Book 12A

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Course 12Book 12A: Entropy and NoncollapsingChapter 5

Reduced Distance and Reduced Volume

A space-time Bishop–Gromov theorem.

25 min read · Updated Oct 3, 2026

Read Perelman I, §6 and §7, then the detailed treatment of L\mathcal L-geodesics and the reduced volume in Kleiner and Lott's notes or in Morgan and Tian's book. Huisken's monotonicity formula (6A.8 Curve Shortening and the First Geometric Flows) and Bishop–Gromov (9B.2 Volume Comparison) are the two models to have in mind.

In this chapter · 9 sections
  1. 5.1Least action and the heat kernel
  2. 5.2L\mathcal LL-length
  3. 5.3The differential inequalities
  4. 5.4The reduced volume
  5. 5.5Bishop–Gromov in space-time
  6. 5.6Perelman's heuristic: Bishop–Gromov in NNN dimensions
  7. 5.7Huisken's formula and the direction of monotonicity
  8. 5.8History
  9. 5.9Exercises

The W\mathcal W-entropy of 12A.3 The 𝓦-Entropy is an integral over a whole manifold, so it controls global things. For the flow with surgery, Perelman needed a monotone quantity attached to a single point of space-time, which sees only the part of the flow that can influence that point. He built it from a new notion of length for curves in space-time, the L\mathcal L-length. Its minimisers give a reduced distance ℓ\ell, and the Gaussian built from ℓ\ell integrates to the reduced volume V~(τ)\tilde V(\tau). The reduced volume is monotone along every Ricci flow, with no curvature assumption, and equals 11 on flat space. It is the Ricci flow's version of the Bishop–Gromov volume ratio, and Perelman found it by applying Bishop–Gromov to a Ricci-flat manifold of very high dimension.

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

  • define the L\mathcal L-length, L\mathcal L-geodesics, the reduced distance and the reduced volume, and compute them on flat space;
  • state the differential inequalities for ℓ\ell and use them to prove that V~\tilde V is nonincreasing in τ\tau;
  • show that ℓ≤n2\ell \leq \frac n2 somewhere at every τ\tau;
  • read the reduced volume against Bishop–Gromov, row by row, and against Huisken's monotonicity formula;
  • describe Perelman's heuristic derivation in N→∞N \to \infty dimensions, and the uses of V~\tilde V.

Least action and the heat kernel

In the world Model Paths, actions and diffusion

A particle diffusing in Rn\mathbb{R}^n from 00 is found near xx after time τ\tau with density (4πτ)−n/2e−∣x∣2/4τ(4\pi\tau)^{-n/2}e^{-|x|^2/4\tau}, the heat kernel. The exponent ∣x∣24τ\frac{|x|^2}{4\tau} is the least value of the action 14∫0τ∣γ˙∣2 ds\frac14\int_0^\tau|\dot\gamma|^2\,ds over paths from 00 to xx in time τ\tau: diffusion is governed, to leading order, by its cheapest paths. Varadhan proved in 1967 that this persists on curved spaces for small times, with −4τlog⁡p(τ,x,y)→d(x,y)2-4\tau\log p(\tau, x, y) \to d(x, y)^2. Peter Li and Shing-Tung Yau used such path lengths, adapted to a linear parabolic equation, to integrate their Harnack inequality in 1986, and Perelman names their paper as the closest precedent for his construction. His reduced distance is the least action for paths in a Ricci flow, with the scalar curvature as a potential, and (4πτ)−n/2e−ℓ(4\pi\tau)^{-n/2}e^{-\ell} plays the part of the heat kernel.

L\mathcal L-length

Work in backward time. Fix a Ricci flow and a base time t0t_0, set τ=t0−t\tau = t_0 - t, so that ∂τg=2Ric⁡\partial_\tau g = 2\operatorname{Ric}, and assume the manifold is closed, or that each g(τ)g(\tau) is complete with uniformly bounded curvature. Fix a base point pp.

Definition 5.1 L\mathcal L-length and reduced distance

For a curve γ:[0,τˉ]→M\gamma : [0, \bar\tau] \to M with γ(0)=p\gamma(0) = p, the L\mathcal L-length is

L(γ)=∫0τˉτ (R(γ(τ),τ)+∣γ˙(τ)∣g(τ)2) dτ.\mathcal L(\gamma) = \int_0^{\bar\tau}\sqrt\tau\,\big(R(\gamma(\tau), \tau) + |\dot\gamma(\tau)|^2_{g(\tau)}\big)\,d\tau.

Let L(q,τˉ)L(q, \bar\tau) be the infimum of L(γ)\mathcal L(\gamma) over curves from pp at τ=0\tau = 0 to qq at τ=τˉ\tau = \bar\tau. The reduced distance is

ℓ(q,τˉ)=L(q,τˉ)2τˉ.\ell(q, \bar\tau) = \frac{L(q, \bar\tau)}{2\sqrt{\bar\tau}}.

The weight τ\sqrt\tau is what makes the theory scale correctly. With s=τs = \sqrt\tau the curve has finite energy near τ=0\tau = 0, and the change of variable turns ∫τ∣γ˙∣2 dτ\int\sqrt\tau|\dot\gamma|^2\,d\tau into 12∫∣γ′(s)∣2 ds\frac12\int|\gamma'(s)|^2\,ds. Perelman's first variation (his (7.1)) gives the L\mathcal L-geodesic equation, with X=γ˙X = \dot\gamma:

∇XX−12∇R+12τX+2Ric⁡(X,⋅)=0.\nabla_XX - \frac12\nabla R + \frac{1}{2\tau}X + 2\operatorname{Ric}(X, \cdot) = 0.

Minimisers exist and are L\mathcal L-geodesics, and τX(τ)\sqrt\tau X(\tau) has a limit v∈TpMv \in T_pM as τ→0\tau \to 0. The L\mathcal L-exponential map sends vv to γv(τˉ)\gamma_v(\bar\tau). Where minimisers are not unique, or LL is not smooth, the inequalities below hold in the barrier sense, as for the distance function (9B.1 Laplacian Comparison).

Example 5.2 Flat space

On Rn\mathbb{R}^n with the static flat metric, R=0R = 0. In s=τs = \sqrt\tau the length is 12∫0τˉ∣γ′(s)∣2 ds\frac12\int_0^{\sqrt{\bar\tau}}|\gamma'(s)|^2\,ds, minimised by the straight line at constant speed ∣x∣/τˉ|x|/\sqrt{\bar\tau}, so L(x,τˉ)=∣x∣22τˉL(x, \bar\tau) = \frac{|x|^2}{2\sqrt{\bar\tau}} and

ℓ(x,τˉ)=∣x∣24τˉ,(4πτˉ)−n/2e−ℓ(x,τˉ)=the heat kernel.\ell(x, \bar\tau) = \frac{|x|^2}{4\bar\tau}, \qquad (4\pi\bar\tau)^{-n/2}e^{-\ell(x, \bar\tau)} = \text{the heat kernel}.

The L\mathcal L-geodesics are γv(τ)=2τ v\gamma_v(\tau) = 2\sqrt\tau\,v (Figure 5.1).

Figure 5.1. L\mathcal L-geodesics from pp in flat R\mathbb{R}, with τ\tau increasing downward: γv(τ)=2τ v\gamma_v(\tau) = 2\sqrt\tau\,v for v=−1,−34,…,1v = -1, -\frac34, \dots, 1 (computed). They leave pp with infinite speed, like Brownian paths at small scales, and the L\mathcal L-exponential map sends vv to γv(τˉ)\gamma_v(\bar\tau). In a curved flow they bend, following the geodesic equation above.

The differential inequalities

Perelman computes the first and second variations of LL exactly as in Riemannian geometry, with one new feature: the time derivative of the metric is 2Ric⁡2\operatorname{Ric}, so Ricci terms appear, and along the curve they combine into Hamilton's Harnack expressions (11B.2 Ancient Solutions and the Harnack Inequality). He collects them in a quantity K=∫0τˉτ3/2H(X) dτK = \int_0^{\bar\tau}\tau^{3/2}H(X)\,d\tau, where HH is Hamilton's trace Harnack expression. Then (his (7.5), (7.6) and (7.10)), in terms of ℓ\ell:

∂τˉℓ=R−ℓτˉ+K2τˉ3/2,∣∇ℓ∣2=−R+ℓτˉ−Kτˉ3/2,Δℓ≤−R+n2τˉ−K2τˉ3/2.\partial_{\bar\tau}\ell = R - \frac{\ell}{\bar\tau} + \frac{K}{2\bar\tau^{3/2}}, \qquad |\nabla\ell|^2 = -R + \frac{\ell}{\bar\tau} - \frac{K}{\bar\tau^{3/2}}, \qquad \Delta\ell \leq -R + \frac{n}{2\bar\tau} - \frac{K}{2\bar\tau^{3/2}}.

The last is the space-time Laplacian comparison, the analogue of Δr≤n−1r\Delta r \leq \frac{n - 1}{r} when Ric⁡≥0\operatorname{Ric} \geq 0 (9B.1 Laplacian Comparison). It is proved with the test fields YY solving ∇XY=−Ric⁡(Y,⋅)+12τY\nabla_XY = -\operatorname{Ric}(Y, \cdot) + \frac{1}{2\tau}Y, which play the part of the linear Jacobi fields srE\frac{s}{r}E in Euclidean space. The unknown KK cancels from the right combinations, and gives (Perelman's (7.13) and (7.14))

∂τˉℓ−Δℓ+∣∇ℓ∣2−R+n2τˉ≥0,2Δℓ−∣∇ℓ∣2+R+ℓ−nτˉ≤0.\partial_{\bar\tau}\ell - \Delta\ell + |\nabla\ell|^2 - R + \frac{n}{2\bar\tau} \geq 0, \qquad 2\Delta\ell - |\nabla\ell|^2 + R + \frac{\ell - n}{\bar\tau} \leq 0.

On flat space both are equalities (Exercise 5.8). No curvature hypothesis is needed anywhere: this is the decisive difference from Bishop–Gromov.

Proposition 5.3 A point where ℓ≤n2\ell \leq \frac n2 (Perelman I, §7.1)

For every τˉ>0\bar\tau > 0 there is a point qq with ℓ(q,τˉ)≤n2\ell(q, \bar\tau) \leq \frac n2.

Proof. Let Lˉ=2τˉL=4τˉℓ\bar L = 2\sqrt{\bar\tau}L = 4\bar\tau\ell. Perelman's (7.5) and (7.10) give ∂τˉLˉ+ΔLˉ≤2n\partial_{\bar\tau}\bar L + \Delta\bar L \leq 2n (Exercise 5.9). At a minimum point of Lˉ(⋅,τˉ)\bar L(\cdot, \bar\tau), ΔLˉ≥0\Delta\bar L \geq 0, so the minimum of Lˉ−2nτˉ\bar L - 2n\bar\tau is nonincreasing (in the barrier sense, on a closed manifold or with the stated bounds). As τˉ→0\bar\tau \to 0, Lˉ(p,τˉ)→0\bar L(p, \bar\tau) \to 0. So min⁡Lˉ(⋅,τˉ)≤2nτˉ\min\bar L(\cdot, \bar\tau) \leq 2n\bar\tau, which is min⁡ℓ≤n2\min\ell \leq \frac n2.

The reduced volume

Definition 5.4 Reduced volume
V~(τˉ)=∫M(4πτˉ)−n/2e−ℓ(q,τˉ) dVg(τˉ)(q).\tilde V(\bar\tau) = \int_M(4\pi\bar\tau)^{-n/2}e^{-\ell(q, \bar\tau)}\,dV_{g(\bar\tau)}(q).

Perelman's own definition omits the factor (4π)−n/2(4\pi)^{-n/2}; with it, V~≡1\tilde V \equiv 1 on flat Rn\mathbb{R}^n, since the integrand is the heat kernel. This normalisation is common in the expositions.

Theorem 5.5 Monotonicity of the reduced volume (Perelman I, §7.1)

Along any Ricci flow as above, V~(τˉ)\tilde V(\bar\tau) is nonincreasing in τˉ\bar\tau. The monotonicity is strict unless the flow is a shrinking gradient soliton.

Proof. Set u=(4πτˉ)−n/2e−ℓu = (4\pi\bar\tau)^{-n/2}e^{-\ell}. Since ∂τˉu=u(−n2τˉ−∂τˉℓ)\partial_{\bar\tau}u = u\big(-\frac{n}{2\bar\tau} - \partial_{\bar\tau}\ell\big) and Δu=u(∣∇ℓ∣2−Δℓ)\Delta u = u(|\nabla\ell|^2 - \Delta\ell),

∂τˉu−Δu+Ru=−u(∂τˉℓ−Δℓ+∣∇ℓ∣2−R+n2τˉ)≤0\partial_{\bar\tau}u - \Delta u + Ru = -u\Big(\partial_{\bar\tau}\ell - \Delta\ell + |\nabla\ell|^2 - R + \frac{n}{2\bar\tau}\Big) \leq 0

by the first inequality above. In backward time the conjugate heat operator is ∂τˉ−Δ+R\partial_{\bar\tau} - \Delta + R, so uu is a subsolution of the conjugate heat equation. Since ∂τˉdV=R dV\partial_{\bar\tau}dV = R\,dV,

ddτˉ∫Mu dV=∫M(∂τˉu+Ru) dV=∫M(∂τˉu−Δu+Ru) dV≤0.\frac{d}{d\bar\tau}\int_Mu\,dV = \int_M(\partial_{\bar\tau}u + Ru)\,dV = \int_M(\partial_{\bar\tau}u - \Delta u + Ru)\,dV \leq 0.

Making this rigorous where ℓ\ell is not smooth needs the barrier sense and care at the cut locus. Perelman's main argument avoids it by working along each L\mathcal L-geodesic: the Jacobian JJ of the L\mathcal L-exponential map satisfies ddτlog⁡J≤n2τ−12τ−3/2K\frac{d}{d\tau}\log J \leq \frac{n}{2\tau} - \frac12\tau^{-3/2}K, so τ−n/2e−ℓ(γ(τ),τ)J(τ)\tau^{-n/2}e^{-\ell(\gamma(\tau), \tau)}J(\tau) is nonincreasing along each geodesic, and integrating over TpMT_pM gives the theorem. Equality along all geodesics forces Ric⁡+∇2ℓ=12τg\operatorname{Ric} + \nabla^2\ell = \frac{1}{2\tau}g, a shrinking gradient soliton.

As τˉ→0\bar\tau \to 0, the flow near (p,t0)(p, t_0) looks Euclidean at the relevant scale τˉ\sqrt{\bar\tau}, and V~(τˉ)→1\tilde V(\bar\tau) \to 1; the expositions prove this limit carefully. So V~≤1\tilde V \leq 1, with 11 the value of flat space. Figure 5.2 computes it for the round shrinking 3-sphere.

Figure 5.2. V~(τˉ)\tilde V(\bar\tau) for the round shrinking S3S^3 of radius 11 at the base time, so that r(τ)2=1+4τr(\tau)^2 = 1 + 4\tau (computed). By symmetry the L\mathcal L-shortest curves run along great circles, which gives ℓ\ell in closed form (Exercise 5.10). V~\tilde V decreases from 11 towards 2πe−3/2≈0.792\sqrt\pi e^{-3/2} \approx 0.79 as τˉ→∞\bar\tau \to \infty: looking further into the past, the flow is seen as the shrinking soliton it is.

Bishop–Gromov in space-time

The reduced volume completes the comparison theory of 9B.1 Laplacian Comparison and 9B.2 Volume Comparison, row by row.

Riemannian geometry, Ric⁡≥0\operatorname{Ric} \geq 0 Ricci flow, backward time τ\tau
length ∫∣γ˙∣ ds\int|\dot\gamma|\,ds L\mathcal L-length ∫τ(R+∣γ˙∣2) dτ\int\sqrt\tau(R + |\dot\gamma|^2)\,d\tau
distance d(p,q)d(p, q), with d24\frac{d^2}{4} as "action" reduced distance ℓ(q,τ)\ell(q, \tau)
geodesics, the exponential map L\mathcal L-geodesics, the L\mathcal L-exponential map
Jacobi fields L\mathcal L-Jacobi fields
Laplacian comparison Δr≤n−1r\Delta r \leq \frac{n - 1}{r} Δℓ≤−R+n2τ−K2τ3/2\Delta\ell \leq -R + \frac{n}{2\tau} - \frac{K}{2\tau^{3/2}}
curvature hypothesis Ric⁡≥0\operatorname{Ric} \geq 0 none: the flow equation supplies it
model: Euclidean space model: the Gaussian soliton (flat Rn\mathbb{R}^n), ℓ=∣x∣24τ\ell = \frac{|x|^2}{4\tau}
volume ratio Vol⁡B(p,r)ωnrn\frac{\operatorname{Vol}B(p, r)}{\omega_nr^n}, nonincreasing in rr reduced volume V~(τ)\tilde V(\tau), nonincreasing in τ\tau
ratio →1\to 1 as r→0r \to 0 V~→1\tilde V \to 1 as τ→0\tau \to 0
equality: Euclidean space equality: shrinking gradient solitons
use: volume doubling, noncollapsing use: noncollapsing that survives surgery

Perelman's heuristic: Bishop–Gromov in NN dimensions

Perelman explains in §6 where the reduced volume comes from. Take a large integer NN and the manifold M~=M×SN×R+\tilde M = M\times S^N\times\mathbb{R}_+ with metric

g~=g(τ)+τ gSN+(N2τ+R)dτ2,\tilde g = g(\tau) + \tau\,g_{S^N} + \Big(\frac{N}{2\tau} + R\Big)d\tau^2,

where g(τ)g(\tau) is the backward Ricci flow and gSNg_{S^N} is the round metric of constant curvature 12N\frac{1}{2N}. He reports that g~\tilde g is Ricci-flat up to errors of order N−1N^{-1}, and that its curvature components reproduce Hamilton's Harnack expressions. For a curve from a point pp at τ=0\tau = 0, orthogonal to the sphere factor, the g~\tilde g-length is

2Nτ(q)+12NL(γ)+O(N−3/2),\sqrt{2N\tau(q)} + \frac{1}{\sqrt{2N}}\mathcal L(\gamma) + O(N^{-3/2}),

so shortest geodesics in M~\tilde M minimise L\mathcal L. The volume of the geodesic sphere of radius 2Nτ\sqrt{2N\tau} about pp, divided by the Euclidean value, is a constant times N−n/2N^{-n/2} times ∫τ−n/2e−ℓ\int\tau^{-n/2}e^{-\ell}, up to O(N−1)O(N^{-1}). Bishop–Gromov for the nearly Ricci-flat M~\tilde M then suggests that this quantity increases as τ\tau decreases. Perelman presents this as a heuristic, and gives the rigorous proof separately in §7. The heuristic explains why no curvature hypothesis is needed: the extra dimensions make the space-time Ricci-flat.

Huisken's formula and the direction of monotonicity

Huisken's monotonicity formula (6A.8 Curve Shortening and the First Geometric Flows) for mean curvature flow integrates the backward heat kernel centred at a space-time point (x0,t0)(x_0, t_0) over the surface:

Θ(t)=∫Mt(4π(t0−t))−n/2e−∣x−x0∣2/4(t0−t) dA,\Theta(t) = \int_{M_t}(4\pi(t_0 - t))^{-n/2}e^{-|x - x_0|^2/4(t_0 - t)}\,dA,

nonincreasing in tt, and constant exactly on self-shrinkers. The reduced volume has the same shape, with the Euclidean distance replaced by ℓ\ell. Perelman points out that the monotonicity runs the other way. In τ=t0−t\tau = t_0 - t, Huisken's Θ\Theta is nondecreasing, while V~\tilde V is nonincreasing. Both are nevertheless constant exactly on the self-similar shrinking solutions, and both are used in the same way: they are evaluated at very small and very large scales and compared, to show that a blow-up looks like a soliton.

Where this goes Uses of the reduced volume

Perelman's §7.3 reproves noncollapsing with V~\tilde V: if a ball is collapsed, V~\tilde V based near it is small at small τ\tau, while at τ\tau comparable to the whole time interval it is bounded below, by using the point where ℓ≤n2\ell \leq \frac n2; monotonicity forbids this. Because the argument needs only local control, it survives surgery (No local collapsing theorem II, §8; 12B.5 Ricci Flow with Surgery for All Time). On κ\kappa-solutions, letting τ→∞\tau \to \infty and rescaling by τ\tau produces the asymptotic soliton (12B.1 κ-Solutions). 12A.6 Pseudolocality turns to the other side of §§8–10: the Harnack inequality for the conjugate heat kernel and pseudolocality.

History

Perelman introduced L\mathcal L-length, reduced distance and reduced volume in §§6–7 of his first preprint (2002). He described the computations as natural modifications of the classical variational theory of geodesics, with Li and Yau's 1986 paper as the closest reference, and noted that Chow and Chu had found the first geometric interpretation of Hamilton's Harnack expressions, using a degenerate metric on M×RM\times\mathbb{R}, to which his construction is in a sense dual. Huisken's formula dates from 1990.

Recall Where we stand

In backward time τ\tau, the L\mathcal L-length ∫τ(R+∣γ˙∣2) dτ\int\sqrt\tau(R + |\dot\gamma|^2)\,d\tau defines a reduced distance ℓ=L/(2τ)\ell = L/(2\sqrt\tau) from a base point; on flat space ℓ=∣x∣24τ\ell = \frac{|x|^2}{4\tau} and (4πτ)−n/2e−ℓ(4\pi\tau)^{-n/2}e^{-\ell} is the heat kernel. Perelman's variational formulas give ∂τℓ−Δℓ+∣∇ℓ∣2−R+n2τ≥0\partial_\tau\ell - \Delta\ell + |\nabla\ell|^2 - R + \frac{n}{2\tau} \geq 0, so (4πτ)−n/2e−ℓ(4\pi\tau)^{-n/2}e^{-\ell} is a subsolution of the conjugate heat equation and the reduced volume V~(τ)\tilde V(\tau) is nonincreasing, for every Ricci flow, strictly unless on a shrinking soliton. Also min⁡ℓ≤n2\min\ell \leq \frac n2, and V~→1\tilde V \to 1 as τ→0\tau \to 0. It is Bishop–Gromov in space-time, found by applying Bishop–Gromov to a nearly Ricci-flat manifold in N→∞N \to \infty dimensions. 12A.6 Pseudolocality completes Book 12A with Perelman's Harnack inequality and pseudolocality.

Exercises

Exercise 5.6 Flat space, in full

On static flat Rn\mathbb{R}^n with p=0p = 0: (a) show that the substitution s=τs = \sqrt\tau turns ∫0τˉτ∣γ˙∣2 dτ\int_0^{\bar\tau}\sqrt\tau|\dot\gamma|^2\,d\tau into 12∫0τˉ∣γ′(s)∣2 ds\frac12\int_0^{\sqrt{\bar\tau}}|\gamma'(s)|^2\,ds; (b) deduce L(x,τˉ)=∣x∣22τˉL(x, \bar\tau) = \frac{|x|^2}{2\sqrt{\bar\tau}} and ℓ=∣x∣24τˉ\ell = \frac{|x|^2}{4\bar\tau}; (c) show V~≡1\tilde V \equiv 1.

Solution

(a) dτ=2s dsd\tau = 2s\,ds and γ˙=γ′(s)2s\dot\gamma = \frac{\gamma'(s)}{2s}, so τ∣γ˙∣2dτ=s∣γ′∣24s22s ds=12∣γ′∣2ds\sqrt\tau|\dot\gamma|^2d\tau = s\frac{|\gamma'|^2}{4s^2}2s\,ds = \frac12|\gamma'|^2ds. (b) Energy is minimised by the constant-speed line, γ(s)=sτˉx\gamma(s) = \frac{s}{\sqrt{\bar\tau}}x, giving 12∣x∣2τˉτˉ\frac12\frac{|x|^2}{\bar\tau}\sqrt{\bar\tau}. Then ℓ=L2τˉ\ell = \frac{L}{2\sqrt{\bar\tau}}. (c) ∫(4πτˉ)−n/2e−∣x∣2/4τˉdx=1\int(4\pi\bar\tau)^{-n/2}e^{-|x|^2/4\bar\tau}dx = 1.

Exercise 5.7 The geodesic equation on flat space

On static flat space the L\mathcal L-geodesic equation reads γ¨+12τγ˙=0\ddot\gamma + \frac{1}{2\tau}\dot\gamma = 0. Solve it with γ(0)=0\gamma(0) = 0, and show that the solutions are γ(τ)=2τ v\gamma(\tau) = 2\sqrt\tau\,v with v=lim⁡τ→0τγ˙(τ)v = \lim_{\tau\to0}\sqrt\tau\dot\gamma(\tau).

Solution

ddτ(τγ˙)=τ(γ¨+12τγ˙)=0\frac{d}{d\tau}(\sqrt\tau\dot\gamma) = \sqrt\tau\big(\ddot\gamma + \frac{1}{2\tau}\dot\gamma\big) = 0, so τγ˙=v\sqrt\tau\dot\gamma = v is constant, γ˙=vτ−1/2\dot\gamma = v\tau^{-1/2}, and γ=2τ v\gamma = 2\sqrt\tau\,v.

Exercise 5.8 The inequalities on flat space

With ℓ=∣x∣24τ\ell = \frac{|x|^2}{4\tau} on flat Rn\mathbb{R}^n (so R=0R = 0, K=0K = 0), check that ∂τℓ−Δℓ+∣∇ℓ∣2+n2τ=0\partial_\tau\ell - \Delta\ell + |\nabla\ell|^2 + \frac{n}{2\tau} = 0 and 2Δℓ−∣∇ℓ∣2+ℓ−nτ=02\Delta\ell - |\nabla\ell|^2 + \frac{\ell - n}{\tau} = 0.

Solution

∂τℓ=−∣x∣24τ2\partial_\tau\ell = -\frac{|x|^2}{4\tau^2}, ∇ℓ=x2τ\nabla\ell = \frac{x}{2\tau}, ∣∇ℓ∣2=∣x∣24τ2|\nabla\ell|^2 = \frac{|x|^2}{4\tau^2}, Δℓ=n2τ\Delta\ell = \frac{n}{2\tau}. First: −∣x∣24τ2−n2τ+∣x∣24τ2+n2τ=0-\frac{|x|^2}{4\tau^2} - \frac{n}{2\tau} + \frac{|x|^2}{4\tau^2} + \frac{n}{2\tau} = 0. Second: nτ−∣x∣24τ2+∣x∣24τ2−nτ=0\frac n\tau - \frac{|x|^2}{4\tau^2} + \frac{|x|^2}{4\tau^2} - \frac n\tau = 0.

Exercise 5.9 The minimum of ℓ\ell

Using ∂τˉL=2τˉR−12τˉL+1τˉK\partial_{\bar\tau}L = 2\sqrt{\bar\tau}R - \frac{1}{2\bar\tau}L + \frac{1}{\bar\tau}K and ΔL≤−2τˉR+nτˉ−1τˉK\Delta L \leq -2\sqrt{\bar\tau}R + \frac{n}{\sqrt{\bar\tau}} - \frac{1}{\bar\tau}K (Perelman's (7.5) and (7.10)), show that Lˉ=2τˉL\bar L = 2\sqrt{\bar\tau}L satisfies ∂τˉLˉ+ΔLˉ≤2n\partial_{\bar\tau}\bar L + \Delta\bar L \leq 2n.

Solution

∂τˉLˉ=Lτˉ+2τˉ∂τˉL=Lτˉ+4τˉR−Lτˉ+2Kτˉ=4τˉR+2Kτˉ\partial_{\bar\tau}\bar L = \frac{L}{\sqrt{\bar\tau}} + 2\sqrt{\bar\tau}\partial_{\bar\tau}L = \frac{L}{\sqrt{\bar\tau}} + 4\bar\tau R - \frac{L}{\sqrt{\bar\tau}} + \frac{2K}{\sqrt{\bar\tau}} = 4\bar\tau R + \frac{2K}{\sqrt{\bar\tau}}. And ΔLˉ=2τˉΔL≤−4τˉR+2n−2Kτˉ\Delta\bar L = 2\sqrt{\bar\tau}\Delta L \leq -4\bar\tau R + 2n - \frac{2K}{\sqrt{\bar\tau}}. Adding gives 2n2n.

Exercise 5.10 Rehearsal: the shrinking sphere

On the round shrinking S3S^3 with g(τ)=(1+4τ)gS3g(\tau) = (1 + 4\tau)g_{S^3} and R=61+4τR = \frac{6}{1 + 4\tau}, let θ\theta be the angle from pp. (a) Explain why L(q,τˉ)=∫0τˉτR dτ+min⁡∫0τˉτ(1+4τ)θ˙2 dτL(q, \bar\tau) = \int_0^{\bar\tau}\sqrt\tau R\,d\tau + \min\int_0^{\bar\tau}\sqrt\tau(1 + 4\tau)\dot\theta^2\,d\tau over θ(0)=0\theta(0) = 0, θ(τˉ)=θ(q)\theta(\bar\tau) = \theta(q). (b) Show that the minimum is θ(q)2/I(τˉ)\theta(q)^2/I(\bar\tau) with I(τˉ)=∫0τˉdττ(1+4τ)=arctan⁡(2τˉ)I(\bar\tau) = \int_0^{\bar\tau}\frac{d\tau}{\sqrt\tau(1 + 4\tau)} = \arctan(2\sqrt{\bar\tau}), and that ∫0τˉτR dτ=3τˉ−32arctan⁡(2τˉ)\int_0^{\bar\tau}\sqrt\tau R\,d\tau = 3\sqrt{\bar\tau} - \frac32\arctan(2\sqrt{\bar\tau}). (c) Show that as τˉ→∞\bar\tau \to \infty, ℓ→32\ell \to \frac32 uniformly, and deduce V~→(4π)−3/2⋅2π2⋅43/2e−3/2=2πe−3/2\tilde V \to (4\pi)^{-3/2}\cdot2\pi^2\cdot4^{3/2}e^{-3/2} = 2\sqrt\pi e^{-3/2}.

Solution

(a) RR depends only on τ\tau, so its integral is the same for every curve; and ∣γ˙∣2≥(1+4τ)θ˙2|\dot\gamma|^2 \geq (1 + 4\tau)\dot\theta^2, with equality along great circles. (b) By Cauchy–Schwarz, θ(q)2=(∫θ˙)2≤∫aθ˙2∫1a\theta(q)^2 = \big(\int\dot\theta\big)^2 \leq \int a\dot\theta^2\int\frac1a with a=τ(1+4τ)a = \sqrt\tau(1 + 4\tau), with equality when θ˙∝1a\dot\theta \propto \frac1a. With s=τs = \sqrt\tau, I=∫0τˉ2 ds1+4s2=arctan⁡(2τˉ)I = \int_0^{\sqrt{\bar\tau}}\frac{2\,ds}{1 + 4s^2} = \arctan(2\sqrt{\bar\tau}) and ∫τR=12∫0τˉs2 ds1+4s2=3s−32arctan⁡2s\int\sqrt\tau R = 12\int_0^{\sqrt{\bar\tau}}\frac{s^2\,ds}{1 + 4s^2} = 3s - \frac32\arctan2s. (c) ℓ=12τˉ(θ2arctan⁡2τˉ+3τˉ−32arctan⁡2τˉ)→32\ell = \frac{1}{2\sqrt{\bar\tau}}\big(\frac{\theta^2}{\arctan2\sqrt{\bar\tau}} + 3\sqrt{\bar\tau} - \frac32\arctan2\sqrt{\bar\tau}\big) \to \frac32, since θ≤π\theta \leq \pi. Then V~≈(4πτˉ)−3/2(4τˉ)3/2∣S3∣e−3/2\tilde V \approx (4\pi\bar\tau)^{-3/2}(4\bar\tau)^{3/2}|S^3|e^{-3/2}, and ∣S3∣=2π2|S^3| = 2\pi^2.

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