Generalized Hamilton—Jacobi Equation and Heat Kernel on Step Two Nilpotent Lie Groups
Abstract
We study geometrically invariant formulas for heat kernels of subelliptic differential 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.
GENERALIZED HAMILTONJACOBI EQUATION AND HEAT KERNEL
ON STEP TWO NILPOTENT LIE GROUPS
OVIDIU CALIN, DERCHEN CHANG, IRINA MARKINA
Abstract. We study geometrically invariant formulas for heat kernels of subelliptic diﬀer
ential operators on two step nilpotent Lie groups and for the Grusin operator in R
2
. We
deduce a general form of t he solution to the HamiltonJacobi equation and its generalized
form in R
n
× R
m
. Using our results, we obtain explicit formulas of the heat kernels for these
diﬀerential operators.
1. Introduction
Let us start with the L ap lace operator on R
n
,
∆ =
1
2
n
X
j=1
∂
2
∂x
2
j
.
It is wellknown that the heat kernel for ∆ is the Gaussian:
P
t
(x, x
0
) =
1
(2πt)
n
2
e
−
x−x
0

2
2t
.
Given a general second order elliptic operator in n dimensional Euclidean sp ace,
∆
X
=
1
2
n
X
j=1
X
2
j
+ lower order term,
where the {X
1
, . . . , X
n
} is a linearly independent set of vector ﬁelds, the heat kernel takes the
form
P
t
(x, x
0
) =
1
(2πt)
n
2
e
−
d
2
(x,x
0
)
2t
a
0
+ a
1
t + a
2
t
2
+ · · ·
.
Here d(x, x
0
) stands for the R iemann ian distance between x and x
0
if the metric is induced
by the orthonormal basis {X
1
, . . . , X
n
}. The a
j
’s are functions of x and x
0
. Note that
∂
∂t
d
2
2t
+
1
2
n
X
j=1
X
j
d
2
2t
2
= 0,
i.e.,
d
2
2t
is a solution of the HamiltonJacobi equation.
2000 Mathematics Subject Classiﬁcation. 53C17, 53C22, 35H20.
Key words and phrases. SubLaplacian, Heat operator, Htype 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
2 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
Now let us move to subelliptic operators. We ﬁrst consider the famous example: Heisenberg
subLaplacian on H
1
(1.1) ∆
X
=
1
2
∂
∂x
1
+ 2x
2
∂
∂y
2
+
1
2
∂
∂x
2
− 2x
1
∂
∂y
2
.
We shall try for a heat kernel in the form
1
t
q
e
−
f
t
· · ·
where h =
f
t
is a solution of the HamiltonJacobi equation
∂h
∂t
+
1
2
∂h
∂x
1
+ 2x
2
∂h
∂y
2
+
1
2
∂h
∂x
2
− 2x
1
∂h
∂y
2
= 0.
In other words,
(1.2)
∂h
∂t
+ H(x, ∇h) = 0,
where
(1.3) H =
1
2
h
ξ
1
+ 2x
2
η
2
+
ξ
2
− 2x
1
η
2
i
=
1
2
ζ
2
1
+ ζ
2
2
is the Hamilton function associated with the subelliptic operator (1.1) and ξ
1
, ξ
2
and η are
dual variable to x
1
, x
2
and y respectively. Usin g the LagrangeChapit method, let us look at
the following equation:
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:
˙x
1
= F
ξ
1
= ξ
1
+ 2x
2
η = ζ
1
,
˙x
2
= F
ξ
2
= ξ
2
− 2x
1
η = ζ
2
,
˙y = F
η
= 2 ˙x
1
x
2
− 2x
1
˙x
2
,
˙
t = F
γ
= 1,
˙
ξ
1
= − F
x
1
− ξ
1
F
h
= 2η ˙x
2
,
˙
ξ
2
= − F
x
2
− ξ
2
F
h
= −2η ˙x
1
,
˙η = − F
y
− γF
h
= 0,
˙γ = − F
t
− γF
h
= 0,
˙
h = ξ · ∇
ξ
F + ηF
η
+ γF
γ
= ξ · ˙x + η ˙y − H
since
˙
t = 1 and γ = −H. With 0 ≤ s ≤ t, one h as
γ(s) =γ = constant,
η(s) =η = constant,
t(s) =s.
Here “constant” means “constant along the bicharacteristic curve”. Furtherm ore,
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
¨x
1
=
˙
ξ
1
+ 2η ˙x
2
= +4η ˙x
2
,
¨x
2
=
˙
ξ
2
− 2η ˙x
1
= −4η ˙x
1
.
(1.4)
HAMILTONJACOBI EQUATION AND HEAT KERNEL 3
Therefore, ¨x
1
˙x
1
+ ¨x
2
˙x
2
= 0, and E =constant.
We need to ﬁnd the classical action integral
S(t) =
Z
t
0
ξ ·
˙
x + η ˙y − H
ds.
Let ﬁnd ξ and x from th e Hamilton system. We obtain
...
x
1
+ 16η
2
˙x
1
= 0,
...
x
2
+ 16η
2
˙x
2
= 0
from (1.4). Hence
˙x
1
(s) = ˙x
1
(0) cos(4ηs) +
¨x
1
(0)
4η
sin(4ηs)
= ˙x
1
(0) cos(4ηs) + ˙x
2
(0) sin(4ηs)
= ζ
1
(0) cos(4ηs) + ζ
2
(0) sin(4ηs)
(1.5)
and
˙x
2
(s) = ˙x
2
(0) cos(4ηs) +
¨x
2
(0)
4η
sin(4ηs)
= ˙x
2
(0) cos(4ηs) − ˙x
1
(0) sin(4ηs)
= − ζ
1
(0) sin(4ηs) + ζ
2
(0) cos(4ηs),
(1.6)
which yields
(1.7) x
1
(s) = x
1
(0) + ζ
1
(0)
sin(4ηs)
4η
+ ζ
2
(0)
1 − cos(4ηs)
4η
and
(1.8) x
2
(s) = x
2
(0) − ζ
1
(0)
1 − cos(4ηs)
4η
+ ζ
2
(0)
sin(4ηs)
4η
.
At s = t one has x
1
(t) = x
1
and x
2
(t) = x
2
, so
1
2
ζ
1
(0) sin(4ηt) +
1
2
ζ
2
(0)
1 − cos(4ηt)
= 2η
x
1
− x
1
(0)
,
−
1
2
ζ
1
(0)
1 − cos(4ηt)
+
1
2
ζ
2
(0) sin(4ηt) = 2η
x
2
− x
2
(0)
,
or,
+ζ
1
(0) cos(2ηt) + ζ
2
(0) sin(2ηt) =
2η
x
1
− x
1
(0)
sin(2ηt)
,
−ζ
1
(0) sin(2ηt) + ζ
2
(0) cos(2ηt) =
2η
x
2
− x
2
(0)
sin(2ηt)
.
(1.9)
Hamilton’s equations give
ξ
2
(s) = − 2ηx
1
(s) +
ξ
2
(0) + 2ηx
1
(0)
= − 2ηx
1
(0) −
1
2
ζ
1
(0) sin(4ηs) −
1
2
ζ
2
(0)
1 − cos(4ηs)
+ ζ
2
(0) + 4ηx
1
(0)
= 2ηx
1
(0) −
1
2
h
ζ
1
(0) sin(4ηs) − ζ
2
(0)
1 + cos(4ηs)
i
,
and
ξ
1
(s) = −2ηx
2
(0) +
1
2
h
ζ
1
(0)
1 + cos(4ηs)
+ ζ
2
(0) sin(4ηs)
i
.
4 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
The above calculations imply
ξ
1
˙x
1
+ ξ
2
˙x
2
= − 2η ˙x
1
(s)x
2
(0) + 2ηx
1
(0) ˙x
2
(s) +
1
2
ζ
2
1
(0) + ζ
2
2
(0)
1 + cos(4ηs)
= − 2η
˙x
1
(s)x
2
(0) − x
1
(0) ˙x
2
(s)
+
1 + cos(4ηs)
E,
and
Z
t
0
ξ ·
˙
x + η ˙y − H
ds = η
h
y − y(0) + 2
x
1
(0)x
2
− x
1
x
2
(0)
+
sin(4ηt)
4η
2
E
i
.
To ﬁnd E we square and add the two equations in (1.9),
E =
1
2
ζ
2
1
(0) +
1
2
ζ
2
2
(0) = 2η
2
x − x
0

2
sin
2
(2ηt )
.
Hence,
S(t) =
Z
t
0
ξ ·
˙
x + η ˙y − H
ds
=η
h
y − y(0) + 2
x
1
(0)x
2
− x
1
x
2
(0)
+ x − x
0

2
cot(2ηt)
i
.
We note that x, y, t, x
0
and η = η(0) are free parameters while y(0) = y(0; x, x
0
, y, η; t) is not.
Therefore, we need to introduce one m ore free variable h(0) such that h(t) = h(0) + S(t) is a
solution of the HamiltonJacobi 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
x, y, ∇
x
h,
∂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
=η y + 2η
x
1
(0)x
2
− x
1
x
2
(0)
+ ηx − x
0

2
cot(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 x
1
(0), x
2
(0) and η.
Before we move fur ther, let us consider a more general situation.
HAMILTONJACOBI EQUATION AND HEAT KERNEL 5
2. Generalized HamiltonJacobi equations
In this section we study the HamiltonJacobi equation which is crucial in th e construction
of the heat ker nel associated with elliptic and subelliptic operators. We ded uce a general
form of the solution to th e HamiltonJacobi equation and its generalized form. We consider
an (n + m)dimensional space R
n
× R
m
. The coordin ates are denoted x = (x
1
, . . . , x
n
) ∈ R
n
and y = (y
1
, . . . , y
m
) ∈ R
m
with dual variables (ξ
1
, . . . , ξ
n
) and (η
1
, . . . , η
m
) 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 th e variables (ξ, η) and has smooth coeﬃcients in (x, y).
We have the f ollow ing nice generalizaition of a result from [11].
Theorem 2.1. Set
(2.1) h(t; x, y, ξ, η) =
m
X
α=1
η
α
(0)y
α
(0) + S(t; x, y, ξ, η)
where
x
j
= x
j
(s; x, y, ξ, η; t), j = 1, . . . , n; y
α
= y
α
(s; x, y, ξ, η; t), α = 1, . . . , m
and
S(t; x, y, ξ, η) =
Z
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 HamiltonJacobi equation:
∂h
∂t
+ H
x, y, ∇
x
h, ∇
y
h
= 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
∂x
j
(t; x, y, ξ, η)
=
Z
t
0
h
n
X
k=1
∂ξ
k
∂x
j
dx
j
ds
+ ξ
k
d
ds
∂x
k
(s; x, y, ξ, η; t)
∂x
j
+
m
X
α=1
∂η
α
∂x
j
dy
α
ds
+ η
α
d
ds
∂y
α
(s; x, · · · ; t)
∂x
j
−
n
X
k=1
∂H
∂ξ
k
∂ξ
k
∂x
j
−
m
X
α=1
∂H
∂η
α
∂η
α
∂x
j
−
n
X
k=1
∂H
∂x
k
∂x
k
(s; x, y, ξ, η; t)
∂x
j
−
m
X
α=1
∂H
∂y
α
∂y
α
(s; x, y, ξ, η; t)
∂x
j
i
ds
=
Z
t
0
d
ds
n
X
k=1
ξ
k
∂x
k
(s; x, y, ξ, η; t)
∂x
j
+
m
X
α=1
η
α
∂y
α
(s; x, y, ξ, η; t)
∂x
j
ds
=
n
X
k=1
ξ
k
(s)
∂x
k
(s; x, y, ξ, η; t)
∂x
j
s=t
s=0
+
m
X
α=1
η
α
(s)
∂y
α
(s; x, y, ξ, η; t)
∂x
j
s=t
s=0
.
It follows that
∂S
∂x
j
(t; x, y, ξ, η) = ξ
j
(t) −
m
X
α=1
η
α
(0)
∂y
α
(0; x, y, ξ, η; t)
∂x
j
.
6 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
Similarly, for β = 1, . . . , m ,
∂S
∂y
β
(t; x, y, ξ, η) = η
β
(t) −
m
X
α=1
η
α
(0)
∂y
α
(0; x, y, ξ, η; t)
∂y
β
.
Moreover,
∂S
∂t
(t; · · · ) =
n
X
k=1
ξ
k
(t; · · · ) ˙x
k
(t; · · · ) +
m
X
α=1
η
α
(t; · · · ) ˙y
α
(t; · · · ) − H
x, y, ξ(t; · · · ), η(t; · · · )
+
n
X
k=1
ξ
k
(s; · · · )
∂x
k
(s; · · · )
∂t
s=t
s=0
+
m
X
α=1
η
α
(s; · · · )
∂y
α
(s; · · · )
∂t
s=t
s=0
.
Diﬀerentiating x
1
= x
1
(t; x, y, ξ, η; t) yields
0 =
d
dt
x
1
(t; x, y, ξ, η; t) = ˙x
1
(t; · · · ) +
∂x
1
(s; x, y, ξ, η; t)
∂t
s=t
.
On the other hand, one has
ξ
k
(s; · · · )
∂x
k
(s; · · · )
∂t
s=t
s=0
= −ξ
k
(t; · · · ) ˙x
k
(t; · · · ), k = 1, . . . , n,
and
η
α
(s; · · · )
∂y
α
(s; · · · )
∂t
s=t
s=0
= −η
α
(t; · · · ) ˙y
α
(t; · · · ) − η
α
(0; · · · )
∂y
α
(0; · · · )
∂t
, α = 1, . . . , n,
therefore,
∂S
∂t
= −H(t; · · · ) −
m
X
α=1
η
α
(0; · · · )
∂y
α
(0; · · · )
∂t
.
It follows that if we set as in the statement of the theorem
h(t; x, y, ξ, η) =
m
X
α=1
η
α
(0)y
α
(0) + S(t; x, y, ξ, η),
then it satisﬁes
∂h
∂x
k
= ξ
k
(t; x, y, ξ, η; t), k = 1, . . . , n
∂h
∂y
α
= η
α
(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, ∇
x
h, ∇
y
h
= 0.
This completes the proof of the theorem.
We note that the d erivation that (2.1) satisﬁes the HamiltonJacobi equation was complete
general, not restriction to H
x, y, ∇
x
h, ∇
y
h
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 are dual variables η
α
(0) instead of y
α
(0).
For the Heisenberg subLaplacian or the Grusin operator, η(0) = η cannot be switched to y(0).
As we know, ˙y = 2( ˙x
1
x
2
− x
1
˙x
2
). From (1.5) – (1.8), one has
˙y = 2
h
˙x
1
x
2
(0) − x
1
(0) ˙x
2
+
1
2
ζ
2
1
(0) + ζ
2
2
(0)
1 − cos(4ηs)
2η
i
,
HAMILTONJACOBI EQUATION AND HEAT KERNEL 7
and
y(s) = 2
x
1
(s)x
2
(0) − x
1
(0)x
2
(s)
+
E
4η
2
4ηs − sin(4ηs)
+ C.
At s = t, one h as x
1
(t) = x
1
, x
2
(t) = x
2
and
y = 2
x
1
x
2
(0) − x
1
(0)x
2
+
E
4η
2
4ηt − sin(4ηt)
+ C.
Hence, one has
y(s) =y − 2
h
x
1
− x
1
(s)
x
2
(0) − x
1
(0)
x
2
− x
2
(s)
i
−
E
4η
2
4η(t − s) − (sin(4ηt) − sin(4ηs))
.
At s = 0,
y(0) = y + 2
x
1
(0)x
2
− x
1
x
2
(0)
+ x − x
0

2
µ(2ηt),
where we set
µ(φ) =
φ
sin
2
φ
− cot φ.
To replace η by y(0), one needs to invert µ,
µ(2ηt) =
y − y(0) + 2
x
1
(0)x
2
− x
1
x
2
(0)
x − x
0

2
.
This is impossible since for most of the values on the right hand side µ
−1
is 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 HamiltonJacobi equation.
Before we go fu rther, we present a scaling property of the solution to the Hamiltonian system
dx
j
ds
=
∂H
∂ξ
j
,
dy
α
ds
=
∂H
∂η
α
,
dξ
j
ds
= −
∂H
∂x
j
,
dη
α
ds
= −
∂H
∂y
α
,
s ∈ [0, t] with the boundary conditions
x(0) = x
0
, x(t) = x, y(t) = y, η(0) = η(0).
Lemma 2.1. One has the following scaling property
x
j
(s; x, x
0
, y, ξ, η(0); t) = x
j
λs; x, x
0
, y, ξ,
η(0)
λ
; λt
, j = 1, . . . , n
y
α
(s; x, x
0
, y, ξ, η(0); t) = y
α
λs; x, x
0
, y, ξ,
η(0)
λ
; λt
, α = 1, . . . , m
ξ
j
(s; x, x
0
, y, ξ, η(0); t) = λξ
j
λs; x, x
0
, y, ξ,
η(0)
λ
; λt
, j = 1, . . . , n
η
α
(s; x, x
0
, y, ξ, η(0); t) = λη
α
λs; x, x
0
, 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.
8 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
Proof. Denote the curve on th e righthand side of (2.2) by {
˜
x(s),
˜
y(s),
˜
ξ(s), ˜η(s)}. Note that
s ∈ (0, t). Then for j = 1, . . . , n
∂ ˜x
j
∂s
= λ ˙x
j
λs; x, x
0
, y, ξ,
η(0)
λ
; λt
= λ
∂H
∂ξ
j
x
1
λs; x, x
0
, y, ξ,
η(0)
λ
; λt
, x
2
λs; x, x
0
, 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 ξ
1
, . . . , ξ
n
and η
1
, . . . , η
m
. Similar
calculations and homogeneity of degree 2 of
∂H
∂x
j
and
∂H
∂y
α
in ξ
1
, . . . , ξ
n
and η
1
, . . . , η
m
yield
∂ ˜y
α
∂s
=
∂H
∂η
α
,
∂
˜
ξ
j
∂s
= −
∂H
∂x
j
,
∂ ˜η
α
∂s
= −
∂H
∂y
α
.
Clearly,
˜x
j
(0) = x
j
0; x, x
0
, y, ξ,
η(0)
λ
; λt
= x
j
(0), ˜x
j
(t) = x
j
λt; x, x
0
, y, ξ,
η(0)
λ
; λt
= x
j
,
for j = 1, . . . , n and
˜y
α
(t) = y
α
(λt; x, x
0
, y, ξ,
η(0)
λ
; λt
= y
α
,
˜η
α
(0) = λη
α
(0; x, x
0
, y, ξ,
η(0)
λ
; λt
= λ
η
α
(0)
λ
= η
α
(0)
for α = 1, . . . , m. The bicharacteristic curves are unique, so th e two sides of (2.2) agree.
Corollary 2.2. One has
h(x, x
0
, y, ξ, η(0); t) = λh
x, x
0
, y, ξ,
η(0)
λ
; λt
.
Proof. In the case of Heisenberg group, the corollary is a direct con s equ en ce of the explicit
formu la (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,
˙x
j
(s; x, x
0
, y, ξ, η(0); t) =
dx
j
ds
(s; x, x
0
, y, ξ, η(0); t)
=
dx
j
ds
λs; x, x
0
, y, ξ,
η(0)
λ
; λt
= λ ˙x
j
λs; x, x
0
, y, ξ,
η(0)
λ
; λt
.
HAMILTONJACOBI EQUATION AND HEAT KERNEL 9
Similar result holds for ˙y
α
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, x
0
, y, ξ,
η(0)
λ
; λt
· λ
˙
x(λs; . . .) +
m
X
α=1
λη
α
(λs; . . .) · λ ˙y
α
(λs; . . .)
− λ
2
H(x(λs; . . .), . . .)
ds
=
1
λ
Z
t
0
λ
2
n
X
k=1
ξ
k
(λs; . . .) ˙x
k
(λs; . . .) + λ
2
m
X
α=1
η
α
(λs; . . .) ˙y
α
(λs; . . .) − λ
2
H(x(λs; . . .), . . .)
d(λs)
= λ
Z
t
0
ξ
s
′
; x, x
0
, y,
η(0)
λ
, λt
·
˙
x(s
′
; . . .) + η(s
′
; . . .) ·
˙
y(s
′
; . . .) − H(x(s
′
; . . .), . . .)
ds
′
= λS
x, x
0
, y, ξ,
η(0)
λ
, λt
.
Also,
m
X
α=1
η
α
(0)y
α
(0; x, x
0
, y, ξ, η(0); t) = λ
m
X
α=1
η
α
(0)
λ
y
α
0; x, x
0
, y, ξ,
η(0)
λ
; λt
and the proof of the corollary is therefore complete.
Set
f(x, x
0
, y, ξ, η(0)) = h(x, x
0
, y, ξ, η(0), t)
t=1
.
Then
Theorem 2.2. f is a solution of the generalized HamiltonJacobi equation
(2.3)
m
X
α=1
η
α
(0)
∂f
∂η
α
(0)
+ H
x, y, ∇
x
f, ∇
y
f
= f.
Proof. By homogeneity property of the function h, one has
h(x, x
0
, y, ξ, η(0), t) =
1
t
h(x, x
0
, y, ξ, tη (0), 1) =
1
t
f(x, x
0
, y, ξ, tη (0)),
so,
(2.4)
∂h
∂t
= −
1
t
2
f +
1
t
m
X
α=1
η
α
(0)
∂f
∂η
α
(0)
on one hand. On the other hand,
(2.5)
∂h
∂t
= −H
x, y, ∇
x
h, ∇
y
h
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.
10 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
2.3. Laplace operator. We start from the Laplace operator ∆ =
P
n
k=1
∂
2
∂x
2
k
in R
n
. The
Hamiltonian function H(ξ) is
H(ξ) =
1
2
n
X
k=1
ξ
2
k
and hence we need to deal with F (ξ, γ) = H + γ = 0. The Hamilton’s s ystem is
˙
x = ξ,
˙
ξ = 0, ˙γ = 0.
with initialb oundary conditions x(0) = x
0
, 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 + x
0
.
Moreover,
x = x(t) = ξ(0)t + x
0
⇒ ξ(0) =
x − x
0
t
and
∂h
∂t
=
1
2
n
X
k=1
ξ
2
k
=
n
X
k=1
(x
k
− x
(0)
k
)
2
2t
2
=
x − x
0

2
2t
2
or,
h(x, x
0
, t) = h(0) +
x − x
0

2
2t
2
t = h(0) +
x − x
0

2
2t
.
Since this is a translation invariant case, we may assume that h(0) = 0. Therefore,
f(x, x
0
) = h(x, x
0
, t)
t=1
=
x − x
0

2
2
gives us the Euclidean action fu nction.
2.4. Grusin operator. We are in R
2
now and th e horizontal vector ﬁelds X
1
, X
2
are given
by
X
1
=
∂
∂x
, and X
2
= x
∂
∂y
.
The Grusin operator is given as follows: ∆
X
=
1
2
∂
∂x
2
+
1
2
x
2
∂
∂y
2
. It is obvious that
∆
X
is elliptic away from the yaxis but degenerate on the yaxis. Since [X
1
, X
2
] =
∂
∂y
,
hence {X
1
, X
2
, [X
1
, X
2
]} spanned the tangent bundle of R
2
everywhere. By H¨ormander’s
theorem [12], ∆
X
is hypoelliptic.
The Hamiltonian function H for the ∆
X
is
(2.6) H(x, y, ξ, η) =
1
2
ξ
2
+
1
2
x
2
η
2
.
The Hamilton system can be obtained as follows;
˙x = H
ξ
= ξ,
˙y = H
η
= ηx
2
,
˙
ξ = − H
x
= −η
2
x,
˙η = − H
y
= 0,
˙
S = ξ ˙x + η ˙y − H.
With 0 ≤ s ≤ t,
η(s) = η(0) = η
0
= constant,
HAMILTONJACOBI EQUATION AND HEAT KERNEL 11
“constant” means “constant along the bicharacteristic curve”. Next,
¨x =
˙
ξ = −xη
2
,
so
¨x + η
2
x = 0.
It follows that
x(s) = A cos(ηs) + B sin(ηs) = x(0) cos(ηs) +
ξ(0)
η
sin(ηs ) = x
0
cos(ηs ) +
ξ(0)
η
sin(ηs).
Hence,
ξ(s) = ˙x(s)
yields
ξ(s) = ξ(0) cos(ηs) − ηx
0
sin(ηs).
We also have
x = x(t) = x
0
cos(ηt) +
ξ(0)
η
sin(ηt),
and
(2.7)
ξ(0)
η
=
x − x
0
cos(ηt)
sin(ηt)
.
Consequently,
x(s) = x(0) cos(ηs) +
x − x
0
cos(ηt)
sin(ηt)
sin(ηs).
The singularities occur at η = η
0
=
kπ
t
when x = ±x
0
; they are η =
(2k+1)π
t
if x = x
0
and
η
0
=
2kπ
t
if x = −x
0
. Next,
˙y(s) =ηx
2
(s)
=η
h
x
0
1
2
+
1
2
cos(2ηs)
+ 2x
0
ξ(0)
η
sin(ηs) cos(ηs) +
ξ(0)
η
2
1
2
−
1
2
cos(2ηs)
i
=
d
ds
n
η
h
x
2
0
2
s +
sin(2ηs)
2η
+
x
0
ξ(0)
η
2
sin
2
(ηs) +
1
2
ξ(0)
η
2
s −
sin(2ηs)
2η
io
=
d
ds
n
η
2
h
x
2
0
+
ξ(0)
η
2
i
s +
1
4
h
x
2
0
−
ξ(0)
η
2
i
sin(2ηs) +
x
0
2
ξ(0)
η
1 − cos(2ηs)
o
.
We replace
ξ(0)
η
by (2.7) and collect terms with x
2
0
:
x
2
0
2
n
ηs +
1
2
sin(2ηs) + ηs
cos
2
(ηt)
sin
2
(ηt)
−
1
2
cos
2
(ηt)
sin
2
(ηt)
sin(2ηt) −
cos(ηt)
sin(ηt)
1 − cos(2ηs)
o
=
x
2
0
2
n
ηs
sin
2
(ηt)
−
1
2
cos
2
(ηt) − sin
2
(ηt)
sin
2
(ηt)
sin(2ηs) −
cos(ηt)
sin(ηt)
1 − cos(2ηs)
o
=
x
2
0
2 sin
2
(ηt)
n
ηs −
1
2
cos(2ηt) sin (2ηs ) + sin(2ηt)
1 − cos(2ηs)
o
=
x
2
0
4 sin
2
(ηt)
n
2ηs −
sin(2ηt) − sin
2η(t − s)
o
.
The terms containing x
2
are:
1
4
x
2
sin
2
(ηt)
2ηs − sin(2ηs)
,
12 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
and the terms with x
0
x are the following:
1
2
2xx
0
sin
2
(ηt)
n
1
2
sin
η(2s − t)
+ sin(ηt)
− η s cos(ηt)
o
.
So,
˙y(s) =
d
ds
n
x
2
0
4 sin
2
(ηt)
h
2ηs −
(sin(2ηt ) − sin
2η(t − s)
)Big]
+
x
2
4 sin
2
(ηt)
2ηs − sin(2ηs)
+
2xx
0
4 sin
2
(ηt)
h
1
2
sin
η(2s − t)
+ sin(ηt)
− η s cos(ηt)
io
.
The action function has the form
S =
Z
t
0
(ξ ˙x + η ˙y − H)ds = η(y − y(0)) +
Z
t
0
(ξ
2
− H)ds.
We ﬁnd ξ
2
as follows
ξ
2
(s) =
ξ
2
(0)
2
1 + cos(2ηs)
− ξ(0)ηx
0
sin(2ηs) +
1
2
η
2
x
2
0
1 − cos(2ηs
)
=
1
2
ξ
2
(0) + η
2
x
2
0

{z }
=H(0)
+
1
2
ξ
2
(0) − η
2
x
2
0
cos(2ηs) − ηx
0
ξ(0) sin(2ηs).
Since H is constant along the bicharacteristic, one has
H = H(0) =
1
2
ξ
2
(0) + η
2
x
2
0
.
Continuing, we obtain the action function
S = η(y − y(0)) +
Z
t
0
(ξ
2
(0) − η
2
x
2
0
)
cos(2ηs)
2
− ηx
0
ξ(0) sin(2ηs)
ds
= η(y − y(0)) +
1
2
ξ
2
(0) − η
2
x
2
0
sin(2ηt)
2η
+ ηx
0
ξ(0)
cos(2ηt) − 1
2η
.
We simplify this
S − η(y − y(0))
=
η
2
2
x − x
0
cos(ηt)
sin(ηt)
2
sin(2ηt)
2η
−
1
2
η
2
x
2
0
sin(2ηt)
2η
+ η
2
x
0
x − x
0
cos(ηt)
sin(ηt)
cos(2ηt) − 1
2η
=
η
4
n
x − x
0
cos(ηt)
sin(ηt)
2
sin(2ηt) − x
2
0
sin(2ηt) − 2x
0
x − x
0
cos(ηt)
sin(ηt)
1 − cos(2ηt)
o
.
(2.8)
In the b racket {· · · } of (2.8), terms involved x
2
0
are
x
2
0
h
cos
2
(ηt)
sin
2
(ηt)
− 1
sin(2ηt) + 2
cos(ηt)
sin(ηt)
(1 − cos(2ηt))
i
= x
2
0
cos(2ηt) s in(2ηt)
sin
2
(ηt)
+ 2
cos(ηt)
sin(ηt)
−
cos(2ηt) sin (2ηt)
sin
2
(ηt)
= 2x
2
0
cot(ηt),
HAMILTONJACOBI EQUATION AND HEAT KERNEL 13
terms involved x
2
are
x
2
sin(2ηt)
sin
2
(ηt)
= 2x
2
cot(ηt),
and terms containing x
0
x are
2xx
0
−
cos(ηt)
sin
2
(ηt)
sin(2ηt) −
1 − cos(2ηt)
sin(ηt)
= − 2xx
0
2 cos
2
(ηt)
sin(ηt)
+ 2 sin(ηt )
= −
4xx
0
sin(ηt)
.
Hence,
{· · · } =2(x
2
+ x
2
0
) cot(ηt) −
4xx
0
sin(ηt)
=
(x + x
0
)
2
+ (x − x
0
)
2
cot(ηt) −
(x + x
0
)
2
− (x − x
0
)
2
sin(ηt)
= (x + x
0
)
2
cot(ηt) −
1
sin(ηt)
+ (x − x
0
)
2
cot(ηt) +
1
sin(ηt)
= (x + x
0
)
2
cos(ηt) − 1
sin(ηt)
+ (x − x
0
)
2
cos(ηt) + 1
sin(ηt)
= − (x + x
0
)
2
tan
ηt
2
+ (x − x
0
)
2
cot
ηt
2
.
Thus S has the following form:
S = η(y − y(0)) −
η
4
h
(x + x
0
)
2
tan
ηt
2
− (x − x
0
)
2
cot
ηt
2
i
.
By Th eorem 2.1, we know that
h(t; x, x
0
, y, η) = ηy(0) + S(t; x, y, η)
= ηy(0) + η(y − y(0)) −
η
4
h
(x + x
0
)
2
tan
η
2
− (x − x
0
)
2
cot
η
2
i
= ηy −
η
4
h
A
2
tan
ηt
2
− B
2
cot
ηt
2
i
is a solution of the HamiltonJacobi equation. Here A = x + x
0
and B = x − x
0
. Now by
Theorem 2.2, the function
f(x, x
0
, y, η) = h(t; x, x
0
, y, η)
t=1
=
1
2
η
2
n
4y − A
2
tan
η
2
+ B
2
cot
η
2
o
is a solution of the generalized HamiltonJacobi equation
η
∂f
∂η
+ H
x, x
0
, y, ∂
x
f, ∂
y
f
= f.
We set
η
2
= eητ,
where τ ∈ R yields the domain of integration and eη is a ﬁxed complex number.
14 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
Lemma 2.5. Suppose f is a smooth f unction of τ ∈ R and
lim
τ→±∞
Re(f)(τ) = ∞
oﬀ the canonical curve x
2
0
+ x
2
= 0. Then eη is pure imaginary.
Proof. Let eη = η
1
+ iη
2
. An elementary calculation yields
f =
1
2
η
1
+ iη
2
τ
n
4y +
sin(2η
1
τ)
(B
2
− A
2
) cosh(2η
2
τ) + (B
2
+ A
2
) cos(2η
1
τ)
cosh
2
(2η
2
τ) − cos
2
(2η
1
τ)
− i
sinh(2η
2
τ)
(B
2
+ A
2
) cosh(2η
2
τ) + (B
2
− A
2
) cos(2η
1
τ)
cosh
2
(2η
2
τ) − cos
2
(2η
1
τ)
o
(i). η
1
= 0, i.e., η ∈ iR. When τ ≈ ±∞,
f ≈
1
2
iη
2
τ
n
4y − i2(x
2
0
+ x
2
) tanh(2η
2
τ)
o
,
and
Re(f) ≈
1
4
(x
2
0
+ x
2
)2η
2
τ tanh(2η
2
τ) → ±∞
as τ → ±∞ as long as x
2
0
+ x
2
6= 0.
(ii). η
2
= 0, that is η ∈ R. Then
f = 2η
1
τy +
1
4
2η
1
τ
sin(2η
1
τ)
h
B
2
− A
2
+ (B
2
+ A
2
) cos(2η
1
τ)
i
is singular in τ ∈ R when x
2
0
+ x
2
6= 0, otherwise
Re(f) = f = 2η
1
τy −→
{z}
τ→±∞
±(sgn(y))∞.
(iii). η
1
6= 0, η
2
6= 0. Here
f ≈
1
2
η
1
+ iη
2
τ
n
4y − i(A
2
+ B
2
) tanh(2η
2
τ)
o
as τ → ±∞, and
Re(f) ≈ 2η
1
τy + (x
2
0
+ x
2
)
η
2
τ
= τ 
2(sgn(τ))η
1
y + (x
2
0
+ x
2
)η
2

and choosing x
0
, x, y so that
2η
1
y > (x
2
0
+ x
2
)η
2

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
eη = −
i
2
.
Then
f = −iτy +
1
2
(x
2
0
+ x
2
)τ coth τ −
τx
0
x
sinh τ
.
HAMILTONJACOBI EQUATION AND HEAT KERNEL 15
2.6. SubLaplace operator on step 2 nilpotent Lie groups. Let M be a simply connected
2step nilpotent Lie group G equipped w ith 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 skewsymmetric 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
2
B(x, x
′
)
and then the exponential map is the identity map. Let {X
1
, . . . , X
n
} be a basis of H and let
{Y
1
, . . . , Y
m
} be a basis of the center [G, G] = C. We assume {X
1
, . . . , X
n
} and {Y
1
, . . . , Y
m
} are
orthonormal, and introduce a left invariant Riemannian metric on the group G in an obvious
way.
We wr ite the vector ﬁelds X
j
, j = 1, . . . , n by:
X
j
=
∂
∂x
j
+
n
X
k=1
m
X
α=1
a
α
jk
x
k
∂
∂y
α
where the a
α
jk
are real numbers and form skewsymm etric matrices
a
α
jk
j,k
, i.e., a
α
jk
= −a
α
kj
.
We are interested in the subLaplacian ∆
X
which can be deﬁned as follows:
∆
X
=
1
2
n
X
j=1
X
2
j
It is easy to see that
(2.9)
X
j
, X
k
= 2
n
X
k=1
m
X
α=1
a
α
jk
∂
∂y
α
.
Lemma 2.7. The operator ∆
X
is hypoelliptic if and only if the rectangular matrix of order
n(n−1)
2
× m with element
a
α
jk
{(j<k),α}
is of rank m (which implies that m ≤
n(n−1)
2
).
Proof. The operator ∆
X
is hypoelliptic when the vector ﬁelds {X
j
}
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 (2.9). If we consider
a
α
jk
as a matrix with ind ices 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 R
n
× R
m
with the following group law:
(2.10)
(x, y) ◦ (x
′
, y
′
) =
x
1
+ x
′
1
, . . . , x
n
+ x
′
n
, y
1
+ y
′
1
+
n
X
j,k=1
a
1
jk
x
′
j
x
k
, . . . , y
m
+ y
′
m
+
n
X
j,k=1
a
m
jk
x
′
j
x
k
.
16 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
It is easy to see that the X
j
are left invariant vector ﬁelds su ch that
X
j
f
(x, y) =
∂
∂x
′
j
f ◦ L
(x,y)
(x
′
, y
′
)
x
′
=0,y
′
=0
where
L
(x,y)
(x
′
, y
′
) = (x, y) ◦ (x
′
, y
′
)
is the left translation by the element (x, y). I n particular, ∆
X
is a left invariant operator for
this group structure (see [1] and [16]).
Let ξ
1
, . . . , ξ
n
be the dual variables of x and η
1
, . . . , η
m
be the dual variables of y. We deﬁne
the symbols ζ
j
of the vector ﬁeld X
j
by
ζ
j
= ξ
j
+
n
X
k=1
m
X
α=1
a
α
jk
x
k
η
α
.
We shall try to ﬁnd a solution of the following equation:
∂h
∂t
+
1
2
n
X
j=1
∂h
∂x
j
+
n
X
k=1
m
X
α=1
a
α
jk
x
k
∂h
∂y
α
2
= 0.
Thus we start with
(2.11)
∂z
∂t
+ H(∇z) = 0,
where H(x, y; ξ, η) is the Hamiltonian function as the full symbol of ∆
X
,
(2.12) H(x, y; ξ, η) =
1
2
n
X
j=1
ξ
j
+
n
X
k=1
m
X
α=1
a
α
jk
x
k
η
α
2
=
1
2
n
X
j=1
ξ
j
+
n
X
k=1
A
kj
(η) · x
k
2
.
Here
A
kj
(η) =
m
X
α=1
a
α
kj
η
α
.
We shall ﬁ nd 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
∗
(R
n
× R
m
) \ {0}. Since A
t
(η) = −A(η), the Hamilton’s system can
be written explicitly as follows:
˙x
j
= H
ξ
j
= ξ
j
−
n
X
k=1
A
jk
(η) · x
k
= ζ
j
, for j = 1, . . . , n
˙y
α
= H
η
α
=
n
X
j=1
n
X
k=1
a
α
jk
x
k
ζ
j
, for α = 1, . . . , m
˙
ξ
j
= − H
x
j
= −
n
X
k=1
A
jk
(η) · ζ
k
=
n
X
k=1
A
kj
(η) · ζ
k
, for j = 1, . . . , n
˙η
α
= − H
y
α
= 0, for α = 1, . . . , m
(2.13)
with the initialboundary conditions such that
(2.14)
x(0) = 0
x(t) = x = (x
1
, . . . , x
n
)
y(t) = y = (y
1
, . . . , y
m
)
η(0) = iτ = i(τ
1
, . . . , τ
m
)
HAMILTONJACOBI EQUATION AND HEAT KERNEL 17
where t ∈ R, x and y are arb itrarily given. With 0 ≤ s ≤ t,
η
α
(s) = η
α
= constant, for α = 1, . . . , m
“constant” means “constant along the bicharacteristic curve”. Also
H =
1
2
n
X
j=1
˙x
2
j
=
1
2
n
X
j=1
ζ
2
= E = energy.
Another way to see that E is constant along the bicharacteristic, note that
¨x
j
=
˙
ζ
j
=
˙
ξ
j
−
n
X
k=1
A
jk
(η) · ˙x
k
= −
n
X
k=1
A
jk
(η) · ζ
k
−
n
X
k=1
A
jk
(η) · ζ
k
= − 2
n
X
k=1
A
jk
(η) · ζ
k
(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 skewsymmetric. It follows that
1
2
n
X
j=1
˙x
2
j
=
1
2
˙
x ·
˙
x = E = energy.
Since
˙
x(s) = e
−2sA(η)
ξ(0), by integrating th e 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 η
α
= η(0) = iτ
α
is pure imaginary, the matrix
iA(τ) is selfadjoint. It follow s that the matrix
isA(τ)
sinh(itA(τ ))
=
1
2πi
Z
γ
λ
sinh(λ)
λ − itA(τ)
−1
dλ
is well deﬁned and invertible for any t ∈ R and τ ∈ R
m
. Here γ is a suitable contour surroun ding
the spectrum of the m atrix itA(τ). The matrix
1
2πi
Z
γ
λ
sinh(λ)
λ − itA(τ)
−1
dλ
has an inverse:
1
2πi
Z
γ
sinh(λ)
λ
λ − itA(τ)
−1
dλ
We write it as
sinh(iA(τ))
iA(τ)
=
∞
X
k=0
(iA(τ))
2k
(2k + 1)!
.
18 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
Then f or any ﬁxed t ∈ R, we have onetoone correspondence between the initial condition ξ(0)
and boundary condition x:
ξ(0) = e
itA(τ)
·
iA(τ)
sinh(itA(τ))
· x, t 6= 0.
Now we may solve the initial value problem:
˙x
j
(s) =
∂H
∂ξ
j
= ξ
j
+ i
P
n
k=1
P
m
α=1
a
α
jk
x
k
τ
α
= ξ
j
+ i
P
k=1
A
kj
(τ)x
k
,
˙
ξ
j
(s) = −
∂H
∂x
j
= −i
P
n
k=1
ξ
k
+ i
P
n
ℓ=1
A
ℓk
(τ)x
ℓ
· A
jk
(τ)
with the initial conditions
(
x(0) = 0
ξ(0) = e
itA(τ)
·
iA(τ)
sinh(itA(τ))
x.
Straightforward computations show that
x(s) = x(s; x, τ, t) = e
i(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)
I − e
−isA(τ)
sinh(isA(τ))
· x
=
e
−isA(τ)
cosh(isA(τ))
·
e
itA(τ)
iA(τ)
sinh(itA(τ))
x
=
e
−isA(τ)
cosh(isA(τ))
· ξ(0).
Hence we obtain solutions for the initialboundary problem (2.13) under the condition (2.14) .
We also have the following solutions for y(s):
y
α
(s) = y
α
(0) +
Z
s
0
n
X
k=1
e
−2iuA(τ)
ξ(0)
k
·
n
X
ℓ=1
a
α
ℓk
x
ℓ
(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 HamiltonJacobi equation. In our case, the fu nction f can be
calculated explicitly.
f(x, y, τ ) = h(x, y, τ, t)
t=1
=
m
X
α=1
η
α
(0)y
α
(0) +
Z
1
0
ξ ·
˙
x + η ·
˙
y − H
ds
= η
0
m
X
α=1
τ
α
y
α
+
Z
1
0
ξ ·
˙
x − H
ds.
Here η
0
is a p ure imaginary number. This choice can be motivated by Lemma 2.5.
Since
ξ ·
˙
x − H =
1
2
hζ, ζi − hζ, Axi,
HAMILTONJACOBI EQUATION AND HEAT KERNEL 19
then
hζ, Axi =
D
ζ,
A(τ)e
2sA(τ)
e
2A(τ)
− I
x
E
−
D
ζ,
A(τ)
e
2A(τ)
− I
x
E
=
1
2
hζ, ζi −
D
2A(τ)e
2sA(τ)
e
2A(τ)
− I
x,
A(τ)
e
2A(τ)
− I
x
E
.
It follows that
ξ ·
˙
x − H =
D
2A(τ)e
2sA(τ)
e
2A(τ)
− I
x,
A(τ)
e
2A(τ)
− I
x
E
=
D
2A(τ) cosh(2sA(τ))
e
2A(τ)
− I
x,
A(τ)
e
2A(τ)
− I
x
E
.
The second equality due to A is skewsymmetric. 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
(2.17) f(x, y, τ ) = − i
m
X
α=1
τ
α
y
α
+
1
2
D
A(τ) coth(A(τ))
x, x
E
.
Using equation (2.17), we may complete the discuss in Section 2.
Example 2.8. When
A =
a
1
0 · · · 0
0 a
2
· · · 0
· · ·
0 0 · · · a
2n
∈ M
2n×2n
, with a
j
= a
j+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
a
k
coth(2a
k
τ)
x
2
k
+ x
2
n+k
.
Example 2.9. In R
4
, the basis of quaternion numbers H = {a + bi + cj + dk : a, b, c, d ∈ R}
can be given by real matrices
M
0
=
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
, M
1
=
0 1 0 0
−1 0 0 0
0 0 0 1
0 0 −1 0
,
M
2
=
0 0 0 −1
0 0 −1 0
0 1 0 0
1 0 0 0
, M
3
=
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
= aM
0
+ bM
1
+ cM
2
+ dM
3
.
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 = I m
1
(q), c = Im
2
(q), d = Im
3
(q), and Im(q) = u = (b, c, d).
20 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
We introd uce the quaternionic Htype group denoted by Q. This group consists of the set
H × R
3
= {[x, y] : x ∈ H, y = (y
1
, y
2
, y
3
) ∈ R
3
}
with the multiplication law deﬁned in (2.10) with [a
α
jk
] = M
α
, α = 1, 2, 3. The horizontal
vector ﬁelds X = (X
1
, X
2
, X
3
, X
4
) of the group Q can be written as follows:
X = ∇
x
+
1
2
M
1
x
∂
∂y
1
+ M
2
x
∂
∂y
2
+ M
3
x
∂
∂y
3
,
with x = (x
1
, x
2
, x
3
, x
4
) and
∇
x
=
∂
∂x
1
,
∂
∂x
2
,
∂
∂x
3
,
∂
∂x
4
.
In this case, the solution for the generalized HamiltonJacobi equation is
f(x, y
1
, y
2
, y
3
, τ
1
, τ
2
, τ
3
) = −i
3
X
α=1
τ
α
y
α
+
x
2
2
τ coth(2τ)
See details in [6] In general multidimensional case, the matrix A can be deﬁned as follows:
A =
P
3
α=1
a
α
1
M
α
0 . . . 0
0
P
3
α=1
a
α
2
M
α
. . . 0
. . .
0 0 . . .
P
3
α=1
a
α
n
M
α
.
In this case we obtain the so called anisotropic quaternion Carnot group considered in [7]. The
complex action is given by
f(x, y, τ ) = −i
X
α
τ
α
y
α
+
1
2
n
X
l=1
x
l

2
τ
l
coth(2τ
l
),
where x
l

2
=
P
3
j=0
x
2
4l−j
, τ 
l
=
P
3
α=1
(a
α
l
)
2
τ
2
α
1/2
. If all a
α
l
, l = 1, . . . , n are equal, we get
the example of multidimensional quaternion Htype group. More information about Htype
groups can be found in [5, 13, 14 , 15].
3. Heat kernel and transport equation
Let us return to the heat kernel. We consider th e subLaplacian
∆
X
=
1
2
n
X
k=1
X
2
k
with X
k
=
∂
∂x
k
+
n
X
j=1
m
X
α=1
a
α
kj
x
j
∂
∂y
α
.
Assume that {X
1
, . . . , X
n
} is an orthonorm al basis of the “horizontal subbundle” on a simply
connected nilpotent 2 step Lie group. The Hamiltonian of the operator ∆
X
is
H(x, y, ξ, η) =
1
2
n
X
k=1
ξ
k
+
n
X
j=1
m
X
α=1
a
α
kj
x
j
η
α
2
.
By Theorem 2.2, the function f associated with H is a solution of the generalized Hamilton
Jacobi equation:
H(x, y, ∇
x
f, ∇
y
f) +
m
X
α=1
τ
α
∂f
∂τ
α
= f(x, y; η
1
, . . . , η
m
).
HAMILTONJACOBI EQUATION AND HEAT KERNEL 21
As we know, the function f depends on free variables η
α
, α = 1, . . . , m. To this end we shall
sum over η
α
, or for convenience τ
α
= tη
α
, α = 1, . . . , m; an extra t can always be absorbed in
the power q which can be determined after we solve the generalized HamiltonJacobi equation.
Thus we write h eat kernel of ∆
X
−
∂
∂t
as following
(3.1) K(x, y; t) = K
t
(x, y) =
1
t
q
Z
R
m
e
−
f (x,y,τ )
t
V (τ)dτ.
Here V is the volume element. To see whether (3.1) is a rep resentation 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
t
q
=
e
−
f (x,y,τ )
t
t
q+2
H(x, y, ∇
x
f, ∇
y
f) − f
−
e
−
f (x,y,τ )
t
t
q+1
∆
X
(f) − q
,
and the eiconal equation (2.3) implies that
∆
X
−
∂
∂t
e
−
f (x,y,τ )
t
V (τ)
t
q
=
e
−
f (x,y,τ )
t
t
q+1
m
X
α=1
τ
α
−
1
t
∂f
∂τ
α
V (τ) −
e
−
f (x,y,τ )
t
t
q+1
∆
X
f − q
V (τ)
= −
e
−
f (x,y,τ )
t
t
q+1
h
m
X
α=1
τ
α
∂V
∂τ
α
+
∆
X
f − q + m
V (τ )
i
+
m
X
α=1
∂
∂τ
α
e
−
f (x,y,τ )
t
τ
α
V (τ )
t
q+1
.
Assuming
e
−
f (u)
t
τ
α
V (τ)
t
q+1
→ 0
as τ
α
→ the en ds of an appropriate contour Γ
α
for α = 1, . . . , m, one has
∆
X
−
∂
∂t
K
t
(x, y)
=
∆
X
−
∂
∂t
n
1
t
q
Z
∪
m
α=1
Γ
α
e
−
f (x,y,τ )
t
V (τ)dτ
= −
1
t
q+1
Z
∪
m
α=1
Γ
α
e
−
f (x,y,τ )
t
h
m
X
α=1
τ
α
∂V
∂τ
α
+
∆
X
f − q + m
V (τ )
i
dτ = 0
if t 6= 0 and
(3.2)
m
X
α=1
τ
α
∂V
∂τ
α
(τ) +
∆
X
f − 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 codimension one form
V dτ =
m
X
ℓ=1
(−1)
ℓ−1
V
ℓ
dτ
1
∧ · · · ∧
c
dτ
ℓ
∧ · · · ∧ dτ
m
22 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
which satisﬁes a socalled “generalized transport equation”:
df ∧ ∆
X
(V ) +
n
X
ℓ=1
X
ℓ
(f)X
ℓ
(dV ) + D(dV ) −
∆
X
f + n − m − 1
dV = 0,
where D(V ) is deﬁned by
D(V ) =
m
X
k=1
τ
k
∂
∂τ
k
(V ) =
m
X
k=1
m
X
ℓ=1
(−1)
ℓ−1
τ
k
∂V
ℓ
∂τ
k
dτ
1
∧ · · · ∧
c
dτ
ℓ
∧ · · · ∧ dτ
m
.
Detailed discussion can be found in Furu tani [9] and Greiner [11].
With f given by (2.17), one has
(3.3) ∆
X
f =
1
2
tr
A(τ) coth(A(τ))
=
1
2
tr
1
2πi
Z
C
λ
cosh(λ)
sinh(λ)
λ − iA(τ)
−1
dλ
.
Then (3.2) becomes
m
X
α=1
τ
α
∂V
∂τ
α
(τ) +
∆
X
f − q + m
V (τ ) = 0 ⇔
m
X
α=1
τ
α
∂V
∂τ
α
(τ) =
q − m −
1
2
tr
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
λ
dW
dλ
=
h
q − m −
1
2
tr
λA(τ) coth(λA(τ))
i
W.
Here we are using the fact that A(τ ) is linear in τ . I t follows that
dW
W
=
q − m
λ
−
1
2
tr
A(τ) coth(λA(τ))
dλ.
Hence,
log W = (q − m)
log λ + log C
−
1
2
log(sinh(A(λτ ))
.
Therefore,
V (τ) =
(det A(τ))
q−m
q
det sinh(A(τ))
.
If we propose the volume element V is real analytic and nonvanish at 0, then we have q =
n
2
+m.
Consequently,
(3.5) P =
A
(2πt)
q
Z
R
m
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.
HAMILTONJACOBI EQUATION AND HEAT KERNEL 23
Theorem 3.2. The equation
P
t
(x, y) =
A
(2πt)
q
Z
R
m
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
− ∆
X
P = 0, t > 0
and
lim
t→0
P (x, y, t) = δ(x)δ(y).
The calculation is long but straightforward. Readers can ﬁnd the proof of this theorem in m any
places, see e.g . , [1, 2, 3, 4, 8, 10]. We skip the pr oof here. Instead, we list some examples.
Example 3.1. The Heisenberg s ubLaplacian: ∆
X
=
1
2
∂
∂x
1
+2x
2
∂
∂y
2
+
1
2
∂
∂x
2
−2x
1
∂
∂y
2
which
is deﬁned as (1.1). The action function is f(x, y) = −iτy + (x
2
1
+ x
2
2
)τ coth(2τ). The volume
is V (τ ) =
2τ
sinh(2τ)
. In this case n = 2, m = 1 and (3.5) has the following expression:
(3.6) P
t
(x, y) =
2
(2πt)
2
Z
+∞
−∞
e
−
f (x,y,τ )
t
τ
sinh(2τ)
dτ.
Example 3.2. The Grusin operator: ∆
G
=
1
2
∂
∂x
2
+
1
2
x
2
∂
∂y
2
. There is no group structure in
this case. However, this operator has connection with the Heisenberg subLaplacian. Let H
1
be the Heisenberg group whose Lie algebra has a basis {X
1
, X
2
, T } with the bracket relation
[X
1
, X
2
] = −4T . As in (1.1),
∆
X
= −
1
2
X
2
1
+ X
2
2
is th e subLaplacian on H
1
. Let N
X
2
= hX
2
i = [{aX
2
}
a∈R
] be a subgroup generated by the
element X
2
. The map ρ : H
1
→ R
2
deﬁned by
ρ : H
1
→ R
2
∼
=
h ∋ g =x
1
X
1
+ x
2
X
2
+ zZ
=(x
1
, x
2
, z) 7→ (u, v) ∈ R
2
where
u = x
1
, v = z +
1
2
x
1
x
2
realizes the projection map
H
1
∼
=
R
3
→ N
X
2
\ H
1
∼
=
R
2
.
In fact, this is a principal bundle and the trivialization is given by the map
N
X
2
× (N
X
2
\ H
1
)
∼
=
R × R
2
∋ (a; u, v) 7→ (x
1
, x
2
, z) ∈ R
3
∼
=
H
1
where
(a; u, v) 7→
u, a, v −
1
2
au
.
So the subLaplacian ∆
X
on H
1
and Gru sin operator ∆
G
commutes each other through the
map ρ:
∆
H
◦ ρ
∗
= ρ
∗
◦ ∆
G
.
The heat kernel P
t
(x, y) ∈ C
∞
(R
+
× H
1
) is given by (3.6). Hence,
Z
+∞
−∞
P
t
(x
1
, x
2
, y), (u, a, v −
1
2
ua)
= P
G
t
((x
1
+ y +
1
2
x
1
x
2
), (u, v))
24 OVIDIU CALIN, DER CHEN CHANG, IRINA MARKINA
that is, the ﬁber integration of the function P
t
(g, h) along the ﬁ ber of the map ρ gives the heat
kernel of the Grusin operator.
P
G
t
((x
0
, 0), (x, y)) =
1
(2πt
3
2
Z
+∞
−∞
e
−
f (x,x
0
,y,τ )
t
s
τ
sinh τ 
dτ
Example 3.3. Step 2 nilpotent Lie group: ∆
X
= −
1
2
P
n
j=1
X
2
j
where
X
j
=
∂
∂x
j
+
n
X
k=1
m
X
α=1
a
α
jk
x
k
∂
∂y
α
with A
(α)
jk
=
a
α
jk
j,k
is a skewsymmetric and orthogonal matrix. The heat kernel is
P
t
(x, y) =
1
(2πt)
n
2
+m
Z
R
m
e
−
2iy·τ −hA(τ ) coth(A(τ ))x,xi
2