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?
Best Answer
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).