Welcome to mathimatikoi.org forum; Enjoy your visit here.

## Show That S is a Sphere- 2

Differential Geometry
Tsakanikas Nickos
Community Team
Articles: 0
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

### Show That S is a Sphere- 2

Let $S$ be a compact connect Riemann surface. If there is a meromorphic function on $S$ with exactly one pole, and that pole is of order $1$, then show that $S$ is biholomorphic to the Riemman Sphere $\hat{\mathbb{C}}$.
Tsakanikas Nickos
Community Team
Articles: 0
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

### Re: Show That S is a Sphere- 2

Comment: A meromorphic function on a Riemann surface $X$ is a holomorphic map from $X$ to the Riemann Sphere $\hat{\mathbb{C}}$ which is not identically equal to $\infty$.

Let $f$ be the given meromorphic function. Then $f \ \colon S \longrightarrow \hat{\mathbb{C}}$ is a non-constant holomorphic map on the compact Riemann surface $S$, so it is proper. This means that the degree $\displaystyle d = d(y) = \sum_{x \in f^{-1}(y)} v_{f}(x)$ of $f$ is well-defined, i.e. independent of the choice of $y \in \hat{\mathbb{C}}$. Let $x$ be the unique pole of $f$. Then we can compute the degree of $f$ using $y = \infty$, and since $x$ is a simple pole of $f$, i.e. the ramification index of $f$ at $x$ is $v_{f}(x) = 1$, we conclude that $d = 1$, as $f$ has a unique pole. By the definition of the degree, it follows that $f$ is bijective (as it assumes every value $y \in \hat{\mathbb{C}}$). Hence, $f$ is biholomorphic, as a holomorphic bijection. This means that $S$ is biholomorphic to the Riemann Sphere.