[proofplan] The Bishop-Gromov inequality is obtained by writing volume in geodesic polar coordinates and comparing the radial Jacobian $J(t,\xi)$ with the model Jacobian $\operatorname{sn}_k(t)^{n-1}$. The endpoint equality of the integrated volume ratios forces the averaged Jacobian deficit to vanish on the whole annulus. Since the normalized radial Jacobian is monotone nonincreasing in each direction before the cut time, vanishing of the averaged deficit implies directional equality for almost every geodesic. Finally, differentiating the equality of the normalized Jacobian and using the trace Riccati equation gives equality in the Ricci lower bound and in the Cauchy-Schwarz step for the shape operator. [/proofplan]

[step:Introduce the radial Jacobian quotient and its monotonicity before the cut locus]Define the model radius \begin{align*} \rho_k:= \begin{cases} \pi/\sqrt{k},&k>0,\\ \infty,&k\leq 0. \end{cases} \end{align*} The updated hypothesis gives $R<\rho_k$. Define the model sine function $\operatorname{sn}_k:(0,\rho_k)\to (0,\infty)$ as the solution of $u''+ku=0$ with $u(0)=0$ and $u'(0)=1$, namely \begin{align*} \operatorname{sn}_k(t)= \begin{cases} \dfrac{\sin(\sqrt{k}t)}{\sqrt{k}},&k>0,\\ t,&k=0,\\ \dfrac{\sinh(\sqrt{-k}t)}{\sqrt{-k}},&k<0. \end{cases} \end{align*} Define $\operatorname{ct}_k:(0,\rho_k)\to \mathbb{R}$ by \begin{align*} \operatorname{ct}_k(t)=\frac{\operatorname{sn}_k'(t)}{\operatorname{sn}_k(t)}. \end{align*} Thus $\operatorname{sn}_k(t)>0$ for every $t\in (0,R)$ and $\operatorname{ct}_k$ is defined on the whole annulus. The model ball volume is the function $V_k:(0,\rho_k)\to (0,\infty)$ given by \begin{align*} V_k(\rho)=\mathcal{H}^{n-1}(S^{n-1})\int_0^\rho \operatorname{sn}_k(t)^{n-1}\,d\mathcal{L}^1(t). \end{align*} Here $\mathcal{H}^{n-1}(S^{n-1})$ denotes the $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) of the unit sphere in $\mathbb{R}^n$. For $\xi \in S_pM$, let $\operatorname{cut}_p(\xi)\in (0,\infty]$ denote the cut time in the initial direction $\xi$, namely the supremum of times $s>0$ for which $t\mapsto \exp_p(t\xi)$ is minimizing on $[0,s]$. Define the unit-speed radial geodesic \begin{align*} \gamma_\xi:[0,\operatorname{cut}_p(\xi))&\to M,\\ t&\mapsto \exp_p(t\xi). \end{align*} On the regular set $0<t<\operatorname{cut}_p(\xi)$, define the normalized Jacobian quotient \begin{align*} q:(0,\infty)\times S_pM&\to [0,\infty),\\ (t,\xi)&\mapsto \begin{cases} \dfrac{J(t,\xi)}{\operatorname{sn}_k(t)^{n-1}},&0<t<\operatorname{cut}_p(\xi),\\ 0,&t\geq \operatorname{cut}_p(\xi). \end{cases} \end{align*} The [radial Jacobian comparison theorem](/page/Radial%20Jacobian%20Comparison) under the Ricci lower bound $\operatorname{Ric}_g\geq (n-1)k\,g$ applies on every interval contained in $(0,\min\{\operatorname{cut}_p(\xi),\rho_k\})$. Its hypotheses are satisfied because $\gamma_\xi$ is a unit-speed minimizing geodesic before $\operatorname{cut}_p(\xi)$ and $\operatorname{sn}_k$ is positive on $(0,R)$. On each such regular interval, the comparison gives the differential inequality \begin{align*} \frac{\partial}{\partial t}\log J(t,\xi)\leq (n-1)\operatorname{ct}_k(t), \end{align*} equivalently \begin{align*} \frac{\partial}{\partial t}\log\left(\frac{J(t,\xi)}{\operatorname{sn}_k(t)^{n-1}}\right)\leq 0. \end{align*} Therefore, for each fixed $\xi$, the map $t\mapsto q(t,\xi)$ is nonincreasing on $(0,R)$ and is smooth on $(0,\min\{R,\operatorname{cut}_p(\xi)\})$.[/step]

[guided]The key object is the quotient comparing the actual polar Jacobian to the model polar Jacobian. First define \begin{align*} \rho_k:= \begin{cases} \pi/\sqrt{k},&k>0,\\ \infty,&k\leq 0. \end{cases} \end{align*} The radius hypothesis gives $R<\rho_k$. Define the model sine function $\operatorname{sn}_k:(0,\rho_k)\to (0,\infty)$ by \begin{align*} \operatorname{sn}_k(t)= \begin{cases} \dfrac{\sin(\sqrt{k}t)}{\sqrt{k}},&k>0,\\ t,&k=0,\\ \dfrac{\sinh(\sqrt{-k}t)}{\sqrt{-k}},&k<0, \end{cases} \end{align*} which is equivalently the solution of $u''+ku=0$ with $u(0)=0$ and $u'(0)=1$. Hence $\operatorname{sn}_k(t)>0$ for $0<t<R$, so division by $\operatorname{sn}_k(t)^{n-1}$ is legitimate throughout the annulus. Define $\operatorname{ct}_k:(0,\rho_k)\to \mathbb{R}$ by $\operatorname{ct}_k(t)=\operatorname{sn}_k'(t)/\operatorname{sn}_k(t)$. For each $\xi\in S_pM$, let $\operatorname{cut}_p(\xi)$ be the supremum of times $s>0$ such that the geodesic $t\mapsto \exp_p(t\xi)$ minimizes distance from $p$ on $[0,s]$. The radial geodesic is the map \begin{align*} \gamma_\xi:[0,\operatorname{cut}_p(\xi))&\to M,\\ t&\mapsto \exp_p(t\xi). \end{align*} Before the cut time, geodesic polar coordinates are smooth, so the volume density $J(t,\xi)$ is smooth and positive. We define \begin{align*} q:(0,\infty)\times S_pM&\to [0,\infty),\\ (t,\xi)&\mapsto \begin{cases} \dfrac{J(t,\xi)}{\operatorname{sn}_k(t)^{n-1}},&0<t<\operatorname{cut}_p(\xi),\\ 0,&t\geq \operatorname{cut}_p(\xi). \end{cases} \end{align*} The extension by $0$ after the cut time is a bookkeeping device: after a geodesic stops minimizing, it no longer contributes to the polar-coordinate representation of metric balls centered at $p$. The [radial Jacobian comparison theorem](/page/Radial%20Jacobian%20Comparison) under $\operatorname{Ric}_g\geq (n-1)k\,g$ states that \begin{align*} t\mapsto \frac{J(t,\xi)}{\operatorname{sn}_k(t)^{n-1}} \end{align*} is nonincreasing on every interval contained in $(0,\min\{\operatorname{cut}_p(\xi),\rho_k\})$. We have checked its two relevant domain hypotheses: $\gamma_\xi$ is minimizing before $\operatorname{cut}_p(\xi)$, and $\operatorname{sn}_k$ is positive for $0<t<R$. More explicitly, on each regular interval the theorem gives \begin{align*} \frac{\partial}{\partial t}\log J(t,\xi)\leq (n-1)\operatorname{ct}_k(t), \end{align*} or equivalently \begin{align*} \frac{\partial}{\partial t}\log\left(\frac{J(t,\xi)}{\operatorname{sn}_k(t)^{n-1}}\right)\leq 0. \end{align*} Setting the quotient equal to $0$ after the cut time preserves monotonicity on $(0,R)$, because the comparison quotient is nonnegative before the cut time. Thus, for every $\xi\in S_pM$, the function $t\mapsto q(t,\xi)$ is nonincreasing on $(0,R)$, and on the regular interval $0<t<\min\{R,\operatorname{cut}_p(\xi)\}$ it is smooth.[/guided]

[step:Convert volume equality into equality of the averaged Jacobian quotient]Let $d\sigma_p$ denote the Riemannian measure on the unit sphere $S_pM\subset T_pM$ induced by the [inner product](/page/Inner%20Product) $g_p$; under any linear isometry $T_pM\cong \mathbb{R}^n$, its total mass is $\mathcal{H}^{n-1}(S^{n-1})$. Define the model area density \begin{align*} w_k:(0,\rho_k)&\to (0,\infty),\\ t&\mapsto \mathcal{H}^{n-1}(S^{n-1})\operatorname{sn}_k(t)^{n-1}, \end{align*} and define the averaged quotient \begin{align*} A:(0,\infty)&\to [0,\infty),\\ t&\mapsto \frac{1}{\mathcal{H}^{n-1}(S^{n-1})}\int_{S_pM} q(t,\xi)\,d\sigma_p(\xi). \end{align*} The cut-time function is measurable and $J$ is smooth on the regular polar-coordinate domain, so the extended quotient $q$ is a nonnegative measurable function on $(0,\infty)\times S_pM$. Hence [Tonelli's Theorem](/page/Tonelli%27s%20Theorem) applies to all iterated integrals below. Since $q(\cdot,\xi)$ is nonincreasing for every $\xi$, integrating the inequality $q(t_1,\xi)\geq q(t_2,\xi)$ for $0<t_1<t_2<R$ over $(S_pM,d\sigma_p)$ gives that $A$ is nonincreasing. The [geodesic polar-coordinate volume formula](/page/Geodesic%20Polar%20Coordinates) gives, for every $\rho\in(0,\rho_k)$, \begin{align*} \operatorname{Vol}_g(B(p,\rho)) &= \int_0^\rho\int_{S_pM} q(t,\xi)\operatorname{sn}_k(t)^{n-1}\,d\sigma_p(\xi)\,d\mathcal{L}^1(t)\\ &= \int_0^\rho A(t)w_k(t)\,d\mathcal{L}^1(t). \end{align*} Also, \begin{align*} V_k(\rho)=\int_0^\rho w_k(t)\,d\mathcal{L}^1(t). \end{align*} Hence \begin{align*} \frac{\operatorname{Vol}_g(B(p,\rho))}{V_k(\rho)} = \frac{\int_0^\rho A(t)w_k(t)\,d\mathcal{L}^1(t)} {\int_0^\rho w_k(t)\,d\mathcal{L}^1(t)}. \end{align*} Because $R<\rho_k$, the weight $w_k(t)$ is positive for every $t\in(0,R)$. We use the following monotone weighted-average fact: if $A$ is nonincreasing on $(0,R)$ and $w_k>0$, then \begin{align*} \rho\mapsto \frac{\int_0^\rho A(t)w_k(t)\,d\mathcal{L}^1(t)}{\int_0^\rho w_k(t)\,d\mathcal{L}^1(t)} \end{align*} is nonincreasing. Indeed, for $0<\rho_1<\rho_2<R$, every value of $A$ on $(0,\rho_1)$ is at least every value of $A$ on $(\rho_1,\rho_2)$ up to null sets, so adding the outer shell can only decrease the weighted average. Let \begin{align*} C=\frac{\operatorname{Vol}_g(B(p,r))}{V_k(r)}=\frac{\operatorname{Vol}_g(B(p,R))}{V_k(R)}. \end{align*} The equality of the averages over $(0,r)$ and $(0,R)$ implies that the weighted average of $A$ over the shell $(r,R)$ is also $C$; this follows by subtracting the identity for the weighted integral over $(0,r)$ from the identity over $(0,R)$. Because $A$ is nonincreasing, for every $s\in(0,r)$ and $t\in(r,R)$ one has $A(s)\geq A(t)$. Suppose there were $s_0\in(0,r)$ with $A(s_0)>C$. Then $A(s)\geq A(s_0)>C$ for every $s\in(0,s_0)$, and since $w_k>0$ on $(0,R)$ the weighted average of $A$ over $(0,r)$ would be strictly larger than $C$. Suppose instead there were $t_0\in(r,R)$ with $A(t_0)<C$. Then $A(t)\leq A(t_0)<C$ for every $t\in(t_0,R)$, and since $w_k>0$ on $(0,R)$ the weighted average of $A$ over $(r,R)$ would be strictly smaller than $C$. Thus $A(s)\leq C\leq A(t)$ for all $s\in(0,r)$ and $t\in(r,R)$, while monotonicity gives the reverse inequality across the cut point. It follows that $A(t)=C$ for every $t\in(0,R)$.[/step]

[guided]The goal of this averaging step is to turn directional comparison into equality of the averaged comparison quotient. The measure $d\sigma_p$ is the spherical measure on $S_pM$ induced by $g_p$, and its total mass is $\mathcal{H}^{n-1}(S^{n-1})$. The function $q$ is nonnegative and measurable on $(0,\infty)\times S_pM$: before the cut time it is the smooth polar Jacobian quotient, and after the cut time it is defined to be $0$. Therefore [Tonelli's Theorem](/page/Tonelli%27s%20Theorem) applies to the nonnegative integrand $q(t,\xi)\operatorname{sn}_k(t)^{n-1}$. Define \begin{align*} A:(0,\infty)&\to [0,\infty),\\ t&\mapsto \frac{1}{\mathcal{H}^{n-1}(S^{n-1})}\int_{S_pM} q(t,\xi)\,d\sigma_p(\xi). \end{align*} Because $q(t_1,\xi)\geq q(t_2,\xi)$ whenever $0<t_1<t_2<R$, integrating this inequality over $S_pM$ gives $A(t_1)\geq A(t_2)$. Thus $A$ is nonincreasing. The [geodesic polar-coordinate volume formula](/page/Geodesic%20Polar%20Coordinates) and [Tonelli's Theorem](/page/Tonelli%27s%20Theorem) give, for $\rho\in(0,\rho_k)$, \begin{align*} \operatorname{Vol}_g(B(p,\rho)) &= \int_0^\rho\int_{S_pM} q(t,\xi)\operatorname{sn}_k(t)^{n-1}\,d\sigma_p(\xi)\,d\mathcal{L}^1(t)\\ &=\int_0^\rho A(t)w_k(t)\,d\mathcal{L}^1(t), \end{align*} where $w_k(t)=\mathcal{H}^{n-1}(S^{n-1})\operatorname{sn}_k(t)^{n-1}$. Also \begin{align*} V_k(\rho)=\int_0^\rho w_k(t)\,d\mathcal{L}^1(t). \end{align*} Hence the Bishop-Gromov ratio is the weighted average of $A$ over $(0,\rho)$ with positive weight $w_k$. Since $A$ is nonincreasing and $w_k>0$ on $(0,R)$, enlarging the interval from $(0,r)$ to $(0,R)$ can only decrease this weighted average. The hypothesis says the two averages are equal. Let \begin{align*} C=\frac{\operatorname{Vol}_g(B(p,r))}{V_k(r)}. \end{align*} Subtracting the weighted integral over $(0,r)$ from the weighted integral over $(0,R)$ shows that the shell average over $(r,R)$ is also $C$. Now monotonicity and positivity of the weight force pointwise equality for the monotone representative. If $A(s_0)>C$ for some $s_0\in(0,r)$, then $A(s)\geq A(s_0)>C$ for every $s\in(0,s_0)$. Since $w_k>0$ on this interval, the weighted average over $(0,r)$ would be larger than $C$, a contradiction. If $A(t_0)<C$ for some $t_0\in(r,R)$, then $A(t)\leq A(t_0)<C$ for every $t\in(t_0,R)$, so the weighted average over $(r,R)$ would be smaller than $C$, again a contradiction. Therefore the monotone representative satisfies \begin{align*} A(t)=C \end{align*} for every $t\in(0,R)$.[/guided]

[step:Promote averaged equality to directional equality for almost every radial geodesic]Let $a,b\in \mathbb{Q}\cap(r,R)$ with $a<b$. Since $q(\cdot,\xi)$ is nonincreasing, \begin{align*} q(a,\xi)-q(b,\xi)\geq 0 \end{align*} for every $\xi\in S_pM$. Integrating over $S_pM$ and using the pointwise equality $A\equiv C$ on $(0,R)$ gives \begin{align*} \int_{S_pM}\bigl(q(a,\xi)-q(b,\xi)\bigr)\,d\sigma_p(\xi) = \mathcal{H}^{n-1}(S^{n-1})\bigl(A(a)-A(b)\bigr) = 0. \end{align*} Thus $q(a,\xi)=q(b,\xi)$ for $d\sigma_p$-almost every $\xi$. Applying this to all rational pairs $a<b$ in $(r,R)$ and intersecting the corresponding full-measure subsets, we obtain a full-measure set $E\subset S_pM$ such that, for every $\xi\in E$, the function $t\mapsto q(t,\xi)$ takes the same value at every rational point of $(r,R)$. Since $q(\cdot,\xi)$ is monotone, equality on the dense set $\mathbb{Q}\cap(r,R)$ implies that $q(\cdot,\xi)$ is constant on $(r,R)$ in the monotone-function sense. Fix $\xi\in E$ and a connected component $I$ of \begin{align*} \{t\in (r,R):t<\operatorname{cut}_p(\xi)\}. \end{align*} On $I$, the function \begin{align*} t\mapsto \frac{J(t,\xi)}{\operatorname{sn}_k(t)^{n-1}} \end{align*} is smooth. Since $\mathbb{Q}\cap I$ is dense in $I$ and the quotient agrees with $q(\cdot,\xi)$ on $I$, continuity implies that it is constant for every $t\in I$.[/step]

[guided]We now convert equality of the average into equality along almost every direction. Choose rational numbers $a,b\in\mathbb{Q}\cap(r,R)$ with $a<b$. Since the directional quotient is nonincreasing, for each $\xi\in S_pM$ we have \begin{align*} q(a,\xi)-q(b,\xi)\geq 0. \end{align*} Integrating this nonnegative function over $(S_pM,d\sigma_p)$ gives \begin{align*} \int_{S_pM}\bigl(q(a,\xi)-q(b,\xi)\bigr)\,d\sigma_p(\xi) = \mathcal{H}^{n-1}(S^{n-1})\bigl(A(a)-A(b)\bigr)=0, \end{align*} because the previous step proved $A\equiv C$ pointwise on $(0,R)$. A nonnegative measurable function with integral zero is zero almost everywhere, so $q(a,\xi)=q(b,\xi)$ for $d\sigma_p$-almost every $\xi$. Apply this argument to all rational pairs $a<b$ in $(r,R)$. Intersecting the corresponding countably many full-measure subsets of $S_pM$ gives a full-measure set $E\subset S_pM$ such that for every $\xi\in E$, the monotone function $t\mapsto q(t,\xi)$ takes the same value at every rational point of $(r,R)$. A monotone function with equal values on a dense set is constant in the monotone-function sense on $(r,R)$. Fix $\xi\in E$ and a connected component $I$ of $\{t\in(r,R):t<\operatorname{cut}_p(\xi)\}$. Since $I$ is an interval, $\mathbb{Q}\cap I$ is dense in $I$. On $I$ the quotient \begin{align*} t\mapsto \frac{J(t,\xi)}{\operatorname{sn}_k(t)^{n-1}} \end{align*} is smooth, hence continuous, and it agrees with $q(\cdot,\xi)$ on $I$. Continuity therefore upgrades equality on $\mathbb{Q}\cap I$ to equality for every $t\in I$.[/guided]

[step:Differentiate the constant quotient to obtain equality of the mean curvature comparison]Fix $\xi\in E$ and a connected component $I$ as above. Let \begin{align*} m:I&\to \mathbb{R},\\ t&\mapsto \frac{\partial}{\partial t}\log J(t,\xi) \end{align*} denote the radial mean curvature, equivalently the trace of the shape operator $S_t$ of the geodesic sphere centered at $p$ at $\gamma_\xi(t)$. Since \begin{align*} \frac{J(t,\xi)}{\operatorname{sn}_k(t)^{n-1}} \end{align*} is constant on $I$, differentiating its logarithm gives \begin{align*} 0 &= \frac{\partial}{\partial t}\log J(t,\xi) - (n-1)\frac{\operatorname{sn}_k'(t)}{\operatorname{sn}_k(t)}\\ &= m(t)-(n-1)\operatorname{ct}_k(t). \end{align*} Therefore \begin{align*} \operatorname{tr}S_t=m(t)=(n-1)\operatorname{ct}_k(t) \end{align*} for every $t\in I$.[/step]

[guided]Fix $\xi\in E$ and a regular connected component $I$. On $I$, the polar Jacobian is positive and smooth, so the logarithmic derivative is defined. Let \begin{align*} m:I&\to \mathbb{R},\\ t&\mapsto \frac{\partial}{\partial t}\log J(t,\xi). \end{align*} This function is the trace of the shape operator $S_t$ of the geodesic sphere centered at $p$ at the point $\gamma_\xi(t)$. The previous step says that \begin{align*} \frac{J(t,\xi)}{\operatorname{sn}_k(t)^{n-1}} \end{align*} is constant on $I$. Taking the logarithm is legitimate because both $J(t,\xi)$ and $\operatorname{sn}_k(t)$ are positive on $I$. Differentiating the logarithm gives \begin{align*} 0 &=\frac{\partial}{\partial t}\log J(t,\xi) -(n-1)\frac{\operatorname{sn}_k'(t)}{\operatorname{sn}_k(t)}\\ &=m(t)-(n-1)\operatorname{ct}_k(t). \end{align*} Therefore \begin{align*} \operatorname{tr}S_t=m(t)=(n-1)\operatorname{ct}_k(t) \end{align*} for every $t\in I$.[/guided]

[step:Use the trace Riccati equation to identify the equality case]On $I$, the [trace Riccati equation](/page/Riccati%20Equation) for the shape operator along $\gamma_\xi$ is \begin{align*} m'(t)+|S_t|^2+\operatorname{Ric}_g(\dot{\gamma}_\xi(t),\dot{\gamma}_\xi(t))=0. \end{align*} The [Cauchy-Schwarz inequality](/page/Cauchy-Schwarz%20Inequality) for the eigenvalues of the symmetric endomorphism $S_t$ gives \begin{align*} |S_t|^2\geq \frac{m(t)^2}{n-1}, \end{align*} with equality if and only if $S_t=\frac{m(t)}{n-1}\operatorname{Id}$ on the tangent space to the geodesic sphere. The Ricci lower bound gives \begin{align*} \operatorname{Ric}_g(\dot{\gamma}_\xi(t),\dot{\gamma}_\xi(t))\geq (n-1)k. \end{align*} Since $m(t)=(n-1)\operatorname{ct}_k(t)$ and $\operatorname{ct}_k$ satisfies \begin{align*} \frac{d}{dt}\bigl((n-1)\operatorname{ct}_k(t)\bigr) + \frac{\bigl((n-1)\operatorname{ct}_k(t)\bigr)^2}{n-1} + (n-1)k = 0, \end{align*} substitution into the trace Riccati equation yields \begin{align*} 0 &= m'(t)+|S_t|^2+\operatorname{Ric}_g(\dot{\gamma}_\xi(t),\dot{\gamma}_\xi(t))\\ &\geq m'(t)+\frac{m(t)^2}{n-1}+(n-1)k\\ &= 0. \end{align*} Both inequalities in the middle line must therefore be equalities. Hence \begin{align*} \operatorname{Ric}_g(\dot{\gamma}_\xi(t),\dot{\gamma}_\xi(t))=(n-1)k \end{align*} and \begin{align*} S_t=\operatorname{ct}_k(t)\operatorname{Id}. \end{align*} This is precisely equality in the radial Riccati comparison inequality along $\gamma_\xi$ on $I$. Since $\xi\in E$ was arbitrary and $d\sigma_p(S_pM\setminus E)=0$, the asserted equality statements hold for almost every radial direction on every regular connected component of the annulus. The construction used only points with $t<\operatorname{cut}_p(\xi)$, so it makes no assertion across the cut locus or outside $(r,R)$.[/step]

[guided]We finish by identifying which inequalities in the Riccati comparison proof have become equalities. Along the regular radial geodesic segment, the [trace Riccati equation](/page/Riccati%20Equation) says \begin{align*} m'(t)+|S_t|^2+\operatorname{Ric}_g(\dot{\gamma}_\xi(t),\dot{\gamma}_\xi(t))=0. \end{align*} Here $S_t$ is the shape operator of the geodesic sphere and $m(t)=\operatorname{tr}S_t$. The [Cauchy-Schwarz inequality](/page/Cauchy-Schwarz%20Inequality) applied to the eigenvalues of the symmetric endomorphism $S_t$ gives \begin{align*} |S_t|^2\geq \frac{m(t)^2}{n-1}, \end{align*} with equality exactly when all eigenvalues are equal, equivalently when \begin{align*} S_t=\frac{m(t)}{n-1}\operatorname{Id} \end{align*} on the tangent space to the geodesic sphere. The Ricci lower bound gives \begin{align*} \operatorname{Ric}_g(\dot{\gamma}_\xi(t),\dot{\gamma}_\xi(t))\geq (n-1)k. \end{align*} From the previous step, $m(t)=(n-1)\operatorname{ct}_k(t)$. The model function satisfies \begin{align*} \frac{d}{dt}\bigl((n-1)\operatorname{ct}_k(t)\bigr) +\frac{\bigl((n-1)\operatorname{ct}_k(t)\bigr)^2}{n-1} +(n-1)k=0. \end{align*} Substituting this value of $m$ into the trace Riccati equation yields \begin{align*} 0 &=m'(t)+|S_t|^2+\operatorname{Ric}_g(\dot{\gamma}_\xi(t),\dot{\gamma}_\xi(t))\\ &\geq m'(t)+\frac{m(t)^2}{n-1}+(n-1)k\\ &=0. \end{align*} A quantity squeezed between the same value $0$ and $0$ must have equality at each intervening inequality. Therefore \begin{align*} \operatorname{Ric}_g(\dot{\gamma}_\xi(t),\dot{\gamma}_\xi(t))=(n-1)k \end{align*} and \begin{align*} S_t=\operatorname{ct}_k(t)\operatorname{Id}. \end{align*} This is precisely equality in the radial Riccati comparison inequality along $\gamma_\xi$ on $I$. Since $\xi$ was chosen from a full-measure subset of $S_pM$, the asserted equality statements hold for almost every direction and every regular connected component of the annulus. The argument never crosses the cut locus and never uses radii outside $(r,R)$.[/guided]