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How is a Riemann surface different from an ordinary surface? I'm studying Riemann surfaces and as I understand they're just differential manifolds of complex dimension one. How is that different from just a surface, so a differential manifold of real dimension two? Is the difference (if any) only algebraic? Or are they topologically different?

Just to be sure: the sphere and the torus are both Riemann surfaces, but they're also normal surfaces right?

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  • $\begingroup$ @NiharKarve the question is different because it focuses on the smoothness or holomorphicity of manifolds, but in the main answer I found "a 1-dimensional complex manifold X (thus X is a 2d-dimensional real manifold, i.e. a topological surface)" which seems to say that no, there is no difference between them. Is that correct? $\endgroup$ Commented Mar 11, 2021 at 10:04
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    $\begingroup$ "thus X is a 2d dimensional real manifold" means that a 1d complex manifold is also a 2d real manifold. The converse is not necessarily true because you would have to define a complex structure on the real manifold $\endgroup$ Commented Mar 11, 2021 at 10:22
  • $\begingroup$ Your link to the description of "normal surfaces" is misleading. The latter were invented for the study of three-dimensional manifolds. I don't think you have this in mind, but "everyday differential-geometric surfaces". $\endgroup$ Commented Mar 11, 2021 at 10:38
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    $\begingroup$ Topologically, every Riemann surface is a surface in the usual sense. The difference lies in the complex structure. $\endgroup$
    – Thorgott
    Commented Mar 11, 2021 at 10:55
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    $\begingroup$ There are surface that is not orientable (say, the Klein bottle), they cannot be given a structure of Riemann surface. $\endgroup$ Commented Mar 11, 2021 at 10:55

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A Riemann surface is just a complex manifold of complex dimension 1. In other words, a Riemann surface is a complex curve.

Now here's the point. A complex curve is of course a real smooth orientable manifold of dimension 2, but the converse is also true.

Any orientable 2-dimensional real smooth manifold $M$ is always going to be a complex manifold. Hence, $M$ always admits the structure of a complex curve. Here's how.

Since $M$ is orientable, it has a volume form $\omega$ which in this case is a nowhere vanishing 2-form. In particular, $\omega$ is closed since $\dim M=2$ and all $k$-forms on $M$ are zero for $k>2$. Hence, $(M,\omega)$ is a symplectic manifold. Its a standard result from symplectic geometry that every symplectic manifold admits an almost complex structure $J$. Now $(M,J)$ is a complex manifold iff $J$ is integrable, that is, its Nijenhuis tensor vanishes. The Nijenhuis tensor associated to $J$ takes two vector fields and outputs another vector field: $$ N(X,Y)=J[X,JY]+J[JX,Y]+[X,Y]-[JX,JY] $$ Let $p\in M$ and let $X$ be any vector field such that $X_p\neq 0$. Since $\dim M=2$, $X_p$ and $JX_p$ both span $T_pM$. Since $N$ is a tensor, $N_p=0$ iff $N(X,JX)=0$. By direct calculation, $$ \begin{align} N(X,JX)&=J[X,J^2X]+J[JX,JX]+[X,JX]-[JX,J^2X]\\ &=-J[X,X]+J[JX,JX]+[X,JX]+[JX,X]\\ &=0+0+[X,JX]-[X,JX]\\ &=0 \end{align} $$ Therefore $N_p=0$. Since $p\in M$ is arbitrary, we have $N\equiv 0$. By the Nijenhuis theorem, $(M,J)$ is a complex manifold.

So the takeaway from all this is that real smooth orientable manifolds of dimension 2 and Riemann surfaces are one and the same. (When I say real manifold I also mean without boundary.)

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    $\begingroup$ Can't you have more than one complex structure on a given real orientable 2d manifold though? $\endgroup$ Commented Mar 11, 2021 at 13:31
  • $\begingroup$ I'm not sure. To answer this question, you have to solve this problem. Suppose $(M,J_1)$ and $(M,J_2)$ are almost complex manifolds with $\dim M=2$. Hence, $J_1$ and $J_2$ are both integrable. Does there always exist a diffeomorphism $\varphi:M\rightarrow M$ such that $d\varphi \circ J_1 =J_2\circ d\varphi$? If yes, then $(M,J_1)$ and $(M,J_2)$ define the same complex structure. $\endgroup$ Commented Mar 11, 2021 at 13:40
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    $\begingroup$ Yes I agree. But assuming you cannot always find such a diffeomorphism... wouldn't it be misleading to say that "real smooth orientable manifolds of dimension 2 and Riemann surfaces are one and the same"? $\endgroup$ Commented Mar 11, 2021 at 13:41
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    $\begingroup$ If a real orientable 2-dim manifold admits distinct complex structures (good question), the more precise statement would be that a Riemann surface is a real orientable 2-dim manifold with a choice of almost complex structure. $\endgroup$ Commented Mar 11, 2021 at 13:46
  • $\begingroup$ There's only one complex structure on the sphere, while there are distinct complex sturctures on other high genus orientable surface. That's essentially uniformization theorem in complex analysis. $\endgroup$ Commented Mar 11, 2021 at 17:15

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