Some more volume comparison results

Let ${(M^n, g)}$ be a Riemannian manifold. We prove two versions of volume comparison results. The first one compares the volume of a geodesic ball with a geodesic ball of the same radius in another Riemannian manifold ${(\overline M^n, \overline g)}$ (not necessarily a space form), under certain assumption on the Ricci curvatures. This is similar to the classical result of Bishop-Gromov in spirit. The second one compares the “exterior region” of a closed hypersurface ${\Sigma}$ with the “exterior region” of a corresponding hypersurface ${\overline \Sigma\subset \overline M}$. Roughly speaking, when the hypersurface ${\Sigma}$ “shrinks” to a point, then this reduces to the first result.

1. Comparing volumes of geodesic balls

Let ${(M^n, g)}$ be a complete Riemannian manifold and ${p\in M}$. The spherical normal coordinates is given by

$\displaystyle (x, t)\mapsto \exp_p(t x)$

for ${(x, t)\in \mathbb{S}^{n-1}(T_pM)\times \mathbb{R}}$ and ${t<\mathrm{inj}(p)}$. Here ${\mathbb{S}^{n-1}(T_pM)=\{v\in T_pM: |v|=1\}}$. Let ${r: M\rightarrow \mathbb{R}}$ be the distance from ${p}$. Let ${B_r(p)=\exp_p (\mathbb{B}(0, r))\subset M}$ be the geodesic ball of radius ${r}$ centered at ${p}$, ${r<\mathrm{inj}(p)}$, the injectivity radius of $p$.

Of course, within the cut locus of ${p}$, ${r}$ is just the coordinate function ${t}$, but we prefer to distinguish these two notions and regard ${r}$ as the distance function and ${t}$ as a parameter for comparison.

Let ${(\overline M^n,\overline g)}$ be another complete Riemannian manifold and ${\overline p\in \overline M}$. The distance from ${\overline p}$ is denoted by ${\overline r}$ and the geodesic ball of radius ${r}$ centered at ${\overline p}$ is denoted by ${B_r(\overline p)}$. We use the spherical normal coordinates to identify the coordinate neighborhood of ${p}$ with that of ${\overline p}$. More precisely, we identify ${\mathbb{S}^{n-1}(T_pM)}$ with ${\mathbb{S}^{n-1}(T_{\overline p}\overline M)}$ by identifying two orthonormal bases ${\{e_i\}_{i=1}^n}$ of ${T_pM}$ and ${\{\overline e_i\}_{i=1}^n}$ of ${T_{\overline p}\overline M}$, then for ${r, we have a diffeomorphism from ${B_r(p)}$ to ${B_r(\overline p)}$ by

$\displaystyle \Phi: \exp_p \left(t\sum_i x^ie_i\right)\mapsto \exp_{\overline p}\left(t\sum_i x^i\overline e_i\right). \ \ \ \ \ (1)$

Under this identification, we can make sense of the following
Ricci curvature condition:

$\displaystyle \mathrm{Ric}(\partial _t, \partial _t)\ge \overline {\mathrm{Ric}}(\partial _t, \partial _t). \ \ \ \ \ (2)$

To state the comparison theorem, we also need to impose another condition on ${\overline M}$. Let ${S_r(p)}$ and ${S_r(\overline p)}$ denote the geodesic sphere of radius ${r}$ centered at ${p}$ and that centered at ${\overline p}$ respectively. We impose the following

Umbilical condition: For all ${r, the sphere ${S_{r}(\overline p)}$ is umbilical at each point. i.e. the second fundamental form ${\overline A}$ of ${S_r(\overline p)}$ is pointwisely proportional to its induced metric at each point ${x}$ (the proportional constant can depend on ${x}$ and ${r}$):

$\displaystyle \overline A=f(x, r)\overline g|_{S_r(\overline p)}. \ \ \ \ \ (3)$

This is for example satisfied if ${\overline g}$ is a warped product metric: ${\overline g= dt^2 + h(x, t)^2 g_S}$, where ${g_S}$ is the round metric on the standard unit sphere. In particular, any space form satisfies this condition.

We can now state the main result:

 Theorem 1 (Volume comparison) Let ${(M^n, g)}$ and ${(\overline M^n, \overline g)}$ be two complete Riemannian manifolds such that the curvature condition (2) and the “umbilical condition” (3) are satisfied. Then for all ${r<\min\{\mathrm{inj}(p), \mathrm{inj}(\overline p)\}}$, $\displaystyle \mathrm{Vol}(B_r(p))\le \mathrm{Vol}(B_r(\overline p)).$ The equality holds if and only if ${B_r(p)}$ is isometric to ${B_r(\overline p)}$.

Let us first focus on the distance function ${r}$ on ${M}$. Let ${\nabla }$ and ${\Delta}$ denote the connection and Laplacian on ${(M,g)}$ respectively. We first state some classic results.

 Lemma 2 (Gauss lemma) Within the cut locus of ${p}$, ${\nabla r=\frac{\partial }{\partial t}.}$

By the Gauss lemma, we immediately get that ${\nabla ^2 r(X, Y)=A(X, Y)}$ on ${S_r(p)}$ and ${\Delta r = H}$, the mean curvature of ${S_r(p)}$. (We use the convention that the second fundamental form of the standard unit sphere ${\mathbb{S}^{n-1}}$ w.r.t. the outward normal ${\nu}$ is positive and its mean curvature ${H}$ w.r.t. ${\nu}$ is ${n-1}$. Also, ${\Delta f=f''}$ on ${\mathbb{R}}$.)

 Lemma 3 (Riccati equation) Within the cut locus of ${p}$, $\displaystyle (\nabla _{\partial _t}\nabla ^2 r) (X, Y)+\left(\nabla ^2 r\right)^2 (X, Y)=-g(R(X, \partial _t) \partial _t, Y).$ Here ${\left(\nabla ^2 r \right)^2(e_i, e_j)=\sum_{k, l=1}^n g^{kl}\nabla ^2 _{ik}r\nabla ^2 _{lj}r }$. (More intrinsically, ${\left(\nabla ^2 r\right) (X, Y)=g(\nabla _X\nabla r, \nabla _Y\nabla r)}$.)

Proof: See the proof of Lemma 12. See also [CLN] p.63 Equation 1.136 or [Petersen] p.265 Equation 2. $\Box$

By taking the trace of the Riccati equation and applying the Cauchy-Schwarz inequality ${(\Delta r)^2 \le (n-1)|\nabla ^2 r|^2}$, we have

 Corollary 4 (Riccati inequality) Within the cut locus of ${p}$, $\displaystyle \frac{\partial }{\partial t}\Delta r +\frac{(\Delta r)^2}{n-1}\le-\mathrm{Ric}(\partial _t, \partial _t).$ The equality holds at the point ${(x, r)}$ if and only if ${S_r(p)}$ is umbilical at this point.

Obviously Corollary 4 is also true on ${(\overline M,\overline g)}$, with “${\le}$” replaced by “${=}$” if the “umbilical condition” (3) holds.

The following maximum principle should be well-known:

 Lemma 5 (Maximum principle) Let ${N}$ be a compact manifold, ${u:N \times (0,T)\rightarrow \mathbb R}$ be a smooth function (${T\le \infty}$) and ${\beta}$ is a locally bounded function on ${[0,T)}$. Suppose ${\displaystyle \limsup_{t\rightarrow 0^+}\max _{x\in N}u(x, t)\le 0}$ and $\displaystyle \frac{\partial u}{\partial t}(x_0,t)\le \beta(t)|u(x_0,t)|$ whenver ${\displaystyle u(x_0,t)= \max _{x\in N}u(x,t)}$. Then ${ u(x,t)\le 0.}$

Proof: Let ${\tau\in (0,T)}$ and suppose ${|\beta|< C}$ on ${[0,\tau]}$. Let ${\delta>0}$ and define ${u_\delta= e^{-Ct}u(x,t)-\delta}$, then ${\displaystyle \limsup_{t\rightarrow 0^+}\max_{x\in N}u_\delta(x,t)<0}$. Suppose there is a first time ${t_1\in (0, \tau]}$ such that ${u_\delta}$ attains ${0}$ at the point ${x_1\in N}$, then ${x_1}$ clearly is the spatial maximum of ${u}$ at time ${t_1}$. Thus,

$\displaystyle \begin{array}{rcl} 0\le \frac{\partial u _\delta}{\partial t} (x_1, t_1) &\le e^{-Ct_1}| \beta(t_1)| |u (x_1,t_1)|- C e^{-Ct_1} u (x_1,t_1)\\ &= (|\beta(t_1)|-C)e^{-Ct_1} u(x_1,t_1)<0 \end{array}$

as ${u(x_1,t_1)>0}$. This contradiction shows that ${u_\delta<0}$ and thus ${u(x,t)\le 0}$ on ${(0,T)}$ by first taking ${\delta\rightarrow 0^+}$ and then ${\tau\rightarrow T}$. $\Box$

We say a function defined on a metric space (e.g. a Riemannian manifold) to be locally Lipschitz if its restriction on any compact set is Lipschitz. For our purpose, we restate the previous result in a form which is more useful for our argument.

 Lemma 6 (Comparison principle) Let ${N}$ be a compact Riemannian manifold. Suppose ${u}$ and ${\overline u: N\times (0, T)\rightarrow \mathbb{R}}$ are smooth functions such that ${\frac{\partial }{\partial t}u(x, t)\le F(u(x, t), x, t)}$ and ${\frac{\partial }{\partial t}\overline u(x,t)=F(\overline u(x,t), x, t)}$ for some locally Lipschitz function ${F}$ on ${\mathbb{R}\times N\times [0,T)}$. Assume that ${\displaystyle \limsup_{t\rightarrow 0^+}\max_{x\in N}(u(x,t)-\overline u(x,t))\le 0}$ and ${u(x, t)-\overline u(x,t)=O(1)}$ as ${t\rightarrow 0^+}$, then $\displaystyle u(x, t)\le \overline u(x,t).$

Proof: It suffices to prove the result on ${N\times (0,\tau)}$ where ${\tau. We can assume ${|u-\overline u|\le C}$ on ${N\times (0,\tau]}$, and that ${|F(y_1, x, t)-F(y_2, x, t)|\le L|y_1- y_2|}$ on ${[-C,C]\times N\times [0,\tau]}$ for some constant ${L}$. Let ${v(x,t)= u(x,t)-\overline u(x,t)}$, then ${\displaystyle \limsup_{t\rightarrow 0^+}\max_{x\in N}v(x,t)\le 0}$ and

$\displaystyle \frac{\partial v }{\partial t} (x_0,t)\le F(u (x_0, t), x_0, t)- F(\overline u(x_0, t), x_0,t)\le L |u(x_0,t)-\overline u(x_0, t)|=L|v(x_0,t)|,$

where ${\displaystyle v(x_0, t)= \max _{x\in N} v(x, t)}$, ${t\in (0,\tau)}$. We can now take ${\beta=L}$ and apply Proposition 5 to conclude that ${u(x,t )\le \overline u (x, t)}$ on ${(0,\tau)}$. $\Box$

Let ${\overline \nabla }$ and ${\overline \Delta}$ denote the connection and Laplacian on ${(\overline M,\overline g)}$ respectively.

 Lemma 7 (Laplacian comparison) If the umbilical condition (3) and ${\mathrm{Ric}(\partial _t, \partial _t)\ge \overline {\mathrm{Ric}}(\partial _t, \partial _t)}$ hold, then ${\Delta r \le \overline \Delta \overline r}$.

Proof: Let ${\sigma=\frac{\Delta r}{n-1}}$ and ${\overline \sigma=\frac{\overline \Delta \overline r}{n-1}}$. Then by Corollary 4,

$\displaystyle \partial _t \sigma+\sigma^2 \le-\frac{\mathrm{Ric}(\partial _t, \partial _t)}{n-1}\le -\frac{\overline {\mathrm{Ric}}(\partial _t, \partial _t)}{n-1}=\partial _t \overline \sigma+\overline \sigma^2.$

Note that ${F(y, x, t): \mathbb{R}\times \mathbb{S}^{n-1}\times [0, l)}$ be defined by ${-\frac{\overline {\mathrm{Ric}}(\partial _t, \partial _t)}{n-1}-y^2}$ is locally Lipschitz. We also have ${\Delta r- \overline \Delta \overline r=H(x, t)-\overline H(x, t)}$, where ${\overline H}$ is the mean curvature of ${S_r(\overline p)}$. It is well-known that both ${H}$ and ${\overline H}$ are of the order ${\frac{n-1}{t}+O(t)}$ as ${t\rightarrow 0^+}$. So ${\sigma - \overline \sigma= O(t)}$, and in particular ${\displaystyle \lim_{t\rightarrow 0^+}(\sigma-\overline \sigma)=0}$, so we can apply Lemma 6 to conclude that ${\Delta r \le \overline \Delta \overline r}$. $\Box$

Let the volume element of ${M}$ and ${\overline M}$ be ${dV}$ and ${d\overline V}$ respectively. Then within ${B_l(p)}$, there is a function ${\lambda(x, t)}$ such that

$\displaystyle dV= \lambda(x, t)dt\wedge dA$

where ${dA}$ is the area element of ${\mathbb{S}^{n-1}}$. Similarly ${d\overline V= \overline \lambda(x, t)dt\wedge dA}$. Note also that ${\lambda(x, r)dA}$ is the area element on ${S_r(p)}$.

As ${\mathcal{L}_{\partial_t}dV=\mathrm{div}(\partial_t)dV=\mathrm{div}(\nabla r)dV=\Delta r dV}$, and ${\mathcal{L}_{\partial_t}dt=\mathcal{L}_{\partial_t}dA=0}$, we have

$\displaystyle \Delta r dV= \mathcal{L}_{\partial_t}(\lambda dt \wedge dA)=(\partial_t\lambda )dt\wedge dA.$

i.e.

$\displaystyle \partial_t \lambda = \lambda\Delta r . \ \ \ \ \ (4)$

Similarly, ${\partial_t \overline \lambda= \overline \lambda \; \overline \Delta \overline r}$.

 Lemma 8 With the same assumption as in Lemma 7, the function ${\frac{\lambda(x, t)}{\overline \lambda(x, t)}}$ is non-increasing and ${\displaystyle \lim_{t\rightarrow 0^+}\frac{\lambda(x, t)}{\overline \lambda(x, t)}=1}$. In particular, ${\lambda(x, t)\le \overline \lambda(x, t)}$.

Proof: Since ${\lambda, \overline \lambda>0}$, by Lemma 7, we have

$\displaystyle \begin{array}{rl} \frac{\partial }{\partial t}\left(\lambda/\overline \lambda\right) =\frac{\overline \lambda \partial _t \lambda-\lambda \partial _t \overline \lambda}{\overline \lambda^2} =\frac{\overline \lambda \lambda(\Delta r - \overline \Delta \overline r)}{\overline \lambda^2}\le 0. \end{array}$

It is easy to see that ${\displaystyle \lim_{t\rightarrow 0^+}\frac{\overline \lambda}{ \lambda}=1}$. $\Box$

We can now prove Theorem 1.
Proof of Theorem 1: By Lemma 8,

$\displaystyle \begin{array}{rl} \mathrm{Vol}(B_r(p))=\int_{\mathbb{B}(0, r)}dV =&\int_0^r \int_{\mathbb{S}^{n-1}}\lambda(x, t)dA(x) dt\\ \le&\int_0^r \int_{\mathbb{S}^{n-1}}\overline \lambda(x, t)dA(x) dt\\ =&\int_{\mathbb{B}(0, r)}d\overline V\\ =&\mathrm{Vol}(B_r(\overline p)). \end{array}$

If the equality holds for some ${r_0}$, then ${\lambda(x, t)=\overline \lambda(x, t)}$ for ${t\le r_0}$, which implies ${\Delta r =\overline \Delta \overline r=(n-1)\sigma(x, t)}$. From the proof of Lemma 7, the equality case of the Riccati inequality (Corollary 4) holds for ${r}$, which implies that ${\nabla ^2 r =\sigma g|_{S_t(p)}}$ for ${t\le r_0}$. In other words, ${\nabla _X \partial _t =\sigma X}$ for any ${X}$ tangential to ${S_t(p)}$.

Now, recall that the diffeomorphism ${\Phi}$ in (1) is defined by ${\exp_p (t\sum x^ie_i)\stackrel{\Phi}\mapsto \exp_{\overline p}(t\sum x^i \overline e_i)}$. To prove that ${\Phi}$ is an isometry, it suffices to show it is an isometry along ${\exp_p(tx)}$ for all ${x\in \mathbb{S}^{n-1}}$. Without loss of generality, assume ${x=e_1}$. (From now on, we ignore the dependence on ${x}$ for any function.) By the canonical identification ${T_V T_pM\cong T_pM}$ for ${V\in T_pM}$, we define the normal Jacobi field ${J_i(t)=(D\exp_p)|_{te_1}(te_i)}$ (${i\ge 2}$), and similarly define ${\overline J_i(t)}$ along ${\exp_{\overline p}(t\overline e_1)}$. It is easy to see that ${D\Phi(J_i(t))=\overline J_i(t)}$. Consider

$\displaystyle \begin{array}{rl} \frac{d}{dt}g(J_i, J_j) =g(\nabla _{\partial _t}J_i, J_j)+g(J_i, \nabla _{\partial _t}J_j) =&g(\nabla _{J_i}\partial _t, J_j)+g(J_i, \nabla _{J_j}\partial _t)\\ =&\sigma g(J_i, J_j)+\sigma g(J_i, J_j)\\ =&2\sigma g(J_i, J_j). \end{array}$

Similarly, ${\frac{d}{dt}\overline g(\overline J_i, \overline J_j)= 2\sigma \overline g(\overline J_i, \overline J_j)}$. Therefore for ${t>0}$,

$\displaystyle \begin{array}{rl} \phi_{ij}'=2\sigma \phi_{ij} \end{array}$

where ${g_{ij}= g(J_i, J_j)}$, ${\overline g_{ij}=\overline g(\overline J_i, \overline J_j)}$ and ${\phi_{ij}=g_{ij}-\overline g_{ij}}$. By multiplying the integrating factor ${\exp\left(\int_\varepsilon^t 2\sigma(s)ds\right)}$ and integrate,

$\displaystyle \begin{array}{rl} \phi_{ij}(t)=\phi_{ij}(\varepsilon) \exp\left(\int_\varepsilon^t 2\sigma(s)ds\right). \end{array}$

It is known that ${\sigma(\varepsilon)=\frac{1}{\varepsilon}+O(\varepsilon)}$, ${g_{ij}(\varepsilon)=\varepsilon^2 \delta_{ij}+O(\varepsilon^4)}$, and ${\overline g_{ij}(\varepsilon)=\varepsilon^2 \delta_{ij}+O(\varepsilon^4)}$ as ${\varepsilon\rightarrow 0}$, so ${\phi_{ij}(t)=O(\varepsilon^2)}$ as ${\varepsilon\rightarrow 0}$. Hence ${g_{ij}(t)-\overline g_{ij}(t)=0}$ by taking ${\varepsilon\rightarrow 0}$. This implies ${\Phi}$ is an isometry. $\Box$

If ${\overline g}$ is a warped product metric of the form

$\displaystyle dt^2 + h(t)^2 g_S \ \ \ \ \ (5)$

around the point ${\overline p}$, where ${g_S}$ is the round metric on the standard unit sphere, then we can get a better result (Theorem 10).

 Lemma 9 Let ${f,g }$ be two continuous functions which are positive for ${t>0}$ and ${\frac{f(t)}{g(t)}}$ is non-increasing for ${t>0}$, then ${\phi(t)=\frac{\int_0^t f(s)ds}{\int_0^t g(s)ds}}$ is non-increasing for ${t>0}$.

Proof: For ${t>0}$,

$\displaystyle \begin{array}{rl} \frac{d}{dt}\left(\frac{\int_0^t f(s)ds}{\int_0^t g(s)ds}\right) =\frac{f(t)\int_0^t g(s)ds - g(t)\int_0^t f(s)ds}{\left(\int_0^t g(s)ds\right)^2} =&\frac{f(t)\int_0^t g(s)ds - g(t)\int_0^t \frac{f(s)}{g(s)}g(s)ds}{\left(\int_0^t g(s)ds\right)^2}\\ \le&\frac{f(t)\int_0^t g(s)ds - g(t) \frac{f(t)}{g(t)}\int_0^t g(s)ds}{\left(\int_0^t g(s)ds\right)^2}\\ =&0. \end{array}$

$\Box$

 Theorem 10 Let ${(M^n, g)}$ and ${(\overline M^n, \overline g)}$ be two complete Riemannian manifolds which satisfy the curvature condition 2 and the “umbilical condition: (3). Suppose ${\overline g}$ is a warped product metric of the form (5) around ${\overline p}$. Then Then for ${r, the following function is non-increasing: $\displaystyle r\mapsto\frac{\mathrm{Vol}(B_r(p))}{\mathrm{Vol}(B_r(\overline p))}.$

Proof: Since ${\overline g=dt^2 + h(t)^2 g_S}$, we have ${d\overline V= \overline \lambda(t) dt\wedge dA= h(t)^{n-1}dt\wedge dA}$. By Lemma 8, ${\frac{\lambda(x,t )}{\overline \lambda(t)}}$ is non-increasing, so

$\displaystyle \begin{array}{rl} \frac{\mathrm{Area}(S_t(p))}{\mathrm{Area}(S_t(\overline p))}=\frac{\int_{\mathbb{S}^{n-1}}\lambda(x, t)dA}{\int_{\mathbb{S}^{n-1}}\overline \lambda(t)dA} =\frac{1}{\mathrm{Area}(\mathbb{S}^{n-1})}\int_{\mathbb{S}^{n-1}}\frac{\lambda(x, t)}{\overline \lambda(t)}dA \end{array}$

is also non-increasing. By Lemma 9, ${\frac{\mathrm{Vol}(B_t(p))}{\mathrm{Vol}(B_t(\overline p))}}$ is also non-increasing. $\Box$

 Example 1 Let ${\overline g=dt^2 + h(t)^2 g_S}$ around ${\overline p}$, then it can be computed that ${\overline {\mathrm{Ric}}(\partial _t, \partial _t)=-(n-1)\frac{h''}{h}}$. So if ${\mathrm{Ric}(\partial _t, \partial _t)\ge -(n-1)\frac{h''}{h}}$, then when ${r<\min\{\mathrm{inj}(p), \mathrm{inj}(\overline p)\}}$, $\displaystyle \mathrm{Vol}(B_r(p))\le \mathrm{Vol}(B_r(\overline p))=\mathrm{Area}(\mathbb{S}^{n-1})\int_0^r h(t)^{n-1}dt.$

2. Comparing volumes of exterior regions

A geodesic ball can be regarded as the “exterior region” of a point, and so Theorem 1 can be regarded as comparing the volumes of two exterior regions of two different points. We now consider the volume of the “exterior region of radius ${r}$” of a hypersurface. The method is the same as in Section 1 with some suitable modification.

Let ${\Sigma}$ be a two-sided hypersurface embedded in ${(M,g)}$ with a unit normal field ${\nu}$ (chosen to be outward whenever this makes sense). The local coordinates of ${M}$ adapted to ${\Sigma}$ is then defined by

$\displaystyle \Sigma\times \mathbb{R}\rightarrow M \ \ \ \ \ (6)$

$\displaystyle (x, t)\mapsto \exp _{x}(t\nu(x)).$

By the inverse function theorem, this defines a local coordinates of ${M}$ around any point in ${\Sigma}$. Let ${r: M\rightarrow \mathbb{R}}$ to be the distance from ${\Sigma}$, i.e. ${\displaystyle r(x)=\inf_{p\in \Sigma}\mathrm{dist}(x, p)}$. Let ${c: \Sigma\rightarrow \mathbb{R}\cup \{\infty\}}$ be the cut function in the outward direction of ${\Sigma}$, i.e. ${c(p)=\sup\{t>0: \mathrm{dist}(\Sigma, \exp_p(t\nu(p)))=t\}\in (0, \infty]}$. Then ${c}$ is continuous and ${r}$ is smooth on ${C(\Sigma)=\bigcup_{p\in \Sigma}\{\exp_p (t\nu(p)): 0; we say a point is within the cut locus of ${\Sigma}$ if it lies in this region.

 Lemma 11 (Gauss lemma) On ${C(\Sigma)}$, ${\nabla r=\frac{\partial }{\partial t}.}$

Proof: Let ${\{x^i\}_{i=1}^{n-1}}$ be a local coordinates of ${\Sigma}$, and extend it by using the coordinates in (6) (such that ${x^n=t}$). As ${\nabla r= \sum_{i=1}^ng^{it}\partial _i}$, it suffices to prove that ${g(\partial_t, \partial _i)=0}$. To see this, consider

$\displaystyle \begin{array}{rl} \partial_t( g( \partial_t, \partial _i)) =g( \nabla _{\partial _t}\partial_t, \partial _i)+ g( \partial_t, \nabla _{\partial _t}\partial _i) =g( \partial_t, \nabla _{\partial _t}\partial _i) =&g( \partial_t, \nabla _{\partial _i} \partial_t)\\ =&\frac{1}{2}\partial _i(g( \partial_t, \partial_t)) =0. \end{array}$

Therefore ${g( \partial _t, \partial _i)}$ is constant and at ${t=0}$, ${g( \partial _t, \partial _i)=0}$. The claim is proved. $\Box$

We define ${\Sigma_t=\{\exp_p(t\nu(p)): p\in \Sigma\}}$. If ${c(\Sigma)=\inf_{p\in \Sigma}c(p)>0}$, then ${\Sigma_t}$ is also a smooth embedded hypersurface for ${0\le t. By the above lemma, for ${t> 0}$, ${\nabla ^2 r}$ is the second fundamental form of ${\Sigma_t}$ and ${\Delta r}$ is the mean curvature of ${\Sigma_t}$ at ${(x,t)}$. Since the second fundamental form varies continuously along ${\Sigma_t}$, ${\displaystyle \lim_{t\rightarrow 0^+}\nabla ^2 r}$ is the second fundamental form of ${\Sigma}$ when restricted to ${T\Sigma}$ and ${\displaystyle \lim_{t\rightarrow 0^+}\Delta r =H}$, the mean curvature of ${\Sigma}$.

 Lemma 12 (Riccati equation) On ${C(\Sigma)}$, $\displaystyle (\nabla _{\partial _t}\nabla ^2 r )(X, Y)+\left(\nabla ^2 r\right)^2 (X, Y)=-g(R(X, \partial _t) \partial _t, Y).$ Here ${\left(\nabla ^2 r\right)^2(e_i, e_j)= \sum_{k, l=1}^ng^{kl}\nabla ^2 _{ik}r\nabla ^2 _{lj}r }$. (More intrinsically, ${\left(\nabla ^2 r\right)^2 (X, Y)=g(\nabla _X\nabla r, \nabla _Y\nabla r)}$.)

Proof: We have

$\displaystyle \begin{array}{rl} (\nabla _{\partial _t}\nabla ^2 r)(X, Y) =&\partial _t\left(\nabla ^2 r(X, Y)\right)-\nabla ^2 r(\nabla _{\partial _t}X, Y)-\nabla ^2 r(X, \nabla _{\partial _t }Y)\\ =&\partial _t\left(g(\nabla _X \nabla r, Y)\right)-g(\nabla _{\nabla _{\partial _t}X}\nabla r, Y)-g(\nabla _X\nabla r, \nabla _{\partial _t}Y)\\ =&g(\nabla _{\partial _t}\nabla _X \partial _t, Y)-g(\nabla _{\nabla _{\partial _t}X}\nabla r, Y)\\ =&g(R(\partial _t, X) \partial _t, Y)+g(\nabla _{[\partial _t, X]}\partial _t, Y)-g(\nabla _{\nabla _{\partial _t}X}\nabla r, Y)\\ =&g(R(\partial _t, X) \partial _t, Y)-g(\nabla _{\nabla _X \nabla r}\nabla r, Y)\\ =&-g(R(X, \partial _t) \partial _t, Y)-g(\nabla _Y\nabla r, \nabla _X \nabla r). \end{array}$

We have used the fact that ${g(\nabla _U \nabla r, V)=g(\nabla _V \nabla r, U)}$ for any ${U,V}$ in the last line. $\Box$

 Corollary 13 (Riccati inequality) On ${C(\Sigma)}$, $\displaystyle \frac{\partial }{\partial t}\Delta r +\frac{(\Delta r)^2}{n-1}\le-\mathrm{Ric}(\partial _t, \partial _t).$ The equality holds at the point ${(x, t)}$ if and only if ${\Sigma_t}$ is umbilical at this point.

Now, recall that ${(\overline M^n, \overline g)}$ is another complete Riemannian manifold with connection ${\overline \nabla }$. Suppose ${\Sigma}$ can be isometrically embedded in ${\overline M}$ by ${\iota: \Sigma\rightarrow \overline M}$ with image ${\overline \Sigma}$, which we assume to be two-sided with a unit (outward) normal field ${\overline \nu}$. Let ${\Omega_r}$ be the region bounded between ${\Sigma}$ and ${\Sigma_r}$, i.e. ${\Omega_r(\Sigma)=\{\exp_p(t\nu(p)): p\in \Sigma, 0\le t< r\}}$ and ${ \Omega_r(\overline \Sigma)=\{\exp_p(t\overline \nu(p)): p\in \overline \Sigma, 0\le t< r\}}$. We can also define a coordinates adapted to ${\overline \Sigma}$ as in (6).

For ${r< l=\min\{c(\Sigma), c(\overline \Sigma)\}}$, we have a diffeomorphism from ${\Omega_r(\Sigma)}$ to ${\Omega_r(\overline \Sigma)}$ by

$\displaystyle \Phi: \exp_p (t\partial _t)\mapsto \exp_{\iota(p)}(t\partial _t). \ \ \ \ \ (7)$

Again, under this identification, we can make sense of the following
Ricci curvature condition:

$\displaystyle \mathrm{Ric}(\partial _t, \partial _t)\ge \overline {\mathrm{Ric}}(\partial _t, \partial _t). \ \ \ \ \ (8)$

As before, we define the

Umbilical condition: For all ${r, the sphere ${\overline \Sigma_r}$ is umbilical at each point. i.e. the second fundamental form ${\overline A}$ of ${\overline \Sigma_r}$ is pointwisely proportional to its induced metric at each point ${x}$ (the proportional constant can depend on ${x}$ and ${r}$):

$\displaystyle \overline A=f(x, r)\overline g|_{\overline \Sigma_r}. \ \ \ \ \ (9)$

We identify ${x}$ with ${\iota(x)}$ and let ${H(x, r)}$ and ${\overline H(x, r)}$ be the mean curvature of ${\Sigma_r}$ and ${\overline \Sigma_r}$ respectively. We also write ${H(x)=H(x, 0)}$ and ${\overline H(x)=\overline H(x, 0)}$.

 Lemma 14 (Laplacian comparison) Suppose the umbilical condition (9) and ${\mathrm{Ric}(\partial _t, \partial _t)\ge \overline {\mathrm{Ric}}(\partial _t, \partial _t)}$ hold. Assume that ${H(x)\le \overline H(x)}$, then ${\Delta r \le \overline \Delta \overline r}$.

Proof: The proof is exactly the same as that of Lemma 7 except that the initial condition is replaced by ${\displaystyle \lim_{r\rightarrow 0^+}(\sigma-\overline \sigma)=\frac{1}{n-1}(H(x)-\overline H(x))\le 0}$, and apply Lemma 6. $\Box$

Let the volume element of ${M}$ and ${\overline M}$ be ${dV}$ and ${d\overline V}$ respectively. Then within ${\Omega_l(\Sigma)}$, there is a function ${\lambda(x, t)}$ such that

$\displaystyle dV= \lambda(x, t)dt\wedge dA$

where ${dA}$ is the area element of ${\Sigma}$. Similarly ${d\overline V= \overline \lambda(x, t)dt\wedge dA}$. As in (4), we have

$\displaystyle \partial_t \lambda = \lambda\Delta r \quad \textrm{and}\quad \partial_t \overline \lambda= \overline \lambda \; \overline \Delta \overline r.$

By the same proof as in Lemma 8, we have

 Lemma 15 With the same assumption as in Lemma 14, the function ${\frac{\lambda(x, t)}{\overline \lambda(x, t)}}$ is non-increasing and ${\lambda(x,0)=\overline \lambda(x,0)}$. In particular, ${\lambda(x, t)\le \overline \lambda(x, t)}$.

We have the following analogue of Theorem 1:

 Theorem 16 (Volume comparison) Let ${(M^n, g)}$ and ${(\overline M^n, \overline g)}$ be two complete Riemannian manifolds. ${\Sigma}$ be a closed hypersurface in ${M}$ which can be isometrically embedded into ${\overline M}$ such that (8) and the “umbilical condition” (9) are satisfied. Assume further that ${H(x)\le \overline H(x)}$. Then for all ${r<\min\{c(\Sigma), c(\overline \Sigma)\}}$, $\displaystyle \mathrm{Vol}(\Omega_r(\Sigma))\le \mathrm{Vol}(\Omega_r(\overline \Sigma)).$ The equality holds if and only if ${\Omega_r(\Sigma)}$ is isometric to ${\Omega_r(\overline \Sigma)}$.

Proof: The proof of the inequality is the same as Theorem 1. If the equality holds for some ${r_0}$, then ${\lambda(x, t)=\overline \lambda(x, t)}$ for ${t\le r_0}$, which implies ${\Delta r =\overline \Delta \overline r=(n-1)\sigma(x, t)}$. From the proof of Lemma 7, the equality case of the Riccati inequality (Corollary 13) holds for ${r}$, which implies that ${\nabla ^2 r =\sigma g|_{S_t(p)}}$ for ${t\le r_0}$. Let ${\{x^i\}_{i=1}^{n-1}}$ be a local coordinates around a point in ${\Sigma}$ (and hence ${\overline \Sigma}$ by identification), we use the coordinates in (6) to extend it to a local coordinates in ${M}$ (and ${\overline M}$). Then for ${g_{ij}=g(\partial _i, \partial _j)}$,

$\displaystyle \begin{array}{rl} \frac{\partial }{\partial t}g_{ij} =\partial _t( g(\partial _i, \partial _j)) =g(\nabla _{\partial _i} \partial _t, \partial _j)+ g(\partial _i,\nabla _ {\partial _j }\partial _t) =2 \nabla ^2 r(\partial _i, \partial _j) =&2 \sigma (x,t)g_{ij}. \end{array}$

Similarly,

$\displaystyle \frac{\partial }{\partial t}\overline g_{ij}=2 \sigma (x,t)\overline g_{ij}.$

Since ${g_{ij}(x, 0)=\overline g_{ij}(x, 0)}$, this implies ${g_{ij}(x, t)=\overline g_{ij}(x,t)}$, and hence ${\Phi}$ in (7) is an isometry. $\Box$

The following is the analogue of Theorem (10):

 Theorem 17 With the same assumptions as in Theorem 16. Suppose ${\overline g}$ is a warped product metric of the form $\displaystyle dt^2+ h(t)^2 dA$ outside ${\overline \Sigma}$, where ${dA}$ is the area element of ${\Sigma}$. Then for ${r<\min\{c(\Sigma), c(\overline \Sigma)\}}$, the following function is non-increasing: $\displaystyle r\mapsto\frac{\mathrm{Vol}(\Omega_r(\Sigma))}{\mathrm{Vol}(\Omega_r(\overline \Sigma))}.$

It is clear that for Theorem 16 and Theorem 17 to hold, it is not really necessary that ${M}$ and ${\overline M}$ to be complete, instead we can assume they to be geodesically complete for geodesics pointing outside ${\Sigma}$ and ${\overline \Sigma}$ respectively.