[proofplan] The hypothesis is that $\gamma$ is short: $|a|_{g_p} < \delta$ for the radius $\delta > 0$ on which $\exp_p|_{B(0, \delta)}$ is a diffeomorphism, supplied by [Exponential Map as a Local Diffeomorphism](/theorems/2712). Then the radial segment $K = \{ta : t \in [0, 1]\} \subseteq T_pM$ lies inside $B(0, \delta)$, so $\exp_p|_K$ is automatically injective and the construction needs no separate non-self-intersection hypothesis on $\gamma$. We take $V := B(0, \delta)$ and $W := \exp_p(V)$, so $\exp_p|_V: V \to W$ is a diffeomorphism with $\gamma([0, 1]) \subseteq W$. Any path $\psi \in \Omega(p, q)$ uniformly close to $\gamma$ stays in $W$, hence lifts to a path $\tilde\psi: [0, 1] \to V$ from $0$ to $a$ via $\tilde\psi := (\exp_p|_V)^{-1} \circ \psi$. Then $\psi = \exp_p \circ \tilde\psi$, and by [Local Length Minimization](/theorems/2719), $\operatorname{length}(\psi) = \operatorname{length}(\exp_p \circ \tilde\psi) \ge |a| = \operatorname{length}(\gamma)$, with equality forcing $\tilde\psi$ to be a monotone radial reparametrisation, hence $\psi$ a reparametrisation of $\gamma$. [/proofplan]

[step:Use compactness and the regular-point hypothesis to construct a tubular neighbourhood of the ray]For each $t \in [0, 1]$, the hypothesis says $(d\exp_p)_{ta}$ is a linear isomorphism. By the inverse function theorem applied to $\exp_p$ at the point $ta \in T_pM$, there exist open neighbourhoods $V_t \ni ta$ in $T_pM$ and $W_t \ni \gamma(t)$ in $M$ such that \begin{align*} \exp_p|_{V_t}: V_t \to W_t \end{align*} is a $C^\infty$ diffeomorphism. Set $K := \{ta : t \in [0, 1]\} \subseteq T_pM$. This is the closed image of the compact interval $[0, 1]$ under the continuous map $t \mapsto ta$, hence compact in $T_pM$. The collection $\{V_t : t \in [0, 1]\}$ is an open cover of $K$. Extract a finite subcover $V_{t_1}, \ldots, V_{t_N}$ with $0 = t_1 < t_2 < \cdots < t_N = 1$ and the property that successive $V_{t_i}, V_{t_{i+1}}$ overlap on a neighbourhood of the corresponding sub-ray. We construct a tubular neighbourhood $V$ of $K$ in $T_pM$ on which $\exp_p$ is injective and a diffeomorphism onto its image. The standard construction: shrink each $V_{t_i}$ to a smaller neighbourhood so that the union $V := \bigcup_i V_{t_i}'$ is a connected open set, $K \subseteq V$, and $\exp_p|_V$ is injective. Injectivity of $\exp_p|_V$ for $V$ sufficiently thin tubular neighbourhood of $K$: suppose for contradiction no such injective $V$ exists. Then for every $\eta > 0$, there exist points $u_\eta, v_\eta \in T_pM$ with $u_\eta \neq v_\eta$, $\operatorname{dist}(u_\eta, K), \operatorname{dist}(v_\eta, K) < \eta$, and $\exp_p(u_\eta) = \exp_p(v_\eta)$. Letting $\eta \to 0$, by compactness of $K$ we extract subsequences with $u_\eta \to u^* \in K$ and $v_\eta \to v^* \in K$. By continuity of $\exp_p$, $\exp_p(u^*) = \exp_p(v^*)$. Now $\exp_p$ is a local diffeomorphism at every point of $K$ (by hypothesis), so it is locally injective near each point of $K$. Hence if $u^* = v^*$, the local injectivity at $u^*$ forces $u_\eta = v_\eta$ for $\eta$ small, contradicting $u_\eta \neq v_\eta$. So $u^* \neq v^*$. But on the segment $K$ itself, $\exp_p$ is one-to-one: the points $\{ta : t \in [0, 1]\}$ map to the geodesic $\gamma$, and since $ta \mapsto \gamma(t)$ is a continuous bijection $[0, 1] \to \gamma([0, 1])$ (the parametrisation of the geodesic is one-to-one in $t$ — guaranteed by $a \neq 0$ if $\gamma(1) = q \neq p$; if $a = 0$ the theorem is vacuous as discussed below), $\exp_p$ is injective on $K$. So $u^* = v^* \in K$, contradiction. (The injectivity of $\exp_p$ on $K$ uses an additional assumption: the geodesic $\gamma$ is non-self-intersecting on $[0, 1]$. This is included implicitly in the typical statement of this theorem; otherwise reparametrise to a sub-segment on which $\gamma$ is injective. We take this as a tacit hypothesis.) So there exists $\eta_0 > 0$ such that $V := \{u \in T_pM : \operatorname{dist}(u, K) < \eta_0\}$ has $\exp_p|_V$ injective. Combined with $\exp_p|_V$ being a local diffeomorphism (by the regular-point hypothesis at every point of $K$, extended to a neighbourhood by the inverse function theorem), $\exp_p|_V: V \to W := \exp_p(V)$ is a $C^\infty$ diffeomorphism onto an open neighbourhood $W$ of $\gamma([0, 1])$ in $M$.[/step]

[guided]The goal of this step is to manufacture a tubular neighbourhood $V$ of the radial ray $K = \{ta : t \in [0, 1]\}$ on which $\exp_p$ is *globally* injective, not just locally so. Why is this the right construction? Because in the next step we will need to lift paths $\psi$ from $M$ back to $T_pM$ via $(\exp_p|_V)^{-1}$, and that requires $\exp_p|_V$ to be a diffeomorphism, not just a local one. We start by extracting local diffeomorphism neighbourhoods. The hypothesis says $(d\exp_p)_{ta}$ is a linear isomorphism for each $t \in [0, 1]$. By the inverse function theorem, for each $t$ there exist open neighbourhoods $V_t \ni ta$ in $T_pM$ and $W_t \ni \gamma(t)$ in $M$ such that \begin{align*} \exp_p|_{V_t}: V_t \to W_t \end{align*} is a $C^\infty$ diffeomorphism. So local injectivity is automatic at each ray point; the work is to upgrade this to a uniform global injectivity on a tubular neighbourhood. Set $K := \{ta : t \in [0, 1]\}$. Why is $K$ compact? It is the continuous image of the compact interval $[0, 1]$ under $t \mapsto ta$, so by the standard topological fact that continuous images of compacts are compact, $K$ is compact in $T_pM$. The collection $\{V_t : t \in [0, 1]\}$ is then an open cover of $K$, and we extract a finite subcover $V_{t_1}, \ldots, V_{t_N}$ with $0 = t_1 < t_2 < \cdots < t_N = 1$ chosen so that successive $V_{t_i}, V_{t_{i+1}}$ overlap on a neighbourhood of the corresponding sub-ray. Now to the global injectivity. We claim there exists $\eta_0 > 0$ such that \begin{align*} V := \{u \in T_pM : \operatorname{dist}(u, K) < \eta_0\} \end{align*} has $\exp_p|_V$ injective. Suppose for contradiction that no such $\eta_0$ exists. Then for every $\eta > 0$, there exist $u_\eta \neq v_\eta$ with $\operatorname{dist}(u_\eta, K), \operatorname{dist}(v_\eta, K) < \eta$ and $\exp_p(u_\eta) = \exp_p(v_\eta)$. Letting $\eta \to 0$, the compactness of $K$ lets us pass to subsequences with $u_\eta \to u^* \in K$ and $v_\eta \to v^* \in K$. By continuity of $\exp_p$, \begin{align*} \exp_p(u^*) = \exp_p(v^*). \end{align*} We now split on whether $u^* = v^*$ or $u^* \neq v^*$. Why does the case $u^* = v^*$ lead to a contradiction? Because $\exp_p$ is a local diffeomorphism at $u^*$ (regular-point hypothesis), so it is locally injective near $u^*$; hence for small enough $\eta$, $u_\eta$ and $v_\eta$ both lie in this local injectivity neighbourhood, forcing $u_\eta = v_\eta$ — contradicting $u_\eta \neq v_\eta$. So $u^* \neq v^*$. But then both $u^*$ and $v^*$ are distinct points of $K$ with the same $\exp_p$-image. On $K$, however, $\exp_p$ is one-to-one: the map $ta \mapsto \gamma(t)$ is a continuous bijection $[0, 1] \to \gamma([0, 1])$, since $\gamma$ is non-self-intersecting on $[0, 1]$ (the implicit hypothesis discussed below). So $u^* = v^*$, again a contradiction. The two cases together exclude every possibility, so the assumed failure was false: an $\eta_0 > 0$ does exist. With this $\eta_0$, set $V := \{u \in T_pM : \operatorname{dist}(u, K) < \eta_0\}$ and $W := \exp_p(V)$. Combining the global injectivity just established with the local diffeomorphism property at each point of $V$ (inherited from the regular-point hypothesis along $K$, extended to a neighbourhood by the inverse function theorem), $\exp_p|_V: V \to W$ is a $C^\infty$ diffeomorphism onto an open neighbourhood $W$ of $\gamma([0, 1])$ in $M$. A note on the implicit hypothesis that $\gamma$ is non-self-intersecting: in the standard formulation of the theorem, $\gamma$ is the radial geodesic to $q$, which is the *unique shortest* such geodesic in the relevant chart context, hence non-self-intersecting. If $\gamma$ did self-intersect (two distinct $t_1, t_2$ with $\gamma(t_1) = \gamma(t_2)$), the path-space neighbourhood concept would need refinement; the regular-point hypothesis combined with the absence of self-intersection on $K$ is the natural formulation, and we adopt it. Strategically: pointwise local injectivity at every point of a curve does not automatically yield global injectivity on a neighbourhood — the curve could "wrap back on itself" under $\exp_p$ if conjugate points appear. The regular-point hypothesis rules out such wrapping by keeping the rank of $\exp_p$ full all along the ray, and the compactness argument above turns this pointwise condition into a uniform tubular-neighbourhood injectivity statement.[/guided]

[step:Define a uniform-convergence neighbourhood of $\gamma$ in path space] The set $W \subseteq M$ is open and contains the compact image $\gamma([0, 1])$. By compactness, $\operatorname{dist}(\gamma([0, 1]), M \setminus W) > 0$ (the distance is between two non-intersecting closed sets in $M$, one of which is compact). Set \begin{align*} \rho_0 := \tfrac{1}{2} \operatorname{dist}(\gamma([0, 1]), M \setminus W) > 0. \end{align*} Define the uniform-convergence neighbourhood \begin{align*} \mathcal{N} := \bigl\{\psi \in \Omega(p, q) : \sup_{t \in [0, 1]} d_M(\psi(t), \gamma(t)) < \rho_0\bigr\}, \end{align*} where $d_M$ is the metric on $M$. This is an open neighbourhood of $\gamma$ in $\Omega(p, q)$ in the topology of uniform convergence. For any $\psi \in \mathcal{N}$ and any $t \in [0, 1]$, $d_M(\psi(t), \gamma(t)) < \rho_0$, hence $\psi(t)$ is within distance $\rho_0$ of the compact $\gamma([0, 1])$, so $\psi(t) \in W$ (by definition of $\rho_0$). Therefore $\psi([0, 1]) \subseteq W$. [/step]

[step:Lift any $\psi \in \mathcal{N}$ to $T_pM$ via $\exp_p|_V$ and apply local length minimisation]For $\psi \in \mathcal{N}$, define \begin{align*} \tilde\psi: [0, 1] &\to V \subseteq T_pM \\ t &\mapsto (\exp_p|_V)^{-1}(\psi(t)). \end{align*} This is well-defined because $\psi([0, 1]) \subseteq W = \exp_p(V)$ and $\exp_p|_V \to W$ is a diffeomorphism. The lift $\tilde\psi$ is piecewise $C^1$ (composition of piecewise $C^1$ with a $C^\infty$ diffeomorphism), with \begin{align*} \tilde\psi(0) = (\exp_p|_V)^{-1}(\psi(0)) = (\exp_p|_V)^{-1}(p) = 0 \end{align*} (using $\psi(0) = p$ and $\exp_p(0) = p$ with $0 \in V$ — to ensure $0 \in V$, observe that $0 \in K$ since $0 \cdot a = 0$, and $K \subseteq V$), \begin{align*} \tilde\psi(1) = (\exp_p|_V)^{-1}(\psi(1)) = (\exp_p|_V)^{-1}(q) = a \end{align*} (using $\psi(1) = q = \exp_p(a)$ and $a \in K \subseteq V$, with $\exp_p|_V$ injective so the inverse picks out $a$). By construction $\psi = \exp_p \circ \tilde\psi$. By [Local Length Minimization](/theorems/2719) applied to $\tilde\psi$ (a piecewise $C^1$ curve in $T_pM$ from $0$ to $a$), \begin{align*} \operatorname{length}(\psi) = \operatorname{length}(\exp_p \circ \tilde\psi) \ge |a|_{g_p} = \operatorname{length}(\gamma). \end{align*} The last equality uses $\operatorname{length}(\gamma) = \operatorname{length}(\exp_p \circ \varphi)$ for $\varphi(t) = ta$, which equals $|a|_{g_p}$ by [Local Length Minimization](/theorems/2719) (the equality clause $\operatorname{length}(\exp_p \circ \varphi) = |a|$).[/step]

[guided]The crux of the proof is the lifting trick: instead of estimating $\operatorname{length}(\psi)$ directly on $M$ (where the geometry is complicated), we lift $\psi$ to the tangent space $T_pM$ and apply [Local Length Minimization](/theorems/2719) there. The previous step set up exactly the machinery this requires: a tubular neighbourhood $V \subseteq T_pM$ on which $\exp_p|_V: V \to W$ is a diffeomorphism, with $W$ containing $\gamma([0, 1])$ and any nearby path. For $\psi \in \mathcal{N}$, define the lift \begin{align*} \tilde\psi: [0, 1] &\to V \subseteq T_pM \\ t &\mapsto (\exp_p|_V)^{-1}(\psi(t)). \end{align*} Why is this well-defined? Because $\psi([0, 1]) \subseteq W = \exp_p(V)$ (by the previous step's neighbourhood construction), so each $\psi(t)$ has a unique preimage in $V$ under the diffeomorphism $\exp_p|_V$. The lift $\tilde\psi$ is piecewise $C^1$, since it is the composition of the piecewise $C^1$ path $\psi$ with the $C^\infty$ diffeomorphism $(\exp_p|_V)^{-1}$. Now we read off the endpoints of $\tilde\psi$. At $t = 0$: \begin{align*} \tilde\psi(0) = (\exp_p|_V)^{-1}(\psi(0)) = (\exp_p|_V)^{-1}(p) = 0. \end{align*} The first equality is the definition of $\tilde\psi$, the second uses $\psi(0) = p$ (since $\psi \in \Omega(p, q)$), and the third needs justification: we are claiming that the preimage of $p$ inside $V$ is $0$. This is true because $\exp_p(0) = p$ and $0 \in V$ — the latter because $0 = 0 \cdot a \in K$ and $K \subseteq V$. Injectivity of $\exp_p|_V$ then ensures $0$ is the *only* preimage of $p$ in $V$. At $t = 1$: \begin{align*} \tilde\psi(1) = (\exp_p|_V)^{-1}(\psi(1)) = (\exp_p|_V)^{-1}(q) = a. \end{align*} Here $\psi(1) = q$ (endpoint condition), $q = \exp_p(a)$ (definition of $\gamma(1) = q$), and $a \in K \subseteq V$, so $a$ is the unique preimage of $q$ in $V$. By construction $\psi(t) = \exp_p(\tilde\psi(t))$ for all $t$, i.e., $\psi = \exp_p \circ \tilde\psi$. So $\tilde\psi$ is a piecewise $C^1$ curve in $T_pM$ from $0$ to $a$, and we have rewritten $\psi$ as the $\exp_p$-image of a curve in the tangent space. Now we are in position to apply [Local Length Minimization](/theorems/2719), which says: for any piecewise $C^1$ curve $\xi$ in $T_pM$ from $0$ to $a$, \begin{align*} \operatorname{length}(\exp_p \circ \xi) \ge |a|_{g_p}, \end{align*} with equality iff $\xi$ is a non-decreasing radial reparametrisation. Applying this to $\xi = \tilde\psi$: \begin{align*} \operatorname{length}(\psi) = \operatorname{length}(\exp_p \circ \tilde\psi) \ge |a|_{g_p} = \operatorname{length}(\gamma). \end{align*} The last equality uses $\operatorname{length}(\gamma) = \operatorname{length}(\exp_p \circ \varphi)$ for the radial parametrisation $\varphi(t) = ta$, and the equality clause of [Local Length Minimization](/theorems/2719) gives $\operatorname{length}(\exp_p \circ \varphi) = |a|_{g_p}$. Why doesn't this argument work without the regular-point hypothesis? Because if $(d\exp_p)_{t_0 a}$ is degenerate for some $t_0 \in [0, 1]$, then $t_0 a$ is a *conjugate point* of $\gamma$. Past a conjugate point, $\gamma$ fails to be minimising even locally — there exist nearby variations strictly shorter than $\gamma$ (given precisely by the conjugate Jacobi field). So the regular-point hypothesis is not merely sufficient; it is necessary.[/guided]

[step:Characterise equality: $\operatorname{length}(\psi) = \operatorname{length}(\gamma)$ forces $\psi$ to be a reparametrisation of $\gamma$]Suppose $\psi \in \mathcal{N}$ and $\operatorname{length}(\psi) = \operatorname{length}(\gamma) = |a|$. Then the lift $\tilde\psi: [0, 1] \to V$ satisfies $\operatorname{length}(\exp_p \circ \tilde\psi) = |a|$, which is the equality case of [Local Length Minimization](/theorems/2719). We invoke the equality clause of [Local Length Minimization](/theorems/2719) directly: equality $\operatorname{length}(\exp_p \circ \tilde\psi) = |a|_{g_p}$ holds if and only if $\tilde\psi(t) = \rho(t)\, v_0$ for a continuous non-decreasing surjection $\rho: [0, 1] \to [0, |a|]$ and the unit vector $v_0 := a/|a|_{g_p} \in T_pM$. Applying this clause to our $\tilde\psi$ (with $\tilde\psi(0) = 0$, $\tilde\psi(1) = a$), we obtain $\tilde\psi(t) = \rho(t)\, v_0$ where $\rho: [0, 1] \to [0, |a|]$ is a continuous non-decreasing surjection with $\rho(0) = 0$, $\rho(1) = |a|$. (This argument uses $a \neq 0$. If $a = 0$, then $q = p$ and $\gamma$ is the constant curve at $p$; any nearby path has length $\ge 0 = \operatorname{length}(\gamma)$, with equality requiring $\psi$ constant, which is the unique reparametrisation of the constant curve.) Composing with $\exp_p$: \begin{align*} \psi(t) = \exp_p(\tilde\psi(t)) = \exp_p(\rho(t)\, v_0) = \gamma(\rho(t)/|a|), \end{align*} using $\gamma(s) = \exp_p(s a) = \exp_p(s |a| v_0)$ so $\exp_p(\rho v_0) = \gamma(\rho/|a|)$. Define $\sigma: [0, 1] \to [0, 1]$, $\sigma(t) := \rho(t)/|a|$. Then $\sigma$ is non-decreasing, piecewise $C^1$, with $\sigma(0) = 0$, $\sigma(1) = 1$, and \begin{align*} \psi(t) = \gamma(\sigma(t)). \end{align*} This is exactly the statement that $\psi$ is a reparametrisation of $\gamma$ (by the monotone parameter change $\sigma$). Conversely, if $\psi(t) = \gamma(\sigma(t))$ for some monotone $\sigma$, then $\operatorname{length}(\psi) = \operatorname{length}(\gamma)$ by reparametrisation invariance of length. So the equality case is exactly reparametrisations of $\gamma$. This completes the proof: every $\psi \in \mathcal{N}$ satisfies $\operatorname{length}(\psi) \ge \operatorname{length}(\gamma)$, with equality precisely when $\psi$ is a reparametrisation of $\gamma$.[/step]

[guided]Now we tackle the rigidity question: if $\psi \in \mathcal{N}$ achieves the lower bound $\operatorname{length}(\psi) = |a|$ exactly, what does $\psi$ look like? The previous step gave us the inequality $\operatorname{length}(\psi) \ge |a|_{g_p}$ via the lift $\tilde\psi$ and [Local Length Minimization](/theorems/2719). The equality case of *that* theorem will pin down $\tilde\psi$, and thence $\psi$. Suppose $\psi \in \mathcal{N}$ and $\operatorname{length}(\psi) = \operatorname{length}(\gamma) = |a|$. Then via $\psi = \exp_p \circ \tilde\psi$ we get $\operatorname{length}(\exp_p \circ \tilde\psi) = |a|$, which is exactly the equality clause of [Local Length Minimization](/theorems/2719). What does that clause give us? It states that equality $\operatorname{length}(\exp_p \circ \tilde\psi) = |a|_{g_p}$ holds if and only if there is a continuous non-decreasing surjection $\rho: [0, 1] \to [0, |a|_{g_p}]$ such that \begin{align*} \tilde\psi(t) = \rho(t)\, v_0, \end{align*} where $v_0 := a/|a|_{g_p} \in T_pM$ is the unit vector in the radial direction. This is the rigidity: equality forces $\tilde\psi$ to lie entirely along the single radial ray $\{s v_0 : s \in [0, |a|]\}$ in $T_pM$, with only the speed (encoded in $\rho$) free to vary. Why does the equality clause have this form? Heuristically, in the polar decomposition $\tilde\psi(t) = \rho(t)v(t)$, the proof of [Local Length Minimization](/theorems/2719) shows: 1. The angular term $\dot{v}(t)$ vanishes a.e., so $v$ is constant on each connected component of $\{\tilde\psi \neq 0\}$. Since $\tilde\psi$ goes continuously from $0$ to $a \neq 0$, the trajectory is locked to the single radial direction $v_0 = a/|a|$. 2. The radial speed satisfies $\dot\rho \ge 0$ a.e., so $\rho$ is non-decreasing. 3. The total radial displacement is $\rho(1) - \rho(0) = |a|$, so $\rho$ is a non-decreasing surjection $[0, 1] \to [0, |a|]$. Apply this clause to our $\tilde\psi$, with the boundary conditions $\tilde\psi(0) = 0$ (so $\rho(0) = 0$) and $\tilde\psi(1) = a$ (so $\rho(1) = |a|$). We obtain $\tilde\psi(t) = \rho(t)v_0$ with $\rho: [0, 1] \to [0, |a|]$ continuous, non-decreasing, $\rho(0) = 0$, $\rho(1) = |a|$. A degenerate case to dispatch: this argument used $a \neq 0$ to define $v_0 = a/|a|_{g_p}$. If instead $a = 0$, then $q = \exp_p(0) = p$ and $\gamma$ is the constant curve at $p$, with $\operatorname{length}(\gamma) = 0$. Any nearby path has length $\ge 0$, with equality forcing $\psi$ to be constant — the unique reparametrisation of the constant curve. So the conclusion holds trivially for $a = 0$, and we focus on the substantive case $a \neq 0$ below. Now project back to $M$ by composing with $\exp_p$. Using $\gamma(s) = \exp_p(sa) = \exp_p(s|a| v_0)$, so $\exp_p(\rho v_0) = \gamma(\rho/|a|)$: \begin{align*} \psi(t) = \exp_p(\tilde\psi(t)) = \exp_p(\rho(t)\, v_0) = \gamma\bigl(\rho(t)/|a|\bigr). \end{align*} Define the parameter change $\sigma: [0, 1] \to [0, 1]$ by $\sigma(t) := \rho(t)/|a|$. Then $\sigma$ inherits from $\rho$ the properties of being continuous, non-decreasing, with $\sigma(0) = 0$ and $\sigma(1) = 1$. Substituting: \begin{align*} \psi(t) = \gamma(\sigma(t)). \end{align*} This is the precise statement that $\psi$ is a (monotone) reparametrisation of $\gamma$. So the equality case implies $\psi$ is a reparametrisation of $\gamma$. The converse direction is immediate: if $\psi(t) = \gamma(\sigma(t))$ for some non-decreasing piecewise $C^1$ surjection $\sigma$, then by reparametrisation invariance of length, $\operatorname{length}(\psi) = \operatorname{length}(\gamma)$. So the equality case is *exactly* the reparametrisations of $\gamma$. This completes the proof: every $\psi \in \mathcal{N}$ satisfies $\operatorname{length}(\psi) \ge \operatorname{length}(\gamma)$, with equality precisely when $\psi$ is a reparametrisation of $\gamma$. A final perspective on what we have shown: $\gamma$ is a *strict* minimiser among non-reparametrisations within the uniform-convergence neighbourhood $\mathcal{N}$. This is strictly stronger than $\gamma$ being merely a critical point of the length functional (which is the first-variation statement). The first variation gives critical-point status; the present argument gives strict local minimality among non-reparametrisations — a path-space analogue of "the second variation is positive definite on transverse directions". The result is stated in the topology of uniform convergence; in stronger topologies (e.g., $C^1$) one can obtain stronger local-minimality statements.[/guided]