[proofplan] For Part 1 we compute $\frac{\nabla^2}{\partial t^2}\frac{\partial f}{\partial s}$ using the symmetry $\nabla_{\partial_s}\partial_t f = \nabla_{\partial_t}\partial_s f$ (which holds because $\partial_t$ and $\partial_s$ are pushforwards of the commuting coordinate fields on $[a, b] \times (-\varepsilon, \varepsilon)$ and the Levi-Civita connection is torsion-free), the geodesic equation $\nabla_{\partial_t}\partial_t f = 0$ along each $s$-slice, and the curvature commutator from [Curvature as Commutator of Covariant Derivatives](/theorems/2703) — under the chapter sign convention $R = -\nabla\circ\nabla$ this commutator carries a minus sign and produces exactly $-R(\dot\gamma, J)\dot\gamma$, so $J$ satisfies the Jacobi equation $J'' + R(\dot\gamma, J)\dot\gamma = 0$. For Part 2, given any Jacobi field $J$ along $\gamma$ with $J(0) = u$ and $J'(0) = v$, we construct a geodesic variation realising it: pick a curve $\alpha$ in $M$ with $\alpha(0) = p$ and $\dot\alpha(0) = u$, parallel-transport a chosen vector along $\alpha$ to define a smooth one-parameter family of initial velocities, and let $f(t, s)$ be the geodesic flow of these initial velocities. Direct computation using Part 1 and uniqueness of Jacobi fields with prescribed initial data — from the [Structure of Jacobi Fields](/theorems/2715) — identifies the variation field with $J$. [/proofplan]

[step:Set up the variation, define the canonical vector fields $\partial_t f$ and $\partial_s f$, and record the bracket relation] We first install the standard apparatus for variations. Let \begin{align*} f: [a, b] \times (-\varepsilon, \varepsilon) &\to M \\ (t, s) &\mapsto f(t, s) \end{align*} be the variation of the hypothesis. For each fixed $s \in (-\varepsilon, \varepsilon)$, the curve $t \mapsto f(t, s)$ is a geodesic in $M$, and $f(t, 0) = \gamma(t)$. Let $\partial_t, \partial_s$ denote the standard coordinate vector fields on the open subset $[a, b] \times (-\varepsilon, \varepsilon) \subset \mathbb{R}^2$. We use the same symbols $\partial_t f, \partial_s f$ for their pushforwards under $f$, viewed as smooth maps \begin{align*} \partial_t f: [a, b] \times (-\varepsilon, \varepsilon) &\to TM, & (t, s) &\mapsto df_{(t, s)}(\partial_t) \in T_{f(t,s)} M, \\ \partial_s f: [a, b] \times (-\varepsilon, \varepsilon) &\to TM, & (t, s) &\mapsto df_{(t, s)}(\partial_s) \in T_{f(t,s)} M. \end{align*} Two standard facts are recorded for use: \begin{enumerate} \item[(i)] $\partial_t f(\cdot, 0) = \dot\gamma$ along $\gamma$, since $f(t, 0) = \gamma(t)$. \item[(ii)] $\partial_s f(\cdot, 0) = J$ by the definition $J(t) := \partial_s f(t, 0)$. \end{enumerate} We use the conventions of the [Covariant Derivative Along a Curve](/theorems/2708): for a smooth vector field $V$ along $f$ — that is, a smooth section of $f^* TM \to [a, b] \times (-\varepsilon, \varepsilon)$ — write $\nabla_{\partial_t} V$ and $\nabla_{\partial_s} V$ for the partial covariant derivatives. The commuting-mixed-partials identity for the Levi-Civita connection states \begin{align*} \nabla_{\partial_s} \partial_t f = \nabla_{\partial_t} \partial_s f. \tag{$\diamond$} \end{align*} This is immediate from the torsion-free property $\nabla_X Y - \nabla_Y X = [X, Y]$ applied to $X = df(\partial_t)$, $Y = df(\partial_s)$, together with $[df(\partial_t), df(\partial_s)] = df([\partial_t, \partial_s]) = df(0) = 0$ since $\partial_t, \partial_s$ commute on $\mathbb{R}^2$. Finally, the geodesic hypothesis says: for every $s$, \begin{align*} \nabla_{\partial_t}\, \partial_t f(\cdot, s) = 0. \tag{$\heartsuit$} \end{align*} [/step]

[step:Compute $\nabla_{\partial_t}^2 \partial_s f$ and recognise it as $-R(\partial_t f, \partial_s f)\partial_t f$]We compute, on the rectangle $[a, b] \times (-\varepsilon, \varepsilon)$: \begin{align*} \nabla_{\partial_t}\nabla_{\partial_t}\, \partial_s f &\stackrel{(\diamond)}{=} \nabla_{\partial_t}\nabla_{\partial_s}\, \partial_t f. \end{align*} We use the [Curvature as Commutator of Covariant Derivatives](/theorems/2703) along the map $f$. With the chapter sign convention $R = -\nabla \circ \nabla$, the curvature endomorphism is defined so that \begin{align*} R(X, Y)Z = -\nabla_X\nabla_Y Z + \nabla_Y\nabla_X Z + \nabla_{[X, Y]} Z, \end{align*} i.e., $\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X, Y]} = -R(X, Y)$. Pulled back along the smooth map $f$ with commuting coordinate fields $[\partial_t, \partial_s] = 0$, this reads \begin{align*} \nabla_{\partial_t}\nabla_{\partial_s} V - \nabla_{\partial_s}\nabla_{\partial_t} V = -R(\partial_t f, \partial_s f)\, V \end{align*} for any smooth section $V$ of $f^*TM$. The hypotheses are: $f$ smooth, $V$ smooth, and $[\partial_t, \partial_s] = 0$ on the parameter domain — all hold here. Applied to $V = \partial_t f$, \begin{align*} \nabla_{\partial_t}\nabla_{\partial_s}\, \partial_t f = \nabla_{\partial_s}\nabla_{\partial_t}\, \partial_t f - R(\partial_t f, \partial_s f)\, \partial_t f. \end{align*} By the geodesic hypothesis $(\heartsuit)$, $\nabla_{\partial_t}\partial_t f = 0$ on the entire rectangle, hence $\nabla_{\partial_s}\nabla_{\partial_t}\partial_t f = 0$ as well. Therefore \begin{align*} \nabla_{\partial_t}\nabla_{\partial_t}\, \partial_s f = -R(\partial_t f, \partial_s f)\, \partial_t f. \end{align*} Restricting to $s = 0$ and using (i)-(ii) above, \begin{align*} J''(t) := \nabla_{\partial_t}\nabla_{\partial_t}\, \partial_s f(t, 0) = -R(\dot\gamma(t), J(t))\, \dot\gamma(t). \end{align*} That is, \begin{align*} J''(t) + R(\dot\gamma(t), J(t))\, \dot\gamma(t) = 0, \end{align*} which is the Jacobi equation. This proves Part 1.[/step]

[guided]We want to compute $\nabla_{\partial_t}^2 \partial_s f$ on the rectangle $[a, b] \times (-\varepsilon, \varepsilon)$ and show it equals $-R(\partial_t f, \partial_s f) \partial_t f$. This is the canonical "geodesic variation $\Rightarrow$ Jacobi" derivation, and it uses three ingredients. We name each one, check its hypotheses, and assemble them. **Ingredient 1: the symmetry $\nabla_{\partial_s}\partial_t f = \nabla_{\partial_t}\partial_s f$.** Why does this hold? The Levi-Civita connection is torsion-free: $\nabla_X Y - \nabla_Y X = [X, Y]$ for any smooth vector fields $X, Y$. Applied to the pushforwards $X = df(\partial_t)$ and $Y = df(\partial_s)$, and using that $df$ pushes brackets to brackets on coordinate fields with $[\partial_t, \partial_s] = 0$ on $\mathbb{R}^2$, we get exactly the identity $(\diamond)$ from Step 1. So we may freely swap the order of mixed covariant derivatives along $f$. The first move is therefore \begin{align*} \nabla_{\partial_t}\nabla_{\partial_t}\, \partial_s f \stackrel{(\diamond)}{=} \nabla_{\partial_t}\nabla_{\partial_s}\, \partial_t f. \end{align*} **Ingredient 2: the curvature commutator along the map $f$.** We apply the [Curvature as Commutator of Covariant Derivatives](/theorems/2703). With the chapter sign convention $R = -\nabla\circ\nabla$, the curvature endomorphism on $M$ satisfies \begin{align*} R(X, Y)Z = -\nabla_X\nabla_Y Z + \nabla_Y\nabla_X Z + \nabla_{[X, Y]} Z, \end{align*} which we rearrange as \begin{align*} \nabla_X\nabla_Y Z - \nabla_Y\nabla_X Z - \nabla_{[X, Y]} Z = -R(X, Y)Z. \end{align*} This identity pulls back along the smooth map $f: [a, b] \times (-\varepsilon, \varepsilon) \to M$. The hypotheses to check are: $f$ is smooth (given), $V$ is a smooth section of $f^*TM$ (we will plug in $V = \partial_t f$, which is smooth), and the parameter coordinate fields commute, $[\partial_t, \partial_s] = 0$ (true on $\mathbb{R}^2$). Under the bracket-vanishing condition the $\nabla_{[\cdot, \cdot]}$ term drops, giving \begin{align*} \nabla_{\partial_t}\nabla_{\partial_s} V - \nabla_{\partial_s}\nabla_{\partial_t} V = -R(\partial_t f, \partial_s f)\, V \end{align*} for any smooth section $V$ of $f^*TM$. Specialising to $V = \partial_t f$ and rearranging: \begin{align*} \nabla_{\partial_t}\nabla_{\partial_s}\, \partial_t f = \nabla_{\partial_s}\nabla_{\partial_t}\, \partial_t f - R(\partial_t f, \partial_s f)\, \partial_t f. \end{align*} **Ingredient 3: the geodesic equation kills the leftover term.** Each $s$-slice $t \mapsto f(t, s)$ is a geodesic by hypothesis, so $\nabla_{\partial_t}\partial_t f = 0$ as a vector field along $f$ on the *entire* rectangle — not just along $\gamma$. Differentiating zero in any direction is still zero, so \begin{align*} \nabla_{\partial_s}\nabla_{\partial_t}\, \partial_t f = 0. \end{align*} This is the step where the geodesic hypothesis $(\heartsuit)$ enters. If $f$ were merely a smooth variation (without the geodesic condition), this term would not vanish and the Jacobi equation would not emerge. **Assembling.** Chaining the three steps: \begin{align*} \nabla_{\partial_t}\nabla_{\partial_t}\, \partial_s f &\stackrel{(\diamond)}{=} \nabla_{\partial_t}\nabla_{\partial_s}\, \partial_t f \quad &&(\text{symmetry of mixed partials}) \\ &= \nabla_{\partial_s}\nabla_{\partial_t}\, \partial_t f - R(\partial_t f, \partial_s f)\, \partial_t f \quad &&(\text{curvature commutator}) \\ &= 0 - R(\partial_t f, \partial_s f)\, \partial_t f \quad &&(\text{geodesic equation}) \\ &= -R(\partial_t f, \partial_s f)\, \partial_t f. \end{align*} **Restricting to $\gamma$.** We now read off the value at $s = 0$. By facts (i)-(ii) from Step 1, $\partial_t f(t, 0) = \dot\gamma(t)$ and $\partial_s f(t, 0) = J(t)$, so the second covariant derivative of $J$ along $\gamma$ is \begin{align*} J''(t) := \nabla_{\partial_t}\nabla_{\partial_t}\, \partial_s f(t, 0) = -R(\dot\gamma(t), J(t))\, \dot\gamma(t). \end{align*} Rearranging, \begin{align*} J''(t) + R(\dot\gamma(t), J(t))\, \dot\gamma(t) = 0, \end{align*} which is precisely the Jacobi equation. This proves Part 1.[/guided]

[step:Construct a geodesic variation realising prescribed initial data $(u, v) \in T_p M \times T_p M$] For Part 2, let $J$ be any Jacobi field along $\gamma$. Set $u := J(0) \in T_p M$ and $v := J'(0) \in T_p M$. We build a geodesic variation $f$ with $f(t, 0) = \gamma(t)$ and $\partial_s f(t, 0) = J(t)$. **Construction of $f$.** Choose a smooth curve \begin{align*} \alpha: (-\varepsilon_0, \varepsilon_0) \to M \end{align*} with $\alpha(0) = p$ and $\dot\alpha(0) = u$. Such an $\alpha$ exists for some $\varepsilon_0 > 0$ (e.g. take $\alpha(s) := \exp_p(s u)$ on a small enough interval; existence of $\exp_p$ near $0$ uses the [Exponential Map as a Local Diffeomorphism](/theorems/2712)). Let $V$ be a smooth vector field along $\alpha$ with \begin{align*} V(0) = \dot\gamma(0), \qquad \frac{\nabla V}{ds}(0) = v. \end{align*} Concretely, parallel-transport $\dot\gamma(0)$ along $\alpha$ to get a parallel vector field $V_0$ along $\alpha$ (so $V_0(0) = \dot\gamma(0)$, $\nabla V_0/ds = 0$), and set $V(s) := V_0(s) + s\, P_\alpha^{0,s}(v)$, where $P_\alpha^{0, s}: T_p M \to T_{\alpha(s)} M$ is the parallel transport along $\alpha$ from time $0$ to time $s$. Then $V$ is smooth, $V(0) = V_0(0) = \dot\gamma(0)$, and \begin{align*} \frac{\nabla V}{ds}(0) = \underbrace{\frac{\nabla V_0}{ds}(0)}_{= 0} + P_\alpha^{0, 0}(v) + 0 = v. \end{align*} Define \begin{align*} f: [a, b] \times (-\varepsilon, \varepsilon) &\to M \\ (t, s) &\mapsto \exp_{\alpha(s)}\bigl((t - a) V(s)\bigr), \end{align*} choosing $\varepsilon \in (0, \varepsilon_0]$ small enough that the exponential is defined on $\{(t, s) : t \in [a, b],\, |s| < \varepsilon\}$. This is possible by the open-domain property of the exponential — the domain of $\exp$ on $TM$ is an open set containing the zero section, and the map $(t, s) \mapsto (\alpha(s), (t - a)V(s)) \in TM$ is continuous with image at $s = 0$ contained in $\{0\} \times \{(b - a)V_0(s)\}$, which lies inside the domain of $\exp$ along $\gamma$ since $\exp_p(t \dot\gamma(0)) = \gamma(t)$ is defined for $t \in [a, b]$. By construction, for each $s$ the curve $t \mapsto f(t, s) = \exp_{\alpha(s)}((t - a) V(s))$ is a geodesic — explicitly, the geodesic with starting point $\alpha(s)$ and initial velocity $V(s)$, by definition of the [exponential map](/page/Exponential%20Map). At $s = 0$, the starting point is $\alpha(0) = p$ and the initial velocity is $V(0) = \dot\gamma(0)$, so $f(t, 0) = \exp_p((t-a)\dot\gamma(0)) = \gamma(t)$. [/step]

[step:Identify the variation field of $f$ with $J$ via uniqueness of Jacobi fields with prescribed initial data]Let $\tilde J(t) := \partial_s f(t, 0)$. By Step 2 applied to $f$, $\tilde J$ is a Jacobi field along $\gamma$. We compute its initial data and conclude $\tilde J = J$ from uniqueness in the [Structure of Jacobi Fields](/theorems/2715). **$\tilde J(0) = u$.** Evaluate at $t = a$: \begin{align*} f(a, s) = \exp_{\alpha(s)}(0) = \alpha(s), \end{align*} so $\partial_s f(a, s) = \dot\alpha(s)$, and $\tilde J(0) = \partial_s f(a, 0) = \dot\alpha(0) = u$, matching $J(0) = u$. (The variation $f$ is defined on $[a, b] \times \cdots$; the parametrisation in the theorem statement uses the same interval, and "$J(0)$" in the conventions of the [Structure of Jacobi Fields](/theorems/2715) corresponds to the value at the left endpoint of the parametrisation. We adopt the same convention here, so $J(0) := J|_{t = a}$.) **$\tilde J'(0) = v$.** We compute, using $(\diamond)$ from Step 1: \begin{align*} \tilde J'(0) &= \frac{\nabla}{\partial t}\Big|_{t = a, s = 0} \partial_s f = \frac{\nabla}{\partial s}\Big|_{s = 0} \partial_t f(a, s). \end{align*} By construction $\partial_t f(a, s) = V(s)$ (the initial velocity of the geodesic $t \mapsto f(t, s)$ at its starting time $t = a$, which equals $V(s)$). Hence \begin{align*} \tilde J'(0) = \frac{\nabla V}{ds}(0) = v. \end{align*} So $\tilde J$ is a Jacobi field with $\tilde J(0) = u$ and $\tilde J'(0) = v$. By the uniqueness assertion in the [Structure of Jacobi Fields](/theorems/2715), $\tilde J = J$. This shows $J$ is the variation field of $f$. Combined with Step 2, the proof is complete: every Jacobi field arises from a geodesic variation (Step 4 construction), and the variation field of any geodesic variation is a Jacobi field (Steps 1-2).[/step]

[guided]We must show that the variation field $\tilde J(t) := \partial_s f(t, 0)$ of the variation built in Step 3 coincides with the given Jacobi field $J$. The strategy has two ingredients: (a) verify that $\tilde J$ has the same initial data $(u, v)$ as $J$, and (b) invoke uniqueness of Jacobi fields with prescribed initial data. Geometrically, $\tilde J$ measures the infinitesimal wiggle of the geodesic $\gamma$ as we perturb the starting point in direction $u$ (via $\alpha$) and the initial velocity in direction $v$ (via $V$); we have to confirm that these wiggles really produce $u$ and $v$ at $t = 0$. **$\tilde J$ is a Jacobi field.** First note that Step 2 applied to the variation $f$ shows $\tilde J$ satisfies the Jacobi equation $\tilde J'' + R(\dot\gamma, \tilde J)\dot\gamma = 0$. So $\tilde J$ is a Jacobi field along $\gamma$, and we only need to compute its initial data. **Computing $\tilde J(0) = u$.** Evaluate the variation at $t = a$. By the definition of the exponential map at the zero vector, $\exp_q(0) = q$ for any $q \in M$, so \begin{align*} f(a, s) = \exp_{\alpha(s)}(0) = \alpha(s). \end{align*} Differentiating in $s$, $\partial_s f(a, s) = \dot\alpha(s)$ as a vector field along $\alpha$. Setting $s = 0$ and using $\dot\alpha(0) = u$ from the construction of $\alpha$: \begin{align*} \tilde J(0) = \partial_s f(a, 0) = \dot\alpha(0) = u. \end{align*} This matches the required $J(0) = u$, so the boundary value at the left endpoint agrees. (A note on indexing: the variation $f$ is parametrised on $[a, b]$ with $f(t, 0) = \gamma(t)$, and "$J(0)$" in the conventions of the [Structure of Jacobi Fields](/theorems/2715) means the value at the left endpoint of the parametrisation. We adopt the same convention, identifying $J(0)$ with $J|_{t = a}$.) **Computing $\tilde J'(0) = v$.** This is the deeper computation, and it is where the symmetry $(\diamond)$ from Step 1 is reused. By definition of $\tilde J$, \begin{align*} \tilde J'(0) = \frac{\nabla}{\partial t}\Big|_{t = a, s = 0} \partial_s f. \end{align*} We swap the order of covariant differentiation using $(\diamond)$ — the same torsion-free identity as in Step 2, applied to the smooth section $\partial_t f$ of $f^*TM$: \begin{align*} \frac{\nabla}{\partial t}\Big|_{t = a, s = 0} \partial_s f = \frac{\nabla}{\partial s}\Big|_{s = 0} \partial_t f(a, s). \end{align*} Now $\partial_t f(a, s)$ is the initial velocity of the $s$-geodesic $t \mapsto f(t, s) = \exp_{\alpha(s)}((t - a) V(s))$ at its starting time $t = a$, which by definition of the exponential map is exactly $V(s)$: \begin{align*} \partial_t f(a, s) = V(s). \end{align*} Therefore \begin{align*} \tilde J'(0) = \frac{\nabla}{\partial s}\Big|_{s = 0} V(s) = \frac{\nabla V}{ds}(0) = v, \end{align*} where the last equality is by the construction of $V$ in Step 3. **Closing via uniqueness.** We now have two Jacobi fields along $\gamma$ — namely $\tilde J$ and the original $J$ — both with initial data $\tilde J(0) = J(0) = u$ and $\tilde J'(0) = J'(0) = v$. The Jacobi equation $J'' + R(\dot\gamma, J)\dot\gamma = 0$ is a second-order linear ODE for sections of $\gamma^*TM$ (working in a parallel frame along $\gamma$ converts it to a linear ODE in $\mathbb{R}^n$), so the [Structure of Jacobi Fields](/theorems/2715) gives unique solvability for any prescribed initial data $(J(0), J'(0))$. Hence \begin{align*} \tilde J = J, \end{align*} i.e., $J$ is the variation field of the geodesic variation $f$ built in Step 3. **Why both wiggles are needed.** Suppose we had defined $V(s)$ to be just the parallel transport $V_0(s)$ of $\dot\gamma(0)$ along $\alpha$, with no correction term. Then $\nabla V/ds(0) = 0$, so the resulting variation would only realise Jacobi fields with $J'(0) = 0$ — half the space of Jacobi fields would be missed. The correction $s \cdot P_\alpha^{0,s}(v)$ is precisely what lets us realise arbitrary $J'(0) = v$. Likewise, if $\alpha$ were constant at $p$, then $\dot\alpha(0) = 0$ and the variation would only realise Jacobi fields with $J(0) = 0$. Both wiggles — of the starting point and of the initial velocity — are essential. **Conclusion.** Combined with Step 2, the proof is complete: every Jacobi field arises as the variation field of some geodesic variation (Steps 3-4), and the variation field of any geodesic variation is a Jacobi field (Steps 1-2).[/guided]