# A fibration of classifying spaces

Let $G$ be a Lie group, $N$ a closed connected normal subgroup. Let $BG$, $BN$, $B(G/N)$ be the classifying spaces of $G,N$ and $G/N$. Is there a fibration $BN\to BG\to B(G/N)$ ?

It seems that such a construction is used in Atiyah, Bott: Yang-Mills equations on Riemann surfaces, formula (9.2), but I can't see how it works.

Thanks a lot!

• A short remark, slightly off-topic, because it's not exactly the setting of the question, but it might be worth a comment anyway: for $G$ a discrete group (and the classifying spaces constructed in as simplicial sets, i.e., as $G$-quotient of $EG_\bullet=G^{\bullet+1}$), there is an alternative argument. The map $BG\to B(G/N)$ is a fibration. This is proved much the same way that $BG$ is fibrant (cf. Goerss-Jardine: Simplicial homotopy, Lemma 3.5). But the fiber (which now is also the homotopy fiber) is obviously $BN$. – Matthias Wendt Oct 5 '14 at 17:09
• As an aside, in this situation you can use the fact that the projection EGxE(G/N) -> E(G/N) is an easy model for EG -> E(G/N), and find that after quotient it is a fiber bundle with fiber (EG)/N, which models BN. – Tyler Lawson Oct 6 '14 at 12:49

This is an edited extract from a book in preparation (Bruner, Catanzaro, May) tentatively titled Characteristic Classes and is therefore overlong for an answer. This is similar to Denis Nardin's answer, but more bundle theoretic; he did refer to an old Memoir of mine, so I thought I'd give an answer. Let $N$ be a closed normal subgroup of a topological group $G$ with quotient group $K$. Let $i\colon N\longrightarrow G$ be the inclusion and $j\colon G\longrightarrow K$ the quotient map. Let $EG \longrightarrow BG$ and $E K \longrightarrow B K$ be universal bundles for $G$ and $K$ and take $EG \longrightarrow EG/ N = B N$ to be the universal bundle for $N$, so that inclusion of orbits gives $Bi\colon BN \longrightarrow BG$ as a bundle with fiber $K$. There is a map $Ej\colon EG \longrightarrow E K$ such that $(Ej)(yg) = (Ej)(y)j(g)$, either by a functorial construction or general principles. Then passage to orbits gives $Bj\colon BG \longrightarrow BK$. By construction this gives a bundle map [this toy does not seem to tex \xymatrix] from the bundle $Bi$ to the bundle $EK\longrightarrow BK$, so that $BN$ is the pullback along $Bj$ of the universal bundle for $K$. Since $EK$ is contractible, this implies that $BN\longrightarrow BG \longrightarrow BK$ is a fibration sequence. [Further details are standard, but I'll supply if wanted.]

By functoriality there is a map $BG\to B(G/N)$. Let $X$ to be its homotopy fiber. Then you can get a fiber sequence

$\Omega X \to G \to G/N \to X \to BG \to B(G/N)$

using the fact that $\Omega BG = G$ for every (nice enough) topological group $G$. Then, since $N$ is the fiber of $G\to G/N$ we have that $\Omega X = N$. So we only need to show that $X$ is connected to conclude that $X=BN$. Consider then the long exact sequence of homotopy groups induced by $X\to BG\to B(G/N)$. It is

$\pi_1(BG)\to \pi_1(B(G/N))\to \pi_0(X)\to \pi_0(BG)=*$

So we just need to show that the map $\pi_1(BG)\to \pi_1(B(G/N))$ is onto. But this is obvious, since this coincides with $\pi_0(G)\to \pi_0(G/N)$, which is certainly onto.

As a final note, I think that a reference for these matters can be the book by May "Classifying spaces and fibrations".

• Thank you so much for your help! Just a short question: why does $\Omega X=N$ imply $X=BN$? Is it a general fact that $\Omega X\simeq\Omega Y$ implies $X\simeq Y$ when $X,Y$ are connected? – Boyu Zhang Oct 5 '14 at 15:22
• A weak equivalence $\Omega X\simeq \Omega Y$ does not necessarily imply the existence of a weak equivalence $X\simeq Y$ (or of the connected components of the base points, for that matter). You need that the weak equivalence is in a suitable sense compatible with the loop space structure, and I think this is not automatic. In the argument of the answer, this is implicit in the fiber sequence argument. – Matthias Wendt Oct 5 '14 at 16:55
• Yeah, sorry I should have made that more explicit. Both $\Omega X$ and $N$ are $E_1$ space in a compatible way because the map $N\to G$ is a map of $E_1$ spaces (in this case, of topological groups) – Denis Nardin Oct 5 '14 at 18:39
• Or maybe more clearly: the equivalence $N\cong \Omega X$ is induced by a map $BN\to X$, given by the fact that we have an explicit model where $BN$ is the fiber of the map $BG\to B(G/N)$ – Denis Nardin Oct 5 '14 at 18:44