Volume Comparison

Laplacian comparison (9B.1 Laplacian Comparison) says that, under a lower Ricci bound, geodesic spheres bend at least as much as in the model space. Integrating that statement over the spheres gives the theorem used more than any other in this part of the Path. Bishop–Gromov volume comparison: if $\operatorname{Ric} \geq (n - 1)k$, the volume of a ball, divided by the volume of the ball of the same radius in the model space, can only decrease as the radius grows. Lower Ricci bounds cap the growth of volume, and they also guarantee that a ball keeps a definite fraction of the volume of any larger concentric ball. The second fact is what makes packing arguments, Gromov's compactness theorem and Perelman's noncollapsing work.

By the end of this chapter you will be able to:

• write the volume form in geodesic polar coordinates and relate its growth to the Laplacian of distance;
• prove the Bishop–Gromov inequality and its absolute form, Bishop's inequality;
• use relative volume comparison for doubling and packing estimates;
• define the asymptotic volume ratio and compute it for cones, cylinders and paraboloids;
• recognise the same monotone-ratio structure in Perelman's reduced volume.

2.1Why hyperbolic space fits trees

2.2Polar coordinates and the volume element

Fix $p$ and use geodesic polar coordinates $(r, \theta)$, $\theta \in S^{n-1} \subset T_pM$, on the domain $\{r < c(\theta)\}$, where $c(\theta)$ is the distance to the cut point in direction $\theta$. The volume form is

with $d\theta$ the round measure on $S^{n-1}$ and $\mathcal A(r, \theta) = r^{n-1} + O(r^{n+1})$. Since $\mathcal A$ is the Jacobian of the exponential map along the radial geodesic, its logarithmic derivative is the mean curvature of the geodesic sphere (Exercise 2.2):

Set $\mathcal A = 0$ for $r \geq c(\theta)$; the cut locus has measure zero, so $\operatorname{Vol}B(p, R) = \int_{S^{n-1}}\int_0^R\mathcal A\,dr\,d\theta$ (the polar formula of 3A.5 Product Measures and Change of Variables on a manifold). In the model space of curvature $k$, $\mathcal A_k(r) = \operatorname{sn}_k(r)^{n-1}$ and $V_k(R) = |S^{n-1}|\int_0^R\operatorname{sn}_k^{n-1}$ (Figure 2.1).

2.3Bishop–Gromov

Equality characterises the model: if $\operatorname{Vol}B(p, R) = V_k(R)$ for some $R$, then $B(p, R)$ is isometric to a ball in the model space. For $k > 0$ this gives Cheng's maximal diameter theorem (1975): if $\operatorname{Ric} \geq (n - 1)g$ and $\operatorname{diam} = \pi$, then $M$ is isometric to the unit sphere (Exercise 2.7).

2.4Doubling, packing and growth

Relative comparison is the useful form. With $\operatorname{Ric} \geq 0$ it says $\operatorname{Vol}B(p, 2r) \leq 2^n\operatorname{Vol}B(p, r)$: the volume measure is doubling, with a constant depending only on the dimension. With $\operatorname{Ric} \geq -(n - 1)$, the same holds for $r \leq 1$ with a slightly larger constant. Two consequences are used again and again.

• Packing. If $\operatorname{Ric} \geq 0$ and the balls $B(x_1, r), \dots, B(x_N, r)$ are disjoint and lie in $B(p, R)$, then $N \leq (\frac{2R}{r})^n$ (Exercise 2.4). Combined with the Vitali covering lemma (3A.6 Modes of Convergence and Differentiation), this bounds how many small balls are needed to cover a large one. Gromov's precompactness theorem (9B.4 Convergence of Manifolds) rests on exactly this count.
• Growth. If $\operatorname{Ric} \geq 0$, then $\operatorname{Vol}B(p, R) \leq \omega_nR^n$: no more than Euclidean. A theorem of Calabi and Yau gives the lower bound for complete noncompact manifolds: $\operatorname{Vol}B(p, R) \geq cR$ for $R \geq 1$ (Exercise 2.5).

The asymptotic volume ratio. If $\operatorname{Ric} \geq 0$ and $M$ is complete and noncompact, the ratio $\frac{\operatorname{Vol}B(p, R)}{\omega_nR^n}$ is nonincreasing, so it has a limit

which does not depend on $p$ (Exercise 2.6). It is $1$ only for $\mathbb{R}^n$. A cone of opening angle less than Euclidean has a positive value below $1$; a cylinder $S^{n-1}\times\mathbb{R}$ and the paraboloid have $\operatorname{AVR} = 0$ (Exercise 2.8). Perelman proved that every $\kappa$-solution, the ancient solutions that model Ricci flow singularities, has asymptotic volume ratio zero (12B.2 The Structure of κ-Solutions). That is one of the facts that make the canonical neighbourhood theorem work.

2.5A pattern that returns: monotone volume ratios

Bishop–Gromov compares a manifold with a model, through a ratio that is monotone in the scale and equal to $1$ exactly for the model. Perelman's reduced volume (12A.5 Reduced Distance and Reduced Volume) has the same structure, in space-time:

	Bishop–Gromov	Perelman's reduced volume
quantity	$\dfrac{\operatorname{Vol}B(p, r)}{\omega_nr^n}$	$\tilde V(\tau) = \displaystyle\int(4\pi\tau)^{-n/2}e^{-\ell}\,dV$
hypothesis	$\operatorname{Ric} \geq 0$	a Ricci flow, backwards in time $\tau$
monotonicity	nonincreasing in $r$	nonincreasing in $\tau$
equality	$\mathbb{R}^n$ (flat)	the Gaussian soliton on $\mathbb{R}^n$
proved from	Laplacian comparison for $r$	Laplacian comparison for the reduced distance $\ell$

The second column is explained in 12A.5 Reduced Distance and Reduced Volume; it is listed here so that, when you meet it, it is a familiar theorem in new clothes.

2.6History

Richard Bishop proved the absolute volume inequality in 1963 (it appears in Bishop and Crittenden's Geometry of Manifolds, 1964); Mikhail Gromov introduced the relative version and its use for compactness around 1980. Shiu-Yuen Cheng proved the maximal diameter theorem in 1975. The linear growth bound is due to Calabi (1975) and Yau (1976).

2.7Exercises

Solution

$\Phi_s$ maps $\exp_p(r\theta)$ to $\exp_p((r + s)\theta)$, so in polar coordinates $\Phi_s^*(\mathcal A\,dr\,d\theta) = \mathcal A(r + s, \theta)\,dr\,d\theta$. Differentiating at $s = 0$: $\partial_r\mathcal A\,dr\,d\theta = \operatorname{div}(\partial_r)\mathcal A\,dr\,d\theta$, and $\operatorname{div}\nabla r = \Delta r$. In $\mathbb{R}^n$: $\partial_r\log r^{n-1} = \frac{n - 1}{r} = \Delta r$.

Solution

For $t \leq R$, $\frac{f(t)}{g(t)} \geq \frac{f(R)}{g(R)}$, so $f(R)g(t) \leq g(R)f(t)$. Integrating in $t$ over $[0, R]$ gives $f(R)G(R) \leq g(R)F(R)$, so $(\frac FG)' \leq 0$.

Solution

$x_i \in B(p, R)$, so for $y \in B(p, R)$, $d(x_i, y) < 2R$. Relative comparison with $k = 0$: $\operatorname{Vol}B(x_i, r) \geq (\frac{r}{2R})^n\operatorname{Vol}B(x_i, 2R) \geq (\frac{r}{2R})^n\operatorname{Vol}B(p, R)$. Summing over the disjoint balls inside $B(p, R)$: $N(\frac{r}{2R})^n\operatorname{Vol}B(p, R) \leq \operatorname{Vol}B(p, R)$.

Solution

By relative comparison at $x$, $\frac{\operatorname{Vol}B(x, R - 1)}{\operatorname{Vol}B(x, R + 1)} \geq (\frac{R - 1}{R + 1})^n$, so the annulus $A = B(x, R + 1)\setminus B(x, R - 1)$ has $\operatorname{Vol}A \leq \big(1 - (\frac{R - 1}{R + 1})^n\big)\operatorname{Vol}B(x, R + 1) \leq \frac{2n}{R + 1}\operatorname{Vol}B(x, R + 1)$. Since $d(p, x) = R$, $B(p, 1) \subset A$, and $B(x, R + 1) \subset B(p, 2R + 1)$. So $\operatorname{Vol}B(p, 1) \leq \frac{2n}{R + 1}\operatorname{Vol}B(p, 2R + 1)$, that is $\operatorname{Vol}B(p, 2R + 1) \geq \frac{R + 1}{2n}\operatorname{Vol}B(p, 1)$: linear growth with $c = \frac{\operatorname{Vol}B(p, 1)}{4n}$, say.

Solution

The inclusions follow from the triangle inequality. Dividing by $\omega_nR^n$ and using $\frac{(R \pm d)^n}{R^n} \to 1$: $\frac{\operatorname{Vol}B(q, R)}{\omega_nR^n}$ is squeezed between quantities converging to $\operatorname{AVR}$ computed at $p$.

Solution

The balls are disjoint by the triangle inequality. Relative comparison at $p$: $\operatorname{Vol}B(p, \frac{\pi}{2}) \geq \frac{V_1(\pi/2)}{V_1(\pi)}\operatorname{Vol}B(p, \pi) = \frac12\operatorname{Vol}M$, since $B(p, \pi)$ is all of $M$ (diameter $\leq \pi$ by Myers). Similarly at $q$. Disjointness forces equality, $\operatorname{Vol}B(p, \frac{\pi}{2}) = \frac12\operatorname{Vol}M$, so the ratio $\frac{\operatorname{Vol}B(p, R)}{V_1(R)}$ is constant on $[\frac{\pi}{2}, \pi]$. Then $\frac{\mathcal A}{\sin^{n-1}r}$ is constant there in every direction, and since it is nonincreasing and tends to $1$ at $0$, in the equality case it is $1$ throughout, which forces the model metric; in particular $\operatorname{Vol}M = |S^n|$.

Solution

(a) $\operatorname{Vol}B(o, R) = |S^{n-1}|\int_0^R(ar)^{n-1}dr = a^{n-1}\omega_nR^n$. (b) $\operatorname{Vol}B(p, R) \leq |S^{n-1}(\rho)|\cdot2R$, linear in $R$, so the ratio with $R^n$ tends to $0$ for $n \geq 2$. (c) $\frac{A}{\pi s^2} \sim \frac{(4\pi/3)\rho^3}{\pi\rho^4} \to 0$.