# Contractible topological groups

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 $n$?

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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. – Ryan Budney Jul 26 '13 at 18:34
@Lars: you probably assume Hausdorff (otherwise the indiscrete topology on any finite group is contractible). – YCor Jul 26 '13 at 19:32
@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) – YCor Jul 26 '13 at 19:33
@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! – Lars Jul 26 '13 at 20:03

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-finite finite-dimensional topological group is a Lie group and being contractible, is homeomorphic to an Euclidean space.

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Irrational line on the torus with induced topology is contractible, but not locally contractible. Or am I missing something? – Victor Protsak Aug 26 '15 at 4:45
No, it is not contractible: the contraction along the line is discontinuous in the topology induced from the torus. – Taras Banakh Aug 26 '15 at 23:08
Thank you, Taras. What is a reference for 'contractible implies locally contractible' in this setting? – Victor Protsak Aug 27 '15 at 5:19
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 $1_G$. – Taras Banakh Aug 27 '15 at 12:06