[proofplan] We prove a biconditional between the geodesic equation $\nabla_{\dot\gamma}\dot\gamma = 0$ and the local length-minimisation property. The forward direction is direct: if $\gamma$ is a unit-speed geodesic and $t \in [0, 1)$, then the segment $\gamma|_{[t, t+\delta]}$ for $\delta$ less than the injectivity radius at $\gamma(t)$ has length less than the radius and lies in a normal-coordinate ball; by [Geodesics Minimize Length Locally](/theorems/2720) part (1), this segment is the unique length-minimiser to its endpoint, so $d(\gamma(t), \gamma(t + \delta)) = \operatorname{length}(\gamma|_{[t, t+\delta]})$. The reverse direction proceeds via [Minimal Geodesics Are Smooth Geodesics](/theorems/2721): the local minimisation hypothesis says $\gamma|_{[t, t+\delta]}$ is a constant-speed minimal curve on each small interval, hence smooth and a geodesic on each such interval, and these patch to give the geodesic equation on $[0, 1]$. Both directions reduce the global statement on $[0, 1]$ to local arguments via covering and patching. [/proofplan] [step:Forward direction setup: assume $\gamma$ is a unit-speed geodesic on $[0, 1]$ and prepare local minimisation] Suppose $\gamma$ is a $C^2$ curve on $[0, 1]$ with $\nabla_{\dot\gamma} \dot\gamma = 0$ (the geodesic equation) and $|\dot\gamma|$ constant. Reduction to unit speed: if $|\dot\gamma| = c \ge 0$, the cases $c = 0$ (constant curve) and $c = 1$ (unit speed) suffice to handle the general case by linear reparametrisation. The constant case ($c = 0$) reduces to: every interval has length $0 = d(\gamma(t), \gamma(t+\delta))$, since $\gamma$ is constant. For $c > 0$, the affine reparametrisation $\bar\gamma(s) := \gamma(s/c)$ has unit speed and is a geodesic (geodesic equation invariant under affine reparametrisation), and the local minimisation property is preserved under reparametrisation. So it suffices to prove the implication for unit-speed $\gamma$. Henceforth assume $|\dot\gamma| \equiv 1$ on $[0, 1]$. Length on $[t, t+\delta]$ is then $\delta$ for any $\delta > 0$ with $t + \delta \in [0, 1]$. [/step] [step:Choose a normal-coordinate ball at $\gamma(t)$ and a small $\delta > 0$ on which $\gamma|_{[t, t+\delta]}$ is contained in the ball] Fix $t \in [0, 1)$. Let $p := \gamma(t) \in M$. By [Exponential Map as a Local Diffeomorphism](/theorems/2712), there exists $\varepsilon = \varepsilon(p) > 0$ such that \begin{align*} \exp_p\big|_{B(0, \varepsilon)}: B(0, \varepsilon) \subseteq T_pM \to U \subseteq M \end{align*} is a diffeomorphism onto an open neighbourhood $U$ of $p$. By continuity of $\gamma$ and openness of $U$, choose $\delta_1 > 0$ with $\gamma([t, t + \delta_1] \cap [0, 1]) \subseteq U$. Set \begin{align*} \delta := \min\bigl(\delta_1, \, \tfrac{1}{2}\varepsilon, \, 1 - t\bigr) > 0. \end{align*} Then $[t, t + \delta] \subseteq [0, 1]$, $\gamma([t, t + \delta]) \subseteq U$, and $\delta \le \tfrac{1}{2}\varepsilon < \varepsilon$. [/step] [step:Apply the local minimisation theorem to identify $\gamma|_{[t, t+\delta]}$ as the unique length-minimizer] The image $\gamma(t + \delta)$ lies in $U$. Set $q := \gamma(t + \delta) \in U$, and let $a := (\exp_p\big|_{B(0, \varepsilon)})^{-1}(q) \in B(0, \varepsilon)$. The unit-speed geodesic $\gamma|_{[t, t+\delta]}$ has $\gamma(t) = p$ and $\dot\gamma(t) \in T_p M$ with $|\dot\gamma(t)| = 1$. By the uniqueness of geodesics with prescribed initial data, $\gamma$ on $[t, t + \delta]$ coincides with the radial geodesic \begin{align*} [t, t + \delta] &\to M \\ s &\mapsto \exp_p((s - t)\, \dot\gamma(t)) \end{align*} on its domain of definition. Setting $s = t + \delta$, $q = \exp_p(\delta\, \dot\gamma(t))$, so $a = \delta\, \dot\gamma(t)$ and $|a| = \delta < \varepsilon$. Hence $\gamma|_{[t, t+\delta]}$, after the affine reparametrisation $u = (s - t)/\delta$ to $u \in [0, 1]$, becomes the radial geodesic $u \mapsto \exp_p(u\, a)$ on $[0, 1]$. By [Geodesics Minimize Length Locally](/theorems/2720) part (1), this radial geodesic is the unique length-minimizing curve from $p$ to $q$, with length $|a| = \delta$. So \begin{align*} d(p, q) = \delta = \operatorname{length}(\gamma|_{[t, t+\delta]}), \end{align*} proving the forward direction. [guided] We have a normal-coordinate ball $\exp_p: B(0, \varepsilon) \to U$ at $p = \gamma(t)$ from the previous step, and the segment $\gamma|_{[t, t+\delta]}$ is contained in $U$ with $\delta < \varepsilon$. The strategy: identify this segment, after a parameter rescaling, with a radial geodesic emanating from $0 \in T_pM$, then invoke [Geodesics Minimize Length Locally](/theorems/2720) to conclude that it minimises length and therefore realises the distance from $p$ to $\gamma(t + \delta)$. Set $q := \gamma(t + \delta) \in U$. Since $\exp_p|_{B(0, \varepsilon)}$ is a diffeomorphism onto $U$, $q$ has a unique preimage in $B(0, \varepsilon)$; we name it \begin{align*} a := \bigl(\exp_p\big|_{B(0, \varepsilon)}\bigr)^{-1}(q) \in B(0, \varepsilon). \end{align*} What is this preimage explicitly? We claim $a = \delta \, \dot\gamma(t)$. To see this, we compare $\gamma$ on $[t, t+\delta]$ with the radial geodesic out of $p$ in the direction $\dot\gamma(t)$. Both curves satisfy the geodesic ODE $\nabla_{\dot c}\dot c = 0$, and both have the same initial data at $s = t$: position $p$ and velocity $\dot\gamma(t)$. By uniqueness of solutions to the geodesic ODE with prescribed initial point and velocity, the two curves coincide on the intersection of their domains. Hence \begin{align*} [t, t + \delta] &\to M \\ s &\mapsto \exp_p\bigl((s - t)\, \dot\gamma(t)\bigr) \end{align*} agrees with $\gamma$ on $[t, t + \delta]$. Setting $s = t + \delta$ gives $q = \exp_p(\delta\, \dot\gamma(t))$, so $a = \delta\, \dot\gamma(t)$ and $|a| = \delta \cdot |\dot\gamma(t)| = \delta < \varepsilon$ (using unit speed). Now reparametrise affinely from $s \in [t, t+\delta]$ to $u = (s-t)/\delta \in [0, 1]$. Under this reparametrisation, the segment becomes the radial geodesic $u \mapsto \exp_p(u\, a)$ on $[0, 1]$, with $|a| = \delta < \varepsilon$. This is exactly the setup of part (1) of the local minimisation theorem. Why did we arrange $\delta \le \varepsilon/2$ in the previous step? Strictly we need $\delta < \varepsilon$ for the radial geodesic to stay in the ball where $\exp_p$ is a diffeomorphism; the factor of $\tfrac{1}{2}$ is just a concrete safe margin. Either bound suffices. By [Geodesics Minimize Length Locally](/theorems/2720) part (1), the radial geodesic $u \mapsto \exp_p(u\, a)$ on $[0, 1]$ is the unique length-minimising piecewise $C^1$ curve from $p$ to $q$, with length equal to $|a| = \delta$. Length is reparametrisation-invariant, so $\operatorname{length}(\gamma|_{[t, t+\delta]}) = \delta$ as well. Therefore \begin{align*} d(p, q) = \delta = \operatorname{length}(\gamma|_{[t, t+\delta]}), \end{align*} which is the forward direction of the theorem. [/guided] [/step] [step:Reverse direction setup: assume the local minimisation property and prepare to apply the smoothness theorem] Now suppose $\gamma: [0, 1] \to M$ is a $C^2$ curve with $|\dot\gamma|$ constant, and that for every $t \in [0, 1)$ there exists $\delta = \delta(t) > 0$ such that \begin{align*} d(\gamma(t), \gamma(t+\delta)) = \operatorname{length}(\gamma|_{[t, t+\delta]}). \tag{$\star$} \end{align*} We must show $\gamma$ satisfies $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[0, 1]$. Reduce to unit speed: as in Step 1, if $|\dot\gamma| = c$ with $c > 0$, replace $\gamma$ by its unit-speed reparametrisation $\bar\gamma$. The hypothesis ($\star$) is preserved (length and distance are reparametrisation-invariant), and the geodesic equation transforms as before. The case $c = 0$ reduces to the constant curve, which satisfies $\nabla_{\dot\gamma}\dot\gamma = 0$ identically. So assume $|\dot\gamma| \equiv 1$. Then $\operatorname{length}(\gamma|_{[t, t+\delta]}) = \delta$, and ($\star$) becomes \begin{align*} d(\gamma(t), \gamma(t+\delta)) = \delta. \end{align*} [/step] [step:Apply the smoothness theorem on each subinterval and patch] Fix $t_0 \in [0, 1)$ and let $\delta_0 := \delta(t_0) > 0$ be the value supplied by ($\star$). Then $\gamma|_{[t_0, t_0 + \delta_0]}$ satisfies \begin{align*} \operatorname{length}(\gamma|_{[t_0, t_0 + \delta_0]}) = \delta_0 = d(\gamma(t_0), \gamma(t_0 + \delta_0)). \end{align*} By definition, this means $\gamma|_{[t_0, t_0 + \delta_0]}$ is a length-minimising piecewise $C^1$ curve from $\gamma(t_0)$ to $\gamma(t_0 + \delta_0)$. Moreover, $\gamma|_{[t_0, t_0 + \delta_0]}$ has constant speed $1$ (inherited from the unit-speed parametrisation of $\gamma$). By [Minimal Geodesics Are Smooth Geodesics](/theorems/2721), every piecewise $C^1$ minimal geodesic of constant speed is a $C^\infty$ geodesic. Apply this to $\gamma|_{[t_0, t_0 + \delta_0]}$: we conclude $\gamma|_{[t_0, t_0 + \delta_0]}$ is $C^\infty$ on $[t_0, t_0 + \delta_0]$ and satisfies $\nabla_{\dot\gamma}\dot\gamma = 0$ on this subinterval. We now propagate this local conclusion to all of $[0, 1]$ by a continuation argument. Define \begin{align*} T := \sup\{t \in [0, 1] : \nabla_{\dot\gamma}\dot\gamma = 0 \text{ on } [0, t]\}. \end{align*} The set is non-empty: taking $t_0 = 0$ above, $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[0, \delta(0)]$, so $T \ge \delta(0) > 0$. Suppose for contradiction that $T < 1$. Apply the local conclusion at $t_0 = T$: there exists $\delta(T) > 0$ such that $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[T, T + \delta(T)]$. Combined with the equation on $[0, T)$ (which holds because $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[0, t]$ for every $t < T$ by definition of $T$ as a supremum), we get $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[0, T + \delta(T)]$, contradicting the definition of $T$. Hence $T = 1$, and $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[0, 1)$. It remains to extend the equation to $t = 1$. Since $\gamma \in C^2$, $\nabla_{\dot\gamma}\dot\gamma$ is continuous on $[0, 1]$. We have shown it vanishes on the dense subset $[0, 1)$. By continuity, it also vanishes at $t = 1$. Hence $\gamma$ satisfies $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[0, 1]$, proving the reverse direction. [guided] We need to upgrade the metric hypothesis ($\star$) — length equals distance on small intervals — into the differential-geometric conclusion that the geodesic equation $\nabla_{\dot\gamma}\dot\gamma = 0$ holds on all of $[0, 1]$. The translation between metric and differential is precisely what [Minimal Geodesics Are Smooth Geodesics](/theorems/2721) supplies, but only locally. The remaining work is a continuation argument that propagates the local conclusion across $[0, 1]$, and a separate continuity step at the right endpoint. **Local conclusion via the smoothness theorem.** Why does the constant-speed hypothesis matter? A length-minimising path can wander at variable speed (e.g. stalling near a point), which destroys the unit-speed parametrisation that the geodesic equation requires. The constant-speed hypothesis pins down the parametrisation up to affine rescaling — exactly what we need. Fix $t_0 \in [0, 1)$ and let $\delta_0 := \delta(t_0) > 0$ be supplied by hypothesis ($\star$). The segment $\gamma|_{[t_0, t_0 + \delta_0]}$ has constant speed $1$ (inherited from the global unit-speed parametrisation of $\gamma$) and satisfies \begin{align*} \operatorname{length}(\gamma|_{[t_0, t_0 + \delta_0]}) = \delta_0 = d(\gamma(t_0), \gamma(t_0 + \delta_0)), \end{align*} so it is a length-minimising piecewise $C^1$ curve of constant speed from $\gamma(t_0)$ to $\gamma(t_0 + \delta_0)$. We verify the hypotheses of theorem 2721: piecewise $C^1$ (yes, even $C^2$ globally), constant speed (yes, speed $1$), length-minimising between its endpoints (yes, by the displayed equation). All three hypotheses are met. By [Minimal Geodesics Are Smooth Geodesics](/theorems/2721), $\gamma|_{[t_0, t_0 + \delta_0]}$ is a $C^\infty$ geodesic, and in particular satisfies $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[t_0, t_0 + \delta_0]$. **Continuation argument: from local to global on $[0, 1)$.** The geodesic equation is local — $\nabla_{\dot\gamma}\dot\gamma|_t$ depends only on the values of $\gamma$ and $\dot\gamma$ in a neighbourhood of $t$ — so once we have the equation on a cover of $[0, 1)$ by intervals $[t_0, t_0 + \delta(t_0)]$, the equation holds on all of $[0, 1)$. We make this precise via a supremum argument. Define \begin{align*} T := \sup\{t \in [0, 1] : \nabla_{\dot\gamma}\dot\gamma = 0 \text{ on } [0, t]\}. \end{align*} The set is non-empty: applying the local conclusion at $t_0 = 0$ gives $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[0, \delta(0)]$, so $T \ge \delta(0) > 0$. Suppose for contradiction that $T < 1$. Then $T \in [0, 1)$, so we can apply the local conclusion at $t_0 = T$: there exists $\delta(T) > 0$ such that $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[T, T + \delta(T)]$. Combined with the equation on $[0, T)$ — which follows from $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[0, t]$ for every $t < T$, by definition of $T$ as a supremum — we obtain $\nabla_{\dot\gamma}\dot\gamma = 0$ on $[0, T + \delta(T)]$. But $T + \delta(T) > T$, contradicting the definition of $T$. Hence $T = 1$, and the geodesic equation holds on $[0, 1)$. **Extending to the right endpoint.** The hypothesis ($\star$) is only stated for $t \in [0, 1)$, so it does not directly give us the equation at $t = 1$. We close the gap by continuity. Since $\gamma \in C^2$, the velocity $\dot\gamma$ is $C^1$ and the covariant acceleration $\nabla_{\dot\gamma}\dot\gamma$ is continuous on $[0, 1]$. We have shown $\nabla_{\dot\gamma}\dot\gamma$ vanishes on $[0, 1)$, which is dense in $[0, 1]$. By continuity, \begin{align*} \nabla_{\dot\gamma}\dot\gamma|_{t=1} = \lim_{t \to 1^-} \nabla_{\dot\gamma}\dot\gamma|_t = 0. \end{align*} Hence $\nabla_{\dot\gamma}\dot\gamma = 0$ on all of $[0, 1]$, completing the reverse direction and the proof. [/guided] [/step]