Page 1

GENERALIZED HAMILTON-JACOBI EQUATION AND HEAT KERNEL

ON STEP TWO NILPOTENT LIE GROUPS

OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

Abstract. We study geometrically invariant formulas for heat kernels of sub-elliptic differ-

ential operators on two step nilpotent Lie groups and for the Grusin operator in R2. We

deduce a general form of the solution to the Hamilton-Jacobi equation and its generalized

form in Rn× Rm. Using our results, we obtain explicit formulas of the heat kernels for these

differential operators.

1. Introduction

Let us start with the Laplace operator on Rn,

∆ =1

2

n

?

j=1

∂2

∂x2

j

.

It is well-known that the heat kernel for ∆ is the Gaussian:

Pt(x,x0) =

1

(2πt)

n

2e−|x−x0|2

2t

.

Given a general second order elliptic operator in n dimensional Euclidean space,

∆X=1

2

n

?

j=1

X2

j+ lower order term,

where the {X1,...,Xn} is a linearly independent set of vector fields, the heat kernel takes the

form

1

(2πt)

Pt(x,x0) =

n

2e−d2(x,x0)

2t

?a0+ a1t + a2t2+ ···?.

Here d(x,x0) stands for the Riemannian distance between x and x0if the metric is induced

by the orthonormal basis {X1,...,Xn}. The aj’s are functions of x and x0. Note that

∂

∂t

?d2

2t

?

+1

2

n

?

j=1

?

Xjd2

2t

?2

= 0,

i.e.,d2

2tis a solution of the Hamilton-Jacobi equation.

2000 Mathematics Subject Classification. 53C17,53C22, 35H20.

Key words and phrases. Sub-Laplacian, Heat operator, H-type groups, action function, volume element.

The first author is partially supported by the NSF grant #0631541.

The second author is partially supported by a Hong Kong RGC competitive earmarked research grant

#600607, a competitive research grant at Georgetown University, and NFR grant #180275/D15.

The third author is supported by NFR grants # 177355/V30, #180275/D15, and ESF Networking Pro-

gramme HCAA.

1

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2OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

Now let us move to subelliptic operators. We first consider the famous example: Heisenberg

sub-Laplacian on H1

∆X=1

2

∂x1

∂y

We shall try for a heat kernel in the form

1

tqe−f

where h =f

tis a solution of the Hamilton-Jacobi equation

∂h

∂t+1

2

∂x1

∂y

In other words,

∂h

∂t+ H(x,∇h) = 0,

where

H =1

2

is the Hamilton function associated with the sub-elliptic operator (1.1) and ξ1, ξ2and η are

dual variable to x1, x2and y respectively. Using the Lagrange-Chapit method, let us look at

the following equation:

(1.1)

?∂

+ 2x2

∂

?2+1

2

?∂

∂x2

− 2x1

∂

∂y

?2.

t···

?∂h

+ 2x2∂h

?2

+1

2

?∂h

∂x2

− 2x1∂h

∂y

?2

= 0.

(1.2)

(1.3)

??ξ1+ 2x2η?2+?ξ2− 2x1η?2?

=1

2

?ζ2

1+ ζ2

2

?

F(x,y,t,h,ξ,η,γ) = γ + H(x,y,ξ,η) = 0.

We shall find the bicharacteristic curves which are solutions to the following Hamilton system:

˙ x1=Fξ1= ξ1+ 2x2η = ζ1,

˙ x2=Fξ2= ξ2− 2x1η = ζ2,

˙ y =Fη= 2˙ x1x2− 2x1˙ x2,

˙t =Fγ= 1,

˙ξ1= − Fx1− ξ1Fh= 2η ˙ x2,

˙ξ2= − Fx2− ξ2Fh= −2η ˙ x1,

˙ η = − Fy− γFh= 0,

˙ γ = − Ft− γFh= 0,

˙h =ξ · ∇ξF + ηFη+ γFγ= ξ · ˙ x + η ˙ y − H

since˙t = 1 and γ = −H. With 0 ≤ s ≤ t, one has

γ(s) =γ = constant,

η(s) =η = constant,

t(s) =s.

Here “constant” means “constant along the bicharacteristic curve”. Furthermore,

H =1

2˙ x2

Another way to see that E is constant along the bicharacteristic, note that

¨ x1=˙ξ1+ 2η ˙ x2= +4η ˙ x2,

¨ x2=˙ξ2− 2η ˙ x1= −4η ˙ x1.

1+1

2˙ x2

2= E = energy.

(1.4)

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HAMILTON-JACOBI EQUATION AND HEAT KERNEL3

Therefore, ¨ x1˙ x1+ ¨ x2˙ x2= 0, and E =constant.

We need to find the classical action integral

S(t) =

?t

0

?ξ · ˙ x + η ˙ y − H?ds.

...

x2+ 16η2˙ x2= 0

Let find ξ and x from the Hamilton system. We obtain

...

x1+ 16η2˙ x1= 0,

from (1.4). Hence

˙ x1(s) = ˙ x1(0)cos(4ηs) +¨ x1(0)

4η

sin(4ηs)

= ˙ x1(0)cos(4ηs) + ˙ x2(0)sin(4ηs)

=ζ1(0)cos(4ηs) + ζ2(0)sin(4ηs)

(1.5)

and

˙ x2(s) =˙ x2(0)cos(4ηs) +¨ x2(0)

4η

sin(4ηs)

=˙ x2(0)cos(4ηs) − ˙ x1(0)sin(4ηs)

= − ζ1(0)sin(4ηs) + ζ2(0)cos(4ηs),

(1.6)

which yields

(1.7)

x1(s) = x1(0) + ζ1(0)sin(4ηs)

4η

+ ζ2(0)1 − cos(4ηs)

4η

and

(1.8)

x2(s) = x2(0) − ζ1(0)1 − cos(4ηs)

4η

+ ζ2(0)sin(4ηs)

4η

.

At s = t one has x1(t) = x1and x2(t) = x2, so

1

2ζ1(0)sin(4ηt) +1

−1

2ζ2(0)?1 − cos(4ηt)?=2η?x1− x1(0)?,

2ζ1(0)?1 − cos(4ηt)?+1

+ζ1(0)cos(2ηt) + ζ2(0)sin(2ηt) =2η?x1− x1(0)?

−ζ1(0)sin(2ηt) + ζ2(0)cos(2ηt) =2η?x2− x2(0)?

Hamilton’s equations give

ξ2(s) = − 2ηx1(s) +?ξ2(0) + 2ηx1(0)?

= − 2ηx1(0) −1

?

and

ξ1(s) = −2ηx2(0) +1

2

2ζ2(0)sin(4ηt) =2η?x2− x2(0)?,

or,

sin(2ηt)

,

sin(2ηt)

.

(1.9)

2ζ1(0)sin(4ηs) −1

ζ1(0)sin(4ηs) − ζ2(0)?1 + cos(4ηs)??

?

2ζ2(0)?1 − cos(4ηs)?+ ζ2(0) + 4ηx1(0)

,

=2ηx1(0) −1

2

ζ1(0)?1 + cos(4ηs)?+ ζ2(0)sin(4ηs)

?

.

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4 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

The above calculations imply

ξ1˙ x1+ ξ2˙ x2= − 2η ˙ x1(s)x2(0) + 2ηx1(0)˙ x2(s) +1

= − 2η?˙ x1(s)x2(0) − x1(0)˙ x2(s)?+?1 + cos(4ηs)?E,

?t

To find E we square and add the two equations in (1.9),

2

?ζ2

1(0) + ζ2

2(0)??1 + cos(4ηs)?

and

0

?ξ · ˙ x + η ˙ y − H?ds = η

?

y − y(0) + 2?x1(0)x2− x1x2(0)?+sin(4ηt)

4η2

E

?

.

E =1

2ζ2

1(0) +1

2ζ2

2(0) = 2η2|x − x0|2

sin2(2ηt).

Hence,

S(t) =

?t

?

0

?ξ · ˙ x + η ˙ y − H?ds

y − y(0) + 2?x1(0)x2− x1x2(0)?+ |x − x0|2cot(2ηt)

We note that x, y, t, x0and η = η(0) are free parameters while y(0) = y(0;x,x0,y,η;t) is not.

Therefore, we need to introduce one more free variable h(0) such that h(t) = h(0) + S(t) is a

solution of the Hamilton-Jacobi equation (1.2).

It reduces to find h(0). To find it we shall substitute S into (1.2). Straightforward compu-

tation shows that

∂h

∂t+ H(x,y,ξ(t),η(t)) = 0

=η

?

.

where

(1.10)

h(t) = η(0)y(0) + S(t),

i.e.,

h(0) = η(0)y(0).

This yields

∂h

∂t+ H

?

x,y,∇xh,∂h

∂y

?

= 0.

We have the following theorem.

Theorem 1.1. We have shown that

h =η(0)y(0) +

=ηy + 2η?x1(0)x2− x1x2(0)?+ η|x − x0|2cot(2ηt)

is a “complete integral” of (1.2) and (1.3), i.e., a solution of (1.2) and (1.3) which depends

on 3 free parameters x1(0), x2(0) and η.

?t

0

?ξ · ˙ x + η ˙ y − H?ds

(1.11)

Before we move further, let us consider a more general situation.

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HAMILTON-JACOBI EQUATION AND HEAT KERNEL5

2. Generalized Hamilton-Jacobi equations

In this section we study the Hamilton-Jacobi equation which is crucial in the construction

of the heat kernel associated with elliptic and sub-elliptic operators. We deduce a general

form of the solution to the Hamilton-Jacobi equation and its generalized form. We consider

an (n + m)-dimensional space Rn× Rm. The coordinates are denoted x = (x1,...,xn) ∈ Rn

and y = (y1,...,ym) ∈ Rmwith dual variables (ξ1,...,ξn) and (η1,...,ηm) respectively. The

roman indices i,j,k,... will vary from 1 to n and the Greek indices α,β,... will vary from 1

to m. As usual, the Hamiltonian function H(x,y,ξ,η) is a homogeneous polynomial of degree

2 in the variables (ξ,η) and has smooth coefficients in (x,y).

We have the following nice generalizaition of a result from [11].

Theorem 2.1. Set

(2.1)

h(t;x,y,ξ,η) =

m

?

α=1

ηα(0)yα(0) + S(t;x,y,ξ,η)

where

xj= xj(s;x,y,ξ,η;t),j = 1,...,n;

yα= yα(s;x,y,ξ,η;t),α = 1,...,m

and

S(t;x,y,ξ,η) =

?t

0

?

ξ(u) · ˙ x(u) + η(u) · ˙ y(u) − H(x(u),y(u),ξ(u),η(u))

?

du.

Then h satisfies the usual Hamilton-Jacobi equation:

∂h

∂t+ H

?

x,y,∇xh,∇yh

?

= 0.

Proof. In order to prove the theorem, we first calculate the partial derivatives of the function

S with respect to all variables explicitly. For j = 1,...,n,

∂S

∂xj(t;x,y,ξ,η)

?t

k=1

n

?

n

?

?t

n

?

It follows that

=

0

?

n

?

∂H

∂ξk

?∂ξk

∂ξk

∂xj

∂xj

dxj

ds

+ ξk

d

ds

∂xk(s;x,y,ξ,η;t)

∂xj

?+

m

?

α=1

?∂ηα

∂xj

dyα

ds

+ ηα

d

ds

∂yα(s;x,··· ;t)

∂xj

?

−

k=1

−

m

?

α=1

∂H

∂ηα

∂ηα

∂xj

−

k=1

∂H

∂xk

∂xk(s;x,y,ξ,η;t)

∂xj

−

m

?

α=1

∂H

∂yα

∂yα(s;x,y,ξ,η;t)

∂xj

?

ds

=

0

d

ds

?

n

?

k=1

ξk∂xk(s;x,y,ξ,η;t)

∂xj

+

m

?

m

?

α=1

ηα∂yα(s;x,y,ξ,η;t)

∂xj

?

ds

=

k=1

ξk(s)∂xk(s;x,y,ξ,η;t)

∂xj

???

s=t

s=0+

α=1

ηα(s)∂yα(s;x,y,ξ,η;t)

∂xj

???

s=t

s=0.

∂S

∂xj(t;x,y,ξ,η) = ξj(t) −

m

?

α=1

ηα(0)∂yα(0;x,y,ξ,η;t)

∂xj

.

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6 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

Similarly, for β = 1,...,m,

∂S

∂yβ(t;x,y,ξ,η) = ηβ(t) −

m

?

α=1

ηα(0)∂yα(0;x,y,ξ,η;t)

∂yβ

.

Moreover,

∂S

∂t(t;···) =

n

?

k=1

ξk(t;···)˙ xk(t;···) +

m

?

α=1

ηα(t;···)˙ yα(t;···) − H?x,y,ξ(t;···),η(t;···)?

s=t

s=0+

α=1

∂t

+

n

?

k=1

ξk(s;···)∂xk(s;···)

∂t

???

m

?

ηα(s;···)∂yα(s;···)

???

s=t

s=0.

Differentiating x1= x1(t;x,y,ξ,η;t) yields

0 =d

dtx1(t;x,y,ξ,η;t) = ˙ x1(t;···) +∂x1(s;x,y,ξ,η;t)

On the other hand, one has

???

ηα(s;···)∂yα(s;···)

∂t

therefore,

∂S

∂t= −H(t;···) −

∂t

???s=t.

ξk(s;···)∂xk(s;···)

∂t

s=t

s=0= −ξk(t;···)˙ xk(t;···),k = 1,...,n,

and

???

s=t

s=0= −ηα(t;···)˙ yα(t;···) − ηα(0;···)∂yα(0;···)

∂t

,α = 1,...,n,

m

?

α=1

ηα(0;···)∂yα(0;···)

∂t

.

It follows that if we set as in the statement of the theorem

h(t;x,y,ξ,η) =

m

?

α=1

ηα(0)yα(0) + S(t;x,y,ξ,η),

then it satisfies

∂h

∂xk

∂h

∂yα

=ξk(t;x,y,ξ,η;t),k = 1,...,n

=ηα(t;x,y,ξ,η;t),α = 1,...,m,

and

∂h

∂t+ H

?

x,y,ξ(t),η(t)

?

= 0⇒

∂h

∂t+ H

?

x,y,∇xh,∇yh

?

= 0.

This completes the proof of the theorem.

?

We note that the derivation that (2.1) satisfies the Hamilton-Jacobi equation was complete

general, not restriction to H

x,y,∇xh,∇yh

that ηα(s) =constant for α = 1,...,m. The action integral S is not a solution of the Hamilton-

Jacobi equation because some of our free parameters are dual variables ηα(0) instead of yα(0).

For the Heisenberg sub-Laplacian or the Grusin operator, η(0) = η cannot be switched to y(0).

As we know, ˙ y = 2(˙ x1x2− x1˙ x2). From (1.5) – (1.8), one has

?

??

being (1.3). In particular we did not assume

˙ y = 2˙ x1x2(0) − x1(0)˙ x2+1

2

?ζ2

1(0) + ζ2

2(0)?1 − cos(4ηs)

2η

?

,

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HAMILTON-JACOBI EQUATION AND HEAT KERNEL7

and

y(s) = 2?x1(s)x2(0) − x1(0)x2(s)?+

At s = t, one has x1(t) = x1, x2(t) = x2and

E

4η2

?4ηs − sin(4ηs)?+ C.

y = 2?x1x2(0) − x1(0)x2

?+

E

4η2

?4ηt − sin(4ηt)?+ C.

Hence, one has

y(s) =y − 2

??x1− x1(s)?x2(0) − x1(0)?x2− x2(s)??

E

4η2

−

?4η(t − s) − (sin(4ηt) − sin(4ηs))?.

y(0) = y + 2?x1(0)x2− x1x2(0)?+ |x − x0|2µ(2ηt),

φ

sin2φ− cot φ.

At s = 0,

where we set

µ(φ) =

To replace η by y(0), one needs to invert µ,

µ(2ηt) =y − y(0) + 2?x1(0)x2− x1x2(0)?

|x − x0|2

.

This is impossible since for most of the values on the right hand side µ−1is a many valued

function [2]. Therefore we must leave η as one of the free parameters which does not permit S

to be a solution of the Hamilton-Jacobi equation.

Before we go further, we present a scaling property of the solution to the Hamiltonian system

dxj

ds

=∂H

∂ξj,

dyα

ds

=∂H

∂ηα,

dξj

ds= −∂H

∂xj,

dηα

ds

= −∂H

∂yα,

s ∈ [0,t] with the boundary conditions

x(0) = x0,

x(t) = x,

y(t) = y,η(0) = η(0).

Lemma 2.1. One has the following scaling property

xj(s;x,x0,y,ξ,η(0);t) =xj

?λs;x,x0,y,ξ,η(0)

?λs;x,x0,y,ξ,η(0)

?λs;x,x0,y,ξ,η(0)

?λs;x,x0,y,ξ,η(0)

λ

;λt?, j = 1,...,n

;λt?, α = 1,...,m

;λt?, j = 1,...,n

;λt?, α = 1,...,m

yα(s;x,x0,y,ξ,η(0);t) =yα

λ

ξj(s;x,x0,y,ξ,η(0);t) =λξj

λ

ηα(s;x,x0,y,ξ,η(0);t) =ληα

λ

(2.2)

for λ > 0, if the two sides of (2.2) stays in the domain of unique solvability of the Hamiltonian

system.

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8 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

Proof. Denote the curve on the right-hand side of (2.2) by {˜ x(s), ˜ y(s),˜ξ(s), ˜ η(s)}. Note that

s ∈ (0,t). Then for j = 1,...,n

∂˜ xj

∂s

=λ˙ xj

?λs;x,x0,y,ξ,η(0)

=λ∂H

∂ξj

=∂H

∂ξj

λ

;λt?

λ

?x1

?˜ x(s), ˜ y(s),˜ξ(s), ˜ η(s)?,

?λs;x,x0,y,ξ,η(0)

;λt?,x2

?λs;x,x0,y,ξ,η(0)

λ

;λt?,...?

since

calculations and homogeneity of degree 2 of∂H

∂H

∂ξj, j = 1...,n, are homogeneous of degree 1 in ξ1,...,ξn and η1,...,ηm. Similar

∂xjand

∂H

∂yαin ξ1,...,ξnand η1,...,ηmyield

∂˜ yα

∂s

=∂H

∂ηα,

∂˜ξj

∂s= −∂H

∂xj,

∂˜ ηα

∂s

= −∂H

∂yα.

Clearly,

˜ xj(0) = xj

?0;x,x0,y,ξ,η(0)

λ

;λt?= xj(0),

˜ xj(t) = xj

?λt;x,x0,y,ξ,η(0)

λ

;λt?= xj,

for j = 1,...,n and

˜ yα(t) =yα(λt;x,x0,y,ξ,η(0)

λ

;λt?= yα,

;λt?= ληα(0)

˜ ηα(0) =ληα(0;x,x0,y,ξ,η(0)

λλ

= ηα(0)

for α = 1,...,m. The bicharacteristic curves are unique, so the two sides of (2.2) agree.

?

Corollary 2.2. One has

h(x,x0,y,ξ,η(0);t) = λh?x,x0,y,ξ,η(0)

λ

;λt?.

Proof. In the case of Heisenberg group, the corollary is a direct consequence of the explicit

formula (1.11) and in this case, η(0) = η is a constant. Here we would like to give a proof

which applies in more general case. We know that for j = 1,...,m,

˙ xj(s;x,x0,y,ξ,η(0);t) =dxj

ds(s;x,x0,y,ξ,η(0);t)

=dxj

ds

?λs;x,x0,y,ξ,η(0)

?λs;x,x0,y,ξ,η(0)

λ

;λt?

;λt?.

=λ˙ xj

λ

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HAMILTON-JACOBI EQUATION AND HEAT KERNEL9

Similar result holds for ˙ yαfor α = 1,...,m. Therefore,

?t

?t

− λ2H(x(λs;...),...)?ds

=1

λ

0

k=1

?t

=λS?x,x0,y,ξ,η(0)

Also,

m

?

and the proof of the corollary is therefore complete.

0

?ξ(s) · ˙ x(s) + η(s) · ˙ y(s) − H(x(s;...),...)?ds

?λξ?λs;x,x0,y,ξ,η(0)

=

0

λ

;λt?· λ˙ x(λs;...) +

m

?

α=1

ληα(λs;...) · λ˙ yα(λs;...)

?t

?λ2

?ξ?s′;x,x0,y,η(0)

n

?

ξk(λs;...)˙ xk(λs;...) + λ2

m

?

α=1

ηα(λs;...)˙ yα(λs;...) − λ2H(x(λs;...),...)?d(λs)

,λt?· ˙ x(s′;...) + η(s′;...) · ˙ y(s′;...) − H(x(s′;...),...)?ds′

=λ

0

λ

λ

,λt?.

α=1

ηα(0)yα(0;x,x0,y,ξ,η(0);t) = λ

m

?

α=1

ηα(0)

λ

yα

?0;x,x0,y,ξ,η(0)

λ

;λt?

?

Set

f(x,x0,y,ξ,η(0)) = h(x,x0,y,ξ,η(0),t)

???t=1.

Then

Theorem 2.2. f is a solution of the generalized Hamilton-Jacobi equation

(2.3)

m

?

α=1

ηα(0)

∂f

∂ηα(0)+ H

?

x,y,∇xf,∇yf

?

= f.

Proof. By homogeneity property of the function h, one has

h(x,x0,y,ξ,η(0),t) =1

th(x,x0,y,ξ,tη(0),1) =1

tf(x,x0,y,ξ,tη(0)),

so,

(2.4)

∂h

∂t= −1

t2f +1

t

m

?

α=1

ηα(0)

∂f

∂ηα(0)

on one hand. On the other hand,

(2.5)

∂h

∂t= −H?x,y,∇xh,∇yh?

from Theorem 2.1. Since (2.4) agrees with (2.5) for all t so we may set t = 1 which yields the

proposition.

?

At the rest of the sectionwe present some examples that revealthe geometrical nature of

functions h and f.

Page 10

10 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

2.3. Laplace operator. We start from the Laplace operator ∆ =?n

Hamiltonian function H(ξ) is

k=1

∂2

∂x2

k

in Rn. The

H(ξ) =1

2

n

?

k=1

ξ2

k

and hence we need to deal with F(ξ,γ) = H + γ = 0. The Hamilton’s system is

˙ x = ξ,

˙ξ = 0,

˙ γ = 0.

with initial-boundary conditions x(0) = x0, x(t) = x. Since˙ξ = 0, it follows that ξ(s) = ξ(0) =

constants, is a constant vector. Then

¨ x =˙ξ = 0⇒

x(s) = ξ(0)s + x0.

Moreover,

x = x(t) = ξ(0)t + x0

⇒

ξ(0) =x − x0

t

and

∂h

∂t=1

2

n

?

k=1

ξ2

k=

n

?

k=1

(xk− x(0)

2t2

k)2

=|x − x0|2

2t2

or,

h(x,x0,t) = h(0) +|x − x0|2

2t2

t = h(0) +|x − x0|2

2t

.

Since this is a translation invariant case, we may assume that h(0) = 0. Therefore,

f(x,x0) = h(x,x0,t)

???t=1=|x − x0|2

2

gives us the Euclidean action function.

2.4. Grusin operator. We are in R2now and the horizontal vector fields X1, X2are given

by

∂

∂x,

X1=

and

X2= x∂

?

∂y.

?2

The Grusin operator is given as follows: ∆X =

1

2

∂

∂x

+1

2x2?

∂

∂y

?2.It is obvious that

∆X is elliptic away from the y-axis but degenerate on the y-axis.

hence {X1,X2,[X1,X2]} spanned the tangent bundle of R2everywhere.

theorem [12], ∆Xis hypoelliptic.

The Hamiltonian function H for the ∆Xis

H(x,y,ξ,η) =1

Since [X1,X2] =

By H¨ ormander’s

∂

∂y,

(2.6)

2ξ2+1

2x2η2.

The Hamilton system can be obtained as follows;

˙ x =Hξ= ξ,

˙ y =Hη= ηx2,

˙ξ = − Hx= −η2x,

˙ η = − Hy= 0,

˙S =ξ ˙ x + η ˙ y − H.

With 0 ≤ s ≤ t,

η(s) = η(0) = η0= constant,

Page 11

HAMILTON-JACOBI EQUATION AND HEAT KERNEL 11

“constant” means “constant along the bicharacteristic curve”. Next,

¨ x =˙ξ = −xη2,

so

¨ x + η2x = 0.

It follows that

x(s) = Acos(ηs) + B sin(ηs) = x(0)cos(ηs) +ξ(0)

η

sin(ηs) = x0cos(ηs) +ξ(0)

η

sin(ηs).

Hence,

ξ(s) = ˙ x(s)

yields

ξ(s) = ξ(0)cos(ηs) − ηx0sin(ηs).

We also have

x = x(t) = x0cos(ηt) +ξ(0)

η

sin(ηt),

and

(2.7)

ξ(0)

η

=x − x0cos(ηt)

sin(ηt)

.

Consequently,

x(s) = x(0)cos(ηs) +x − x0cos(ηt)

sin(ηt)

sin(ηs).

The singularities occur at η = η0=kπ

η0=2kπ

t

if x = −x0. Next,

˙ y(s) =ηx2(s)

?

=d

ds

2

=d

ds

2

twhen x = ±x0; they are η =

(2k+1)π

t

if x = x0and

=ηx0

?1

η

2+1

?x2

?η

ξ(0)

η

by (2.7) and collect terms with x2

?

?

x2

0

2sin2(ηt)

2

x2

0

4sin2(ηt)

The terms containing x2are:

2cos(2ηs)?+ 2x0ξ(0)

?s +sin(2ηs)

0+?ξ(0)

η

sin(ηs)cos(ηs) +?ξ(0)

?+x0ξ(0)

s +1

4

η

η

?2?1

2−1

2cos(2ηs)??

???

?1 − cos(2ηs)??

?

0

2η

?2?

η2

0−?ξ(0)

sin2(ηs) +1

2

?ξ(0)

η

?2?s −sin(2ηs)

2

η

2η

?

x2

η

?

x2

?2?

0:

sin(2ηs) +x0

ξ(0)

.

We replace

x2

2

0

ηs +1

2sin(2ηs) + ηscos2(ηt)

cos2(ηt) − sin2(ηt)

sin2(ηt)

?

?

sin2(ηt)−1

2

cos2(ηt)

sin2(ηt)sin(2ηt) −cos(ηt)

sin(2ηs) −cos(ηt)

sin(ηt)

sin(ηt)

?1 − cos(2ηs)??

?1 − cos(2ηs)??

=x2

0

2

ηs

sin2(ηt)−1

2

=

ηs −1

?cos(2ηt)sin(2ηs) + sin(2ηt)?1 − cos(2ηs)???

2ηs −?sin(2ηt) − sin?2η(t − s)???

=

.

1

4

x2

sin2(ηt)

?2ηs − sin(2ηs)?,

Page 12

12 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

and the terms with x0x are the following:

1

2

2xx0

sin2(ηt)

?1

2

?sin?η(2s − t)?+ sin(ηt)?− ηscos(ηt)

?

.

So,

˙ y(s) =d

ds

?

x2

0

4sin2(ηt)

x2

4sin2(ηt)

2xx0

4sin2(ηt)

?

2ηs −?(sin(2ηt) − sin?2η(t − s)??)Big]

?2ηs − sin(2ηs)?

?1

+

+

2

?sin?η(2s − t)?+ sin(ηt)?− η scos(ηt)

??

.

The action function has the form

S =

?t

0

(ξ ˙ x + η ˙ y − H)ds = η(y − y(0)) +

?t

0

(ξ2− H)ds.

We find ξ2as follows

ξ2(s) =ξ2(0)

2

?ξ2(0) + η2x2

=H(0)

?1 + cos(2ηs)?− ξ(0)ηx0sin(2ηs) +1

0

??

H = H(0) =1

2η2x2

0

?1 − cos(2ηs)

?

=1

2

?

?

?

+1

2

?ξ2(0) − η2x2

0

?cos(2ηs) − ηx0ξ(0)sin(2ηs).

Since H is constant along the bicharacteristic, one has

2

?ξ2(0) + η2x2

0

?.

Continuing, we obtain the action function

S =η(y − y(0)) +

?t

0

?ξ2(0) − η2x2

?(ξ2(0) − η2x2

0)cos(2ηs)

2

− ηx0ξ(0)sin(2ηs)?ds

+ ηx0ξ(0)cos(2ηt) − 1

=η(y − y(0)) +1

2

0

?sin(2ηt)

2η

2η

.

We simplify this

S − η(y − y(0))

=η2

2

=η

4

?x − x0cos(ηt)

??x − x0cos(ηt)

sin(ηt)

?2sin(2ηt)

?2sin(2ηt) − x2

2η

−1

2η2x2

0

sin(2ηt)

2η

+ η2x0x − x0cos(ηt)

sin(ηt)

cos(2ηt) − 1

2η

?1 − cos(2ηt)??

sin(ηt)

0sin(2ηt) − 2x0x − x0cos(ηt)

sin(ηt)

.

(2.8)

In the bracket {···} of (2.8), terms involved x2

??cos2(ηt)

=x2

0

0are

x2

0

sin2(ηt)− 1

?cos(2ηt)sin(2ηt)

=2x2

0cot(ηt),

?

sin(2ηt) + 2cos(ηt)

sin(ηt)(1 − cos(2ηt))

+ 2cos(ηt)

?

?

sin2(ηt)

sin(ηt)−cos(2ηt)sin(2ηt)

sin2(ηt)

Page 13

HAMILTON-JACOBI EQUATION AND HEAT KERNEL 13

terms involved x2are

x2sin(2ηt)

sin2(ηt)= 2x2cot(ηt),

and terms containing x0x are

2xx0

?

−cos(ηt)

sin2(ηt)sin(2ηt) −1 − cos(2ηt)

?2cos2(ηt)

4xx0

sin(ηt).

sin(ηt)

?

?

= − 2xx0

sin(ηt)

+ 2sin(ηt)

= −

Hence,

{···} =2(x2+ x2

0)cot(ηt) −

4xx0

sin(ηt)

=?(x + x0)2+ (x − x0)2?cot(ηt) −(x + x0)2− (x − x0)2

=(x + x0)2?

=(x + x0)2cos(ηt) − 1

sin(ηt)

= − (x + x0)2tan?ηt

Thus S has the following form:

?

By Theorem 2.1, we know that

sin(ηt)

cot(ηt) −

1

sin(ηt)

+ (x − x0)2cos(ηt) + 1

?

+ (x − x0)2?

cot(ηt) +

1

sin(ηt)

?

sin(ηt)

2

?+ (x − x0)2cot?ηt

2

?.

S = η(y − y(0)) −η

4

(x + x0)2tan?ηt

2

?− (x − x0)2cot?ηt

2

??

.

h(t;x,x0,y,η) =ηy(0) + S(t;x,y,η)

=ηy(0) + η(y − y(0)) −η

4

?

(x + x0)2tan?η

?− B2cot?ηt

2

?− (x − x0)2cot?η

2

??

=ηy −η

4

?

A2tan?ηt

22

??

is a solution of the Hamilton-Jacobi equation. Here A = x + x0and B = x − x0. Now by

Theorem 2.2, the function

???t=1=1

η∂f

∂η+ H

f(x,x0,y,η) = h(t;x,x0,y,η)

2

η

2

?

4y − A2tan?η

2

?+ B2cot?η

2

??

is a solution of the generalized Hamilton-Jacobi equation

?

x,x0,y,∂xf,∂yf

?

= f.

We set

η

2= ? ητ,

where τ ∈ R yields the domain of integration and ? η is a fixed complex number.

Page 14

14 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

Lemma 2.5. Suppose f is a smooth function of τ ∈ R and

lim

τ→±∞Re(f)(τ) = ∞

off the canonical curve x2

0+ x2= 0. Then ? η is pure imaginary.

?

− isinh(2η2τ)?(B2+ A2)cosh(2η2τ) + (B2− A2)cos(2η1τ)?

(i). η1= 0, i.e., η ∈ iR. When τ ≈ ±∞,

f ≈1

2iη2τ

and

Re(f) ≈1

4(x2

as τ → ±∞ as long as x2

(ii). η2= 0, that is η ∈ R. Then

f = 2η1τy +1

4 sin(2η1τ)

is singular in τ ∈ R when x2

Proof. Let ? η = η1+ iη2. An elementary calculation yields

2

f =1

?η1+ iη2

?τ

4y +sin(2η1τ)?(B2− A2)cosh(2η2τ) + (B2+ A2)cos(2η1τ)?

cosh2(2η2τ) − cos2(2η1τ)

cosh2(2η2τ) − cos2(2η1τ)

?

?

4y − i2(x2

0+ x2)tanh(2η2τ)

?

,

0+ x2)2η2τ tanh(2η2τ) → ±∞

0+ x2?= 0.

2η1τ

?

B2− A2+ (B2+ A2)cos(2η1τ)

?

0+ x2?= 0, otherwise

Re(f) = f = 2η1τy −→

????

τ→±∞

±(sgn(y))∞.

(iii). η1?= 0, η2?= 0. Here

f ≈1

2

?η1+ iη2

?τ

?

4y − i(A2+ B2)tanh(2η2τ)

?

as τ → ±∞, and

Re(f) ≈2η1τy + (x2

=|τ|?2(sgn(τ))η1y + (x2

2η1y > (x2

0+ x2)??η2τ??

0+ x2)|η2|?

and choosing x0,x,y so that

0+ x2)|η2|

we have

lim

τ→±∞Re(f) = ±∞

which we do not want. This complete the proof of Lemma (2.5).

?

Following the tradition, we shall choose

? η = −i

2(x2

2.

Then

f = −iτy +1

0+ x2)τ coth τ −

τx0x

sinh τ.

Page 15

HAMILTON-JACOBI EQUATION AND HEAT KERNEL 15

2.6. Sub-Laplace operator on step 2 nilpotent Lie groups. Let M be a simply connected

2-step nilpotent Lie group G equipped with a left invariant metric. Let G be its Lie algebra

and it is identified with the group G by the exponential map:

exp : G → G.

We assume

G = [G,G] ⊕ [G,G]⊥= C ⊕ [G,G]⊥= C ⊕ H,

where H and C are vector spaces over R with an skew-symmetric bilinear form

B : H × H → C

such that B(H,H) = C. The group law is given by

(H ⊕ C) × (H ⊕ C) → H ⊕ C

with

(x,y) ∗ (x′,y′) =

?x + x′,y + y′+1

2B(x,x′)?

and then the exponential map is the identity map. Let {X1,...,Xn} be a basis of H and let

{Y1,...,Ym} be a basis of the center [G,G] = C. We assume {X1,...,Xn} and {Y1,...,Ym} are

orthonormal, and introduce a left invariant Riemannian metric on the group G in an obvious

way.

We write the vector fields Xj, j = 1,...,n by:

Xj=

∂

∂xj

+

n

?

k=1

m

?

α=1

aα

jkxk

∂

∂yα

where the aα

We are interested in the sub-Laplacian ∆Xwhich can be defined as follows:

jkare real numbers and form skew-symmetric matrices?aα

∆X=1

jk

?

j,k, i.e., aα

jk= −aα

kj.

2

n

?

j=1

X2

j

It is easy to see that

(2.9)

?Xj,Xk

?= 2

n

?

k=1

m

?

α=1

aα

jk

∂

∂yα.

Lemma 2.7. The operator ∆X is hypoelliptic if and only if the rectangular matrix of order

n(n−1)

2

jk

× m with element?aα

Proof. The operator ∆X is hypoelliptic when the vector fields {Xj}n

bracket generating condition. This implies that we can recover all the

relations (2.9). If we consider

jk

(j,k) where j < k, this means that this matrix should have rank m.

?

{(j<k),α}is of rank m (which implies that m ≤n(n−1)

2

).

j=1satisfy the “first”

∂

∂yαfrom the

n(n−1)

2

?aα

?

as a matrix with indices α = 1,...,m and the couples

?

We may define a Lie group structure on Rn× Rmwith the following group law:

(2.10)

?

(x,y) ◦ (x′,y′) =

x1+ x′

1,...,xn+ x′

n,y1+ y′

1+

n

?

j,k=1

a1

jkx′

jxk,...,ym+ y′

m+

n

?

j,k=1

am

jkx′

jxk

?

.

Page 16

16 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

It is easy to see that the Xjare left invariant vector fields such that

?Xjf?(x,y) =

where

L(x,y)(x′,y′) = (x,y) ◦ (x′,y′)

is the left translation by the element (x,y). In particular, ∆Xis a left invariant operator for

this group structure (see [1] and [16]).

Let ξ1,...,ξnbe the dual variables of x and η1,...,ηmbe the dual variables of y. We define

the symbols ζjof the vector field Xjby

n

?

We shall try to find a solution of the following equation:

n

?

Thus we start with

∂z

∂t+ H(∇z) = 0,

where H(x,y;ξ,η) is the Hamiltonian function as the full symbol of ∆X,

n

?

Here

?

We shall find the bicharacteristic curves which are solutions to the corresponding Hamilton’s

system. The solutions define a one parameter family of symplectic isomorphism of the (punc-

tures) cotangent bundle T∗(Rn×Rm)\{0}. Since At(η) = −A(η), the Hamilton’s system can

be written explicitly as follows:

n

?

n

?

˙ξj= − Hxj= −

k=1

˙ ηα= − Hyα= 0,

for α = 1,...,m

∂

∂x′

j

?f ◦ L(x,y)

?(x′,y′)

???x′=0,y′=0

ζj= ξj+

k=1

m

?

α=1

aα

jkxkηα.

∂h

∂t+1

2

j=1

?∂h

∂xj

+

n

?

k=1

m

?

α=1

aα

jkxk

∂h

∂yα

?2

= 0.

(2.11)

(2.12)

H(x,y;ξ,η) =1

2

j=1

?ξj+

n

?

k=1

m

?

α=1

aα

jkxkηα

?2=1

2

n

?

j=1

?ξj+

n

?

k=1

Akj(η) · xk

?2.

Akj(η) =

m

α=1

aα

kjηα.

˙ xj=Hξj= ξj−

k=1

n

?

?

Ajk(η) · xk= ζj,

for j = 1,...,n

˙ yα=Hηα=

j=1

k=1

aα

jkxkζj,

for α = 1,...,m

n

Ajk(η) · ζk=

n

?

k=1

Akj(η) · ζk,

for j = 1,...,n

(2.13)

with the initial-boundary conditions such that

(2.14)

x(0) = 0

x(t) = x = (x1,...,xn)

y(t) = y = (y1,...,ym)

η(0) = iτ = i(τ1,...,τm)

Page 17

HAMILTON-JACOBI EQUATION AND HEAT KERNEL17

where t ∈ R, x and y are arbitrarily given. With 0 ≤ s ≤ t,

ηα(s) = ηα= constant,

for α = 1,...,m

“constant” means “constant along the bicharacteristic curve”. Also

H =1

2

n

?

j=1

˙ x2

j=1

2

n

?

j=1

ζ2= E = energy.

Another way to see that E is constant along the bicharacteristic, note that

¨ xj=˙ζj=˙ξj−

n

?

k=1

Ajk(η) · ˙ xk

= −

n

?

?

k=1

Ajk(η) · ζk−

n

?

k=1

Ajk(η) · ζk

= − 2

n

k=1

Ajk(η) · ζk

(2.15)

for j = 1,...,n. Hence

(2.16)

¨ x =˙ζ =˙ξ + A(η)˙ x = −2A(η)ζ.

Therefore,

¨ x · ˙ x = −2A(η)ζ · ζ = 0

since A is skew-symmetric. It follows that

1

2

n

?

j=1

˙ x2

j=1

2˙ x · ˙ x = E = energy.

Since ˙ x(s) = e−2sA(η)ξ(0), by integrating the equation

A(η)˙ x(s) = A(η)e−2sA(η)ξ(0),

one has

A(η)x(s) = −1

2

?

e−2sA(η)− I

?

ξ(0)

where I is the n × n identity matrix. Since ηα= η(0) = iταis pure imaginary, the matrix

iA(τ) is self-adjoint. It follows that the matrix

?

is well defined and invertible for any t ∈ R and τ ∈ Rm. Here γ is a suitable contour surrounding

the spectrum of the matrix itA(τ). The matrix

?

has an inverse:

1

2πi

γ

λ

We write it as

sinh(iA(τ))

iA(τ)

isA(τ)

sinh(itA(τ))=

1

2πi

γ

λ

sinh(λ)

?

λ − itA(τ)

?−1dλ

1

2πi

γ

λ

sinh(λ)

?

?

λ − itA(τ)

?−1dλ

?−1dλ

?

sinh(λ)

λ − itA(τ)

=

∞

?

k=0

(iA(τ))2k

(2k + 1)!.

Page 18

18OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

Then for any fixed t ∈ R, we have one-to-one correspondence between the initial condition ξ(0)

and boundary condition x:

ξ(0) = eitA(τ)·

iA(τ)

sinh(itA(τ))· x,t ?= 0.

Now we may solve the initial value problem:

with the initial conditions

˙ xj(s) =

˙ξj(s) =

∂H

∂ξj= ξj+ i?n

−∂H

k=1

?m

α=1aα

ξk+ i?n

jkxkτα= ξj+ i?

k=1Akj(τ)xk,

· Ajk(τ)

∂xj= −i?n

?

k=1

?

ℓ=1Aℓk(τ)xℓ

?

x(0) = 0

ξ(0) = eitA(τ)·

iA(τ)

sinh(itA(τ))x.

Straightforward computations show that

x(s) =x(s;x,τ,t) = ei(t−s)A(τ)sinh(isA(τ))

sinh(itA(τ))· x

ξ(s) =ξ(s;x,τ,t)

sinh(itA(τ))· eitA(τ)?

=

e−isA(τ)cosh(isA(τ))

?

=

iA(τ)

I − e−isA(τ)sinh(isA(τ))

?

x

· x

??

?

·

?

eitA(τ)

iA(τ)

sinh(itA(τ))

?

=

e−isA(τ)cosh(isA(τ))· ξ(0).

Hence we obtain solutions for the initial-boundary problem (2.13) under the condition (2.14).

We also have the following solutions for y(s):

?s

k=1

Again by Theorem 2.2, the function

yα(s) = yα(0) +

0

n

?

??e−2iuA(τ)ξ(0)?

k·

n

?

ℓ=1

aα

ℓkxℓ(u)

?

du,α = 1,...,m.

f(x,y,τ) = h(x,y,τ,t)

???t=1

is a solution of the generalized Hamilton-Jacobi equation. In our case, the function f can be

calculated explicitly.

???t=1

=

ηα(0)yα(0) +

f(x,y,τ) =h(x,y,τ,t)

m

?

α=1

?1

?ξ · ˙ x − H?ds.

0

?ξ · ˙ x + η · ˙ y − H?ds

=η0

m

?

α=1

ταyα+

?1

0

Here η0is a pure imaginary number. This choice can be motivated by Lemma 2.5.

Since

ξ · ˙ x − H =1

2?ζ,ζ? − ?ζ,Ax?,

Page 19

HAMILTON-JACOBI EQUATION AND HEAT KERNEL 19

then

?ζ,Ax? =

?

2?ζ,ζ? −

ζ,A(τ)e2sA(τ)

e2A(τ)− I

=1

x

?

−

?

ζ,

A(τ)

e2A(τ)− Ix

A(τ)

e2A(τ)− Ix

?

?2A(τ)e2sA(τ)

?

e2A(τ)− I

x,

?

.

It follows that

ξ · ˙ x − H =

?2A(τ)e2sA(τ)

e2A(τ)− I

x,

A(τ)

e2A(τ)− Ix

=

?2A(τ)cosh(2sA(τ))

e2A(τ)− I

x,

A(τ)

e2A(τ)− Ix

?

.

The second equality due to A is skew-symmetric. Now we can integrate from s = 0 to s = 1

to obtain

?1

It follows that

m

?

Using equation (2.17), we may complete the discuss in Section 2.

0

?ξ · ˙ x − H?ds =1

2

??A(τ)coth(A(τ))?x,x

ταyα+1

2

?

.

(2.17)

f(x,y,τ) = −i

α=1

??A(τ)coth(A(τ))?x,x

?

.

Example 2.8. When

A =

a1

0

0

a2

···

···

···

···

0

0

00

a2n

∈ M2n×2n,

with aj= aj+n,j = 1,...,n,

i.e., the group is an anisotropic Heisenberg group. In this case, m = 1 and

f(x,y,τ) = −iτy + τ

n

?

k=1

akcoth(2akτ)?x2

k+ x2

n+k

?.

Example 2.9. In R4, the basis of quaternion numbers H = {a + bi + cj + dk : a,b,c,d ∈ R}

can be given by real matrices

We have

The number a is called the real part and denoted by a = Re(q). The vector u = (b,c,d) is the

imaginary part of q. We use the notations

M0=

1

0

0

0

0

1

0

0

0

0

1

0

0

0

0

1

,

M1=

01

0

0

0

0

0

0

0

0

1

0

−1

0

0−1

,

0

1

0

0

M2=

0

0

0

1

0

0

1

0

0−1

−10

0

0

0

0

,

M3=

0

0

1

0

0

0

0

−1

0

0

0−1

.

q =

ab

a

c

−d

−c

−c

−bd

b

a

d

c

a

−d

−b

= aM0+ bM1+ cM2+ dM3.

b = Im1(q),c = Im2(q),d = Im3(q),

andIm(q) = u = (b,c,d).

Page 20

20OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

We introduce the quaternionic H-type group denoted by Q. This group consists of the set

H × R3= {[x,y] : x ∈ H, y = (y1,y2,y3) ∈ R3}

with the multiplication law defined in (2.10) with [aα

vector fields X = (X1,X2,X3,X4) of the group Q can be written as follows:

X = ∇x+1

2

∂y1

with x = (x1,x2,x3,x4) and

?∂

In this case, the solution for the generalized Hamilton-Jacobi equation is

jk] = Mα, α = 1,2,3. The horizontal

?

M1x∂

+ M2x∂

∂y2

+ M3x∂

∂y3

?

,

∇x=

∂x1,

∂

∂x2,

∂

∂x3,

∂

∂x4

?

.

f(x,y1,y2,y3,τ1,τ2,τ3) = −i

3

?

α=1

ταyα+|x|2

2

|τ|coth(2|τ|)

See details in [6] In general multidimensional case, the matrix A can be defined as follows:

In this case we obtain the so called anisotropic quaternion Carnot group considered in [7]. The

complex action is given by

A =

?3

α=1aα

0

1Mα

0

...

...

...

...

0

0

?3

α=1aα

2Mα

00

?3

α=1aα

nMα

.

f(x,y,τ) = −i

?

α=1(aα

α

ταyα+1

2

n

?

?1/2. If all aα

l=1

|xl|2|τ|lcoth(2|τ|l),

where |xl|2=?3

groups can be found in [5, 13, 14, 15].

j=0x2

4l−j, |τ|l=??3

l)2τ2

α

l, l = 1,...,n are equal, we get

the example of multidimensional quaternion H-type group. More information about H-type

3. Heat kernel and transport equation

Let us return to the heat kernel. We consider the sub-Laplacian

∆X=1

2

n

?

k=1

X2

k

with

Xk=

∂

∂xk

+

n

?

j=1

m

?

α=1

aα

kjxj

∂

∂yα.

Assume that {X1,...,Xn} is an orthonormal basis of the “horizontal subbundle” on a simply

connected nilpotent 2 step Lie group. The Hamiltonian of the operator ∆Xis

n

?

By Theorem 2.2, the function f associated with H is a solution of the generalized Hamilton-

Jacobi equation:

H(x,y,ξ,η) =1

2

k=1

?

ξk+

n

?

j=1

m

?

α=1

aα

kjxjηα

?2.

H(x,y,∇xf,∇yf) +

m

?

α=1

τα

∂f

∂τα

= f(x,y;η1,...,ηm).

Page 21

HAMILTON-JACOBI EQUATION AND HEAT KERNEL 21

As we know, the function f depends on free variables ηα, α = 1,...,m. To this end we shall

sum over ηα, or for convenience τα= tηα, α = 1,...,m; an extra t can always be absorbed in

the power q which can be determined after we solve the generalized Hamilton-Jacobi equation.

Thus we write heat kernel of ∆X−∂

∂tas following

(3.1)

K(x,y;t) = Kt(x,y) =1

tq

?

Rme−f(x,y,τ)

t

V (τ)dτ.

Here V is the volume element. To see whether (3.1) is a representation of the heat kernel we

apply the heat operator to K and take it across the integral.

?

∆X−∂

∂t

?e−f(x,y,τ)

t

tq

=e−f(x,y,τ)

tq+2

t

?H(x,y,∇xf,∇yf) − f?−e−f(x,y,τ)

t

tq+1

?∆X(f) − q?,

and the eiconal equation (2.3) implies that

?

∆X−∂

∂t

?e−f(x,y,τ)

m

?

? m

t

V (τ)

tq

=e−f(x,y,τ)

tq+1

t

α=1

τα

?

−1

t

∂f

∂τα

?

V (τ) −e−f(x,y,τ)

t

tq+1

?∆Xf − q?V (τ)

?

= −e−f(x,y,τ)

t

tq+1

?

α=1

τα∂V

∂τα

+?∆Xf − q + m?V (τ)+

m

?

α=1

∂

∂τα

?e−f(x,y,τ)

t

ταV (τ)

tq+1

?

.

Assuming

e−f(u)

t ταV (τ)

tq+1

→ 0

as τα→ the ends of an appropriate contour Γαfor α = 1,...,m, one has

?

=

∂ttq

∪m

α=1Γα

1

tq+1

∪m

α=1Γα

∆X−∂

?

∂t

?

??1

Kt(x,y)

∆X−∂

?

e−f(x,y,τ)

e−f(x,y,τ)

t

V (τ)dτ?

τα∂V

∂τα

= −

?

t

? m

?

α=1

+?∆Xf − q + m?V (τ)

?

dτ = 0

if t ?= 0 and

(3.2)

m

?

α=1

τα∂V

∂τα(τ) +?∆Xf − q + m?V (τ) = 0.

The equation (3.2) is called the first order transport equation.

Remark 3.1. Here we have made a crucial assumption on the volume element, i.e., V does

not depend on the space variables x and y. That simplify the transport equation significantly.

Under a more general situation a function V will found among co-dimension one form

V dτ =

m

?

ℓ=1

(−1)ℓ−1Vℓdτ1∧ ··· ∧?

dτℓ∧ ··· ∧ dτm

Page 22

22OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

which satisfies a so-called “generalized transport equation”:

df ∧ ∆X(V ) +

n

?

ℓ=1

Xℓ(f)Xℓ(dV ) + D(dV ) −?∆Xf + n − m − 1?dV = 0,

where D(V ) is defined by

D(V ) =

m

?

k=1

τk

∂

∂τk(V ) =

m

?

k=1

m

?

ℓ=1

(−1)ℓ−1τk∂Vℓ

∂τkdτ1∧ ··· ∧?

dτℓ∧ ··· ∧ dτm.

Detailed discussion can be found in Furutani [9] and Greiner [11].

With f given by (2.17), one has

(3.3)∆Xf =1

2tr?A(τ)coth(A(τ))?=1

2tr

?1

2πi

?

C

λcosh(λ)

sinh(λ)

?λ − iA(τ)?−1dλ

?

.

Then (3.2) becomes

m

?

m

?

α=1

τα∂V

∂τα(τ) +?∆Xf − q + m?V (τ) = 0 ⇔

τα∂V

∂τα(τ) =

α=1

?

q − m −1

2tr?A(τ)coth(A(τ))??

V.

(3.4)

Fix τ and define for 0 ≤ λ ≤ 1

W(λ) = V (λτ).

Hence, (3.4) reduces to

λdW

dλ

=

?

q − m −1

2tr?λA(τ)coth(λA(τ))??

2tr?A(τ)coth(λA(τ))??

W.

Here we are using the fact that A(τ) is linear in τ. It follows that

?q − m

Hence,

logW = (q − m)?log λ + log C?−1

Therefore,

dW

W

=

λ

−1

dλ.

2log(sinh(A(λτ))?.

V (τ) =

(detA(τ))q−m

??det sinh(A(τ))?.

If we propose the volume element V is real analytic and non-vanish at 0, then we have q =n

Consequently,

A

(2πt)q

2+m.

(3.5)

P =

?

Rme−f(x,y,τ)

t

V (τ)dτ,

where f is given by (3.3) and

V (τ) =

(detA(τ))

??det sinh(A(τ))?

n

2

where the branch is taken to be V (0) = 1. Finally we can write down the second main results

on this paper.

Page 23

HAMILTON-JACOBI EQUATION AND HEAT KERNEL 23

Theorem 3.2. The equation

Pt(x,y) =

A

(2πt)q

?

Rme−f(x,y,τ)

t

V (τ)dτ,

represents the heat kernel for ∆Xif and only if q =n

2+ m, in which case A = 1.

We clearly have

∂P

∂t− ∆XP = 0, t > 0

and

lim

t→0P(x,y,t) = δ(x)δ(y).

The calculation is long but straightforward. Readers can find the proof of this theorem in many

places, see e.g., [1, 2, 3, 4, 8, 10]. We skip the proof here. Instead, we list some examples.

Example 3.1. The Heisenberg sub-Laplacian: ∆X=1

is defined as (1.1). The action function is f(x,y) = −iτy + (x2

is V (τ) =

sinh(2τ). In this case n = 2, m = 1 and (3.5) has the following expression:

2

?

∂

∂x1+2x2∂

∂y

?2+1

2)τ coth(2τ). The volume

2

?

∂

∂x2−2x1∂

∂y

?2

which

1+ x2

2τ

(3.6)

Pt(x,y) =

2

(2πt)2

?+∞

?

−∞

e−f(x,y,τ)

t

τ

sinh(2τ)dτ.

2x2?

Example 3.2. The Grusin operator: ∆G=1

this case. However, this operator has connection with the Heisenberg sub-Laplacian. Let H1

be the Heisenberg group whose Lie algebra has a basis {X1,X2,T} with the bracket relation

[X1,X2] = −4T. As in (1.1),

∆X= −1

2

is the sub-Laplacian on H1. Let NX2= ?X2? = [{aX2}a∈R] be a subgroup generated by the

element X2. The map ρ : H1→ R2defined by

ρ : H1→ R2∼= h ∋ g =x1X1+ x2X2+ zZ

=(x1,x2,z) ?→ (u,v) ∈ R2

2

∂

∂x

?2+1

∂

∂y

?2. There is no group structure in

?X2

1+ X2

2

?

where

u = x1,v = z +1

2x1x2

realizes the projection map

H1∼= R3→ NX2\ H1∼= R2.

In fact, this is a principal bundle and the trivialization is given by the map

NX2× (NX2\ H1)∼= R × R2∋ (a;u,v) ?→ (x1,x2,z) ∈ R3∼= H1

where

(a;u,v) ?→

?u,a,v −1

2au?.

So the sub-Laplacian ∆X on H1and Grusin operator ∆Gcommutes each other through the

map ρ:

∆H◦ ρ∗= ρ∗◦ ∆G.

The heat kernel Pt(x,y) ∈ C∞(R+× H1) is given by (3.6). Hence,

?+∞

−∞

Pt

?(x1,x2,y),(u,a,v −1

2ua)?= PG

t((x1+ y +1

2x1x2),(u,v))

Page 24

24 OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA

that is, the fiber integration of the function Pt(g,h) along the fiber of the map ρ gives the heat

kernel of the Grusin operator.

?+∞

PG

t((x0,0),(x,y)) =

1

(2πt?3

2

−∞

e−f(x,x0,y,τ)

t

?

|τ|

sinh |τ|dτ

Example 3.3. Step 2 nilpotent Lie group: ∆X= −1

2

?n

aα

jkxk

j=1X2

?∂

jwhere

Xj=

∂

∂xj

+

n

?

k=1

? m

?

α=1

∂yα

with A(α)

jk=?aα

jk

?

j,kis a skew-symmetric and orthogonal matrix. The heat kernel is

?

Here Ajk(τ) =?m

?

Pt(x,y) =

1

(2πt)

n

2+m

Rme−2iy·τ−?A(τ) coth(A(τ))x,x?

2t

?

det

A(τ)

sinh(A(τ))dτ.

α=1aα

jkτα. In particular, if A(α)

jksatisfies further assumption: A(α)

jkA(γ)

jk+

A(γ)

jkA(α)

jk= 0, i.e., the group in a H-type group. Then the heat kernel has the following form:

Pt(x,y) =

1

(2πt)

n

2+m

Rme−2iy·τ+|x|2|τ| coth(|τ|)

2t

?

|τ|

sinh |τ|

?n

2dτ.

Acknowledgement. Part of this article is based on a lecture presented by the first author

during the International Conference on “New Trends in Complex and Harmonic Analysis”

which was held on May 7-11, 2007 at Voss, Norway. The first author thanks the organizing

committee, especially Professor Alexander Vasiliev for his invitation. He would also like to

thank all the colleagues at the Mathematics Department of University of Bergen for the warm

hospitality during his visit to Norway. We also would like to thank Professor Peter Griener

and Professor Kenro Furutani for many inspired conversations on this project.

References

[1] Beals R., Gaveau B., and Greiner P. C. The Green function of model step two hypoelliptic operators and

the analysis of certain tangential Cauchy-Rieman complexes. Advances in Math. 121 (1996), 288–345.

[2] Beals R., Gaveau B., and Greiner P. C. Hamilton-Jacobi theory and the heat kernel on Heisenberg groups.

J. Math. Pures Appl. 79 (2000), no. 7, 633–689.

[3] Calin O., Chang D. C., and Greiner P. C. On a step 2(k + 1) sub-Riemannian manifold. J. Geom. Anal.,

14 (2004), no. 1, 1–18

[4] Calin O., Chang D. C., and Greiner P. C. Geometric Analysis on the Heisenberg Group and Its Gener-

alizations, to be published in AMS/IP series in advanced mathematics, International Press, Cambridge,

Massachusetts, 2007. 244 pp.

[5] Cowling M., Dooley A. H., Kor´ anyi A., and Ricci F. H-type groups and Iwasawa decompositions. Adv.

Math. 87 (1991), no. 1, 1–41.

[6] Chang D. C., Markina I. Geometric analysis on quaternion H-type groups. J. Geom. Anal. 16 (2006), no.

2, 266–294.

[7] Chang D. C., Markina I. Anisotropic quaternion Carnot groups: geometric analysis and Green’s function,

to appear in Advanced in Applied Math., (2007).

[8] Chang D. C., Markina I. Quaternion H-type group and differential operator ∆λ, to appear in Science in

China, Series A: Mathematics, (2007).

[9] Furutani K. Heat kernels of the sub-Laplacian and the Laplacian on nilpotent Lie groups, Analysis, Geom-

etry and Topology of Elliptic Operators, World Scientific (2006), 185–226.

Page 25

HAMILTON-JACOBI EQUATION AND HEAT KERNEL 25

[10] Gaveau B. Principe de moindre action, propagation de la chaleur et estim´ ees sous elliptiques sur certains

groupes nilpotents, Acta Math. 139 (1977), no. 1-2, 95–153.

[11] Greiner P. On H¨ ormander operators and non-holonomic geometry, Fields Institute Communications, 52

(2008).

[12] H¨ ormander L. Hypoelliptic second order differential equations. Acta Math. 119 (1967) 147–171.

[13] Kaplan A. Fundamental solutions for a class of hypoelliptic PDE generated by composition of quadratics

forms. Trans. Amer. Math. Soc. 258 (1980), no. 1, 147–153.

[14] Kaplan A. On the geometry of groups of Heisenberg type. Bull. London Math. Soc. 15 (1983), no. 1, 35–42.

[15] Kor´ anyi A. Geometric properties of Heisenberg-type groups. Adv. in Math. 56 (1985), no. 1, 28–38.

[16] Nagel A., Ricci F., and Stein E.M. Singular integrals with flag kernels and analysis on quadratic CR

manifolds Jour. Func. Anal. 181 (2001), 29–181.

Department of Mathematics, Eastern Michigan University, Ypsilanti, MI, 48197, USA

E-mail address: ocalin@emunix.emich.edu

Department of Mathematics, Georgetown University, Washington D.C. 20057, USA

Department of Mathematics, National Tsing Hua University, Hsinchu, Taiwan 30013, ROC

E-mail address: chang@georgetown.edu

Department of Mathematics, University of Bergen, Johannes Brunsgate 12, Bergen 5008, Nor-

way

E-mail address: irina.markina@uib.no

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