Timeline for Is every locally compact connected homogeneous metric space a manifold cross a continuum?
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11 events
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| Sep 22, 2019 at 15:47 | comment | added | Taras Banakh | @JamesHanson Yes, see degruyter.com/view/j/forum.2015.27.issue-4/forum-2013-0033/… Also the same result was independently proved by Hoffman: ivv5hpp.uni-muenster.de/u/lkram_01/publ/39.pdf | |
| Sep 22, 2019 at 15:16 | comment | added | James E Hanson | @TarasBanakh Do you have a source for that? | |
| Sep 22, 2019 at 15:01 | comment | added | Taras Banakh | @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. | |
| Sep 22, 2019 at 14:28 | comment | added | James E Hanson | @TarasBanakh I know that locally compact connected Abelian groups are always $\mathbb{R}^n \times K$ for some compact group $K$. This is discussed in this blog post. I don't know what happens in the non-Abelian case. | |
| Sep 22, 2019 at 14:26 | comment | added | James E Hanson | @D.S.Lipham I don't think a topologically homogeneous metric space can always be given a homogeneous metric, can it? For instance the Hilbert cube is topologically homogeneous but I doubt it has a homogeneous metric. | |
| Sep 22, 2019 at 13:47 | comment | added | Taras Banakh | And what about locally compact topological groups? Is it true that any locally compact topological group decomposes into the product of a Lie group and a compact group? | |
| Sep 22, 2019 at 12:05 | comment | added | Taras Banakh | By the way, what happens at the level of locally compact abelian groups? In this case the Pontryagin duality theory can give some counterexamples. | |
| Sep 22, 2019 at 12:04 | comment | added | Taras Banakh | This is continuation of my comment. Now it remains to prove (or disprove) that the fibration $G/H\to G/KH$ is trivial. | |
| Sep 22, 2019 at 12:02 | comment | added | Taras Banakh | 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 $HK/H$. | |
| Sep 22, 2019 at 6:29 | comment | added | D.S. Lipham | 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. | |
| Sep 22, 2019 at 3:43 | history | asked | James E Hanson | CC BY-SA 4.0 |