Definition and basic properties of heat

kernels III, Constructions

Zhiqin Lu,Department of Mathematics,UC Irvine, Irvine CA 92697

April 27, 2010

TheoremLet M be a complete Riemannian manifold. Then there is aheat kernel

H(x, y, t) ∈ C∞(M ×M × R+),

such that

(e∆tf)(x) =

∫M

H(x, y, t)f(y)dy

satisfying

1 H(x, y, t) = H(y, x, t),

2 limt→0+

H(x, y, t) = δx(y);

3 (∆− ∂∂t

)H = 0;

4 H(x, y, t) =∫MH(x, z, t− s)H(z, y, s)dz for any

0 < s ≤ t.

The proof composed of two steps

1 Construct the paramatrix;

2 Using the paramatrix to construct the heat kernel.

We only formally construct the heat kernel.

The k-simplex ∆k is the following subset of Rk

{(t1, · · · , tk) | 0 ≤ t1 ≤ · · · ≤ tk ≤ 1}.

For t > 0, we write t∆k for the rescaled simplex

{(t1, · · · , tk) | 0 ≤ t1 ≤ · · · ≤ tk ≤ t}.

We only formally construct the heat kernel.

The k-simplex ∆k is the following subset of Rk

{(t1, · · · , tk) | 0 ≤ t1 ≤ · · · ≤ tk ≤ 1}.

For t > 0, we write t∆k for the rescaled simplex

{(t1, · · · , tk) | 0 ≤ t1 ≤ · · · ≤ tk ≤ t}.

We assume that U(x, y, t) be a function on M ×M ×R+ suchthat limt→0+ U(x, y, t) = δx(y) be the the Dirac function. Let

R(x, y, t) =dU(x, y, t)

dt−∆xU(x, y, t),

where ∆x means the Laplace operator for the variable x. Notethat formally, any function g(x, y, t) defines one-parameterfamily of operators Gt by the formula

Gtf(x) =

∫M

g(x, y, t)f(y)dy.

We use the Rt, Ut to denote the corresponding families ofoperators with respect to the functions R(x, y, t) andU(x, y, t), respectively. For any k ≥ 1, define the operator

Qkt =

∫t∆k

Ut−tkRtk−tk−1· · ·Rt2−t1Rt1dt1 · · · dtk,

and Q0t = Ut. Let

R(k)(s) =

∫s∆k−1

Rs−tk−1· · ·Rt2−t1Rt1dt1 · · · dtk−1,

and R(0)(s) = 0.

Since the derivative of the integral of the form∫ t0a(t− s)b(s)ds is equal to∫ t

0

da

dt(t− s)b(s)ds+ a(0)b(t),

we have (∂

∂t−∆

)Qkt = R(k+1)(t) +R(k)(t).

As a result, we have(∂

∂t−∆

) ∞∑k=0

(−1)kQkt = 0.

We wish to find the the following form of the fundamentalsolution of the heat equation:

U(x, y, t) ∼ (4πt)−n2 e−d

2(x,y)/4t

{∑i≥0

φi(x, y)ti

}

where d(x, y) is the distance function on the Riemannianmanifold. U(x, y, t) should satisfy

1 limt→0+ U(x, y, t) = δx(y), where δx(y) is the Diracfunction at x;

2 For any N , limt→0+ (∆− ∂∂t

)U(x, y, t) = O(tN).

The function U(x, y, t) is called the paramatrix of the heatkernel.

We pick a normal coordinate system (y1, · · · , yn), and letr = d(x, y) be the Riemannian distance. We identify aneighborhood of y to a small ball of Ty(M) by the exponentialmap. Under this map, the coordinates of x can be written as(x1, · · · , xn). On the other hand, let (θ1, · · · , θn−1) be acoordinate system on Sn−1, then (r, θ1, · · · , θn−1) gives acoordinate system at y also, and this coordinate system iscalled the polar coordinates.

Let ψ(r) be a function of r, and let g = det(sij). Then wehave

∆ψ =d2ψ

dr2+

(d log

√g

dr

)dψ

dr,

∆(φψ) = φ∆ψ + ψ∆φ+ 2dφ

dr

dψ

dr.

We let

ψ =1

(4πt)n/2e−

r2

4t ,

φ = φ0 + φ1t+ · · ·+ φN tN ,

where φj = φj(x, y) are smooth functions on M ×M , and

uN = ψφ =1

(4πt)n/2e−

r2

4t

N∑i=0

φiti.

Then(∆− ∂

∂t

)uN = φ

(∆ψ − ∂ψ

∂t

)+ ψ

(∆φ− ∂φ

∂t

)+ 2

∂φ

∂r

dψ

dr

Since

∆ψ − ∂ψ

∂t=d log

√g

dr

dψ

dr,

dψ

dr= − r

2tψ.

we have(∆− ∂

∂t

)uN =

ψ

t

N∑k=0

[∆φk−1 −

(k +

r

2

d log√g

dr

)φk − r

dφkdr

]tk.

Thus in order to find the paramatrix, we set

rdφkdr

+

(k +

r

2

d log√g

dr

)φk = ∆φk−1

for k = 0, · · · , N , where we let φ−1 = 0. The solutions of theabove ODEs are

φ0(x, y) = g−14 (x);

φk(x, y) = g−14 (x)r(x, y)−k

×∫ r(x,y)

0

rk−1(∆φk−1)

(rx

r(x, y)

)g

(rx

r(x, y)

) 14

dr.

Thus we have (∆− ∂

∂t

)uN =

ψ

t(∆φN)tN .

From the above, we proved that

LemmaThere is a unique formal solution U(x, y, t) of the heatequation (

∆− ∂

∂t

)U(x, y, t) = 0

of the form

uN = ψφ =1

(4πt)n/2e−

r2

4t

N∑i=0

φiti.

Let η be a smooth function such that η = 1 for t < 1 andη = 0 for t > 2. Let

p(x, y) = η

(2r(x, y)

δ

)be the cut-off function, where δ is the injectivity radius. Thenfor any N , we consider the functionuN(x, y, t) = p(x, y)uN(x, y, t). We shall prove that

LemmaFor any N sufficiently large, we have

1 limt→0+ uN(x, y, t) = δx(y);

2 The kernel RN(x, y, t) =(∆y − ∂

∂t

)uN(x, y, t) satisfies

the estimate

||RN(x, y, t)||Cl ≤ C(l)tN−l/2−1.

Using the above estimate, we know that for N � 0, we have

||RN(x, y, t)||Cl ≤ Ctα.

It follows that

||R(k)(s)||Cl ≤ Ckα!

(α + k)!.

Since ∑k

Ckα!

(α + k)!< +∞,

our formal construction is convergent to the heat kernel.

TheoremThe ∆p on Lp(M) is well defined as the infinitesimalgenerator of the heat semi-group.

A short-cut

The following equation is true

H(x, y, t) = U(x, y, t)−∫ t

0

e∆x(t−s)(∂

∂s−∆x

)U(x, y, s)ds

Two problems

1 The self-adjoint extension of ∆ has to be assumed first;

2 The convergence of U(x, y, t).

A short-cut

The following equation is true

H(x, y, t) = U(x, y, t)−∫ t

0

e∆x(t−s)(∂

∂s−∆x

)U(x, y, s)ds

Two problems

1 The self-adjoint extension of ∆ has to be assumed first;

2 The convergence of U(x, y, t).

Let u be a positive harmonic function on M .

Defineφ = |∇ log u|.Then we have

φi =1

u|∇u|ujuji −

|∇u|u2

ui

∆φ = φii =1

u|∇u|∑

u2ji +

1

u|∇u|ujujii

− 2

u2|∇u|uiujuij −

1

u|∇u|3∑i

(∑j

ujuji

)2

+ 2φ3

Let u be a positive harmonic function on M .Defineφ = |∇ log u|.

Then we have

φi =1

u|∇u|ujuji −

|∇u|u2

ui

∆φ = φii =1

u|∇u|∑

u2ji +

1

u|∇u|ujujii

− 2

u2|∇u|uiujuij −

1

u|∇u|3∑i

(∑j

ujuji

)2

+ 2φ3

Let u be a positive harmonic function on M .Defineφ = |∇ log u|.Then we have

φi =1

u|∇u|ujuji −

|∇u|u2

ui

∆φ = φii =1

u|∇u|∑

u2ji +

1

u|∇u|ujujii

− 2

u2|∇u|uiujuij −

1

u|∇u|3∑i

(∑j

ujuji

)2

+ 2φ3

We assume at a point uj = 0 for j > 1. Then we have

φ1 =1

u|∇u|u1u11 −

|∇u|u2

u1

∆φ =1

u|∇u|∑

u2ji +

1

u|∇u|ujujii

− 2u1

u2u11 −

1

u|∇u|∑i

u21i + 2φ3

We have

1

u|∇u|ujujii ≥

1

u|∇u|uj(∆u)j −Kφ

and

1

u|∇u|∑

u2ji −

1

u|∇u|∑i

u21i

≥ 1

u|∇u|∑j≥2

u2jj ≥

1

(n− 1)u|∇u|u2

11

We have

1

u|∇u|ujujii ≥

1

u|∇u|uj(∆u)j −Kφ

and

1

u|∇u|∑

u2ji −

1

u|∇u|∑i

u21i

≥ 1

u|∇u|∑j≥2

u2jj ≥

1

(n− 1)u|∇u|u2

11

Thus we have

∆φ ≥ 2φ3 − 2u1

u2u11 +

1

(n− 1)u|∇u|u2

11 −Kφ

We have

φ3 − |∇u|u2

u11 = −∇φ∇uu

Thus we have

∆φ ≥ A∇φ∇uu

+ (2 + A)φ3 − (2 + A)u1

u2u11 +

1

(n− 1)u|∇u|u2

11 −Kφ

If we choose A close to 2 enough, we can find an ε > 0 suchthat

∆φ ≥ A∇φ∇uu

+1

2εφ3 −Kφ.

Thus we have

∆φ ≥ 2φ3 − 2u1

u2u11 +

1

(n− 1)u|∇u|u2

11 −Kφ

We have

φ3 − |∇u|u2

u11 = −∇φ∇uu

Thus we have

∆φ ≥ A∇φ∇uu

+ (2 + A)φ3 − (2 + A)u1

u2u11 +

1

(n− 1)u|∇u|u2

11 −Kφ

If we choose A close to 2 enough, we can find an ε > 0 suchthat

∆φ ≥ A∇φ∇uu

+1

2εφ3 −Kφ.

Thus we have

∆φ ≥ 2φ3 − 2u1

u2u11 +

1

(n− 1)u|∇u|u2

11 −Kφ

We have

φ3 − |∇u|u2

u11 = −∇φ∇uu

Thus we have

∆φ ≥ A∇φ∇uu

+ (2 + A)φ3 − (2 + A)u1

u2u11 +

1

(n− 1)u|∇u|u2

11 −Kφ

If we choose A close to 2 enough, we can find an ε > 0 suchthat

∆φ ≥ A∇φ∇uu

+1

2εφ3 −Kφ.

The Bakry-Emery geometry was introduced to study thediffusion processes. For a Riemannian manifold (M, g) and asmooth function φ on M , the Bakry-Emery manifold is a triple(M, g, φ), where the measure on M is the weighted measuree−φdVg. The Bakry-Emery Ricci curvature is defined to be

Ric∞ = Ric + Hess(φ),

and the Bakry-Emery Laplacian is

∆φ = ∆−∇φ · ∇.

The operator can be extended as a self-adjoint operator withrespect to the weighted measure e−φdVg.

The B-E Laplacian

We have ∫∆φfge

−φ = −∫∇f∇g · e−φ

Therefore, it can be extended as a self-adjoint ellipticoperator. Hodge theorem is also valid for this new operator.

The B-E Laplacian

We have ∫∆φfge

−φ = −∫∇f∇g · e−φ

Therefore, it can be extended as a self-adjoint ellipticoperator.

Hodge theorem is also valid for this new operator.

The B-E Laplacian

We have ∫∆φfge

−φ = −∫∇f∇g · e−φ

Therefore, it can be extended as a self-adjoint ellipticoperator. Hodge theorem is also valid for this new operator.

Theorem (Brighton)Let f be a bounded function and let u be a f -harmonicfunction.

That is∆u−∇f∇u = 0

Then if M is a Ricci nonnegative manifold and u is positive, umust be a constant.

Theorem (Brighton)Let f be a bounded function and let u be a f -harmonicfunction.That is

∆u−∇f∇u = 0

Then if M is a Ricci nonnegative manifold and u is positive, umust be a constant.

Define a function U(x, y) on M × R+.

U(x, y) = u(x)− 1

2y2∇f∇u

Then∆U = 0

at (x, 0).Let Φ = |∇ logU |. Then we have

∆Φ ≥ AΦ3 + C∇Φ∇UU

for A > 0 and ∆ the Laplacian on the n+ 1 dimensionalmanifold M × R+.

Define a function U(x, y) on M × R+.

U(x, y) = u(x)− 1

2y2∇f∇u

Then∆U = 0

at (x, 0).Let Φ = |∇ logU |. Then we have

∆Φ ≥ AΦ3 + C∇Φ∇UU

for A > 0 and ∆ the Laplacian on the n+ 1 dimensionalmanifold M × R+.

Define a function U(x, y) on M × R+.

U(x, y) = u(x)− 1

2y2∇f∇u

Then∆U = 0

at (x, 0).

Let Φ = |∇ logU |. Then we have

∆Φ ≥ AΦ3 + C∇Φ∇UU

for A > 0 and ∆ the Laplacian on the n+ 1 dimensionalmanifold M × R+.

Define a function U(x, y) on M × R+.

U(x, y) = u(x)− 1

2y2∇f∇u

Then∆U = 0

at (x, 0).Let Φ = |∇ logU |. Then we have

∆Φ ≥ AΦ3 + C∇Φ∇UU

for A > 0 and ∆ the Laplacian on the n+ 1 dimensionalmanifold M × R+.