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Suppose I have a chain complex $C$ with differential $D$ and filtration $F$. Suppose further that I can decompose $D$ by its action on the filtration, i.e. there are maps $D = D_1 + D_2 + \cdots$ so that $D_k: F^iC \to F^{i+k}C$. (In every example I am working with, the filtration is bounded.)

Now consider the spectral sequence $\{E^r, d^r\}$ induced by the filtration on $C$. Is there an easy relationship between $d^r$ and $D_r$?

I know (say, from ncatlab) that $d^r$ is the restriction of $D$ to $r$-almost cycles, i.e. "elements in filtering degree $p$ on which $D$ decreases the filtering degree to $p-r$." Can we say that $d^r$ is the map induced by $D_r$ on the appropriate quotient?

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Here is some motivation. McCleary (and probably Weibel) makes it very clear that knowledge of $E^r$ and $d^r$ is sufficient to determine $E^{r+1}$ but not $d^{r+1}$. The situation I've described above seems ideals for calculating the differentials. –  Adam Saltz Apr 2 '13 at 15:37

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The answer is no! One can see this by computing small examples in which $D = D_1 + D_2$ but the differential on $E^3$ is non-trivial. It is true that $d^r$ is determined by $E_1, \ldots, E_r$ and $D_1, \ldots, D_r$.

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