Splitting of a Short-Exact Sequence of Lie Groups

Question

Let $G$ be a Lie group (possibly disconnected). Consider the natural short-exact sequence $$1\rightarrow G_0\rightarrow G\rightarrow\pi_0(G)\rightarrow 1,$$ where $G_0$ is the identity component of $G$, and $\pi_0(G)\cong G/G_0$ is the component group. Under what conditions does this sequence admit a splitting, so that $G$ is a semidirect product of $G_0$ and $\pi_0(G)$?. Useful answers might include (but need not be limited to) some types of Lie group $G$ for which a splitting exists. I would appreciate any and all references and suggestions.

Thanks!

Answer 1

See pages 177 to 190 of:

Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008. (pdf)

Answer 2

There is a crossed module associated to this situation, which will be trivial if $G \to \pi_0(G)$ splits on the level of gropus. Thus an obstruction for the existence of a splitting is the non-triviality of the crossed module.

In some more detail, the crossed module is given by the universal covering $\tilde{G_0}\to G_0$, composed with the inclustion $G_0\to G$. The action is given by lifting the conjugation action of $G$ on $G_0$ to $\tilde{G_0}$ (note that each automorphism of $G_0$ lifts in a unique way to its simply connected cover). This defines the smooth crossed module $\tilde{G_0} \to G$. This has an obstruction class in the group cohomology $H^3(\pi_0(G),\pi_1(G))$, where the action of $\pi_0(G)$ on $\pi_1(G)$ in induced by the conjugation in $G$.

The purpose of the obstruction class is twofold. On one hand, it is an obstruction for the existence of the splitting. On the other hand, it allows you to reduce the question to simply connected $G_0$ in case that the obstruction vanishes, since then there exists a $H$ with $H_0$ simply connected and $\pi_0(H)=\pi_0(G)$ and $G\to\pi_0(G)$ splits if $H\to \pi_0(H)$ does so. Details on the latter arguing can be found in Theorem III.8 of http://arxiv.org/abs/math/0504295 (I hope that I am not overseeing some smoothness issues here).

Answer 3

Having consulted Peter Michor's book and a few other references, I would like to describe a few of my conclusions. We note that $G_0\rightarrow G\rightarrow\pi_0(G)$ is a principal $G_0$-bundle over the discrete group $\pi_0(G)$. Hence, the bundle is trivial, and there is a (non-canonical) global section $s:\pi_0(G)\rightarrow G$ satisfying $s(e)=e$ (yielding a manifold isomorphism $G_0\times\pi_0(G)\cong G$). However, it may not be possible to choose $s$ to be a morphism of Lie groups, since it is known that $G$ need not be a semidirect product of $G_0$ and $\pi_0(G)$. So, we may endow the manifold product $G_0\times\pi_0(G)$ with the Lie group structure for which our above isomorphism is actually an isomorphism of Lie groups.