I'm sorry for the quantity of questions I'm asking, but I would like to solve once and for all many doubts I have on equivalent definitions of the moduli space of Riemann surfaces.

Definition 1: The Moduli space of Riemann surfaces of genus $g$ is

$\mathcal{M}_g:=\{$ Riemann surfaces of genus $g$ }/isomorfisms $\qquad$ ($\mathcal{M}_g$ here is just as a set)

I know that, for $g\ge 2$, we have this equivalent definition of $\mathcal{M}_g$:

Definition 2: $\mathcal{M}_g:=\{$hyperbolic metrics on $S_g$}/isometry $\qquad S_g$ is the oriented topological surface of genus $g$.

I've been told that, for $g\ge 2$, definition 1 and 2 are equivalent because of the Uniformization theorem, which says that a simply connected Riemann surface has conformal class of curvature 1 or 0 or -1. For what I've understood, for any Riemann surface $X$ of genus $g\ge 2$ we consider the universal cover $\widetilde{X}$ which is of conformal type -1 (is the hyperbolic plane $\mathbb{H}$) and the hyperbolic metric descends on $X$.

Question 1: Why is the conformal type of $\widetilde{X}$ -1? Why can't it be spheric or flat?

Question 2: Given an isomorphism class of Riemann surfaces, (i.e. a complex structure, up to isomorphism) how do I get (or compute, if you prefer) the hyperbolic metric? Given the hyperbolic metric, how do I get the complex structure?

I've also found this other definition for $g\ge 2$:

Definition 3: $\mathcal{M}_g:=\{$hyperbolic metrics on $S_g$}/orientation preserving diffeomorphisms

which brings me directly to the next question:

Question 3: Why are definition 2 and 3 equivalent? I.e why is it the same to consider hyperbolic metrics up to isometries or up to just orientation preserving diffeomorphisms?

Now I want to consider the case $g=1$. I've found this definition:

Definition 4: $\mathcal{M}_1:=\{$flat metrics on $S_1\}/$action of $\mathbb{R}^+\times Diffeo^+(S_1)$

I guest this too must be a conseguence of the Uniformization theorem, but I can't really get it:

Question 4: Why can't the universal cover of a Riemann surface of genus 1 be spheric or hyperbolic?

I know that the Riemann surface $X_\tau:=\mathbb{C}/(\mathbb{Z}+\tau\mathbb{Z})$, $\tau\in \mathbb{H}$, is isomorphic to $X_\eta$, $\eta\in\mathbb{H}$ if and only if there are integer numbers $a,b,c,d$ with $ad-bc=1$ such that $\eta=\frac{a\tau +b}{c\tau +d}$

Question 5: Given an isomorphism class of Riemann surfaces of genus 1, (and so by the above observation, an orbit of $\mathbb{H}$ by the action of $PSL(2,\mathbb{Z})$) how do I get (or compute, if you prefer) the flat metric? Given the flat metric, how do I get the complex structure?

Thank you very much!


Question 1: We know that from the uniformization theorem, the Riemann surface is covered by either one of these:

$$\tilde M = \mathbb S^2 , \mathbb C, \ \text{or }\mathbb H.$$

Then for any Riemann surface $M$, $M$ is biholomorphic to $\tilde M/\Gamma$, where $\Gamma$ is a subgroup of bilomorphism on $\tilde M$ and each $F\in \Gamma$ has no fixed points.

Hence the only Riemann surface that is covered by $\mathbb S^2$ is $\mathbb S^2$ itself: the reason is that all bihomorphisms (indeed, all orientation-preserving map) on $\mathbb S^2$ must have a fixed point. This forces $\Gamma = \{I\}$ and $M \cong \mathbb S^2$.

For $\mathbb C$, you can check that all biholomorphism $F:\mathbb C \to \mathbb C$ are rotations and translations. As rotations must have fixed points, all Riemann surfaces covered by $\mathbb C$ are biholomorphic to $\mathbb C/\Gamma$, where $\Gamma$ is a group of translation on $\mathbb C^2$. Then one can see that $M$ has to be a torus.

In particular, the above discussion shows that any surfaces of genus $\ge 2$ cannot be covered by $\mathbb S^2$ or $\mathbb C$. Thus by the uniformization theorem, they are covered by $\mathbb H$. (I haven't excluded the possibility that some genus one Riemann surfaces is covered by $\mathbb H$, which is your question 4: that can be ruled out after introducing the metric)

Question 2 and 5 First we show that given a Riemann surface, one can construct a metric with constant curvature.

Fix a Riemann surface $M$, we know that it is biholomorphic to $\tilde M/\Gamma$, where $\Gamma$ is a subgroup of biholomorphism. If we happen to find a metric on $\tilde M$ so that all biholomorphisms (without fixed points) are oriented isometries, then the metric on $\tilde M$ "descends" to $\tilde M/\Gamma$ and thus to $M$.

Indeed we have such a metric: For $\mathbb C$, we use the usual Euclidean metric. It is obvious that all translations are isometries. For $\mathbb H$, one can check that all biholomophisms are of the form

$$ z\mapsto \frac{az+b}{cz+d}, \ \text{where }a, b, c, d\in \mathbb R, ad-bc >0.$$

One can show that (see here) if we introduce the metric

$$ds^2 = \frac{1}{y^2} (dx^2 + dy^2),$$

then all such biholomorphisms are isometries. One can also calculate that this metric has constant curvature $-1$.

To sum up, if a Riemann surface $M$ is covered by $\mathbb C$ (resp. $\mathbb H$), then $M$ can be given a metric of constant curvature $0$ (resp. $-1$) so that $\mathbb C \to M$ (resp. $\mathbb H \to M$) is a local isometry.

Question 4 Now we can see that a genus one Riemann surfaces cannot be covered by $\mathbb H$: if not, it has a metric with $-1$ curvature. Then by the Gauss Bonnet theorem,

$$0 = 2-2g = \frac{1}{2\pi} \int_M K d\mu = - \frac{1}{2\pi} \text{Area}(M) \neq 0,$$

which is a contradiction.

Back to Question 2 and 5 Now we answer the other part of the question 2 and 5: Given a metric $g$ of constant curvature on $M$, how to construct on $M$ a complex structure?

Let $\tilde M$ be an universal cover of $M$. Then as $\pi : \tilde M \to M$ is a local diffeomorphism, one can pullbacks the metric to the universal cover $\tilde M$. Thus $M \cong \tilde M/\Gamma$, where $\Gamma$ is a group of isometries on $\tilde M$.

However, all simply connected Rieamnnian surfaces with constant curvature are classified. They are nothing but $\mathbb S^2, \mathbb C$ and $\mathbb H$. (with the obvious metrics). Thus one can define a complex structure on $M$ by that of $\tilde M$.

To sum up a little bit, we have:

  • Given a Riemann surfaces, we can construct a metric of constant curvature on $M$.

  • Given a surface with Riemannian metric $g$ of constant curvature, one can construct a complex structure.

One can check that this process are inverse to each other, thus

$\{$all Riemann surfaces$\} \leftrightarrow \{$all oriented Riemannian surfaces with constant curvature$\}$

As we have introduced the metric in a way that biholomorphism without fixed points corresponds to oriented isomtries, we have the equivalence of definition one and two.

Question 3 I am a bit skeptical about even the definition, given a hyperbolic metric $h$, and an arbitrary oriented diffeomorphism $f$, $f^*h$ might not be a hyperbolic metric.

  • 1
    $\begingroup$ To your skepticism of Question 3, if $f:M\to M$ is a diffeomorphism of smooth manifolds, and $M$ carries a Riemannian metric $g$, then by definition $f$ is an isometry mapping $(M,f^*g)\to (M,g)$. $\endgroup$ – Neal Sep 3 '16 at 6:07

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