Perspectives and open problems in geometric analysis: spectrum of Laplacian

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1 Perspectives and open problems in geometric analysis: spectrum of Laplacian Zhiqin Lu ay 2, 2010 Abstract 1. Basic gradient estimate; different variations of the gradient estimates; 2. The theorem of Brascamp-Lieb, Barkey-Émery Riemannian geometry, relation of eigenvalue gap with respect to the first Neumann eigenvalue; the Friedlander-Solomayak theorem, 3. The definition of the Laplacian on L p space, theorem of Sturm, 4. Theorem of Wang and its possible generalizations. 1 Gradient estimate of the first eigenvalue Let be an n-dimensional Riemannian manifold with or without boundary. Let the metric ds 2 be represented by ds 2 = g ij dx i dx j, where (x 1,, x n ) are local coordinates. Let = 1 g x i (g ij g x j ) be the Laplace operator, where (g ij ) = g 1 ij, g = det(g ij). The operator acts on smooth functions. If, then we may define one of the following two boundary conditions: 1. Dirichlet condition: f = Neumann condition: the manifold. f n = 0, where n is the outward normal vector of 1

2 By the elliptic regularity, if is compact, then the spectrum of consists of eigenvalues λ 1 λ 2 λ k + of finite multiplicity. By the variational principal, we have the following Poincaré inequality f 2 λ 1 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 be a compact manifold without boundary. Let d be the diameter of. Assume that the Ricci curvature of is non-negative. Then we have the following estimate λ 1 π2 4d 2. Proof. Let u be the first eigenfunction such that max u 2 = 1. f 2 Let g(x) = 1 2 ( u 2 + (λ 1 + ε)u 2 ), where ε > 0. The function g is a smooth function. Let x 0 be the maximal point of g. Then at x 0 we have u j u ji + (λ 1 + ε)uu i = 0 (1) and 0 u 2 ji + u j u jii + (λ 1 + ε) u 2 + (λ 1 + ε)u u. Using the Ricci identity, we have Thus we have u j u jii = u j ( u) j + Ric( u) u j ( u) j. 0 u 2 ji + u j ( u) j + (λ 1 + ε) u 2 + (λ 1 + ε)u u. (2) 2

3 Suppose that at the maximum point of g(x), u 0. Then we have u 2 ji u 4 ( i,j u j u i u ij ) 2 by the Cauchy inequality. Using the first order condition, we conclude u 2 ji (λ 1 + ε) 2 u 2. Putting the above inequality into, we get 0 ε u 2 + ε(λ 1 + ε)u 2 which is not possible. Thus at the maximum point, we must have u = 0. Therefore we have From the above estimate, we get g(x) 1 2 (λ 1 + ε) max u 2 = 1 2 (λ 1 + ε). u 2 1 u 2 λ 1 + ε. Since ε is arbitrary, we let it go to zero and obtain u 2 1 u 2 λ 1. By changing the sign of u, we may assume that max u = 1. Let u(p) = 1 for p. Since u = 0, there is a point q such that u(q) = 0. Let σ(t) be the minimal geodesic line connecting q and p. Consider the function By the above inequality, we get arcsin u(σ(t)) (arcsin u(σ(t))) λ 1 σ (t) Integrating the above inequality along the geodesic line, we get and the theorem is proved. π 2 λ 1 d Several extensions of the above method can be obtained when 1 The manifold has boundary; 2 The Ricci curvature has a lower bound. 3

4 We first address the Neumann boundary condition. Lemma 1. If 0 and is convex, then g(x) doesn t attain its maximum on unless at the point u = 0. Proof. Let x 0 such that g(x) attains the maximum at x 0. Then g(x 0 ) n 0, where n is the outward unit normal vector. By the definition of g(x) and the = 0, we have u fact u n 0. n Let h ij be the second fundamental form. Then j u j 0 u j u j n = h iju i u j 0. If the equality is true, then we must have u(x 0 ) = 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 x 2 + y 2 = R 2. 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 R 2 θ 2 As a result, u = cos θ, sin θ are the two eigenfunctions with the first eigenvalue 1/R 2. If u = cos θ, then with the induced Riemannian metric, u 2 = 1 R 2 sin2 θ. Thus we have and thus which is not sharp. g(θ) = 1 2R 2, u 2 1 u 2 λ 1, The problem is that in general, we don t know whether the first eigenfunction is always symmetric. ore precisely, if we assume that 1 = sup u > inf u = k 1 4

5 we don t know whether k = 1. From Li-Yau s basic estimate, we can improve the estimate λ 1 π 2 /4d 2 to λ 1 ( π 2 + arcsin k)2 d 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 ũ = u 1 k 2. 1+k 2 Then using the standard gradient estimate, we get where ũ 2 λ 1 (1 + a)(1 ũ 2 ) (3) a = 1 k 1 + k. Now the function ũ is symmetric: max ũ = min ũ = 1. Using the same method, we get π 2 π2 λ 1 (1 + a)d2 2d 2 Zhong-Yang [12] took one more step and proved the following result. Theorem 2 (Zhong-Yang). Let be a compact Riemannian manifold with non-negative Ricc curvature. Then λ 1 π2 d 2. 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 If such function ϕ exists, then we have ũ 2 λ 1 (1 + aϕ(arcsin u))(1 ũ 2 ). (4) λ1 d π 2 π 2 dθ 1 + aϕ(θ) π by the convexity of the function 1 1+x, which implies the optimal inequality. 5

6 To prove the inequality (4), we use the maximal principle. At the point x 0 such that the equality of (4) holds, we have ϕ(arcsin u) ũ ũ 1 ũ 2 ϕ (arcsin u) (1 ũ2 )ϕ (arcsin u). We define a function { ( 4 ψ(θ) = π (θ + cos θ sin θ) 2 sin θ) cos 2 θ, ψ( π 2 ) = 1, ψ( π 2 ) = 1 θ ( π 2, π 2 ). Then a straightforward computation gives { ψ (θ) 0 ψ sin θ + sin θ cos θψ 1 2 cos2 θψ = 0. Using the maximum principle we get ϕ(arcsin u) ψ(arcsin u). Since ψ is an odd function, the theorem is proved. Remark 1. Recently, Hang-Wang [3] proved that, in fact, λ 1 > π2 d 2 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 for some constant K > 0. Then Ric() (n 1)K λ 1 π2 (n 1)K. 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 be a compact Riemannian manifold without boundary. Assume that Ric() (n 1)K for K > 0. Then λ 1 where d is the diameter of. 1 (n 1)d 2 exp( [1 + (1 + 4(n 1)2 d 2 K) 1/2 ]), 6

7 Proof. Let u be the normalized first eigenfunction. That is Let β > 1. Consider 1 = sup u > inf u 1. G(x) = u 2 (β u) 2. Let x 0 be the maximum point of G(x). Then G(x 0 ) = 0, G(x 0 ) 0. Since we have G(x)(β u) 2 = u 2, Thus at x 0, we have That is G(β u) G (β u) 2 + G (β u) 2 = u 2. 0 u 2 G (β u) 2 = 2 u 2 ij + 2 u i u ijj 2G[(β u)( u) + u 2 ] = 2u 2 ij + 2u i ( u i ) i + 2Ric( u, u) 2G[λ 1 u(β u) + u 2 ]. u 2 ij λ 1 u 2 (n 1)K u 2 G(λ 1 u(β u) + u 2 ) 0. We choose a local coordinate system at x 0 such that u j = 0 (j = 2,, n), u 1 = u. Then u 1 0 (or otherwise, G(x 0 ) = 0 which is not possible). From G(x 0 ) = 0, we have { u11 = u 2 (β u) 1. u 1i = 0, i 1 Using the following trick ( n n u 2 ij u 2 ii 1 n ) 2 u ii = 1 n 1 n 1 ( u u 11) 2 i j=2 i=2 i=2 1 = n 1 (λu + u 11) 2 = 1 n 1 (λ2 u 2 + 2λuu 11 + u 2 11) we have u (n 1) 1 n 1 λ2 u 2, 1 u 4 2(n 1) (β u) 2 λ2 u 2 n 1 (λ 1 + (n 1)K) u 2 u 2 u λ 1 0. β u 7

8 Let α = u(β u) 1. Then Thus which gives α 1 β u 1 β (n 1) G2 λ2 n 1 α2 (λ 1 + (n 1)K)G λ 1 Gα 0, ( ) λβ G(x) G(x 0 ) 4(n 1) + (n 1)K. β 1 Let l be the geodesic line connecting x 1 and x 2, where u(x 1 ) = 0, u(x 2 ) = sup u = 1. Then we have log or in other words [ ( )] 1/2 β β 1 u γ β u βλ1 4(n 1) + (n 1)K d, β 1 λ 1 β 1 β [ 1 4(n 1)d 2 ( log ) 2 β (n 1)K] β 1 Choosing β 0 such that the right side above maximized, we proved the theorem. The optimal estimate, in this direction, was obtained by Yang: Theorem 4. Let be a compact Riemannian manifold. Then Ric() (n 1)K, (K > 0), d = diam() λ 1 π2 d 2 exp( C nkd 2 ) where C n = n 1 for n > 2 and C n = 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 be a compact Riemannian manifold. Assume that d is the diameter of the manifold and Ric() (n 1)K > 0. for K > 0. Then λ 1 nk. 8

9 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() (n 1)K > 0 the first eigenvalue λ 1, with respect to the closed, the Neumann, or the Dirichlet Laplacian satisfies λ 1 π2 + (n 1)K. d2 Here is, we assume that is convex. Note that by yer s theorem, we always have π 2 /d 2 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 λ 1 π2 d (n 1)K, 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 if the manifold has convex boundary. λ 1 π2 d (n 1)K, 4 Ling [5] was able to improve the above estimate into λ 1 π2 d (n 1)K 100 Further improvements are possible, see Ling-Lu [6] for example. We end this lecture by making the following Conjecture 2. Let be a compact Ricci flat Riemannian manifold such that λ 1 π2 d 2 < ε for a sufficiently small ε > 0. Then = S 0, where 0 is a Ricci flat compact Riemannian manifold. 9

10 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 R n. Let u be the first Dirichlet eigenfunction with the eigenvalue λ 1. 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 u. Thus we have u 2 u 2 u 2. u 2 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(x 0 ) = 0 at a point x 0 in the interior of Ω. Let u(x) = p N (x) + O(x N+ε ) be the Taylor s expansion of the eigenfunction at x 0, where p N (x) is the polynomial of degree N. From the equation u = λ 1 u, we have p 2 (x) = 0. Since u(x) 0, we must have p 2 (x) 0, a contradiction to the maximal principal. Lemma 3. Using the above notations, we have on Ω, the boundary of Ω. u n < 0 Proof. This follows from the strong maximum principle. By the above lemma, we have u n 0 10

11 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 get p 2 0 and p 2 = 0. Since p 2 is harmonic, by the strong maximum principle, p 2 / n < 0 unless p 2 = 0. But if p 2 0 then u 0. This completes the proof. We now consider the heat flow we have u t = u + λ 1u, u Ω = 0. Lemma 4. For any smooth initial function u 0 > 0, the flow exists and converges to the first eigenfunction. Proof. Let u 1 = a j f j (x), where f j (x) is the eigenfunction of the j-th eigenvalue. Then the solution of the equation is u(t, x) = a j e (λj λi)t f j (x) Obviously, we have lim u(t, x) = a 1f 1 (x). t If we choose u 0 such that a 1 0, then the flow converges to the first eigenfunction. f 1 (x) > 0 by the above lemmas. By our choice of u 0 a 1 = u 0 f 1 (x) > 0. The proof is complete. Lemma 5. Let w be any smooth positive function on Ω such that Ω w Ω = 0 and w Ω 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(x 1,, x n ) = 0 to get the function If Ω is convex, then x n = x n (x 1,, x n 1 ). 2 x n x i x j > 0, 11

12 is a positive definite matrix. Using the chain rule, the above inequality is equivalent to w ij + w inw j w n wn 2 + w jnw i wn 2 w iw j w nn wn 3 > 0. where 1 i, j n 1. However, if we allow i or j to be n, then as the n n matrix, we still have w ij + w inw j w n wn 2 + w jnw i wn 2 w iw j w nn wn 3 0. oreover, for any (a 1,, a n ), if a j w j = 0, we have w ij a i a j ε a 2 for some positive ε > 0. A generic vector has the form a + µb, where b = (w 1,, w n ). For w small enough, we have w ij (a i + µb i )(a j + µb j ) + 1 w µ 2 b 4 > 0. Thus 2 log w, whose matrix entries are is negative definite for w small enough. w ij w w iw j w 2, Remark 2. The above proof is purely elementary. We can use differential geometry to give another proof. Assume e 1,, e n are the local frame fields at the boundary point such that e n is normal to the boundary and the rest are tangent to the boundary. Then for 1 i, j n 1, we have 2 (e i, e j )w = e i (e j (w)) e i e j w = h ij w n, where h ij is the second fundamental form, and n field. Thus w n 0. To prove w ij w w iw j w 2 is the outward normal vector is negative definite, we write V = V 1 + µe n, where V 1 is tangent to Ω. Then there is a constant C such that 2 log w(v, V ) w 1 w 4 n (π(v 1, V 1 ) + Cµ V 1 + Cµ 2 ) w 2 µ 2 w n. Since w n 0, for points sufficient close to the boundary, w is small enough. Using the Cauchy inequality, we can prove the above negativeness. 12

13 Now we begin to prove the theorem: We will choose a function u 0 such that u 0 > 0 and log u 0 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 u 0, we first pick up any smooth function w > 0 on Ω with w Ω = 0, w Ω 0. Let u 0 = we C x 2 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 u 0 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 t = u + λ 1u. The flow of ϕ is ϕ t = ϕ ϕ 2 λ 1. (5) By the maximum principle, if T is the first time the matrix 2 ϕ is generated, then there is an x 0 and a direction i such that ϕ ii = 0 and for other j s, ϕ jj 0. oreover, at x 0, ϕ iik = 0, ϕii t 0 and ϕ ii 0. Differentiating (5) by i, i, we have 0 ϕ ii t = ϕ ii 2ϕ k ϕ kii 2ϕ 2 ki 2ϕ 2 ki. By the convexity, ϕ 2 ki ϕ iiϕ kk = 0. Thus ϕ ki = 0 for k 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 R n. Let Ω be a bounded domain in R n with smooth boundary. Let λ 1, λ 2 be the Dirichlet first two eigenvalues. Since λ 1 must be simple (of multiplicity one) we have λ 2 λ 1 > 0. 13

14 The question we would like to answer is that, how to get the lower bound estimate of λ 2 λ 1? Let ϕ 1, ϕ 2 be the eigenfunctions with respect to λ 1, λ 2. We set ϕ = ϕ 2 ϕ 1. Then a straightforward computation gives ϕ + log ϕ 1 ϕ = λϕ, where λ = λ 2 λ 1. oreover, since ϕ 1 Ω = 0, we must have ϕ n = 0. Ω We have a notion of Bakry-Émery Ricci tensor. Let f = ϕ 2 1 Then the Bakry-Émery Laplacian is defined as f = + log f which is a self-adjoint operator with respect to the volume form ϕ 2 1dV. The Bakry-Émery Ricci tensor is defined as 2 log f By the Brascamp-Lieb theorem, the Barkey-Emery Ricci curvature is nonnegative. We make the following conjecture: Conjecture 3. The method of gradient estimates can be generalized to the Bakry-Émery case without additional difficulties. The above conjecture, if true, will give a unified proof between the estimation of the first eigenvalue and gap estimate. We can go one more step further. Since the Barkey-Emery Ricci tensor comes from the setting (, ds 2, e f dv ), which can be considered as the limit of the wrap product e f dr 2 + ds 2 we make the following definition. Let Ω ε = {(x, y) x Ω, 0 y εϕ 2 1(x)}. Let µ ε be the first Neumann eigenvalue. Then 14

15 Conjecture 4. Using the above notations, we have Similarly, we make the following λ 2 λ 1 = lim ε 0 µ ε. Conjecture 5. Let (µ ε ) k be the k-th Neumann eigenvalue of Ω ε. Then We can prove the following Theorem 7. lim ε 0 (µ ε) k = λ k+1 λ 1. λ 2 λ 1 lim ε 0 µ ε. Proof. We first note that h n Ω ε = 0 on the part y = εϕ 1 (x) 2 can be written as h(ε ϕ 2 1, 1) = 0. We define Ũε as follows Ũ ε = ϕ + y 2 log ϕ 1 ϕ. Then a straightforward computation gives { Ũ ε + λũε = O(ε 2 ) = O(ε 2 ) Ωε Ũε n. From the above, we have Let α be a number such that Ω ε Ũ ε = O(ε 2 ). Ω ε (Ũε α) = 0. Then α = O(ε). By the variational principle, we have Ω µ ε ε Ũε 2 Ω ε (Ũε α). 2 However, since We have and the theorem is proved. µ ε Ũε 2 Cε. Ω ε Ω ε λũ ε 2 + O(ε). Ω ε Ũε 2 15

16 Remark 3. We consider the wrap product e f dr 2 + d 2 over Ω ε, 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 R n. Then λ 2 λ 1 π2 d 2. Proof. We consider the domain Ω ε and let U ε be the first Neumann eigenfunction. By what we proved in the last lecture, we have µ ε π2 d 2 ε where d ε is the diameter of Ω ε. Since d ε 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 be an m-dimensional domain in R m. Let be star-shaped. Let R be the radius of the largest ball centered at p contained in and let R 0 be the smallest ball centered at p containing. Then there is a constant C, depending only on m, such that η 1 C Rm R m+2 0 Let U ε be the first Neumann eigenfunction of Ω ε. Then asymptotically we can write U ε ϕ + y 2 log ϕ 1 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

17 As is well-known, the eigenfunctions of the sector are of the form f(r) sin θ α where f(r) is the so-called Bessel function If we set f + 1 r f + 1 r 2 (r2 1 )f = λf α2 g(r) = rf(r) Then the corresponding equation becomes g (r) + 1 r 2 ( 1 α 2 1 )g = λg 4 When α 0, we take the following renomorization: set 1 r = α 2/3 x. Then we have 1 r 2 1 2xα2/3 Let λ = λ + 1 α. Then we have 2 ( ( 1 1 g (r) + r 2 α 2 1 ) 1α ) 4 2 g = λg and if α 0, we have This is the Airy s function. g + 2xg = λg 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 { C+ x x > 0 h(x) = C x x < 0 where the choice of, C +, C are so that h( a) = h(b) = 0. For any positive ε > 0, let Ω ε = {(x, y) x I, 0 y εh(x)} Theorem 10 (Friedlander-Solomyak). Let α = 2/3. Let l j (ε) be the Dirichlet eigenvalues of Ω ε. Let ) µ j = lim ε (l 2α j (ε) π2 ε 0 2 ε 2. 17

18 exists. where where Then {µ j } are eigenvalues of the Schrödinger operator H on L 2 (R), q(x) = H = d2 dx 2 + q(x) { 2π 2 3 C + x x > 0 2π 2 3 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 be a triangle and let d be the diameter of. Then d 2 (λ 2 λ 1 ) + 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 Friedlander et.la. Question. Let h(x) > 0 be a piecewise smooth function on a bounded domain Ω of R n. Let Ω ε = {(x, y) x Ω, 0 y εh(x)} Then what is the asymptotical behavior of the Dirichlet (Neumann) eigenvalues of Ω ε. 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 R n. Then λ 2 λ 1 3π2 d 2 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 Schrödinger 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 eigenvalues determine to triangle. The result of Chang-Deturck gives partial answer to the above conjecture: 18

19 Theorem 12. There exists N = N(λ 1, λ 2 ) such that λ 1,, λ N determines the triangle. Unfortunately, if α 0, N = N(λ 1, λ 2 ) +. 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 1, and we use (α, b) as coordinates. Conjecture 8. Let ξ = α 2/3. Define two functions P (ζ, b) = λ 2 /λ 1 Q(ζ, b) = (λ 3 λ 2 )/(λ 2 λ 1 ) Then P, Q are analytic functions on [0, ε) [ 1 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 is invertible. (ζ, b) (P, Q) Since the limit of the Laplacian is the 1-d Schrödinger operator, we must first solve the problem of hearing the Schrödinger operator. Gelfand-Levitan theory doesn t apply directly here. 3 The L p -spectrum of the Laplacian 3.1 The Laplacian on L p space Definition 1. A one-parameter semi-group on a complex Banach space B is a family T t of bounded linear operators, where T t : B B parameterized by real numbers t 0 and satisfies the following relations: 1. T 0 = 1; 2. If 0 s 1 t < +, then T s T t = T s+t 3. The map t 1 f T t f from [0, + ) B to B is jointly continuous. 19

20 The (infinitesimal) generator Z of a one-parameter semi-group T t is defined by Zf = lim t 0 + t 1 (T t f f) 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. oreover, we have Lemma 1. The subspace Dom(Z) is dense in B, and is invariant under T t in the sense that T t (Dom(Z)) Dom(Z) for all t 0. oreover for all f Dom(Z) and t 0. Proof. If f B, we define T t Zf = ZT t f f t = t 0 T x fdx The above integration exists in the following sense: since T x f 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 t 0 T x f dx is convergent. But this follows easily from the joint continuity in the definition of semi-group. T x f must be uniformly bounded for small x. We compute lim h 0 h 1 (T h f t f t ) + { t+h } t = lim h 1 T x fdx h 1 T x fdx h 0 + h 0 { t+h } h = lim h 1 T x fdx h 1 T x fdx h 0 + = T t f f Therefore, f t Dom(Z) and t Z(f t ) = T t f f Since t 1 f t 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 0 20

21 Lemma 2. The generator Z is a closed operator. Proof. We first observe that T t f f = t 0 T x Zfdx if f Dom(Z). To see this, we consider the function r(t) = T t f f t 0 T xzfdx. Obviously we have r(0) = 0, and r (t) 0. Thus r(t) 0. Using the above formula, we have T f f f = lim n (T tf n f n ) = lim n By the Lebegue theorem, the above limit is equal to Thus we have and therefore f Dom(Z), Zf = g. t 0 T x gdx lim t 1 (T t f f) = g t 0 + t 0 T x Zf n dx Lemma 3. If B is a Hilbert space, then Z must be densely defined and selfadjoint. Let be a manifold of dimension n, not necessarily compact or complete. The semi-group can formally be defined as e t ore precisely, the following result is true Theorem 13. Let be a manifold, then there is a heat kernel H(x, y, t) C ( R t ) such that satisfying (T t f)(x) = H(x, y, t)f(y)dy 1. H(x, y, t) = H(y, x, t). 2. lim t 0 + H(x, y, t) = δ x (y). 3. ( t )H = 0. 21

22 4. H(x, y, t) = 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 L 2 (). In particular, we don t need to assume to be complete. The infinitesimal generator on L 2 () is in fact the Dirichlet Laplacian. We let p denote the Laplacian on L p space. With this notation, for most of the theorems in linear differential geometry, the completeness assumption can be removed. Example 1. Let f Dom( 2 ) such that f L 2 () and f = 0. Then f has to be a constant. When is a complete manifold, the above is a theorem of Yau. However, it is interesting to see that even when 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 and if is an (n 1)-dimensional manifold, then A 2 is the Dirichlet Laplacian. B. If = R n {0}. Then if f L 2 (), f = 0 and f Dom( 2 ). Then f(0) = 0 and f must be bounded near 0. By the removable singularity theorem, f extends to a harmonic function on R n, which must be a constant. C. 2 is particularly useful on moduli spaces, where it is very difficult to describe the boundary 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 L 2 -eigenvalues on R n. That is, for any λ R, if f +λf = 0 and f L 2 (R n ), then we have f 0. The above well-known result was generalized by Escobar, who proved that if has a rotational symmetric metric, then there is no L 2 -eigenvalue. Let be the Laplacian on a complete non-compact manifold. 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 = 0 22 λde

23 The heat kernel is defined as e t f(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 1 There exists an L 2 function f 0 such that 2 The multiplicity of λ is finite. f + λf = 0 3 In a neighborhood of λ 1 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 discussion, 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 σ ess ( ). Using the above definition, λ σ ess ( ), if either 1 λ is an eigenvalue of infinite multiplicity, or 2 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 G ε 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 (A λi)f a f 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

24 Theorem 16. Using the above notations, define λ 0 = inf f 2 f C0 () f 2 and λ ess = sup K inf f C0 (\K) f 2 f 2 where K is a compact set running through an exhaustion of the manifold. Then λ 0 and λ ess are the least lower bound of σ( ) and σ ess ( ), respectively. Corollary 1. If λ 0 < λ ess, then λ 0 is an eigenvalue of with finite dimensional eigenspace. In this case, λ 0 is called the ground state. In the following, we give a non-trivial application of the above principle. Theorem 17 (Lin-Lu). Let be a complex complete surface embedded in R 3. Assume that is not totally geodesic, but asymptotically flat in the sense that the second fundamental form goes to zero at infinity. Define Ω = {y R 3 d(y, ) 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 is asymptotically flat, at infinity As a result Ω [ a, a] λ ess = π2 4a 2 Thus the main difficulty in the proof of the above theorem is to prove λ 0 < π2 4a 2 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 0, K < and being asymptotically flat, the ground state exists. following conjecture to give the complete picture. Thus we make the 24

25 Conjecture 9. Let be a complete, no-totally geodesic, and asymptotically embedded surface in R 3. Let Ω be defined as before. Let K be the Gauss curvature. If K < + 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 be a complete Riemannian manifold. We say that the volume (, g) grows uniformly sub-exponentially, if for any ε > 0, there is a constant C < such that for all r > 0 and all x, we have v(b r (x)) Ce εr v(b 1 (X)) Theorem 18 (Sturm). If the volume of (, g) grows uniformly and sub-exponentially, then the spectrum σ( p ) of p acting on L p () 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 (, g) grows uniformly and sub-exponentially, then for any ε > 0 e εd(x,y) (v(b 1 (x)) 1 2 v(b1 (y)) 1 2 dv(y) < sup x Proof. We take r = d(x, y). Then since we must have B 1 (y) B r+1 (x) v(b 1 (y)) 1 2 v(br+1 (x)) 1 2 Ce 1 2 (r+1) v(b 1 (x))

26 for any x. Thus the integration in the lemma is less than C e εr e 1 2 ε(r+1) v(b 1 (x)) 1 dy We let f(r) = v( B r (x)) and F (r) = r f(t)dt. Then up to a constant, the 0 above expression is less than (v(b 1 (x))) 1 e 1 2 εr f(r)dr By the volume growth assumption f(r) Cr n 1 vb 1 (x), the lemma follows. In fact, the assertion is true if 1 The Ricci curvature of has a lower bound; 0 2 sup e εd(x,y) (v(b 1 (x)) 1 2 v(b1 (y)) 1 2 dv(y) < x 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 L p (). Let 1 p + 1 q = 1. By dualization ( q ζ) 1 is bounded in L q (). By the interpolation 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 estimates. Let ζ ρ( 2 ). Then ( 2 ζ) 1 is bounded from L 2 L 2. In order to prove that the operator is bounded on L p, we need to prove that it has a kernel g(x, y) such that ( 2 ζ) n f = g(x, y)f(y)dy Lemma 5. If g(x, y) satisfies sup x g(x, y) dy C Then ( 2 ζ) n is a bounded operator on L p (). 26

27 Proof. This is essentially Hölder inequality: ( p g(x, y)f(y)dy) dx ( ) p g 1 1 q g p fdy dx C 1 q ( C 1+ 1 q ) 1 q g gf p dx g(x, y)f p (y)dydx f p (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 we get ( 2 ζ) n C ( 2 ζ ) n C + 1 provided that ε is small enough. Let dist(ζ, σ( p )) be the distance to the spectrum of p, then we have Thus C + 1 lim ( 2 ζ) nm 1 m dist(ζ, σ( p )) n m dist(ζ 1, σ( p )) (C + 1) 1 n Since ζ is arbitrary, we have dist(ζ, σ( p )) > δ > 0. 4 On the essential spectrum of complete noncompact manifold Let 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, the Ricci curvature satisfies Ric() δ(n) 1 r 2 where r(x), the distance from x to q is sufficiently large. J-P. Wang [10] proved the following theorem: 27

28 Theorem 19. Let be the complete non-compact Riemannian manifold defined above. Then the spectrum of the Laplacian p acting on the space L p () is [0, ) for all p [1, ). Corollary 2. Let be a complete manifold with non-negative Ricci curvature, then the L 2 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 be a complete non-compact Riemannian manifold, and let Ric() 0. Then there is a constant C = C(n, v(b 1 (p))) such that v(b p (R)) C(n, v(b 1 (p)))r. Proof. Let p be a fixed point. Let ρ be the distance function with respect to p, Let R > 0 be a large number. Fixing x 0 B R (p). By the Laplacian comparison theorem, we have ρ 2 2n. It follows that for any ϕ C0 (), ϕ 0, we have ϕ ρ 2 2n ϕ. (6) We choose a standard cut-off function ϕ = ψ(ρ(x)), where 1 0 t R 1 1 ψ(t) = 2 (R + 1 t) R 1 t R t R + 1. By the Stoke s theorem ϕ ρ 2 = 2 ρ ϕ ρ = 2 ψ ρ. By the definition of ψ, the right hand side of the above is equal to ρ (R 1)v(B R+1 (x 0 )\B R 1 (x 0 )). B R+1 (x 0)\B R 1 (x 0) Combining the above equation with, we have (R 1)v(B R+1 (x 0 ) B R 1 (x 0 )) 2n ϕ 2nv R+1 (x 0 ). Obviously, we have B 1 (p) B R+1 (x 0 )\B R 1 (x 0 ). 28

29 Thus we have Since B 2(R+1) (p) B R+1 (x 0 ), we have or in other word, 2nv R+1 (x 0 ) (R 1)v 1 (p). 2nv 2(R+1) (p) (R 1)v 1 (p). v 2(R+1) (p) R 1 2n v 1(p). 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: ρ n + n ρ B(R)\K B(R)\K CR n CR n. B(R) Thus we can also estimate ρ from below. B(R)\K ρ n + Using the above observation, Wang computed the L 1 -spectrum. Using the theorem of Sturm, all L p -spectrum, in particular the L 2 -spectrum we are interested, are the same. It is possible to compute the L 2 -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 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 f 2 C f 2. 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 ( could be of infinite volume or finite volume), Wang s theorem is not. In what follows, we shall construct an example that all L p -spectrum are the same of finite volume, L 1 -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 = dr 2 + f(r) 2 dθ 2, r > 1 K ρ n 29

30 where f(r) = 1 r α for some α large. The Gauss curvature of g is f /f = α(α + 1) 1 doesn t satisfy Wang s assumption. r 2. Thus the manifold We prove that the volume of grows uniformly and sub-exponentially. To see this, we observe that for any point (x) v(b 1 (x)) C r(x) α for some constant C > 0. On the other hand, if α > 1, the volume of manifold is finite. Thus εr C(ε) v(b r (x)) C e r α for any r 0. Thus the manifold satisfies the assumption of Sturm and as a result, all L p - spectrum of are the same. We compute the L 1 -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 We have We have the following estimate g = ψ( r k )ei λr. g = ψe i λr + 2 ψ e i λr + ψ e i λr. ψ e i λr L 1 C (V (4k) V (k)). K where V (r) is the volume of the manifold of radius r. A straightforward computation gives ψ e i λr L 1 C k α. By the same reason ψe i λr L 1 C k α + C k B(4k)\B(k) r. Since ds 2 = dr 2 + f(r) 2 dθ 2, we have r α r. 30

31 Thus Finally B(4k)\(k) r = 4k k α r α+1 dr C k α. ψ e i λr + λg L 1 = λψe i λr r L 1 C k α. On the other hand g L 1 Thus if k is sufficiently large 3k 2k 1 V (3k) V (2k) C 1 k α 1. g + λg L 1 ε g L 1 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 be a compact Riemannian manifold and let be the universal cover of. We assume that is non-compact, then It would be interesting to ask λ 0 ( ) = 0 π 1 () is amenable. Conjecture 10. Using the same assumptions as above. Then σ ess ( ) = [0, ). In the case when π 1 () = Z n, the above conjecture is true. Lemma 8. Suppose = T n, = R n. Then σ ess ( ) = [0, ) 31

32 Here the metric on is an arbitrary metric. Proof. Let N be any finite cover of. Let λ be an eigenvalue of N. Then λ σ ess ( ) In fact, let ρ be a cut-off function. Since Then on If ρ, ρ are small, then f + λf = 0 on N (ρf) + λρf L 2 f ρ L 2 + α ρ f L 2 (fρ) + λρf L 2 ε fρ L 2 If the result is not true, since σ ess ( ) 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 such that Let k, l be integers such that f + λf L 2 < ε f L 2 g + µg L 2 < ε g L 2 k f 2 + l g 2 k f 2 + l g 2 (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 curvature, generalizing Wang s result. References [1] H. J. Brascamp and E. H. Lieb, On extensions of the Brunn-inkowski and Prékopa- Leindler theorems, including inequalities for log concave functions, and with an application to the diffusion equation, J. Functional Analysis 22 (1976), no. 4, R (56 #8774) [2] Robert Brooks, The fundamental group and the spectrum of the Laplacian, Comment. ath. Helv. 56 (1981), no. 4, , DOI /BF R (84j:58131) [3] Fengbo Hang and Xiaodong Wang, A remark on Zhong-Yang s eigenvalue estimate, Int. ath. Res. Not. IRN 18 (2007), Art. ID rnm064, 9. R (2008m:53083) 32

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