Local flow boxes, adapted charts and the Frobenius criterion for rest distributions

Flow box and adapted charts: the flat model with its inertial atlas is a smooth 4-manifold (The affine atlas makes the Minkowski model a smooth 4-manifold diffeomorphic to R^4) and $X$ is a smooth vector field on the open set $\mathrm{\Omega }$, so [F1] applied on this manifold gives coordinates $(u^0,u^1,u^2,u^3)$ on an open neighbourhood $U$ of $p$ with $X=\mathrm{\partial }\mathrm{/}\mathrm{\partial }{u}^0$. In such a chart the coordinate representation of a curve $\gamma$ in $U$ satisfies ${\gamma }^{\mathrm{\prime }}=X\circ \gamma$ if and only if $(u^a\circ \gamma {)}^{\mathrm{\prime }}={\delta }_0^a$ for every $a$, i.e. if and only if $u^1,u^2,u^3$ are constant and $u^0\circ \gamma$ is affine; hence the integral curves of $X$ in $U$ are exactly the $u^0$-coordinate lines, up to reparametrization. If $n$ is an observer congruence, then $n$ is nowhere zero by [F6], so applying the flow box to $X=n$ produces a chart with $n=\mathrm{\partial }\mathrm{/}\mathrm{\partial }{u}^0$ in which $u^1,u^2,u^3$ are constant along every integral curve of $n$ contained in $U$; this is precisely an $n$-adapted chart in the sense of Charts, tetrads, observer congruences and adapted charts in affine Minkowski space clause (d).

Frobenius criterion: the one-form $\alpha$ is smooth by hypothesis and nowhere zero, and $\ker \alpha$ is a hyperplane distribution by definition; [F2] gives immediately that $\ker \alpha$ is integrable on $\mathrm{\Omega }$ if and only if $\alpha \wedge d\alpha =0$ on $\mathrm{\Omega }$.

The exterior derivative of the metric dual (iii)(a): let $X,Y$ be smooth vector fields on $\mathrm{\Omega }$. By [F4], $d\alpha (X,Y)=X(\alpha (Y))-Y(\alpha (X))-\alpha ([X,Y])$; here $\alpha (Y)=n_b{Y}^b$ and $\alpha (X)=n_b{X}^b$, and with [F3] the product rule gives $X(n_b{Y}^b)=({\mathrm{\nabla }}_X{n}_b)Y^b+n_b{\mathrm{\nabla }}_X{Y}^b$ and $Y(n_b{X}^b)=({\mathrm{\nabla }}_Y{n}_b)X^b+n_b{\mathrm{\nabla }}_Y{X}^b$, while torsion-freeness gives $\alpha ([X,Y])=n_b({\mathrm{\nabla }}_X{Y}^b-{\mathrm{\nabla }}_Y{X}^b)$. Substituting, the terms $n_b{\mathrm{\nabla }}_X{Y}^b$ and $n_b{\mathrm{\nabla }}_Y{X}^b$ cancel and $d\alpha (X,Y)=({\mathrm{\nabla }}_X{n}_b)Y^b-({\mathrm{\nabla }}_Y{n}_b)X^b=({\mathrm{\nabla }}_a{n}_b-{\mathrm{\nabla }}_b{n}_a)X^a{Y}^b$; since $\mathrm{\nabla }n$ is a tensor, this is the component identity $d{\alpha }_{ab}={\mathrm{\nabla }}_a{n}_b-{\mathrm{\nabla }}_b{n}_a=2{\mathrm{\nabla }}_{[a}{n}_{b]}$.

The horizontal form of $d\alpha$ (iii)(b), first part: let $X,Y$ satisfy $\alpha (X)=\alpha (Y)=0$ at every point, so that $X,Y$ take values in the pointwise rest spaces by [F6]. By step 1.3, $d\alpha (X,Y)=2{\mathrm{\nabla }}_{[a}{n}_{b]}{X}^a{Y}^b$, and by [F5] ${\mathrm{\nabla }}_a{n}_b=-n_a{a}_b+{\theta }_{ab}+{\omega }_{ab}$; substituting the decomposition, the acceleration terms contribute $(-n_a{a}_b+n_b{a}_a)X^a{Y}^b=-\alpha (X)a^{\mathrm{♭}}(Y)+\alpha (Y)a^{\mathrm{♭}}(X)=0$, because $\alpha (X)=\alpha (Y)=0$; the $\theta$ contribution is $({\theta }_{ab}-{\theta }_{ba})X^a{Y}^b=0$ because $\theta$ is symmetric; and the $\omega$ contribution is $({\omega }_{ab}-{\omega }_{ba})X^a{Y}^b=2{\omega }_{ab}{X}^a{Y}^b$ because $\omega$ is antisymmetric. Hence $d\alpha (X,Y)=2\omega (X,Y)$ for horizontal $X,Y$.

The wedge condition (iii)(b), equivalence: let $p\in \mathrm{\Omega }$ and let $(v_0,v_1,v_2,v_3)$ be a basis of $V$ with $v_0:=n(p)$ and $v_1,v_2,v_3\in n(p{)}^{\perp }$. At $p$ one has $\alpha (v_0)=\eta (n,n)=-1$ and $\alpha (v_i)=0$ for $i\ge 1$ by [F6]; every $u\in V$ decomposes as $u=-\alpha (u)n(p)+h(u)$ with $h(u)\in n(p{)}^{\perp }$ by [F6]. Since $\alpha \wedge d\alpha$ is a three-form, it is determined by its values on basis triples: $(\alpha \wedge d\alpha )(v_0,v_i,v_j)=-d\alpha (v_i,v_j)=-2\omega (v_i,v_j)$ by alternation and step 2.1, $(\alpha \wedge d\alpha )(v_i,v_j,v_k)=0$ because all three arguments are horizontal and $\alpha$ vanishes on them, and every triple with a repeated argument vanishes by alternation. Hence $\alpha \wedge d\alpha$ vanishes at $p$ if and only if $\omega (v_i,v_j)=0$ for all $i,j\ge 1$, which by horizontality of $\omega$ is equivalent to $\omega (p)=0$. Passing to the open set $U\subseteq \mathrm{\Omega }$: $\alpha \wedge d\alpha =0$ on $U$ if and only if $\omega =0$ on $U$ by the pointwise statement, and $\omega =0$ on $U$ if and only if $\ker \alpha$ is integrable on $U$ by steps 2.1 and 1.2, since $\ker \alpha$ is integrable exactly when $\alpha \wedge d\alpha =0$ there.

Integral hypersurfaces are rest slices (iii)(c): suppose $\ker \alpha$ is integrable on $U$ and let $S\subseteq U$ be an integral hypersurface, so that $T_{q}S=\ker {\alpha }_q=\{v:{\alpha }_q(v)=0\}$ for every $q\in S$. Since ${\alpha }_q(v)=\eta (n(q),v)$, we have $T_{q}S=n(q{)}^{\perp }$ by [F6]; hence $n(q)$ is $\eta$-orthogonal to $T_{q}S$, and the restriction of $\eta$ to $T_{q}S=n(q{)}^{\perp }$ is positive definite by [F6]. Thus each integral hypersurface is a rest slice in the tangent-space sense; it need not equal an affine pointwise hyperplane. No global foliation is asserted.

Conclusion: step 1.1 proves (i), including the existence of local adapted charts for an observer congruence with $n=\mathrm{\partial }\mathrm{/}\mathrm{\partial }{u}^0$; step 1.2 proves (ii) as a direct application of the codimension-one Frobenius criterion; step 1.3 proves (iii)(a); steps 2.1 and 3.1 prove the horizontal identity $d\alpha (X,Y)=2\omega (X,Y)$ and the equivalence of $\alpha \wedge d\alpha =0$, $\omega =0$ and integrability of the rest distribution on an open set; step 4.1 proves (iii)(c); and the scope clause (iv) is respected throughout, since every statement is local on $\mathrm{\Omega }$ and none of the arguments uses a global foliation, completeness, or a variable-rank or non-smooth distribution.