5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Belated comment: adding a tag 'lie-groups' would be appropriate here? $\endgroup$ Apr 16, 2011 at 17:50
  • $\begingroup$ and what if we consider a topological group which is not locally euclidean? $\endgroup$
    – agt
    Apr 16, 2011 at 18:12
  • 2
    $\begingroup$ @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
    Apr 17, 2011 at 8:22
  • $\begingroup$ @Stephen: thanks a lot for your attention. $\endgroup$
    – agt
    Apr 18, 2011 at 6:07

3 Answers 3

10
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thanks a lot Andreas, this is exactly what I was looking for. $\endgroup$ Apr 16, 2011 at 17:36
  • $\begingroup$ Corollary: Any continuous Galois representation on Euclidean space has finite image. $\endgroup$ Apr 16, 2011 at 21:10
3
$\begingroup$

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 $G$.

$\endgroup$
9
  • $\begingroup$ Hi Giuseppe, but $H$ is not necessarily a topological manifold. So are you saying that $H$ is a $0$-dimensional manifold so therefore discrete? $\endgroup$ Apr 16, 2011 at 16:22
  • 2
    $\begingroup$ 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. $\endgroup$ Apr 16, 2011 at 16:42
  • $\begingroup$ So what is the definition of a 0-dimensional manifold? $\endgroup$ Apr 16, 2011 at 16:55
  • 1
    $\begingroup$ @Hugo: exactly what it sounds like: a (second-countable Hausdorff etc.) space which is locally homeomorphic to a point, hence discrete. $\endgroup$ 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. $\endgroup$ Apr 16, 2011 at 17:35
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Apr 26, 2011 at 14:54
  • $\begingroup$ Well, I suppose I'm appealing to van Dantzig's theorem... $\endgroup$ Apr 26, 2011 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.