[proofplan] We write the Riemannian volume of $B_g(p,R)$ in geodesic polar coordinates around $p$, with each radial integral stopped at the cut time. The Ricci lower bound gives a Riccati inequality for the mean curvature of geodesic spheres, and comparison with the model solution $\operatorname{sn}_k'(t)/\operatorname{sn}_k(t)$ shows that the radial Jacobian divided by $\operatorname{sn}_k(t)^{n-1}$ is nonincreasing along each ray. Averaging this pointwise monotonicity over the unit tangent sphere gives monotonicity of the radial area ratio. A one-dimensional weighted-average lemma then converts monotonicity of the area ratio into monotonicity of the ball-volume ratio. [/proofplan] [step:Define the model density and the polar-coordinate data around $p$] Fix $p \in M$. Let \begin{align*} S_pM := \{v \in T_pM : g_p(v,v)=1\} \end{align*} be the unit tangent sphere at $p$, and let $\sigma_p$ denote the Riemannian hypersurface measure on $S_pM$. Thus $\sigma_p(S_pM)=\omega_{n-1}$. For each $\theta \in S_pM$, define the unit-speed geodesic \begin{align*} \gamma_\theta:[0,\infty) &\to M \\ t &\mapsto \exp_p(t\theta), \end{align*} which is defined for all $t \geq 0$ because $(M,g)$ is complete. Define the cut time \begin{align*} c:S_pM &\to (0,\infty] \\ \theta &\mapsto \sup\{s>0:\gamma_\theta|_{[0,s]} \text{ is minimizing}\}. \end{align*} On the set \begin{align*} \mathcal{U}_p := \{(t,\theta)\in (0,\infty)\times S_pM : 0<t<c(\theta)\}, \end{align*} the map \begin{align*} \Phi_p:\mathcal{U}_p &\to M\setminus(\{p\}\cup \operatorname{Cut}(p)) \\ (t,\theta) &\mapsto \exp_p(t\theta) \end{align*} is a smooth polar-coordinate parametrization. Let \begin{align*} J_p:\mathcal{U}_p &\to (0,\infty) \end{align*} denote its Riemannian Jacobian, characterized by \begin{align*} \Phi_p^*(d\operatorname{vol}_g) = J_p(t,\theta)\,d\mathcal{L}^1(t)\,d\sigma_p(\theta). \end{align*} Let $\operatorname{Cut}(p) \subset M$ denote the cut locus of $p$, namely the set of endpoints $\gamma_\theta(c(\theta))$ for directions $\theta \in S_pM$ with $c(\theta)<\infty$, together with the usual closure of these cut endpoints. We use the standard [geodesic polar-coordinate theorem](/page/Geodesic%20Polar%20Coordinates), including that $\Phi_p$ is a smooth parametrization on $\mathcal{U}_p$ and that $\operatorname{Cut}(p)$ has $\operatorname{vol}_g$-measure zero. Define the model radial area density \begin{align*} a_k:I_k &\to (0,\infty) \\ t &\mapsto \omega_{n-1}\operatorname{sn}_k(t)^{n-1}. \end{align*} Since $t \in I_k$, we have $\operatorname{sn}_k(t)>0$. [/step] [step:Compare the radial Jacobian with the model Jacobian before the cut time] Fix $\theta \in S_pM$. We prove that the function \begin{align*} q_\theta:(0,c(\theta))\cap I_k &\to (0,\infty) \\ t &\mapsto \frac{J_p(t,\theta)}{\operatorname{sn}_k(t)^{n-1}} \end{align*} is nonincreasing. On the interval $(0,c(\theta))$, the radial distance from $p$ is smooth along the ray and the polar-coordinate map has no cut singularity in the direction $\theta$. Therefore the standard [first variation formula for the polar Jacobian](/page/First%20Variation%20Formula) and the [Riccati equation for geodesic spheres](/page/Riccati%20Equation) apply on every compact subinterval of $(0,c(\theta))$. Let $S_\theta(t):\dot{\gamma}_\theta(t)^\perp \to \dot{\gamma}_\theta(t)^\perp$ be the shape operator of the geodesic sphere centered at $p$ at the point $\gamma_\theta(t)$, computed with respect to the outward radial unit normal $\dot{\gamma}_\theta(t)$. Define the radial mean curvature function \begin{align*} h_\theta:(0,c(\theta)) &\to \mathbb{R} \\ t &\mapsto \operatorname{tr} S_\theta(t). \end{align*} The [first variation formula](/theorems/2728) for the polar Jacobian gives \begin{align*} \frac{\partial}{\partial t}\log J_p(t,\theta)=h_\theta(t). \end{align*} The Riccati equation for $S_\theta(t)$ gives \begin{align*} h_\theta'(t)+\operatorname{tr}(S_\theta(t)^2)+\operatorname{Ric}_g(\dot{\gamma}_\theta(t),\dot{\gamma}_\theta(t))=0. \end{align*} Since $S_\theta(t)$ is symmetric on the $(n-1)$-dimensional [inner product space](/page/Inner%20Product%20Space) $\dot{\gamma}_\theta(t)^\perp$, Cauchy-Schwarz for its eigenvalues gives \begin{align*} \operatorname{tr}(S_\theta(t)^2)\geq \frac{h_\theta(t)^2}{n-1}. \end{align*} The Ricci hypothesis and $|\dot{\gamma}_\theta(t)|_g=1$ give \begin{align*} \operatorname{Ric}_g(\dot{\gamma}_\theta(t),\dot{\gamma}_\theta(t))\geq (n-1)k. \end{align*} Therefore \begin{align*} h_\theta'(t)+\frac{h_\theta(t)^2}{n-1}+(n-1)k \leq 0. \end{align*} Define the model mean curvature \begin{align*} h_k:I_k &\to \mathbb{R} \\ t &\mapsto (n-1)\frac{\operatorname{sn}_k'(t)}{\operatorname{sn}_k(t)}. \end{align*} Because $\operatorname{sn}_k''(t)+k\operatorname{sn}_k(t)=0$, the function $h_k$ satisfies \begin{align*} h_k'(t)+\frac{h_k(t)^2}{n-1}+(n-1)k=0. \end{align*} The initial asymptotics of geodesic polar coordinates and of the model density are \begin{align*} J_p(t,\theta)=t^{n-1}(1+O(t^2)), \qquad \operatorname{sn}_k(t)^{n-1}=t^{n-1}(1+O(t^2)), \end{align*} as $t\downarrow 0$, and hence \begin{align*} h_\theta(t)-h_k(t)=O(t). \end{align*} Set \begin{align*} w_\theta:(0,c(\theta))\cap I_k &\to \mathbb{R} \\ t &\mapsto h_\theta(t)-h_k(t). \end{align*} Subtracting the equality for $h_k$ from the inequality for $h_\theta$ gives \begin{align*} w_\theta'(t)+\frac{h_\theta(t)+h_k(t)}{n-1}w_\theta(t)\leq 0. \end{align*} For $0<\varepsilon<t<c(\theta)$ with $t\in I_k$, multiply this inequality by the positive integrating factor \begin{align*} E_{\varepsilon,t} := \exp\left( \int_\varepsilon^t \frac{h_\theta(s)+h_k(s)}{n-1}\,d\mathcal{L}^1(s) \right). \end{align*} We obtain \begin{align*} w_\theta(t)E_{\varepsilon,t}\leq w_\theta(\varepsilon). \end{align*} The same initial asymptotics give \begin{align*} \frac{h_\theta(s)+h_k(s)}{n-1}=\frac{2}{s}+O(s) \end{align*} as $s\downarrow 0$. Hence, for fixed $t$, there are constants $C_t>0$ and $\varepsilon_t>0$ such that \begin{align*} E_{\varepsilon,t}\geq C_t\varepsilon^{-2} \end{align*} whenever $0<\varepsilon<\varepsilon_t$. Since the preceding inequality is equivalent to \begin{align*} w_\theta(t)\leq \frac{w_\theta(\varepsilon)}{E_{\varepsilon,t}}, \end{align*} and $w_\theta(\varepsilon)=O(\varepsilon)$, the right-hand side tends to $0$ as $\varepsilon\downarrow 0$. Thus \begin{align*} h_\theta(t)\leq h_k(t) \end{align*} for every $t\in(0,c(\theta))\cap I_k$. Finally, \begin{align*} \frac{\partial}{\partial t}\log q_\theta(t) = \frac{\partial}{\partial t}\log J_p(t,\theta) - (n-1)\frac{\operatorname{sn}_k'(t)}{\operatorname{sn}_k(t)} = h_\theta(t)-h_k(t) \leq 0. \end{align*} Therefore $q_\theta$ is nonincreasing. [guided] Fix a direction $\theta \in S_pM$. The goal along this single geodesic is to compare the actual polar volume density $J_p(t,\theta)$ with the model density $\operatorname{sn}_k(t)^{n-1}$. The correct quantity is their quotient, because the model space form has radial Jacobian exactly $\operatorname{sn}_k(t)^{n-1}$ in each direction. Define \begin{align*} q_\theta:(0,c(\theta))\cap I_k &\to (0,\infty) \\ t &\mapsto \frac{J_p(t,\theta)}{\operatorname{sn}_k(t)^{n-1}}. \end{align*} We prove that $q_\theta$ is nonincreasing by differentiating its logarithm. On the interval $(0,c(\theta))$, the point $\gamma_\theta(t)$ lies before the cut time, so the radial distance from $p$ is smooth along this ray and the polar-coordinate parametrization has no cut singularity in the direction $\theta$. This is precisely the setting in which the standard [first variation formula for the polar Jacobian](/page/First%20Variation%20Formula) and the [Riccati equation for geodesic spheres](/page/Riccati%20Equation) apply. Let \begin{align*} S_\theta(t):\dot{\gamma}_\theta(t)^\perp &\to \dot{\gamma}_\theta(t)^\perp \end{align*} be the shape operator of the geodesic sphere centered at $p$ at $\gamma_\theta(t)$, with outward unit normal $\dot{\gamma}_\theta(t)$. Define \begin{align*} h_\theta:(0,c(\theta)) &\to \mathbb{R} \\ t &\mapsto \operatorname{tr} S_\theta(t). \end{align*} The polar Jacobian changes at the rate given by the mean curvature: \begin{align*} \frac{\partial}{\partial t}\log J_p(t,\theta)=h_\theta(t). \end{align*} The shape operator satisfies the Riccati equation along the radial geodesic: \begin{align*} h_\theta'(t)+\operatorname{tr}(S_\theta(t)^2)+\operatorname{Ric}_g(\dot{\gamma}_\theta(t),\dot{\gamma}_\theta(t))=0. \end{align*} Now we use exactly the two hypotheses that allow comparison. First, $S_\theta(t)$ is symmetric on an $(n-1)$-dimensional [inner product](/page/Inner%20Product) space, so Cauchy-Schwarz applied to its eigenvalues gives \begin{align*} \operatorname{tr}(S_\theta(t)^2)\geq \frac{h_\theta(t)^2}{n-1}. \end{align*} Second, the geodesic is unit-speed, so $g(\dot{\gamma}_\theta(t),\dot{\gamma}_\theta(t))=1$, and the Ricci lower bound gives \begin{align*} \operatorname{Ric}_g(\dot{\gamma}_\theta(t),\dot{\gamma}_\theta(t))\geq (n-1)k. \end{align*} Substituting these two inequalities into the Riccati equation yields \begin{align*} h_\theta'(t)+\frac{h_\theta(t)^2}{n-1}+(n-1)k \leq 0. \end{align*} The model comparison function is \begin{align*} h_k:I_k &\to \mathbb{R} \\ t &\mapsto (n-1)\frac{\operatorname{sn}_k'(t)}{\operatorname{sn}_k(t)}. \end{align*} Since $\operatorname{sn}_k''+k\operatorname{sn}_k=0$, direct differentiation gives \begin{align*} h_k'(t)+\frac{h_k(t)^2}{n-1}+(n-1)k=0. \end{align*} Thus $h_\theta$ is a subsolution of the scalar Riccati equation solved by $h_k$. To compare the two, define \begin{align*} w_\theta:(0,c(\theta))\cap I_k &\to \mathbb{R} \\ t &\mapsto h_\theta(t)-h_k(t). \end{align*} Subtracting the equality for $h_k$ from the inequality for $h_\theta$ gives \begin{align*} w_\theta'(t)+\frac{h_\theta(t)+h_k(t)}{n-1}w_\theta(t)\leq 0. \end{align*} The singularity at $t=0$ is harmless because both actual and model polar densities have the same Euclidean first-order behavior: \begin{align*} J_p(t,\theta)=t^{n-1}(1+O(t^2)), \qquad \operatorname{sn}_k(t)^{n-1}=t^{n-1}(1+O(t^2)). \end{align*} Consequently \begin{align*} w_\theta(t)=O(t) \end{align*} as $t\downarrow 0$. For $0<\varepsilon<t<c(\theta)$ with $t\in I_k$, multiply the differential inequality for $w_\theta$ by the positive integrating factor \begin{align*} E_{\varepsilon,t} := \exp\left( \int_\varepsilon^t \frac{h_\theta(s)+h_k(s)}{n-1}\,d\mathcal{L}^1(s) \right). \end{align*} This gives \begin{align*} w_\theta(t)E_{\varepsilon,t}\leq w_\theta(\varepsilon). \end{align*} We still have to justify the limit as $\varepsilon\downarrow 0$, because the integrating factor depends on $\varepsilon$. The same initial asymptotics imply \begin{align*} \frac{h_\theta(s)+h_k(s)}{n-1}=\frac{2}{s}+O(s) \end{align*} as $s\downarrow 0$. Therefore, for fixed $t$, there are constants $C_t>0$ and $\varepsilon_t>0$ such that \begin{align*} E_{\varepsilon,t}\geq C_t\varepsilon^{-2} \end{align*} whenever $0<\varepsilon<\varepsilon_t$. From \begin{align*} w_\theta(t)E_{\varepsilon,t}\leq w_\theta(\varepsilon) \end{align*} we get \begin{align*} w_\theta(t)\leq \frac{w_\theta(\varepsilon)}{E_{\varepsilon,t}}. \end{align*} Since $w_\theta(\varepsilon)=O(\varepsilon)$, the right-hand side is $O(\varepsilon^3)$ and tends to $0$. Therefore \begin{align*} h_\theta(t)\leq h_k(t) \end{align*} for every $t\in(0,c(\theta))\cap I_k$. Now differentiate the logarithm of the quotient: \begin{align*} \frac{\partial}{\partial t}\log q_\theta(t) &= \frac{\partial}{\partial t}\log J_p(t,\theta) - (n-1)\frac{\operatorname{sn}_k'(t)}{\operatorname{sn}_k(t)} \\ &= h_\theta(t)-h_k(t) \\ &\leq 0. \end{align*} Since $q_\theta(t)>0$, the inequality for the logarithmic derivative proves that $q_\theta$ is nonincreasing along the radial geodesic before the cut time. [/guided] [/step] [step:Average the pointwise comparison over all radial directions] Define the actual radial area density \begin{align*} a:I_k &\to [0,\infty) \\ t &\mapsto \int_{S_pM}\mathbb{1}_{\{t<c(\theta)\}}J_p(t,\theta)\,d\sigma_p(\theta). \end{align*} For each $t\in I_k$, define \begin{align*} b_t:S_pM &\to [0,\infty) \\ \theta &\mapsto \mathbb{1}_{\{t<c(\theta)\}} \frac{J_p(t,\theta)}{\operatorname{sn}_k(t)^{n-1}}. \end{align*} If $0<s<t$ and $t<c(\theta)$, then $s<c(\theta)$ and the previous step gives \begin{align*} \frac{J_p(t,\theta)}{\operatorname{sn}_k(t)^{n-1}} \leq \frac{J_p(s,\theta)}{\operatorname{sn}_k(s)^{n-1}}. \end{align*} If $t\geq c(\theta)$, then $b_t(\theta)=0\leq b_s(\theta)$. Hence \begin{align*} b_t(\theta)\leq b_s(\theta) \end{align*} for every $\theta\in S_pM$ whenever $0<s<t$ in $I_k$. Since $\operatorname{sn}_k(t)>0$ on $I_k$, \begin{align*} \frac{a(t)}{a_k(t)} &= \frac{1}{\omega_{n-1}} \int_{S_pM} \mathbb{1}_{\{t<c(\theta)\}} \frac{J_p(t,\theta)}{\operatorname{sn}_k(t)^{n-1}} \,d\sigma_p(\theta) \\ &= \frac{1}{\omega_{n-1}}\int_{S_pM} b_t(\theta)\,d\sigma_p(\theta). \end{align*} Averaging the pointwise inequality $b_t\leq b_s$ over $S_pM$ gives \begin{align*} \frac{a(t)}{a_k(t)} \leq \frac{a(s)}{a_k(s)} \end{align*} whenever $0<s<t$ in $I_k$. Thus $a/a_k$ is nonincreasing on $I_k$. [/step] [step:Convert monotonicity of area densities into monotonicity of ball volumes] Define \begin{align*} A:I_k &\to (0,\infty) \\ R &\mapsto \operatorname{vol}_g(B_g(p,R)), \end{align*} and \begin{align*} A_k:I_k &\to (0,\infty) \\ R &\mapsto V_k^n(R). \end{align*} By the polar-coordinate formula and the fact that the cut locus has $\operatorname{vol}_g$-measure zero, \begin{align*} A(R)=\int_0^R a(t)\,d\mathcal{L}^1(t). \end{align*} By definition of the model ball volume, \begin{align*} A_k(R)=\int_0^R a_k(t)\,d\mathcal{L}^1(t). \end{align*} We now use the following one-dimensional monotone-density lemma. [claim:Weighted averages of a nonincreasing density are nonincreasing] Let $0<R_1<R_2$ be numbers in $I_k$. Let $\alpha:I_k\to (0,\infty)$ be locally integrable and let $q:I_k\to[0,\infty)$ be nonincreasing. Define the accumulated weighted functions \begin{align*} F:I_k &\to [0,\infty) \\ R &\mapsto \int_0^R q(t)\alpha(t)\,d\mathcal{L}^1(t), \end{align*} and \begin{align*} G:I_k &\to (0,\infty) \\ R &\mapsto \int_0^R \alpha(t)\,d\mathcal{L}^1(t). \end{align*} Then \begin{align*} \frac{F(R_2)}{G(R_2)}\leq \frac{F(R_1)}{G(R_1)}. \end{align*} [/claim] [proof] Set \begin{align*} m_1:=\frac{F(R_1)}{G(R_1)}. \end{align*} Since $q$ is nonincreasing, for $0<s<R_1<t<R_2$ we have $q(t)\leq q(s)$. Averaging this inequality first in $s$ over $(0,R_1)$ with weight $\alpha(s)\,d\mathcal{L}^1(s)$ gives \begin{align*} q(t)\leq m_1 \end{align*} for every $t\in(R_1,R_2)$ at which $q(t)$ is defined. Therefore \begin{align*} \int_{R_1}^{R_2} q(t)\alpha(t)\,d\mathcal{L}^1(t) \leq m_1\int_{R_1}^{R_2}\alpha(t)\,d\mathcal{L}^1(t). \end{align*} Adding $F(R_1)=m_1G(R_1)$ to both sides gives \begin{align*} F(R_2) \leq m_1G(R_1)+m_1\int_{R_1}^{R_2}\alpha(t)\,d\mathcal{L}^1(t) = m_1G(R_2). \end{align*} Since $G(R_2)>0$, division by $G(R_2)$ gives \begin{align*} \frac{F(R_2)}{G(R_2)} \leq m_1 = \frac{F(R_1)}{G(R_1)}. \end{align*} [/proof] Apply the lemma with the maps \begin{align*} \alpha:I_k &\to (0,\infty) \\ t &\mapsto a_k(t), \end{align*} and \begin{align*} q:I_k &\to [0,\infty) \\ t &\mapsto \frac{a(t)}{a_k(t)}. \end{align*} The previous step proves that $q$ is nonincreasing, and $a_k(t)>0$ for every $t\in I_k$. Since $a(t)=q(t)a_k(t)$, the lemma gives, for $0<r<R$ in $I_k$, \begin{align*} \frac{A(R)}{A_k(R)} \leq \frac{A(r)}{A_k(r)}. \end{align*} [guided] At this point the geometric work is finished. We know that the ratio of radial area densities \begin{align*} q:I_k &\to [0,\infty) \\ t &\mapsto \frac{a(t)}{a_k(t)} \end{align*} is nonincreasing by the pointwise radial Jacobian comparison and averaging over $S_pM$ from the previous step. The remaining question is purely one-dimensional: why does a nonincreasing density ratio imply that the accumulated-volume ratio is also nonincreasing? Define the ball-volume maps \begin{align*} A:I_k &\to (0,\infty) \\ R &\mapsto \operatorname{vol}_g(B_g(p,R)), \end{align*} and \begin{align*} A_k:I_k &\to (0,\infty) \\ R &\mapsto V_k^n(R). \end{align*} The [geodesic polar-coordinate theorem](/page/Geodesic%20Polar%20Coordinates) gives \begin{align*} A(R)=\int_0^R a(t)\,d\mathcal{L}^1(t), \end{align*} because the polar parametrization covers the ball outside the cut locus and the cut locus has $\operatorname{vol}_g$-measure zero. The model identity is \begin{align*} A_k(R)=\int_0^R a_k(t)\,d\mathcal{L}^1(t). \end{align*} Since $a(t)=q(t)a_k(t)$, the quotient $A(R)/A_k(R)$ is the $a_k(t)\,d\mathcal{L}^1(t)$-weighted average of $q$ over $(0,R)$: \begin{align*} \frac{A(R)}{A_k(R)} = \frac{\int_0^R q(t)a_k(t)\,d\mathcal{L}^1(t)} {\int_0^R a_k(t)\,d\mathcal{L}^1(t)}. \end{align*} Fix $0<r<R$ in $I_k$ and set \begin{align*} m_r:= \frac{\int_0^r q(t)a_k(t)\,d\mathcal{L}^1(t)} {\int_0^r a_k(t)\,d\mathcal{L}^1(t)}. \end{align*} Because $q$ is nonincreasing, every later value $q(t)$ with $r<t<R$ is bounded above by the weighted average of the earlier values on $(0,r)$. Therefore \begin{align*} \int_r^R q(t)a_k(t)\,d\mathcal{L}^1(t) \leq m_r\int_r^R a_k(t)\,d\mathcal{L}^1(t). \end{align*} Adding the identity \begin{align*} \int_0^r q(t)a_k(t)\,d\mathcal{L}^1(t) = m_r\int_0^r a_k(t)\,d\mathcal{L}^1(t) \end{align*} to this inequality gives \begin{align*} \int_0^R q(t)a_k(t)\,d\mathcal{L}^1(t) \leq m_r\int_0^R a_k(t)\,d\mathcal{L}^1(t). \end{align*} Since $a_k(t)>0$ on $I_k$, the denominator is positive, so division gives \begin{align*} \frac{A(R)}{A_k(R)} \leq m_r = \frac{A(r)}{A_k(r)}. \end{align*} This proves the desired monotonicity of the ball-volume ratio. [/guided] [/step] [step:Conclude the Bishop-Gromov inequality for metric balls] For every $0<r<R$ with $r,R\in I_k$, the preceding step gives \begin{align*} \frac{\operatorname{vol}_g(B_g(p,R))}{V_k^n(R)} = \frac{A(R)}{A_k(R)} \leq \frac{A(r)}{A_k(r)} = \frac{\operatorname{vol}_g(B_g(p,r))}{V_k^n(r)}. \end{align*} Since $p\in M$ was arbitrary, the function \begin{align*} R\mapsto \frac{\operatorname{vol}_g(B_g(p,R))}{V_k^n(R)} \end{align*} is nonincreasing on $I_k$ for every $p\in M$. This is exactly the asserted Bishop-Gromov volume comparison. [/step]