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

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Universal property of the finite-dimensional exterior power

Statement

Let $A : V^{k} \to W$ be an alternating $k$ -linear map into a real vector space $W$ . Then there is a unique linear map

\tilde{A} : \overset{k}{⋀} V \to W

such that

\tilde{A} (v_{1} \land \dots \land v_{k}) = A (v_{1}, \dots, v_{k})

for all $v_{1}, \dots, v_{k} \in V$ .

Facts & Assumptions

Given: An alternating $k$ -linear map $A : V^{k} \to W$ .

[F1]

The $k$ th exterior power is the dual space $⋀^{k} V = {Alt}^{k} (V)^{*}$ , and $v_{1} \land \dots \land v_{k}$ is the evaluation functional $ω \mapsto ω (v_{1}, \dots, v_{k})$ (The finite-dimensional exterior power of vectors).

[L1]

The wedges of a basis form a basis of ${Alt}^{k} (V)$ , so the corresponding decomposable $k$ -vectors span $⋀^{k} V$ by duality (Wedge monomials in a dual basis form a basis).

Proof

technique · direct

1.1

Choose a basis $e_{1}, \dots, e_{n}$ of $V$ . Define $\tilde{A}$ on the spanning set of decomposable wedges by $\tilde{A} (e_{i_{1}} \land \dots \land e_{i_{k}}) : = A (e_{i_{1}}, \dots, e_{i_{k}})$ for $i_{1} < \dots < i_{k}$ , and extend linearly. This is possible because [L1] gives a basis indexed by those increasing tuples.

L1givenchooseconstruct

2.1

For arbitrary $v_{1}, \dots, v_{k}$ , expand each $v_{j}$ in the chosen basis. Multilinearity of $A$ and of the wedge, together with alternation on both sides, reduce both expressions to the same signed sum over increasing $k$ -tuples. Hence $\tilde{A} (v_{1} \land \dots \land v_{k}) = A (v_{1}, \dots, v_{k})$ .

F1step 1.1algebra

3.1

If $L : ⋀^{k} V \to W$ is another linear map with the same property, then $L$ and $\tilde{A}$ agree on every decomposable basis wedge, hence on all of $⋀^{k} V$ by linearity and [L1]. So $\tilde{A}$ is unique.

L1step 2.1

4.1

Therefore every alternating $k$ -linear map factors uniquely through $⋀^{k} V$ .

step 1.1step 2.1step 3.1∎

Depends on

Used by

Functoriality of finite-dimensional exterior powers Proposition

Dependency tree · two levels

6 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)