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CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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De rham cohomology of punctured euclidean space

Statement

Under countable choice, Rn∖{0} has cohomology R in degrees 0,n−1 only for n≥2. For n=1 it has R2 in degree zero only, and for n=0 all groups vanish.

Facts & Assumptions

Given: A nonnegative integer n and countable choice.

[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]

De rham cohomology is smooth homotopy invariant: A smooth homotopy equivalence induces an isomorphism of de Rham graded real algebras.

[F3]

De rham cohomology of a finite disjoint union is the direct sum: For a finite disjoint union M=∐j=1mMj, restrictions give HdRk(M)≅⨁j=1mHdRk(Mj).

[F4]

Poincare lemma for differential forms on star shaped domains: 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.

Proof

technique · direct
1.1F1F2given

For n≥2, set r(x)=x/∣x∣ and i:Sn−1↪Rn∖{0}. Then ri=id⁡ and F(x,t)=((1−t)+t/∣x∣)x is a smooth homotopy from the identity to ir. Its scalar coefficient is strictly positive for 0≤t≤1, so it never reaches zero. Smooth homotopy invariance and the sphere computation give the groups asserted.

2.1F3F4given∎

For n=1, the two half-lines are star-shaped and connected, so each has only H0=R; their finite disjoint union gives R2. For n=0 the punctured space is empty, all its form spaces are zero, and every cohomology group is zero.

Source locator

Lee, Corollary 17.23, p.451, with the low-dimensional cases computed explicitly.

Depends on

Used by

Dependency tree · two levels

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Sources