Is the continuous part of the symmetry group of a compact subset $\mathbb{R}^3$ isomorphic to $SO(k)$?

Question

Let $K$ be a non-empty compact subset of $\mathbb{R}^3$. Let $G= \{f \in \mathrm{Isom}~\mathbb{R}^3 | f(K) = K\}$, $G'$ be the identity connected component of $G$. Is it true that $G'$ is isomorphic (as Lie group) $SO(k)$ for some $k \in \{1, 2, 3\}$?

P.S. Related my question on MSE, there I asked about compact sets in $\mathbb{R}^n$, for arbitrary n the answer is no, see also comments here

@markvs I mean "there is k", of course (it is clear that in this case, for three-dimensional groups - SO (3), one-dimensional - SO (2), zero-dimensional - SO (1)) — Arshak Aivazian, Jan 07 '22 at 20:22
A closed connected subgroup of $\mathrm{SO}(3)$ is either trivial, the whole group, or conjugate to the standard upper left copy of $\mathrm{SO}(2)$. — YCor, Jan 09 '22 at 17:10
@YCor I understand that this is exactly what needs to be proved (but the comment that the statement is true is certainly useful). — Arshak Aivazian, Jan 09 '22 at 17:40

score 1 · Accepted Answer · answered Jan 09 '22 at 17:39

1

It turns out there are already two answers to this question on MSE: All connected closed subgroups in $SO(3)$ and Connected lie subgroup of $SO(3)$

answered Jan 09 '22 at 17:39

Arshak Aivazian

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Is the continuous part of the symmetry group of a compact subset $\mathbb{R}^3$ isomorphic to $SO(k)$?

1 Answers1