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Given an extension of groups, say $$0 \to H \stackrel{i}{\to} G \stackrel{q}{\to} G/H \to 0$$ and a $H$-principal fiber bundle $P \to X$, one can use induction to obtain bundles with fibers $G$ and $G/H$, say $P_G \to X$ and $P_{G/H} \to X$.

Under which assumptions does one obtain an extension of principal bundles?

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    $\begingroup$ $P_{G/H}$ is a trivial $(G/H)$-bundle. $\endgroup$ – abx Aug 29 '16 at 19:08
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You set $P_H=P$, $P_G=P \times^H G$ and $P_{G/H}=(P_G)/H=P \times^H (G/H)$, so you get obvious bundle maps $P_H \to P_G \to P_{G/H}$. Is that what you mean by extension of bundles?

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  • $\begingroup$ I mean a short exact sequence. $\endgroup$ – far Aug 29 '16 at 21:10
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    $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Aug 29 '16 at 22:51
  • $\begingroup$ 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? $\endgroup$ – far Aug 30 '16 at 10:09
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    $\begingroup$ 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. $\endgroup$ – Ben McKay Aug 30 '16 at 10:58
  • $\begingroup$ is $P_G/G=P/H$? (answering construction of map (1.2) on page 548 of paper "Atiyah-Singer III") $\endgroup$ – hänsel Feb 6 '18 at 18:14

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