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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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Poincare lemma for differential forms on star shaped domains

Statement

Every closed smooth k-form on a star-shaped open domain is exact for k≥1. For centre 0, one primitive is ηx(v1,…,vk−1)=∫01tk−1ωtx(x,v1,…,vk−1) dt.

Facts & Assumptions

Given: A domain U star-shaped about c, and a closed k-form ω with k≥1.

[F1]

Radial contraction of a star shaped domain: For an open U⊆Rn star-shaped about a specified c∈U, the radial contraction is F:U×[0,1]→U, F(x,t)=c+t(x−c). def-star-shaped-open-subset-of-rn says exactly that each displayed value lies in U. The coordinate expression is polynomial, so its restriction is smooth up to both endpoints; F(x,0)=c and F(x,1)=x. The centre is part of the data, so U is nonempty. For n=0 the unique nonempty domain is a point and the formula is constant.

[F2]

De rham homotopy formula for a smooth homotopy: If F:M×[0,1]→N is smooth up to the endpoints and Ft(x)=F(x,t), then F1∗−F0∗=d(KF∗)+(KF∗)d.

Proof

technique · direct
1.1F1F2given

Take the radial homotopy from the constant map to the identity. Its time-zero pullback on positive-degree forms vanishes because the differential of the constant map is zero. The homotopy formula and dω=0 give ω=d(KF∗ω), so η=KF∗ω is a smooth primitive.

2.1F2step 1.1∎

For c=0, dF(x,t)∂t=x and dF(x,t)(v,0)=tv. The contraction coefficient therefore equals tk−1ωtx(x,v1,…,vk−1). Integrating gives the displayed formula. When k=1 the factor is 1, including t=0; for k>1 the integrand is smooth and vanishes there. Degrees exceeding the dimension have zero form and zero primitive.

Source locator

Lee, Theorem 17.14, p.447; the explicit primitive follows by evaluating the interval operator.

Depends on

Dependency tree · 2 levels

Sources