240B CLASS NOTE. Contents. 1. Integration on manifolds.

Transcription

1 240B CLASS NOTE ZHIQIN LU Contents 1. Integration on manifolds The extension of the Levi-Civita connection Covariant derivatives The Laplace operator Self-adjoint extension of the Laplace operator The Poincaré Lemma de Rham Theorem Formal Hodge Theorem The Hodge theorem Proof of the Hodge Theorem ore about elliptic regularity Semigroups and their generators Heat kernel Variational characterization of spectrum Poincaré inequality and Sobolev inequality De Giorgi-Nash-oser Estimates Aleksandrov maximum principle Krylov-Safonov Estimates Integration on manifolds. Let be an n-dimensional manifold and let ω be an n-form on with compact support. The integration of ω over can be defined in the following way. Let {U i } be a locally finite cover of and let {ρ i } be the partition of unity subordinating to the cover. That is (1) ρ i 0; (2) supp ρ i U i for all i; (3) i ρ i = 1. Date: January 1st, The author is partially supported by the NSF award DS

2 2 ZHIQIN LU Then we can define ω = j U j ρ j ω, where the right hand side is a sum of integrations on Euclidean space, hence well defined. Exercise 1. Prove that the definition of integral is independent to the choices of cover and the partition of unity. The definition of integration can be generalized to manifold with boundary. A manifold with boundary is a topological space such that (1) It can be covered by {U j }, where U j is either the unit ball of R n, or the upper half of the unit ball of R n defined by {(x 1,, x n ) x 2 i < 1, x n 0}. (2) The transition functions are smooth (up to the boundary). Let be a manifold with boundary. Let be the boundary of. Then we have (1) is a smooth manifold; (2) is a manifold without boundary. That is = 0, or 2 = 0. The theorem of partition of unity can be generalized into the following setting. Theorem 1. Let {U i } I be a locally finite cover of. Then there exists a partition of unity subordinating to the cover {ρ i }. oreover, let Then J = {j I U j = }. {U j } j J, ρ j Uj is a partition of unity subordinating to the corresponding cover. Exercise 2. Prove the above theorem. A manifold is called orientable, if there is a cover {U i } on such that if x 1 j,, xn j are the coordinates on U j, then we have det (x 1 j,, xn j ) (x 1 i,, xn i ) > 0 for any U i U j. Such a set of coordinate charts is called an orientation of the manifold. Exercise 3. If is an orientable manifold with boundary, prove that is also an orientable manifold.

3 240B CLASS NOTE 3 Theorem 2 (Stokes Theorem). Let be an orientable manifold with the given orientation. Then there is a natural orientation such that for any (n 1) forms on. dω = ω Proof. Let {U i } be a locally finite cover and let {ϕ i } be the partition of unity subordinating to the cover. Since d ρ j = 0, we have dω = ρ j dω = d(ρ j ω), j I U j j I U j ω = ρ j ω. U j j J and The Stokes theorem follows from the following statements: (1) U j d(ρ j ω) = 0, j J (2) U j d(ρ j ω) = U j {x n =0} ρ jω, j J. In the first case, we may assume that U j is the unit ball of R n, and supp (ρ j ω) U j. Write ρ j ω = ( 1) i f i dx 1 dx ˆ i dx n, where ˆ means omit the term. Then f d(ρ j ω) = dx 1 dx n. U j x i Since f j vanishes on the boundary of U j, we have U j d(ρ j ω) = 0 by the Newton-Lebinitz theorem. The proof of the second assertion is similar. We assume that U j = {(x 1,, x n ) x 2 j < 1, x n 0}. Then we have j d(ρ j ω) = n i=1 f i x i dx 1 dx n.

4 4 ZHIQIN LU If i n, then U j f i x i dx 1 dx n = 0 With the appropriate orientation, we have f n dx 1 dx n = U j x n U j {x n =0} f n dx 1 dx n 1. Exercise 4. Provided the details in the last part of the proof of Stokes Theorem. Lemma 1. On an orientable Riemannian manifold, the n-form det(g ij )dx 1 dx n is globally defined. Definition 1. Let f be a smooth function with compact support in a Riemannian manifold. Then f dv = f det(g ij )dx 1 dx n. i Exercise 5. Let X = a i x i be a vector field of. The divergence of the vector field X is defined to be a i div X = x i ai log det(g kl ). x i Prove that div XdV = The extension of the Levi-Civita connection. Let be a Riemannian manifold and let be the Levi-Civita connection. We extend to all tensor fields as follows. Definition 2. Let X, Y be vector fields. (1) If f C (), then X f = X f ; (2) X Y is defined as in the usual Levi-Civita connection; (3) If ω is a one-form, we use to define the one form X ω; ( X ω)y = X(ω(Y)) ω( X Y)

5 240B CLASS NOTE 5 (4) In general, let {e 1,, e n } be a local frame of and let ω 1,, ω n be the dual frame. Then if We define T = a i1 i p j 1 j q e i1 e ip ω j1 ω jq, X T = X(a i1 i p j 1 j q )e i1 ω jq p + a i1, i p j 1 j q e i1 X e is ω jq + s=1 q a i1, i p j 1 j q e i1 X ω jt ω jq. t=1 Using the above notations, we have Theorem 3. X ds 2 = 0, where ds 2 is the Riemannian metric. and Proof 1. Using the local coordinates: we have Thus we have x k = Γ l x ki i x l x k dx j = Γ j kl dx l x k g ij dx i dx j = g ij x k dx i dx j g ij Γ i kl dx ldx j g ij Γ j kl dx idx l = 0. Proof 2. Using the intrinsic characterization of the Riemannian metric ( X ds 2 )(Y, Z) = X( Y, Z ) X Y, Z Y, X Z = 0. We also have the following result which well justifies our definition of the curvature operator. Theorem 4. Let d 2 be the operator defined by the composition T T T Then it is the curvature operator. and Proof. We have T T T = Γ j dx x ki k, i x j 2 = Γ j ki dx k dx l x i x l x j T Λ 2 (). + Γ j ki Γm lj dx k dx l Γ j x ki Γk lm dx m dx l. m x j

6 6 ZHIQIN LU Thus we have d 2 j ki = Γ x i x l x j dx k dx l + Γ j ki Γm lj dx k dx l Γ j x ki Γk lm dx m dx l. m x j Obviously, the last term of the above equation is zero. Thus after changing indices, we have d 2 Γ j ki = x i Γ m x ki Γm lm l dx k dx l. x j By a straightforward computation, we get d 2 = 1 x i 2 R imklg mj dx k dx l. x k As showed above, it is quite tedious to use local coordinates. The natural frame ( ), x 1 x n may not be our best choice. For the rest of the lecture notes, in the most cases, we use orthonormal frames. Let e 1,, e n be local orthonormal frame and let ω 1,, ω n be the dual frame. Then the Riemannian metric can be written as ds 2 = ω 1 ω 1 + ω n ω n, where the tensor product is understood as the symmetric tensor product. We write X e j = ω ij (X)e i, and R ijkl = ek el e j el ek e j [ek,e l ]e j, e i. The Cartan s formulas are { dωj + ω i ω ij = 0, dω ij + ω is ω sj = 1 2 R ijklω k ω l. Here ω ij = ω ji. Exercise 6. Verify the above formulas. Exercise 7. Prove that given ω j, there are unique ω ij with ω ij = ω ji satisfying the first Cartan equation. Exercise 8. Prove that at any point p, one can choose an orthonormal frame such that at p, the connection matrix ω ij is zero.

7 240B CLASS NOTE 7 Exercise 9. Prove that, if the curvature is zero on a neighborhood, then on that neighborhood, one can choose an orthonormal frame such that the connection matrix is identically zero on that neighborhood. 3. Covariant derivatives. The notation of connection on tensors field can be applied to co-tensors and yield the following Lemma 2. Let Then ek η = e k(a i1 i p ) η = a i1 i p ω i1 ω ip. s a i1 r sth i p ω r is (e k ) ω i 1 ω ip. Proof. A straightforward computaton. Because of the above result, we make the following definition: Definition 3. We define a i1, i p,k = e k (a i1,,i p ) a i1, r i p ω r is (e k ) s sth and call it the covariant derivative of the coeficients a i1 i p. By the above definition, we have We have the following Theorem 5. We have η = a i1 i p,k ω k ω i1 ω ip. a i1 i p,k,l a i1 i p,l,k = a i1, r i p R ris kl. sth Proof. To prove the above result, we introduce some notations. We let I = (i 1,, i p ). Then we have 1 2 (a I,k,l a I,l,k ) ω k ω l ω I = a I,k,l ω k ω l ω I = d 2η. Using the above notations, we can re-write the definition of covariant derivatives as following η = da I a Ir ω ris, where I r = (i 1 r i p ). sth

8 8 ZHIQIN LU We compute ( ) d 2η = d ( η)i,k ω k ω I (1) = ( ) d( η) I,k ( η) Ir,kω ris ( η) I,r ω rk ωk ω I. The first and the third term of the above can be consolidated because d( η) I,k ω k ( η) I,r ω rk ω k = d(( η) I,k ω k ) by the first Cartan s formula. Thus we have d( η) I,k ω k ( η) I,r ω rk ω k = d(da Ir a Ir ω ris ) = da Ir ω ris a Ir dω ris. Combining the above equation with (1), we have d 2η = ( da Ir ω ris a Ir dω ris ) ω I ( η) Ir,kω k ω ris ω I = a Ir (dω ris + ω rl ω lis ) ω I + a i1 µ i p ω µit ω ris ω I. For fixed s t, we have a i1 µ r tth sth Thus we have and the lemma is proved. s t i p ω µit ω ris ω I + a i1 µ r tth sth tth r sth d 2η = 1 2 a I r R ris kl ω k ω l ω I, i p ω µis ω rit ω I = 0. Let f be a smooth function. The following notations are often used We have f i = e i ( f ); f ij = f i,j ; f ijk = f i,j,k. Corollary 1 (Ricci identity). Using the above notations, we have f ij = f ji ; f ijk f ikj = f r R rijk. In particular, we have the following useful Ricci identity: f iji f iij = (Ric) rj f r. i Proof. We only need to prove f ij = f ji. To see this, we compute i f ij ω j ω i = (d f i f r ω ri ) ω i = d f i ω i f r ω ri ω i. By the Cartan s formula, the right hand side of the above is d( f i ω i ) = dd f = 0.

9 240B CLASS NOTE 9 4. The Laplace operator. Let be a compact orientable Riemannian manifold and let Λ p () be the vector space of p forms. As in the last section, we use ω 1,, ω n to be the local orthonormal frame, and ω 1 ω n is the volume form. We write a p-form in the following expression η = a i1 i p ω i1 ω ip, where we assume the coefficients are skew-symmetric. That is, we assume that a σ(i1 ) σ(i p ) = ( 1) sgn σ a i1 i p for any permutation σ of {1,, p}. The L 2 inner product is defined as (2) (ω, η) = p! a i1 i p b i1 i p ω 1 ω n. We remark that in the above definition, we only require one of the coefficients to be skew-symmetric, which is useful in the following computation. Exercise 10. Prove the following: if η = b i1 i p ω i1 ω ip, where η = b i1 i p are not assumed to be skew-symmetric. Then we still have (ω, η) = p! a i1 i p b i1 i p ω 1 ω n. Lemma 3. We have dω = ( 1) p a i1 i p,k ω i1 ω ip ω k. Since the coefficients above are not skew symmetric, we may also rewrite the expression as p dω = ( 1)p p + 1 a i 1 i p,k a i1 k i p,i s ω i1 ω ip ω k. s=1 sth Proof. The proof is through a straightforward computation: p dω = da i1 i p ω i1 ω ip + ( 1) s 1 a i1 i p ω i1 dω is ω ip. s=1 Using Cartan s formula dω is = ω i ω iis, we get the first formula. The second formula is through skew-symmetrization. Lemma 4. We have δω = ( 1) p p a i1 i p,i p ω i1 ω ip 1.

10 10 ZHIQIN LU Proof. Let be a (p 1)-form. We verify that We have η = b i1 i p 1 ω i1 ω ip 1 (δω, η) (ω, dη) = 0. (δω, η) = ( 1) p p! a i1 i p 1 i p,i p b i1 i p 1, and (ω, dη) = ( 1) p 1 p! a i1 i p b i1 i p 1,i p. define an (n 1) form α such that n α = ( 1) k 1 a i1 i p 1 kb i1 i p 1 ω 1 ˆω k ω n. Then we have k=1 dα = (a i1 i p 1 i p,i p b i1 i p 1 + a i1 i p b i1 i p 1,i p ) ω 1 ω n. The lemma follows from the Stokes theorem dα = 0. Theorem 6 (Weizenböck formula). We have p 1 ω = a i 1 i p,k,k + p i p 1 kr ris i p k + p a i1 i p 1 r(ric) rip ω i 1 ω ip. and s=1 a i1 r sth Proof. By the above two lemmas we have p δdω = a i 1 i p,k,k + s=1 a i1 k sth i p,i s,k ω i 1 ω ip, dδω = p a i1 i p,i p,kω i1 ω ip 1 ω k. By changing the indices and using the skew-symmetry of a i1 i p in the above, we get and δdω = ( a i1 i p,k,k + pa i1 i p 1 k,i p,k) ωi1 ω ip, dδω = p a i1 k,k,i p ω i1 ω ip 1 ω ip. The theorem follows by applying Theorem 5. Define the raw Laplacian on a p form to be ω = k k ω.

11 Then the Weizenböck formula can be written as 240B CLASS NOTE 11 ω = ω + E(ω), where E is a 0-th order differential operator depending on the curvature. We would like to list the following special cases of the above theorem as exercises. Exercise 11. If f is a smooth function, then Exercise 12. If ω is a one-form, then = f,k,k = f kk. ω = ω + a r (Ric) rk ω k. 5. Self-adjoint extension of the Laplace operator. In this section, we assume that is a compact manifold. Let L 2 (Λ p ()) be the metric space completion of Λ p () under the inner product product defined in (2). Exercise 13. Prove that we can identify L 2 (Λ p ()) to be the space of p-forms ω, where ω = a i1 i p ω i1 ω ip, such that a i1 i p are locally L 2 integrable functions. The reason we would like to use L 2 (Λ p ()) in stead of Λ p () is, of course, that L 2 (Λ p ()) is complete, allowing many applications in analysis. Unfortunately, it is not possible to extend to L 2 (Λ p ()) as a symmetric linear operator. In what follows, we briefly explain the reason. First, the Laplace operator is not a bounded operator on Λ p (). We can give counterexamples even in one-dimensional case: let f be a smooth function on [0, 1]. Then there doesn t exist a constant C such that 1 0 ( f (t)) 2 dt C 1 0 ( f (t)) 2 dt. Second, like most differential operators, the Laplace operator is a closedgraph operator. That is, if η j η and η j η in L 2 and η Λ p (), then we must have η = η. To see this, we consider any smooth p form ω: We have η η, ω = lim j η j, ω lim j η j, ω = 0. Thus we have η = η. If could be extended to a linear operator of L 2 (Λ p ()), then by the Closed Graph Theorem, could have been bounded, which is a contradiction. Because of the above result, the best we can do is to extend the Laplacian operator into a densely defined self-adjoint operator. Of course, Λ p () is dense in L 2 (Λ p ()). But for such an operator (i.e, and its domain Dom( )), we don t have the so-called spectral theorem, which we introduce below.

12 12 ZHIQIN LU Definition 4. Let H be a Hilbert space, Given a densely defined linear operator A on H, its adjoint A is defined as follows: (1) The domain of A consists of vectors x in H such that y x, Ay is a bounded linear functional, where y Dom( ); (2) By the Riesz Representation Theorem for linear functionals, if x is in the domain of A, there is a unique vector z in H such that x, Ay = z, y for any y Dom( ). This vector z is defined to be A x. It can be shown that the dependence of z on x is linear. If A = A (which implies that Dom (A ) = Dom (A)), then A is called self-adjoint. For a (densely-defined) self-adjoint operator A, we have the Spectral Theorem. That is, there is a spectral measure such that A = + λde. For the rest of this section, we define the self-adjoint extension of the Laplace operator. Define H 1 () be the Sobolev space of the completion of the vector space 0 Λ p () under the norm η 1 = η 2 dv + η 2 dv. We define the quadratic form Q on H 1 () by 0 ( ) Q(ω, η) = dω, dη + δω, δη dv for any ω, η H 1 (). Then we define 0 Dom( ) = { ϕ H 1 0 () ψ Λ p (), f L 2 (Λ p ()), s.t. Q(ϕ, ψ) = ( f, ψ) }. Lemma 5. Using the above notations, we have Dom( ) = Dom( ). Proof. We first observe that for any ψ Λ p () and ϕ H 1 (), we have 0 Q(ϕ, ψ) = (ϕ, ψ). Using this result, the proof goes as follows: for any ϕ Dom( ), the functional ψ ( ψ, ϕ) = Q(ψ, ϕ) = ( f, ψ) is a bounded functional. Thus ϕ Dom( ). On the other hand, if ϕ Dom( ), then the functional ψ ( ψ, ϕ) is bounded. By the Riesz representation theorem, there is

13 240B CLASS NOTE 13 a unique f L 2 (Λ()) such that ( ψ, ϕ) = ( f, ψ). Q(ϕ, ψ) = ( ψ, ϕ) = ( f, ψ). Exercise 14. Provide the details of the proof of the above lemma. From the above discussion, we have proved that Theorem 7. The Laplace operator has a self-adjoint extension. In general, the self-adjoint extension is not unique. Thus we must have Example 1. In the above definition, that we require y Dom (A)) happens to be crucial. Consider the following example: let be a compact orientable manifold with smooth boundary. Let C () be the space of smooth functions whose supports 0 are within the interior of, and let C () be the space of smooth functions (smooth up to the boundary). We can define two Laplacians: 1 = (, C 0 ()) and 2 = (, C ()). We can prove that C () Dom The Poincaré Lemma. In this section, we prove the following Theorem 8. Let U be the unit ball of the original point of R n. Let ω be a p-form on U with p 1. Then there is a p 1 form η such that ω = dη. Proof. If ω is a 1-form, the proof is straightforward. Let f (x) = ω, l where l is a curve connecting the original point to x. Then d f = ω. In general, we use the math induction. We write ω = ω 1 + dx n ω 2, where ω 1, ω 2 are forms without the factor dx n. Let ω = a i1 i p 1 (x 1,, x n )dx i1 dx ip 1. Define η = a i1 i p 1 (x 1,, x n )dx n dx i1 dx ip 1. Then d(ω dη) = 0 and ω dη has not dx n factor. As a result, the coefficients of ω dη are independent to x n. Thus the theorem follows from the inductive assumption. The following theorem is a nonlinear generalization of the above result:

14 14 ZHIQIN LU Theorem 9. Let U be the unit tube of the original point of R n. Let ω is a skewsymmetric matrix valued 1-form on U. Assume that Then on U, the equation is solvable. dω + ω ω = 0. dg + ωg = 0 Proof. We assume that g(0) = 1. Assume that for r > 0, we can define g(x 1,, x r, 0,, 0) such that dg + ωg = 0 on {x r+1 = x n = 0}. We define g(x 1,, x r+1, 0,, 0) by the following ODE: { g x r+1 + ω r+1 g = 0 g(x 1,, x r, 0,, 0) were defined. To complete the proof, we need to prove that for any k r, we have (3) g x k + ω k g = 0. Taking the derivative with respect to x r+1, we get Using the ODE, we get 2 g + ω k g g + ω k. x r+1 x k x r+1 x r+1 ω r+1 g g ω r+1 + ω k g g + ω k. x k x k x r+1 x r+1 Using the fact that dω + ω ω = 0, we can prove that the expression in (3) satisfies the ODE with the zero initial value, hence must be zero. 7. de Rham Theorem. The differential operator d : Λ p () Λ p+1 () satisfies the relation d 2 = 0. Using this, we can define the de Rham cohomology groups as follows: Definition 5. Let Z k () = {ω Λ k () dω = 0}, B k () = {dη η Λ k 1 ()}. Then B k () Z k (), and the quotient spaces H k DR () = Zk ()/B k (), are called the de Rham cohomology groups. We have the following result

15 Theorem 10 (de Rham). For any k 0, we have 240B CLASS NOTE 15 H k DR () = Hk sing (), where H k () is the singular cohomology groups. sing Proof. There are many fancy proofs of the above result. The proof provided below is the best one (and can be made rigorous). Let S k () be the vector spaces generated by all dimension k submanifolds of with boundaries. The map : S k () S k 1 () sends those submanifolds into their boundaries. Obviously, 2 = 0. That is, the boundary of the boundary of a submanifold has no boundary. We can define the homology groups as for all k 0. We assume that H k () = {V V = 0, V S k()} { W W S k+1 ()} H k sing () = (H k()), which is a combinatorial result and the proof can be found in any book in algebraic topology. There is a natural non-degenerated coupling between Λ k () and S k () by the integration Λ k () S k () R, (ω, V) ω. By the Stokes theorem, the above coupling descends to the map H k DR () H k() R. To complete the proof, we need to prove that the above coupling is nondegenerated. What we need to prove is that, for any p-form ω such that dω = 0 and for any p-dimensional submanifold K without boundary K ω = 0, then there is a p 1 form η such that ω = dη. Let U, V be two open sets and suppose that the theorem is proved for U, V. Let ω = dη 1 on U and ω = dη 2 on V. Then on U V, d(η 1 η 2 ) = 0. We need to prove that for any p 1 dimensional submanifold L in U V, if L (η 1 η 2 ) = 0, then there is a ξ such that η 1 η 2 = dξ. Thus η 1 dξ = η 2 provides the required solution. Exercise 15. Provide the details of the proof. V

16 16 ZHIQIN LU 8. Formal Hodge Theorem. The de Rham Theorem introduced in the last section is among the very fundamental results of compact manifolds. However, in terms of the concrete computation of the cohomology groups, they are not very efficient. The elements of the groups are classes of differential forms. The Hodge Theorem provides the way to pick up the best possible candidates in the classes. It asserts that, among the class of the form ω + dη, there is a unique representative, called the harmonic form, which is better than other representatives of the class in the following sense. We consider the following variational problem for a fixed element [a] of Λ p (): finding an η 0 such that (4) η 0 2 = inf η [a] η, η dv g. X By definition, each η [a] can be represented by η = ξ 0 + dξ 1. If the space of ξ 1 were of finite dimensional, then since η 2 0, we could have found a ξ such that 1 ξ 0 + dξ 1 2 = inf η [a] η 2, and the variational problem (4) would have been solved. Now let s assume that there is a unique p form η 0 that solves problem (4). Let ξ 1 be an arbitrary p 1 form. Then we have the inequality Since ε is arbitrary, we have Then η 0 + εdξ 1 2 η 0 2. (η 0, dξ 1 ) = 0. (5) δη 0 = 0. Then from (5), η 0 is harmonic, i.e., η 0 = 0. Thus the solution of the problem (4) must be harmonic. For any element η in a fixed cohomologyical class [a], we have the following Hodge decomposition η = η 0 + dη 1, where η 0 is the harmonic form, and η 1 is a p 1 form. In general, for arbitrary p form η, we have the following Hodge decomposition: η = η 0 + dξ 1 + δξ 2, where η 0, ξ 1, and ξ 2 are smooth p, p 1, and p + 1 forms, respectively, and η 0 is harmonic.

17 240B CLASS NOTE The Hodge theorem. Unfortunately, it is far from trivial that the variational problem (4) can be solved. The PDE theory, especially the elliptic regularity theory kicks in here to make the above arguments rigid. Theorem 11 (real Hodge Theorem). Let be a compact orientable Riemannian manifold. Let p 0. Define H p () = {ϕ Λ p ()) ϕ = 0}. Then (1) dim H p () < + ; (2) Let η be an arbitrary smooth p-form. Then we have η = η 0 + dη 1 + δη 2, where η 0 is a harmonic p-form; η 1 and η 2 are (p 1) and (p + 1) forms, respectively, (3) H p DR () = H p () for p Proof of the Hodge Theorem. The proof of the (real) Hodge theorem heavily depends on the analysis of the Laplacian. The three basic PDE tools we are going to use are the Sobolev Lemma, the Rellich Lemma, and the Gårding inequality. For the first two lemmas, we need to introduce the Sobolev s-norms. The proof of the Gårding inequality depends on the Weitzenböck formula. Let be a compact Riemannian manifold. X = U α be a finite cover such that each (U α, {x 1 α,, x n α}) is a real local coordinate system. Let {ρ α } be the partition of unity subordinating to the cover {U α }. Let S be a smooth function. For any nonnegative integer s, define the Sobolev norm of S to be S 2 s = D K α(ρ α S) 2 dv, α K s where K = (k 1,, k n ) is the multiple index; K = k i ; and k 1+ +k n D K αs = (x 1 S. α) k 1 (x n α ) k n Obviously, the definition of the Sobolev norms depend on the choice of the cover {U α } and the partition of unity {ρ α }. Exercise 16. prove that the Sobolev norms s, defined by different covers and partitions of unity, are equivalent. In particular, the norm 0 is equivalent to the L 2 inner product in (2). There is a way to generalize the notation of Sobolev norms s from integers s to any nonnegative numbers s, using pseudo-differential operators, or using the elementary definition as follows:

18 18 ZHIQIN LU We first define the Sobolev s-norms for ρ α S on each U α. Since U α is a coordinate patch, we can assume, without loss of generality, that U α is a part of the torus T = T n = R n /Z n. Thus the section ρ α S can be extended as a smooth C r -valued function on T. Let ρ α S = ξ Z n S ξ e i ξ,x be the Fourier expansion of S α. We define ρ α S 2 s = (1 + ξ 2 ) s (ρ α S) ξ 2. ξ Z n The Sobolev s-norm of S is defined as S 2 s = ρ α S 2 s. α Define H p s () to be the completion of the smooth p forms under the norm s. As before, it is independent of the choice of the cover and the partition of unity. Theorem 12 (Sobolev Lemma). Using the above notations, we have s H p s () = Λ p (). Proof. The theorem being local, we assume that S is a C r -valued smooth function of T. By definition, for any s > 0, we have ξ Z n (1 + ξ 2 ) s S ξ 2 C <. Let K = (k 1,, k n ) be a multiple index. We consider the Fourier expansion of D K S (i) K ξ k 1 1 ξk n n S ξ e i ξ,x. By definition, if S s H p s (), then ξ Z n ξ k1 1 ξk n n 2 S ξ 2 < + for any K. Thus D K S is well-defined using the Fourier expansion and thus S Λ p (). In order to prove the Rellich Lemma, we give the following definition of compact operators. Definition 6. Let B 1, B 2 be two Banach spaces. Let A : B 1 B 2 be a linear operator from B 1 to B 2. A is a compact operator, if the image of the unit ball of B 1 under T is sequential compact in B 2. In other word, if {x i } is a bounded sequence of B 1, then a subsequence of {Ax i } converges to some point of B 2.

19 Theorem 13 (Rellich Lemma). For s > r, the inclusion is compact. 240B CLASS NOTE 19 H p s () H p r () Proof. Similar to the previous theorem, we can work on the torus T. Assume that {u k } is a bounded sequence of H p s (T). We wish to prove that a subsequence of {u k } will converge in the r norm. Let {(u k ) ξ } be the Fourier coefficients of u k for k 1. Then (6) (1 + ξ 2 ) s (u k ) ξ 2 C ξ Z n for some constant C independent of k. For any fixed ξ, a subsequence of (1 + ξ 2 ) s/2 (u k ) ξ converges. By the standard diagonalization, we can find a subsequence of {u k }, which we still denote as {u k } for abusing of notations, such that for each ξ, (1 + ξ 2 ) s/2 (u k ) ξ is convergent. We prove that {u k } is a Cauchy sequence under the normal r. For any ε > 0, we choose R large so that (7) 1 (1 + R 2 < ε. ) s r We consider all ξ with ξ R. There are only finitely many such ξ s. By the assumption on the sequence {u k }, there is an N > 0, such that (1 + ξ 2 ) r (u k ) ξ (u l ) ξ 2 < ε ξ R for k, l > N. Thus we have u k u l 2 r ε + Using (6) and (7), we have and the theorem is proved. (1 + ξ 2 ) r (u k ) ξ (u l ) ξ 2. ξ >R u k u l 2 s < (1 + 2C)ε, Exercise 17. Let be a compact manifold. Define two Banach spaces of. Let C 0 () be the space of continuous functions with the maximum norm, and let C 1 () be the space of C 1 () functions with the norm max( f + f ). Prove the Arzelá-Ascoli Theorem: the inclusion map is a compact operator. ι : C 1 () C 0 ()

20 20 ZHIQIN LU The standard method in proving the above theorem, like in that of the Rellich Lemma, is to use the diagonal subsequence method, which is used everywhere and is a kind of boring. In the following, we give a different proof, using the Tychonoff Theorem. We consider F(), the space of all bounded functions with bound 1 on. By the identification F() = [ 1, 1], x and by the Tychonoff Theorem, we know that F() is compact under the product topology. On the other hand, on C 0 (), we can define the so-called compact-open topology, whose subbase composed of the set of the form C(K, U) = { f F(K) U}, where K is a compact subset of and U is an open set of R. We can prove that when we restrict to the space C 1 (), the product topology is the same as the compact-open topology. On the other hand, on C 0 (), compact-open topology defines the same topology as in the Banach space C 0 (), and the theorem is proved. Exercise 18. Provide the details. Next we shall introduce the Gårding inequality. Let and let ϕ, ψ Γ(X, Λ p ()), D(ϕ, ψ) = (ϕ, ψ) + (dϕ, dψ) + (δϕ, δψ) = (ϕ, (1 + )ψ). Then D(ϕ, ψ) defines an inner product on Λ p (). Apparently we have the inequality D(ϕ, ϕ) C ϕ 2 1 for some constant C. The Gårding inequality states that the inverse inequality is also true so that the norms D(ϕ, ϕ) and ϕ 1 are equivalent. Theorem 14 (Gårding Inequality). For ϕ Λ p (), we have (8) ϕ 2 1 CD(ϕ, ϕ), for some constant C > 0. Proof. The proof of the inequality depends on the Weitzenböck formula, Theorem 6. Let ϕ Λ p (). Then (9) ϕ = ϕ + E(ϕ), where E is a zero-th differential operator. Using integration by parts, we get ( ϕ, ϕ) ϕ 2 C(ϕ, ϕ).

21 It follows that D(ϕ, ϕ) (ϕ, ϕ) ( ϕ, ϕ) 2C 2C 240B CLASS NOTE 21 ϕ (ϕ, ϕ), 2 and the inequality is proved. Proof of the Hodge Theorem. For ϕ Λ p (), we define a linear functional l(ψ) = (ϕ, ψ) for ψ Λ p (). With respect to the inner product D(ϕ, ψ), l is bounded: l(ψ) ϕ 0 D(ψ, ψ). Thus by the Riesz representation theorem, there is an η such that (1) (ϕ, ψ) = (η, (1 + )ψ) for any ψ Γ(X, A p,q (E)); (2) η 2 CD(η, η) < +. (Gårding inequality) 1 Let A : Λ p () Λ p () be the linear operator defined by sending ϕ Λ p () to the unique η defined above. Then since η is actually in H p (), the operator A is a compact operator by the Rellich Lemma. On the other hand, it is not hard to see that A is a self-adjoint operator 1. According to the spectral theorem for compact, self-adjoint operators, there is a Hilbert-space decomposition Λ p () = m E(ρ m ), where ρ m are the eigenvalues of A and E(ρ m ) are the finite-dimensional eigenspaces corresponding to the eigenvalue ρ m. Since A is one-to-one, all ρ m 0. We claim that H p () = E(1), where H p () is the space of harmonic forms. To see this, we assume that ϕ H p (), and Aϕ = ϕ. By the Weitzenböck formula, is an elliptic operator. Using the Sobolev Lemma and Schauder estimate, we can prove that ϕ is smooth. Thus (1 + )ϕ = ϕ and ϕ is harmonic. Since E(1) is of finite dimensional, we have dim H p () < +. This proves the first assertion of the theorem. To prove the second assertion of the theorem, we note that the biggest possible eigenvalue of A is 1. By the property of compact operators, there is a gap between the eigenvalue 1 and the rest of the eigenvalues. Thus if ϕ H p () and if ϕ is smooth, we have ϕ 0 ε ϕ 0 1 Since A is a bounded operator, in order to prove that A is self-adjoint, we just need to verify the equality (Aϕ, ψ) = (ϕ, Aψ) for smooth ϕ and ψ, which follows easily from integration by parts.

22 22 ZHIQIN LU for some ε > 0. Thus we can define the inverse operator G on H p (), called the Green s operator, such that G = 1 H p (). If we compare G with the operator A (they have the same eigenspaces), we shall see that G is also a compact operator because G C A for some constant C. Furthermore, by the elliptic regularity, G maps smooth p forms to smooth p forms. Using this fact, the Hodge decomposition is given by ϕ = η 0 + dδgϕ + δdgϕ. This completes the proof of the Hodge Theorem. 11. ore about elliptic regularity. Let be a compact Riemnnian manifold and let ϕ H 1 () be a p-form. We say ϕ is a weak solution of the equation for ψ a C p-form, if ( + I)ϕ = ψ (1) ϕ H 1 (); (2) for any η a smooth p-form, we have (ϕ, ( + I)η) = (ψ, η). In this section, we prove the following result. Theorem 15. Let ϕ be a weak solution of the above equation, then ϕ must be smooth. Let ρ be a smooth function. Then by the Weizenböck formula, we have (10) ( + I)(ρη) = ρ( + I)η + J(η), where J is some first order differential operator (depending on ρ) on the space of p-forms. Thus we have (11) (ρϕ, ( + I)ξ) = (ϕ, ρξ) (ϕ, J(ξ)). Now we specify our choice of ρ. Let x be a fixed point. We choose a local coordinate system (U, (x 1,, x n )). We choose a smooth function ρ such that the support of ρ is within U. Without loss of generality, we may assume that ρϕ, η are p-forms of the torus T n. Let (, ) 0 be the inner product on Λ p (T n ) induced by the flat Riemannian metric and let 0 be the corresponding norm. Then there is a constant C > 0 such that C 1 0 C 0. That is, the norm is equivalent to the L 2 norm induced by the Riemannian metric of. Let ρϕ ξ Z n a ξ e i x,ξ

23 240B CLASS NOTE 23 be the Fourier expansion of ρϕ, where a ξ are actually p-forms. Let R, s be large real numbers and let η R = a ξ e i x,ξ (1 + ξ 2 ) s Define ξ <R 0 = We use the following observation n 2 x 2 j=1 j 0 e i x,ξ = ξ 2 e i x,ξ. Then we have (12) A ξ 2 (1 + ξ 2 ) s+2 = (η R, ( 0 + I)η R ) 0 C η R 2 H 1 (). ξ <R By the Gårding inequlity, we have (13) a ξ 2 (1 + ξ 2 ) s+2 C(η R, ( + I)η R ). ξ <R Since ρϕ is in L 2, a ξ 2 is convergent. We assume that a ξ 2 < 1. Then (14) (η R, ( + I)η R ) (ρϕ, ( + I)η R ) + C A ξ 2 (1 + ξ 2 ) s+2. Using the equation (11), we have. ξ <R (ρϕ, ( + I)η R ) (ρϕ, η R ) + (ϕ, J(η R ). Using Cauchy inequality, we have (15) (ρϕ, ( + I)η R ) C a ξ 2 (1 + ξ 2 ) s+2. Using (13), (14), we have ξ <R a ξ 2 (1 + ξ 2 ) s+2 C(ϕ, ( + I)η k ) ξ <R + 2 a ξ 2 ξ >R a ξ 2 (1 + ξ 2 ) s+2 In summary, we proved that a ξ 2 (1 + ξ 2 ) s+2 C a ξ 2 (1 + ξ 2 ) s+2, ξ <R ξ <R

24 24 ZHIQIN LU and hence a ξ 2 (1 + ξ 2 ) s+2 C. ξ <R Thus ρϕ is smooth and the theorem is proved. Exercise 19. Provide the details of the above proof. In particular, (1) Prove (10); (2) Prove (14) in details: why there is a constant C such that ( + I)η R C ( 0 + I)η R 0? (3) (tricky) Why there is a constant C such that (ϕ, J(η R )) C a ξ 2 (1 + ξ 2 ) s+2? ξ <R (4) Using the Sobolev lemma and the Rellich lemma, prove the following: if for all s > 0 ξ Z n a ξ 2 (1 + ξ 2 ) s are convergent, then ρϕ is smooth. 12. Semigroups and their generators. In this section, we define abstract semigroup and prove some of the basic properties. Definition 7. A one-parameter semigroup 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, t <, then (3) The map T s T t = T s+t. t, f T t f from [0, ) B to B is jointly continuous. The (infinitesimal) generator Z of a one-parmeter semigroup T t is defined by Z f = 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

25 240B CLASS NOTE 25 Lemma 6. 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 T t Z f = ZT t f for all f Dom(Z) and t 0. Proof. If f B, we define f t = t 0 T x f dx. 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 conditional convergence. Thus in order to prove the convergence of the Riemann sums, we only need to verify that t T x f dx 0 is convergent. But this follows easily from the joint continuity in the definition of the semigroup: 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 0 + h 1 T x f dx h 1 T x f dx h 0 t+h h = lim h 0 + h 1 T x f dx h 1 T x f dx = 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 Lemma 7. The generator Z is a closed operator. Proof. We first observe that T t f f = t 0 T x Z f dx for f Dom(Z). To see this, we consider the function r(t) = T t f f t 0 T xz f dx. Obviously we have r(0) = 0 and r (t) 0. Thus r(t) 0. 0

26 26 ZHIQIN LU Using the above formula, we have T t f f = lim n (T t f n f n ) = lim n t By the Lebegue theorem, the above limit is equal to t T x gdx. Thus we have and therefore, f Dom(Z), Z f = g. 0 lim t 0 t 1 (T t f f ) = g, + 0 T x Z f n dx. Remark 1. If Z is not a bounded operator on Dom(Z), it is not possible to extend Z to the whole banach space B because otherwise since Z is closed, Z has to be bounded by the closed graph theorem. 13. Heat kernel. The fact that the Laplace operator is only a linear operator on smooth functions is not enough for most applications in geometry and analysis. We shall extend the operator to Banach spaces. L p () are the good Banach spaces in our mind. In this section, we construct the heat kernel, hence the semi-group, and we shall define the Laplace operator as the infinitesimal generator of the semi-group. For the sake of simplicity, we shall only consider the Laplace operator on functions. oreover, the sign convention is that on the Euclidean space, = 2. The semigroup is formally defined as T x 2 t = e t. j The main result of this section is the following Theorem 16. Let be a complete Riemannian manifold, then there is a heat kernel such that satisfying H(x, y, t) C ( R + ), (T t f )(x) = H(x, y, t) f (y)dy (1) H(x, y, t) = H(y, x, t), (2) lim H(x, y, t) = δ x(y); t 0 + (3) ( )H = 0; t (4) H(x, y, t) = H(x, z, t s)h(z, y, s)dz for any 0 < s t.

27 240B CLASS NOTE 27 Before proving the theorem, we first formally construct the heat kernel. This formal construction also outline the proof of the theorem. The k-simplex k is the following subset of R k {(t 1,, t k ) 0 t 1 t k 1}. For t > 0, we write t k for the rescaled simplex {(t 1,, t k ) 0 t 1 t k t}. We assume that U(x, y, t) be a function on R + such that lim t 0 + U(x, y, t) = δ x (y) be the the Dirac function. Let du(x, y, t) R(x, y, t) = x U(x, y, t), dt where x means the Laplace operator for the variable x. Note that formally, any function g(x, y, t) defines one-parameter family of operators G t by the formula G t f (x) = g(x, y, t) f (y)dy. We use the R t, U t to denote the corresponding families of operators with respect to the functions R(x, y, t) and U(x, y, t), respectively. For any k 1, define the operator Q k t = t k U t tk R tk t k 1 R t2 t 1 R t1 dt 1 dt k, and Q 0 t = U t. Let R (k) (s) = R s tk 1 R t2 t 1 R t1 dt 1 dt k 1, s k 1 and R (0) (s) = 0. Since the derivative of the integral of the form t 0 a(t s)b(s)ds is equal to t da (t s)b(s)ds + a(0)b(t), 0 dt we have ( ) t Q k t = R(k+1) (t) + R (k) (t). As a result, we have ( ) t ( 1) k Q k t = 0. k=0 For R n, the fundamental solution of the heat equation ( t )u = 0 is 1 r2 e 4t (4πt) n/2,

28 28 ZHIQIN LU where r = d(x, y) is the Euclidean distance of x and y. We wish to find the the following form of the fundamental solution of the heat equation: (16) U(x, y, t) (4πt) n 2 e d 2 (x,y)/4t ϕ i (x, y)t i where d(x, y) is the distance function on the Riemannian manifold. U(x, y, t) should satisfy (1) lim t 0 + U(x, y, t) = δ x (y), where δ x (y) is the Dirac function at x; (2) For any N, lim t 0 + ( t )U(x, y, t) = O(tN ). The function U(x, y, t) is called the paramatrix of the heat kernel. We pick a normal coordinate system (y 1,, y n ), and let r = d(x, y) be the Riemannian distance. We identify a neighborhood of y to a small ball of T y () by the exponential map. Under this map, the coordinates of x can be written as (x 1,, x n ). On the other hand, let (θ 1,, θ n 1 ) be a coordinate system on S n 1, then (r, θ 1,, θ n 1 ) gives a coordinate system at y also, and this coordinate system is called the polar coordinates. Exercise 20. Let ds 2 be the Riemannian metric. Then we can write That is, prove that r i 0 n 1 ds 2 = dr 2 + r 2 s ij (x)dθ i dθ j. i,j=1 is orthogonal to any θ j. Let ψ(r) be a function of r, and let g = det(s ij ). Then we have ( ) ψ = d2 ψ d log g dψ dr 2 + dr dr, We let (ϕψ) = ϕ ψ + ψ ϕ + 2 dϕ dr ψ = 1 r2 e 4t (4πt) n/2, ϕ = ϕ 0 + ϕ 1 t + + ϕ N t N, dψ dr. where ϕ j = ϕ j (x, y) are smooth functions on, and u N = ψϕ = 1 r2 e 4t (4πt) n/2 N ϕ i t i. Then ( ) ( u N = ϕ ψ ψ ) ( + ψ ϕ ϕ t t t i=0 ) + 2 ϕ dψ r dr

29 240B CLASS NOTE 29 Since ψ ψ t = d log g dr dψ dr = r 2t ψ. dψ dr, we have ( t ) u N = ψ t N [ ( ϕ k 1 k + r d log ) g ϕ k r dϕ ] k t k. 2 dr dr k=0 Thus in order to find the paramatrix, we set r dϕ ( k dr + k + r d log ) g ϕ k = ϕ k 1 2 dr for k = 0,, N, where we let ϕ 1 = 0. The solutions of the above ODEs are (17) ϕ 0 (x, y) = g 1 4 (x); ϕ k (x, y) = g 1 4 (x)r(x, y) k Thus we have ( r(x,y) u N = ψ t t ( ϕ N)t N. From the above, we proved that 0 ) ( ) ( ) 1 r k 1 rx rx 4 ( ϕ k 1 ) g dr. r(x, y) r(x, y) Lemma 8. There is a unique formal solution U(x, y, t) of the heat equation ( ) t U(x, y, t) = 0 of the form (16) such that ϕ i (x, y) are defined in (17). Let η be a smooth function such that η = 1 for t < 1 and η = 0 for t > 2. Let ( ) 2r(x, y) p(x, y) = η δ be the cut-off function, where δ is the injectivity radius 2. Then for any N, we consider the function u N (x, y, t) = p(x, y)u N (x, y, t). We shall prove that Lemma 9. For any N sufficiently large, we have (1) lim t 0 + u N (x, y, t) = δ x (y); (2) The kernel R N (x, y, t) = ( y t) un (x, y, t) satisfies the estimate R N (x, y, t) C l C(l)t N l/ For the sake of simplicity, we assume the existence of global injectivity radius.

30 30 ZHIQIN LU Using the above estimate, we know that for N 0, we have It follows that Since R N (x, y, t) C l Ct α. R (k) (s) C l Ck α! (α + k)!. k C k α! (α + k)! < +, our formal construction is convergent to the heat kernel. Theorem 17. The p on L p () is well defined as the infinitesimal generator of the heat semi-group. If we assume that has been extended as a self-adjoint operator, then we have Theorem 18. Using the above notations, we have H(x, y, t) = U(x, y, t) t 0 e (t s) ( t x ) U(x, y, s)ds. 14. Variational characterization of spectrum. Let be the Laplace operator on functions of a manifold. As in the last section, we make the convention that the operator is a negative operator. Let λ be a complex number, (λi ) 1 is called the resolvent of. If for some λ, (λi ) 1 doesn t exist, or it does exist but is an unbounded operator, then we call λ spectrum point of. We use σ( ) or Spec ( ) to denote the set. Theorem 19. Using the above notations, we have Spec ( ). Proof. This is a general fact about the spectrum of any linear operator on a Banach space. We assume that is densely defined in a Banach space B. Assume that Spec ( ) =. Let f B and let l be abounded linear functional on B, then l((λi ) 1 f ) is a bounded holomorphic function of C, which has to be a constant. ore over, it must be the zero function. Since l and f are arbitrary, we conclude that (λi ) 1 = 0 which is not possible for all λ. This is a contradiction.

31 240B CLASS NOTE 31 When is a compact manifold, the best possible things happen: by the Hodge theorem, the spectrum of is made from eigenvalues. ore precisely, we have such that for each λ j, the space 0 = λ 0 < λ 1 λ 2 λ k +, E j = { f f = λ j f } is not trivial and is of finite dimensional. By the above result, we can prove the following variational (or min-max) principal. Theorem 20. We have f 2 Ω (18) λ k = inf Ω f 2 where ϕ j are the eigenfunctions of λ j. and f Ω = 0, f 0 = Proof. Let { f α } be a sequence such that f α ϕ j = 0, j < k, f α 2 f 2 α λ k. Ω f ϕ j, j < k, If we normalize f α such that f 2 α = 1, then the sequence { f α } is bounded in H 1 (). Therefore, there is an f H 1 () such that f α f in the weak sense. In particular, for any ϕ smooth, we have We have f ϕ f α ϕ f ϕ. f α ϕ f α L 2 ϕ L 2 λ k ϕ L 2. Since smooth functions are dense in H 1 (), we have f 2 lim inf k f α 2, which is known as the Fatou Lemma. On the other hand, by the Rellich Lemma, f α f strongly in L 2 (). Thus we have f 2 = 1,

32 32 ZHIQIN LU hence f 2 λ k = f 2 Let ϕ be a smooth function such that ϕϕ j = 0 Then for any ε, we must have f + ε ϕ 2 f 2 ( f + εϕ)2 Since this is true for any ε, we must have f = λ k f f 2 in the weak sense. By the elliptic regularity, f has to be smooth. The following result was used in my recent paper with J. Rowlett. Lemma 10. For k 1, let ξ 0,, ξ k 1 be a nontrivial orthogonal set with respect to the weighted L 2 measure; that is for i j and ξ i 0. Then we have k λ j j=2 ξ i ξ j ϕ 2 1 = 0 Ω k 1 j=0 Ω ξ j 2 ϕ 2 1 Ω ξ j 2 ϕ 2 1 Proof. Clearly, we may assume that all ξ j have unit L 2 norm with respect to the weighted measure. By Theorem 2, for 0 j k 1, let ξ j = a ji ψ i i=1 be the Fourier expansion of ξ j with respect to {ψ i } and the weighted measure. Then we have k 1 k 1 (19) ξ j 2 ϕ 2 1 = λ i a 2 ji. j=0 By the normalization of ξ j, we have j=1 a 2 ij Ω j=0 i=1 = 1, i = 0, 1,..., k 1..

33 Since ξ i are orthogonal, we may write where Then Ω 240B CLASS NOTE 33 k 1 ψ i = b ij ξ j + ψ i, j=0 ψ i ψ i ϕ2 1 = 0, b ij = Ω k 1 ψ 2 i ϕ2 1 = 1 = b 2 ij + ψ i 2 j=0 for all i 1. Thus we have k k 1 λ i (1 a 2 ji ) λ k i=2 k 1 = λ k Thus we have j=0 j=0 i=k+1 ψ i ξ j ϕ 2 1 = a ji. Ω k 1 k 1 = 1 b 2 ij = a 2 ji, j=0 k k 1 k 1 (1 a 2 ji ) = λ k (1 i=2 j=0 k 1 a 2 ji + λ k a 2 j1 λ k k λ i i=2 j=0 k 1 j=0 k i=2 k 1 k 1 λ i a 2 ji + j=0 i=k+1 j=0 i=k+1 j=0 λ i a 2 ji. λ i a 2 ji, j=0 k a 2 ji ) λ k which by (19) completes the proof of the lemma. Unlike in the case of compact manifold, in general, a complete noncompact 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 + λ f = 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. We will not study the L 2 -eigenvalues in this section. Let be the Laplace operator on a complete non-compact manifold. naturally extends to a self-adjoint densely defined operator on L 2 (), which we still denote as for the sake of simplicity. The pure point spectrum of are those λ R such that (1) there exists an L 2 function f 0 such that f + λ f = 0. (2) the multiplicity of λ is finite; (3) in a neighborhood of λ, it is the only spectrum point. 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. i=2

34 34 ZHIQIN LU 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. Theorem 21. 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 (A f λ f ε f, f ) < 0. Theorem 22. 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 ε Dom(A) for which (A λi) f a f for all f G ε. Using the above result, we give the following variational characterization of the lower bound of spectrum and the lower bound of essential spectrum. Theorem 23. Using the above notations, define f 2 λ 0 = f, 2 and λ ess = sup K inf f C 0 () inf f C 0 (\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. Proof. We prove the formula for λ ess. The formula for λ 0 is similar. Let λ ess = inf σ ess ( ). If σ ess ( ) =, we define λ ess = +. When λ ess = +, we prove that λ ess is also infinity. Assume not, then there is a constant C, such that for any compact set K, there is a function f C (\K), we have 0 f 2 f 2 < C. By inductively choosing f, we can make sure that the support of these functions are disjoint. Thus the functions span an infinite dimensional space G with ( f C f, f ) < 0, which contradicts to the fact that the essential spectrum is an empty set. The proof of the general case is similar. We fist prove that λ ess λ ess By definition, for ε > 0, (, λ ess + ε) σ ess ( ).

35 240B CLASS NOTE 35 Therefore we can find infinite dimensional space V such that for any f V ( f (λ ess + ε) f, f ) < 0. In what follows, we shall prove that, using the cut-off functions, there exist infinitely many elements in V such that their supports are disjoint. We assume that K is a compact set. Let K be a large ball containing K. Let ρ be the cut-off function such that ρ = 1 on K but ρ = 0 outside K. We claim that for any ε > 0, there is an f V with f 2 = 1 but ρ 2 f 2 < ε. If the above is not true, then there is an ε 0 > 0 such that for any f V, ρ 2 f 2 ε 0, Since the set f V is of infinite dimensional, the set ρ f is of infinite dimensional as well, otherwise the above inequality is not valid. Thus we can find an orthogonal basis ρ 2 f i f j = 0, if i j, while we still keeping f 2 = 1. We consider i (ρ f i ) 2 2 ρ 2 f 2 i + 2 By f i 2 λ ess + ε, we have Thus we have (ρ f i) 2 2C + 2(λ ess + ε) (ρ f i) 2 ε 0 f i 2. for any i. This is a contradiction because on the compact set K, the are only finitely many eigenvalues (counting multiplicity) below a fixed number. Now we can prove our theorem. For all ε > 0. We can find an f with f 2 = 1 but ρ 2 f 2 < ε. Considering ρ 1 = 1 ρ, we have From (ρ 1 f ) 2 = ρ 2 1 f f 2 ρ 1 2 Cε, ρ 1 f ρ 1 f + f 2 ρ 1 2. and 2 ρ 1 f ρ 1 f = 1 2 ρ 2 1 f 2 Cε,

36 36 ZHIQIN LU we get and (ρ 1 f ) 2 λ ess + ε + 3Cε, ρ 2 1 f 2 1 ε. By definition, we get λ ess λ ess + ε + 3Cε. 1 ε and thus λ ess λ ess. The other direction is easier to prove. It is particularly interesting to get the lower bound estimate for the essential spectrum because of the following theorem. Theorem 24 (Variational principal). Suppose λ 0 < λ ess, then λ 0 is en eigenvalue of with finite dimensional eigenspace. That is, there exists an L 2 function f 0, such that which is a very strong result. f = λ 0 f, 15. Poincaré inequality and Sobolev inequality. Let be a compact manifold without boundary (we call such a manifold closed). Then by the Hodge theorem, for any function f such that f = 0, we have f 2 C f 2 for a positive constant C. For manifold with boundary, we have similar versions of Poincaré inequalities with respect to the boundary conditions. The other fundamental inequality is the Sobolev inequality Sobolev inequality. Assume that is a compact manifold with boundary, then there is a constant C such that (20) C ( f ) n 1 n n n 1 f for any smooth functions satisfies the Dirichlet or Neumann boundary conditions. The Soboleve inequality is equivalent to the so-called isoperimetric inequality:

37 240B CLASS NOTE 37 Isoperimetric inequality. Let Ω be a domain of which is relatively compact. Then there is a constant C such that (21) C(Vol (Ω)) n 1 n Vol ( Ω). To prove the equivalence, we first assume the Sobolev inequality (20). We take the function 1, x Ω, d(x, Ω) ε, d(x, Ω) f ε (x) = ε, x Ω, d(x, Ω) ε, 0 otherwise Using the Soboleve inequality on f ε and lettting ε 0, we get the isoperimetric inequality. On the other hand, we have the following co-area formula Theorem 25 (Co-area formula). Let be a compact manifold with boundary. f H 1 (). Then for any nonnegative function g on, we have ( ) g g = dσ. f { f =σ} Exercise 21. Proof the above co-area formula. For the sake of simplicity, we assume that f 0 and we assume that the isoperimetric inequality (21) is valid. By the co-area formula, we have f = Area ( f = σ)dσ. 0 At the same time we have n n n f n 1 = Vol ( f n 1 > λ)dλ = Vol ( f > σ)σ n 1 n dσ. 0 n 1 0 Using the isoperimetric inequality, we have f C Vol ( f > σ) n 1 n dσ. 0 Therefore, we just need to prove that ( Vol ( f > σ) n 1 n dσ C Let 0 F(σ) = Vol ( f > σ), ϕ(t) = ψ(t) = t 0 ( t 0 0 F(σ) n 1 n dσ, ) n 1 Vol ( f > σ)σ n 1 1 n dσ. ) n 1 F(σ)σ n 1 1 n dσ. Then ϕ(0) = ψ(0). It is not hard to see that ϕ (t) n n 1 ψ( ). n n 1 ψ (t). Thus ϕ( )