## Why is the mapping class group of hyperbolic manifolds finite?

Hi! I'm trying to understand why a hyperbolic n-manifold has finite mapping class group if $n \geq 3$. In books I'm reading it's said it's a consequence of Mostow's rigidity theorem: "If M and N are complete hyperbolic manifolds with finite total volume, any isomorphism of fundamental groups is realized by a unique isometry."

A corollary of this is that: " If M is hyperbolic (complete, with finite total volume) and $n \geq 3$, then Out($\pi_{1}(M)$) is a finite group, isomorphic to the group of isometries of M ".

But how could this could solve my problem? I mean, I know there is Dehn-Nielsen Theorem which states that Out($\pi_{1}(M)$) is isomorphic to MCG(M), but I know this to be true only in dimension 2...what can I say in dimension (at least) 3? Thank you.

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Hyperbolic manifolds have contractible universal covers (ie hyperbolic space), so they are $K(\pi,1)$'s. It is standard that there is a bijection between self-homotopy equivalences of $K(\pi,1)$'s and outer automorphisms of $\pi$. – Andy Putman May 30 2011 at 15:09
I know there is Dehn-Nielsen Theorem which states that $Out(\pi_1(M))$ is isomorphic to $MCG(M)$, but I know this to be true only in dimension 2...what can I say in dimension (at least) 3?'' Taking away the hypothesis that $M$ is hyperbolic turns this into a much more interesting question (in my humble opinion). An attempt at rephrasing it is here: mathoverflow.net/questions/66484/… – Dave Futer May 30 2011 at 19:31
@Lor: notice the difference between Agol's response and my own. Depending on how you define "mapping class group" the question you ask is quite different. My response was assuming "mapping class group" meant $\pi_0 HomotopyEquivalences(M)$. Ian's response is for $\pi_0 Homeo(M)$. I don't believe Ian's response applies to $\pi_0 Diff(M)$ provided the dimension of $M$ is large. – Ryan Budney May 30 2011 at 19:36

In dimension three, this was proven by Gabai, Meyerhoff, and N. Thurston. This was proven orginally for Haken 3-manifolds by Hatcher. Gabai extended this to hyperbolic 3-manifolds satisfying a certain technical condition, which was then verified for all closed hyperbolic 3-manifolds in the above paper. Gabai extended this result to prove that $Diff(M) \simeq Isom(M)$.

The analogous result in higher dimensions was proved by Farrell and Jones (see Theorem 5, I think this is only for dimension $>5$, but this isn't explicitly stated). Proofs are given here. I don't think dimensions $4$ or $5$ have been worked out.

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As Andy mentions, the isometry group $Isom(M)$ is isomorphic to $\pi_0 HomotopyEquiv(M)$ by Mostow rigidity. The isomorphism between $\pi_0 HomotopyEquiv(M)$ and $Out(\pi_1 M)$ is true for any $K(\pi,1)$-space, I believe this appears in Hatcher's Algebraic Topology book, but it's essentially the same as the argument you've seen for surfaces -- try comparing the two.

To prove that the isometry group of a finite-volume hyperbolic manifold is finite, there's a variety of ways. For example, consider the shortest geodesics in the manifold -- they have to be permuted by the isometry group, and then consider the stabilizer of that action. You've got a few special cases to consider but that's a start.

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