Given a closed, oriented surface $\Sigma$ of genus greater than 1, let $Mod(\Sigma)$ denote the mapping class group of orientation preserving diffeomorphisms of $\Sigma$ up to isotopy. Given any finite subgroup $F\subset Mod(\Sigma),$ the affirmative solution to the Nielsen Realization problem states that there exists a hyperbolic metric $h$ on $\Sigma$ such that $F$ is isomorphic to the isometry group of $h.$ Now, by Hurwitz's Theorem, the order of $F$ is bounded above by $84(g-1).$ I've always wondered if there was a topological/algebraic method to deduce a uniform bound on the order of finite subgroups of the mapping class group without going through the above mentioned argument. If $Mod(\Sigma)$ were a linear group over a field of characteristic zero, I suppose this would follow immediately from Selberg's Lemma, but I vaguely recall hearing that $Mod(\Sigma)$ has no faithful, finite dimensional linear representations.

So, my question is: is there a way to see that finite subgroups of the mapping class group have a uniform bound on their order without using the above argument, and if so how sharp can one make the bounds in this manner.

I've long been curious about this question and finally remembered to post it on mathoverflow. As always, thanks for your input.

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    $\begingroup$ "I vaguely recall hearing that Mod(Σ) has no faithful, finite dimensional linear representations." --- Actually, it's a major open problem whether or not mapping class groups are linear. $\endgroup$ – Dan Petersen Aug 31 '14 at 20:38
  • $\begingroup$ Thank you Dan, a google search drew up some notes of Benson Farb stating exactly this. $\endgroup$ – Andy Sanders Aug 31 '14 at 21:22

You can also use Serre's theorem which says that kernel of the natural homomorphism from the mapping class group of $\Sigma$ to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$ is torsion free, and therefore every finite subgroup injects to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$. But that gives a polynomial bound of degree $g^2$ compared to $84(g-1)$.

  • $\begingroup$ Thank you Lee. This is certainly the style of observation I had in mind. $\endgroup$ – Andy Sanders Aug 31 '14 at 15:52
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    $\begingroup$ @AndySanders: My pleasure. And I even fixed the typos. $\endgroup$ – Lee Mosher Aug 31 '14 at 15:54
  • $\begingroup$ I thought that Serre's result still uses Nielsen's realization for cyclic groups (due to Nielsen-Fenchel). Am I wrong ? $\endgroup$ – BS. Aug 31 '14 at 16:39
  • $\begingroup$ Well, I'm willing to give Nielsen realization for cyclic groups, there's no hyperbolic geometry there. $\endgroup$ – Andy Sanders Aug 31 '14 at 16:40

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