[proofplan] We build explicit test fields along $\gamma$ by multiplying a parallel orthonormal frame perpendicular to $\dot{\gamma}$ by the scalar sine function that vanishes at both endpoints. Parallelism makes the covariant derivative term exactly the derivative of the scalar factor, while the curvature terms sum to the Ricci curvature in the geodesic direction. The Ricci lower bound then gives an upper estimate by an elementary trigonometric integral, and the assumption $\ell > \pi/\sqrt{K}$ makes that estimate strictly negative. [/proofplan] [step:Choose a parallel orthonormal frame perpendicular to the geodesic] Since $\gamma$ is unit-speed, $|\dot{\gamma}(0)|=1$. Choose an [orthonormal basis](/page/Orthonormal%20Basis) \begin{align*} e_1,\dots,e_{n-1} \end{align*} of the orthogonal complement \begin{align*} \dot{\gamma}(0)^\perp = \{w \in T_{\gamma(0)}M : g_{\gamma(0)}(w,\dot{\gamma}(0))=0\}. \end{align*} For each $i \in \{1,\dots,n-1\}$, let \begin{align*} E_i: [0,\ell] &\to TM \\ t &\mapsto E_i(t) \in T_{\gamma(t)}M \end{align*} be the unique parallel vector field along $\gamma$ satisfying $E_i(0)=e_i$, so \begin{align*} \nabla_{\dot{\gamma}}E_i(t)=0 \end{align*} for every $t \in [0,\ell]$. Metric compatibility of the Levi-Civita connection gives \begin{align*} \frac{d}{dt}g_{\gamma(t)}(E_i(t),E_j(t)) = g_{\gamma(t)}(\nabla_{\dot{\gamma}}E_i(t),E_j(t)) + g_{\gamma(t)}(E_i(t),\nabla_{\dot{\gamma}}E_j(t)) = 0. \end{align*} Thus $g_{\gamma(t)}(E_i(t),E_j(t))=\delta_{ij}$ for all $t$. Since $\gamma$ is a geodesic, \begin{align*} \nabla_{\dot{\gamma}}\dot{\gamma}(t)=0, \end{align*} and therefore \begin{align*} \frac{d}{dt}g_{\gamma(t)}(E_i(t),\dot{\gamma}(t)) = 0. \end{align*} Because $g_{\gamma(0)}(E_i(0),\dot{\gamma}(0))=0$, we have \begin{align*} g_{\gamma(t)}(E_i(t),\dot{\gamma}(t))=0 \end{align*} for every $t \in [0,\ell]$. [/step] [step:Define sine test fields that vanish at both endpoints] Define the smooth scalar function \begin{align*} f: [0,\ell] &\to \mathbb{R} \\ t &\mapsto \sin\left(\frac{\pi t}{\ell}\right). \end{align*} For each $i \in \{1,\dots,n-1\}$, define the smooth vector field along $\gamma$ \begin{align*} V_i: [0,\ell] &\to TM \\ t &\mapsto f(t)E_i(t) \in T_{\gamma(t)}M. \end{align*} Since $f(0)=f(\ell)=0$, each $V_i$ satisfies $V_i(0)=V_i(\ell)=0$. Since each $E_i(t)$ is perpendicular to $\dot{\gamma}(t)$, each $V_i(t)$ is perpendicular to $\dot{\gamma}(t)$. [guided] The endpoint condition suggests using a scalar factor that is zero at $0$ and $\ell$. The first Dirichlet sine function is the natural choice because its derivative and its square have exactly computable integrals. We define \begin{align*} f: [0,\ell] &\to \mathbb{R} \\ t &\mapsto \sin\left(\frac{\pi t}{\ell}\right). \end{align*} Then $f(0)=0$ and $f(\ell)=0$. For each $i \in \{1,\dots,n-1\}$, define \begin{align*} V_i: [0,\ell] &\to TM \\ t &\mapsto f(t)E_i(t) \in T_{\gamma(t)}M. \end{align*} This is a smooth vector field along $\gamma$ because $f$ is smooth and $E_i$ is smooth. The endpoint values are \begin{align*} V_i(0)=f(0)E_i(0)=0, \qquad V_i(\ell)=f(\ell)E_i(\ell)=0. \end{align*} The perpendicularity condition is also preserved by scalar multiplication: \begin{align*} g_{\gamma(t)}(V_i(t),\dot{\gamma}(t)) = f(t)g_{\gamma(t)}(E_i(t),\dot{\gamma}(t)) = 0 \end{align*} for every $t \in [0,\ell]$. [/guided] [/step] [step:Compute the summed index form in terms of Ricci curvature] For each $i$, the product rule for covariant differentiation along $\gamma$ gives \begin{align*} \nabla_{\dot{\gamma}}V_i(t) = f'(t)E_i(t)+f(t)\nabla_{\dot{\gamma}}E_i(t) = f'(t)E_i(t). \end{align*} Since $|E_i(t)|=1$, \begin{align*} |\nabla_{\dot{\gamma}}V_i(t)|^2 = (f'(t))^2. \end{align*} Also, \begin{align*} g_{\gamma(t)}(R(V_i(t),\dot{\gamma}(t))\dot{\gamma}(t),V_i(t)) = f(t)^2 g_{\gamma(t)}(R(E_i(t),\dot{\gamma}(t))\dot{\gamma}(t),E_i(t)). \end{align*} Because $E_1(t),\dots,E_{n-1}(t),\dot{\gamma}(t)$ is an orthonormal basis of $T_{\gamma(t)}M$ and the term with $\dot{\gamma}(t)$ contributes zero, the Ricci trace gives \begin{align*} \sum_{i=1}^{n-1} g_{\gamma(t)}(R(E_i(t),\dot{\gamma}(t))\dot{\gamma}(t),E_i(t)) = \operatorname{Ric}_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t)). \end{align*} Hence \begin{align*} \sum_{i=1}^{n-1} I_\gamma(V_i,V_i) = \int_0^\ell \left( (n-1)(f'(t))^2 - f(t)^2\operatorname{Ric}_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t)) \right) \, d\mathcal{L}^1(t). \end{align*} [/step] [step:Use the Ricci lower bound to obtain a strict negative upper estimate] Since $\gamma$ is unit-speed, \begin{align*} g_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t))=1. \end{align*} The Ricci lower bound therefore gives \begin{align*} \operatorname{Ric}_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t)) \geq (n-1)K \end{align*} for every $t \in [0,\ell]$. Since $f(t)^2 \geq 0$, the previous identity implies \begin{align*} \sum_{i=1}^{n-1} I_\gamma(V_i,V_i) \leq (n-1) \int_0^\ell \left( (f'(t))^2 - K f(t)^2 \right) \, d\mathcal{L}^1(t). \end{align*} Now \begin{align*} f'(t) = \frac{\pi}{\ell}\cos\left(\frac{\pi t}{\ell}\right), \end{align*} so \begin{align*} \int_0^\ell (f'(t))^2 \, d\mathcal{L}^1(t) &= \left(\frac{\pi}{\ell}\right)^2 \int_0^\ell \cos^2\left(\frac{\pi t}{\ell}\right) \, d\mathcal{L}^1(t), \\ \int_0^\ell f(t)^2 \, d\mathcal{L}^1(t) &= \int_0^\ell \sin^2\left(\frac{\pi t}{\ell}\right) \, d\mathcal{L}^1(t). \end{align*} Using \begin{align*} \int_0^\ell \cos^2\left(\frac{\pi t}{\ell}\right) \, d\mathcal{L}^1(t) = \int_0^\ell \sin^2\left(\frac{\pi t}{\ell}\right) \, d\mathcal{L}^1(t) = \frac{\ell}{2}, \end{align*} we get \begin{align*} \sum_{i=1}^{n-1} I_\gamma(V_i,V_i) &\leq (n-1) \left( \left(\frac{\pi}{\ell}\right)^2\frac{\ell}{2} - K\frac{\ell}{2} \right) \\ &= \frac{(n-1)\ell}{2} \left( \frac{\pi^2}{\ell^2} - K \right). \end{align*} The hypothesis $\ell > \pi/\sqrt{K}$ is equivalent to $\pi^2/\ell^2 < K$, and therefore the final quantity is strictly negative. Thus \begin{align*} \sum_{i=1}^{n-1} I_\gamma(V_i,V_i)<0. \end{align*} [/step]