# 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 $\xi^{*}\in T_{a}X$. 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 $(\overline{\Omega}_{\nu},\partial\Omega_{n}u)$ 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 $H^{1}(Y_{\nu},\mathbb{R})=0$.

## References

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

Title proof of the uniformization theorem ProofOfTheUniformizationTheorem 2013-03-22 15:37:50 2013-03-22 15:37:50 Simone (5904) Simone (5904) 14 Simone (5904) Proof msc 30F20 msc 30F10