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I have recently been learning about spectral sequences, not in the context of any problem, but mostly out of curiosity. There are two questions that have occurred to me which I've been unable to answer, even after computing a few examples.

For the purposes of both questions, suppose I have two filtered chain complexes: $0=C_{0}\subset C_{1}\subset\ldots\subset C_{n}$ and $0=K_{0}\subset K_{1}\subset\ldots\subset K_{m}$. For the sake of these questions one can assume that the modules making up the complexes are finite dimensional vector spaces.

  1. If $n=m$ and there exists $q$, such that $0<q<n$, and for all $p\neq q$ $K_{p}=C_{p}$, then what is the relationship between the spectral sequences associated to each filtration? To put it another way, if I change a filtration at one level, how does the associated spectral sequence change?

  2. If $m<n$ and there exist $m$ integers $0\leq i_{0}<i_{1}<\ldots<i_{m} \leq n$ such that $K_{k}=C_{i_{k}}$, then how are the spectral sequences of the two complexes related. That if I forget certain $C_{i}$ to form a new (shorter ) filtration for the same complex, how does the spectral sequence change?

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    $\begingroup$ The spectral sequences can be pretty much totally unrelated. For example, with (1) consider $C_0=0$ and $C_i=C_n$ otherwise -- the trivial filtration. The gap between your 2nd spectral sequence and the 1st is identical to the gap between your 2nd spectral sequence and the object you're studying by filtering. In general if one filtration is a subfiltration of the other you would have a natural map between the two spectral sequences. But it could of course be much worse than that. It's probably helpful to think about the extension problem for abelian groups. $\endgroup$ – Ryan Budney Sep 18 '14 at 18:38
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    $\begingroup$ For (1), the parts above q+1 and below q-1 respectively are definitely isomorphic in these spectral sequences. So I'd disagree with "totally unrelated". $\endgroup$ – Achim Krause Mar 18 '15 at 2:44
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    $\begingroup$ Note that he only wants to change the filtration very locally, whereas you seem to assume more general modifications. $\endgroup$ – Achim Krause Mar 18 '15 at 2:45
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You might want to look at persistent homology, which contains the same information as the dimensions of the vector spaces appearing in the spectral sequence. It is known that persistent homology doesn't change much if one makes small changes to the filtration ("stability theorem"). However "smallness" there isn't the same as in your case. For your case, the parts of the spectral sequence not going "across" the modification would be obviously unchanged, but the rest could be quite messed up.

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This probably isn't a Good Answer to your question, as it addresses neither 1) nor 2), and Ryan Budney's comment looks like a good exercise; nonetheless, you mention not currently having an application in mind, and I think it only to the good to mention a particular spectral sequence which is fairly typical of the immediate applications of spectral sequences in topology; I will call it the Serre spectral sequence, though other names might be attached as well;

Theorem Given a reasonable fibration $$ F \to E \to B $$ in which the base $B$ and the fiber $F$ are connected, in which $\Pi_1 B$ acts trivially on $H^*(F)$, there is a spectral sequence with page $$ E_2^{p,q} \simeq H^p(B;H^q(F)) $$ and converging to (an extension problem for) $H^{p+q}(E)$.

And the point of mentioning this is:

Theorem(ish) one can construct this spectral sequence in such a way that the filtered chain complexes you start with have the same underlying filtered groups, whatever particular compatible $E$ you're actually trying to study; the differences are in the differentials. More: the differences are in the part of the differentials of filtration degree at least $2$

which is why the spectral sequence groups on page $E_2$ don't mention the total space $E$.

A better answer should go into why "lots of groups/filtration layers in common" isn't the right kind of similarity for filtered chain complexes — you will want to think about chain homotopy operators and filtered versions of that; but as far as scope/variability, Ryan's remarks look good, and fibrations are a good family of examples to keep in mind.

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    $\begingroup$ Now it occurs to me that when you write "C_i = K_i", you might already mean "as chain complexes", which ... Ryan's looks much more helpful, there. $\endgroup$ – Jesse C. McKeown Sep 18 '14 at 22:20

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