The Lorentzian Gauss lemma and the radial maximum of proper time

(Two-parameter family.) Let $r\mapsto v(r)$ be a smooth curve in $W_p$ and put $F(s,r):={\exp }_p(sv(r))$ for $s\in [0,1]$, so that $F(0,r)=p$ and $F(1,r)={\exp }_p(v(r))$. Set $T:={\mathrm{\partial }}_{s}F$, $J:={\mathrm{\partial }}_{r}F$. For each fixed $r$, $s\mapsto F(s,r)$ is the radial geodesic with tangent $v(r)$ at $s=0$; hence $D_{s}T=0$ and $g(T,T)=g_p(v(r),v(r))$ for all $s$ by [L1]. By [L2], $D_{s}J=D_{r}T$.

(The lift of a curve.) Let $\alpha$ be as in (b) and define $v(t):={\exp }_p^{-1}(\alpha (t))$ for $t\in (0,1]$, $v(0):=0$; by [L6], $v$ is piecewise $C^1$ with $\alpha ={\exp }_p\circ v$ and ${\alpha }^{\mathrm{\prime }}=d({\exp }_p{)}_v(v^{\mathrm{\prime }})$, and $v(t)=t{\alpha }^{\mathrm{\prime }}(0)+o(t)$. Write $Q(t):=g_p(v(t),v(t))$ and $R(t):=\sqrt{-Q(t)}$ on $\{Q<0\}$.

(First possible exit from the future timelike cone.) Let $T^{\ast }:=\sup \{t\in [0,1]:v(s)\text{ is future timelike for every }s\in (0,t]\}$; this is positive by [L6] and continuity, because $v(t)=t{\alpha }^{\mathrm{\prime }}(0)+o(t)$ and ${\alpha }^{\mathrm{\prime }}(0)$ is future timelike. The vector $v(t)$ is future timelike for every $0<t<T^{\ast }$. If $T^{\ast }<1$, openness of the future timelike cone rules out $v(T^{\ast })$ being timelike, since that would extend the defining interval past $T^{\ast }$. If $T^{\ast }=1$ and $v(1)$ is future timelike, the desired cone assertion already holds at the endpoint. In every remaining case, continuity and closure of the future timelike cone imply that $v(T^{\ast })$ is either a nonzero future causal vector or zero; after excluding the timelike case it is either future null (case A) or zero (case B). With signature $(-,+,+,+)$, $Q(T^{\ast })=g_p(v(T^{\ast }),v(T^{\ast }))\le 0$, with equality in case A; the case B value is also $0$.

(Gauss computation.) By [L2] and step 1.1, ${\mathrm{\partial }}_{s}g(T,J)=g(D_{s}T,J)+g(T,D_{s}J)=g(T,D_{r}T)=\frac{1}{2}{\mathrm{\partial }}_{r}g(T,T)=g_p(v(r),v^{\mathrm{\prime }}(r)),$ using symmetry of $g$ in the middle term and step 1.1 in the last. Since $J(0,r)={\mathrm{\partial }}_r{\exp }_p(0)=0$, integrating the constant-in-$s$ right-hand side from $s=0$ to $s=1$ gives $g_{{\exp }_p(v(r))}(d({\exp }_p{)}_{v(r)}(v(r)),d({\exp }_p{)}_{v(r)}(v^{\mathrm{\prime }}(r)))=g_p(v(r),v^{\mathrm{\prime }}(r))$. Every pair $(v,w)$ with $v\in W_p$ is of the form $(v(0),v^{\mathrm{\prime }}(0))$ for the smooth family $r\mapsto v+rw$, so the identity holds for all such pairs; this proves (a).

(Case A is impossible.) In case A, $Q<0$ on $(0,T^{\ast })$ and $Q(T^{\ast })=0$, so its one-sided left derivative satisfies $Q^{\mathrm{\prime }}(T^{\ast -})\ge 0$; if $T^{\ast }$ is a corner of the piecewise curve, use the left tangent and left derivative. On the smooth pieces the Gauss lemma of step 2.1 gives $Q^{\mathrm{\prime }}(t)=2g_p(v,v^{\mathrm{\prime }})=2g_{\alpha (t)}(T(t),{\alpha }^{\mathrm{\prime }}(t))$, where $T(t)=d({\exp }_p{)}_{v(t)}(v(t))$ is the radial tangent. For each $t<T^{\ast }$, the radial geodesic $s\mapsto {\exp }_p(sv(t))$ has constant timelike speed by [L1]; its tangent begins in the future timelike component and stays there by continuity and the two-component classification [L4]. Thus $T(t)$ is future timelike. Since $v(T^{\ast })$ is a nonzero future null vector, $d({\exp }_p{)}_{v(T^{\ast })}$ is invertible and $T(T^{\ast })={\lim }_{t\uparrow T^{\ast }}T(t)$ is nonzero, null by [L1], and future-directed by [L4]. Let $s:=\sqrt{-g({\alpha }^{\mathrm{\prime }}(T^{\ast -}),{\alpha }^{\mathrm{\prime }}(T^{\ast -}))}>0$ and $O:={\alpha }^{\mathrm{\prime }}(T^{\ast -})\mathrm{/}s$; then $O$ is an observer. By [L3] applied to observer $O$ and future null vector $T(T^{\ast })$, $-g(O,T(T^{\ast }))>0$. Therefore $Q^{\mathrm{\prime }}(T^{\ast -})=2g(T(T^{\ast }),{\alpha }^{\mathrm{\prime }}(T^{\ast -}))=2sg(T(T^{\ast }),O)<0$, contradicting $Q^{\mathrm{\prime }}(T^{\ast -})\ge 0$. The argument also applies when $T^{\ast }=1$, using the left endpoint tangent of the final piece.

(Case B is impossible.) In case B, $v(t)$ is future timelike for $t<T^{\ast }$ and $v(T^{\ast })=0$, so $R>0$ on $(0,T^{\ast })$ and $R(t)\to 0$ as $t\uparrow T^{\ast }$. On each smooth piece, by [L5] and the Gauss lemma, $R^{\mathrm{\prime }}=-g_p(v,v^{\mathrm{\prime }})\mathrm{/}R=-g_{\alpha }(\widehat{T},{\alpha }^{\mathrm{\prime }})$, where $\widehat{T}:=T\mathrm{/}R$ is future unit timelike by [L1] and the future-directedness of $v$. Put $s:=\sqrt{-g({\alpha }^{\mathrm{\prime }},{\alpha }^{\mathrm{\prime }})}>0$ and $O:={\alpha }^{\mathrm{\prime }}\mathrm{/}s$, a future unit timelike observer. Applying claim 2 of [L3] to observers $\widehat{T},O$ gives $-g(\widehat{T},O)\ge 1$, hence $R^{\mathrm{\prime }}=-g(\widehat{T},{\alpha }^{\mathrm{\prime }})\ge s>0$. Thus $R$ is strictly increasing on each smooth piece of $(0,T^{\ast })$ and, by continuity across the finitely many corners, $R(t)>R(\delta )$ whenever $0<\delta <t<T^{\ast }$. Taking $t\uparrow T^{\ast }$ yields $R(T^{\ast })\ge R(\delta )>0$, contradicting $R(T^{\ast })=0$. This argument also applies when $T^{\ast }=1$.

(Length inequality.) By steps 1.2, 1.3, 3.1 and 3.2, $v(t)$ is future timelike for every $t\in (0,1]$, so $R>0$ on $(0,1]$. On each smooth piece put $s:=\sqrt{-g({\alpha }^{\mathrm{\prime }},{\alpha }^{\mathrm{\prime }})}>0$, $O:={\alpha }^{\mathrm{\prime }}\mathrm{/}s$, and $\widehat{T}:=T\mathrm{/}R$; both $O$ and $\widehat{T}$ are observers. Claim 2 of [L3] gives $-g(\widehat{T},O)\ge 1$, hence $R^{\mathrm{\prime }}=-g(\widehat{T},{\alpha }^{\mathrm{\prime }})\ge s$. Integrating over the smooth pieces from $\delta$ to $1$ and adding gives $R(1)-R(\delta )\ge {\int }_{\delta }^{1}s(t)dt$. Letting $\delta \downarrow 0$, continuity of $v$ gives $R(\delta )\to 0$, and the integral over $(0,\delta )$ tends to $0$ because $s$ is bounded on the finitely many pieces with finite one-sided limits. Therefore $L_g(\alpha )\le R(1)=\sqrt{-g_p(v_q,v_q)}$. Dividing by $c$ and using Proper time is well defined and reparametrization invariant; the four-velocity is its proper-time tangent gives the proper-time statement.

(Equality case.) If equality holds in step 4.1, the continuous nonnegative function $R^{\mathrm{\prime }}-s$ on each smooth open piece has zero integral over $(0,1)$, so it vanishes identically there. By the equality case of claim 2 in [L3], applied to the observers $\widehat{T}=T\mathrm{/}R$ and $O={\alpha }^{\mathrm{\prime }}\mathrm{/}s$, one has $O=\widehat{T}$ and hence ${\alpha }^{\mathrm{\prime }}=s\widehat{T}$ on every smooth piece. Since $d({\exp }_p{)}_v$ is invertible for $v\in W_p$, ${\alpha }^{\mathrm{\prime }}=d({\exp }_p{)}_v(v^{\mathrm{\prime }})$ and $T=d({\exp }_p{)}_v(v)$ imply $v^{\mathrm{\prime }}=(s\mathrm{/}R)v$. Also $R^{\mathrm{\prime }}=s$ on each piece, so $(v\mathrm{/}R{)}^{\mathrm{\prime }}=v^{\mathrm{\prime }}\mathrm{/}R-v{R}^{\mathrm{\prime }}\mathrm{/}{R}^2=0$ there; continuity across the finitely many corners makes $v\mathrm{/}R$ the same constant throughout $(0,1]$. Put $v_0:=v(1)=v_q\in W_p$ and $\rho (t):=R(t)\mathrm{/}R(1)$ for $t>0$, with $\rho (0):=0$. Then $\rho (1)=1$, $v(t)=\rho (t)v_0$, and $\rho$ is piecewise $C^1$ and strictly increasing because $R^{\mathrm{\prime }}=s>0$ on every smooth piece. At $t=0$, put $a:={\alpha }^{\mathrm{\prime }}(0)$. Since ${\exp }_p^{-1}$ is smooth with $d({\exp }_p^{-1}{)}_p=\mathrm{id}$ and the first piece of $\alpha$ is $C^1$ up to $0$, $v^{\mathrm{\prime }}(t)\to a$ from the right; also $v(t)=ta+o(t)$ by [L6]. Thus $R(t)\mathrm{/}t\to \sqrt{-g_p(a,a)}>0$ and $R^{\mathrm{\prime }}(t)=-g_p(v,v^{\mathrm{\prime }})\mathrm{/}R\to \sqrt{-g_p(a,a)}$, so $\rho$ and its derivative extend continuously from the right to $0$. Consequently $\alpha (t)={\exp }_p(\rho (t)v_0)$, which is the equality form in (b). Conversely, for any equality-form curve from the Statement, ${\alpha }^{\mathrm{\prime }}={\rho }^{\mathrm{\prime }}d({\exp }_p{)}_{\rho v_0}(v_0)$ has norm $\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }\sqrt{-g_p(v_0,v_0)}$; its length is $\rho (1)\sqrt{-g_p(v_0,v_0)}=\sqrt{-g_p(v_q,v_q)}$, so equality holds. This proves (b) and completes the proof.