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De rham cohomology of a contractible smooth manifold

Statement

Assume countable choice. A nonempty contractible smooth manifold $M$ has $H_{d R}^{0} (M) ≅ R$ and $H_{d R}^{k} (M) = 0$ for every $k > 0$ .

Facts & Assumptions

Given: A nonempty contractible smooth manifold $M$ and countable choice.

[F1]

De rham cohomology is continuous homotopy invariant on smooth manifolds: Assume countable choice ${A C}_{ω}$ . Continuously homotopic smooth maps induce equal de Rham maps. Continuous homotopy equivalences between smooth manifolds induce inverse de Rham graded algebra maps via smooth representatives, independently of those representatives.

[F2]

Zero and out of range de rham cohomology: $H_{d R}^{k} (M) = 0$ if $k < 0$ or $k > \dim M$ . If $M = \emptyset$ , its cohomology vanishes in every degree.

[F3]

Zero th de rham cohomology is locally constant functions: $H_{d R}^{0} (M)$ is the algebra of locally constant real functions. For nonempty connected $M$ it is canonically $R$ .

[F4]

Nullhomotopic maps and contractible spaces: Let $f : X \to Y$ be continuous. The map $f$ is nullhomotopic if there is a point $y_{0} \in Y$ such that $f$ is homotopic to the constant map $c_{y_{0}} : X \to Y$ , $c_{y_{0}} (x) = y_{0}$ (def-homotopy-relative-and-path-homotopy). A nonempty topological space $X$ is contractible if every continuous map $f : X \to Y$ to every topological space $Y$ is nullhomotopic. This definition separates the property of the space from the particular map ${id}_{X}$ . The next corollary proves that it is equivalent to the familiar condition that the identity map be nullhomotopic.

Proof

technique · direct

1.1

Apply contractibility to the identity map to obtain a continuous homotopy from ${id}_{M}$ to a constant $c_{p}$ for some $p \in M$ . Let $a : M \to {p}$ and $b : {p} \to M$ be the unique map and inclusion. Then $a b = {id}_{{p}}$ and $b a = c_{p} ≃ {id}_{M}$ , so these are continuous homotopy inverses.

F4given

2.1

Continuous homotopy invariance identifies the de Rham groups with those of the point. Its zero-degree functions are precisely $R$ , and all positive-degree form spaces vanish by dimension. Hence its positive-degree cohomology vanishes and its degree-zero cohomology is $R$ , giving the asserted groups of $M$ .

F1F2F3step 1.1∎

Source locator

Lee, Theorem 17.13, pp.446–447; continuous smoothing is supplied by the preceding fully local corollary.

Depends on

Used by

De rham cohomology of the circle from mayer vietoris Example

Dependency tree · two levels

21 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)