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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The wedge product is alternating and bilinear

Statement

If $α \in {Alt}^{k} (V)$ and $β \in {Alt}^{ℓ} (V)$ , then $α \land β$ is alternating of degree $k + ℓ$ , and the wedge product is bilinear in $(α, β)$ .

Facts & Assumptions

Given: Alternating covectors $α, α^{'} \in {Alt}^{k} (V)$ , $β, β^{'} \in {Alt}^{ℓ} (V)$ , and scalars $a, b$ .

[F1]

The wedge product is the normalized alternation of the tensor product, equivalently the signed shuffle sum (The wedge product of alternating covectors).

Proof

technique · direct

1.1

By [F1], $α \land β$ is obtained by applying the alternation operator to $α \otimes β$ . Alternation produces an alternating multilinear form, so $α \land β \in {Alt}^{k + ℓ} (V)$ .

F1given

1.2

Both tensor product and alternation are linear in each argument, so [F1] gives $(a α + b α^{'}) \land β = a (α \land β) + b (α^{'} \land β),$ and similarly in the second slot.

F1givenalgebra

2.1

Therefore the wedge product is alternating and bilinear.

step 1.1step 1.2∎

Depends on

The wedge product of alternating covectors

Used by

The wedge product is associative and graded commutative Theorem

Dependency tree · two levels

4 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, 2nd ed. (standard reference, not scraped)