2

I understand there are many questions on here that show give an explicit map to show that $SU(2)$ is a double cover of $SO(3)$ (seevia quaternions).

I am trying and use the fact that $SU(2)$ is a double cover of $SO(3)$ to write $SU(2)$ as some fibre bundle. But I seem to encounter some contradictions. I want to know what is incorrect. Here are my ideas:

(1) Since the Lie algebras of $SO(3)$ and $SU(2)$ are isomorphic. They have isomorphic connected components.

(2) Covering spaces can always be thought of as discrete fibre bundles over the base space.

(3) But $SU(2)$ is simply connected.

I think (2) is wrong.

(4) If two is wrong what is the correct way to think about covering spaces. As far as I understand $SU(2)$ has one connected component, and $SO(3)$ has two connected components. So it seems like maybe $SO(3)$ should be a double cover of $SU(2)$?

Mark Murray
  • 820
  • 2
    (1) is wrong. The correspondence in Lie's theorem is between simply-connected (connected) Lie groups and Lie algebras. (2),(3) are correct. (4) is wrong. SO(3) is connected because you can rotate the axis of rotation and rotate some more. – user10354138 Jul 28 '20 at 08:37
  • Wow the ones I was most sure of were the incorrect ones – Mark Murray Jul 28 '20 at 08:56

1 Answers1

3

(1) Is wrong. The kernel of $SU(2)\to SO(3)$ is $\{\pm I\}$. The element $-I$ is central in $SU(2)$, but $SO(3)$ has trivial center, so they cannot be isomorphic. (Both $SU(2)$ and $SO(3)$ are connected.)

(2) True.

(3) True.

(4) It's wrong to say $SO(3)$ has two connected component.

It's true to say $SO(3)$ is the identity component of $O(3)$, the unique other component being comprised of reflections (and "improper reflections").

anon
  • 151,657
  • Is there a nice way to think of $SU(2)$ as a discrete fibre bundle then? How can a discrete fibre bundle be connected? – Mark Murray Jul 28 '20 at 09:02
  • 2
    @MarkMurray There is an analogous situation in fewer dimensions. Namely, the double covering $S^1\to S^1$ given by $z\mapsto z^2$. It may be easier to think about the infinite covering $\Bbb R\to S^1$: imagine $\Bbb R$ as an infinite helix being projected onto a plane - the helix and the circle projection are both connected, and each fiber is a copy of $\Bbb Z$ (countably infinitely many isolated points). It is perhaps more rewarding to think of $S^1\to S^1$ as going from a plain rubber band to a twice-looped rubber band, since double covers of $SO(n)$ generalize this idea. – anon Jul 28 '20 at 09:07
  • Excellent example thank you, I'm guessing the only way to think of $SU(2)$ as a fibre bundle is then to think of it as the set $SO(3)\times Z_2$ but with the topology from $SU(2)$ – Mark Murray Jul 28 '20 at 09:40
  • I don't know how you would think of $SU(2)$ as the set $SO(3)\times\Bbb Z_2$. – anon Jul 28 '20 at 09:50