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Let $A$ be a dg algebra, and $x, z \in A$ cocycles. Let's consider the maps $$ A \to A \oplus A \to A$$ given by $y \mapsto (xy,yz)$ and $(u,v) \mapsto uz-xv$, respectively. We think of this as defining a double complex with three nonzero columns. (Here we need either to assume that $x,z$ are of degree $0$, or shift the grading in the second and third column.) The spectral sequence of a double complex will have an $E_1$-page given by $H(A) \to H(A) \oplus H(A) \to H(A)$. The $E_2$-page will have as its first column the kernel $$ \{ y \in H(A) : xy = yz = 0\},$$ and the $E_2$-differential gives a map to the third column, i.e. the cokernel $$ H(A)/\left( x H(A) + H(A)z\right).$$ This map is exactly the triple Massey product $\langle x,y,z\rangle$.

Question Is there a generalization of this construction to higher Massey products? I.e. can one write down a double complex with $n$ columns, such that the $E_{n-1}$-differential could have been used as an alternate definition of the $n$-fold Massey product?

To clarify what I'm after, there are of course many spectral sequences in which the differentials are known to be given by Massey products. What is special with the above construction is that it also produces the right indeterminacy: the triple Massey product $\langle x,y,z\rangle$ is in general well defined precisely as an element of $ H(A)/\left( x H(A) + H(A)z\right),$ which is also what you get from the spectral sequence. One tricky thing about trying to cook up a similar thing for higher products is that there is no similarly simple description of the indeterminacy for the higher products.

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    $\begingroup$ One might speculate that the correct "generic" higher analog might be modeled after associahedra: e. g. with the very next step, while $d^2=0$ in the horizontal direction is caused by $(xy)z=x(yz)$, with four variables there are five instances of this, and the next level complex might be caused by the fact that the appropriate linear combination of these five instances is a boundary. If all this makes sense, one probably should get what is usually called a multicomplex, with the sequence of differentials $d_n$ such that $d_{n+1}d_{n+1}$ is $d_n$-nullhomotopic. $\endgroup$ Jan 25, 2018 at 23:38

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This question is rather old (and maybe you have found an answer to this), but https://arxiv.org/abs/1507.04691 gives an answer to what you were looking for, I think (page 16, at the middle): if $(A,d)$ is dg the differentials of the Eilenberg--Moore s.s. (look at the bar construction as a double complex) "are" the Massey products and the usual higher ones defined partially on $H^1(A)^{\otimes n}$ are given by a differential on $E_{n-1}$, concretely $E_{n-1}^{n,n} \longrightarrow E_{n-1}^{1,2}$.

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    $\begingroup$ Hi Pedro, thanks a lot! I think I knew this when I asked the question, but if so I don't know why I didn't write it there. Unfortunately the implied indeterminacy in the Eilenberg-Moore SS is larger than the one for the Massey products, so this SS does not precisely produce the correct indeterminacies. $\endgroup$ Aug 23, 2021 at 9:47
  • $\begingroup$ @DanPetersen Ah, okay. Is the indeterminacy in the EMSS larger because in your case you actually pick representative cocycles to define your spectral sequence, but in the EMSS, there is an extra indeterminacy of the lift of each pair of classes $[x],[z]$? I suppose the indeterminacy is a bit more complicated than this, but I just wanted to understand exactly what the issue is. :) $\endgroup$
    – Pedro
    Aug 23, 2021 at 10:55
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    $\begingroup$ Yes - if you consider the differential in the EMSS corresponding to the triple product, then its target is $HA/(HA \cdot HA)$. In general the implied indeterminacy in the higher operation obtained from the EMSS is independent of the choice of elements being multiplied, unlike the classical Massey product. $\endgroup$ Aug 23, 2021 at 11:25

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