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

Someone asked me this question, and I was embarrassed to not know the answer: is the volume of Moduli space with respect to the Teichmuller metric finite? The answer is "yes" when we replace Teichmuller metric with Weil-Petersson metric, but the geometry of the two spaces is quite different.

share|cite|improve this question
Thanks! I don't suppose any actual estimates on the volume are available... – Igor Rivin Feb 20 '11 at 18:16
up vote 9 down vote accepted

The answer is YES, the volume of the moduli space is finite with respect to the Teichmuller metric.

The reason is the theorem of Royden, that the Kobayashi metric on Teich(S) coincides with the Teichmuller metric, and the fact that the moduli space $M(S)$ associated to S has a nice compactification $\overline{M(S)}$, the Deligne-Mumford compactification.

The argument goes as follows:

For a stable curve Z in $\overline{M(S)}$ with k nodes, you can find a neighborhood U of Z such that U is locally $\Delta^n /G$, where $\Delta$ is the unit disc in $\mathbb{C}$ and $G$ is a finite group. Then $U\cap M(S)$ is locally isomorphic to $((\Delta^{*})^{k} \times \Delta^{n-k} )/ G$.

The volume of $(\Delta^{*})^{k} \times \Delta^{n-k} $ near the origin is finite in the Kobayashi metric. Since inclusions contract the Kobayashi metric it follows that there is a small neighborhood V of $Z \in \overline{M(S)}$ such that volume of $V\cap M(S)$ is finite. The result now follows by compactness of $\overline{M(S)}$.

You can look at Curt McMullen's paper : for more details and references. (Proof of Theorem 8.1)

share|cite|improve this answer

The answer is again yes. See the proof of theorem 8.1 in Curtis McMullen's paper "The moduli space of Riemann surfaces is Kahler hyperbolic".

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.