ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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Reversing orientation negates the Hodge star while keeping the metric fixed

Example

Fix the standard inner product on $R^{3}$ and compare the two orientations: the standard one, represented by the ordered basis $(e_{1}, e_{2}, e_{3})$ , and the reversed one, represented by $(e_{2}, e_{1}, e_{3})$ . For the standard orientation the Hodge star satisfies $⋆_{+} (e_{1} \land e_{2}) = e_{3}$ ; for the reversed orientation the unit volume form is $e_{2} \land e_{1} \land e_{3} = - e_{1} \land e_{2} \land e_{3}$ , and the Hodge star satisfies $⋆_{-} (e_{1} \land e_{2}) = - e_{3}$ . The metric is the same in both computations; only the orientation changed, and the star changed sign.

Facts & Assumptions

Given: The standard inner product on $R^{3}$ and the two ordered bases $(e_{1}, e_{2}, e_{3})$ and $(e_{2}, e_{1}, e_{3})$ .

[L1]

Two ordered bases lie in the same orientation class exactly when their change-of-basis isomorphism has positive determinant (A real linear isomorphism preserves or reverses orientation according to the sign of its determinant).

[L2]

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

[L3]

The Hodge star is an isometry for the Gram pairing (The Hodge star is an isometry and satisfies $⋆^{2} = (- 1)^{k (n - k)}$ on $Λ^{k} V$ ).

[L4]

The oriented unit volume form is $ω = b_{1} \land \dots \land b_{n}$ for a positively oriented orthonormal basis (The oriented unit volume form).

Verification

technique · direct

1.1

The change of basis from $(e_{1}, e_{2}, e_{3})$ to $(e_{2}, e_{1}, e_{3})$ is a transposition with determinant $- 1$ , so by [L1] the two bases represent the two opposite orientation classes.

L1algebra

1.2

By [L4], the unit volume forms are $ω_{+} = e_{1} \land e_{2} \land e_{3}$ and $ω_{-} = e_{2} \land e_{1} \land e_{3} = - ω_{+}$ .

L4algebra

2.1

By the characterizing relation of [L2] and its uniqueness clause, $α \land ⋆_{-} β = ⟨ α, β ⟩ ω_{-} = - ⟨ α, β ⟩ ω_{+} = α \land (- ⋆_{+} β)$ for all $α$ , so $⋆_{-} = - ⋆_{+}$ on every degree.

L2step 1.2

3.1

In particular $⋆_{+} (e_{1} \land e_{2}) = e_{3}$ by [L2], and step 2.1 gives $⋆_{-} (e_{1} \land e_{2}) = - e_{3}$ ; the metric used in both computations is the same standard inner product, and [L3] records that each star is an isometry for it.

L2L3step 2.1

4.1

Steps 1.1 through 3.1 show that reversing the orientation negates the Hodge star while the metric stays fixed.

step 1.1step 2.1step 3.1∎

Depends on

Used by

FALSE: an inner product determines an orientation False statement
FALSE: Hodge star needs only the vector-space structure False statement

Dependency tree · two levels

13 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, §2.4 (standard reference, not scraped)