[proofplan] Let $I_k\subset(0,a]$ denote the comparison interval from the statement, so $\operatorname{sn}_k(t)>0$ for every $t\in I_k$. We compare the given Jacobi field with the model Jacobi field in the simply [connected space](/page/Connected%20Space) form of constant sectional curvature $k$, restricted to the same parameter values in $I_k$. The model field is $\widetilde J(t)=\operatorname{sn}_k(t)E(t)$, where $E$ is a parallel unit vector field perpendicular to the model geodesic, so its length is exactly $\operatorname{sn}_k(t)$ and its initial covariant derivative has norm $1$. The two inequalities then follow from the general Rauch comparison theorem: under the upper curvature bound on $M$, the ratio $|J|/|\widetilde J|$ is nondecreasing; under the lower curvature bound and the no-conjugate-points hypothesis, the comparison is reversed. [/proofplan] [step:Construct the model Jacobi field in the space form of curvature $k$] Let $(\widetilde M_k,\widetilde g_k)$ denote the simply connected complete Riemannian manifold of constant sectional curvature $k$, and let $R^{\widetilde M_k}$ denote the Riemann curvature tensor of $(\widetilde M_k,\widetilde g_k)$, with the Jacobi equation written in the convention $\nabla_t\nabla_tY+R^{\widetilde M_k}(Y,\dot{\widetilde\gamma})\dot{\widetilde\gamma}=0$ for a vector field $Y$ along $\widetilde\gamma$. Let \begin{align*} \widetilde\gamma:[0,a]\to \widetilde M_k \end{align*} be a unit-speed geodesic. Choose a vector $\widetilde v\in T_{\widetilde\gamma(0)}\widetilde M_k$ with $|\widetilde v|_{\widetilde g_k}=1$ and $\widetilde v\perp \dot{\widetilde\gamma}(0)$, and let \begin{align*} E:[0,a]\to T\widetilde M_k \end{align*} be the parallel vector field along $\widetilde\gamma$ satisfying $E(0)=\widetilde v$. Define \begin{align*} \widetilde J:[0,a]\to T\widetilde M_k,\qquad t\mapsto \operatorname{sn}_k(t)E(t). \end{align*} Since $E$ is parallel and $\operatorname{sn}_k$ satisfies \begin{align*} \operatorname{sn}_k''(t)+k\operatorname{sn}_k(t)=0,\qquad \operatorname{sn}_k(0)=0,\qquad \operatorname{sn}_k'(0)=1, \end{align*} we have \begin{align*} \nabla_t\nabla_t\widetilde J(t) = \operatorname{sn}_k''(t)E(t) = -k\operatorname{sn}_k(t)E(t). \end{align*} Because every sectional curvature of $\widetilde M_k$ equals $k$ and $E(t)\perp\dot{\widetilde\gamma}(t)$, the curvature term satisfies \begin{align*} R^{\widetilde M_k}(\widetilde J(t),\dot{\widetilde\gamma}(t))\dot{\widetilde\gamma}(t) = k\operatorname{sn}_k(t)E(t). \end{align*} Thus \begin{align*} \nabla_t\nabla_t\widetilde J(t) + R^{\widetilde M_k}(\widetilde J(t),\dot{\widetilde\gamma}(t))\dot{\widetilde\gamma}(t) =0, \end{align*} so $\widetilde J$ is a Jacobi field. Moreover \begin{align*} \widetilde J(0)=0,\qquad \nabla_t\widetilde J(0)=\operatorname{sn}_k'(0)E(0)=\widetilde v, \end{align*} and therefore $|\nabla_t\widetilde J(0)|_{\widetilde g_k}=1$. Finally, for every $t\in I_k$, \begin{align*} |\widetilde J(t)|_{\widetilde g_k} = \operatorname{sn}_k(t)|E(t)|_{\widetilde g_k} = \operatorname{sn}_k(t), \end{align*} because $E$ is parallel and $\operatorname{sn}_k(t)>0$ on $I_k$. [guided] The comparison field should be the Jacobi field in the constant-curvature model with the same initial size data as $J$. Let $(\widetilde M_k,\widetilde g_k)$ be the simply connected complete space form of sectional curvature $k$, and let $R^{\widetilde M_k}$ denote its Riemann curvature tensor. We use the sign convention in which a vector field $Y$ along $\widetilde\gamma$ is Jacobi precisely when $\nabla_t\nabla_tY+R^{\widetilde M_k}(Y,\dot{\widetilde\gamma})\dot{\widetilde\gamma}=0$. Let \begin{align*} \widetilde\gamma:[0,a]\to \widetilde M_k \end{align*} be a unit-speed geodesic. Choose a unit vector $\widetilde v\in T_{\widetilde\gamma(0)}\widetilde M_k$ with $\widetilde v\perp\dot{\widetilde\gamma}(0)$, and let \begin{align*} E:[0,a]\to T\widetilde M_k \end{align*} be the parallel vector field along $\widetilde\gamma$ determined by $E(0)=\widetilde v$. Now define \begin{align*} \widetilde J:[0,a]\to T\widetilde M_k,\qquad t\mapsto \operatorname{sn}_k(t)E(t). \end{align*} This is the correct model field because $\operatorname{sn}_k$ is the scalar solution of the Jacobi equation in constant curvature $k$. Since $E$ is parallel, $\nabla_tE=0$, so differentiating gives \begin{align*} \nabla_t\nabla_t\widetilde J(t) = \operatorname{sn}_k''(t)E(t). \end{align*} By the defining differential equation for $\operatorname{sn}_k$, \begin{align*} \operatorname{sn}_k''(t)+k\operatorname{sn}_k(t)=0, \end{align*} hence \begin{align*} \nabla_t\nabla_t\widetilde J(t) = -k\operatorname{sn}_k(t)E(t). \end{align*} In the space form $\widetilde M_k$, every sectional curvature equals $k$. Since $E(t)$ is perpendicular to $\dot{\widetilde\gamma}(t)$ and $\widetilde J(t)=\operatorname{sn}_k(t)E(t)$, the curvature term in the Jacobi equation is \begin{align*} R^{\widetilde M_k}(\widetilde J(t),\dot{\widetilde\gamma}(t))\dot{\widetilde\gamma}(t) = k\operatorname{sn}_k(t)E(t). \end{align*} Adding the two displayed identities gives \begin{align*} \nabla_t\nabla_t\widetilde J(t) + R^{\widetilde M_k}(\widetilde J(t),\dot{\widetilde\gamma}(t))\dot{\widetilde\gamma}(t) =0. \end{align*} Thus $\widetilde J$ is a Jacobi field along $\widetilde\gamma$. The initial conditions also match the field $J$. Since $\operatorname{sn}_k(0)=0$ and $\operatorname{sn}_k'(0)=1$, \begin{align*} \widetilde J(0)=0,\qquad \nabla_t\widetilde J(0)=E(0)=\widetilde v, \end{align*} so $|\nabla_t\widetilde J(0)|_{\widetilde g_k}=1$. Finally, parallel transport preserves length, so $|E(t)|_{\widetilde g_k}=1$ for every $t$. On $I_k$ we have $\operatorname{sn}_k(t)>0$, and therefore \begin{align*} |\widetilde J(t)|_{\widetilde g_k} = \operatorname{sn}_k(t)|E(t)|_{\widetilde g_k} = \operatorname{sn}_k(t). \end{align*} This identifies the comparison length exactly. [/guided] [/step] [step:Apply Rauch comparison under the upper curvature bound] Assume \begin{align*} K_M(\dot{\gamma}(t),J(t))\le k \end{align*} for every $t\in I_k$ with $J(t)\neq0$. The comparison field $\widetilde J$ constructed above satisfies \begin{align*} K_{\widetilde M_k}(\dot{\widetilde\gamma}(t),\widetilde J(t))=k \end{align*} for every $t\in I_k$, and \begin{align*} J(0)=\widetilde J(0)=0,\qquad |\nabla_tJ(0)|=|\nabla_t\widetilde J(0)|_{\widetilde g_k}=1. \end{align*} We invoke the [Rauch Comparison Theorem](/theorems/TEMP-6) on each subinterval of $I_k$ on which $J$ does not vanish. Its hypotheses are met as follows: both geodesics are unit speed; both fields are transverse Jacobi fields; the initial values vanish; the initial derivative norms agree; the comparison field $\widetilde J$ is nonzero on $I_k$ because $|\widetilde J(t)|_{\widetilde g_k}=\operatorname{sn}_k(t)>0$ there; and the curvature inequality is $K_M(\dot{\gamma}(t),J(t))\le K_{\widetilde M_k}(\dot{\widetilde\gamma}(t),\widetilde J(t))$ wherever $J(t)\neq0$. Applying that theorem with $M$ as the lower-curvature side and $\widetilde M_k$ as the upper-curvature side, the function \begin{align*} r:I_k\cap\{t:J(t)\neq0\}\to\mathbb{R},\qquad t\mapsto \frac{|J(t)|}{|\widetilde J(t)|_{\widetilde g_k}} \end{align*} is nondecreasing on every such subinterval. Since both Jacobi fields vanish at $0$ and have initial derivative of norm $1$, \begin{align*} \lim_{t\downarrow0}\frac{|J(t)|}{|\widetilde J(t)|_{\widetilde g_k}}=1. \end{align*} Hence $r(t)\ge1$ for every $t\in I_k$ for which $J(t)\neq0$. Using $|\widetilde J(t)|_{\widetilde g_k}=\operatorname{sn}_k(t)$ gives \begin{align*} |J(t)|\ge \operatorname{sn}_k(t). \end{align*} If $J$ vanished somewhere in $I_k$, let $\tau\in I_k$ be its first zero after $0$. On the open interval $(0,\tau)$ the field $J$ is nonzero, so the Rauch comparison conclusion just proved gives \begin{align*} |J(t)|\ge \operatorname{sn}_k(t) \end{align*} for every $t\in(0,\tau)$. Passing to the limit as $t\uparrow\tau$ and using continuity of $J$ and of $\operatorname{sn}_k$ gives \begin{align*} 0=|J(\tau)|\ge \operatorname{sn}_k(\tau)>0, \end{align*} which is impossible because $\tau\in I_k$. Therefore $J$ has no such zero on $I_k$, and the inequality holds for every $t\in I_k$. [guided] The only delicate point in the upper-curvature case is that the ratio \begin{align*} r(t)=\frac{|J(t)|}{|\widetilde J(t)|_{\widetilde g_k}} \end{align*} is meaningful only where both Jacobi fields are nonzero. The model field is nonzero on $I_k$ because \begin{align*} |\widetilde J(t)|_{\widetilde g_k}=\operatorname{sn}_k(t)>0. \end{align*} The field $J$ might a priori have a later zero, so we first apply Rauch on each interval where $J$ is nonzero. Assume first that $J(t)\neq0$ on a subinterval of $I_k$. The hypotheses of the [Rauch Comparison Theorem](/theorems/TEMP-6) are satisfied on that subinterval: both geodesics are unit-speed, both fields are transverse Jacobi fields, both vanish at $0$, their initial derivative norms agree, and the curvature inequality is \begin{align*} K_M(\dot\gamma(t),J(t))\le K_{\widetilde M_k}(\dot{\widetilde\gamma}(t),\widetilde J(t))=k. \end{align*} Therefore Rauch gives that \begin{align*} r(t)=\frac{|J(t)|}{|\widetilde J(t)|_{\widetilde g_k}} \end{align*} is nondecreasing there. Since both fields vanish at $0$ and have initial derivative norm $1$, \begin{align*} \lim_{t\downarrow0}r(t)=1. \end{align*} Thus $r(t)\ge1$, and using the model identity $|\widetilde J(t)|_{\widetilde g_k}=\operatorname{sn}_k(t)$ gives \begin{align*} |J(t)|\ge \operatorname{sn}_k(t) \end{align*} wherever the ratio is defined. Now rule out the possible later zero. Suppose $J$ vanished somewhere in $I_k$, and let $\tau\in I_k$ be the first zero after $0$. Then $J(t)\neq0$ for $0<t<\tau$, so the preceding Rauch argument gives \begin{align*} |J(t)|\ge \operatorname{sn}_k(t) \end{align*} for every $t\in(0,\tau)$. Letting $t\uparrow\tau$ and using continuity gives \begin{align*} 0=|J(\tau)|\ge \operatorname{sn}_k(\tau)>0, \end{align*} which is impossible because $\tau\in I_k$. Hence $J$ is nonzero on $I_k$, and the upper-curvature comparison proves \begin{align*} |J(t)|\ge\operatorname{sn}_k(t) \end{align*} for every $t\in I_k$. [/guided] [/step] [step:Apply Rauch comparison under the lower curvature bound with no conjugate points] Assume \begin{align*} K_M(\dot{\gamma}(t),J(t))\ge k \end{align*} for every $t\in I_k$ with $J(t)\neq0$, and assume that $\gamma(0)$ has no conjugate point along $\gamma|_{I_k}$. Since $|\nabla_tJ(0)|=1$, the Jacobi field $J$ is not the zero Jacobi field. The no-conjugate-points hypothesis therefore implies that this nonzero Jacobi field along $\gamma$ vanishing at $0$ cannot vanish again on $I_k$; in particular, $J(t)\neq0$ for every $t\in I_k$. Now the space form is the lower-curvature side: \begin{align*} K_{\widetilde M_k}(\dot{\widetilde\gamma}(t),\widetilde J(t))=k \le K_M(\dot{\gamma}(t),J(t)). \end{align*} The initial data again satisfy \begin{align*} \widetilde J(0)=J(0)=0,\qquad |\nabla_t\widetilde J(0)|_{\widetilde g_k}=|\nabla_tJ(0)|=1. \end{align*} We invoke the [Rauch Comparison Theorem](/theorems/TEMP-6). Its hypotheses are met as follows: both geodesics are unit speed; both fields are transverse Jacobi fields; the initial values vanish; the initial derivative norms agree; the field $J$ is nonzero on $I_k$ by the no-conjugate-points argument above; the field $\widetilde J$ is nonzero on $I_k$ because $|\widetilde J(t)|_{\widetilde g_k}=\operatorname{sn}_k(t)>0$; and the curvature inequality is $K_{\widetilde M_k}(\dot{\widetilde\gamma}(t),\widetilde J(t))\le K_M(\dot{\gamma}(t),J(t))$. Applying that theorem with $\widetilde M_k$ as the lower-curvature side and $M$ as the upper-curvature side, the function \begin{align*} q:I_k\to\mathbb{R},\qquad t\mapsto \frac{|\widetilde J(t)|_{\widetilde g_k}}{|J(t)|} \end{align*} is nondecreasing. Since the initial derivative norms agree, \begin{align*} \lim_{t\downarrow0}\frac{|\widetilde J(t)|_{\widetilde g_k}}{|J(t)|}=1. \end{align*} Therefore $q(t)\ge1$ for every $t\in I_k$, equivalently \begin{align*} |J(t)|\le |\widetilde J(t)|_{\widetilde g_k}. \end{align*} Using the model length identity $|\widetilde J(t)|_{\widetilde g_k}=\operatorname{sn}_k(t)$ yields \begin{align*} |J(t)|\le \operatorname{sn}_k(t) \end{align*} for every $t\in I_k$. [guided] In the lower-curvature case the no-conjugate-points hypothesis is what makes the ratio comparison globally meaningful on $I_k$. Since $|\nabla_tJ(0)|=1$, the Jacobi field $J$ is not identically zero. If $J(\tau)=0$ for some $\tau\in I_k$, then $\tau$ would be a conjugate time to $\gamma(0)$ along $\gamma$, contradicting the hypothesis. Hence \begin{align*} J(t)\neq0 \end{align*} for every $t\in I_k$. The model field is also nonzero on $I_k$ because \begin{align*} |\widetilde J(t)|_{\widetilde g_k}=\operatorname{sn}_k(t)>0. \end{align*} Now the space form is the lower-curvature side: \begin{align*} K_{\widetilde M_k}(\dot{\widetilde\gamma}(t),\widetilde J(t))=k\le K_M(\dot\gamma(t),J(t)). \end{align*} The [Rauch Comparison Theorem](/theorems/TEMP-6) applies because both geodesics are unit-speed, both fields are transverse Jacobi fields, both vanish at $0$, their initial derivative norms agree, and both fields are nonzero on $I_k$. It gives that \begin{align*} q(t)=\frac{|\widetilde J(t)|_{\widetilde g_k}}{|J(t)|} \end{align*} is nondecreasing on $I_k$. The initial derivative norms both equal $1$, so near $0$ both lengths have the same first-order behaviour, and therefore \begin{align*} \lim_{t\downarrow0}q(t)=1. \end{align*} Since $q$ is nondecreasing, $q(t)\ge1$ for every $t\in I_k$. This is exactly \begin{align*} |J(t)|\le |\widetilde J(t)|_{\widetilde g_k}. \end{align*} Finally, substituting the model length identity gives \begin{align*} |J(t)|\le \operatorname{sn}_k(t) \end{align*} for every $t\in I_k$. [/guided] [/step] [step:Combine the two comparisons to obtain the stated inequalities] The first comparison proves that an upper bound \begin{align*} K_M(\dot{\gamma}(t),J(t))\le k \end{align*} forces the Jacobi field in $M$ to have length at least that of the space-form Jacobi field: \begin{align*} |J(t)|\ge\operatorname{sn}_k(t). \end{align*} The second comparison proves that a lower bound \begin{align*} K_M(\dot{\gamma}(t),J(t))\ge k \end{align*} together with the absence of conjugate points forces the reverse length inequality: \begin{align*} |J(t)|\le\operatorname{sn}_k(t). \end{align*} Both inequalities hold precisely on the interval $I_k$ where the model length $\operatorname{sn}_k(t)$ is positive. This proves the theorem. [/step]