4
$\begingroup$

Let $\pi:E\rightarrow B$ be a (smooth) principle bundle with structure group $G$.

By definition, there exists a reduction of the structure group to a subgroup $H<G$, if there exists a global section $s:B\rightarrow E/H$.

Now, clearly, if $G$ acts freely and fiber-preserving on $E$, then so does any subgroup $H<G$. And if $G$ acts transitively on $E$, so does $H$ on $E/H$... so $E/H\rightarrow B$ is a principle $H$-bundle, right?

But then, there is the following fact:

If a principle bundle admits a global section, then it is already trivial.

Question So doesn't the definition of a reduction imply that $E/H\rightarrow B$ is a trivial bundle?

$\endgroup$
1
  • 1
    $\begingroup$ A good example to work out is the frame bundle to $S^2$ and it's reduction to the orthogonal frame bundle (aka the unit tangent bundle to the sphere). $\endgroup$ Commented May 22 at 23:18

1 Answer 1

9
$\begingroup$

$E/H \to B$ is a fiber bundle with fiber $G/H$, it is not a principal $H$-bundle (note, $H$ acts trivially on $E/H$). Therefore, the existence of a section does not imply that the bundle is trivial.

$\endgroup$

You must log in to answer this question.

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