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Feb 6, 2018 at 18:41 comment added Ben McKay @hänsel: Yes: both are canonically diffeomorphic to $X$.
Feb 6, 2018 at 18:14 comment added hänsel is $P_G/G=P/H$? (answering construction of map (1.2) on page 548 of paper "Atiyah-Singer III")
Aug 30, 2016 at 10:58 comment added Ben McKay I didn't notice this: $H$ doesn't act trivially on $G/H$ by left translation, so $H$ doesn't act trivially on $P_{G/H}$. So $P_{G/H}$ is a bundle whose fibers are diffeomorphic to $G/H$, and has a $G$-action, but is not principal, and has no $G/H$-action.
Aug 30, 2016 at 10:09 comment added far I am assuming that $0 \to H \to G \to G/H \to 0$ is a short exact sequence of groups, so that $G/H$ is a group and the maps in the sequence are group homomorphisms. By functoriality I would expect the $G/H$ bundle to be a principal bundle as well. Am I missing something?
Aug 29, 2016 at 22:51 comment added Qiaochu Yuan What is a short exact sequence of bundles? Note that even when $G/H$ is a group, the action of $G$ on it is not an action by group homomorphisms, so the associated bundle with fiber $G/H$ is not a bundle of groups.
Aug 29, 2016 at 21:10 comment added far I mean a short exact sequence.
Aug 29, 2016 at 19:07 history answered Ben McKay CC BY-SA 3.0