Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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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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Closed forms have zero boundary integral

Statement

Assume ACω. For oriented Mn with boundary, n≥1, if η∈Ωcn−1(M) is closed, then ∫∂Mj∗η=0. When M is compact, no separate support assumption on the smooth closed form is needed.

Facts & Assumptions

[F1]

The general Stokes theorem: Assume ACω. Let M be an oriented smooth n-manifold with boundary, n≥1, and let η∈Ωcn−1(M). With j:∂M↪M and the outward-normal-first orientation, ∫Mdη=∫∂Mj∗η. 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.

1.1F1

Closedness says dη=0. General Stokes identifies the boundary integral with ∫Mdη, which is the integral of the zero top form and hence zero. An empty boundary is included.

2.1step 1.1algebra∎

For compact M every closed support is compact. Thus the same argument applies to every smooth closed (n−1)-form. For n=1 it gives the signed sum of boundary values of a locally constant function; for the zero form all terms vanish.

Depends on

Used by

Dependency tree · two levels

7 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