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 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 @ p @ ∆ = p (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: ¬. Dirichlet condition: fj@M = 0. @f ­. Neumann condition: @n j@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 ! +1 of finite multiplicity. By the variational principal, we have the following Poincar´einequality Z Z 2 2 jrfj > 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) = (jruj2 + (λ + ")u2); 2 1 where " > 0. The function g is a smooth function. Let x0 be the maximal point of g. Then at x0 we have ujuji + (λ1 + ")uui = 0 (1) and 2 2 0 > uji + ujujii + (λ1 + ")jruj + (λ1 + ")u∆u: Using the Ricci identity, we have ujujii = uj(∆u)j + Ric(ru) > uj(∆u)j: Thus we have 2 2 0 > uji + uj(∆u)j + (λ1 + ")jruj + (λ1 + ")u∆u: (2) 2 Suppose that at the maximum point of g(x), ru 6= 0. Then we have 2 −4 X 2 uji > jruj ( ujuiuij) i;j by the Cauchy inequality. Using the first order condition, we conclude 2 2 2 uji > (λ1 + ") u : Putting the above inequality into ~, we get 2 2 0 > "jruj + "(λ1 + ")u which is not possible. Thus at the maximum point, we must have ru = 0. Therefore we have 1 1 g(x) (λ + ") max u2 = (λ + "): 6 2 1 2 1 From the above estimate, we get jruj2 λ + ": 1 − u2 6 1 Since " is arbitrary, we let it go to zero and obtain jruj2 λ : 1 − u2 6 1 By changing the sign of u, we may assume that max u = 1. Let u(p) = 1 for p 2 M. Since R u = 0, there is a point q 2 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 j(arcsin u(σ(t))) j 6 λ1jσ (t)j 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 ¬ 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 maximum on @M unless at the point ru = 0. Proof. Let x0 2 @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 ru(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 jruj2 = sin2 θ: R2 Thus we have 1 g(θ) = ; 2R2 and thus jruj2 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 jruej 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 esti- mate (3), we suspect that there is an odd function '(arcsin u) such that 2 2 jruej 6 λ1(1 + a'(arcsin u))(1 − ue ): (4) If such function ' exists, then we have π p Z 2 dθ λ1d > > π π p − 2 1 + a'(θ) by the convexity of the function p 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 '(arcsin u) u − u 1 − u2'0(arcsin u) + (1 − u2)'00(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 ; 2 ) (θ) = π π : ( 2 ) = 1; (− 2 ) = −1 Then a straightforward computation gives 0 (θ) > 0 0 1 2 00 : − sin θ + sin θ cos θ − 2 cos θ = 0 Using the maximum principle we get '(arcsin u) 6 (arcsin u). Since 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 jruj2 G(x) = : (β − u)2 Let x0 be the maximum point of G(x). Then rG(x0) = 0; ∆G(x0) 6 0: Since G(x)(β − u)2 = jruj2; we have ∆G(β − u)2 + 2rGr(β − u)2 + G∆(β − u)2 = ∆jruj2: Thus at x0, we have 2 2 0 > ∆jruj − G∆(β − u) X 2 X 2 = 2 uij + 2 uiuijj − 2G[(β − u)(−∆u) + jruj ] 2 = 2uij + 2ui(∆ui)i 2 + 2Ric(ru; ru) − 2G[λ1u(β − u) + jruj ]: That is 2 2 2 2 uij − λ1jruj − (n − 1)Kjruj − G(λ1u(β − u) + jruj ) 6 0: We choose a local coordinate system at x0 such that uj = 0 (j = 2; ··· ; n), u1 = jruj. Then u1 6= 0 (or otherwise, G(x0) = 0 which is not possible). From rG(x0) = 0, we have u = −|∇uj2(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 jruj4 λ2u2 jruj2u − − (λ + (n − 1)K)jruj2 − λ 0: 2(n − 1) (β − u)2 n − 1 1 1 β − u 6 7 Let α = u(β − u)−1.