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<l=\min\{\mathrm{inj}(p), \mathrm{inj}(\overline p)\}}, 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<l}, 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<T}. 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<l}, 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<t<c(p)\}}; 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<c(\Sigma)}. 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<l}, 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.

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