1. Jul 29, 2014

### 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!

2. Jul 30, 2014

### 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"

3. Jul 31, 2014

### Matterwave

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

4. Jul 31, 2014

### micromass

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