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A compactly supported primitive has zero total derivative integral
Statement
Assume . If is oriented and boundaryless, , and , then . In particular, on a compact such manifold every exact smooth top form has zero integral. The compact-support assumption is on the primitive , not merely on .
Facts & Assumptions
The general Stokes theorem: Assume . Let be an oriented smooth -manifold with boundary, , and let . With and the outward-normal-first orientation, 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.
View as a manifold with empty boundary. General Stokes applies to the compactly supported primitive and gives , including the zero primitive.
If is compact, the closed support of any smooth primitive is a compact subset of , so the first conclusion applies to every exact top form. This holds for n=1 as well; no negative-degree form or dimension-zero Stokes assertion is used.
Depends on
Used by
Dependency tree · two levels
6 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 Corollary 16.13, p.414 (compact-support version from Theorem 16.11) (standard reference, not scraped)