Is every locally compact, Hausdorff, locally path-connected topological group $G$ locally Euclidean? (That would imply of course also being a Lie group.) Is it true when countable basis is assumed? I wasn't able to find a discussion of this question in the literature on topological groups and the Hilbert 5th problem.

EDIT: As YCor rightly pointed out, let's assume that $G$ is of finite topological dimension. The general theme of my question is to identify a minimal set of conditions on a topological group making it into a Lie group.

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    $\begingroup$ You probably want finite-dimensional, or something. Otherwise, any infinite product of positive-dimensional compact connected Lie groups will provide a counterexample. For instance, infinite-dimensional tori $(\mathbf{R}/\mathbf{Z})^X$, $X$ infinite, satisfy the assumption (for $X$ countable this is second countable). $\endgroup$
    – YCor
    Commented Mar 7, 2021 at 22:24
  • $\begingroup$ If you browse Annals of Math in the years 1940-1955 you find a considerable literature on the subject. $\endgroup$
    – YCor
    Commented Mar 7, 2021 at 22:25
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    $\begingroup$ Actually, even more is true and every locally compact finite dimensional group is locally a topological product of Cantor set and $R^n$. $\endgroup$ Commented Mar 8, 2021 at 1:57
  • $\begingroup$ @MoisheKohan That's an interesting fact. Would you know a reference to it? $\endgroup$
    – Adam
    Commented Mar 8, 2021 at 4:31
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    $\begingroup$ In the case of compact (Hausdorff) groups, this is due to Pontryagin, see the reference here. The general case follows from the approximation property (of l.c. groups by Lie groups). I will write a proof when I have more time. $\endgroup$ Commented Mar 8, 2021 at 12:18

2 Answers 2


Under the additional assumption of finite topological dimension pointed by YCor in the comments to the OP, the answer is yes, see e.g. Theorem 10, pp. 120 of the paper of K. Whittington, Local connectedness in topological groups, Topology and its Applications 180 (2015) 111-123. In fact, according to the same Theorem the hypothesis of local compactness is not needed and follows from the others. The second axiom of countability is also not needed.

Edit: in due time, the hypotheses needed for a positive solution to Hilbert's 5th problem when the assumption of finite topological dimension is allowed to fail were a major theme in Per Enflo's PhD thesis (1970).

Edit 2: Implication (2) $\Rightarrow$ (3) of Theorem 10 ibid. is a literal citation of Theorem 4.10.1, page 185 of the book by D. Montgomery and L. Zippin, Topological Transformation Groups (Wiley Interscience, 1955), see Theorem 2 in Moishe Kohan's answer.


This is an addendum to Pedro's answer:

Theorem 1. Suppose that $G$ is a locally compact finite dimensional (Hausdorff) topological group. Then $G$ is locally homeomorphic to the product of a totally disconnected space and ${\mathbb R}^n$.

A reference for this result (that should be better known) is

Montgomery, Deane; Zippin, Leo, Topological transformation groups, Mineola, NY: Dover Publications (ISBN 978-0-486-82449-9). xi, 289 p. (2018). ZBL1418.57024.

section 4.9.3. (The actual theorem says a bit more, namely that locally, at $e$, $G$ is a product of a local Lie group and a totally disconnected group.

In the same book you will also find Theorem 4.10.1 answering your original question:

Theorem 2. If $G$ is a locally compact, locally connected and finite dimensional topological group, then $G$ is a Lie group (possibly non-2nd countable).

Of course, this is a consequence of Theorem 1.

  • $\begingroup$ Granted that the proof of Whittington's Theorem 10 uses Theorem 4.10.1 from Montgomery-Zippin (the latter is the former's implication (2) => (3)). However, the former is stronger in the sense that local compactness needs not to be assumed and is rather a consequence. $\endgroup$ Commented Mar 9, 2021 at 16:42
  • $\begingroup$ @PedroLauridsenRibeiro I would have to read Whittington's paper to understand what these parts are. $\endgroup$ Commented Mar 9, 2021 at 18:00
  • $\begingroup$ Did you forget to assume finite dimension in Theorem 1? $\endgroup$ Commented Mar 9, 2021 at 20:50
  • $\begingroup$ @GeraldEdgar: Yes, I did. Thanks! $\endgroup$ Commented Mar 9, 2021 at 21:00

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