7
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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.) $\endgroup$ Nov 16, 2010 at 17:31

2 Answers 2

12
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Nov 16, 2010 at 17:28
  • 3
    $\begingroup$ 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. $\endgroup$ Nov 17, 2010 at 19:54
8
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.