Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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The standard volume form generates top cohomology of a sphere

Example

Under countable choice, for n≥1 the form ω=∑i=1n+1(−1)i−1xi dx1∧⋯∧dxi^∧⋯∧dxn+1∣Sn generates Hn(Sn).

Facts & Assumptions

Given: Assume countable choice. The outward orientation on the unit sphere.

[F1]

De rham cohomology of spheres: Assume countable choice. For n≥1, HdRk(Sn) is R in degrees 0,n and zero otherwise. For S0 it is R2 in degree zero and zero otherwise.

[F2]

Nonzero total integral obstructs exactness on a closed manifold: Let Mn be compact, oriented, and boundaryless, n≥1. A smooth top form ω with ∫Mω≠0 is not exact. In particular every positive smooth top form on a nonempty such M is not exact.

Verification

technique · direct
1.1givenalgebra

For tangent vectors v1,…,vn, expansion along the first column gives ωx(v1,…,vn)=det⁡(x,v1,…,vn). The outward orientation is precisely the convention that this determinant is positive on positive tangent bases. Since x is a nonzero normal to the tangent space, the determinant is nonzero on every tangent basis; thus ω is smooth, positive and nowhere zero.

2.1F1F2step 1.1∎

The sphere is nonempty, compact, oriented and boundaryless, so the positive-top-form clause of the integral obstruction theorem makes ω nonexact. It is closed by top degree. The sphere computation gives a one-dimensional Hn, and its nonzero class therefore generates it.

Source locator

Lee, Theorem 17.21, pp.450–451, and Proposition 16.28, p.422, positivity of volume integration; the proof verifies nonexactness by the stated Stokes supplier.

Depends on

Used by

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Sources