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Smoothly embed a genus g surface in $\mathbb{R}^3$, and pick a normal vector pointing "out" of the surface at each point. Then on each tangent plane, I have a map which rotates the tangent plane 90 degrees in the direction given by the right hand rule. This gives an almost complex structure on the manifold, and every almost complex structure is integrable in real dimension 2, so this defines a complex structure on the surface.

Question 1: Does the above paragraph make sense? I am at the point where I think I can make grammatically correct sentences using the words above, but I am still not sure about all of the logical interconnections.

Say the first paragraph is correct. Then for each smooth embedding of the g-holed torus into $\mathbb{R}^3$ I get a complex structure on that surface. But I know that there are many different complex structures on the g-holed torus (and that the moduli space of such curves is 3g-3 complex dimensional for g>1).

Question 2,3,4: How many of these complex structures can I get through different embeddings? If I give you actual equations for an embedding, can you compute which point in the moduli space I am determining? In the particular case of the torus, can you tell me which $\tau$ corresponds to a given embedding?

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    $\begingroup$ Your first paragraph looks good to me! $\endgroup$
    – Paul Reynolds
    Sep 13, 2012 at 21:36
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    $\begingroup$ Please do not use the phrase "g-holed torus"! It's called a surface of genus g. (Even if you insist on using the word "hole", then a torus has one hole, so the phrase "g-holed torus" is nonsense). $\endgroup$
    – John Pardon
    Sep 13, 2012 at 21:49
  • $\begingroup$ It seems an almost identical question has been asked before by Igor Rivin: mathoverflow.net/questions/53999/… $\endgroup$
    – John Pardon
    Sep 13, 2012 at 21:50
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    $\begingroup$ Q1: yes, this is right. It doesn't even need to be 3-dimensional space, you can do this for embeddings (even immersions) into $\mathbb{R}^n$, as long as you also pick an orientation of the surface. The point is that the Riemannian metric on $\mathbb{R}^n$ pulls back to one on the embedded surface, and along with the orientation this lets you define "rotation by 90 degrees". In the special case of $\mathbb{R}^3$, the "outwards pointing normal vector" gives a canonical orientation: in higher dimensions this must be given additionally. $\endgroup$
    – Oscar Randal-Williams
    Sep 13, 2012 at 21:53
  • $\begingroup$ @unknown: I agree question 2 is the same as Igor Rivin's, but not the others. $\endgroup$
    – Steven Gubkin
    Sep 13, 2012 at 22:24

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Question 1: Looks good to me.

Question 2: is a duplicate of this question (the answer is: every conformal structure can be so realized).

Question 3/4. There are algorithms to compute the conformal structure given a surface, due to Sasha Bobenko and his collaborators, which work by discretizing the surface, and then computing periods. Looking at the list of arxiv preprints, I am not seeing something directly on this topic, so this may be in preparation, but you can write to him.

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  • $\begingroup$ Very cool. I actually asked this question because I want to make some 3D models illustrating a lot of Riemann surface theory, and making sure I could represent all the complex structures was the first step. Now I will have to see if I can animate them... $\endgroup$
    – Steven Gubkin
    Sep 13, 2012 at 22:22
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Your first paragraph does not only make sense but is a very important observation for studying the differential geometry of surfaces in 3-space: For special surfaces classes one can use the induced Riemann surface structure to reduce the differential geometric problem to an algebraic geometric one. Let me just mention some examples

  1. The most obvious one is about minimal surfaces in $\mathbb R^3.$ The (conformal) immersion is harmonic, and is determined (up to euclidean motions) by a pair of holomorphic spinors. Further restriction to special behavior at the ends led to an algebraic geometric problem.

  2. Willmore spheres turn out to be (after a moebius transformation) minimal spheres with planar ends. They are all given by special contact curves in $\mathbb CP^3.$ (see the surfaces papers of Bryant).

  3. CMC tori: Their Gauss map is harmonic. This turns out to be an integrable system. There exists an associated family of flat $SL(2,\mathbb C)$-connections, which is completely deterimend by a (special) compact Riemannn surface together with some additional spectral data, see Hitchin (Harmonic maps froma torus to $S^3$) and Pinkall Sterling (all CMC tori in $\mathbb R^3$).

Q2,3,4: For $g\geq 2$ I am not aware of any explicit embedding, nevertheless there are implicitely constructed surfaces, and for some of them one can compute the complex structure explicitely, see for example Lawson's minimal surfaces.

For tori, it depends (of course) how your formula for your embedding looks like. But there are many examples (even of geometric interesting tori) where you can compute the modulus $\tau$ explicitely. I think the easiest way to obtain all complex structures on a 2-torus through emebddings in $\mathbb R^3$ is given by a work of Pinkall (Hopf tori in S^3): The preimage of a (simple, closed) curve in $S^2$ under the Hopf fibration is a torus, and its modulus can be computeted easily in terms of the enclosed area and of the length of the curve. (After stereographic projection, which is conformal, you get a torus in $\mathbb R^3$.)

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Question 1 makes sense, and can be stated without "almost complex structure", just in classical terms. There is a Riemannian metric on your surface, induced from $R^3$. The distance between two points is the length of a rope on the surface stretched stretched between these two points. The length is measured as usual in $R^3$. Every Riemannian metric on a surface defines a complex structure. This is the same structure you are talking about.

I agree with Igor that every complex structure arises in this way.

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  • $\begingroup$ You also need orientability, which OP, I think, implicitly assumes when he says "genus g surface". $\endgroup$
    – Misha
    Sep 14, 2012 at 1:59

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