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Chart independence of the metric volume form, its value on orthonormal frames, and the unit normal and density of a non-null hyperplane

Statement

Let $(M, g, N)$ be a time-oriented smooth Lorentzian four-manifold (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model) whose underlying manifold carries an orientation (Oriented smooth manifolds and oriented charts), and let $w$ and the local forms ${v o l}_{g, U}$ be as in The metric volume form, current flux through 3-faces and finite-chain charges clause (a).

(a) Chart independence. Let $(U, x)$ and $(V, y)$ be oriented charts with $x = φ \circ y$ on the overlap of their domains, where $φ$ is the smooth transition map; write $g_{a b}$ and $g_{a b}^{'}$ for the metric components in the two charts and $w = \sqrt{∣ \det (g_{a b}) ∣}$ , $w^{'} = \sqrt{∣ \det (g_{a b}^{'}) ∣}$ . Then

$\det (g_{a b}^{'}) = (\det D φ)^{2} \det (g_{a b} \circ φ), w^{'} = (\det D φ) (w \circ φ),$

and the local forms agree on the overlap,

$φ^{*} (w d x^{0} \land d x^{1} \land d x^{2} \land d x^{3}) = w^{'} d y^{0} \land d y^{1} \land d y^{2} \land d y^{3} .$

Consequently there is a unique smooth nowhere-vanishing $4$ -form ${v o l}_{g}$ on $M$ — the metric volume form — whose restriction to the domain of every oriented chart $(U, x)$ is $w d x^{0} \land d x^{1} \land d x^{2} \land d x^{3}$ ; it is a smooth section with nowhere-zero values and positive on the given orientation.

(b) Value on orthonormal frames. For every $p \in M$ and every positively oriented ordered basis $(e_{0}, e_{1}, e_{2}, e_{3})$ of $T_{p} M$ with $g_{p} (e_{A}, e_{B}) = diag (- 1, 1, 1, 1)_{A B}$ ,

${v o l}_{g, p} (e_{0}, e_{1}, e_{2}, e_{3}) = 1.$

(c) Non-null hyperplanes: normal and density. Let $p \in M$ , let $H \subseteq T_{p} M$ be a $3$ -dimensional subspace that is nondegenerate for $g_{p}$ , and let $v_{1}, v_{2}, v_{3}$ be an ordered basis of $H$ . Then $H^{⊥}$ is one-dimensional and nondegenerate and $H = (H^{⊥})^{⊥}$ . Let $u \in H^{⊥} ∖ {0}$ and write $ν : = u / \sqrt{∣ g_{p} (u, u) ∣}$ , so that $ε : = g_{p} (ν, ν) \in {- 1, 1}$ . If $δ \in {- 1, 1}$ is the sign for which $(δ ν, v_{1}, v_{2}, v_{3})$ is a positively oriented basis of $T_{p} M$ , then $ν^{*} : = δ ν$ is the unit normal to $H$ adapted to the ordered basis $(v_{1}, v_{2}, v_{3})$ ; it satisfies $g_{p} (ν^{*}, ν^{*}) = ε$ , $g_{p} (ν^{*}, h) = 0$ for every $h \in H$ , and it is the unique vector with these two properties for which $(ν^{*}, v_{1}, v_{2}, v_{3})$ is positively oriented. Moreover

${v o l}_{g, p} (ν^{*}, v_{1}, v_{2}, v_{3}) = \sqrt{∣ \det (g_{p} (v_{i}, v_{j})) ∣} .$

Facts & Assumptions

Given: A time-oriented smooth Lorentzian four-manifold $(M, g, N)$ with an orientation, oriented charts $(U, x)$ , $(V, y)$ with transition $φ$ , a point $p \in M$ , a three-dimensional nondegenerate subspace $H \subseteq T_{p} M$ with ordered basis $v_{1}, v_{2}, v_{3}$ , and the objects $w$ , ${v o l}_{g, U}$ of The metric volume form, current flux through 3-faces and finite-chain charges clause (a).

[L1]

The metric components transform as $g_{a b}^{'} (y) = \partial_{a} φ^{c} (y) \partial_{b} φ^{d} (y) g_{c d} (φ (y))$ (Chart independence and the transformation law of a smooth covariant 2-tensor field, Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model).

[L2]

For square matrices of one size, $\det (A B) = \det A \det B$ and $\det (A^{T}) = \det A$ ; the determinant of $diag (- 1, 1, 1, 1)$ is $- 1$ (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L3]

Pullback of forms is functorial and commutes with the wedge product; in coordinates, $φ^{*} (d x^{0} \land \dots \land d x^{3}) = \det (D φ) d y^{0} \land \dots \land d y^{3}$ for a smooth map of open subsets of $R^{4}$ , by the coordinate expression of pullback and of the wedge product (The pullback of a differential form, The wedge product of differential forms, Local coordinate expression for a differential form). Oriented charts of an oriented manifold have orientation-preserving transitions, so $\det D φ > 0$ (Oriented smooth manifolds and oriented charts).

[L4]

At every point there is a $g_{p}$ -orthonormal ordered basis $E_{0}, E_{1}, E_{2}, E_{3}$ with $g_{p} (E_{A}, E_{B}) = diag (- 1, 1, 1, 1)_{A B}$ ; exchanging two of its spacelike vectors preserves orthonormality and the position of the negative entry and reverses its orientation, so such a basis may be chosen positively oriented (Every Lorentz form admits an ordered basis with matrix diag(-1,1,1,1), Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model, Alternating $k$ -linear maps).

[L5]

$⊥$ is the $g_{p}$ -orthogonal complement of eta-orthogonality and orthogonal complements in a Lorentz vector space: $S^{⊥} = {v : g_{p} (v, s) = 0 \forall s \in S}$ is a linear subspace, $S \subseteq (S^{⊥})^{⊥}$ , and $S^{⊥} = (span S)^{⊥}$ .

[L6]

Choose a basis $h_{1}, h_{2}, h_{3}$ of $H$ and extend it to a basis $h_{1}, h_{2}, h_{3}, e_{4}$ of the four-dimensional space $T_{p} M$ . Any $λ \in H^{*}$ extends to a functional $\tilde{λ} \in T_{p} M^{*}$ by assigning, for example, $\tilde{λ} (e_{4}) = 0$ . In this basis the matrix of $g_{p}$ is invertible because $g_{p}$ is nondegenerate; hence the linear map $T_{p} M \to T_{p} M^{*}$ , $v \mapsto g_{p} (v, \cdot)$ , is represented by an invertible matrix and is bijective. Thus $\tilde{λ} = g_{p} (v, \cdot)$ for a unique $v$ (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model clause (a)); this proves the extension-and-representation fact used below by finite-dimensional basis extension and an invertible matrix.

Proof

technique · direct; transform the metric determinant and the coordinate volume form between oriented charts, evaluate the result on orthonormal frames, and reduce the hyperplane statement to a finite coordinate computation

1.1L1L2L3algebra

(Determinant transformation.) On the overlap, [L1] reads $g^{'} = D φ^{T} (g \circ φ) D φ$ in matrix form. By [L2], $\det g^{'} = \det (D φ)^{2} \det (g \circ φ)$ . Since $\det D φ > 0$ by [L3] and $w, w^{'} > 0$ by definition, taking square roots gives $w^{'} = (\det D φ) (w \circ φ)$ .

1.2L5L6algebra

(The orthogonal complement is one-dimensional.) Let $L : T_{p} M \to H^{*}$ be $L (v) = g_{p} (v, \cdot) ∣_{H}$ . Then $\ker L = H^{⊥}$ . For each $λ \in H^{*}$ , [L6] constructs an extension $\tilde{λ} \in T_{p} M^{*}$ and the invertible matrix of $g_{p}$ gives a unique $u \in T_{p} M$ with $\tilde{λ} = g_{p} (u, \cdot)$ ; hence $L (u) = λ$ and $L$ is surjective. Rank-nullity gives $\dim H^{⊥} = 4 - \dim H^{*} = 1$ . Applying the same basis-extension and nondegenerate-matrix argument to the one-dimensional space $H^{⊥}$ shows $\dim (H^{⊥})^{⊥} = 3$ .

1.3L5algebra

(Coordinate computation.) Choose a positively oriented $g_{p}$ -orthonormal basis $E_{0}, \dots, E_{3}$ with $g_{p} (E_{A}, E_{B}) = diag (- 1, 1, 1, 1)_{A B}$ as in [L4] and write vectors as their coordinate columns. Let $c$ be the coordinate column of $ν^{*}$ and let $w_{i}$ be the coordinate columns of $v_{i}$ . Since $g_{p} (ν^{*}, v_{i}) = 0$ , the matrix $η = diag (- 1, 1, 1, 1)$ satisfies $w_{i}^{T} η c = 0$ for $i = 1, 2, 3$ , and $c^{T} η c = ε$ . Let $W$ be the $4 \times 3$ matrix with columns $w_{1}, w_{2}, w_{3}$ and $M$ the $4 \times 4$ matrix with columns $c, w_{1}, w_{2}, w_{3}$ . Then $M^{T} η M = (\begin{array}{cc} ε & 0 \\ 0 & Q \end{array}), Q = (g_{p} (v_{i}, v_{j}))_{i, j = 1}^{3} .$

2.1step 1.1L3algebra

(Agreement of local forms.) By [L3], $φ^{*} (d x^{0} \land \dots \land d x^{3}) = \det (D φ) d y^{0} \land \dots \land d y^{3}$ . Multiplying by $w \circ φ$ and substituting step 1.1 gives $φ^{*} (w d x^{0} \land \dots \land d x^{3}) = w^{'} d y^{0} \land \dots \land d y^{3}$ , the displayed identity. In particular the two local forms have the same value at every point of the overlap under the identification by $φ$ .

2.2step 1.2L5algebra

(Nondegeneracy and double complement.) If $u \in H \cap H^{⊥}$ , then $g_{p} (u, h) = 0$ for every $h \in H$ , and nondegeneracy of the restriction of $g_{p}$ to $H$ forces $u = 0$ . Hence $H \cap H^{⊥} = {0}$ , the sum $H + H^{⊥} = T_{p} M$ is direct by dimension, and the restriction of $g_{p}$ to $H^{⊥}$ is nondegenerate: if $u \in H^{⊥}$ satisfies $g_{p} (u, u^{'}) = 0$ for all $u^{'} \in H^{⊥}$ , then $g_{p} (u, \cdot)$ vanishes on $H$ and on $H^{⊥}$ and hence on $T_{p} M$ , so $u = 0$ . Finally $H \subseteq (H^{⊥})^{⊥}$ by [L5] and both have dimension $3$ by steps 1.2 and the definition of $H$ , so $H = (H^{⊥})^{⊥}$ .

2.3step 1.3L2algebra

( $M$ is invertible and a block identity.) The columns of $M$ form a basis of $T_{p} M$ : the $w_{i}$ are linearly independent because $H$ has dimension $3$ , and $c \notin span (w_{1}, w_{2}, w_{3})$ since $c$ is $η$ -orthogonal to that span while $c^{T} η c = ε \neq 0$ . Hence $\det M \neq 0$ , and $Q$ is invertible because $H$ is nondegenerate. Taking determinants in step 1.3 with [L2] and expanding the block determinant in the first coordinate gives $\det (M^{T} η M) = \det M^{2} \det η = - \det M^{2}$ and $\det (M^{T} η M) = ε \det Q$ , so $\det M^{2} = - ε \det Q = ∣ \det Q ∣$ . Therefore $∣ \det M ∣ = \sqrt{∣ \det Q ∣}$ .

3.1step 2.1L3given

(Gluing.) Each local form is smooth and nowhere vanishing on its chart domain because $w, w^{'} > 0$ are smooth and $d x^{0} \land \dots \land d x^{3}$ is a nowhere-vanishing smooth $4$ -form. By step 2.1 the local forms agree on every overlap, so they define a unique smooth section ${v o l}_{g}$ of $⋀^{4} T^{*} M$ with the stated local expressions; the transition identity also shows that ${v o l}_{g}$ evaluates positively on the positively oriented frames of the orientation at every point.

3.2step 2.2L4algebra

(Existence and uniqueness of the adapted normal.) Let $u$ span $H^{⊥}$ , put $ν = u / \sqrt{∣ g_{p} (u, u) ∣}$ ; then $g_{p} (ν, ν) \in {- 1, 1}$ , a value depending only on $H$ because replacing $u$ by a nonzero multiple does not change it. Choose $δ \in {- 1, 1}$ so that $(δ ν, v_{1}, v_{2}, v_{3})$ is positively oriented; such a sign exists and is unique because the two choices give opposite orientations and the orientation is a choice of one of the two rays of $⋀^{4} T_{p} M$ . Uniqueness of $ν^{*} = δ ν$ as a vector with $g_{p} (ν^{*}, ν^{*}) = \pm 1$ , $ν^{*} ⊥ H$ and $(ν^{*}, v_{1}, v_{2}, v_{3})$ positively oriented is immediate: any such vector is a nonzero multiple of $u$ , the last condition fixes its sign, and the unit condition fixes its modulus.

4.1step 3.1L2L4algebra

(Value on orthonormal frames.) Fix $p$ and a positively oriented $g_{p}$ -orthonormal basis $e_{0}, \dots, e_{3}$ as in [L4], take an oriented chart at $p$ , and let $E = [e_{A}^{i}]$ be the matrix whose columns are the coordinate components of the $e_{A}$ in the chart basis. Orthonormality is $E^{T} G E = diag (- 1, 1, 1, 1)$ with $G = (g_{i j} (p))$ , so [L2] gives $\det (E)^{2} \det G = - 1$ , that is, $∣ \det E ∣ = 1 / w (p)$ . Positivity of the orientation of $(e_{A})$ and of the chart gives $\det E > 0$ , hence $\det E = 1 / w (p)$ . By the local expression of step 3.1 and the defining property of wedge products, ${v o l}_{g, p} (e_{0}, \dots, e_{3}) = w (p) \det E = 1$ .

5.1step 4.1step 3.2step 2.3algebra∎

(Value of the volume form.) By step 4.1 applied to the basis $E_{A}$ , the coordinate expression of ${v o l}_{g}$ from step 3.1 gives ${v o l}_{g, p} (ν^{*}, v_{1}, v_{2}, v_{3}) = \det M$ , the determinant of the coordinate matrix of the four vectors. By construction of $ν^{*}$ in step 3.2 the tuple is positively oriented, so $\det M > 0$ ; by step 2.3, $\det M = \sqrt{∣ \det Q ∣} = \sqrt{∣ \det (g_{p} (v_{i}, v_{j})) ∣}$ . This is the displayed formula and completes the proof.

Remarks

Orientation versus time orientation. The construction uses the orientation of $M$ only; the time orientation enters neither ${v o l}_{g}$ nor the sign of the unit normal. Replacing the orientation by its opposite replaces $ν^{*}$ by $- ν^{*}$ and ${v o l}_{g}$ by $- {v o l}_{g}$ , leaving $ι_{J} {v o l}_{g}$ unchanged up to the corresponding sign convention on chains.
Null hyperplanes. A three-dimensional subspace $H$ on which $g_{p}$ is degenerate has $H^{⊥} \subseteq H$ ; it has no unit normal and the density identity above fails, which is why null faces require the three-form itself (The metric volume form, current flux through 3-faces and finite-chain charges clause (c)).

Chart independence of the metric volume form, its value on orthonormal frames, and the unit normal and density of a non-null hyperplane

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources