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Let $f: E\rightarrow B$ be a Kan fibration between pointed connected Kan complexes with fibre the Eilenberg-MacLane space $\mathrm{K}(M, n), n\geq 2, M$ an abelian group. Assume $f$ induces an isomorphism on $\pi_1$ with common value $G$, and so $G$ acts naturally on $\pi_n\mathrm{K}(M, n)=M$ hence on (the simplicial abelian group) $\mathrm{K}(M, n)$. Let $\tilde{B}$ be the universal cover of $B$, on which $\pi_1B=G$ acts.

Is there a fibration equivalence $\tilde{B}\times_G\mathrm{K}(M, n)\sim E$ over $B$?

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    $\begingroup$ The question becomes a lot simpler if you consider the special case $\pi_1(E) = \pi_1(B) = 0$. Then you are asking if for every fibration with fiber $K(M,n)$ is trivial (i.e. $E \sim B \times K(M,n)$). $\endgroup$ – Najib Idrissi May 27 '18 at 22:40
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    $\begingroup$ Good point to give counterexample! $\endgroup$ – Lao-tzu May 29 '18 at 5:51
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No. If this were the case then there would be a section $s: B \to E$ to $f$ induced by the $G$-equivariant map $\widetilde{s}:\widetilde{B} \to \widetilde{B} \times {\rm K}(M,n)$ sending $x$ to $(x,0)$ (where $0 \in {\rm K}(M,n)$ is the neutral element, which is fixed by the action of $G$). However, there exists fibrations with fibers ${\rm K}(M,n)$ for $n \geq 2$ which do not admit a section, such as the quaternionic Hopf fibration $S^7 \to S^4$. In general fibrations $E \to B$ with fiber ${\rm K}(M,n)$ with a fixed $\pi_1(B)$-action on $M$ are classified by the cohomology group with local coefficients $H^{n+1}(B,M)$, where the ones of the form you describe correspond to the $0$ element.

Edit:

As pointed in the remarks below, the quaternionic Hopf fibration is indeed not a counter-example for what the OP is asking since its fibers are $S^3$, which is not an Eilenberg-MacLane space. One way to fix this is to take the relative Postnikov truncation (which still doesn't have a section). Alternatively, there are probably other more natural examples.

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  • $\begingroup$ Thanks a lot! My question is in fact stupid, $\tilde{B}\times_G\mathrm{K}(M, n)$ of course represents only one fibration, given the $G$-action on $M$. Could you indicate where can I find your last general statement which I don't figure out now? $\endgroup$ – Lao-tzu May 27 '18 at 9:01
  • $\begingroup$ I think you are using a result of Robinson about cohomology with twisted coefficients. $\endgroup$ – Lao-tzu May 27 '18 at 9:30
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    $\begingroup$ How does the Hopf fibration have fibres an Eilenberg-Mac Lane space? Sp(1) has highly nontrivial homotopy groups... $\endgroup$ – David Roberts May 27 '18 at 11:09
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    $\begingroup$ @Lao-tzu It is a classical result in obstruction theory, I don't know who proved it first. See for example chapter VI in Goerss, Jardine Simplicial Homotopy Theory. $\endgroup$ – Denis Nardin May 27 '18 at 11:12
  • $\begingroup$ @DavidRoberts, right! added an edit to the answer accordingly. $\endgroup$ – Yonatan Harpaz May 27 '18 at 13:05

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