Timeline for To which extent can one recover a manifold from its group of homeomorphisms
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Mar 29, 2012 at 18:12 | comment | added | Ben Wieland | My comment was about the homotopy category. However, the LES of homotopy groups of Homeo(M,p) -> Homeo(M) -> M and shows that the kernel is the quotient of the fundamental group by an abelian group. The aspherical assumption and comparison with the similar sequence for Aut(M) shows that the abelian group is contained in the center. | |
Mar 29, 2012 at 16:59 | comment | added | Vitali Kapovitch | @Ben Wieland are you talking about the mapping class groups in the homotopy category? Then I understand why what you say holds but I don't know why this would be true for the mapping class group in the homeomorphism category which I think is what Misha wanted to know. Is it known/true there? I can see that it would follow if we knew that $\pi_0(Homeo(M))\to \pi_0(Aut(M))$ is an isomorphism. I believe this is known for many (all?) aspherical 3-manifolds but for dim>3 one needs the Borel conjecture (for onto) and that homotopic homeomorphisms of $M$ are istotopic (for 1-1). | |
Mar 29, 2012 at 16:39 | comment | added | Ben Wieland | Yes, the groups are the automorphism group and its outer quotient, so the kernel is the group modulo its center. (I originally excluded center so that the components are contractible. In general, they are the classifying space of the center.) | |
Mar 28, 2012 at 17:41 | comment | added | Misha | @Ben Wieland: In the case of mapping class groups surfaces, kernel of the homomorphism $Mod(S,p)\to Mod(S)$ is $\pi_1(S)$. Here $Mod(S,p)$ is the "pointed" mapping class group. Do you know if the same holds in higher dimensions? If so, then one can recover $\pi_1(M)$ from the two mapping class groups. | |
Mar 28, 2012 at 4:58 | comment | added | Vitali Kapovitch | that's a nice observation. I was only thinking about the simply connected case which I think is more difficult here but your example certainly shows that in the non simply connected case the homotopy type of the monoid of self equivalences of $M$ does not determine the homotopy type of $M$. I still wonder whether it's true in the simply connected case. | |
Mar 28, 2012 at 4:41 | comment | added | Ben Wieland | If the manifold is aspherical and the fundamental group has trivial center, then the monoid of homotopy self-equivalences has contractible components and the set of components is the outer automorphism group of the fundamental group. This is often trivial. (Maybe this is a silly example, since the base-point preserving version yields the automorphism group, which might recover the group.) | |
Mar 28, 2012 at 3:58 | history | answered | Vitali Kapovitch | CC BY-SA 3.0 |