Path connected subgroups of SO(3),

[Tinyboss]

Hi, I'm giving a talk tomorrow morning, and I'd like to use the following fact: a path-connected subgroup of SO(3) consists of either a) only the identity, b) all the rotations about a single axis, or c) all of SO(3).

Unfortunately, I can't for the life of me find where I read it, and I'm not 100% sure I'm remembering it correctly. Can anyone provide a reference, a quick proof, or tell me it's wrong? Thanks!

 

[micromass]

I probably saw this too late to be helpful for your talk, but the fact is true. The easiest proof is by applying Yamabe's theorem which says that any path-connected subgroup of $SO(3)$ corresponds to a subalgebra of $\mathfrak{so}(3)$. More precisely, if $H$ is a path-connected subgroup of $SO(3)$, then there is a Lie-subgroup $\mathfrak{h}$ of $\mathfrak{so}(3)$, such that $\mathfrak{h} = \langle \textrm{exp}\mathfrak{h}\rangle$.

So if $\mathfrak{h}$ has dimension $0$, then you get the identity. If it has dimension $1$, then you have the rotations with fixed axis. If it has dimensions $3$ then it is the entire group. It cannot have dimension $2$.

For more information, see the excellent book by Hilgert and Neeb: "Structure and Geometry of Lie Groups"

 

[Matterwave]

I probably saw this too late to be helpful for your talk, but the fact is true. The easiest proof is by applying Yamabe's theorem which says that any path-connected subgroup of $SO(3)$ corresponds to a subalgebra of $\mathfrak{so}(3)$. More precisely, if $H$ is a path-connected subgroup of $SO(3)$, then there is a Lie-subgroup $\mathfrak{h}$ of $\mathfrak{so}(3)$, such that $\mathfrak{h} = \langle \textrm{exp}\mathfrak{h}\rangle$.

So if $\mathfrak{h}$ has dimension $0$, then you get the identity. If it has dimension $1$, then you have the rotations with fixed axis. If it has dimensions $3$ then it is the entire group. It cannot have dimension $2$.

For more information, see the excellent book by Hilgert and Neeb: "Structure and Geometry of Lie Groups"

I really thought the proof would be easier than this. This fact seemed very intuitive to me.

 

[micromass]

I really thought the proof would be easier than this. This fact seemed very intuitive to me.

There is likely an elementary, direct proof. But this is the fastest way to prove it.

 

[blue_raver22]

Hi there, thank you for your question. I am familiar with the fact you mentioned about path-connected subgroups of SO(3). This is known as the "Axis-Angle Theorem" and it states that any path-connected subgroup of SO(3) must be either the identity, a subgroup of rotations about a single axis, or the entire group SO(3). This theorem was first proved by Sophus Lie in the 19th century and has been a fundamental result in the study of rotations and symmetries in mathematics and physics.

A quick proof of this theorem can be done by considering the fact that SO(3) is a compact, connected, and simply connected Lie group. This means that any subgroup of SO(3) must also be compact, connected, and simply connected. Since any compact, connected, and simply connected Lie group is either the identity, a single rotation about an axis, or the entire group, the Axis-Angle Theorem follows.

I hope this helps with your talk tomorrow morning. Good luck!