[proofplan] We write the polar Jacobian as the determinant of a Jacobi tensor along the radial geodesic $\gamma(r)=\exp_p(r\theta)$. Its logarithmic derivative is the trace of the radial shape operator, and the matrix Riccati equation turns the Ricci lower bound into a scalar differential inequality for $J_p(r,\theta)^{1/(n-1)}$. The model function $\operatorname{sn}_k$ satisfies the corresponding equality. A Wronskian comparison then shows that $J_p(r,\theta)^{1/(n-1)}/\operatorname{sn}_k(r)$ is nonincreasing, and raising to the power $n-1$ gives the desired monotonicity. [/proofplan]

[step:Represent the polar Jacobian by transverse Jacobi fields] Let \begin{align*} \gamma: [0,c(\theta)) &\to M,\\ r &\mapsto \exp_p(r\theta) \end{align*} be the unit-speed radial geodesic. Choose an [orthonormal basis](/page/Orthonormal%20Basis) $(e_1,\dots,e_{n-1})$ of the orthogonal complement $\theta^\perp \subset T_pM$, and let $(E_1,\dots,E_{n-1})$ be the parallel orthonormal frame along $\gamma$ determined by \begin{align*} E_i(0)=e_i,\qquad \nabla_{\dot\gamma}E_i=0. \end{align*} For each $i \in \{1,\dots,n-1\}$, let \begin{align*} Y_i: [0,c(\theta)) &\to TM \end{align*} be the Jacobi field along $\gamma$ determined by \begin{align*} Y_i(0)=0,\qquad \nabla_{\dot\gamma}Y_i(0)=e_i. \end{align*} For $0<r<c(\theta)$, define the Jacobi tensor \begin{align*} A(r): \theta^\perp &\to \dot\gamma(r)^\perp \end{align*} by \begin{align*} A(r)e_i=Y_i(r). \end{align*} Since $r<c(\theta)$, the radial exponential map is nonsingular in the direction $\theta$ at radius $r$, so $A(r)$ is invertible. With respect to the parallel frames $(e_i)$ and $(E_i(r))$, its determinant is positive, and \begin{align*} J_p(r,\theta)=\det A(r). \end{align*} The Jacobi initial conditions imply the expansion \begin{align*} A(r)=rI+O(r^3) \end{align*} as $r\downarrow 0$, where $I:\theta^\perp\to\theta^\perp$ is the identity map. Hence \begin{align*} J_p(r,\theta)=r^{n-1}+O(r^{n+1}). \end{align*} [/step]

[step:Derive the scalar differential inequality from the Riccati equation]Define the radial shape operator \begin{align*} S(r): \dot\gamma(r)^\perp &\to \dot\gamma(r)^\perp,\\ v &\mapsto \nabla_{\dot\gamma}\bigl(A(r)A(r)^{-1}v\bigr) \end{align*} equivalently, in the parallel frame, $S(r)=A'(r)A(r)^{-1}$. Define its trace \begin{align*} m:(0,c(\theta)) &\to \mathbb{R},\\ r &\mapsto \operatorname{tr}S(r). \end{align*} By differentiating the determinant, \begin{align*} m(r)=\frac{\partial}{\partial r}\log J_p(r,\theta). \end{align*} Let \begin{align*} R_r:\dot\gamma(r)^\perp &\to \dot\gamma(r)^\perp,\\ v &\mapsto R(v,\dot\gamma(r))\dot\gamma(r) \end{align*} be the curvature endomorphism along $\gamma$. The Jacobi equation gives the matrix Riccati equation \begin{align*} S'(r)+S(r)^2+R_r=0. \end{align*} Taking traces gives \begin{align*} m'(r)+\operatorname{tr}(S(r)^2)+\operatorname{Ric}_g(\dot\gamma(r),\dot\gamma(r))=0. \end{align*} Since $S(r)$ is self-adjoint on the Euclidean space $\dot\gamma(r)^\perp$ of dimension $n-1$, the [Cauchy-Schwarz inequality](/theorems/432) applied to its eigenvalues gives \begin{align*} \operatorname{tr}(S(r)^2)\geq \frac{m(r)^2}{n-1}. \end{align*} Because $\gamma$ has unit speed, the Ricci lower bound gives \begin{align*} \operatorname{Ric}_g(\dot\gamma(r),\dot\gamma(r))\geq (n-1)k. \end{align*} Therefore \begin{align*} m'(r)+\frac{m(r)^2}{n-1}+(n-1)k\leq 0. \end{align*}[/step]

[guided]The purpose of this step is to convert the tensorial Jacobi equation into a one-dimensional inequality. The Jacobi tensor $A(r)$ records how nearby radial geodesics separate from $\gamma$. Its logarithmic determinant measures infinitesimal volume growth, so the natural quantity to differentiate is \begin{align*} m(r)=\frac{\partial}{\partial r}\log J_p(r,\theta). \end{align*} Since $J_p(r,\theta)=\det A(r)$ and $A(r)$ is invertible for $0<r<c(\theta)$, the determinant differentiation formula gives \begin{align*} \frac{\partial}{\partial r}\log J_p(r,\theta) = \operatorname{tr}\bigl(A'(r)A(r)^{-1}\bigr). \end{align*} Thus $m(r)=\operatorname{tr}S(r)$, where $S(r)=A'(r)A(r)^{-1}$ is the radial shape operator. The Jacobi equation for $A$ is \begin{align*} A''(r)+R_rA(r)=0, \end{align*} where \begin{align*} R_r:\dot\gamma(r)^\perp &\to \dot\gamma(r)^\perp,\\ v &\mapsto R(v,\dot\gamma(r))\dot\gamma(r). \end{align*} Differentiating $S=A'A^{-1}$ gives \begin{align*} S' = A''A^{-1}-A'A^{-1}A'A^{-1} = A''A^{-1}-S^2. \end{align*} Using $A''=-R_rA$, we obtain \begin{align*} S'+S^2+R_r=0. \end{align*} Taking the trace of this identity gives \begin{align*} m'(r)+\operatorname{tr}(S(r)^2)+\operatorname{tr}R_r=0. \end{align*} The trace of $R_r$ over the orthogonal complement $\dot\gamma(r)^\perp$ is exactly $\operatorname{Ric}_g(\dot\gamma(r),\dot\gamma(r))$, so \begin{align*} m'(r)+\operatorname{tr}(S(r)^2)+\operatorname{Ric}_g(\dot\gamma(r),\dot\gamma(r))=0. \end{align*} Now we estimate the two non-derivative terms. Since $S(r)$ is self-adjoint on the $(n-1)$-dimensional [inner product space](/page/Inner%20Product%20Space) $\dot\gamma(r)^\perp$, let $\lambda_1(r),\dots,\lambda_{n-1}(r)$ denote its real eigenvalues. Then \begin{align*} m(r)=\sum_{i=1}^{n-1}\lambda_i(r), \qquad \operatorname{tr}(S(r)^2)=\sum_{i=1}^{n-1}\lambda_i(r)^2. \end{align*} Applying Cauchy-Schwarz in $\mathbb{R}^{n-1}$ to the vectors $(\lambda_1(r),\dots,\lambda_{n-1}(r))$ and $(1,\dots,1)$ gives \begin{align*} m(r)^2 = \left(\sum_{i=1}^{n-1}\lambda_i(r)\right)^2 \leq (n-1)\sum_{i=1}^{n-1}\lambda_i(r)^2 = (n-1)\operatorname{tr}(S(r)^2). \end{align*} Hence \begin{align*} \operatorname{tr}(S(r)^2)\geq \frac{m(r)^2}{n-1}. \end{align*} Finally, $\gamma$ is unit speed, so $g(\dot\gamma(r),\dot\gamma(r))=1$. The Ricci lower bound therefore gives \begin{align*} \operatorname{Ric}_g(\dot\gamma(r),\dot\gamma(r))\geq (n-1)k. \end{align*} Substituting both estimates into the trace Riccati identity yields \begin{align*} m'(r)+\frac{m(r)^2}{n-1}+(n-1)k\leq 0. \end{align*}[/guided]

[step:Convert the logarithmic inequality into a comparison for the normalized root] Define \begin{align*} z:(0,c(\theta)) &\to (0,\infty),\\ r &\mapsto J_p(r,\theta)^{1/(n-1)}. \end{align*} Then \begin{align*} \frac{z'(r)}{z(r)}=\frac{m(r)}{n-1}. \end{align*} Using the scalar inequality from the previous step, \begin{align*} \frac{z''(r)}{z(r)} = \frac{m'(r)}{n-1}+\frac{m(r)^2}{(n-1)^2} \leq -k. \end{align*} Thus \begin{align*} z''(r)+kz(r)\leq 0. \end{align*} The Jacobi expansion $J_p(r,\theta)=r^{n-1}+O(r^{n+1})$ gives \begin{align*} z(r)=r+O(r^3), \qquad z'(r)=1+O(r^2) \end{align*} as $r\downarrow 0$. [/step]

[step:Compare with the model solution by a Wronskian argument] Let \begin{align*} s:(0,\rho_k) &\to (0,\infty),\\ r &\mapsto \operatorname{sn}_k(r), \end{align*} where $\rho_k=\infty$ if $k\leq 0$ and $\rho_k=\pi/\sqrt{k}$ if $k>0$. By the definition of $\operatorname{sn}_k$, \begin{align*} s''(r)+ks(r)=0, \qquad s(r)=r+O(r^3), \qquad s'(r)=1+O(r^2) \end{align*} as $r\downarrow 0$. For $0<r<\min\{c(\theta),\rho_k\}$, define the Wronskian-type function \begin{align*} W:(0,\min\{c(\theta),\rho_k\}) &\to \mathbb{R},\\ r &\mapsto z'(r)s(r)-z(r)s'(r). \end{align*} Differentiating and using $z''+kz\leq 0$ and $s''+ks=0$, we obtain \begin{align*} W'(r) = z''(r)s(r)-z(r)s''(r) \leq -kz(r)s(r)+kz(r)s(r) = 0. \end{align*} The expansions of $z$ and $s$ at $0$ give \begin{align*} \lim_{r\downarrow 0}W(r)=0. \end{align*} Therefore $W(r)\leq 0$ for every $0<r<\min\{c(\theta),\rho_k\}$. Since $s(r)>0$ on this interval, \begin{align*} \frac{\partial}{\partial r}\left(\frac{z(r)}{s(r)}\right) = \frac{z'(r)s(r)-z(r)s'(r)}{s(r)^2} = \frac{W(r)}{s(r)^2} \leq 0. \end{align*} Thus $r\mapsto z(r)/s(r)$ is nonincreasing. [/step]

[step:Raise the root comparison to obtain the Bishop Jacobian monotonicity] By the definition of $z$ and $s$, \begin{align*} \left(\frac{z(r)}{s(r)}\right)^{n-1} = \frac{J_p(r,\theta)}{\operatorname{sn}_k(r)^{n-1}}. \end{align*} The exponent $n-1$ is positive because $n\geq 2$, and the function $t\mapsto t^{n-1}$ is increasing on $(0,\infty)$. Since $z(r)/s(r)$ is nonincreasing and positive on \begin{align*} 0<r<\min\{c(\theta),\rho_k\}, \end{align*} the function \begin{align*} r\longmapsto \frac{J_p(r,\theta)}{\operatorname{sn}_k(r)^{n-1}} \end{align*} is nonincreasing on the same interval. This is precisely the claimed interval, including the restriction $r<\pi/\sqrt{k}$ when $k>0$. [/step]