## Return to Answer

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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 $((\Delta^{})^{k} ((\Delta^{*})^{k} \times \Delta^{n-k} )/ G$.

The volume of $(\Delta^{})^{k} (\Delta^{*})^{k} \times \Delta^{n-k}$ near the origin is finite in the Kobayashi metric. Since inclusion contracts 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 : http://www.math.harvard.edu/~ctm/papers/home/text/papers/kahler/kahler.pdf for more details and references. (Proof of Theorem 8.1)

1

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 inclusion contracts 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 : http://www.math.harvard.edu/~ctm/papers/home/text/papers/kahler/kahler.pdf for more details and references. (Proof of Theorem 8.1)
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