A volume comparison result

Let {(M^{n+1},g)} be a complete Riemannian manifold with non-negative Ricci curvature. Suppose we have a geodesic ball {B} in {(M,g)} whose boundary has the same surface area as a Euclidean ball (not necessarily of the same radius), we want to compare the volumes of these two balls.

Note that the usual comparison theorem says that the volume of a geodesic ball is bounded from above (often by the volume of a ball in a space form with the same radius) given that {\mathrm{Ric}} is bounded from below. In this note, we investigate a dual problem: instead of comparing the volumes of balls with the same radius, we compare the balls with the same surface area. Naturally, the main ingredient is a comparison theorem of Bishop.

We use the following notations. Denote the geodesic ball and geodesic sphere in {(M,g)} centered at {p} with radius {r} by {B(p,r)} and {S(p,r)} respectively. Let {p\in M} be fixed and define {A(r)=\mathrm{Area}(S(p,r))} and {V(r)= \mathrm{Vol}(B(p,r))}. Similarly, we define {\overline A(r)=\mathrm{Area}(\mathbb{S}(r))} and {\overline V(r)=\mathrm{Vol}(\mathbb{B}(r))}, where {\mathbb{S}(r)=\{x\in \mathbb{R}^{n+1}: |x|=r\}} and {\mathbb{B}(r)=\{x\in \mathbb{R}^{n+1}: |x|\le r\}}. Let {\omega_n=\overline A(1)}, we define

\displaystyle \overline r(t)=\left(\frac{t}{\omega_n}\right)^{\frac{1}{n}},

so that the Euclidean ball of radius {\overline r(A)} has surface area {A}.

Theorem 1 Let {(M^{n+1},g)} be a complete Riemannian manifold with {\mathrm{Ric}\ge 0}. Suppose {B} is a geodesic ball such that its boundary has area {A_0}, then {\mathrm{Vol}(B)\ge \overline V(\overline r(A_0))}. The equality holds if and only if {B} is isometric to a Euclidean ball.  

From now on, we fix a geodesic ball {B=B(p, r_0)} in {(M,g)} and let {A_0} be the area of {\partial B}. Let {r_1=\min \{r\le r_0: A(r)=A_0\}}. Note that {A(r)\le A_0} for {r\le r_1}. Obviously, to prove Theorem 1, it suffices to show {V(r_1)\ge \overline V(\overline r(A_0))}.

For {0\le t\le A_0}, we define

\displaystyle r(t)=\max\{r\in [0,r_1]: A(r)=t\}.

It is easy to see that {r(t)} is increasing (and is upper semi-continuous). Indeed, if {t_1<t_2\le A_0}, then by the continuity of {A(r)}, there exists {r(t_1)<r'\le r_1} such that {A(r')=t_2}. Therefore {r(t_2)>r(t_1)}.

We also have the simple properties that {A(r(t))=t \textrm{ and }r(A(\rho))\ge \rho.}

Lemma 2 There exists a non-decreasing function {\tau(t)\ge1} such that for {0\le t\le A_0},

\displaystyle r(t)= \tau(t)\overline r (t).

 

Proof: By dividing {\overline A} and {A} by the same constant, we can assume {\overline A(r)= r^n}, and regard {\overline A} and {A} as functions of {r^n} (i.e. {\overline A=\mathrm{id}}). In the same way we normalize {r} and {\overline r} so that {\overline r^n(t)=t}. By a comparison theorem of Bishop (cf. [BC] p.256 Cor. 3 or [SY] Theorem 1.2), there exists a non-increasing function {s=s(r)\le 1} such that

\displaystyle  \begin{array}{rl}  A(r^n)=s(r)r^n. \end{array}

Let {t\le A_0} be fixed and substitute {r^n=r^n(t)} in the above equation,

\displaystyle  t=A(r^n(t))= s(r(t))r^n(t).

So

\displaystyle r^n (t)=\frac{1}{s(r(t))}t=\frac{1}{s(r(t))}\overline r^n(t).

Clearly {\frac{1}{s(r(t))}\ge1 } is non-decreasing (as {r(t)} is increasing). The result follows. \Box

As a consequence, we have:

Lemma 3 The function {r-\overline r} is non-decreasing for {0\le t\le A_0}. In particular, { r(A_0)-r(t)\ge\overline r(A_0)-\overline r(t)}.  

Proof of Theorem 1: Recall that {B=B(p,r_0)}, {A_0=\mathrm{Area}(\partial B)}, {r_1=\min \{r\le r_0: A(r)=A_0\}}, and that {r(t)=\max\{r\in [0,r_1]: A(r)=t\}} for {t\le A_0}. By the layercake integral formula and Lemma 3,

\displaystyle  \begin{array}{rl}  \mathrm{Vol}(B)\ge V(r_1) =\int_0^{r_1}A(r)dr =&\int_0^{A_0}|\{r\in [0,r_1]: A(r)\ge t\}|dt\\ \ge&\int_0^{A_0}(r(A_0)-r(t))dt\\\ \ge&\int_0^{A_0}(\overline r(A_0)-\overline r(t))dt\\ =&\int_0^{\overline r(A_0)}\overline A(r)dr\\ =&\overline V(\overline r(A_0)). \end{array}

By the equality case of the comparison theorem of Bishop, {\mathrm{Vol}(B)=\overline V(\overline r(A))} if and only if {B} is isometric to a Euclidean ball. \Box

Question: Can Theorem 1 be generalized to a Riemannian manifold with smooth boundary?

Advertisements
This entry was posted in Geometry, Inequalities. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s