# Finite subgroups of mapping class groups

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 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. Aug 31, 2014 at 20:38
• Thank you Dan, a google search drew up some notes of Benson Farb stating exactly this. Aug 31, 2014 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 $84(g-1)$.