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The question is pretty much in the title. It is a classical fact that a filtered dga gives rise to a multiplicative spectral sequence. It is claimed in Remark 4.1 of https://arxiv.org/pdf/1410.6728.pdf that this generalizes to filtered $A_\infty$- algebras(see the page 27 of the same for this definition). One can define a notion of filtered $A_N$-algebras for $N \in \mathbb{N}^{\geq 0}$, $N \geq 3$ where $m_i$ are only defined for $i \leq N$ and the $A_{\infty}$ equations hold for $i \leq N$ (so the multiplication is still associative on the level of homology).

Question: Is the natural spectral sequence associated to a filtered $A_N$ algebra multiplicative?

I would guess the answer is yes, but could not find a reference for this fact and there seems to be some subtleties in the literature according to this earlier post:

Multiplicative structure on spectral sequence

Motivation: It seems like this might be an efficient way to prove that some spectral sequences that arise "in nature" are multiplicative, say by equipping the relevant (co) chain complex with an $A_3$ structure.

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    $\begingroup$ It seems to me this should follow from the fact that the filtering condition for $A_\infty$-algebras is simply the condition that the coderivation $b : BA\to BA$ on the bar construction on $A$ preserves the induced filtration on $BA$. Then all your known formalism works. $\endgroup$
    – Pedro
    Commented Jan 26, 2018 at 11:44
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    $\begingroup$ Are you sure in your definition of $A_N$-algebras to only have the $A_\infty$-equations for $i\le N$? I mean, a natural way to truncate $A_\infty$ is to say that operations of arity greater than $N$ vanish, but that would give equations for $i\le 2N-1$. Do you have any meaningful examples of your $A_N$-algebras where the equations from $N+1$ to $2N-1$ do not hold? $\endgroup$
    – Vladimir Dotsenko
    Commented Jan 26, 2018 at 11:47
  • $\begingroup$ Perhaps I was not sufficiently clear in my explanation since I thought it was a standard definition. I am assuming that operations of arity N+1 and higher are not defined. So (with suitable sign conventions) for N=3, I would have m_1 is a differential (Eq. 1), m_2 satisfies Leibnitz rule (Eq. 2), multiplication is associative up to a homotopy m_3 (Eq. 3). This seems like a very natural structure. Examples typically extend to $A_\infty$ structure, however that extension may require some extra work. The question is whether that work is needed for constructing a mult. spectral seq. $\endgroup$
    – algebrachallenged
    Commented Jan 26, 2018 at 12:44
  • $\begingroup$ P.S. I gather examples that you are asking about can be constructed using homotopy theory by taking suitable versions of chains on spaces considered here: mathoverflow.net/questions/248416/… and equipping them with a Pontryagin product. $\endgroup$
    – algebrachallenged
    Commented Jan 26, 2018 at 12:49

1 Answer 1

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For the $d_r$-differentials to be derivations, i.e., to satisfy the Leibniz rule $d_r(x \cdot y) = d_r(x) \cdot y \pm x \cdot d_r(y)$ with $x \cdot y = m_2(x \otimes y)$, it is enough to have a filtered differential graded $A_2$-algebra. This follows from Sections 7 and 8 of Massey's 1954 paper "Products in Exact Couples", since the pairing $m_2 \colon A \otimes A \to A$ is a filtration-preserving chain map.

Assuming the Leibniz rule, if the pairing of $E_1$-terms (or $E_2$-terms) is associative and unital, then so are the induced pairings of $E_r$-terms for all greater $r$.

Aside: The "subtleties in the literature" concern exact couples that do not come from filtered chain complexes, e.g., those that arise from the homotopy groups of sequences of spectra. If a pairing of sequences is only defined in the stable homotopy category, then it is not generally clear that there will be a corresponding product in the spectral sequence.

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  • $\begingroup$ Thanks for your answer. Just to fill in between lines, the answer to the question is "yes" (at least assuming one omits "unital" from the definition of multiplicative spectral sequence) because it follows from the axiom of filtered $A_3$ algebra structure on the first page will be associative. $\endgroup$
    – algebrachallenged
    Commented Jan 27, 2018 at 23:22
  • $\begingroup$ In the case of $A_2$ (or $A_3$...) algebra, assuming the filtration is bounded, then I take it that it is still true that we have convergence of the multiplicative spectral sequence (i.e. the possibly non-associative multiplication on the $E_\infty$ page is the associated graded of the induced multplication on homology of $A$)? $\endgroup$
    – algebrachallenged
    Commented Jan 28, 2018 at 21:48

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