Local length comparison for a conjugate-free geodesic

Set-up and exponential parametrization of $\gamma$. [F1, F9, F10, F11, given] Put $X:=(b-a)v\in T_{p}M$ and $\psi (t):=(t-a)v$ for $t\in [a,b]$, so $\psi$ is affine, $\psi (a)=0$ and $\psi (b)=X$. By [F9] the maximal geodesic ${\gamma }_{p,v}$ has an open interval $I_{p,v}\ni 0$ and, by uniqueness in [F9], $s\mapsto {\gamma }_{p,v}(s)$ and $s\mapsto \gamma (a+s)$ agree for $s\in [0,b-a]$; in particular $t-a\in I_{p,v}$ for every $t\in [a,b]$. Hence by [F1] each $\psi (t)=(t-a)v$ lies in ${\mathcal{E}}_p$ and ${\exp }_p(\psi (t))={\gamma }_{p,v}(t-a)=\gamma (t)(a\le t\le b)\mathrm{.}$ Finally [F11] with [F10] gives $\mathrm{\mid }\dot{\gamma }\mathrm{\mid }\equiv \mathrm{\mid }v{\mathrm{\mid }}_g$, so $L_g(\gamma )=(b-a)\mathrm{\mid }v{\mathrm{\mid }}_g=\mathrm{\mid }X{\mathrm{\mid }}_g$.

The differential of ${\exp }_p$ is invertible along the segment. [F3, F4, F5, F17, step 1.1, given] Fix $t\in (a,b]$, put $\tau :=t-a>0$ and $\sigma (s):={\exp }_p(s\tau v)$ for $s\in [0,1]$, which by [F1] and step 1.1 equals $\gamma (a+s\tau )$. Let $w\in \ker d({\exp }_p{)}_{\tau v}$ and let $J$ be the Jacobi field along $\sigma$ with $J(0)=0$, $D_{s}J(0)=w$; by [F4] it exists uniquely, is nonzero when $w\mathrm{\ne }0$, and $J(1)=d({\exp }_p{)}_{\tau v}(w)=0$. If $w\mathrm{\ne }0$, [F5] realizes $J$ as the variation field of a smooth variation $F:(-\eta ,\eta )\times [0,1]\to M$ by affinely parametrized geodesics with $F(0,s)=\sigma (s)$, and $u\mapsto (u-a)\mathrm{/}\tau$ maps $[a,t]$ affinely onto $[0,1]$; hence $\tilde{F}(r,u):=F(r,(u-a)\mathrm{/}\tau )$ is a smooth variation of $\gamma {\mathrm{\mid }}_{[a,t]}$ whose longitudinal curves are affinely parametrized geodesics by [F5], and its variation field $\tilde{J}(u)=J((u-a)\mathrm{/}\tau )$ is a Jacobi field along $\gamma {\mathrm{\mid }}_{[a,t]}$ by [F5], with $\tilde{J}(a)=J(0)=0$, $\tilde{J}(t)=J(1)=0$ and $\tilde{J}\mathrm{\ne }0$. By [F5] that would make $\gamma (a)$ and $\gamma (t)$ conjugate along $\gamma {\mathrm{\mid }}_{[a,t]}$, contrary to the hypothesis. Hence $\ker d({\exp }_p{)}_{\tau v}=\{0\}$. Both $T_{\tau v}(T_{p}M)$ and $T_{\gamma (t)}M$ have the same finite dimension, so [F17] makes $d({\exp }_p{)}_{\tau v}$ invertible. At $t=a$ the differential $d({\exp }_p{)}_{0_p}$ is invertible by [F3].

Local diffeomorphism branches over a finite cover of the segment. [F1, F6, F8, step 2.1, given] For every $s\in [a,b]$, [F6] applies to the smooth map ${\exp }_p$ at $\psi (s)\in {\mathcal{E}}_p$, whose differential is invertible by step 2.1. Consider all pairs $(s,\rho )$ with $s\in [a,b]$, $\rho >0$, and ${\exp }_p$ a diffeomorphism on $B(\psi (s),4\rho )$ onto its open image. Such a pair exists for each $s$ by [F6], and the family of all smaller balls $B(\psi (s),\rho )$ from these pairs covers the compact set $\psi ([a,b])$ (compact as the continuous image of $[a,b]$ by [F8]). Choose a finite subcover with pairs $(s_j,{\rho }_j)$, put $x_j:=\psi (s_j)$ and $V_j:=B(x_j,4{\rho }_j)$, and write $O_j:={\exp }_p(V_j)$ and ${\beta }_j:=({\exp }_p{\mathrm{\mid }}_{V_j}{)}^{-1}:O_j\to V_j$. This makes only the finite subcover choice, not a choice of one inverse neighbourhood for every $s$.

A finite strip partition. [F7, F8, step 3.1, given] The sets $W_j:={\psi }^{-1}(B(x_j,{\rho }_j))$ form an open cover of the compact metric space $[a,b]$. By [F7] let $\delta >0$ be a Lebesgue number, and choose a finite partition $a=t_0<t_1<\cdots <t_N=b$ with $t_i-t_{i-1}<\delta \mathrm{/}(1+\mathrm{\mid }v{\mathrm{\mid }}_g)$; then $\mathrm{diam}([t_{i-1},t_i])=t_i-t_{i-1}<\delta$, so $[t_{i-1},t_i]$ lies in one $W_{j(i)}$. Thus for each $i$ there is $j(i)\in \{1,\dots ,m\}$ with $\psi ([t_{i-1},t_i])\subseteq B(x_{j(i)},{\rho }_{j(i)})\subseteq V_{j(i)}\mathrm{.}$ This uses only finitely many selections.

Closeness thresholds. [F8, step 3.1, step 4.1, given] For each $i$, the compact set $\gamma ([t_{i-1},t_i])={\exp }_p(\psi ([t_{i-1},t_i]))$ is contained in the open set $O_{j(i)}$ by step 4.1. The family of all half-radius balls $B(q,R\mathrm{/}2)$ with $q$ in this compact set, $R>0$ and $B(q,R)\subseteq O_{j(i)}$ covers it because $O_{j(i)}$ is open. Choose a finite subcover $B(\gamma (u_k),R_k\mathrm{/}2)$. Their corresponding full-radius balls $B(\gamma (u_k),R_k)$ lie in $O_{j(i)}$. Put ${\eta }_1:={\min }_i{\min }_k{R}_k\mathrm{/}2>0$. For each $i<N$, openness of $O_{j(i)}$ and continuity of ${\beta }_{j(i)}$ at $\gamma (t_i)$, with ${\beta }_{j(i)}(\gamma (t_i))=\psi (t_i)$, supply ${\mu }_i>0$ such that $B(\gamma (t_i),{\mu }_i)\subseteq O_{j(i)}$ and every $q$ in that ball satisfies $\mathrm{\mid }{\beta }_{j(i)}(q)-\psi (t_i){\mathrm{\mid }}_g<3{\rho }_{j(i+1)}$; put ${\eta }_2:={\min }_{i<N}{\mu }_i>0$ (the empty minimum being $+\mathrm{\infty }$, and ${\eta }_2=+\mathrm{\infty }$ when $N=1$). Set $\epsilon :=\min ({\eta }_1,{\eta }_2,1)>0$.

Pull-back of a nearby curve. [F6, F10, step 4.1, step 5.1, given] Let $\bar{\gamma }:[a,b]\to M$ be continuous and piecewise $C^1$ with $\bar{\gamma }(a)=\gamma (a)$, $\bar{\gamma }(b)=\gamma (b)$ and ${\sup }_t{d}_g(\bar{\gamma }(t),\gamma (t))<\epsilon$. For $t\in [t_{i-1},t_i]$, choose $k$ with $\gamma (t)\in B(\gamma (u_k),R_k\mathrm{/}2)$ as in step 5.1; then, by the triangle inequality for $d_g$ [F16], $d_g(\bar{\gamma }(t),\gamma (u_k))\le d_g(\bar{\gamma }(t),\gamma (t))+d_g(\gamma (t),\gamma (u_k))<\epsilon +R_k\mathrm{/}2\le R_k,$ so $\bar{\gamma }(t)\in O_{j(i)}$. Hence ${\phi }_i:={\beta }_{j(i)}\circ \bar{\gamma }$ is defined and continuous on $[t_{i-1},t_i]$ and piecewise $C^1$ there (as ${\beta }_{j(i)}$ is smooth and $\bar{\gamma }$ is piecewise $C^1$), with ${\exp }_p({\phi }_i(t))=\bar{\gamma }(t)$. At an interior subdivision point $t_i$, $i<N$: with $q:=\bar{\gamma }(t_i)$ one has $d_g(q,\gamma (t_i))<\epsilon \le {\mu }_i$, so $\mathrm{\mid }{\phi }_i(t_i)-\psi (t_i){\mathrm{\mid }}_g<3{\rho }_{j(i+1)}$; since $\psi (t_i)\in B(x_{j(i+1)},{\rho }_{j(i+1)})$, this gives ${\phi }_i(t_i)\in B(x_{j(i+1)},4{\rho }_{j(i+1)})=V_{j(i+1)}$, and so does ${\phi }_{i+1}(t_i)$; both are mapped to $q$ by ${\exp }_p$, which is injective on $V_{j(i+1)}$, hence ${\phi }_i(t_i)={\phi }_{i+1}(t_i)$. Therefore the formulas $\phi {\mathrm{\mid }}_{[t_{i-1},t_i]}:={\phi }_i$ define a single continuous piecewise $C^1$ curve $\phi :[a,b]\to T_{p}M$ with ${\exp }_p(\phi (t))=\bar{\gamma }(t)(a\le t\le b),\phi (t)\in V_{j(i)}\text{ for }t\in [t_{i-1},t_i]\mathrm{.}$ Moreover $\phi (a)={\beta }_{j(1)}(\gamma (a))={\beta }_{j(1)}({\exp }_p(0))=0$ and $\phi (b)={\beta }_{j(N)}(\gamma (b))={\beta }_{j(N)}({\exp }_p(X))=X$, because $0\in B(x_{j(1)},{\rho }_{j(1)})\subseteq V_{j(1)}$ and $X\in B(x_{j(N)},{\rho }_{j(N)})\subseteq V_{j(N)}$ by step 4.1.

Pointwise Gauss-lemma comparison. [F2, F12, step 6.1, given] Let $\sigma >0$, put $r:=\mathrm{\mid }\phi {\mathrm{\mid }}_g$ and $f_{\sigma }:=\sqrt{r^2+{\sigma }^2}=\sqrt{g_p(\phi ,\phi )+{\sigma }^2}$, which is piecewise $C^1$ on $[a,b]$ with $f_{\sigma }^{\mathrm{\prime }}=\frac{g_p(\phi ,{\phi }^{\mathrm{\prime }})}{f_{\sigma }}=\frac{g_p(\phi ,{\phi }^{\mathrm{\prime }})}{\sqrt{\mathrm{\mid }\phi {\mathrm{\mid }}_g^2+{\sigma }^2}}$ by [F12]. At every interior point $t$ of a smooth piece of $\phi$, [F12] applied to ${\exp }_p\circ \phi =\bar{\gamma }$ gives $\dot{\bar{\gamma }}(t)=d({\exp }_p{)}_{\phi (t)}({\phi }^{\mathrm{\prime }}(t))$. If $\phi (t)=0$, then $f_{\sigma }^{\mathrm{\prime }}(t)=0$ and the inequality below is trivial. Otherwise decompose ${\phi }^{\mathrm{\prime }}(t)=a\phi (t)+w$ with $a:=g_p(\phi (t),{\phi }^{\mathrm{\prime }}(t))\mathrm{/}\mathrm{\mid }\phi (t){\mathrm{\mid }}_g^2$ and $w:={\phi }^{\mathrm{\prime }}(t)-a\phi (t)$, so that $g_p(w,\phi (t))=0$ by positive definiteness and symmetry of the inner product $g_p$ [F16]. By [F2] the images of $w$ and of the radial direction $\phi (t)$ are orthogonal in $T_{\bar{\gamma }(t)}M$, and $\mathrm{\mid }d({\exp }_p{)}_{\phi (t)}(\phi (t)){\mathrm{\mid }}_g=\mathrm{\mid }\phi (t){\mathrm{\mid }}_g$; hence $\mathrm{\mid }\dot{\bar{\gamma }}(t){\mathrm{\mid }}_g^2=\mathrm{\mid }d({\exp }_p{)}_{\phi (t)}({\phi }^{\mathrm{\prime }}(t)){\mathrm{\mid }}_g^2=a^2\mathrm{\mid }\phi (t){\mathrm{\mid }}_g^2+\mathrm{\mid }d({\exp }_p{)}_{\phi (t)}(w){\mathrm{\mid }}_g^2\ge \frac{g_p(\phi (t),{\phi }^{\mathrm{\prime }}(t){)}^2}{\mathrm{\mid }\phi (t){\mathrm{\mid }}_g^2}\ge f_{\sigma }^{\mathrm{\prime }}(t{)}^2,$ because $\mathrm{\mid }\phi (t){\mathrm{\mid }}_g^2\le \mathrm{\mid }\phi (t){\mathrm{\mid }}_g^2+{\sigma }^2$. Thus $\mathrm{\mid }\dot{\bar{\gamma }}(t){\mathrm{\mid }}_g\ge \mathrm{\mid }{f}_{\sigma }^{\mathrm{\prime }}(t)\mathrm{\mid }(a<t<b)$ at all points where the two functions are differentiable, and by continuity the inequality extends to each closed smooth piece.

Lower bound for the length, and its sub-interval form. [F10, F11, F13, step 6.1, step 7.1, given] Take the common finite refinement of the strip partition $a=t_0<\cdots <t_N=b$ and an admissible piecewise-$C^1$ subdivision of $\bar{\gamma }$. On each refined closed subinterval, both $\phi$ and $f_{\sigma }$ are continuous and $C^1$ in the interior, with one-sided derivatives at its ends. Step 7.1, monotonicity [F13], and Newton--Leibniz [F13] there give the speed integral at least the absolute change of $f_{\sigma }$. Summing first within each original strip and using the triangle inequality gives ${\int }_{t_{i-1}}^{t_i}\mathrm{\mid }\dot{\bar{\gamma }}{\mathrm{\mid }}_g\ge {\int }_{t_{i-1}}^{t_i}\mathrm{\mid }{f}_{\sigma }^{\mathrm{\prime }}\mathrm{\mid }\ge \mathrm{\mid }{f}_{\sigma }(t_i)-f_{\sigma }(t_{i-1})\mathrm{\mid }\mathrm{.}$ Subdivision independence [F10] identifies the sum of speed integrals with $L_g(\bar{\gamma })$. Summing over $i$ and using the triangle inequality, $L_g(\bar{\gamma })\ge \mathrm{\mid }{f}_{\sigma }(b)-f_{\sigma }(a)\mathrm{\mid }=\sqrt{\mathrm{\mid }X{\mathrm{\mid }}_g^2+{\sigma }^2}-\sigma \mathrm{.}$ This holds for every $\sigma >0$; the right-hand side increases to $\mathrm{\mid }X{\mathrm{\mid }}_g$ as $\sigma \downarrow 0$ and never exceeds $\mathrm{\mid }X{\mathrm{\mid }}_g$, so $L_g(\bar{\gamma })\ge \mathrm{\mid }X{\mathrm{\mid }}_g=L_g(\gamma ),$ which is the asserted inequality. The same computation applied to a sub-interval $[u,w]\subseteq [t_{i-1},t_i]$ of a single strip (with $r=\mathrm{\mid }\phi {\mathrm{\mid }}_g$ and using $f_{\sigma }\to r$ as $\sigma \downarrow 0$) gives, for all $a\le u\le w\le b$ after splitting at the finitely many $t_i$ and adding, $L_g(\bar{\gamma }{\mathrm{\mid }}_{[u,w]})\ge \mathrm{\mid }r(w)-r(u)\mathrm{\mid }\mathrm{.}$

Equality case, I: the radius is bounded, nondecreasing, and has no internal excursion. [F10, step 6.1, step 8.1, given] Assume from now on that $L_g(\bar{\gamma })=\mathrm{\mid }X{\mathrm{\mid }}_g$, and recall $r(a)=\mathrm{\mid }\phi (a){\mathrm{\mid }}_g=0$ and $r(b)=\mathrm{\mid }X{\mathrm{\mid }}_g$ from step 6.1. (i) $r\le \mathrm{\mid }X{\mathrm{\mid }}_g$ on $[a,b]$: if $r(u)>\mathrm{\mid }X{\mathrm{\mid }}_g$, then by the sub-interval bound of step 8.1 and additivity [F10], $L_g(\bar{\gamma })=L_g(\bar{\gamma }{\mathrm{\mid }}_{[a,u]})+L_g(\bar{\gamma }{\mathrm{\mid }}_{[u,b]})\ge r(u)+(r(u)-\mathrm{\mid }X{\mathrm{\mid }}_g)>\mathrm{\mid }X{\mathrm{\mid }}_g,$ a contradiction. (ii) $r$ is nondecreasing: if $s<s^{\mathrm{\prime }}$ with $r(s)>r(s^{\mathrm{\prime }})$, then by the sub-interval bound of step 8.1, (i) and additivity, $L_g(\bar{\gamma })\ge r(s)+(r(s)-r(s^{\mathrm{\prime }}))+(\mathrm{\mid }X{\mathrm{\mid }}_g-r(s^{\mathrm{\prime }}))=\mathrm{\mid }X{\mathrm{\mid }}_g+2(r(s)-r(s^{\mathrm{\prime }}))>\mathrm{\mid }X{\mathrm{\mid }}_g,$ a contradiction. (iii) No component $(u,w)$ of the open set $\{r>0\}$ has $w<b$: for such a component $r(u)=r(w)=0$ (or $u=a$) and, picking $x\in (u,w)$ with $r(x)>0$, $L_g(\bar{\gamma })\ge 0+r(x)+r(x)+\mathrm{\mid }X{\mathrm{\mid }}_g>\mathrm{\mid }X{\mathrm{\mid }}_g$ by the sub-interval bound of step 8.1 applied to $[a,u]$, $[u,x]$, $[x,w]$, $[w,b]$ and additivity [F10], again a contradiction. Hence, if $X\mathrm{\ne }0$, the set $\{r>0\}$ is a single interval $(\alpha ,b]$ with $\alpha \in [a,b)$, and $r=0$ on $[a,\alpha ]$.

Equality case, II: the pull-back is radial. [F2, F13, F14, step 6.1, step 7.1, step 9.1, given] Assume $X\mathrm{\ne }0$, so by step 9.1(iii) $\{r>0\}=(\alpha ,b]$. On $(\alpha ,b]$ put $e:=\phi \mathrm{/}r$, a continuous piecewise $C^1$ map into the unit sphere of $(T_{p}M,g_p)$. Recall from step 7.1 the decomposition ${\phi }^{\mathrm{\prime }}=a\phi +w$ with $w={\phi }^{\mathrm{\prime }}-a\phi$ perpendicular to $\phi$ and $\mathrm{\mid }\dot{\bar{\gamma }}{\mathrm{\mid }}_g^2=(\frac{g_p(\phi ,{\phi }^{\mathrm{\prime }})}{\mathrm{\mid }\phi {\mathrm{\mid }}_g}{)}^2+\mathrm{\mid }d({\exp }_p{)}_{\phi }(w){\mathrm{\mid }}_g^2=\mathrm{\mid }{r}^{\mathrm{\prime }}{\mathrm{\mid }}^2+\mathrm{\mid }d({\exp }_p{)}_{\phi }(w){\mathrm{\mid }}_g^2\mathrm{.}$ Suppose $w(t_0)\mathrm{\ne }0$ at some interior point $t_0$ of a smooth piece contained in $(\alpha ,b]$; then $g_1:=\mathrm{\mid }d({\exp }_p{)}_{\phi (t_0)}(w(t_0)){\mathrm{\mid }}_g>0$ because $d({\exp }_p{)}_{\phi (t_0)}$ is injective (step 6.1 places $\phi (t_0)$ in $V_{j(i)}$, where ${\exp }_p$ is a diffeomorphism), and by continuity there is a closed interval $J\subseteq (\alpha ,b)\cap [t_{i-1},t_i]$ on which $\mathrm{\mid }d({\exp }_p{)}_{\phi }(w){\mathrm{\mid }}_g\ge g_1\mathrm{/}2$. There the continuous function $H:=\mathrm{\mid }\dot{\bar{\gamma }}{\mathrm{\mid }}_g-\mathrm{\mid }{r}^{\mathrm{\prime }}\mathrm{\mid }$ satisfies $H=\sqrt{\mathrm{\mid }{r}^{\mathrm{\prime }}{\mathrm{\mid }}^2+\mathrm{\mid }d({\exp }_p{)}_{\phi }(w){\mathrm{\mid }}_g^2}-\mathrm{\mid }{r}^{\mathrm{\prime }}\mathrm{\mid }>0,$ so ${\int }_{J}H>0$ by [F14]. On the other hand, $\mathrm{\mid }{f}_{\sigma }^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }{r}^{\mathrm{\prime }}\mathrm{\mid }$ wherever $r>0$, so for every $\sigma >0$, step 7.1 gives $0\le H\le \mathrm{\mid }\dot{\bar{\gamma }}{\mathrm{\mid }}_g-\mathrm{\mid }{f}_{\sigma }^{\mathrm{\prime }}\mathrm{\mid }$ on $J$ and $\mathrm{\mid }\dot{\bar{\gamma }}{\mathrm{\mid }}_g-\mathrm{\mid }{f}_{\sigma }^{\mathrm{\prime }}\mathrm{\mid }\ge 0$ on the whole strip, so [F13] yields ${\int }_{J}H\le {\int }_{t_{i-1}}^{t_i}(\mathrm{\mid }\dot{\bar{\gamma }}{\mathrm{\mid }}_g-\mathrm{\mid }{f}_{\sigma }^{\mathrm{\prime }}\mathrm{\mid })\le L_g(\bar{\gamma })-\mathrm{\mid }{f}_{\sigma }(b)-f_{\sigma }(a)\mathrm{\mid }=\mathrm{\mid }X{\mathrm{\mid }}_g-(\sqrt{\mathrm{\mid }X{\mathrm{\mid }}_g^2+{\sigma }^2}-\sigma )\le \sigma ,$ contradicting ${\int }_{J}H>0$ after $\sigma \downarrow 0$. Hence $w\equiv 0$ on $(\alpha ,b]$, so ${\phi }^{\mathrm{\prime }}=a\phi$ and $e^{\mathrm{\prime }}=0$ on the interiors of the pieces; by continuity $e$ is constant on $(\alpha ,b]$. Its value is $e_0=\frac{\phi (b)}{\mathrm{\mid }\phi (b){\mathrm{\mid }}_g}=\frac{X}{\mathrm{\mid }X{\mathrm{\mid }}_g}=\frac{v}{\mathrm{\mid }v{\mathrm{\mid }}_g},$ and $\phi (t)=r(t)e_0$ for $t\in (\alpha ,b]$, while $\phi \equiv 0$ on $[a,\alpha ]$. Consequently $r(t)=g_p(\phi (t),e_0)$ for every $t\in [a,b]$, so $r$ is piecewise $C^1$ with $r\ge 0$ nondecreasing by step 9.1(ii).

Equality case, III: the monotone reparametrization. [F1, F9, F11, step 1.1, step 9.1, step 10.1, given] Assume first $X\mathrm{\ne }0$ and define $\tau (t):=a+\frac{g_p(\phi (t),e_0)}{\mathrm{\mid }v{\mathrm{\mid }}_g}=a+\frac{r(t)}{\mathrm{\mid }v{\mathrm{\mid }}_g}(a\le t\le b)\mathrm{.}$ Then $\tau$ is continuous, piecewise $C^1$ (as $\phi$ is), nondecreasing (by step 10.1), and $\tau (t)\in [a,b]$ for all $t$ (by step 9.1(i) and $\mathrm{\mid }X{\mathrm{\mid }}_g=(b-a)\mathrm{\mid }v{\mathrm{\mid }}_g$), with $\tau (a)=a$ and $\tau (b)=a+\mathrm{\mid }X{\mathrm{\mid }}_g\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}_g=b$. Since $r(t)\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}_g\in [0,b-a]\subset I_{p,v}$ by step 1.1, [F1] gives ${\exp }_p(\phi (t))={\exp }_p(r(t)e_0)={\exp }_p\left(\frac{r(t)}{\mathrm{\mid }v{\mathrm{\mid }}_g}v\right)={\gamma }_{p,v}\left(\frac{r(t)}{\mathrm{\mid }v{\mathrm{\mid }}_g}\right)=\gamma (a+\frac{r(t)}{\mathrm{\mid }v{\mathrm{\mid }}_g})=\gamma (\tau (t)),$ so $\bar{\gamma }=\gamma \circ \tau$ as required. If $X=0$ (that is $v=0$ and $\gamma$ constant), then $\mathrm{\mid }X{\mathrm{\mid }}_g=0=L_g(\gamma )$ and $L_g(\bar{\gamma })=0$, so $\mathrm{\mid }\dot{\bar{\gamma }}{\mathrm{\mid }}_g\equiv 0$ on every smooth piece by [F14] and $\bar{\gamma }$ is constant with value $\bar{\gamma }(a)=p=\gamma$; thus $\bar{\gamma }=\gamma \circ \tau$ for every admissible $\tau$, for instance $\tau =\mathrm{id}$. This proves the forward direction of the equality clause in all cases.

Converse direction, and the boundary audit. [A1, F5, F15, step 11.1, given] Conversely, if $\bar{\gamma }=\gamma \circ \tau$ for a continuous nondecreasing surjection $\tau :[a,b]\to [a,b]$, piecewise $C^1$, with $\tau (a)=a$, $\tau (b)=b$, then [F15] applies to the piecewise $C^1$ curve $\gamma$ and gives $\gamma \circ \tau$ piecewise $C^1$ with $L_g(\gamma \circ \tau )=L_g(\gamma )$; note that $\bar{\gamma }$ automatically has the two fixed endpoint values $\gamma (\tau (a))=\gamma (a)$ and $\gamma (\tau (b))=\gamma (b)$. This proves the reverse direction of the equality clause. Boundaries and choice. Dimensions zero: $T_{p}M=\{0\}$, $\gamma$ is constant and the argument of step 11.1 with $X=0$ applies. Constant geodesics ($v=0$): $\gamma$ is constant and verbatim the same case applies; the equality clause holds because $\bar{\gamma }$ is then constant and every admissible $\tau$ factors it, which is [F15] applied to constant $\gamma$. Degenerate interval: the statement assumes $a<b$; a singleton interval carries no affine geodesic with $a<b$ and is not an instance. Endpoints: all derivatives at $a$ and $b$ are one-sided, the strips include the endpoints, and $\phi (a)=0$, $\phi (b)=X$ are used as values only. Choice: exactly the inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ is spent, through [F5] and the exponential-map suppliers; the finite cover, the finite partition, the finitely many threshold radii and the junction conditions are finite selections, and no field or direction is chosen from an infinite family.