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Suppose that $X \subset \mathbb{R}^n$ is compact, homogeneous and contractible (and thus connected). Does $X$ have to be a point?

I couldn't think of a non-trivial example, and there isn't a counterexample in the plane. The homogeneous planar continua have been classified (point, circle, pseudo-arc, circle of pseudo-arcs) and the only contractible one is a point. Maybe there is some twisty sort of example in three or four dimensions, though?

Does it become true if $\dim(X) = n $ and is embeddable in $\mathbb{R}^{n+1}$?

By homogeneous I mean for any $x, y \in X$ there is a homeomorphism $f$ of $X$ with $f(x) = y$. By contractible I mean the identity map on $X$ is homotopic to a constant map. For example the circle is homogeneous but not contractible, and the closed disc is contractible but not homogeneous.

John Samples
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    What's the definition of homogeneous? – qualcuno Feb 14 '21 at 23:53
  • @guidoar $\mathrm{Homeo}(X)$ acts transitively on $X$. In other words for all $x,y\in X$ there is an homeomorphism $f\colon X\to X$ with $f(x)=y$. – Alessandro Codenotti Feb 14 '21 at 23:55
  • Ah, right. Thanks :) – qualcuno Feb 14 '21 at 23:56
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    An easy observation: there are such continua in $\Bbb R^\Bbb N$, for example $[0,1]^\Bbb N$. (Also "connected" is redundant in your list of properties, being implied by "contractible") – Alessandro Codenotti Feb 14 '21 at 23:58
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    @AlessandroCodenotti $[0,1]^N$ is not homogeneous. No homeomorphism maps a boundary point to an interior point. – Paul Frost Feb 15 '21 at 00:02
  • Suppose that $X$ is a contractible CW complex. If $\dim X = 1$, then $X$ is a tree. Serre proves in Trees that any action of a finite group will have fixed points. That any finite $2$-dimension $G$-complex (with $G$ finite) admits fixed points is the Casacuberta-Dicks conjecture. They proved it originally for $X$ acyclic and $G$ solvable. So even though this is not exactly what you asked for, the problem seems hard (maybe the fact that we can work with the whole group $\mathsf {Homeo}(X)$ makes things easier, I don't know). – qualcuno Feb 15 '21 at 00:07
  • @AlessandroCodenotti You are right, I didn't properly see that you wrote $\mathbb N$; I read $N$. – Paul Frost Feb 15 '21 at 00:10
  • Ok, I added the relevant definitions and removed the superfluity of "connected." Is there a way to adjust Bing's "two drill bits" space, perhaps? I forget the exact properties of that space. – John Samples Feb 15 '21 at 00:35
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    Another easy observation: if $X$ contractible is relaxed to $\pi_n(X)=0$ for $n\geq 1$ then the pseudo-arc is a (connected) example – Alessandro Codenotti Feb 15 '21 at 04:51
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    For those of you (like me) who where looking for an example which is a manifold, this is not possible: a compact simply-connected manifold is orientable so it has $\mathbb Z$ for top homology. Hence, by homogeneity, no neighborhood of any point is homeomorphic to a disk, thus the "local topology" of a possible counterexample must be pretty nasty. – Adam Chalumeau Feb 15 '21 at 15:07
  • Nice observation @AdamChalumeau – John Samples Feb 15 '21 at 19:45
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    Fun fact: if $G$ is a compact topological group and $G$ is contractible, then $G$ is the trivial group. This rules out such objects from being counterexamples. – Tyrone Feb 15 '21 at 20:49
  • I also just realized that the homogeneous planar continua are known to be the point, the circle, the pseudo-arc and the circle of pseudo-arcs, so that case is actually known. Going to edit the post. – John Samples Feb 15 '21 at 20:53
  • What about $S^{\infty}$? – JohannesPauling Feb 23 '21 at 16:20
  • It's infinite-dimensional, right? – John Samples Feb 23 '21 at 18:46
  • @JohnSamples yes : ) – JohannesPauling Feb 24 '21 at 04:54

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