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PostPosted: 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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PostPosted: Sun Mar 13, 2016 9:28 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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