# Finding cocycles that square to zero

Suppose $x$ is a chosen class in the singular cohomology (integer coefficients) of a space $X$. I'm thinking primarily of classes of odd degree on a simply connected space. What are necessary conditions (besides $x^2=0$) for the existence of a cocycle representing $x$ whose cup-square equals zero as a cocycle? Sufficient conditions?

Take your pick of the precise form of the question: you can fix a cochain model for cup products before or after choosing $x$, or even allow a DGA quasi-isomorphic to the singular cochains on $X$.

You may feel inclined to mutter "Steenrod square" or "Massey product" - but which, and why?

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Tim, have you found more stringent necessary conditions? I'd love to know... –  Mariano Suárez-Alvarez Jan 3 '10 at 23:21
No, I haven't - only vanishing of the Massey powers. A nice test example would be the space $SU(N)$, whose cohomology is an exterior algebra on generators in degrees $3,5,\dots,2N-1$. A surprising little paper of Karoubi ("Stabilizing and commuting cochains", C. R. Acad. Sci. Paris Ser. I Math. 333 (2001), no. 8, 769-771) shows that there's a functorial DGA model for cohomology in which one can find commuting representatives for a countable set of commuting cohomology classes. –  Tim Perutz Jan 7 '10 at 14:04

## 1 Answer

The triple product $\langle x,x,x\rangle$ has to contain zero.

Indeed, if $a$, $b$, $c$ are odd cohomology classes such that $ab=0$ and $bc=0$, to compute the triple product $\langle a, b, c\rangle$, one picks representative cocycles $\alpha$, $\beta$ and $\gamma$, then picks cochains $\delta$ and $\eta$ such that $\alpha\beta=d\delta$ and $\beta\gamma=d\eta$, and then observes that $\tau=\alpha\eta+\delta\gamma$ is a cocycle. Then $\tau$ is a representative of $\langle a,b,c\rangle$ in an appropriate quotient of the cohomology group which contains the class of $\tau$.

In your case, suppose we can represent the class $x$ by a cocycle $\xi$ such that $\xi^2=0$. Then if we take $a=b=c=x$, we can take $\alpha=\beta=\gamma=\xi$ and $\delta=\eta=0$, so that $\tau=0$, that is, $0\in \langle x,x,x\rangle$.

In fact, all Massey products $\langle x,x,\dots,x\rangle$ ("Massey powers"?) have to be zero, by a similar computation---see the book by McCleary on spectral sequences, chapter 8, for a speedy description of these.

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Yes, the Massey product $\langle x,x,x \rangle\in H^*(X)/(x)$ must vanish, and likewise the higher powers. When $Ann(x)$ is the ideal $(x)$, as it sometimes is, this seems to be automatic: $x\langle x,x,x\rangle= \pm \langle x^2,x,x\rangle=0$. Similarly for the higher powers (cf. McCleary). –  Tim Perutz Dec 13 '09 at 2:58