Explicit example of recovering a manifold from its group of homeomorphisms

Question

Due to this MO post, it is known that

one can recover a closed connected manifold from its group of homeomorphisms uniquely.

I want to see this fact by some concrete examples. Fore example what is the $n$-dimensional closed and connected manifold that its homeomorphisms group $\mathrm{Homeo}(M)$ is $\mathrm{SL}(k,\Bbb R)$ for some appropriate $k$?

Answer 1

To complete the discussion:

1. The group $G=Homeo_+({\mathbb R})$ contains a finitely generated subgroup $\Gamma$ which is not isomorphic to any matrix group:

B.Farb, J.Franks, Groups of homeomorphisms of one-manifolds, I: actions of nonlinear groups, https://arxiv.org/pdf/math/0107085.pdf.

Here $Homeo_+$ denotes the subgroup of orientation-preserving homeomorphisms.

2. The group $Homeo_+({\mathbb R})$ is isomorphic to the group of $Homeo_+(I)$, where $I$ is the unit interval. (Use the fact that $I$ is homeomorphic to $[-\infty, \infty]$.) The group $Homeo_+(I)$ embeds naturally in the group of homeomorphisms of the cube $I^n= I\times I^{n-1}$ fixing $\partial I\times I^{n-1}$ pointwise:

Extend each homeomorphism by the identity to $I^{n-1}$.

Let $G$ denote the image of this embedding of $Homeo_+(I)$ in $Homeo(I^n)$.

Define the following equivalence relation $\sim$ on $I^n$: For each $q\in \partial I^{n-1}$, the product $I\times \{q\}$ is one equivalence class. The rest of the equivalence classes are singletons. Then quotient space $I^n/\sim$ is homeomorphic to the closed $n$-dimensional ball $B^n$. The continuous action of $G$ on $I^n$ descends to a continuous action of $G$ on $B^n=I^n/\sim$, fixing $\partial B^n$ pointwise.

3. From (2) one sees that if $M$ is any manifold of dimension $n\ge 1$, then $G$ embeds in $Homeo(M^n)$: Embed $B^n$ in $M^n$ and extend the action of $G$ on $B^n$ as above by the identity to the rest of $M$.

4. From (1) & (3), one concludes that if $M$ is a manifold of dimension $\ge 1$, then $Homeo(M)$ cannot be isomorphic (as an abstract group) to any group of matrices.

5. To conclude: there is no manifold of dimension $\ge 1$ whose group of homeomorphisms is isomorphic to $GL(k, {\mathbb R})$.