Why is the mapping class group of hyperbolic manifolds finite? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:34:02Z http://mathoverflow.net/feeds/question/66459 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66459/why-is-the-mapping-class-group-of-hyperbolic-manifolds-finite Why is the mapping class group of hyperbolic manifolds finite? Lor 2011-05-30T14:57:54Z 2011-05-30T19:27:02Z <p>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."</p> <p>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 ".</p> <p>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.</p> http://mathoverflow.net/questions/66459/why-is-the-mapping-class-group-of-hyperbolic-manifolds-finite/66477#66477 Answer by Ryan Budney for Why is the mapping class group of hyperbolic manifolds finite? Ryan Budney 2011-05-30T18:21:14Z 2011-05-30T18:21:14Z <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/66459/why-is-the-mapping-class-group-of-hyperbolic-manifolds-finite/66485#66485 Answer by Ian Agol for Why is the mapping class group of hyperbolic manifolds finite? Ian Agol 2011-05-30T19:27:02Z 2011-05-30T19:27:02Z <p>In dimension three, this was proven by <a href="http://www.ams.org/mathscinet-getitem?mr=1973051" rel="nofollow">Gabai, Meyerhoff, and N. Thurston</a>. 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 $Diff(M) \simeq Isom(M)$. </p> <p>The analogous result in higher dimensions was proved by <a href="http://www.ams.org/journals/bull/1988-19-01/S0273-0979-1988-15640-6/home.html" rel="nofollow">Farrell and Jones</a> (see Theorem 5, I think this is only for dimension $>5$, but this isn't explicitly stated). Proofs are given <a href="http://www.ams.org/mathscinet-getitem?mr=973309" rel="nofollow">here</a>. I don't think dimensions $4$ or $5$ have been worked out. </p>