Intuition behind moduli space of curves

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For a genus g compact smooth surface $M$, an algebraic structure is the same as a complex structure is the same as a conformal structure. So the moduli space of smooth curves should be the same as the moduli space of conformal structures on $M$. A conformal structure is an equivalence class of Riemmanian metrics that give the same angle measurements.

If I embed $M$ in $\mathbb{R}^3$, I get a metric on $M$. This gives me a conformal structure, hence a point in the moduli space. The moduli space is known to have complex dimension 3g-3 (except for g=0, where the moduli space is a point, and g=1, where the moduli space is 1-dimensional).

My question is: can we visualize the 6g-6 real dimensions as deformations of the embedding of $M$ in $\mathbb{R}^3$. In particular, how can we see that small deformations of a 2-sphere are conformally equivalent to the original 2-sphere.

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Great question! – Sam Nead Dec 4 2009 at 17:23

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Although the space of deformations is 6g-6 for g>1, the machinery used to prove that is quite different from the machinery used to prove that all 2-spheres are conformally equivalent. In particular, the latter is one branch of the uniformization theorem and intuition about how that applies to the specific case you ask about, 2-spheres embedded in 3D, probably won't give intuition about the g>1 case without quite a bit of further work. – Dan Piponi Dec 4 2009 at 19:01
Is it even clear what subspace of the space of conformal structures can be realized by surfaces embedded in $\mathbb R^3$? Do all hyperbolic metrics admit conformal embeddings in $\mathbb R^3$? – Ryan Budney Dec 5 2009 at 3:28

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My intuition is that if you wrinkle the surface, you can basically decouple its conformal class from the macroscopic geometry of the embedding. It might not be easy to develop this idea, and I also have a vague recollection that there has been a paper on this. – Greg Kuperberg Dec 5 2009 at 3:38

5 Answers

There is one thing that you can "see" when you have a surface embedded in $R^3$. Namelly, the moduli space of conformal (complex) structures on a surface of genus $g$, $g>0$ is non-compact. If you take in $R^3$ a surface that has a very long cylinder inside of it, such that the circle generated by it on the surface is not contractible, then in the moduli space of complex curves you get a point "close to its boundary". The longer the cylinder close you get to the boundary. En example will be a long and thin torus.

Considering such pictures you will be able to viualise at least points of the moduli space that are the most close to the boundary in the case of arbirtary genus.

Over thing you could destinguish are complex curves that admit an anti-holomorphic involution - just take a surface in $R^3$ symmetric with respect the $x,y$ plane.

As for small deformation of the sphere, unfotrunatelly I don't think it will be any easier to "see" that it is the same as on any other sphere. You will need PDE to prove it. Even to prove that it has a holomorphic sturcture you will need to know that for any metric on a 2-dimesnional surface you have local isotermal coordinates, where it looks as $f(x,y)(dx^2+dy^2)$.

ADDED.

By Pogorelov's theorem every sphere with positive Gauss curvature can be emdedded isometrically in R^3. So a sphere close to a unite sphere is just a sphere whose curvature is approximatively 1. Knowing this does not make it any easier for you to know that the conformal structure on this sphere is standard, according to what I understand.

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I think sigfpe's comment is correct -- the formula $3g-3$ is only true for $g \geq 2$, and however you can "visualize" or understand these $3g-3$ dimensions for $g \geq 2$ probably won't work or help for $g=0,1$.

I'm not sure if this will help, but one way (the only way that I know, but there are others) to "visualize" the $6g-6$ dimensions is using the Fenchel-Nielsen coordinates. Roughly it works like this: You can take your genus $g \geq 2$ surface and cut it along $3g-3$ circles to get a bunch of pairs of pants ($2g-2$ of them). Then $3g-3$ of the $6g-6$ dimensions come from the lengths of the circles; the remaining $3g-3$ dimensions come from how much we "twist" when we identify the pants along the circles. A key fact which makes this work is that, for each specification of the lengths of the boundary circles, there is a unique pair of pants (unique complex structure). If you google "Fenchel-Nielsen coordinates" I am sure that you will find lots of papers with nice pictures illustrating this, and explaining this in more detail.

Anyway, you'll notice that you can't do this for $g=0,1$. A sphere can't be cut into pairs of pants, nor can a torus.

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This is related to Kevin's answer but goes back to the 19th century. A Riemann surface is planar if a simple closed curve separates it. A Riemann surface of genus g becomes planar after cutting g handles along simple closed curves. Then the classical uniformaization theorem states that it is equivalent to a sphere with 2g slits parallel to the "x axis". Then the location of the slits is determine by the 2g centers (4g real coordinates) and the 2g lengths (2g real coordinates). This give 6g real coordinates, which are then reduced by the 6 real dimensions of the conformal automorphisms of the sphere. Thus Riemann surfaces of genus g have 6g-6 real parameters, plus the number of parameters of the conformal automorphisms of the generic surface of genus g, which adds 6 for g=0 and 2 for g=1. (Alan Mayer explained it this way in a colloquium at Brandeis about 1967.)

As usual, induction is the easiest way to see anything, and here one degenerates a Riemann surface to one of lower genus by acquiring a node (shrinking off a loop). Then the lower genus Riemann surface has 3(g-1)-3 complex parameters. the two identified points have 2 parameters, and we add one more for the degeneration direction. (Such degeneration methods occur in Wirtinger's book on theta functions in 1895.)

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(EDIT 1: Replaced hand-waving argument in third paragraph with a hopefully less incorrect version)

(EDIT 2: Added final paragraph about obtaining all conformal deformations for surfaces other than sphere.)

I think it is possible to see the infinitesimal rigidity of the sphere, even if it does involve a PDE as Dmitri says. I think you can also try and see if for other embedded surfaces, all infinitesimal deformations of conformal structure are accounted for by deformations of the embedding in a similar way.

For the case of S2, what you want is to do is take a normal vector field V (i.e. infinitesimal change of embedding) and produce a tangent vector field X such that flowing along X gives the same infinitesimal change in conformal structure as flowing along V. This should amount to solving a linear PDE, so as Dmitri says a PDE is definitely involved, but probably not as hard as proving the existence of isothermal coordinates (which from memory is non-linear). For the standard embedding of S2 there can't be too many choices for this linear differential operator given that it has to respect the SO(3)-symmetry.

I guess we're looking for a first-order equivariant linear operator from normal vector fields to tangent vector fields. If we identify normal fields with functions then two possible candidates are to take X=grad V or X to be the Hamiltonian flow generated by V. I can't think of any others and probably it's possible to prove these are the only such ones. (Assuming it's elliptic, the symbol of the operator must be an SO(3)-equivariant isomorphism from T*S2 to TS2 and there can't be too many choices! Using the metric leads to grad and using the area form leads to the Hamiltonian flow.) Then you just have to decide which one to use.

For the case of a general embedded surface $M$, you can ask "is it possible to obtain all deformations of conformal structure by deforming the embedding into R3?" To answer this we can again think of a normal vector field as a function V on the surface. There is a second-order linear differential operator $$D\colon C^\infty(M) \to \Omega^{0,1}(T)$$ which sends a normal vector field to the corresponding infinitesimal change of conformal structure. This operator will factor through the hessian with a homomorphism from `$T^* \otimes T^*$` to `$T^{*0,1}\otimes T^{1,0}$`. The operator $D$ will not be onto, but what we want to know is whether every cohomology class in $H^{0,1}(T)$ has a representative in the image of $D$. At least, this is how I would try and approach the question; I'm sure there are other methods.

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You should have a look into the paper "Constrained Willmore Surfaces" by Bohle, Pinkall, Peters. There they give the formula (easy to derive) that the infinitesimal change of the complex structure $\cdot J$ by an infinitesimal normal variation $\cdot f=u N$ (N being the unit normal of the surface) is given by $2uA^\circ J,$ where $A^\circ$ is the trace free part of the Weingarten operator. Especially, in the case of the round sphere, you see that an infinitesimal normal variation does not change the complex structure.

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