Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09
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.

First integral homology and conjugation coinvariants

Statement

Naturally H1(G;Z)Gab. If RF, conjugation induces an F/R-action on H1(R;Z) and H1(R;Z)F/RR/[F,R], where [F,R] is generated by frf1r1. Derived homology carries the inherited DC and supplied-resolution convention.

Facts & Assumptions

Given: G is a group; in the second assertion R is normal in F; coefficients are trivial integers.

[F1]

Normalized diagonal bars compute homology (Diagonal bar coinvariants compute group homology).

[F2]

Inhomogeneous normalized chains kill an entry equal to 1 (The normalized homogeneous bar complex).

Proof

1.1

In degree one all boundaries to degree zero are zero with trivial coefficients. Degree two has d[xy]=[y][xy]+[x], with [1]=0. Thus H1 is the abelian group generated by symbols [x] subject precisely to [xy]=[x]+[y]. The assignment [x]x[G,G] induces a map to Gab; conversely x[x] is a group homomorphism to an abelian group, kills every commutator, and factors through Gab. The maps are inverse on generators and commute with group homomorphisms.

F1F2algebra
2.1

For R the isomorphism sends conjugation by f to r[R,R]frf1[R,R]. Conjugation by an element of R is trivial on this quotient, so the action factors through F/R. Taking coinvariants adds exactly the relations frf1r1=1. Since [R,R][F,R], the resulting quotient is R/[F,R]. If R=1 it is zero, and if R=F it is Fab.

step 1.1algebra

Depends on

Used by

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources