# Lie groups and NSS+LC group

1. Let $G$ be a locally compact group without small subgroups. Is $G$ a "finite" dimensional Lie group? (i.e, $G$ is not infinite dimensional Lie group.)
2. Are Lie groups precisely the locally Euclidean topological groups?

I need exact answer and exact references for the above questions. Thanks a lot for your answers.

• What do you mean by small subgroups? For question 2, you also need the topology to be countably generated. – user1688 Feb 11 '13 at 15:37
• A topological group $G$ has no small subgroups if there is a neighborhood $U$ of the identity such that no nontrivial subgroup of $G$ is contained in $U$. By Gleason-Yamabe-Montgomery-Zippin, a locally compact group with no small subgroups is isomorphic to a real analytic Lie group (of finite dimension, with an arbitrary number of components). – YCor Feb 11 '13 at 21:26

## 2 Answers

Yes to both question (at least for separable groups):

MR0379739 Reviewed Montgomery, Deane; Zippin, Leo Topological transformation groups. Reprint of the 1955 original. Robert E. Krieger Publishing Co., Huntington, N.Y., 1974. xi+289 pp.

MR0049204 Reviewed Montgomery, Deane; Zippin, Leo Small subgroups of finite-dimensional groups. Ann. of Math. (2) 56, (1952). 213–241.

• Separability is, of course, needed, otherwise you can take an uncountable group with discrete topology. – Misha Feb 11 '13 at 17:01
• Depending on courage, taste, and (bad?) judgement one could allow Lie groups with uncountably many connected components. – Peter Michor Feb 11 '13 at 20:11
• to sustain Peter's comment: an uncountable discrete group is certainly a Lie group, is finite-dimensional (indeed 0-dimensional), and has no small subgroups. If there is a separability issue, it might only incurs at a local level, but actually I don't think it's relevant, i.e. I think Peter's answer is correct without separability assumption. – YCor Feb 11 '13 at 21:21
• Peter: A Lie group is supposed to be a manifold; the latter is always (to the best of my knowledge) assumed to be 2nd countable. Incidentally, I thought that you also need to assume that the group is Hausdorff, but maybe this could be derived from "no small subgroups" assumption... – Misha Feb 12 '13 at 3:38
• @Misha: I think the most usual convention is to assume that connected components are second countable, or equivalently that manifolds are paracompact. – YCor Jul 30 '13 at 21:56

Yes to both questions, assuming that the group $G$ is locally compact and Hausdorff. In such a group one can always find an open subgroup $H$ which is isomorphic to $(K\times L)/\Gamma$, where $K$ is compact, $L$ is a $1$-connected Lie group and $\Gamma$ is a discrete subgroup of $K\times L$. This reduces both questions (or assumptions) to the compact group $K$. No topological countability assumptions are needed.