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In a previous question, I learned that there exist infinitely many non biholomorphic Riemann surfaces homeomorphic to the torus.

Is it also true for the sphere?

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Seirios
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    No it's not. In fact the complex structure is unique. (Think of it as $\mathbb{C}\mathbb{P}^1$) For a proof, do you know Riemann-Roch? –  Mar 03 '13 at 18:44
  • Any textbook on Riemann surfaces should thoroughly discuss such issues. – Qiaochu Yuan Mar 03 '13 at 19:54
  • @Sanchez: No, I don't know Riemann-Roch. In fact, I just begin to study Riemann surfaces, so if there is no elementary argument, a little explanation with a reference (for more details) will be appreciated. – Seirios Mar 03 '13 at 20:56
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    Alternatively, use the uniformisation theorem and Riemann–Hurwitz. – Zhen Lin Mar 03 '13 at 21:41
  • @ZhenLin: Why is Riemann-Hurwitz useful? The uniformatisation theorem seems to be sufficient, no? – Seirios Mar 06 '13 at 12:23
  • The sphere is a surface of genus $0$, so the Hurwitz formula implies there can be no unramified coverings of it by a compact surface of positive genus. Hence the uniformisation theorem implies any Riemann surface that is topologically a sphere must be $\mathbb{C P}^1$. – Zhen Lin Mar 06 '13 at 12:51
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    @ZhenLin: Let $S$ be a Riemann surface homeomorphic so $\mathbb{S}^2$. So $S$ is simply connected, and according to the uniformization theorem, $S$ is either the unit disk, the sphere or the complex plane. But neither the unit disk nor the complex plane are homeomorphic to the sphere, so we can conclude without Riemann-Hurwitz, no? – Seirios Mar 07 '13 at 15:01

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As stated in comments, the answer turns out to be an immediate consequence of the uniformization theorem:

Theorem: Every simply connected Riemann surface is biholomorphic to the open unit disk, the complex plane, or the Riemann sphere.

So if $S$ is a Riemann surface homeomorphic to $\mathbb{S}^2$, it is simply connected, so biholomorphic to the open unit disk, the complex plane or the Riemann sphere. But a biholomorphism is a homeomorphism and neither the open disk nor the complex plane are homeomorphic to $\mathbb{S}^2$. Therefore, $S$ is biholomorphic to the Riemann sphere.

Seirios
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