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 ∂  √ ∂  ∆ = √ 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 ∂  √ ∂  ∆ = √ 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   λ1  ..   .  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 ∈ B. Then we have the Fourier expansion

∞ 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 ∞ 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 ∞ 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 ∞ 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 ∞ X 2 2 ∆f = (−k ak cos kx − k bk sin kx) k=1 Then we can write ∞ 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 ∞ 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 ∞ 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 ∞ X 2 2 ∆ = 0 · P0 − (k Pk + k Qk) k=1 2 2 ∞ 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 ∞ ∆ = − λ dE, 0 where E is the so-called spectral measure. In general, we can write Z ∞ ∆ = − λ dE, 0 where E is the so-called spectral measure. 2 2 ∞ 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

∞ 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

∞ 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. ∞ 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

∞ 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

∞ 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

∞ 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 ∈ 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 ∈ 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 ||η||1 = |η| dVM + |∇η| dVM . M M

1 We define the quadratic form Q on H0 (M) by Z Q(ω, η) = (hdω, dηi) dVM M

1 for any ω, η ∈ H0 (M). Then we define

 1 Dom(∆) = φ ∈ H0 (M) | ∀ψ ∈ C∞(M), ∃f ∈ L2(M), s.t. Q(φ, ψ) = −(f, ψ) . Proof. ∞ 1 We first observe that for any ψ ∈ C (M) and φ ∈ H0 (M), we have Q(φ, ψ) = −(φ, ∆ψ). Using this result, the proof goes as follows: for any φ ∈ Dom(∆), the functional ψ 7→ (∆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 7→ (∆ψ, φ) is bounded. By the Riesz representation theorem, there is a unique f ∈ L2(M) such that (∆ψ, φ) = (f, ψ). Thus we must have Q(φ, ψ) = −(∆ψ, φ) = −(f, ψ).

From the above discussion, we have proved that Theorem The Laplace operator has a self-adjoint extension. Definition A one-parameter semigroup of operators on a complex Banach space B is a family Tt of bounded linear operators, where Tt : B → B parameterized by real numbers t ≥ 0 and satisfies the following relations:

1 T0 = 1; 2 If 0 ≤ s, t < ∞, then

TsTt = Ts+t.

3 The map t, f → Ttf from [0, ∞) × B to B is jointly continuous. Even though e−∆t are all bounded operator, the kernel doesn’t exist in general. Definition of operator kernel Let A be an operator on L2(M). If there is a function A(x, y) such that Z Af(x) = A(x, y)f(y)dy

for all functions f, then we call A(x, y) is the kernel of the operator. By the above definition, the kernel of an operator doesn’t exist in general. For example, let B be a Banach space, and let I be the identity map. Then the kernel of I doesn’t exist.

The family of bounded operators

e∆t forms a semi-group. Definition of operator kernel Let A be an operator on L2(M). If there is a function A(x, y) such that Z Af(x) = A(x, y)f(y)dy

for all functions f, then we call A(x, y) is the kernel of the operator. By the above definition, the kernel of an operator doesn’t exist in general. For example, let B be a Banach space, and let I be the identity map. Then the kernel of I doesn’t exist.

The family of bounded operators

e∆t forms a semi-group. Even though e−∆t are all bounded operator, the kernel doesn’t exist in general. By the above definition, the kernel of an operator doesn’t exist in general. For example, let B be a Banach space, and let I be the identity map. Then the kernel of I doesn’t exist.

The family of bounded operators

e∆t forms a semi-group. Even though e−∆t are all bounded operator, the kernel doesn’t exist in general. Definition of operator kernel Let A be an operator on L2(M). If there is a function A(x, y) such that Z Af(x) = A(x, y)f(y)dy for all functions f, then we call A(x, y) is the kernel of the operator. The family of bounded operators

e∆t forms a semi-group. Even though e−∆t are all bounded operator, the kernel doesn’t exist in general. Definition of operator kernel Let A be an operator on L2(M). If there is a function A(x, y) such that Z Af(x) = A(x, y)f(y)dy for all functions f, then we call A(x, y) is the kernel of the operator. By the above definition, the kernel of an operator doesn’t exist in general. For example, let B be a Banach space, and let I be the identity map. Then the kernel of I doesn’t exist. Definition of heat kernel Let M be a complete Riemannian manifold. Then there exists heat kernel H(x, y, t) ∈ C∞(M × M × R+) such that Z (e∆tf)(x) = H(x, y, t)f(y)dy M for any L2 function f. The heat kernel satisfies 1 H(x, y, t) = H(y, x, t)

2 lim H(x, y, t) = δx(y) t→0+ 3 ∂ (∆ − ∂t )H = 0 4 R H(x, y, t) = M H(x, z, t − s)H(x, y, s)dz, t > s > 0 An example

Let M be a compact manifold. Let fj(x) be an orthonormal basis of eigenfunctions. Let

∆fj = −λjfj

Then we an write the heat kernel as a series

∞ X −λj t H(x, y, t) = e fj(x)fj(y) j=0 Consider the equation ( ∂f = ∆f ∂t f(0, x) = φ(x),

where φ is the initial value function.Then Z ∞ X −λj t f(t, x) = H(x, y, t)φ(y)dy = e ajfj(x), M j=1 R where aj = M φ(x)fj(x)dx. λ1t Conclusion: If a1 6= 0, then e f(t, x) → a1f1(x) 6= 0

Let Ω be a bounded domain of Rn with smooth boundary. Then Z ∞ X −λj t f(t, x) = H(x, y, t)φ(y)dy = e ajfj(x), M j=1 R where aj = M φ(x)fj(x)dx. λ1t Conclusion: If a1 6= 0, then e f(t, x) → a1f1(x) 6= 0

Let Ω be a bounded domain of Rn with smooth boundary. Consider the equation ( ∂f = ∆f ∂t f(0, x) = φ(x), where φ is the initial value function. λ1t Conclusion: If a1 6= 0, then e f(t, x) → a1f1(x) 6= 0

Let Ω be a bounded domain of Rn with smooth boundary. Consider the equation ( ∂f = ∆f ∂t f(0, x) = φ(x), where φ is the initial value function.Then Z ∞ X −λj t f(t, x) = H(x, y, t)φ(y)dy = e ajfj(x), M j=1 R where aj = M φ(x)fj(x)dx. Let Ω be a bounded domain of Rn with smooth boundary. Consider the equation ( ∂f = ∆f ∂t f(0, x) = φ(x), where φ is the initial value function.Then Z ∞ X −λj t f(t, x) = H(x, y, t)φ(y)dy = e ajfj(x), M j=1 R where aj = M φ(x)fj(x)dx. λ1t Conclusion: If a1 6= 0, then e f(t, x) → a1f1(x) 6= 0 1 Construct a log-concave function φ(x) with a1 6= 0; 2 Prove that along the flow, the log-concavity is preserved.

A theorem of Brascamp-Lieb

Theorem [Brascamp-Lieb] Let Ω be a bounded convex domain with smooth boundary in n R . Let f1(x) > 0 be the first eigenfunction. Then log f1(x) is a concave function.

Ideas of the proof 2 Prove that along the flow, the log-concavity is preserved.

A theorem of Brascamp-Lieb

Theorem [Brascamp-Lieb] Let Ω be a bounded convex domain with smooth boundary in n R . Let f1(x) > 0 be the first eigenfunction. Then log f1(x) is a concave function.

Ideas of the proof

1 Construct a log-concave function φ(x) with a1 6= 0; A theorem of Brascamp-Lieb

Theorem [Brascamp-Lieb] Let Ω be a bounded convex domain with smooth boundary in n R . Let f1(x) > 0 be the first eigenfunction. Then log f1(x) is a concave function.

Ideas of the proof

1 Construct a log-concave function φ(x) with a1 6= 0; 2 Prove that along the flow, the log-concavity is preserved. 2 2 Let φ = ϕe−C|x| . Then for sufficient large C, φ is log-concave.

Construction of the log-concavity function

1 Since Ω is a convex domain, the local defining function ϕ(x) is log-concave neat the boundary. Construction of the log-concavity function

1 Since Ω is a convex domain, the local defining function ϕ(x) is log-concave neat the boundary. 2 2 Let φ = ϕe−C|x| . Then for sufficient large C, φ is log-concave. Maximum Principle

Let g = log f. Then ∂g = ∆g + |∇g|2. ∂t For any i, we have

∂(−gii) X = ∆(−g ) + 2g g + 2 g2 ∂t ii kii k ki k