Fiber bundles of $G$-spaces

Question

So if $G$ is a topological group and $H,\ J$ are closed subgroups such that $H\lhd J$, then the principal bundle $G/H\to G/J$ is trivial iff it has a global section.

I have questions about the general case:

1. If $H$ and $J$ are closed subgroups of $G$ such that $H<J$ (not necessarily normal), then is the bundle $G/H\to G/J$, $\ \ gH\mapsto gJ$ trivial iff it has a global section?
2. If $G$ has a closed normal subgroup $N$ that acts transitively on the base $G/J$ and trivially on the fiber $J/H$. Is the bundle $G/H\to G/J$ trivial?

Answer 1

For 1, of course if the bundle is trivial it has a global section. But the converse need not hold.

Consider the chain of subgroups $SO(2n)\subseteq SO(2n+1)\subseteq SO(2n+2)$. Then the bundle in 1 is the unit tangent bundle $T^1S^{2n+1}\rightarrow S^{2n+1}$. A section of this bundle is essentially a non-vanishing vector field on $S^{2n+1}$, so exists for all $n$. On the other hand, Adams showed this bundle is trivial only for $n=0,1,3$.

For 2, I'm not sure what you're asking. How does $N$ act on $J/H$?

Answer 2

To answer your first question, the fiber bundle being trivial is actually a more general result about principle fiber bundles. The group structure doesn't matter since you essentially only need an arbitrary manifold as your base.

Suppose $(P,\pi, M, G)$ is a principle fiber bundle where $P\times G\to P$ is a right action and $\pi:P \to M$ is projection. Then $P$ is trivial if and only if there exists a smooth global section.

Proof: The first direction is trivial (if $P$ is trivial then it is a product bundle $P = M\times G$). Conversely, suppose there is a smooth global section $\sigma:M\to P$. Then we can define a diffeomorphism $\Phi:P \to M\times G$ as $$ \Phi(p) \;\; =\;\; \left (\pi(p), \hat{g}\right ) $$ where $\hat{g} \in G$ is the unique group element that has $\sigma(\pi(p)) \cdot \hat{g} = p$. Such a group element is unique since the action of $G$ is required to be free. $\Phi$ is clearly smooth and also has a smooth inverse $\Phi^{-1}:M\times G\to P$ given by $$ \Phi^{-1}(x,g) \;\; =\;\; \sigma(x)\cdot g. $$ $$\tag*{$\Box$}$$