MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.