LemmaStatement: AI-generatedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Alternating face boundaries square to zero

PDF p. 4: Smooth comparison with Bloch cycles

Statement

Let $C_{n}$ be abelian groups for $n \geq 0$ and let $d_{i}^{n} : C_{n} \to C_{n - 1}$ , $0 \leq i \leq n$ , satisfy $d_{i}^{n - 1} d_{j}^{n} = d_{j - 1}^{n - 1} d_{i}^{n}$ for $i < j$ and $n \geq 2$ . Put $C_{n} = 0$ for $n < 0$ , $\partial_{n} = 0 : C_{n} \to C_{n - 1}$ for every $n \leq 0$ , and $\partial_{n} = \sum_{i = 0}^{n} (- 1)^{i} d_{i}^{n}$ for $n \geq 1$ . Then $\partial_{n - 1} \partial_{n} = 0$ for every integer $n$ , and $H_{n} (C) = \ker \partial_{n} / im \partial_{n + 1}$ is defined and vanishes for $n < 0$ . This applies to the admissible cycle face maps for smooth finite-type schemes over fields.

Facts & Assumptions

Given: The groups and face identities in the Statement.

Goal: Prove the chain identity and justify the homology quotient.

[F1]

The admissible cycle face maps satisfy these identities. Admissible face restrictions and their cycle identities

[F2]

A chain complex has consecutive differential composites zero; its homology is cycles modulo boundaries. Chain complex in an abelian category, Homology object of a chain complex

Proof

1.1givenconstructalgebra

For $n \geq 2$ , expand $\partial_{n - 1} \partial_{n} = \sum_{a = 0}^{n - 1} \sum_{b = 0}^{n} (- 1)^{a + b} d_{a}^{n - 1} d_{b}^{n}$ . Pair $(a, b)$ with $a < b$ to $(b - 1, a)$ . This is a bijection onto the pairs whose first coordinate is at least the second, with inverse $(c, d) \mapsto (d, c + 1)$ .

2.1givenstep 1.1algebra

The two maps in each pair agree by the face identity, while their signs are opposite. All terms cancel. For $n = 1$ the composite is zero because $\partial_{0} = 0$ ; for $n \leq 0$ it is zero because $\partial_{n}$ is the explicitly defined zero map.

3.1F1F2step 2.1∎

Therefore boundaries lie in kernels, and [F2] defines the displayed homology quotient. For $n < 0$ the group itself is zero. Substitution of [F1] gives the cycle-complex application, completing the proof.

Depends on

Used by · 1

Bloch cycle complexes and higher Chow groups

Dependency tree · 3 levels

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