PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Zero and out of range de rham cohomology

Statement

$H_{d R}^{k} (M) = 0$ if $k < 0$ or $k > \dim M$ . If $M = \emptyset$ , its cohomology vanishes in every degree.

Facts & Assumptions

Given: The de Rham complex with zero out-of-range terms.

[F1]

De rham cohomology: The real de Rham cohomology is $H_{d R}^{k} (M) = Z^{k} (M) / B^{k} (M)$ , with $Z^{k}, B^{k}$ as in def-closed-and-exact-differential-forms. This is def-cohomology-object-of-a-cochain-complex in real vector spaces. Only a closed form $ω$ represents a class $[ω]$ . For closed forms $ω, ω'$ , equality $[ω] = [ω']$ means precisely $ω' - ω = d η$ for some $(k - 1)$ -form $η$ . Addition and real scalar multiplication are induced by those of forms. All groups on the empty manifold are zero.

Proof

technique · direct

1.1F1given

If $k < 0$ or $k > \dim M$ , the cycle space is a subspace of $Ω^{k} (M) = 0$ , so $Z^{k} = B^{k} = 0$ . Its quotient is therefore zero.

2.1F1given∎

On the empty manifold there is just one section of each form bundle, namely the zero section. Thus in every degree the quotient is again $0 / 0 = 0$ , proving the empty clause as well.

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., Chapter 17, pp.441–443, Proposition 17.2 and Corollary 17.3; local quotient calculations below.

Depends on

De rham cohomology

Used by

De rham cohomology of a point Example
De rham cohomology of a contractible smooth manifold Theorem

Dependency tree · two levels

3 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)