# To which extent can one recover a manifold from its group of homeomorphisms

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 $M$?

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.

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).

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.

Another reference in the smooth case, sent to me by Beson Farb is the book by Augustin Banyaga, "The structure of classical diffeomorphism groups."

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This has really interesting applications in theoretical physics! For group actions on the discrete structures embedded in a manifold determined by causal dynamical triangulations (or other approaches to Quantum Gravity), connecting the topology to the group action is very important. –  Samuel Reid Mar 28 '12 at 6:03
@Samuel: any references for the things you mention? –  Igor Rivin Mar 29 '12 at 1:49

Answer is: Yes, one can recover $M$ if it is a compact manifold. See J. V. Whittaker: On Isomorphic groups and homeomorphic spaces, Annals of Math 1963.

EDIT 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

And references therein...

ANOTHER EDIT

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.

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Great answer, Igor! –  Misha Mar 28 '12 at 1:58
Thanks for these references! –  Samuel Reid Mar 28 '12 at 6:04
The paper by Whittacker can be found here webpages.ursinus.edu/nscoville/Whittaker%201963.pdf. It has the more general result that every isomorphism of abstract groups $\mathrm{Aut}(X) \cong \mathrm{Aut}(Y)$ is the conjugation of some homeomorphism $X \cong Y$, when $X,Y$ are compact manifolds. –  Martin Brandenburg Mar 28 '12 at 7:07
This result was generalized to the $C^r$-setting by Filipkiewicz in Isomorphisms between diffeomorphism groups. The article by Tomasz Rybicki mentioned by Igor, which can be found here ams.org/journals/proc/1995-123-01/S0002-9939-1995-1233982-7/… tries to enhance these results. –  Martin Brandenburg Mar 28 '12 at 7:13
Very erudite answer, Igor: thanks! –  Georges Elencwajg Mar 28 '12 at 9:37

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.

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 the rational homotopy type then in contrast with the diffeomorphism case the answer is actually NO.

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 $H_\ast(M,\mathbb Q)$.

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).

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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.) –  Ben Wieland Mar 28 '12 at 4:41
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. –  Vitali Kapovitch Mar 28 '12 at 4:58
@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. –  Misha Mar 28 '12 at 17:41
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.) –  Ben Wieland Mar 29 '12 at 16:39
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. –  Ben Wieland Mar 29 '12 at 18:12

For the smooth case, the result is in:

• Takens, F. (1979). Characterization of a differentiable structure by its group of diffeomorphisms. Bol. Soc. Brasil. Mat., 10, 17–25. MR552032

For completeness, Takens' theorem is:

Theorem 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.

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that's a much easier question though. The original question doesn't assume that the isomorphism between the diffeomorphism groups comes from a bijection of the underlying manifolds. –  Vitali Kapovitch Mar 28 '12 at 15:13
Results that Igor and Martin quote give complete answer in the smooth case. –  Misha Mar 28 '12 at 17:43
Misha: If you do the reference chase then the results that Igor and Martin quote use this result so this one is the original that sparked all of the others. I thought that worth mentioning. Vitali: I hadn't picked up on that point. However, when I posted then this question already had an accepted answer so I wasn't trying to provide an answer as such, rather I just thought this might be a useful addition to what was already there. –  Loop Space Mar 29 '12 at 6:55

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 $G$.

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).

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Okay, but there's no way you can extract $G_x$ given only $G$ as a topological group. –  Qiaochu Yuan Mar 28 '12 at 0:31
@Qiaochu, according to Igor's answer, you actually can! (in the compact case) It'd be cool to know how. –  Mariano Suárez-Alvarez Mar 28 '12 at 1:34