Suppose $\{(E_r,d_r)\}$ for $r>0$ forms a spectral sequence over a field $\mathbb{F}$, i.e. for any $r$, $(E_r,d_r)$ is a chain complex over $\mathbb{F}$ and $E_{r+1}=H(E_r,d_r)$. For simplicity, suppose $E_N=E_{N+1}=\cdots=E_{\infty}$ for some fixed number $N$. I have the following questions:

  1. Can we construct a filtered chain complex $(C,d)$ such that the induced spectral sequence is $\{(E_r,d_r)\}$?
  2. Can we determine $(C,d)$ up to some equivalence (like filtered quasi-isomorphism)?
  3. Can we relax the assumption to obtain the similar results?
  4. Are there any good references?

Indeed, the spectral sequence in my mind is from the unrolled spectral sequence in


There are many similar questions, which gives some potential counterexamples when we replace $\mathbb{F}$ by $\mathbb{Z}$ or without the assumption about the convergence at some finite page $E_N$.

Complete invariant of filtered chain complexes under chain homotopy equivalence

Functoriality of filtered spectral sequences

isomorphic spectral sequences => quasi-isomorphic filtered chain complexes?


1 Answer 1

  1. Yes, since you're working over a field, you can decompose your spectral sequence into a big direct sum of

a) permanent cycles, and

b) for each r>1, chains supporting a d_r, and the cycle hit by that d_r.

It is straightforward to represent each such summand by a simple filtered chain complex, so do so, and then take a direct sum. The resulting filtered chain complex has its spectral sequence isomorphic to the one you began with. I don't know a written reference for this, but it is an old idea, and it is not difficult to write out the details.

  1. I think this question is addressed by the more general results in "Model category structures and spectral sequences" by Cirici, Santander, Livernet, and Whitehouse: https://arxiv.org/pdf/1805.00374.pdf

  2. If I recall correctly, you can replace your ground field with any commutative ring without changing the answer to #1. But I am only going from memory here!

  3. See answers to #1 and #2.

You mentioned unrolled exact couples. Let me mention that question #2 becomes much, much more subtle if you work with exact couples rather than filtered chain complexes, but at least question #1 still has the same answer (for basically the same reasons) in that setting.

  • $\begingroup$ Just in case other, more general references for #4 could also be helpful: after Cartan and Eilenberg's book, the next good thing to read when it comes to algebraic manipulation of the filtered chain complexes "underneath" a spectral sequence is probably Deligne's "Theorie de Hodge II". For similar ideas in the setting of exact couples, Boardman's "Conditionally convergent spectral sequences" is excellent. Sorry if these references are too basic and already well-known to you. $\endgroup$
    – user164898
    May 11, 2021 at 20:49

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