8
$\begingroup$

Let $M=G/H$ be a homogeneous manifold, with $G$ connected Lie group. Suppose that $\widetilde{M}$ is a covering of $M$.

QUESTION: is there a general prescription to obtain a Lie group $\widetilde{G}$, starting from $G$, in such a way that $\widetilde{M}=\widetilde{G}/\widetilde{H}$?

Using the case of $M=G=S^1$ and $H=1$ as a toy model (and, for instance, $\widetilde{M}=\widetilde{G}=\mathbb{{R}}$) we see that $\widetilde{G}$ has two remarkable properties:

1) it contains the group $\Gamma=\mathbb{{Z}}$ of "gauge symmetries" of $\widetilde{M}\to M$;

2) its factor by $\Gamma$ returns the original group $G$.

So, I guess that the two properties above are enough to characterise $\widetilde{G}$ but I'm not able to prove it. I'm sure it's a well-known result, but I can't find any reference (I could not get which book is this "Bredon" mentioned here: lifting group action). In the case that my guess is correct, I'd like to understand if there is a constructive way to obtain $\widetilde{G}$, e.g., by realising the Lie algebra $\frak{g}$ as vector fields on $M$, lifting them to $\widetilde{M}$, and then take the group generated by their flows.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

If $G$ is connected, take its Lie algebra, acting as vector fields on $M$. Lift the vector fields by the covering map. There is a unique connected Lie group $\tilde{G}$ acting on $\tilde{M}$ whose Lie algebra has this action, by a theorem of Dick Palais: http://en.wikipedia.org/wiki/Lie%E2%80%93Palais_theorem

Improve this answer
$\endgroup$
8
  • $\begingroup$ Indeed. This is more or less what I suggested in my last lines. I'm aware of Lie-Palais theorem, though you need to require some extra topological conditions from $\widetilde{M}$. Still, I'd like to see some universal property characterising $\widetilde{G}$, like, e.g., "it is the unique group admitting $G$ as a factor by covering transformations", or something like that, and/or some algebraic way to construct it out of the available data! $\endgroup$
    – Giovanni Moreno
    Commented Sep 29, 2014 at 15:28
  • $\begingroup$ The only extra condition you need on $\tilde{M}$ is that it is connected. $\endgroup$
    – Ben McKay
    Commented Sep 29, 2014 at 15:39
  • $\begingroup$ The fundamental group of $G$ lies inside the universal covering group of $G$. The group $\Gamma$ you want is the subgroup of $\pi_1(G)$ which maps trivially to $\pi_1(M)$ when a loop in $G$ carries a base point of $M$ around. $\endgroup$
    – Ben McKay
    Commented Sep 29, 2014 at 15:43
  • $\begingroup$ Don't you need $\widetilde{M}$ to be compact? $\endgroup$
    – Vít Tuček
    Commented Sep 29, 2014 at 20:35
  • 1
    $\begingroup$ @G_infinity: you cannot always embed every Lie algebra into the Lie algebra of vector fields on every manifold. For example, you cannot embed the Lie algebra of $SU(2)$ into the vector fields on the circle, or else $SU(2)$ would have a nontrivial action on the circle, and by compactness would have a compact orbit, so the circle itself, with stabilizer of a point a closed subgroup of codimension 1, i.e. dimension 2. But the Lie algebra of $SU(2)$ is cross product, so you can picture the cross product on a plane in 3-dimensions: no invariant plane. So no subgroup of dimension 2. $\endgroup$
    – Ben McKay
    Commented Sep 30, 2014 at 9:57
3
$\begingroup$

The Palais theorem assumes that the manifold $\tilde M$ is compact. The positive answer gives proposition 6 of the Onishchik book "Topology of transitive transformation groups". It states : For any action of a Lie group $G$ on a manifold and any covering $\pi : N \to M$ there is an action of the universal cover $\tilde{G}$ on $N$ which cover the action of $G$ on $M$, i.e. such that the projection $\pi : N \to M$ is $\tilde{G}$-equivariant.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Welcome to Math Overflow, @Dmitri Alekseevsky! $\endgroup$
    – Mikhail Borovoi
    Commented Dec 29, 2015 at 18:19
  • $\begingroup$ It seems that Palais doesn't assume compactness of any manifold in his book. He assumes completeness of a finite dimensional Lie algebra of vector fields. The wikipedia entry simplifies the presentation of the result, by assuming compactness of the manifold, in order to ensure completeness of the vector fields. $\endgroup$
    – Ben McKay
    Commented Apr 30, 2017 at 13:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .