# Classifying space for fibrations with Eilenberg-MacLane space fibers and nontrivial fundamental group actions

Let $$A$$ be an abelian group and let $$n \geq 2$$. For any connected CW complex $$X$$, it is standard that a fibration $$f\colon E \rightarrow X$$ whose fibers are homotopy equivalent to a $$K(A,n)$$ is fiberwise homotopy equivalent to the pull-back of the loop-space fibration over a $$K(A,n+1)$$ if and only if $$\pi_1(X)$$ acts trivially on the fibers of $$f$$ (up to homotopy equivalence).

Question: Consider the more general set of fibrations $$f\colon E \rightarrow X$$ whose fibers are homotopy equivalent to a $$K(A,n)$$, but with a possibly nontrivial monodromy action, up to fiber homotopy equivalence. Is this functor representable?

From the first paragraph, if it is representatable by a space $$Z$$ then the universal cover of $$Z$$ is a $$K(A,n+1)$$.

Of course, there are many variants here (e.g. of the definition of a fibration, or the equivalence relation on the set), and I'd be interested in any of these.

• Do the hypotheses of Brown representability not apply? – John Greenwood Jan 7 at 4:17
• @JohnGreenwood: It's not clear to me. The issue is that you have to be able to glue fibrations that are isomorphic on a subcomplex, and since the gluing map is just a fiberwise homotopy equivalence (and thus not a homeomorphism) this seems subtle. – Tina Jan 7 at 4:19
• (if we could assume that these fibrations were actually locally trivial fiber bundles and isomorphisms were fiber bundle isomorphisms, then this would not be a problem; however, I don't know if this is possible). – Tina Jan 7 at 4:21
• My mistake! Gluing fibrations seems tricky...what if you glue with the homotopy equivalence anyway, and then force the result to be a fibration? – John Greenwood Jan 7 at 5:16

Mark Grant's excellent answer already resolves the question. However, let me sketch how this arises as a special case of the more general problem of classifying fibrations with a given fiber.

For any space $$X$$, the homotopy automorphisms $$\operatorname{hAut}(X)$$ are defined as the self-homotopy equivalences of $$X$$ (topologized with the compact-open topology; note that they form a union of path components of the space of all self-maps of $$X$$). They form a group-like monoid, so there is a connected pointed space $$B\operatorname{hAut}(X)$$ such that there is an equivalence of $$A_\infty$$-spaces $$\operatorname{hAut}(X)\simeq \Omega B\operatorname{hAut}(X)$$. In fact, $$B\operatorname{hAut}(X)$$ can be built as the geometric realization of the simplicial space $$B(*,\operatorname{hAut}(X),*) = * \leftarrow \operatorname{hAut}(X) \Leftarrow \operatorname{hAut}(X)\times\operatorname{hAut}(X) \Lleftarrow \dots$$. Since $$\operatorname{hAut}(X)$$ acts on $$X$$ from the left, we can also build the simplicial space $$B(*,\operatorname{hAut}(X),X) = X\leftarrow \operatorname{hAut}(X)\times X\Leftarrow\dots$$, and the map $$X\to *$$ induces a fibration $$X\to E_X\to B\operatorname{hAut}(X)$$. This is the universal fibration with fiber $$X$$, that is, for every fibration $$X\to F\to Y$$ there is a unique homotopy class of maps $$f: Y\to B\operatorname{hAut}(X)$$ such that $$F\simeq f^*E_X$$.

If $$X = K(A,n)$$ is an Eilenberg-MacLane space, the grouplike monoid $$\operatorname{hAut}(X)$$ can be described quite explicitly: In this case, $$X$$ can also be given the structure of a grouplike monoid with identity $$e$$, so that the map $$\operatorname{hAut}(X)\to X, f\mapsto f(e)$$ has a homotopy right inverse given by sending $$x$$ to (left, say) translation by $$x$$. Thus there is a homotopy equivalence $$\operatorname{hAut}(X)\simeq \operatorname{hAut}_*(X)\times X$$, where $$\operatorname{hAut}_*(X)$$ is the submonoid of self homotopy equivalences that preserve the basepoint (note, however, that this is not compatible with the monoid structure). Now it is straightforward to show that $$\operatorname{hAut}_*(K(A,n))\simeq K(\operatorname{Aut}(A),0)$$ is homotopy equivalent to a discrete space, and there is a retraction $$\operatorname{hAut}(X) \to \pi_0(\operatorname{hAut}(X))\cong \operatorname{Aut}(A)$$. All in all, we obtain an equivalence of grouplike monoids $$\operatorname{hAut}(K(A,n))\simeq \operatorname{Aut}(A)\ltimes K(A,n)$$ It follows that $$B\operatorname{hAut}(K(A,n))$$ has exactly two nonvanishing homotopy groups, namely $$\pi_1(\operatorname{hAut}(K(A,n)))\cong \operatorname{Aut}(A)$$ and $$\pi_{n+1}(\operatorname{hAut}(K(A,n)))\cong A$$, with the evident action of $$\pi_1$$ on $$\pi_{n+1}$$. In particular, a map $$f:Y\to B\operatorname{hAut}(K(A,n))$$ is determined by $$\rho\in H^1(Y;\operatorname{Aut}(A))$$, which determines a local system $$A_\rho$$ on $$Y$$, and a cohomology class $$[f]\in H^{n+1}(Y;A_\rho)$$.

• As a side note, for bundles whose fibers are $K(G,1)$ with $G$ is a nonabelian groupthere there is a similar analysis. But then $\pi_0(hAut)$ turns out to be the outer automorphism group $Out(G)$ and $\pi_1(hAut)$ is the center $Z(G)$. – Tyler Lawson Jan 7 at 17:00
• There's also work of Wirth, summarised in this paper joint with Stasheff: arxiv.org/abs/math/0609220 on homotopy locally trivial fibrations. – David Roberts Jan 8 at 6:03
• So is it right to think that, morally, the difference between fiber bundles and fibrations is the structure "group" being an actual group versus a group-up-to-homotopy? – John Greenwood Jan 8 at 19:08
• This is wonderful! Do you know a good place to read about this? (ps: sorry for only replying now -- I am not on the internet very often). – Tina Jan 11 at 1:44
• (pps: I'll accept your answer once I get a reference.) – Tina Jan 12 at 14:32

Denote $$\pi=\pi_1(X)$$ and fix the monodromy action $$\rho:\pi\to \operatorname{Aut}(A)$$. There is a generalized Eilenberg-Mac Lane space $$L_\rho(A,n+1)$$, whose only non-trivial homotopy groups are $$\pi_1(L_\rho(A,n+1))=\pi$$ and $$\pi_{n+1}(L_\rho(A,n+1))=A$$ and such that the action of $$\pi_1$$ on $$\pi_{n+1}$$ is given by $$\rho$$. The construction appears in Section 7 of

Gitler, Samuel, Cohomology operations with local coefficients, Am. J. Math. 85, 156-188 (1963). ZBL0131.38006.

Essentially, $$L_\rho(A,n+1)$$ is the Borel construction $$E\pi\times_\rho K(A,n+1)$$ done with simplicial sets, so that everything is sufficiently natural (and the action of $$\pi$$ on $$A$$ gives an action of $$\pi$$ on $$K(A,n)$$ for all $$n$$). The path loop fibration is equivariant, giving a fibration sequence $$K(A,n)\to E\pi\times_\rho PK(A,n+1) \to E\pi\times_\rho K(A,n+1)=L_\rho(A,n+1).$$ I would bet that fibrations over $$X$$ with fibre $$K(A,n)$$ and monodromy $$\rho$$ are classified by maps (Edit: from $$X$$) into $$L_\rho(A,n+1)$$ (Edit: inducing the identity on $$\pi_1$$) using this construction.

• Arbitrary maps? Maybe rather only those inducing isomorphism on $\pi_1$, or something like that? – მამუკა ჯიბლაძე Jan 7 at 21:18
• You're right, of course. For example, pulling back by the trivial map would give a fibration with trivial monodromy. In fact, the monodromy of the pullback should (I think) be the pullback of the monodromy, so we'd want the map to induce the identity on $\pi_1$; see my edit. – Mark Grant Jan 8 at 9:53
• Bertram's answer is more general, and probably does a better job of answering the quesiton asked. – Mark Grant Jan 8 at 9:54

One place this was done but in greater generality was by Blakers, Annals of Matyh.2nd series, 49 (2) 428-461. He defines what he calls "group systems". These are now called "crossed complexes". An account is in the book Nonabelian Algebraic Topology (EMS, 2011). A particular case $$C$$ is where the crossed complex has $$C_0$$ a singleton, $$C_1$$ is a group, say $$G$$, for some $$n>1$$, $$C_n$$ is a $$G$$-module, and otherwise $$C_n$$ is trivial.

Any crossed complex $$C$$ has a classifying space $$BC$$ whose simplicial definition in the case $$C_0$$ is a point is due to Blakers. Any filtered space, $$X_*$$, and in particular any, CW-complex $$X$$ with skeletal filtration $$X_*$$, has a fundamental crossed complex $$\Pi X_*$$, (also due to Blakers). In particular the cubical nerve $$NC$$ of $$C$$ can be defined in dimension $$n$$ by $$(NC)_n = Crs(\Pi I^n_*, C),$$ where $$I^n$$ is the standard $$n$$-cube. (The simplicial version of this was published with P.J. Higgins as Math. Proc. Camb. Phil. Soc. 110 (1991) 95-120.)

(I confess I rather suspect the homotopy classification theorems of our book have to be restricted to $$X$$ finite dimensional or $$BC$$ having only a finite number of non trivial homotopy groups,)