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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
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@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
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