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Hodge decomposition of two-forms in four dimensions

Statement

Let $V$ be a four-dimensional real vector space with an orientation and inner product $g$ , and let $W$ be a finite-dimensional real inner-product space. Define the Hodge operator $*_{g}$ on $Λ^{2} V^{*}$ by $α \land *_{g} β = ⟨ α, β ⟩_{g} {v o l}_{g}$ , and extend it $W$ -linearly to $Λ^{2} V^{*} \otimes W$ . Give this tensor product the inner product induced by those on $Λ^{2} V^{*}$ and $W$ . Then $*_{g}^{2} = I$ , $*_{g}$ is an isometry, and its $+ 1$ and $- 1$ eigenspaces are orthogonal. Every $Φ \in Λ^{2} V^{*} \otimes W$ has the unique decomposition

Φ = Φ^{+} + Φ^{-}, Φ^{\pm} : = \frac{Φ \pm *_{g} Φ}{2}, *_{g} Φ^{\pm} = \pm Φ^{\pm} .

If the inner product on $V$ is rescaled to $\hat{g} = λ^{2} g$ for any $λ > 0$ , then the Hodge operator on two-forms is unchanged: $*_{\hat{g}} = *_{g}$ .

Statement references

Tong, Lecture 1 §1.1.1, equations (1.6)–(1.8)

Facts & Assumptions

Given: The oriented four-dimensional inner-product space $V$ , the finite-dimensional inner-product space $W$ , and the Hodge operator and induced tensor-product inner product specified in the Statement.

[A1]

The oriented space $V$ has a positively oriented orthonormal coframe, and $W$ has an orthonormal basis; these are finite-dimensional inner-product spaces.

Proof

1.1givenA1algebra

Choose a positively oriented orthonormal coframe $e^{1}, e^{2}, e^{3}, e^{4}$ of $V^{*}$ . The defining wedge identity gives $*_{g} e^{12} = e^{34}$ , $*_{g} e^{13} = - e^{24}$ , $*_{g} e^{14} = e^{23}$ , $*_{g} e^{23} = e^{14}$ , $*_{g} e^{24} = - e^{13}$ , and $*_{g} e^{34} = e^{12}$ . Thus $*_{g}$ permutes the orthonormal basis of $Λ^{2} V^{*}$ with signs, so it is an isometry; applying it twice to each of the six basis vectors gives $*_{g}^{2} = I$ . The same conclusions hold after tensoring each basis vector with any member of an orthonormal basis of $W$ .

2.1givenA1step 1.1algebra

For $Φ^{+} = \frac{1}{2} (Φ + *_{g} Φ)$ and $Φ^{-} = \frac{1}{2} (Φ - *_{g} Φ)$ , the identity $*_{g}^{2} = I$ from step 1.1 gives $*_{g} Φ^{+} = Φ^{+}$ , $*_{g} Φ^{-} = - Φ^{-}$ , and $Φ = Φ^{+} + Φ^{-}$ . If $*_{g} α = α$ and $*_{g} β = - β$ , the isometry property from step 1.1 gives $⟨ α, β ⟩ = ⟨ *_{g} α, *_{g} β ⟩ = - ⟨ α, β ⟩$ , hence $⟨ α, β ⟩ = 0$ . The decomposition is unique because the intersection of the two eigenspaces is zero.

3.1givenA1algebra∎

Under $\hat{g} = λ^{2} g$ , the induced inner product on $Λ^{2} V^{*}$ scales by $λ^{- 4}$ because each covector factor scales by $λ^{- 2}$ , while the oriented volume form scales by $λ^{4}$ . Hence for all $α, β \in Λ^{2} V^{*}$ , $⟨ α, β ⟩_{\hat{g}} {v o l}_{\hat{g}} = ⟨ α, β ⟩_{g} {v o l}_{g}$ . The defining wedge identity therefore gives $α \land *_{\hat{g}} β = α \land *_{g} β$ for every $α$ ; nondegeneracy of the wedge pairing between two-forms in dimension four implies $*_{\hat{g}} β = *_{g} β$ . The identity extends $W$ -linearly.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

David Tong, TASI Lectures on Solitons: Instantons, Monopoles, Vortices and Kinks