Worked flat slab: a rational temporal function with Cauchy levels, complete flow and orthogonal product

(The gradient is past timelike and $X(T)=1$.) By [L3], $\mathrm{\nabla }T=-T^{\mathrm{\prime }}{\mathrm{\partial }}_0$ with $T^{\mathrm{\prime }}>0$ by [L2]. Hence $g(\mathrm{\nabla }T,\mathrm{\nabla }T)=-T^{\mathrm{\prime }2}<0$, so $\mathrm{\nabla }T$ is timelike, and its direction is $-{\mathrm{\partial }}_0$, which is past-directed because ${\mathrm{\partial }}_0=N$ is future-directed [L4]; thus $T$ is a temporal function in the sense of Chronology, causality, strong causality, global hyperbolicity and time functions clause (e). Define $X:=\mathrm{\nabla }T\mathrm{/}g(\mathrm{\nabla }T,\mathrm{\nabla }T)={\mathrm{\partial }}_0\mathrm{/}{T}^{\mathrm{\prime }}$. Then $X$ is future-directed timelike and $X(T)=T^{\mathrm{\prime }}(t)\cdot dt(X)=T^{\mathrm{\prime }}\cdot \frac{1}{T^{\mathrm{\prime }}}=1$.

(The levels are Cauchy.) The level sets of $T$ are exactly the coordinate slices $x^{-1}(\{t\}\times {\mathbb{R}}^3)$ with $0<t<1$: $T$ depends only on $t$ and is strictly increasing [L2]. For an inextendible future causal curve, [L5] gives strictly increasing time coordinate and the spatial increment bound. If its future time limit were $b<1$, its spatial coordinates would be Cauchy and converge in ${\mathbb{R}}^3$, giving an interior slab point limit because $b\ge t(\gamma (s_0))>0$. This contradicts future-inextendibility. Dually its past time limit cannot exceed zero, so the time range is exactly $(0,1)$. Continuity and strict increase then give exactly one parameter at every coordinate slice. The slice chart and positive spatial metric make each slice a spacelike embedded hypersurface; since $T$ takes each real value on exactly one slice, every inextendible causal curve meets every level of $T$ exactly once. Consequently $T$ takes every real value (each level is nonempty) and $T\to +\mathrm{\infty }$ along every inextendible future-directed causal curve and $T\to -\mathrm{\infty }$ along every inextendible past-directed one: a future-directed causal curve meets the level $\{T=c\}$ for every $c$, and $T$ is strictly increasing along it by [L4] and the gradient computation, so it cannot turn back.

(The flow is complete and explicit.) By [L2] let $u\mapsto \sigma (u):=T^{-1}(u)$ be the smooth inverse of $T$. Fix $(t_0,z_0)$ and put $\gamma (s):=x^{-1}(\sigma (s+T(t_0)),z_0)$. Since $\sigma$ takes values in $(0,1)$, the curve stays in $M$ for every $s\in \mathbb{R}$; it is smooth; and by the chain rule and [L2] its $t$-coordinate satisfies $\frac{d}{ds}\sigma (s+T(t_0))={\sigma }^{\mathrm{\prime }}(s+T(t_0))=\frac{1}{T^{\mathrm{\prime }}(\sigma (s+T(t_0)))}$. Hence the chart velocity of $\gamma$ is $(1\mathrm{/}{T}^{\mathrm{\prime }}(t),0,0,0)=X(\gamma (s))$ in the chart components, so $\gamma$ is an integral curve of $X$ defined on all of $\mathbb{R}$. The maximal integral curve of $X$ through any point therefore has domain $\mathbb{R}$ (an integral curve defined on a larger interval cannot exist because $\gamma$ covers the whole maximal interval: the uniqueness theorem identifies all integral curves through a common point on overlaps), so $X$ is complete and its flow $\mathrm{\Phi }:\mathbb{R}\times M\to M$ is global and smooth (Through each point there is a unique maximal integral curve, Complete vector fields, A vector field is complete if and only if its flow is global, Local and global flows generated by a vector field).

(The orthogonal product form.) Put $\beta :=-1\mathrm{/}g(\mathrm{\nabla }T,\mathrm{\nabla }T)=1\mathrm{/}{T}^{\mathrm{\prime }2}>0$ and $h_T:=g+\beta dT\otimes dT$. By [L3], $\beta dT\otimes dT=(1\mathrm{/}{T}^{\mathrm{\prime }2})(T^{\mathrm{\prime }2}dt\otimes dt)=dt\otimes dt$, so $h_T={\delta }_{ij}d{x}^i\otimes d{x}^j$ on the spatial coordinates: it is the Euclidean metric on $\ker dT=\mathrm{span}({\mathrm{\partial }}_1,{\mathrm{\partial }}_2,{\mathrm{\partial }}_3)$, positive definite and smooth, and $g=-dt\otimes dt+h_T$. Moreover $X={\mathrm{\partial }}_0\mathrm{/}{T}^{\mathrm{\prime }}$ satisfies $h_T(X,\cdot )=0$ because $dt(X)\mathrm{\ne }0$ is compensated exactly as in the general computation $h_T(X,\cdot )=g(X,\cdot )+\beta dT=-\beta dT+\beta dT=0$. The flow map $\mathrm{\Psi }:\mathbb{R}\times {\mathrm{\Sigma }}_{1\mathrm{/}2}\to M$, $\mathrm{\Psi }(s,\cdot )=x^{-1}(\sigma (s+T(1\mathrm{/}2)),\cdot )$, is a bijection because $\sigma \circ T={\mathrm{i}\mathrm{d}}_{(0,1)}$ and $T\circ \sigma ={\mathrm{i}\mathrm{d}}_{\mathbb{R}}$, and it is a diffeomorphism since both maps are smooth; its pullback of $g$ is $-(\beta \circ \mathrm{\Psi })d{s}^2+h$ with $h$ the Euclidean metric on ${\mathrm{\Sigma }}_{1\mathrm{/}2}$.

(Summary.) The rational function $T$ is a temporal function on the globally hyperbolic slab whose levels are the Cauchy slices, whose normalized gradient field $X={\mathrm{\partial }}_0\mathrm{/}{T}^{\mathrm{\prime }}$ is complete with explicit flow $s\mapsto \sigma (s+T(t_0))$, and whose metric splits orthogonally as $g=-\beta dT\otimes dT+h_T$ with $\beta >0$ and $h_T$ Riemannian on the levels. This realises the conclusions (a)-(f) of Smooth Cauchy temporal splitting: a global temporal function with Cauchy levels and an orthogonal complete flow in an explicit model.