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Rank-one tangent changes, signature and Hodge transfer

Statement

Let $V$ be an oriented four-dimensional real vector space with a Lorentz form $g$ of signature $(-, +, +, +)$ , meaning a nondegenerate symmetric bilinear form with one negative and three positive directions. Let $T \in V$ , $W \in V^{*}$ , and $D, d > 0$ satisfy $W (T) = d - D$ . Define $A = I + D^{- 1} T \otimes W$ , where $(T \otimes W) (v) = W (v) T$ , and $g^{'} = (D / 4) A^{*} g$ , where $(A^{*} g) (u, v) = g (A u, A v)$ . Give the domain of $A$ the transported orientation: a basis is positive exactly when its image under $A$ is positive.

Then $\det A = d / D$ , $A^{- 1} = I - d^{- 1} T \otimes W$ , and $g^{'}$ has the same inertia as $g$ . For every real or complex two-form $F$ , with $(A^{*} F) (u, v) = F (A u, A v)$ , the corresponding metric Hodge stars satisfy $⋆_{g^{'}} (A^{*} F) = A^{*} (⋆_{g} F)$ . For any two such forms $F, H$ , their full complex-bilinear contractions satisfy $(A^{*} F)_{a b} (A^{*} H)_{g^{'}}^{a b} = (4 / D)^{2} F_{a b} H_{g}^{a b}$ , where both indices are raised with the indicated inverse metric and no complex conjugation or factor of one half is used.

Statement references

Orientation of a finite-dimensional real vector space

Complex self-dual forms and metric-dependent contraction

For A∈ Mₙ(R) and columns u,v over a commutative ring, det(A+uv^T)=det(A)+v^Tadj(A)u

Sylvester's law of inertia: every real symmetric form is congruent to diag(Iₚ,-I_q,0ᵣ), and (p,q,r) is unique

Lorentzian exterior calculus and Hodge star

Conformal Hodge scaling in every degree

Facts & Assumptions

Given: The oriented Lorentz space, $T$ , $W$ , $D$ and $d$ in the statement.

Goal: Establish the determinant, inverse, inertia, Hodge transfer and contraction formulas in the statement.

[A1]

Work in the ZFC background, including the choice assumptions inherited by the geometric suppliers. The Axiom of Choice

[F1]

The rank-one determinant identity gives $\det (B + u v^{T}) = \det B + v^{T} adj (B) u$ over a commutative ring. For A∈ Mₙ(R) and columns u,v over a commutative ring, det(A+uv^T)=det(A)+v^Tadj(A)u

[F2]

A real symmetric form has basis-independent inertia under congruence. Sylvester's law of inertia: every real symmetric form is congruent to diag(Iₚ,-I_q,0ᵣ), and (p,q,r) is unique

[F3]

Algebraic pullback composes each covariant slot with the given linear map; inverse metrics raise indices, and the full two-form contraction is complex bilinear. Complex self-dual forms and metric-dependent contraction

[F4]

The metric Hodge star is uniquely characterized by $α \land ⋆_{g} β = ⟨ α, β ⟩_{g} {vol}_{g}$ , with the exterior pairing induced by the inverse metric and the metric volume determined by the orientation. Lorentzian exterior calculus and Hodge star

[F5]

In four dimensions a positive conformal rescaling leaves the Hodge star on two-forms unchanged. Conformal Hodge scaling in every degree

[F6]

An orientation of a finite-dimensional real vector space is an equivalence class of ordered bases, where two bases are equivalent when their change-of-basis isomorphism has positive determinant. Orientation of a finite-dimensional real vector space

[F7]

A linear isomorphism of an oriented real vector space of positive dimension preserves its orientation exactly when its determinant is positive. A real linear isomorphism preserves or reverses orientation according to the sign of its determinant

[F8]

Linear pullback respects tensor products and permutations of covariant slots. Linear pullback respects tensor products and permutations

[F9]

The wedge product of a $k$ -covector and an $ℓ$ -covector is $(k + ℓ)! / (k! ℓ!)$ times the alternation of their tensor product. The wedge product of alternating covectors

Proof

1.1givenF1algebra

Put $K = T \otimes W$ . For every $v \in V$ , $K^{2} v = W (v) W (T) T = (d - D) K v$ . Thus $(I + D^{- 1} K) (I - d^{- 1} K) = I + (D^{- 1} - d^{- 1} - (d - D) / (D d)) K = I$ , and the same calculation in the opposite order gives $A^{- 1} = I - d^{- 1} K$ . In any basis, [F1] with $B = I$ , $u = D^{- 1} T$ and $v^{T} = W$ gives $\det A = 1 + D^{- 1} W (T) = d / D > 0$ .

2.1step 1.1F2F6F7algebra

Since $A$ is invertible and $D / 4 > 0$ , a basis in which $g$ is diagonal with one negative and three positive entries pulls back by $A^{- 1}$ to a basis in which $g^{'}$ has those entries multiplied by $D / 4$ . Hence $g^{'}$ has the same inertia as $g$ , independently of basis by [F2]. By [F7], $A$ preserves the original orientation because $\dim V = 4$ and $\det A > 0$ . Hence the transported positive-basis class in the sense of [F6] agrees with the original orientation on $V$ .

3.1step 1.1step 2.1F3F4F8F9algebra

Write $h = A^{*} g$ . The map $A$ is an orientation-preserving isometry from $(V, h)$ with transported orientation to $(V, g)$ . For two-forms $α, β$ , the induced inverse-metric pairing satisfies $⟨ A^{*} α, A^{*} β ⟩_{h} = ⟨ α, β ⟩_{g}$ : in a basis this follows by substituting $h^{- 1} = A^{- 1} g^{- 1} A^{- T}$ into the two raised slots. Also ${vol}_{h} = A^{*} {vol}_{g}$ , as both sides take value $1$ on a transported positively oriented $h$ -orthonormal basis. By [F8] and [F9], slotwise pullback commutes with the normalized alternating sum defining the wedge: it is linear, respects tensor products and commutes with every permutation of the slots. Thus $A^{*} (α \land γ) = (A^{*} α) \land (A^{*} γ)$ for alternating covectors $α, γ$ . Pulling back the identity in [F4] therefore gives $(A^{*} α) \land A^{*} (⋆_{g} β) = ⟨ A^{*} α, A^{*} β ⟩_{h} {vol}_{h}$ . Since every two-form is an $A^{*} α$ , uniqueness in [F4] yields $⋆_{h} (A^{*} β) = A^{*} (⋆_{g} β)$ .

4.1step 3.1F3F4F5A1algebra∎

Regard $V$ as an affine manifold with constant metric $h$ , a constant time orientation and the transported orientation. Applying [F5] to the constant positive factor $D / 4$ gives $⋆_{g^{'}} = ⋆_{h}$ on two-forms, so step 3.1 proves the Hodge transfer. Moreover $g^{' - 1} = (4 / D) h^{- 1}$ , and the full contraction uses two inverse-metric factors. The pairing calculation in step 3.1, with the factor of two converting the exterior pairing to the full two-form contraction, therefore gives $(A^{*} F)_{a b} (A^{*} H)_{g^{'}}^{a b} = (4 / D)^{2} F_{a b} H_{g}^{a b}$ . Real linearity extends both identities to complex forms by complex bilinearity.

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