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De rham cohomology of punctured euclidean space

Statement

Under countable choice, $R^{n} ∖ {0}$ has cohomology $R$ in degrees $0, n - 1$ only for $n \geq 2$ . For $n = 1$ it has $R^{2}$ 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 $n \geq 1$ , $H_{d R}^{k} (S^{n})$ is $R$ in degrees $0, n$ and zero otherwise. For $S^{0}$ it is $R^{2}$ 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 $M = ∐_{j = 1}^{m} M_{j}$ , restrictions give $H_{d R}^{k} (M) ≅ ⨁_{j = 1}^{m} H_{d R}^{k} (M_{j})$ .

[F4]

Poincare lemma for differential forms on star shaped domains: Every closed smooth $k$ -form on a star-shaped open domain is exact for $k \geq 1$ . For centre $0$ , one primitive is $η_{x} (v_{1}, \dots, v_{k - 1}) = \int_{0}^{1} t^{k - 1} ω_{t x} (x, v_{1}, \dots, v_{k - 1}) d t$ .

Proof

technique · direct

1.1

For $n \geq 2$ , set $r (x) = x / ∣ x ∣$ and $i : S^{n - 1} ↪ R^{n} ∖ {0}$ . Then $r i = id$ and $F (x, t) = ((1 - t) + t / ∣ x ∣) x$ is a smooth homotopy from the identity to $i r$ . Its scalar coefficient is strictly positive for $0 \leq t \leq 1$ , so it never reaches zero. Smooth homotopy invariance and the sphere computation give the groups asserted.

F1F2given

2.1

For $n = 1$ , the two half-lines are star-shaped and connected, so each has only $H^{0} = R$ ; their finite disjoint union gives $R^{2}$ . For $n = 0$ the punctured space is empty, all its form spaces are zero, and every cohomology group is zero.

F3F4given∎

Source locator

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

Depends on

Used by

De rham cohomology of punctured three space Example

Dependency tree · two levels

13 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

John M. Lee, Introduction to Smooth Manifolds, second edition (standard reference, not scraped)