Smooth Cauchy temporal splitting: a global temporal function with Cauchy levels and an orthogonal complete flow

(Strip functions.) For every integer $j$ apply [L1] with the quadruple $a=j-2$, $a^{\mathrm{\prime }}=j-1$, $c^{\mathrm{\prime }}=j+1$, $c=j+2$, obtaining a smooth ${\tau }_j:M\to [-1,1]$ with ${\tau }_j=-1$ on $\{t\le j-2\}$, ${\tau }_j=+1$ on $\{t\ge j+2\}$ and $\mathrm{\nabla }{\tau }_j$ past causal or zero everywhere and past timelike on the inner strip $\{j-1\le t\le j+1\}$. The countably many applications are made with ${\mathrm{A}\mathrm{C}}_{\omega }$.

(The locally finite sum $T$.) Define $T:={\tau }_0+{\sum }_{j\ge 1}({\tau }_{-j}+{\tau }_j)\mathrm{.}$ The series is locally finite: choose $R>\mathrm{\mid }t(x)\mathrm{\mid }$ at a point $x$. On a neighborhood of $x$ one has $\mathrm{\mid }t\mathrm{\mid }<R$; if $j>R+2$, then $t<R<j-2$, so ${\tau }_j=-1$ and (because $t\ge -R>-j+2$) ${\tau }_{-j}=+1$ identically there, whence ${\tau }_{-j}+{\tau }_j=0$ on that neighborhood. By A locally finite sum of smooth functions is smooth the sum $T$ is smooth, and $\mathrm{\nabla }T=\mathrm{\nabla }{\tau }_0+{\sum }_{j\ge 1}(\mathrm{\nabla }{\tau }_{-j}+\mathrm{\nabla }{\tau }_j)$ with only finitely many nonzero terms near any point. Every term is past causal or zero, so every partial sum is past causal or zero and so is the limit [L5]; at a point with $t(x)\in [k-1,k+1]$ for some integer $k$ the summand ${\tau }_k$ has past timelike gradient and all others past causal or zero, so $\mathrm{\nabla }T(x)$ is past timelike by [L5]. Since ${\bigcup }_{k\in \mathbb{Z}}[k-1,k+1]=\mathbb{R}$, the gradient $\mathrm{\nabla }T$ is past timelike everywhere; in particular $g(\mathrm{\nabla }T,\mathrm{\nabla }T)<0$ and $g(\mathrm{\nabla }T,N)>0$, so $T$ is a temporal function and (a) holds.

(The order bounds.) For every integer $k\ge 0$ one has $T\ge 2k+1$ on $\{t\ge k+2\}$ and $T\le -2k-1$ on $\{t\le -k-2\}$. Indeed on $\{t\ge k+2\}$ we have ${\tau }_0=+1$, ${\tau }_{-j}=+1$ for every $j\ge 1$ (since $t\ge k+2\ge 2\ge -j+2$), and ${\tau }_j=+1$ for $1\le j\le k$ (since $t\ge k+2\ge j+2$), while ${\tau }_j\ge -1$ always; hence $T\ge 1+2k$. The past bound is the same computation with the signs reversed.

(The gradient field.) Define $X:=\mathrm{\nabla }T\mathrm{/}g(\mathrm{\nabla }T,\mathrm{\nabla }T)$, smooth because the denominator is a nowhere-vanishing smooth function, and future-directed timelike because it is a negative multiple of the past-directed timelike $\mathrm{\nabla }T$ [L5]. Moreover $X(T)=dT(X)=g(\mathrm{\nabla }T,X)=g(\mathrm{\nabla }T,\mathrm{\nabla }T)\mathrm{/}g(\mathrm{\nabla }T,\mathrm{\nabla }T)=1$.

(Divergence along causal curves.) Let $\gamma :I\to M$ be an inextendible future-directed causal curve. By [L2], $t(\gamma (s))\to +\mathrm{\infty }$ as $s\to \sup I$; given an integer $k\ge 0$, for all sufficiently large $s$ one has $t(\gamma (s))\ge k+2$, hence $T(\gamma (s))\ge 2k+1$ by step 3.1. As $k$ is arbitrary, $T(\gamma (s))\to +\mathrm{\infty }$; the past-directed case is dual. This is (c).

(Levels are Cauchy hypersurfaces.) By [L4] and step 2.1, $T$ is a time function: $T(\gamma (a))<T(\gamma (b))$ for every future causal curve not constant on $[a,b]$. Let $c\in \mathbb{R}$. Every inextendible causal curve meets $S_c:=T^{-1}(c)$ in at least one parameter value: for a future-directed curve, $s\mapsto T(\gamma (s))$ is continuous with limits $-\mathrm{\infty }$ at the past end (dual of step 4.1) and $+\mathrm{\infty }$ at the future end (step 4.1), so [L8] gives a parameter with value $c$; the past-directed case is dual. It meets $S_c$ at most once, because two distinct parameter values would give two causally related distinct points with equal $T$-value, contradicting strict monotonicity. Hence $S_c$ is a Cauchy set; in particular $T$ is surjective. Since $dT\mathrm{\ne }0$ everywhere (the gradient is timelike), $S_c$ is an embedded hypersurface by [L6], and its tangent space at $p$ is $\ker d{T}_p=(\mathrm{\nabla }{T}_p{)}^{\perp }$, which is positive definite because $\mathrm{\nabla }{T}_p$ is timelike [L6]. Hence $S_c$ is a spacelike embedded hypersurface and a Cauchy set, that is, a Cauchy hypersurface; this is (b).

(Completeness of $X$.) Let $\gamma :(\alpha ,\beta )\to M$ be a maximal integral curve of $X$ with $\gamma (0)=z$. By the chain rule, $\frac{d}{ds}T(\gamma (s))=dT(X)(\gamma (s))=1$, so $T(\gamma (s))=T(z)+s$ throughout its domain. Suppose $\beta <+\mathrm{\infty }$ and choose $s_0\in (\alpha ,\beta )$. Define the smooth increasing bijection $\theta :[0,\mathrm{\infty })\to [s_0,\beta )$ by $\theta (u)=\beta -(\beta -s_0)e^{-u}$ and set $\widehat{\gamma }(u)=\gamma (\theta (u))$. This is a future causal curve. The escape lemma [L8] says the maximal flow line leaves every compact set as $s\to \beta$, so $\gamma (s)$ has no limit there; therefore $\widehat{\gamma }$ has no future endpoint limit and is future-inextendible. Apply the one-sided extension Equivalence of the absolutely-continuous and continuous causal conventions, and one-sided inextendible extensions clause (d) to obtain an inextendible future causal curve ${\widehat{\gamma }}^{\ast }$ which agrees with $\widehat{\gamma }$ on $[0,\mathrm{\infty })$. By step 4.1, $T({\widehat{\gamma }}^{\ast }(u))\to +\mathrm{\infty }$ as $u\to \mathrm{\infty }$. But on this same tail $T(\widehat{\gamma }(u))=T(z)+\theta (u)\to T(z)+\beta <+\mathrm{\infty }$, a contradiction. Hence $\beta =+\mathrm{\infty }$; the time-reversed argument gives $\alpha =-\mathrm{\infty }$. Therefore every maximal integral curve is defined on $\mathbb{R}$, so $X$ is complete by [L7]; its flow is global and smooth.

(The flow diffeomorphism.) Define $\mathrm{\Psi }:\mathbb{R}\times S_0\to M$ by $\mathrm{\Psi }(s,z):=\mathrm{\Phi }(s,z)$. For $z\in S_0$ the curve $s\mapsto \mathrm{\Phi }(s,z)$ is an integral curve of $X$ with $\mathrm{\Phi }(0,z)=z$, so $T(\mathrm{\Phi }(s,z))=T(z)+s=s$ by step 5.2; hence $\mathrm{\Psi }$ takes values in $M$ with $T\circ \mathrm{\Psi }(s,z)=s$. It is surjective: given $p\in M$, the point $z:=\mathrm{\Phi }(-T(p),p)$ satisfies $T(z)=T(p)-T(p)=0$, so $z\in S_0$, and the composition law of the global flow gives $\mathrm{\Psi }(T(p),z)=\mathrm{\Phi }(T(p),\mathrm{\Phi }(-T(p),p))=\mathrm{\Phi }(0,p)=p$. It is injective: if $\mathrm{\Phi }(s,z)=\mathrm{\Phi }(s^{\mathrm{\prime }},z^{\mathrm{\prime }})$, applying $T$ gives $s=s^{\mathrm{\prime }}$, and then $z=\mathrm{\Phi }(-s,\mathrm{\Phi }(s,z))=\mathrm{\Phi }(0,z^{\mathrm{\prime }})=z^{\mathrm{\prime }}$ by the inverse-time law. The inverse $p\mapsto (T(p),\mathrm{\Phi }(-T(p),p))$ is smooth because $T$ and $\mathrm{\Phi }$ are smooth. Hence $\mathrm{\Psi }$ is a diffeomorphism, which is (e).

(The orthogonal product.) Put $\beta :=-1\mathrm{/}g(\mathrm{\nabla }T,\mathrm{\nabla }T)$, a smooth positive function. Since $g(\mathrm{\nabla }T,\mathrm{\nabla }T)<0$ everywhere, $\beta >0$; and $X=\mathrm{\nabla }T\mathrm{/}g(\mathrm{\nabla }T,\mathrm{\nabla }T)=-\beta \mathrm{\nabla }T$, so $g(X,\cdot )=-\beta g(\mathrm{\nabla }T,\cdot )=-\beta dT$. Consequently $g(X,X)=-\beta$ and $g(X,v)=0$ for every $v\in \ker dT$. Define the smooth symmetric tensor field $h_T:=g+\beta dT\otimes dT$. Then $h_T(X,\cdot )=g(X,\cdot )+\beta dT=-\beta dT+\beta dT=0$, while for $v,w\in \ker dT$ one has $h_T(v,w)=g(v,w)$; since $\ker dT=(\mathrm{\nabla }T{)}^{\perp }$ is positive definite by [L6], $h_T$ restricts to a Riemannian metric on $\ker dT$. This proves the identity $g=-\beta dT\otimes dT+h_T$ and the orthogonal decomposition of (f). Finally, differentiating $\mathrm{\Psi }$ and using $d\mathrm{\Psi }({\mathrm{\partial }}_s)=X$, $d\mathrm{\Psi }({\mathrm{\partial }}_{z}v)=(d{\mathrm{\Phi }}_s{)}_{z}v$ with $(d{\mathrm{\Phi }}_s{)}_{z}v\in \ker dT$ by step 6.1, one obtains ${\mathrm{\Psi }}^{\ast }g({\mathrm{\partial }}_s,{\mathrm{\partial }}_s)=g(X,X)=-\beta \circ \mathrm{\Psi }$, ${\mathrm{\Psi }}^{\ast }g({\mathrm{\partial }}_s,{\mathrm{\partial }}_{z}v)=g(X,(d{\mathrm{\Phi }}_s{)}_{z}v)=0$, and ${\mathrm{\Psi }}^{\ast }g({\mathrm{\partial }}_{z}v,{\mathrm{\partial }}_{z}w)=h_{(s,z)}(v,w)$; the last expression is positive definite because $(d{\mathrm{\Phi }}_s{)}_z$ is a linear isomorphism onto $\ker d{T}_{\mathrm{\Phi }(s,z)}$. This is (f), and it completes the proof.