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It is well-known that the universal cover $\tilde G$ of a connected Lie group $G$ has a Lie group structure such that the covering projection $\tilde G\to G$ is a Lie group morphism. Of course $\tilde G$ might not be linear even though $G$ is, but this is not the point here.

My question is: assume that $G$ is a not necessarily connected Lie group. Does there exist a Lie group $\tilde G$ and an onto Lie group morphism $\tilde G\to G$ whose restriction to the identity component of $\tilde G$ is the universal cover of the identity component of $G$?

I assume that the answer is "no" in general, but I could not find any counter-example.

@Jim: Of course, the terminology "universal cover" would have been inappropriate even though such a cover existed (which as you and André pointed out, is not the case).

I came to this question from some other direction. Namely, the $PSL_2(R)$ action on $RP^1$ lifts to a $\widetilde{PSL_2(R)}$ action on the universal cover of $RP^1$, and this action extends to an action of a two-sheeted cover of $PGL_2(R)$. It is tempting to denote this cover by $\widetilde{PGL_2(R)}$, and I wondered whether such a construction was standard.

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The Pin groups of Atiyah-Bott-Shapiro (covering orthogonal groups) are the most natural examples of what you are looking for, but I'm not at all sure about the existence of such a construction for arbitrary non-connected Lie groups. (Calling it a "universal cover" as in your header might be overkill, since that is such a standard term: usually a universal cover is itself simply connected, in particular connected.) – Jim Humphreys Nov 16 '10 at 17:31

2 Answers 2

up vote 12 down vote accepted

The group $Pin_-(2)$ is an example of what you're looking for.
It can be described explicitly as a subgroup of the group of unit quaternions:
$Pin_-(2)=$ { $a+bi| a^2+b^2=1$ } $\cup$ { $cj+dk| c^2+d^2=1$ } $\subset \mathbb H^\times$.

Its main interesting properties are:
- The conjugation action of $\pi_0$ on its Lie algebra is non-trivial.
- All the elements of the non-identity component have a non-trivial square.

There is no Lie group that is diffeomorphic to $\mathbb Z/2\times \mathbb R$ and that shares those properties.

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Thank you André! This is the group I was originally testing, but for some reason I thought that it were no counter-example. I have to check your last assertion in order to credit your answer. – Andrei Moroianu Nov 16 '10 at 17:28
I first learned about $Pin_-(2)$ from my Berkeley advisor Allen Knutson. He had asked me the following question: "how many group structures are there on the disjoint union of two circles?". The answer is three. – André Henriques Nov 17 '10 at 19:54

Expanding on Andre's answer, there is an obstruction class in $H^3(\pi_0(G),\pi_1(G,e))$ (due to R.L. Taylor, Covering groups of non connected topological groups, Proc. Amer. Math. Soc. 5, pp753-768, 1954) to the existence of a universal covering space. There is a University of Wales thesis by Mucuk with the main results contained in this paper by Brown and Mucuk which detail when you can get a universal covering space that acts as you want.

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