Teichmuller volume of moduli space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:45:45Z http://mathoverflow.net/feeds/question/56034 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56034/teichmuller-volume-of-moduli-space Teichmuller volume of moduli space Igor Rivin 2011-02-20T02:59:00Z 2011-02-20T06:00:05Z <p>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.</p> http://mathoverflow.net/questions/56034/teichmuller-volume-of-moduli-space/56039#56039 Answer by Gjergji Zaimi for Teichmuller volume of moduli space Gjergji Zaimi 2011-02-20T05:33:45Z 2011-02-20T05:33:45Z <p>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".</p> http://mathoverflow.net/questions/56034/teichmuller-volume-of-moduli-space/56041#56041 Answer by Stergios for Teichmuller volume of moduli space Stergios 2011-02-20T05:55:00Z 2011-02-20T06:00:05Z <p>The answer is YES, the volume of the moduli space is finite with respect to the Teichmuller metric.</p> <p>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. </p> <p>The argument goes as follows: </p> <p>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 $((\Delta^{*})^{k} \times \Delta^{n-k} )/ G$. </p> <p>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 $\overline{M(S)}$.</p> <p>You can look at Curt McMullen's paper : <a href="http://www.math.harvard.edu/~ctm/papers/home/text/papers/kahler/kahler.pdf" rel="nofollow">http://www.math.harvard.edu/~ctm/papers/home/text/papers/kahler/kahler.pdf</a> for more details and references. (Proof of Theorem 8.1)</p>