PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-12

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E zero is the associated graded complex

Statement

In the specified quotient model, $(E^{0}, d^{0})$ is the associated graded complex: $E_{p, q}^{0} = {gr}_{p} C_{p + q}$ .

Facts & Assumptions

Given: A filtered chain complex with the specified initial-page quotient model.

[F1]

The initial page is $F_{p} C_{n} / F_{p - 1} C_{n}$ (R page of the spectral sequence of a filtered complex).

[F2]

At $r = 0$ , the page differential is characterized on quotient representatives by $[x] \mapsto [d_{n} x]$ (The filtered differential induces d r on the r page).

[F3]

The associated graded complex has the unique differential satisfying $d_{p, q}^{0} π_{p, n} = π_{p, n - 1} (d_{n} ∣_{F_{p} C_{n}})$ (The associated graded of a filtered complex is a bigraded complex), as supplied by quotient-complex descent (The differential descends to a quotient complex).

Proof

technique · direct

1.1F1

The defining numerator and denominator of [F1] are $Z^{0} = F_{p} C_{n}$ and $B^{0} = F_{p - 1} C_{n}$ . Their quotient is exactly the chosen model of ${gr}_{p} C_{n}$ .

2.1F2F3step 1.1∎

By [F2], the initial-page differential sends every quotient representative $[x]$ to $[d_{n} x]$ , so its composite with the quotient projection is $π_{n - 1} d_{n}$ . By [F3], the associated-graded quotient differential is the unique map with that same composite. The quotient projection is epic, hence the two maps agree. With $n = p + q$ , their common bidegree is $(0, - 1)$ .

Source notes

Weibel, Chapter 5, Construction 5.4.6 and Lemma 5.4.7, pp.133–134; Sharifi, Theorem 4.2.3, pp.91–92. Increasing homological indices are used here.

E zero is the associated graded complex

Statement

Facts & Assumptions

Proof

Source notes

Depends on

Dependency tree · 2 levels

Sources