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Let $X$ be a compact Riemann surface. Prove that if $X$ is isomorphic to $\mathbb{P}^1$, then $X$ admits a meromorphic function $f$ that has a single pole and that this pole has multiplicity one.

Since $X$ and $\mathbb{P}^1$ are isomorphic, then there exists a biholomorphic function $f$ between them. This $f$ happens to be meromorphic with only one pole. However, I can't seem to be able to prove that this pole must have multiplicity one. All I got so far is, assume that it has multiplicity greater than one, $n$, then every point in $\mathbb{P}^1$ must have multiplicity $n$. Can someone help me out with where to go next?

Zev Chonoles
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user44322
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1 Answers1

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Hint: The integer $n$ is the number of points in $X$ (counting multiplicity) in the preimage of an arbitrary point in $\mathbf{P}^{1}$.

Andrew D. Hwang
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  • Since $f$ is biholomorphic then the multiplicity of every point in X in the preimage of an arbitrary point in $\mathbb{P}^1$ is $n$. So we got that the multiplicity of every point in $f$ and $f^{-1}$ are both $n$. Where do I go from there? – user44322 Feb 22 '14 at 21:51
  • Let $z$ denote an element of $\mathbf{P}^{1}$. How many points are in $f^{-1}({z})$, given that $f$ is a biholomorphism? – Andrew D. Hwang Feb 22 '14 at 22:50
  • Only 1 and that point should have multiplicity $n$ – user44322 Feb 22 '14 at 22:57
  • Just asking: What exactly is your definition of "multiplicity"? – Andrew D. Hwang Feb 22 '14 at 23:06
  • There is a theorem that says every holomorphic mapping between two Riemann Surfaces, when composed with charts can be written as $z^k$. This $k$ here is the multiplicity. – user44322 Feb 22 '14 at 23:28
  • Ah. Let's suppose there exists a point of $X$ where $f$ has multiplicity $k > 1$. Can you use your definition to show $f$ is not a biholomorphism? (You'll need to think about the behavior of $z \mapsto z^{k}$ in small neighborhoods of $0$.) – Andrew D. Hwang Feb 23 '14 at 00:14
  • If $f$ is a bijection then composing it with charts is also a bijection. Thus, $z^k$ must be a bijection which is not true. – user44322 Feb 23 '14 at 07:10