Does there exist a Hausdorff topological group which is contractible and of finite covering dimension but which is not homeomorphic to $\mathbb{R}^n$ for some $n$?
5
-
$\begingroup$ By the Yamabe theorem / Hilbert's 5th problem this reduces to the question of if such a group can have "small subgroups". I imagine the answer to that is no, and so your question would have the answer no as well. But off the top of my head I don't see a proof. $\endgroup$– Ryan BudneyCommented Jul 26, 2013 at 18:34
-
$\begingroup$ @Lars: you probably assume Hausdorff (otherwise the indiscrete topology on any finite group is contractible). $\endgroup$– YCorCommented Jul 26, 2013 at 19:32
-
3$\begingroup$ @Ryan: you seem to implicitly assume the group locally compact. In this case the answer is indeed negative: if $G$ is a connected LC-group and $K$ a compact subgroup then $G/K$ is contractible iff it is homeomorphic to a Euclidean space. Reference: arxiv.org/abs/1104.1820 (Arch. der Math 2012) $\endgroup$– YCorCommented Jul 26, 2013 at 19:33
-
$\begingroup$ @YvesCornulier : Yes, I did mean to assume Hausdorff. I'll add that assumption. Thank you very much for the reference in the case where the group is locally compact! $\endgroup$– LarsCommented Jul 26, 2013 at 20:03
-
$\begingroup$ I think that I have an example (I still need to check its dimension). $\endgroup$– Wlod AACommented Aug 2, 2017 at 2:34
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
8
If a topological group is contractible, then it is locally contractible (using the group operation produce a contraction which does not move the unit of the group). By a classical result of [A. Gleason, R. Palais, On a class of transformation groups, Amer. J. Math. 79 (1957), 631–648], a locally path-connected finite-dimensional topological group is a Lie group and being contractible, is homeomorphic to an Euclidean space.
answered Aug 25, 2015 at 21:53
-
$\begingroup$ Irrational line on the torus with induced topology is contractible, but not locally contractible. Or am I missing something? $\endgroup$ Commented Aug 26, 2015 at 4:45
-
$\begingroup$ No, it is not contractible: the contraction along the line is discontinuous in the topology induced from the torus. $\endgroup$ Commented Aug 26, 2015 at 23:08
-
1$\begingroup$ I do not know. This is just a simple exercise: if $h:[0,1]\times G\to G$ is a contraction of a topological group $G$ with $h(\{1\}\times G)=\{1_G\}$, then the homotopy $l:[0,1]\times G\to G$ defined by $l(t,x)=h(t,x)\cdot h(t,1_G)^{-1}$ does not move the unit $1_G$ of the group and hence witnesses that $G$ is locally contractible at $1_G$. $\endgroup$ Commented Aug 27, 2015 at 12:06
-
1$\begingroup$ @Wlod AA But for topological groups "your" local contractibility is equivalent to "mine". $\endgroup$ Commented Aug 1, 2017 at 18:30
-
1$\begingroup$ @freakish This follows from the Cartan-Iwasawa-Malcev theorem (en.wikipedia.org/wiki/Maximal_compact_subgroup) which implies that every connected Lie group $G$ is homeomorphic to $M\times \mathbb R^n$ for some maximal compact subgroup $M$ of $G$ and some $n$. The contractibility og $G$ implies the contractibility of the compact group $M$, and contractible compact groups are singletons, see e.g. link.springer.com/content/pdf/10.1007/BF01238544.pdf $\endgroup$ Commented Oct 7, 2021 at 20:50