To which extent can one recover a manifold from its group of homeomorphisms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:17:28Z http://mathoverflow.net/feeds/question/92422 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92422/to-which-extent-can-one-recover-a-manifold-from-its-group-of-homeomorphisms To which extent can one recover a manifold from its group of homeomorphisms Misha 2012-03-27T23:35:52Z 2012-03-29T13:30:31Z <p>Question. Suppose that $M$ is a closed connected topological manifold and $G$ is its group of homeomorphisms (with compact-open topology). Does $G$ (as a topological group) uniquely determine $M$? </p> <p>One can ask the same question where we regard $G$ as an abstract group (ignoring topology), replace topological category by smooth category (here one can equip $G=Diff(M)$ with a finer structure of a Frechet manifold), varying degree of smoothness, dropping compactness assumption, recovering $M$ up to homotopy, etc. </p> <p>I do not know how to answer any of these questions. I do not even know if one can recover the dimension of $M$ from its group of homeomorphisms. In low dimensions, or assuming that $M$ has a locally-symmetric Riemannian metric, and if $dim(M)$ is given, I know few things. For instance, among 2-dimensional manifolds one can recover $M$ from $G$ since $G/G_0$ is the mapping class group $Mod(M)$ of $M$ and one can tell the genus of $M$ from maximal rank of free abelian subgroups of $Mod(M)$. Same for, say, closed hyperbolic manifolds with non-isomorphic isometry groups. However, given, for instance, two closed hyperbolic 3-manifolds $M_1, M_2$ with trivial isometry groups, I do not know how to distinguish $M_i$'s by, say, $Homeo(M_i)$ (the problem reduces to a question about homeomorphism groups of the unit ball commuting with $\pi_1(M_i)$, $i=1,2$, but I do not see how to solve it). </p> <p>Update: Results quoted by Igor and Martin give the complete answer in topological and smooth category in the strongest possible form (much more than I expected!). Positive answer is also known in the symplectic category, but, apparently, is open for contact manifolds and their groups of contactomorphisms. </p> <p>Another reference in the smooth case, sent to me by Beson Farb is the book by Augustin Banyaga, "The structure of classical diffeomorphism groups." </p> http://mathoverflow.net/questions/92422/to-which-extent-can-one-recover-a-manifold-from-its-group-of-homeomorphisms/92426#92426 Answer by Craig Westerland for To which extent can one recover a manifold from its group of homeomorphisms Craig Westerland 2012-03-28T00:30:17Z 2012-03-28T00:30:17Z <p>I suppose that the answer to this question depends upon how much information you are willing to allow yourself to extract from $G$. Since your manifold is connected, $G$ acts transitively upon it, and so if $x \in M$ is any point, and $G_x$ the stabiliser of $x$ in $G$, then there is a homeomorphism $G/G_x \cong M$. So $M$ can be completely reconstructed from $G$.</p> <p>To be fair, though, this presupposes that you have a very good understanding of $G$ and its subgroups, perhaps more than is reasonable. Furthermore, this is not the sort of information that is preserved by passing to the mapping class group (e.g., a surface is clearly not homeomorphic a quotient of its mapping class group).</p> http://mathoverflow.net/questions/92422/to-which-extent-can-one-recover-a-manifold-from-its-group-of-homeomorphisms/92427#92427 Answer by Igor Rivin for To which extent can one recover a manifold from its group of homeomorphisms Igor Rivin 2012-03-28T00:30:26Z 2012-03-29T01:43:23Z <p>Answer is: Yes, one can recover $M$ if it is a compact manifold. See <a href="http://dl.dropbox.com/u/5188175/whit.pdf" rel="nofollow">J. V. Whittaker: On Isomorphic groups and homeomorphic spaces, Annals of Math 1963.</a></p> <p><strong>EDIT</strong> Actually, one knows a lot more, see, for example Tomasz Rybicki Journal: Proc. Amer. Math. Soc. 123 (1995), 303-310. MSC: Primary 58D05; Secondary 17B66, 22E65, 57R50 MathSciNet review: 1233982 </p> <p>And references therein...</p> <p><strong>ANOTHER EDIT</strong></p> <p>A quite different proof of a stronger theorem (actually a large set of theorems) than Whittaker's (actually, Whittaker's paper seems to be rather badly written) is given by Matatyahu Rubin in Rubin, Matatyahu(3-SFR) On the reconstruction of topological spaces from their groups of homeomorphisms. Trans. Amer. Math. Soc. 312 (1989), no. 2, 487–538. </p> http://mathoverflow.net/questions/92422/to-which-extent-can-one-recover-a-manifold-from-its-group-of-homeomorphisms/92438#92438 Answer by Vitali Kapovitch for To which extent can one recover a manifold from its group of homeomorphisms Vitali Kapovitch 2012-03-28T03:58:01Z 2012-03-28T03:58:01Z <p>This is more of a longish comment rather but I'd like to point out that while Igor's reference in principle gives a complete answer, actually reading off any specific information about $M$ (such as its dimension) from some topological invariants of $Diff(M)$ is likely hard. </p> <p>Moreover, if one relaxes the categories somewhat then the answer to Misha's question can even be negative! Specifically, one can ask if the homotopy type of the monoid of self homotopy equivalences of $M$ determines $M$ up to homotopy type. I actually don't know the answer to this but if one relaxes the category even further and looks at <em>the rational homotopy type</em> then in contrast with the diffeomorphism case the answer is actually <strong>NO</strong>.</p> <p>Specifically, it's rather easy to compute that the rational homotopy type of the identity component $Aut(M)$ of the monoid of self homotopy equivalences of an equal rank biquotient of Lie groups $M=G//H$ that satisfies Halperin's conjecture (which says that in this case $H^\ast (M,\mathbb Q)$ has no negative degree derivations) is determined by rational homotopy and homology groups of $M$. In this case $Aut(M)$ is rationally equivalent to a product of finitely many odd dimensional spheres and one can write an explicit (if somewhat ugly) formula for the dimensions of the spheres that show up in terms of $\pi_\ast(M)\otimes \mathbb Q$ and $H_\ast(M,\mathbb Q)$.</p> <p>But there are plenty of examples of such biquotients in dimensions above 5 which have distinct rational types but the same rational homotopy and homology. For example, one can take $G//T$ where $G$ is a simply connected Lie group and $T\le G\times G$ is a torus of the same rank as rank $G$. All such biquotients satisfy Halperin's conjecture so the formula I mention above applies. It is then clear that rational homology and homotopy groups of $G//T$ are completely determined by $G$ but the rational type of $G//T$ can be different depending on the embedding $T\to G\times G$. There are infinitely many such examples already in dimension 6 of the form $(S^3\times S^3\times S^3)//T^3$. Still, in this case one can read off for example, the dimension of $M$ from the knowledge of the rational homotopy groups of $Aut(M)$ but I don't know how to get such formula for a general closed simply connected manifold $M$ (and I'm not even sure if it's possible).</p> http://mathoverflow.net/questions/92422/to-which-extent-can-one-recover-a-manifold-from-its-group-of-homeomorphisms/92449#92449 Answer by Andrew Stacey for To which extent can one recover a manifold from its group of homeomorphisms Andrew Stacey 2012-03-28T11:22:37Z 2012-03-28T11:22:37Z <p>For the smooth case, the result is in:</p> <ul> <li>Takens, F. (1979). Characterization of a differentiable structure by its group of diffeomorphisms. <em>Bol. Soc. Brasil. Mat.</em>, <em>10</em>, 17&ndash;25. <a href="http://www.ams.org/mathscinet-getitem?mr=552032" rel="nofollow">MR552032</a></li> </ul> <p>and the answer is "Yes".</p> <p>For completeness, Takens' theorem is:</p> <p><strong>Theorem</strong> Let $\Phi \colon M_1 \to M_2$ be a bijection between two smooth $n$-manifolds such that $\lambda \colon M_2 \to M_2$ is a diffeomorphism iff $\Phi^{-1} \circ \lambda \circ \Phi$ is a diffeomorphism. Then $\Phi$ is a diffeomorphism.</p>