[proofplan] Part 1 is a direct calculation: differentiate $E(\gamma_s) = \tfrac12\int_0^T g(\dot\gamma_s, \dot\gamma_s)\, d\mathcal{L}^1(t)$ under the integral sign, use metric compatibility of the Levi-Civita connection to convert $\partial_s g(\dot\gamma_s, \dot\gamma_s)$ into $2g(\nabla_s \dot\gamma_s, \dot\gamma_s)$, swap covariant derivatives via vanishing torsion ($\nabla_s \dot\gamma = \nabla_t Y$), and integrate by parts to push $\nabla_t$ off $Y$ onto $\dot\gamma$. Part 2 follows from Part 1: if the boundary term and the integral both vanish for all admissible $Y$, the fundamental lemma of the calculus of variations forces $\nabla_t \dot\gamma = 0$. Part 3 is a side-by-side computation: under the unit-speed normalisation of $\gamma_0$ and constant-speed assumption on $\gamma_s$, $E$ and $\ell$ have proportional first derivatives at $s = 0$. Part 4 follows by reparametrising a critical point of $\ell$ to constant speed and applying Part 2 to the resulting critical point of $E$. [/proofplan]

[step:Differentiate the energy integral and exchange differentiation with the integral]Let $H : [0, T] \times (-\varepsilon, \varepsilon) \to M$ be the smooth variation, $\gamma_s := H(\cdot, s)$, and $Y(t) := \frac{\partial H}{\partial s}(t, 0)$ the variational vector field. Both partial derivatives $\partial H/\partial t$ and $\partial H/\partial s$ are smooth vector fields along $H$. The energy is \begin{align*} E : (-\varepsilon, \varepsilon) &\to \mathbb{R} \\ s &\mapsto \frac{1}{2}\int_0^T g(\dot\gamma_s(t), \dot\gamma_s(t)) \, d\mathcal{L}^1(t). \end{align*} The integrand $(t, s) \mapsto g(\dot\gamma_s(t), \dot\gamma_s(t))$ is smooth on the compact rectangle $[0, T] \times [-\varepsilon/2, \varepsilon/2]$, so we may differentiate under the integral sign: \begin{align*} \frac{d}{ds} E(\gamma_s) = \frac{1}{2}\int_0^T \frac{\partial}{\partial s} g(\dot\gamma_s(t), \dot\gamma_s(t)) \, d\mathcal{L}^1(t). \end{align*}[/step]

[guided]We need to compute $\frac{d}{ds} E(\gamma_s)|_{s=0}$. To set up the calculation, recall that the variation supplies a smooth two-parameter map $H : [0, T] \times (-\varepsilon, \varepsilon) \to M$, with $\gamma_s := H(\cdot, s)$ and variational field $Y(t) := \frac{\partial H}{\partial s}(t, 0)$. The energy of the $s$-th variation curve is the kinetic-energy integral \begin{align*} E : (-\varepsilon, \varepsilon) &\to \mathbb{R} \\ s &\mapsto \frac{1}{2}\int_0^T g(\dot\gamma_s(t), \dot\gamma_s(t)) \, d\mathcal{L}^1(t). \end{align*} We want to differentiate this in $s$ and evaluate at $s = 0$. The natural move is to push the $s$-derivative through the integral sign — but this is a non-trivial step that requires justification. The general Leibniz rule for differentiation under the integral states \begin{align*} \frac{d}{ds}\int_0^T f(t, s) \, d\mathcal{L}^1(t) = \int_0^T \frac{\partial f}{\partial s}(t, s) \, d\mathcal{L}^1(t), \end{align*} provided that $f$ is continuous in both variables and $\partial f/\partial s$ is continuous in both variables on a compact rectangle containing the relevant slice. Here the relevant integrand is $f(t, s) = \tfrac12 g(\dot\gamma_s(t), \dot\gamma_s(t))$. Smoothness of $H$ implies smoothness of $\partial_t H$, hence smoothness of $g(\partial_t H, \partial_t H)$ on the compact rectangle $[0, T] \times [-\varepsilon/2, \varepsilon/2]$ — both hypotheses are satisfied. Note also that the interval of integration $[0, T]$ does not depend on $s$, so there are no boundary-motion terms. Applying Leibniz: \begin{align*} \frac{d}{ds} E(\gamma_s) = \frac{1}{2}\int_0^T \frac{\partial}{\partial s} g(\dot\gamma_s(t), \dot\gamma_s(t)) \, d\mathcal{L}^1(t). \end{align*} The remaining work is to compute the inner $s$-derivative, which is what the next step does using metric compatibility.[/guided]

[step:Use metric compatibility to expand $\partial_s g(\dot\gamma_s, \dot\gamma_s)$] Let $\nabla$ denote the Levi-Civita connection on $(M, g)$. Write $\nabla_s := \nabla_{\partial_s H}$ and $\nabla_t := \nabla_{\partial_t H}$ for covariant differentiation along the $s$- and $t$-coordinate curves of the variation $H$ — these are well-defined operators on vector fields along $H$ by the [Covariant Derivative Along a Curve](/theorems/2708) extended to maps from a rectangle. Metric compatibility of $\nabla$ states $X g(V, W) = g(\nabla_X V, W) + g(V, \nabla_X W)$. Applied with $X = \partial_s$ and $V = W = \dot\gamma_s = \partial_t H$: \begin{align*} \frac{\partial}{\partial s} g(\dot\gamma_s, \dot\gamma_s) = 2 g(\nabla_s \dot\gamma_s, \dot\gamma_s). \end{align*} [/step]

[step:Use the vanishing of torsion to swap $\nabla_s \dot\gamma_s$ with $\nabla_t Y$ at $s = 0$]For a smooth map $H : [0, T] \times (-\varepsilon, \varepsilon) \to M$, the symmetry-of-second-derivatives identity for the pullback covariant derivative on $H^* TM$ states \begin{align*} \nabla_s \partial_t H = \nabla_t \partial_s H. \end{align*} This is the correct identity: there is no bracket term, because the coordinate vector fields $\partial_s$ and $\partial_t$ on the parameter rectangle $[0,T] \times (-\varepsilon, \varepsilon)$ satisfy $[\partial_s, \partial_t] = 0$, and the Levi-Civita connection is torsion-free. Hence \begin{align*} \nabla_s \dot\gamma_s = \nabla_t \frac{\partial H}{\partial s}. \end{align*} Specialising to $s = 0$, where $\partial H/\partial s|_{s = 0} = Y(t)$: \begin{align*} \nabla_s \dot\gamma_s \big|_{s = 0} = \nabla_t Y(t). \end{align*}[/step]

[guided]After Step 2 we have $\frac{d}{ds} E(\gamma_s) = \int_0^T g(\nabla_s \dot\gamma_s, \dot\gamma_s)\, d\mathcal{L}^1(t)$. The expression $\nabla_s \dot\gamma_s$ is awkward: it differentiates the $t$-velocity in the $s$-direction. We want to convert it into something involving $\nabla_t Y$, because eventually we plan to integrate by parts in the $t$-variable to obtain a clean first variation formula. So the question is: can we swap $\nabla_s$ and $\nabla_t$ when applied to the variation? The answer comes from the torsion-freeness of the Levi-Civita connection. Recall that the torsion of a connection $\nabla$ is the tensor \begin{align*} T(X, Y) := \nabla_X Y - \nabla_Y X - [X, Y]. \end{align*} The Levi-Civita connection is the unique torsion-free metric-compatible connection on $(M, g)$, so $T \equiv 0$. Why does this help us? For a smooth two-parameter map $H : [0, T] \times (-\varepsilon, \varepsilon) \to M$, the coordinate fields $\partial_t H$ and $\partial_s H$ are sections of the pullback bundle $H^* TM$. The Lie bracket of the parameter-space vector fields $\partial_t$ and $\partial_s$ on the rectangle $[0,T] \times (-\varepsilon, \varepsilon)$ vanishes — this is just the equality of mixed partials $\partial_t \partial_s = \partial_s \partial_t$ on the flat parameter space. Pulling the torsion identity back to $H^* TM$ and using $[\partial_t, \partial_s] = 0$ gives the symmetry-of-second-derivatives identity for the pullback covariant derivative: \begin{align*} \nabla_s \partial_t H = \nabla_t \partial_s H. \end{align*} There is no bracket correction term precisely because the parameter-space bracket vanishes. Specialising the left side to $\partial_t H = \dot\gamma_s$ gives \begin{align*} \nabla_s \dot\gamma_s = \nabla_t \frac{\partial H}{\partial s}. \end{align*} This is the key swap: it converts the $s$-derivative of the velocity into a $t$-derivative of the variation, exactly what we need for the integration-by-parts step in Step 4. Now evaluate at $s = 0$. By definition of the variational field, $\partial H/\partial s|_{s = 0} = Y(t)$, so \begin{align*} \nabla_s \dot\gamma_s \big|_{s = 0} = \nabla_t Y(t), \end{align*} the covariant derivative of $Y$ along $\gamma$. Without torsion-freeness, the right-hand side would carry an extra $T(\partial_t H, \partial_s H)$ term and the entire computation would collapse — this is precisely the place in the proof where the torsion-free axiom of Levi-Civita is consumed.[/guided]

[step:Combine and integrate by parts]Combining Steps 1, 2, 3 and evaluating at $s = 0$: \begin{align*} \frac{d}{ds} E(\gamma_s)\Big|_{s = 0} = \int_0^T g\!\left(\nabla_t Y(t), \dot\gamma(t)\right) d\mathcal{L}^1(t). \end{align*} Apply the product rule via metric compatibility once more to the function $t \mapsto g(Y(t), \dot\gamma(t))$: \begin{align*} \frac{d}{dt} g(Y(t), \dot\gamma(t)) = g(\nabla_t Y(t), \dot\gamma(t)) + g(Y(t), \nabla_t \dot\gamma(t)). \end{align*} Solving for $g(\nabla_t Y, \dot\gamma)$: \begin{align*} g(\nabla_t Y(t), \dot\gamma(t)) = \frac{d}{dt} g(Y(t), \dot\gamma(t)) - g(Y(t), \nabla_t \dot\gamma(t)). \end{align*} Integrate over $[0, T]$ using the fundamental theorem of calculus on each smooth segment of the piecewise $C^1$ curve $\gamma$. Let $0 = t_0 < t_1 < \cdots < t_N = T$ be the breakpoints of $\gamma$. The variational field $Y$ is smooth on $[0,T]$, but the velocity $\dot\gamma$ may have one-sided jumps $\dot\gamma(t_i^+) - \dot\gamma(t_i^-)$ at each interior breakpoint $t_i$. Applying FTC on each segment $[t_{i-1}, t_i]$ and summing: \begin{align*} \int_0^T g(\nabla_t Y, \dot\gamma) \, d\mathcal{L}^1(t) = g(Y, \dot\gamma)\Big|_0^T - \sum_{i=1}^{N-1} g\!\left(Y(t_i),\, \dot\gamma(t_i^+) - \dot\gamma(t_i^-)\right) - \int_0^T g(Y, \nabla_t \dot\gamma) \, d\mathcal{L}^1(t). \end{align*} This yields the first variation formula in Part 1: \begin{align*} \frac{d}{ds} E(\gamma_s)\Big|_{s = 0} = -\int_0^T g\!\left(Y(t), \nabla_t \dot\gamma(t)\right) d\mathcal{L}^1(t) + g(Y(t), \dot\gamma(t))\Big|_0^T - \sum_{i=1}^{N-1} g\!\left(Y(t_i),\, \dot\gamma(t_i^+) - \dot\gamma(t_i^-)\right). \end{align*}[/step]

[guided]Combining the previous steps we have \begin{align*} \frac{d}{ds} E(\gamma_s)\Big|_{s = 0} = \int_0^T g\!\left(\nabla_t Y(t), \dot\gamma(t)\right) d\mathcal{L}^1(t). \end{align*} This is progress, but the $\nabla_t Y$ in the integrand is problematic: a critical-point analysis requires us to pick out the coefficient of $Y$ itself and apply the fundamental lemma of the calculus of variations. So we want to **move the $\nabla_t$ off of $Y$ onto $\dot\gamma$** — this is the Riemannian analogue of integration by parts. The tool, used a second time in this proof, is metric compatibility — but now applied along the $t$-direction. Differentiating the function $t \mapsto g(Y(t), \dot\gamma(t))$ along $\gamma$, metric compatibility gives the Riemannian Leibniz rule: \begin{align*} \frac{d}{dt} g(Y(t), \dot\gamma(t)) = g(\nabla_t Y(t), \dot\gamma(t)) + g(Y(t), \nabla_t \dot\gamma(t)). \end{align*} This is structurally identical to the product rule from one-variable calculus, with $\nabla_t$ replacing the ordinary derivative — and indeed metric compatibility is exactly the statement that $\nabla$ acts as a derivation on the inner product. Solve for the term we want to eliminate: \begin{align*} g(\nabla_t Y(t), \dot\gamma(t)) = \frac{d}{dt} g(Y(t), \dot\gamma(t)) - g(Y(t), \nabla_t \dot\gamma(t)). \end{align*} Now we integrate over $[0, T]$. The first term on the right is a total derivative, so by the fundamental theorem of calculus it integrates to a boundary term $g(Y, \dot\gamma)|_0^T$. The second term is what we wanted: the inner product of $Y$ with $\nabla_t\dot\gamma$, the geodesic-acceleration vector. A subtlety: $\gamma$ is only piecewise $C^1$, so $\nabla_t \dot\gamma$ may have jumps at the interior breakpoints $0 = t_0 < t_1 < \cdots < t_N = T$. We cannot apply FTC across a breakpoint because $\dot\gamma$ may not be continuous there. Instead we apply FTC segment-by-segment on each smooth piece $[t_{i-1}, t_i]$ and sum: \begin{align*} \int_0^T g(\nabla_t Y, \dot\gamma) \, d\mathcal{L}^1(t) = g(Y, \dot\gamma)\Big|_0^T - \sum_{i=1}^{N-1} g\!\left(Y(t_i),\, \dot\gamma(t_i^+) - \dot\gamma(t_i^-)\right) - \int_0^T g(Y, \nabla_t \dot\gamma) \, d\mathcal{L}^1(t). \end{align*} The variational field $Y$ is smooth on all of $[0, T]$ (it comes from differentiating a smooth $H$), so the $Y$ values at adjacent endpoints of consecutive segments agree — that's why the interior contributions collapse to a single $Y(t_i)$ in each summand. The jumps that survive are the jumps in $\dot\gamma$ itself: at each interior breakpoint $t_i$, the FTC produces a contribution $-g(Y(t_i), \dot\gamma(t_i^+) - \dot\gamma(t_i^-))$, capturing the failure of $\dot\gamma$ to be continuous. Substituting back into the energy derivative yields the first variation formula for Part 1: \begin{align*} \frac{d}{ds} E(\gamma_s)\Big|_{s = 0} = -\int_0^T g\!\left(Y(t), \nabla_t \dot\gamma(t)\right) d\mathcal{L}^1(t) + g(Y(t), \dot\gamma(t))\Big|_0^T - \sum_{i=1}^{N-1} g\!\left(Y(t_i),\, \dot\gamma(t_i^+) - \dot\gamma(t_i^-)\right). \end{align*} Reading the right-hand side: the integral term is the geodesic equation contracted against the test field $Y$, the boundary term records the endpoint behaviour, and the discrete sum records the corner kinks of the piecewise curve. Each is a separate obstruction to criticality, which is exactly what we will exploit in Step 5.[/guided]

[step:Critical points of $E$ among endpoint-fixing variations are geodesics] For Part 2, we characterise the critical points. An endpoint-fixing variation has $H(0, s) = \gamma(0)$ and $H(T, s) = \gamma(T)$ for all $s$, so $Y(0) = Y(T) = 0$ and the boundary term in the first variation formula vanishes. The first variation reduces to \begin{align*} \frac{d}{ds} E(\gamma_s)\Big|_{s = 0} = -\int_0^T g\!\left(Y(t), \nabla_t \dot\gamma(t)\right) d\mathcal{L}^1(t). \end{align*} $(\Leftarrow)$ If $\gamma$ is a geodesic, $\nabla_t \dot\gamma \equiv 0$ on $[0, T]$, so the integrand vanishes and the first variation is $0$ for every variation. Thus $\gamma$ is a critical point. $(\Rightarrow)$ Suppose $\gamma$ is a critical point: the first variation vanishes for every endpoint-fixing variation. We claim $\nabla_t \dot\gamma \equiv 0$. [claim:The vanishing of $\int_0^T g(Y, \nabla_t \dot\gamma) \, d\mathcal{L}^1 = 0$ for all $Y$ vanishing at the endpoints forces $\nabla_t \dot\gamma \equiv 0$] [proof] We prove this on each smooth segment $(t_{i-1}, t_i)$ of $\gamma$. Suppose, for contradiction, that $\nabla_t \dot\gamma(t^*) \neq 0$ for some $t^* \in (t_{i-1}, t_i)$. By continuity of $\nabla_t \dot\gamma$ on $(t_{i-1}, t_i)$, there is a closed sub-interval $[a, b] \subset (t_{i-1}, t_i)$ with $t^* \in (a, b)$ on which $\nabla_t \dot\gamma$ is continuous and non-vanishing. **Construction of the test field.** Let $\eta \in C_c^\infty(0, T)$ be a smooth bump function with $\eta \ge 0$, $\operatorname{supp} \eta \subseteq [a, b]$, and $\eta(t^*) = 1$. Let $P_t : T_{\gamma(t^*)} M \to T_{\gamma(t)} M$ denote parallel transport along $\gamma$ from $\gamma(t^*)$ to $\gamma(t)$. Define \begin{align*} Y(t) := \eta(t) \cdot P_t\!\left(\nabla_t \dot\gamma(t^*)\right), \end{align*} so $Y$ is smooth along $\gamma$, vanishes outside $[a,b]$, and so vanishes at endpoints and near every breakpoint. **Sign of the integral.** Parallel transport is an isometry, so the function $t \mapsto g(P_t(\nabla_t \dot\gamma(t^*)), \nabla_t \dot\gamma(t))$ is continuous on $[a, b]$ and equals $|\nabla_t \dot\gamma(t^*)|_g^2 > 0$ at $t = t^*$. By shrinking $[a,b]$ around $t^*$ if necessary, we may assume this inner product remains $\ge \tfrac12 |\nabla_t \dot\gamma(t^*)|_g^2 > 0$ throughout $[a,b]$. Hence \begin{align*} \int_0^T g(Y, \nabla_t \dot\gamma) \, d\mathcal{L}^1(t) \ge \frac{1}{2}|\nabla_t \dot\gamma(t^*)|_g^2 \int_a^b \eta\, d\mathcal{L}^1(t) > 0, \end{align*} contradicting the vanishing of $\int_0^T g(Y, \nabla_t\dot\gamma)\, d\mathcal{L}^1$ implied by the criticality assumption (since $Y$ vanishes at endpoints and near breakpoints, all boundary and jump terms drop out). Realise $Y$ as the variational field of $H(t, s) := \exp_{\gamma(t)}(s Y(t))$, well-defined and smooth for $|s|$ small. Therefore $\nabla_t \dot\gamma \equiv 0$ on every smooth segment of $\gamma$. **Continuity at breakpoints.** With $\nabla_t \dot\gamma = 0$ on each segment now established, the first variation formula reduces, for any admissible $Y$ with $Y(0) = Y(T) = 0$, to \begin{align*} 0 = \frac{d}{ds} E(\gamma_s)\Big|_{s=0} = -\sum_{i=1}^{N-1} g\!\left(Y(t_i),\, \dot\gamma(t_i^+) - \dot\gamma(t_i^-)\right). \end{align*} For each interior breakpoint $t_i$ choose a smooth $Y$ along $\gamma$ with $Y(t_i) = \dot\gamma(t_i^+) - \dot\gamma(t_i^-)$ and $Y(t_j) = 0$ for $j \ne i$ (constructed by parallel-transporting the jump from $\gamma(t_i)$ along $\gamma$ and multiplying by a bump function supported near $t_i$). The criticality condition gives $|\dot\gamma(t_i^+) - \dot\gamma(t_i^-)|_g^2 = 0$, hence $\dot\gamma(t_i^+) = \dot\gamma(t_i^-)$. So $\dot\gamma$ is continuous at every breakpoint. Combined with $\nabla_t \dot\gamma = 0$ on each segment, the geodesic ODE propagates smoothness across each $t_i$, so $\gamma$ is a smooth geodesic on $[0, T]$. [/proof] [/claim] This proves Part 2: critical points are geodesics, and conversely. [/step]

[step:Energy and length first variations agree under unit speed and constant-speed assumption] For Part 3, assume $|\dot\gamma_s(t)|_g$ is constant in $t$ for each fixed $s$, and $|\dot\gamma_0(t)|_g \equiv 1$. Compute the length: \begin{align*} \ell(\gamma_s) = \int_0^T |\dot\gamma_s(t)|_g \, d\mathcal{L}^1(t) = T \cdot c(s), \end{align*} where $c(s) := |\dot\gamma_s(t)|_g$ is the constant-in-$t$ speed of $\gamma_s$, with $c(0) = 1$. The energy is \begin{align*} E(\gamma_s) = \frac{1}{2}\int_0^T |\dot\gamma_s(t)|_g^2 \, d\mathcal{L}^1(t) = \frac{T}{2} c(s)^2. \end{align*} Differentiating in $s$ and evaluating at $s = 0$: \begin{align*} \frac{d}{ds} \ell(\gamma_s)\Big|_{s = 0} &= T \cdot c'(0), \\ \frac{d}{ds} E(\gamma_s)\Big|_{s = 0} &= T \cdot c(0) \cdot c'(0) = T \cdot c'(0). \end{align*} The two are equal: \begin{align*} \frac{d}{ds} E(\gamma_s)\Big|_{s = 0} = \frac{d}{ds} \ell(\gamma_s)\Big|_{s = 0}. \end{align*} [/step]

[step:Critical points of $\ell$ are reparametrisations of geodesics]For Part 4, suppose $\gamma : [0, T] \to M$ is a critical point of $\ell$ on $\Omega(\gamma(0), \gamma(T))$, meaning $\frac{d}{ds}\ell(\gamma_s)|_{s = 0} = 0$ for every endpoint-fixing variation. We show $\gamma$ is a reparametrisation of a geodesic. Assume $\gamma$ is regular, i.e., $\dot\gamma(t) \neq 0$ for all $t$. (If $\gamma$ has zero velocity on a set of positive measure, the standard convention is to reparametrise so that the zero-velocity portions are removed; this changes neither $\ell$ nor the critical-point property, since $\ell$ is invariant under monotone reparametrisation.) Define the reparametrisation $\tilde\gamma : [0, \ell(\gamma)] \to M$ by arc length: $\tilde\gamma(s(t)) := \gamma(t)$ where $s(t) := \int_0^t |\dot\gamma|_g \, d\mathcal{L}^1$. Since $|\dot\gamma|_g > 0$ everywhere, $s : [0, T] \to [0, \ell(\gamma)]$ is a smooth diffeomorphism, and $\tilde\gamma$ is smooth with $|\dot{\tilde\gamma}|_g \equiv 1$. We claim $\tilde\gamma$ is a critical point of $\ell$ on $[0, \ell(\gamma)]$. Length is invariant under smooth reparametrisation, so for any smooth variation $\tilde H$ of $\tilde\gamma$ we may pull back to a variation $H$ of $\gamma$ (compose with the inverse of $s$ on each fixed slice), and $\ell(\tilde\gamma_s) = \ell(\gamma_s)$ for every $s$. Differentiating in $s$, criticality of $\gamma$ implies criticality of $\tilde\gamma$. **Direct reduction of length-criticality to energy-criticality.** Let $L = \ell(\gamma)$ and let $\tilde\gamma : [0, L] \to M$ be the arc-length reparametrisation, so $|\dot{\tilde\gamma}|_g \equiv 1$. Fix any smooth endpoint-fixing variation field $\tilde Y$ along $\tilde\gamma$. Decompose $\tilde Y$ into orthogonal and tangential components with respect to the unit-speed velocity: \begin{align*} \tilde Y(t) = \tilde Y^\perp(t) + f(t)\, \dot{\tilde\gamma}(t), \end{align*} where $f(t) := g(\tilde Y(t), \dot{\tilde\gamma}(t))$ and $\tilde Y^\perp$ is pointwise orthogonal to $\dot{\tilde\gamma}$. Since $\tilde Y$ vanishes at endpoints, $f(0) = f(L) = 0$. A direct computation (analogous to Part 1, with the same metric-compatibility and torsion-free arguments applied to $\ell$) gives, when $\tilde\gamma$ has unit speed, \begin{align*} \frac{d}{ds}\ell(\tilde\gamma_s)\Big|_{s=0} = -\int_0^L g(\tilde Y, \nabla_t \dot{\tilde\gamma})\, d\mathcal{L}^1(t) + g(\tilde Y, \dot{\tilde\gamma})\Big|_0^L. \end{align*} The tangential piece $f \dot{\tilde\gamma}$ contributes nothing: $g(f \dot{\tilde\gamma}, \nabla_t \dot{\tilde\gamma}) = \tfrac{f}{2} \partial_t g(\dot{\tilde\gamma}, \dot{\tilde\gamma}) = 0$ since $|\dot{\tilde\gamma}|_g \equiv 1$, and the boundary term $g(f\dot{\tilde\gamma}, \dot{\tilde\gamma})|_0^L = (f \cdot 1)|_0^L = f(L) - f(0) = 0$. So the length first variation along $\tilde Y$ equals the same expression along $\tilde Y^\perp$, namely \begin{align*} -\int_0^L g(\tilde Y^\perp, \nabla_t \dot{\tilde\gamma})\, d\mathcal{L}^1(t). \end{align*} This equals the first variation of $E$ along $\tilde Y^\perp$ by Part 1 (with $\tilde Y^\perp$ vanishing at endpoints). Hence: length-criticality of $\tilde\gamma$ over endpoint-fixing variations is equivalent to energy-criticality of $\tilde\gamma$ over endpoint-fixing variations. By Part 2, this forces $\nabla_t \dot{\tilde\gamma} = 0$, so $\tilde\gamma$ is a geodesic, and $\gamma$ is the reparametrisation of $\tilde\gamma$ via the arc-length parameter.[/step]

[guided]The strategy for Part 4: starting from a critical point $\gamma$ of length, reparametrise to unit speed to obtain $\tilde\gamma$, show that length-criticality of $\tilde\gamma$ reduces to energy-criticality of $\tilde\gamma$, and then invoke Part 2 to conclude $\tilde\gamma$ is a geodesic. We carry this out in detail. Suppose $\gamma : [0, T] \to M$ is a regular critical point of $\ell$ on $\Omega(\gamma(0), \gamma(T))$. Why does the regularity assumption matter? Because we need to invert the arc-length function $s(t) = \int_0^t |\dot\gamma|_g\, d\mathcal{L}^1$. By the inverse function theorem, $s$ is a smooth diffeomorphism iff $s'(t) = |\dot\gamma(t)|_g > 0$ for all $t$, which is precisely the regularity hypothesis. (For non-regular curves the standard convention is to delete the zero-velocity portions; this leaves $\ell$ unchanged because $\ell$ is invariant under monotone reparametrisation, so criticality is preserved.) Setting $L := \ell(\gamma)$, define the arc-length reparametrisation \begin{align*} \tilde\gamma : [0, L] &\to M \\ \sigma &\mapsto \gamma(s^{-1}(\sigma)), \end{align*} which is smooth and satisfies $|\dot{\tilde\gamma}|_g \equiv 1$ by the chain rule. Does the criticality of $\gamma$ for $\ell$ carry over to $\tilde\gamma$? Yes, because $\ell$ is reparametrisation-invariant: for any smooth variation $\tilde H$ of $\tilde\gamma$, we may pull back via $s$ to get a variation $H$ of $\gamma$, and $\ell(\tilde\gamma_s) = \ell(\gamma_s)$ for every $s$. Differentiating, criticality of $\gamma$ transfers to criticality of $\tilde\gamma$. Now we want to compare the first variation of $\ell$ at $\tilde\gamma$ to the first variation of $E$. The cleanest tool is to decompose any variation field along $\tilde\gamma$ into pieces along and perpendicular to the velocity. Let $\tilde Y$ be any smooth endpoint-fixing variation field along $\tilde\gamma$ and write \begin{align*} \tilde Y(t) = \tilde Y^\perp(t) + f(t)\, \dot{\tilde\gamma}(t), \end{align*} where $f(t) := g(\tilde Y(t), \dot{\tilde\gamma}(t))$ extracts the tangential component and $\tilde Y^\perp$ is the orthogonal residual. Since $\tilde Y$ vanishes at the endpoints, so does its tangential coefficient: $f(0) = f(L) = 0$. A direct computation, repeating the metric-compatibility and torsion-freeness arguments from Step 4 but applied to the integrand of $\ell$ instead of $E$ and using $|\dot{\tilde\gamma}|_g \equiv 1$ to drop the $|\dot{\tilde\gamma}|_g$ factor in the length integrand, gives the unit-speed first variation formula for length: \begin{align*} \frac{d}{ds}\ell(\tilde\gamma_s)\Big|_{s=0} = -\int_0^L g(\tilde Y, \nabla_t \dot{\tilde\gamma})\, d\mathcal{L}^1(t) + g(\tilde Y, \dot{\tilde\gamma})\Big|_0^L. \end{align*} The tangential piece of $\tilde Y$ contributes nothing to this expression — and this is the technical heart of the reduction. The integral contribution of the tangential piece is \begin{align*} g(f \dot{\tilde\gamma}, \nabla_t \dot{\tilde\gamma}) = \tfrac{f}{2} \partial_t g(\dot{\tilde\gamma}, \dot{\tilde\gamma}) = 0, \end{align*} because $g(\dot{\tilde\gamma}, \dot{\tilde\gamma}) \equiv 1$ has zero $t$-derivative. The boundary contribution of the tangential piece is \begin{align*} g(f\dot{\tilde\gamma}, \dot{\tilde\gamma})\big|_0^L = (f \cdot 1)\big|_0^L = f(L) - f(0) = 0, \end{align*} because $f$ vanishes at the endpoints. So along $\tilde Y$ the first variation of length equals the first variation along the perpendicular component: \begin{align*} \frac{d}{ds}\ell(\tilde\gamma_s)\Big|_{s=0} = -\int_0^L g(\tilde Y^\perp, \nabla_t \dot{\tilde\gamma})\, d\mathcal{L}^1(t). \end{align*} But this is precisely the first variation of energy along $\tilde Y^\perp$ (using Part 1 with the $Y^\perp$ vanishing at endpoints — both endpoint values of $\tilde Y^\perp$ inherit the zero from $\tilde Y$). Hence length-criticality of $\tilde\gamma$ over endpoint-fixing variations is **equivalent** to energy-criticality of $\tilde\gamma$ over the same variations. Apply Part 2 to conclude $\nabla_t \dot{\tilde\gamma} \equiv 0$, so $\tilde\gamma$ is a (smooth) geodesic on $[0, L]$. Composing with the arc-length function $s : [0, T] \to [0, L]$ gives $\gamma = \tilde\gamma \circ s$, exhibiting $\gamma$ as a reparametrisation of a geodesic. This completes Part 4 and the proof of the theorem.[/guided]