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I am following these lectures on principal fiber bundles. Here, a principal fiber bundle is defined as a fiber bundle of which total space $P$ has a right free action of some Lie group and which is bundle isomorphic to the bundle $P\longrightarrow P/G$, with the canonical quotient space projection map. The lecturer goes on to deduce from this a few properties of principal bundles, such as the fibers being diffeomorphic to the group solely from the fact the action is free, the action conserving the fibers and so on and so forth.

My problem is that recently I started to complement these lectures with other books, and in most of the literature principal bundles are defined with (seemingly) more conditions. In them, a principal fiber bundle is defined in a similar setting but with a fibre preserving simply transitive right action (so, not only free but surjective), and, additionally, a G-equivariant atlas (that is, with bundle charts "commuting" with the action). So I'd like some help in obtaining properties from this definition from the first one I have.

  1. Firstly: the lecturer of the first definition seems to ignore that $G$ be compact in order for the orbits to be diffeomorphic to it. Does the first definition nonetheless somehow make up for it? How would I go about showing it? I tried to show that the orbit map is a diffeo by composing it with the bundle isomorphism maps but couldn't really see if that works.
  2. Secondly, how do I prove that there exists a G-equivariant bundle atlas? My first guess was composing the bundle charts in the bundle atlas of the principal bundle with bundle charts of the second bundle, via the isomorphism maps, but I can't see how those interact with the action since the compatibility conditions only make mention of the projections. I would suppose information about the action is in the map which projects to the orbit space, but I could not extract it.
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1 Answer 1

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similar setting but with a fibre preserving simply transitive right action

Well, for any group $G$ acting on a set $P$, if you look consider the fibers of the quotient set $P/G$, these are by definition the orbits of the group. So, transitivity of the action on the fibers is automatic, meaning one only needs to impose freeness, if you decide to introduce the concept this way.

Other times, people might introduce the definition by allowing for a possibly different base manifold $B$ which is not $P/G$ set theoretically. But really, the choice of set is not what matters but rather how the group action $G$ plays with the projection map $P\to B$, i.e what we require is, roughly speaking, a surjective map $\pi:P\to B$ such that the action of $G$ on $P$ restricts, for each $b\in B$, to a transitive action on the fiber $\pi^{-1}(\{b\})$ (which is equivalent to fixing a point $p$ in this fiber and saying this fiber equals the orbit of $p$ under $G$).

So, as far as definitions are concerned, they may look different, but that is only superficial.


Question 1.

You do not need compactness of the group. Take a look at the following smooth orbit-stabilizer theorem. The definition of a principal bundle in the first sense requires that

  • $G$ be a Lie group which acts smoothly on the right on a smooth manifold $P$.
  • The orbit manifold $P/G$ must exist, i.e there must exist a smooth structure on $P/G$ such that the underlying topology is the quotient topology for the projection $\pi:P\to P/G$, and this map $\pi$ must be a smooth submersion.
  • $G$ must act freely on $P$.

What the lecturer doesn’t emphasize is the second bullet point. Just FYI: A necessary and sufficient condition (I think this is called Godement’s criterion) is that the orbit relation \begin{align} \mathcal{R}:=\{(x,y)\in P\times P\,:\, \text{$x,y$ lie in the same $G$-orbit}\} \end{align} must be a closed subset and a smooth embedded submanifold of $P\times P$. This is somewhat difficult to check in practice, so we typically settle for a simpler sufficient condition, namely that $G$ acts freely and properly (and properness will trivially be satisfied if $G$ is compact).

So, now by closedness of the orbit relation, it implies each orbit is closed (it is the preimage of a singleton (which is closed since we have a Hausdorff topology on $P/G$) under the projection $\pi:P\to P/G$), so by the result in the link, we have that for each $p\in P$, the orbit $\mathcal{O}_p$ is an embedded submanifold of $P$ diffeomorphic to the quotient $G/\text{Stab}(p)$, and this latter set is $G/\{e\}$ since the action is free, so trivially diffeomorphic to $G$.


Question 2.

First what you do is show that locally there always exist smooth sections of $\pi:P\to P/G$. This is a simple consequence of the fact that $\pi$ is a surjective submersion. So, fix a smooth local section $\phi:U\subset P/G\to \pi^{-1}(U)\subset P$. Since we have a free action, we see that the map $\Phi:U\times G\to \pi^{-1}(U)$, $(x,g)\mapsto \phi(x)\cdot g$ is injective (surjectivity is obvious), so it is bijective, and it is certainly smooth and it preserves base points. All that remains is showing that the inverse is smooth. This follows by an inverse-function theorem type of argument so I omit it.

So, $\Phi$ is now a diffeomorphism $U\times G\to \pi^{-1}(U)$ which preserves the base points. On the domain, we have an obvious right action of $G$ namely $(x,s)\cdot g:=(x,sg)$ where $sg$ is the usual group product. On the target, $\pi^{-1}(U)$ also has a natural action of $G$ by restricting the one on $P$. Observe that we have for each $(x,s)\in U\times G$ and $g\in G$, \begin{align} \Phi((x,s)\cdot g):=\Phi(x,sg):=\phi(x)\cdot (sg) = [\phi(x)\cdot s]\cdot g=\Phi(x,s)\cdot g, \end{align} so we have an equivariance condition satisfied by $\Phi$.

To summarize starting from the first definition, the way you get an equivariant bundle trivialization is by taking a local section (which is a way to ‘lift’ the base manifold into the bundle), and then you act by $G$ to move up and down the fiber (freeness and fiberwise-transitivity then make this all bijective).

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  • $\begingroup$ This is such a good and complete answer, thank you. I'll need to go over it slowly. A short preliminary question though: I can see how the action is simply transitive on the fibers of $P/G$, but can't see how it is also transitive on the fibers of $M$. I tried proving with help of the isomorphism maps, but at the last step I don't know how these maps interaction with the action so I can't invert everything. $\endgroup$ Commented Jan 19 at 15:46
  • $\begingroup$ @LourencoEntrudo What is $M$? $\endgroup$
    – peek-a-boo
    Commented Jan 19 at 16:34
  • $\begingroup$ sorry, the base space of the principal bundle $P$ $\endgroup$ Commented Jan 19 at 16:41
  • $\begingroup$ If you start with an arbitary bundle with arbitary base, then you must demand by assumption the action is transitive on fibers. That’s what I said in the beginning. If you simply start with a manifold $P$ and an action on it, then the action will trivially be transitive in the fibers of $P/G$. $\endgroup$
    – peek-a-boo
    Commented Jan 19 at 16:43
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    $\begingroup$ @LourencoEntrudo I think I see where your confusion is now. The lecturer merely says $P\to M$ must be isomorphic as a bundle to $P\to P/G$, but here I think it should be a little more restrictive by saying the isomorphism must be of the form $(\text{id}_P,f)$ for some diffeomorphism $f:M\to P/G$. Because the point is that we want the fibers to be equal to the orbits (so that the group action of $P$ restricts to a transitive action on the fibers). $\endgroup$
    – peek-a-boo
    Commented Jan 19 at 19:30

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