PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Orientation reversal changes the integral sign

Statement

Let $- M$ have the opposite orientation on every component of an oriented smooth manifold $M$ . For every compactly supported top form, $\int_{- M} ω = - \int_{M} ω$ , in all dimensions.

Facts & Assumptions

[F1]

Independence of atlas, partition and refinement: The compact-support integral on an oriented manifold is independent of the chart cover, coordinate maps, subordinate partition, and refinement. If $U \subset M$ is open and contains $supp ω$ , with its restricted orientation, then $\int_{U} ω ∣_{U} = \int_{M} ω$ .

Proof

Given: The objects and hypotheses in the statement above.

1.1

Compute both integrals with the same chart cover and partition, as independence permits. In positive dimension each chart sign changes from $σ$ to $- σ$ , with its coefficient and Riemann integral unchanged.

2.1

In dimension zero each point sign changes from $ε$ to $- ε$ . Factoring $- 1$ out of either finite sum proves the formula; for empty support or the zero form it reads $0 = - 0$ .

step 1.1algebra∎

Depends on

Independence of atlas, partition and refinement

Used by

False: top-form integration needs no orientation False statement
Change of variables on oriented manifolds Theorem

Dependency tree · two levels

4 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

Lee Proposition 16.6(b), pp.407–408 (standard reference, not scraped)