# Generalized Hamilton—Jacobi Equation and Heat Kernel on Step Two Nilpotent Lie Groups

Chapter (PDF Available) · February 2009with23 Reads
DOI: 10.1007/978-3-7643-9906-1_3
In book: Analysis and Mathematical Physics, pp.49-76
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 HAMILTON-JACOBI EQUATION AND HEAT KERNEL
ON STEP TWO NILPOTENT LIE GROUPS
OVIDIU CALIN, DER-CHEN CHANG, IRINA MARKINA
Abstract. We study geometrically invariant formulas for heat kernels of sub-elliptic diﬀer-
ential operators on two step nilpotent Lie groups and for the Grusin operator in R
2
. We
deduce a general form of t he solution to the Hamilton-Jacobi 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
n
,
=
1
2
n
X
j=1
2
x
2
j
.
It is well-known that the heat kernel for is the Gaussian:
P
t
(x, x
0
) =
1
(2πt)
n
2
e
|xx
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 Hamilton-Jacobi equation.
2000 Mathematics Subject Classiﬁcation. 53C17, 53C22, 35H20.
Key words and phrases. Sub-Laplacian, Heat operator, H-type groups, action function, volume element.
The ﬁrst author is partially supported by the NSF grant #0631541.
The second author is partially supported by a Hong Kong RGC competitive earmarked research grant
#600607, a competitive research grant at Georgetown University, and NFR grant #180275/D15.
The third author is supported by NFR grants # 177355/V30, #180275/D15, and ESF Networking Pro-
gramme HCAA.
1
2 OVIDIU CALIN, DER -CHEN CHANG, IRINA MARKINA
Now let us move to subelliptic operators. We ﬁrst consider the famous example: Heisenberg
sub-Laplacian 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 Hamilton-Jacobi 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 sub-elliptic operator (1.1) and ξ
1
, ξ
2
and η are
dual variable to x
1
, x
2
and y respectively. Usin g the Lagrange-Chapit 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)
HAMILTON-JACOBI 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 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
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)
+ η|x 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.
HAMILTON-JACOBI EQUATION AND HEAT KERNEL 5
2. Generalized Hamilton-Jacobi equations
In this section we study the Hamilton-Jacobi equation which is crucial in th e construction
of the heat ker nel associated with elliptic and sub-elliptic operators. We ded uce a general
form of the solution to th e Hamilton-Jacobi 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 Hamilton-Jacobi 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 Hamilton-Jacobi 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
α
(0).
For the Heisenberg sub-Laplacian 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
,
HAMILTON-JACOBI 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 Hamilton-Jacobi 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
,
j
ds
=
H
x
j
,
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 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
˜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
.
HAMILTON-JACOBI 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 Hamilton-Jacobi 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, (0), 1) =
1
t
f(x, x
0
, y, ξ, (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 initial-b 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 y-axis but degenerate on the y-axis. Since [X
1
, X
2
] =
y
,
hence {X
1
, X
2
, [X
1
, X
2
]} spanned the tangent bundle of R
2
everywhere. By 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,
HAMILTON-JACOBI EQUATION AND HEAT KERNEL 11
“constant” means constant along the bicharacteristic curve”. Next,
¨x =
˙
ξ =
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
=
2
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),
HAMILTON-JACOBI 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 Hamilton-Jacobi 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 Hamilton-Jacobi 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
+
2
. An elementary calculation yields
f =
1
2
η
1
+
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
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
+
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 = y +
1
2
(x
2
0
+ x
2
)τ coth τ
τx
0
x
sinh τ
.
HAMILTON-JACOBI EQUATION AND HEAT KERNEL 15
2.6. Sub-Laplace operator on step 2 nilpotent Lie groups. Let M be a simply connected
2-step nilpotent Lie group G equipped 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 skew-symmetric bilinear form
B : H × H C
such that B(H, H) = C. The group law is given by
(H C) × (H C) H C
with
(x, y) (x
, y
) =
x + x
, y + y
+
1
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 skew-symm etric matrices
a
α
jk
j,k
, i.e., a
α
jk
= a
α
kj
.
We are interested in the sub-Laplacian
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(n1)
2
× m with element
a
α
jk
{(j<k)}
is of rank m (which implies that m
n(n1)
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(n1)
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 eld 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.
(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 initial-boundary conditions such that
(2.14)
x(0) = 0
x(t) = x = (x
1
, . . . , x
n
)
y(t) = y = (y
1
, . . . , y
m
)
η(0) = = i(τ
1
, . . . , τ
m
)
HAMILTON-JACOBI 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 skew-symmetric. 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) =
α
is pure imaginary, the matrix
isA(τ)
sinh(itA(τ ))
=
1
2πi
Z
γ
λ
sinh(λ)
λ itA(τ)
1
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
has an inverse:
1
2πi
Z
γ
sinh(λ)
λ
λ itA(τ)
1
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 xed t R, we have one-to-one 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
=1
A
ℓk
(τ)x
· 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(ts)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 initial-boundary 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
=1
a
α
ℓk
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 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,
HAMILTON-JACOBI 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 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
(2.17) 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 ) = 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 H-type 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 Hamilton-Jacobi 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(2|τ|)
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(2|τ|
l
),
where |x
l
|
2
=
P
3
j=0
x
2
4lj
, |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
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 sub-Laplacian
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
).
HAMILTON-JACOBI 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
=
α
, 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 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 )
t
V (τ).
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 )
t
t
q
=
e
f (x,y )
t
t
q+2
H(x, y,
x
f,
y
f) f
e
f (x,y )
t
t
q+1
X
(f) q
,
and the eiconal equation (2.3) implies that
X
t
e
f (x,y )
t
V (τ)
t
q
=
e
f (x,y )
t
t
q+1
m
X
α=1
τ
α
1
t
f
τ
α
V (τ)
e
f (x,y )
t
t
q+1
X
f q
V (τ)
=
e
f (x,y )
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 )
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 )
t
V (τ)
=
1
t
q+1
Z
m
α=1
Γ
α
e
f (x,y )
t
h
m
X
α=1
τ
α
V
τ
α
+
X
f q + m
V (τ )
i
= 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 co-dimension one form
V =
m
X
=1
(1)
1
V
1
· · ·
c
· · ·
m
22 OVIDIU CALIN, DER -CHEN CHANG, IRINA MARKINA
which satisﬁes a so-called “generalized transport equation”:
df
X
(V ) +
n
X
=1
X
(f)X
(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
=1
(1)
1
τ
k
V
τ
k
1
· · ·
c
· · ·
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
.
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
=
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))
qm
q
det sinh(A(τ))
.
If we propose the volume element V is real analytic and non-vanish at 0, then we have q =
n
2
+m.
Consequently,
(3.5) P =
A
(2πt)
q
Z
R
m
e
f (x,y )
t
V (τ),
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.
HAMILTON-JACOBI EQUATION AND HEAT KERNEL 23
Theorem 3.2. The equation
P
t
(x, y) =
A
(2πt)
q
Z
R
m
e
f (x,y )
t
V (τ),
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
t0
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 ub-Laplacian:
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) = y + (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 )
t
τ
sinh(2τ)
.
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 sub-Laplacian. 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 sub-Laplacian on H
1
. Let N
X
2
= hX
2
i = [{aX
2
}
aR
] 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 sub-Laplacian
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 )
t
s
|τ|
sinh |τ |
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 skew-symmetric 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