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 Post subject: Show That S is a Sphere- 2Posted: Wed Feb 03, 2016 9:12 pm
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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 $\hat{\mathbb{C}}$.

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 Post subject: Re: Show That S is a Sphere- 2Posted: Sun Mar 13, 2016 9:28 pm
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Joined: Tue Nov 10, 2015 8:25 pm
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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.

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