# 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 $\mathrm{\Omega}\subset \u2102$ with ${H}^{1}(\mathrm{\Omega},\mathbb{R})=0$ is either equal to $\u2102$ 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 $\u2102$.

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 ${H}^{1}({Y}_{\nu},\mathbb{R})=0$. 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 ${\overline{\mathrm{\Omega}}}_{\nu}$ 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 ${\mathrm{\Phi}}_{\nu}:{\mathrm{\Omega}}_{\nu}\to \u2102$. 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 $\u2102\to \u2102$ 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 $\mathrm{\Phi}:X\to \u2102$.

Proof of the ”fact”: Let us first compute the cohomology with compact support ${H}_{c}^{1}({\mathrm{\Omega}}_{\nu},\mathbb{R})$. Let $u$ be a closed $1$-form with compact support in ${\mathrm{\Omega}}_{\nu}$. By Poincaré duality ${H}_{c}^{1}(X,\mathbb{R})=0$, so $u=df$ for some ”test” function $f\in \mathcal{D}(X)$. As $df=0$ on a neighborhood^{} of $X\setminus {\mathrm{\Omega}}_{\nu}$ 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 ${H}_{c}^{1}({\mathrm{\Omega}}_{\nu},\mathbb{R})$ and we get ${H}_{c}^{1}({\mathrm{\Omega}}_{\nu},\mathbb{R})={H}^{1}({\mathrm{\Omega}}_{\nu},\mathbb{R})=0$. The exact sequence of the pair $({\overline{\mathrm{\Omega}}}_{\nu},\partial {\mathrm{\Omega}}_{n}u)$ yelds

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

thus ${H}^{0}(\partial {\mathrm{\Omega}}_{\nu},\mathbb{R})=\mathbb{R}$. 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

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

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 |
---|---|

Canonical name | ProofOfTheUniformizationTheorem |

Date of creation | 2013-03-22 15:37:50 |

Last modified on | 2013-03-22 15:37:50 |

Owner | Simone (5904) |

Last modified by | Simone (5904) |

Numerical id | 14 |

Author | Simone (5904) |

Entry type | Proof |

Classification | msc 30F20 |

Classification | msc 30F10 |