The phase coordinate u = x^0 - x^3 is null and profile functions are flat-harmonic and constant along its transport curves

(The covector and the transport field.) By [F2], $n_a={\mathrm{\partial }}_{a}u$ has components $(1,0,0,-1)$, so $n\mathrm{\ne }0$ and ${\eta }^{ab}{n}_a{n}_b=-1+1=0$; its spatial part has Euclidean norm $1$ by [F3]. The vector $T={\mathrm{\partial }}_0+{\mathrm{\partial }}_3$ is nonzero with $\eta (T,T)=0$ and $\eta (T,N)=-1<0$, hence future null by The flat model spacetime: the Minkowski metric on R^4 and its inertial chart clause (c). Its metric dual is $(-1,0,0,1)=-n$, so $n(T)=0$. Multiplication by $\omega$ gives the stated components of $k$ and $k^{\mathrm{♯}}=\omega T$. For $\omega \mathrm{\ne }0$, $\eta (\omega T,\omega T)=0$ and $\eta (\omega T,N)=-\omega$, proving the future/past alternatives; for $\omega =0$ the vector is zero, excluded from the null class by Timelike, null, spacelike and causal vectors of a Lorentz vector space. This proves (a).

(Profiles are transported and flat-harmonic.) Fix $x$. By [F1], ${\mathrm{\partial }}_a\phi (x)=f^{\mathrm{\prime }}(u(x)){\mathrm{\partial }}_{a}u(x)$, so by step 1.1, $T\phi (x)={\mathrm{\partial }}_0\phi (x)+{\mathrm{\partial }}_3\phi (x)=f^{\mathrm{\prime }}(u(x))(1+(-1))=0$. For the d'Alembertian, [F1] applied to $f^{\mathrm{\prime }}$ gives ${\mathrm{\partial }}_0^2\phi =f^{\mathrm{\prime }\mathrm{\prime }}(u(x))$ and ${\mathrm{\partial }}_3^2\phi =f^{\mathrm{\prime }\mathrm{\prime }}(u(x))({\mathrm{\partial }}_{3}u{)}^2=f^{\mathrm{\prime }\mathrm{\prime }}(u(x))$, since ${\mathrm{\partial }}_3^{2}u=0$, while ${\mathrm{\partial }}_1\phi ={\mathrm{\partial }}_2\phi =0$; hence $\mathrm{\square }\phi (x)=-f^{\mathrm{\prime }\mathrm{\prime }}(u(x))+f^{\mathrm{\prime }\mathrm{\prime }}(u(x))=0$ by [F2]. This proves (b).

(Planar perturbations.) For each fixed pair $(a,b)$, the component $h_{ab}$ is the composition of the smooth function $u$ with the smooth function $F_{ab}$; hence $h$ has smooth components in the inertial chart and, by the transformation law for covariant two-tensors in the one available chart [F3], defines a smooth symmetric field. Applying [F1] to each component gives ${\mathrm{\partial }}_c{h}_{ab}=F_{ab}^{\mathrm{\prime }}(u){\mathrm{\partial }}_{c}u=F_{ab}^{\mathrm{\prime }}(u)n_c$; applying the same computation to the derivative profile $F_{ab}^{\mathrm{\prime }}$ gives ${\mathrm{\partial }}_c{\mathrm{\partial }}_d{h}_{ab}=F_{ab}^{\mathrm{\prime }\mathrm{\prime }}(u)n_c{n}_d$, so $\mathrm{\square }{h}_{ab}=F_{ab}^{\mathrm{\prime }\mathrm{\prime }}(u){\eta }^{cd}{n}_c{n}_d=F_{ab}^{\mathrm{\prime }\mathrm{\prime }}(u)(n^d{n}_d)=0$ by step 1.1. For the same reason $({\mathrm{\partial }}_0+{\mathrm{\partial }}_3)h_{ab}=F_{ab}^{\mathrm{\prime }}(u)(1-1)=0$. If the profile is TT, then the components $h_{a0}$ and $h_{a3}$ vanish for every $a$, and ${\eta }^{ab}{h}_{ab}=F_{11}+F_{22}=0$; for the divergence, the TT condition gives $F_{a0}^{\mathrm{\prime }}=F_{a3}^{\mathrm{\prime }}=0$, and with $n_1=n_2=0$, $n_0=1$, $n_3=-1$ the sum ${\eta }^{bc}{\mathrm{\partial }}_b{h}_{ac}={\eta }^{00}{F}_{a0}^{\mathrm{\prime }}{n}_0+{\eta }^{33}{F}_{a3}^{\mathrm{\prime }}{n}_3=-F_{a0}^{\mathrm{\prime }}-F_{a3}^{\mathrm{\prime }}$ vanishes. This proves (c).

(Transport along $T$.) The integral curve of $T$ through $x$ is $\sigma \mapsto x+\sigma T=(x^0+\sigma ,x^1,x^2,x^3+\sigma )$, because the derivative of this curve is $T$ by [F2] and $T$ is constant. Along it, $u$ takes the constant value $(x^0+\sigma )-(x^3+\sigma )=x^0-x^3=u(x)$; hence $h_{ab}$ is constant along it by step 2.2. In the coordinate reparametrization $t_c=x^0\mathrm{/}c$ the curve has $d{t}_c\mathrm{/}d\sigma =1\mathrm{/}c$ and $d{x}^3\mathrm{/}d\sigma =1$, so $\mathrm{\mid }d{x}^3\mathrm{/}d{t}_c\mathrm{\mid }=c$. This proves (d).

(Conclusion.) Steps 1.1, 2.1, 2.2 and 3.1 prove clauses (a)-(d) from the explicit chart functions and the cited differential calculus; no field equation, no choice principle and no physical assumption is used.