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Let $M$ be a $2$-dimensional orientable manifold. Is $M$ a Riemann surface? If it is true, how can I show it? Thank you very much.

joseabp91
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  • It is a $1$-dimensional complex manifold (in complex sense). – joseabp91 Dec 02 '18 at 19:27
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    See: https://math.stackexchange.com/a/907476/534616 – K B Dave Dec 02 '18 at 19:27
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    Not quite a duplicate of that one because the surface isn't required to be compact – K B Dave Dec 02 '18 at 19:29
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    I do not see why they answer my question. – joseabp91 Dec 02 '18 at 19:40
  • @joseabp91 Isn't it the definition of the Riemann surface?(See here:https://en.wikipedia.org/wiki/Riemann_surface. In fact,dimension-two differentiable manifold naturally admits an almost complex structure and it is intergrable.Then it determines a complex structure (By Newlander-Nierenberg's theorem) . – Invariance Sep 22 '19 at 12:11
  • By the way,any $n$-dimension complex is an oriented $2n$-dimension differentiable manifold shows the equivalence of the definition. – Invariance Sep 22 '19 at 12:17

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