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In oriented Euclidean three-space, the cross product is $⋆ (u \land v)$

Statement

In $R^{3}$ with the standard inner product and the standard orientation (the class of the ordered basis $(e_{1}, e_{2}, e_{3})$ ), define the cross product in the oriented orthonormal basis by

$u \times v : = (u_{2} v_{3} - u_{3} v_{2}, u_{3} v_{1} - u_{1} v_{3}, u_{1} v_{2} - u_{2} v_{1}) .$

Then $u \times v = ⋆ (u \land v)$ . Equivalently, with the unit volume form $ω = e_{1} \land e_{2} \land e_{3}$ , one has $u \times v = ι_{v} ι_{u} ω$ , and $u \times v$ is orthogonal to both $u$ and $v$ .

Facts & Assumptions

Given: Vectors $u, v \in R^{3}$ , the oriented orthonormal basis $(e_{1}, e_{2}, e_{3})$ , and the unit volume form $ω$ .

[L1]

In an oriented orthonormal basis, $⋆ e_{I} = ε_{I} e_{I^{c}}$ , and the star is characterized by $α \land ⋆ β = ⟨ α, β ⟩ ω$ (The Hodge star exists uniquely and is given by the complementary-basis formula in an oriented orthonormal basis).

[L2]

On $Λ^{2} R^{3}$ , one has $⋆^{2} = id$ because $(- 1)^{2 \cdot 1} = 1$ (The Hodge star is an isometry and satisfies $⋆^{2} = (- 1)^{k (n - k)}$ on $Λ^{k} V$ ).

[L3]

The interior product contracts by $ι_{v} (v_{1} \land \dots \land v_{k}) = \sum_{r = 1}^{k} (- 1)^{r - 1} ⟨ v, v_{r} ⟩ v_{1} \land \dots \land \hat{v_{r}} \land \dots \land v_{k}$ (Interior product is the adjoint of exterior multiplication by a vector).

Proof

technique · direct

1.1

Every bilinear alternating map $B : R^{3} \times R^{3} \to R^{3}$ is determined by its values on the three pairs $(e_{1}, e_{2}), (e_{2}, e_{3}), (e_{3}, e_{1})$ : expanding in coordinates, $B (u, v) = \sum_{i < j} (u_{i} v_{j} - u_{j} v_{i}) B (e_{i}, e_{j})$ .

algebra

1.2

By [L1], $⋆ (e_{1} \land e_{2}) = e_{3}$ , $⋆ (e_{2} \land e_{3}) = e_{1}$ , and $⋆ (e_{3} \land e_{1}) = e_{2}$ ; the map $(u, v) \mapsto ⋆ (u \land v)$ is bilinear and alternating because the wedge is and $⋆$ is linear.

L1algebra

1.3

The coordinate cross product is bilinear and alternating, and satisfies $e_{1} \times e_{2} = e_{3}$ , $e_{2} \times e_{3} = e_{1}$ , $e_{3} \times e_{1} = e_{2}$ .

algebra

2.1

By step 1.1, the two bilinear alternating maps of steps 1.2 and 1.3 agree on the three generating pairs, hence agree everywhere: $⋆ (u \land v) = u \times v$ .

step 1.1step 1.2step 1.3

3.1

The interior-product form: by [L3], $ι_{u} ω = ⟨ u, e_{1} ⟩ e_{2} \land e_{3} - ⟨ u, e_{2} ⟩ e_{1} \land e_{3} + ⟨ u, e_{3} ⟩ e_{1} \land e_{2}$ , and applying $ι_{v}$ to that expansion collects the $e_{1}, e_{2}, e_{3}$ coordinates $u_{2} v_{3} - u_{3} v_{2}$ , $u_{3} v_{1} - u_{1} v_{3}$ , $u_{1} v_{2} - u_{2} v_{1}$ , so $ι_{v} ι_{u} ω = u \times v = ⋆ (u \land v)$ by step 2.1.

L3step 2.1algebra

3.2

Orthogonality: by the defining relation of [L1] and the square law of [L2], $⟨ u, ⋆ (u \land v) ⟩ ω = u \land ⋆ (⋆ (u \land v)) = u \land u \land v = 0$ , and $ω \neq 0$ , so $⟨ u, u \times v ⟩ = 0$ ; the same argument with $v$ gives $⟨ v, u \times v ⟩ = 0$ .

L1L2step 2.1algebra

4.1

Steps 2.1, 3.1 and 3.2 establish the three equivalent descriptions and the orthogonality.

step 2.1step 3.1step 3.2∎

Depends on

Used by

The cross product is recovered from ⋆(u∧ v) in ℝ³ Example

Dependency tree · two levels

12 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

Reyer Sjamaar, Manifolds and Differential Forms, §2.5 (standard reference, not scraped)

In oriented Euclidean three-space, the cross product is ⋆(u∧v)⋆(u∧v)