# Expressing a term in $BH\to BG \to B(G/H)$ as a function of the other two. Classifying spaces and fibrations.

Let $$G$$ be a Lie compact group and $$H be a normal subgroup. Then we have an $$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 $$B(Aut(BH))$$.

• 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 $BH$ – Max Apr 16 at 16:53