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A metric space $(X,d)$ is isometrically homogeneous if its isometry group acts transitively on points, i.e., for every $x,y \in X$ there is an isometry $\varphi:X\to X$ with $\varphi(x) = y$. I'd like to know an example of a compact isometrically homogeneous metric space which is not a manifold (a space with finitely many points counts as a 0-dimensional manifold).

Googling a bit I've discovered enough recent literature on this general subject to be sure there must be classical examples known to experts, but I haven't managed to find them written down. For example, Theorem 1.2 of this paper implies:

A compact isometrically homogeneous metric space is a finite-dimensional manifold if and only if it is locally contractible.

So equivalently, I'd like an example of a compact isometrically homogeneous metric space which is not locally contractible.

Added: Pete and Neil both gave very nice answers. I'm accepting Neil's since, as Pete points out, it essentially contains Pete's answer as a special case.

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2 Answers 2

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Take $X=\prod_{n=0}^\infty S^1$ with $d(x,y)=\sum_n|x_n-y_n|/2^n$. Then the metric topology is the same as the product topology, which is compact by Tychonov. There is an obvious group structure by pointwise multiplication, and multiplication by any fixed element is an isometry, so the space is isometrically homogeneous. It is path-connected but not locally contractible.

More generally, I guess you can take any sequence of compact isometrically homogeneous spaces $X_n$, rescale the metric so that $d(x,y)\leq 2^{-n}$ for all $x,y\in X_n$, and then take $X=\prod_nX_n$ with $d(x,y)=\sum_nd(x_n,y_n)$.

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  • $\begingroup$ @Neil: nice answer. I guess you can't take quite any sequence of spaces $X_n$ (what if each has a single point?). On the other hand, if you take each $X_n$ to be a $p$-point discrete space, you recover my example (at least up to homeomorphism). $\endgroup$
    – Pete L. Clark
    Commented Feb 24, 2011 at 15:40
  • $\begingroup$ It's interesting that, like Pete's example of the $p$-adic integers, your product of shrinking circles is also like a manifold in some sufficiently generalized sense. $\endgroup$
    – Mark Meckes
    Commented Feb 24, 2011 at 15:43
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The ring of $p$-adic integers $\mathbb{Z}_p$ with its standard metric seems to be an example of what you want.

Of course this space is a "$p$-adic analytic manifold", so you may want to see another example. E.g., how about a space which satisfies all of your properties and is also connected?

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  • $\begingroup$ I haven't had a reason to get to know the $p$-adic metric before, and I'm having some trouble seeing the homogeneity. Can you give me a hint about that? (The Wikipedia page is of limited helpfulness, as it discusses several perspectives on $\mathbb{Q}_p$, but doesn't clarify their relationships, and $\mathbb{Z}_p$ and the $|\cdot|_p$ are never discussed from the same perspective.) $\endgroup$
    – Mark Meckes
    Commented Feb 24, 2011 at 15:39
  • $\begingroup$ As for your comment that $\mathbb{Z}_p$ is a kind of manifold, I'm quite happy to have an example which is not a finite-dimensional Riemannian manifold. $\endgroup$
    – Mark Meckes
    Commented Feb 24, 2011 at 15:40
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    $\begingroup$ @Mark: the $p$-adic metric is induced by a norm, so is therefore homogeneous for the same reasons as in usual functional analysis: for every $x,y,z \in \mathbb{Z}_p$, $d_p(x+y,x+z) = |(x+y)-(x+z)|_p = |y-z|_p = d_p(y,z)$. (BTW, maybe a moral here is that all mathematicians should know at least a little bit about the $p$-adic numbers, as of course we all know about $\R$ and $\C$.) $\endgroup$
    – Pete L. Clark
    Commented Feb 24, 2011 at 15:46
  • $\begingroup$ Thanks. Since in my personal experience a "norm" is always over $\mathbb{R}$ or $\mathbb{C}$ that simply didn't occur to me. I happily stipulate your moral. $\endgroup$
    – Mark Meckes
    Commented Feb 24, 2011 at 15:50
  • $\begingroup$ This is the example I would have offered, too, but the total disconnectedness of $\mathbb{Z}_p$ held me back, since it looked as if Mark was looking for something connected. Beyond that, I'd say, Mark, that norms aren't "over" anything but rather have their values somewhere. In the case of $\mathbb{Z}_p$, the values are a discrete multiplicative semigroup of the reals, but that's okay. $\endgroup$
    – Lubin
    Commented Feb 24, 2011 at 16:31

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