The open flat slab: an incomplete timelike geodesic, a Cauchy slice and global hyperbolicity

(The slab model.) By [L2] the set $M$ is an open submanifold of $A$ with the restricted chart $x$, and the restricted fields $g=\eta {\mathrm{\mid }}_M$ and $N={\mathrm{\partial }}_0{\mathrm{\mid }}_M$ are smooth; the components of $g$ in the restricted chart are the constants $\mathrm{diag}(-1,1,1,1)$, and $g(N,N)=\eta ({\mathrm{\partial }}_0,{\mathrm{\partial }}_0)=-1<0$, so $N$ is a time orientation of $(M,g)$ and $(M,g,N)$ is a time-oriented smooth Lorentzian four-manifold. Applying [L3] to this chart gives the vanishing of the Christoffel symbols and the curvature of the metric-selected connection and the geodesic character of coordinate curves. Since $M$ is open in $A$ and $g$ restricts $\eta$, the notions of locally absolutely continuous curve, velocity and pointwise causal character agree for curves in $M$ whether read in $(M,g)$ or in $(A,\eta )$. This proves (a).

(Monotonicity and Lipschitz bounds.) Let $u:I\to M$ be a future causal curve of $(M,g)$. By (a) it is a future causal curve of $(A,\eta )$, so [L4] applies: $x^0\circ u$ is strictly increasing and $\mathrm{\mid }z(u(t^{\mathrm{\prime }}))-z(u(t)){\mathrm{\mid }}_{\mathrm{e}}\le x^0(u(t^{\mathrm{\prime }}))-x^0(u(t))$ for $t<t^{\mathrm{\prime }}$ in $I$, and $x^0\circ u$ is locally absolutely continuous. This is the quantitative input used below.

(Cone characterization and convexity in the slab.) By [L1] and $g=\eta {\mathrm{\mid }}_M$, the future cones at points of $M$ are those of the chart. Since $M$ is convex in the coordinates of $x$, the straight segment between $p,q\in M$ lies in $M$. If $p=q$, reflexivity gives $q\in J_M^{+}(p)$ and $q-p=0\in \overline{C^{+}}$; this case uses no constant causal curve. For $p\mathrm{\ne }q$, if $q-p\in C^{+}$ (respectively $\overline{C^{+}}$) the straight path $s\mapsto p+s(q-p)$ is a future timelike (respectively causal) curve in $M$ from $p$ to $q$, so $q\in I_M^{+}(p)$ (respectively $q\in J_M^{+}(p)$). Conversely a future causal curve in $M$ is one in $A$, and [L5] gives $q-p\in \overline{C^{+}}$ when $q\in J_M^{+}(p)$; for $q\in I_M^{+}(p)$ the strict cone membership is the same statement with $C^{+}$. Finally, [L6] gives $C^{+}+\overline{C^{+}}\subseteq C^{+}$ and $\overline{C^{+}}+\overline{C^{+}}\subseteq \overline{C^{+}}$.

(The vertical geodesic, its maximal domain and inextendibility.) Let $\gamma (s):=x^{-1}(s,z_0)$ for $s\in (0,1)$. Its coordinate path is $y(s)=(s,z_0)$, so $\dot{y}=(1,0,0,0)$ and $\ddot{y}=0$; by [L3] and [L8] it is a smooth geodesic with affine parameter $s$, and its tangent ${\gamma }^{\mathrm{\prime }}(s)={\mathrm{\partial }}_0$ satisfies $g({\gamma }^{\mathrm{\prime }},{\gamma }^{\mathrm{\prime }})=-1<0$, so $\gamma$ is a future timelike curve. Suppose a geodesic $\tilde{\gamma }:(a,b)\to M$ with $a<1\mathrm{/}2<b$ restricts to $\gamma$ on $(0,1)$ and has $b>1$. In the chart, both coordinate paths satisfy the first-order geodesic system of [L8] with the same initial data at $s=1\mathrm{/}2$, namely $(y,\dot{y})=((1\mathrm{/}2,z_0),(1,0,0,0))$, because the Christoffel symbols vanish; by uniqueness in [L8] the coordinate paths agree on $(a,b)$, so $\tilde{\gamma }(s)=x^{-1}(s,z_0)$ for $s\in (1\mathrm{/}2,b)$. But for $s\in (1,b)$ the point $(s,z_0)$ does not lie in $x(M)=(0,1)\times {\mathbb{R}}^3$, so $\tilde{\gamma }$ is not defined there — a contradiction. The same argument at the past end excludes $a<0$. Hence no geodesic with these data extends the parameter interval beyond $(0,1)$. If $\gamma (s)$ converged to a point $q\in M$ as $s\to 1$, then continuity of the chart coordinate $x^0$ on $M$ would give $1={\lim }_{s\to 1}{x}^0(\gamma (s))=x^0(q)<1$, a contradiction; hence $\gamma$ is future-inextendible in the sense of clause (d) of Causal curves, chronological and causal reachability, and inextendibility, and its affine parameter is bounded above by $1$. Thus $(M,g,N)$ admits an inextendible future timelike geodesic with bounded affine domain: it is timelike geodesically incomplete. This proves (b).

(Range of an inextendible causal curve.) Let $u:I\to M$ be an inextendible causal curve (future-inextendible and past-inextendible). Suppose first that $u$ is future-inextendible, and put $T:=\sup (x^0\circ u)(I)\le 1$, finite because $x^0<1$ on $M$. If $T<1$, then $x^0\circ u$ is increasing with limit $T$, and by the Lipschitz bound of step 2.1 the spatial curve $z\circ u$ is Cauchy as $t\to \sup I$: for $t,t^{\mathrm{\prime }}$ close to $\sup I$ one has $\mathrm{\mid }z(u(t^{\mathrm{\prime }}))-z(u(t)){\mathrm{\mid }}_{\mathrm{e}}\le \mathrm{\mid }{x}^0(u(t^{\mathrm{\prime }}))-x^0(u(t))\mathrm{\mid }\to 0$. By [L9] it converges to some $z_{\mathrm{\infty }}\in {\mathbb{R}}^3$; hence $u(t)\to q:=x^{-1}(T,z_{\mathrm{\infty }})$ as $t\to \sup I$, and $q\in M$ because $T<1$ and $T\ge x^0(u(t_0))>0$ for any fixed $t_0\in I$. This point limit contradicts future-inextendibility directly by [L7]. Hence $T=1$. The past argument is the time reverse and gives $\inf (x^0\circ u)(I)=0$; so $x^0\circ u$ is a strictly increasing continuous function on the interval $I$ whose image has infimum $0$ and supremum $1$ and contains neither endpoint (all values lie in $(0,1)$). Therefore its image is exactly the interval $(0,1)$, and for every $c\in (0,1)$ there is exactly one $t\in I$ with $x^0(u(t))=c$: existence by surjectivity, uniqueness by strict monotonicity.

(Strong causality.) Let $p\in M$ and let $U\subseteq M$ be an open neighborhood of $p$. Choose $0<\epsilon <\min (x^0(p),1-x^0(p))$ so small that the open set $V:=\{q\in M:p-\epsilon e_0{\ll }_{M}q{\ll }_{M}p+\epsilon e_0\}$ is contained in $U$; this is possible because $V$ shrinks to $\{p\}$ as $\epsilon \downarrow 0$ (in the chart it is an open double cone around $p$ intersected with $M$). By the choice of $\epsilon$, the points $p\pm \epsilon e_0$ lie in $M$. Let $\gamma :[a,b]\to M$ be a future causal curve with $\gamma (a),\gamma (b)\in V$ and let $a<t<b$. Since $\gamma {\mathrm{\mid }}_{[a,t]}$ and $\gamma {\mathrm{\mid }}_{[t,b]}$ are future causal curves, step 2.2 gives $\gamma (t)-\gamma (a)\in \overline{C^{+}}$ and $\gamma (b)-\gamma (t)\in \overline{C^{+}}$; since $\gamma (a)\in V$, step 2.2 gives $\gamma (a)-(p-\epsilon e_0)\in C^{+}$, and by [L6] $\gamma (t)-(p-\epsilon e_0)=(\gamma (a)-(p-\epsilon e_0))+(\gamma (t)-\gamma (a))\in C^{+}$. Similarly $(p+\epsilon e_0)-\gamma (b)\in C^{+}$ and $(p+\epsilon e_0)-\gamma (t)=((p+\epsilon e_0)-\gamma (b))+(\gamma (b)-\gamma (t))\in C^{+}$. By the convexity and cone characterization of step 2.2, $p-\epsilon e_0{\ll }_M\gamma (t){\ll }_{M}p+\epsilon e_0$, that is $\gamma (t)\in V$. Hence no future causal curve leaves $V$ and later re-enters it, and $(M,g,N)$ is strongly causal.

(Compactness of the causal diamonds.) Let $p,q\in M$. By the cone characterization of step 2.2, $J_M^{+}(p)\cap J_M^{-}(q)=\{r\in M:r-p\in \overline{C^{+}},q-r\in \overline{C^{+}}\}\mathrm{.}$ The corresponding set $K:=\{r\in {\mathbb{R}}^4:r-p\in \overline{C^{+}},q-r\in \overline{C^{+}}\}$ in the chart is closed (the cone is closed) and bounded with $x^0(r)\in [x^0(p),x^0(q)]\subset (0,1)$; hence $K$ is a compact subset of ${\mathbb{R}}^4$ by [L9] and, being contained in the open box $x(M)$, it is a compact subset of $x(M)\cong M$. Therefore the diamond $x^{-1}(K)=J_M^{+}(p)\cap J_M^{-}(q)$ is compact in $M$.

(The Cauchy slice.) The set ${\mathrm{\Sigma }}_{1\mathrm{/}2}=x^{-1}(\{1\mathrm{/}2\}\times {\mathbb{R}}^3)$ has the smooth slice chart $y=(x^1,x^2,x^3,x^0-1\mathrm{/}2)$, with image ${\mathbb{R}}^3\times \{0\}$ inside $y(M)={\mathbb{R}}^3\times (-1\mathrm{/}2,1\mathrm{/}2)$; hence it is an embedded three-dimensional submanifold by Embedded submanifolds and slice charts; its tangent space is spanned by ${\mathrm{\partial }}_1,{\mathrm{\partial }}_2,{\mathrm{\partial }}_3$ with $g({\mathrm{\partial }}_i,{\mathrm{\partial }}_j)={\delta }_{ij}>0$, so it is spacelike. By step 3.1 every inextendible causal curve of $(M,g)$ has $x^0$-range exactly $(0,1)$ and meets the level $x^0=1\mathrm{/}2$ in exactly one parameter value; by clause (c) of Spacelike hypersurfaces, acausal sets, Cauchy sets, domains of dependence and Cauchy horizons this says that ${\mathrm{\Sigma }}_{1\mathrm{/}2}$ is a Cauchy set, and with the first part a Cauchy hypersurface. This proves (c).

(Conclusion.) Part (a) is step 1.1, part (b) is step 2.3, part (c) is step 4.1. By steps 3.2 and 3.3 the model is strongly causal and all its causal diamonds are compact, which is global hyperbolicity in the sense of clause (c) of Chronology, causality, strong causality, global hyperbolicity and time functions; this is (d). The same model is timelike geodesically incomplete by (b), so in this class global hyperbolicity does not imply geodesic completeness.

(The converse non-implication.) A metric is geodesically complete here when every maximal affinely parametrized geodesic has domain $\mathbb{R}$. For the cylinder of [L10], fix any $p\in C$ and $w\in T_{p}C$, choose a lift $a\in {\mathbb{R}}^4$ with $q_C(a)=p$, and put $v=(d{q}_C{\mathrm{\mid }}_a{)}^{-1}w$, using the canonical coordinate identification $T_a{\mathbb{R}}^4={\mathbb{R}}^4$. The smooth curve $\beta (s)=q_C(a+sv)$ is defined for all $s\in \mathbb{R}$ and has initial data $(p,w)$. Near each $s_0$, take a neighborhood of $a+s_{0}v$ on which $q_C$ is a diffeomorphism; the quotient chart $x\circ (q_C{\mathrm{\mid }}_U{)}^{-1}$ gives constant metric components $\eta$ by [L10], so its Christoffel symbols vanish by [L3], and the coordinate path of $\beta$ is $a+sv$. It satisfies the geodesic equation [L8]. Thus $\beta$ is a geodesic on $\mathbb{R}$ through every initial datum, including $w=0$; local uniqueness [L8], continued through overlapping charts, makes any geodesic with these data its restriction. The cylinder is geodesically complete. It is not strongly causal: $\alpha (0)\mathrm{\ne }\alpha (L\mathrm{/}2)$, so choose an open neighborhood $U$ of $\alpha (0)$ excluding $\alpha (L\mathrm{/}2)$; for every $V$ with $\alpha (0)\in V\subseteq U$, the causal segment $\alpha {\mathrm{\mid }}_{[0,L]}$ has both endpoints in $V$ but leaves $V$. This violates the strong-causality definition, hence the cylinder is not globally hyperbolic. Together with step 5.1, this proves both non-implications asserted in the Remarks.