Analysis on Manifolds - Aristotle University of...

Analysis on Manifolds

Michel MARIAS

February 2005

Contents

I Heat Kernel Estimates 10.1 Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1 Introduction 5

2 Preliminaries 112.1 The Laplacian in local coordinates . . . . . . . . . . . . 11

3 Gaussian estimates of the heat kernel 153.1 The gradient estimate of Li and Yau . . . . . . . . . . . 15

3.1.1 Proof of the Gradient estimate . . . . . . . . . . . 163.1.2 End of proof of the gradient estimate . . . . . . . 19

3.2 Parabolic Harnack Inequality . . . . . . . . . . . . . . . 223.3 Upper bounds of the heat kernel . . . . . . . . . . . . . . 24

3.3.1 The diagonal upper bound of the heat kernel whenRic (M) ≥ 0 . . . . . . . . . . . . . . . . . . . . . 26

3.4 The off-diagonal gaussian upper bound . . . . . . . . . . 273.5 Proof of the upper gaussian estimate . . . . . . . . . . . 34

4 Heat kernel bounds on compact manifolds 41

5 Applications of heat kernel bounds 435.1 Estimates of the Green function . . . . . . . . . . . . . . 435.2 Liouville Theorem . . . . . . . . . . . . . . . . . . . . . . 475.3 Lichnerowicz Theorem . . . . . . . . . . . . . . . . . . . 49

iii

iv CONTENTS

6 Estimates of the eigenvalues on compact manifolds 516.1 The Laplace operator on a compact manifold . . . . . . . 526.2 The min-max principle . . . . . . . . . . . . . . . . . . . 546.3 The Polya conjecture . . . . . . . . . . . . . . . . . . . . 556.4 Some important inequalities . . . . . . . . . . . . . . . . 56

6.4.1 Equivalence of Sobolev and isoperimetric inequal-ities . . . . . . . . . . . . . . . . . . . . . . . . . 57

6.4.2 A lower bound of the first eigenvalue . . . . . . . 596.4.3 Upper bounds for the first eigenvalue . . . . . . . 62

6.5 Estimates of higher eigenvalues via the heat kernel . . . . 626.5.1 The Polya conjecture . . . . . . . . . . . . . . . . 626.5.2 Heat kernel expansion on a compact manifold . . 64

6.6 Lower estimates of higher eigenvalues . . . . . . . . . . . 64

7 Appendix 1. Basic facts from Riemannian Geometry 717.1 The metric tensor . . . . . . . . . . . . . . . . . . . . . . 717.2 The Curvature . . . . . . . . . . . . . . . . . . . . . . . 73

7.2.1 The curvature of a surface . . . . . . . . . . . . . 737.2.2 The curvature of a manifold . . . . . . . . . . . . 767.2.3 Ricci curvature . . . . . . . . . . . . . . . . . . . 77

7.3 Divergence theorem and Green’s fomulas . . . . . . . . . 787.4 Appendix 2 . . . . . . . . . . . . . . . . . . . . . . . . . 79

Part I

Heat Kernel Estimates

1

0.1. Preface 3

0.1 Preface

This is the first writing of the first part of a graduate course I shallgive in the Department of Mathematics of the AUTH in the spring fall2005. In these lectures we present the heat kernel bounds on a completeRiemannian manifold and the spectrum of the Laplacian on a compactRiemannian manifold. The material comes mainly from the books ofSchoen and Yau [15] and Davies [3].

In the second part, not written yet!, we shall present some aspects ofHarmonic Analysis on manifolds, and eventually on graphs, where theresults of Part I and mainly the heat kernel bounds plays the centralrole.

Thessaloniki, February 2005.

Chapter 1

Introduction

Let M be an n−dimensional Riemann manifold. Let (gij (x)) be theRiemannian metric of M in local coordinates. Let us denote by d (x, y)the Riemannian distance on M and by dm its Riemannian measure.We have

dm (x) =√

g (x)dx1 · · · dxn,

where g (x) = det (gij (x)), (see Section 2 for precise definitions). Letalso B (x, r) be the ball with center x and radius r. We shall denote by|B (x, r)| its volume with respect to the measure dm.

The Laplace operator ∆ of M is a second order differential operatorwhich on C∞ functions is written as

∆f =1√g

∑i,j

∂i

(√ggij∂jf

),

where (gij (x)) is the inverse matrix of (gij (x)).The heat kernel

pt (x, y) , t > 0, x, y ∈ M,

of M , is the fundamental solution of the heat equation:

∂tpt (x, y) = ∆xpt (x, y) , p0 (x, y) = δx (y) ,

5

6 1. Introduction

where δx is the Dirac measure at x ∈ M .

In the particular case of Rn, pt (x, y) is the well known Gauss kernel:

pt (x, y) =e−‖x−y‖2/4t

(4πt)n/2.

Unfortunately (!), Rn is one of the very few cases we have a close formula

of the heat kernel. In the case of manifolds, is hopeless to look for aclose formula since the Laplacian is a second order differential operatorwith variable coefficients.

In the past 20 years, great efforts have been made in order to un-derstand the behavior of the heat kernel in various geometric settings,for example on Riemann manifolds, Lie groups, and their discrete ana-logues. Today, we know quite well the heat kernel on Riemann mani-folds with non negative Ricci curvature, and on Lie groups of polyno-mial volume growth. The main theorem we shall prove in these lecturesis the following result of Li and Yau, [7].

Theorem 1 Let M be an n−dimensional, complete, non-compact Rie-mannian manifold such that Ric (M) ≥ 0 and let pt (x, y) be its heatkernel. Then, there are positive constants C1, c2, C3 and c4 such that

C1e− d(x,y)2

c2t∣∣B (x,√

t)∣∣ ≤ pt (x, y) ≤ C3

e− d(x,y)2

c4t∣∣B (x,√

t)∣∣ , (1.1)

for all t > 0 and all x, y ∈ M .

If M = Rn, then ∣∣∣B (x,

√t)∣∣∣ = cnt

n2 ,

and (1.1) gives that

pt (x, y) ∼ t−n2 e−

d(x,y)2

ct ,

7

that is the Gauss kernel modulo multiplicative constants. Because of

the exponential factor e−d(x,y)2

ct , the estimates (1.1) are known as thelower and the upper Gaussian estimates of the heat kernel.

If Ric (M) ≥ −K2, the Yau-Li estimate is valid only for t ≤ 1. Thebest result for (constant) negative curvature till now is this of Daviesand Mandouvalos, [4].

In the setting of Lie groups, Varopoulos and his team have ob-tained sharp results in the case when the group is of polynomial volumegrowth, [17].

I shall now explain why the heat kernel is a fundamental tool inAnalysis. In what follows I shall use many heuristics.

In the setting of Rn, to study the Laplace operator as an unbounded

operator on L2 (Rn), we use the Fourier transform which allow us towrite

∆f (ξ) = −‖ξ‖2 f (ξ) ,

for all f ∈ C∞0 (Rn). This formula allow us to solve some important

differential equations in Rn. For example, the wave equation

∂2t wt (x) = ∆wt (x) , w0 (x) = δx, ∂tw0 (x) = 0,

in the Fourier transform variables is written as

wt (ξ) = cos t ‖ξ‖and the solution wt (x) is found by inverting the above Fourier trans-form.

Also, by means of the Fourier transform, we find out the heat kernel ofR

n. In any case, the Fourier transform is a basic tool in doing Analysison R

n. For example, Harmonic Analysis on Rn depends heavily on the

Fourier transform, (cf. [16]).

But, let us return to the heat equation. For any f ∈ C∞0 (Rn) and

s > 0, we set

Tsf (x) =

∫ ∞

0

ps (x, y) f (y) dy, (1.2)

8 1. Introduction

where

ps (x, y) =e−‖x−y‖2/4t

(4πt)n/2

is the Gauss kernel.It easy to check (cf. for example [11]), that Ts, s > 0, is a semigroup

of continuous operators on L2 (Rn). Semigroup means that

(Tr Ts) f = Tr+sf.

Using (1.2), we can see that the infinitesimal generator of the heatsemigroup Ts, is precisely the Laplacian ∆, i.e. for any f ∈ C∞

0 (Rn)

Tsf − f

s−→s→0

−∆f,

in the L2 sense. This is the reason we write

Tsf = e−s∆.

In the setting of a manifold, the Fourier transform is no more avail-able, and the study of the Laplacian on L2 (M) goes through the spectraltheorem:

∆ =

∫ ∞

λ0

λdEλ,

where dEλ is the spectral measure of M , (cf. [12], Vol. 1)Again, unfortunately, the spectral measure dEλ of a manifold is

known only in very few cases. But, the spectral theorem allow us todefine the operator

m (∆) =

∫ ∞

λ0

m (λ) dEλ (1.3)

as a bounded operator on L2 (M) for any bounded Borel function m (λ)on the real line. In particular, if

m (λ) = e−sλ, s > 0,

9

formula (1.3), allow us to define the heat semigroup of the manifold M :

e−s∆ =

∫ ∞

λ0

e−sλdEλ (1.4)

as a bounded operator on L2 (M).

So, the heat semigroup e−s∆, appears as the Laplace transform of thespectral measure and if one knows the Laplace transform of a measure,then by inversion, he also knows the measure. Thus, a very good way tostudy the spectral measure of a manifold, is to study the heat semigroupe−s∆.

Further, a highly non-trivial result, tell us that the heat semigroupe−s∆ has a C∞ kernel , (cf. [3], p. 149) and that this kernel is preciselythe heat kernel of M :

e−s∆f (x) =

∫M

ps (x, y) f (y) dm (x) .

In conclusion, if one knows the heat kernel, he also knows the heatsemigroup e−s∆ and also the spectral measure dEλ, which is the inverseLaplace transform of e−s∆. Thus, in the sense we described above, theheat kernel is a substitute of the Fourier transform in the setting ofmanifolds.

A natural question is the following. Why in formula (1.3), we have

preferred e−sλ to e−i√

λs. Using e−i√

λs, we have the Fourier transformof the spectral measure:

e−is√

∆ =

∫ ∞

λ0

e−is√

λdEλ.

The reason is that the kernel of the wave operator e−is√

∆ is singular,and the study of the wave equation (which is a hyperbolic one) is muchmore harder than the study of the heat equation (a parabolic one).

10 1. Introduction

In the present notes I shall give, in Chapter 1, the proof of theparabolic Harnack inequality and then the proof of the upper gaus-sian estimate of the heat kernel. I shall follow the approach of Li andYau. The main references are the paper of Li and Yau [7] and the booksof Shoen and Yau [15] and Davies [3]. I hope that in an other occasionI will have the opportunity to present the approach of N. Varopouloswhich gives the heat kernel estimates in various geometric settings.

Chapter 2

Preliminaries

In this section we fix the notation we will use in these lectures. Moredetails on the basic facts of Riemannian geometry we shall need aregiven in the Appendix.

2.1 The Laplacian in local coordinates

Let (gij (x)) be the Riemannian metric of M in local coordinates. Theinduced Riemannian distance is denoted by d (x, y) and the integral ofa function f ∈ C∞

0 (M) in local coordinates is given by∫M

f (x) dm (x) =

∫M

f (x)√

g (x)dx,

whereg (x) = det (gij (x)) .

Thus the Riemannian measure dm (x) in local coordinates is given by

dm (x) =√

g (x)dx =√

det (gij (x))dx.

In the sequel, very often we shall denote the measure dm (x) simply bydx.

11

12 2. Preliminaries

If f ∈ C∞0 (M), its gradient ∇f is the vector field with coordinates

given by

(∇f)i (x) =∑

j

gij (x) ∂jf (x) , (2.1)

where (gij (x)) is the inverse matrix of (gij (x)).If

ξ (x) = (ξ1 (x) , . . . , ξn (x)) ,

is a vector field written in local coordinates, then its divergence div ξ (x)is a scalar function and it is defined by the validity of the formula∫

M

〈∇f (x) , ξ (x)〉x dm (x) =

∫M

f (x) div ξ (x) dm (x) , (2.2)

where〈∇f (x) , ξ (x)〉x =

∑i,j

gij (x) (∇f)i (x) ξj (x)

is the scalar product on the tangent space Tx (M) of M at x.Let us suppose that f has support in a coordinate neighborhood,

then ∫M

〈∇f (x) , ξ (x)〉x dm (x)

=

∫M

∑i,k

gik (x) (∇f)i (x) ξk (x)√

g (x)dx

=

∫M

∑i,k

gik (x)

(∑j

gij (x) ∂jf (x)

)ξk (x)

√g (x)dx

=

∫M

∑j

∂jf (x)∑

k

ξk (x)∑

i

gik (x) gij (x)√

g (x)dx

But ∑i

gik (x) gij (x) = δkj,

2.1. The Laplacian in local coordinates 13

so that∫M

〈∇f (x) , ξ (x)〉x dm (x) =

∫M

∑j

∂jf (x) ξj (x)√

g (x)dx

= −∫

M

f (x)∑

j

∂j

(ξj (x)

√g (x)

)dx

which implies that

div ξ (x) = − 1√g (x)

∑j

∂j

(√g (x)ξj (x)

). (2.3)

The Laplacian ∆ is defined by

∆f = div∇f.

From (2.1) and (2.3), it follows that in local coordinates, the Laplacianis written as

∆f =1√g

∑i,j

∂i

(√ggij∂jf

).

From this formula it follows that for any f ∈ C∞0 (M),

〈−∆f, f〉 = −∫

M

div∇f (x) f (x) dm (x)

=

∫M

〈∇f (x) ,∇f (x)〉x dm (x) ≥ 0,

and we see that −∆ is a symmetric and positive operator on C∞0 (M).

14 2. Preliminaries

Chapter 3

Gaussian estimates of theheat kernel

In this Chapter we shall present the Li-Yau approach of proving Gaus-sian estimates of the heat kernel on a complete Riemannian manifold.

The first step is the proof of the parabolic gradient estimate whichallow us to prove the parabolic Harnack inequality for positive solutionsof the heat equation. Then, from the Harnack inequality we derive theon-diagonal upper bound of the heat kernel and finally the off-diagonalupper bound.

3.1 The gradient estimate of Li and Yau

In this Section we shall describe the Li-Yau approach of proving theparabolic Harnack inequality for positive solutions of the heat equationon a complete manifold M with Ric (M) ≥ −K2, (cf. [7]). The basicgradient estimate of Li-Yau is the following

Theorem 2 (Gradient estimate of Li-Yau) Let M be an n-dimensionalcomplete manifold with Ric (M) ≥ −K2, then any positive solution

15

16 3. Gaussian estimates of the heat kernel

ut (x) of the heat equation

∂tut (x) = ∆ut (x) ,

satisfies the following estimate

‖∇ut‖2

(ut)2 − α

∂tut

ut

≤ nα2

2

1

t+

K2

2 (α − 1)

, (3.1)

for all x ∈ M , t ∈ [0, T ] and α > 1.

If Ric (M) ≥ 0, then we can take α = 1 and T = ∞.

Remark 3 Let u be a positive harmonic function i.e. a solution of theLaplace equation

∆u (x) = 0

for all x ∈ M . Since u (x) is time-independent, it is also a solutionof the heat equation: ∆u (x) = 0 = ∂tu (x). Therefore, the Li-Yauestimate applies and gives by choosing for example t = 2 and α = 2

‖∇u‖2

u2≤ n

1 + K2

. (3.2)

The above estimate is the elliptic version of (3.1), (see [15], p.23).

3.1.1 Proof of the Gradient estimate

For the proof of the gradient estimate we need several Lemmata.

Lemma 4 If f : M −→ R is smooth, then

∆‖∇f‖2 ≥ 2

n(∆f)2 + 2 〈∇ (∆f) ,∇f〉 − 2K2 ‖∇f‖2 . (3.3)

3.1. The gradient estimate of Li and Yau 17

Proof. Since we have to verify the above inequality (3.3) only at apoint a, we choose normal coordinates with center the point a, so that

gij = δij = gij.

With this choice of coordinates

∆f (a) =∑

i

∂iif (a) , ‖∇f (a)‖2 =∑

i

(∂if (a))2 .

(in the sequel, for simplicity, we drop a)This implies that

∆‖∇f‖2 =

∑i

∂ii

∑j

(∂jf)2 = 2∑

i

∂i

∑j

∂i∂jf∂jf

= 2∑i,j

∂i∂i∂jf∂jf + (∂i∂jf)2 .

Now, in changing the order of derivation, the Ricci curvature tensorappears:

∂i∂i∂jf = ∂j∂i∂if + Rij∂jf.

Also (∑i

∂iif

)2

=∑

i

(∂iif)2 + 2∑i=j

∂iif∂jjf

≤∑

i

(∂iif)2 +∑i=j

(∂iif)2 + (∂jjf)2

≤ n∑i,j

(∂ijf)2 .

Therefore

∆‖∇f‖2 ≥ 2

∑i,j

∂j∂i∂if∂jf + Rij∂jf∂jf +2

n

(∑i

∂iif

)2

≥ 2 〈∇ (∆f) ,∇f〉 − 2K2 ‖∇f‖2 +2

n(∆f)2 ,

since Ric (M) ≥ −K2.


Lemma 5 Let u > 0 be a smooth solution of the heat equation

∂tu (x) = ∆u (x) , x ∈ M , t ∈ [0, T ] ,

and putF (x, t) = t

‖∇f‖2 − α∂tf

, (3.4)

where f = log u and α ∈ R. Then

∆F − ∂tF + 2 〈∇f,∇F 〉 + t−1F (3.5)

≥ 1

t

[2

n

‖∇f‖2 − ∂tf2 − 2K2 ‖∇f‖2

].

Proof. Let us denote by A (F ) the LHS of (3.5), i.e.

A (F ) = ∆F − ∂tF + 2 〈∇f,∇F 〉 + t−1F.

SinceF (x, t) = t

‖∇f‖2 − α∂tf

,

it follows that

A (F ) = A (t ‖∇f‖2)− αAA (∂tf) .

Since ∂tu = ∆u, it follows that the function

f = log u

satisfies∆f + ‖∇f‖2 = ∂tf. (3.6)

So, ifG = t∂tf,

then

A (G) = ∆G − ∂tG + 2 〈∇f,∇G〉 + t−1G

= t∆∂tf − ∂tf − t ∆∂tf + 2 〈∇f,∇∂tf〉+2t 〈∇f,∇∂tf〉 + ∂tf = 0.


So, the LHS of (3.5) equals to

A (t ‖∇f‖2)= t∆ ‖∇f‖2 − ∂t

(t ‖∇f‖2)+ 2t

⟨∇f,∇‖∇f‖2⟩+ ‖∇f‖2

= t∆ ‖∇f‖2 − ‖∇f‖2 − 2t 〈∇f,∇∂tf〉 + 2t⟨∇f,∇‖∇f‖2⟩+ ‖∇f‖2

(3.3)

≥ t

2

n(∆f)2 + 2 〈∇ (∆f) ,∇f〉 − 2K ‖∇f‖2

−2t 〈∇f,∇∂tf〉 + 2

⟨∇f,∇‖∇f‖2⟩(3.6)= t

2

n(∆f)2 − 2K ‖∇f‖2

which leads at once to the RHS of (3.5).

3.1.2 End of proof of the gradient estimate

To prove the Li-Yau estimate, it suffices to show that the function

F (x, t) = t‖∇f‖2 − α∂tf

,

defined by (3.4) satisfies

F (x, t) ≤ nα2

2

(1 +

K2t

2 (α − 1)

)(3.7)

for all x ∈ M , t ∈ [0, T ] and α > 1.Indeed, since

f = log u,

it follows

F (x, t) = t‖∇f‖2 − α∂tf

= t

‖∇ut‖2

(ut)2 − α

∂tut

ut

,

and the inequality (3.7) implies

‖∇ut‖2

(ut)2 − α

∂tut

ut

≤ nα2

2

1

t+

K2

2 (α − 1)

,


for all x ∈ M , t ∈ [0, T ] and α > 1, i.e. the gradient estimate.

We shall prove (3.7) in two steps.

Step 1. We assume that M is a compact manifold without boundary.

Since F is a continuous function and M is compact, there is a point(x0, s0) at which F takes its maximum value. If F (x0, s0) ≤ 0, then wehave nothing to prove. If F (x0, s0) > 0, then s0 > 0 since F (x, 0) = 0.

Next, we observe that since (x0, s0) is a local maximum, then by themaximum principle (na koitaxtei)

∇xF (x0, s0) = 0, ∂tF (x0, s0) ≥ 0 and ∆F (x0, s0) ≤ 0. (3.8)

From (3.5) and (3.8) we get that

s−20 F (x0, s0) (3.9)

≥[

2

n

‖∇f (x0, s0)‖2 − ∂tf (x0, s0)2 − 2K2 ‖∇f (x0, s0)‖2

].

Let us now set

m =‖∇f (x0, s0)‖2

F (x0, s0).

Then m ≥ 0 and since

F (x, t) = t‖∇f‖2 − α∂tf

,

we get that

F (x0, s0) = s0 mF (x0, s0) − α∂tf (x0, s0) ,

which implies that

∂tf (x0, s0) =F (x0, s0)

αs0

ms0 − 1 . (3.10)


From (3.9) and (3.10) and since α > 1, it follows that

s−20 F (x0, s0) + 2K2 ‖∇f (x0, s0)‖2

≥ 2

n

‖∇f (x0, s0)‖2 − ∂tf (x0, s0)2

=2

n

mF (x0, s0) − F (x0, s0)

αs0

ms0 − 12

≥ 2F (x0, s0)2

n

(m − ms0 − 1

αs0

),

i.e.

s−20 F (x0, s0) + 2K2mF (x0, s0) ≥ 2F (x0, s0)

2

n

(m − ms0 − 1

αs0

)or

F (x0, s0) ≤ nα2

2

1 + 2K2ms20

1 + (α − 1) ms02 =nα2

2

1 + Cv

(1 + v)2 ,

where

v = (α − 1) ms0 ≥ 0 and C = 2K2 s0

(α − 1)> 0.

Finally, v ≥ 0 implies that

1 + Cv ≤ (1 + v)2

(1 +

C

4

),

so

F (x, t) ≤ F (x0, s0) ≤ nα2

2

(1 +

C

4

)≤ nα2

2

(1 +

K2

2 (α − 1)

).

This completes the proof of Step 1.

Step 2. We assume now that M is a complete and non-compact man-ifold with Ric (M) ≥ −K2.


Step 2 is in fact a very technical lemma and the interested readercan consult the book [3], p. 160, Lemma 5.3.4, for a detailed proof.The problem is that we can no longer assume that F has a maximumin M × [0, T ]. The idea which allow us to overcome this difficulty is thefollowing. Let φ be a smooth function on M with support in B (a,R)such that:

0 ≤ φ (x) ≤ φ (a) = 1,

for all x ∈ M and

‖∇φ (x)‖2 ≤ εφ (x) and ∆φ (x) ≥ −δ,

where ε and δ are positive and small as we wish.

Step 2 will follow by applying Step 1 to the function φF which issupported on B (a,R) and consequently φF admits a maximum.

Remark 6 The construction of a function φ as above has its own in-terest and is given in the Appendix.

3.2 Parabolic Harnack Inequality

One of the important consequences of the Li-Yau estimate is the fol-lowing

Theorem 7 (Parabolic Harnack Inequality) Let M be an n-dimensional,complete non-compact Riemannian manifold such that Ric (M) ≥ −K2

and let us (x) be a positive solution of the heat equation on M × [0, T ].Then there is an α > 1 such that

us (x) ≤ us+t (y)

(s + t

t

)αn2

eαd(x,y)2

4s enαK2s4(α−1) ,

for all s, t > 0 such that s + t ≤ T and all x, y ∈ M .

3.2. Parabolic Harnack Inequality 23

Proof. By replacing u by u + ε, we can always assume that u > 0.Set

f = log u

and the Li-Yau estimates says that

‖∇f‖2 − α∂tf ≤ nα2

2t+

nαK2

4 (α − 1). (3.11)

Let

β =αd (x, y)2

4s2+

nαK2

4 (α − 1).

andφ (λ) = u (γ (λ) , λ) λ

αn2 eβλ, λ ∈ [t, t + s] ,

where γ (λ) is geodesic between x and y and which satisfies

∂γ

∂λ(λ) =

d

s.

It easy to check that to prove the theorem, it suffices to show thatthe function λ −→ φ (λ) is increasing.

We have that

log φ (λ) = log u (γ (λ) , λ) +nα

2log λ + βλ

= f (γ (λ) , λ) +nα

2log λ + βλ.

By the chain rule

∂λ log φ (λ) = ∇f (γ (λ) , λ)∂γ

∂λ(λ) +

∂f

∂λ(λ) +

nα

2λ+ β

and by the Li-Yau estimate (3.11)

∂f

∂λ≥ 1

α‖∇f‖2 − nα

2λ− nαK2

4 (α − 1).


Therefore

∂λ log φ (λ) ≥ ∇f (γ (λ) , λ)d

s+

1

α‖∇f‖2 − nα

2λ− nαK2

4 (α − 1)

+nα

2λ+

αd2

4s2+

nαK2

4 (α − 1)

≥ −‖∇f (γ (λ) , λ)‖ d

s+

1

α‖∇f‖2 +

αd2

4s2≥ 0,

which implies that

φ′ (λ) = φ (λ) ∂λ log φ (λ) ≥ 0

and λ −→ φ (λ) is increasing.

Corollary 8 (Parabolic Harnack inequality for Ric (M) ≥ 0) Letus (x) be a positive solution of the heat equation on M × [0, T ], then

us (x) ≤ us+t (y)

(s + t

t

)n2

ed(x,y)2

4s ,


Proof. We put K = 0 and then let α → 1.

3.3 Upper bounds of the heat kernel

In this Section we give some very important applications of the Harnackinequality we proved in Corollary 8 concerning the heat kernel, i.e.the fundamental solution of the heat equation ∂t − ∆, on a completemanifold M .

We recall that the fundamental solution pt (x, y) of the heat equationsatisfies:

(∂t − ∆x) pt (x, y) = 0,

3.3. Upper bounds of the heat kernel 25

for all t > 0, x, y ∈ M , and

limt→0

pt (x, y) = δx (y) .

Further, if for example f ∈ C∞0 (M), then the function

Pt (f) (x) =

∫M

pt (x, y) f (y) dy

is a solution of the heat equation

(∂t − ∆x) (Ptf) (x) = f (x) .

The existence of the heat kernel is highly a non trivial result. Thefollowing can be found in [15], p. 94.

Theorem 9 (existence of the heat kernel) Let M a complete Rie-mannian manifold, then the fundamental solution pt (x, y) is a positiveand C∞ function of t > 0, x, y ∈ M , and satisfies:

pt (x, y) = pt (y, x) , (Symmetry)

For all s ≤ t

pt (x, y) =

∫M

pt−s (x, z) ps (z, y) dz, (semi-group property).

Further, Yau in [19], proved that if Ric (M) ≥ −K2, then the heatsemigroup is Markov i.e.

Pt (1) (x) =

∫M

pt (x, y) dy = 1,

for all t > 0, x ∈ M .


3.3.1 The diagonal upper bound of the heat kernelwhen Ric (M) ≥ 0

Let pt (x, y) be the heat kernel of M . We recall pt (x, y) is a positivesolution of the heat equation on R × M :

∂tpt (x, z) = ∆xpt (x, z) ,

for all t ≥ 0 and x, z ∈ M . Note that the Laplacian acts on the firstvariable.

Theorem 10 (Diagonal upper bound of the heat kernel) Thereis a positive constant c depending only upon n such that

pt (x, x) ≤ c∣∣B (x,√

t)∣∣ ,

for all t > 0 and x ∈ M .

Proof. Corollary 8 implies that

pt (x, z) ≤ pt+s (y, z)

(s + t

s

)n2

ed(x,y)2

4t ,

for all t, s > 0 and x, y, z ∈ M . Let us choose z = x and then integrateover y ∈ B (x, r)

|B (x, r)| pt (x, x) =

∫B(x,r)

pt (x, x) dy

≤∫

B(x,r)

pt+s (y, x)

(s + t

s

)n2

ed(x,y)2

4t dy

≤(

s + t

s

)n2

er2

4t

∫B(x,r)

pt+s (y, x) dy. (3.12)

3.4. The off-diagonal gaussian upper bound 27

But pt (y, x) is the kernel of the semigroup e−t∆, so∫B(x,r)

pt+s (y, x) dy =

∫M

pt+s (y, x)1B(x,r) (y) dy

=[e−(t+s)∆1B(x,r)

](x) . (3.13)

Further, the semigroup e−t∆ is a contraction semigroup on the Lp (M)for all p ≥ 1, [3], p. 22. This gives[

e−(t+s)∆1B(x,r)

](x) ≤ sup

y∈B(x,r)

[e−(t+s)∆1B(x,r)

](y)

=∥∥e−(t+s)∆1B(x,r)

∥∥∞ ≤ ∥∥1B(x,r)

∥∥∞ = 1. (3.14)

Putting together (3.12), (3.13) and (3.14) we get that

|B (x, r)| pt (x, x) ≤(

s + t

s

)n2

er2

4t .

The Theorem follows upon putting s = t = r2.

3.4 The off-diagonal gaussian upper bound

In this section we shall apply the parabolic Harnack inequality to derivethe off diagonal gaussian upper bound of the heat kernel which is themain result of these lectures.

We have seen that if Ric (M) ≥ −K2 and if us (x) is a positivesolution of the heat equation on M × [0, T ], then us (x) satisfies theparabolic Harnack inequality: there is an α > 1 such that

us (x) ≤ us+t (y)

(s + t

t

)αn2

eαd(x,y)2

4s enαK2s4(α−1) , (3.15)


We need the following corollary of the above Harnack inequality(3.15).


Corollary 11 With the same assumptions as above we have the fol-lowing mean value inequality

us (x) ≤(∮

B(x,R)

us+t (y)p dm (y)

) 1p(

s + t

t

)αn2

eαR2

4s enαK2s4(α−1) , (3.16)

for all p > 0, s, t > 0 such that s + t ≤ T and all x, y ∈ M , where∮B(x,R)

u =1

V (x,R)

∫B(x,R)

u.

Proof. Let us+t (y0) the minimal value of us+t (y) in the ball B (x,R).Then

us+t (y0) ≤(∮

B(x,R)

us+t (y)p dm (y)

) 1p

.

Next apply (3.15) for the points x and y0:

us (x) ≤ us+t (y0)

(s + t

t

)αn2

eαd(x,y0)2

4s enαK2s4(α−1)

≤(∮

B(x,R)

us+t (y)p dm (y)

) 1p(

s + t

t

)αn2

eαR2

4s enαK2s4(α−1) .

Let us set

g (x, t) =−d2 (x, y)

(1 + δ) T − t≤ 0.

Bearing in mind that

‖∇xd (x, y)‖2 = 1 a.e.

we can see that g (x, t) satisfies:

1

4‖∇g‖2 + ∂tg = 0

for any t ≤ (1 + δ) T .

The following Lemma plays an important role in the estimates.


Lemma 12 Let M be a complete, noncompact Riemannian manifoldand let u (s, x) be an L2 solution of the heat equation with initial datau0 (x): ⎧⎨⎩

∆u (s, x) = ∂su (s, x) ,

u (s, x) = u0 (x) .(3.17)

Assume further that g (x, t) ∈ C1 (M × R+) and satisfies

1

4‖∇g‖2 + ∂tg = 0, g ≤ 0. (3.18)

Then for all R > 0, T > 0 and y ∈ M we have

∫B(y,R)

eg(x,T )

2 u (T, x)2 dm (x) ≤∫

M

eg(x,0)

2 u0 (x)2 dm (x) . (3.19)

Proof. Let ϕ ∈ C∞0 (M) satisfy

ϕ (x) =

⎧⎨⎩1, x ∈ B (y,R)

0, x /∈ B (y,R + k) ,

‖∇ϕ‖ ≤ C

k, 0 ≤ ϕ ≤ 1,

where C and k are positive constants. Further, for any ε > 0, set

g =g

ε.


Since u (s, x) is a solution of the heat equation we have that

0 = 2

∫ T

0

∫M

ϕ (x)2 eg(s,x)u (s, x) (∆ − ∂s) u (s, x) dm (x) ds

= 2

∫ T

0

∫M

ϕ (x)2 eg(s,x)u (s, x) div∇u (s, x) dm (x) ds

− 2

∫ T

0

∫M

ϕ (x)2 eg(s,x)u (s, x) ∂su (s, x) dm (x) ds

= −2

∫ T

0

∫M

⟨∇ (ϕ (x)2 eg(s,x)u (s, x)),∇u (s, x)

⟩dm (x) ds

−∫ T

0

∫M

ϕ (x)2 eg(s,x)∂su (s, x)2 dm (x) ds

= I1 + I2.

But

I1 = −4

∫ T

0

∫M

ϕ (x) eg(s,x)u (s, x) 〈∇ϕ (x) ,∇u (s, x)〉 dm (x) ds

− 2

∫ T

0

∫M

ϕ (x)2 eg(s,x)u (s, x) 〈∇g (s, x) ,∇u (s, x)〉 dm (x) ds

− 2

∫ T

0

∫M

ϕ (x)2 eg(s,x)u (s, x) ‖∇u‖2 (s, x) dm (x) ds.

Integrating by parts

I2 = −∫ T

0

∫M

ϕ (x)2 eg(s,x)∂su (s, x)2 dm (x) ds

= −∫

M

ϕ (x)2 [eg(s,x)u (s, x)2]T0

dm (x) ds+

−∫ T

0

∫M

ϕ (x)2 u (s, x)2 eg(s,x)∂sg (s, x) dm (x) ds.


Now, by the Cauchy-Schwarz inequality

4ϕ (x) eg(s,x)u (s, x) 〈∇ϕ (x) ,∇u (s, x)〉= 4eg(s,x) 〈u (s, x)∇ϕ (x) , ϕ (x)∇u (s, x)〉

= 4eg(s,x)

⟨(ε − 2

ε

) 12

u (s, x)∇ϕ (x) ,

(ε

ε − 2

) 12

ϕ (x)∇u (s, x)

⟩

≤ 2eg(s,x)

ε − 2

εu (s, x)2 ‖∇ϕ‖2 (x) +

ε

ε − 2ϕ (x)2 ‖∇u‖2 (s, x)

.

Similarly

2ϕ (x)2 eg(s,x)u (s, x) 〈∇g (s, x) ,∇u (s, x)〉

= 2ϕ (x)2 eg(s,x)

⟨√ε

2u (s, x)∇g (s, x) ,

2√ε∇u (s, x)

⟩≤ ϕ (x)2 eg(s,x)

ε

4u (s, x)2 ‖∇g (s, x)‖2 +

4

ε‖∇u (s, x)‖2

.

Using the above computations we get that

0 = I1 + I2 ≤ 2ε

ε − 2

∫ T

0

∫M

eg(s,x)u (s, x)2 ‖∇ϕ‖2 (x) dm (x) ds

+

∫ T

0

∫M

eg(s,x)ϕ (x)2[ε4‖∇g‖2 (s, x) + ∂sg (s, x)

]u (s, x)2 dm (x) ds

−∫

M

ϕ (x)2 [eg(s,x)u (s, x)2]T0

dm (x) ds.

Further, the assumptions on ∇ϕ and the fact that

ε

4‖∇g‖2 (s, x) + ∂sg (s, x) =

1

4ε‖∇g‖2 (s, x) +

1

ε∂sg (s, x) = 0


imply that∫M

ϕ (x)2 [eg(s,x)u (s, x)2]T0

dm (x) ds

≤ 2ε

ε − 2

C2

k2

∫ T

0

∫M

eg(s,x)u (s, x)2 dm (x) ds

+

∫ T

0

∫M

eg(s,x)ϕ (x)2[ε4‖∇g‖2 (s, x) + ∂sg (s, x)

]u (s, x)2 dm (x) ds

≤ 2ε

ε − 2

C2

k2

∫ T

0

∫M

eg(s,x)u (s, x)2 dm (x) ds.

Noting that u ∈ L2 and g (s, x) ≤ 0, we can led k → ∞ and notingthat ϕ (x) = 1 on B (y,R) we get that∫

B(y,R)

[eg(s,x)u (s, x)2]T

0dm (x) ds ≤ 0

i.e. ∫B(y,R)

eg(s,T )

ε u (T, x)2 dm (x) ds ≤∫

B(y,R)

eg(s,0)

ε u (0, x)2 dm (x) ds

=

∫B(y,R)

eg(s,0)

ε u0 (x)2 dm (x) ds.

The Lemma follows by taking ε = 2.

Remark 13 From the Lemma above, it follows that if u0 (x) ≡ 0, thenu (t, x) ≡ 0 for all t > 0. This shows that the heat equation (3.17) withinitial data u0 ∈ L2 has at most one solution.

Lemma 14 Let pt (x, y) be the heat kernel of a complete Riemannianmanifold M . For any ρ > 0 and T > 0. Set

F (y, t) =

∫M\B(x,ρ)

pT (x, z) pt (z, y) dz.


Then for any δ > 0 and R > 0,∫B(x,R)

F (y, (1 + δ) T )2 dy ≤ eR2

2δT e−ρ2

2(1+2δ)T

∫M\B(x,ρ)

pT (x, z)2 dz.

(3.20)

Proof. By the definition of F (y, t), F (y, t) is a solution of the heatequation with initial value

F (y, 0) =

∫M\B(x,ρ)

pT (x, z) p0 (z, y) dz

=

∫M\B(x,ρ)

pT (x, z) δz (y) dz

=

pT (x, y) if y ∈ M\B (x, ρ)

0 if y ∈ B (x, ρ) .

Moreover, F (y, 0) ∈ L2. Therefore we can apply Lemma 12 with thechoice

g (x, t) =−r2 (x, y)

(1 + δ) T − t

and get that for any t ≤ (1 + 2δ) T∫B(x,R)

e−r2(x,y)

2(1+δ)T−t F (y, t)2 dy ≤∫

M

e−r2(x,y)2(1+δ)T F (y, 0)2 dy.

If we choose t = (1 + δ) T we have that

e−R2

δT

∫B(x,R)

F (y, (1 + δ) T )2 dy ≤∫

M\B(x,ρ)

e−r2(x,y)2(1+δ)T pT (x, y)2 dy

and the proof is complete.

Remark 15 Let us note that by the definition of F (y, t) and (3.20) itfollows that∫

B(x,R)

F (y, (1 + δ) T )2 dy ≤ eR2

δT e−ρ2

2(1+δ)T

∫M\B(x,ρ)

pT (x, y)2 dy

= eR2

δT e−ρ2

2(1+δ)T F (y, T ) .


If we choose ρ = 0, i.e. when

F (y, t) =

∫M

pT (x, z) pt (z, y) dz,

then Lemma 14 says that for any δ > 0, T > 0 and R > 0∫B(x,R)

F (y, (1 + δ) T )2 dy ≤ eR2

δT F (y, T ) . (3.21)

3.5 Proof of the upper gaussian estimate

We are now in position to prove one of the main result of this chapter

Theorem 16 (Li-Yau upper Gaussian estimate) Let M be a com-plete Riemannian manifold with Ric (M) ≥ −K2. Let pt (x, y) be theheat kernel of M , i.e. the fundamental solution of ∂tu (x, t) = ∆u (x, t).Then there are positive constants C (δ, n) and C1 (n) such that

pt (x, y) ≤ C (δ, n)

V(x,√

t) 1

2 V(y,√

t) 1

2

e−d(x,y)2

(4+δ)t eC1(n)δK2t, (3.22)

for all t > 0 and x, y ∈ M .

Before proceed to the proof we make the following interesting re-marks.

Remark 17 1. C (δ, n) −→ +∞ as δ → 0.2. If K2 = 0 i.e. if Ric (M) ≥ 0, then (3.22) is written as

pt (x, y) ≤ C (δ, n)e−

d(x,y)2

(4+δ)t

V(x,√

t) 1

2 V(y,√

t) 1

2

, (3.23)


3.5. Proof of the upper gaussian estimate 35

3. If Ric (M) ≥ 0, then by the Bishop comparison theorem

V (x, r)

V (x, s)≤(r

s

)n

,

for all r ≥ s. So that

V(x,√

t)

V(y,√

t) ≤ V

(y,√

t + d (x, y))

V(y,√

t) ≤

(√t + d (x, y)√

t

)n

,

which combined with (3.23) yields

pt (x, y) ≤ C (δ, n)e−

d(x,y)2

(4+δ)t

V(x,√

t) 1

2 V(x,√

t) 1

2

(1 +

d (x, y)√t

)n/2

≤ C2e−

d(x,y)2

ct

V(x,√

t) , (3.24)

since

C (δ, n) e−d(x,y)2

(4+δ)t

(1 +

d (x, y)√t

)n/2

≤ C2e− d(x,y)2

ct

for some positive constants C2 and c. Similarly, we obtain the symmet-ric estimate of (3.24):

pt (x, y) ≤ C2e−

d(x,y)2

ct

V(y,√

t) , (3.25)


Proof of the upper bound. The proof is given in two steps.Step 1. We assume that d (x, y)2 ≤ 4t.Let

F (y, t) =

∫M



Then by (3.21), for any δ > 0, T > 0 and R > 0∫B(x,R)

F (y, (1 + δ) T )2 dy ≤ eR2

δT F (y, T ) . (3.26)

Further F (y, t) is a positive solution of the heat equation and conse-quently it satisfies the mean-value Harnack inequality (3.16) with

t = T, t + s = (1 + δ) T and p = 2

i.e.

F (x, T ) ≤(∮

B(x,R)

F (y, (1 + δ) T )2 dm (y)

) 12

(1 + δ)αn2 e

αR2

4δT enαK2δT2(α−1) .

If we combine it with (3.26) we get that

F (x, T ) ≤ 1√V (x,R)

eR2

2δT F (y, T )12 (1 + δ)

αn2 e

αR2

4δT enαK2δT2(α−1)

i.e.

F (x, T ) ≤ 1

V (x,R)e

R2

δT (1 + δ)αn eαR2

2δT enαK2δT(α−1) .

If we choose R2 = 2T we get

F (x, T ) ≤ 1

V(x,√

2T)e

1+αδ (1 + δ)αn e

nαK2δT(α−1) (3.27)

= C (n, α, δ)e

nαK2δT(α−1)

V(x,√

2T) ,

whereC (n, α, δ) = e

1+αδ (1 + δ)αn .

Now, observe first that

F (x, T ) =

∫M

pT (x, y)2 dm (y) ,


and secondly that pt (x, y) satisfies the semigroup property:

pt+s (x, z) =

∫M

pt (x, y) ps (y, z) dm (y) .

Combining these with (3.27) we get that

pt (x, y) =

∫M

p t2(x, y) p t

2(y, z) dm (y)

≤(∫

M

p t2(x, y)2 dm (y)

) 12(∫

M

p t2(y, z)2 dm (y)

) 12

≤ C (n, α, δ)e

nαK2δt(α−1)

V(x,√

t) 1

2 V(y,√

t) 1

2

which competes the proof of step 1.

Step 2. We assume that d (x, t)2 > 4t.In this case we set

Fρ (y, t) =

∫M\B(x,ρ)


By Lemma 14 we have that for all δ > 0 and R > 0,∫B(x,R)

Fρ (y, (1 + δ) T )2 dy

≤ eR2

2δT e−ρ2

2(1+2δ)T

∫M\B(x,ρ)

pT (x, z)2 dz

= eR2

2δT e−ρ2

2(1+2δ)T Fρ (x, T ) . (3.28)

Now, the function Fρ (y, t) as a positive solution of the heat equationsatisfies the mean-value Harnack inequality (3.16) with t = T , t + s =(1 + δ) T and p = 2:

Fρ (x, T ) ≤(∮

B(x,R)

Fρ (y, (1 + δ) T )2 dm (y)

) 12

(1 + δ)αn2 e

αR2

4δT enαK2δT2(α−1)


i.e.

Fρ (x, T )2 ≤ (1 + δ)αn eαR2

2δT enαK2δT(α−1)

V (x,R)

∫B(x,R)

Fρ (y, (1 + δ) T )2 dm (y) .

Combining with (3.28) we get that

Fρ (x, T )2 ≤ (1 + δ)αn

V (x,R)e

(1+α)R2

2δT enαK2δT(α−1) e−

ρ2

2(1+2δ)T Fρ (x, T ) .

With R2 = (1 + δ)−1 T , this becomes

Fρ (x, T ) ≤ (1 + δ)αn

V

(x,√

(1 + δ)−1 T

)enαK2δT(α−1) e−

ρ2

2(1+2δ)T . (3.29)

Let us now apply the mean-value Harnack inequality to the heat kernelpt (x, y) on the ball B (y, d (x, y) − ρ) which is contained in M\B (x,R).If we choose T = (1 + δ) t, we have that

pt (x, y) ≤(∮

B(y,d(x,y)−ρ)

pT (x, ξ)2 dξ

) 12(

T

t

)nα2

eα(d−ρ)2

4(T−t) eαnK2

2(α−1)(T−t)

≤ (1 + δ)αn2

V (y, d (x, y) − ρ)12

eα(d−ρ)2

4δt eαnK2

2(α−1)δt

(∫M\B(x,R)

pT (x, ξ)2 dξ

) 12

≤ (1 + δ)αn2

V (y, d (x, y) − ρ)12

eα(d−ρ)2

4δt eαnK2

2(α−1)δtFρ (x, T )

12

≤ C (α, δ)

V(x,√

t) 1

2 V (y, d (x, y) − ρ)12

eα(d−ρ)2

4δt eαnK2

2(α−1)δ(2+δ)te−

ρ2

4(1+2δ)(1+δ)t ,


where the last inequality follows from the estimate (3.29) of Fρ (x, T ).Choosing ρ such that d − ρ =

√t we get

pt (x, y) ≤ C (α, δ)e

αt4δt e

αnK2

2(α−1)δ(2+δ)te−

(d(x,y)−√t)2

4(1+2δ)(1+δ)t

V(x,√

t) 1

2 V(y,√

t) 1

2

≤ C (α, δ)

V(x,√

t) 1

2 V(y,√

t) 1

2

eC1K2δte−d(x,y)2

4(1+2δ)(1+δ)t ,

where in the last inequality we have used the Schwarz inequality

ρ2 =(d (x, y) −√

t)2

≥ d (x, y)2

(1 + δ)− t

δ.

The Harnack inequality gives also a lower gaussian bound of the heatkernel, in the case when Ric (M) ≥ 0, which is of the same order as theupper bound. We state the result without proof. The interest readercan found the proof in [15], p. 172.

Theorem 18 (Lower gaussian bound of the heat kernel) Let Mbe a complete, noncompact Riemannian manifold with Ric (M) ≥ 0,then its heat kernel pt (x, y) satisfies

pt (x, y) ≥ c (n, ε)

V(x,√

t)e− d(x,y)2

(4−ε)t . (3.30)

Chapter 4

Heat kernel bounds oncompact manifolds

The Harnack inequality on complete noncompact manifolds also holdson compact manifolds and the proof is exactly the same.

Theorem 19 Let M be an n−dimensional, complete compact Rieman-nian manifold possibly with boundary. If ∂M = ∅ we assume that ∂Mis convex. Assume that Ric (M) ≥ −K2. If us (x) be a positive solutionof the heat equation on M × [0,∞] and

∂us

∂ν

∣∣∣∣∂M

= 0,

(if ∂M = ∅). Then there is an α > 1 such that

us (x) ≤ us+t (y)

(s + t

t

)αn2

eαd(x,y)2

4s enαK2s4(α−1) ,

for all s, t > 0 such that s + t > 0 and all x, y ∈ M .

The above Harnack inequality gives an upper gaussian estimate forthe heat kernel of M with Neumann boundary conditions. The proofis exactly the same as in the case of noncompact manifolds.

41

42 4. Heat kernel bounds on compact manifolds

Theorem 20 Let M be an n−dimensional, complete compact Rieman-nian manifold possibly with boundary. If ∂M = ∅ we assume that ∂Mis convex. Assume that Ric (M) ≥ −K2. Let pt (x, y) be the heat ker-nel with Neumann boundary conditions. Then there is a constant c > 0such that

pt (x, y) ≤ ce−d(x,y)2/ctecK2t

V(x,√

t)1/2

V(y,√

t)1/2

,


Remark 21 If K = 0, then using the volume comparison theorems,the above heat kernel satisfies:

pt (x, y) ≤ ce−d(x,y)2/ct

V(x,√

t)


As for the lower bound of pt (x, y), we have

Theorem 22 Let M be an n−dimensional, complete compact Rieman-nian manifold possibly with boundary. If ∂M = ∅ we assume that ∂Mis convex. Assume that Ric (M) ≥ 0. Let pt (x, y) be the heat kernelwith Neumann boundary conditions. Then there is a constant c > 0such that

pt (x, y) ≥ ce−d(x,y)2/ct

V(x,√

t)


Chapter 5

Applications of heat kernelbounds

We shall finish this chapter with some important applications of thegaussian bounds of the heat kernel.

5.1 Estimates of the Green function

We shall in fact give the estimates of the Green function of a completeRiemannian manifold with Ric (M) ≥ 0.

Let us first recall that the Green function G (x, y) of M is defined by

G (x, y) =

∫ ∞

0

pt (x, y) dt, x = y.

if the integral above converges. If for example M = R2, then

G (x, y) =

∫ ∞

0

e−‖x−y‖2

4t

4πtdt

43

44 5. Applications of heat kernel bounds

and it is easy to check that the integral above diverges. In fact

G (x, y) ≥∫ ∞

1

e−‖x−y‖2

4t

4πtdt

andg (t) = e−

at , a > 0,

is increasing. Thus g (t) ≥ g (1) for all t > 1. So,

G (x, y) ≥ e−‖x−y‖2

4

4π

∫ ∞

1

dt

t= +∞.

The situation is quite different in the case of Rn, n ≥ 3:

G (x, y) =

∫ ∞

0

e−‖x−y‖2

4t

(4πt)n/2dt =

∫ ∞

0

e−‖x−y‖2

4u

(4π)n/2u

n2du

u2

=1

(4π)n/2

∫ ∞

0

e−‖x−y‖2

4uu

n2−2du

‖x−y‖24

u=w=

1

(4π)n/2

∫ ∞

0

e−w

(4w

‖x − y‖2

)n2−2

4

‖x − y‖2dw

=1

(4π)n/2

(4

‖x − y‖2

)n2−1 ∫ ∞

0

e−wwn2−2dw

=1

(4π)n/2Γ(n

2− 1)( 4

‖x − y‖2

)n2−1

=Γ(

n2− 1)

4πn/2 ‖x − y‖n−2 ,

where for the computation of the integral∫ ∞

0

e−wwn2−2dw

5.1. Estimates of the Green function 45

we have used the definition of the function Γ.

It can be checked that G (x, y) is positive and that G (x, y) is thefundamental solution of the Laplace equation i.e.

∆G (x, y) = −δx (y) . (5.1)

From (5.1) it follows that if for example f ∈ C∞0 (M), then

G (f) (x) =

∫M

G (x, y) f (y) dm (y)

satisfies the Poisson equation

∆G (f) (x) = f (x) .

The following estimates of G (x, y) are a consequence of the gaussianbounds of the heat kernel.

Theorem 23 Let M be a complete, noncompact Riemannian manifoldwith Ric (M) ≥ 0, then there are positive constants c1 (n) and c2 (n)such that

c1 (n)

∫ ∞

d(x,y)2

dt

V(x,√

t) ≤ G (x, y) ≤ c2 (n)

∫ ∞

d(x,y)2

dt

V(x,√

t) .

Proof. We have that

G (x, y) =

∫ ∞

0

pt (x, y) dt

=

∫ d2

0

pt (x, y) dt +

∫ ∞

d2

pt (x, y) dt.

The upper gaussian bound of the heat kernel implies that∫ ∞

d2

pt (x, y) dt ≤ c (n)

∫ ∞

d2

e−d(x,y)2/ct

V(x,√

t) dt

≤ c (n)

∫ ∞

d2

dt

V(x,√

t) .


To finish the proof we will show that there is c′ (n) > 0 such that∫ d2

0

pt (x, y) dt ≤ c′ (n)

∫ ∞

d2

pt (x, y) dt.

By the upper bound again, we have that∫ d2

0


∫ d2

0

e−d(x,y)2/ct

V(x,√

t) dt

s= d4

t= c (n)

∫ ∞

d2

e−s

cd2

V(x, d2√

s

) d4

s2ds

= c (n)

∫ ∞

d2

V (x,√

s)

V(x, d2√

s

) 1

V (x,√

s)e−

scd2

d4

s2ds.

Butd2

√s≤ √

s

and by the Bishop comparison theorem we get that

V (x,√

s)

V(x, d2√

s

) ≤(√

sd2√

s

)n

=( s

d2

)n

.

This gives∫ d2

0


∫ ∞

d2

( s

d2

)n e−s

cd2

V (x,√

s)

d4

s2ds

≤ c (n)

∫ ∞

d2

( s

d2

)n 1

V (x,√

s)

(d2

s

)n−2d4

s2ds

= c (n)

∫ ∞

d2

ds

V (x,√

s)

and the proof of upper bound of the Green function is complete.

5.2. Liouville Theorem 47

The lower bound of G (x, y) follows directly by the lower bound ofthe heat kernel:

G (x, y) =

∫ ∞

0

pt (x, y) dt ≥ G (x, y) ≥ c (n)

∫ ∞

d2

e−d(x,y)2/ct

V(x,√

t) dt

≥ c (n) e−1/c

∫ ∞

d2

dt

V(x,√

t) .

Remark 24 A slightly weaker version of this result was proved first byN. Th. Varopoulos, (cf. [18]).

5.2 Liouville Theorem

In this Section we give some immediate consequences of the Li-Yauestimate to potential theory. M is as usual an n−dimensional, completemanifold with Ric (M) ≥ −K2.

The first result gives an upper bound of eigenvalues associated topositive eigenfunctions.

Theorem 25 Let v be a positive solution of the eigenvalue equation

−∆v = λv.

Then

λ ≤ nK2

4,

and‖∇v‖2

v2≤ nK2

2− λ +

(nK2

4− λ

)1/2 (nK2

)1/2. (5.2)

Proof. If we putu (t, x) = v (x) e−λt,


then u (t, x) is a positive solution of the heat equation on R+ × M . Sothe Li-Yau estimate applies and give

‖∇v‖2

v2+ λα ≤ nα2

2

1

t+

K2

2 (α − 1)

,

for all x ∈ M , t > 0 and α > 1. Putting β = α − 1 and letting t → ∞,we get

0 ≤ ‖∇v‖2

v2≤ −λ (β + 1) +

nK2

4β(β + 1)2 (5.3)

for all β > 0. Thus

λ (β + 1) ≤ nK2

4β(β + 1)2

or

λ ≤ nK2

4β(β + 1) ≤ nK2

4

since the function g (x) = 1 + 1x

is decreasing on (0,∞) and f (∞) = 1.To prove (5.2), we choose the β which minimizes the RHS of (5.3).

Inequality (5.2) allow us to prove that there are no nonconstant pos-itive harmonic functions on M if Ric (M) ≥ 0. This is a generalizationof the well known Liouville theorem.

Corollary 26 (Liouville Theorem) If Ric (M) ≥ 0, then there areno positive harmonic functions on M other that the constants.

Proof. A positive harmonic function satisfies (5.2) with λ = 0 andK = 0, i.e.

‖∇v‖2

v2≤ 0 ⇒ ‖∇v‖2 = 0,

which by the mean value theorem implies that v = c.

Remark 27 When Ric (M) ≤ 0, then in general there exists manynontrivial harmonic functions. See for example [8, 9, ?].

5.3. Lichnerowicz Theorem 49

5.3 Lichnerowicz Theorem

The Lichnerowicz theorem (1958) gives lower bounds of the smallesteigenvalue λ1 of the Laplacian on a complete manifold without bound-ary. It is the first result in this direction i.e. when curvature conditionsare take in account.

Theorem 28 Let M be a complete n−dimensional Riemannian man-ifold without boundary and such that

Ric M ≥ (n − 1) k.

Let λ1 be the smallest eigenvalue of the Laplacian i.e. λ1 is the inf ofλ ≥ 0 for which there is a function fλ in C∞ ∩ L2 such that

∆fλ = −λfλ,

thenλ1 ≥ nk

Proof. Let fλ1 = f be an eigenfunction with eigenvalue λ1. Thenby Lemma 4

∆‖∇f‖2 ≥ 2

n(∆f)2 + 2 〈∇ (∆f) ,∇f〉 − 2K2 ‖∇f‖2 .

Integrating on M the above inequality we get

0 =

∫∆‖∇f‖2

≥ 2

n

∫(∆f)2 + 2

∫〈∇ (λ1f) ,∇f〉 − 2 (n − 1) k

∫‖∇f‖2

=2

n

∫λ2

1f2 + 2λ1

∫f 2 + 2 (n − 1) k

∫(∆f) f

=

(2

nλ2

1 + 2λ1 + 2 (n − 1) kλ1

)∫f 2.


Since ∫f 2 = 0

we get (2

nλ2

1 + 2λ1 + 2 (n − 1) kλ1

)≤ 0

from which we deduce the bound of λ1.

Chapter 6

Estimates of the eigenvalueson compact manifolds

In this chapter we shall study the spectrum of the Laplacian on a com-pact manifold. For that, the estimates of the heat kernel plays animportant role.

The classical theorem of Hodge (cf. [13], p.32), states that if M isa compact, connected and oriented Riemannian manifold, then thereis an orthonormal basis of L2 (M) consisting of eigenfunctions of theLaplacian −∆. All the eigenvalues

0 = λ0 < λ1 ≤ λ2 ≤ · · · ≤ λn ≤ · · ·

are positive expect that 0 is an eigenvalue with multiplicity one. Eacheigenvalue has finite multiplicity and the eigenvalues accumulate onlyat infinity.

The study of the spectrum of the Laplacian is very important andhas a long history. The spectrum, which is an analytic object, is relatedto the geometry of the manifold. To illustrate this astonishing fact wejust mention the celebrated Weyl formula :

51

52 6. Estimates of the eigenvalues on compact manifolds

Let M ⊂ Rn be a connected compact domain with a smooth bound-

ary and let (λj) the eigenvalues of the Laplacian with Dirichlet data(see below for precise definitions). Set

N (λ) = # λj ≤ λ .

ThenN (λ) ∼ ωn

(2π)n V ol (M) λn/2, as λ → ∞, (6.1)

where ωn is the volume of the unit sphere in Rn.

Thus, if we can estimate the eigenvalues, then we shall have a es-timate of N (λ) and thus, by the Weyl formula we shall estimate thevolume of M which of course is a geometric object.

A second important problem related to the Weyl formula, is thePolya conjecture. To explain it we need first to define the Laplacianwith Dirichlet or Neumann data.

6.1 The Laplace operator on a compact

manifold

Let M be a complete and compact manifold with boundary ∂M (∂Mmay be empty).

Let us first define properly the Laplace operator as an unboundedoperator on L2 (M). For that we define the following norm on C∞ (M):

‖|ϕ|‖ :=

∫M

‖ϕ‖2 +

∫M

‖∇ϕ‖2 < ∞.

We denote by H1 the completion of C∞ (M) with respect to theabove norm. The space H1 is a Hilbert space, it is of course a subspaceof L2 (M) and it is known as the Sobolev space H1.

We also denote by H10 the completion of C∞

0 (M) with respect to theabove norm. The space H1

0 is a Hilbert space.

6.1. The Laplace operator on a compact manifold 53

From the general theory of Sobolev spaces (cf. Dwse anafora), itis known that if M is complete, then

H1 = H10 .

The space H1 will be the domaine of the Laplacian.

For the study of the spectrum of the Laplacian we shall consider thefollowing cases.

First case: ∂M = ∅. In this case ∆ is selfadjoint elliptic operator onH1. By the spectral theory of selfadjoint operators (Dwse anafora),we know that ∆ has discrete eigenvalues

0 = λ0 < λ1 ≤ λ2 ≤ · · · ≤ λn ≤ · · ·and the corresponding eigenfunctions ϕi satisfying

∆ϕi = −λiϕi

belongs in C∞∩H1 and can be chosen so that ϕi forms an orthonor-mal basis of H1.

Second case: ∂M = ∅. In this case we specify some boundary con-ditions so that ∆ is selfadjoint. Usually we have two kinds of boundaryconditions.

Dirichlet boundary conditions : In this case we impose the con-dition that the eigenfunctions ϕi belongs in C∞ ∩ H1 and vanisheson the boundary ∂M . So, they are solutions of the Dirichlet problem

∆ϕi = −λiϕi, ϕi|∂M = 0.

With this choice, ∆ is selfadjoint on H1 and dom ∆ = H1.The corresponding eigenvalues satisfy

0 < λ1 ≤ λ2 ≤ · · · ≤ λn ≤ · · ·


and ϕi forms an orthonormal basis of H1.

Neumann boundary conditions : In this case dom (∆) = H1, itseigenfunctions ϕi satisfy

∂νϕi|∂M = 0,

they are C∞ and forms an orthonormal basis of H1. The correspondingeigenvalues satisfy

0 = λ0 < λ1 ≤ λ2 ≤ · · · ≤ λn ≤ · · ·

6.2 The min-max principle

In the spectral theory of ∆ the min-max principle plays an importantrole. It can be formulated as follows.

For simplicity, let D be:

• If ∂M = ∅, then

D =

f ∈ H1 :

∫M

f = 0

,

• If ∂M = ∅, and Dirichlet boundary condition is posed, then

D = H1,

• If ∂M = ∅, and Neumann boundary condition is posed, then

D =

f ∈ H1 :

∫M

f = 0

.

6.3. The Polya conjecture 55

In all the cases above, ∆ is selfadjoint on D and we can find anorthonormal basis ϕi of D with

∆ϕi = −λiϕi,

ϕi ∈ C∞ ∩ D such that

λ1 = inff∈D

∫M‖∇f (x)‖2 dx∫

M|f (x)|2 dx

(6.2)

and

λj = inff∈D

∫M‖∇f (x)‖2 dx∫

M|f (x)|2 dx

:

∫M

f (x) ϕi (x) dx = 0, 1 ≤ i ≤ j − 1

.

(6.3)

6.3 The Polya conjecture

A second celebrated conjecture is the Polya conjecture which assertsthat the kth eigenvalue λk (resp. µk) for the Dirichlet problem (resp.for the Neumann problem) of an bounded domain M of R

n satisfies

λk ≥ c (n)

(k

vol (M)

)2/n

(resp.

µk ≤ c (n)

(k

vol (M)

)2/n

) with

c (n) = 4π2(ωn−1

n

)−2/n

.

Finally, an other important problem is the estimate of the lowereigenvalue λ1 of the Dirichlet problem. These estimates have nice appli-cations in geometry (see below for the geometric implications of Cheng’stheorem). For more details see the survey article of Yau in [15], p. 113.


In the sequel we shall study some important aspects of the spectrumof the Laplacian on a compact manifold. We start with the relation ofthe eigenvalues and especially of λ1 with

6.4 Some important inequalities

It follows from (6.2) that for any c ≤ λ1 the following inequality holds

c

∫M

|f (x)|2 dx ≤∫

M

‖∇f (x)‖2 dx, f ∈ D, (6.4)

i.e. λ1 is the biggest constant for which the above inequality holds.This type of inequality is called Poincare inequality and it is one of themost important inequalities in the theory of P.D.E.

Remark 29 Saloff-Coste proved in [14] that on a Riemannian man-ifold the conjunction of the “doubling volume property”: ∃c > 0 suchthat for all x ∈ M and r > 0,

V (x, 2r) ≤ cV (x, r) (6.5)

and of the following Poincare inequality∫B(x0,r)

|f(x) − fB|2 dx ≤ Cr2

∫B(x0,2r)

‖∇f(x)‖2 dx, (6.6)

where is fB is the mean value of f on B (x0, r), is equivalent with thevalidity of the gaussian upper and lower bounds of the heat kernel:

pt (x, y) ∼ e−d(x,y)2/ct

V(x,√

t) .

This fact is valid also on graphs (the skeleton of manifolds) where thereis no the notion of curvature (see [5]).

6.4. Some important inequalities 57

An other important inequality is the Sobolev inequality : Let M be andimensional compact manifold with boundary, then there is a constantc > 0 such that(∫

M

|f (x)| nn−1 dx

)n−1n

≤ c

∫M

‖∇f (x)‖ dx, f ∈ D. (6.7)

If M is noncompact, then the Sobolev inequality may not valid. Weshall see below that its validity is related to the volume growth of M .

6.4.1 Equivalence of Sobolev and isoperimetric in-equalities

The isoperimetric inequality is the following:Let M be a Riemannian manifold. Then there is a constant c > 0

such that for any domain Ω with compact closure in M

vol (Ω)n

n−1 ≤ cvol (∂Ω) . (6.8)

Theorem 30 The Sobolev and the isoperimetric inequality are equiv-alent.

Proof. To prove that Sobolev implies the isoperimetric we considerthe function

fε (x) =

⎧⎨⎩1, if x ∈ Ω and d (x, ∂Ω) ≥ ε,

d(x,∂Ω)ε

, if x ∈ Ω and d (x, ∂Ω) < ε,0 otherwise.

Applying (6.7) to fε and letting ε → 0 we get the isoperimetric inequal-ity.

To prove that (6.8) implies (6.7), we needThe co-area formula. Let M be compact manifold with boundary.Then for any nonnegative function g on M , we have∫

M

g (x) dx =

∫ +∞

−∞

(∫x∈M :f(x)=α

g (x)

‖∇f (x)‖dσ (x)

)dα, (6.9)


where dσ is the measure of the “surface” x ∈ M : f (x) = α. (nakoitaxtei) (For the proof see [6])

Let us return to our proof. For simplicity, we assume that f ≥ 0 andwe apply (6.9) with g = ‖∇f (x)‖. We get∫

M

‖∇f (x)‖ dx =

∫ +∞

0

(∫x∈M :f(x)=α

dσ (x)

)dα

=

∫ +∞

0

area x ∈ M : f (x) = α dα

which combined with the isoperimetric inequality gives:

∫M

‖∇f (x)‖ dx =

∫ +∞

0

area x ∈ M : f (x) = α dα

≥ c

∫ +∞

0

(vol x ∈ M : f (x) > α)n−1n dα.

Since∫M

|f (x)| nn−1 dx =

∫ ∞

0

vol

x : |f (x)| nn−1 > λ

dλ

=n

n − 1

∫ ∞

0

vol x : |f (x)| > λλ1

n−1 dλ

it suffices to prove that

ϕ (α) :=

∫ +∞

0

(vol x ∈ M : f (x) > α)n−1n dα (6.10)

≥ c

(∫ ∞

0

vol x : |f (x)| > αα1

n−1 dα

)n−1n

:= ψ (α) .

But

ϕ (0) = ψ (0)


and

ϕ′ (t) ≥(

n

n − 1

)n−1n

ψ′ (t) .

Integrating the above inequality we get that

ϕ (∞) ≥(

n

n − 1

)n−1n

ψ (∞)

which is (6.10).

Remark 31 Let us now mention an important fact. The link betweenthe Sobolev and isoperimetric inequalities and the volume growth. Wetreat the simple case of R

n.Observe that

∂t |B (x, t)| = voln−1 (∂B (x, t)) .

Now, apply the isoperimetric inequality for the ball B (x, t). We get

|B (x, t)| nn−1 ≤ cvol (∂B (x, t)) = c∂t |B (x, t)|

which implies that|B (x, t)| ≥ ctn. (6.11)

The connection between Sobolev and volume growth is studied in thecontext of nilpotent Lie groups by N. Th. Varopoulos. He proved, usinghis own heat kernel estimates, that (6.11) is in fact equivalent withSobolev, (cf. [17]).

6.4.2 A lower bound of the first eigenvalue

In this section we shall obtain a lower bound of the first eigenvalue λ1

of the Laplacian with Dirichlet data, by using only the co-area formula.For that we need the Cheeger constant hD: If M is a compact man-

ifold with ∂M = ∅, the Cheeger constant is defined by

hD = inf

vol (∂Ω)

volΩ: Ω ⊂⊂ M

.


Theorem 32 (Cheeger) Let M be a compact manifold with ∂M = ∅and λ1 the first eigenvalue of the Laplacian with Dirichlet data. Then

λ1 ≥ 1

4h2

D.

Proof. Let ϕ1 be an eigenfunction with respect to the eigenvalue λ1.Then, by applying a strong maximum principle, one can see that ϕ1

does not change sign (cf. [2], p.160) and we may assume that ϕ1 > 0.Now, the Green formula for C2 compactly supported functions h, f ,

(cf. [2], p. 144) asserts that∫M

h∆f + 〈∇h,∇f〉 =

∫∂M

h 〈ν,∇f〉 (6.12)

where ν is the outward unit normal of ∂M .If we apply (6.12) with f = h = ϕ1, the r.h.s is 0 and we get∫

M

ϕ1∆ϕ1 +

∫M

‖∇ϕ1‖2 = 0.

But

∆ϕ1 = −λ1ϕ1,

so,

λ1

∫M

ϕ21 =

∫M

‖∇ϕ1‖2 (6.13)

Let us observe that if there is a constant c0 > 0 such that∫M

‖∇f‖ ≥ c0

∫M

f (6.14)

for all C∞ f with f |∂M = 0, then

λ1 ≥ 1

4c20. (6.15)


Indeed, (6.14) with f = ϕ21 gives

c0

∫M

ϕ21 ≤∫

M

∥∥∇ϕ21

∥∥ = 2

∫M

|ϕ1| ‖∇ϕ1‖C.S≤ 2

(∫M

|ϕ1|2)1/2(∫

M

‖∇ϕ1‖2

)1/2

i.e.c20

4

∫M

ϕ21 ≤∫

M

‖∇ϕ1‖2 (6.13)= λ1

∫M

ϕ21

which implies that λ1 ≥ 14c20.

Next, to complete the proof, we shall prove that (6.14) is valid withc0 = hD. By the co-area formula∫

M

‖∇f‖ =

∫ +∞

−∞

(∫f=α

1

)dα

=

∫ +∞

−∞area f = α dα

=

∫ +∞

−∞

area f = αvol f > α vol f > α dα

≥ infα

area f = αvol f > α

∫ +∞

−∞vol f > α dα

= infα

area f = αvol f > α

∫M

|f |

≥ hD

∫M

|f | .

Remark 33 An analogue of Theorem 32 is valid in the case of the firsteigenvalue with Neumann data, (cf. [15], p.91)


6.4.3 Upper bounds for the first eigenvalue

Let V (k, r) be the volume of the ball with radius r in the space of con-stat Ricci curvature (n − 1) k. S.Y. Cheng proved a comparison theo-rem for heat kernels. If fact he proved that if M is an n−dimensinalcomplete manifold with Ric (M) ≥ (n − 1) k, and let pt (x, y) andEt (x, y) the heat kernel on M and on the space of constat Ricci curva-ture (n − 1) k. Then

pt (x, x) ≥ Et (x, x) .

Using the above comparison of the heat kernels one can prove the fol-lowing theorem (cf. [15], p. 104)

Theorem 34 (Cheng) Let M be a compact manifold without bound-ary, and Ric (M) ≥ (n − 1) k, then

λ1 (M) ≤ λ1

(V

(k,

d

2m

))where

d = diam (M) .

Remark 35 Cheng’ s theorem implies that a compact manifold, whoseRicci curvature is not less that (n − 1) and whose diameter is π, isisometric to Sn (0, 1).

6.5 Estimates of higher eigenvalues via the

heat kernel

6.5.1 The Polya conjecture

The Polya conjecture deals with the eigenvalues of the Laplacian on acompact domain of R

n with both Dirichlet or Neumann data.

6.5. Estimates of higher eigenvalues via the heat kernel 63

Let Ω be a bounded domain of Rn and let

0 < λ1 ≤ λ2 ≤ · · · ≤ λn ≤ · · ·

be the eigenvalues of the Laplacian with Dirichlet data:

∆ϕj = −λjϕj, ϕj|∂M = 0.

Using the tauberian theorem and the heat kernel, H. Weyl in 1912proved the asymptotic formula

λk ∼ cn

(k

vol (Ω)

)2/n

, as k → ∞,

where

cn = (2π)2(ωn−1

n

)−2/n

.

Based on this formula, Polya conjectured in 1960 that

λk ≥ cn

(k

vol (Ω)

)2/n

, for all k.

In the case n = 2, Polya proved that the conjecture holds true forsome special domains of R

2. E. Lieb proved in 1980 that there is aconstant cn smaller than cn such that

λk ≥ cn

(k

vol (Ω)

)2/n

, for all k.

The most recent result (???) is the

Theorem 36 (Li-Yau) (cf. [15], p. 118)

λk ≥ n

n + 2cn

(k

vol (Ω)

)2/n

, for all k.


6.5.2 Heat kernel expansion on a compact mani-fold

The following about the heat kernel on a compact manifold is well-known, (cf. [1, 13]).

Theorem 37 Let M be a compact Riemannian manifold, ϕj be anorthonormal basis of L2 (M) consisting of eigenfunctions, λj the cor-responding eigenvalues, then

pt (x, y) =∑

e−λjtϕj (x) ϕj (y) .

In particular ∑e−λjt =

∫M

pt (x, x) dx

6.6 Lower estimates of higher eigenvalues

To start with we shall prove, by using the heat kernel and the Sobolevinequality, the following

Theorem 38 Let M be a compact Riemannian manifold without bound-ary. Then there is a c > 0 such that

λk ≥ c

(k

vol (M)

)2/n

, for all k.

Proof. The proof is based on the Cheng-Li’s method, (cf. ???). Letpt (x, y) be the heat kernel of M . Then

pt (x, y) =

∫ps (x, z) pt−s (z, y) dz

for 0 < s ≤ t.

6.6. Lower estimates of higher eigenvalues 65

In particular

p2t (x, x) =

∫pt (x, z)2 dz.

Now

∂t

∫pt (x, z)2 dz = 2

∫pt (x, z) ∂tpt (x, z) dz

= 2

∫pt (x, z) ∆pt (x, z) dz

= −2

∫‖∇pt (x, z)‖2 dz

Sobolev≤ −2c

(∫pt (x, z)

2nn−2 dz

)n−22n

.

Also, since∫

pt (x, y) dy ≤ 1 for all t > 0 and x ∈ M , we get∫pt (x, z)2 dz =

∫pt (x, z)

2nn+2 pt (x, z)

4n+2 dz

Holder≤(∫

pt (x, z)2n

n+2n+2n−2 dz

)n−2n+2(∫

pt (x, z)4

n+2n+2

4 dz

) 4n+2

≤(∫

pt (x, z)2n

n−2 dz

)n−2n+2(∫

pt (x, z) dz

) 4n+2

≤(∫

pt (x, z)2n

n−2 dz

)n−2n+2

.

Putting together the above two estimates we get

∂t

∫pt (x, z)2 dz ≤ −2c

(∫pt (x, z)2 dz

) 2+nn

= −2cp2t (x, x)2+n

n .


Integrating the above inequality and bearing in mind that pt (x, x) → ∞as t → 0, we obtain (pws???) that

p2t (x, x) =

∫pt (x, z)2 dz ≤

(4ct

n

)−n/2

.

Recall now, that ∫p2t (x, x) dx =

∑e−2λjt.

This imply ∑e−2λjt ≤

(4ct

n

)−n/2

vol (M) .

But,0 < λ1 ≤ λ2 ≤ · · · ≤ λk,

so,

ke−λk ≤k∑

j=0

e−2λjt ≤∑

e−2λjt

≤(

4ct

n

)−n/2

vol (M) .

If we choose t such that 2λkt = n2

we get

ke−λk ≤(

c

n

n

λk

)−n/2

vol (M)

i.e.

λk ≥ c

n

(k

vol (M)

)2/n

.

Next, using the estimates of the heat kernel we shall prove the fol-lowing


Theorem 39 Let M be an n−dimensional compact Riemannian man-ifold possibly with boundary and with Ric (M) ≥ 0. We assume that∂M is convex. Denote its Neumann eigenvalues by

0 = µ0 < µ1 ≤ µ2 ≤ · · · ≤ µk ≤ · · ·and its Dirichlet eigenvalues by

0 < λ1 < λ2 ≤ · · · ≤ λk ≤ · · ·(if ∂M is empty, no boundary conditions), then

λk ≥ c1 (n)

d2k2/n, µk ≥ c2 (n)

d2(k + 1)2/n , for all k,

where c1 (n) and c2 (n) are positive constants depending only on n andd = diam (M).

Proof. Since the Dirichlet heat kernel is smaller than the Neumannheat kernel (na exeigithei) the proof for the Neumann eigenvaluesgives also the case of the Dirichlet eigenvalues.

Let pt (x, y) be the Neumann heat kernel. Then by Theorem 19

pt (x, y) ≤ c (n)

V(x,√

t) .

So, ∫M

pt (x, x) dx =∑

e−µjt ≤ c (n)

∫M

dx

V(x,√

t) .

But for√

t ≥ d,

V(x,√

t)

= vol (M) ,

and for√

t ≤ d, by the Bishop comparison theorem

V (x, d)

V(x,√

t) ≤ ( d√

t

)n


which implies that

1

V(x,√

t) ≤ 1

vol (M)

(d√t

)n

Thus ∑e−µjt ≤ c (n)

(d√t

)n

, if t ≤ d2

1, if t ≥ d2.

Fixing any k and taking the sum of the first k + 1 terms we get

(k + 1) e−µkt ≤ c (n)

(d√t

)n

, if t ≤ d2

1, if t ≥ d2.

Therefore

(k + 1) ≤ c (n) inft

eµkt(

d√t

)n

, if t ≤ d2

eµkt, if t ≥ d2

≤ c (n) min

eµkd2

, inf0<t≤d2

eµkt

(d√t

)n. (6.16)

It is easy to check that

inf0<t≤d2

eµkt

(d√t

)n

= eµkt

(d√t

)n

∣∣∣t= n2µk

(6.17)

=

(2e

n

)n/2

(d√

µk)n .

Also,

eµkd2 ≥ k + 1

c (n)

i.e.

µk ≥ 1

d2log

k + 1

c (n),


there are only finitely many µk such that

µk <n

2d2.

For these µk one can choose c′ (n) such that

µk ≥ 1

d2log

k + 1

c (n)≥ c′ (n)

d2(k + 1)2/n

Now, for the µk’ that satisfy

µk ≥ n

2d2

we settk =

n

2µk

≤ d2

and hence by (6.16) and (6.17)(2e

n

)n/2

(d√

µk)n ≥ k + 1

c (n). (6.18)

From (6.18) it follows that we can choose c′′ (n) such that

µk ≥ c′′ (n)

d2(k + 1)2/n ,

and the proof of the estimate for the µk’ is complete.Replacing µk by λk, we get the result for the Dirichlet problem.

Chapter 7

Appendix 1. Basic facts fromRiemannian Geometry

In this chapter we shall present some basic facts from Riemannian Ge-ometry useful for Analysts that hate the kingdom of indices and so on!!!The material come for [13].

7.1 The metric tensor

The first thing we have to do on a differential manifold, if we want todo “geometry” is to be able to measure lengths of curves.

If M is a smooth surface of R3, we can measure the length of a curve

γ : [0, 1] −→ M

by the usual formula

l (γ) =

∫ 1

0

‖γ′ (t)‖ dt.

So, to measure the length of γ, the basic ingredient is to measure thelength of the vector γ′ (t) which belongs in the tangent space Tγ(t)M .

71

72 7. Appendix 1. Basic facts from Riemannian Geometry

Of course, by the same formula we can measure the length of a curveon any manifold embedded in R

n.Thus in order to measure lengths of curves in any manifold we need

to introduce a method of measuring lengths of tangent vectors. Forthat we need the

Definition 40 A Riemannian manifold is a smooth manifold with afamily of smoothly varying positive definite inner products gx on TxMfor each x ∈ M . The family g = (gx) is called a Riemannian metric.

Given a Riemannian metric on M , we can set the length of the curve

γ : [0, 1] −→ M

to be

l (γ) =

∫ 1

0

gx (γ′ (t) , γ′ (t)) dt.

(of course x = γ (t)).To compute the metric g, we must be able to analyze it in local

coordinates. If (x1, · · · , xn) are local coordinates near x and X,Y ∈TxM , then

X =∑

ai∂i and Y =∑

bi∂i.

Thus

gx (X,Y ) = gx

(∑ai∂i,

∑bi∂i

)=∑

aibjgx (∂i, ∂j)

and gx is determined by the symmetric and positive definite matrix

gij (x) = gx (∂i, ∂j) .

We writeg =

∑i,j

gijdxi ⊗ dxj.

7.2. The Curvature 73

Remark 41 Lethij (x) = gx (∂i, ∂j)

be the matrix of the metric g in an another coordinate chart with co-ordinates (y1, . . . , yn). Then one can show that on the overlap of thecharts we have that

gij =∑k,l

∂xk

∂yi

∂xl

∂yj

hkl.

Now we can measure lengths of curves, it seems plausible to set thedistance between any two points of M to be the length of the shortestpath between them:

d (x, y) = infγ piecewise C∞ l (γ) : γ (0) = x, γ (1) = y .

7.2 The Curvature

7.2.1 The curvature of a surface

There is no better place to begin the discussion of curvature than withGauss’ solution of the question: when a piece of surface in R

3 is flat.By flat we mean of course, that there is a isometry of the piece of

the surface to a region in the standard plane. Our intuition is thatthis should be possible iff the piece is not curved in some sense. Gauss’work gives a precise meaning to this notion of curvature.

Locally, the given surface M is oriented and hence has an outwardpointing unit vector nx at every point x ∈ M . Consider the Gauss map

Mν−→ S2 (0, 1)

x ν (x) = nx.

As test cases, consider first M = S2 (0, r), the sphere of radius r. LetA be the ball of S2 (0, r) with center the north pole and radius ε, thenone can see that

area (ν (A)) =c (ε)

r2.


Now, if A is any region in the two plane, then

area (ν (A)) = 0.

Thus the more curved the surface, the greater the area of A.

Thus the curvature of the surface at x must be something like

“ limA→x

area (ν (A))

area (A)”.

But

area (ν (A)) =

∫A

det dν (x)

and if A is too small, then

area (ν (A)) ∼ det dν (x) area (A) .

Thus the curvature of the surface at x can be defined as det dν (x).

Definition 42 The Gaussian curvature Kx of a surface of R3 at x is

given by det dν (x).

At this point it is reasonable to say that a piece A of surface is flat if

Kx = 0 for every x ∈ A.

Let us then suppose that A is flat and let (x1, x2) be a coordinate chartaround x. Let ϕ be an isometry from A to a region B of R

2:

Aϕ−→ B

(x1, x2) ϕ (x1, x2) = (y1, y2) .

So (y1, y2) can be considered as coordinates of B and the conditionthat ϕ is an isometry is that the inner product in B with coordinates(y1, y2) is the same as the inner product of A with coordinates (x1, x2):∑

i,j=1,2

gijdxi ⊗ dxj = dy1 ⊗ dy1 + dy2 ⊗ dy2. (7.1)


Thus we want to find the functions y1 (x1, x2) , y2 (x1, x2) satisfying(7.1). If we write

dy1 =∂y1

∂x1dx1 +

∂y1

∂x2dx2

and

dy2 =∂y2

∂x1dx1 +

∂y2

∂x2dx2,

we see that (7.1) is a system of 3 nonlinear first order partial differentialequations of two unknown functions y1 (x1, x2) , y2 (x1, x2).

In order to solve it uniquely we need a third equation, the integrabilitycondition. Gauss found that the integrability condition is the following:

∂Γ122

∂x1− ∂Γ1

21

∂x2+∑s=1,2

Γs

22Γ1s1 − Γs

21Γ1s2

= 0, (7.2)

(the relation is valid around x), where Γijk are the Christoffel symbols ,

Γijk (x) =

1

2gij (x)

∂gkl

∂xj(x) +

∂gjl

∂xk(x) − ∂gjk

∂xl(x)

and (gij (x)) = (gjl (x))−1.

Here we shall note that although Gauss has given an analytic solutionof the isometry mapping problem, it is not obvious how it is related toour geometric intuition that this solution is in fact determined by theGaussian curvature defined above. This is in fact the content of theGauss’ celebrated Theorema Egregium which states that the Gaussiancurvature is given by

1

det (g)

(∂Γ1

22

∂x1− ∂Γ1

21

∂x2+∑s=1,2

Γs

22Γ1s1 − Γs

21Γ1s2

), (7.3)

i.e. the curvature is defined directly off the metric tensor of the surface.


Theorem 43 (Theorema Egregium) A surface in R3 admits an isom-

etry from a neighborhood of a point x in the flat plane iff (7.2) is validin that neighborhood. More over the Gaussian curvature at x is givenby (7.3).

7.2.2 The curvature of a manifold

After Gauss, Riemann posed and solved the question when a piece of aRiemannian manifold is isometric with a region of R

n with the standardflat metric.

Riemann found the integrability condition for the system∑i,j=1,n

gijdxi ⊗ dxj =∑

i,j=1,n

δijdyi ⊗ dyj, (7.4)

the natural generalization of the system (7.1). To set the notation,define the Riemann curvature tensor of type (1, 3) to be

R (x) =n∑

i,j,k,l=1

Rijkl (x) ∂xi

⊗ dxj ⊗ dxl ⊗ dxk

where

Rijkl (x) =

∂Γijk

∂xl− ∂Γi

jl

∂xk+

n∑s=1

ΓsjkΓ

isl − Γs

jlΓisk (7.5)

and the Christoffel symbols are defined as above.

Remark 44 It is not obvious that Rijkl (x) are the components of a ten-

sor, although a long computation verifies that the Rijkl (x) do transform

correctly.

Theorem 45 (Riemann) The system (7.4) has a solution in the neigh-borhood of a point x iff R ≡ 0 in a neighborhood of x.

Remark 46 One can show that two manifolds are locally isometric iffthe curvature tensors agree in a certain technical sense. Thus the Rie-mann curvature tensor completely determines the Riemannian geometrylocally.


7.2.3 Ricci curvature

In practice the Riemann curvature tensor is not much of use. Aftertaking account of the symmetries of R, (cf. [13], p. 62), we still haven (n − 1) /2 independent components of R on an n−manifold. This istoo much and this the reason we introduce the Ricci curvature whichis a first simplification of the curvature tensor. The Ricci tensor Ricit is a type (0, 2) tensor given by

Ric =n∑

i,j=1

Rijdxi ⊗ dxj,

whereRij =

∑k

Rkijk.

The Ricci tensor is of the same type as the metric tensor, so we cancompare then. Namely, we write

Ric (M) ≥ K

ifRic (w,w) ≥ Kg (w,w)

for all 1− forms w ∈ TM .

The Ricci tensor, while weaker than the curvature tensor, still con-trols some crucial rough geometric properties of a Riemannian manifold.It is of basic importance that the growth of the balls is controlled by theRicci tensor, (cf. [15], p.11).

Theorem 47 (Bishop comparison theorem) Let M be an n-dimensionalcomplete Riemannian manifold with Ric (M) ≥ 0. Then for any x ∈ Mand R > 0

|B (x,R)| ≤ |Sn−1 (0, 1)|Rn,

where Sn−1 (0, 1) is the unit sphere in Rn.


Also, for any x ∈ M and r > s > 0,

|B (x, r)||B (x, s)| ≤

(r

s

)n

. (7.6)

Theorem 48 In particular, if we take r = 2s in (7.6), we get

|B (x, 2s)| ≤ 2n |B (x, s)| . (7.7)

Property (7.7) is known as the doubling volume property and its rolein doing analysis on a manifold is really crucial.

For the lower bound of |B (x,R)| on a complete noncompact manifoldwith Ric (M) ≥ 0 we only have that for R ≥ 1,

|B (x,R)| ≥ c (x) R.

For more details on this questions of comparison results, see [15],Chapter 1.

Ricci tensor and partial derivatives

Finally, we notice that the Ricci tensor appears in change of order ofpartial derivatives. For example in the course of the proof of the Li-Yaugradient estimate, we shall make use of the following formula

∂i∂i∂jf = ∂j∂i∂if + Rij∂jf.

7.3 Divergence theorem and Green’s fo-

mulas

In this Section we give the Divergence theorem and Green’s fomulas onan n−dimensional Riemann manifold M . For the proofs see [2], p.142.

7.4. Appendix 2 79

Theorem 49 (Green’s formulas I) Let f ∈ C2 (M) and h ∈ C1 (M)with at least one of them compactly supported. Then∫

M

h∆f + 〈∇f,∇h〉 dx = 0.

If both are C2, then ∫M

h∆f − f∆h dx = 0.

Theorem 50 (Divergence Theorem) Let M be oriented, Ω a do-main in M with smooth boundary ∂Ω and let ν be the outward unitnormal of ∂Ω. Then for any compactly supported C1 vector field X∫

Ω

div X =

∫∂Ω

〈X, ν〉

Theorem 51 (Green’s formulas II) Let M and Ω be as in the di-vergence theorem above, and let f ∈ C2 (M) and h ∈ C1 (M) with atleast one of them compactly supported. Then∫

Ω

h∆f + 〈∇f,∇h〉 dx =

∫∂Ω

h 〈ν,∇f〉 .

If both are C2, then∫Ω

h∆f − f∆h dx =

∫∂Ω

h 〈ν,∇f〉 − f 〈ν,∇h〉 .

7.4 Appendix 2

We give the construction of the function φ we used in the proof of theLi-Yau estimate.

We need first the following theorem of Calabi (cf. [3], p. 153).


Theorem 52 (Calabi) Assume that Ric (M) ≥ −nK2. If

ψ : [0, 1] −→ R,

is a C2 function andf (x) = ψ (d (x, x0))

for some x0 fixed, then

∆f (x) ≤⎧⎨⎩

ψ′′ (d) + ψ (d) nK2 coth K2d, if K2 > 0,

ψ′′ (d) + ψ (d) nd−1, if K = 0,

in the weak sense, for all x ∈ M .

We note that because of problems with the cut locus, the functionx −→ d (x, x0) is C∞ only for x = x0 and for x close to x0. This is thereason we have the Calabi inequality in the weak sense.

Lemma 53 There is a smooth function φ : M −→ R supported on theball B (a,R) such that

0 ≤ φ (x) ≤ φ (a) = 1,

for all x ∈ M and

‖∇φ (x)‖2 ≤ εφ (x) and ∆φ (x) ≥ −δ,

where ε and δ are positive and small as we wish.

Proof. Let

ψ (ρ) =

⎧⎨⎩ (ρ − 1)2 , if ρ ∈ [0, 1] ,

0 if ρ ≥ 1,

and set

φn (x) = ψ

(d (x, a)2

n

).

7.4. Appendix 2 81

Then by the chain rule

∇φn (x) = ψ′(

d (x, a)2

n

)∇(

d (x, a)2

n

)

= ψ′(

d (x, a)2

n

)2d (x, a)∇d (x, a)

n.

So

‖∇φn (x)‖2 = ψ′(

d (x, a)2

n

)24d (x, a)2 ‖∇d (x, a)‖2

n2.

Now‖∇d (x, a)‖ ≤ 1???

(dieukrinise!), so

‖∇φn (x)‖2 ≤ ψ′(

d (x, a)2

n

)24d (x, a)2

n2

≤ 4d (x, a)2

n2

⎧⎪⎨⎪⎩4(

d(x,a)2

n− 1)2

, if d (x, a)2 ≤ n,

0 if d (x, a)2 > n

≤ 16n

n2=

16

n.

Also,

∆φn (x) = ψ′′(

d (x, a)2

n

)4d (x, a)2

∥∥∇d (x, a)2∥∥2

n2

+ ψ′(

d (x, a)2

n

)∆(d (x, a)2)

n

= 24d (x, a)2

∥∥∇d (x, a)2∥∥2

n2+ ψ′

(d (x, a)2

n

)∆(d (x, a)2)

n.


By Calabi’s theorem one can see that

∆(d (x, a)2) ≤ c (d (x, a) + 1) .

So, if d (x, a)2 ≤ n,

∆φn (x) ≥ −∣∣∣∣∣ψ′(

d (x, a)2

n

)∣∣∣∣∣ c (d + 1)

n≥ c

n

for n big enough.

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Index

Bishop comparison theorem, 78

Cheeger constant, 59Christoffel symbols, 75co-area formula, 57

Dirichlet boundary conditions, 53divergence, 12Divergence Theorem, 79

Gaussian curvature, 74gradient, 12Green function, 43Green’s formulas, 79

Laplace equation, 45Laplacian, 13Lichnerowicz theorem, 49Liouville theorem, 48lower gaussian bound, 39

Neumann boundary conditions,54

parabolic Harnack inequality, 22Poincare inequality, 56Poisson equation, 45Polya conjecture, 55

Ricci curvature, 77Riemann curvature tensor, 76riemannian measure, 11

Sobolev inequality, 57

Theorema Egregium, 76

upper gaussian estimate, 34

Weyl formula, 51

85

Analysis on Manifolds - Aristotle University of...

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