# 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 $n$?

• 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
• I think that I have an example (I still need to check its dimension). – Wlod AA Aug 2 '17 at 2:34

• 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