Definition and basic properties of heat kernels I, An introduction Zhiqin Lu, Department of Mathematics, UC Irvine, Irvine CA 92697 April 23, 2010 1 What is the heat kernel? 2 Why it is so difficult to understand the heat kernel? 3 The basic properties of the heat kernel. In this lecture, we will answer the following questions: 2 Why it is so difficult to understand the heat kernel? 3 The basic properties of the heat kernel. In this lecture, we will answer the following questions: 1 What is the heat kernel? 3 The basic properties of the heat kernel. In this lecture, we will answer the following questions: 1 What is the heat kernel? 2 Why it is so difficult to understand the heat kernel? In this lecture, we will answer the following questions: 1 What is the heat kernel? 2 Why it is so difficult to understand the heat kernel? 3 The basic properties of the heat kernel. In this lecture, we will answer the following questions: 1 What is the heat kernel? 2 Why it is so difficult to understand the heat kernel? 3 The basic properties of the heat kernel. S-T. Yau, R. Schoen Differential Geometry Academic Press, 2006 N. Berline, E. Getzler, M. Vergne Heat kernels and Dirac operators Springer 1992 E.B. Davies, One-parameter semi-groups Academic Press 1980 E.B. Davies, Heat kernels and spectral theory Cambridge University Press, 1990 The Laplace operator is defined as 1 X @ p @ ∆ = p gij g ; g @xi @xj ij −1 where (g ) = (gij) , g = det(gij). The basic settings: Let M be a Riemannian manifold with the Riemannian metric 2 ds = gijdxi dxj: The basic settings: Let M be a Riemannian manifold with the Riemannian metric 2 ds = gijdxi dxj: The Laplace operator is defined as 1 X @ p @ ∆ = p gij g ; g @xi @xj ij −1 where (g ) = (gij) , g = det(gij). 2 The semi-groupof operators; 3 The existence of heat kernel. At least three questions have to be addressed: 1 The Laplace operator as a densely defined self-adjoint operator; 3 The existence of heat kernel. At least three questions have to be addressed: 1 The Laplace operator as a densely defined self-adjoint operator; 2 The semi-groupof operators; At least three questions have to be addressed: 1 The Laplace operator as a densely defined self-adjoint operator; 2 The semi-groupof operators; 3 The existence of heat kernel. Finite dimensional case Let A be a positive definite matrix. Then A can be diagonalized. That is, up to a similar transformation by an orthogonal matrix, A is similar to the matrix 0 1 λ1 B .. C @ . A λn Let Ei be the eigenspace with respect to the eigenvalue λi and n let Pi : R ! Ei be the orthogonal projection. Then we can write n X A = λiPi i=1 Infinite dimensional case, an example Let B be the space of L2 periodic functions on [−π; π]. Let f 2 B. Then we have the Fourier expansion 1 a0 X f(x) ∼ + (a cos kx + b sin kx): 2 k k k=1 If f is smooth, then the above expansion is convergent to the function. If f is smooth, then 1 X 2 2 ∆f = (−k ak cos kx − k bk sin kx) k=1 Define Pkf = ak cos kx Qkf = bk sin kx Then we can write 1 X 2 2 ∆ = 0 · P0 − (k Pk + k Qk) k=1 The Laplace operator on one dimensional is @2 ∆ = @x2 Define Pkf = ak cos kx Qkf = bk sin kx Then we can write 1 X 2 2 ∆ = 0 · P0 − (k Pk + k Qk) k=1 The Laplace operator on one dimensional is @2 ∆ = @x2 If f is smooth, then 1 X 2 2 ∆f = (−k ak cos kx − k bk sin kx) k=1 Then we can write 1 X 2 2 ∆ = 0 · P0 − (k Pk + k Qk) k=1 The Laplace operator on one dimensional is @2 ∆ = @x2 If f is smooth, then 1 X 2 2 ∆f = (−k ak cos kx − k bk sin kx) k=1 Define Pkf = ak cos kx Qkf = bk sin kx The Laplace operator on one dimensional is @2 ∆ = @x2 If f is smooth, then 1 X 2 2 ∆f = (−k ak cos kx − k bk sin kx) k=1 Define Pkf = ak cos kx Qkf = bk sin kx Then we can write 1 X 2 2 ∆ = 0 · P0 − (k Pk + k Qk) k=1 2 2 1 Let L (M) be the space of L functions. The space C0 (M) is the space of smooth functions on M with compact support. In general, we can write Z 1 ∆ = − λ dE; 0 where E is the so-called spectral measure. In general, we can write Z 1 ∆ = − λ dE; 0 where E is the so-called spectral measure. 2 2 1 Let L (M) be the space of L functions. The space C0 (M) is the space of smooth functions on M with compact support. It is symmetric.That is Z Z ∆f · g = f · ∆g 1 However, C0 (M) is not a Banach space. We would like to pick L2(M), the space of L2 functions. Question Can we extend the Laplacian onto L2(M)? The Laplacian ∆ is defined on the space of smooth functions with compact support. That is Z Z ∆f · g = f · ∆g 1 However, C0 (M) is not a Banach space. We would like to pick L2(M), the space of L2 functions. Question Can we extend the Laplacian onto L2(M)? The Laplacian ∆ is defined on the space of smooth functions with compact support.. It is symmetric. 1 However, C0 (M) is not a Banach space. We would like to pick L2(M), the space of L2 functions. Question Can we extend the Laplacian onto L2(M)? The Laplacian ∆ is defined on the space of smooth functions with compact support.. It is symmetric.That is Z Z ∆f · g = f · ∆g We would like to pick L2(M), the space of L2 functions. Question Can we extend the Laplacian onto L2(M)? The Laplacian ∆ is defined on the space of smooth functions with compact support.. It is symmetric.That is Z Z ∆f · g = f · ∆g 1 However, C0 (M) is not a Banach space. Question Can we extend the Laplacian onto L2(M)? The Laplacian ∆ is defined on the space of smooth functions with compact support.. It is symmetric.That is Z Z ∆f · g = f · ∆g 1 However, C0 (M) is not a Banach space. We would like to pick L2(M), the space of L2 functions. The Laplacian ∆ is defined on the space of smooth functions with compact support.. It is symmetric.That is Z Z ∆f · g = f · ∆g 1 However, C0 (M) is not a Banach space. We would like to pick L2(M), the space of L2 functions. Question Can we extend the Laplacian onto L2(M)? 1 The Laplacian is an unbounded operator; 2 Like most differential operator, it is a closed graph operator; 3 By the Closed Graph Theorem, if ∆ can be extended, then it must be a bounded operator, a contradiction. Answer: No! 2 Like most differential operator, it is a closed graph operator; 3 By the Closed Graph Theorem, if ∆ can be extended, then it must be a bounded operator, a contradiction. Answer: No! 1 The Laplacian is an unbounded operator; 3 By the Closed Graph Theorem, if ∆ can be extended, then it must be a bounded operator, a contradiction. Answer: No! 1 The Laplacian is an unbounded operator; 2 Like most differential operator, it is a closed graph operator; Answer: No! 1 The Laplacian is an unbounded operator; 2 Like most differential operator, it is a closed graph operator; 3 By the Closed Graph Theorem, if ∆ can be extended, then it must be a bounded operator, a contradiction. Self-adjoint densely defined operator Definition 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 7! hx; Ayi is a bounded linear functional, where y 2 Dom(∆); 2 If x is in the domain of A∗, there is a unique vector z in H such that hx; Ayi = hz; yi for any y 2 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. 1 Define H0 (M) be the Sobolev space of the completion of the vector space Λp(M) under the norm sZ sZ 2 2 jjηjj1 = jηj dVM + jrηj dVM : M M 1 We define the quadratic form Q on H0 (M) by Z Q(!; η) = (hd!; dηi) dVM M 1 for any !; η 2 H0 (M). Then we define 1 Dom(∆) = φ 2 H0 (M) j 8 2 C1(M); 9f 2 L2(M); s:t: Q(φ, ) = −(f; ) : Proof. 1 1 We first observe that for any 2 C (M) and φ 2 H0 (M), we have Q(φ, ) = −(φ, ∆ ): Using this result, the proof goes as follows: for any φ 2 Dom(∆), the functional 7! (∆ ; φ) = −Q( ; φ) = (f; ) is a bounded functional. Thus φ 2 Dom(∆∗). On the other hand, if φ 2 Dom(∆∗), then the functional 7! (∆ ; φ) is bounded. By the Riesz representation theorem, there is a unique f 2 L2(M) such that (∆ ; φ) = (f; ).