\Omega }}_{\nu },ℝ)=0$. The exact sequence of the pair $({\overline{\mathrm{\Omega }}}_{\nu },\partial {\mathrm{\Omega }}_{n}u)$ yelds

$$ℝ=H^0({\overline{\mathrm{\Omega }}}_{\nu },ℝ)\to H^0(\partial {\mathrm{\Omega }}_{\nu },ℝ)\to H^1({\overline{\mathrm{\Omega }}}_{\nu },\partial {\mathrm{\Omega }}_{\nu };ℝ)\simeq H_c^1({\mathrm{\Omega }}_{\nu },ℝ)=0,$$

thus $H^0(\partial {\mathrm{\Omega }}_{\nu },ℝ)=ℝ$. Finally, the Mayer-Vietoris sequence applied to small neighborhoods of the two copies of ${\overline{\mathrm{\Omega }}}_{\nu }$ in $Y_{\nu }$ gives an exact sequence

$$H^0{({\overline{\mathrm{\Omega }}}_{\nu },ℝ)}^{\oplus 2}\to H^0(\partial {\mathrm{\Omega }}_{\nu },ℝ)\to H^1(Y_{\nu },ℝ)\to H^1{({\overline{\mathrm{\Omega }}}_{\nu },ℝ)}^{\oplus 2}=0$$

where the first map is onto. Hence $H^1(Y_{\nu },ℝ)=0$.