Perspectives and open problems in geometric

analysis: spectrum of Laplacian

Zhiqin Lu

May 2, 2010

Abstract

1. Basic gradient estimate; different variations of the gradient esti-mates;

2. The theorem of Brascamp-Lieb, Barkey-Emery Riemannian geom-etry, relation of eigenvalue gap with respect to the first Neumanneigenvalue; the Friedlander-Solomayak theorem,

3. The definition of the Laplacian on Lp space, theorem of Sturm,

4. Theorem of Wang and its possible generalizations.

1 Gradient estimate of the first eigenvalue

Let M be an n-dimensional Riemannian manifold with or without boundary.Let the metric ds2 be represented by

ds2 =∑

gijdxidxj ,

where (x1, · · · , xn) are local coordinates. Let

∆ =1√g

∑ ∂

∂xi(gij√g∂

∂xj)

be the Laplace operator, where (gij) = g−1ij , g = det(gij).

The operator ∆ acts on smooth functions. If ∂M 6= ∅, then we may defineone of the following two boundary conditions:

¬. Dirichlet condition: f |∂M = 0.

­. Neumann condition: ∂f∂n |∂M = 0, where n is the outward normal vector of

the manifold ∂M .

1

By the elliptic regularity, if M is compact, then the spectrum of ∆ consistsof eigenvalues

λ1 6 λ2 6 · · · 6 λk → +∞

of finite multiplicity.

By the variational principal, we have the following Poincare inequality∫|∇f |2 > λ1

∫f2

To our special interests, we would like to give “computable” lower boundestimates of the first eigenvalue. Here by “computable” we mean the geometricquantities like the diameter, the bounds of the curvature, etc, that are readilyavailable.

Li-Yau [4] discovered the method of gradient estimates to give “computable”lower bounds of the first eigenvalue. The prototype of the estimates is as follows:

Theorem 1 (Li-Yau). Let M be a compact manifold without boundary. Let dbe the diameter of M . Assume that the Ricci curvature of M is non-negative.Then we have the following estimate

λ1 >π2

4d2.

Proof. Let u be the first eigenfunction such that

maxu2 = 1.

Let

g(x) =1

2(|∇u|2 + (λ1 + ε)u2),

where ε > 0.

The function g is a smooth function. Let x0 be the maximal point of g.Then at x0 we have

ujuji + (λ1 + ε)uui = 0 (1)

and0 > u2

ji + ujujii + (λ1 + ε)|∇u|2 + (λ1 + ε)u∆u.

Using the Ricci identity, we have

ujujii = uj(∆u)j +Ric(∇u) > uj(∆u)j .

Thus we have

0 > u2ji + uj(∆u)j + (λ1 + ε)|∇u|2 + (λ1 + ε)u∆u. (2)

2

Suppose that at the maximum point of g(x), ∇u 6= 0. Then we have

u2ji > |∇u|−4(

∑i,j

ujuiuij)2

by the Cauchy inequality. Using the first order condition, we conclude

u2ji > (λ1 + ε)2u2.

Putting the above inequality into ~, we get

0 > ε|∇u|2 + ε(λ1 + ε)u2

which is not possible. Thus at the maximum point, we must have ∇u = 0.Therefore we have

g(x) 61

2(λ1 + ε) maxu2 =

1

2(λ1 + ε).

From the above estimate, we get

|∇u|2

1− u26 λ1 + ε.

Since ε is arbitrary, we let it go to zero and obtain

|∇u|2

1− u26 λ1.

By changing the sign of u, we may assume that maxu = 1. Let u(p) = 1 forp ∈ M . Since

∫u = 0, there is a point q ∈ M such that u(q) = 0. Let σ(t) be

the minimal geodesic line connecting q and p. Consider the function

arcsinu(σ(t))

By the above inequality, we get

|(arcsinu(σ(t)))′| 6√λ1|σ′(t)|

Integrating the above inequality along the geodesic line, we get

π

26√λ1d

and the theorem is proved.

Several extensions of the above method can be obtained when

¬ The manifold has boundary;

­ The Ricci curvature has a lower bound.

3

We first address the Neumann boundary condition.

Lemma 1. If ∂M 6= 0 and ∂M is convex, then g(x) doesn’t attain its maximumon ∂M unless at the point ∇u = 0.

Proof. Let x0 ∈ ∂M such that g(x) attains the maximum at x0. Then

∂g(x0)

∂n6 0,

where ~n is the outward unit normal vector. By the definition of g(x) and thefact ∂u

∂n = 0, we have

uj∂uj∂n

6 0.

Let hij be the second fundamental form. Then

0 > uj∂uj∂n

= hijuiuj > 0.

If the equality is true, then we must have ∇u(x0) = 0.

The next question is to sharpen the Li-Yau estimates. Even for the unitcircle, Li-Yau estimate is not sharp.

Let’s consider the circle x2 +y2 = R2. Let the parameter, or the coordinate,of the circle be

x = R cos θ, y = R sin θ

Then the Laplace operator is1

R2

∂2

∂θ2

As a result, u = cos θ, sin θ are the two eigenfunctions with the first eigenvalue1/R2. If u = cos θ, then with the induced Riemannian metric,

|∇u|2 =1

R2sin2 θ.

Thus we have

g(θ) =1

2R2,

and thus|∇u|2

1− u26 λ1,

which is not sharp.

The problem is that in general, we don’t know whether the first eigenfunctionis always symmetric. More precisely, if we assume that

1 = supu > inf u = −k > −1

4

we don’t know whether k = 1. From Li-Yau’s basic estimate, we can improvethe estimate λ1 > π2/4d2 to

λ1 > (π

2+ arcsin k)2d−2.

The above inequality is essentially useless because we know nothing about k.However, using a simple trick, we can double the estimate of Li-Yau:

Take

u =u− 1−k

21+k

2

.

Then using the standard gradient estimate, we get

|∇u|2 6 λ1(1 + a)(1− u2) (3)

where

a =1− k1 + k

.

Now the function u is symmetric: max u = −min u = 1. Using the samemethod, we get

λ1 >π2

(1 + a)d2>

π2

2d2

Zhong-Yang [12] took one more step and proved the following result.

Theorem 2 (Zhong-Yang). Let M be a compact Riemannian manifold withnon-negative Ricc curvature. Then

λ1 >π2

d2.

The estimate is called “optimal” in the sense that for 1 dimensional manifold,the lower bound is achieved. We shall soon see that the estimate, in general, isfar from being optimal.

The basic idea of the proof is still the maximum principle. From the esti-mate (3), we suspect that there is an odd function ϕ(arcsinu) such that

|∇u|2 6 λ1(1 + aϕ(arcsinu))(1− u2). (4)

If such function ϕ exists, then we have

√λ1d >

∫ π2

−π2

dθ√1 + aϕ(θ)

> π

by the convexity of the function 1√1+x

, which implies the optimal inequality.

5

To prove the inequality (4), we use the maximal principle. At the point x0

such that the equality of (4) holds, we have

ϕ(arcsinu) 6 u− u√

1− u2ϕ′(arcsinu) +1

2(1− u2)ϕ′′(arcsinu).

We define a function

ψ(θ) =

( 4π (θ + cos θ sin θ)− 2 sin θ) cos−2 θ, θ ∈ (−π2 ,

π2 )

ψ(π2 ) = 1, ψ(−π2 ) = −1.

Then a straightforward computation givesψ′(θ) > 0ψ − sin θ + sin θ cos θψ′ − 1

2 cos2 θψ′′ = 0.

Using the maximum principle we get ϕ(arcsinu) 6 ψ(arcsinu). Since ψ is anodd function, the theorem is proved.

Remark 1. Recently, Hang-Wang [3] proved that, in fact,

λ1 >π2

d2

unless the manifold is of one dimensional.

The Li-Yau-Zhong-Yang estimate is still effective when the Ricci curvatureis not “too negative”. Namely, let

Ric(M) > −(n− 1)K

for some constant K > 0. Then

λ1 >π2

d2− (n− 1)K.

Thus as long as the right hand side of the above is positive, the estimate iseffective.

When K is very negative, we need to modify the basic gradient estimate.The following theorem belongs to Li-Yau.

Theorem 3. Let M be a compact Riemannian manifold without boundary.Assume that

Ric(M) > −(n− 1)K

for K > 0. Then

λ1 >1

(n− 1)d2exp(−[1 + (1 + 4(n− 1)2d2K)1/2]),

where d is the diameter of M .

6

Proof. Let u be the normalized first eigenfunction. That is

1 = supu > inf u > −1.

Let β > 1. Consider

G(x) =|∇u|2

(β − u)2.

Let x0 be the maximum point of G(x). Then

∇G(x0) = 0,∆G(x0) 6 0.

SinceG(x)(β − u)2 = |∇u|2,

we have

∆G(β − u)2 + 2∇G∇(β − u)2 +G∆(β − u)2 = ∆|∇u|2.

Thus at x0, we have

0 > ∆|∇u|2 −G∆(β − u)2

= 2∑

u2ij + 2

∑uiuijj − 2G[(β − u)(−∆u) + |∇u|2]

= 2u2ij + 2ui(∆ui)i

+ 2Ric(∇u,∇u)− 2G[λ1u(β − u) + |∇u|2].

That is

u2ij − λ1|∇u|2 − (n− 1)K|∇u|2 −G(λ1u(β − u) + |∇u|2) 6 0.

We choose a local coordinate system at x0 such that uj = 0 (j = 2, · · · , n),u1 = |∇u|. Then u1 6= 0 (or otherwise, G(x0) = 0 which is not possible). From∇G(x0) = 0, we have

u11 = −|∇u|2(β − u)−1

u1i = 0, i 6= 1.

Using the following trick

n∑i·j=2

u2ij >

n∑i=2

u2ii >

1

n− 1

(n∑i=2

uii

)2

=1

n− 1(∆u− u11)2

=1

n− 1(λu+ u11)2 =

1

n− 1(λ2u2 + 2λuu11 + u2

11)

>u2

11

2(n− 1)− 1

n− 1λ2u2,

we have

1

2(n− 1)

|∇u|4

(β − u)2− λ2u2

n− 1− (λ1 + (n− 1)K)|∇u|2 − λ1

|∇u|2uβ − u

6 0.

7

Let α = u(β − u)−1. Then

α 61

β − u6

1

β − 1.

Thus1

2(n− 1)G2 − λ2

n− 1α2 − (λ1 + (n− 1)K)G− λ1Gα 6 0,

which gives

G(x) 6 G(x0) 6 4(n− 1)

(λβ

β − 1+ (n− 1)K

).

Let l be the geodesic line connecting x1 and x2, where u(x1) = 0, u(x2) =supu = 1. Then we have

logβ

β − 16∫γ

|∇u|β − u

6

[4(n− 1)

(βλ1

β − 1+ (n− 1)K

)]1/2

d,

or in other words

λ1 >β − 1

β

[1

4(n− 1)d2

(log

β

β − 1

)2

− (n− 1)K

]

Choosing β0 such that the right side above maximized, we proved the theorem.

The optimal estimate, in this direction, was obtained by Yang:

Theorem 4. Let M be a compact Riemannian manifold.

Ric(M) > −(n− 1)K, (K > 0), d = diam(M)

Then

λ1 >π2

d2exp(−CnKd2)

where Cn =√n− 1 for n > 2 and Cn =

√2 for n = 2.

The case when the Ricci curvature is positive is also very interesting. Thefollowing theorem of Lichnerowicz is well known.

Theorem 5. Let M be a compact Riemannian manifold. Assume that d is thediameter of the manifold and

Ric(M) > −(n− 1)K > 0.

for K > 0. Thenλ1 > nK.

8

In seeking the common generalization of the above theorem and the Zhong-Yang estimate, Peter Li (see [11]) proposed the following conjecture.

Conjecture 1. For a compact manifold with Ric(M) > (n − 1)K > 0 thefirst eigenvalue λ1, with respect to the closed, the Neumann, or the DirichletLaplacian satisfies

λ1 >π2

d2+ (n− 1)K.

Here is ∂M 6= ∅, we assume that ∂M is convex.

Note that by Myer’s theorem, we always have π2/d2 > K. Thus the conjec-ture, if true, will give a common generalization of the result of Lichnerowicz’sand the one obtained by the gradient estimate.

In this direction, D-G Yang [11] proved that the first Dirichlet eigenvalue ofthe Laplacian satisfies

λ1 >π2

d2+

1

4(n− 1)K,

if the manifold has weakly convex boundary. He also proved that the first closedeigenvalue and the first Neumann eigenvalue of the Laplacian satisfies

λ1 >π2

d2+

1

4(n− 1)K,

if the manifold has convex boundary.

Ling [5] was able to improve the above estimate into

λ1 >π2

d2+

31

100(n− 1)K

Further improvements are possible, see Ling-Lu [6] for example.

We end this lecture by making the following

Conjecture 2. Let M be a compact Ricci flat Riemannian manifold such that

λ1 −π2

d2< ε

for a sufficiently small ε > 0. Then M = S′ ×M0, where M0 is a Ricci flatcompact Riemannian manifold.

9

2 Spectrum gap of the first two eigenvalues

2.1 Heat flow proof of a theorem of Brascamp-Lieb

The following result was first proved by Brascamp-Lieb [1]. In Singer-Wong-Yau-Yau [9], a simplified proof was given. In this subsection, we give a heatflow proof.

Theorem 6. Let Ω be a bounded convex domain of Rn. Let u be the firstDirichlet eigenfunction with the eigenvalue λ1. Then (up to a sign), u is positiveand log u is concave.

We begin by the following lemmas.

Lemma 1. Up to a sign, u > 0.

Proof. Otherwise, we may use |u| in place of u, From Kato’s inequality, wehave

|∇|u|| 6 |∇u|.

Thus we have ∫|∇|u||2∫u2

6

∫|∇u|2∫u2

.

By the variational characterizing of the first eigenvalue, we know that the rightside of the above is minimized. Thus the equality must hold and |u| is an eigen-function of the first eigenvalue.

The multiplicity of the first eigenfunction must be one. Thus up to a sign,the eigenfunction must be non-negative.

Lemma 2. u > 0 inside Ω.

Proof. If not, assume that u(x0) = 0 at a point x0 in the interior of Ω. Let

u(x) = pN (x) +O(xN+ε)

be the Taylor’s expansion of the eigenfunction at x0, where pN (x) is the polyno-mial of degree N . From the equation ∆u = −λ1u, we have ∆p2(x) = 0. Sinceu(x) > 0, we must have p2(x) > 0, a contradiction to the maximal principal.

Lemma 3. Using the above notations, we have

∂u

∂n< 0

on ∂Ω, the boundary of Ω.

Proof. This follows from the strong maximum principle. By the above lemma,we have

∂u

∂n6 0

10

If ∂u∂n = 0, then since the function u vanishes on the boundary, we have ∇u =

0 at the point. Using the Taylor’s expansion for the boundary point, we getp2 > 0 and ∆p2 = 0. Since p2 is harmonic, by the strong maximum principle,∂p2/∂n < 0 unless p2 = 0. But if p2 ≡ 0 then u ≡ 0. This completes the proof.

We now consider the heat flow

∂u

∂t= ∆u+ λ1u, u|∂Ω = 0.

we have

Lemma 4. For any smooth initial function u0 > 0, the flow exists and convergesto the first eigenfunction.

Proof. Letu1 =

∑ajfj(x),

where fj(x) is the eigenfunction of the j-th eigenvalue. Then the solution of theequation is

u(t, x) =∑

aje−(λj−λi)tfj(x)

Obviously, we havelimt→∞

u(t, x) = a1f1(x).

If we choose u0 such that a1 6= 0, then the flow converges to the first eigen-function. f1(x) > 0 by the above lemmas. By our choice of u0

a1 =

∫Ω

u0f1(x) > 0.

The proof is complete.

Lemma 5. Let w be any smooth positive function on Ω such that

w|∂Ω = 0 and ∇w|∂Ω 6= 0

Then near the boundary of Ω, logw is concave.

Proof. Since w is a smooth function vanishes on the boundary, it can be viewedas the defining function of Ω. By the implicit function theorem, we solve theequation

w(x1, · · · , xn) = 0

to get the functionxn = xn(x1, · · · , xn−1).

If Ω is convex, then∂2xn∂xi∂xj

> 0,

11

is a positive definite matrix. Using the chain rule, the above inequality is equiv-alent to

−wijwn

+winwjw2n

+wjnwiw2n

− wiwjwnnw3n

> 0.

where 1 6 i, j 6 n − 1. However, if we allow i or j to be n, then as the n × nmatrix, we still have

−wijwn

+winwjw2n

+wjnwiw2n

− wiwjwnnw3n

> 0.

Moreover, for any (a1, · · · , an), if∑ajwj = 0, we have

−wijaiaj > ε|a|2

for some positive ε > 0. A generic vector has the form a + µb, where b =(w1, · · · , wn). For w small enough, we have

−wij(ai + µbi)(aj + µbj) +1

w|µ|2|b|4 > 0.

Thus ∇2 logw, whose matrix entries are

wijw− wiwj

w2,

is negative definite for w small enough.

Remark 2. The above proof is purely elementary. We can use differentialgeometry to give another proof. Assume e1, · · · , en are the local frame fieldsat the boundary point such that en is normal to the boundary and the rest aretangent to the boundary. Then for 1 6 i, j 6 n− 1, we have

∇2(ei, ej)w = ei(ej(w))−∇eiejw = hij∂w

∂n,

where hij is the second fundamental form, and ∂∂n is the outward normal vector

field. Thus ∂w∂n 6 0. To prove

wijw− wiwj

w2

is negative definite, we write V = V1 + µen, where V1 is tangent to ∂Ω. Thenthere is a constant C such that

∇2 logw(V, V ) 6 w−1

∣∣∣∣∂w∂n∣∣∣∣ (π(V1, V1) + Cµ‖V1‖+ Cµ2)− w−2µ2

∣∣∣∣∂w∂n∣∣∣∣4 .

Since ∂w∂n 6= 0, for points sufficient close to the boundary, w is small enough.

Using the Cauchy inequality, we can prove the above negativeness.

12

Now we begin to prove the theorem: We will choose a function u0 suchthat u0 > 0 and log u0 is concave. Then we shall prove that the log-concavity ispreserved under the heat flow. The theorem thus follows from the above lemmas.

To construct the required u0, we first pick up any smooth function w > 0on Ω with w|∂Ω = 0, ∇w|∂Ω 6= 0. Let

u0 = we−C∑x2j

for a constant C > 0 sufficiently large. By the above lemma, the functionis log-concave near the boundary. Away from the boundary, since w > δ > 0for some constant δ > 0, we can choose C large enough so that u0 is log-concave.

Using the matrix version of the maximum principle, we can prove that theflow keeps the log-concavity. Let ϕ = log u, where u is the solution of the heatequation

∂u

∂t= ∆u+ λ1u.

The flow of ϕ is∂ϕ

∂t= ∆ϕ− |∇ϕ|2 − λ1. (5)

By the maximum principle, if T is the first time the matrix ∇2ϕ is generated,then there is an x0 ∈M and a direction i such that

−ϕii = 0

and for other j’s, −ϕjj > 0. Moreover, at x0, ϕiik = 0, ∂ϕii∂t 6 0 and ∆ϕii > 0.

Differentiating (5) by i, i, we have

0 >∂ϕii∂t

= ∆ϕii − 2ϕkϕkii − 2ϕ2ki > −2ϕ2

ki.

By the convexity, ϕ2ki 6 ϕiiϕkk = 0. Thus ϕki = 0 for k 6= i. The theorem

follows from the strong maximum principle.

This completes the proof of the Brascamp-Lieb theorem.

2.2 Gap of the first two eigenvalues

For the sake of simplicity, we only consider bounded smooth domain in Rn.

Let Ω be a bounded domain in Rn with smooth boundary. Let λ1, λ2 be theDirichlet first two eigenvalues. Since λ1 must be simple (of multiplicity one) wehave

λ2 − λ1 > 0.

13

The question we would like to answer is that, how to get the lower bound esti-mate of λ2 − λ1?

Let ϕ1, ϕ2 be the eigenfunctions with respect to λ1, λ2. We set

ϕ =ϕ2

ϕ1.

Then a straightforward computation gives

∆ϕ+ ∂∇ logϕ1∇ϕ = −λϕ,

where λ = λ2 − λ1. Moreover, since ϕ1|∂Ω = 0, we must have

∂ϕ

∂n

∣∣∣∣∂Ω

= 0.

We have a notion of Bakry-Emery Ricci tensor. Let

f = ϕ21

Then the Bakry-Emery Laplacian is defined as

∆f = ∆ +∇ log f∇

which is a self-adjoint operator with respect to the volume form ϕ21dV .

The Bakry-Emery Ricci tensor is defined as

−∇2 log f

By the Brascamp-Lieb theorem, the Barkey-Emery Ricci curvature is non-negative.

We make the following conjecture:

Conjecture 3. The method of gradient estimates can be generalized to theBakry-Emery case without additional difficulties.

The above conjecture, if true, will give a unified proof between the estima-tion of the first eigenvalue and gap estimate.

We can go one more step further. Since the Barkey-Emery Ricci tensorcomes from the setting (M,ds2, efdV ), which can be considered as the limit ofthe wrap product

efdr2 + ds2

we make the following definition.

LetΩε = (x, y)|x ∈ Ω, 0 6 y 6 εϕ2

1(x).Let µε be the first Neumann eigenvalue. Then

14

Conjecture 4. Using the above notations, we have

λ2 − λ1 = limε→0

µε.

Similarly, we make the following

Conjecture 5. Let (µε)k be the k-th Neumann eigenvalue of Ωε. Then

limε→0

(µε)k = λk+1 − λ1.

We can prove the following

Theorem 7.λ2 − λ1 > lim

ε→0µε.

Proof. We first note that ∂h∂n |∂Ωε = 0 on the part y = εϕ1(x)2 can be written

as∇h(ε∇ϕ2

1,−1) = 0.

We define Uε as follows

Uε = ϕ+ y2∇ logϕ1∇ϕ.

Then a straightforward computation gives∆Uε + λUε = O(ε2)

∂Uε∂n

∣∣∣∂Ωε

= O(ε2).

From the above, we have ∫Ωε

Uε = O(ε2).

Let α be a number such that ∫Ωε

(Uε − α) = 0.

Then α = O(ε). By the variational principle, we have

µε 6

∫Ωε|∇Uε|2∫

Ωε(Uε − α)2

.

However, since ∫Ωε

U2ε = Cε.

We have

µε 6

∫ΩελU2

ε∫ΩεU2ε

+O(ε).

and the theorem is proved.

15

Remark 3. We consider the wrap product

efdr2 + d2

over Ωε, we see the relation between two settings. This also gives the relationbetween the Barkey-Emery geometry with respect to the ordinary Riemanniangeometry.

Some applications.

Theorem 8 (Singer-Wong-Yau-Yau, Yu-Zhong). Let Ω be a convex domain inRn. Then

λ2 − λ1 >π2

d2.

Proof. We consider the domain Ωε and let Uε be the first Neumann eigenfunc-tion. By what we proved in the last lecture, we have

µε >π2

d2ε

where dε is the diameter of Ωε. Since dε → d, the theorem is proved.

Question: How to recover the recent result of Yau? We are going to usethe following result of Chen-Li.

Theorem 9. Let M be an m-dimensional domain in Rm. Let M be star-shaped.Let R be the radius of the largest ball centered at p ∈M contained in M and letR0 be the smallest ball centered at p containing M . Then there is a constant C,depending only on m, such that

η1 > CRm

Rm+20

Let Uε be the first Neumann eigenfunction of Ωε. Then asymptotically wecan write

Uε ∼ ϕ+ y2∇ logϕ1∇ϕ

In general, power series method can be used to prove the conjecture.

Remark 4. Conjecture 4 was proved by Lu-Rowlett [7].

Appendix: Eigenvalues of collapsing domain

It is important to study the asymptotic behavior of eigenvalues when a domaincollapses. In this appendix, we give some preliminary results in this direction.

We begin with the following observation. Consider the sectorWhat is the asymptotic behavior of the Dirichlet eigenvalues when α→ 0?

16

As is well-known, the eigenfunctions of the sector are of the form

f(r) sinθ

α

where f(r) is the so-called Bessel function

f ′′ +1

rf ′ +

1

r2(r2 − 1

α2)f = −λf

If we setg(r) =

√rf(r)

Then the corresponding equation becomes

−g′′(r) +1

r2(

1

α2− 1

4)g = λg

When α → 0, we take the following renomorization: set 1 − r = α2/3x. Thenwe have

1

r2− 1 ∼ 2xα2/3

Let λ = λ+ 1α2 . Then we have

−g′′(r) +

(1

r2

(1

α2− 1

4

)− 1

α2

)g = λg

and if α→ 0, we have−g′′ + 2xg = λg

This is the Airy’s function.

Friedlander and Solomyak was able to generalize the above result in the fol-lowing setting:

Let h(x) > 0 be a piecewise linear function defined on [−a, b], where a, b > 0.We assume that

h(x) =

M − C+x x > 0M − C−x x < 0

where the choice of M,C+, C− are so that h(−a) = h(b) = 0.

For any positive ε > 0, let

Ωε = (x, y)|x ∈ I, 0 6 y 6 εh(x)

Theorem 10 (Friedlander-Solomyak). Let α = 2/3. Let lj(ε) be the Dirichleteigenvalues of Ωε. Let

µj = limε→0

ε2α

(lj(ε)−

π2

M2ε2

).

17

exists. Then µj are eigenvalues of the Schrodinger operator H on L2(R),where

H = − d2

dx2+ q(x)

where

q(x) =

2π2M−3C+x x > 02π2M−3C−x x < 0

Note that if C+ = C−, thenH turns to the harmonic oscillator. If C+ = +∞,then it turns to the above discussed case.

Theorem 11 (Lu-Rowlett). Let M be a triangle and let d be the diameter ofM . Then

d2(λ2 − λ1)→ +∞

if the triangle collapses. In other words, the “gap” function is a proper functionon the moduli space of triangles.

The original proof of the above result is independent to the work of Fried-lander et.la.

Question. Let h(x) > 0 be a piecewise smooth function on a boundeddomain Ω of Rn. Let

Ωε = (x, y)|x ∈ Ω, 0 6 y 6 εh(x)

Then what is the asymptotical behavior of the Dirichlet (Neumann) eigenvaluesof Ωε.

The question is very important in answering the following conjecture of Vanden Berg and Yau.

Conjecture 6. Let Ω be a convex bounded domain in Rn. Then

λ2 − λ1 >3π2

d2

The conjecture is asymptotically optimal for thin rectangles.

In some sense, the result of Friedlander gives the compactification of theLaplacians on the moduli space. Namely, if α→ 0, then the Laplace operatorstend to the Schrodinger operator defined above.

As an application, we relate the result to the following conjecture.

Conjecture 7. Does it exist a number N such that the first N Dirichlet eigen-values determine to triangle.

The result of Chang-Deturck gives partial answer to the above conjecture:

18

Theorem 12. There exists N = N(λ1, λ2) such that λ1, · · · , λN determinesthe triangle.

Unfortunately, if α→ 0, N = N(λ1, λ2)→ +∞.

In order to solve this “hearing the shape of a triangle” problem, we considerthe following parametrization of the moduli space of a triangle when one of theangle is smallwhere b 6 1, and we use (α, b) as coordinates.

Conjecture 8. Let ξ = α2/3. Define two functions

P (ζ, b) = λ2/λ1

Q(ζ, b) = (λ3 − λ2)/(λ2 − λ1)

Then P,Q are analytic functions on [0, ε)× [ 12 , 1].

Note that by the result of Friedlander-Solomyak, we know that

P (ζ, b) = 1 + aζ +O(ζ)

Q(ζ, b) = 1 + o(1)

The hearing problem is implied by proving

(ζ, b) 7→ (P,Q)

is invertible.

Since the limit of the Laplacian is the 1-d Schrodinger operator, we mustfirst solve the problem of hearing the Schrodinger operator. Gelfand-Levitantheory doesn’t apply directly here.

3 The Lp-spectrum of the Laplacian

3.1 The Laplacian on Lp space

Definition 1. A one-parameter semi-group on a complex Banach space B is afamily Tt of bounded linear operators, where Tt : B → B parameterized by realnumbers t > 0 and satisfies the following relations:

¬. T0 = 1;

­. If 0 6 s1t < +∞, thenTsTt = Ts+t

®. The mapt1f 7→ Ttf

from [0,+∞)×B to B is jointly continuous.

19

The (infinitesimal) generator Z of a one-parameter semi-group Tt is definedby

Zf = limt→0+

t−1(Ttf − f)

The domain Dom(Z) of Z being the set of f for which the limit exists. It isevident that Dom(Z) is a linear space. Moreover, we have

Lemma 1. The subspace Dom(Z) is dense in B, and is invariant under Tt inthe sense that

Tt(Dom(Z)) ⊂ Dom(Z)

for all t > 0. MoreoverTtZf = ZTtf

for all f ∈ Dom(Z) and t > 0.

Proof. If f ∈ B, we define

ft =

∫ t

0

Txfdx

The above integration exists in the following sense: since Txf is a continuousfunction of x, we can define the integration as the limit of the correspondingRiemann sums. In a Banach space, absolute convergence implies the conditionalconvergence. Thus in order to prove the convergence of the Riemann sums, weonly need to verify that ∫ t

0

‖Txf‖dx

is convergent. But this follows easily from the joint continuity in the definitionof semi-group. ‖Txf‖ must be uniformly bounded for small x.

We compute

limh→0+

h−1(Thft − ft)

= limh→0+

h−1

∫ t+h

h

Txfdx− h−1

∫ t

0

Txfdx

= limh→0+

h−1

∫ t+h

t

Txfdx− h−1

∫ h

0

Txfdx

= Ttf − f

Therefore, ft ∈ Dom(Z) and

Z(ft) = Ttf − f

Since t−1ft → f in norm as t→ 0+, we see that Dom(Z) is dense in B.

The generator Z, in general, is not a bounded operator. However, we canprove the following

20

Lemma 2. The generator Z is a closed operator.

Proof. We first observe that

Ttf − f =

∫ t

0

TxZfdx

if f ∈ Dom(Z). To see this, we consider the function r(t) = Ttf − f −∫ t0TxZfdx. Obviously we have r(0) = 0, and r′(t) ≡ 0. Thus r(t) ≡ 0.

Using the above formula, we have

Tff − f = limn→∞

(Ttfn − fn) = limn→∞

∫ t

0

TxZfndx

By the Lebegue theorem, the above limit is equal to∫ t

0

Txgdx

Thus we havelimt→0+

t−1(Ttf − f) = g

and therefore f ∈ Dom(Z), Zf = g.

Lemma 3. If B is a Hilbert space, then Z must be densely defined and self-adjoint.

Let M be a manifold of dimension n, not necessarily compact or complete.The semi-group can formally be defined as

e∆t

More precisely, the following result is true

Theorem 13. Let M be a manifold, then there is a heat kernel

H(x, y, t) ∈ C∞(M ×M × Rt)

such that

(Ttf)(x) =

∫M

H(x, y, t)f(y)dy

satisfying

¬. H(x, y, t) = H(y, x, t).

­. limt→0+ H(x, y, t) = δx(y).

®. (∆− ∂∂t )H = 0.

21

¯. H(x, y, t) =∫MH(x, z, t− s)H(z, y, s)dz.

In [Getzler], the above theorem was proved. One of the feature of the abovetheorem is that the proof is independent to the fact that ∆ can be extended asa densely defined self-adjoint operator on L2(M). In particular, we don’t needto assume M to be complete. The infinitesimal generator on L2(M) is in factthe Dirichlet Laplacian.

We let ∆p denote the Laplacian on Lp space. With this notation, for most ofthe theorems in linear differential geometry, the completeness assumption canbe removed.

Example 1. Let f ∈ Dom(∆2) such that f ∈ L2(M) and ∆f = 0. Then f hasto be a constant.

When M is a complete manifold, the above is a theorem of Yau. However,it is interesting to see that even when M is incomplete, the above result is stilltrue, and the proof is exactly the same as the original proof of Yau.

Examining some special cases of the above setting is interesting.

A©. If ∂M 6= ∅ and if ∂M is an (n − 1)-dimensional manifold, then A2 is theDirichlet Laplacian.

B©. If M = Rn − 0. Then if f ∈ L2(M), ∆f = 0 and f ∈ Dom(∆2). Thenf(0) = 0 and f must be bounded near 0. By the removable singular-ity theorem, f extends to a harmonic function on Rn, which must be aconstant.

C©. ∆2 is particularly useful on moduli spaces, where it is very difficult todescribe the boundary.

3.1.1 Variational characterization of spectrum

Unlike in the case of compact manifold, in general, a complete manifold doesn’tadmit any pure point spectrum. For example, there are no L2-eigenvalues onRn. That is, for any λ ∈ R, if ∆f+λf = 0 and f ∈ L2(Rn), then we have f ≡ 0.

The above well-known result was generalized by Escobar, who proved thatif M has a rotational symmetric metric, then there is no L2-eigenvalue.

Let ∆ be the Laplacian on a complete non-compact manifold M . By theargument in the previous section, ∆ naturally extends to a self-adjoint denselydefined operator, which we still denote as ∆ for the sake of simplicity.

It is well-known that there is a spectrum measure E such that

−∆ =

∫ ∞0

λdE

22

The heat kernel is defined as

e∆tf(x) =

∫H(x, y, t)f(y)dy

and the Green’s function is defined as

G(x, y) =

∫ ∞0

H(x, y, t)dt

The pure-point spectrum of ∆ are these λ ∈ R such that

¬ There exists an L2 function f 6= 0 such that

∆f + λf = 0

­ The multiplicity of λ is finite.

® In a neighborhood of λ1 it is the only spectrum point.

We define

ρ(∆) = y ∈ R|(∆− y)−1 is a bounded operator

and we define σ(∆) = R−ρ(∆) to be the spectrum of ∆. From the above discus-sion, σ(∆) decomposes as the union of pure point spectrum, and the so-calledessential spectrum, which is, by definition, the complement of the pure-pointspectrum.

The set of the essential spectrum is denoted as σess(∆). Using the abovedefinition, λ ∈ σess(∆), if either

¬ λ is an eigenvalue of infinite multiplicity, or

­ λ is the limiting point of σ(∆).

The following theorems in functional analysis are well-known. For reference,see Donnelly.

Theorem 14. A necessary and sufficient condition for the interval (−∞, λ) tointersect the essential spectrum of an self-adjoint densely defined operator A isthat, for all ε > 0, there exists an infinite dimensional subspace Gε ⊂ Dom(A),for which

(Af − λf − εf, f) < 0

Theorem 15. A necessary and sufficient condition for the interval (λ−a, λ+a)to intersect the essential spectrum of A is that there exists an infinite dimen-sional subspace G ⊂ Dom(A) for which ‖(A− λI)f‖ 6 a‖f‖ for all f ∈ G.

Using the above result, we give the following variational characterization ofthe lower bound of spectrum and the lower bound of essential spectrum.

23

Theorem 16. Using the above notations, define

λ0 = inff∈C∞0 (M)

∫M|∇f |2∫Mf2

and

λess = supK

inff∈C∞0 (M\K)

∫M|∇f |2∫Mf2

where K is a compact set running through an exhaustion of the manifold. Thenλ0 and λess are the least lower bound of σ(∆) and σess(∆), respectively.

Corollary 1. If λ0 < λess, then λ0 is an eigenvalue of M with finite dimen-sional eigenspace.

In this case, λ0 is called the ground state.

In the following, we give a non-trivial application of the above principle.

Theorem 17 (Lin-Lu). Let M be a complex complete surface embedded in R3.Assume that M is not totally geodesic, but asymptotically flat in the sense thatthe second fundamental form goes to zero at infinity. Define

Ω = y ∈ R3|d(y,M) 6 a

for a small positive number a > 0. Then Ω is a 3-d manifold with boundary.The Dirichlet Laplacian of Ω has a ground state.

Sketch of the proof: Since M is asymptotically flat, at infinity

Ω ≈M × [−a, a]

As a result

λess =π2

4a2

Thus the main difficulty in the proof of the above theorem is to prove

λ0 <π2

4a2

which can be obtained by careful analysis of the Gauss and the mean curvatures.

Remark 5. Exner et al. proved that under the condition∫K 6 0,

∫|K| <∞

and M being asymptotically flat, the ground state exists. Thus we make thefollowing conjecture to give the complete picture.

24

Conjecture 9. Let M be a complete, no-totally geodesic, and asymptoticallyembedded surface in R3. Let Ω be defined as before. Let K be the Gauss curva-ture. If ∫

M

|K| < +∞

then the ground state exists.

The difficulty of the above conjecture is that even the surface is asymptoti-cally flat, we still don’t known the long-range behavior of the surface.

3.2 On the theorem of Sturm

Let Mbe a complete Riemannian manifold. We say that the volume (M, g)grows uniformly sub-exponentially, if for any ε > 0, there is a constant C <∞such that for all r > 0 and all x ∈M , we have

v(Br(x)) 6 Ceεrv(B1(X))

Theorem 18 (Sturm). If the volume of (M, g) grows uniformly and sub-exponentially,then the spectrum σ(∆p) of ∆p acting on Lp(M) is independent of p ∈ [1,∞).In particular, it is a subset of the real line.

One feature of the concept “uniformly and sub-exponentially” is that it isself-dual. Take the following example: A hyperbolic space is not “uniformly andsub-exponentially”. On the other side, let Γ be a discrete group acting on thehyperbolic space H, such that Γ\H has finite volume. Since the infinity of Γ\Hare cusps, it is still not “uniformly and sub-exponentially”.

A manifold with non-negative Ricci curvature satisfies the assumption thatthe volume grows “uniformly and sub-exponentially”. However, for such a man-ifold, its volume is infinite. It doesn’t has the finite volume counterpart.

The proof of Sturm’s theorem depends on the heat kernel estimates. Webegin with the following

Lemma 4. If the volume of (M, g) grows uniformly and sub-exponentially, thenfor any ε > 0

supx∈M

∫M

e−εd(x,y)(v(B1(x))−12 v(B1(y))−

12 dv(y) <∞

Proof. We take r = d(x, y). Then since

B1(y) ⊂ Br+1(x)

we must have

v(B1(y))−12 > v(Br+1(x))−

12 > Ce−

12 (r+1)v(B1(x))−

12

25

for any x. Thus the integration in the lemma is less than

C

∫M

e−εre12 ε(r+1)v(B1(x))−1dy

We let f(r) = v(∂Br(x)) and F (r) =∫ r

0f(t)dt. Then up to a constant, the

above expression is less than

(v(B1(x)))−1

∫ ∞0

e−12 εrf(r)dr

By the volume growth assumption f(r) 6 Crn−1vB1(x), the lemma follows.

In fact, the assertion is true if

¬ The Ricci curvature of M has a lower bound;

­

supx∈M

∫M

e−εd(x,y)(v(B1(x))−12 v(B1(y))−

12 dv(y) <∞

The hard part is to prove σ(∆p) ⊂ σ(∆2) for all p ∈ [1,∞]. If this is done,then it is easy to prove

σ(∆2) ⊂ σ(∆p)

as follows:

Let ξ ∈ ρ(∆p). Then (∆p − ξ)−1 is a bounded operator on Lp(M). Let1p + 1

q = 1. By dualization (∆q − ζ)−1 is bounded in Lq(M). By the interpola-

tion theorem, (∆2 − ζ)−1 is bounded and this ζ ∈ ρ(∆2).

In order to prove σ(∆p) ⊂ σ(∆2), or ρ(∆2) ⊂ ρ(∆p), we need some esti-mates. Let ζ ∈ ρ(∆2). Then

(∆2 − ζ)−1

is bounded from L2 → L2. In order to prove that the operator is bounded onLp, we need to prove that it has a kernel g(x, y) such that

(∆2 − ζ)−nf =

∫g(x, y)f(y)dy

Lemma 5. If g(x, y) satisfies

supx∈M

∫M

|g(x, y)|dy 6 C

Then (∆2 − ζ)−n is a bounded operator on Lp(M).

26

Proof. This is essentially Holder inequality:∫M

(∫M

g(x, y)f(y)dy

)pdx

6∫M

(∫M

g1q g

1p fdy

)pdx

6∫M

(∫M

g

) 1q

·∫gfpdx

6 C1q

∫M

∫M

g(x, y)fp(y)dydx

6 C1+ 1q

∫M

fp(y)dy

If we assume that σ(∆p) is a no-where dense set in C, then we have

Lemma 6. If (∆2 − ζ)−n is bounded, so is (∆2 − ζ)−1.

Proof. For any ε, let ζ ′ ∈ ρ(∆ρ) and |ζ − ζ ′| < ε. Then from

‖(∆2 − ζ)−n‖ 6 C

we get‖(∆2 − ζ ′)−n‖ 6 C + 1

provided that ε is small enough. Let dist(ζ ′, σ(∆p)) be the distance to the spec-trum of ∆p, then we have

C + 1 > limm→∞

‖(∆2 − ζ)−nm‖ 1m > dist(ζ ′, σ(∆p))

−n

Thus

dist(ζ ′, σ(∆p)) >1

(C + 1)1n

Since ζ ′ is arbitrary, we have dist(ζ, σ(∆p)) > δ > 0.

4 On the essential spectrum of complete non-compact manifold

Let M be a complete non-compact manifold. We assume that there exists asmall constant δ(n) > 0, depending only on n such that for some point q ∈M ,the Ricci curvature satisfies

Ric(M) > −δ(n)1

r2

where r(x), the distance from x to q is sufficiently large. J-P. Wang [10] provedthe following theorem:

27

Theorem 19. Let M be the complete non-compact Riemannian manifold de-fined above. Then the spectrum of the Laplacian ∆p acting on the space Lp(M)is [0,∞) for all p ∈ [1,∞).

Corollary 2. Let M be a complete manifold with non-negative Ricci curvature,then the L2 essential spectrum of the Laplacian is [0,+∞).

By the Bishop volume comparison theorem, we know that for any completenon-compact manifold with non-negative Ricci curvature, the volume growthis at most polynomial. In general, it is not correct to have the lower boundestimate. However, we have the following:

Theorem 20. Let M be a complete non-compact Riemannian manifold, andlet Ric(M) > 0. Then there is a constant C = C(n, v(B1(p))) such that

v(Bp(R)) > C(n, v(B1(p)))R.

Proof. Let p ∈ M be a fixed point. Let ρ be the distance function with respectto p, Let R > 0 be a large number. Fixing x0 ∈ ∂BR(p). By the Laplaciancomparison theorem, we have

∆ρ2 6 2n.

It follows that for any ϕ ∈ C∞0 (M), ϕ > 0, we have∫M

ϕ∆ρ2 6 2n

∫M

ϕ. (6)

We choose a standard cut-off function ϕ = ψ(ρ(x)), where

ψ(t) =

1 0 6 t 6 R− 112 (R+ 1− t) R− 1 6 t 6 R+ 10 t > R+ 1

.

By the Stoke’s theorem∫M

ϕ∆ρ2 = −2

∫M

ρ∇ϕ∇ρ = −2

∫M

ψ′ρ.

By the definition of ψ, the right hand side of the above is equal to∫BR+1(x0)\BR−1(x0)

ρ > (R− 1)v(BR+1(x0)\BR−1(x0)).

Combining the above equation with ~, we have

(R− 1)v(BR+1(x0)−BR−1(x0)) 6 2n

∫ϕ 6 2nvR+1(x0).

Obviously, we haveB1(p) ⊂ BR+1(x0)\BR−1(x0).

28

Thus we have2nvR+1(x0) > (R− 1)v1(p).

Since B2(R+1)(p) ⊃ BR+1(x0), we have

2nv2(R+1)(p) > (R− 1)v1(p).

or in other word,

v2(R+1)(p) >R− 1

2nv1(p).

What Wang observed was the following inverse Laplacian comparison the-orem: We don’t have a lower bound for the Laplacian. However, we have thefollowing: ∫

B(R)\K|∆ρ| 6

∫B(R)\K

∂n+

∫B(R)\K

|∂n−∆ρ|

6 CRn −∫∂B(R)

∂ρ

∂n+

∫∂K

∂ρ

∂n

6 CRn.

Thus we can also estimate∫

∆ρ from below.

Using the above observation, Wang computed the L1-spectrum. Using thetheorem of Sturm, all Lp-spectrum, in particular the L2-spectrum we are inter-ested, are the same.

It is possible to compute the L2-spectrum directly, but that would be moreor less the same as repeating the proof of Sturm’s theorem. In fact, we can geta little more information than Wang’s theorem provided.

Lemma 7. Let M be a complete non-compact manifold with non-negative Riccicurvature. Let B(R) be a very large ball of radius R. Let λ be a Dirichleteigenvalue and let f be its eigenfunction of B(R). Then there is a constantC > 0 such that ∫

B(R)·B(R−1)

f2 6 C

∫B(R)

f2.

For the rest of this lecture we are seeking possible extensions of Wang’stheorem. While we observe that Sturm’s theorem is self-dual (M could be ofinfinite volume or finite volume), Wang’s theorem is not. In what follows, weshall construct an example that all Lp-spectrum are the same of finite volume,L1-spectrum computable, but doesn’t satisfy the assumption of Wang.

The manifold we construct is of 2 dimensional rotational symmetric outsidea compact set, and the Riemannian metric g can be written as

g = dr2 + f(r)2dθ2, r > 1

29

where f(r) = 1rα for some α large.

The Gauss curvature of g is −f ′′/f = −α(α + 1) 1r2 . Thus the manifold

doesn’t satisfy Wang’s assumption.

We prove that the volume of M grows uniformly and sub-exponentially. Tosee this, we observe that for any point (x)

v(B1(x)) >C

r(x)α

for some constant C > 0. On the other hand, if α > 1, the volume of manifoldis finite. Thus

v(Br(x)) 6 C 6 eεrC(ε)

rα

for any r 0.

Thus the manifold satisfies the assumption of Sturm and as a result, all Lp-spectrum of M are the same.

We compute the L1-spectrum concretely. Following Wang, we pick up alarge number k. Let ψ be a cut-off function whose support is in [1, 4], and isidentically 1 on [2, 3]. Consider the function

g = ψ(r

k)ei√λr.

We have∆g = ∆ψei

√λr + 2∇ψ∇ei

√λr + ψ∆ei

√λr.

We have the following estimate

‖∇ψ∇ei√λr‖L1 6

C

K(V (4k)− V (k)).

where V (r) is the volume of the manifold of radius r. A straightforward com-putation gives

‖∇ψ∇ei√λr‖L1 6

C

kα.

By the same reason

‖∆ψei√λr‖L1 6

C

kα+C

k

∫B(4k)\B(k)

|∆r|.

Since dS2 = dr2 + f(r)2dθ2, we have

|∆r| 6 α

r.

30

Thus ∫B(4k)\(k)

|∆r| =∫ 4k

k

α

rα+1dr 6

C

kα.

Finally

‖ψ∆ei√λr + λg‖L1 = ‖

√λψei

√λr∆r‖L1 6

C

kα.

On the other hand

‖g‖L1 >∫ 3k

2k

1 > V (3k)− V (2k) >C1

kα−1.

Thus if k is sufficiently large

‖∆g + λg‖L1 6 ε‖g‖L1

Thus there should be a finite volume version of Wang’s theorem.

We end the lecture by some speculations of the essential spectrum.

Definition 2. A discrete group G is called amenable, if there is a measure suchthat

1. The measure is a probability measure;

2. The measure is finitely additive;

3. The measure is left-invariant: given a subset A and an element g of G,the measure of A equals to the measure of gA.

In one sentence, G is amenable if it has finitely-additive left-invariant prob-ability measure.

The following theorem of R. Brooks [2] is remarkable:

Theorem 21. ([Brooks] Let M be a compact Riemannian manifold and let M

be the universal cover of M . We assume that M is non-compact, then

λ0(M) = 0⇔ π1(M) is amenable.

It would be interesting to ask

Conjecture 10. Using the same assumptions as above. Then

σess(M) = [0,∞).

In the case when π1(M) = Zn, the above conjecture is true.

Lemma 8. Suppose M = Tn, M = Rn. Then

σess(M) = [0,∞)

31

Here the metric on M is an arbitrary metric.

Proof. Let N be any finite cover of M . Let λ be an eigenvalue of N . Then

λ ∈ σess(M)

In fact, let ρ be a cut-off function. Since

∆f + λf = 0 on N

Then on M‖∆(ρf) + λρf‖L2 5 ‖f∆ρ‖L2 + ‖α∇ρ∇f‖L2

If |∇ρ|, |∆ρ| are small, then

‖∆(fρ) + λρf‖L2 6 ε‖fρ‖L2

If the result is not true, since σess(∆) is a closed set, there is an interval(a, b) such that for any N , there is no eigenvalues in (a, b).

We prove this by contradiction. Let λ, µ be two consecutive eigenvalues suchthat λ < a and µ > b. By the above argument, we can find a C∞0 function on

M such that

‖∆f + λf‖L2 < ε‖f‖L2

‖∆g + µg‖L2 < ε‖g‖L2

Let k, l be integers such that

k∫|∇f |2 + l

∫|∇g|2

k∫f2 + l

∫g2

∈ (a, b)

Then by repeating f k-times and g l-times we are done.

Remark 6. Recently, Lu-Zhou [8] proved that the essential spectrum is [0,+∞)for any complete non-compact manifold with asymptotic nonnegative Ricci cur-vature, generalizing Wang’s result.

References

[1] H. J. Brascamp and E. H. Lieb, On extensions of the Brunn-Minkowski and Prekopa-Leindler theorems, including inequalities for log concave functions, and with an ap-plication to the diffusion equation, J. Functional Analysis 22 (1976), no. 4, 366–389.MR0450480 (56 #8774)

[2] Robert Brooks, The fundamental group and the spectrum of the Laplacian, Comment.Math. Helv. 56 (1981), no. 4, 581–598, DOI 10.1007/BF02566228. MR656213 (84j:58131)

[3] Fengbo Hang and Xiaodong Wang, A remark on Zhong-Yang’s eigenvalue estimate, Int.Math. Res. Not. IMRN 18 (2007), Art. ID rnm064, 9. MR2358887 (2008m:53083)

32

[4] Peter Li and Shing Tung Yau, Estimates of eigenvalues of a compact Riemannian mani-fold, Geometry of the Laplace operator (Proc. Sympos. Pure Math., Univ. Hawaii, Hon-olulu, Hawaii, 1979), Proc. Sympos. Pure Math., XXXVI, Amer. Math. Soc., Providence,R.I., 1980, pp. 205–239. MR573435 (81i:58050)

[5] Jun Ling, The first eigenvalue of a closed manifold with positive Ricci curvature, Proc.Amer. Math. Soc. 134 (2006), no. 10, 3071–3079, DOI 10.1090/S0002-9939-06-08332-8.MR2231634 (2007d:58057)

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