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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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  • $\begingroup$ Tim, have you found more stringent necessary conditions? I'd love to know... $\endgroup$ Jan 3 '10 at 23:21
  • $\begingroup$ 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. $\endgroup$
    – Tim Perutz
    Jan 7 '10 at 14:04
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    $\begingroup$ If you are willing to weaken your ask to "There exists an $A_\infty$-homomorphism from the exterior algebra on a degree 3 generator $f: \Bbb Z[3] \to C^*(X;\Bbb Z)$ with $[f(1)] = x_3$ for $[x_3]$ a chosen cohomology class", then this does exist for $SU(n)$. The data of an $A_\infty$-homomorphism is more or less precisely what the vanishing of the higher "Massey powers" guarantees: first an element $x_5 \in C^*$ with $dx_5 = x_3^2$, then an element $x_7 \in C^*$ with $dx_7 = x_3 x_5 + x_5 x_3$... You probably already know all of this but perhaps this weaker answer is enough for your needs. $\endgroup$
    – mme
    May 11 at 2:25
  • $\begingroup$ Since cochains are an E-infinity algebra, $x$ has odd degree and you're happy about quasi-iso replacements, your question is kind of equivalent to the following one: given an E-infinity álgebra $A$ and an odd cycle $x\in H_*(A)$, is there a quasi-isomorphic E-Infinity algebra $B$ such that $x$ is represented by a cochain $y\in B$ such that the sub-E-infinity algebra generated by $y$ is strictly commutative? $\endgroup$ Jul 11 at 3:28
  • $\begingroup$ Moreover, if there is a chain e-infinity operad whose category of algebras happens to be left proper, then it is equivalent to ask whether the sub-E-infinity algebra of $a$ generated by a representative of $x$ is rectifiable. if there is not, this is weaker. all this implies that all steenrod operations, massey products etc. vanish on $x$, and in general all primary, secondary and higher order operations on $x$, but even that wouldn't suffice. $\endgroup$ Jul 11 at 3:30
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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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  • $\begingroup$ 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). $\endgroup$
    – Tim Perutz
    Dec 13 '09 at 2:58
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I don't know what your application is, so the below may not be sufficient, but it was fun to think about.


The issue we run into here is that being (respresentable by) a square-zero cocycle is not a homotopical condition, and is certainly not invariant under quasi-isomorphisms. For instance, $C^*(S^3)$ contains no square-zero 3-cocycles representing top cohomology: if $\sigma:\Delta^3 \to S^3$ is a singular 3-simplex with $\sigma$ constant on $\partial \Delta^3$, on which $\phi(\sigma) = 1$, then we can always find a 6-simplex $\overline \sigma$ with $\overline \sigma_{[0, 3]} = \sigma$ and $\overline \sigma_{[3, 6]} = \sigma$. (Choose a projection $\Delta^6 \to \Delta^3 \vee \Delta^3$ and compose with the map $\sigma \vee \sigma$, which makes sense because $\sigma$ is constant on the boundary.) Then $(\phi \cup \phi)(\overline \sigma) = 1$. Of course, if you instead work with the quasi-isomorphic dg-algebra of simplicial cochains, every representative is square-zero because $C^6_\Delta(S^3) = 0$.

A more homotopical condition is being coherently square zero (in the literature, being a 'twisting element', though this is a very special case of a general theory). If $[x_1]$ is a cohomology class of odd degree $2n+1$, then a coherently square-zero extension is a sequence of cochains $x_2, x_3, \cdots$ with $|x_i| = 2ni + 1$, so that $$dx_k + \sum_{\substack{i+j = k \\ i,j \geq 1}} x_i x_j = 0.$$

There are two natural ways to come up with this formula.

First: consider the dg-algebra $\Bbb Z[2n+1]$, which is a copy of $\Bbb Z$ in degree 3 with zero differential and zero product. Then a dg-algebra homomorphism $\Bbb Z[2n+1] \to C^*(X;\Bbb Z)$ is precisely a square-zero cocycle. When we want more "homotopical" conditions, we learn to generalize from dg-algebra maps to $A_\infty$ maps. And the above is (maybe up to sign, I didn't check) the same thing as an $A_\infty$-homomorphism $\Bbb Z[2n+1] \to C^*(X;\Bbb Z)$.

Second: you might be looking for a "twisted differential" on the cochain complex $C^*(X;\Bbb Z) \otimes \Bbb Z[T, T^{-1}]$ which incorporates cup-product with $x_1$. The first formula you would guess is $d_{tw} = d + L_1 T$, where $L_1(y) = x_1 y$. The conditon $d_{tw}^2 = 0$ is equivalent to the demand that $dx_1 = 0$ and $x_1^2 = 0$.

When you give up on finding $x_1^2 = 0$, you might try to correct for this by adding higher terms. You could then guess that the right differential is $d_{tw} = d + L_1 T + L_2 T^2 + \cdots$ where $L_m(y) = x_m y$. The claim that $d_{tw}^2 = 0$ is now precisely the formulae I wrote above.


(There is also a notion of homotopy of these. A non-trivial but not too difficult fact is that if $A$ and $B$ are algebras equipped with a cup-1 product satisfying the Hirsch formula, and if $f: A \to B$ is a quasi-isomorphism of dg-algebras which preserves the cup-1 product, then $f$ induces a bijection on the sets of twisting sequences as above up to homotopy.)


Theorem. Suppose $X$ is a topological space with $H^*(X;\Bbb Z)$ is torsion-free. Then if $c \in H^{2n+1}(X;\Bbb Z)$ is an odd class, there is always a coherently square-zero extension $(x_1, x_2, \cdots) \in \prod_{i \geq 1} C^{2ni+1}(X;\Bbb Z)$ with $[x_1] = c$.

More generally --- as in Mariano's nice answer --- all we need to demand is that all Massey powers $\langle c, \cdots, c\rangle$ is well-defined and equal to zero.

Proof: Construct this by induction.

Choose $x_1$ to be any cocycle with $[x_1] = c$. Inductively suppose that we have constructed $x_1, \cdots, x_m$ with $$dx_k + \sum_{\substack{i+j = k \\ i,j \geq 1}} x_i x_j = 0$$ for all $k \leq m$. Then by explicit computation the term $$\sum_{\substack{i+j = m + 1 \\ i,j \geq 1}} x_i x_j$$ is a cocycle, and referring to Kraines, "Massey higher products", Section 3, you see that this is a representative for the so-called Massey power $\langle c\rangle^{m+1}$.

By Kraines, "Massey higher products", Theorem 15, these vanish in rational cohomology; by naturality of Massey products under change-of-coefficients and injectivity of the map $H^*(X;\Bbb Z) \to H^*(X;\Bbb Q)$, it follows that these vanish integrally. Thus there exists an $x_{m+1}$ so that $$dx_{m+1} + \sum_{\substack{i+j = m + 1 \\ i,j \geq 1}} x_i x_j = 0,$$ completing the induction.


Notice that the simplest case of this is that $c^2 = 0$ in rational cohomology because $c^2$ is 2-torsion. The Massey powers are a sequence of higher torsion obstructions to such an extension.

One can further analyze the obstructions to such sequences being homotopic, but I will not do so here. A special case is that if $x_1^2 = y_1^2 = 0$ in your dga with cup-1 products satisfying the Hirsch formula, then if you have torsion-free cohomology then in fact $(x_1, 0, \cdots)$ and $(y_1, 0, \cdots)$ are homotopic as twisting sequences if and only if $[x_1] = [y_1]$. To manage the higher cases one has to also ask that certain higher cohomology classes are equal, where the higher cohomology classes are made out of iterated cup-1 products of lower ones.


In particular you can extend every odd cohomology class on $SU(n)$ to a twisting sequence. (If you're trying to extend the 3-dimensional generator and you work with simplicial cochains, I can prove that you can find a representative with $x_k = 0$ for $k \geq n$, but this is a little tedious.)

The argument I outlined above that you will never be able to construct an honest square-zero cocycle in the singular chains on $S^3$ works just as well to show you won't be able to do that in the singular chains on $SU(n)$. I don't think you can do it in simplicial chains on $SU(n)$ for $n > 2$, either, but I haven't thought about that so much.

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  • $\begingroup$ Could you provide a reference for the 'coherently square zero' idea? $\endgroup$
    – Ryan
    Jul 10 at 19:18
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    $\begingroup$ @Ryan This is a simplified version of a concept which appears in the context of twisted tensor products. I do not know a reference for the simpler concept, but I've added some remarks to help inspire it: I think they may be more useful for you than the literature, which works towards a different idea. The name you want is "twisting element" or "twisting cochain" or "Maurer-Cartan equation" (though this is often in a different context where one works over the rationals). $\endgroup$
    – mme
    Jul 10 at 19:29
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    $\begingroup$ The original reference is Ed Brown's paper "Twisted Tensor Products I", but you can find a definition of twisting cochain near the end of May's "Simplicial objects...". In the context of twisting cochains/elements on $A_\infty$-algebras over field, you can see this paper, where when you forget all the higher multiplications you recover what I've written above. But as you can see all of these references target generalizations of what I've written above, focused on different applications. $\endgroup$
    – mme
    Jul 10 at 19:30
  • $\begingroup$ Yes, your motivation was quite helpful, many thanks $\endgroup$
    – Ryan
    Jul 10 at 21:28
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    $\begingroup$ @FernandoMuro The argument in Lemma 16 of his paper shows that over the rationals they are indeed defined without indeterminacy (and it just uses the cup-1 product and the Hirsch formula). This is special for powers of odd classes. It's because our cohomology is torsion-free that they are also thus well-defined and vanish over the integers. $\endgroup$
    – mme
    Jul 11 at 3:44

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