RECAP on classification of bundles

We want to classify $G$-principal bundles over $X$ (smooth manifold, G compact Lie). These are in 1-1 correspondence with homotopy classes of maps $[X,BG]$ (where $EG\to BG$ is the universal bundle as usual). If $BG \simeq K(\pi,n)$ then it's easy: $$[X, BG]\leftrightarrow H^n(X,\pi),$$ therefore there is a cohomology class that gives us the classification (e.g. the 1st Chern class for the frame bundles of complex line bundles). In general, $BG$ can be more complicated, in any case $BG$ has a Postnikov tower which induces a factorization of the classifying map $f\in [X, BG]$ in $(f_i)_i$ $\require{AMScd}$ \begin{CD} \vdots@. \vdots\\ @| @VVV \\ X@>>f_2> P_2(BG)\\ @| @VVp_2V \\ X @>>f_1> P_1(BG)@.\simeq K(\pi_1(BG),1) \end{CD}

The homotopy type of $f$ is given by the homotopy type of the $f_i$s. Since $P_1(BG)\simeq K(\pi_1(BG),1)$ $f_1$ is given by a cohomology class in $H^1(X, \pi_1(BG))$. However, not any choice of $f_2$ works, because it must lift $f_1$. From this answer* I learned that there is a Cartesian diagram $\require{AMScd}$ \begin{CD} X@>>f_2> P_{2}(BG) @>>> K(\pi_0G,1)\\ @| @VVp_2V @VVV\\ X @>>f_1> P_1(BG) @>>> K(\pi_2 G,3)_{h\pi_0G} \end{CD}

1)Explanation/references for this? I was expecting the second column to be something like $K(\pi_2(BG),2)\to K(\pi_1(BG),1)$, it reminds me of principal fibrations.

2)How to see that these lifts are parametrized by $H^{2}(X,\pi_{2}(BG))$ cohomology with local coefficients twisted by $f_1\in H^1(X,\pi_1(BG))$? Obstruction theory tells the necessary conditions to lift $f_1$ but not how many lifts there are.

In the end we get that the principal bundle is classified by $f_1\sim \alpha_1 \in H^1(X,\pi_1(BG))$ and a sequence of cohomology classes $\alpha_k \in H^{k}(X, \pi_k(BG)) $ in the cohomology with local coefficients twisted by $\alpha_1$.

3) How to compute them? Is there any example for say $G=O(2)$? Any link with invariant polynomials in $\mathfrak{g}$ or the Weyl algebra of $\mathfrak{g}$?

This is essentially Denis Nardin's answer. In his comment Nardin, says another interesting thing if $G=O(n)$, then $\alpha_1 = 0 $ iff the bundle is orientable, $\alpha_1, \alpha_2 = 0$ iff the bundle is spin and so on climbing the Whitehead tower of $O(n)$ $$O(n)\leftarrow SO(n)\leftarrow Spin(n)\leftarrow String(n)\leftarrow ...$$

4)Is this true for any Whitehead tower of groups? Does this implicitly say that the Postnikov tower of $SO(n)$ $(Spin(n))$ is the one of $O(n)$ without the first (second) term?

BG as a twisted product

If $\pi_1(BG)$acts on $\pi_{n+1}(BG)$ trivially then

  1. the Postnikov tower gives us an expression for $BG$ in terms of a twisted product of Eilenberg-MacLane spaces $BG \simeq K(\pi_1(BG),1)\times_{k_1} K(\pi_2(BG),2)\times_{k_2} \dots$
  2. There is no need of local coefficients for the $\alpha_k$ above.

In the same answer*, Mark Grant says that in the case of $G=O(2)$:

there is a fibration $$ K(\mathbb{Z},2)\to E\mathbb{Z}/2\times_{\mathbb{Z}/2} K(\mathbb{Z},2)\to B\mathbb{Z}/2 $$ given by the twisting of $w_1$ on the universal $SO(2)$ bundle, and this fibration agrees up to homotopy with the fibration $$BSO(2)\to BO(2)\to BO(1).$$


5) Can you explain this in the more general setting of a fibration $F\to E\to B$? Also I do not understand $E\mathbb{Z}/2\times_{\mathbb{Z}/2} K(\mathbb{Z},2)$, what is the $\mathbb{Z}/2$ action on $K(\mathbb{Z},2)$? Grant says $w_1$ is involved but I cannot imagine how. (I know that in general $w_1 \in H^1(X, \pi_1(BO(2))$ gives me an action of $\pi_1(X)$ on $\pi_n(BO(2))$). How does this relate to the Whitehead tower above?

*Classification of $O(2)$-bundles in terms of characteristic classes

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    $\begingroup$ In the second diagram, the bottom right corner should be $K(\pi_2G, 3)_{h\pi_0G}$, so that the fiber of the vertical map over it is equivalent to $K(\pi_2 G, 2)$. $\endgroup$ Apr 14, 2019 at 14:52
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    $\begingroup$ The tower in Mark Grant's answer is not the Whitehead tower, but the Postnikov tower (which is fairly degenerate, since $BO(2)$ is a 2-type so that $P_2(BO(2))=BO(2)$ and $P_1(BO(2))=B\mathbb{Z}/2=BO(1)$). I'll see if I can write an answer, but I'd have to say that I don't understand most of your questions $\endgroup$ Apr 14, 2019 at 15:13
  • $\begingroup$ Also, the reason why the homotopy quotient is there is precisely that this is not a principal fibration, so local coefficients appear $\endgroup$ Apr 14, 2019 at 15:19
  • $\begingroup$ @DenisNardin Thanks, many of the above questions came out trying to understand your previous answer. Sorry if they are messy, so is my understanding of the subject. In particular question 2) asks about why you state in your answer that "the possible choices are parametrized by a class in $H^{n+3}(X,\pi_{n+1}G)$". As far as I know (very little) obstruction theory tells us if we can lift but -gives us a cohomological obstruction to lifting- but it does not tell how many possible lifts we have. $\endgroup$ Apr 14, 2019 at 15:47
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    $\begingroup$ @WarlockofFiretopMountain Obstruction theory tells you absolutely how many lifts there are! In fact it tells you the homotopy type of the space of lifts :). I'll try to write something, without repeating too much what Charles has already written. BTW I think I screwed up the indexing in my old answer (what's new, heh? :)), I think I fixed it now, so you might want to double check. $\endgroup$ Apr 14, 2019 at 15:48

1 Answer 1


I'll try to answer question 1. Unfortunately, I know of no especially convenient reference for the case of understanding general Postnikov towers; however, everything I say below is "well known".

In a Postikov tower for $X$, the map $p=p_n\colon P_n\to P_{n-1}$ is, by definition, a fibration with fiber equivalent to $K(A,n)$, where $A=\pi_n X$. The idea is that there exists an (essentially unique) homotopy pullback square $\require{AMScd}$ \begin{CD} P_{n} @>>> E_F\\ @VpVV @VVqV\\ P_{n-1} @>>> B_F \end{CD} where $q$ is the universal example of a fibration with fiber equivalent to $F:=K(A,n)$. This universal fibration is rarely ever a principal bundle.

Here is the formula for $q$: Let $\def\Aut{\mathrm{Aut}}\def\Map{\mathrm{Map}}$ $\Aut(F)\subseteq \Map(F,F)$ be the union of all components of the mapping space which contain homotopy equivalences. Then $\Aut(F)$ is a topological monoid which is "grouplike" (i.e., $\pi_0\Aut(F)$ is a group), and which acts on $F$. Then the map $$ F_{h\Aut(F)} \to (*)_{h\Aut(F)}=B\Aut(F)$$ is the universal $F$-fibration.

For $F=K(A,n)$, it turns out you can identify $\Aut(F)$ very explicitly. (I'm going to assume $n\geq 2$ here.) The construction of Eilenberg-MacLane $A\mapsto K(A,n)$ spaces lifts to a functor $$(\text{abelian groups})\to (\text{topological abelian groups}).$$ Therefore we get a topological group $K(A,n)\rtimes \Aut(A)$ acting on $K(A,n)$ (the $K(A,n)$ acts on itself by left-translation, and $\Aut(A)$ is the discrete automorphism group of $A$.) It turns out (by a computation) that this group is equivalent (up to homotopy) to the topological monoid $\Aut(K(A,n))$, so $$B_{K(A,n)} \approx B\Aut(K(A,n)) \approx B\bigl( K(A,n)\rtimes \Aut(A)\bigr) \approx BK(A,n)_{h\Aut(A)}\approx K(A,n+1)_{h\Aut(A)},$$ and $$E_{K(A,n)} \approx K(A,n)_{h\bigl(K(A,n)\rtimes\Aut(A)\bigr)} \approx (*)_{h\Aut(A)} \approx B\Aut(A).$$ So the desired pullback square has the form \begin{CD} P_{n} @>>> B\Aut(A)\\ @VpVV @VVqV\\ P_{n-1} @>>> K(A,n+1)_{h\Aut(A)} \end{CD} In practice, when you have a Postnikov tower of a space like $BG$, the action of the fundamental group $\pi_1 BG=\pi_0G$ on $A=\pi_nBG$ determines a homomorphism $\pi_0G\to \Aut(A)$, and you can "restrict" along this homomorphism to get a pullback square \begin{CD} P_{n} @>>> B\pi_0G\\ @VpVV @VVqV\\ P_{n-1} @>>> K(A,n+1)_{h\pi_0G} \end{CD}

  • $\begingroup$ The reference I know is the one I gave in my old post, the paper by Blanc, Dwyer and Goerss The realization space of a ∏-algebra. $\endgroup$ Apr 14, 2019 at 15:45
  • $\begingroup$ @DenisNardin Thanks! Though I kinda want a reference appropriate for someone who is just now learning what a Postnikov tower is. $\endgroup$ Apr 14, 2019 at 15:52
  • $\begingroup$ Unfortunately the fact that they had to write the theory themselves makes me suspect that there's no elementary exposition yet. $\endgroup$ Apr 14, 2019 at 16:12
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    $\begingroup$ If a group $K$ acts on itself by left translation, then $K_{hK}\approx *$. In general there's an equivalence of the form $X_{h(K\rtimes H)}\approx (X_{hK})_{hH}$. $\endgroup$ Apr 14, 2019 at 18:31
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    $\begingroup$ The composite $\def\Aut{\mathrm{Aut}}$ $P_{n-1}\to B\Aut(K(A,n))\to B\Aut(A)$ factors through a map $P_1\to B\Aut(A)$. Since $P_1\approx B\pi_0G$, you can pullback everything along this map. That gives the final square I drew. $\endgroup$ Apr 14, 2019 at 18:33

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