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De Rham integration cochain
Definition
For a smooth manifold , possibly with boundary, and , define On this means , a finite sum. In negative degrees is the zero map. Forms and the de Rham complex at a boundary use The de Rham complex and pullback extend to manifolds with boundary. No orientation of is required: the domain simplex has the specified orientation. The name integration cochain map is justified by De Rham integration is a cochain map ↗.
Facts & Assumptions
Integral of a form over a smooth singular simplex assigns an extension-independent real number to a form and smooth simplex, with degree-zero evaluation.
Smooth singular chain and cochain complexes identifies smooth cochains with real functions on the supplied simplex set, evaluated by finite sums.
The de Rham complex and pullback extend to manifolds with boundary supplies , its degree conventions and the boundaryless specialization.
Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in gives linearity of Riemann integrals, hence of the Jordan integrals in [F1] by zero extension.
Verification
Given: A smooth manifold , an integer and a form .
Definition [F1] gives one uniquely determined value on every smooth -simplex. For each finite chain, distributing its coefficients in the displayed sum shows that this function extends to a real linear functional. Two such functionals agreeing on all simplices agree on every finite chain. By [F2] this defines a unique element of the cochain space. No choice of a vector-space basis or of simplex extensions is made.
For forms of degree and real , alternating evaluation in the definition of pullback gives . For , the Jordan coefficient integral is linear by [F4]; for , evaluation is linear directly. Thus on every simplex, and step 1.1 makes a real linear map.
For the value at a point simplex is exactly the function value there; for it is the integral of the pulled-back one-form over the closed oriented interval. Degenerate simplices remain in the supplied set and receive their actual integrals, rather than being removed. If , [F3] gives zero source and the map is zero, even though the smooth cochain space may be nonzero. Negative degrees are zero on both sides. On an empty manifold there are no simplices and the only cochain is zero. These conventions include the zero form and zero chain and require no choice.
Depends on
Used by
- De Rham integration cochain on a smooth path Example
- The de Rham map on the angular form Example
- De Rham integration respects wedge and cup in cohomology Lemma
- The de Rham map is an isomorphism on convex coordinate domains Lemma
- Naturality of the de Rham map Proposition
- De Rham integration is a cochain map Theorem
Dependency tree · two levels
31 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
- Peter S. Park, Proof of de Rham's Theorem (standard reference, not scraped)