Let $G$ be a compact Lie group with a compatible biinvariant metric $d$. The hyperspace $K(G)$ of nonempty compact subsets of $G$ is a compact metric space with the Hausdorff metric, and it is easy to check that subgroups of $G$ form a closed subspace in $K(G)$, hence we may talk about the (compact) space of closed subgroups of $G$. Let us denote this space by $\mathbf{K}(G)$.

General question:

(1) Does anyone know any source that may help exploring spaces of the form $\mathbf{K}(G)$?

I have a conjecture:

(2) For a compact connected Lie group $G$ the following are equivalent:

a) $G$ is a limit point in $\mathbf{K}(G)$ (that is, it can be approximated by proper closed subgroups).

b) The circle group is a quotient of $G$.

Is it true? ( b)$\implies$a) is easy, take inverse images of finite subgroups of the circle group by the quotient map.)

For (1) I have found only the papers of Fischer and Gartside: On the space of subgroups of a compact group I and On the space of subgroups of a compact group II.

They mostly deal with arbitrary compact $G$ or profinite $G$, not Lie groups.

For (2) I found the MO question Approximating Lie groups by finite groups, which says that only compact abelian Lie groups can be approximated by finite subgroups (it refers to a paper of A. M. Turing, Finite Approximations to Lie Groups).

  • 1
    $\begingroup$ These questions can be tricky; one can generalize this to a decent topology on the space of closed subgroups of any topological group, called the Chabauty space. Chabauty space on $\mathbb{R}^2$ is a $4$-sphere, but the topology of the chabauty space of $\mathbb{R}^n$ is not precisely known for $n>2$. There are works by e.g. Haettel and de la Harpe-Kleptsyn-de Cornulier on this. $\endgroup$ Apr 25, 2022 at 13:50
  • $\begingroup$ Somebody commented here in April but later they deleted the comment. It contained extremely useful information for us: a paper of Mongomery and Zippin Together with Nicolas Tholozan's answer it helped us in our work with the space of closed subgroups. $\endgroup$
    – chj
    Jun 10, 2022 at 15:03

1 Answer 1


Assume a sequence of subgroups $G_n$ converges to $G$. Up to extraction, we can assume that $\mathrm{Lie}(G_n)$ converges to a Lie subalgebra $\mathrm{Lie}(H)$. Since the adjoint action of $G_n$ preserves $\mathrm{Lie}(G)$, by passing to the limit we get that $\mathrm{Lie}(H)$ is an ideal of $\mathrm{Lie}(G)$.

Now a bit of (elementary ?) Lie theory should give you that $\mathrm{Lie}(G_n) = \mathrm{Lie}(H)$ for $n$ large enough. Let $H$ be the connected subgroup with Lie algebra $\mathrm{Lie}(H)$. One concludes that $G_n/H$ is a sequence of discrete groups approximating $G/H$. By your second reference, $G/H$ is abelian, which proves your conjecture.

  • $\begingroup$ What does "Up to extraction" mean? $\endgroup$
    – LSpice
    May 25, 2022 at 18:04
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    $\begingroup$ Taking a subsequence. (Maybe that's a "frenchism".) $\endgroup$
    – Nicolast
    May 26, 2022 at 7:03
  • $\begingroup$ @NicolasTholozan Thank you very much for your help! We have not studied Lie theory before so it took some time and research for us to understand your argument. Probably we will use this as a lemma in a paper we are writing. $\endgroup$
    – chj
    Jun 10, 2022 at 14:49

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