Elements of finite order in mapping class groups of high dimensional manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:24:24Z http://mathoverflow.net/feeds/question/77868 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77868/elements-of-finite-order-in-mapping-class-groups-of-high-dimensional-manifolds Elements of finite order in mapping class groups of high dimensional manifolds unknown (google) 2011-10-11T21:35:02Z 2011-10-12T10:16:25Z <p>Let $M$ be a manifold with boundary. Consider the following groups:</p> <p>(1) $\pi_0(\operatorname{Diff}(M,\partial M))$.</p> <p>(2) $\pi_0(\operatorname{Homeo}(M,\partial M))$.</p> <p>(3) $\pi_0(\operatorname{HomEq}(M,\partial M))$.</p> <p>That is, isotopy (resp. isotopy and homotopy) classes of diffeomorphisms (resp. homeomorphisms and homotopy equivalences) relative to $\partial M$.</p> <blockquote> <p>I would like to find conditions on $M$ which guarantee that one of the groups (1)/(2)/(3) has no element of finite order.</p> </blockquote> <p>My motivation for this is to generalize from the case $\dim M=2$, when as long as $\partial M\ne\varnothing$, none of the groups (1)/(2)/(3) has an element of finite order (This is left as an exercise for the reader. Hint: an element of finite order in $\operatorname{MCG}(\Sigma_g)$ fixes some hyperbolic structure).</p> <p>Of course, if $\dim M=2$, then (1)=(2)=(3). This question is really about finding a proper generalization of the result for $\dim M=2$ to higher dimensions, so I'm leaving it open as to which of (1)/(2)/(3) this question is really about. We remark that (1) seems unlikely to be the right group to consider; for instance when $(M,\partial M)=(D^n,S^{n-1})$ and $n>4$, then it is the group of exotic spheres in dimension $n+1$.</p> http://mathoverflow.net/questions/77868/elements-of-finite-order-in-mapping-class-groups-of-high-dimensional-manifolds/77873#77873 Answer by Ryan Budney for Elements of finite order in mapping class groups of high dimensional manifolds Ryan Budney 2011-10-11T22:20:46Z 2011-10-11T22:20:46Z <p>Here's one case where you can answer (3). </p> <p>Let $M$ be a compact hyperbolic manifold of dimension $\geq 3$. By Mostow rigidity, $HomEq(M)$ has the same homotopy-type as its isometry group, a finite group. By design, $Isom(M) \simeq Out(\pi_1 M)$. This latter isomorphism is because $M$ is a $K(\pi,1)$ space. </p> <p>So $\pi_0 HomEq(M)$ contains elements of finite order if and only if $Out(\pi_1 M)$ does, if and only if this hyperbolic manifold has symmetries. </p> http://mathoverflow.net/questions/77868/elements-of-finite-order-in-mapping-class-groups-of-high-dimensional-manifolds/77913#77913 Answer by Jeffrey Giansiracusa for Elements of finite order in mapping class groups of high dimensional manifolds Jeffrey Giansiracusa 2011-10-12T10:16:25Z 2011-10-12T10:16:25Z <p>In the case of simply connected 4-manifolds, a famous theorem of Michael Friedman asserts that $\pi_0 Homeo(M)$ is isomorphic to the group of automorphisms $Aut(Q)$ of the intersection quadratic form $Q$ on the middle homology. This is an arithmetic group, and hence it contains a finite index subgroup that is torsion free. However, $Aut(Q)$ will almost always have torsion, except possibly in some low-rank cases.</p> <p>One way of thinking about this is: For surfaces, the map from MCG to $Aut(Q)=Sp_{2g}(\mathbb{Z})$ (sending a homeomorphism to the induced automorphism of homology) has a huge kernel, namely the Torelli group. But in dimension 4 under the simply connected hypothesis this map is an isomorphism. </p>