Gr13 Convex Normal Neighborhoods — Examples

(The cone function.) For $q\in A$ write $v:=q-p^{\mathrm{\prime }}$, so $v=(v^0,v_{\mathrm{v}\mathrm{e}\mathrm{c}})$ and $Q(q):=-\eta (v,v)=(v^0{)}^2-\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mathrm{\mid }}^2$. By [L1], $Q(q)>0$ exactly when $v$ is timelike, $Q(q)=0$ exactly when $v$ is null (or zero), and the sign of $v^0$ together with [L1] identifies the future and past sides: $\{Q>0,v^0>0\}$ is the chronological future $I^{+}(p^{\mathrm{\prime }})$ of the flat model and $\{Q>0,v^0<0\}$ its chronological past.

(The gradient of $Q$.) In the chart components, $dQ=2v^{0}d{x}^0-2{\sum }_i{v}^{i}d{x}^i$, and the vector field $W:=-2v^0{\mathrm{\partial }}_0-2{\sum }_i{v}^i{\mathrm{\partial }}_i$ satisfies $\eta (W,w)=2v^0{w}^0-2{\sum }_i{v}^i{w}^i=dQ(w)$ for all $w$, so $\mathrm{\nabla }Q=W=-2v$ by [L4]. Consequently $\mathrm{\nabla }Q$ is past-directed timelike exactly on $\{Q>0,v^0>0\}$, namely on $I^{+}(p^{\mathrm{\prime }})$, and it vanishes nowhere on the zero set of $Q$ except at the vertex $q=p^{\mathrm{\prime }}$.

(The smoothed bump is smooth and its gradient is past timelike where it is positive.) Define $f(q)=\phi (Q(q))$ when $v^0(q)>0$, and $f(q)=0$ when $v^0(q)\le 0$. Near every nonvertex point with $v^0=0$, one has $Q<0$, so this prescription is locally zero. On the future half-space it is the smooth composite; it extends smoothly across the null cone by flatness of $\phi$. At the vertex, every derivative on the future side is a finite sum of polynomial factors times derivatives of $\phi (Q)$, all tending to zero faster than any power of $\mathrm{\mid }v\mathrm{\mid }$; the zero extension is therefore smooth there as well. Thus $f$ is smooth everywhere [L3]. Moreover $f\ge 0$ and $f>0$ exactly on $I^{+}(p^{\mathrm{\prime }})$ by step 1.1. Where $f>0$ one has $\mathrm{\nabla }f={\phi }^{\mathrm{\prime }}(Q)\mathrm{\nabla }Q$ by the chain rule; since $Q>0$ there and ${\phi }^{\mathrm{\prime }}(Q)=\frac{1}{Q^2}{e}^{-1\mathrm{/}Q}>0$, the gradient $\mathrm{\nabla }f$ is a positive multiple of the past-directed timelike $\mathrm{\nabla }Q$ of step 2.1, hence is past timelike on $\{f>0\}$.

(The normalized bump.) Since $p-p^{\mathrm{\prime }}=(\delta ,0,0,0)$ is future timelike, $Q(p)={\delta }^2>0$ and $f(p)>0$; thus $h_p=\chi f\mathrm{/}f(p)$ is smooth and compactly supported for every smooth compactly supported $\chi :A\to [0,1]$. To realize the full truncated-gradient assertion, put $K=\{q:0\le v^0(q)\le \delta ,\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}(q)\mathrm{\mid }\le v^0(q)\}$, the compact set $J^{+}(p^{\mathrm{\prime }})\cap \{x^0\le 0\}$, and choose a precompact open ball $C$ containing K. By [L5] choose $\chi$ with support compactly in C and $\chi =1$ on a neighborhood of K. Then $h_p(p)=1$, and continuity gives a neighborhood with $h_p>1\mathrm{/}2$. If $h_p(q)>0$ and $x^0(q)\le 0$, future-branch positivity puts q in K, hence in the cutoff plateau; there $\mathrm{\nabla }{h}_p=\mathrm{\nabla }f\mathrm{/}f(p)$ is past timelike by step 3.1. At points with $x^0\le 0$ and $h_p=0$ the gradient vanishes by flatness on the future null boundary and by the zero definition outside the future cone. Thus the compact support, normalized value, truncated gradient and nonzero-gradient locus properties of the cone-bump construction hold. For an arbitrary other cutoff the gradient assertion remains limited to its plateau.

(Summary.) The explicit data $(Q,f,h_p)$ realise the cone-bump lemma at $p^{\mathrm{\prime }}$ in the flat model: $Q$ is the exponential cone function, $f$ the smoothing of the indicator of the chronological future of $p^{\mathrm{\prime }}$, and $h_p$ the normalized compactly supported bump with $h_p(p)=1$ and past-timelike gradient in the cutoff plateau where $h_p>0$.

(The two legs are future null.) The displacements are $q-p=(1,1,0,0)$ and $r-q=(1,-1,0,0)$; both satisfy $v^0=\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }=1>0$, hence both are future null by [L1], and the corresponding straight paths are future causal curves. Therefore $p\le q$ and $q\le r$, and consequently $p\le r$.

(The total displacement is future timelike.) The displacement $r-p=(2,0,0,0)$ satisfies $v^0=2>0=\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }$, hence is future timelike by [L1]. In particular $r\mathrm{\ne }p$.

(Contrast: the parallel case stays null.) For $r^{\mathrm{\prime }}=x^{-1}(2,2,0,0)$ the second displacement $r^{\mathrm{\prime }}-q=(1,1,0,0)$ is a positive multiple of the first, the broken path is the straight null path from $p$ to $r^{\mathrm{\prime }}$ (the two legs are collinear), and $r^{\mathrm{\prime }}-p=(2,2,0,0)$ satisfies $v^0=\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }=2$, so it is future null: in the parallel case the composition of two null legs remains null, and no strict chronology follows. This shows why the non-parallel corner is exactly what produces the timelike total displacement.

(Strict chronology by the straight path and by an explicit curve.) The displacement $r-p=(2,0,0,0)$ satisfies $v^0=2>0=\mathrm{\mid }{v}_{\mathrm{v}\mathrm{e}\mathrm{c}}\mathrm{\mid }$, so it is future timelike by [L1]; hence the straight path $\sigma (s):=x^{-1}(x(p)+s(r-p))$, $s\in [0,1]$, has the constant future timelike velocity $r-p$ and is a smooth future timelike curve from $p$ to $r$, so $p\ll r$ [L1, L2]. As an independent certificate, $\gamma (s):=x^{-1}(s,\frac{1}{4}\sin (\pi s),0,0)$ on $[0,2]$ is smooth with chart velocity $(1,\frac{\pi }{4}\cos (\pi s),0,0)$ of spatial norm at most $\pi \mathrm{/}4<1$, hence future timelike everywhere by [L1]; it runs from $\gamma (0)=p$ to $\gamma (2)=r$. Since $\gamma$ is a future causal curve carrying a timelike piece, clause (c) of the push-up rule The push-up rule: p << q and q <= r imply p << r, together with the timelike-piece criterion equally yields $p\ll r$, in agreement with the direct computation.

(Summary.) For the non-parallel null legs the two legs give $p\le q\le r$ while the total displacement is future timelike, and $p\ll r$ is certified both by the straight path and by the explicit timelike curve $\gamma$; for parallel null legs the total displacement stays null. This exhibits exactly the strict output that the push-up rule records for a causal connection with a timelike piece.

(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.

(The function is temporal with Cauchy levels.) The coordinate $t$ is smooth with past timelike gradient by [L2]. For any two-sided inextendible future causal curve $\gamma :I\to M$, [L4] makes $t\circ \gamma$ strictly increasing with limits $a,b$ in $[0,1]$. If $b<1$, its spatial coordinates are Cauchy at the future end by the increment bound, hence converge in ${\mathbb{R}}^3$; as $b\ge t(\gamma (s_0))>0$, this gives a point limit $(b,z_{\mathrm{\infty }})$ inside the slab, contradicting inextendibility. Dually $a>0$ would give a past point limit, so $a=0,b=1$. Continuity and strict increase give exactly one parameter at every $c\in (0,1)$. Thus each $t^{-1}(c)$ is Cauchy; its slice chart and positive spatial metric make it a spacelike embedded hypersurface.

(The maximal integral curve leaves the slab at finite parameter.) Take $c=1\mathrm{/}2$, so the curve $\gamma (s):=x^{-1}(1\mathrm{/}2+s,0,0,0)$. By [L3] it is the maximal integral curve of $X$ through $p_0=\gamma (0)$, and its maximal domain is $(-1\mathrm{/}2,1\mathrm{/}2)$. At $s\to 1\mathrm{/}2-$ the curve converges in the ambient flat model to the boundary point $x^{-1}(1,0,0,0)\mathrm{\notin }M$ and similarly at the lower end, so it cannot be extended within $M$; its domain is not all of $\mathbb{R}$. Therefore $X={\mathrm{\partial }}_0$ is not complete, and its flow is not global.

(Conclusion.) The flat slab is globally hyperbolic, the coordinate time is a temporal function with Cauchy levels, and yet its normalized gradient field is incomplete. Hence the refuted statement is false: completeness of $X=\mathrm{\nabla }T\mathrm{/}g(\mathrm{\nabla }T,\mathrm{\nabla }T)$ is not a consequence of being a temporal function on a globally hyperbolic manifold, and the completeness clause of Smooth Cauchy temporal splitting: a global temporal function with Cauchy levels and an orthogonal complete flow is a property of the function constructed there.