Local Contractibility of Homeomorphism Group of $\mathbb{R}^n$

Question

For this question, let $X = \mathbb{R}^n$, and let $Aut(X)$ be the group of self-homeomorphisms with the compact-open topology.

Then I know that $Aut(X)$ is both locally path-connected, and locally contractible. I have a decent sense of the implications of local path-connectedness: Two homeomorphisms which are sufficiently close in the compact-open topology can be joined by an isotopy of $X$.

But I don't have a good sense of what local contractibility means for this group. I've read that spaces very similar to $X$, for example Euclidean Neighborhood Retracts, don't necessarily have locally contractible homeomorphism groups. Why would that be?

What's the geometric interpretation of local contractibility for these spaces? How do you use it?

Thanks! Sorry it's a bit of a 'soft question.'

Answer 1

https://mathoverflow.net/questions/388706/do-locally-contractible-path-connected-groups-have-accessible-bases/388966?noredirect=1#comment990788_388966

Ok, it turns out that $Aut(\mathbb{R}^2)$ is locally contractible in the strong sense after all. Y'all were making me so paranoid! But it was a fun reference-hunt and makes me want to go back and study some of that old school retract theory stuff, I always tried to avoid it.

As per the answer in the MO thread, the space is locally homeomorphic to separable Hilbert space, so it satisfies everything you could ever want.

Answer 2

According to Černavskiĭ, the local contractibility of $Homeo(M)$ for a noncompact topological manifold $M$ implies that $M$ could be the interior of a compact manifold with boundary. Of course this is true when $M=\mathbb{R}^n$. For a statement see p.8, Th.2, of his paper

A. Černavskiĭ, Local Contractibility of the Group of Homeomorphisms of a Manifold, 1969 Math. USSR Sb. 8 287.

Notes: If $M$ is a metrisable topological manifold, then $Homeo(M)$ in the compact-open topology is a metrisable topological group. If $M$ is compact, then $Homeo(M)$ is locally contractible. If $M$ is noncompact, then $Homeo(M)$ may fail to be locally contractible (cf. Kirby-Edwards). The implication is that $Homeo(M)$ for compact $M$ may be an ANR. This is known to be true in dimensions $\leq 2$ (cf. Yagasaki) however it is an open problem for dimensions $\geq3$.

Answer 3

I can tell you about interesting applications of this result. This is an important result if you care about classification of higher-dimensional topological manifolds, for instance, relating the TOP and PL categories. To see how , see section 8 in

Rudyak, Yuli, Piecewise linear structures on topological manifolds, Hackensack, NJ: World Scientific (ISBN 978-981-4733-78-6/hbk; 978-981-4733-80-9/ebook). xxii, 106 p. (2016). ZBL1356.57003.

A preliminary form, with some typos, you can find on the archive here. I find this much more readable than Kirby and Siebenmann.

An example: Suppose that $f: T^n\to T^n$ is a homeomorphism homotopic to the identity. Then there is a finite covering map $M\to T^n$ such that $f$ lifts to a homeomorphism $M\to M$ which is isotopic to the identity. This, of course, uses only local path-connectivity of $Homeo(T^n)$. In order to get a sense of what local contractibility means, you should work with a family of homeomorphisms parameterized by your favorite compact manifold. Then local contractibility implies that (if the entire family moves all points just a tiny bit) then all these homeomorphisms can be simultaneously isotopied to the identity (as a family). For maps of the torus as above, this would mean that there is a finite covering such that the family of lifts is isotopic to $id_M$ as a family.

Answer 4

So, someone pointed out to me that there's a complication. Consider the following definition, which I'll call semi-locally contractible:

$X$ is semi-locally contractible at the point $x$ if for every neighborhood $U$ of $x$, there is a neighborhood $V \subset U$ of $x$ such that $V$ deforms to $x$ in $U$.

It turns out that this is the definition of "locally contractible" that all of these papers from the 50's/60's/70's were using; I was under the impression that this definition was equivalent for these groups, but I think I was wrong. It's sometimes referred to as the "geometric topologist's local contractibility." So it may not be possible to say the following: A family of homeomorphisms sufficiently close to the identity (or some other homeomorphism $f$) can be simultaneously isotopied to the identity (or $f$ respectively).

However, it's not a priori obvious that $Aut(\mathbb{R}^2) = G$ doesn't also satisfy local contractibility in the strong sense, i.e. that every point has a local basis of contractible neighborhoods. I'm wondering if the following collections of functions shed light on the issue, here.

We'll consider homeomorphisms close to the identity map of the plane.

First family: Let $x_n = (0, \frac{1}{n}) \in \mathbb{R}^2$, and inductively define pairwise-disjoint discs $D_n$ centered at $x_n$ none containing $0$. For visualization's sake, note that $D_n \rightarrow 0$. For a fixed $n$, we will define a collection $\mathcal{F}_n$ of functions $f_k^n$ equal to the identity outside of $D_n$:

These functions pinch a pair of points closer and closer to the center of $D_n$. It seems impossible to simultaneously isotopy all of them back to the identity. Instead, we will need to take a smaller neighborhood in $G$ that doesn't contain the tail of any family $\mathcal{F}_n$.

Second family: Let $J$ be a circle in the plane around the center point $x$, and let $\mathcal{F}$ denote the family of plane homeomorphisms such that $|f(j) - j| < \epsilon$ for all $f \in \mathcal{F}$ and all $j \in J$. In particular, it contains the periodic, radial oscillations of the plane centered at $x$, i.e. for every straight line $L_\theta$ through $x$ it linearly maps $L_\theta$ to itself (orientation-preserving) by $f_\theta$. If we parametrize the angles $L_\theta$ by $\pi$ we can let the derivative of $f_n^m$ along $L_\theta$ be $1+\frac{1}{n} \cdot sin(2m\theta)$ (for $n > 2$ to avoid some self-intersection complications).

Then for any $\epsilon$-nbhd of the identity in $G$, it will contain the family $\mathcal{F}_n = \lbrace f_n^m$ $|$ $m \in \mathbb{N} \rbrace$ for some $n$ but I think there is also no contraction of the whole family simultaneously.

Am I on the right track, or am I missing the idea here?