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Interior product is the adjoint of exterior multiplication by a vector

Statement

For $v \in V$ , the interior product $ι_{v}$ of Interior product on the exterior algebra is the adjoint of exterior multiplication $m_{v} (α) = v \land α$ . On a decomposable wedge,

$ι_{v} (v_{1} \land \dots \land v_{k}) = \sum_{r = 1}^{k} (- 1)^{r - 1} ⟨ v, v_{r} ⟩ v_{1} \land \dots \land \hat{v_{r}} \land \dots \land v_{k},$

where $\hat{v_{r}}$ means $v_{r}$ is omitted. Equivalently, $ι_{v}$ is the unique linear operator with $ι_{v} (1) = 0$ , $ι_{v} (w) = ⟨ v, w ⟩$ for $w \in V$ , and the graded derivation rule $ι_{v} (w \land α) + w \land ι_{v} (α) = ⟨ v, w ⟩ α$ , established on the following page item as the anticommutation identity.

Facts & Assumptions

Given: A finite-dimensional real inner product space $V$ , a vector $v$ , and a degree $k \geq 1$ .

[L1]

The interior product is the adjoint $ι_{v} = m_{v}^{*}$ , characterized by $⟨ ι_{v} α, β ⟩ = ⟨ α, v \land β ⟩$ (Interior product on the exterior algebra).

[L2]

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

[L3]

A linear map between finite-dimensional inner product spaces has a unique adjoint (Every linear map between finite-dimensional inner product spaces has a unique adjoint).

[L4]

A finite-dimensional inner product space has an orthonormal basis (Every finite-dimensional real or complex inner product space has an orthonormal basis).

[L5]

A linear map out of $Λ^{k} V$ is determined by its values on decomposable wedges (Exterior powers represent alternating multilinear maps and are unique up to unique isomorphism).

Proof

technique · direct

1.1L1L2L3

The map $m_{v} : Λ^{k - 1} V \to Λ^{k} V$ is linear by [L2], so [L3] supplies its unique adjoint; by [L1] that adjoint is $ι_{v}$ with $⟨ ι_{v} α, β ⟩ = ⟨ α, v \land β ⟩$ .

1.2L2L4algebra

Choose an orthonormal basis $(e_{1}, \dots, e_{n})$ by [L4], write $v = \sum_{j} v_{j} e_{j}$ , and let $I = {i_{1} < \dots < i_{k}}$ . Then $⟨ ι_{v} e_{I}, e_{J} ⟩ = ⟨ e_{I}, v \land e_{J} ⟩ = \sum_{j} v_{j} ⟨ e_{I}, e_{j} \land e_{J} ⟩$ , and the pairing is nonzero only when $J = I ∖ {i_{r}}$ for some $r$ and $j = i_{r}$ , in which case $e_{j} \land e_{J} = (- 1)^{r - 1} e_{I}$ .

2.1step 1.2L1

By step 1.2, $ι_{v} e_{I} = \sum_{r = 1}^{k} (- 1)^{r - 1} v_{i_{r}} e_{I ∖ {i_{r}}}$ , since the $e_{J}$ form an orthonormal basis of $Λ^{k - 1} V$ and the pairing is nondegenerate by [L1].

3.1L2L5step 2.1

The assignment $Φ_{v} (v_{1}, \dots, v_{k}) : = \sum_{r = 1}^{k} (- 1)^{r - 1} ⟨ v, v_{r} ⟩ v_{1} \land \dots \land \hat{v_{r}} \land \dots \land v_{k}$ is multilinear (each term is multilinear) and alternating (when $v_{a} = v_{b}$ , the two terms $r = a$ and $r = b$ cancel by the block sign of [L2], and the remaining terms have a repeated entry), so by [L5] there is a unique linear map agreeing with $Φ_{v}$ on decomposables; on the basis wedge $e_{I}$ it equals step 2.1, hence it is $ι_{v}$ .

4.1step 1.1step 3.1∎

Steps 1.1 and 3.1 prove the adjoint description and the displayed contraction formula; evaluating at $k = 0, 1$ gives $ι_{v} (1) = 0$ and $ι_{v} (w) = ⟨ v, w ⟩$ , the two boundary clauses of the characterization.

Depends on

Used by

Cited to discharge well-definedness by Interior product on the exterior algebra.

Dependency tree · two levels

25 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

Albert Chern, Geometric Fluid Dynamics notes, Interior Products (standard reference, not scraped)