MHB What is the Homology Spectral Sequence for a Chain Complex with Filtration?

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    2015
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The discussion revolves around computing the homology spectral sequence for a chain complex defined by the filtration \( F_m := \sum_{n = 0}^m C_n \). Participants are encouraged to engage with the problem of the week (POTW), but no responses have been provided. The original poster has shared their solution for reference. The focus remains on understanding the implications of the filtration on the spectral sequence. Overall, the thread highlights a lack of community interaction on this mathematical topic.
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Here is this week's POTW:

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Let $\mathcal{C} = \{C_n\}_{n\ge 0}$ be a chain complex. Compute the homology spectral sequence associated with the filtration $F_m := \sum_{n = 0}^m C_n$, $m\ge 0$.
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No one answered this week's problem. You can read my solution below.
I'll let $d$ represent the differential of the chain complex. Since $E_n^0 \approx F_n/F_{n-1} \approx C_n$ and $d(F_n) \subseteq F_{n-1}$, then the map $d^0_n : F_n/F_{n-1} \to F_n/F_{n-1}$ is zero, $(E^0,E_n^0,d^0)$ is identified with $(\mathcal{C},C_n,0)$, and $E_n^1 \approx E_n^0 \approx C_n$. The map $d_n^1 : F_n/F_{n-1} \to F_{n-1}/F_{n-2}$ reduces to $d_n : C_n \to C_{n-1}$, and so $(E^1,E_n^1,d^1)$ is identified with $(\mathcal{C},C_n,d_n)$. We have $E_n^2 \approx H_n(\mathcal{C})$ and $d^2_n$ is zero, so $(E^2, E_n^2, d^2)$ is identified with $(H_*(\mathcal{C}), H_n(C), 0)$. All higher order $d^r$ are zero, $E_n^r \approx H_n(\mathcal{C})$ for $2 \le r < \infty$. Finally, $E_n^\infty \approx \sum_{m = 0}^n H_m(\mathcal{C})/\sum_{m = 0}^{n-1} H_m(\mathcal{C}) \approx H_n(\mathcal{C})$.