12
$\begingroup$

What is a topological characterization of the class of spaces that have the form $G/H$ for a locally compact, Hausdorff group $G$ and a closed subgroup $H$ ?

Such a space $X=G/H$ necessarily satisfies the following conditions:

1) $X$ is locally compact, Hausdorff.

2) $X$ is (topologically) homogeneous (i.e., for any points $x,y\in X$ there exists a homeomorphism $\phi$ of $X$ such that $\phi(x)=y$).

3) $X$ satisfies the Suslin condition (i.e., every collection of non-empty, disjoint, open subsets of $X$ is countable; this follows from the Haar-measure on $G$).

Are these conditions sufficient? Maybe this is easier if one restricts attention to the class of separable, metric spaces.

Note that homogeneity for $X$ means exactly that the homeomorphism-group Homeo(X) acts transitively on $X$. If $X$ is also locally compact, separable and metric, then Homeo(X) is a separable, complete metric space (in the compact-open topology), and it follows that $X$ is homeomorphic to $G/G_x$, where $G_x$ is the stabilizer subgroup (also called isotopy group). See for instance the following articles for these results:

Effros (1965) # Transformation groups and C*-algebras [Ann. Math. (2) 81]

Ungar (1975) # On all kinds of homogeneous spaces [Trans. AMS 212]

The question now is under which conditions there exists a locally compact subgroup of Homeo(X) which still acts transitively on $X$.

The motivation for this question is to clarify the notion of "homogeneous space". Sometimes in the literature, by a homogeneous space it is not understood a (topological) homogeneous space but a coset space $G/H$ where $G$ is usually even assumed to be a Lie group.

$\endgroup$
4
  • 3
    $\begingroup$ These conditions are not sufficient, the Sierpinski carpet, etc, are not homogeneous spaces of LC-groups (although compact, metrizable and homogeneous). The only 1-dimensional compact connected spaces occuring as homogeneous spaces of LC-groups are solenoids (projective limits of circles). Except the circle, these spaces are not path-connected. Also, all connected compact manifolds are homogeneous, but in dimension $\ge 2$ most of them are not homogeneous under a locally compact group (it is not hard to check that this implies homogeneous under a connected Lie group) $\endgroup$
    – YCor
    May 6, 2013 at 17:25
  • 1
    $\begingroup$ Condition 3) is not necessary unless you restrict to compact G´s. $\endgroup$ May 6, 2013 at 19:23
  • $\begingroup$ @Ramiro: indeed, probably (3) should be replaced by: $X$ is a disjoint union of clopen subsets with the Suslin condition. $\endgroup$
    – YCor
    May 6, 2013 at 23:07
  • $\begingroup$ @Yves: Thank you for the examples. I also agree with you (and Ramiro) that the Suslin condition is too strong and should be relaxed to what you suggest. $\endgroup$ May 7, 2013 at 8:28

1 Answer 1

2
$\begingroup$

This problem is considered in the recent papers by Hofmann, Kramer (http://arxiv.org/pdf/1301.5114.pdf) and Antonyan, Dobrowolski [Locally contractible coset spaces, Forum Mathematicum. 27:4 (2015), 2157–2175]. According to these papers, for a locally compact group $G$ and a closed subgroup $H$ in $G$ the homogeneous space $X=G/H$ is an Euclidean manifold if either $X$ is finite-dimensional and locally connected or $X$ contains a non-empty open set contractible in $X$. This result implies that the Hilbert and Menger cubes are not coset spaces of locally compact groups (in spite of the fact that they are topologically homogeneous).

Concerning non-metrizable compact coset spaces of locally compact groups, I think that all such spaces should be supercompact and Dugunji compact (at least this is true for compact topological groups, see http://arxiv.org/abs/1010.3329 and http://iopscience.iop.org/article/10.1070/SM1990v067n02ABEH002098/pdf).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.