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Computing form integrals by finite parametrizations
Statement
Let , let be oriented, and let . For let be bounded open Jordan domains and continuous and smooth up to the boundary in target coordinates: near each parameter point, a target coordinate representative extends smoothly to a Euclidean neighborhood. Suppose is an orientation-preserving diffeomorphism onto an open , the are pairwise disjoint, and . Then An empty family is allowed when the support is empty. No nonsingularity of on , and no -valued extension across a genuine target boundary, is assumed.
Facts & Assumptions
Integration on an oriented embedded submanifold: Let be an oriented embedded smooth -submanifold, with boundary allowed. For a smooth -form on such that has compact support on , define . If is an orientation-preserving diffeomorphism, this equals . Compact support is required on itself.
Linearity and additivity of the form integral: For compactly supported smooth top forms on an oriented and , Also , where ranges over connected components with their restricted orientations; only finitely many meet .
A map sends a compact set of content zero to a set of content zero: Let . Then if is on an open with values in and is compact with content zero, then is compact and has content zero. Content zero and nullity are those of def-null-and-content-zero-in-rn.
Additivity of the integral over finitely many Jordan pieces that fill a Jordan set up to content zero: Let , let be bounded and Jordan measurable, let , and let be bounded Jordan measurable sets such that has content zero whenever and such that has content zero. Let be bounded, Riemann integrable over and Riemann integrable over each . Then
Change of variables for an injective map on a compact Jordan set: Let , let be open, let be injective and , and suppose is invertible for every . Let be compact and Jordan measurable. For a bounded function , the following are equivalent: 1. is Riemann integrable on ; 2. is Riemann integrable on . When either condition holds,
Proof
Given: The objects and hypotheses in the statement above.
First record boundary control. Compactness and continuity give and : a limit of interior image points has a convergent parameter subsequence, and an interior parameter limit has image in . Each is compact of content zero. Cover it by finitely many parameter neighborhoods with smooth coordinate extensions. Intersect smaller closed neighborhoods with and apply the null-image lemma on each extension domain. Thus is content zero in every fixed relatively compact target chart, after finite localization. No derivative rank condition is used here.
By a finite chart partition of the compact support and linearity, it suffices to consider a form supported compactly inside a small chart whose coordinate domain is a bounded rectangle or half-rectangle and whose chart extends past its artificial edges. Such charts come from restricting a larger chart; the Euclidean boundary of has content zero. Put . The boundaries of the bounded lie in together with the chart images of , so are Jordan measurable. The localized coefficient is bounded, zero near artificial edges, and Riemann integrable, including the genuine face.
For that localized coefficient, outside . The disjoint overlap only on null boundaries after closure. Apply finite almost-partition additivity to the pieces and (whose integral is zero since vanishes there except on those boundaries). Consequently the signed chart integral is .
Fix and write on . This is a diffeomorphism onto . To justify substitution despite possible singularities at parameter boundary, let be the coefficient of on ; it extends continuously to the compact and is bounded, say by . Choose a finite union of grid cubes, with disjoint interiors, covering all but a collar of of arbitrarily small volume. Images of that collar have arbitrarily small chart volume as well: finitely many smooth coordinate extensions have bounded derivatives and are Lipschitz on smaller convex neighborhoods; a cube of side maps into a cube of side at most , so total covering volume increases by at most a fixed factor. Such collars exist because has content zero.
On the compact part , the nonzero support of lies in a compact subset of , since the localized form is supported inside . Subdivide or cover this compact part by finitely many cubes compactly contained in , splitting overlaps along their faces. On each such compact Jordan piece, the published substitution theorem applies to : it is injective, , and has invertible derivative on the surrounding open subset of . Pieces where the form vanishes contribute zero. Add these equalities. The omitted integrals on the parameter side are bounded by times collar volume; on the image side they are bounded by times the image-collar covering volume. Let those bounds tend to zero. This proves , with the sign supplied by orientation preservation.
Sum over and then over the finite chart localization. Empty support gives only zero coefficients; for the same collar estimate uses intervals and point boundaries. Degenerate Jacobians at boundary points are harmless because substitution was used only on compact subsets of the diffeomorphism domains.
Depends on
- Integration on an oriented embedded submanifold
- Linearity and additivity of the form integral
- A $C^1$ map sends a compact set of content zero to a set of content zero
- Additivity of the integral over finitely many Jordan pieces that fill a Jordan set up to content zero
- Change of variables for an injective $C^1$ map on a compact Jordan set
Used by
- Agreement of general and classical surface Stokes Corollary
- General Stokes agrees with both planar Green formulas Corollary
- An exact top form with nonzero integral on a disk Example
- Change of variables on an oriented circle Example
- Green circulation and flux on a disk Example
- Surface Stokes on a graph disk Example
- The angular period and the obstruction to bounding Example
- Volume-form divergence on the Euclidean ball Example
- Agreement with classical Gauss flux in Euclidean space Proposition
- Orientation-free density integration and its properties Theorem
Dependency tree · two levels
41 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 16.8 and proof, pp.408–409 (standard reference, not scraped)