CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29

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A bivector in $R^{4}$ need not be decomposable

Statement refuted

Every bivector $α \in Λ^{2} R^{4}$ is decomposable, that is, of the form $α = u \land v$ .

Facts & Assumptions

Given: The standard basis of $R^{4}$ and the bivector $α = e_{1} \land e_{2} + e_{3} \land e_{4}$ .

[L1]

The wedge $v_{1} \land \dots \land v_{k}$ is a basis element of $Λ^{k} R^{4}$ , hence nonzero, for independent vectors (In a finite-dimensional vector space, a decomposable wedge is nonzero exactly when its vectors are linearly independent).

[L2]

The wedge product is associative and satisfies $v \land w = - w \land v$ for vectors (Exterior multiplication is well defined, graded, associative, unital, and graded-commutative).

Counterexample

technique · direct

1.1L2algebra

If $α = u \land v$ were decomposable, then by [L2], $α \land α = u \land v \land u \land v = - u \land u \land v \land v = 0$ .

1.2L1L2algebra

For the displayed $α$ , compute $α \land α = e_{1} \land e_{2} \land e_{3} \land e_{4} + e_{3} \land e_{4} \land e_{1} \land e_{2} = 2 e_{1} \land e_{2} \land e_{3} \land e_{4}$ , which is the nonzero basis vector of $Λ^{4} R^{4}$ by [L1].

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 contradict each other, so the bivector $α$ is not decomposable.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Keith Conrad, Exterior Powers (standard reference, not scraped)

A bivector in R4R4 need not be decomposable