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.

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The degree-zero homology quotient

PDF p. 13: CH3 identification and degree seven vanishing

Statement

Under the hypotheses of Degree-zero Bloch cycles are ordinary cycles, there is a canonical identification $H_{0} z^{r} (X, ∙) = Z^{r} (X) / B^{r} (X)$ , where $B^{r} (X)$ is the subgroup in Boundaries in degree zero.

Facts & Assumptions

Given: $k$ , $X$ and $r$ as in the statement.

Goal: Compute the degree-zero homology from the chain complex.

[F1]

The differential $\partial_{0}$ is zero and homology is the kernel of the outgoing differential modulo the image of the incoming differential. Bloch cycle complexes and higher Chow groups

[F2]

The degree-zero group is $Z^{r} (X)$ and the incoming image is $B^{r} (X)$ . Degree-zero Bloch cycles are ordinary cycles, Boundaries in degree zero

Proof

1.1F1

Since $\partial_{0} = 0$ , its kernel is all of the degree-zero chain group.

2.1F1F2step 1.1∎

Dividing that kernel by the incoming image and applying the identifications in [F2] yields $Z^{r} (X) / B^{r} (X)$ .

Depends on

Used by · 1

Degree zero and negative homology of the cycle complex

Dependency tree · 3 levels

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