(Unstable) cohomological operations are in bijection with cohomology classes of Eilenberg–MacLane spaces. Rings $H^\bullet(K(\mathbb Z;n);\mathbb Z)$ can be computed step by step by considering spectral sequences of Serre fibrations $K(\mathbb Z;n-1)\to\ast\to K(\mathbb Z;n)$.

But how to extract an explicit description of an operation from such computation?


  • Consider LSSS of fibration $\mathbb CP^\infty\cong\Omega K(\mathbb Z;3)\to\ast\to K(\mathbb Z;3)$: let $t$ be the generator of $H^2(\mathbb CP^\infty)$ and let $u\in H^3(K(\mathbb Z;3);\mathbb Z)$ be the fundamental class; then $d_3(t^3)=3t^2u$ and $d_5(t^2u)$ gives a 3-torsion element in $H^8(K(\mathbb Z;3);\mathbb Z)$.
  • The corresponding cohomological operation is the following «Massey cube»: if $a$ is an integral $3$-cocycle, $[a,a]=0\in H^6$ (where $[-,-]$ is the suppercommutator), so $[a,a]=db$ (where $b$ is some cochain); define $\langle a\rangle^3:=[a,b]\in H^8$.

So concrete question is: suppose $\gamma=d_i(\alpha\beta)$ (in LHSS for $K(\mathbb Z;n-1)\to\ast\to K(\mathbb Z;n)$; $\alpha\in H^\bullet(K(\mathbb Z;n-1))$, $\beta,\gamma\in H^\bullet(K(\mathbb Z;n))$) and we already know opeations corresponding to $\alpha$ and $\beta$; how to construct the operation corresponding to $\gamma$?

  • 2
    $\begingroup$ Notice that this is not strictly a Massey cube: it is rather $\langle a,2a,a\rangle$, but one can check that there is no indeterminacy, so instead of a set we get a single class. The paper I mentioned by Kraines does deal with cubes $\langle a,a,a\rangle$. $\endgroup$ – Mariano Suárez-Álvarez Jun 30 '13 at 1:38
  • $\begingroup$ In my opinion, Eilenberg and MacLane work is better to understand the homology of these spaces. The Serre spectral sequence is more complicated. $\endgroup$ – Fernando Muro Jun 30 '13 at 10:06
  • $\begingroup$ @Mariano Well, yes, this operation is $(a,2a,a)$ — so perhaps I should write $\langle 2a\rangle^3$ or $-\langle a\rangle^3$ (in 3-torsion $4=1$ and $8=-1$). Anyway, it's a Massey cube up to a sign. $\endgroup$ – Grigory M Jun 30 '13 at 16:52
  • $\begingroup$ @GrigoryM, but a priori this is defined on integral cohomology, so $2$ is not $-1$. (Kraines does work module three) $\endgroup$ – Mariano Suárez-Álvarez Jun 30 '13 at 17:54

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