How is a Riemann surface different from an ordinary surface?

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?

• @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? Commented Mar 11, 2021 at 10:04
• "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 Commented Mar 11, 2021 at 10:22
• 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". Commented Mar 11, 2021 at 10:38
• Topologically, every Riemann surface is a surface in the usual sense. The difference lies in the complex structure. Commented Mar 11, 2021 at 10:55
• There are surface that is not orientable (say, the Klein bottle), they cannot be given a structure of Riemann surface. Commented Mar 11, 2021 at 10:55

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.)

• Can't you have more than one complex structure on a given real orientable 2d manifold though? Commented Mar 11, 2021 at 13:31
• 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. Commented Mar 11, 2021 at 13:40
• 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"? Commented Mar 11, 2021 at 13:41
• 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. Commented Mar 11, 2021 at 13:46
• 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. Commented Mar 11, 2021 at 17:15