Let $(C,\partial)$ and $(C',\partial')$ be chain complexes of $R$-modules where $R$ is a (commutative) ring. Let $F$ and $F'$ be finite filtrations of $C$ and $C'$ respectively, i.e., $$\varnothing = F_0C \subset F_1C \subset \ldots \subset F_nC = C$$ and similarly for $F'$. There exist spectral sequences associated to $F$ and $F'$, let's call them $E_{\ast,\ast}^\ast$ and ${E'}_{\ast,\ast}^\ast$.

It is quite easy to see that an isomorphism from $F$ to $F'$ induces an isomorphism from $E$ to $E'$. What is completely unclear to me (and sorry if this is not a research level question but I couldn't find an answer) is the converse, namely:

Does an isomorphism from $E$ to $E'$ imply a (quasi-)isomorphism between $F$ and $F'$?

More precisely, if two filtrations give rise to isomorphic spectral sequences, what is the strongest statement that can be made about them? Does this statement depend on the finiteness of the filtrations, on the nature of the ring $R$, on convergence of the spectral sequences, etc?

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    $\begingroup$ In other words, is the map from the filtered derived category to the category of spectral sequences full? I doubt it, although I don't have a counterexample. $\endgroup$ Jun 20, 2012 at 17:56
  • $\begingroup$ I realized you're only asking for isomomorphisms, nevertheless, it feels too strong. $\endgroup$ Jun 20, 2012 at 17:58
  • $\begingroup$ Donu, I agree that it seems strong. But the question is natural: in the filtered derived category, does the equivalence $C \sim C'$ iff $E(C)$ is isomorphic to $E(C')$ go by another name? $\endgroup$ Jun 20, 2012 at 18:09
  • $\begingroup$ Well, a morphism of filtered complexes which induces an isomorphism at the $E_1$ level (which implies it all the way down) is the same thing as a filtered quasi-isomorphism = isomorphism in the filtered derived category. Although this may not answer your question. $\endgroup$ Jun 20, 2012 at 18:29

2 Answers 2


Say your two filtered chain complexes are concentrated in degree zero. Then the spectral sequences degerate, and your questions become: If you have two filtered abelian groups and an isomorphism between the associated graded modules, can you deduce that the abelian groups are isomorphic? The answer is no; you can take $$ 0 \subset 2 \mathbb{Z} \subset \mathbb{Z} $$ and $$ 0 \subset \mathbb{Z} \subset \mathbb{Z} \oplus \mathbb{Z}/2. $$ As Ralph says, you usually need something on the chain level, unless you're in very degenerate cases (where the spectral sequence determines the isomorphism type of the object in the derived category).


What can be said is that if $C,C'$ have degree-wise finite filtrations and if $f: C \to C'$ is a filtration preserving chain map such that there is $r$ such that $E^r(f): E^r_{ij}(C) \to E^r_{ij}(C')$ is an isomorphism for all $i,j$ then $f$ is a quasi-isomorphm. This can be found in Brown: Cohomology of Groups, VII, Prop. 2.6.

Special cases where $E^\infty(C) \cong E^\infty(C')$ implies $H_\ast(C) \cong H_\ast(C')$ include:

  1. The spectral sequence degenerates, e.g. $E^\infty_{ij}=0$ if $j \neq 0$ because in this case $H^i(C)=E^\infty_{i,0}$.

  2. If $C,C'$ are complexes of modules over a ring $R$ and $E^\infty(C)$ consists of projective $R$-modules. In this case there is no extension problem and we have $H^n(C)=\oplus_{i+j=n}E^\infty_{ij}(C)$.

Here it is essential to consider the $E^\infty$-term since it can happen that $E^r(C) \cong E^r(C')$ but $E^{r+1}(C) \not\cong E^{r+1}(C')$. As an example take the LHS spectral sequence of the group extension

$$0 \to \mathbb{Z}/2 \to G \to \mathbb{Z}/2 \oplus \mathbb{Z}/2 \to 0.$$

Both, the dihedral group $D_8$ and the quaternion group $Q_8$ fit into this extension. Hence the spectral sequences have the same $E_2$-term but their $E_3$-terms differ.

  • $\begingroup$ Thank you for the answer, I've just taken a look at Brown. As you have said, the result already presupposes the existence of a filtered chain map. Do you know if it is possible to induce such an $f$ from the given isomorphism of spectral sequences? In any case, if no one makes a stronger statement than yours soon then I will accept this answer. $\endgroup$ Jun 20, 2012 at 18:03
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    $\begingroup$ I have to think about it. From a philosophical/practical point of view it's almost always worse in homological algebra if one hasn't maps on chain level to start with. Don't care about accepting my answer. $\endgroup$
    – Ralph
    Jun 20, 2012 at 18:11
  • $\begingroup$ One might be able to come up with an «$\infty$-thing», though, if one does not start with a chain level map (over a field, at least) $\endgroup$ Jun 20, 2012 at 19:21

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