Convergent causal ends are extendible, maximal geodesics have no limit, and Cauchy sets are acausal

(Endpoint absolute continuity and extension.) Let $\gamma$ and $x$ be as in (a). By [L9] choose a Lorentz-orthonormal basis $e_0,e_1,e_2,e_3$ at $x$, with $e_0$ future-directed. An invertible linear change of any chart centred at $x$ gives coordinates $z$ whose coordinate basis at $x$ is this basis, hence $g_{ab}(x)=\mathrm{diag}(-1,1,1,1)$. After shrinking to a coordinate ball there is $C<\mathrm{\infty }$ such that $\mathrm{\mid }v{\mathrm{\mid }}_{\mathrm{e}}\le C{v}^0$ for every nonzero future causal vector based in that ball. To prove this, otherwise there are base points $q_n\to x$ and coordinate unit future causal vectors $v_n$ with $v_n^0<1\mathrm{/}2$. Pass to a convergent subsequence of the unit vectors. Its unit limit $v$ satisfies $g_x(v,v)\le 0$ and $g_x(v,N_x)\le 0$; the latter is strict by [L9], so $v$ is future causal. Thus $v^0>0$ and $\mathrm{\mid }{v}_s\mathrm{\mid }\le v^0$ in the chosen orthonormal basis, giving $v^0\ge 1\mathrm{/}\sqrt{2}$, a contradiction. The coordinate $z^0$ is therefore strictly increasing on future causal curves in this ball, and supplies the local strict temporal coordinate. Choose $a<b$ so the tail of $\gamma$ stays in this chart, write $y=z\circ \gamma$, and extend $y(b)=z(x)$. Put $\tau =y^0$; on each compact subinterval of $[a,b)$, the AC chain rule and FTC give ${\tau }^{\mathrm{\prime }}>0$ and $\mathrm{\mid }{y}^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{e}}\le C{\tau }^{\mathrm{\prime }}$ almost everywhere and make $\tau$ strictly increasing. Its continuous endpoint extension is bounded and nondecreasing on $[a,b]$, so [L10] gives ${\tau }^{\mathrm{\prime }}\in L^1[a,b]$. Consequently each component of $y^{\mathrm{\prime }}$ is integrable on $[a,b]$. For $t<b$ the local FTC gives $y(t)=y(a)+{\int }_a^t{y}^{\mathrm{\prime }}$ componentwise; letting $t\uparrow b$ gives the same identity at $b$, since the integral of an L1 function is continuous and $y(t)\to y(b)$. Thus the endpoint extension of $y$ is AC on $[a,b]$ by [L10], and retains its original nonzero future causal velocity almost everywhere. Choose a future timelike $w\in T_{x}M$; [L4] supplies a short geodesic $\beta (s)={\exp }_x(sw)$ with $\beta (0)=x$, whose tangents are future timelike by [L3]. Concatenate the now AC closed tail with $\beta$ by [L7] and leave the earlier part of $\gamma$ unchanged. This defines the required future causal $\tilde{\gamma }$ on $I\cup [b,b+\epsilon )$ and proves (a). The past case reverses time.

(Set-up for the maximal geodesic.) Suppose, for contradiction, that $\gamma :={\gamma }_{p,v}$ with $v\mathrm{\ne }0$ converges to some $x\in M$ at one of its ends. Treat the upper end $b:=\sup I_{p,v}$, the past case being the time reverse of the same argument. Choose a chart $(U,y)$ around $x$ with $y(x)=0$; by convergence there is $t_1<b$ with $\gamma ([t_1,b))\subseteq K$ for a compact $K\subseteq U$. Put $K^{\mathrm{\prime }}=y(K)$, a compact subset of the open coordinate domain $y(U)\subseteq {\mathbb{R}}^4$. Write $y(t):=y(\gamma (t))$ and $w(t):=y^{\mathrm{\prime }}(t)$ for $t\in [t_1,b)$; the curve is smooth and $w(t)\mathrm{\ne }0$ for all $t$, by uniqueness, including when $v$ is null: if ${\gamma }^{\mathrm{\prime }}(t_{\ast })=0$ at any interior parameter, its geodesic initial data at $t_{\ast }$ agree with the constant geodesic at $\gamma (t_{\ast })$. Local uniqueness of the chart geodesic ODE [L2], [L6] makes them agree near $t_{\ast }$. The agreement set is closed and locally open in the connected geodesic interval, so they agree throughout, contradicting $v\mathrm{\ne }0$. Put $\rho (t):=\mathrm{\mid }w(t){\mathrm{\mid }}_{\mathrm{e}}>0$, $u(t):=w(t)\mathrm{/}\rho (t)\in S^3$, and $\sigma (t):={\int }_{t_1}^t\rho$, a strictly increasing $C^1$ function with ${\sigma }^{\mathrm{\prime }}=\rho >0$. Let ${\sigma }_{\mathrm{\infty }}:={\lim }_{t\to b}\sigma (t)\in (0,\mathrm{\infty }]$ and let $t(\sigma )$ be the inverse of $\sigma$ on $[0,{\sigma }_{\mathrm{\infty }})$. By [L2], $w^{\mathrm{\prime }}=-\mathrm{\Gamma }(y)(w,w)$ componentwise.

(A maximal solution whose graph stays in a compact set has no finite endpoint.) Let $O\subseteq {\mathbb{R}}^n$ be open, let $F:O\to {\mathbb{R}}^n$ be smooth, and let $z:[0,{\tau }_{\mathrm{\infty }})\to O$ be a maximal solution of $z^{\mathrm{\prime }}=F(z)$ whose graph is contained in a compact set $K\subseteq O$. If ${\tau }_{\mathrm{\infty }}<\mathrm{\infty }$, then $z$ extends to a solution on a strictly larger interval: indeed, by compactness of $K$ and [L5], finitely many cylinders cover $K$ and there is a common $h>0$ such that every initial condition in $K$ admits a solution on an interval of length $h$ contained in $O$; decreasing $h$ if necessary so that $h<{\tau }_{\mathrm{\infty }}$, and starting from $z({\tau }_{\mathrm{\infty }}-h\mathrm{/}2)$ and invoking uniqueness [L6] extends $z$ past ${\tau }_{\mathrm{\infty }}$, contradicting maximality. Hence ${\tau }_{\mathrm{\infty }}=\mathrm{\infty }$.

(The normalized equations in both parameters.) Put $Q(y,u):=\mathrm{\Gamma }(y)(u,u)$ and $a(y,u):=⟨u,Q(y,u){⟩}_{\mathrm{e}}$. In the affine parameter $t$, differentiating $\rho =\mathrm{\mid }w{\mathrm{\mid }}_{\mathrm{e}}$ and $u=w\mathrm{/}\rho$ gives $\frac{d\rho }{dt}=-{\rho }^{2}a(y,u)$ and $\frac{du}{dt}=-\rho P_{u}Q(y,u)$, where for unit $u$, $P_u=I-u\otimes u$ is the Euclidean orthogonal projection onto $u^{\perp }$. Since $d\sigma \mathrm{/}dt=\rho$, in the parameter $\sigma$ the fields $Y(\sigma )=y(t(\sigma ))$, $U(\sigma )=u(t(\sigma ))$, and $r(\sigma )=\rho (t(\sigma ))$ satisfy $\frac{dY}{d\sigma }=U$, $\frac{dU}{d\sigma }=-P_{U}Q(Y,U)$, and $\frac{dr}{d\sigma }=-ra(Y,U)$. For ODE continuation extend $P_U$ off the unit sphere by $I-U\otimes U$; this defines a smooth field on the open coordinate domain times ${\mathbb{R}}^4$. The unit sphere is invariant: $q=\mathrm{\mid }U{\mathrm{\mid }}_{\mathrm{e}}^2-1$ satisfies $q^{\mathrm{\prime }}=2a(Y,U)q$, so the zero initial value forces $q=0$. Thus the normalized solution remains unit-speed in $\sigma$.

(Finite normalized length forces a finite affine end and extension.) Suppose ${\sigma }_{\mathrm{\infty }}<\mathrm{\infty }$. On the original normalized interval $(Y,U)$ lies in the compact set $y(K)\times S^3$ inside the coordinate ODE domain. Its maximal ODE extension from the initial data therefore passes beyond ${\sigma }_{\mathrm{\infty }}$ by step 1.3 and uniqueness [L6]. Choose $\delta >0$ so this extension on $[0,{\sigma }_{\mathrm{\infty }}+\delta ]$ still has coordinate values in the chart; by the sphere invariance of step 2.1 its direction remains unit. The continuous coefficient $a(Y,U)$ is bounded there, by $C$ say. Its speed is $r(\sigma )=\rho (t_1)\exp (-{\int }_0^{\sigma }a(Y,U)ds)$, so $0<\rho (t_1)e^{-C({\sigma }_{\mathrm{\infty }}+\delta )}\le r(\sigma )\le \rho (t_1)e^{C({\sigma }_{\mathrm{\infty }}+\delta )}<\mathrm{\infty }$. In particular, on the original interval $t(\sigma )=t_1+{\int }_0^{\sigma }r(s{)}^{-1}ds$ has a finite limit $t_1+{\int }_0^{{\sigma }_{\mathrm{\infty }}}{r}^{-1}$; since this is the inverse of $\sigma (t)$, that limit equals $b$, so this case explicitly excludes $b=+\mathrm{\infty }$. On the extended interval this strictly increasing smooth affine-time reconstruction attains values greater than $b$. Its inverse gives $dy\mathrm{/}dt=rU$ and $d(rU)\mathrm{/}dt=rd(rU)\mathrm{/}d\sigma =-r^{2}Q(Y,U)$ by step 2.1, hence the coordinate geodesic equation [L2]. The reconstructed curve agrees with $\gamma$ on the old tail by uniqueness, glues with its earlier part, and extends the affine geodesic past $b$. This contradicts maximality [L1].

(Second case: infinite ${\sigma }_{\mathrm{\infty }}$.) Suppose ${\sigma }_{\mathrm{\infty }}=\mathrm{\infty }$. Then $t(\sigma )\to b$ as $\sigma \to \mathrm{\infty }$, and by convergence of $\gamma$ one has $Y(\sigma )=y(t(\sigma ))\to 0$ in the chart. By step 2.1, $\mathrm{\mid }{U}^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{e}}=\mathrm{\mid }{P}_U\mathrm{\Gamma }(Y)(U,U){\mathrm{\mid }}_{\mathrm{e}}\le C$ on the compact set $K^{\mathrm{\prime }}\times S^3$. Let $\sigma >0$ and $L>0$; the mean value bound [L8] gives ${\mid U(\sigma )-\frac{1}{L}{\int }_{\sigma }^{\sigma +L}U\mid }_{\mathrm{e}}\le \frac{CL}{2}$, while ${\mid \frac{1}{L}{\int }_{\sigma }^{\sigma +L}U\mid }_{\mathrm{e}}=\frac{1}{L}\mathrm{\mid }Y(\sigma +L)-Y(\sigma ){\mathrm{\mid }}_{\mathrm{e}}\le \frac{1}{L}(\mathrm{\mid }Y(\sigma +L){\mathrm{\mid }}_{\mathrm{e}}+\mathrm{\mid }Y(\sigma ){\mathrm{\mid }}_{\mathrm{e}})$. Choose $L$ with $CL<1$ and then $\sigma$ so large that the last expression is less than $\frac{1}{2}$; then $\mathrm{\mid }U(\sigma ){\mathrm{\mid }}_{\mathrm{e}}<\frac{1}{2}+\frac{1}{2}=1$, contradicting $\mathrm{\mid }U{\mathrm{\mid }}_{\mathrm{e}}=1$. Hence ${\sigma }_{\mathrm{\infty }}=\mathrm{\infty }$ is impossible.

(Conclusions.) Steps 3.1 and 3.2 show that a maximal geodesic with $v\mathrm{\ne }0$ cannot converge at either end, which is (b); the past end is the time reverse of the same argument. For (c), let $\gamma :[t_0,t_1]\to M$ be future causal. Choose a future timelike $w_1\in T_{\gamma (t_0)}M$ and a future timelike $w_2\in T_{\gamma (t_1)}M$. By [L1] there are maximal geodesics ${\gamma }_1:={\gamma }_{\gamma (t_0),w_1}$ and ${\gamma }_2:={\gamma }_{\gamma (t_1),w_2}$; both are future-directed and smooth on their domains by [L3], and they contain $\gamma (t_0)$ at parameter $0$ and $\gamma (t_1)$ at parameter $0$ respectively. Concatenating the restriction of ${\gamma }_1$ to $(\inf I_1,0]$, the curve $\gamma$ (after shifting the parameter), and the restriction of ${\gamma }_2$ to $[0,\sup I_2)$ gives a future causal curve $\widehat{\gamma }:J\to M$ by [L7], where $J$ is the corresponding interval, and there is a strictly increasing affine identification of $[t_0,t_1]$ with a subinterval of $J$ carrying $\gamma$ to $\widehat{\gamma }$. By (b) the maximal geodesics do not converge at their outer ends, so the two outer ends of $\widehat{\gamma }$ have no limits, at finite or infinite endpoints alike; this is exactly inextendibility in Causal curves, chronological and causal reachability, and inextendibility clause (d). If $p\mathrm{\ne }q$ and $p\le q$, a future causal curve from $p$ to $q$ exists by the definition of $\le$, and the construction applies to it; the past case is the time reverse. This proves (c).

(Cauchy sets are acausal.) Let $S$ be a Cauchy set in the sense of Spacelike hypersurfaces, acausal sets, Cauchy sets, domains of dependence and Cauchy horizons clause (c). Suppose $p,q\in S$ are distinct and $p\le q$. Choose a future causal curve $\gamma :[t_0,t_1]\to M$ with $\gamma (t_0)=p$ and $\gamma (t_1)=q$, and let $\widehat{\gamma }:J\to M$ and the affine identification $\rho$ be as in (c). Then $\widehat{\gamma }$ is an inextendible causal curve and $\rho (t_0)\mathrm{\ne }\rho (t_1)$, since $\rho$ is strictly increasing and $t_0<t_1$ while $\widehat{\gamma }(\rho (t_0))=p\mathrm{\ne }q=\widehat{\gamma }(\rho (t_1))$; thus the parameter set ${\widehat{\gamma }}^{-1}[S]$ contains the two distinct elements $\rho (t_0)$ and $\rho (t_1)$, contradicting the defining property of a Cauchy set. The relation $q\le p$ is excluded symmetrically by the past version of (c). Hence no two distinct points of $S$ are causally related, that is, $S$ is acausal; this is (d) and completes the proof.