I am trying to understand better the connectedness, simply-connectedness, and disconnectedness of Lie groups $L$ have anything to do with giving a nontrivial finite extension such that $G/N=L$ with $G$ is the finite extension of the Lie group $L$ by a finite group $N$ (as a normal subgroup of $G$). The nontrivial finite extension means that $G$ is not the product group as $G \neq N \times L$.

It looks to me that:

(i) If $L$ is connected but not simply-connected, there is always a nontrivial finite extension.

(ii) If $L$ is connected and simply-connected, there is no nontrivial finite extension.

(iii) If $L$ is disconnected, there are still examples with nontrivial finite extensions, such as the disconnected $L=O(n)$ Lie group, we can find $Pin^{\pm}(n)/\mathbb{Z}_2=O(n)$.

Questions: Are there any more nontrivial finite extension examples of a disconnected $L$ lie groups? (other than $Pin^{\pm}(n)/\mathbb{Z}_2=O(n)$ I gave.)

Do we require that $\pi_1(L)\neq 0$, even though $\pi_0(L)\neq 0$ (disconnected) is still OK?

What are general statements one can make about the disconnected Lie groups and their finite extension existence or not?


There are two issues here: First any two connected components of a Lie group are diffeomorphic (via left translations). Hence it is no problem that $\pi_1(L)$ "sees" only the connected component of the identity.

Second, it is tricky to say what "non-trivial" should mean in the disconnected case. For example, take the extension $Spin(n)\to SO(n)$ by $\mathbb Z_2$. Then choose any finite group $F$ and put $L:=SO(n)\times F$ and $G:=Spin(n)\times F$. Then clearly $G\to L$ is again an extension by $\mathbb Z_2$ and both $G$ and $L$ have $|F|$ connected components. If you consider this as non-trivial then there are too many examples. On the other hand, finding a definition which makes this extension "trivial" seems hard to me.

As I said in a reply to an earlier question I think that right way to study this is to first look for connected extensions of the connected component $L_0$ of the identity of $L$. Having found a "good" extension try to build up an extension of all of $L$.

  • $\begingroup$ Thanks, +1, maybe I will ask further issues later. $\endgroup$ – wonderich May 10 '17 at 3:26

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