PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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The associated graded of a filtered complex is a bigraded complex

Statement

Let $(C, d, F)$ be an increasingly filtered chain complex in an abelian category, and let $π_{p, n} : F_{p} C_{n} \to F_{p} C_{n} / F_{p - 1} C_{n}$ be the quotient projections. The family ${gr}_{p}^{F} C_{p + q}$ has the canonical differential $d_{p, q}^{0}$ uniquely characterized, for $n = p + q$ , by $d_{p, q}^{0} π_{p, n} = π_{p, n - 1} (d_{n} ∣_{F_{p} C_{n}}) .$ This differential has bidegree $(0, - 1)$ and satisfies $(d^{0})^{2} = 0$ . The family with this differential is the associated graded complex.

Facts & Assumptions

Given: A filtered chain complex in an abelian category.

[F1]

The graded quotients exist as subquotients (Associated graded quotients are well defined subquotients).

[F2]

$F_{p - 1} C$ and $F_{p} C$ are subcomplexes (Filtered chain complex).

[F3]

The differential of a complex descends to its quotient by a subcomplex, with square zero (The differential descends to a quotient complex).

Proof

technique · direct

1.1F1F2F3

The inclusion $F_{p - 1} C ↪ F_{p} C$ is a chain map of subcomplexes by [F2]. Apply [F3] to this inclusion: it gives $d^{0} : F_{p} C_{n} / F_{p - 1} C_{n} \to F_{p} C_{n - 1} / F_{p - 1} C_{n - 1}$ , uniquely satisfying $d^{0} π_{n} = π_{n - 1} d_{n}$ . The quotients are those supplied by [F1].

2.1F3step 1.1algebra∎

Writing $n = p + q$ , the target has complementary degree $n - 1 - p = q - 1$ ; its filtration degree is still $p$ . Moreover $(d^{0})^{2} π_{n} = π_{n - 2} d_{n - 1} d_{n} = 0$ , and the quotient map $π_{n}$ is epic. Hence $(d^{0})^{2} = 0$ , including zero graded pieces.

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.

The associated graded of a filtered complex is a bigraded complex

Statement

Facts & Assumptions

Proof

Source notes

Depends on

Dependency tree · 2 levels

Sources