[proofplan] We reduce the Jacobi equation $J'' + R(\dot\gamma, J)\dot\gamma = 0$ along $\gamma$ to a linear ODE on $\mathbb{R}^n$ using a parallel orthonormal frame $\{X_i(t)\}_{i=1}^n$ with $X_1 = \dot\gamma/|\dot\gamma|$. Writing $J(t) = \sum_i a_i(t) X_i(t)$ converts the Jacobi equation into a linear second-order ODE system for the coefficients $a_i: [0, L] \to \mathbb{R}$ on a compact interval, to which the standard existence-and-uniqueness theorem applies — yielding existence, uniqueness, and dimension $2n$. The longitudinal solution $J(t) = kt\dot\gamma(t)$ is verified by direct substitution using $\dot\gamma$ parallel and $R(\dot\gamma, \dot\gamma) = 0$. Orthogonality is preserved because the function $t \mapsto g(J(t), \dot\gamma(t))$ satisfies a homogeneous second-order ODE with vanishing initial data, hence vanishes identically. Reparametrisation invariance follows from the chain rule applied to the Jacobi ODE under $t \mapsto \lambda t$. [/proofplan]

[step:Set up a parallel orthonormal frame along $\gamma$ and rewrite the Jacobi equation as a linear ODE system]Pick an orthonormal basis $e_1, \ldots, e_n$ of $T_p M$ with $e_1 = \dot\gamma(0)/|\dot\gamma(0)|$ (assuming $\dot\gamma(0) \ne 0$, since $\gamma$ is a geodesic; the constant-speed property from [Geodesics Have Constant Speed](/theorems/2709) ensures $\dot\gamma(t) \ne 0$ for all $t$). For each $i$, let \begin{align*} X_i: [0, L] &\to TM \\ t &\mapsto X_i(t) \in T_{\gamma(t)} M \end{align*} be the parallel transport of $e_i$ along $\gamma$, defined by the parallel transport ODE $\nabla_{dt} X_i = 0$ with $X_i(0) = e_i$ — see the [Covariant Derivative Along a Curve](/theorems/2708). Parallel transport along $\gamma$ is a $g$-isometry $T_p M \to T_{\gamma(t)} M$, so $\{X_i(t)\}_{i=1}^n$ is a $g_{\gamma(t)}$-orthonormal frame for $T_{\gamma(t)} M$ at every $t$. Moreover, $X_1(t) = \dot\gamma(t)/|\dot\gamma(0)|$ for all $t$: both sides are parallel along $\gamma$ (the left by construction, the right because $\nabla_{dt} \dot\gamma = 0$ since $\gamma$ is a geodesic, so $\nabla_{dt}(\dot\gamma/|\dot\gamma(0)|) = 0$) and they coincide at $t = 0$. A vector field $J$ along $\gamma$ has a unique expansion \begin{align*} J(t) = \sum_{i=1}^n a_i(t)\, X_i(t), \qquad a_i: [0, L] \to \mathbb{R}, \quad a_i \in C^\infty([0, L]). \end{align*} Since each $X_i$ is parallel ($\nabla_{dt} X_i = 0$), the Leibniz rule for $\nabla_{dt}$ gives \begin{align*} J'(t) &:= \frac{\nabla J}{dt}(t) = \sum_{i=1}^n \dot a_i(t)\, X_i(t), \\ J''(t) &:= \frac{\nabla^2 J}{dt^2}(t) = \sum_{i=1}^n \ddot a_i(t)\, X_i(t). \end{align*} The Jacobi equation $J'' + R(\dot\gamma, J)\dot\gamma = 0$ thus reads \begin{align*} \sum_i \ddot a_i(t)\, X_i(t) + \sum_j a_j(t)\, R(\dot\gamma(t), X_j(t))\dot\gamma(t) = 0. \end{align*} Define the smooth coefficient functions $A_{ij}: [0, L] \to \mathbb{R}$ by \begin{align*} A_{ij}(t) := g_{\gamma(t)}\bigl(R(\dot\gamma(t), X_j(t))\dot\gamma(t),\, X_i(t)\bigr). \end{align*} Smoothness of $A_{ij}$ follows from smoothness of $g$, $R$, $\dot\gamma$, and $X_j$. Taking the inner product of the Jacobi equation with $X_i(t)$ and using $g(X_i, X_k) = \delta_{ik}$ yields the linear second-order ODE system \begin{align*} \ddot a_i(t) + \sum_{j=1}^n A_{ij}(t)\, a_j(t) = 0, \qquad i = 1, \ldots, n. \tag{$\ast$} \end{align*}[/step]

[guided]The Jacobi equation $J'' + R(\dot\gamma, J)\dot\gamma = 0$ is intrinsically a coordinate-free linear second-order ODE for vector fields along $\gamma$. To apply the classical existence-and-uniqueness theorem for ODEs we need scalar coefficients on a fixed vector space. The standard device is to **parallel-transport an orthonormal frame** along $\gamma$ and expand $J$ in it: this gives a global frame for the bundle $\gamma^*TM$ over $[0, L]$, identifying it as $[0, L] \times \mathbb{R}^n$, and turns vector-valued ODEs into matrix ODEs. Concretely, pick an orthonormal basis $e_1, \ldots, e_n$ of $T_p M$ with $e_1 = \dot\gamma(0)/|\dot\gamma(0)|$. (This is possible because the Gram-Schmidt procedure on any basis containing $\dot\gamma(0)$ produces such an orthonormal basis; we use that $\dot\gamma(0) \ne 0$, which holds for any geodesic with $\dot\gamma \not\equiv 0$ combined with the constant-speed property of [Geodesics Have Constant Speed](/theorems/2709).) Define \begin{align*} X_i: [0, L] &\to TM, \\ t &\mapsto X_i(t) \in T_{\gamma(t)} M \end{align*} to be the parallel transport of $e_i$ along $\gamma$, the unique solution of $\nabla_{dt} X_i = 0$ with $X_i(0) = e_i$ from the [Covariant Derivative Along a Curve](/theorems/2708). **Why does $\{X_i(t)\}$ remain an orthonormal frame at every $t$?** Because the Levi-Civita connection is metric-compatible: for any parallel vector fields $V, W$ along $\gamma$, $\frac{d}{dt} g(V, W) = g(\nabla_{dt} V, W) + g(V, \nabla_{dt} W) = 0$, so $g(V, W)$ is constant. Hence $g(X_i(t), X_j(t)) = g(e_i, e_j) = \delta_{ij}$ for all $t$. **Why is $X_1(t)$ proportional to $\dot\gamma(t)$?** Since $\gamma$ is a geodesic, $\nabla_{dt}\dot\gamma = 0$, so $\dot\gamma$ is parallel along $\gamma$. Hence $\dot\gamma/|\dot\gamma(0)|$ is parallel and equals $e_1$ at $t = 0$; by uniqueness of parallel transport, $X_1(t) = \dot\gamma(t)/|\dot\gamma(0)|$. (The denominator is constant in $t$ because $|\dot\gamma|$ is constant on a geodesic.) With the frame in hand, every vector field $J$ along $\gamma$ has a unique expansion in this basis: \begin{align*} J(t) = \sum_{i=1}^n a_i(t)\, X_i(t), \qquad a_i: [0, L] \to \mathbb{R}, \quad a_i \in C^\infty([0, L]). \end{align*} Uniqueness comes from $\{X_i(t)\}$ being a basis at each $t$; smoothness of $a_i$ comes from smoothness of $J$ and of $X_i$. The whole point of using a **parallel** frame is that the covariant derivative now passes through the expansion: differentiating term-by-term and using $\nabla_{dt} X_i = 0$, \begin{align*} J'(t) := \frac{\nabla J}{dt}(t) = \sum_i \dot a_i(t)\, X_i(t) + \sum_i a_i(t) \cdot \underbrace{\nabla_{dt} X_i(t)}_{= 0} = \sum_i \dot a_i(t)\, X_i(t). \end{align*} Differentiating once more in exactly the same way (the $X_i$ are still parallel, so the second term vanishes again): \begin{align*} J''(t) := \frac{\nabla^2 J}{dt^2}(t) = \sum_{i=1}^n \ddot a_i(t)\, X_i(t). \end{align*} Without parallelism we would pick up extra $\nabla_{dt} X_i$ terms — the whole reduction would fail. This is the structural reason for choosing a parallel frame. Substitute these expansions into the Jacobi equation $J'' + R(\dot\gamma, J)\dot\gamma = 0$: \begin{align*} \sum_i \ddot a_i(t)\, X_i(t) + \sum_j a_j(t)\, R(\dot\gamma(t), X_j(t))\dot\gamma(t) = 0. \end{align*} Now take the $g$-inner product with each $X_i(t)$ and use orthonormality $g(X_i, X_k) = \delta_{ik}$. The first sum collapses to $\ddot a_i(t)$. The second produces the smooth coefficient matrix \begin{align*} A_{ij}(t) := g_{\gamma(t)}\bigl(R(\dot\gamma(t), X_j(t))\dot\gamma(t),\, X_i(t)\bigr), \end{align*} which is smooth in $t$ because $g$, $R$, $\dot\gamma$, and $X_j$ are smooth (in particular, the curvature tensor is smooth on $M$). The resulting scalar system \begin{align*} \ddot a_i(t) + \sum_{j=1}^n A_{ij}(t)\, a_j(t) = 0, \qquad i = 1, \ldots, n \end{align*} is exactly the system $(\ast)$. We have reduced the Jacobi equation — a coordinate-free vector-valued ODE on a manifold — to a linear second-order ODE system on a compact interval, which is the form to which classical ODE theory applies.[/guided]

[step:Apply the linear ODE existence-and-uniqueness theorem to $(\ast)$ to obtain $J$ from initial data $(u, v)$] Write $u = \sum_i u_i e_i$ and $v = \sum_i v_i e_i$ in the chosen basis of $T_p M$. The initial conditions $J(0) = u$, $J'(0) = v$ translate to \begin{align*} a_i(0) = u_i, \qquad \dot a_i(0) = v_i, \qquad i = 1, \ldots, n. \end{align*} The system $(\ast)$ is a linear second-order ODE system on $[0, L]$ with smooth coefficients $A_{ij} \in C^\infty([0, L])$. Equivalently, setting $b_i := \dot a_i$, it becomes the first-order linear system \begin{align*} \frac{d}{dt} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 & I \\ -A(t) & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix}, \qquad \begin{pmatrix} a(0) \\ b(0) \end{pmatrix} = \begin{pmatrix} u_i \\ v_i \end{pmatrix}, \end{align*} where $a, b \in \mathbb{R}^n$ and $A(t) = (A_{ij}(t)) \in \mathbb{R}^{n \times n}$ is smooth. The coefficient matrix is continuous on the compact interval $[0, L]$, so by the existence and uniqueness theorem for linear ODEs, the system admits a unique global solution $(a, b): [0, L] \to \mathbb{R}^{2n}$ for every initial datum $(u_1, \ldots, u_n, v_1, \ldots, v_n) \in \mathbb{R}^{2n}$. Setting $J(t) := \sum_i a_i(t) X_i(t)$ recovers a unique smooth Jacobi field along $\gamma$ with $J(0) = u$ and $J'(0) = v$. The map $T_p M \times T_p M \to \{\text{Jacobi fields along } \gamma\}$ sending $(u, v) \mapsto J$ is linear (since $(\ast)$ is linear in $(a, b)$ and the initial-data assignment is linear) and bijective (existence-uniqueness). The space of Jacobi fields therefore has dimension $\dim(T_p M \times T_p M) = 2n$. [/step]

[step:Verify the longitudinal solution $J(t) = kt\dot\gamma(t)$ by substitution] Suppose $J(0) = 0$ and $J'(0) = k\dot\gamma(0)$. Define \begin{align*} \tilde J: [0, L] &\to TM \\ t &\mapsto kt\dot\gamma(t) \in T_{\gamma(t)} M. \end{align*} We verify that $\tilde J$ is a Jacobi field with the same initial conditions, then invoke uniqueness from the previous step to conclude $J = \tilde J$. **Initial conditions.** $\tilde J(0) = 0 = J(0)$. For the derivative, \begin{align*} \frac{\nabla \tilde J}{dt}(t) = k\dot\gamma(t) + kt \cdot \underbrace{\nabla_{dt} \dot\gamma}_{= 0} = k\dot\gamma(t), \end{align*} since $\gamma$ is a geodesic. At $t = 0$ this gives $\tilde J'(0) = k\dot\gamma(0) = J'(0)$. **Jacobi equation.** Differentiating once more, \begin{align*} \frac{\nabla^2 \tilde J}{dt^2}(t) = \frac{\nabla}{dt}\bigl(k\dot\gamma(t)\bigr) = 0. \end{align*} For the curvature term, \begin{align*} R(\dot\gamma, \tilde J)\dot\gamma = R(\dot\gamma, kt\dot\gamma)\dot\gamma = kt\, R(\dot\gamma, \dot\gamma)\dot\gamma = 0, \end{align*} using $\mathbb{R}$-linearity of $R$ in the second slot and the antisymmetry $R(X, X) = 0$ (which is the first symmetry from the [Symmetries of the Riemann Curvature Tensor](/theorems/2704)). Hence $\tilde J'' + R(\dot\gamma, \tilde J)\dot\gamma = 0 + 0 = 0$, so $\tilde J$ satisfies the Jacobi equation. By uniqueness of Jacobi fields with prescribed initial data (Step 2), $J(t) = \tilde J(t) = kt\dot\gamma(t)$. [/step]

[step:Show that orthogonality to $\dot\gamma$ is preserved by the Jacobi flow via a scalar ODE for $g(J, \dot\gamma)$]Suppose $J(0)$ and $J'(0)$ are both $g_p$-orthogonal to $\dot\gamma(0)$. Define \begin{align*} \varphi: [0, L] &\to \mathbb{R} \\ t &\mapsto g_{\gamma(t)}(J(t), \dot\gamma(t)). \end{align*} We compute $\ddot\varphi$ and show $\varphi \equiv 0$. By metric compatibility of $\nabla$ and $\nabla_{dt} \dot\gamma = 0$, \begin{align*} \dot\varphi(t) = g\bigl(\nabla_{dt} J,\, \dot\gamma\bigr) + g\bigl(J,\, \nabla_{dt} \dot\gamma\bigr) = g(J', \dot\gamma). \end{align*} Differentiating again, again using metric compatibility and $\nabla_{dt}\dot\gamma = 0$, \begin{align*} \ddot\varphi(t) = g(J'', \dot\gamma) + g(J', \nabla_{dt} \dot\gamma) = g(J'', \dot\gamma). \end{align*} Substituting the Jacobi equation $J'' = -R(\dot\gamma, J)\dot\gamma$, \begin{align*} \ddot\varphi(t) = -g\bigl(R(\dot\gamma, J)\dot\gamma,\, \dot\gamma\bigr). \end{align*} By the pair symmetry $g(R(X, Y)Z, W) = g(R(Z, W)X, Y)$ from the [Symmetries of the Riemann Curvature Tensor](/theorems/2704), \begin{align*} g\bigl(R(\dot\gamma, J)\dot\gamma,\, \dot\gamma\bigr) = g\bigl(R(\dot\gamma, \dot\gamma) \dot\gamma,\, J\bigr) = 0, \end{align*} where the last equality uses $R(\dot\gamma, \dot\gamma) = 0$ from the antisymmetry in the first pair. Hence $\ddot\varphi \equiv 0$ on $[0, L]$. Since $\varphi$ has $\ddot\varphi \equiv 0$, $\varphi(t) = \varphi(0) + t\dot\varphi(0)$. By hypothesis, $\varphi(0) = g_p(J(0), \dot\gamma(0)) = 0$ and $\dot\varphi(0) = g_p(J'(0), \dot\gamma(0)) = 0$, so $\varphi \equiv 0$. This proves $g(J(t), \dot\gamma(t)) = 0$ for all $t$, i.e., $J(t) \perp \dot\gamma(t)$. The space of Jacobi fields with $J(0), J'(0) \perp \dot\gamma(0)$ corresponds via Step 2 to pairs $(u, v) \in (\dot\gamma(0))^\perp \times (\dot\gamma(0))^\perp$, which has dimension $2(n - 1)$.[/step]

[guided]The clean way to show that orthogonality is preserved is to introduce a scalar test function and derive a closed ODE for it. Define \begin{align*} \varphi: [0, L] &\to \mathbb{R} \\ t &\mapsto g_{\gamma(t)}(J(t), \dot\gamma(t)). \end{align*} The strategy: show that the Jacobi equation, combined with the symmetries of the Riemann tensor, forces $\ddot\varphi \equiv 0$. Then $\varphi$ is at most affine in $t$, and vanishing initial data $\varphi(0) = \dot\varphi(0) = 0$ gives $\varphi \equiv 0$ — which is precisely the statement that $J(t) \perp \dot\gamma(t)$ for all $t$. **First derivative.** Metric compatibility of the Levi-Civita connection gives the product rule $\frac{d}{dt} g(V, W) = g(\nabla_{dt} V, W) + g(V, \nabla_{dt} W)$ for vector fields along $\gamma$. Applied to $V = J$, $W = \dot\gamma$, with $\nabla_{dt}\dot\gamma = 0$ (the geodesic equation kills the second term): \begin{align*} \dot\varphi(t) = g\bigl(\nabla_{dt} J,\, \dot\gamma\bigr) + g\bigl(J,\, \nabla_{dt}\dot\gamma\bigr) = g(J'(t), \dot\gamma(t)). \end{align*} **Second derivative.** Differentiate again, using the same product rule and $\nabla_{dt}\dot\gamma = 0$ once more: \begin{align*} \ddot\varphi(t) = g(J''(t), \dot\gamma(t)) + g(J'(t), \nabla_{dt}\dot\gamma(t)) = g(J''(t), \dot\gamma(t)). \end{align*} **Use Jacobi.** This is where the Jacobi equation enters: $J'' = -R(\dot\gamma, J)\dot\gamma$. Substituting, \begin{align*} \ddot\varphi(t) = -g\bigl(R(\dot\gamma(t), J(t))\dot\gamma(t),\, \dot\gamma(t)\bigr). \end{align*} **Why does this vanish?** This is the geometric heart of the argument and uses *both* curvature symmetries from the [Symmetries of the Riemann Curvature Tensor](/theorems/2704). Recall the $(0,4)$-curvature $R(X, Y, Z, W) := g(R(X, Y)Z, W)$. Theorem 2704 gives us - **pair symmetry:** $R(X, Y, Z, W) = R(Z, W, X, Y)$; - **antisymmetry in the first pair:** $R(X, X, Z, W) = 0$. Apply the pair symmetry with $X = \dot\gamma$, $Y = J$, $Z = \dot\gamma$, $W = \dot\gamma$: \begin{align*} g\bigl(R(\dot\gamma, J)\dot\gamma,\, \dot\gamma\bigr) = R(\dot\gamma, J, \dot\gamma, \dot\gamma) = R(\dot\gamma, \dot\gamma, \dot\gamma, J) = 0, \end{align*} where the last equality uses the antisymmetry in the first pair (the first two slots are identically $\dot\gamma, \dot\gamma$). Hence $\ddot\varphi \equiv 0$ on $[0, L]$. **Conclude $\varphi \equiv 0$.** From $\ddot\varphi \equiv 0$ we integrate twice to get $\varphi(t) = \varphi(0) + t\dot\varphi(0)$. The hypotheses now do their job: $J(0) \perp \dot\gamma(0)$ gives $\varphi(0) = g_p(J(0), \dot\gamma(0)) = 0$, and $J'(0) \perp \dot\gamma(0)$ gives $\dot\varphi(0) = g_p(J'(0), \dot\gamma(0)) = 0$. Hence $\varphi \equiv 0$, which means $g(J(t), \dot\gamma(t)) = 0$ for all $t \in [0, L]$, i.e., $J(t) \perp \dot\gamma(t)$ everywhere. **Dimension count.** The map $(u, v) \mapsto J$ from Step 2 restricts to a linear isomorphism between pairs $(u, v)$ with $u, v \in (\dot\gamma(0))^\perp$ and Jacobi fields perpendicular everywhere — both directions hold: ($\Rightarrow$) is what we just proved; ($\Leftarrow$) if $J \perp \dot\gamma$ everywhere, then in particular $g(J(0), \dot\gamma(0)) = 0$ and $\frac{d}{dt}\big|_0 g(J, \dot\gamma) = g(J'(0), \dot\gamma(0)) = 0$. The dimension of $(\dot\gamma(0))^\perp \subset T_p M$ is $n - 1$, so the total Jacobi-perpendicular subspace has dimension $2(n - 1)$.[/guided]

[step:Establish reparametrisation invariance via the chain rule on the Jacobi equation] Suppose $\tilde\gamma(t) = \gamma(\lambda t)$ for $\lambda \ne 0$. Then $\dot{\tilde\gamma}(t) = \lambda \dot\gamma(\lambda t)$. Let $J$ be a Jacobi field along $\gamma$, and define \begin{align*} \tilde J: [0, L/\lambda] &\to TM \\ t &\mapsto J(\lambda t) \in T_{\gamma(\lambda t)} M = T_{\tilde\gamma(t)} M. \end{align*} By the chain rule for the covariant derivative along a reparametrised curve, \begin{align*} \frac{\nabla \tilde J}{dt}(t) = \lambda\, \frac{\nabla J}{dt}(\lambda t), \qquad \frac{\nabla^2 \tilde J}{dt^2}(t) = \lambda^2\, \frac{\nabla^2 J}{dt^2}(\lambda t). \end{align*} Substituting the Jacobi equation $J'' + R(\dot\gamma, J)\dot\gamma = 0$ at the point $\lambda t$, \begin{align*} \frac{\nabla^2 \tilde J}{dt^2}(t) = -\lambda^2\, R(\dot\gamma(\lambda t), J(\lambda t))\dot\gamma(\lambda t). \end{align*} Using $\mathbb{R}$-bilinearity of $R$ in its first and third slots, \begin{align*} \lambda^2\, R(\dot\gamma(\lambda t), J(\lambda t))\dot\gamma(\lambda t) = R(\lambda \dot\gamma(\lambda t),\, J(\lambda t))\, \lambda \dot\gamma(\lambda t) = R(\dot{\tilde\gamma}(t), \tilde J(t))\, \dot{\tilde\gamma}(t). \end{align*} Hence \begin{align*} \frac{\nabla^2 \tilde J}{dt^2}(t) + R(\dot{\tilde\gamma}(t), \tilde J(t))\, \dot{\tilde\gamma}(t) = 0, \end{align*} so $\tilde J$ satisfies the Jacobi equation along $\tilde\gamma$. This shows the map $J \mapsto J \circ (\lambda \cdot)$ takes Jacobi fields along $\gamma$ to Jacobi fields along $\tilde\gamma$, and therefore the property of being a Jacobi field is independent of the parametrisation in the stated sense. [/step]

Combining the four steps yields the existence and uniqueness with $2n$-dimensional solution space (Steps 1-2), the longitudinal formula $J(t) = kt\dot\gamma(t)$ (Step 3), the orthogonality preservation with $2(n-1)$-dimensional perpendicular subspace (Step 4), and the reparametrisation invariance (Step 5). The proof is complete.