1

What are the closed subgroups of $SO(3)$? I think that it is a textbook question but I haven't studied it anywhere. I haven't studied Lie theory. I am thinking about the question from a basic abstract algebra(my knowledge is Dummit and Foote) and Topology(Munkres) point of view.

I don't want a detailed answer if you would like to write it. I appreciate that. If you give me some ideas and hints in steps. That will help me a lot.

Ri-Li
  • 9,038
  • 3
    Related: https://math.stackexchange.com/q/2931074/96384, https://math.stackexchange.com/q/4114618/96384, https://math.stackexchange.com/q/2130157/96384. – Torsten Schoeneberg Apr 19 '22 at 00:45
  • 1
    The finite subgroups, and the copies of SO(2) and O(2) by fixing an axis. – anon Apr 19 '22 at 01:12
  • @runway44 Can you give some hints to follow? – Ri-Li Apr 19 '22 at 01:37
  • HINT: Every element of $SO(3)$ has $1$ as an eigenvalue. – Ted Shifrin Apr 19 '22 at 01:44
  • Okay, I will try to prove it. Then it implies that $SO(3)$ fixes the eigenvector, which means it always fixes at least one axis(not necessarily the x,y,z but with rotation we can assume x,y,z). Am I right? What next? – Ri-Li Apr 19 '22 at 02:07
  • @Ri-Li . Do the rotations that share the same rotation axis form a closed subgroup? – Kurt G. Apr 19 '22 at 06:48

1 Answers1

2

It's shown in https://math.stackexchange.com/a/2931125/32766 that the identity component $G_0$ of such a closed subgroup $G$ is either trivial, all of $\mathrm{SO}(3)$ or the stabilizer of some $v \in S^2$. Since $G$ is compact, $[G: G_0]$ is finite.

So in the first case $G$ is a finite group, and its a classical result that these are either cyclic (fixing some $v$), dihedral (fixing some line $\mathbb{R}v$ or equivalently $\{\pm v\}$ as a set), or the (oriented) symmetry group of some platonic solid. If $G_0 = \mathrm{SO}(3)$ then of course $G = G_0 = \mathrm{SO}(3)$.

If $G_0 = \operatorname{Stab}(v)$ then for any $g \in G$ we must have $\operatorname{Stab}(gv) = gG_0 g^{-1} = G_0$, so $gv = \pm v$. So either $G = G_0$ or $G = \operatorname{Stab}(\{\pm v\})$ with $[G: G_0 ] = 2$.

ronno
  • 11,390