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

1.1

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.

F1given

2.1

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.

F1given∎

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)