Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Divergence theorem for a volume form

Statement

Assume ACω. Let Mn be oriented with boundary, n1, let μ be a positive smooth volume form, and let X be a compactly supported smooth vector field. Then M(divμX)μ=Mj(ιXμ), with outward-normal-first orientation. For compact M every smooth X is allowed.

Facts & Assumptions

[F1]

Divergence as an exterior derivative: For a positive volume form μ and smooth vector field X on an oriented smooth n-manifold, n1, with boundary allowed, d(ιXμ)=(divμX)μ. No tangency assumption on X at the boundary is needed.

[F2]

The general Stokes theorem: Assume ACω. Let M be an oriented smooth n-manifold with boundary, n1, and let ηΩcn1(M). With j:MM and the outward-normal-first orientation, Mdη=Mjη. An empty boundary contributes zero; in dimension one its integral is a finite signed sum of point values.

Proof

Given: The objects and hypotheses in the statement above.

1.1

The form ιXμ is smooth and has support contained in suppX, hence compact. Its derivative is (divμX)μ by the divergence-form identity.

F1given
2.1

Apply general Stokes to that compactly supported (n1)-form. This gives the stated formula and ensures the boundary restriction is compactly supported. On compact M the support of every smooth X is compact; an empty boundary yields zero, as does X=0. For n=1 the right side is a signed sum of contraction values.

F2step 1.1

Depends on

Used by

Dependency tree · two levels

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Sources