4
$\begingroup$

Suppose that $(X,d)$ is a locally compact connected homogeneous metric space, where by homogeneous I mean that for any $x_0,x_1 \in X$ there exists an isometry $f:X\rightarrow X$ such that $f(x_0)=x_1$. Does there necessarily exist a connected manifold $M$ and a continuum (compact connected metric space) $K$ such that $X$ and $M\times K$ (with the max metric) are bi-uniformly equivalent, i.e. there exists a bijection $g:M\times K \rightarrow X$ such that $g$ and $g^{-1}$ are both uniformly continuous. If so can $M$ and $K$ be taken to be homogeneous themselves?

When I looked for an answer to this question the only thing I could find was this paper of Berestovskii, which, if I'm understanding it correctly, gives a positive answer with the additional assumption that the space has a geodesic metric.

$\endgroup$
10
  • $\begingroup$ What about (pseudo-arc)$\times$(pseudo-arc minus a point)? Or (pseudo-arc)$^2$ minus a point? Probably not the product of a continuum and a manifold. But are they homogeneous? My intuition is that the second example is, because (circle)$^2$ minus a point is homeomorphic to $\mathbb R^2$. Maybe an easier example to consider is (solenoid)$^2$ minus a point. $\endgroup$ Sep 22, 2019 at 6:29
  • $\begingroup$ It seems that you locally compact homogeneous space $(X,d)$ is uniformly homeomorphic to the homogeneous space $G/H$ of some locally compact group $G$ by a closed subgroup $H$. Being locally compact (and maybe something else like $\sigma$-compact), the group $G$ contains a compact normal subgroup $K$ such that $G/K$ is a Lie group, so a manifold. Then $HK$ is a closed subgroup in $G$ and $G/HK$ seems to be a manifold. Therefore, $(X,d)$ admits a perfect map onto a manifold $M$, whose fiber is homeomorphic to the compact homogeneous space $HK/H$. $\endgroup$ Sep 22, 2019 at 12:02
  • $\begingroup$ This is continuation of my comment. Now it remains to prove (or disprove) that the fibration $G/H\to G/KH$ is trivial. $\endgroup$ Sep 22, 2019 at 12:04
  • $\begingroup$ By the way, what happens at the level of locally compact abelian groups? In this case the Pontryagin duality theory can give some counterexamples. $\endgroup$ Sep 22, 2019 at 12:05
  • 1
    $\begingroup$ @JamesHanson You are right: not every topologically homogeneous space (form example the Hilbert cube) can be endowed with a homogenous metric. It is known that locally contractible locally compact homogeneous metric spaces are finite-dimensional manifolds. $\endgroup$ Sep 22, 2019 at 15:01

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy