GENERALIZED HAMILTON-JACOBI EQUATION AND HEAT KERNEL ON STEP TWO NILPOTENT LIE GROUPS

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 nilpotent Lie groups. Using our results, we obtain explicit formulas of the heat kernels for the sub-Laplacian on H-type groups satisfying the J2 _{condition and Grusin}

operator in R2_.

1. Introduction Recall that the Laplace operator on Rn,

∆ = 1 2 n X j=1 ∂2 ∂x2 j

has the heat kernel

Pt(x, x0) = 1

(2πt)n2

e−|x−x0|

2 2t .

Given a general second order elliptic operator in n dimensions, ∆X = 1₂

n

X

j=1

X_j2+ · · ·

where the {X1, . . . , Xn} is a linearly independent set of vector fields and + · · · stands for lower

order terms, the heat kernel takes the form

Pt(x, x0) = 1

(2πt)n2

e−d2(x,x0)2t ¡a₀+ a₁t + a₂t2+ · · ·¢.

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

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

∂ ∂t d2 2t + 1 2 n X j=1 ³ Xjd 2 2t ´₂ = 0, i.e., d2

2t is a solution of the Hamilton-Jacobi equation.

Now let us move to subelliptic operators. We first study the famous example: Heisenberg sub-Laplacian on H1 (1.1) ∆X = 1₂³ ∂_∂x 1 + 2x2_∂y∂ ´₂ + 1 2 ³ ∂ ∂x₂ − 2x1 ∂ ∂y ´₂ .

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 The author is partially supported by a Hong Kong RGC competitive earmarked research grant #600607 and a competitive research grant at Georgetown University.

The third author is supported by grants of the Norwegian Research Council # 177355/V30 and #180275/D15.

We shall try for a heat kernel in the form 1 tqe− f t · · · where h = f_t is a solution of ∂h ∂t + 1 2 ³ ∂h ∂x1 + 2x2 ∂h ∂y ´₂ +1 2 ³ ∂h ∂x2 − 2x1 ∂h ∂y ´₂ = 0. Thus we start with

(1.2) ∂h ∂t + H(x, ∇h) = 0, where (1.3) H = 1 2 h¡ ξ1+ 2x2η ¢₂ +¡ξ2− 2x1η ¢₂i = 1 2 £ ζ₁2+ ζ₂2¤,

where ξ1, ξ2 and η are dual to x1, x2 and y respectively. Using the Lagrange-Chapit method,

let us look at the following equation:

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

where ξ = ∇_xh = ¡∂h ∂x1,∂x2∂h

¢

, η = ∂h

∂y and γ = ∂h∂t. We shall find the bicharacteristic curves

which are solutions to the following system:

˙x1 = Fξ1 = ξ1+ 2x2η = ζ1, ˙x2 = Fξ2 = ξ2− 2x1η = ζ2, ˙y = F_η = 2 ˙x₁x₂− 2x₁˙x₂, ˙t = F_γ= 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,

γ(s) =γ = −H = constant, η(s) =η = −H = constant,

t(s) =s,

constant meaning “constant along the bicharacteristic curve”. Also

H = 1 2˙x 2 1+ 1 2˙x 2 2= E = energy.

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.

(1.4)

Therefore, ¨x1˙x1+ ¨x2˙x2= 0, and E =constant. Next from (1.4), we obtain

..._x

Hence ˙x1(s) = ˙x1(0) cos(4ηs) + x¨1_4η(0)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) + x¨2_4η(0)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) = x₁ and x₂(t) = x₂, so

1 2ζ1(0) sin(4ηt) + 1 2ζ2(0) ¡ 1 − cos(4ηt)¢= 2η¡x1− x1(0) ¢ − 1 2ζ1(0) ¡ 1 − cos(4ηt)¢+1 2ζ2(0) sin(4ηt) = 2η ¡ x2− x2(0) ¢ , or, ζ1(0) cos(2ηt) + ζ2(0) sin(2ηt) = 2η¡x1− x1(0) ¢ sin(2ηt) − ζ1(0) sin(2ηt) + ζ2(0) cos(2ηt) = 2η¡x2− x2(0) ¢ sin(2ηt) (1.9)

Hamilton’s equations give

ξ2(s) = − 2ηx1(s) + ¡ ξ2(0) + 2ηx1(0) ¢ = − 2ηx1(0) − 1₂ζ1(0) sin(4ηs) − 1₂ζ2(0) ¡ 1 − cos(4ηs)¢+ ζ2(0) + 4ηx1(0) = 2ηx1(0) − 1₂ h ζ1(0) sin(4ηs) − ζ2(0) ¡ 1 + cos(4ηs)¢i, and ξ1(s) = −2ηx1(0) +1₂ h ζ1(0) ¡ 1 − cos(4ηs)¢+ ζ2(0) sin(4ηs) i .

We need to find the classical action integral

S(t) =

Z _t

0

¡

ξ · ˙x + η ˙y − H¢ds.

The above calculations imply

ξ1˙x1+ ξ2˙x2 = − 2η ˙x1(s)x2(0) + 2ηx1(0) ˙x2(s) +1₂ ¡ ζ₁2(0) + ζ₂2(0)¢¡1 + cos(4ηs)¢ = − 2η¡˙x1(s)x2(0) + x1(0) ˙x2(s) ¢ +¡1 + cos(4ηs)¢E,

and _Z t 0 ¡ ξ · ˙x + η ˙y − H¢ds = η h y − y(0) + 2¡x1(0)x2− x1x2(0) ¢ +sin(4ηt) 4η2 E i .

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

E = 1 2ζ 2 1(0) + 1 2ζ 2 2(0) = 2η2 |x − x0|2 sin2(2ηt). Hence, S(t) = Z _t 0 ¡ ξ · ˙x + η ˙y − H¢ds =η h y − y(0) + 2¡x1(0)x2− x1x2(0) ¢ + |x − x0|2cot(2ηt) i .

We note that x, y, t, x0 and 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

where

(1.10) h(t) = η(0)y(0) + S(t), i.e., h(0) = η(0)y(0).

This yields ∂h ∂t + H ³ x, y, ∇xh,∂h_∂y ´ = 0. We have the following theorem.

Theorem 1.1. We have shown that

h =η(0)y(0) + Z _t 0 ¡ ξ · ˙x + η ˙y − H¢ds =ηy + 2η¡x₁(0)x₂− x₁x₂(0)¢+ η|x − x₀|2cot(2ηt) (1.11)

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.

Before we move further, let us consider a more general situation. 2. Generalized Hamilton-Jacobi equations

We consider an (n + m)-dimensional space Rn_{× R}m_{. The coordinates are denoted x =}

(x1, . . . , xn) ∈ Rnand y = (y1, . . . , ym) ∈ Rm. The roman indices i, j, k, . . . will vary from 1 to

n and the Greek indices α, β, . . . will vary from 1 to m.

We have the following theorem. Theorem 2.1. Set (2.1) h(t; x, y, ξ, η) = m X 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(s; x, y, ξ, η) =

Z _s

0

³

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 ³ 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) = Z _t 0 h_Xn k=1 ¡ ∂ξ_k ∂x_j dxj ds + ξk d ds ∂x_k(s; x, y, ξ, η; t) ∂x_j ¢ + m X α=1 ¡ ∂ηα ∂x_j dyα ds + ηα d ds ∂yα(s; x, · · · ; t) ∂x_j ¢ − n X k=1 ∂H ∂ξk ∂ξ_k ∂xj − m X α=1 ∂H ∂ηα ∂η_α ∂xj − n X k=1 ∂H ∂xk ∂xk(s; x, y, ξ, η; t) ∂xj − m X α=1 ∂H ∂yα ∂yα(s; x, y, ξ, η; t) ∂xj i ds = Z _t 0 d ds ³_Xn k=1 ξk∂xk(s; x, y, ξ, η; t)_∂x j + m X α=1 ηα∂yα(s; x, y, ξ, η; t)_∂x j ´ ds = n X k=1 ξ_k(s)∂xk(s; x, y, ξ, η; t) ∂x_j ¯ ¯ ¯s=t s=0+ m X α=1 ηα(s)∂yα(s; x, y, ξ, η; t)_∂x j ¯ ¯ ¯s=t s=0. It follows that ∂S ∂xj(t; x, y, ξ, η) = ξj(t) − m X α=1 ηα(0)∂yα(0; x, y, ξ, η; t)_∂x j . Similarly, for β = 1, . . . , m, ∂S ∂y_β(t; x, y, ξ, η) = ηβ(t) − m X α=1 ηα(0)∂yα(0; x, y, ξ, η; t)_∂y β , β = 1, . . . , m. Moreover, ∂S ∂t(t; · · · ) = n X k=1 ξk(t; · · · ) ˙xk(t; · · · ) + m X α=1 ηα(t; · · · ) ˙yα(t; · · · ) − H ¡ x, y, ξ(t; · · · ), η(t; · · · )¢ + n X k=1 ξ_k(s; · · · )∂xk(s; · · · ) ∂t ¯ ¯ ¯s=t s=0+ m X α=1 ηα(s; · · · )∂yα(s; · · · )_∂t ¯ ¯ ¯s=t s=0. Differentiating x1= x1(t; x, y, ξ, η; t) yields 0 = d dsx1(t; x, y, ξ, η; t) = ˙x1(t; · · · ) + ∂x1(t; x, y, ξ, η; t) ∂t ¯ ¯ ¯ s=t.

On the other hand, one has

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

and η_α(s; · · · )∂yα(s; · · · ) ∂t ¯ ¯ ¯s=t s=0 = −ηα(t; · · · ) ˙yα(t; · · · ) − ηα(0; · · · ) ∂y_α(0; · · · ) ∂t , α = 1, . . . , n, therefore, ∂S ∂t = −H(t; · · · ) − m X α=1 ηα(0; · · · )∂yα(0; · · · )_∂t . It follows that h(t; x, y, ξ, η) = m X α=1 ηα(0)yα(0) + S(t; x, y, ξ, η), satisfies ∂h ∂xk = ξk(t; x, y, ξ, η; t), k = 1, . . . , n ∂h ∂y_α = ηα(t; x, y, ξ, η; t), α = 1, . . . , m, and ∂h ∂t + H ³ x, y, ξ(t), η(t) ´ = 0 ⇒ ∂h ∂t + H ³ 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

³

x, y, ∇xh, ∇yh

´

being (1.3). In particular we did not assume that η_α(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 is a dual variable 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), (1.7), (1.6), (1.8), one has

˙y = 2 h ˙x1x2(0) − x1(0) ˙x2+ 1₂ ¡ ζ₁2(0) + ζ₂2(0)¢ 1 − cos(4ηs) 2η i , and y(s) = 2¡x1(s)x2(0) − x1(0)x2(s) ¢ + E 4η2 ¡ 4ηs − sin(4ηs)¢+ C. At s = t, one has x1(t) = x1, x2(t) = x2 and

y = 2¡x1x2(0) − x1(0)x2 ¢ + E 4η2 ¡ 4ηt − sin(4ηt)¢+ C. Hence, one has

y(s) =y − 2h¡x1− x1(s) ¢ x2(0) − x1(0) ¡ x2− x2(s) ¢i − E 4η2 £ 4η(t − s) − (sin(4ηt) − sin(4ηs))¤. At s = 0, y(0) = y + 2¡x₁(0)x₂− x₁x₂(0)¢+ |x − x₀|2µ(2ηt), where we set µ(φ) = φ sin2φ− cot 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−1 _{is a many valued}

function. 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.

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)_λ ; λt ¢ , j = 1, . . . , n yα(s; x, x0, y, ξ, η(0); t) = yα ¡ λs; x, x0, y, ξ, η(0) λ ; λt ¢ , α = 1, . . . , m ξ_j(s; x, x₀, y, ξ, η(0); t) = λξ_j¡λs; x, x₀, y, ξ,η(0) λ ; λt ¢ , j = 1, . . . , n η_α(s; x, x₀, y, ξ, η(0); t) = λη_α¡λs; x, x₀, y, ξ,η(0) λ ; λt ¢ , α = 1, . . . , m (2.2)

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

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 ; λt ¢ = λ∂H ∂ξj ¡ x₁¡λs; x, x₀, y, ξ,η(0) λ ; λt ¢ , x₂¡λs; x, x₀, y, ξ,η(0) λ ; λt ¢ , . . .¢ =∂H ∂ξ_j ¡ ˜ x(s), ˜y(s), ˜ξ(s), ˜η(s)¢, since ∂H_∂ξ

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

calculations yield ∂ ˜yα ∂s = ∂H ∂ηα , ∂ ˜ξj ∂s = − ∂H ∂xj , ∂ ˜ηα ∂s = − ∂H ∂yα ,

and clearly for j = 1, . . . , n, ˜ xj(0) = xj ¡ 0; x, x0, y,η(0) λ ; λt ¢ = 0, x˜j(t) = xj ¡ λt; x, x0, y,η(0) λ ; λt ¢ = xj, and for α = 1, . . . , m ˜ yα(t) = yα(λt; x, x0, y, ξ,η(0)_λ ; λt ¢ = yα, ˜ ηα(0) = ληα(0; x, x0, y, ξ,η(0)_λ ; λt ¢ = ληα(0) λ = ηα(0).

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) =dx_dsj(s; x, x0, y, ξ, η(0); t) =dxj ds ¡ λs; x, x0, y, ξ,η(0)_λ ; λt ¢ = λ ˙xj ¡ λs; x, x0, y, ξ,η(0)_λ ; λt ¢ .

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

Z _t 0 £ ξ(s) · ˙x(s) + η(s) · ˙y(s) − H(x(s; . . .), . . .)¤ds = Z _t 0 £ λξ¡λs; x, x0, y,η(0)_λ ; λt ¢ · λ ˙x(λs; . . .) + m X α=1 ληα(λs; . . .) · λ ˙yα(λs; . . .) − λ2H(x(λs; . . .), . . .)¤ds =1 λ Z _t 0 £ λ2 n X k=1 ξk(λs; . . .) ˙xk(λs; . . .) + λ2 m X α=1 ηα(λs; . . .) ˙yα(λs; . . .) − λ2H(x(λs; . . .), . . .) ¤ d(λs) = λ Z _t 0 £ ξ¡s0; x, x0, y,η(0)_λ , λt ¢ · ˙x(s0; . . .) + η(s0; . . .) · ˙y(s0; . . .) − H(x(s0; . . .), . . .)¤ds0 = λS¡x, x₀, y, ξ,η(0) λ , λt ¢ . Also, m X α=1 ηα(0) ˙yα(0; x, x0, y, ξ, η(0); t) = λ m X α=1 η_α(0) λ yα ¡ 0; x, x0, y, ξ, η(0) λ ; λt ¢

and the proof of the corollary is therefore complete. ¤

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 X α=1 ηα(0)_∂η∂f α(0) + H ³ 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, ∂h ∂t = − 1 t2f + 1 t m X α=1 ηα(0)_∂η∂f α(0)

on one hand. On the other hand,

∂h ∂t = −H ¡ x, y, ∇xh, ∇yh ¢ .

The two right hand sides agree for all t so we may as well set t = 1 which yields the proposition. ¤ 2.3. Laplace operator. The Laplace operator in Rn _{is ∆ =} Pn

k=1 ∂ 2 ∂x2 k. The Hamiltonian function H(ξ) is H(ξ) = 1 2 n X k=1 ξ_k2

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, a constant vector. Then

¨ x = ˙ξ = 0 ⇒ x(s) = ξ(0)s + x0. Moreover, x = x(t) = ξ(0)t + x0 ⇒ ξ(0) = x − x_t 0 and ∂h ∂t = 1 2 n X k=1 ξ2_k= n X k=1 (x_k− x(0)_k )2 2t2 = |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 R2 _{now and the horizontal vector fields X}

1, X2 are given

by

X1 = _∂x∂ , and X2 = x_∂y∂ .

The Grusin operator is given as follows: ∆X = 1₂

³ ∂ ∂x ´₂ + 1₂x2³∂ ∂y ´₂ . It is obvious that ∆X is elliptic away from the y-axis but degenerate on the y-axis. Since [X1, X2] = _∂y∂,

hence {X1, X2, [X1, X2]} spanned the tangent bundle of R2 everywhere. By H¨ormander’s

theorem [16], ∆X is hypoelliptic.

The Hamiltonian function H for the ∆X is

(2.4) H(x, y, ξ, η) = 1

2ξ

2₊1

2x

The Hamilton system can be obtained as follows; ˙x = H_ξ = 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,

constant meaning “constant along the bicharacteristic curve”. Next, ¨ x = ˙ξ = −xη2, so ¨ x + η2x = 0. It follows that

x(s) = A cos(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 (2.5) 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_t when x = ±x0; they are η = (2k+1)π_t if x = x0 and

η0= 2kπ_t if x = −x0. Next, ˙y(s) =ηx2(s) =η h x0¡ 1₂+1₂cos(2ηs) ¢ + 2x0ξ(0)_η sin(ηs) cos(ηs) +¡ ξ(0)_η ¢₂¡ 1 2 − 1 2cos(2ηs) ¢i =d ds n ηh x20 2 ¡ s + sin(2ηs) 2η ¢ +x0ξ(0) η2 sin 2_{(ηs) +} 1 2 ¡ ξ(0) η ¢₂¡ s −sin(2ηs) 2η ¢io =d ds n η 2 h x2₀+¡ ξ(0) η ¢₂i s +1 4 h x2₀−¡ ξ(0) η ¢₂i sin(2ηs) + x0 2 ξ(0) η ¡ 1 − cos(2ηs)¢o.

We replace ξ(0)_η by (2.5) and collect terms with x2 0: x2 0 2 n ηs +1 2sin(2ηs) + ηs cos2_(ηt) sin2(ηt) − 1 2 cos2_(ηt) sin2(ηt) sin(2ηt) − cos(ηt) sin(ηt) ¡ 1 − cos(2ηs)¢o =x20 2 n _ηs sin2_(ηt)− 1 2 cos2_{(ηt) − sin}2_(ηt) sin2_(ηt) sin(2ηs) − cos(ηt) sin(ηt) ¡ 1 − cos(2ηs)¢o = x20 2 sin2_(ηt) n ηs −1 2 £

cos(2ηt) sin(2ηs) + sin(2ηt)¡1 − cos(2ηs)¢¤o

= x

2 0

4 sin2(ηt) n

2ηs −£sin(2ηt) − sin¡2η(t − s)¢¤o.

Also the x2 _terms:

1 4 x2 sin2(ηt) ¡ 2ηs − sin(2ηs)¢,

and the x0x terms:

1 2 2xx0 sin2(ηt) n 1 2 £

sin¡η(2s − t)¢+ sin(ηt)¤− η s cos(ηt)

o . So, ˙y(s) = d ds n _x2 0 4 sin2_(ηt) h

2ηs −£(sin(2ηt) − sin¡2η(t − s)¢¤)Big] + x2 4 sin2(ηt) ¡ 2ηs − sin(2ηs)¢ + 2xx0 4 sin2(ηt) h 1 2 ¡

sin¡η(2s − t)¢+ sin(ηt)¢− η s cos(ηt)

io . Next, S = Z _t 0

(ξ ˙x + η ˙y − H)ds = η(y − y(0)) + Z _t 0 (ξ2− H)ds. Now, ξ2(s) =ξ2(0) 2 ¡ 1 + cos(2ηs)¢− ξ(0)ηx0sin(2ηs) +1₂η2x20 ¡ 1 − cos(2ηs₎¢ = 1 2 £ ξ2(0) + η2x2₀¤ | {z } =H(0) +1 2 £ ξ2(0) − η2x2₀¤cos(2ηs) − ηx0ξ(0) sin(2ηs).

Since H is constant along the bicharacteristic, one has

H = H(0) = 1

2 £

ξ2(0) + η2x2₀¤.

Continuing one has

S = η(y − y(0)) + Z _t 0 £ (ξ2(0) − η2x2₀)cos(2ηs) 2 − ηx0ξ(0) sin(2ηs) ¤ ds = η(y − y(0)) + 1 2 ¡ ξ2(0) − η2x2₀¢ sin(2ηt) 2η + ηx0ξ(0) cos(2ηt) − 1 2η .

We simplify this S − η(y − y(0)) =η2 2 ³ x − x₀cos(ηt) sin(ηt) ´_2sin(2ηt) 2η − 1 2η 2_x2 0 sin(2ηt) 2η + η 2_x 0 x − x₀cos(ηt) sin(ηt) cos(2ηt) − 1 2η =η 4 n³ x − x₀cos(ηt) sin(ηt) ´₂

sin(2ηt) − x2₀sin(2ηt) − 2x₀x − x0cos(ηt) sin(ηt)

¡

1 − cos(2ηt)¢o.

(2.6)

In the bracket {· · · } of (2.6), terms involved x2₀ are

x2₀h³ cos2(ηt) sin2(ηt) − 1 ´ sin(2ηt) + 2cos(ηt) sin(ηt)(1 − cos(2ηt)) i = x2₀³ cos(2ηt) sin(2ηt) sin2(ηt) + 2 cos(ηt) sin(ηt) − cos(2ηt) sin(2ηt) sin2(ηt) ´ = 2x2₀cot(ηt), terms involved x2 _are

x2sin(2ηt)

sin2(ηt) = 2x

2_cot(ηt),

and terms involved x0x are

2xx0 ³ − cos(ηt) sin2_(ηt)sin(2ηt) − 1 − cos(2ηt) sin(ηt) ´ = − 2xx0³ 2 cos 2_(ηt) sin(ηt) + 2 sin(ηt) ´ = − 4xx0 sin(ηt). Hence, {· · · } =2(x2+ x2₀) cot(ηt) − 4xx0 sin(ηt) =£(x + x0)2+ (x − x0)2 ¤ cot(ηt) −(x + x0) 2_{− (x − x} 0)2 sin(ηt) = (x + x0)2 ³ cot(ηt) − 1 sin(ηt) ´ + (x − x0)2 ³ cot(ηt) + 1 sin(ηt) ´

= (x + x0)2cos(ηt) − 1_sin(ηt) + (x − x0)2cos(ηt) + 1_sin(ηt)

= − (x + x0)2tan ¡ ηt 2 ¢ + (x − x0)2cot ¡ ηt 2 ¢ .

Thus S has the following form:

S = η(y − y(0)) −η 4 h (x + x0)2tan¡ ηt₂ ¢ − (x − x0)2cot¡ ηt₂ ¢i .

By Theorem 2.1, we know that

h(t; x, x0, y, η) = ηy(0) + S(t; x, y, η)

= ηy(0) + η(y − y(0)) −η 2 h (x + x0)2tan¡ η₂ ¢ − (x − x0)2cot¡ η₂ ¢i = ηy − η 2 h A2tan¡ ηt 2 ¢ − B2cot¡ ηt 2 ¢i

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

Theorem 2.2, the function

f (x, x₀, y, η) = h(t; x, x₀, y, η) ¯ ¯ ¯ t=1= 1 2 η 2 n 4y − A2tan¡ η 2 ¢ + B2cot¡ η 2 ¢o is a solution of the generalized Hamilton-Jacobi equation

η∂f ∂η + H ³ x, x0, y, ∂xf, ∂yf ´ = f. We set η 2 = eητ,

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

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 eη is pure imaginary.

Proof. Let eη = η₁+ iη₂. An elementary calculation yields

f =1 2 ¡ η1+ iη2 ¢ τ n 4y + sin(2η1τ ) £ (B2− A2) cosh(2η2τ ) + (B2+ A2) cos(2η1τ ) ¤ cosh2(2η2τ ) − cos2(2η1τ ) − isinh(2η2τ ) £ (B2+ A2) cosh(2η2τ ) + (B2− A2) cos(2η1τ ) ¤ cosh2(2η2τ ) − cos2(2η1τ ) o

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

f ≈ 1 2iη2τ n 4y − i2(x2₀+ x2) tanh(2η2τ ) o , and Re(f ) ≈ 1 4(x 2 0+ x2)2η2τ tanh(2η2τ ) → ±∞ as τ → ±∞ as long as x2 0+ x2 6= 0.

(ii). η2 = 0, that is η ∈ R. Then

f = 2η1τ y +1_4sin(2η2η1τ 1τ ) h B2− A2+ (B2+ A2) cos(2η1τ ) i is singular in τ ∈ R when x2 0+ x26= 0, otherwise Re(f ) = f = 2η1τ y −→_|{z} τ →±∞ ±(sgn(y))∞. (iii). η1 6= 0, η2 6= 0. Here f ≈ 1 2η1τ ¡ η₁+ iη₂¢τ n 4y − i(A2+ B2) tanh(2η₂τ ) o as τ → ±∞, and Re(f ) ≈ 2η₁τ y + (x2₀+ x2)¯¯η₂τ¯¯ = |τ |£2(sgn(τ ))η1y + (x20+ x2)|η2| ¤ and choosing x0, x, y so that

we have

lim

τ →±∞Re(f ) = ±∞

which we do not want. This complete the proof of the Lemma. ¤ Following tradition we shall choose

e η = −i 2. Then f = −iτ y + 1 2(x 2 0+ x2)τ coth τ − τ x₀x sinh τ.

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 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) ∗ (x0, y0) = ¡x + x0, y + y0+1 2B(x, x

0₎¢

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 denote the vector fields Xj, j = 1, . . . , n by the formulas:

Xj = _∂x∂ j + n X k=1 m X α=1 aα_jkxk_∂y∂ α where the aα

jk are real numbers. After a change of coordinates, we may assume that the

matrices£aα jk

¤

j,k are skew-symmetric, i.e., aαjk = −aαkj. We are interested in the sub-Laplacian

∆X which can be defined as follows:

∆X = 1₂ n

X

j=1

X_j2

It is easy to see that

(2.7) £Xj, Xk ¤ = 2 n X k=1 m X 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 × m with element £ aα jk ¤

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

Proof. The operator ∆X is hypoelliptic when the vector fields {Xj}n_j=1 satisfy the “first”

bracket generating condition. This implies that we can recover all the _∂y∂_α from the n(n−1)₂ relations (2.7). If we consider £aα

jk

¤

as a matrix with indices α = 1, . . . , m and the couples (j, k) where j < k, this means that this matrix should have rank m. ¤

We may define a Lie group structure on Rn_{× R}m _{with the following group law:}

(2.8) (x, y)◦(x0, y0) = ³ x1+x01, . . . , xn+x0n, y1+y10+ n X j,k=1 a1_jkx0_jxk, . . . , ym+y0m+ n X j,k=1 am_jkx0_jxk ´ .

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

¡ Xjf ¢ (x, y) = ∂ ∂x0 j ¡ f ◦ L_(x,y)¢(x0, y0) ¯ ¯ ¯ x0_=0,y0₌₀ where L_(x,y)(x0, y0) = (x, y) ◦ (x0, y0)

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

this group structure (see [2] and [20]).

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

the symbols ζ_j of the vector field X_j by

ζj = ξj+ n X k=1 m X α=1 aα_jkxkηα.

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

∂h ∂t + 1 2 n X j=1 ³ ∂h ∂xj + n X k=1 m X α=1 aα_jkxk_∂y∂h α ´₂ = 0.

Thus we start with

(2.9) ∂z

∂t + H(∇z) = 0,

where H(x, y; ξ, η) is the Hamiltonian function as the full symbol of ∆_X, (2.10) H(x, y; ξ, η) = 1 2 n X j=1 ¡ ξj + n X k=1 m X α=1 aα_kjxkηα ¢₂ = 1 2 n X j=1 ¡ ξj+ n X k=1 Akj(η) · xk ¢₂ . Here Akj(η) = m X α=1 aα_kjηα.

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_{× R}m_{) \ {0}. The Hamilton’s system can be written explicitly}

as follows: ˙xj = Hξj = ξj− n X k=1 Ajk(η) · xk= ζj, for j = 1, . . . , n ˙yα= Hηα = n X j=1 n X k=1 aα_jkxkζj, for α = 1, . . . , m ˙ξj = − Hxj = − n X k=1 A_kj(η) · ζ_k= n X k=1 A_jk(η) · ζ_k, for j = 1, . . . , n ˙η_α= − H_yα = 0, for α = 1, . . . , m (2.11)

with the initial-boundary conditions such that

(2.12)          x(0) = 0 x(t) = x = (x1, . . . , xn) y(t) = y = (y1, . . . , ym) η(0) = iη = i(η₁, . . . , η_m) where t ∈ R, x and y are arbitrarily given. With 0 ≤ s ≤ t,

ηα(s) = ηα= constant, for α = 1, . . . , m

constant meaning “constant along the bicharacteristic curve”. Also

H = 1 2 n X j=1 ˙x2_j = 1 2 n X j=1 ζ2 = E = energy.

Another way to see that E is constant along the bicharacteristic, note that ¨ xj = ˙ζj = ˙ξj+ n X k=1 Ajk(η) · ˙xk = − n X k=1 Akj(η) · ζk+ n X k=1 Ajk(η) · ζk = 2 n X k=1 A_jk(η) · ζ_k= 2 n X k=1 A_jk(η) · ζ_k (2.13)

for j = 1, . . . , n since At_{(η) = −A(η). Hence}

(2.14) 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 X j=1 ˙x2_j = 1 2˙x · ˙x = E = energy. Since ˙x(s) = e−2sA(η)_{ξ(0), by integrating the equation}

one has A(η)x(s) = −1 2 ³ e−2sA(η)− I ´ ξ(0)

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

iA(τ ) is self-adjoint. It follows that the matrix isA(τ ) sinh(itA(τ )) = 1 2πi Z C λ sinh(λ) ³ λ − itA(τ ) ´₋₁ dλ

is well defined and invertible for any t ∈ R and τ ∈ Rm. Here C is a suitable contour surrounding the spectrum of the matrix itA(τ ). The matrix

1 2πi Z C λ sinh(λ) ³ λ − itA(τ ) ´₋₁ dλ has an inverse: 1 2πi Z C sinh(λ) λ ³ λ − itA(τ ) ´₋₁ dλ We write it as sinh(iA(τ )) iA(τ ) = ∞ X k=0 (iA(τ ))2k (2k + 1)!.

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 6= 0. Now we may solve the initial value problem:

   ˙x_j(s) = ∂H ∂ξj = ξj+ i P_n k=1 P_m α=1aαkjxkτα = ξj+ i P k=1Akj(τ )xk, ˙ξj(s) = −_∂x∂H_j = −i P_n k=1 ³ ξ_k+ iPn_{`=1</sub>A<sub>`k}(τ )x_` ´ · A_jk(τ ) with the initial conditions ₍

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)

= iA(τ )

sinh(itA(τ )) · e

itA(τ )³_{I − e}−isA(τ )_{sinh(isA(τ ))}´_{· x}

= ³ e−isA(τ )cosh(isA(τ )) ´ · ³ eitA(τ ) iA(τ ) sinh(itA(τ )) ´ x = ³ e−isA(τ )cosh(isA(τ )) ´ · ξ(0).

Hence we obtain solutions for the initial-boundary problem (2.11) under the condition (2.12). We also have the following solutions for y(s):

yα(s) = yα(0) + Z _s 0 n X k=1 ³¡ e−2iuA(τ )ξ(0)¢_k· n X `=1 aα<sub>`k</sub>x_`(u) ´ du, α = 1, . . . , m.

Again by Theorem 2.2, the function 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. f (x, y, τ ) = h(x, y, τ, t) ¯ ¯ ¯ t=1 = m X α=1 ηα(0)yα(0) + Z ₁ 0 ¡ ξ · ˙x + η · ˙y − H¢ds = η0 m X α=1 ταyα+ Z ₁ 0 ¡ ξ · ˙x − H¢ds.

Here η0 is a pure imaginary number as we have seen in Lemma 2.5. Since

ξ · ˙x − H = 1 2hζ, ζi − hζ, Axi, then hζ, Axi = D ζ,A(τ )e2sA(τ ) e2A(τ )_{− I} x E − D ζ, A(τ ) e2A(τ )_{− I}x E =1 2hζ, ζi − D 2A(τ)e2sA(τ ) e2A(τ )_{− I} x, A(τ ) e2A(τ )_{− I}x E . It follows that ξ · ˙x − H =D 2A(τ)e2sA(τ ) e2A(τ )_{− I} x, A(τ ) e2A(τ )_{− I}x E =D 2A(τ) cosh(2sA(τ)) e2A(τ )_{− I} x, A(τ ) e2A(τ )_{− I}x E .

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

to obtain _Z 1 0 ¡ ξ · ˙x − H¢ds = 1 2 D¡ A(τ ) coth(A(τ ))¢x, x E . It follows that (2.15) f (x, y, τ ) = −i m X α=1 ταyα+1₂ D¡ A(τ ) coth(A(τ ))¢x, x E .

Using equation (2.15), we may complete the discuss in section 1.

Example 2.8. When A =     a1 0 · · · 0 0 a2 · · · 0 · · · 0 0 · · · 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 ) = −iτ y + τ

n

X

k=1

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

M0 =     1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1     , M1=     0 1 0 0 −1 0 0 0 0 0 0 1 0 0 −1 0     , M2=     0 0 0 −1 0 0 −1 0 0 1 0 0 1 0 0 0     , M3=     0 0 −1 0 0 0 0 1 1 0 0 0 0 −1 0 0     . We have q =     a b −d −c −b a −c d d c a b c −d −b a     = aM0+ bM1+ cM2+ dM3.

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

b = Im₁(q), c = Im₂(q), d = Im₃(q), and Im(q) = u = (b, c, d).

We introduce the quaternionic H-type group denoted by Q. This group consists of the set H × R3 = {[x, z] : x ∈ H, z = (z1, z2, z3) ∈ R3}

with the multiplication law defined in (2.8). The horizontal vector fields X = (X1, X2, X3, X4)

of the group Q can be written as follows:

X = ∇x+ 1₂ ³ M1x_∂z∂ 1 + M2x ∂ ∂z2 + M3x ∂ ∂z3 ´ , with x = (x1, x2, x3, x4) and ∇x=³ ∂_∂x 1, ∂ ∂x2, ∂ ∂x3, ∂ ∂x4 ´ .

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

f (x, z1, z2, z3, τ1, τ2, τ3) = −i 3 X α=1 ταzα+|x| 2 2 |τ | coth(2|τ |). In general, the matrix A can be defined as follows:

A =     P₃ α=1aα1Mα 0 . . . 0 0 P3_α=1aα 2Mα . . . 0 . . . 0 0 . . . P3_α=1aα nMα     .

In this case we obtain the so called anisotropic quaternion Carnot group considered in [10]. If all aα

l, l = 1, . . . , n are equal, we get the example of quaternion H-type group (see [8, 9, 17]).

In this case, the complex action is given by

f (x, y, τ ) = −iX α ταyα+1₂ n X l=1 |xl|2|τ |lcoth(2|τ |l), where |xl|2= P₃ j=0x24l−j, |τ |l= ¡ P₃ α=1(aαl)2τα2 ¢_1/2 .

3. Heat kernel and transport equation Let us return to the heat kernel. We consider the sub-Laplacian

∆X = 1₂ n X k=1 X_k2 with X_k = ∂ ∂x_k + n X j=1 m X α=1 aα_kjxj_∂y∂ α .

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

connected nilpotent Lie group. The Hamiltonian of the operator ∆X is

H(x, y, ξ, η) = 1 2 n X k=1 ³ ξk+ n X j=1 m X α=1 aα_kjxjηα ´₂ .

By Theorem 2.2, f is a solution of the generalized Hamilton-Jacobi equation:

H(x, y, ∇xf, ∇yf ) + m X α=1 τα_∂τ∂f α = f (x, y; η1, . . . , ηm).

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− _∂t∂ as following

(3.1) K(x, y; t) = Kt(x, y) = _t1_q

Z

Rm

e−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.

³ ∆X− _∂t∂´ e −f (x,y,τ )_t tq = e−f (x,y,τ )t tq+2 ¡ 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 ³ ∆X −_∂t∂ ´ e−f (x,y,τ )_{t V (τ )} tq =e− f (x,y,τ ) t V (τ ) tq+1 m X α=1 τ_α ³ −1 t ∂f ∂τα ´ V (τ ) − e− f (x,y,τ ) t V (τ ) tq+1 ¡ ∆_Xf − q¢V (τ ) = − e −f (x,y,τ )_{t V (τ )} tq+1 h_Xm α=1 τα_∂τ∂V α +¡∆Xf − q + m ¢ V (τ ) i + m X α=1 ∂ ∂τ_α ³ e−f (x,y,τ )_{t τ} αV (τ ) tq+1 ´ . Assuming e−f (u)t τ_αV (τ ) tq+1 → 0

as τ_α→ the ends of an appropriate contour Γ_α for α = 1, . . . , m, one has ³ ∆_X − ∂ ∂t ´ K_t(x, y) = ³ ∆_X − ∂ ∂t ´n 1 tq Z ∪m α=1Γα e−f (x,y,τ )t V (τ )dτ ª = − 1 tq+1 Z ∪m α=1Γα e−f (x,y,τ )t h_Xm α=1 τα_∂τ∂V α + ¡ ∆Xf − q + m ¢ V (τ ) i dτ = 0

if t 6= 0 and (3.2) m X α=1 τα_∂τ∂V α (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

X

`=1

(−1)`−1V`dτ1∧ · · · ∧ cdτ`∧ · · · ∧ dτm

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

df ∧ ∆_X(V ) + n X `=1 X<sub>`</sub>(f )X<sub>`</sub>(dV ) + D(dV ) −¡∆<sub>X</sub>f + n − m − 1¢dV = 0, where D(V ) is defined by D(V ) = m X k=1 τk<sub>∂τ</sub>∂ k(V ) = m X k=1 m X `=1 (−1)`−1τk∂V<sub>∂τ</sub>` kdτ1∧ · · · ∧ cdτ`∧ · · · ∧ dτm

Detailed discussion can be found in Furutani [12] and Greiner [14]. With f given by (2.15), one has

(3.3) ∆Xf = 1₂tr ¡ A(τ ) coth(A(τ ))¢= 1 2tr ³ 1 2πi Z C λcosh(λ) sinh(λ) ¡ λ − iA(τ )¢−1dλ ´ . Then (3.2) becomes m X α=1 τα_∂τ∂V α (τ ) +¡∆Xf − q + m ¢ V (τ ) = 0 ⇔ m X α=1 τα_∂τ∂V α(τ ) = ³ 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 = h q − m −1 2tr ¡

λA(τ ) coth(λA(τ ))¢iW.

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

dW W = ³ q − m λ − 1 2tr ¡ A(τ ) coth(λA(τ ))¢´dλ. Hence,

log W = (q − m)¡log λ + log C¢−1

2log(sinh(A(λτ )) ¢ . Therefore, V (τ ) = _q¡(det A(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 = n₂+m. Consequently, (3.5) P = A (2πt)q Z Rm e−f (x,y,τ )t V (τ )dτ,

where f is given by (3.3) and

V (τ ) = (det A(τ ))

n

2

q¡

det sinh(A(τ ))¢

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

Theorem 3.2. The equation

Pt(x, y) = _(2πt)A _q

Z

Rm

e−f (x,y,τ )t V (τ )dτ, represents the heat kernel for ∆_X if 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., [2, 4, 5, 11, 13]. We skip the proof here. Instead, we list some examples here.

Example 3.1. The Heisenberg sub-Laplacian: ∆X = 1₂

³ ∂ ∂x1+2x2∂y∂ ´₂ +1₂ ³ ∂ ∂x2−2x1∂y∂ ´₂ which is defined as (1.1). The action function is f (x, y) = −iτ y + (x2

1+ x22)τ coth(2τ ). The volume

is V (τ ) = _{sinh(2τ )}2τ . In this case n = 2, m = 1 and (3.5) has the following expression: (3.6) Pt(x, y) = _(2πt)2 ₂

Z _+∞

−∞

e−f (x,y,τ )t τ

sinh(2τ )dτ.

Example 3.2. The Grusin operator: ∆G = 1₂

³ ∂ ∂x ´₂ +1₂x2³∂ ∂y ´₂

. There is no group structure in 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 ¡

X₁2+ X₂2¢

is the sub-Laplacian on H1. Let NX2 = hX2i = [{aX2}a∈R] be a subgroup generated by the

element X2. The map ρ : H → R2 defined by

ρ : H → R2 ∼= h 3 g =x1X1+ x2X2+ zZ

=(x1, x2, z) 7→ (u, v) ∈ R2

where

u = x1, v = z + 1₂x1x2

realizes the projection map

In fact, this is a principal bundle and the trivialization is given by the map NX2× (NX2 \ H) ∼= R × R2 3 (a; u, v) 7→ (x1, x2, z) ∈ R3 ∼= H where (a; u, v) 7→ ¡u, a, v − 1 2au ¢ .

So the sub-Laplacian ∆_X on H₁ and Grusin operator ∆_G commutes each other through the map ρ:

∆_H◦ ρ∗= ρ∗◦ ∆_G.

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

Z _+∞

−∞

Pt(x1, x2, y), (u, a, v − 1₂ua)) = PtG((x1+ y +1₂x1x2), (u, v))

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.

P_tG((x0, 0), (x, y)) = 1 (2πt¢32 Z _+∞ −∞ e−f (x,x0,y,τ)t s |τ | sinh |τ |dτ

Example 3.3. Step 2 nilpotent Lie group: ∆X = −1₂

P_n j=1Xj2 where Xj = _∂x∂ j + n X k=1 ³_Xm α=1 aα_jkx_k´ ∂ ∂y_α with A(α)_jk =£aα jk ¤

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

Pt(x, y) = 1

(2πt)n2+m

Z

Rm

e−2iy·τ −hA(τ ) coth(A(τ ))x,xi2t

s

det A(τ )

sinh(A(τ ))dτ. Here A_jk(τ ) = Pm_α=1aα

jkτα. In particular, if A(α)jk satisfies further assumption: A (α) jk A

(γ) jk +

A(γ)_jk A(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)n2+m Z Rme −2iy·τ +|x|2|τ | coth(|τ |)_2t ³ |τ | sinh |τ | ´n 2 dτ.

Acknowledgment. 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] Baez J. C. The octonions. Bull. Amer. Math. Soc. (N.S.) 39 (2002), no. 2, 145–205.

[2] 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.

[3] Beals R., Gaveau B., and Greiner P. C. Complex Hamiltonian mechanics and parametrices for subelliptic

Laplacians, I, II, III. Bull. Sci. Math., 21 (1997), no. 1–3, 1–36, 97–149, 195–259.

[4] 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.

[5] 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

[6] Calin O., Chang D. C., and Greiner P. C. Real and complex Hamiltonian mechanics on some

subRieman-nian manifolds. Asian J. Math., 18, No. 1 (2004), 137–160.

[7] 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.

[8] 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.

[9] Chang D. C., Markina I. Geometric analysis on quaternion H-type groups. J. Geom. Anal. 16 (2006), no. 2, 266–294.

[10] Chang D. C., Markina I. Anisotropic quaternion Carnot groups: geometric analysis and Green’s function, to appear in Advanced in Applied Math., (2007).

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

China, Series A: Mathematics, (2007).

[12] Furutani K. Heat kernels of the sub-Laplacian and the Laplacian on nilpotent Lie groups, Analysis, Ge-ometry and Topology of Elliptic Operators, World Scientific (2006), 185–226.

[13] 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.

[14] Greiner P. On H¨ormander operators and non-holonomic geometry, to appear in Fields Institute Commu-nications, (2007).

[15] G¨urlebeck K., Spr¨ossig W. Quaternionic and Clifford Calculus for Physicists and Engineers John Wiley and Sons, Chichester, 1997. 371 pp.

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

[17] 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.

[18] Kaplan A. On the geometry of groups of Heisenberg type. Bull. London Math. Soc. 15 (1983), no. 1, 35–42. [19] Kor´anyi A. Geometric properties of Heisenberg-type groups. Adv. in Math. 56 (1985), no. 1, 28–38. [20] 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.

[21] Porteous I. R. Clifford algebras and the classical groups. Cambridge Studies in Advanced Mathematics, 50. Cambridge University Press, Cambridge, 1995. 295 pp.

[22] Ricci F. The spherical transform on harmonic extensions of H-type groups. Differential geometry (Turin, 1992). Rend. Sem. Mat. Univ. Politec. Torino, 50 (1992), no. 4, 381–392 (1993).

[23] Stein E. M. Harmonic analysis: real-variable methods, orthogonality, and oscillatory integrals. Princeton Mathematical Series, 43. Monographs in Harmonic Analysis, III. Princeton University Press, Princeton, 1993, 695 pp.

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

E-mail address: [email protected]

Department of Mathematics, Georgetown University, Washington D.C. 20057, USA Department of Mathematics, National Tsing Hua University, Hsinchu, Taiwan 30013, ROC

E-mail address: [email protected]

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