TITLE: Can we construct a filtered chain complex from a spectral sequence?
QUESTION [5 upvotes]: 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:

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

Indeed, the spectral sequence in my mind is from the unrolled spectral sequence in
https://ncatlab.org/nlab/show/conditional+convergence
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?

REPLY [3 votes]: 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.

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

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!

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.