Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (gpt-6.1-sol)
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.

An exact top-degree current has zero total charge on a closed manifold

Statement

If X is a compact oriented smooth n-manifold without boundary and T is a degree-(n−1) de Rham current on X, then ⟨dT,1⟩=0. Thus a top-degree current identity can be paired with the constant test function only after every term has been defined as a degree-n current with the same pairing convention. This conclusion determines no physical source coefficient or normalization.

Facts & Assumptions

Given: The closed oriented manifold X, a degree-(n−1) current T, and the constant function 1∈Ω0(X).

[F1]

De Rham currents of compact oriented source cycles defines ⟨dT,η⟩=(−1)k+1⟨T,dη⟩ for a degree-k current and sets d to zero outside current degrees 0,…,n.

[F2]

The current differential is continuous and squares to zero verifies that this differential is a current in the stated degree.

Proof

Given: The current differential sign and the smooth exterior derivative on functions.

1.1F1F2givencasesalgebra

If n=0, then T has degree −1 and is zero by the out-of-range current convention in [F1], so the conclusion holds. If n≥1, then k=n−1 is an allowed degree and, for the test function 1, [F1] gives ⟨dT,1⟩=(−1)(n−1)+1⟨T,d1⟩=(−1)n⟨T,0⟩=0, since the exterior derivative of a constant function is zero. The pairing is well-typed by [F2].

2.1F1step 1.1given∎

By [F1], a degree-n current acts on Ω0(X), so equality of degree-n currents permits evaluation on 1 term by term. If any proposed Bianchi term has not been defined in this same dual space, the displayed pairing is not defined and the calculation in step 1.1 cannot be applied to it. The argument is mathematical only and supplies no value for any physical coefficient.

Depends on

Dependency tree · 2 levels

Sources