1.2. ‘FOURIER TRANSFORM’ FC ON THE ISOTROPIC CONE C 5
see Remark 2.4.10. In general, these operators P1,...,Pn satisfy the following
properties (see Theorem 2.4.1):
P1 PiPj = PjPi for any 1 i, j n.
P2 Pj R[x,

∂x
]C
for any 1 j n.
P3 The induced differential operators Pj|C on C0 ∞(C) extend to self-adjoint oper-
ators on the Hilbert space L2(C).
P4 (P1 2 + · · · + Pn1 2 Pn1+1 2 · · · Pn)|C 2 = 0.
P5 The Lie algebra generated by xi, Pi (1 i n) contains the vector fields
E, Xij (1 i j n).
From now on, we simply write Pj for Pj|C . Thus, we have commuting self-
adjoint, second-order differential operators P1,...,Pn on
L2(C).
I. Todorov drew
our attention that the fundamental differential operators P1,...,Pn appeared also
in [5, (3.4)] under the name “interior derivatives”.
We are brought naturally to the following:
Problem 1.1.1. 1) Find joint eigenfunctions of the differential operators
P1,...,Pn on the isotropic cone C.
2) Given a function f on C, find an explicit expansion formula of f into joint
eigenfunctions of P1,...,Pn.
1.2. ‘Fourier transform’ FC on the isotropic cone C
In this book, we shall give a solution to Problem 1.1.1 by introducing a unitary
operator FC on
L2(C).
To elucidate the operator FC, let us consider first much simpler operators
pj :=

−1

∂xj
(1 j n)
in place of Pj. Then, p1,...,pn form a commuting family of differential operators
which extend to self-adjoint operators on
L2(Rn).
Analogously to Problem 1.1.1,
consider the question of finding the explicit eigenfunction expansion for the op-
erators p1,...,pn. Then, as is well-known, this is done by using the (Euclidean)
Fourier transform F FRn on
Rn.
In what follows, we normalize FRn as
(1.2.1) FRn u(ξ) :=
1
(2π)
n
2 Rn
u(x)e

−1 x,ξ
dx,
where x, ξ =
∑n
i=1
xiξi and dx = dx1 · · · dxn. We note that the signature of the
power here is opposite from the usual convention. Obviously, the kernel
k(x, ξ) :=
1
(2π)
n
2
e

−1 x,ξ
of the Fourier transform FRn is real analytic on the direct product space Rn × Rn.
We recall the following key properties of the Euclidean Fourier transform:
F1 pj k(x, ξ) = ξj k(x, ξ).
F2 k(x, ξ) = k(ξ, x).
F3 FRn (C0
∞(Rn))

C∞(Rn)

L2(Rn).
F4 FRn extends to a unitary operator on L2(Rn).
F5 FRn xj = pj FRn ,
FRn pj = −ξj FRn .
F6 (FRn 2 u)(x) = u(−x), FRn 4 = id.
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