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A principal $G$-bundle has a cross section iff it is trivial (e.g. Husemoller's Fibre Bundles, 3rd ed., 8.3 in chapter 4).

A principal $G$-bundle is in particular a fiber bundle with fiber $G$.

My question: does there exist a group $G$ and a non-trivial principal $G$-bundle $p:E\rightarrow B$ that does have a cross section when considered as a mere fiber-bundle?

If so, I would be glad to see a simple example. Thanks!

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    $\begingroup$ By "fiber bundle", you mean locally trivial bundle with typical fiber $G$? I think you do not get more cross-sections just by ignoring the free right action of $G$ on the fiber bundle. $\endgroup$
    – Claudio Gorodski
    Commented Jun 25, 2012 at 11:36
  • $\begingroup$ Yes, locally trivial with fiber G. $\endgroup$
    – Shlomi A
    Commented Jun 25, 2012 at 11:46
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    $\begingroup$ I agree with Claudio. Specifically, I see no difference between the notions of a section of a principal bundle and a section of the same bundle considered as a mere fiber bundle. In either case, a section is just a map from the base to the total space which, followed by the projection, yields the identity map of the base. $\endgroup$
    – Andreas Blass
    Commented Jun 25, 2012 at 13:11
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    $\begingroup$ Maybe the question should have been phrased without the word "principal"? In other words, one could ask "Are there non-trivial fiber bundles whose fibers are homeomorphic to a fixed topological group $G$, but which do admit sections?" The answer is certainly "yes". For instance, every (real, say) vector bundle has a zero section, and the fibers are topological groups. Of course vector bundles are not principal bundles, since the transition functions are linear maps, whereas a principal bundle for the additive group $\mathbb{R}^n$ would have translations as its transition maps. $\endgroup$
    – Dan Ramras
    Commented Jun 25, 2012 at 18:33
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    $\begingroup$ The word "principal" in the question is intentional. Another way of putting it is whether applying the forgetful functor (from principal $G$-bundles over $B$ to fiber-bundles over $B$) to a non-trivial principal bundle might yield a fiber bundle with a global cross section. I tend to agree with Claudio's answer. Are sections in both categories indeed the same? $\endgroup$
    – Shlomi A
    Commented Jun 26, 2012 at 8:47

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Let $p\colon E \to B$ be a principal $G$-bundle and $s\colon B \to E$ a global section. Then the map $F\colon B\times G \to E$, $F(x,g) = s(x)\cdot g$ is a global trivialization of $E$. Here the dot denotes the right action of the group $G$. The map $F$ is surjective, because the action is transitive on fibres, and it is injective because the action is free. Continuity/smoothness is the same as the continuity/smoothness of your section $s$ and the action of the group $G$.

There's no special requierement in the definition of a section of a principal $G$-bundle, it's still the section of the underlying fibre bundle. But once you have a global section, you just use the $G$-action to trivialize the bundle.

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  • $\begingroup$ Hi Oldřich. Thank you for your answer. You must have meant $F(x,g)=s(x)\cdot g$. I'm not sure that I got your answer. My question was whether it is possible that a principal $G$-bundle does not have a section, but does have one when considered just a fiber bundle? $\endgroup$
    – Shlomi A
    Commented Jun 27, 2012 at 8:33
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    $\begingroup$ Yes, $F(x,g)=s(x)⋅g$, sorry about that, I've corrected it. So maybe you should ask this: Is the definition of a section of a principal bundle different from the definition of a section of a general fibre bundle? You should find out that the answer is: no. Let $p\colon E\to B$ be a fibre bundle, then a section of $p$ is a mapping $s\colon B\to E$ such that $p\circ s=\text{id}_{B}$. And a principal bundle is first of all a fibre bundle, plus some additional structure. A section of a principal bundle is the same as a section of the underlying fibre bundle. $\endgroup$
    – Oldřich Spáčil
    Commented Jun 27, 2012 at 13:23

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