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

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Nonzero total integral obstructs exactness on a closed manifold

Statement

Let $M^{n}$ be compact, oriented, and boundaryless, $n \geq 1$ . A smooth top form $ω$ with $\int_{M} ω \neq 0$ is not exact. In particular every positive smooth top form on a nonempty such $M$ is not exact.

Facts & Assumptions

[F1]

A compactly supported primitive has zero total derivative integral: If $M^{n}$ is oriented and boundaryless, $n \geq 1$ , and $η \in Ω_{c}^{n - 1} (M)$ , then $\int_{M} d η = 0$ . In particular, on a compact such manifold every exact smooth top form has zero integral. The compact-support assumption is on the primitive $η$ , not merely on $d η$ .

[F2]

Positivity of the oriented integral: Let $ω \in Ω_{c}^{n} (M)$ be nonnegative on the positive determinant ray of an oriented smooth manifold. Then $\int_{M} ω \geq 0$ , and $ω \neq 0$ implies $\int_{M} ω > 0$ .

Proof

Given: The objects and hypotheses in the statement above.

1.1

If $ω = d η$ , compactness of $M$ makes the primitive compactly supported. The exact-integral vanishing result gives $\int_{M} ω = 0$ . Thus a nonzero integral excludes exactness.

2.1

For a positive form on nonempty $M$ , positivity implies it is nonzero, and its integral is strictly positive. Apply the preceding implication. On the empty manifold every integral is zero, so the nonzero-integral hypothesis cannot hold; the positive-form conclusion explicitly assumed nonempty $M$ .

F2step 1.1∎

Depends on

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Sources

Lee Corollary 16.13; Merry Corollary 27.2 proof (nonexactness consequence without cohomology terminology) (standard reference, not scraped)