Let $G$ be a Lie compact group and $H<G$ be a normal subgroup. Then we have an $H$-principal bundle $$H\overset{i}{\to} G \overset{\pi}{\to}G/H.$$ The classifying space construction is functorial and we get maps $$ BH \overset{Bi}{\to} BG \overset{B\pi}{\to} B(G/H)$$

1) Is this a fibration in general?

We know that $EH = EG$ and we have a fibration $G/H\to BH\to BG$ and that $$BH= EG/H = EG\times_G G/H.$$ Therefore we can express $BH$ in terms of $EG$.

2) Can we express $BG$ in terms of $EH,BH,E(G/H),B(G/H)$?

3) Can we express $B(G/H)$ in terms of $EH,BH,EG,BG$?

An example: when $G = H\rtimes G/H$ (for example $O(2)=SO(2) \rtimes \mathbb{Z}/2)$ then $$ B(H\rtimes G/H)) = E(G/H)\times_{G/H} BH$$

Therefore I expect that maybe if we can express $G$ with some generalized semidirect product (or a sequence of them) maybe there is hope to answer question 2). At least I expect that if $BH\to BG \to B(G/H)$ is a fibration then $BG$ should be classified by some map in $B(Aut(BH))$.

  • $\begingroup$ I think it's not a fibration in general, even for discrete groups. In this case, however, it is homotopic to a fibration (that is, there is a fibration $E\to B$ with $B$ homotopic to $B(G/H)$, $E$ to $BG$ and the fiber to $BH$ $\endgroup$ – Max Apr 16 at 16:53

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