# Local triviality of principal bundles

Suppose I define a principal $G$-bundle as a map $\pi: P \to M$ with a smooth right action of $G$ on $P$ that acts freely and transitively on the fibers of $\pi$. Does it follow that $P$ is locally isomorphic to $M \times G$ with the obvious right action of $G$ on $M \times G$? Let's suppose $M$ is a manifold.

I know that fiber bundles over a contractible set are trivial and a manifold is locally contractible, but I believe this statements refers to locally trivial fiber bundles and so will not apply to this case.

A related question is: if we have a fibration such that the base space is contractible and all fibers are homeomorphic, does it follow that the fibration is just the product of the base with the fiber?

Thanks!

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You should maybe say for any $q\in M$ there is a neighborhood $U$ such that $\pi^{-1}(U)\simeq U\times G$. Being "locally isomorphic" to $M\times G$ is sort of awkward wording since $M\times G$ is "global". Also you use the term "locally trivial fiber bundles" to distinguish from "fiber bundles", but to my knowledge, all definitions of fiber bundles include being locally trivial. It wouldn't be a "bundle" otherwise. – Matt Feb 25 '11 at 18:28
(cont) Lastly, in your setup, by definition you have that the fibers are isomorphic to $G$, so you only need to check local triviality, since what it is locally trivially isomorphic to must be $U\times G$ otherwise the fibers wouldn't be the right thing. – Matt Feb 25 '11 at 18:31
@Matt: google.com/… – Eric O. Korman Feb 25 '11 at 18:35
It is a theorem of Ehresmann's that the two notions coincide: en.wikipedia.org/wiki/Ehresmann%27s_theorem To that end you might be interested in the following: mathoverflow.net/questions/17826/… – Sean Tilson Feb 27 '11 at 13:56
I could be missing something, but doesn't Ehresmann's Theorem say that you need a proper submersion $f:P \to M$ in order to conclude that $f$ fibers $P$ locally trivially over $M$? The 'proper' part says that the inverse image of every compact set is compact, and this won't be true of the projection map $\pi$ unless the fiber is compact. So it seems to me that you only get local triviality of the bundle for free if the fiber is compact? – Idempotent Dec 19 '15 at 23:12

For the first question - yes, at least if you suppose $P$ is a smooth manifold and, say, $G$ is a Lie group. For the principal $G$-bundles with your definition, just as with the regular one, being trivial is the same as admitting a section (in the section is $s$, map $s(x)$ to ($x$, unit of $G$), and use the action to define the rest of the trivializing map). Now to construct a section locally, near $x_0$ take any $s(x_0)$ in the fiber above $x_0$. Pick an auxiliary Riemann metric near take the orthogonal subspace to the tangent of the fiber at $s(x)$. Exponential map will give you a section locally.

In other categories you would need to construct the section $s$ in a different manner. I think this can be done in the category of topological manifolds. Not sure about more general cases, but it would seem ok. Maybe for CW complexes you can go cell by cell?

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nice answer, thanks! – Eric O. Korman Feb 25 '11 at 20:53
(I know this question is old!) maybe this was implicit for both of you, but I'm pretty certain you need the right $G$ action on $P$ to be proper : $\mathbb{R}$ acts freely and smoothly on the torus $\mathbb{R}/\mathbb{Z}\times \mathbb{R}/\mathbb{Z}$ through $([x],[y])\cdot t=([x+t],[y+\alpha t])$ with $\alpha$ irrational, yet gives not rise to a principal $G$ bundle. Of course there is no problem when $G$ is compact. – Olivier Bégassat Aug 13 '11 at 0:39
Max, If I am not mistaken, your proof assumes that the fibre is a smooth submanifold of the smooth manifold P. As pointed by @OlivierBégassat this need not be the case if the action is not proper. Could any of you please give me a reference to a proof in the CW category with appropriate hypothesis ? Thanks. – user90041 Apr 1 at 15:48
@user90041: See my answer to math.stackexchange.com/questions/560371/… – studiosus Jun 5 at 5:51