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So the following statement seems to be obvious but I don't see how to prove it:

Q: How does one prove that a closed totally disconnected subgroup of a connected real Lie group is discrete?

Note that it is essential to take into account the group structure since the Cantor set is a closed totally disconnected subset of $\mathbf{R}$ which is not discrete.

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  • $\begingroup$ Belated comment: adding a tag 'lie-groups' would be appropriate here? $\endgroup$
    – Jim Humphreys
    Commented Apr 16, 2011 at 17:50
  • $\begingroup$ and what if we consider a topological group which is not locally euclidean? $\endgroup$
    – agt
    Commented Apr 16, 2011 at 18:12
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    $\begingroup$ @Giuseppe: If the group isn't locally Euclidean, then the subgroup needn't be discrete. For example, let $G$ be a direct product of infinitely many circle groups. The circle group has a subgroup of order $2$, so $G$ has a subgroup that is a direct product of infinitely many $2$-element discrete groups. This subgroup is closed and totally disconnected but not discrete. $\endgroup$
    – Stephen S
    Commented Apr 17, 2011 at 8:22
  • $\begingroup$ @Stephen: thanks a lot for your attention. $\endgroup$
    – agt
    Commented Apr 18, 2011 at 6:07

3 Answers 3

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Let's just give a quick argument:

Let $G$ be a Lie group (with Lie algebra $\mathbb g$) and $H \subset G$ be a closed subgroup. If $H$ is not discrete, then there exists a sequence $(g_n)_{n \in \mathbb N}$ of elements of $H$, which converges to $1_G$. Write $g_n = \exp( \alpha_n \xi_n)$ (for $n$ large enough), where $\xi_n \in \mathbb g$ are unit vectors (with respect to some innerproduct) and $\alpha_n \in \mathbb R_{> 0}$. Since $g_n \to 1_G$, we get $\alpha_n \to 0$. Let $\delta \in \mathbb R$ and find integers $k_n(\delta)$, such that $|\alpha_n \cdot k_n -\delta|< \alpha_n$.

Let $\xi$ be some accumulation point of the set $\lbrace \xi_n \mid n \in \mathbb N \rbrace$. Now, it is easy to see that $$g_n^{k_n(\delta)} = \exp(\alpha_n k_n(\delta) \xi_n) \to \exp(\delta \xi)$$ on a subsequence. This implies that the whole 1-parameter subgroup generated by $\xi$ lies in $H$. In particular, $H$ is not totally disconnected.

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  • $\begingroup$ Thanks a lot Andreas, this is exactly what I was looking for. $\endgroup$
    – Hugo Chapdelaine
    Commented Apr 16, 2011 at 17:36
  • $\begingroup$ Corollary: Any continuous Galois representation on Euclidean space has finite image. $\endgroup$
    – Kevin Ventullo
    Commented Apr 16, 2011 at 21:10
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0)The solution by Gleason, Montgomery, and Zippin of the fifth Hilbert's problem says that any òpcally euclidean topological group admits a compatible analytical structure.

1)By Cartan--Von Neumann theorem on closed subgroups, if $G$ is a Lie group (i.e. smooth group manifold) and $H$ is a closed subgroup of $G$, then $H$ is an embedded Lie subgroup of $G$, so its connected components w.r.t. the subspace topology and its connected components w.r.t. this Lie group structure are the same.

2)The connected components of a topological manifold are open.

Now 0),1) and 2) should be conclusive for more than your question, i.e.:

If $G$ is a locally euclidean topological group and $H$ is a totally disconnected, closed subgroup of $G$ then $H$ is discrete topological subgroup of $G$.

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  • $\begingroup$ Hi Giuseppe, but $H$ is not necessarily a topological manifold. So are you saying that $H$ is a $0$-dimensional manifold so therefore discrete? $\endgroup$
    – Hugo Chapdelaine
    Commented Apr 16, 2011 at 16:22
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    $\begingroup$ Cartan's theorem implies that $H$ is necessarily a smooth submanifold of $G$. So if it's totally disconnected, it must be 0-dimensional, i.e. discrete. $\endgroup$
    – Brad Hannigan-Daley
    Commented Apr 16, 2011 at 16:42
  • $\begingroup$ So what is the definition of a 0-dimensional manifold? $\endgroup$
    – Hugo Chapdelaine
    Commented Apr 16, 2011 at 16:55
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    $\begingroup$ @Hugo: exactly what it sounds like: a (second-countable Hausdorff etc.) space which is locally homeomorphic to a point, hence discrete. $\endgroup$
    – Qiaochu Yuan
    Commented Apr 16, 2011 at 17:33
  • $\begingroup$ Well take $G=H=\mathbf{Z}_p$, then $H$ is not discrete. The result that you claim in $0)$ probably applies to topological groups which have a topological real manifold structure. $\endgroup$
    – Hugo Chapdelaine
    Commented Apr 16, 2011 at 17:35
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To complement what has been said earlier: a Lie group $G$ has no small subgroup, i.e. there exists a neighborhood $U$ of the identity $e\in G$ such that the only subgroup of $G$ contained in $U$, is $\{e\}$. [To see it, let $B$ be a small ball around $0$ in the Lie algebra of $G$, such that the exponential map $\exp$ induces a diffeomorphism between $B$ and its image. Set $U=\exp(B/2)$. Then for every $x\in U\backslash\{e\}$, there exists $p>1$ such that $x^p\notin U$.]

Let then $H$ be a closed subgroup of $G$. If $H$ is totally disconnected, then there is a basis of neighborhoods of $e$ in $H$, consisting of open subgroups of $H$. So $H\cap U=\{e\}$, meaning that $H$ is discrete.

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  • $\begingroup$ So how do you show the existence of such a basis of neighborhoods at $e$ in $H$. Definitely you need $H$ to be closed but I don't see how to use it. $\endgroup$
    – Hugo Chapdelaine
    Commented Apr 26, 2011 at 14:54
  • $\begingroup$ Well, I suppose I'm appealing to van Dantzig's theorem... $\endgroup$
    – Alain Valette
    Commented Apr 26, 2011 at 16:11

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