MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hello everyone,

Let $S_g$ be a compact orientable surface of genus $g \geq 2$, and let $\mathcal{A}$ be the set of $\mathcal{C}^{\infty}$ Riemanniann metric on $S_g$ endowed with the topology of uniform convergence.

Let $h$ be in $\mathcal{A}$. Since $g \geq 2$, $S_g$ is covered by $D^2 \simeq \mathbb{R}^2$, and $h$ can be pulled back on $D^2$ to $\tilde{h}$ in such a way that the projection $(D^2, \tilde{h}) \longrightarrow (S_g , h)$ is a local isometry, and $\pi_1(S_g) \simeq \Gamma \subset \text{Iso}((D^2,\tilde{h}))$ acts on $(D_2, \tilde{h})$ by isometry, so $S_g = D^2 / \Gamma$

My question is, how big is $\Gamma$ in $ \text{Iso}((D^2,\tilde{h}))$ ? For example if one restricts $\mathcal{A}$ to a relevant subset that can be endowed with a good probability measure (to be defined), can one say if $\Gamma = \text{Iso}((D^2,\tilde{h}))$ with probability $1$ ? Or if $[\Gamma :\text{Iso}((D^2,\tilde{h}))] < \infty $? More generally, can one compare $\Gamma$ and $ \text{Iso}((D^2,\tilde{h}))$ for a randomly chosen metric on $S_g$ in any way? I

t's clear for example that in the hyperbolic case, $\Gamma$ is very small in $\textbf{PSl}_2(\mathbb{R})$, but in the general case, one can expect that $\text{Iso}(D^2,\tilde{h})$ is small since the geometry on a random Riemannian manifold has very few symmetries.

Maybe this question is very classic, in that case I would be very gratefull if anyone could give me references on the topic. Thanks for the future answers !

share|cite|improve this question
The earlier MO question, "Random manifolds," contains some related information and references: – Joseph O'Rourke Jun 12 '13 at 13:10
up vote 5 down vote accepted

If you take any reasonable probability measure on $\mathcal{A}$, then with probability $1$ there should be a single point of maximum curvature ; then the isometry group of $\tilde h$ must preserve a $\Gamma$-orbit, namely the set of the maximum curvature points in the universal covering. You should also be able to single out one direction at the maximum curvature point by the same kind of consideration, showing that indeed the isometry group above is reduced to $\Gamma$.

To make all this precise, you would have to define your random metrics, but the same kind of reasoning should work for generic metric in a reasonable topology.

share|cite|improve this answer

It follows from Theorem 1.3 of the paper of Farb and Weinberger, "Isometries, rigidity and universal covers", MR2456886 that either $\text{Iso}((D^2,\tilde{h}))$ is discrete in which case it contains $\Gamma$ with finite index, or the metric $h$ has constant negative curvature in which case, after scaling the metric, we may take $(D^2,\tilde h)$ to be the Poincare metric and $\Gamma$ to be a Fuchsian group.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.