Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The angular form generates the first de rham cohomology of the circle

Example

Under countable choice, α=(xdyydx)/(2π)S1 has period one and its class generates HdR1(S1).

Facts & Assumptions

Given: Assume countable choice. The counterclockwise oriented unit circle and the displayed one-form.

[F1]

De rham cohomology of spheres: Assume countable choice. For n1, HdRk(Sn) is R in degrees 0,n and zero otherwise. For S0 it is R2 in degree zero and zero otherwise.

[F3]

A nonzero period obstructs exactness and bounding: Let SM be an oriented compact boundaryless embedded k-submanifold, k1, and let ω be a closed smooth k-form on M. If Sω0, then ω is not exact on M, and S cannot be the induced oriented boundary of a compact embedded (k+1)-submanifold of M.

Verification

technique · direct
1.1

The form is smooth and closed because two-forms on a one-manifold vanish. For γ(t)=(cost,sint), 0t2π, substitution gives γα=dt/(2π), hence S1α=1.

givenalgebra
2.1

The circle is compact, oriented, boundaryless and embedded, and the form is closed, so its nonzero period obstructs exactness. Since the sphere theorem gives dimH1(S1)=1, this nonzero class is a basis.

F1F3step 1.1

Source locator

Lee, angular form (17.1), p.441, and Theorem 17.21, pp.450–451; the period is calculated explicitly.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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