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Every Lie group is the semidirect product of a connected Lie group and a discrete group.

I think the component of the identity could be useful.

ಠ_ಠ
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Gsanm
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1 Answers1

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Warning. Apparently the proof is not correct. I will update it later.

For every topological group $G$ with identity component $G^0$ it is well-known and easy to prove that $G^0$ is a normal, in fact characteristic, closed subgroup. If $G$ is locally connected, $G^0$ is open, which implies that the quotient group $G/G^0$ is discrete. The Lie algebra of $G$ is isomorphic to the Lie algebra of $G/G^0 \times G^0$, hence these Lie groups are locally isomorphic. It is not hard to see that the isomorphisms glue to a global isomorphism, since the decomposition into a product of a connected and a discrete Lie group is essentially unique.

By the way, the corresponding statement for topological groups is wrong.

Martin Brandenburg
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  • What about, for example, $O(2n+1)$? As a manifold, it's diffeomorphic to $SO(2n+1) \times \mathbb{Z}/2$, but not as a group (since $O(2n+1)$ has trivial center but $SO(2n+1)\times \mathbb{Z}/2$ doesn't). Or am I missing something stupid? – Jason DeVito - on hiatus Feb 28 '13 at 00:13
  • Why does $O(2n+1)$ has trivial center? If your counterexample is correct, you should add it as an answer (or "anti-answer" ;)). – Martin Brandenburg Feb 28 '13 at 01:04
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    Ha! I got it backwards. $-I \in O(2n+1)$, and in fact $O(2n+1)\cong SO(2n+1)\times\mathbb{Z}/2$ (as Lie groups). On the other hand, $O(2n)$ is not Lie isomorphic to $SO(2n)\times\mathbb{Z}/2$. For example, when $n=1$, $SO(2)\times\mathbb{Z}/2$ is abelian, but generally reflections and rotations don't commute. (I'm not going to add this as an aswer because the OP has now added a correction asking for a semi-direct product instead of direct product. I'm not sure how to prove it.) – Jason DeVito - on hiatus Feb 28 '13 at 03:25