Classification of fibrations for classifying spaces $B^2\mathbb{Z}_2$ and $BSO(3)$ or $BO(3)$

Question

I am interested in knowing what can we say about the classification of fibrations for classifying spaces $B^2M \equiv B^2\mathbb{Z}_2$ and $BG \equiv BSO(3)$ or $BO(3)$.

Here we can take either:

1. $B^2M=B^2\mathbb{Z}_2$ as fibers, and $BG=BSO(3)$ (or $BO(3)$) as base manifold.

or

2. $BG=BSO(3)$ (or $BO(3)$) as fibers, and $B^2M=B^2\mathbb{Z}_2$ as base manifold.

For example,

(A) We consider the Postinikov classes $\omega \in H^3(BG,M)$. This classifies the fibrations of $$ B^2M \hookrightarrow BG_\text{{new}} \to BG. $$ If I understand correctly, the distinct $BG_\text{{new}}$ are determined by the classes of cocycles $\omega \in H^3(BG,M)$.

(B) On the other hand, we have $1\to \mathbb{Z}_2 \to O(3) \to SO(3) \to 1$, this gives rise $B\mathbb{Z}_2 \to BO(3) \to BSO(3) \to B^2\mathbb{Z}_2,$ $$ BO(3) \to BSO(3) \to B^2\mathbb{Z}_2, $$ which seems to suggest a fibration of $BO(3)$ over $B^2\mathbb{Z}_2$

(C) On the other hand, we further have $1\to \mathbb{Z}_2 \to SU(2) \to SO(3) \to 1$, this gives rise $B\mathbb{Z}_2 \to BSU(2) \to BSO(3) \to B^2\mathbb{Z}_2,$ $$ BSU(2)\to BSO(3) \to B^2\mathbb{Z}_2, $$ which seems to suggest a fibration of $BSU(2)$ over $B^2\mathbb{Z}_2$

My questions:

(a) Do (B) and (C) above really suggest a valid fibration of classifying space fibered over another classifying spaces? How do we classify them? Are there similar classifications like Postinikov classes?

(b) The first example (A) and the later (B) and (C) look very different from each other. Are there any relations or generalizations to relate each of them?

(c) In general, given $B^2M \equiv B^2\mathbb{Z}_2$ and $BG \equiv BSO(3)$ or $BO(3)$, what are other possible fibrations between them (these classifying spaces)?

Answer 1

Firstly, yes, your examples are all correct. However, in example~(B) we just have $O(3)=\{\pm I\}\times SO(3)$ as groups, so your fibration is just the product of $$BSO(3)\xrightarrow{1}BSO(3)\to 1$$ with the path-loop fibration $$ B\mathbb{Z}/2\to 1\to B^2\mathbb{Z}/2 . $$
The same happens with $BSO(m)$ and $BO(m)$ whenever $m$ is odd. When $m$ is even, there is no normal subgroup of order two in $O(m)$, so we cannot do an analogous thing. However, for any $m$ we have an extension $SO(m)\to O(m)\to \mathbb{Z}/2$ giving a fibration $BSO(m)\to BO(m)\to B\mathbb{Z}/2$, the other way around from your example.

As Dylan mentioned, fibrations with fibre $F$ are classified by maps to $B\text{hAut}(F)$. If $F$ has a group structure then it acts on itself by translation, giving a map $F\to\text{hAut}(F)$. In this context we have a homeomorphism $\text{Map}(F,F)=F\times\text{Map}_*(F,F)$ and similarly $\text{hAut}(F)=F\times\text{hAut}_*(F)$, where $\text{hAut}_*(F)$ is the space of based weak equivalences from $F$ to itself. If $F=K(\mathbb{Z}/2,d)=B^d(\mathbb{Z}/2)$ then for $k>0$ we have $\pi_k(\text{Map}_*(F,F))=[\Sigma^kF,F]=H^d(\Sigma^kF,\mathbb{Z}/2)=0$ (because $\Sigma^kF$ is $(d+k-1)$-connected). Moreover, in this case we have $\pi_0(\text{Map}_*(F,F))=H^d(F;\mathbb{Z}/2)=\{0,1\}$, with the $0$-component consisting of nullhomotopic maps, and the $1$-component consisting of weak equivalences. From this we find that the translation map $F\to\text{hAut}(F)$ is a weak equivalence in this case (as well as being a homomorphism of topological groups). This gives $B\text{hAut}(F)\simeq K(\mathbb{Z}/2,d+1)$, so fibrations with fibre $F$ and base $B$ are classified by maps $B\to K(\mathbb{Z}/2,d+1)$, or in other words by Postnikov invariants in $H^{d+1}(B;\mathbb{Z}/2)$. None of this relies on $B$ being the classifying space of a Lie group, which is a bit of a distraction from the real issues here.

Whenever you have a fibration as above, with total space $E$ say, you get a longer sequence $$ \dotsb \to \Omega^2B\to K(\mathbb{Z}/2,d-1)\to \Omega E \to \Omega B \to K(\mathbb{Z}/2,d-1) \to E \to B \to K(\mathbb{Z}/2,d)$$ in which any three adjacent terms give a fibration. In particular, we have some fibrations with Eilenberg-MacLane spaces as the base, and others with Eilenberg-MacLane spaces as the fibre. All of the examples that we have discussed arise in this way, for suitable choices of $B$ and $d$.

Given a compact Lie group $G$, we could in principle classify fibrations with fibre $BG$ using $B\text{hAut}(BG)$, which is similar to $\text{Map}(BG,BG)$. If $G$ is an $n$-torus then $BG=K(\mathbb{Z}^n,2)$ and$\text{hAut}(BG)\simeq BG\times GL_n(\mathbb{Z})$ by an argument similar to that given above, so this is not too hard to understand. However, even in the simplest nonabelian case of $G=SU(2)$, the set $[BG,BG]$ is uncountable and has no tractable structure, so this approach is not very useful. We can instead try to produce extensions $G\to P\to Q$ of topological groups, for various $Q$, and then take classifying spaces to produce fibrations $BG\to BP\to BQ$. This will not produce all $BG$-fibrations over $BQ$, but it will produce a class of such fibrations that one can reasonably hope to analyse.