False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29

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FALSE: $Λ^{k} V$ is canonically a subspace of $V^{\otimes k}$ over every field

Statement

For every field $F$ , every $F$ -vector space $V$ , and every $k \geq 0$ , the $k$ th exterior power $Λ^{k} V$ is canonically a linear subspace of the $k$ -fold tensor power $V^{\otimes k}$ : the antisymmetrization map

$Alt (v_{1} \land \dots \land v_{k}) : = \sum_{σ \in S_{k}} sgn (σ) v_{σ (1)} \otimes \dots \otimes v_{σ (k)}$

is a well-defined injective linear section of the quotient map $π : V^{\otimes k} \to Λ^{k} V$ , for every field $F$ .

Facts & Assumptions

Given: A field $F$ , a vector space $V$ with $\dim V \geq k \geq 2$ , the quotient map $π$ , and the antisymmetrization $Alt$ .

[L1]

The exterior power is the quotient $Λ^{k} V = V^{\otimes k} / W_{k}$ with quotient map $π$ (The $k$ th exterior power as the tensor-power quotient by repeated-vector relations).

[L2]

Permuting the entries of a wedge multiplies it by the permutation sign (Exterior multiplication is well defined, graded, associative, unital, and graded-commutative).

Refutation

technique · direct

1.1

By [L1], $Λ^{k} V$ is a quotient of $V^{\otimes k}$ , and the structure map $π$ is a surjection with nonzero kernel: for $\dim V \geq 2$ and $k \geq 2$ , extend a nonzero vector $v$ to a basis $(v, e_{2}, \dots, e_{n})$ ; then the pure tensor $v \otimes v \otimes e_{3} \otimes \dots \otimes e_{k} \in V^{\otimes k}$ (with the tail omitted when $k = 2$ ) is nonzero yet $π$ of it is $v \land v \land e_{3} \land \dots \land e_{k} = 0$ because the first two entries repeat, so the canonical construction presents $Λ^{k} V$ as a quotient, not a subspace.

1.2

For the formula-defined antisymmetrization, [L2] gives

$π (Alt (v_{1} \land \dots \land v_{k})) = \sum_{σ} sgn (σ) v_{σ (1)} \land \dots \land v_{σ (k)} = \sum_{σ} (sgn σ)^{2} v_{1} \land \dots \land v_{k} = k! v_{1} \land \dots \land v_{k} .$

[L2, algebra]

2.1

Over a field whose characteristic divides $k!$ , step 1.2 gives $π \circ Alt = k! \cdot id = 0$ , while $Λ^{k} V \neq 0$ by the hypothesis $\dim V \geq k$ ; a section must satisfy $π \circ s = id$ , so $Alt$ is not a section over such a field. The concrete witness is $F = F_{2}$ , $V = F_{2}^{2}$ , $k = 2$ : $Alt (e_{1} \land e_{2}) = e_{1} \otimes e_{2} + e_{2} \otimes e_{1}$ and $π$ of that is $2 e_{1} \land e_{2} = 0$ .

step 1.1step 1.2

3.1

Step 2.1 gives a field $F$ , a vector space $V$ , and a degree $k$ for which the displayed formula is not a section of $π$ , so the universal claim "for every field" is false.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

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

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

FALSE: ΛkVΛkV is canonically a subspace of V⊗kV⊗k over every field