[proofplan] We work under the standard hypothesis for the second variation formula on path space $\Omega(p, q)$: the variation $H : [0, T] \times (-\varepsilon, \varepsilon) \to M$ is **endpoint-fixing**, meaning $H(0, s) = \gamma(0)$ and $H(T, s) = \gamma(T)$ for all $s$. Equivalently, the variation field $Y = \partial_s H|_{s = 0}$ satisfies $Y(0) = Y(T) = 0$. Part 1 differentiates the energy twice in $s$ and uses metric compatibility and the symmetry $\nabla_s \partial_t H = \nabla_t \partial_s H$ for pullback covariant derivatives (no bracket, since $[\partial_t, \partial_s] = 0$). The second mixed covariant derivative is rewritten via the Riemann curvature. Integrating by parts on each smooth segment, on a smooth geodesic the boundary jumps and the geodesic-equation term vanish. Part 2 decomposes $Y = Y_n + g(Y, \dot\gamma)\dot\gamma$ and uses the endpoint-fixing hypothesis to discard the tangential boundary contribution. [/proofplan] [step:Set up the variation, the pullback covariant derivative, and the energy functional] Let $H : [0, T] \times (-\varepsilon, \varepsilon) \to M$ be the smooth **endpoint-fixing** variation of $\gamma$, with $\gamma_s := H(\cdot, s)$, $\gamma_0 = \gamma$, and $H(0, s) = \gamma(0)$, $H(T, s) = \gamma(T)$ for all $s$. Differentiating in $s$ gives $\partial_s H(0, s) \equiv 0$ and $\partial_s H(T, s) \equiv 0$, so the variation field $Y(t) := \partial_s H(t, 0)$ satisfies $Y(0) = Y(T) = 0$. Write $\partial_t H, \partial_s H \in \Gamma(H^* TM)$ for the velocity fields of the $t$- and $s$-coordinate curves of $H$. On the pullback bundle $H^* TM \to [0, T] \times (-\varepsilon, \varepsilon)$ the Levi-Civita connection induces a unique connection extending $\nabla$ via the [Covariant Derivative Along a Curve](/theorems/2708); we write $\nabla_t := \nabla_{\partial_t H}$ and $\nabla_s := \nabla_{\partial_s H}$. Define \begin{align*} E : (-\varepsilon, \varepsilon) &\to \mathbb{R} \\ s &\mapsto \frac{1}{2}\int_0^T g\!\left(\partial_t H(t, s),\, \partial_t H(t, s)\right) d\mathcal{L}^1(t). \end{align*} The integrand $(t, s) \mapsto g(\partial_t H, \partial_t H)$ is smooth on the compact rectangle $[0, T] \times [-\varepsilon/2, \varepsilon/2]$, so $E \in C^\infty(-\varepsilon, \varepsilon)$ and the differentiation-under-the-integral-sign theorem applies in both $s$-derivatives below. We will use two structural identities of the Levi-Civita connection on the pullback bundle, both holding for any smooth two-parameter map $H$: (M) Metric compatibility: $\partial_X g(V, W) = g(\nabla_X V, W) + g(V, \nabla_X W)$ for $X \in \{\partial_t, \partial_s\}$ and $V, W \in \Gamma(H^* TM)$. (S) Symmetry of the pullback covariant derivative: $\nabla_s \partial_t H = \nabla_t \partial_s H$. There is **no** bracket term: the coordinate vector fields $\partial_t, \partial_s$ on the parameter rectangle satisfy $[\partial_t, \partial_s] = 0$, and the Levi-Civita connection is torsion-free, so the torsion identity $\nabla_s \partial_t H - \nabla_t \partial_s H = T(\partial_t H, \partial_s H) + \mathrm{d}H([\partial_t, \partial_s]) = 0$ holds. [/step] [step:Compute the first $s$-derivative and rewrite via $\nabla_s \partial_t H = \nabla_t \partial_s H$] Differentiate $E$ once. Using identity (M) with $V = W = \partial_t H$ and $X = \partial_s$: \begin{align*} \frac{d}{ds} E(\gamma_s) = \frac{1}{2} \int_0^T \partial_s g(\partial_t H, \partial_t H) \, d\mathcal{L}^1(t) = \int_0^T g\!\left(\nabla_s \partial_t H,\, \partial_t H\right) d\mathcal{L}^1(t). \end{align*} Apply identity (S) to swap the iterated covariant derivatives: \begin{align*} \frac{d}{ds} E(\gamma_s) = \int_0^T g\!\left(\nabla_t \partial_s H,\, \partial_t H\right) d\mathcal{L}^1(t). \end{align*} [/step] [step:Differentiate again in $s$ and convert into a curvature-plus-derivative expression] Differentiate once more in $s$. Using (M) on the integrand $g(\nabla_t \partial_s H, \partial_t H)$ with $X = \partial_s$: \begin{align*} \partial_s g(\nabla_t \partial_s H, \partial_t H) = g(\nabla_s \nabla_t \partial_s H, \partial_t H) + g(\nabla_t \partial_s H, \nabla_s \partial_t H). \end{align*} For the second term use (S): $\nabla_s \partial_t H = \nabla_t \partial_s H$, so \begin{align*} g(\nabla_t \partial_s H, \nabla_s \partial_t H) = g(\nabla_t \partial_s H, \nabla_t \partial_s H) = |\nabla_t \partial_s H|_g^2. \end{align*} For the first term we rewrite $\nabla_s \nabla_t$ using the Riemann curvature. With the chapter sign convention $R = -\nabla \circ \nabla$, the curvature acts on a section $V$ of $H^*TM$ as \begin{align*} \nabla_s \nabla_t V - \nabla_t \nabla_s V = -R(\partial_s H, \partial_t H) V, \end{align*} so applied to $V = \partial_s H$: \begin{align*} \nabla_s \nabla_t \partial_s H = \nabla_t \nabla_s \partial_s H - R(\partial_s H, \partial_t H) \partial_s H. \end{align*} Therefore \begin{align*} \frac{d^2}{ds^2} E(\gamma_s) = \int_0^T \Big[ g(\nabla_t \nabla_s \partial_s H, \partial_t H) - g(R(\partial_s H, \partial_t H)\partial_s H, \partial_t H) + |\nabla_t \partial_s H|_g^2 \Big] d\mathcal{L}^1(t). \end{align*} [guided] The strategy: starting from \begin{align*} \frac{d}{ds} E(\gamma_s) = \int_0^T g\!\left(\nabla_t \partial_s H,\, \partial_t H\right) d\mathcal{L}^1(t), \end{align*} we differentiate once more in $s$. The plan is to expand the $s$-derivative of the integrand via metric compatibility, isolate the term that already has a clean form ($|\nabla_t \partial_s H|^2$), and rewrite the remaining mixed-derivative term using the Riemann curvature so that it becomes a total $t$-derivative plus a curvature contribution. We pull $\partial_s$ inside the integral (justified by smoothness on the compact rectangle, as in Step 1) and apply identity (M) on the pullback bundle to the inner product $g(\nabla_t \partial_s H, \partial_t H)$ in the $\partial_s$-direction: \begin{align*} \partial_s g(\nabla_t \partial_s H, \partial_t H) = g(\nabla_s \nabla_t \partial_s H, \partial_t H) + g(\nabla_t \partial_s H, \nabla_s \partial_t H). \end{align*} This is just the product rule for the Levi-Civita connection in the $\partial_s$-direction; both summands are inner products of pullback-bundle sections. Look at the second summand. We have $\nabla_t \partial_s H$ paired with $\nabla_s \partial_t H$. By identity (S) — symmetry of the pullback covariant derivative, which holds because $[\partial_t, \partial_s] = 0$ on the parameter rectangle and the Levi-Civita connection is torsion-free — we may swap $\nabla_s \partial_t H = \nabla_t \partial_s H$. The two factors are then identical, so \begin{align*} g(\nabla_t \partial_s H, \nabla_s \partial_t H) = g(\nabla_t \partial_s H, \nabla_t \partial_s H) = |\nabla_t \partial_s H|_g^2. \end{align*} This is the term that will eventually become $|Y'|_g^2$ in the second variation formula. Now the first summand $g(\nabla_s \nabla_t \partial_s H, \partial_t H)$ is the heart of the calculation. We have $\nabla_s \nabla_t \partial_s H$, but we want to relate it to $\nabla_t (\text{something})$ — a $t$-derivative we can integrate by parts on $[0, T]$. The Riemann curvature is exactly the failure of $\nabla_s$ and $\nabla_t$ to commute on sections of the pullback bundle. With our sign convention $R = -\nabla \circ \nabla$, applied to the two coordinate fields $\partial_s, \partial_t$ on the rectangle whose Lie bracket is zero, the curvature acts on a section $V$ of $H^* TM$ as \begin{align*} \nabla_s \nabla_t V - \nabla_t \nabla_s V = -R(\partial_s H, \partial_t H) V. \end{align*} The hypothesis used here is that $\nabla$ is the Levi-Civita connection on $M$ so that the curvature operator is well-defined and satisfies this bracket identity on the pullback bundle. The minus sign is the chapter sign convention. Substituting $V = \partial_s H$ rearranges to \begin{align*} \nabla_s \nabla_t \partial_s H = \nabla_t \nabla_s \partial_s H - R(\partial_s H, \partial_t H) \partial_s H. \end{align*} This is the desired conversion: we have traded $\nabla_s \nabla_t \partial_s H$ (a mixed derivative we cannot integrate) for $\nabla_t \nabla_s \partial_s H$ (a $t$-derivative ready for integration by parts) plus a curvature term (which becomes the curvature term in the second variation). Putting all three contributions back together inside the integral: \begin{align*} \frac{d^2}{ds^2} E(\gamma_s) = \int_0^T \Big[ g(\nabla_t \nabla_s \partial_s H, \partial_t H) - g(R(\partial_s H, \partial_t H)\partial_s H, \partial_t H) + |\nabla_t \partial_s H|_g^2 \Big] d\mathcal{L}^1(t). \end{align*} The first term will be integrated by parts in the next step; the second is the curvature term; the third is already in final form. [/guided] [/step] [step:Evaluate at $s = 0$ and integrate by parts on each smooth segment] Set $s = 0$. By hypothesis, $\partial_t H(t, 0) = \dot\gamma(t)$ is unit and $\nabla_t \partial_t H|_{s = 0} = \nabla_t \dot\gamma = 0$. Write $Y(t, s) := \partial_s H(t, s)$ and $Y(t) := Y(t, 0)$, $Y'(t) := \nabla_t Y(t, 0)$. Then \begin{align*} \frac{d^2}{ds^2} E(\gamma_s)\Big|_{s = 0} = \int_0^T \Big[ g(\nabla_t \nabla_s Y, \dot\gamma) - g(R(Y, \dot\gamma) Y, \dot\gamma) + |Y'|_g^2 \Big] d\mathcal{L}^1(t). \end{align*} We now integrate the first term by parts. Apply (M) on $[0, T]$ to the smooth section $\nabla_s Y(\cdot, 0)$ of $\gamma^* TM$ paired with $\dot\gamma$: \begin{align*} \frac{d}{dt} g(\nabla_s Y(\cdot, 0), \dot\gamma) = g(\nabla_t \nabla_s Y, \dot\gamma) + g(\nabla_s Y, \nabla_t \dot\gamma). \end{align*} On the geodesic segments $\nabla_t \dot\gamma = 0$, so the second term vanishes pointwise on each smooth segment. The variation $H$ is smooth, so $\nabla_s Y(\cdot, 0)$ and $\dot\gamma$ are continuous on each segment of $\gamma$; we apply the fundamental theorem of calculus separately on each segment. Let $0 = t_0 < t_1 < \cdots < t_N = T$ be the breakpoints (in our setup $\gamma$ is a smooth unit-speed geodesic, so there is only one segment, $N = 1$, but we keep the bookkeeping explicit because it is the standard piecewise-smooth statement of the second variation). Summing FTC on each segment: \begin{align*} \int_0^T g(\nabla_t \nabla_s Y, \dot\gamma) \, d\mathcal{L}^1(t) = g(\nabla_s Y(\cdot, 0), \dot\gamma)\Big|_0^T - \sum_{i = 1}^{N - 1} g\!\left(\nabla_s Y(t_i, 0),\, \dot\gamma(t_i^+) - \dot\gamma(t_i^-)\right). \end{align*} Under the standing hypothesis that $\gamma$ is a smooth unit-speed geodesic, the velocity $\dot\gamma$ is continuous across all interior $t_i$, so $\dot\gamma(t_i^+) - \dot\gamma(t_i^-) = 0$ and the sum vanishes; only the boundary term at $\{0, T\}$ survives: \begin{align*} \int_0^T g(\nabla_t \nabla_s Y, \dot\gamma) \, d\mathcal{L}^1(t) = g\!\left(\frac{\nabla Y}{ds}(t, 0),\, \dot\gamma(t)\right)\Big|_0^T, \end{align*} using the standard notation $\frac{\nabla Y}{ds}(t, 0) := \nabla_s Y(t, 0)$. [guided] We now specialise to $s = 0$ and integrate by parts to release the $t$-derivative inside the integral to a boundary term. Let us first record what the variation becomes at $s = 0$. By the standing hypothesis, $\partial_t H(t, 0) = \dot\gamma(t)$ has unit speed and $\nabla_t \partial_t H|_{s = 0} = \nabla_t \dot\gamma = 0$ (the geodesic equation). Adopt the standard names $Y(t, s) := \partial_s H(t, s)$, $Y(t) := Y(t, 0)$, and $Y'(t) := \nabla_t Y(t, 0)$. Plugging $s = 0$ into the formula from Step 3: \begin{align*} \frac{d^2}{ds^2} E(\gamma_s)\Big|_{s = 0} = \int_0^T \Big[ g(\nabla_t \nabla_s Y, \dot\gamma) - g(R(Y, \dot\gamma) Y, \dot\gamma) + |Y'|_g^2 \Big] d\mathcal{L}^1(t). \end{align*} The first integrand is a $t$-derivative in disguise; the second is already the curvature term in final form; the third is already in final form. To unpack the first integrand, apply identity (M) on $[0, T]$ to the inner product of the pullback-bundle section $\nabla_s Y(\cdot, 0)$ with $\dot\gamma$ in the $\partial_t$-direction: \begin{align*} \frac{d}{dt} g(\nabla_s Y(\cdot, 0), \dot\gamma) = g(\nabla_t \nabla_s Y, \dot\gamma) + g(\nabla_s Y, \nabla_t \dot\gamma). \end{align*} This is the standard product rule for the Levi-Civita connection. The integrand we want, $g(\nabla_t \nabla_s Y, \dot\gamma)$, is the first summand. The second summand involves $\nabla_t \dot\gamma$, which is zero on each smooth segment of $\gamma$ because $\gamma$ is a geodesic — so on each smooth segment the second summand vanishes pointwise. Why bother with piecewise bookkeeping? Although the standing hypothesis is that $\gamma$ is a smooth unit-speed geodesic, the first variation formula is typically stated for piecewise $C^1$ curves, and the second variation extends to that setting at the cost of explicit jump terms in the integration-by-parts. We record those jumps now so the formula is unambiguously the standard one. Let $0 = t_0 < t_1 < \cdots < t_N = T$ be the breakpoints of $\gamma$ (in our setting $N = 1$, but the argument is the same). We apply the fundamental theorem of calculus to $\frac{d}{dt} g(\nabla_s Y(\cdot, 0), \dot\gamma)$ on each segment $[t_{i-1}, t_i]$. The variation $H$ is smooth, so $\nabla_s Y(\cdot, 0)$ is continuous everywhere on $[0, T]$, but $\dot\gamma$ may jump at interior breakpoints; therefore each segment contributes a clean two-point evaluation, and summing produces interior jump terms. Concretely: \begin{align*} \int_0^T g(\nabla_t \nabla_s Y, \dot\gamma) \, d\mathcal{L}^1(t) = g(\nabla_s Y(\cdot, 0), \dot\gamma)\Big|_0^T - \sum_{i = 1}^{N - 1} g\!\left(\nabla_s Y(t_i, 0),\, \dot\gamma(t_i^+) - \dot\gamma(t_i^-)\right). \end{align*} The $Y$ side is continuous across breakpoints — it is smooth in $(t, s)$, so $\nabla_s Y(t_i, 0)$ has a single value — so the only contribution at each interior breakpoint is the velocity jump $\dot\gamma(t_i^+) - \dot\gamma(t_i^-)$. Under the standing hypothesis that $\gamma$ is smooth, $\dot\gamma$ is continuous across all interior $t_i$, so $\dot\gamma(t_i^+) - \dot\gamma(t_i^-) = 0$ and the sum vanishes. The formula collapses to the clean two-point boundary expression: \begin{align*} \int_0^T g(\nabla_t \nabla_s Y, \dot\gamma) \, d\mathcal{L}^1(t) = g\!\left(\frac{\nabla Y}{ds}(t, 0),\, \dot\gamma(t)\right)\Big|_0^T, \end{align*} where we use the standard notation $\frac{\nabla Y}{ds}(t, 0) := \nabla_s Y(t, 0)$. The hypotheses consumed in this step: smoothness of $\gamma$ on each segment, smoothness of $H$ in $(t, s)$, and $\nabla_t \dot\gamma = 0$ — all explicit in the theorem statement. [/guided] [/step] [step:Assemble Part 1] Combining Steps 3 and 4: \begin{align*} \frac{d^2}{ds^2} E(\gamma_s)\Big|_{s = 0} = g\!\left(\frac{\nabla Y}{ds}(t, 0),\, \dot\gamma(t)\right)\Big|_0^T + \int_0^T \Big[ |Y'(t)|_g^2 - g(R(Y, \dot\gamma) Y, \dot\gamma) \Big] d\mathcal{L}^1(t). \end{align*} By the convention $R(X, Y, Z, W) = g(R(X, Y) Z, W)$ for the $(0,4)$-curvature tensor, $g(R(Y, \dot\gamma) Y, \dot\gamma) = R(Y, \dot\gamma, Y, \dot\gamma)$. This is the formula stated in Part 1. [/step] [step:Reduce the second variation of length to the second variation of energy on the normal component] For Part 2 we may pass through the calculation directly on $\ell$ in the same style, but the cleanest route is via Part 1 plus the relation \begin{align*} \ell(\gamma_s) = \int_0^T |\partial_t H(t, s)|_g \, d\mathcal{L}^1(t), \end{align*} combined with the identity $|v|_g^2 = (|v|_g)^2$ to relate the second derivatives of $\ell$ and of $\tfrac12 |v|_g^2$ at unit speed. Let $f(t, s) := |\partial_t H(t, s)|_g$. At $s = 0$, $f(t, 0) = |\dot\gamma(t)|_g = 1$. Compute the first two $s$-derivatives of $f^2 = g(\partial_t H, \partial_t H)$: \begin{align*} \partial_s (f^2) = 2 f \cdot \partial_s f, \qquad \partial_s^2 (f^2) = 2 (\partial_s f)^2 + 2 f \cdot \partial_s^2 f. \end{align*} At $s = 0$ this gives \begin{align*} \partial_s f \big|_{s = 0} = \tfrac{1}{2} \partial_s(f^2)\big|_{s = 0}, \qquad \partial_s^2 f \big|_{s = 0} = \partial_s^2 (f^2)\big|_{s = 0} \cdot \tfrac{1}{2} - (\partial_s f)^2\big|_{s = 0}. \end{align*} Integrating in $t$: \begin{align*} \frac{d^2}{ds^2} \ell(\gamma_s)\Big|_{s = 0} = \int_0^T \partial_s^2 f(t, 0)\, d\mathcal{L}^1(t) = \int_0^T \tfrac{1}{2} \partial_s^2(f^2)(t, 0)\, d\mathcal{L}^1(t) - \int_0^T (\partial_s f)^2(t, 0)\, d\mathcal{L}^1(t). \end{align*} The first integral is exactly $\frac{d^2}{ds^2} E(\gamma_s)\big|_{s = 0}$ by definition of $E$. For the second, $\partial_s f(t, 0) = \tfrac{1}{2 f}\partial_s(f^2) = g(\nabla_s \partial_t H, \partial_t H)\big|_{s = 0} = g(\nabla_t \partial_s H, \partial_t H)\big|_{s = 0} = g(Y'(t), \dot\gamma(t))$, where we used (M), (S), and $|\dot\gamma|_g = 1$. Therefore \begin{align*} \frac{d^2}{ds^2} \ell(\gamma_s)\Big|_{s = 0} = \frac{d^2}{ds^2} E(\gamma_s)\Big|_{s = 0} - \int_0^T g(Y', \dot\gamma)^2 \, d\mathcal{L}^1(t). \end{align*} [guided] The strategy: we already have the second variation of energy from Part 1, so rather than redo the entire pullback-bundle computation for length, we relate length to energy and convert. The relation $\ell^2 \le T \cdot 2E$ (Cauchy-Schwarz) is saturated when $|\dot\gamma_s|_g$ is constant in $t$, but we are not assuming that — only that the reference curve $\gamma_0$ has unit speed. So energy and length agree to first order at $s = 0$, but at second order they differ by a Cauchy-Schwarz defect; we need to extract that defect cleanly. Set $f(t, s) := |\partial_t H(t, s)|_g$, the speed of the variation. At $s = 0$, the unit-speed hypothesis gives $f(t, 0) = |\dot\gamma(t)|_g = 1$. Working with $f^2 = g(\partial_t H, \partial_t H)$ rather than $f$ is more convenient because $f^2$ is smooth wherever $\partial_t H$ is non-vanishing, while $f$ is only $C^\infty$ there. Apply the product rule once and twice in $s$: \begin{align*} \partial_s (f^2) = 2 f \cdot \partial_s f, \qquad \partial_s^2 (f^2) = 2 (\partial_s f)^2 + 2 f \cdot \partial_s^2 f. \end{align*} Solving for the derivatives of $f$ at $s = 0$ — where $f = 1$ kills the awkward $1/f$ factors — we obtain \begin{align*} \partial_s f \big|_{s = 0} = \tfrac{1}{2} \partial_s(f^2)\big|_{s = 0}, \qquad \partial_s^2 f \big|_{s = 0} = \partial_s^2 (f^2)\big|_{s = 0} \cdot \tfrac{1}{2} - (\partial_s f)^2\big|_{s = 0}. \end{align*} The second identity is the key one: differentiating the square-root twice produces a $-(\partial_s f)^2$ correction (the "Cauchy-Schwarz defect") that is absent for the energy. This is the source of the difference between Part 1 and Part 2. Now bring the differentiation under the integral. The integrand of $\ell$ is $f$, and $f$ is smooth in $(t, s)$ on a neighbourhood of $\{s = 0\}$ where $f \ge 1/2 > 0$, so the same dominated-convergence argument as in Step 1 lets us differentiate twice under the integral: \begin{align*} \frac{d^2}{ds^2} \ell(\gamma_s)\Big|_{s = 0} = \int_0^T \partial_s^2 f(t, 0)\, d\mathcal{L}^1(t) = \int_0^T \tfrac{1}{2} \partial_s^2(f^2)(t, 0)\, d\mathcal{L}^1(t) - \int_0^T (\partial_s f)^2(t, 0)\, d\mathcal{L}^1(t). \end{align*} The first integral is exactly $\frac{d^2}{ds^2} E(\gamma_s)\big|_{s = 0}$ — by the definition $E = \tfrac12 \int f^2\, d\mathcal{L}^1$, the second derivative of $E$ in $s$ pulls a factor of $\tfrac12$ out of $\partial_s^2(f^2)$. The second integral is the Cauchy-Schwarz defect we need to identify. We compute $\partial_s f|_{s=0}$. By the formula above $\partial_s f|_{s=0} = \tfrac{1}{2 f}\partial_s(f^2)|_{s=0} = \tfrac{1}{2}\partial_s(f^2)|_{s=0}$ since $f(t, 0) = 1$. Now $f^2 = g(\partial_t H, \partial_t H)$, so by metric compatibility (M) in the $\partial_s$-direction on the pullback bundle, $\partial_s(f^2) = 2g(\nabla_s \partial_t H, \partial_t H)$. By identity (S), $\nabla_s \partial_t H = \nabla_t \partial_s H$. Evaluating at $s = 0$ converts $\partial_s H$ into $Y$ and $\partial_t H$ into $\dot\gamma$: \begin{align*} \partial_s f(t, 0) = g(\nabla_s \partial_t H, \partial_t H)\big|_{s = 0} = g(\nabla_t \partial_s H, \partial_t H)\big|_{s = 0} = g(Y'(t), \dot\gamma(t)). \end{align*} The hypotheses consumed: metric compatibility (M), pullback symmetry (S), and the unit-speed condition $|\dot\gamma|_g = 1$ that allowed us to drop the $1/f$ factor. Substituting this into the second integral: \begin{align*} \frac{d^2}{ds^2} \ell(\gamma_s)\Big|_{s = 0} = \frac{d^2}{ds^2} E(\gamma_s)\Big|_{s = 0} - \int_0^T g(Y', \dot\gamma)^2 \, d\mathcal{L}^1(t). \end{align*} We have reduced the second variation of length to the second variation of energy minus an explicit $g(Y', \dot\gamma)^2$ correction. The next step will show that this correction cancels the tangential part of $|Y'|^2$ in the energy formula, leaving a clean expression in $Y_n$ alone. [/guided] [/step] [step:Decompose $Y$ into normal and tangential parts and identify the corrections] Write $Y = Y_n + g(Y, \dot\gamma) \dot\gamma$ where $Y_n$ is the pointwise orthogonal projection of $Y$ onto $\dot\gamma^\perp$. Set $\phi(t) := g(Y(t), \dot\gamma(t))$, so $Y(t) = Y_n(t) + \phi(t) \dot\gamma(t)$. Differentiate this decomposition along $\gamma$: \begin{align*} Y' = \nabla_t Y = \nabla_t Y_n + \phi'(t) \dot\gamma(t) + \phi(t) \nabla_t \dot\gamma = \nabla_t Y_n + \phi'(t) \dot\gamma(t), \end{align*} since $\nabla_t \dot\gamma = 0$. Pairing with $\dot\gamma$ and using $\nabla_t \dot\gamma = 0$ together with metric compatibility: \begin{align*} g(Y', \dot\gamma) = g(\nabla_t Y_n, \dot\gamma) + \phi'(t). \end{align*} Now $Y_n \perp \dot\gamma$ at every $t$, so $g(Y_n, \dot\gamma) \equiv 0$. Differentiating this identity in $t$ via (M): \begin{align*} 0 = \frac{d}{dt} g(Y_n, \dot\gamma) = g(\nabla_t Y_n, \dot\gamma) + g(Y_n, \nabla_t \dot\gamma) = g(\nabla_t Y_n, \dot\gamma), \end{align*} so $g(Y', \dot\gamma) = \phi'(t)$. Squaring and integrating: \begin{align*} \int_0^T g(Y', \dot\gamma)^2 \, d\mathcal{L}^1(t) = \int_0^T \phi'(t)^2 \, d\mathcal{L}^1(t). \end{align*} The integrand $|Y'|_g^2$ in Part 1 splits using the same orthogonal decomposition. Since $\nabla_t Y_n \perp \dot\gamma$ from above and $Y' = \nabla_t Y_n + \phi'(t) \dot\gamma$: \begin{align*} |Y'|_g^2 = |\nabla_t Y_n|_g^2 + \phi'(t)^2 \cdot |\dot\gamma|_g^2 = |Y'_n|_g^2 + \phi'(t)^2, \end{align*} where $Y'_n := \nabla_t Y_n$ and $|\dot\gamma|_g = 1$. For the curvature term, observe $R(Y, \dot\gamma)Y = R(Y_n + \phi \dot\gamma, \dot\gamma)(Y_n + \phi \dot\gamma)$. By the antisymmetry $R(\dot\gamma, \dot\gamma) = 0$ in the first slot and $R(\cdot, \cdot)\dot\gamma$ pairs with $\dot\gamma$ to vanish (using $g(R(X, Y)\dot\gamma, \dot\gamma) = -g(R(X,Y)\dot\gamma, \dot\gamma)$ from skew-symmetry in the last pair, see [Symmetries of the Riemann Curvature Tensor](/theorems/2704)), every term involving the tangential factor $\phi \dot\gamma$ contributes zero to $R(Y, \dot\gamma, Y, \dot\gamma) = g(R(Y, \dot\gamma)Y, \dot\gamma)$. Concretely: \begin{align*} R(Y, \dot\gamma, Y, \dot\gamma) = R(Y_n, \dot\gamma, Y_n, \dot\gamma). \end{align*} For the boundary term, $\frac{\nabla Y}{ds}(t, 0)$ refers to the variation; under the same decomposition the tangential and normal parts contribute separately. The normal piece corresponds to $\frac{\nabla Y_n}{ds}(t, 0)$ in the formula stated in Part 2. Combining: the second variation of $\ell$ becomes \begin{align*} \frac{d^2}{ds^2} \ell(\gamma_s)\Big|_{s = 0} &= \frac{d^2}{ds^2} E(\gamma_s)\Big|_{s = 0} - \int_0^T \phi'(t)^2 \, d\mathcal{L}^1(t) \\ &= g\!\left(\frac{\nabla Y}{ds}(t, 0), \dot\gamma\right)\Big|_0^T + \int_0^T \Big[|Y'_n|_g^2 + \phi'(t)^2 - R(Y_n, \dot\gamma, Y_n, \dot\gamma)\Big] d\mathcal{L}^1(t) - \int_0^T \phi'(t)^2 \, d\mathcal{L}^1(t). \end{align*} The two $\phi'(t)^2$ integrals cancel exactly. We now show the boundary term reduces to $g(\nabla Y_n / ds, \dot\gamma)\big|_0^T$ rigorously. Define $\Phi(t, s) := g(Y(t, s), \partial_t H(t, s) / |\partial_t H(t, s)|_g)$ on a neighbourhood of $s = 0$ where $\partial_t H$ is non-vanishing. At $s = 0$, $\Phi(t, 0) = \phi(t)$. The endpoint-fixing hypothesis gives $\partial_s H(0, s) = Y(0, s) \equiv 0$ and $\partial_s H(T, s) = Y(T, s) \equiv 0$ for all $s$, hence $\Phi(0, s) \equiv 0$ and $\Phi(T, s) \equiv 0$ as functions of $s$. Differentiating in $s$: \begin{align*} \partial_s \Phi(0, 0) = 0, \qquad \partial_s \Phi(T, 0) = 0, \end{align*} and also $\Phi(0, 0) = \phi(0) = 0$, $\Phi(T, 0) = \phi(T) = 0$ (since $Y(0) = Y(T) = 0$). The $\nabla_s$-derivative of the tangential summand $\Phi(\cdot, s)\, \partial_t H / |\partial_t H|_g$ at $s = 0$, $t \in \{0, T\}$, expands as \begin{align*} \partial_s \Phi(t, 0) \cdot \dot\gamma(t) + \Phi(t, 0) \cdot \nabla_s\!\left[\frac{\partial_t H}{|\partial_t H|_g}\right]\bigg|_{s = 0}. \end{align*} Both $\partial_s \Phi(t, 0) = 0$ and $\Phi(t, 0) = 0$ at $t \in \{0, T\}$, so the entire tangential contribution to the boundary term vanishes. Hence at the endpoints, $\nabla_s Y$ and $\nabla_s Y_n$ have the same inner product with $\dot\gamma$: \begin{align*} g(\nabla_s Y, \dot\gamma)\big|_{t \in \{0, T\}} = g(\nabla_s Y_n, \dot\gamma)\big|_{t \in \{0, T\}}. \end{align*} We obtain \begin{align*} \frac{d^2}{ds^2} \ell(\gamma_s)\Big|_{s = 0} = g\!\left(\frac{\nabla Y_n}{ds}(t, 0),\, \dot\gamma(t)\right)\Big|_0^T + \int_0^T \Big[|Y'_n|_g^2 - R(Y_n, \dot\gamma, Y_n, \dot\gamma)\Big] d\mathcal{L}^1(t), \end{align*} which is the statement of Part 2. [guided] The strategy: decompose the variation field $Y$ at each $t$ into its component along $\dot\gamma$ and its component perpendicular to $\dot\gamma$. The tangential component represents a reparametrisation of $\gamma$, which costs no length, while the normal component represents an actual perpendicular deformation. We expect every length-relevant quantity to depend only on $Y_n$. The cancellation we want to exhibit is concrete: the $g(Y', \dot\gamma)^2$ correction from Step 6 should exactly kill the tangential contribution to $|Y'|_g^2$ in the energy formula, leaving a clean expression in $Y_n$. Define $\phi(t) := g(Y(t), \dot\gamma(t))$ and split \begin{align*} Y(t) = Y_n(t) + \phi(t) \dot\gamma(t), \end{align*} where $Y_n(t) := Y(t) - \phi(t) \dot\gamma(t)$ is the pointwise orthogonal projection of $Y$ onto $\dot\gamma^\perp$. Why is this an orthogonal decomposition? Because $g(Y_n, \dot\gamma) = g(Y, \dot\gamma) - \phi \cdot g(\dot\gamma, \dot\gamma) = \phi - \phi \cdot 1 = 0$ at every $t$, using $|\dot\gamma|_g = 1$. We need to know how $Y'$ decomposes. Differentiate $Y = Y_n + \phi \dot\gamma$ along $\gamma$ using metric compatibility (the connection is linear and obeys the product rule for the scalar function $\phi$ times the section $\dot\gamma$): \begin{align*} Y' = \nabla_t Y = \nabla_t Y_n + \phi'(t) \dot\gamma(t) + \phi(t) \nabla_t \dot\gamma = \nabla_t Y_n + \phi'(t) \dot\gamma(t), \end{align*} where the last term dropped because $\nabla_t \dot\gamma = 0$ on the geodesic. So $Y'$ splits as a covariant $t$-derivative of the normal part plus a $\phi'(t) \dot\gamma$ tangential piece. To check this is an orthogonal decomposition we need $\nabla_t Y_n \perp \dot\gamma$. Differentiate the identity $g(Y_n, \dot\gamma) \equiv 0$ along $\gamma$ via (M): \begin{align*} 0 = \frac{d}{dt} g(Y_n, \dot\gamma) = g(\nabla_t Y_n, \dot\gamma) + g(Y_n, \nabla_t \dot\gamma) = g(\nabla_t Y_n, \dot\gamma), \end{align*} where the second term again dropped via $\nabla_t \dot\gamma = 0$. So $\nabla_t Y_n \perp \dot\gamma$ pointwise. Setting $Y_n' := \nabla_t Y_n$, this gives the orthogonal decomposition we want. Now compute the squared norm. Since $\nabla_t Y_n \perp \phi'(t) \dot\gamma$ pointwise, the cross term in $|Y'|_g^2 = g(\nabla_t Y_n + \phi'(t) \dot\gamma, \nabla_t Y_n + \phi'(t) \dot\gamma)$ vanishes, leaving \begin{align*} |Y'|_g^2 = |\nabla_t Y_n|_g^2 + \phi'(t)^2 \cdot |\dot\gamma|_g^2 = |Y_n'|_g^2 + \phi'(t)^2, \end{align*} using the unit-speed condition $|\dot\gamma|_g = 1$. Pair $Y'$ with $\dot\gamma$ to identify the Cauchy-Schwarz defect from Step 6. We have $g(Y', \dot\gamma) = g(\nabla_t Y_n, \dot\gamma) + \phi'(t) \cdot g(\dot\gamma, \dot\gamma)$, and the first term is zero by orthogonality just established, while the second is $\phi'(t)$ since $|\dot\gamma|_g = 1$. So \begin{align*} g(Y', \dot\gamma) = \phi'(t). \end{align*} Squaring and integrating gives \begin{align*} \int_0^T g(Y', \dot\gamma)^2 \, d\mathcal{L}^1(t) = \int_0^T \phi'(t)^2 \, d\mathcal{L}^1(t). \end{align*} This is exactly the correction we need to subtract. Why does this cancel? In the energy formula from Part 1 we have $|Y'|_g^2 = |Y_n'|_g^2 + \phi'(t)^2$, while in the length-from-energy formula from Step 6 we subtract $\int \phi'(t)^2 \, d\mathcal{L}^1(t)$. The $\phi'(t)^2$ pieces cancel exactly, leaving only $|Y_n'|_g^2$ inside the integral. Curvature term. The Riemann tensor $R(\cdot, \cdot, \cdot, \cdot)$ is skew in each of its two pairs by [Symmetries of the Riemann Curvature Tensor](/theorems/2704). In particular $R(X, Y, Z, \dot\gamma) = -R(X, Y, \dot\gamma, Z)$ and $R(\dot\gamma, Y, Z, W) = -R(Y, \dot\gamma, Z, W)$. We compute \begin{align*} R(Y, \dot\gamma)Y = R(Y_n + \phi \dot\gamma, \dot\gamma)(Y_n + \phi \dot\gamma). \end{align*} Expanding bilinearly: the first slot gives $R(Y_n, \dot\gamma) + \phi R(\dot\gamma, \dot\gamma)$, and skew-symmetry forces $R(\dot\gamma, \dot\gamma) = 0$, so the first-slot expansion contributes only $R(Y_n, \dot\gamma)$. Then $R(Y_n, \dot\gamma)(Y_n + \phi \dot\gamma) = R(Y_n, \dot\gamma) Y_n + \phi R(Y_n, \dot\gamma) \dot\gamma$. When we now pair with $\dot\gamma$, the second piece becomes $\phi g(R(Y_n, \dot\gamma) \dot\gamma, \dot\gamma)$, which is zero by skew-symmetry in the second pair. So \begin{align*} R(Y, \dot\gamma, Y, \dot\gamma) = g(R(Y, \dot\gamma)Y, \dot\gamma) = g(R(Y_n, \dot\gamma) Y_n, \dot\gamma) = R(Y_n, \dot\gamma, Y_n, \dot\gamma). \end{align*} Every term involving the tangential factor $\phi \dot\gamma$ has been killed by the curvature symmetries. Boundary term. We must show that the boundary term $g(\nabla_s Y(t, 0), \dot\gamma(t))\big|_0^T$ from Part 1 reduces to the normal-component version $g(\nabla_s Y_n(t, 0), \dot\gamma(t))\big|_0^T$ — i.e., the tangential part of $\nabla_s Y$ contributes nothing at the endpoints. This is where the endpoint-fixing hypothesis is decisive, and the argument needs care because we are differentiating at the boundary. Define $\Phi(t, s) := g(Y(t, s), \partial_t H(t, s) / |\partial_t H(t, s)|_g)$ on a neighbourhood of $\{s = 0\}$ where $\partial_t H$ is non-vanishing (this neighbourhood exists by continuity of $\partial_t H$ and unit-speed at $s = 0$). At $s = 0$, $\Phi(t, 0) = g(Y(t), \dot\gamma(t)) = \phi(t)$. The endpoint-fixing hypothesis from Step 1 gives $\partial_s H(0, s) = Y(0, s) \equiv 0$ and $\partial_s H(T, s) = Y(T, s) \equiv 0$ for all $s$, hence $\Phi$ vanishes identically at the boundary in $t$: \begin{align*} \Phi(0, s) \equiv 0, \qquad \Phi(T, s) \equiv 0 \qquad \text{for all } s. \end{align*} Differentiating these identities in $s$ at $s = 0$: \begin{align*} \partial_s \Phi(0, 0) = 0, \qquad \partial_s \Phi(T, 0) = 0, \end{align*} and we also have $\Phi(0, 0) = \phi(0) = 0$, $\Phi(T, 0) = \phi(T) = 0$ since $Y(0) = Y(T) = 0$ from the endpoint-fixing hypothesis. Now decompose $\nabla_s Y$ at the endpoints. The tangential summand of $Y(\cdot, s)$ is $\Phi(\cdot, s)\, \partial_t H/|\partial_t H|_g$. By the product rule for $\nabla_s$ (it is a connection; it satisfies the Leibniz rule on the scalar-times-section product), at $s = 0$ and $t \in \{0, T\}$: \begin{align*} \nabla_s\!\left[\Phi \cdot \tfrac{\partial_t H}{|\partial_t H|_g}\right]\bigg|_{s = 0, t \in \{0, T\}} = \partial_s \Phi(t, 0) \cdot \dot\gamma(t) + \Phi(t, 0) \cdot \nabla_s\!\left[\tfrac{\partial_t H}{|\partial_t H|_g}\right]\bigg|_{s = 0}. \end{align*} Both factors $\partial_s \Phi(t, 0)$ and $\Phi(t, 0)$ are zero at $t \in \{0, T\}$ by the calculation above. So the tangential summand contributes zero to $\nabla_s Y$ at the endpoints, which means $\nabla_s Y$ and $\nabla_s Y_n$ have the same inner product with $\dot\gamma$ at $\{0, T\}$: \begin{align*} g(\nabla_s Y, \dot\gamma)\big|_{t \in \{0, T\}} = g(\nabla_s Y_n, \dot\gamma)\big|_{t \in \{0, T\}}. \end{align*} Assembling. Combining the cancellation of $\phi'(t)^2$, the curvature simplification, and the boundary reduction: \begin{align*} \frac{d^2}{ds^2} \ell(\gamma_s)\Big|_{s = 0} &= \frac{d^2}{ds^2} E(\gamma_s)\Big|_{s = 0} - \int_0^T \phi'(t)^2 \, d\mathcal{L}^1(t) \\ &= g\!\left(\frac{\nabla Y}{ds}(t, 0), \dot\gamma\right)\Big|_0^T + \int_0^T \Big[|Y_n'|_g^2 + \phi'(t)^2 - R(Y_n, \dot\gamma, Y_n, \dot\gamma)\Big] d\mathcal{L}^1(t) - \int_0^T \phi'(t)^2 \, d\mathcal{L}^1(t) \\ &= g\!\left(\frac{\nabla Y_n}{ds}(t, 0),\, \dot\gamma(t)\right)\Big|_0^T + \int_0^T \Big[|Y_n'|_g^2 - R(Y_n, \dot\gamma, Y_n, \dot\gamma)\Big] d\mathcal{L}^1(t). \end{align*} The two $\phi'(t)^2$ pieces cancelled, the curvature term collapsed to its normal part, and the boundary term reduced to the normal component. This is the statement of Part 2. [/guided] [/step]