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Lorentzian exterior calculus and Hodge star

Statement

Let $(M, g, N, o)$ have the oriented, time-oriented Lorentzian structure and volume form of Oriented Lorentzian fields, observers and coordinate conventions. On decomposable $k$ -covectors define the induced exterior pairing by

$⟨ α_{1} \land \dots \land α_{k}, β_{1} \land \dots \land β_{k} ⟩_{g} : = \det (g^{- 1} (α_{i}, β_{j})),$

and extend it bilinearly. For $0 \leq k \leq 4$ , its coordinate expression is the pointwise exterior metric on $k$ -covectors

$⟨ α, β ⟩_{g} : = \frac{1}{k!} α_{a_{1} \dots a_{k}} β^{a_{1} \dots a_{k}},$

where indices are raised by $g^{- 1}$ . There is a unique smooth bundle map $⋆_{g} : Ω^{k} (M) \to Ω^{4 - k} (M)$ characterized by

$α \land ⋆_{g} β = ⟨ α, β ⟩_{g} {v o l}_{g}$

at every point. In a positively oriented $g$ -orthonormal coframe $(e^{0}, e^{1}, e^{2}, e^{3})$ , put $ϵ_{0} = - 1$ and $ϵ_{1} = ϵ_{2} = ϵ_{3} = 1$ . For an increasing index set $I \subseteq {0, 1, 2, 3}$ , let $I^{c}$ be its increasing complement, let $s (I, I^{c})$ be the sign of the permutation $(I, I^{c})$ to $(0, 1, 2, 3)$ , and set $ϵ_{I} : = \prod_{i \in I} ϵ_{i}$ . Then

$⋆_{g} e^{I} = ϵ_{I} s (I, I^{c}) e^{I^{c}} .$

Consequently, on $k$ -forms,

$⋆_{g}^{2} = (- 1)^{k (4 - k) + 1} id .$

In particular $⋆_{g}^{2} = - id$ on two-forms. For a two-form $F$ , the component formula in a positively oriented chart is

$(⋆_{g} F)_{a b} = \frac{1}{2} ({v o l}_{g})_{a b c d} F^{c d} .$

The exterior derivative in The exterior derivative by the invariant vector-field formula has the local coordinate formula of The local coordinate formula for the exterior derivative, satisfies $d^{2} = 0$ , and obeys the graded product rule of The exterior derivative is a graded derivation.

Facts & Assumptions

Given: The smooth oriented Lorentzian structure, its metric volume form, and smooth differential forms of degrees $0$ through $4$ .

[F1]

The oriented Lorentzian manifold has signature $(-, +, +, +)$ , and each point has a neighborhood with a positively oriented smooth orthonormal tetrad. The metric volume form has value $+ 1$ on a positively oriented orthonormal tetrad. (Oriented Lorentzian fields, observers and coordinate conventions, Local frame fields, coordinate frames and tetrad fields, The metric volume form, current flux through 3-faces and finite-chain charges, Chart independence of the metric volume form, its value on orthonormal frames, and the unit normal and density of a non-null hyperplane)

[F2]

A smooth $k$ -form is a smooth section of $⋀^{k} T^{*} M$ and has a unique local expansion in increasing coordinate wedges. (A smooth differential $k$ -form, Local coordinate expression for a differential form)

[F3]

The wedge product is the pointwise alternating product of covectors and differential forms form an associative graded-commutative algebra. (The wedge product of differential forms, Differential forms form a graded commutative algebra)

[F4]

The exterior derivative is defined by its invariant vector-field formula; it is a smooth form, and in coordinates it is $(d α)_{a_{0} \dots a_{k}} = (k + 1) \partial_{[a_{0}} α_{a_{1} \dots a_{k}]}$ . (The exterior derivative by the invariant vector-field formula, The invariant exterior-derivative formula is $C^{\infty}$ -multilinear, The local coordinate formula for the exterior derivative)

[F5]

Smooth mixed partial derivatives commute, and the exterior derivative is an $R$ -linear degree-one graded derivation of the wedge algebra. (Clairaut--Schwarz theorem for continuous second partial derivatives, The exterior derivative is a graded derivation)

[F6]

The inverse metric raises covector slots. In an orthonormal coframe, the determinant pairing of decomposable wedges is diagonal, with diagonal entry the product of their metric signs; its component expression is the normalized contraction displayed in the Statement, since each increasing index set occurs $k!$ times in the full antisymmetric sum. (The inverse metric, musical isomorphisms and index raising)

Proof

1.1F1F2F3F6algebraconstruct

The determinant is alternating multilinear in each list, so it defines a bilinear pairing on exterior powers. In a positively oriented orthonormal coframe, it makes the basis wedges $e^{I}$ orthogonal with squared norm $ϵ_{I}$ ; in the full component contraction this same diagonal value occurs $k!$ times, canceled by the factor $1 / k!$ . Define $⋆_{g} e^{I} = ϵ_{I} s (I, I^{c}) e^{I^{c}}$ and extend linearly. If $J \neq I$ has the same size as $I$ , then $e^{J} \land e^{I^{c}} = 0$ because an index repeats. If $J = I$ , then $e^{I} \land e^{I^{c}} = s (I, I^{c}) {v o l}_{g}$ , so $e^{I} \land ⋆_{g} e^{I} = ϵ_{I} {v o l}_{g}$ . Bilinearity proves the characterizing identity for all forms in the coframe. The complement wedges pair each basis vector with exactly one basis vector, so this wedge pairing is nondegenerate and the characterization determines $⋆_{g}$ uniquely.

2.1step 1.1F1F2algebra

The characterizing identity uses only $g$ , the orientation, and ${v o l}_{g}$ . Any other positively oriented orthonormal coframe therefore defines the same map by uniqueness. In each smooth tetrad neighborhood the displayed matrix of $⋆_{g}$ is constant in the smooth coframe, so its coordinate coefficients are smooth; the locally defined maps agree on overlaps and give a global smooth bundle map.

2.2step 1.1algebra

Applying the basis formula twice gives $⋆_{g}^{2} e^{I} = ϵ_{I} ϵ_{I^{c}} s (I, I^{c}) s (I^{c}, I) e^{I}$ . The product $ϵ_{I} ϵ_{I^{c}}$ equals $- 1$ because exactly one of the four $ϵ_{i}$ is negative. Reversing the two blocks of sizes $k$ and $4 - k$ takes $k (4 - k)$ transpositions, so $s (I, I^{c}) s (I^{c}, I) = (- 1)^{k (4 - k)}$ . Their product is $(- 1)^{k (4 - k) + 1}$ , proving the square formula for each basis element and hence for every $k$ -form.

2.3step 1.1F1F6algebra

For a two-form $F = \frac{1}{2} F_{c d} e^{c} \land e^{d}$ , the basis formula sends each $e^{c} \land e^{d}$ to its oriented complementary wedge with the product of its metric signs. In components this is $(⋆_{g} F)_{a b} = \frac{1}{2} ({v o l}_{g})_{a b c d} F^{c d}$ : the factor $1 / 2$ counts each antisymmetric pair once, and raising both indices includes the negative sign in the timelike slot.

3.1F4F5algebra∎

In a chart, [F4] gives $(d α)_{a_{0} \dots a_{k}} = (k + 1) \partial_{[a_{0}} α_{a_{1} \dots a_{k}]}$ . Applying it twice expresses each coefficient of $d^{2} α$ as an alternating sum of second partial derivatives; terms paired by exchanging the first two derivative indices cancel because those mixed partials agree by [F5]. Thus $d^{2} α = 0$ in every chart and hence globally. The graded product rule is exactly the established identity in [F5].

Remarks

The positive-definite Hodge formula would omit the extra signature sign. It is not a supplier for this item: the single negative direction contributes $ϵ_{I} ϵ_{I^{c}} = - 1$ in step 2.2. No metric field equation, Maxwell postulate, global potential, or physical interpretation is used.

The geometric suppliers Oriented Lorentzian fields, observers and coordinate conventions and Local frame fields, coordinate frames and tetrad fields record ${A C}_{ω}$ as a background axiom: every countable family of nonempty sets has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )). This is an assumption of their development, not a property of a smooth structure. The finite coframe construction and uniqueness-based patching above require no additional choice principle.

Lorentzian exterior calculus and Hodge star

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

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