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Smoothness of the rest-space family and its Frobenius integrability criterion

Statement

Let $(M, g, N)$ be a time-oriented smooth Lorentzian four-manifold, let $U \subseteq M$ be a nonempty open subset, and let $n$ be an observer congruence on $U$ (Observer congruences, their observer curves and adapted charts). Let $α : = n^{♭} \in Ω^{1} (U)$ be the smooth $1$ -form $α_{q} (v) = g_{q} (n_{q}, v)$ (The inverse metric, musical isomorphisms and index raising, A smooth differential $k$ -form), and let $D_{q} : = \ker α_{q} = {v \in T_{q} M : g_{q} (n_{q}, v) = 0} = n_{q}^{⊥}$ be the rest space of the congruence at $q$ (Observers, rest spaces and the spatial projection at an event). Then:

$α$ is nowhere zero on $U$ , and $q \mapsto D_{q}$ is a smooth rank- $3$ distribution on $U$ (Smooth distributions on a manifold).
$D$ is integrable (Integrable distributions) if and only if $α \land d α = 0$ on $U$ (The exterior derivative by the invariant vector-field formula, The wedge product of differential forms).
If $α \land d α = 0$ on $U$ , then every point of $U$ has a chart $(u^{0}, u^{1}, u^{2}, u^{3})$ in which $D = span (\partial_{1}, \partial_{2}, \partial_{3})$ ; equivalently, the rest spaces are tangent to the level hypersurfaces of $u^{0}$ .

Facts & Assumptions

Given: A time-oriented smooth Lorentzian four-manifold $(M, g, N)$ , a nonempty open $U \subseteq M$ , an observer congruence $n$ on $U$ , the one-form $α = n^{♭}$ and the family $D_{q} = \ker α_{q}$ .

[L1]

The congruence $n$ is a smooth vector field on $U$ with $g_{q} (n_{q}, n_{q}) = - 1$ for all $q \in U$ , so $n_{q} \neq 0$ (Observer congruences, their observer curves and adapted charts).

[L2]

A vector field on a chart domain is smooth exactly when its coordinate components are smooth (Smoothness of a vector field is equivalent to smooth coordinate components).

[L3]

In a chart the components $g_{a b} (q) = g_{q} (\partial_{a} ∣_{q}, \partial_{b} ∣_{q})$ of $g$ are $C^{\infty}$ (Smooth covariant 2-tensor fields on a smooth manifold).

[L4]

The flat map is $♭_{q} (v) = g_{q} (v, \cdot)$ , so $α_{q} (v) = g_{q} (n_{q}, v)$ ; the components of $α$ in a chart are $α_{a} = \sum_{b} g_{a b} n^{b}$ (The inverse metric, musical isomorphisms and index raising).

[L5]

Finite sums and products of $C^{\infty}$ functions on chart domains are $C^{\infty}$ ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition).

[L6]

A family of tangent subspaces is a smooth rank- $k$ distribution exactly when every point has a neighborhood carrying smooth vector fields $X_{1}, \dots, X_{k}$ that are pointwise independent and span the subspaces (A smooth distribution is exactly a locally framed constant-rank family of tangent spaces); smooth distributions are defined as rank- $k$ smooth subbundles of $T M$ , under the axiom ${A C}_{ω}$ (Smooth distributions on a manifold, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L7]

For a nowhere-zero one-form $α$ , the hyperplane distribution $\ker α$ is integrable if and only if $α \land d α = 0$ (The codimension-one Frobenius criterion).

[L8]

A rank- $k$ smooth distribution is integrable if and only if it is involutive (Involutive distributions), and in that case every point has a coordinate neighborhood in which the distribution equals $span (\partial_{1}, \dots, \partial_{k})$ (Frobenius local coordinate theorem).

Proof

technique · direct

1.1L1L2L3L4L5given

In a chart $(U^{'}, x)$ contained in $U$ , the components of $α$ are $α_{a} = \sum_{b} g_{a b} n^{b}$ by [L4], where $g_{a b}$ are $C^{\infty}$ by [L3] and the components $n^{b}$ of the smooth field $n$ are $C^{\infty}$ by [L2]; hence by [L5] the functions $α_{a}$ are $C^{\infty}$ , so $α$ is a smooth $1$ -form. Moreover $α_{q} (n_{q}) = g_{q} (n_{q}, n_{q}) = - 1 \neq 0$ for every $q \in U$ by [L1] and [L4], so $α$ is nowhere zero.

2.1step 1.1L2L4algebra

Fix $q_{0} \in U$ and a chart $(U^{'}, x)$ around $q_{0}$ ; since $α_{q_{0}} \neq 0$ by step 1.1, some component $α_{a} (q_{0})$ is nonzero, and after relabelling the coordinates we may assume $α_{0} (q_{0}) \neq 0$ . On the open subset $V \subseteq U^{'}$ where $α_{0} \neq 0$ define, for $i = 1, 2, 3$ , the vector fields $X_{i} : = α_{0} \partial_{i} - α_{i} \partial_{0}$ ; their components are $C^{\infty}$ by step 1.1, so they are smooth by [L2]. They are tangent to $D$ , because $α (X_{i}) = α_{0} α_{i} - α_{i} α_{0} = 0$ . They are pointwise independent on $V$ , since the coefficient of $\partial_{i}$ in $X_{i}$ is $α_{0} \neq 0$ . Finally they span $D_{q}$ at every $q \in V$ : if $v = \sum_{a} v^{a} \partial_{a}$ satisfies $α (v) = α_{0} v^{0} + \sum_{i \geq 1} α_{i} v^{i} = 0$ , then $v^{0} = - (\sum_{i \geq 1} α_{i} v^{i}) / α_{0}$ , so $v = \sum_{i \geq 1} (v^{i} / α_{0}) X_{i}$ .

2.2step 1.1L7given

Applying [L7] to the nowhere-zero smooth one-form $α$ of step 1.1 gives: $\ker α = D$ is integrable if and only if $α \land d α = 0$ . This proves claim 2.

3.1step 2.1L6given

Step 2.1 exhibits, near each point of $U$ , three smooth pointwise-independent fields spanning the $D_{q}$ ; by [L6] the family $q \mapsto D_{q}$ is therefore a smooth rank- $3$ distribution on $U$ . This proves claim 1.

3.2step 2.2L8given

Assume $α \land d α = 0$ on $U$ ; by step 2.2 the distribution $D$ is integrable. Then [L8] gives, near every point of $U$ , a chart $(u^{0}, u^{1}, u^{2}, u^{3})$ in which $D = span (\partial_{1}, \partial_{2}, \partial_{3})$ ; the slices with $u^{0}$ constant are integral hypersurfaces tangent to the rest spaces. This proves claim 3.

Smoothness of the rest-space family and its Frobenius integrability criterion

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources