Why does «Massey cube» of an odd element lie in 3-torsion?

Question

The cup product is supercommutative, i.e the supercommutator $[-,-]$ is trivial at the cohomology level — but not at the cochain level, which allows one to produce various cohomology operations.

The simplest (in some sense) of such (integral) operations is the following «Massey cube». Suppose $a$ is an integral $k$-cocycle, $k$ is odd; $[a,a]=0\in H^{2k}$, so $[a,a]=db$ (where $b$ is some cochain); define $\langle a\rangle^3:=[a,b]\in H^{3k-1}$ (clearly this is a cocycle; it doesn’t depend on choice of $b$, since by supercommutativity $[a,b’-b]=0\in H^{3k-1}$ whenever $d(b-b’)=0$).

The question is,

why $\langle a\rangle^3$ lies in 3-torsion?

For $k=3$, for example, this is true since $H^8(K(\mathbb Z,3);\mathbb Z)=\mathbb Z/3$, but surely there should be a more direct proof? (Something like Jacobi identity, maybe?)

@Drew If by "this" you mean $\langle a\rangle^3$, then that's exactly my question. — Grigory M, Jun 28 '13 at 15:23
I might suggest using the operations and boundary formulas for the sequence operad in McClure-Smith's paper on multivariable cochains: their operation $\langle 121\rangle$ is the degree one cup-1 product expressing supercommutativity, and so you might be able to write down a pretty explicit element in terms of three-variable operations of degree two whose boundary is $\langle a,a,a\rangle$. — Tyler Lawson, Jun 29 '13 at 19:39
(The paper mentioned by Tyler Lawson is Multivariable cochain operations and little n-cubes, I guess) — Grigory M, Jun 30 '13 at 14:52
@TylerLawson Thank you! (In the end I've written a description in terms of cup-1 product, but I've found it using your advice; in particular, $a\cup_1 a\cup_1 a$ is just another name for something like $\langle31213\rangle(a,a,a)$.) — Grigory M, Jul 01 '13 at 19:19

score 5 · Accepted Answer · answered Jul 01 '13 at 19:13

Recall that $d(x\cup_1y)=[x,y]\pm dx\cup_1 y\pm x\cup_1dy$.

In particular, in the definition from the question one can take $b=a\cup_1a$. So $\langle a\rangle^3=[a,a\cup_1a]$.

Now $d((a\cup_1a)\cup_1a)=[a,a\cup_1a]+(d(a\cup_1 a))a=\langle a\rangle^3+[a,a]\cup_1 a$. Now by Hirsch formula $a^2\cup_1a=a(a\cup_1a)+(a\cup_1a)a=\langle a\rangle^3$. So $$3\langle a\rangle^3=d(a\cup_1a\cup_1a)=0\in H^{3k-1}.$$

Why does «Massey cube» of an odd element lie in 3-torsion?

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