PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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Differential forms form a graded commutative algebra

Statement

The graded vector space

Ω^{*} (M) : = ⨁_{k \geq 0} Ω^{k} (M)

with the wedge product is an associative graded-commutative algebra.

Facts & Assumptions

Given: Differential forms $α, β, γ$ of homogeneous degrees $k, ℓ, m$ .

[F1]

The wedge product of forms is defined pointwise from the wedge product of alternating covectors (The wedge product of differential forms).

[L1]

The fibrewise wedge product is associative and graded commutative (The wedge product is associative and graded commutative).

[L2]

Tensor-field smoothness can be checked on coordinate components (Smoothness of a tensor field is equivalent to smooth coordinate components).

Proof

technique · direct

1.1

At each point $p \in M$ , [F1] and [L1] give $((α \land β) \land γ)_{p} = α_{p} \land (β_{p} \land γ_{p})$ and $(α \land β)_{p} = (- 1)^{k ℓ} (β \land α)_{p} .$

F1L1given

1.2

The local coefficient functions of $α \land β$ are polynomial expressions in the local coefficients of $α$ and $β$ , so [L2] shows that wedge products remain smooth.

F1L2givenalgebra

2.1

Therefore $Ω^{*} (M)$ is closed under wedge and inherits associativity and graded commutativity pointwise from [L1].

step 1.1step 1.2∎

Depends on

Used by

The wedge product is not commutative False statement
Pullback of forms is smooth functorial and preserves wedges Proposition

Cited to discharge well-definedness by The wedge product of differential forms.

Dependency tree · two levels

9 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

Will J. Merry, Differential Geometry (standard reference, not scraped)