Lectures on Ricci Curvaturemath.ucsd.edu/~benchow/lcct/BamlerJiangRicciBookChap1-2...2020/06/18 · the Ricci curvature on M. Here {e i}n i=1 denotes a local orthonormal frame. In

Lectures on Ricci Curvature

Richard Bamler

Wenshuai JiangWorking version: June 18, 2020

Contents

Preface i

Notation and Symbols iii

Chapter 1. Basics in Riemannian Geometry 11.1. Introduction 11.2. Lohkamp’s construction 31.3. Regularity of distance functions 3

1.3.1. The cut locus and injectivity domain 31.3.2. Upper Hessian bounds and the existence of the weak Laplacian 6

1.4. The Laplacian comparison theorem 111.5. Distance functions and the geometry of their level sets 12

1.5.1. Second fundamental form and mean curvature of geodesicspheres 13

1.5.2. The Jacobi and Riccati equations 131.5.3. Lower Ricci curvature bound comparison with model spaces 151.5.4. Laplacian Comparison Theorem, II 161.5.5. Myers’s Theorem 16

1.6. Bishop–Gromov volume comparison theorem 161.6.1. A local version of the volume comparison theorem 161.6.2. The Bishop–Gromov volume comparison theorem 181.6.3. Bishop–Gromov relative volume comparison theorem 191.6.4. Volume and diameter rigidity 221.6.5. Further comparison results 231.6.6. Further application of Bishop–Gromov 24

1.7. Cheeger–Gromoll splitting 241.8. Notes and commentary 241.9. Exercises 24

Chapter 2. Metric Spaces and Gromov–Hausdorff Distance 252.1. Hausdorff distance 252.2. Gromov–Hausdorff distance 272.3. Gromov’s precompactness theorem 33

3

4 CONTENTS

2.4. Length spaces and the pointed Gromov–Hausdorff distance 372.5. Basic Geometric measure theory 39

2.5.1. Covering Lemma 392.5.2. Hausdorff measure 41

2.6. Notes and commentary 482.7. Exercises 48

Chapter 3. Analysis on manifolds with lower Ricci curvature bounds 513.1. Segment inequality 513.2. Segment inequality (Richard’s version) 533.3. Integral estimates along families of geodesics (Richard’s version) 543.4. Geodesic flow and integral estimates (Wenshuai’s version) 563.5. Poincare and Sobolev inequality (Richard’s version) 583.6. Poincare and Sobolev inequality (Wenshuai’s version 613.7. Cheng–Yau gradient estimate 713.8. Li-Yau heat kernel estimate 743.9. Cheeger-Colding’s cut-off function 74

Chapter 4. Almost GH-splitting and δ-splitting maps 75

Chapter 5. Volume stability 77

Chapter 6. Almost volume cone implies almost metric cone 79

Chapter 7. Quantitative Stratification 81

Chapter 8. Singularity and Regularity of bounded Ricci curvature 838.1. ε-regularity 838.2. Lp-estimates of Riemann curvature and Lp-Hessian estimates of

harmonic functions 848.3. codimension four singularity 84

Chapter 9. Singularity and Regularity of lower Ricci curvature 85

Chapter 10. Geometric and analytic aspects of—Ricci curvature 8710.1. Definitions 8710.2. Proofs 8710.3. section quoted on p. 69 8810.4. labelled: sec: heat kernel 8810.5. Gaussian estimates for the heat kernel 8810.6. Liouville type theorem 9110.7. Existence of cutoff function 9310.8. Almost splitting implies existence of a splitting map 95

Preface

This is a working version of a book by by Richard Bamler and Wenshuai Jiangarising from their notes on Cheeger–Colding theory.

The University of Sydney has a seminar on Cheeger–Colding theory given byWenshuai Jiang: University of Sydney Differential Geometry Seminar Series

We will go through these notes in the UCSD reading seminar on Cheeger–Colding theory and Perelman’s L-geometry. Seminar time: 5 pm Tuesdays andThursdays, starting June 23, 2020.

Please email any suggestions or lists of typos to [email protected] . Thanks!

i

https://math.berkeley.edu/~rbamler/

https://person.zju.edu.cn/en/wsjiang

http://www.maths.usyd.edu.au/s/scnitm/haotianw-DifferentialGeometrySemin-011

http://www.math.ucsd.edu/~benchow/cc-seminar.html

Notation and Symbols

:= defined to be equal to

∂t∂

∂t0n the origin in Rn

A(p, r1, r2) annulus

As second fundamental form of the level set f = sA(t, θ) ... superceded by J(t, θ) ?

b∗ see (1.7)

B(p, r) open ball of radius r centered at p

B(p, r) closed ball of radius r centered at p

B(A, r) (tubular) r-neighborhood of A

∇ covariant derivative or gradient

∇2 Hessian

c a sufficiently small positive constant

C a sufficiently large positive constant

Cp cut locus in M of p

C⋃p∈Mp × Cp

C(Z) see (2.2)

Ckc Ck with compact support

C∞0 smooth functions with compact support

CapX(r) capacity of X with respect to the scale r

cnk see (1.16)

χA characteristic function of AD∂t covariant derivative with respect to t along a map

dH Hausdorff distance

Dt(A) set of t-density points of A

Dp injectivity domain of expp

iii

iv NOTATION AND SYMBOLS

D⋃p∈M Dp ⊂ TM

dGH Gromov–Hausdorff distance

dH Hausdorff distance

deg2 mod 2 degree; see (11.55)

dg volume form of g

dµ4b measure defining the weak Laplacian of b

t disjoint union

diam diameter

div divergence

expp exponential map at p

Ff see (3.1)

g Riemannian metric

geucl Euclidean metric on Rn

γv constant speed geodesic with γ′v(0) = v

γv geodesic with γ′v(0) = v

H mean curvature of a level set

H-distance Hausdorff distance

Hk mean curvature of geodesic sphere in model space; see (1.18)

Ht(A) t-Hausdorff measure of A

inj(x) injectivity radius estimate at xfflaverage; see (3.34)

Jx Jacobian of expxJ∗p see (1.44)

J(t) Jacobi field

K(x, y, t) heat kernel

4 Laplacian

L length

L Lie derivative

M smooth manifold

(M, g) Riemannian manifold with metric g

Mnk model space of constant curvature k

M space of isometry classes of compact metric spaces

Mx see (11.385)

NOTATION AND SYMBOLS v

M(n, k) the class of manifolds defined by (2.41)

M(n, k,D) the class of manifolds defined by (2.28)

M(n, k, v,D) the class of manifolds defined by (1.48)

µ4b see (1.6)

nωn volume of unit (n− 1)-sphere in Rn

osc oscillation: sup− infrp(x) distance function d(x, p)rRm(p) ...

rGH,n(p) ...

R scalar curvature

Ric Ricci tensor

Rm Riemann curvature tensor

ρt(x, y) heat kernel [different than Richard’s notation]

R≥0 set of non-negative real numbers

R+ set of positive real numbers

R(X) regular set of X

Rη(X) see (11.355)

Ss shape operator of the level set f = sS(X) singular set of X

Sk(X) see (11.353)

Skη(X) see (11.354)

sec sectional curvatures

Skr sublevel set of rGH,k; see (10.68)

SM unit tangent bundle of M

snk see (1.15)

supp support

TM tangent bundle of M

T ‖M tangent bundle of the level sets of a function

∆xyz triangle ...

vk(r) volume of r-ball in model space; see (1.33)

vol volume

|Ω| the volume of a measurable set Ωωn volume of unit Euclidean n-ball

vi NOTATION AND SYMBOLS

ωt see (2.52)

Z+ set of nonnegative integers

CHAPTER 1

Basics in Riemannian Geometry

For the requisite Riemannian geometry, see Chapters 1, 2, and 5 of the bookby Cheeger and Ebin. Some of the material therein is reviewed in this chapter.

1.1. Introduction

Consider an n-dimensional Riemannian manifold (Mn, g) and denote by

Ric(X,Y ) =n∑i=1

Rm(ei, X, Y, ei)

the Ricci curvature on M . Here eini=1 denotes a local orthonormal frame.In this book we will analyze the following conditions on the Ricci curvature:

Einstein condition: Ric = λg, where λ is a constant, called the Einstein equation.Lower Ricci curvature bound: Ric ≥ λg.Two-sided Ricci curvature bound: −λg ≤ Ric ≤ λg.

An upper Ricci curvature bound is not very interesting from a geometric oranalytic point of view, due to the following results of Lohkamp:

THEOREM 1.1 (Lohkamp, 1994, [26]). Let (Mn, g) be a compact Riemannianmanifold of dimension n ≥ 3 and K > 0 a constant. Then there is a sequence ofRiemannian metrics gi on M such that Ricgi < −Kgi for all i and

(M, gi)i→∞−−−→ (M, g)

in the Gromov–Hausdorff sense.

For more details on Gromov–Hausdorff convergence see section 10.1. Notethat there is a similar result in the case in which M is non-compact.

THEOREM 1.2 (Lohkamp, 1994, [27]). In Theorem 1.1 we can even choose gisuch that

gii→∞−−−→ g

uniformly as tensors.

We will give an intuition on the proof of Theorem 1.1 in section 10.2.Consider again the conditions listed in the beginning of this section. It turns out

that the lower Ricci curvature condition will be the most fruitful of the conditions

1

https://bookstore.ams.org/cdn-1,590,737,037,146/chel-365-h/~~FreeAttachments/chel-365-h-prev.pdf

2 1. BASICS IN RIEMANNIAN GEOMETRY

mentioned above. An additional upper Ricci curvature bound (resulting in a two-sided bound) can only be used in very specific — but important — settings. Itwill turn out that the (sharp) Einstein condition will hardly have any geometric oranalytic improvements over the two-sided Ricci curvature bound discussed in thisbook. It will only turn out to be useful regarding regularity questions. However,sometimes we will assume the Einstein condition in lieu of a more general andsufficient two-sided Ricci curvature bound in order to simplify some of our proofs.

Let us now discuss in a broad sense, what type of questions we will consider inthis book. Given a condition on the Ricci curvature, as mentioned above, we mayask:

• What geometric properties are implied by such a condition?• What topological restriction does such a condition impose on the under-

lying manifold?• Are there specific, precise geometric bounds (such as bounded volume or

diameter)?• For which geometries does equality occur for such bounds? Results of

this type are called rigidity results.• Are geometries for which we observe almost equality close to geometries

for which exact equality holds? Results of this type are called almostrigidity or stability results.• What are the analytic properties of spaces with bounded Ricci curvature?

Can we find uniform constants for analytic bounds such as: isoperimetricinequalities, Poincare and Sobolev inequalities, heat kernel estimates, andgradient estimates for harmonic functions?• How do spaces with bounded Ricci curvature degenerate? What is the

structure of the limit space?

Perhaps one of the most intriguing application of the theory of spaces withbounded Ricci curvature is the following integral curvature bound for two-sidedRicci curvature bounds:

THEOREM 1.3 (Jiang and Naber, 2016 [20]). For any dimension n and v > 0there is a constant C = C(n, v) <∞ such that the following holds:

Let (M, g) be a compact n-dimensional Riemannian manifold p ∈ M a pointand assume that

volB(p, 1) ≥ v and − g ≤ Ric ≤ g on M.

Then

(1.1)ˆB(p,1)

|Rm|2 < C.

1.3. REGULARITY OF DISTANCE FUNCTIONS 3

The proof of this bound combines many of the geometric and analytic resultsthat we will discuss in this book.

1.2. Lohkamp’s construction

The purpose of this section is to convey some of the ideas that were used in theproof of Theorem 1.1. We will sketch the proof of the following simplified result:

THEOREM 1.4. There is a metric g on R3 such that

(a) g = geucl on R3 \K for some compact K ⊂ R3,(b) Ricg ≤ 0 on R3, and(c) Ric < 0 at one point.

Sketch of the proof of Theorem 1.4.

1.3. Regularity of distance functions

In this book, we will frequently employ analytic arguments involving distancefunctions. We will now explain how we will deal with issues arising from the factthat such distance functions are not smooth everywhere.

1.3.1. The cut locus and injectivity domain. Let (M, g) be a complete Rie-mannian manifold, and let p ∈ M be a point. For every tangent vector v ∈ TpMat p, let γv : R→M be the (constant speed) geodesic with γ′v(0) = v. Note that

γav(t) = γv(at), a, t ∈ R.

The exponential map at p is defined as

(1.2) expp : TpM −→M, v 7−→ γv(1).

The Hopf–Rinow theorem implies that

expp(TpM) = M.

DEFINITION 1.5. The cut locus Cp ⊂M of p is defined by

M \ Cp = x ∈M : there is a minimizing geodesic γ : [0, l + δ]→M ,

δ > 0, such that γ(0) = p and γ(l) = x .

Define moreover the injectivity domain of expp by

(1.3) Dp := v ∈ TpM : γv : [0, 1 + δ]→M is minimizing for some δ > 0.

Lastly, set

(1.4) C :=⋃p∈Mp × Cp ⊂M ×M, D :=

⋃p∈M

Dp ⊂ TM.


By an arc-length geodesic we mean a geodesic that is parametrized by arclength, that is, has unit speed. We now recall the following result regarding the cutlocus and injectivity domain.

THEOREM 1.6 (Properties of the cut locus and injectivity domain). For anyp ∈M the following holds:

(a) x ∈ M \ Cp if and only if there exists a unique minimizing arc-lengthgeodesic γ : [0, l] → M with γ(0) = p, γ(l) = x, and p, x are notconjugate along γ.

(b) (x, y) ∈ C⇐⇒ (y, x) ∈ C.(c) D is open and C is closed.(d) expp |Dp : Dp →M \ Cp is a diffeomorphism.

PROOF. Let us first prove the “only if” part of assertion (a). Let x ∈M − Cp.So there is a minimizing geodesic γ : [0, l + δ] → M with γ(0) = p, γ(l) = x,and δ > 0. Assume that there is a minimizing arc-length geodesic σ : [0, l] → M

with σ(0) = p and σ(l) = x. We want to prove that σ = γ|[0,l]. To see this, let γbe the concatenation of σ and γ|[l,l+δ]. Then `(γ) = l+ δ = `(γ) = d(p, γ(l+ δ)).So γ is a length-minimizing curve and must therefore be a smooth geodesic. Soσ′(l) = γ′(l), from which it follows that σ = γ|[0,l]. So γ|[0,l] is the uniqueminimizing geodesic between p and x. The fact that p and x are not conjugatealong γ follows by a standard argument involving the second variation formula.

Next, we show that D is open in TM . To see this, we argue by contradiction.Assume that there is some lv ∈ Dp ⊂ D with |v| = 1, l ≥ 0 that is not in theinterior of D. Set x := γv(l). So we can find a sequence of vectors livi ∈ TM \D,|vi| = 1, li ≥ 0 such that livi → lv. Denote by pi := π(vi) the basepoints of thesevectors. Fix a sequence of positive numbers δi → 0 and set x′i := γvi(li).

Note that pi → p and x′i → x as i → ∞. Since livi 6∈ Dpi , there is an arc-length geodesic minimizing geodesic σi : [0, l′i] → M between pi and x′i. So l′i <li+δi. On the other hand, li = d(pi, x′i) ≥ d(p, x)−d(p, pi)−d(x, x′i)→ d(p, x).So limi→∞ l

′i = l.

Let ui := σ′i(0). So σi = γui |[0,l′i]. By compactness, and after passing to asubsequence, we may assume that ui → u ∈ TpM , |u| = 1. Then σi = γui |[0,l′i]converges to γu|[0,l] uniformly. So γu|[0,l] is a minimizing geodesic and

γu(l) = limi→∞

γui(l′i) = limi→∞

x′i = x.

It follows that γu|[0,l] is a minimizing geodesic between p and x. Using the partof assertion (a) that we have already proven, we know that γv|[0,l] is the uniqueminimizing geodesic between p and x. So γu|[0,l] = γv|[0,l] and therefore u = v.So it follows that vi → v and therefore that γui |[0,l′i] → γv|[0,l] as i→∞.


By the part of assertion (a) that we have already established, we also know thatp and x are not conjugate along γ. So the differential (d expp)lv is non-degenerate.So if we consider the map F : TM → M × M , w 7→ (p, exp(w)), then thedifferential dFlv is non-degnerate as well. So, by the Inverse Function Theorem, Fis a local diffeomorphism at lv. As (li + δi)vi and l′ivi both converge to lv and

F((li + δi)vi

)= (pi, x′i) = F

(l′iui

),

it follows that (li + δi)vi = l′iui for large i, which contradicts l′i < li. This showsthat D is open.

The fact that C is closed can be seen as follows. Consider again the map Ffrom the last paragraph and observe that by definition F (D) = M \ C. Also,by the part of assertion (a) that we have already established, we know that thedifferential d expp is non-degenerate on Dp for all p ∈ M . So, as argued in thelast paragraph, we know that dF is non-degenerate on D. It follows that F is anopen map and therefore M \ C is open. So C is closed, which finishes the proof ofassertion (c).

To see assertion (d), observe that Dp = D ∩ TpM is open and p × Cp =C∩p×M is closed. So it remains to show that expp |Dp is injective. This followsimmediately from the direction of assertion (a) that we have already established.

Let us now prove the “if” direction of assertion (a). To see this, let us inspectthe proof of the fact that D is open. It turns out that the only way that we haveused the property lv ∈ Dp ⊂ D was using the “only if” direction of assertion (a).So the proof also works in the case in which we only assumed that γv|[0,l] is theunique minimizing geodesic between p and x and that p and x are not conjugatealong γv|[0,l]. So that for all v ∈ TpM , |v| = 1, l ≥ 0 the following holds: If γv|[0,l]is the unique arc-length geodesic minimizing geodesic between its endpoints andif its endpoints are not conjugate to one another, then lv ∈ Dp, and consequentlyexpp(lv) = γv(l) ∈ M \ Cp. This prove assertion (a) and finishes the proof of thetheorem.

As corollaries, we obtain:

COROLLARY 1.7. The distance function rp(x) := d(x, p) is smooth on M \(Cp ∪ p).

COROLLARY 1.8. The injectivity radius satisfies inj(x) = d(x,Cx). This im-plies that the injectivity radius is a continuous function.

We also recall:

THEOREM 1.9. Cp has measure zero.


PROOF. Let x ∈ Cp ∩ B(p,R) for some R > 0. There is a v ∈ TpM ,v ∈ B(0p, R) such that γv|[0,1] is a minimizing geodesic between p and x =expp(v) = γv(1). So (1 + δ)−1v ∈ Dp for any δ > 0. This shows that

Cp ∩B(p,R) ⊂ expp( ⋂δ>0

((1 + δ)

(Dp ∩B(0p, R)

)\(Dp ∩B(0p, R)

))).

Since Dp ∩B(0p, R) is star-shaped, we know that

Dp ∩B(0p, R) ⊂ (1 + δ)(Dp ∩B(0p, R)

).

So it is easy to see that⋂δ>0

((1 + δ)

(Dp ∩B(0p, R)

)\(Dp ∩B(0p, R)

))has measure zero. It follows that Cp ∩ B(p,R) and therefore Cp have measurezero.

1.3.2. Upper Hessian bounds and the existence of the weak Laplacian. Ifh ∈ C∞(U), where U is an open subset of M , then its Hessian is ∇2h = ∇∇hand its Laplacian is 4h = tr(∇2h), where ∇ denotes the Riemannian covariantderivative. Denote ∇2

v,wh = (∇2h)(v, w). Then4h =∑ni=1∇2

ei,eih, where eiis any orthonormal frame field on U .

By Corollary 1.7, the Laplacian 4rp is well-defined on M \ (Cp ∪ p). Itturns out that we can make sense of this Laplacian on M \ p if we allow4rp tobe a signed measure of locally finite variation.

To this end, we define:

DEFINITION 1.10 (Upper Hessian bound). Let p ∈ M and v ∈ TpM . Acontinuous function b ∈ C0(M) is said to satisfy ∇2

v,vb < C in the barrier sense(at p) if there is a neighborhood p ∈ U ⊂ M and a barrier ϕ ∈ C2(U) such thatb ≤ ϕ on U and b(p) = ϕ(p) and ∇2

v,vϕ(p) < C.

Similarly, we can define lower Hessian bounds.Let us note one important observation: Assume that b is a-Lipschitz, i.e.,

|b(x)− b(y)| ≤ ad(x, y), x, y ∈M,

for some a ≥ 0 and consider a barrier ϕ for b at p ∈ U . Then we claim that|∇ϕ(p)| ≤ a. This can be seen as follows: If ∇ϕ(p) = 0, then there is nothing toshow. So assume that ∇ϕ(p) 6= 0 and set v := 1

|∇ϕ(p)|∇ϕ(p). Choose a geodesicσ : (−ε, ε)→M with σ′(0) = v. Then t 7→ ϕ(σ(t)) is a-Lipschitz. The derivativeof this function at 0 is equal to 〈v,∇ϕ(p)〉 = |∇ϕ(p)|, which proves our claim.


LEMMA 1.11. Let F : (0,∞)→ R be a non-decreasing function of regularityC2. Set b(x) := F (d(p, x)) for some p ∈ M . Then for any compact subsetK ⊂ M \ p there is a CK such that we have ∇2

v,vb < CK in the barrier sensefor all z ∈ K and v ∈ TzM , |v| = 1.

Analogously, if F is non-increasing instead of non-decreasing, then we obtaina lower Hessian bound on b.

PROOF. Choose C such that |sec| < C on a tubular 1-neighborhood B(K, 1)around K and choose

0 < r0 < min1, d(p,K), (10C)−1

such that r0 < inj(x) for all x ∈ K. Note that inj(x) := d(x,Cx) is continuous,since C is closed.

Let now z ∈ K and choose a minimizing arc-length geodesic

γ : [0, l]→M

from p to z. Set q := γ(l − r0) and

ϕ(x) := F(d(p, q) + d(q, x)

).

Then ϕ(z) = b(z) and by the triangle we have b ≤ ϕ everywhere. So ϕ is a barrierfor b at z. By a standard Jacobi field estimate (see for example Theorem 10.1) andthe injectivity radius bound, we obtain that

∇2v,vϕ ≤ ∇2

v,vd(p, ·) · F ′ +(∇vd(p, ·)

)2F ′′ ≤ r−1

0 F ′ + |F ′′|.

This proves the desired bound.

We can now show the existence of the weak Laplacian.

PROPOSITION 1.12 (Existence of weak Laplacian). Let (Mn, g) be a Rie-mannian manifold and consider a function b : M → R that is locally Lipschitzand whose Hessian is locally uniformly bounded from above in the barrier sense.

Then the gradient ∇b exists almost everywhere and for any vector field Z ∈C1c (M ;TM) we have

(1.5)ˆM〈Z,∇b〉dg = −

ˆM

(divZ)b dg.

Moreover, there is a unique signed measure µ4b on M0 of locally finite totalvariation such that for any compactly supported ϕ ∈ C2

c (M0) we have(1.6)

Mb(x)4ϕ(x)dg(x) = −

ˆM〈∇b(x),∇ϕ(x)〉dg(x) =

ˆMϕ(x)dµ4b(x).

Also, µ4b can be expressed as the difference of its positive and negative part,µ4b = (µ4b)+ − (µ4b)−. The positive part (µ4b)+ is absolutely continuous


with respect to dg. This implies in particular that µ4b can be decomposed as thesum µ4b,a.c. + µ4b,sing of an absolutely continuous and a singular part that ispurely non-positive.

Lastly, dµ4b = (4b)dg wherever b is C2.

PROOF. Let U ⊂ M be an open subset that is relatively compact in M andon which we can find coordinates (x1, . . . , xn) : U → V ⊂ Rn. Moreover, letU ′ ⊂ U be an open subset such that U ′ ⊂ U , and let U ′′ ⊂ U ′ be an open subsetsuch that U ′′ ⊂ U ′. We will first construct µ4b on U ′′ such that (1.6) holds for allϕ ∈ C2

c (U ′′).Express b = b(x1, . . . , xn) in terms of the coordinates (x1, . . . , xn). We claim

that for some sufficiently large constant C1 <∞ the function

(1.7) b∗(x1, . . . , xn) := b(x1, . . . , xn)− C1((x1)2 + · · ·+

(xn)2)

is concave in the sense that for any segment σ : [0, 1]→ Rn

σ(t) =((1− t)x1 + ty1, . . . , (1− t)xn + tyn

)whose image is contained in V we have

(1.8) b∗(σ(t)) ≥ (1− t) · b∗(σ(0)) + t · b∗(σ(1)).

To see this, observe first that ∇2b∗ < −C∗(C1)g in the barrier sense, whereC∗(C1) → ∞ as C1 → ∞. Now consider such a segment σ and let us showthat (1.8) holds whenever C1 is chosen large enough depending on some constantsthat don’t depend on σ. The proof is by contradiction. Assume that (1.8) wasviolated for some t ∈ [0, 1]. Then we can choose t0 ∈ [0, 1] such that

b∗(σ(t0))− (1− t0) · b∗(σ(0))− t0 · b∗(σ(1))

is minimal. Obviously, t0 ∈ (0, 1). Let ϕ be a barrier for b′ at σ(t0) with∇2σ′(t0),σ′(t0)ϕ < −C

∗|σ′(t0)|2. Then for t near t0 we have

b∗(σ(t0))− (1− t0) · b∗(σ(0))− t0 · b∗(σ(1))≤ b∗(σ(t))− (1− t0) · b∗(σ(0))− t · b∗(σ(1))≤ ϕ(σ(t))− (1− t) · b∗(σ(0))− t · b∗(σ(1)),

with equality for t = t0. So t0 is a local minimum for

t 7→ ϕ(σ(t))− (1− t) · b∗(σ(0))− t · b∗(σ(1)).

Therefore,

0 = (ϕ σ)′(t0) = ∇σ′(t0)ϕ+ b∗(σ(0))− b∗(σ(1)),


and

0 ≤ (ϕ σ)′′(t0)(1.9)

= ∇2σ′(t),σ′(t)ϕ+ 〈∇σ′(t0)σ

′(t0),∇ϕ〉

< −C∗|σ′(t0)|2 − C ′|σ′(t0)|2 · |∇ϕ(σ(t0))|

for some C ′ <∞ that only depends on coordinate chart (x1, . . . , xn). Since |∇ϕ|is bounded by the Lipschitz constant of ϕ, it follows that the right-hand side of(1.9) can be chosen negative if C1 is chosen large enough, giving us the desiredcontradiction. So b∗ is in fact concave.

By convoluting b∗ with a kernel, we can construct local smoothings b∗ε ∈C∞(U ′), ε > 0 of b∗ on U ′ such that the following holds for some uniform con-stant C∗2 <∞:

(i) limε→0

b∗ε = b∗ uniformly on U ′.

(ii)∂b

∂xi→ ∂b∗

∂xiin L1(U ′) as ε→ 0.

(iii)∣∣∣∣∂b∗∂xi

∣∣∣∣ < C∗2 and b∗ε is concave on U ′. As b∗ε is smooth, this implies that

(∂2b∗ε∂xi∂xj

)ni,j=1

≤ 0

on U ′ for all ε > 0.

Using the fact that items (ii) and (iii) imply∣∣∣∣∇2∂

∂xi, ∂

∂xjb∗ε −

∂2b∗ε∂xi∂xj

∣∣∣∣ =∣∣∣∣⟨∇ ∂

∂xi

∂∂xj

,∇b∗ε⟩∣∣∣∣,

we conclude that

(ii’) ∇b∗ε → ∇b∗ in L1(U ′) as ε→ 0.(iii’) |∇b∗ε| < C∗∗2 and ∇2b∗ε < C∗∗2 on U ′ for all ε > 0, for some uniform

C∗∗2 <∞.

Set now

bε(x1, . . . , xn) := b∗ε(x1, . . . , xn) + C1((x1)2 + · · ·+ (xn)2).

Then the following holds for some uniform C2 <∞:

(i) limε→0 bε = b uniformly on U ′,(ii) ∇bε → ∇b in L1(U ′) as ε→ 0.

(iii) |∇bε| < C2 and4bε < C2 on U ′ for all ε > 0. Note that in the followingwe will only need an upper bound on the Laplacian of bε.


We will now work with this family of smoothings of b. We first conclude thatfor any Z ∈ C1

c (U ′;TM0),ˆM〈Z,∇b〉dg = lim

ε→0

ˆM〈Z,∇bε〉dg = − lim

ε→0

ˆM

(divZ)bεdg = −ˆM

(divZ)bdg.

This shows (1.5) on U ′ and by setting Z = ∇ϕ, we obtain that the first part of (1.6)holds for all ϕ ∈ C2

c (U ′).Next, observe that for any ϕ ∈ C2

c (U ′) we have(1.10)ˆMb(x)4ϕ(x)dg(x) = lim

ε→0

ˆMbε(x)4ϕ(x)dg(x) = lim

ε→0

ˆM4bε(x)ϕ(x)dg(x).

Using the upper bound on the Laplacian of bε and (1.10), we find that wheneverϕ ≥ 0, we have

(1.11)ˆMb(x)4ϕ(x)dg(x) ≤ C2

ˆMϕ(x)dg(x) ≤ C2|U | ·max

Mϕ.

Let us now derive a lower bound for the integral on the left-hand side. To do this,we first choose and fix now a cutoff function ψ ∈ C2

c (U ′) such that 0 ≤ ψ ≤ 1everywhere and ψ ≡ 1 on U ′′. Set

C3 :=∣∣∣∣ ˆ

Mb(x)4ψ(x)dg(x)

∣∣∣∣.Then, whenever ϕ ∈ C2

c (U ′′) and ϕ ≥ 0, we have with A := maxM ϕ that

ˆMb(x)4ϕ(x)dg(x) = A

ˆMb(x)4ψ(x)dg(x)−

ˆMb(x)4

(Aψ(x)− ϕ(x)

)dg(x)

(1.12)

≥ AˆMb(x)4ψ(x)dg(x)− C2|U |max

M

(Aψ − ϕ) = −

(C3 + C2|U |

)maxM

ϕ.

Combining (1.11) and (1.12) implies that the functional

H : C2c (U ′′) −→ R, ϕ 7−→

ˆMb(x)4ϕ(x)dg(x)

satisfies

(1.13) |H(ϕ)| ≤ C4 maxM

ϕ

for all ϕ ∈ C2c (U ′′) with ϕ ≥ 0, where C4 < ∞ is a uniform constant. By

approximation, this implies that H can be extended to C0c (U ′′) and that (1.13)

for all ϕ ∈ C0c (U ′′). So, by the Riesz–Markov representation theorem, there is

a unique signed measure µ4b of finite total variation on U ′′ such that for all ϕ ∈C2c (U ′′) we have

(1.14)Û ′′b(x)4ϕ(x)dg(x) =

Û ′′ϕdµ4b.

1.4. THE LAPLACIAN COMPARISON THEOREM 11

The fact that (µ4b)+ is absolutely continuous with respect to dg can be seenas follows: By the first inequality of (1.11) and (1.14), we getˆ

U ′ϕdµ4b ≤ C2

Û ′′ϕ(x)dg(x).

So, again by the Riesz–Markov representation theorem, dµ4b + C2dg is a non-negative measure. It follows that d(µ4b)− ≤ C2dg.

Repeating the argument above on the elements U ′′i of an open cover U =U ′′i i∈I of M yields a signed measure µ4b of locally finite total variation onM0 such that (1.6) holds for all ϕ ∈ C2

c (U ′′i ). Using a partition of unity, we canthen establish (1.5) for all Z ∈ C1

c (M0;TM0) and (1.6) for all ϕ ∈ C2c (M0).

1.4. The Laplacian comparison theorem

Let us define the following sine-type function, which characterizes the growthof Jacobi fields in spaces of constant curvature k:

snk(t) =

1√−k sinh(t

√−k) if k < 0,

t if k = 0,1√k

sin(t√k) if k > 0.

(1.15)

We have that the associated cosine-type function

cnk(t) =

cosh(t

√−k) if k < 0,

1 if k = 0,cos(t

√k) if k > 0

(1.16)

satisfies

(1.17) sn′k = cnk and cn′k = −k snk .

Define the cotangent-type function

(1.18) Hk(t) = (n− 1)cnk(t)snk(t)

.

We can now derive the following estimate:

THEOREM 1.13 (Laplacian comparison for increasing functions of distance).Let (M, g) be a complete Riemannian manifold with Ric ≥ (n − 1)k for somek ∈ R, and let p ∈M . Let F : (0,∞)→ R be a smooth, non-decreasing function.Then the function b(x) := F (d(x, p)) satisfies

(1.19) dµ4b ≤(Hk(d(x, p))F ′(d(x, p)) + F ′′(d(x, p))

)dg

on M \ p. If F is non-increasing, then the reverse estimate holds.


In particular, if F (x) = x, i.e., b(x) = d(x, p), then we have

(1.20) dµ4b ≤ Hk(d(x, p))dg.

PROOF. By Proposition 1.12, the positive part (µ4b)+ is absolutely continu-ous with respect to dg. As such, it suffices to show that (1.19) holds away from thecut locus Cp, which has measure 0. As x 7→ d(x, p) is smooth on M \ (Cp ∪ p),we can check easily that away from the cut locus

4b = 4d(·, p)F ′(d(·, p)) + |∇d(·, p)|2F ′′(d(·, p)).

The claim now follows from the fact that |∇d(·, p)|2 = 1 and

4d( ·, p) ≤ (n− 1)sn′k(d(x, p))snk(d(x, p)) .

For the last bound see [30, Lemma 34].

THEOREM 1.14 (Laplacian comparison for distance, I). Let rp = d(p, ·). IfRic ≥ (n− 1)k on M , then

(1.21) 4rp ≤ Hk(rp) = (n− 1) cnk(rp)snk(rp)

.

As we shall see in the next section, the right-hand side is equal to 4Mnkd(p, ·),

where Mnk is the model space of constant curvature k.

1.5. Distance functions and the geometry of their level sets

Let (M, g) be a Riemannian manifold, and let f ∈ C∞(M). Initially, we willassume the smoothness of f for simplicity. We say that f is a distance functionif |∇f | ≡ 1. Set N := ∇f . We use this notation since N is a normal vector fieldof the level sets of f . For example, given p ∈ M , f(x) = rp(x) = d(x, p) is adistance function whose level sets are call geodesic spheres.

LEMMA 1.15. We have ∇NN = 0. Moreover, the trajectories of N are of theform t 7→ exp(tNp) and are geodesics.

PROOF. For any v ∈ TpM ,

0 = v(|∇f |2

)= 2〈∇v∇f,∇f〉 = 2∇2

v,∇ff = 2〈∇∇f∇f, v〉.

1.5. DISTANCE FUNCTIONS AND THE GEOMETRY OF THEIR LEVEL SETS 13

1.5.1. Second fundamental form and mean curvature of geodesic spheres.For each s ∈ R, the level set Σs := f−1(s) ⊂ M is a regular hypersurface, andN |Σs is a unit normal vector field. The shape operator is defined by

(1.22) Ssv = ∇vN.

Observe that 〈Ssv,N〉 = 0. The second fundamental form is defined by

(1.23) As = 〈Ss · , ·〉.

If v, w ∈ TΣs, then As(v, w) = As(w, v). The mean curvature is the traceHs = trSs, where the traces over TM and TΣs are equal since∇NN = 0.

View H : M → R. We have that S and A are tensors on

T ‖M = v ∈ TM : 〈v,N〉 = 0 = ker df =⋃s∈R

TΣs ⊂ TM.

Recall that for a vector field V , its divergence is defined by

divV =n∑i=1〈∇eiV, ei〉,

where eini=1 is an orthonormal frame.

LEMMA 1.16. We have

(1.24) A = ∇2f |T ‖M , H = 4f = divN.

That is, the second fundamental form and mean curvature of the level sets of f aregiven by the Hessian and Laplacian of f , respectively.

PROOF. We compute that

A(u, v) = 〈∇u∇f, v〉 = ∇2u,vf.

This formula and its trace yield the lemma. Note that∇N,v = 〈∇NN, v〉 = 0.

1.5.2. The Jacobi and Riccati equations. Let γ : (a, b)→M be a trajectoryof N , that is, γ′(t) = N (γ(t)). Pick t0 ∈ (a, b) and v ∈ T

‖γ(t0)M . Choose a

smooth path σ : (−ε, ε)→M such that σ(0) = γ(t0) and σ′(0) = v. Set

γs(t) := exp((t− t0)Nσ(s)

).

Then γs(t) is a trajectory of N , and hence is a geodesic, so that

J(t) := ∂γs∂s

∣∣∣∣s=0

(t)

is a Jacobi field along γ with J(t0) = v and 〈J(t), γ′(t)〉 = 0.For a vector field or tensor field V (t) along γ(t), let V ′(t) = (∇NV )(t).


LEMMA 1.17. Let S = St. Then the Jacobi field evolves along γ by

S (J(t)) = J ′(t).


S (J(t)) = ∇J(t)∇f = D

∂s

∂γ

∂t

∣∣∣∣s=0

= D

∂t

∂γ

∂s

∣∣∣∣s=0

= J ′(t).

Let J ′′(t) = ∇N (∇NJ)(t). The Jacobi equation is

(1.25) J ′′(t) = R(N, J)N(t).

LEMMA 1.18 (Riccati equation). The shape operator evolves along γ by

(1.26) ∇NS + S2 +RN = 0,

where RN = R( ·, N)N .

PROOF. We compute, using the Jacobi equation and (1.17), that

−R(J(t), γ′(t)

)γ′(t) = J ′′(t)

= D

∂t(S(J(t)))

= S′ (J(t)) + S(J ′(t)

)= S′ (J(t)) + S2 (J(t)) .

Since, for any given t ∈ (a, b), J(t) may be chosen arbitrarily, the lemma follows.

LEMMA 1.19. The mean curvature evolves along γ by

(1.27) ∂NH + 1n− 1H

2 + Ric(N,N) ≤ 0,

with equality if and only if S = 1n−1H idT ‖M . Here, ∂NH is the directional

derivative N(H).

PROOF. Using the Riccati equation, we compute that

0 = trT ‖M(S′ + S2 +R(· , N)N

)= ∂NH + trT ‖M (S2) + Ric(N,N)

≥ ∂NH + (trS)2 + Ric(N,N),

sinceρ2

1 + · · ·+ ρ2n−1 ≥

1n− 1 (ρ1 + · · ·+ ρn−1)2 ,

where the ρi are the eigenvalues of the shape operator S and where equality holdsif and only if all of the ρi are equal.

1.5. DISTANCE FUNCTIONS AND THE GEOMETRY OF THEIR LEVEL SETS 15

1.5.3. Lower Ricci curvature bound comparison with model spaces. Aconsequence of Lemma 1.19 is:

LEMMA 1.20. Along the trajectory γ of N , if Ric ≥ (n− 1)k, then

(1.28) H ′ + 1n− 1H

2 + (n− 1)k ≤ 0.

Equality, at a point, holds if and only if S = 1n−1H idT ‖M and Ric(N,N) =

(n − 1)k at that point. Equality along γ implies that S ≡ 1n−1H idT ‖M and

RN ≡ k idT ‖M along γ.

From now on, f(x) := d(p, x), where p ∈ M , and γ : [0, `] → M withγ(0) = p. In this case, we have explicit comparisons and the model space hasconstant curvature k.

LEMMA 1.21. If (M, g) = Mnk , the model space of constant curvature k, then

(1.29) H ′ + 1n− 1H

2 + (n− 1)k = 0.

PROOF. We have on Mnk that H(t) = Hk(t) and that(

(n− 1)cnksnk

)′= −(n− 1)k snk

snk− (n− 1)cn2

k

sn2k

.

LEMMA 1.22 (Comparison principle). If ϕ,ψ ∈ C1 ((0, `)) satisfy

ϕ′ + 1n− 1ϕ

2 + (n− 1)k ≤ 0,(1.30)

ψ′ + 1n− 1ψ

2 + (n− 1)k = 0,(1.31)

and if limt0 ψ(t) = ∞, then ϕ ≤ ψ. Moreover (equality case), if ϕ(t) = ψ(t)for some t ∈ (0, `), then ϕ ≡ ψ on (0, t].

PROOF. Assume first that ϕ(t0) ≤ ψ(t0). Then we have the following stringof inequalities, each implying the next,

(ϕ− ψ)′ + 1n− 1(ϕ− ψ)(ϕ+ ψ) ≤ 0,(

(ϕ− ψ)(t) exp(

1n− 1

ˆ t

t0

(ϕ+ ψ)))′≤ 0,

ϕ(t) ≤ ψ(t) for all t ≥ t0.

Set ϕε(t) := ϕ(t+ε). Then ϕε(t0) ≤ ψ(t0) for sufficiently small t0, which impliesthat ϕε(t) ≤ ψ(t) for all t ∈ (0, `). By taking ε → 0, we obtain ϕ(t) ≤ ψ(t) fort ∈ (0, `).


1.5.4. Laplacian Comparison Theorem, II.

THEOREM 1.23 (Laplacian comparison for distance, II). Let γ : [0, `] → M

be a minimal arc-length geodesic emanating from p. Along the trajectory γ of N ,if Ric ≥ (n− 1)k, then

(1.32) H(t) ≤ Hk(t) = (n− 1) cnk(t)snk(t)

.

If equality holds for some t ∈ (0, `), then equality holds on (0, t]. Moreover, ifk > 0, then ` ≤ π√

k.

PROOF. Inequality (1.32) and its equality case characterization follow fromthe comparison principle and Lemma 1.21 since limt0Hk(t) =∞.

Now suppose that k > 0. We have limt→ π√

k

Hk(t) = −∞. Since γ being minimal

implies that H(t) is finite on (0, `), (1.32) implies that ` ≤ π√k

.

1.5.5. Myers’s Theorem.

COROLLARY 1.24 (Myers’s Theorem). If Ric ≥ (n− 1)k > 0 on M , then

diamM ≤ π√k.

PROOF. Every minimal geodesic on M has length at most π√k

.

1.6. Bishop–Gromov volume comparison theorem

One of the main tools in the study of spaces with lower Ricci curvature boundsis the volume comparison result due to Bishop and Gromov. For any k ∈ R definethe function

(1.33) vk(r) := nωn

ˆ r

0

(snk(t)

)n−1dt,

where ωn is the volume of unit ball in n-dimensional Euclidean space. Note thatvk(r) is equal to the volume of an r-ball in a model space of constant sectionalcurvature k; see (1.39) below.

1.6.1. A local version of the volume comparison theorem. Let ωg denotethe volume form of g and let L denote the Lie derivative.

LEMMA 1.25.

LNωg = Hωg.

1.6. BISHOP–GROMOV VOLUME COMPARISON THEOREM 17


LNωg = (divN)ωg = Hωg.

Indeed, if ei is an orthonormal frame field, then

(LNωg)(e1, . . . , en) =n∑i=1

ωg(e1, . . . , ei−1,∇eiN, ei+1, . . . , en)

=n∑i=1〈∇eiN, ei〉.

Consider the diffeomorphism

expp |Dp : Dp →M \ Cp,

where Dp is defined as in (1.3) and Cp ⊂ M is the cut locus of p. Given p ∈ M ,we have exponential coordinates defined by its inverse

(1.34) (expp |Dp)−1 : M − Cp → Dp ⊂ TpM ∼= Rn.

Let Jp : Dp → R be the Jacobian of the diffeomorphism expp |Dp . Let (r, θ),where r > 0 and θ ∈ Sn−1, be the spherical coordinates of points rθ in Dp−0n.Define λ : Dp → R by

(1.35) ωg = λ dr ∧ dvoln−1,

where dvoln−1 is the volume form on Sn−1 and where, by abuse of notation, ωgdenotes the pullback of ωg by expp |Dp to Dp. We have that

λ(r, θ) = rn−1 Jp(r, θ)

and that

(1.36) ∂rλ = Hλ.

We first state a more local version of the volume comparison result.

PROPOSITION 1.26 (Bishop–Gromov, local version). Let (M, g) be a com-plete Riemannian manifold with Ric ≥ (n − 1)k for some k ∈ R and let p ∈ M

be a point. Thenλ(r, θ)

snn−1k (r)

is non-increasing in r and λ(r, θ) ≤ snn−1k (r), where

we also require that 0 < r < π√k

. In other words, for any v ∈ TpM , |v| = 1, thequantity

t 7→ Jp(tv) · tn−1

(snk(t))n−1

is non-increasing in t as long as tv ∈ Dp.



(1.37)∂

∂rlog

(λ(r, θ)

snn−1k (r)

)= ∂rλ

λ(r, θ)− (n− 1)cnk(r)

snk(r)≤ 0,

where we used (1.36) and (1.32). That λ(r, θ) ≤ snn−1k (r) follows from this mono-

tonicity and the fact that limr0

λ(r, θ)rn−1 = 1.

Now

(1.38) |B(p, r)| =ˆ r

0

ˆSn−1

λ(r, θ)dθdr, where λ :≡ 0 on TpM −Dp.

On the model space Mnk , we have

Jk(r, θ) := JMnk (r, θ) = snn−1

k (r).

Thus, on Mnk , the r-ball volume is independent of p:

(1.39) vk(r) := |Bk(p, r)| = nωn

ˆ r

0snn−1k (r′)dr′,

where Bk = BMnk .

1.6.2. The Bishop–Gromov volume comparison theorem. The following isthe Bishop–Gromov volume comparison theorem.

THEOREM 1.27 (Bishop and Gromov). If Ric ≥ (n− 1)k, then the quantity

(1.40)|B(p, r)|vk(r)

is non-increasing in r, as long as 0 < r ≤ π√k

when k > 0. Moreover,

(1.41) |B(p, r)| ≤ vk(r).

If |B(p, r)| = vk(r) (equality case), then sec ≡ k at p.

PROOF. By (1.37) and the mean value theorem,

(1.42) λ(r, θ)ˆ r

0snn−1k (r′)dr′ ≤ snn−1

k (r)ˆ r

0λ(r′, θ)dr′.

By differentiating (1.38) and then integrating (1.42) over Sn−1, we have that(d

dr|B(p, r)|

) ˆ r

0snn−1k (r′)dr′ =

ˆSn−1

(λ(r, θ)dθ

ˆ r

0snn−1k (r′)dr′

)≤ snn−1

k (r)ˆ r

0

ˆSn−1

λ(r′, θ)dr′

= v′k(r)nωn

|B(p, r)|,


where we also used (1.39). This says thatd

dr

( |B(p, r)|vk(r)

)≤ 0. On the other hand,

limr→0

|B(p, r)|vk(r)

= limr→0

|B(p,r)|rn

vk(r)rn

= ωnωn

= 1.

Therefore,|B(p, r)|vk(r)

≤ 1.

If we have equality for some r > 0, that is,|B(p, r)|vk(r)

= 1, thenRN ≡ k idT ‖MonB(p, r)∩ (M − Cp − p). This implies thatR(u, v, v, u) = k for u, v ∈ TpMwith |u| = |v| = 1 and 〈u, v〉 = 0.

1.6.3. Bishop–Gromov relative volume comparison theorem. GeneralizingTheorem 1.27, we have the annulus (a.k.a. relative) version of the Bishop–Gromovvolume comparison theorem, which is the following:

THEOREM 1.28 (Relative volume comparison). Let (M, g) be a Riemannianmanifold with Ric ≥ (n− 1)k for some k ∈ R and let p ∈M be a point. Supposethat a collection of radii satisfy 0 ≤ r1 ≤ r2 and 0 ≤ r′1 ≤ r′2 with r1 ≤ r′1 and0 < r2 ≤ r′2, where we also require that r′2 <

π√k

when k > 0. Then

(1.43)|A(p, r1, r2)|vk(r2)− vk(r1) ≥

|A(p, r′1, r′2)|vk(r′2)− vk(r′1) .

Here,

A(p, r1, r2) := x ∈M : r1 < d(p, x) < r2

is an annulus centered at p and vk(r2)− vk(r1) = |AMnk (p, r1, r2)|.

PROOF. Recall that

expp |Dp : Dp −→M \ Cp

is a (bijective) diffeomorphism and that Cp has measure zero. Recall also thatJp : Dp → R is the Jacobian of expp, and set

J∗p(v) := Jp(v)χDp(v).

By Proposition 1.26, the quantity

(1.44) J∗p(v) · |v|n−1(snk(|v|)

)n−1

is non-increasing along radial lines. For any r > 0 we have

|A(p, r1, r2)| =Â(0,r1,r2)∩Dp

Jp(v)dv =Â(0,r1,r2)

J∗p(v)dv.


So, in order to verify the monotonicity of (1.40), it suffices to check that

1vk(r2)− vk(r1)

Â(0,r1,r2)

J∗p(v)dv

is non-increasing in r1 and r2. For this, it is enough to show that for any v ∈ TpM ,|v| = 1, the quantity

1vk(r2)− vk(r1)

ˆ r2

r1

J∗p(tv)tn−1dt

is non-increasing in r1 and r2. The rest follows via Fubini’s theorem. To see themonotonicity of this quantity in r1, observe that its derivative in r1 can be boundedas follows:

− 1vk(r2)− vk(r1) Jp(r1v)rn−1

1 + v′k(r1)(vk(r2)− vk(r1))2

ˆ r2

r1

J∗p(tv)tn−1dt

= nωn(vk(r2)− vk(r1))2

ˆ r2

r1

(−(

snk(t))n−1 J∗p(r1v)rn−1

1

+(

snk(r1))n−1 J∗p(tv)tn−1

)≤ 0.

Similarly, the derivative in r2 can be bounded as follows

1vk(r2)− vk(r1) J∗p(r2v)rn−1

2 − v′k(r2)(vk(r2)− vk(r1))2

ˆ r2

r1

J∗p(tv)tn−1dt

= nωn(vk(r2)− vk(r1))2

ˆ r2

r1

((snk(t)

)n−1 J∗p(r2v)rn−12

+(

snk(r2))n−1 J∗p(tv)tn−1

)≤ 0.

This shows the monotonicity of (1.43).

Note that Theorem 1.28 implies

(1.45)|A(p, r1, r2)|vk(r2)− vk(r1) ≤

|B(p, r2)|vk(r2) ≤ |B(p, r1)|

vk(r1) .

Lastly, we will analyze the case in which the quantity (1.40) changes only verylittle in r. In this case, we can derive an L1-bound on the Laplacian of the distancefunction. For this purpose, define the function

Fk(r) := 1 + (n− 1)ˆ r

1

dt

(snk(t))n−1 .

Note that Fk is increasing in r and

F0(r) = −r2−n,


which is, up to a multiplicative constant, the Green’s kernel in Euclidean space.For general k ∈ R, it is not hard to check that

(n− 1)sn′k(r)snk(r)

F ′k(r) + F ′′k (r) = 0.

So Fk is characterizes the Green’s kernel in the model space of constant curvaturek. Using Theorem 1.13, we obtain moreover that on any Riemannian manifoldwith Ric ≥ k, we have

dµ4(Fkd( · ,p)) ≤ 0.We can now estimate the mass of dµ4(Fkd( · ,p)) in terms of the decrease of thequantity (1.40).

PROPOSITION 1.29. Let (M, g) be a complete Riemannian manifold with Ric ≥(n− 1)k for some non-positive k ≤ 0, let p ∈M be a point, and let 0 < r1 < r2.Set b(x) := Fk(d(x, p)). Then dµ4b ≤ 0 and

−Â(p,r1,r2)

dµ4b ≤ C( |B(p, r1)|

vk(r1) − |B(p, 2r2)|vk(2r2)

).

Here C is a universal constant.

PROOF. Let

H(r) :=

vk(r)vk(r1) if 0 ≤ r ≤ r1,

1 if r1 ≤ r ≤ r2,vk(2r2)− vk(r)vk(2r2)− vk(r2) if r2 ≤ r ≤ 2r2,

0 if 2r2 ≤ r.

and note that H : [0,∞)→ [0, 1] is continuous. Define

f(x) := H(d(x, p)).

We can then compute that, using (1.33),

−ˆB(p,2r2)

fdµ4b =ˆB(p,2r2)

∇f · ∇b dg

=ˆB(p,2r2)

n− 1(snk(d(x, p)))n−1 H

′(d(x, p))dg(x)

= n(n− 1)ωnvk(r1)

ˆB(p,r1)

(snk(d(x, p)))n−1

(snk(d(x, p)))n−1 dg(x)

− n(n− 1)ωnvk(2r2)− vk(r2)

Â(p,r2,2r2)

(snk(d(x, p)))n−1

(snk(d(x, p)))n−1 dg(x)

= n(n− 1)ωn( |B(p, r1)|

vk(r1) − |A(p, r2, 2r2)|vk(2r2)− vk(r2)

).


The last term can be estimated as follows:

|B(p, r1)|vk(r1) − |A(p, r2, 2r2)|

vk(2r2)− vk(r2)

= vk(2r2)vk(2r2)− vk(r2)

|B(p, r1)|vk(r1) − vk(r2)

vk(2r2)− vk(r2)|B(p, r1)|vk(r1) − |A(p, r2, 2r2)|

vk(2r2)− vk(r2)

≤ vk(2r2)vk(2r2)− vk(r2)

|B(p, r1)|vk(r1) − |B(p, r2)|

vk(2r2)− vk(r2) −|A(p, r2, 2r2)|vk(2r2)− vk(r2)

≤ vk(2r2)vk(2r2)− vk(r2)

( |B(p, r1)|vk(r1) − |B(p, 2r2)|

vk(2r2)

).

Since we assumed that k ≤ 0, we have, by volume comparison, that

(2r2)n

vk(2r2) ≤rn2

vk(r2) =⇒ vk(r2) ≤ 2−nvk(2r2).

So

|B(p, r1)|vk(r1) − |A(p, r2, 2r2)|

vk(2r2)− vk(r2) ≤1

1− 2−n( |B(p, r1)|

vk(r1) − |B(p, 2r2)|vk(2r2)

).

The claim now follows from the facts that dµ4b ≤ 0 and f ≡ 1 on A(p, r1, r2).

1.6.4. Volume and diameter rigidity.

COROLLARY 1.30. Let (M, g) be a complete Riemannian manifold with Ric ≥(n− 1)k > 0. Then

volM ≤ volMnk ,

with equality if and only if (M, g) = Mnk .

PROOF. By Myers’s Theorem, diamM ≤ π√k

. Thus

(1.46) volM =∣∣∣∣B (p, π√

k

)∣∣∣∣ ≤ vk ( π√k

)=∣∣∣∣Bk (p, π√

k

)∣∣∣∣ = volMnk .

If volM = volMnk , then equality holds in (1.46) for all p ∈ M , which implies

that sec ≡ k for all p. It then follows that (M, g) = Mnk , whereas the converse is

evident.

As an application, we have:

THEOREM 1.31 (Diameter rigidity). Let (M, g) be a complete Riemannianmanifold with Ric ≥ (n− 1)k > 0. If diamM = π√

k, then (M, g) = Mn

k .

PROOF. Without loss of generality, we may assume that k = 1. Then, bythe compactness of M , there exist p, q ∈ M such that d(p, q) = π. For any


0 < r < π√k

, B(p, r) ∩ B(q, π − r) = ∅. Therefore, by the Bishop–Gromovvolume comparison theorem,

|B(p, r)|v1(r) ≥ volM

v1(π) ≥|B(p, r)|v1(π) + |B(q, π − r)|

v1(π) .

On the other hand, by the annulus version of the Bishop–Gromov volume compar-ison theorem,

|B(p, r)|v1(r) ≥ |B(q, π − r)|

v1(π)− v1(r) = |B(q, π − r)|v1(π − r) ,

so thatvolMv1(π) = lim

r→π|B(p, r)|v1(r) ≥ lim sup

r→π

|B(q, π − r)|v1(π − r) = 1.

The theorem now follows from the equality case of the Bishop–Gromov volumecomparison theorem.

1.6.5. Further comparison results.

THEOREM 1.32. If k1 ≤ sec ≤ k2, then for all x ∈ M − Cp and v ∈ TxMwith 〈v,∇r〉 = 0 and |v| = 1,

cnk2(r)snk2(r) ≤ ∇

2u,vr ≤

cnk1(r)snk1(r) .

This theorem follows from:

LEMMA 1.33 (Comparison principle for matrices). If Φ ∈ C1((0, `),Rm×m,sym

),

Ψ ∈ C1 ((0, `)),

Φ′ + Φ2 + kIm≤ 0 (case 1),≥ 0 (case 2),

Ψ′ + Ψ2 + k = 0,

limt→0 Ψ(t) =∞ (case 1), and limt→0 λ1(Φ(t)) =∞ (case 2), where λ1 denotesthe smallest eigenvalue, then

Φ(t) ≤ Ψ(t)Im (case1),≥ Ψ(t)Im (case2).

PROOF. Assume that Φ(t0) ≤ Ψ(t0)Im,≥ Ψ(t0)Im.

Fix t1 > t0. Since

(Φ−ΨIm)′ + (Φ−ΨIm)(Φ + ΨIm) ≤ 0,≥ 0,

we have for 0 < t ≤ t1 that((Φ−ΨIm)(t)e∓Ct

)′ ≤ 0,≥ 0


for C sufficiently large. Therefore,

Φ(t1)−Ψ(t1)Im≤ 0,≥ 0.

The remainder of the argument is analogous to the scalar comparison principle.

THEOREM 1.34. If sec ≤ k and 0 < r < inj(p), then

(1.47) |B(p, r)| ≥ vk(r)

and|B(p, r)|vk(r)

is non-decreasing in r.

We leave it to the reader to consider the example of the torus.

1.6.6. Further application of Bishop–Gromov. Let(1.48)

M(n, k, v,D) := (Mn, g) : Ric ≥ (n− 1)k, volM > v, diamM < D.

THEOREM 1.35 (Anderson [1]). There are only finitely many isomorphismclasses for π1(M), where M ∈M(n, k, v,D).

PROOF. See Petersen’s book [30, Theorem 64].

1.7. Cheeger–Gromoll splitting

Fill in.

1.8. Notes and commentary

For excellent treatises on Riemannian geometry and geometric analysis, seeCheeger and Ebin [8], Li [22], Petersen [30], and Schoen and Yau [32].

1.9. Exercises

CHAPTER 2

Metric Spaces and Gromov–Hausdorff Distance

The main purpose of this chapter is to introduce some basic knowledge ofGromov–Hausdorff distance.

2.1. Hausdorff distance

DEFINITION 2.1 (Metric space). Let X be a set, and let d : X ×X → R≥0 bea function. We say (X, d) is a metric space with metric d if the function d satisfies:

(1) For all x, y ∈ X , d(x, y) = d(y, x) ≥ 0 and d(x, y) = 0 if and only ifx = y.

(2) (Triangle inequality) For all x, y, z ∈ X , d(x, y) ≤ d(x, z) + d(y, z).

EXAMPLE 2.2. Euclidean space Rn, with the standard norm, is a metric space.A Riemannian manifold is a metric space.

A metric space (X, d) is totally bounded if for any ε > 0 there exists a finite ε-net, i.e., a finite number of points x1, . . . , xN inX such thatX =

⋃Ni=1B(xi, ε).

DEFINITION 2.3. A map bewteen metric spaces Φ : (X1, d1) → (X2, d2) isan isometry if d2(Φ(x),Φ(y)) = d1(x, y) for all x, y ∈ X1. Two metric spaces(X1, d1) and (X2, d2) are isometric if there exists an isometry between them. Inthis case, we write X1 ∼= X2 or (X1, d1) = (X2, d2).

Let us now define the Hausdorff distance in a metric space:

DEFINITION 2.4. Let (X, d) be a complete metric space, and let Z,W ⊂ X

be two bounded closed subsets. Define their Hausdorff distance (H-distance) by:

dH(Z,W ) := inf ε > 0 : Z ⊂ B(W, ε) and W ⊂ B(Z, ε),(2.1)

where B(Z, r) := x ∈ X : d(x, Z) < r is the r-neighborhood of Z. IfZ = p, then we call B(p, r) := B(p, r) an r-ball centered at p.

EXAMPLE 2.5. Let Z = [0, 1] ⊂ R, and letW = [1, 2] ⊂ R, with the standardmetric. Then dH(Z,W ) = 1.

Let

(2.2) C(Z) :=A ⊂ Z : A compact and A 6= ∅

.

25

26 2. METRIC SPACES AND GROMOV–HAUSDORFF DISTANCE

LEMMA 2.6. (C(Z), dH) is a metric space.

PROOF. Exercise 2.1.

We remark that in the lemma above we can also replace C(Z) by the set of allnon-empty closed subsets of Z.

THEOREM 2.7. (C(Z), dH) is complete if and only if (Z, dZ) is complete.

PROOF. Let us first prove the “only if” direction, that is, we assume that(C(Z), dH) is complete and claim that (Z, dZ) must be complete as well. To seethis, note that for any x, y ∈ Z we have dH(x, y) = dZ(x, y). So, if xi∞i=1is a Cauchy sequence in Z, then the sequence of singleton sets xi∞i=1 is aCauchy sequence in C(Z). Therefore this sequence has a limit A ∈ C(Z), mean-ing that limi→∞ dH(xi, A) = 0. We claim that A consists of only one point. Tosee this, consider a1, a2 ∈ A. Then

dZ(a1, a2) ≤ dZ(a1, xi) + dZ(xi, a2) ≤ 2dH(A, xi)→ 0.

So dZ(a1, a2) = 0 and therefore a1 = a2. So A = a and it follows thatlimi→∞ dZ(xi, a) = 0.

Next, we prove the “if” direction. So assume that (Z, dZ) is complete and con-sider a Cauchy sequence Ai∞i=1 in C(Z). We claim that this sequence converges.As a Cauchy sequence with a convergent subsequence is convergent, we may passto a subsequence and assume without loss of generality that dH(Ai, Ai+1) < 2−i

for all i.We now claim that this sequence converges to

A∞ := z ∈ Z : for all r > 0 we have B(x, r) ∩Ai 6= ∅ for infinitely many i.

Fix some ε > 0. We first show that Ai is contained in a small tubular neigh-borhood of A∞ for large i. Choose N large enough such that dH(Ai, Aj) < ε

for all i, j ≥ N . Let z ∈ A∞. By the definition of A∞, we can find an i ≥ N

and an y ∈ Ai such that dZ(y, z) < ε. Given this y, for all j ≥ N we canuse dH(Ai, Aj) < ε to find an x ∈ Aj such that dZ(x, y) < ε. Therefore,dZ(x, z) < 2ε. It follows that A∞ ⊂ B2ε(Aj) for all j ≥ N .

Next, we show that Ai is contained in a small tubular neighborhood of A∞ forlarge i. Let j be large enough such that 2−j < ε and consider a point x ∈ Aj .Then, since dH(Ai, Ai+1) < 2−i for all i, we can find a sequence xj := x ∈Aj , xj+1 ∈ Aj+1, . . . such that d(xi, xi+1) < 2−i for i ≥ j. So xi∞i=j is aCauchy sequence and has a limit x∞ ∈ Z with d(xj , x∞) < 2−j + 2−j−1 + · · · <2ε. By the definition of A∞ we must have x∞ ∈ A∞. So, in conclusion, we havethat Aj ⊂ B(A∞, 2ε) for sufficiently large j.

2.2. GROMOV–HAUSDORFF DISTANCE 27

The last two paragraphs showed that limi→∞ dH(Ai, A∞) = 0. In order toshow that Ai converges in C(Z), we now need to show that A∞ ∈ C(Z). Theargument in the previous paragraph implies that A∞ 6= ∅ and by definition of A∞,it is clear that A∞ is closed. So it remains to show that A∞ is totally bounded.To see this, let ε > 0 and choose i large enough such that dH(Ai, A∞) < ε. AsAi is totally bounded, we can find a finite ε-net z1, . . . , zn ⊂ Ai. Pick pointsz′1, . . . , z′n ⊂ A∞ with dZ(zk, z′k) < ε. We claim that z′1, . . . , z′n is a 3ε-netfor A∞. To see this, consider a point y′ ∈ A∞. Choose y ∈ Ai with dZ(y, y′) < ε

and a k with dZ(y, zk) ≤ ε. Then

dZ(y′, z′k) ≤ dZ(y′, y) + dZ(y, zk) + dZ(zk, z′k) < 3ε,

as desired.

DEFINITION 2.8 (Isometric embedding). We say that (X, dX) isometricallyembeds in (Y, dY ), written as (X, dX) → (Y, dY ), if there exists an injective mapι : X → Y such that dX(x1, x2) = dY (ι(x1), ι(x2)) for all x1, x2 ∈ X .

LEMMA 2.9. Let (X, dX) and (Y, dY ) be two compact metric spaces. Thenthere exists a compact metric space (Z, dZ) such that X and Y both isometricallyembed in Z.

PROOF. Assume that diam(X),diam(Y ) ≤ D. LetZ = XtY be the disjointunion of X and Y . Define a “metric” on Z by:

• dZ(x1, x2) = dX(x1, x2) for all x1, x2 ∈ X .• dZ(y1, y2) = dY (y1, y2) for all y1, y2 ∈ Y .• dZ(x, y) = D for all x ∈ X and y ∈ Y .

We may check that dZ is indeed a metric on Z, in particular, that the triangleinequality holds. That X and Y both isometrically embed in Z is clear.

2.2. Gromov–Hausdorff distance

Let (X, dX) and (Y, dY ) be two compact metric spaces; we will define theirGromov–Hausdorff distance.

DEFINITION 2.10 (GH-distance). Let (X, dX) and (Y, dY ) be two compactmetric spaces. Their Gromov–Hausdorff (GH) distance is defined by

dGH((X, dX), (Y, dY )

):= inf

(Z,dZ)dH(X,Y ),(2.3)

where the infimum is taken over all (Z, dZ) such that X and Y both isometricallyembed in Z.

REMARK 2.11. By Lemma 2.9, a metric space (Z, dZ) such as in (2.3) alwaysexists. This means that the definition is well-defined.


REMARK 2.12. Often, we just write dGH(X,Y ), suppressing the metrics dXand dY in our notation.

REMARK 2.13. Note that an isometric embedding is a map φ : X → Z suchthat dZ(φ(x), φ(y)) = dX(x, y) for all x, y ∈ X . So, for example, the Riemannianisometric embedding S2 → R3 is not an isometric embedding in this sense.

REMARK 2.14. LetX = [0, 1] ⊂ R and Y = [1, 2] ⊂ R, each with the naturalinduced metric from R. Then dGH(X,Y ) = 0. Comparing with the Hausdorffdistance of X and Y , we see that the Gromov–Hausdorff distance is independentof the locations ofX and Y , while the Hausdorff distance depends on the locationsof X and Y .

EXAMPLE 2.15. Consider the torus X1 = S1(1) × S1(r), whose meridianshave lengths 2π and 2πr, and the circle X2 = S1. If we consider Z = X1,ι1 = idX1 and ι2 being an isometric embedding S1 → S1 × S1(r), then it is notnot hard to see from (2.3) that dGH(X1, X2) ≤ πr.

This is not the best possible isometric embedding. For example, consider themetric space Y = S1(r)t∗ where the distance between ∗ and any point in S1(r)is equal to π

2 r. Let Z = S1(1)× Y , let ι1 be the isometric embedding of S1(1)×S1(r) whose image is S1(1)× (Y \ ∗) and let ι2 be the isometric embedding ofS1(1) whose image is S1(1)× ∗. Then (2.3) implies that dGH(X1, X2) ≤ π

2 r.

LEMMA 2.16. If X1, X2 are isometric, then dGH(X1, X2) = 0.

PROOF. Let Φ : X1 → X2 be an isometry and, in the definition of GH dis-tance, consider the case of (Z, dZ) = (X2, d2), the embedding Φ : X1 → Z, andthe embedding idX2 : X2 → Z in (2.3).

While the definition of the Gromov–Hausdorff distance above seems very cleanfrom an abstract point of view, it is somewhat unhandy for actual computations.The following lemma shows that we can restrict our attention to a smaller collec-tion of metric spaces (Z, dZ).

LEMMA 2.17. For any ε > 0 there is a metric d12 on the disjoint unionX1tX2such that d12|Xk = dk, for k = 1, 2, and dH,d12(X1, X2) < dGH(X1, X2) + ε,where the Hausdorff distance is taken with respect to the metric d12.

Therefore,

dGH(X1, X2) = infdH,d′12

(X1, X2) : d′12 on X1 tX2, d′12|Xk = dk, k = 1, 2.

PROOF. Given any metric space (Z, dZ) and isometric embeddings ιk : Xk →Z, k = 1, 2, we can define d12(x1, x2) := dZ(ι1(x1), ι2(x2)) + ε for for x1 ∈ X1


and x2 ∈ X2, and d12(xk, x′k) := dk(xk, x′k) for xk, x′k ∈ Xk, k = 1, 2. It is nothard to see that d12 is a metric on X1 tX2.

For the last claim, observe that (Z = X1 tX2, d12) with the obvious embed-dings ιk : Xk → X1 tX2 are possibilities in (2.3).

LEMMA 2.18. The Gromov–Hausdorff distance dGH satisfies the triangle in-equality. That is, for any three metric spaces (X1, d1), (X2, d2), (X3, d3) we have

dGH(X1, X3) ≤ dGH(X1, X2) + dGH(X2, X3).

PROOF. The idea of the proof will be to construct a metric dZ on the spaceX1tX2tX3 that contains isometric copies of X1, X2, X3, and for which X1 andX3 are close in the Hausdorff distance. To do this, fix some ε > 0 and let d12 andd23 be the metrics on X1 tX2 and X2 tX3, respectively, from Lemma 2.17. Wenow define

dZ(x1, x2) := d12(x1, x2), dZ(x2, x3) := d23(x2, x3)

anddZ(xk, x′k) := dk(xk, x′k),

whenever xk, x′k ∈ Xk. Lastly, we define

dZ(x1, x3) := infy∈X2

(d12(x1, y) + d23(y, x3)

).

It is not difficult to check that dZ is a metric and that dH(X1, X2) ≤ dGH(X1, X2)+ε and dH(X2, X3) ≤ dGH(X2, X3) + ε in (Z, dZ). Therefore

dGH(X1, X3) ≤ dH(X1, X3) ≤ dH(X1, X2) + dH(X2, X3)≤ dGH(X1, X2) + dGH(X2, X3) + 2ε.

Letting ε→ 0 yields the desired inequality.

Next, we present another useful way to characterize the geometric closeness oftwo metric spaces.

DEFINITION 2.19. A map f : X1 → X2 between two metric spaces (X1, d1)and (X2, d2) is called an ε-isometry if

(1) |d2(f(x), f(y))− d1(x, y)| ≤ ε for all x, y ∈ X1 and(2) d2(z, f(X1)) ≤ ε for all z ∈ X2.

Sometimes we also call such map an ε-Gromov–Hausdorff approximated map(ε-GH map).

DEFINITION 2.20. Let (X, dX) and (Y, dY ) be two compact metric spacessuch that dGH(X,Y ) ≤ ε. Assume that h : X → R and g : Y → R are two


functions. We say |g − h| ≤ ε in the Gromov–Hausdorff sense if there exists anε-GH map f : X → Y such that

supx∈X|h− g f |(x) ≤ ε.(2.4)

Geometric closeness with respect to the ε-isometry characterization is equiva-lent to Gromov–Hausdorff closeness, as the following lemma shows:

LEMMA 2.21. For two metric spaces (X1, d1) and (X2, d2) the following aretrue:

(1) If dGH(X1, X2) < r, then there is a 2r-isometry f : X1 → X2.(2) If there is an r-isometry f : X1 → X2, then dGH(X1, X2) < 2r.

PROOF. For part (1), we can use Lemma 2.17 to find a metric d12 on X1 tX2such that

(2.5) dH(X1, X2) < r.

So, for any x ∈ X1, there is a y ∈ X2 such that d12(x, y) < r. Set f(x) = y.Doing this for all x ∈ X1 yields a map f : X1 → X2 with the property thatd12(f(x), x) < r. Now, for all x, x′ ∈ X we have

d2(f(x), f(x′)) ≤ d12(f(x), x) + d12(x, x′) + d12(x′, f(x′)) < d12(x, x′) + 2r.

Moreover, for any y′ ∈ X2 we can use (2.5) to find a y ∈ X1 such that d12(y, y′) <r. Then

d2(y′, f(X1)) ≤ d2(y′, f(y)) ≤ d12(y′, y) + d12(y, f(y)) < 2r.

For part (2), consider the following metric dZ on Z := X1 tX2: For xk, x′k ∈Xk, k = 1, 2, we set dZ(xk, x′k) := dk(xk, x′k), and for x1 ∈ X1 and x2 ∈ X2 weset

dZ(x1, x2) := d2(f(x1), x2) + ε.

It is again not hard to verify that dZ is a metric and that

dGH(X1, X2) ≤ dH(X1, X2) < 2r.

The following lemma plays a role in the proof of Gromov’s precompactnesstheorem. This lemma also gives us a rough idea of how to estimate the Gromov–Hausdorff distance.

LEMMA 2.22. Let (X = x1, . . . , xN, dX) and (Y = y1, . . . , yN, dY ) betwo finite metric spaces with distances dX(xi, xj) = dij and dY (yi, yj) = hij .Assume that

|dij − hij | ≤ ε for each pair (i, j), where 1 ≤ i, j ≤ N.(2.6)


Then

dGH((X, dX), (Y, dY )

)≤ ε.(2.7)

PROOF. Let Z = X t Y be the disjoint union of X and Y . Define a metric onZ by

• dZ(x1, x2) = dX(x1, x2) for all x1, x2 ∈ X .• dZ(y1, y2) = dY (y1, y2) for all y1, y2 ∈ Y .• dZ(xi, yj) = ε+ minkdik + hkj.

It is easy to check that dZ is a metric on Z. Since dZ(xi, yi) = ε, this implies that

dGH(X,Y ) ≤ ε.

Lastly, we mention another notion of geometric closeness.

LEMMA 2.23. For two totally bounded metric spaces (X1, d1) and (X2, d2)the following are true:

(1) If dGH(X1, X2) < r, then there are finite 3r-nets x1, . . . , xn ⊂ X1and y1, . . . , yn ⊂ X2 such that |d1(xi, xj) − d2(yi, yj)| < 2r for alli, j.

(2) If there are r-nets x1, . . . , xn ⊂ X1 and y1, . . . , yn ⊂ X2 such that|d1(xi, xj)− d2(yi, yj)| < 2r for all i, j, then dGH(X1, X2) < 4r.

PROOF. For part (1), we use Lemma 2.21 to find a 2r-isometry f : X1 → X2.Next, let x1, . . . , xn ⊂ X1 be an r-net. Then it is not hard to see that y1 :=f(x1), . . . , yn := f(xn) ⊂ X2 is a 3r-net with the desired properties.

For part (2), view x1, . . . , xn and y1, . . . , yn as metric spaces with theinduced metric and note that

dGH(X1, X2) ≤ dGH(X1, x1, . . . , xn

)+ dGH

(x1, . . . , xn, y1, . . . , yn

)+ dGH

(y1, . . . , yn, X2

)< 4r,

where we have used Lemma 2.22 in the last inequality.

As an application of ε-isometries, we prove the following lemma, which illus-trates that the Gromov–Hausdorff distance is a very weak distance.

LEMMA 2.24. If (X, dX) is totally bounded, then there is a sequence of finitemetric spaces (Y1, d1), (Y2, d2), . . . such that limi→∞ dGH(X,Yi) = 0.

PROOF. For any i choose a 1i -net x1, . . . , xni ⊂ X and consider the space

Yi := x1, . . . , xni with the restricted metric. Then the inclusion map Yi → X isa 1i -isometry.


Using the notion of ε-isometries, we can finally show the definiteness of theGromov–Hausdorff distance.

LEMMA 2.25. If (X1, d1) and (X2, d2) are compact and if dGH(X1, X2) = 0,then X1 ∼= X2.

PROOF. Since X1 is compact, it is separable and we can find a dense andcountable subset x1, x2, . . . ⊂ X1. By assumption we can choose 1

i -isometriesfi : X1 → X2. After passing to a subsequence via a diagonal argument, we canassume that h(xk) := limi→∞ fi(xk) exists for all k. Then d2(h(xk), h(xk′)) =limi→∞ d2(fi(xk), fi(xk′)) = d1(xk, xk′). So h : x1, x2, . . . → X2 is an iso-metric embedding and it can be seen easily that its image is dense in X2. So wecan extend h to an isometric embedding H : X1 → X2. Since X1 is compact, itsimage is closed. So, as the image of h is dense in X2, the map H is surjective.

EXAMPLE 2.26. Consider the following two (non-compact!) metric spaces:Let X1 be the union of isometric copies of intervals of the form [0, l] for all l ∈Q ∩ (0, 1], where we identify all intervals at 0. Similarly, let X2 be the unionof such intervals [0, l], but for l ∈ (

√2 + Q) ∩ (0, 1]. Then dGH(X1, X2) = 0

and X1, X2 are complete but X1 6∼= X2. This demonstrates the necessity of thecompactness assumption in the lemma above.

We now present the main result of this section. For this, denote by M the spaceof isometry classes of compact metric spaces. Then we have the following:

THEOREM 2.27. (M, dGH) is a complete metric space.

PROOF. It only remains to prove completeness. To this end, let

(X1, d1), (X2, d2), . . . ∈M

be a Cauchy sequence. Our goal will be to show that this sequence converges inM. As a Cauchy sequence with a convergent subsequence converges, we may passto a subsequence and henceforth assume that we have

dGH(Xi, Xi+1) < 2−i.

We now construct a metric space (Z, dZ) that contains the spaces Xi as iso-metric copies and in which these spaces form a Cauchy sequence with respect tothe Hausdorff distance. Then we use (2.7) to deduce convergence.

To do this, we apply Lemma 2.17 to find metrics di,i+1 on XitXi+1 such thatdi,i+1|Xi = di and di,i+1|Xi+1 = di+1 and

dH,di,i+1(Xi, Xi+1) < 2−i.

2.3. GROMOV’S PRECOMPACTNESS THEOREM 33

Set

Z := X1 tX2 t · · ·

and define dZ as follows: For any x, x′ ∈ Xi we set dZ(x, x′) := di(x, x′) and forxi ∈ Xi and xj ∈ Xj , i < j, let dZ(xi, xj) be the infimum of

di,i+1(xi, yi+1) + di+1,i+2(yi+1, yi+2) + · · ·+ dj−1,j(yj−1, xj)

over all yi+1 ∈ Xi+1, . . . , yj−1 ∈ Xj−1. It is straightforward to check that dZ isa metric on Z and that dZ |XitXi+1 = di,i+1. So X1, X2, . . . is a Cauchy sequencein Z.

Let now (Z, dZ), Z ⊂ Z, be the completion of (Z, dZ). Then X1, X2, . . . isstill a Cauchy sequence in Z, and by Theorem 2.7 this sequence has a limit, i.e.there is a compact X∞ ⊂ Z such that limi→∞ dH(Xi, X∞) = 0. It follows thatlimi→∞ dGH(Xi, X∞) = 0, which finishes the proof.

2.3. Gromov’s precompactness theorem

In this section, we will introduce and prove Gromov’s precompactness theo-rem. First, we introduce the notion of capacity.

DEFINITION 2.28 (Capacity). Let (X, d) be a compact metric space, and letr > 0. The capacity of X at the scale r is defined as the maximum number ofdisjoint r2 -balls in X , that is,

CapX(r) := max #B(x, r/2) : B(x, r/2) ⊂ X are pairwise disjoint.(2.8)

In other words, the capacity of X at the scale r is the largest number of r2 -balls

one can “pack” in X .

REMARK 2.29. If X is compact, then CapX(r) <∞ for any r > 0.

REMARK 2.30. If CapX(r) ≤ N , then X can be covered by at most N manyr-balls.

LEMMA 2.31. Let (X, dX) and (Y, dY ) be two compact metric spaces. Assumethat dGH(X,Y ) ≤ ε/10. Then, for all r > 0,

CapX(r + ε

2

)≤ CapY (r).(2.9)

In particular, if (Xi, di)→ (X, d), then we have

limi→∞

CapXi(r) = CapX(r).(2.10)


PROOF. Assume that CapX(r + ε/2) = N and that B(xi, (2r + ε)/4), i =1, . . . , N is a collection of pairwise disjoint balls. Since dGH(X,Y ) ≤ ε/10,there exists yi, i = 1, . . . , N ⊂ Y such that

dGH(yi, i = 1, . . . , N, xi, i = 1, . . . , N) ≤ ε/6.(2.11)

This implies that dY (yi, yj) ≥ dX(xi, xj)− ε/3 > r. Hence

CapY (r) ≥ N.

THEOREM 2.32 (Gromov’s Precompactness). Let C be a collection of compactmetric spaces. Then the following are equivalent:

(1) C is a precompact subset of (M, dGH), i.e., for any sequence in C thereexists a convergent subsequence with limit in M.

(2) There exists a map N : (0, 1) → Z+ such that CapX(r) ≤ N(r) for allX ∈ C and 0 < r < 1.

PROOF. (1) ⇒ (2): Assume that C is precompact. Then, for any ε > 0, thereexists a finite subset X1, . . . , XKε of C such that for any X ∈ C there existsXi ∈ X1, . . . , XKε such that

dGH(Xi, X) ≤ ε/10.(2.12)

Let

N(ε) := maxj=1,...,Kε

CapXj (ε/10).(2.13)

By Lemma 2.31 and (2.12), we conclude by letting r = ε/10 for any X ∈ C that

CapX(ε) ≤ CapX(3ε/5) ≤ CapXi(ε/10) ≤ N(ε),(2.14)

which proves (2).(2) ⇒ (1): For any ε ∈ (0, 1) and X ∈ C, since CapX(ε) ≤ N(ε) := Nε,

there exists a finite set

Xε := x1X , . . . , x

NεX ⊂ X

such that

X ⊂ B(x1X , . . . , x

NεX , ε).(2.15)

This means that

dGH((x1

X , . . . , xNεX , dX

), (X, dX)

)≤ ε,(2.16)

and so diam(X) ≤ 2εNε. Let us now define a map f : C→ R(Nε−1)Nε/2 by

f : X 7→ (eij(X)) ∈ R(Nε−1)Nε/2,(2.17)

2.3. GROMOV’S PRECOMPACTNESS THEOREM 35

where

eij(X) = dX(xiX , xjX).

Since diam(X) ≤ 2εNε, the image f(C) is a bounded subset in R(Nε+1)Nε/2.Hence f(C) is totally bounded. Thus there exists finite many X1, . . . , XKε ∈ Csuch that for any X ∈ C there exists Xi such that

dR(Nε−1)Nε/2(f(Xi), f(X)) ≤ ε,(2.18)

where dR(Nε+1)Nε/2 is the maximum norm in Euclidean space, that is, dRk(x, y) =maxki=1 |xi − yi|. Therefore, by Lemma 2.22, we have that

dGH((Xε, dX), (Xε

i , dXi))≤ 2ε.(2.19)

By the triangle inequality for the GH-distance, we have

dGH(Xi, X) ≤ dGH((Xε, dX), (X, dX)

)+ dGH

((Xε

i , dXi), (Xi, dXi))

+ dGH((Xε, dX), (Xε

i , dXi))

≤ 5ε.

This means that C is a totally bounded subset of (M, dGH). Hence C is a precom-pact subset.

DEFINITION 2.33. Let µ be a measure on a topological space X . We say thata subset A ⊂ X is µ-measurable if

µ(C) = µ(C ∩A) + µ(C \A)(2.20)

for any subset C ⊂ X . We say that µ is a Borel measure if every Borel set is µ-measurable. A metric measure space is a triple (X, d, µ), where (X, d) is a metricspace and µ is a Borel measure.

Let (X, d, µ) be a metric measure space with Borel measure µ such that 0 <µ(X) <∞. We say that µ is a doubling measure with constant κ > 0 if

µ(B(x, r)) ≥ κµ(B(x, 2r)) for all B(x, r) ⊂ X.(2.21)

LEMMA 2.34. Let (X, d, µ) be a compact metric measure space with boundeddiam(X) ≤ D, and let µ be a doubling measure with constant κ > 0. Then, forall r ≤ D,

CapX(r) ≤ C(κ)(D

r

)α(κ).(2.22)

Moreover, if µ satisfies

µ(B(x, r)) ≥ κ(r

R

)nµ(B(x,R))


for all B(x, r) ⊂ B(x,R) ⊂ X , then

CapX(r) ≤ κ−12n(r

D

)−n.(2.23)

PROOF. For anyB(x, r) ⊂ X , we have by the doubling measure property that

µ(B(x, r/2)) ≥ κµ(B(x, r)) ≥ κ2µ(B(x, 2r)) ≥ κN+1µ(B(x, 2Nr)).(2.24)

Let the integer N be such that 2D > 2Nr ≥ D. Therefore, we have

κN+1 ≥ κ2(D

r

)log κ/ log 2.(2.25)

Hence

µ(B(x, r/2)) ≥ κ2(D

r

)log κ/ log 2µ(X).(2.26)

This implies

CapX(r) ≤ κ−2(D

r

)− log κ/ log 2.(2.27)

Therefore we have proved (2.21). Inequality (2.23) follows directly as above byletting R = D. Hence we have finished the proof of the lemma.

Given constants n ∈ Z+, k ∈ R, and D ∈ R+, denote

(2.28) M(n, k,D) := (Mn, g) : Ric ≥ (n− 1)k and diamM ≤ D,

where (Mn, g) is a compact Riemannian n-manifold. By the volume comparisontheorem for manifolds with a Ricci curvature lower bound, we have the followingcorollary of Gromov’s precompactness theorem.

THEOREM 2.35. The collection of Riemannian manifolds M(n, λ,D) is a pre-compact subset of (M, dGH).

PROOF. Let µM = vol. By volume comparison, if (M, g) ∈M(n, λ,D), thenwe have for all B(x, r) ⊂M and r ≤ D that

µM (B(x, r)) ≥ µM (B(x, 2r)) vk(r)vk(2r)

≥ κ(n, λ,D)µM (B(x, 2r)),(2.29)

where vk(r) is the r-ball volume in the space form with curvature k, as defined in(1.33). Now the theorem follows directly from Gromov’s Theorem 2.32.

REMARK 2.36. We should notice that the limit of manifolds with respect toGromov–Hausdorff distance is a metric space, which may not be a manifold any-more. The main goal of this book is to study the structure of the limit space of asequence of manifolds with lower Ricci curvature bounds.

2.4. LENGTH SPACES AND THE POINTED GROMOV–HAUSDORFF DISTANCE 37

2.4. Length spaces and the pointed Gromov–Hausdorff distance

Let (X, d, x) be a pointed metric space with x ∈ X . In this section we willdefine Gromov–Hausdorff convergence for noncompact, locally compact metricspaces. Let us begin with the definition of pointed Gromov–Hausdorff distance.

DEFINITION 2.37. Let (X, dX , x) and (Y, dY , y) be two compact pointed met-ric spaces. Define their pointed Gromov–Hausdorff distance by

dpGH((X, dX , x), (Y, dY , y)

):= inf

(Z,dZ)dH(X,Y ) + dZ(x, y),(2.30)

where the infimum is taken over all possible (Z, dZ) such that X and Y both iso-metrically embed in Z.

EXAMPLE 2.38. Let (X = [0, 1], 0) and (Y = [0, 1], 1/2) be two pointedmetric spaces. Then dpGH(X,Y ) = 1/2.

Let (X, d) be a path-connected metric space, and let γ : [0, 1] → X be acontinuous curve. Define its length by

L[γ] := sup0=t0<t1<···<tN=1

N−1∑i=0

d(γ(ti), γ(ti+1)).(2.31)

From the definition, for any 0 < t < 1,

d(γ(0), γ(t)) + d(γ(t), γ(1)) ≤ L[γ].(2.32)

We say that a path-connected metric space (X, d) is a length space if for anyx0, x1 ∈ X there exists a continuous curve γ connecting x0 and x1 such that

d(x0, x1) = L[γ].(2.33)

LEMMA 2.39. Let (Y, d) be a complete metric space. Then the following areequivalent:

(1) Y is a length space.(2) For any y1, y2 ∈ Y , there exists a midpoint y3 of y1, y2, i.e.,

d(y1, y3) = d(y2, y3) = d(y1, y2)/2.(2.34)

PROOF. Assume (1), that Y is a length space. For any y1, y2 ∈ Y , there thenexists a curve γ connecting y1 and y2 such that

d(y1, y2) = L[γ].(2.35)

Let y3 ∈ γ be such that d(y1, y3) = d(y1, y2)/2. By the definition of the length ofa curve and by the triangle inequality, we have

d(y1, y2) ≤ d(y1, y3) + d(y3, y2) ≤ L[γ] = d(y1, y2).(2.36)


This implies that y3 is a midpoint of y1, y2.Assume (2). Let us now prove that Y is a length space. For any y0, y1 ∈ Y ,

by (2) there exists a midpoint of y0, y1; denote it by y1/2. Applying (2) to bothy0, y1/2 and y1/2, y1, we obtain midpoints y1/4 and y3/4 of y0, y1/2 and y1/2, y1,respectively. Inductively, we can construct yi/2k for all 0 ≤ i ≤ 2k and all k ≥ 1such that yi/2k is a midpoint of y(i+1)/2k , y(i−1)/2k . Denote

T = i/2k : 0 ≤ i ≤ 2k, k ≥ 1 ⊂ [0, 1],

which is dense in [0, 1]. It is not hard to check that for any t, s ∈ T, we have

d(yt, ys) = |t− s|d(y0, y1).(2.37)

In particular, for any t0 < t1 < · · · < tN with ti ∈ T ,

d(yt0 , ytN ) =N−1∑i=0

d(yti , yti+1).(2.38)

Define γ : T → Y by γ(t) = yt. By (2.37) and noting that Y is complete, we canextend γ to a continuous curve γ : [0, 1]→ Y . Then it follows directly that

d(y0, y1) = L[γ].(2.39)

Hence we have proved Y is a length space.

LEMMA 2.40. Let (Xi, di) be a sequence of length spaces with (Xi, di) →(X, d) in the Gromov–Hausdorff sense. Then (X, d) is a (path-connected) lengthspace.

PROOF. This follows directly from Lemma 2.39 since midpoints converge tomidpoints.

REMARK 2.41. Comparing with the Gromov–Hausdorff limit of length spaces,generally, the limit of path-connected metric spaces may not be path-connectedagain. For example, consider the sequence of metric spaces

Xi = (x, sin(x−1)) ∈ R2 : i−1 ≤ x ≤ 1

for i ≥ 1, with the metrics induced by their inclusions in the metric space R2.Clearly, the Xi are not length spaces, and their limit is not path-connected.

DEFINITION 2.42. Let (Xi, di, xi), i = 1, . . . be a sequence of completelength spaces, and let (X, d, x) be a complete, locally compact, length space.We say that (Xi, di, xi) converges to (X, d, p) in the pointed Gromov–Hausdorffsense if for any r > 0,

dpGH((B(xi, r), xi), (B(x, r), x))→ 0.(2.40)

2.5. BASIC GEOMETRIC MEASURE THEORY 39

REMARK 2.43. For a general complete metric space, one can use Gromov–Hausdorff approximated maps to define the pointed Gromov–Hausdorff distance(see []). However, since we will focus on complete manifolds which are lengthspaces, the above definition is good enough for our applications.

By using Gromov’s precompactness theorem and a diagonal argument, we canprove:

THEOREM 2.44. Let

(2.41) M(n, k) := complete n-manifold (Mn, g, p) : Ric ≥ (n− 1)k.

Then, for any sequence (Mni , gi, pi) ∈ M(n, k), there exists a subsequence which

converges in the pointed Gromov–Hausdorff sense to a length space.

2.5. Basic Geometric measure theory

2.5.1. Covering Lemma.

DEFINITION 2.45 (Finely covers). Let (X, d) be a metric space, and let A ⊂X . We say that a covering B = B(y, r) of A finely covers A if for any a ∈ Athere exists a sequence B(a, ri) ∈ B such that ri → 0.

LEMMA 2.46 (5-times covering). Let (X, d) be a metric space, and letA ⊂ X .Assume that B = B(y, r) is a covering of A with uniformly bounded diameter.Then there exists a pairwise disjoint countable subcollection B′ ⊂ B such that

A ⊂⋃

B′∈B′5B′,(2.42)

where 5B′ = B(x, 5r) for B′ = B(x, r).Moreover, if we assume that B finely coversA, then, for any finite subcollection

B1, B2, . . . , BN := B′′ ⊂ B′, the following holds:

A ⊂(N⋃i=1

Bi

)∪

⋃B∈B′\B′′

5B

.(2.43)

PROOF. Let r = supdiam(B)/2 : B ∈ B. Denote

Bj = B ∈ B : r2−j−1 < diam(B)/2 ≤ r2−j.

Then

B =∞⋃j=0

Bj .(2.44)

Let us now define an increasing sequence B′j inductively. Let B′1 ⊂ B1 be amaximal pairwise disjoint subcollection of B1. Assume that B′k is defined. Define


B′k+1 to be a maximal pairwise disjoint subcollection of B′k ∪ Bk+1 such thatB′k ⊂ B′k+1. We now show that the collection B′ =

⋃k≥1 B

′k satisfies the lemma.

Let us first check (2.42). For any x ∈ A, since B is a covering of A, wemay define rx = supdiam(B)/2 : x ∈ B ∈ B. Assume that x is such thatr2−j−1 < rx ≤ r2−j . Then there exists B ∈ Bj such that x ∈ B. From theconstruction, there exists some B′ ∈ B′j such that B ∩ B′ 6= ∅. In particular,diam(B′) ≥ r2−j ≥ diam(B)/2. Thus B ⊂ 5B′. So we have proved (2.42).

To see (2.43), for any finite subcollection B1, B2, . . . , BN := B′′ ⊂ B′ andx ∈ A \

⋃Ni=1Bi, since X \

⋃Ni=1Bi is open and since B finely covers A, there

exists B ∈ Bj such that x ∈ B and B ∩ Bi = ∅ for all i = 1, · · · , N . By theconstruction of B′j there exists B′ ∈ B′j such that B′ ∩ B 6= ∅. As above, we seethat B ⊂ 5B′. Hence we have proved (2.43).

LEMMA 2.47. Let (X, d, µ) be a metric measure space with doubling Borelmeasure µ satisfying

µ(B(x, r)) ≥ κµ(B(x, 2r)) for all x ∈ X, r ≤ 1.(2.45)

Let B = B(x, r) : x ∈ A finely cover a subset A with 0 < µ(A) < ∞. Then,for any ε > 0, there exists a pairwise disjoint subcollection B′ ⊂ B such thatB′ ⊂ B(p,R) for any B′ ∈ B′ and

µ

A \ ⋃B∈B′

B

≤ ε.(2.46)

PROOF. Since µ is Borel, there exists an open subset D such that A ⊂ D withµ(D) ≤ 2µ(A). Since B finely covers A and since D is open, we can choosea subcover B1 ⊂ B that finely covers A and satisfies that for each B ∈ B1,B ⊂ D. Applying Lemma 2.46 to B1, there exists a pairwise disjoint countableB′ ⊂ B1 such that for any finite subcollection B1, B2, . . . , BN := B′′ ⊂ B′, thefollowing holds

A ⊂(N⋃i=1

Bi

)∪

⋃B∈B′\B′′

5B

.(2.47)

Since B′ ⊂ D for B′ ∈ B′, the countable sum∑B′∈B′

µ(B′) ≤ µ(D) <∞.(2.48)

Choose finite B′′ ⊂ B′ such that∑B∈B′\B′′

µ(5B) ≤∑

B∈B′\B′′C(κ)µ(B) ≤ ε.(2.49)


We get from (2.47) that

A \⋃

B∈B′′B ⊂

⋃B′∈B′\B′′

5B′.(2.50)

Hence

µ

A \ ⋃B∈B′′

B

≤ ∑B′∈B′\B′′

µ(5B′) ≤ ε.(2.51)

Thus we have finished the proof.

2.5.2. Hausdorff measure. In this subsection, we will define Hausdorff mea-sure and Hausdorff dimension for a metric space, and discuss some basic propertiesof Hausdorff measure. For any t ≥ 0, set

(2.52) ωt = πt/2

Γ(t2 + 1

) ,where Γ is the Gamma function Γ(a) =

´∞0 xa−1e−xdx for a > 0. In particular,

ωt is equal to the volume of the unit ball in Rt if t is an integer.

DEFINITION 2.48 ((Spherical) Hausdorff measure). Let (X, d) be a metricspace, and let A ⊂ X . Given t ≥ 0, the (spherical) t-Hausdorff measure of A isdefined by

Ht(A) := limδ→0

Htη(A),(2.53)

where

Htδ(A) = inf

∑ωtr

ti : A ⊂

⋃B(xi, ri) and ri ≤ δ

.(2.54)

REMARK 2.49. The general Hausdorff measure is defined by using generalsubsets to cover A instead of balls. However, both Hausdorff measures (generalHausdorff and spherical Hausdorff) are equivalent. In this book, we will use Defi-nition 2.48 for the Hausdorff measure.

DEFINITION 2.50. We say that a measure µ on a topological space X is Borelregular if every Borel subset is measurable and if for each each subset A ⊂ X

there exists a Borel subset B such that A ⊂ B and µ(B) = µ(A).

From the definition it is easy to check that Ht and Htδ are (outer) measures:

µ

( ∞⋃i=1

Ai

)≤∞∑i=1

µ(Ai),(2.55)

and µ(∅) = 0 for µ = Ht or Htδ.

Furthermore, it is easy to check that

Htδ(A ∪B) = Ht

δ(A) + Htδ(B), for d(A,B) > 2δ.(2.56)


Hence

Ht(A ∪B) = Ht(A) + Ht(B), for d(A,B) > 0.(2.57)

Thus, by Caratheodory’s Criterion, Ht is a Borel measure, i.e. every Borel set isHt measurable. On the other hand, from the definition, Ht is Borel regular. Indeed,by definition, for eachA and δ > 0 there exists a covering B = B(xi, ri), ri ≤ δof A such that ∑

ωtrti − δ ≤ Ht

δ(A) ≤ Ht(A).(2.58)

Denote Uδ =⋃iB(xi, ri), which is open. From the definition of Ht

δ, we get

Htδ(A) ≤ Ht

δ(Uδ) ≤∑

ωtrti .(2.59)

Therefore, we have

Ht(A) ≥ Htδ(Uδ)− δ.(2.60)

Let U =⋂

0<δ<1 Uδ. Then U is Borel and A ⊂ U . Moreover, since Ht and Htδ

are measures, we have

Ht(U) ≥ Ht(A) ≥ Htδ(Uδ)− δ ≥ Ht

δ(U)− δ.(2.61)

Letting δ → 0, we obtain Ht(U) = Ht(A). Summarizing all of the above, hencewe have proved:

LEMMA 2.51. The Hausdorff measure Ht is a Borel regular measure for allt ≥ 0.

REMARK 2.52. In the construction of Uδ in (2.59), Ht(Uδ) may not be finite.Even when a measure µ is regular and satisfies µ(A) <∞, there may not exist anopen subset U such that A ⊂ U and µ(U) < µ(A) + ε. To see this, it suffices tochoose A = 0n ⊂ Rn and µ = H0.

Let us now give the definition of Hausdorff dimension.

DEFINITION 2.53 (Hausdorff dimension). Let (X, d) be a metric space, andlet A ⊂ X . The Hausdorff dimension of A is defined by

dimH A := dimA := inft ≥ 0 : Ht(A) = 0 = supt ≥ 0 : Ht(A) =∞.(2.62)

REMARK 2.54. It is an easy exercise to show that

inft ≥ 0 : Ht(A) = 0 = supt ≥ 0 : Ht(A) =∞

for any A ⊂ X . See Exercise 2.6.


The point a ∈ A is called a t-density point if

lim supr→0

Ht∞(A ∩B(a, r))

ωtrt≥ 2−t.(2.63)

Denote the set of t-density points of A by Dt(A). Then

LEMMA 2.55. Let (X, d) be metric space, let t ≥ 0, and let A ⊂ X be an Ht

measurable subset. Then

Ht(A \Dt(A)) = 0,(2.64)

and if Ht(A) <∞, then for Ht-a.e. a ∈ A we have

lim supr→0

Ht(A ∩B(a, r))ωtrt

≤ 1.(2.65)

PROOF. For any ε > 0, define

Dε :=a ∈ A : lim sup

r→0

Ht∞(A ∩B(a, r))

ωtrt< (1− ε)2−t

.(2.66)

Then A \Dt(A) =⋃∞i=1D2−i . To prove (2.64), it suffices to prove Ht(Dε) = 0

for each ε > 0. By definition, for each a ∈ Dε, there exists ra > 0 such that for allr ≤ ra,

Ht∞(A ∩Br(a))

ωtrt< (1− ε)2−t.(2.67)

Let us argue by contradiction. Assume that Ht(Dε) > 0. Denote

Dε,i := a ∈ Dε : ra < i−1.

Then

Dε =∞⋃i=1

Dε,i.(2.68)

Since Ht(Dε) > 0, for some Dε,i0 := D′ we have that Ht(D′) > 0. For any δ <i−10 /10, by the definition of Ht

δ, there exists a finite collection B0 = B1, . . . , BNwith diam(Bi) ≤ 2δ and such that D′ ⊂

⋃B∈B0 B and

Htδ(D′) ≥

∑B∈B0

ωt

(diam(B)2

)t− δ.(2.69)

For eachB ∈ B0, let x ∈ B∩D′ and considerB′ = B(x, s) with s = diam(B) ≤2δ < i−1

0 . Then B ⊂ B′. On the other hand, by definition of Htδ and Ht

∞ we havefor s ≤ δ that

Htδ(B(x, s) ∩A) = Ht

∞(B(x, s) ∩A) < (1− ε)2−tωtst,(2.70)


where we have used (2.67) in the last inequality. Hence∑B∈B0

ωt

(diam(B)2

)t= 2−t

∑B∈B0

ωtdiam(B)t

≥ 11− ε

∑B∈B0

Htδ(B′ ∩A)

≥ 11− ε

∑B∈B0

Htδ(B ∩A)

≥ 11− εH

tδ(D′),

where we have used D′ ⊂⋃B∈B0 B for the last inequality. Therefore, combining

this with (2.69), we arrive at

Htδ(D′) ≥

11− εH

tδ(D′)− δ.(2.71)

Letting δ → 0, we get Ht(D′) = 0, which is a contradiction. Thus we have provedHt(Dε) = 0 and hence Ht(A \Dt(A)) = 0.

Let us now prove (2.65). Denote

A0 =x ∈ A : lim sup

r→0


> 1.

It suffices to prove Ht(A0) = 0. For any ε > 0, define

Aε :=x ∈ A : lim sup

r→0


> 1 + 2ε.(2.72)

Then

A0 =∞⋃i=1

A2−i .(2.73)

To prove (2.65), it suffices to show Ht(Aε) = 0 for each ε > 0. By definition, foreach a ∈ Aε, there exists a sequence of ra,i > 0 satisfying limi→0 ra,i = 0 and

Ht(A ∩B(a, ra,i))ωtrta,i

> 1 + 2ε.(2.74)

By shrinking the balls a little bit, there exists a sequence of closed balls B(a, sa,i)with sa,i → 0 such that

Ht(A ∩ B(a, sa,i))ωtsta,i

> 1 + ε.(2.75)

Therefore, for each δ > 0, there exists a finely cover B of Aε such that eachB(a, r) ∈ B satisfies (2.75) and r ≤ δ. Since A is Ht measurable and Ht is Borelregular, the measure Ht|A is also Borel regular (see Exercise 2.7). Therefore, there


exists an open subset D such that Aε ⊂ D and Ht((D \Aε)∩A) ≤ δ. By Lemma2.46, there exists a pairwise disjoint countable subcollection B′ ⊂ B such that forany finite subcollection B1, B2, . . . , BN := B′′ ⊂ B′, the following holds

Aε ⊂N⋃i=1

Bi ∪⋃

B∈B′\B′′5B,(2.76)

and B′ ⊂ D for each B′ ∈ B′. This implies that

∑B∈B′

Ht5δ(5B ∩A) ≤

∑B∈B′

ωt

(5 diam(B)2

)t

≤ 5t∑B∈B′

ωt

(diam(B)2

)t

≤ 5t

1 + ε

∑B∈B′

Ht(A ∩B),

where we have used the trivial upper bound by definition

Ht5δ(5B ∩A) ≤ ωt

(5 diam(B)2

)tsince diam(5B)/2 ≤ 5δ, and by (2.75) for the last inequality. Since B′ are pair-wise disjoint, we get

∑B∈B′

Ht5δ(5B ∩D) ≤ 5t

1 + εHt(D ∩A) <∞.(2.77)

Hence there exists a finite subcollection B′′ ⊂ B′ such that∑B∈B′\B′′

Ht5δ(5B ∩A) ≤ δ.(2.78)

Thus we get from (2.76) that

Ht5δ

Aε \ ⋃B∈B′′

B

≤ δ.(2.79)

In particular, we have by (2.75) and

Ht5δ(B ∩A) ≤ ωt

(diam(B)2

)t


for diam(B) ≤ 10δ from the definition of Ht5δ that

Ht5δ(Aε) ≤

∑B∈B′′

Ht5δ(B ∩A) + δ(2.80)

≤∑B∈B′′

ωt

(diam(B)2

)t+ δ

≤ 11 + ε

∑B∈B′′

Ht(A ∩B) + δ

≤ 11 + ε

Ht(D ∩A) + δ

≤ 11 + ε

(Ht(Aε) + δ) + δ.

Letting δ → 0, we get

Ht(Aε) ≤1

1 + εHt(Aε),(2.81)

which implies Ht(Aε) = 0. Thus we have proved (2.65).

LEMMA 2.56. Let (Xi, di, pi)→ (X, d, p) and t ≥ 0. Then

lim supi→∞

Ht∞(Xi) ≤ Ht

∞(X),(2.82)

and for any δ > 0

lim supi→∞

Ht2δ(Xi) ≤ Ht

δ(X).(2.83)

REMARK 2.57. In general, we will not have lim supHt(Xi) ≤ Ht(X) dueto any collapsing limit. For example, Xi = 0, i−1 ⊂ R and X = 0. ThenH0(Xi) = 2 and H0(X) = 1.

PROOF. It suffices to prove (2.83). Inequality (2.82) follows from (2.83). Bythe intersection with a bounded ball B(p,R), without loss of generality we mayassume that diamX < ∞ and that Ht

δ(X) = a < ∞. For any ε > 0, choose afinite covering B(xα, rα) of X such that rα < δ and∑

rtαωt ≤ a+ ε.(2.84)

For i sufficiently large, there exists an associated covering B(xiα, (1 + ε)rα) ofXi. This implies that

Ht2δ(Xi) ≤ (1 + ε)t

∑rtαωt ≤ (a+ ε)(1 + ε)t.(2.85)

In particular, we have

lim supi→∞

Ht2δ(Xi) ≤ (a+ ε)(1 + ε)t.(2.86)

By letting ε→ 0, we have proved (2.83).


We will use the following lemma to study noncollapsed Ricci limit spaces,where the Ahlfors regularity assumption holds.

LEMMA 2.58. Suppose that a sequence of metric spaces (Xi, di) satisfies theAhlfors regularity assumption that there exist constants 0 < λ ≤ Λ such thatλrn ≤ Hn(B(x, r)) ≤ Λrn for any B(x, r) ⊂ Xi with r ≤ diam(Xi, di). If(Xi, di)→ (X, d), then for any x ∈ X and r ≤ diam(X, d), we have

C0(n, λ,Λ)rn ≤ Hn(B(x, r)) ≤ C1(n, λ,Λ)rn,(2.87)

and for any r > s,

Hn(B(x, r) \B(x, s)) ≤ C(n, λ,Λ)(r − s)rn−1.(2.88)

REMARK 2.59. This implies that dimX = n and Hn(∂B(x, r)) = 0.

REMARK 2.60. Even under the Ahlfors regularity assumption, one still cannot get Hausdorff measure convergence; see also the example in Remark 2.57.

PROOF OF LEMMA 2.58. Assume xi → x, with xi ∈ Mi. By the Ahlforsregularity assumption for Hn, we get for any s > 0 and r > 0 that (see alsoLemma 2.34)

CapB(xi,2r)(s) ≤ C(n, λ,Λ)s−nrn.(2.89)

Since B(xi, 2r)→ B(x, 2r) and by Lemma 2.31, we get

CapB(x,2r)(s) ≤ C(n, λ,Λ)s−nrn.(2.90)

This implies by the definition of Hns that

Hns (B(x, r)) ≤ C(n, λ,Λ)s−nrnωnsn ≤ C(n, λ,Λ)rn.(2.91)

Letting s → 0, we have Hn(B(x, r)) ≤ C1(n, λ,Λ)rn. For another direction, weuse Lemma 2.56 to get for any δ > 0 that

Hnδ (B(x, r)) ≥ lim

i→∞Hn

2δ(B(xi, r)).(2.92)

Let us show a uniform lower bound for Hn2δ(B(xi, r)). Actually, for any ε > 0, by

the definition of Hn2δ there exists a covering B(yi, ri), ri ≤ 2δ, i = 1, . . . , N of

B(xi, r) such that

Hn2δ(B(xi, r)

)≥

N∑i=1

ωnrni − ε.(2.93)


Noting that Hn(B(yi, ri)) ≤ Λrni , we get

Hn2δ(B(xi, r)

)≥ C(n,Λ)

N∑i=1

Hn(B(yi, ri))− ε(2.94)

≥ C(n,Λ)Hn(B(xi, r))− ε≥ C(n,Λ, λ)rn − ε.

Letting ε→ 0, we get Hn2δ(B(xi, r)

)≥ C(n,Λ, λ)rn. By (2.92), we arrive at

Hnδ (B(x, r)) ≥ C(n, λ,Λ)rn.(2.95)

Letting δ → 0, we obtain the lower bound Hn(B(x, r) ≥ C(n, λ,Λ)rn, whichfinishes the proof of (2.87).

To prove (2.88), by the Ahlfors regularity of Hn, we have

CapA(xi,r−2s,r+s)(10(r − s)) ≤ C(n, λ,Λ)(r − s)−n+1rn−1.(2.96)

Since xi → x and by Lemma 2.31, we obtain

CapA(x,r−2s,r+s)(10(r − s)) ≤ C(n, λ,Λ)(r − s)−n+1rn−1.(2.97)

This implies that

Hn(A(x, r − 2s, r + s)) ≤ C(n, λ,Λ)(r − s)−n+1rn−1

× maxy∈A(x,r−2s,r+s)

Hn(B(y, 20(r − s))

≤ C(n, λ,Λ)(r − s)rn−1,

where we have used (2.87) in the last inequality. Thus we have finished the proofof (2.88).

2.6. Notes and commentary

Excellent references for geometric measure theory are the books by Federer[16] and Leon Simon [33].

2.7. Exercises

EXERCISE 2.1. Prove Lemma 2.6.

EXERCISE 2.2. Let (X, d) be a compact metric space. Denote, as in (2.2),C(Z) := A ⊂ X : A is closed. Prove: (C(Z), dH) is a compact metric space.

EXERCISE 2.3. Compute dGH(S1(r1), S1(r2)), where S1(r) carries the arc-length metric.

2.7. EXERCISES 49

EXERCISE 2.4. Let (Xi, di) and (X, d) be compact metric spaces and sup-pose that (Xi, di)→ (X, d). Assume that fi : Xi → R is a sequence of uniformlybounded and uniformly Lipschitz functions. Prove: There exists a Lipschitz func-tion f : X → R and a subsequence f ′i of fi such that f ′i → f in the Gromov–Hausdorff sense.

EXERCISE 2.5. Let (Xi, dXi) and (Yi, dYi) be a sequence of compact metricspaces. Assume that the product spaces Xi × Yi converge to a compact metricspace (Z, dZ). Prove: Z is isometric to a product space.

EXERCISE 2.6. Let (X, d) be a metric space, and let A ⊂ X . Prove thatinft ≥ 0 : Ht(A) = 0 = supt ≥ 0 : Ht(A) =∞.

EXERCISE 2.7. Let (X, d) be a metric space, and let A ⊂ X be Ht measur-able. Prove that the restricted measure Ht|A is Borel regular.

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Index

Ahlfors regularity assumption, 47annulus, 19

ball, r-, 25barrier sense, 6Bishop–Gromov volume comparison

theorem, 18Borel measure, 35Borel regular measure, 41

capacity, 33comparison principle, 15cut locus, 3

density pointt-, 43

diameter rigidity theorem, 22distance function, 12

exponential coordinates, 17exponential map, 3

geodesicarc-length, 4

geodesic sphere, 12GH map, ε-, see also Gromov–Hausdorff

approximated map, ε-Gromov–Hausdorff approximated map, ε-,

29Gromov–Hausdorff distance

pointed, 37Gromov–Hausdorff sense

ε-close, 30

H-distance, see also Hausdorff distanceHarnack inequality

differential, 89Hausdorff dimension, 42Hausdorff distance, 25heat kernel, 88Hessian, 6Hopf–Rinow theorem, 3

injectivity domain, 3isometrically embeds, 27isometry, ε-, 29

Jacobi equation, 14Jacobi field, 13

Laplacian comparison theorem, 12length, 37length space, 37

maximum norm, 35mean curvature, 13metric measure space, 35metric space, 25model space

of constant curvature k, 15Myers’s Theorem, 16

neighborhood, r-, 25net, ε-, 25

Peter–Paul inequality, 72

Riccati equation, 14

second fundamental form, 13shape operator, 13

totally bounded

205

206 INDEX

metric space, 25

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