Perspectives and Open Problems in Geometric Analysis: Spectrum of Laplacian

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 estimates; 2. The theorem of Brascamp-Lieb, Barkey-Emery´ Riemannian geometry, relation of eigenvalue gap with respect to the first Neumann eigenvalue; 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

2 X ds = gijdxidxj, where (x1, ··· , xn) are local coordinates. Let

1 X ∂ √ ∂ ∆ = √ (gij g ) g ∂xi ∂xj

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

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

x. Dirichlet condition: f|∂M = 0.

∂f y. Neumann condition: ∂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 ∆ consists of eigenvalues λ1 6 λ2 6 ··· 6 λk → +∞ of ﬁnite multiplicity.

By the variational principal, we have the following Poincar´einequality Z Z 2 2 |∇f| > λ1 f

To our special interests, we would like to give “computable” lower bound estimates of the ﬁrst eigenvalue. Here by “computable” we mean the geometric quantities like the diameter, the bounds of the curvature, etc, that are readily available.

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 d be the diameter of M. Assume that the Ricci curvature of M is non-negative. Then we have the following estimate

π2 λ . 1 > 4d2 Proof. Let u be the ﬁrst eigenfunction such that

max u2 = 1.

Let 1 g(x) = (|∇u|2 + (λ + ε)u2), 2 1 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) and 2 2 0 > uji + ujujii + (λ1 + ε)|∇u| + (λ1 + ε)u∆u. Using the Ricci identity, we have

ujujii = uj(∆u)j + Ric(∇u) > uj(∆u)j. Thus we have

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

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

2 −4 X 2 uji > |∇u| ( ujuiuij) i,j by the Cauchy inequality. Using the ﬁrst order condition, we conclude

2 2 2 uji > (λ1 + ε) u .

Putting the above inequality into ~, we get

2 2 0 > ε|∇u| + ε(λ1 + ε)u which is not possible. Thus at the maximum point, we must have ∇u = 0. Therefore we have 1 1 g(x) (λ + ε) max u2 = (λ + ε). 6 2 1 2 1 From the above estimate, we get

|∇u|2 λ + ε. 1 − u2 6 1 Since ε is arbitrary, we let it go to zero and obtain

|∇u|2 λ . 1 − u2 6 1 By changing the sign of u, we may assume that max u = 1. Let u(p) = 1 for p ∈ M. Since R 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

arcsin u(σ(t))

By the above inequality, we get

0 p 0 |(arcsin u(σ(t))) | 6 λ1|σ (t)| Integrating the above inequality along the geodesic line, we get π p λ d 2 6 1 and the theorem is proved.

Several extensions of the above method can be obtained when x The manifold has boundary;

y 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 maximum on ∂M unless at the point ∇u = 0.

Proof. Let x0 ∈ ∂M such that g(x) attains the maximum at x0. Then ∂g(x ) 0 0, ∂n 6 where ~n is the outward unit normal vector. By the deﬁnition of g(x) and the ∂u fact ∂n = 0, we have ∂u u j 0. j ∂n 6

Let hij be the second fundamental form. Then ∂u 0 u j = h u u 0. > j ∂n ij i j >

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 unit circle, 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 is 1 ∂2 R2 ∂θ2 As a result, u = cos θ, sin θ are the two eigenfunctions with the ﬁrst eigenvalue 1/R2. If u = cos θ, then with the induced Riemannian metric, 1 |∇u|2 = sin2 θ. R2 Thus we have 1 g(θ) = , 2R2 and thus |∇u|2 λ , 1 − u2 6 1 which is not sharp.

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

1 = sup u > inf u = −k > −1

4 we don’t know whether k = 1. From Li-Yau’s basic estimate, we can improve 2 2 the estimate λ1 > π /4d to π λ ( + arcsin k)2d−2. 1 > 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 − 1−k u = 2 . e 1+k 2 Then using the standard gradient estimate, we get

2 2 |∇ue| 6 λ1(1 + a)(1 − ue ) (3) where 1 − k a = . 1 + k Now the function ue is symmetric: max ue = − min ue = 1. Using the same method, we get π2 π2 λ 1 > (1 + a)d2 > 2d2 Zhong-Yang [12] took one more step and proved the following result. Theorem 2 (Zhong-Yang). Let M be a compact Riemannian manifold with non-negative Ricc curvature. Then

π2 λ . 1 > 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, is far from being optimal.

The basic idea of the proof is still the maximum principle. From the estimate (3), we suspect that there is an odd function ϕ(arcsin u) such that

2 2 |∇ue| 6 λ1(1 + aϕ(arcsin u))(1 − ue ). (4) If such function ϕ exists, then we have

π p Z 2 dθ λ1d > > π π p − 2 1 + aϕ(θ) by the convexity of the function √ 1 , which implies the optimal inequality. 1+x

5 To prove the inequality (4), we use the maximal principle. At the point x0 such that the equality of (4) holds, we have p 1 ϕ(arcsin u) u − u 1 − u2ϕ0(arcsin u) + (1 − u2)ϕ00(arcsin u). 6 e e e 2 e We deﬁne a function 4 −2 π π ( π (θ + cos θ sin θ) − 2 sin θ) cos θ, θ ∈ (− 2 , 2 ) ψ(θ) = π π . ψ( 2 ) = 1, ψ(− 2 ) = −1 Then a straightforward computation gives

0 ψ (θ) > 0 0 1 2 00 . ψ − sin θ + sin θ cos θψ − 2 cos θψ = 0

Using the maximum principle we get ϕ(arcsin u) 6 ψ(arcsin u). Since ψ is an odd function, the theorem is proved.

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

π2 λ > 1 d2 unless the manifold is of one dimensional. The Li-Yau-Zhong-Yang estimate is still eﬀective when the Ricci curvature is not “too negative”. Namely, let

Ric(M) > −(n − 1)K for some constant K > 0. Then π2 λ − (n − 1)K. 1 > d2 Thus as long as the right hand side of the above is positive, the estimate is eﬀective.

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 1 λ exp(−[1 + (1 + 4(n − 1)2d2K)1/2]), 1 > (n − 1)d2 where d is the diameter of M.

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

1 = sup u > inf u > −1. Let β > 1. Consider |∇u|2 G(x) = . (β − u)2

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

∇G(x0) = 0, ∆G(x0) 6 0. Since G(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

2 2 0 > ∆|∇u| − G∆(β − u) X 2 X 2 = 2 uij + 2 uiuijj − 2G[(β − u)(−∆u) + |∇u| ] 2 = 2uij + 2ui(∆ui)i 2 + 2Ric(∇u, ∇u) − 2G[λ1u(β − u) + |∇u| ]. That is

2 2 2 2 uij − λ1|∇u| − (n − 1)K|∇u| − G(λ1u(β − u) + |∇u| ) 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 u = −|∇u|2(β − u)−1 11 . u1i = 0, i 6= 1 Using the following trick

n n n !2 X X 1 X 1 u2 u2 u = (∆u − u )2 ij > ii > n − 1 ii n − 1 11 i·j=2 i=2 i=2 1 1 = (λu + u )2 = (λ2u2 + 2λuu + u2 ) n − 1 11 n − 1 11 11 u2 1 11 − λ2u2, > 2(n − 1) n − 1 we have 1 |∇u|4 λ2u2 |∇u|2u − − (λ + (n − 1)K)|∇u|2 − λ 0. 2(n − 1) (β − u)2 n − 1 1 1 β − u 6

7 Let α = u(β − u)−1. Then 1 1 α . 6 β − u 6 β − 1 Thus 1 λ2 G2 − α2 − (λ + (n − 1)K)G − λ Gα 0, 2(n − 1) n − 1 1 1 6 which gives

λβ G(x) G(x ) 4(n − 1) + (n − 1)K . 6 0 6 β − 1

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

Z 1/2 β |∇u| βλ1 log 6 6 4(n − 1) + (n − 1)K d, β − 1 γ β − u β − 1 or in other words " # β − 1 1 β 2 λ log − (n − 1)K 1 > β 4(n − 1)d2 β − 1

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 π2 λ exp(−C Kd2) 1 > d2 n √ √ 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. The following theorem of Lichnerowicz is well known. Theorem 5. Let M be a compact Riemannian manifold. Assume that d is the diameter of the manifold and

Ric(M) > −(n − 1)K > 0. for K > 0. Then λ1 > 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 the ﬁrst eigenvalue λ1, with respect to the closed, the Neumann, or the Dirichlet Laplacian satisﬁes π2 λ + (n − 1)K. 1 > d2 Here is ∂M 6= ∅, we assume that ∂M is convex.

2 2 Note that by Myer’s theorem, we always have π /d > K. Thus the conjecture, if true, will give a common generalization of the result of Lichnerowicz’s and the one obtained by the gradient estimate.

In this direction, D-G Yang [11] proved that the first Dirichlet eigenvalue of the Laplacian satisfies π2 1 λ + (n − 1)K, 1 > d2 4 if the manifold has weakly convex boundary. He also proved that the first closed eigenvalue and the first Neumann eigenvalue of the Laplacian satisfies

π2 1 λ + (n − 1)K, 1 > d2 4 if the manifold has convex boundary.

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

π2 31 λ + (n − 1)K 1 > d2 100 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

π2 λ − < ε 1 d2

0 for a suﬃciently small ε > 0. Then M = S × M0, where M0 is a Ricci ﬂat compact 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 heat flow proof.

Theorem 6. Let Ω be a bounded convex domain of Rn. Let u be the ﬁrst Dirichlet eigenfunction with the eigenvalue λ1. Then (up to a sign), u is positive and 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, we have |∇|u|| 6 |∇u|. Thus we have R |∇|u||2 R |∇u|2 . R u2 6 R u2 By the variational characterizing of the ﬁrst eigenvalue, we know that the right side of the above is minimized. Thus the equality must hold and |u| is an eigenfunction 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)

N be the Taylor’s expansion of the eigenfunction at x0, where p (x) is the polyno- 2 mial of degree N. From the equation ∆u = −λ1u, we have ∆p (x) = 0. Since 2 u(x) > 0, we must have p (x) > 0, a contradiction to the maximal principal. Lemma 3. Using the above notations, we have ∂u < 0 ∂n on ∂Ω, the boundary of Ω. Proof. This follows from the strong maximum principle. By the above lemma, we have ∂u 0 ∂n 6

10 ∂u If ∂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 get 2 2 2 p > 0 and ∆p = 0. Since p 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 = ∆u + λ u, u| = 0. ∂t 1 ∂Ω we have

Lemma 4. For any smooth initial function u0 > 0, the ﬂow exists and converges to the ﬁrst eigenfunction. Proof. Let X u1 = ajfj(x), where fj(x) is the eigenfunction of the j-th eigenvalue. Then the solution of the equation is X −(λj −λi)t u(t, x) = aje fj(x) Obviously, we have lim u(t, x) = a1f1(x). t→∞

If we choose u0 such that a1 6= 0, then the ﬂow converges to the ﬁrst eigenfunction. f1(x) > 0 by the above lemmas. By our choice of u0 Z a1 = u0f1(x) > 0. Ω 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, log w is concave. Proof. Since w is a smooth function vanishes on the boundary, it can be viewed as the deﬁning function of Ω. By the implicit function theorem, we solve the equation w(x1, ··· , xn) = 0 to get the function xn = xn(x1, ··· , xn−1). If Ω is convex, then ∂2x n > 0, ∂xi∂xj

11 is a positive deﬁnite matrix. Using the chain rule, the above inequality is equiv- alent to wij winwj wjnwi wiwjwnn − + 2 + 2 − 3 > 0. wn wn wn wn where 1 6 i, j 6 n − 1. However, if we allow i or j to be n, then as the n × n matrix, we still have

wij winwj wjnwi wiwjwnn − + 2 + 2 − 3 > 0. wn wn wn wn P Moreover, for any (a1, ··· , an), if ajwj = 0, we have

2 −wijaiaj > ε|a| for some positive ε > 0. A generic vector has the form a + µb, where b = (w1, ··· , wn). For w small enough, we have 1 −w (a + µb )(a + µb ) + |µ|2|b|4 > 0. ij i i j j w Thus ∇2 log w, whose matrix entries are w w w ij − i j , w w2 is negative definite for w small enough. Remark 2. The above proof is purely elementary. We can use differential geometry to give another proof. Assume e1, ··· , en are the local frame fields at the boundary point such that en is normal to the boundary and the rest are tangent to the boundary. Then for 1 6 i, j 6 n − 1, we have ∂w ∇2(e , e )w = e (e (w)) − ∇e e w = h , i j i j i j ij ∂n

∂ where hij is the second fundamental form, and ∂n is the outward normal vector ∂w ﬁeld. Thus ∂n 6 0. To prove w w w ij − i j w w2 is negative deﬁnite, we write V = V1 + µen, where V1 is tangent to ∂Ω. Then there is a constant C such that 4 2 −1 ∂w 2 −2 2 ∂w ∇ log w(V,V ) w (π(V1,V1) + CµkV1k + Cµ ) − w µ . 6 ∂n ∂n

∂w Since ∂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 such that u0 > 0 and log u0 is concave. Then we shall prove that the log-concavity is preserved 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 > 0 on Ω with w|∂Ω = 0, ∇w|∂Ω 6= 0. Let

−C P x2 u0 = we j for a constant C > 0 suﬃciently large. By the above lemma, the function is log-concave near the boundary. Away from the boundary, since w > δ > 0 for 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 the flow keeps the log-concavity. Let ϕ = log u, where u is the solution of the heat equation ∂u = ∆u + λ u. ∂t 1 The flow of ϕ is ∂ϕ = ∆ϕ − |∇ϕ|2 − λ . (5) ∂t 1 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

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

Diﬀerentiating (5) by i, i, we have

∂ϕ 0 ii = ∆ϕ − 2ϕ ϕ − 2ϕ2 −2ϕ2 . > ∂t ii k kii ki > ki

2 By the convexity, ϕki 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.

n Let Ω be a bounded domain in R with smooth boundary. Let λ1, λ2 be the Dirichlet ﬁrst two eigenvalues. Since λ1 must be simple (of multiplicity one) we have λ2 − λ1 > 0.

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

Let ϕ1, ϕ2 be the eigenfunctions with respect to λ1, λ2. We set ϕ ϕ = 2 . ϕ1 Then a straightforward computation gives

∆ϕ + ∂∇ log ϕ1∇ϕ = −λϕ, where λ = λ2 − λ1. Moreover, since ϕ1|∂Ω = 0, we must have

∂ϕ = 0. ∂n ∂Ω We have a notion of Bakry-Emery´ Ricci tensor. Let

2 f = ϕ1 Then the Bakry-Emery´ Laplacian is deﬁned as ∆f = ∆ + ∇ log f∇

2 which is a self-adjoint operator with respect to the volume form ϕ1dV .

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 the Bakry-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 tensor comes from the setting (M, ds2, ef dV ), which can be considered as the limit of the wrap product ef dr2 + ds2 we make the following deﬁnition.

Let 2 Ωε = {(x, y)|x ∈ Ω, 0 6 y 6 εϕ1(x)}.

Let µε be the ﬁrst Neumann eigenvalue. Then

14 Conjecture 4. Using the above notations, we have

λ2 − λ1 = lim µε. ε→0 Similarly, we make the following

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

k lim(µε) = λk+1 − λ1. ε→0 We can prove the following Theorem 7. λ2 − λ1 lim µε. > ε→0 ∂h 2 Proof. We ﬁrst note that ∂n |∂Ωε = 0 on the part y = εϕ1(x) can be written as 2 ∇h(ε∇ϕ1, −1) = 0.

We deﬁne Ueε as follows

2 Ueε = ϕ + y ∇ log ϕ1∇ϕ. Then a straightforward computation gives

( 2 ∆Ueε + λUeε = O(ε ) . ∂Ueε 2 ∂n = O(ε ) ∂Ωε From the above, we have Z 2 Ueε = O(ε ). Ωε Let α be a number such that Z (Ueε − α) = 0. Ωε Then α = O(ε). By the variational principle, we have

R 2 |∇Uε| Ωε e µε . 6 R 2 (Uε − α) Ωε e However, since Z 2 Ueε = Cε. Ωε We have R λU 2 Ωε eε µε 6 + O(ε). R U 2 Ωε eε and the theorem is proved.

15 Remark 3. We consider the wrap product

ef dr2 + d2 over Ωε, we see the relation between two settings. This also gives the relation between the Barkey-Emery geometry with respect to the ordinary Riemannian geometry. Some applications. Theorem 8 (Singer-Wong-Yau-Yau, Yu-Zhong). Let Ω be a convex domain in Rn. Then π2 λ − λ . 2 1 > d2

Proof. We consider the domain Ωε and let Uε be the ﬁrst Neumann eigenfunction. By what we proved in the last lecture, we have π2 µε > 2 dε 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 use the 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 let R0 be the smallest ball centered at p containing M. Then there is a constant C, depending only on m, such that Rm η1 > C m+2 R0

Let Uε be the ﬁrst Neumann eigenfunction of Ωε. Then asymptotically we can write 2 Uε ∼ ϕ + y ∇ log ϕ1∇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 domain collapses. In this appendix, we give some preliminary results in this direction.

We begin with the following observation. Consider the sector What 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 θ f(r) sin α where f(r) is the so-called Bessel function 1 1 1 f 00 + f 0 + (r2 − )f = −λf r r2 α2

If we set √ g(r) = rf(r) Then the corresponding equation becomes 1 1 1 −g00(r) + ( − )g = λg r2 α2 4 When α → 0, we take the following renomorization: set 1 − r = α2/3x. Then we have 1 − 1 ∼ 2xα2/3 r2 1 Let λ = λe + α2 . Then we have 1 1 1 1 −g00(r) + − − g = λge r2 α2 4 α2 and if α → 0, we have −g00 + 2xg = λg This is the Airy’s function.

Friedlander and Solomyak was able to generalize the above result in the following setting:

Let h(x) > 0 be a piecewise linear function deﬁned on [−a, b], where a, b > 0. We assume that M − C x x > 0 h(x) = + M − 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 Dirichlet eigenvalues of Ωε. Let

2 2α π µj = lim ε lj(ε) − . ε→0 M 2ε2

17 2 exists. Then {µj} are eigenvalues of the Schr¨odinger operator H on L (R), where d2 H = − + q(x) dx2 where 2 −3 2π M C+x x > 0 q(x) = 2 −3 2π M C−x x < 0

Note that if C+ = C−, then H 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 of M. Then 2 d (λ2 − λ1) → +∞ if the triangle collapses. In other words, the “gap” function is a proper function on 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 bounded domain 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) eigenvalues of Ωε.

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

Conjecture 6. Let Ω be a convex bounded domain in Rn. Then 3π2 λ − λ 2 1 > d2 The conjecture is asymptotically optimal for thin rectangles.

In some sense, the result of Friedlander gives the compactification of the Laplacians on the moduli space. Namely, if α → 0, then the Laplace operators tend to the Schrödingeroperator 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 eigenvalues 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 determines the triangle.

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

In order to solve this “hearing the shape of a triangle” problem, we consider the following parametrization of the moduli space of a triangle when one of the angle is small where 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)

1 Then P,Q are analytic functions on [0, ε) × [ 2 , 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 Schrödingeroperator, we must first solve the problem of hearing the Schrödingeroperator. Gelfand-Levitan theory 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 a family Tt of bounded linear operators, where Tt : B → B parameterized by real numbers t > 0 and satisﬁes the following relations:

x. T0 = 1;

y. If 0 6 s1t < +∞, then TsTt = Ts+t

z. The map t1f 7→ Ttf from [0, +∞) × B to B is jointly continuous.

19 The (inﬁnitesimal) generator Z of a one-parameter semi-group Tt is deﬁned by −1 Zf = lim t (Ttf − f) t→0+ The domain Dom(Z) of Z being the set of f for which the limit exists. It is evident 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 in the sense that Tt(Dom(Z)) ⊂ Dom(Z) for all t > 0. Moreover TtZf = ZTtf for all f ∈ Dom(Z) and t > 0. Proof. If f ∈ B, we deﬁne Z t ft = Txfdx 0

The above integration exists in the following sense: since Txf is a continuous function of x, we can deﬁne the integration as the limit of the corresponding Riemann sums. In a Banach space, absolute convergence implies the conditional convergence. Thus in order to prove the convergence of the Riemann sums, we only need to verify that Z t kTxfkdx 0 is convergent. But this follows easily from the joint continuity in the deﬁnition of semi-group. kTxfk must be uniformly bounded for small x. We compute

−1 lim h (Thft − ft) h→0+ ( Z t+h Z t ) −1 −1 = lim h Txfdx − h Txfdx h→0+ h 0 ( Z t+h Z h ) −1 −1 = lim h Txfdx − h Txfdx h→0+ t 0

= Ttf − f

Therefore, ft ∈ Dom(Z) and

Z(ft) = Ttf − f

−1 + Since t ft → 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 can prove the following

20 Lemma 2. The generator Z is a closed operator. Proof. We ﬁrst observe that

Z t Ttf − f = TxZfdx 0 if f ∈ Dom(Z). To see this, we consider the function r(t) = Ttf − f − R t 0 0 TxZfdx. Obviously we have r(0) = 0, and r (t) ≡ 0. Thus r(t) ≡ 0. Using the above formula, we have

Z t Tf f − f = lim (Ttfn − fn) = lim TxZfndx n→∞ n→∞ 0 By the Lebegue theorem, the above limit is equal to

Z t Txgdx 0 Thus we have −1 lim t (Ttf − f) = g t→0+ 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

∞ t H(x, y, t) ∈ C (M × M × R ) such that Z (Ttf)(x) = H(x, y, t)f(y)dy M satisfying x. H(x, y, t) = H(y, x, t).

y. limt→0+ H(x, y, t) = δx(y).

∂ z. (∆ − ∂t )H = 0.

21 R {. H(x, y, t) = M H(x, z, t − s)H(z, y, s)dz. In [Getzler], the above theorem was proved. One of the feature of the above theorem is that the proof is independent to the fact that ∆ can be extended as a densely deﬁned self-adjoint operator on L2(M). In particular, we don’t need to assume M to be complete. The inﬁnitesimal generator on L2(M) is in fact the Dirichlet Laplacian.

p We let ∆p denote the Laplacian on L space. With this notation, for most of the theorems in linear diﬀerential geometry, the completeness assumption can be removed.

2 Example 1. Let f ∈ Dom(∆2) such that f ∈ L (M) and ∆f = 0. Then f has to 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 still true, 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 the Dirichlet Laplacian.

n 2 B . If M = R − {0}. Then if f ∈ L (M), ∆f = 0 and f ∈ Dom(∆2). Then f(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 a constant.

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

3.1.1 Variational characterization of spectrum Unlike in the case of compact manifold, in general, a complete manifold doesn’t admit any pure point spectrum. For example, there are no L2-eigenvalues on Rn. 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 that if M has a rotational symmetric metric, then there is no L2-eigenvalue.

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

It is well-known that there is a spectrum measure E such that Z ∞ −∆ = λdE 0

22 The heat kernel is deﬁned as Z e∆tf(x) = H(x, y, t)f(y)dy and the Green’s function is deﬁned as Z ∞ G(x, y) = H(x, y, t)dt 0

The pure-point spectrum of ∆ are these λ ∈ R such that x There exists an L2 function f 6= 0 such that

∆f + λf = 0

y The multiplicity of λ is ﬁnite.

z In a neighborhood of λ1 it is the only spectrum point. We deﬁne

−1 ρ(∆) = {y ∈ R|(∆ − y) 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-called essential spectrum, which is, by deﬁnition, the complement of the pure-point spectrum.

The set of the essential spectrum is denoted as σess(∆). Using the above definition, λ ∈ σess(∆), if either x λ is an eigenvalue of infinite multiplicity, or y 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) to intersect the essential spectrum of an self-adjoint densely defined operator A is that, 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 dimensional subspace G ⊂ Dom(A) for which k(A − λI)fk 6 akfk for all f ∈ G. Using the above result, we give the following variational characterization of the lower bound of spectrum and the lower bound of essential spectrum.

23 Theorem 16. Using the above notations, deﬁne

R 2 M |∇f| λ0 = inf f∈C∞(M) R 2 0 M f and R 2 M |∇f| λess = sup inf f∈C∞(M\K) R 2 K 0 M f where K is a compact set running through an exhaustion of the manifold. Then λ0 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 dimensional 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 that the second fundamental form goes to zero at infinity. Define

3 Ω = {y ∈ R |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 π2 λ = ess 4a2 Thus the main diﬃculty in the proof of the above theorem is to prove

π2 λ < 0 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 Z Z K 6 0, |K| < ∞ and M being asymptotically ﬂat, the ground state exists. Thus we make the following conjecture to give the complete picture.

24 Conjecture 9. Let M be a complete, no-totally geodesic, and asymptotically embedded surface in R3. Let Ω be deﬁned as before. Let K be the Gauss curvature. If Z |K| < +∞ M then the ground state exists.

The diﬃculty of the above conjecture is that even the surface is asymptotically 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

εr v(Br(x)) 6 Ce v(B1(X)) Theorem 18 (Sturm). If the volume of (M, g) grows uniformly and sub-exponentially, p then the spectrum σ(∆p) of ∆p acting on L (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 is self-dual. Take the following example: A hyperbolic space is not “uniformly and sub-exponentially”. On the other side, let Γ be a discrete group acting on the hyperbolic space H, such that Γ\H has ﬁnite volume. Since the inﬁnity of Γ\H are cusps, it is still not “uniformly and sub-exponentially”.

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

The proof of Sturm’s theorem depends on the heat kernel estimates. We begin with the following Lemma 4. If the volume of (M, g) grows uniformly and sub-exponentially, then for any ε > 0 Z −εd(x,y) − 1 − 1 sup e (v(B1(x)) 2 v(B1(y)) 2 dv(y) < ∞ x∈M M Proof. We take r = d(x, y). Then since

B1(y) ⊂ Br+1(x) we must have

− 1 − 1 − 1 (r+1) − 1 v(B1(y)) 2 > v(Br+1(x)) 2 > Ce 2 v(B1(x)) 2

25 for any x. Thus the integration in the lemma is less than Z −εr 1 ε(r+1) −1 C e e 2 v(B1(x)) dy M R r We let f(r) = v(∂Br(x)) and F (r) = 0 f(t)dt. Then up to a constant, the above expression is less than Z ∞ −1 − 1 εr (v(B1(x))) e 2 f(r)dr 0

n−1 By the volume growth assumption f(r) 6 Cr vB1(x), the lemma follows. In fact, the assertion is true if x The Ricci curvature of M has a lower bound; y Z −εd(x,y) − 1 − 1 sup e (v(B1(x)) 2 v(B1(y)) 2 dv(y) < ∞ x∈M M

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:

−1 p Let ξ ∈ ρ(∆p). Then (∆p − ξ) is a bounded operator on L (M). Let 1 1 −1 q p + q = 1. By dualization (∆q − ζ) is bounded in L (M). By the interpola- −1 tion theorem, (∆2 − ζ) is bounded and this ζ ∈ ρ(∆2).

In order to prove σ(∆p) ⊂ σ(∆2), or ρ(∆2) ⊂ ρ(∆p), we need some estimates. Let ζ ∈ ρ(∆2). Then −1 (∆2 − ζ) is bounded from L2 → L2. In order to prove that the operator is bounded on Lp, we need to prove that it has a kernel g(x, y) such that Z −n (∆2 − ζ) f = g(x, y)f(y)dy

Lemma 5. If g(x, y) satisﬁes Z sup |g(x, y)|dy 6 C x∈M M

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

26 Proof. This is essentially H¨olderinequality: Z Z p g(x, y)f(y)dy dx M M p Z Z 1 1 6 g q g p fdy dx M M 1 Z Z q Z p 6 g · gf dx M M Z Z 1 p 6 C q g(x, y)f (y)dydx M M Z 1+ 1 p 6 C q f (y)dy M

If we assume that σ(∆p) is a no-where dense set in C, then we have −n −1 Lemma 6. If (∆2 − ζ) is bounded, so is (∆2 − ζ) .

0 0 Proof. For any ε, let ζ ∈ ρ(∆ρ) and |ζ − ζ | < ε. Then from

−n k(∆2 − ζ) k 6 C we get 0 −n k(∆2 − ζ ) k 6 C + 1 0 provided that ε is small enough. Let dist(ζ , σ(∆p)) be the distance to the spectrum of ∆p, then we have

−nm 1 0 −n C + 1 lim k(∆2 − ζ) k m dist(ζ , σ(∆p)) > m→∞ > Thus 0 1 dist(ζ , σ(∆p)) > 1 (C + 1) n 0 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 a small constant δ(n) > 0, depending only on n such that for some point q ∈ M, the Ricci curvature satisﬁes 1 Ric(M) −δ(n) > r2 where r(x), the distance from x to q is suﬃciently large. J-P. Wang [10] proved the following theorem:

27 Theorem 19. Let M be the complete non-compact Riemannian manifold de- p ﬁned above. Then the spectrum of the Laplacian ∆p acting on the space L (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 complete non-compact manifold with non-negative Ricci curvature, the volume growth is at most polynomial. In general, it is not correct to have the lower bound estimate. However, we have the following:

Theorem 20. Let M be a complete non-compact Riemannian manifold, and let 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 respect to p, Let R > 0 be a large number. Fixing x0 ∈ ∂BR(p). By the Laplacian comparison theorem, we have 2 ∆ρ 6 2n. ∞ It follows that for any ϕ ∈ C0 (M), ϕ > 0, we have Z Z 2 ϕ∆ρ 6 2n ϕ. (6) M M We choose a standard cut-off function ϕ = ψ(ρ(x)), where  1 0 6 t 6 R − 1  1 ψ(t) = 2 (R + 1 − t) R − 1 6 t 6 R + 1 .  0 t > R + 1 By the Stoke’s theorem Z Z Z ϕ∆ρ2 = −2 ρ∇ϕ∇ρ = −2 ψ0ρ. M M M By the definition of ψ, the right hand side of the above is equal to Z ρ > (R − 1)v(BR+1(x0)\BR−1(x0)). BR+1(x0)\BR−1(x0)

Combining the above equation with ~, we have Z (R − 1)v(BR+1(x0) − BR−1(x0)) 6 2n ϕ 6 2nvR+1(x0).

Obviously, we have B1(p) ⊂ BR+1(x0)\BR−1(x0).

28 Thus we have 2nvR+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, R − 1 v (p) v (p). 2(R+1) > 2n 1 What Wang observed was the following inverse Laplacian comparison theorem: We don’t have a lower bound for the Laplacian. However, we have the following: Z Z Z |∆ρ| 6 ∂n + |∂n − ∆ρ| B(R)\K B(R)\K B(R)\K Z Z n ∂ρ ∂ρ 6 CR − + ∂B(R) ∂n ∂K ∂n n 6 CR . Thus we can also estimate R ∆ρ from below.

Using the above observation, Wang computed the L1-spectrum. Using the theorem 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 more or less the same as repeating the proof of Sturm’s theorem. In fact, we can get a little more information than Wang’s theorem provided. Lemma 7. Let M be a complete non-compact manifold with non-negative Ricci curvature. Let B(R) be a very large ball of radius R. Let λ be a Dirichlet eigenvalue and let f be its eigenfunction of B(R). Then there is a constant C > 0 such that Z Z 2 2 f 6 C f . B(R)·B(R−1) B(R) For the rest of this lecture we are seeking possible extensions of Wang’s theorem. While we observe that Sturm’s theorem is self-dual (M could be of infinite volume or finite volume), Wang’s theorem is not. In what follows, we shall 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 outside a compact set, and the Riemannian metric g can be written as

g = dr2 + f(r)2dθ2, r > 1

29 1 where f(r) = rα for some α large.

00 1 The Gauss curvature of g is −f /f = −α(α + 1) r2 . Thus the manifold doesn’t satisfy Wang’s assumption.

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

C v(B (x)) 1 > r(x)α for some constant C > 0. On the other hand, if α > 1, the volume of manifold is ﬁnite. Thus C(ε) v(B (x)) C eεr r 6 6 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 a large number k. Let ψ be a cut-oﬀ function whose support is in [1, 4], and is identically 1 on [2, 3]. Consider the function r √ g = ψ( )ei λr. k

We have √ √ √ ∆g = ∆ψei λr + 2∇ψ∇ei λr + ψ∆ei λr. We have the following estimate

√ i λr C k∇ψ∇e k 1 (V (4k) − V (k)). L 6 K where V (r) is the volume of the manifold of radius r. A straightforward computation gives √ i λr C k∇ψ∇e k 1 . L 6 kα By the same reason

√ Z i λr C C 1 k∆ψe kL 6 α + |∆r|. k k B(4k)\B(k)

Since dS2 = dr2 + f(r)2dθ2, we have α |∆r| . 6 r

30 Thus Z Z 4k α C |∆r| = α+1 dr 6 α . B(4k)\(k) k r k Finally √ √ √ i λr i λr C kψ∆e + λgk 1 = k λψe ∆rk 1 . L L 6 kα On the other hand Z 3k C1 1 kgkL > 1 > V (3k) − V (2k) > α−1 . 2k k Thus if k is suﬃciently large

k∆g + λgkL1 6 εkgkL1 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 such that 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 probability measure.

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

Theorem 21. ([Brooks] Let M be a compact Riemannian manifold and let Mf be the universal cover of M. We assume that Mf is non-compact, then

λ0(Mf) = 0 ⇔ π1(M) is amenable. It would be interesting to ask Conjecture 10. Using the same assumptions as above. Then

σess(Mf) = [0, ∞).

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

Lemma 8. Suppose M = T n, Mf = Rn. Then

σess(Mf) = [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(Mf)

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

∆f + λf = 0 on N

Then on Mf k∆(ρf) + λρfkL2 5 kf∆ρkL2 + kα∇ρ∇fkL2 If |∇ρ|, |∆ρ| are small, then

k∆(fρ) + λρfkL2 6 εkfρkL2

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 such ∞ that λ < a and µ > b. By the above argument, we can ﬁnd a C0 function on Mf such that

k∆f + λfkL2 < εkfkL2

k∆g + µgkL2 < εkgkL2

Let k, l be integers such that

k R |∇f|2 + l R |∇g|2 ∈ (a, b) k R f 2 + l R g2

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 curvature, generalizing Wang’s result.

References

[1] H. J. Brascamp and E. H. Lieb, On extensions of the Brunn-Minkowski and Pr´ekopa- Leindler theorems, including inequalities for log concave functions, and with an application 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 manifold, 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) [6] Jun Ling and Zhiqin Lu, Bounds of Eigenvalues on Riemannian Manifolds, Trends in Partial Differential Equations, ALM, 10, Higher Education Press and International Press Beijing-Boston, 2009, pp. 241–264. [7] Zhiqin Lu and Julie Rowlett, preprint. [8] Zhiqin Lu and Detang Zhou, preprint. [9] I. M. Singer, B. Wong, S. T. Yau, and S. S.-T. Yau, An estimate of the gap of the first two eigenvalues in the Schrödingeroperator, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 12 (1985), no. 2, 319–333. MR829055 (87j:35280) [10] Jiaping Wang, The spectrum of the Laplacian on a manifold of nonnegative Ricci curvature, Math. Res. Lett. 4 (1997), no. 4, 473–479. MR1470419 (98h:58194) [11] DaGang Yang, Lower bound estimates of the first eigenvalue for compact manifolds with positive Ricci curvature, Pacific J. Math. 190 (1999), no. 2, 383–398. MR1722898 (2001b:53039) [12] Jia Qing Zhong and Hong Cang Yang, On the estimate of the first eigenvalue of a compact Riemannian manifold, Sci. Sinica Ser. A 27 (1984), no. 12, 1265–1273. MR794292 (87a:58162)