Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-23 (gpt-6-sol)
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The Hodge star exists uniquely and is given by the complementary-basis formula in an oriented orthonormal basis

Statement

Let V be an oriented finite-dimensional real inner product space of dimension n. For every degree 0≤k≤n there is a unique linear map ⋆:ΛkV→Λn−kV satisfying

α∧⋆β=⟨α,β⟩ ωfor all α,β∈ΛkV,

where ω is the oriented unit volume form of The oriented unit volume form. If (e1,…,en) is a positively oriented orthonormal basis and I={i1<⋯<ik} has complement Ic={i1c<⋯<in−kc}, let εI be the sign of the permutation (i1,…,ik,i1c,…,in−kc)↦(1,…,n); then

⋆eI=εI eIc.

Facts & Assumptions

Given: An oriented n-dimensional real inner product space V, its unit volume form ω, a degree k, and a positively oriented orthonormal basis (e1,…,en).

[L1]

The Hodge star is characterized by α∧⋆β=⟨α,β⟩ω (The Hodge star on an oriented finite-dimensional real inner-product space).

[L3]

The unit volume form is ω=e1∧⋯∧en for the positively oriented orthonormal basis, with ⟨ω,ω⟩=1 (The oriented unit volume form).

[L4]

The wedges eI form a basis of ΛkV and the wedges eIc a basis of Λn−kV (Increasing-index wedges of a basis form a basis of ΛkV).

Proof

technique · direct
1.1L3L4

By [L3] and [L4], the positively oriented orthonormal basis exists and yields the wedge bases (eI)I of ΛkV and (eJ)J of Λn−kV.

2.1step 1.1

Define a linear map ⋆:ΛkV→Λn−kV by ⋆eI:=εI eIc on the basis of step 1.1.

3.1L2L3step 1.1step 2.1

For subsets I,J of sizes k and k, one has eJ∧eIc=0 unless J=I: if J≠I some index occurs twice, and if J=I then eI∧eIc=εI e1∧⋯∧en=εIω by the definition of εI and [L3]. The Gram pairing of [L2] on the orthonormal basis satisfies ⟨eJ,eI⟩=1 for J=I and 0 otherwise, so eJ∧⋆eI=εI eJ∧eIc=⟨eJ,eI⟩ω for all I,J.

4.1step 3.1algebra

Both sides of the relation of [L1] are bilinear in (α,β), so step 3.1 extends from basis wedges to the identity α∧⋆β=⟨α,β⟩ω for all α,β∈ΛkV.

5.1L3L4step 3.1step 4.1

Uniqueness: if ⋆′ also satisfies [L1], fix β and write δ:=⋆β−⋆′β=∑∣J∣=n−kcJeJ in the basis of [L4]. Then eJc∧δ=0 for every J by subtracting the two defining identities. All terms except cJeJc∧eJ vanish because they repeat a basis vector, while eJc∧eJ=±ω≠0. Hence every cJ=0 and δ=0. This argument also covers k=0,n and n=0, where the relevant bases each have one empty or full wedge.

6.1step 2.1step 4.1step 5.1∎

Steps 2.1 and 4.1 prove existence with the complementary-basis formula, and step 5.1 proves uniqueness.

Depends on

Used by

Cited to discharge well-definedness by The Hodge star on an oriented finite-dimensional real inner-product space.

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Sources