Hermite Operator on the Heisenberg Group

Hermite Operator on the Heisenberg Group·

Ovidiu Calin1, Der-Chen Chang2, and Jingzhi Tie3

Department of MathematicsEastern Michigan UniversityYpsilanti, MI [email protected]

2 Department of MathematicsGeorgetown UniversityWashington, DC [email protected]

3 Department of MathematicsUniversity of GeorgiaAthens, GA [email protected]

Dedicated to Professor Carlos Berenstein on his 60th birthday.

Summary. In this article, we first introduce a new geometric method based on multipliers tocompute heat kernels for operators with potentials. Using the heat kernel, we may computethe fundamental solution for the Hermite operator with a singularity at an arbitrary point onthe Heisenberg group. As a consequence, one may obtain the fundamental solutions for thesub-Laplacian in Hn and the Grusin operator in ffi.n.

1 Introduction

The Heisenberg group and its sub-Laplacian are at the crossroads of many domains of analysis and geometry: nilpotent Lie group theory, hypoelliptic secondorder partial differential equations, strongly pseudoconvex domains in complexanalysis, probability theory of degenerate diffusion process, sub-Riemannian geometry, control theory and semiclassical analysis of quantum mechanics, see, e.g.,[BGGr, BGr, CCGr, CGr, cn, CT2].

The Heisenberg group is the simplest nilpotent Lie group with underlying manifold JR.2n+l and multiplicative law

* The second author is partially supported by a William Fulbright Research grant and acompetitive research grant at Georgetown University.

38 O. CaHn, D.-C. Chang, and J. Tie

The nonisotropic dilation [Xl, ... , X2n, t] f---+ [OXI, ... , OX2n, 02t ] for 0 > 0 definesan automorphism on the group Hn .

The Heisenberg Lie algebra is the vector space of left invariant vector fields withthe usual Lie bracket

[X, Y] = XY - YX

with the following basis:

a aXJ'+n = -- + 2aJ·xJ·-, J' = 1, ... , n,

aXj+n at

and T = ft. Here aj = aj+n for j = 1, ... , n. It is easy to see that Xl, ... , X2nsatisfy the Heisenberg uncertainty principle:

Hence Hn is a noncommutative Lie group of step 2, i.e., XI, ... ,X2n and their firstbrackets yield the tangent bundle THn . The sub-Laplacian on Hn is defined by

. a L:n2 2L).. = lA- - (X" +X.+ )at J J n

j=l

a 2n [ a2

a2

] n (a a ) a= iA- - '""' - + 4a2x~- + 2'""' aj Xj-- - Xj+n- -.at L...J ax~ J J at2 L...J aX"+ ax' at

j=l J j=l J n J

This operator is a sum of squares of 2n "horizontal" vector fields, and it is thereforenot elliptic, although it is hypoelliptic (see Hormander [HI]), i.e., the solution u ofL).u = f is smooth whenever f E COO(Hn ) provided A 1= LJ=1 (2kj + l)aj,where (kl, , kn) E (z+)n (see [BCT] and [CTl]). It is easy to see that the vectorfields Xl, , X2n and T = ft and the operator L).. are left-invariant with respectto the Heisenberg translation. The operator La arises naturally when the operatorDb = aba; + a;ab acts on (0, q)-forms on the group Hn . Here 3b is the tangential

Cauchy-Riemann operator and a; is the Hilbert space adjoint of abo Moreover, thegroup Hn can be identified as the boundary of the Siegel upper space:

which plays a very important role in analysis on strongly pseudoconvex domains.Here we are interested in the fundamental solution of the Hermite operator

Hermite Operator on the Heisenberg Group 39

2n 2n

Ha =a+ L(,BJxJ - X]) + i).T = L:). +a+ L ,BJxJ}=1 }=1

in Hn . More precisely, we are looking for a distribution Ka (x, y) such that

[ " + %; (P]X] - xl) + ;),T] K.(x, Y) ~ 8(x - yl. (1)

Since the operator Ha is not invariant under the Euclidean group action or theHeisenberg group action, we have to compute the fundamental solution with singularity at any point y. One way to achieve this goal is to take the Fourier transform withrespect to the t-variable (which is the center ofthe group) and reduce the operator asa Hermite operator in JR2n with parameters a, ). and the dual variable r of t. Using theresults for Hermite operator (see, e.g., [B] and [CT3]) and inverse Fourier transform,one may solve the problem.

Here we shall use the Hamilton-Jacobi theory to find the heat kernel for the heatequation with potentials. The idea is to write down the Euler-Lagrange system ofequations for the Lagrangian and to characterize the system qualitatively from theconservation law point of view. In general, these systems cannot be solved explicitly.For simple equations, one may characterize the solutions by finding the first integralsof motions. We will demonstrate this method by solving the operator with potentialV (x) = ,B2x 2. Then we shall give some examples and comments for general potentialsat the end of Section 2. For more details, readers can consult the book [Ce]o

2 Heat kernel for the Hermite operator in 1R.2

We start with the Hermite operator

where). E JR+ is a nonnegative real parameter. We associate the Hamiltonian functionas half of the principal symbol

(2)

The Hamiltonian system is

. 2X = H~ = I; and I; = - Hx = ). x.

As we are interested in finding the geodesic between the points XQ, x E JR, x(s) willsatisfy the boundary problem

x=).2x withx(O)=xQ,x(t)=x.

40 O. Calin, D.-C. Chang, and J. Tie

The conservation of energy law is

where E is the energy constant. This can be used to obtain an ODE for the solution xes):

dx J dx- = 2E + A2X2 ===} = ds.ds J2E + A2X2

Integrating between s = 0 and s = t, with x(O) = Xo and x(t) = x, yields

lx

--;==d=u:::::;;:::::;;: = t {:::::::} 1v d v = At,XQ J2E + A2u2 VQ JI+V2

with v = & and Vo = jIT. Integrating yields

which is equivalent to

v = sinh(sinh-1(vo) + At)

{:::::::} v = vocosh(At) + cosh(sinh-1(vo» sinh(At)

{:::::::} v= Vo cosh(At) + J1 + v5 sinh(At).

Hence

AX AXo~= ~cosh(At)+

v2E v2E

and

It follows thatA(X - xo cosh(At» / -; 2--s-inh-(A-t-)-- = V2E + A Xo'

Solving for E yields

A2(x - xo cosh(At»2 2 22E = sinh(At)2 - A Xo

A2(x2 - 2xxo cosh(At) + x5 COSh(At)2 - x5 sinh(At)2)= sinh(At)2

A2(x2 + x5 - 2xxo cosh(At»= ----"----------;;------sinh(at)2

(3)


Proposition 1. The energy along a geodesic derived from the Hamiltonian (2) between the points Xo and x is

).2(x2 + x5 - 2xxo cosh(M))E = 2 sinh(at)2 .

Making Xo = 0, we obtain the following result.

Corollary 1. The energy along a geodesic derived from the Hamiltonian (2) joiningthe origin and x is given by

).2x 2

E- ~- 2 sinh(M)2 .

We note that if we take the limit a -+ 0 in (3), we obtain the Euclidean energy

. .).2t2 (x 2 + x5 - 2xxo cosh(M))hm E = hm --~----'<.----;;:----

'----+0 '----+0 sinh(M)2 2t2

2.1 The action function

Let S = S(xo, x, t) be the action with initial point Xo and final point x, within time t.The action satisfies the Hamilton-Jacobi equation

a,s + H(VS) = O.

One may note that

1 1 1H(~, x) = _(~2 - ).2x 2) = _x2 - _).2x 2 = E,

222

and hence a,s = - E. Using (3) yields

as ).2(x2 +X5 - 2xxocosh(M))

at 2 sinh().t)2

). 2 2 a a 1= -(x + xo)- coth(M) - ).xxo---,---

2 at at sinh(M)

a [). 2 2 ).xxo ]= - -(x +xo)coth().t) - . .at 2 SInh().t)

Hence we arrived at the action

a [2 2 2xxo ]S(xo, x, t) = - (x + xo) coth(M) - -.--2 slnh(M)). 1- . [(x 2 + x5) cosh().t) - 2xxol2 slnh(M)

One may note easily that

1· S (x - xo)21m = ,

'----+0 2t

which is the Euclidean action.

(5)

42 O. Calin, D.-C. Chang, and 1. Tie

Lemma 1. We have

(i) (axS)2 = A2X 2 + 2E,

(ii) a;s = Acoth(M).

Proof

(i) Differentiating in (5) with respect to the x-variable yields

Aaxs =. (x cosh(M) - xo), (6)

smh(M)

2 A2(x 2cosh2(M) + xJ - 2xxo cosh(M»)(axS) = 2

sinh (M)

A2(x2 + x 2 sinh2(M) + xJ - 2xxo cosh(M)

sinh2(M)

2 2 A2(x2+xJ-2xxocosh(M))=ax+----"------=-----

sinh2 (M)

(ii) Again, differentiating in (6) with respect to the x-variable yields

Aa;s=. cosh(M) = Acoth(M).

slnh(M)o

We shall look for a fundamental solution of the heat equation which has a representation as follows.

K (xo, x, t) = V (t)ekS(xo,x,t) , (7)

where V (t) will satisfy a volume function equation and k is a real constant. Lemma Iprovides

atK = V'(t)ekS + V(t)kekSatS

= ekS(V'(t) - kV(t)E).

Hence

axekS = kekSaxS,

a;ekS = k2ekS (axS)2 + kekSa;S

= kekS [k(axg)2 + a;S]

= kekS[k(A2x 2 + 2E) + Acoth(M)].

We shall find the heat kernel using a multiplier method. Let

P = at - a; +exA2x 2, (8)

where ex is a real multiplier, which will be determined such that P K (xo, x, t) = 0 forany t > 0:

Hennite Operator on the Heisenberg Group 43

P K(xo, x, t) = ekS(V'(t) - kEV(t))

- kekS (k()..2x 2 + 2E) + )..coth(M))V(t)a)..2x 2ekS V (t)

= ekSV(t) [VI(t) _ kE _ k2()..2x 2 + 2E) - kA coth(M) + a)..2x 2]V(t)

= ekSV(t) [VI(t) _ kE _ k2)..2x 2 - 2k2E + a)..2x 2 - ka coth(M)]V(t)

= ekSV(t) [VI(t) _ kE(2k + 1) + (a - k2))..2x 2 - kacoth(M)].V(t)

In order to eliminate the middle two terms in the brackets, we choose k = -! and

a = ~. Let f3 = ~ > O. Then the operator (8) becomes

(9)

and

P K (xo, x, t) = K (xo, x, t) (VI (t) + f3 coth(2f3 t)) .V(t)

We shall choose V(t) such that

V'(t)-- = -f3 coth(2f3t) , t > O.V(t)

Integrating yields

1 . CIn V(t) = --In(slllh(2f3t)) ==> V(t) = -r=====

2 y'sinh(2f3t)

Using the action (5), the fundamental solution formula (7) becomes

K ( t) - C - ¥ sinh(12fJt) [(x2+x5) cosh(2tlt)-2xxoJxo, x, - ey'sinh(2f3t)

,..-----

_~ 2f3t e-;t,'Slll~r{fJt)[(x2+X5)cOSh(2tlt)-2xxoJ- y'2f3t sinh (2f3 t) .

We shall find the constant C investigating the limit case f3 -+ 0, when the operator

(9) becomes the usual one-dimensional heat operator at - a;. As si~1;tlt) -+ 1, theabove fundamental solution becomes

C 1 ( )2K(xo x t) '" --e 4t x-xo f3 -+ O.

"y'2f3t '

By comparison with the fundamental solution for the usual heat operator, which is

1 1 ( )2--e4t x-xo.J4rrt '

we find C = j:f;. We arrive at the following result.


Theorem 1. Let f3 ~ O. The fundamental solution for the operator P = at - a; +f32 x 2 is

1 2 Rt I 2fJt (2 2 hK( t) . P e- 4t sinh(2fJt) [x +xo) cos (2,Bt)-2xxolxo, x, = r:t=

V 4m slllh(2f3t)

1

v0Ji

f3 fJ(x 2+xij) cosh(2fJt)-2fJxxo. e 2siDb(2fJtj

slllh(2f3t)t > O.

The computations are similar in the case when f3 = -iy. Using cosh(iyt) =cos(yt) and sinh(2iyt) = i sin(2yt), we obtain a dual theorem.

Theorem 2. Let y ~ O. The fundamental solution for the operator P = at - a; y2x 2 is

1K(xo, x, t) = r:t=

V47Tt. 2y t e- -it sin~~~t) [(x2+xij) cos(2yt)-2xxol

slll(2yt)

t > O.y(x2+xij) cos(2ytj-2yxxO

2sin(2yt)y

. eslll(2yt)

1=

v0Ji

2.2 The harmonic oscillator ch - L:j=l (a;j ± 1 }xJ)

Consider the Hermite operator in jRn

where Aj ~ 0 for j = 1, ... , n. The associated Hamiltonian is

with the Hamiltonian system

Xj = H~j = ~j and ~j = -Hxj = A]Xj, j = 1, ... ,n.

The geodesic x(s) starting at Xo = (x?, ... x~) and having the final point x =(Xl, ... , xn ) satisfies the equations

j = I,. .n.

As in the one-dimensional case, we have the law of conservation of energy

xJ(s) - A]XJ(S) = 2Ej, j = 1, ... , n,


where Ej is the energy constant for the jth component. The total energy, which isthe Hamiltonian, is given by

Proposition 1 yields

and hence

The action between XQ and x in time t satisfies the equation fr S = - E or

Hence we shall choose

LetA· 1

Sj=-.L. [(x;+(xJ)2)cosh(Ajt)-2xjxJ]. (11)2 smh(Ajt)

Then S = S! + ... + Sn and 3xj S = 3xj Sj. Then Lemma 1 yields

n n n n

L(3xj S)2 = L(3xj Sj)2 = L(A]X; + 2Ej) = LA]X; + 2E.j=! j=! j=! j=!

Hencen n n

L 3;j S = L 3;j Sj = LAj cothO· jt).j=! j=! j=!

We shall look for a kernel of the form

K(xQ, x, t) = V (t)ekS(xo,x,t) , k E R (12)


A computation similar to the one-dimensional case yields

~K = ekS(V'(t) - kEV(t)),at

and

and hence

!:l.nekS = kekS

jk t[A]X] + 2Ej + Aj COth(Ajt)]} .J=l

In order to find the kernel for the heat operator, we employ the multiplier methodagain. We shall consider the parabolic operator

n

Pn = at - ~)a;j - ajA]X]),j=l

where a is a multiplier subject to be found later. Then

PnK = ekS[V'(t) - kEV(t)]

- kekS Ik (tA]X] + 2Ej) + tAj coth(Aj) V(t)l J=l J=l J

n

+ 'LajA]X]V(t)ekS

j=l

= ekSV(t) [~~) - kE(1 + 2k) + t(aj - k2 )A]X] - k tAj COth(Ajt)l( ) j=l j=l J

kS [V'(t) na ]=e Vet) -- + - coth(at) ,Vet) 2

where we choose k = -! and al = ... = an = i. Let f3j = ~ ~ 0 and chooseV (t) satisfying

V'(t) n-- = - 'LAj coth(2f3jt), t > O.Vet) j=l

Integrating yields Vet) = flnJ'-l . l~j • Hence the fundamental solution for the- 8mh (2f3jt)

operator Pn = at - L:J=l (a;j - A]X]) expressed in the form (12) is

Hennite Operator on the Heisenberg Group 47

When f3 j ---* 0, j = 1, ... , n, we should obtain the kernel of the heat operator

at - L:J=1 a;j' which is

I e- lr Ix-xol2 0/2 ' t> .

(4rrt)n

By comparison, we obtain the value

Theorem 3. Let f3 j ::: 0 for j = 1, ... , n. The fundamental solution for the operator

Pn = at - L:J=1 (a;j - f3JxJ) is

for t > O.

Theorem 3 recovers a result in [CGr] (see also [H2, Chapter 6]). In a similar wayas in the one-dimensional case, choosing f3 j = -iYj yields the following result.

Theorem 4. Let Yj ::: 0 for j = 1, ... , n. Then the fundamental solution for theoperator P = at - L:J=1 (a;j + yJxJ) is

( )

1/21 n 2Yjt

K(xQ, x, t) =(4rrt)n/2 Dsin(2Yjt)

{1 ~ 2y·t 2 }

X exp -- L.,. j [(xJ + (x7) ) cos(2Yjt) - 2XjX7]4t. sm(2Yj,t)

j=1

48 o. Calin, D.-C. Chang, and J. Tie

fort> o.

Remarks.

(1) We should point out here that the method we used in this section is based onHamilton-Jacobi theory which provides deep insight into the geometry inducedby the operator. Using this method, we not only can construct the heat kernelfor the Hermite operator. We can also construct fundamental solutions for moregeneral operators. For example, we can find the fundamental solution for the heatequation with quartic potential, i.e.,

2 1 4 4P = at - a - -).. x with)" :::: O.x 4

This is a quite different behavior than the quadratic potential case (the Hermiteoperator), where there is only one energy and one solution between two givenpoints. However, there are infinitely many energies associated to the quartic case.This makes the operator P very difficult to invert.

However, given any two points XQ and x and a time t > 0, there is a sequence ofenergies En = En (XQ, x, t) which have explicit representations. For each energywe associate an action Sn = Sn(xQ, x, t), which satisfies the Hamilton-Jacobiequation

We can show that

S ""' ~ (nK)4 ~ asn --+ 00.n 3 2a t3

For each action Sn we associate a volume function Vn.Finally, we can write downthe asymptotic expansion of the kernel

The constants en should be chosen such that

for any compact supported function q;. Using this method, we may handle evenmore general potentials. Readers may consult [CC, Chapter 10].

(2) This method can be used to solve a more general operator such as


The fundamental solution for the operator P is given by

(13)

This equation is derived from the famous Euler equation of compressible fluids(see, e.g., [L] and [LZ]).

(3) If one replaces t by it, the heat operator becomes a SchrOdinger operator. Thepropagator for the Schr6dinger operator P= ihat + !h2ax - !,B2x 2 is

K(x, xo, t, to) =

where t - to > O.

.h R. i [ 2 2 2xxo ]I P e'Iii fJ(x +xo ) cot(fJ(t-to))- sin(!J(t to))

4n sin(a(t - to» ,

3 Hermite operator on the Heisenberg group

In this section, we will calculate the fundamental solution Ka for the Hermite operatoron the Heisenberg group. First, note that for j = 1, ... , n,

It follows that

Here a j = a j +n for j = 1, ... , n.Since the operator X] +X]+n is invariant under rotations in the (x j , x j+n) plane,

so the fundamental solution Ka is also invariant under rotations in the (x j, Xj+n) plane(see [CT2] and [H2]). Therefore, we may assume that Ka(x, t; y, s) is independentof the rotational variables. It follows that

aKa aKaXj---Xj+n--=O, j=l, ... ,n.

aXj+n aXj


Denote2n 2n [ a2 a2 ] a

R a = a + Lf3JxJ - L -2 +4aJxJ-z - iA-.. 1 . 1 aX. at atJ= J= J

Taking the partial Fourier transform of Ha with respect to the t-variable, one has

2n 2n[2 ]~ 22 a 222R a = ~ f3·x. - ~ - -4a·x·T +a +hL.J J J L.J ax~ J Jj=l j=l J

2n [ 2 ]= a + AT + ~ (4a~T2 + f32)x~ - ~L.J J J J ax~j=l J

2n [ a2

]= a+~ A2x 2 - - .L.J J J ax~j=l J

This is exactly the Hermite operator on 1R2n with parameter Aj = J 4aJT2 + f3J and

a = a + Ar. Recall that aj = aj+n for j = I, ... , n. Hence the exceptional set ofR a is

A =I(a, A) : a + AT = - tJ4a;T2 + f3;(2k j + I); k E (Z+)2n}.J=l

Now using Theorem 3 (see also [CC, Chapter 4] and [CT3, Theorem 2.1]), one canderive the fundamental solution of the Hermite operator by integration of the heatkernel.

Theorem 5. Fora 1- {- LJ=l Aj(2kj+ 1), k = (kl, ... , kn) E (z+)n}, the Hermite

operator Ha = a - ~ + LJ=l A]X; has fundamental solution

where Ps(x, y) is defined as in Theorem 3.

~ Using Theorem 5, we know that for (a, A) 1- A, the fundamental solution forR a is

~ 1 100[ 2n A' ] !Ka(x, T; y) = -- e-(a+Ar)S Tl. J

(2rr)n 0 j=l smh(2Ajs)

I 2n [2 ]}A' (x' - y.)xexp -L--.L .J J + (xJ +y;)tanh(Ajs) ds.

. 2 slnh(2AJ'S)J=l


Next, we can find the fundamental solution for the operator 'Ha with singularpoint (y, 0) by taking the inverse Fourier transform with respect to r. We note that

Aj = J4a;r 2 + f3; and it depends on r:

I /00 irt ~Ka(x, t; y, 0) = - e Ka(x, r)dr2n -00

1 /00 iOO [ 2n A ] i= eitr e-(a+Ar)s j(2n)n+1 -00 0 Dsinh(2Ajs)

I 2n [2 ]}A' (x' - y.)xexp -L:---.l.. .) ) + (x; +y;)tanh(Ajs) dsdr

. 2 slnh(2A)'S))=1

I

1

/

00 iOO [ 2n (4a~r2 + R~) i ] 2:- itr-(a+Ar)s n ) p)

- (2n)n+1 -00 0 e j=1 sinh(2s(4a;r2 + f3;)i)

{

2n (4a2r2+f3~)i [ (x._y.)2x exp - L:)) ))

j=1 2 sinh(2s(4a;r2 + f37)i)

+ (x; + y;) tanh(s(4a;r 2 + f37) i)]} dsdr.

The fundamental solution at any point (y, s) is

Ka(x, t; y, s) = Ka(x, t - s; y, 0).

The above formula is very complicated and cannot be simplified in the general case.

3.1 Fundamental solution for the isotropic Heisenberg sub-Laplacian

It is very difficult to integrate the above integrals for arbitrary f3 j and a j' It can besimplified in some special cases. For example, if f3 j = 0 for all j and ex = O. We canfind the fundamental solution of the sub-Laplacian LA with singular point at (0, 0):

Here

n

y(x,s) = L:aj(x7+x7+n)coth(ajs)j=1

andn

a'v(s) = n ) .

sinh(a 's)j=1 )


This is the fundamental solution with singularity at the origin. Consequently, thefundamental solution with a singularity at an arbitrary point can be obtained by theHeisenberg translation.

When Qj = 1 for all j = 1, ... , n, then we may obtain an exact form for thekernel. In this case, the kernel will be

A(n - I)! 100 1 e- 4s

K).(x,t) = +1 n ds8nn

-00 [sinh(s)]n (L]:1 x;coth(s) - it)

A(n - 1)! 100 e- 4S

= ~8nn+1 -00 (lxl 2 cosh(s) - it sinh(s))n '

where Ixl2 = L]:l x;' Denote

r = (lxl4 + t2)~ and e- irp = r-2(lxI2 - it)

with cp E (- ~, ~). Using the identity

cosh(s + icp) = cosh(s) cos cp + i sinh(s) sin cp,

one hasA

(n - I)! 100 e- 4S

K).(x, t) = ds.8nn+1 -00 [r 2 cosh(s + icp)]n

Changing the contour, the formula (14) becomes

A

(n - 1)! f(n) i!"rp100 e- 4S

Ka(x, t) = 2 e 4 ds.8nn+1 r n -00 [cosh(s)]n

The above integral can be evaluated as follows:

(14)

From the above formula, we know that the kernel can be extended from IRe().,) I < nto the region C \ fa, where

fa = {±(n + 2k) : k E Z+}.

This coincides with the result obtained by Folland and Stein [FSt] except for a constantfactor, since our vector fields have different constants in front of it. In particular, when)., = 0, one has

Ko(x, t) = Cn (lxl2 + it)-J. (Ixe - it)-J. = Cn (lxl4 + t 2)-I.

In fact, the above result can be generalized to the nonisotropic case by using Laguerrecalculus; see, e.g., [B] and [BGr].


3.2 Connection with the Grusin operator

Consider the operator

1 [( a a )2 (a a )2]£ = - - + 2X2 - + - - 2Xl -2 aXl at aX2 at

on the Heisenberg group. The fundamental solution for £ with singularity at theorigin is

1 /00 csch(2r)drK(Xl,x2,t;0,0,0) = 2" 2 2 .

n -00 (Xl + x2 ) coth(2r) - it

SetXl = X, X2 = z,

Then the operator £ transforms to

Ll = ~ [(~)2 + (x~ + ~)2] .2 ax ay az

The operator Ll is translation invariant in y and z. Hence it suffices to have thesingularity at (xo, 0, 0).

(xo, 0, 0)-1 . (X, z, 2XlX2 - 4y) = (X - Xo, z, 2(xo + x)z - 4y).

Therefore,

i n-2csch(2r)drK(x, z, y; Xo, 0, 0) = .

lR [(x - xO)2 + z2] coth(2r) - i[2(xo + x)z - 4y]

Note that a parametrix for !::>.G may be obtained from a parametrix for Ll, using theHadamard method of descent, by integrating the parametrix for!::>. with respect to z.Recall

i d).. 2n sgn(a)2 - , ).. E JR,

lR a).. + bA + c v'4ac - b2

if a =1= °and a)..2 + b)" + c =1= 0. Hence

i 2[n2 sinh(2r) cosh(2r)]-lj2drK(x, Xo, y) =

lR J(x - xO)2 coth(2r) + 4iy + (x + xO)2 tanh(2r)

2 i [sinh(r) cosh(r)]-lj2dr-- (l~- n lR J(x - xO)2 coth(r) + 4iy + (x + XO)2 tanh(r) .

This is the fundamental solution for the step 2 Grusin operator

1 2 2LlG = 2(Xl + X2 ),

where Xl = ax and X2 = xay • For further discussion, see [CCGrKl] and [CCGrK].

54 O. Calin, D.-C. Chang, and 1. Tie

Acknowledgments Part of this article is based on the lecture presented by the second authorat the conference on "A Celebration of Carlos Berenstein's Mathematics: Harmonic Analysis,Signal Processing and Complexity," which was held on May 17-29,2004 at George MasonUniversity, Fairfax, VA. We would like to thank Professor Daniele Struppa for the invitation.We would also like to thank the organizing committee for the warm hospitality during theconference.

References

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[CCGrK] O. Calin, D. C. Chang, P. Greiner, and Y. Kannai, Heat kernels for highly degenerateGrusin operators, in preparation, 2004.

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[CTl] D. C. Chang and J. Tie, Estimates for spectral projection operators of the subLaplacian on the Heisenberg group, J. Anal. Math., 71 (1997), 315-347.

[CT2] D. C. Chang and J. Tie, Estimates for powers of sub-Laplacian on the non-isotropicHeisenberg group, J. Geom. Anal., 10 (2000), 653-678.

[CT3] D. C. Chang and J. Tie, A note on Hermite and subelliptic operators, Acta Math.Sinica, 2004, to appear.

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[LZ] Y. Zeng and T. P. Liu, Large Time Behavior ofSolutions for General QuasilinearHyperbolic-Parabolic Systems of Conservation Laws, Memoirs of the AmericanMathematical Society 125, American Mathematical Society, Providence, RI, 1997.

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Hermite Operator on the Heisenberg Group