Polynomials Associated with Dihedral Groups
There is a commutative algebra of differential-difference operators, with two parameters, associated to any dihedral group with an even number of reflections. The intertwining operator relates this algebra to the algebra of partial derivatives. This paper presents an explicit form of the action of t...
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Dunkl, C.F. 2019-02-16T08:35:28Z 2019-02-16T08:35:28Z 2007 Polynomials Associated with Dihedral Groups / C.F. Dunkl // Symmetry, Integrability and Geometry: Methods and Applications. — 2007. — Т. 3. — Бібліогр.: 10 назв. — англ. 1815-0659 2000 Mathematics Subject Classification: 33C45; 33C80; 20F55 https://nasplib.isofts.kiev.ua/handle/123456789/147808 There is a commutative algebra of differential-difference operators, with two parameters, associated to any dihedral group with an even number of reflections. The intertwining operator relates this algebra to the algebra of partial derivatives. This paper presents an explicit form of the action of the intertwining operator on polynomials by use of harmonic and Jacobi polynomials. The last section of the paper deals with parameter values for which the formulae have singularities. en Інститут математики НАН України Symmetry, Integrability and Geometry: Methods and Applications Polynomials Associated with Dihedral Groups Article published earlier |
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Polynomials Associated with Dihedral Groups |
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Polynomials Associated with Dihedral Groups Dunkl, C.F. |
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Polynomials Associated with Dihedral Groups |
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Polynomials Associated with Dihedral Groups |
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Polynomials Associated with Dihedral Groups |
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Polynomials Associated with Dihedral Groups |
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polynomials associated with dihedral groups |
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Dunkl, C.F. |
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Dunkl, C.F. |
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2007 |
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English |
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Symmetry, Integrability and Geometry: Methods and Applications |
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Інститут математики НАН України |
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Article |
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There is a commutative algebra of differential-difference operators, with two parameters, associated to any dihedral group with an even number of reflections. The intertwining operator relates this algebra to the algebra of partial derivatives. This paper presents an explicit form of the action of the intertwining operator on polynomials by use of harmonic and Jacobi polynomials. The last section of the paper deals with parameter values for which the formulae have singularities.
|
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1815-0659 |
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https://nasplib.isofts.kiev.ua/handle/123456789/147808 |
| citation_txt |
Polynomials Associated with Dihedral Groups / C.F. Dunkl // Symmetry, Integrability and Geometry: Methods and Applications. — 2007. — Т. 3. — Бібліогр.: 10 назв. — англ. |
| work_keys_str_mv |
AT dunklcf polynomialsassociatedwithdihedralgroups |
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2025-11-24T16:26:16Z |
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2025-11-24T16:26:16Z |
| _version_ |
1850482758082625536 |
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 3 (2007), 052, 19 pages
Polynomials Associated with Dihedral Groups
Charles F. DUNKL
Department of Mathematics, University of Virginia, Charlottesville, VA 22904-4137, USA
E-mail: cfd5z@virginia.edu
URL: http://www.people.virginia.edu/∼cfd5z
Received February 06, 2007; Published online March 23, 2007
Original article is available at http://www.emis.de/journals/SIGMA/2007/052/
Abstract. There is a commutative algebra of differential-difference operators, with two
parameters, associated to any dihedral group with an even number of reflections. The
intertwining operator relates this algebra to the algebra of partial derivatives. This paper
presents an explicit form of the action of the intertwining operator on polynomials by use
of harmonic and Jacobi polynomials. The last section of the paper deals with parameter
values for which the formulae have singularities.
Key words: intertwining operator; Jacobi polynomials
2000 Mathematics Subject Classification: 33C45; 33C80; 20F55
1 Introduction
The dihedral group of type I2 (2s) acts on R2, contains 2s reflections and the rotations through
angles of mπ
s for 1 ≤ m ≤ 2s − 1, and is of order 4s, where s is a positive integer. It is the
symmetry group of the regular 2s-gon and has two conjugacy classes of reflections (the mirrors
passing through midpoints of pairs of opposite edges and those joining opposite vertices). There
is an associated commutative algebra of differential-difference (“Dunkl”) operators with two
parameters, denoted by κ0, κ1. It is convenient to use complex coordinates for R2, that is,
z = x1 + ix2, z = x1 − ix2. Notations like f (z) will be understood as functions of z, z;
except that f (z, z) will be used to indicate the result of interchanging z and z. Let N, N0, Q
denote the sets of positive integers, nonnegative integers and rational numbers, respectively. Let
ω = eiπ/s, then the reflections in the group are (z, z) 7→ (zωm, zω−m), 0 ≤ m < 2s and the
rotations are (z, z) 7→ (zωm, zω−m, ), 1 ≤ m < 2s. Note that f (zωm) is the abbreviated form of
f (zωm, zω−m). The differential-difference operators are defined by
Tf (z) :=
∂
∂z
f (z) + κ0
s−1∑
j=0
f (z)− f
(
zω2j
)
z − zω2j
+ κ1
s−1∑
j=0
f (z)− f
(
zω2j+1
)
z − zω2j+1
,
T f (z) :=
∂
∂z
f (z)− κ0
s−1∑
j=0
f (z)− f
(
zω2j
)
z − zω2j
ω2j − κ1
s−1∑
j=0
f (z)− f
(
zω2j+1
)
z − zω2j+1
ω2j+1,
for polynomials f (z). (The second formula implicitly uses the relation − ωm
z−zωm = 1
z−zω−m .) The
key fact is that T and T commute. The explicit action of T and T on monomials is given by
Tzazb = aza−1zb + s
b(a−b−1)/sc∑
j=0
(
κ0 + (−1)j κ1
)
za−1−jszb+js, (1.1)
Tzazb = bzazb−1 − s
b(a−b)/sc∑
j=1
(
κ0 + (−1)j κ1
)
za−jszb−1+js, (1.2)
mailto:cfd5z@virginia.edu
http://www.people.virginia.edu/~cfd5z
http://www.emis.de/journals/SIGMA/2007/052/
2 C.F. Dunkl
for a ≥ b; the relations remain valid when both (z, z) and
(
T, T
)
are interchanged. The Laplacian
is 4TT . These results are from [2, Section 3]. The harmonic polynomials and formulae (1.1)
and (1.2) also appear in Berenstein and Burman [1, Section 2]. The aim of this paper is to find
an explicit form of the intertwining operator V . This is the unique linear transformation that
maps homogeneous polynomials to homogeneous polynomials of the same degree and satisfies
TV f (z) = V
∂
∂z
f (z) , TV f (z) = V
∂
∂z
f (z) , V 1 = 1.
The operator was defined for general finite reflection groups in [4]. Rösler [8] proved that V is
a positive operator when κ0, κ1 > 0; this roughly means that if a polynomial f satisfies f (y) ≥ 0
for all y with ‖y‖ < R (for some R) then V f (y) ≥ 0 on the same set. The present paper does
not shed light on the positivity question since the formulae are purely algebraic. In Section 5 the
special case − (κ0 + κ1) ∈ N is considered in more detail. These values of (κ0, κ1) are apparent
singularities in the expressions for V zazb which are found in Section 4. The book by Y. Xu and
the author [7] is a convenient reference for the background of this paper.
In a way, to find V zazb only requires to solve a set of equations involving V zjzk for 0 ≤ j ≤ a,
0 ≤ k ≤ b. This can be implemented in computer algebra for small a, b but it is not really an
explicit description. For example, by direct computation we find that
V z2 =
(κ0 + κ1 + 1) z2 + (κ0 − κ1) z2
(2κ0 + 1) (2κ1 + 1) (2κ0 + 2κ1 + 1)
, s = 2,
V z2 =
2z2
(sκ0 + sκ1 + 1) (sκ0 + sκ1 + 2)
, s > 2.
The idea is to find the harmonic expansion of V zazb; suppose f (z) is (real-) homogeneous of
degree n then there is a unique expansion f (z) =
bn/2c∑
j=0
(zz)j fn−2j (z) where fn−2j is homo-
geneous of degree n − 2j and is harmonic, that is, TTfn−2j = 0, for 0 ≤ j ≤ n/2. There is
some more information easily available for the expansion of V zazb. Let n = a + b and suppose
V zazb =
n∑
j=0
cjz
n−jzj for certain coefficients cj . Because V commutes with the action of the
group we deduce that
V
(
(ωz)a (ωz)b
)
= ωa−b
n∑
j=0
cjz
n−jzj =
n∑
j=0
ωn−2jcjz
n−jzj ;
thus cj 6= 0 implies n− 2j ≡ a− b mod (2s) or j ≡ b mod s. Further
V
(
zazb
)
=
n∑
j=0
cjz
n−jzj ,
so it will suffice to determine V zazb for a ≥ b. We will use the Poisson kernel to calculate
the polynomials denoted Kn (x, y) := V x
(
1
n! (x1y1 + x2y2)
n) (see [5, p. 1219]), where y ∈ R2
and V x acts on the variable x. Thus V xn−j
1 xj
2 is j! (n− j)! times the coefficient of yn−j
1 yj
2
in Kn (x, y). This is adapted to complex coordinates by setting w = y1 + iy2, in which case
x1y1 + x2y2 = 1
2 (zw + zw).
2 The Poisson kernel
Actually it is only the series expansion of this kernel that is used. For now we assume κ0, κ1 ≥ 0.
The measure on the circle T :=
{
eiθ : −π < θ ≤ π
}
associated to the group I2 (2s) and the
Polynomials Associated with Dihedral Groups 3
operators T , T is
dµ
(
eiθ
)
:=
1
2B
(
κ0 + 1
2 , κ1 + 1
2
) (sin2 sθ
)κ0
(
cos2 sθ
)κ1 dθ.
Suppose g is a function of t = cos 2sθ then∫
T
g (t (θ)) dµ
(
eiθ
)
=
2−κ0−κ1
B
(
κ0 + 1
2 , κ1 + 1
2
) ∫ 1
−1
g (t) (1− t)κ0−1/2 (1 + t)κ1−1/2 dt.
The inner product in L2 (T, µ) is
〈f, g〉 :=
∫
T
f (z) g (z)dµ (z)
and ‖f‖ := 〈f, f〉1/2. Throughout the polynomials under consideration have real coefficients
so that g (z, z) = g (z, z). By the group invariance of µ the integral
∫
T zazbdµ (z)is real-valued
when a ≡ b mod (2s) and vanishes otherwise. There is an orthogonal decomposition L2 (T, µ) =
∞∑
n=0
⊕Hn; for n > 0 each Hn is of dimension two and consists of the polynomials in z, z (real-)
homogeneous of degree n and annihilated by TT (the harmonic property), while H0 consists of
the constant functions. The Poisson kernel is the reproducing kernel for harmonic polynomials
(for more details see [3, 5]). Xu [10] investigated relationships between harmonic polynomials,
the intertwining operator and the Poisson kernel for the general reflection group. The paper
of Scalas [9] concerns boundary value problems for the dihedral groups. The projection of the
kernel onto Hn is denoted by Pn (z, w) and satisfies∫
T
Pn (z, w) g (w) dµ (w) = g (z)
for each polynomial g ∈ Hn. There is a formula for Pn in terms of
{
Kn−2j : 0 ≤ j ≤ n
2
}
(see [5,
p. 1224]) which can be inverted. In the present case
Pn (z, w) =
bn/2c∑
j=0
(γ0)n
j! (2− n− γ0)j
2n−2j (zzww)j Kn−2j (z, w) ,
where γ0 = sκ0 + sκ1 + 1. The inverse relation is
Kn (z, w) = 2−n
bn/2c∑
j=0
1
j! (γ0)n−j
(zzww)j Pn−2j (z, w) . (2.1)
This is a consequence of the following:
Proposition 1. Suppose there are two sequences {ξn : n ∈ N0} and {ηn : n ∈ N0} in a vector
space over Q (γ0) where γ0 is transcendental, then
ξn =
bn/2c∑
j=0
(γ0)n
j! (2− n− γ0)j
ηn−2j , n ∈ N0,
if and only if
ηn =
bn/2c∑
j=0
1
j! (γ0)n−j
ξn−2j , n ∈ N0.
4 C.F. Dunkl
Proof. Consider the matrices A and B defined by ξn =
∑
j Ajnηj , ηn =
∑
j Bjnξj ; these
matrices are triangular and the diagonal entries are nonzero, hence they are nonsingular. It
suffices to show B is a one-sided inverse of A; this is actually finite-dimensional linear algebra,
since one can truncate to the range 0 ≤ n, j ≤ M for any M ∈ N. Indeed
bn/2c∑
j=0
(γ0)n
j! (2− n− γ0)j
bn/2−jc∑
i=0
1
i! (γ0)n−2j−i
ξn−2j−2i
=
bn/2c∑
k=0
ξn−2k
(γ0)n
k! (γ0)n−k
k∑
j=0
(−k)j (1− n− γ0 + k)j
j! (2− n− γ0)j
=
bn/2c∑
k=0
ξn−2k
(γ0)n (1− k)k
k! (γ0)n−k (2− n− γ0)k
= ξn;
using the substitution i = k − j we obtain (γ0)n−2j−i = (γ0)n−k−j =
(−1)j(γ0)n−k
(1−n−γ0+k)j
and 1
i! =
(−1)j (−k)j
k! ; the sum over j is found by the Chu–Vandermonde formula. �
Set ξn = Pn(z,w)
(zzww)n/2 and ηn = 2nKn(z,w)
(zzww)n/2 for n ∈ N0 to prove equation (2.1).
Suppose for each n ∈ N there exist a basis {hn1, hn2} and a biorthogonal basis {gn1, gn2}
for Hn with real coefficients in z, z (so hn1 (z, z) = hn1 (z, z), for example). Thus 〈hni, gnj〉 =
δij/λni,with structural constants λni. Then
Pn (z, w) =
2∑
i=1
λnihni (z, z) gni (w,w) . (2.2)
Once this is made sufficiently explicit we can compute Kn (z, w) and V zn−jzj . The description of
harmonic polynomials is in terms of the case s = 1 (corresponding to the group I2 (2) = Z2×Z2).
In terms of Jacobi polynomials the polynomials annihilated by T are:
f2n
(
reiθ
)
:= r2nP
(κ0− 1
2
,κ1− 1
2)
n (cos 2θ) +
i
2
(
r2 sin 2θ
)
r2n−2P
(κ0+ 1
2
,κ1+ 1
2)
n−1 (cos 2θ) , (2.3)
f2n+1
(
reiθ
)
:=
(
n + κ0 +
1
2
)
r cos θ r2nP
(κ0− 1
2
,κ1+ 1
2)
n (cos 2θ) (2.4)
+ i
(
n + κ1 +
1
2
)
r sin θ r2nP
(κ0+ 1
2
,κ1− 1
2)
n (cos 2θ) ;
where the subscript indicates the degree of homogeneity, (clearly fn is a polynomial with real
coefficients in z, z; cos 2θ =
(
z2 + z2
)
/ (2zz) and i
2
(
r2 sin 2θ
)
= 1
4
(
z2 − z2
)
). The real and
imaginary parts form a basis for the harmonic polynomials. Specifically let
f0
n (z) := Re fn (z) , f1
n (z) := i Im fn (z) .
This implies that both f0
n and f1
n have real coefficients in z, z and f0
n (z, z) = f0
n (z, z) , f1
n (z, z) =
−f1
n (z, z). When s > 1 and 1 ≤ t < s it is known [2, p. 182] that
{
ztfn (zs) , ztfn (zs)
}
is
an orthogonal basis for Hns+t for n ≥ 0. Henceforth we denote hns+t,1 (z) = gns+t,1 (z) =
ztfn (zs) = hns+t,2 = gns+t,2 and λns+t,1 = λns+t,2 = ‖fn‖−2. The integral
〈
ztfn (zs) , ztfn (zs)
〉
reduces to the case s = 1 and t = 0. When s ≥ 1
{
f0
n (zs) , f1
n (zs)
}
is an orthogonal basis
for Hns and zsfn−1 (zs) is orthogonal to fn (zs). By orthogonality ‖fn‖2 =
∥∥f0
n
∥∥2 +
∥∥f1
n
∥∥2 and
Polynomials Associated with Dihedral Groups 5
the latter two norms are standard Jacobi polynomial facts. The associated structural constants
are denoted by labeled λ’s. Thus
λ0
2n :=
∥∥f0
2n
∥∥−2 =
n! (κ0 + κ1 + 1)n (κ0 + κ1 + 2n)(
κ0 + 1
2
)
n
(
κ1 + 1
2
)
n
(κ0 + κ1 + n)
, (2.5)
λ1
2n :=
∥∥f1
2n
∥∥−2 =
(n− 1)! (κ0 + κ1 + 1)n (κ0 + κ1 + 2n)(
κ0 + 1
2
)
n
(
κ1 + 1
2
)
n
, (2.6)
λ2n := ‖f2n‖−2 =
n! (κ0 + κ1 + 1)n(
κ0 + 1
2
)
n
(
κ1 + 1
2
)
n
; (2.7)
and
λ0
2n+1 :=
∥∥f0
2n+1
∥∥−2 =
n! (κ0 + κ1 + 1)n (κ0 + κ1 + 2n + 1)(
n + κ0 + 1
2
) (
κ0 + 1
2
)
n+1
(
κ1 + 1
2
)
n+1
, (2.8)
λ1
2n+1 :=
∥∥f1
2n+1
∥∥−2 =
n! (κ0 + κ1 + 1)n (κ0 + κ1 + 2n + 1)(
n + κ1 + 1
2
) (
κ0 + 1
2
)
n+1
(
κ1 + 1
2
)
n+1
, (2.9)
λ2n+1 := ‖f2n+1‖−2 =
n! (κ0 + κ1 + 1)n(
κ0 + 1
2
)
n+1
(
κ1 + 1
2
)
n+1
. (2.10)
From this point on we no longer need the measure µ on the circle. Only the algebraic expressions
are used. The condition κ0, κ1 ≥ 0 is replaced by the requirement that none of −κ0+ 1
2 , −κ1+ 1
2 ,
−s (κ0 + κ1) equal a positive integer. The exceptional case − (κ0 + κ1) ∈ N is taken up in the
last section. In the next section we compute the structural constants for the biorthogonal bases
{fn (zs) , fn (zs)} and {zsfn−1 (zs) , zsfn−1 (zs)} for Hns (see [3, p. 461]. It is easier to carry this
out with material developed in the next section.
3 Expressions for coefficients
This is a detailed study of the coefficients of fn (z) in terms of powers of z, z. The expressions
are in the form of a single sum of hypergeometric 3F2-type, and can not be simplified any
further. For a polynomial f in z, z define c (f ; a, b) to be the coefficient of zazb in f , that is,
f (z, z) =
∑
a,b≥0
c (f ; a, b) zazb.
Since we restrict to polynomials with real coefficients the equation c
(
f (z); a, b
)
= c (f ; b, a) is
valid. Further c (f (zs) ; as, bs) = c (f ; a, b). Recall
Kn (z, w) :=
1
2nn!
V z ((zw + zw)n) ,
thus V zn−jzj is 2nj! (n− j)! times the coefficient of wjwn−j in Kn (z, w). To adapt the notation
from equation (2.2) for P0 set h01 = g01 = λ01 = 1 and h02 = g02 = λ02 = 0. Then
Kn (z, w) = 2−n
bn/2c∑
j=0
1
j! (sκ0 + sκ1 + 1)n−j
(3.1)
× (zzww)j
2∑
i=1
λn−2j,ihn−2j,i (z, z) gn−2j,i (w,w) .
6 C.F. Dunkl
Proposition 2. For 0 ≤ m ≤ n,
V
(
zn−mzm
)
= m! (n−m)!
bn/2c∑
j=0
1
j! (sκ0 + sκ1 + 1)n−j
(3.2)
× (zz)j
2∑
i=1
λn−2j,ic (gn−2j,i;n−m− j, m− j) hn−2j,i (z, z) .
The nonzero terms appear at increments (in j) of 2s. We start by finding c
(
f0
n;n− j, j
)
and
c
(
f1
n;n− j, j
)
. This is straightforward and will serve as motivation for introducing a specific
useful 3F2-series. Consider f0
2n (z) and recall that
P (α,β)
n (t) =
(α + 1)n
n! 2F1
(
−n, n + α + β + 1
α + 1
;
1− t
2
)
.
When z = reiθ we have 1
2 (1− cos 2θ) = − (z − z)2 / (4zz) so
r2nP
(κ0− 1
2
,κ1− 1
2)
n (cos 2θ)
=
(
κ0 + 1
2
)
n
n!
n∑
l=0
2l∑
i=0
(−n)l (n + κ0 + κ1)l (2l)!
l!
(
κ0 + 1
2
)
l
i! (2l − i)!
2−2lzn+l−izn−l+i (−1)l+i
=
(
κ0 + 1
2
)
n
n!
n∑
j=−n
(−1)j zn+jzn−j
n−j∑
i=max(−2j,0)
(−n)j+i (n + κ0 + κ1)j+i
(
1
2
)
j+i
i!
(
κ0 + 1
2
)
j+i
(2j + i)!
;
(substituting l = i + j, so 0 ≤ i + j ≤ n and 0 ≤ i ≤ 2i + 2j are the ranges of the summation)
by the (z, z)-symmetry it suffices to consider j ≥ 0. Thus
c
(
f0
2n;n + j, n− j
)
=
(
κ0 + 1
2
)
n
(−n)j (n + κ0 + κ1)j
(
1
2
)
j(
κ0 + 1
2
)
j
(2j)!n!
(−1)j
×
n−j∑
i=0
(j − n)i (n + κ0 + κ1 + j)i
(
1
2 + j
)
i
i!
(
κ0 + 1
2 + j
)
i
(2j + 1)i
=
(n + κ0 + κ1)j
(
κ0 + 1
2 + j
)
n−j
22jj! (n− j)! 3F2
(
j − n, n + κ0 + κ1 + j, j + 1
2
κ0 + 1
2 + j, 2j + 1
; 1
)
;
this used (2j)! = 22jj!
(
1
2
)
j
and
(
κ0 + 1
2
)
n
/
(
κ0 + 1
2
)
j
=
(
κ0 + 1
2 + j
)
n−j
. The sum, which
appears to be a mysterious combination of the parameters, actually has a nice form revealing
more useful information.
Definition 1. For n ∈ N0 and parameters a, b, c1, c2 let
En (a, b; c1, c2) :=
(a)n (c2)n
n! (c1 + c2)n
3F2
(
−n, b, c1
1− n− a, 1− c2 − n
; 1
)
=
1
n! (c1 + c2)n
n∑
j=0
(−n)j
j!
(a)n−j (b)j (c1)j (c2)n−j .
Observe the symmetry En (a, b; c1, c2) = (−1)n En (b, a; c2, c1). This follows from manipula-
tions such as (a)n−j = (−1)j (a)n / (1− n− a)j . The following transformation is relevant to the
calculation of coefficients.
Polynomials Associated with Dihedral Groups 7
Proposition 3. For n ∈ N0 and parameters a, b, c1, c2
En (a, b; c1, c2) =
(a + c1)n
n! 3F2
(
−n, n + a + b + c1 + c2 − 1, c1
a + c1, c1 + c2
; 1
)
.
Proof. Use the transformation
3F2
(
−n, A,B
C,D
; 1
)
=
(D −B)n
(D)n
3F2
(
−n, C −A,B
C, 1 + B −D − n
; 1
)
.
First set A = b, B = c1, C = 1− n− a,D = 1− n− c2 then
(a)n (c2)n
n! (c1 + c2)n
3F2
(
−n, b, c1
1− n− a, 1− c2 − n
; 1
)
=
(a)n
n! 3F2
(
−n, 1− n− a− b, c1
1− n− a, c1 + c2
; 1
)
.
Set A = 1− n− a− b, B = c1, C = c1 + c2, D = 1− n− a to obtain the stated formula. In the
calculation the reversal such as (1− n− a)n = (−1)n (a)n is used several times. �
We arrive at a pleasing formula:
c
(
f0
2n;n + j, n− j
)
=
(n + κ0 + κ1)j
22jj!
En−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
.
It is useful because it clearly displays the result of setting one or both parameters equal to zero (or
a negative integer). That is En (κ0, 0; c1, c2) = (κ0)n(c2)n
n!(c1+c2)n
and n ≥ 1 implies En (0, 0; c1, c2) = 0.
(When κ0 = κ1 = 0 the polynomial f0
2n is a multiple of the Chebyshev polynomial of the
first kind, that is f0
2n (z) = (n)n
n!22n
(
z2n + z2n
)
, a fact obvious from the definition of f0
2n.) The
remaining basis polynomials can all be expressed in terms of the function E.
Proposition 4. For n ∈ N
f0
2n (z) =
n∑
j=1
(
zn+jzn−j + zn−jzn+j
) 1
22jj!
× (n + κ0 + κ1)j En−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
+ znznEn
(
κ0, κ1;
1
2
,
1
2
)
,
f1
2n (z) =
n∑
j=1
(
zn+jzn−j − zn−jzn+j
) 1
22j (j − 1)!
× (n + κ0 + κ1 + 1)j−1 En−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
.
Proof. The expansion for f0
2n has already been determined. Next i
2r2 sin 2θ = 1
4
(
z2 − z2
)
so(
i
2
r2 sin 2θ
)
r2n−2P
(κ0+ 1
2
,κ1+ 1
2)
n−1 (cos 2θ) =
(
κ0 + 3
2
)
n−1
(n− 1)!
×
n−1∑
l=0
(1− n)l (n + κ0 + κ1 + 1)l
l!
(
κ0 + 3
2
)
l
2−2l−2 (−1)l (z + z) (z − z)2l+1 (zz)n−1−l ,
and
2−2l−2 (z + z) (z − z)2l+1 (zz)n−1−l
= 2−2l−2
2l+2∑
i=0
(2l + 1)! (2l + 2− 2i)
i! (2l + 2− i)!
(−1)i zn+l+1−izn−l−1+i
8 C.F. Dunkl
=
2l+2∑
i=0
l!
(
1
2
)
l+1
(l + 1− i)
i! (2l + 2− i)!
(−1)i zn+l+1−izn−l−1+i;
substitute l = j+i−1. By the symmetry f1
2n (z) = −f1
2n (z) it suffices to find c
(
f1
2n;n + j, n− j
)
for 1 ≤ j ≤ n. Indeed
c
(
f1
2n;n + j, n− j
)
=
j
(
κ0 + 3
2
)
n−1
(1− n)j−1 (n + κ0 + κ1 + 1)j−1
(
1
2
)
j(
κ0 + 3
2
)
j−1
(2j)! (n− 1)!
(−1)j−1
×
n−j∑
i=0
(j − n)i (n + κ0 + κ1 + j)i
(
1
2 + j
)
i
i!
(
κ0 + 1
2 + j
)
i
(2j + 1)i
=
(n + κ0 + κ1 + 1)j−1
22j (j − 1)!
En−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
.
This completes the proof. �
Proposition 5. For n ∈ N0
f0
2n+1 (z) =
(
n + κ0 +
1
2
) n∑
j=0
(
zn+1+jzn−j + zn−jzn+1+j
) 1
22j+1j!
× (n + κ0 + κ1 + 1)j En−j
(
κ0, κ1; j +
1
2
, j +
3
2
)
,
f1
2n+1 (z) =
(
n + κ1 +
1
2
) n∑
j=0
(
zn+1+jzn−j − zn−jzn+1+j
) 1
22j+1j!
× (n + κ0 + κ1 + 1)j En−j
(
κ0, κ1; j +
3
2
, j +
1
2
)
.
Proof. The second equation is straightforward:
f1
2n+1 (z) =
1
2
(
n + κ1 +
1
2
)
(z − z) r2nP
(κ0+ 1
2
,κ1− 1
2)
n (cos 2θ) =
(
n + κ1 +
1
2
) (
κ0 + 3
2
)
n
n!
×
n∑
l=0
2l+1∑
i=0
(−n)l (n + κ0 + κ1 + 1)l (2l + 1)!
l!
(
κ0 + 3
2
)
l
i! (2l + 1− i)!
2−2l−1zn+1+l−izn−l+i (−1)l+i
=
(
n + κ1 +
1
2
) (
κ0 + 3
2
)
n
n!
n∑
j=0
(
zn+1+jzn−j − zn−jzn+1−j
)
× (−1)j
(−n)j (n + κ0 + κ1 + 1)j
(
1
2
)
j+1(
κ0 + 3
2
)
j
(2j + 1)!
n−j∑
i=0
(j − n)i (n + κ0 + κ1 + j + 1)i
(
3
2 + j
)
i
i!
(
κ0 + 3
2 + j
)
i
(2j + 2)i
,
(substituting l = j + i for 0 ≤ j ≤ n) thus
c
(
f1
2n+1;n + 1 + j, n− j
)
= −c
(
f1
2n+1;n− j, n + 1 + j
)
=
(
n + κ1 +
1
2
) (n + κ0 + κ1 + 1)j
j!22j+1
En−j
(
κ0, κ1; j +
3
2
, j +
1
2
)
.
Note that (2j + 1)! = 22j+1j!
(
1
2
)
j+1
and
(
κ0 + 3
2
)
n
/
(
κ0 + 3
2
)
j
=
(
κ0 + 3
2 + j
)
n−j
. For f0
2n+1
reverse the parameters, that is,
P
(κ0− 1
2
,κ1+ 1
2)
n (cos 2θ) = (−1)n P
(κ1+ 1
2
,κ0− 1
2)
n (− cos 2θ) ,
Polynomials Associated with Dihedral Groups 9
and note 1
2 (1 + cos 2θ) = (z + z)2 / (4zz) and r cos θ = 1
2 (z + z). Thus
f0
2n+1 (z) = (−1)n
(
n + κ0 +
1
2
) (
κ1 + 3
2
)
n
n!
×
n∑
l=0
2l+1∑
i=0
(−n)l (n + κ0 + κ1 + 1)l (2l + 1)!
l!
(
κ1 + 3
2
)
l
i! (2l + 1− i)!
2−2l−1zn+1+l−izn−l+i
= (−1)n
(
n + κ0 +
1
2
) (
κ1 + 3
2
)
n
n!
n∑
j=0
(
zn+1+jzn−j + zn−jzn+1−j
)
×
(−n)j (n + κ0 + κ1 + 1)j
(
1
2
)
j+1(
κ1 + 3
2
)
j
(2j + 1)!
n−j∑
i=0
(j − n)i (n + κ0 + κ1 + j + 1)i
(
3
2 + j
)
i
i!
(
κ1 + 3
2 + j
)
i
(2j + 2)i
,
thus
c
(
f0
2n+1;n + 1 + j, n− j
)
= c
(
f0
2n+1;n− j, n + 1 + j
)
=
(
n + κ0 +
1
2
) (n + κ0 + κ1 + 1)j
j!22j+1
(−1)n−j En−j
(
κ1, κ0; j +
3
2
, j +
1
2
)
.
The symmetry relation Em (b, a; c2, c1) = (−1)m Em (a, b; c1, c2) finishes the computation. �
To find the coefficients of fn we use contiguity relations satisfied by Em.
Lemma 1. For m ∈ N0 and parameters a, b, c
(m + a + c) Em (a, b; c, c + 1)− (m + b + c) Em (a, b; c + 1, c) (3.3)
= 2 (m + 1) Em+1 (a, b; c, c) ,
(m + a + c) Em (a, b; c, c + 1) + (m + b + c) Em (a, b; c + 1, c) (3.4)
=
m + 2c + 1
2c + 1
(m + a + b + 2c) Em (a, b; c + 1, c + 1) .
Proof. We compute the coefficient of (b)j for 0 ≤ j ≤ m + 1 in the two identities. Note that
(m + b + c) (b)j = (b)j+1 + (m + c− j) (b)j , then replace j by j − 1 for the first term. The
coefficient of (b)j in (m + b + c) Em (a, b; c + 1, c) is
1
m! (2c + 1)m j!
×
{
(−m)j (m + c− j) (a)m−j (c)m−j (c + 1)j + j (−m)j−1 (a)m+1−j (c)m+1−j (c + 1)j−1
}
=
(−m)j−1
m! (2c + 1)m j!
(a)m−j (c)m+1−j (c + 1)j−1 {(−m + j − 1) (c + j) + j (a + m− j)} .
The coefficient of (b)j in the left side of (3.3) is
(−m)j−1
m! (2c + 1)m j!
(a)m−j (c)j (c + 1)m−j
× {(m + a + c) (−m + j − 1)− (−m + j − 1) (c + j)− j (a + m− j)}
=
(−m)j−1
m! (2c + 1)m j!
(a)m−j (c)j (c + 1)m−j (a + m− j) (−m− 1)
=
2 (−1−m)j
m! (2c)m+1 j!
(a)m+1−j (c)j (c)m+1−j .
10 C.F. Dunkl
This proves equation (3.3). For the right side of (3.4) the coefficient of (b)j is found similarly as
before ((m + a + b + 2c) (b)j = (m + a + 2c− j) (b)j +(b)j+1, and so on). The coefficient of (b)j
in the left side is
(−m)j−1
m! (2c + 1)m j!
(a)m−j (c)j (c + 1)m−j {(m + a + 2c) (−m + j − 1) + j (a− 1)} ,
and in the right side
m + 2c + 1
m! (2c + 1) (2c + 2)m
{
(−m)j
j!
(m + a + 2c− j) (a)m−j (c + 1)j (c + 1)m−j
+
(−m)j−1
j!
j (a)m+1−j (c + 1)j−1 (c + 1)m+1−j
}
=
(−m)j−1
m! (2c + 1)m j!
(a)m−j (c + 1)j−1 (c + 1)m−j
× {(−m + j − 1) (m + a + 2c− j) (c + j) + j (a + m− j) (c + m + 1− j)} ;
the expression in {·} equals c (m + a + 2c) (−m + j − 1) + cj (a− 1) which proves (3.4). �
Proposition 6. For n ∈ N0
f2n (z) =
n∑
j=1
(
(n + κ0 + κ1 + j) zn+jzn−j + (n + κ0 + κ1 − j) zn−jzn+j
)
× 1
22jj!
(n + κ0 + κ1 + 1)j−1 En−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
+ En
(
κ0, κ1;
1
2
,
1
2
)
znzn,
f2n+1 (z) =
n+1∑
j=1
(
(n + 1 + j) zn+jzn+1−j + (n + 1− j) zn−jzn+1+j
)
× 1
22jj!
(n + κ0 + κ1 + 1)j En+1−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
+ (n + 1) En+1
(
κ0, κ1;
1
2
,
1
2
)
znzn+1.
Proof. Recall fn = f0
n + f1
n. For 0 ≤ j ≤ n from Proposition 4 we find
c (f2n;n + j, n− j) = c
(
f0
2n;n + j, n− j
)
+ c
(
f1
2n;n + j, n− j
)
=
(n + κ0 + κ1 + 1)j−1
22jj!
((n + κ0 + κ1) + j) En−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
,
c (f2n;n− j, n + j) = c
(
f0
2n;n + j, n− j
)
− c
(
f1
2n;n + j, n− j
)
=
(n + κ0 + κ1 + 1)j−1
22jj!
((n + κ0 + κ1)− j) En−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
.
It remains to compute c (f2n+1;n + 1 + j, n− j) and c (f2n+1;n− j, n + 1− j) for 0 ≤ j ≤ n.
Write the arguments as
(
n + 1
2 + ε
(
j + 1
2
)
, n + 1
2 − ε
(
j + 1
2
))
with ε = ±1. Then, by Proposi-
tion 5,
c
(
f2n+1;n +
1
2
+ ε
(
j +
1
2
)
, n +
1
2
− ε
(
j +
1
2
))
Polynomials Associated with Dihedral Groups 11
= c
(
f0
2n+1;n + 1 + j, n− j
)
+ εc
(
f1
2n+1;n + 1 + j, n− j
)
=
(n + κ0 + κ1 + 1)j
j!22j+1
{(
n + κ0 +
1
2
)
En−j
(
κ0, κ1; j +
1
2
, j +
3
2
)
+ ε
(
n + κ1 +
1
2
)
En−j
(
κ0, κ1; j +
3
2
, j +
1
2
)}
.
When ε = 1 by (3.4) we obtain
c (f2n+1;n + 1 + j, n− j) =
(n + κ0 + κ1 + 1)j+1
(j + 1)!22j+2
(n + j + 2) En−j
(
κ0, κ1; j +
3
2
, j +
3
2
)
,
and when ε = −1 by (3.3) we obtain
c (f2n+1;n− j, n + 1− j) =
(n + κ0 + κ1 + 1)j
j!22j
(n− j + 1) En−j+1
(
κ0, κ1; j +
1
2
, j +
1
2
)
.
The stated formula for f2n+1 uses c (f2n+1;n + j, n + 1− j) explicitly (j is shifted by 1). �
For Hns with n > 0 we intend to use both the orthogonal basis
{
f0
n (zs) , f1
n (zs)
}
as well as
the biorthogonal bases
{
fn (zs) , fn (zs)
}
and
{
zsfn−1 (zs) , zsfn−1 (zs)
}
. For the latter we need
the value of νn := 〈fn (zs) , zsfn−1 (zs)〉. Instead of doing the integral directly we use the two
formulae for Pns (z, w), that is,
Pns (z, w) = λ0
nf0
n (zs) f0
n (ws) + λ1
nf1
n (zs) f1
n (ws)
= ν−1
n (fn (zs) wsfn−1 (ws) + fn (zs) wsfn−1 (ws)) .
From the coefficients of wns in the equation we obtain
λ0
nc
(
f0
n; 0, n
)
f0
n (zs) + λ1
nc
(
f1
n;n, 0
)
f1
n (zs) = ν−1
n c (fn−1;n− 1, 0) fn (zs) .
But fn = f0
n + f1
n so by the linear independence of
{
f0
n, f1
n
}
there are two equations for cn (one
is redundant). Thus
νn =
c (fn−1;n− 1, 0)
λ0
nc (f0
n; 0, n)
=
c (fn−1;n− 1, 0)
λ1
nc (f1
n;n, 0)
.
The calculation has two cases depending on n being even or odd:
ν2n =
2
(
κ0 + 1
2
)
n
(
κ1 + 1
2
)
n
(n− 1)! (κ0 + κ1 + 1)n−1 (κ0 + κ1 + 2n)
, n ≥ 1, (3.5)
ν2n+1 =
2
(
κ0 + 1
2
)
n+1
(
κ1 + 1
2
)
n+1
n! (κ0 + κ1 + 1)n (κ0 + κ1 + 2n + 1)
, n ≥ 0. (3.6)
4 The intertwining operator
We describe V zazb for a ≥ b. It is helpful to consider the representations of I2 (2s) since V
commutes with the group action on polynomials. Since zz is invariant it suffices to consider
(zz)b za−b, or zm. The residue of m mod2s is the determining factor. Suppose m ≡ j mod2s and
j 6= 0, s. The representation of I2 (2s) on span {zm, zm} is irreducible and isomorphic to the one
on span
{
zj , zj
}
if 1 ≤ j < s, and to the one on span
{
z2s−j , z2s−j
}
if s < j < 2s. If m ≡ 0 mod 2s
then span {zm, zm} is the direct sum of the identity and determinant representations (on C1 and
12 C.F. Dunkl
C
(
z2s − z2s
)
respectively). If m ≡ smod2s then span {zm, zm} is the direct sum of the two
representations realized on C (zs − zs) and C (zs + zs) (these are relative invariants). Recall
Pm (z, w) =
2∑
i=1
λmihmi (z, z) gmi (w,w) and equation (3.2) shows that the nonzero terms in the
expansion of V zazb occur only when the condition
c (ga+b−2j,i; a− j, b− j) 6= 0 (4.1)
is satisfied. If m ≡ 0 mod s then gmi (w,w) is a polynomial in ws, ws thus (4.1) is equivalent to
a− j ≡ b− j ≡ 0 mod s, in particular a ≡ b mod s. In this case suppose a = us + r ≥ b = vs + r
with 0 ≤ r < s. Set b−j = (v − k) s then j = ks+r, a−j = (u− k) s, a+b−2j = (u + v − 2k) s,
and 0 ≤ k ≤ v ≤ u. We see that the nonzero terms occur for P(u+v−2k)s with 0 ≤ k ≤ v.
If m ≡ t mod s and 1 ≤ t < s then gm1 (w,w) = wtf(m−t)/s (ws, ws) and (4.1) implies
a−j ≡ t mod s, b−j ≡ 0 mod s; further gm2 (w,w) = gm1 (w,w) and (4.1) implies a−j ≡ 0 mod s,
b− j ≡ t mod s.
Theorem 1. Suppose a − b ≡ t mod s, 1 ≤ t < s and a > b. Let b = vs + r with v ≥ 0 and
0 ≤ r < s and a = us + r + t, then
V
(
zazb
)
= a!b!
v∑
k=0
1
(ks + r)! (sκ0 + sκ1 + 1)a+(v−k)s
λu+v−2k
× c (fu+v−2k;u− k, v − k) (zz)ks+r ztfu+v−2k (zs)
+ a!b!
v∑
k=1−b(r+t)/sc
1
((k − 1) s + r + t)! (sκ0 + sκ1 + 1)b+(u−k+1)s
λu+v+1−2k
× c (fu+v+1−2k; v − k, u− k + 1) (zz)(k−1)s+r+t zs−tfu+v+1−2k (zs) .
Proof. Since 0 < a − b = (u− v) s + t we have u ≥ v. For the first part of the series,
corresponding to i = 1 in Pns+t let b− j = (v − k) s with k ≤ v; then j = b− (v − k) s = ks + r,
implying k ≥ 0. Further a − j = (u− k) s + t, a + b − 2j = (u + v − 2k) s + t = a + b −
2r − 2ks (and a + b − j = a + (b− j) = a + (v − k) s). Also c
(
ztfu+v−2k (zs) ; a− j, b− j
)
=
c (fu+v−2k;u− k, v − k) . This proves the first part. For the second part, with i = 2 in Pns−t
let b − j = (v − k) s + (s− t), thus k ≤ v. Then j = (k − 1) s + r + t. The requirement j ≥ 0
implies 1− k ≤ r+t
s , that is k ≥ 1−
⌊
r+t
s
⌋
(if 0 ≤ r + t < s then k ≥ 1, otherwise s ≤ r + t < 2s
and k ≥ 0). Also a − j = (u− k + 1) s and a + b − 2j = (u + v + 1− 2k) s + (s− t) (and
a+b−j = b+(a− j) = b+(u− k + 1) s). In this case we use c
(
zs−tfu+v+1−2k (zs) ; a− j, b− j
)
=
c (fu+v+1−2k; v − k, u− k + 1). �
Note that the degrees of fm have the same parity as u + v in the first sum, and the opposite
in the second sum. By Proposition 6 we can find the coefficients explicitly. If u+v is even then
c (fu+v−2k;u− k, v − k) =
1
2u−v
(
u−v
2
)
!
(
u + v
2
− k + κ0 + κ1
)
u−v
2
× Ev−k
(
κ0, κ1;
u− v + 1
2
,
u− v + 1
2
)
,
and
c (fu+v+1−2k; v − k, u− k + 1) =
(v − k + 1)
2u−v
(
u−v
2
)
!
(
u + v
2
− k + κ0 + κ1 + 1
)
u−v
2
× Ev−k+1
(
κ0, κ1;
u− v + 1
2
,
u− v + 1
2
)
.
Polynomials Associated with Dihedral Groups 13
If u + v is odd then
c (fu+v−2k;u− k, v − k) =
(u− k + 1)
2u−v+1
(
u−v+1
2
)
!
(
u + v + 1
2
− k + κ0 + κ1
)
u−v+1
2
× Ev−k
(
κ0, κ1;
u− v
2
+ 1,
u− v
2
+ 1
)
,
and
c (fu+v+1−2k; v − k, u− k + 1) =
(v − k + 1)
2u−v+1
(
u−v+1
2
)
!
(
u + v + 3
2
− k + κ0 + κ1
)
u−v−1
2
× Ev−k
(
κ0, κ1;
u− v
2
+ 1,
u− v
2
+ 1
)
.
Theorem 2. Suppose a ≡ b mod s, and a ≥ b. Let a = us + r ≥ b = vs + r with 0 ≤ r < s and
v ≥ 0. If a > b then
V
(
zazb
)
= a!b!
v∑
k=0
1
(ks + r)! (sκ0 + sκ1 + 1)b+(u−k)s
ν−1
u+v−2k
× (zz)ks+r {c (fu+v−2k−1;u− k − 1, v − k) fu+v−2k (zs)
+ c (fu+v−2k−1; v − k − 1, u− k) fu+v−2k (zs)},
V
(
1
2
(
zazb − zbza
))
= a!b!
v∑
k=0
1
(ks + r)! (sκ0 + sκ1 + 1)b+(u−k)s
λ1
u+v−2k
× c
(
f1
u+v−2k;u− k, v − k
)
(zz)ks+r f1
u+v−2k (zs) .
If a ≥ b then
V
(
1
2
(
zazb + zbza
))
= a!b!
v∑
k=0
1
(ks + r)! (sκ0 + sκ1 + 1)b+(u−k)s
λ0
u+v−2k
× c
(
f0
u+v−2k;u− k, v − k
)
(zz)ks+r f0
u+v−2k (zs) .
Proof. The three different expansions for zazb, 1
2
(
zazb − zbza
)
and 1
2
(
zazb + zbza
)
use the
bases
{
fj , fj
}
,
{
f1
j
}
and
{
f0
j
}
respectively. Suppose a = us + r ≥ b = vs + r with 0 ≤ r < s.
Set b − j = (v − k) s then j = ks + r, a − j = (u− k) s, a + b − 2j = (u + v − 2k) s, and
0 ≤ k ≤ v ≤ u. Consider the case a > b, that is, u > v. For arbitrary m ≥ 1 the basis
{fm (zs, zs) , fm (zs, zs)} for Hsm has the biorthogonal set {zsfm−1 (zs, zs) , zsfm−1 (zs, zs)} and
c (zsfn1+n2−1 (zs, zs) ;n1s, n2s) = c (fn1+n2−1;n1 − 1, n2) ,
c (zsfn1+n2−1 (zs, zs) ;n1s, n2s) = c (fn1+n2−1;n2 − 1, n1) .
The constants νm are given in equations (3.5) and (3.6). This demonstrates the first series. The
remaining two follow from Proposition 3.2. �
Observe that in the series for V
(
zazb
)
the lowest-degree term with k = v < u reduces to
one summand since c (fu−v−1;−1, u− v) = 0. Each term in V
(
zazb − zbza
)
is of the same
representation type, C (zs − zs) when a− b ≡ smod2s or C
(
z2s − z2s
)
when a− b ≡ 0 mod 2s.
Similarly each term in V
(
zazb + zbza
)
is of the representation type C (zs + zs) or C1 (depending
on the parity of a−b
s ). The coefficients can be found from Propositions 4 and 5.
14 C.F. Dunkl
For a > b consider zazb as (zz)b times the (ordinary) harmonic polynomial za−b. The fact
that V
(
zazb
)
is L2 (T, µ)-orthogonal to Hn for n < a− b, equivalently, that the above series for
V
(
zazb
)
contain no terms involving Hn with n < a−b (that is, a term like cn (zz)(a+b−n)/2 pn (z)
with pn ∈ Hn), is a special case of a result of Xu [10]. This paper also has formulae for V z2m
when s = 2, that is, the group I2 (4).
5 Singular values
The term “singular values” refers to the set Ks of pairs (κ0, κ1) ∈ C2 for which V is not defined
on all polynomials in z, z. Let
K0 :=
{
(κ0, κ1) ∈ C2 : {κ0, κ1} ∩
(
−1
2 − N0
)
6= ∅
}
,
(at least one of κ0, κ1 is in
{
−1
2 ,−3
2 , . . .
}
). It was shown by de Jeu, Opdam and the author [6,
p. 248] that Ks = K0 ∪
{
(κ0, κ1) : κ0 + κ1 = − j
s , j ∈ N, j
s /∈ N
}
. To illustrate how the singular
values appear in the formulae for V consider V z2ns+1 (for s > 1, n ≥ 1) which has only one term
in the formula from Theorem 1. In particular
c
(
V z2ns+1; 2ns + 1, 0
)
=
(2ns + 1)! (κ0 + κ1 + 1)2n (n + κ0 + κ1 + 1)n
24nn!
(
κ0 + 1
2
)
n
(
κ1 + 1
2
)
n
(sκ0 + sκ1 + 1)2ns+1
.
The denominator vanishes for κ0, κ1 = −1
2 ,−3
2 , . . . ,−2n−1
2 and κ0+κ1 = −k
s for 1 ≤ k ≤ 2ns+1.
There appear to be singularities at κ0 + κ1 = −k for 1 ≤ k ≤ 2n but the term (κ0 + κ1 + 1)2n
in the numerator cancels these zeros. The same cancellation occurs for arbitrary V
(
zazb
)
in
a more complicated way. The formula for Kn (z, w) has the factors (s (κ0 + κ1) + 1)n−j in the
denominators thus the individual terms can have simple poles at κ0 + κ1 = −k
s for k ∈ N.
We will show directly that the singularities at κ0 + κ1 = −m are removable when Kn (z, w) is
expressed as a quotient of polynomials in κ0, κ1. It turns out that the terms with poles can be
paired in such a way that the sum of each pair has a removable singularity. The pairs correspond
to {Pk, P2sm−k} for certain values of k.
Throughout we assume that (κ0, κ1) /∈ K0.
We use an elementary algebraic result: suppose a rational function F (α, β) (with coefficients
in the ring Q [z, z, w, w]) vanishes for a countable set of values {α = 0, β = rn : n ∈ N0} (which
are not poles) then F (α, β) is divisible by α; indeed the numerator of F (0, β) is a polynomial
in β vanishing at all β = rn hence is zero. This result will be applied with α = κ0 + κ1 + m,
β = κ0 − κ1.
Most of the section concerns the proof of the following result: let κ0 + κ1 = −m then
PN (z, w) = 0 for N > 2sm and PN (z, w) + (zzww)N−sm P2sm−N (z, w) = 0 for 0 ≤ N ≤ 2sm.
The Poisson kernels Pn were described in equation (2.2). There are a number of cases, roughly
corresponding to the representations of I2 (2s).
Proposition 7. Suppose − (κ0 + κ1) = m ∈ N then
f0
2n (z) =
(
κ0 + 1
2
)
n
(m− n)!(
κ0 + 1
2
)
m−n
n!
(zz)2n−m f0
2m−2n (z) , 0 ≤ n ≤ m,
f1
2n (z) =
(
κ0 + 1
2
)
n
(m− n− 1)!(
κ0 + 1
2
)
m−n
(n− 1)!
(zz)2n−m f1
2m−2n (z) , 1 ≤ n ≤ m− 1.
Proof. The argument uses the Jacobi polynomials directly. Recall z=reiθ. Then for 0 ≤ n ≤ m
f0
2n (z) = r2n
(
κ0 + 1
2
)
n
n! 2F1
(
−n, n−m
κ0 + 1
2
;
1− cos 2θ
2
)
,
Polynomials Associated with Dihedral Groups 15
f0
2m−2n (z) = r2m−2n
(
κ0 + 1
2
)
m−n
(m− n)! 2F1
(
− (m− n) , (m− n)−m
κ0 + 1
2
;
1− cos 2θ
2
)
,
while for 1 ≤ n ≤ m− 1
f1
2n (z) = ir2n sin 2θ
(
κ0 + 3
2
)
n−1
(n− 1)! 2F1
(
1− n, n−m + 1
κ0 + 3
2
;
1− cos 2θ
2
)
,
f1
2m−2n (z) = ir2m−2n sin 2θ
(
κ0 + 3
2
)
m−n−1
(m− n− 1)! 2F1
(
− (m− n− 1) , 1− n
κ0 + 1
2
;
1− cos 2θ
2
)
.
This proves the formulae. �
Proposition 8. Suppose − (κ0 + κ1) = m ∈ N and 0 ≤ n < m then
f0
2n+1 (z) =
(
κ0 + 1
2
)
n+1
(m− n− 1)!(
κ0 + 1
2
)
m−n
n!
(zz)2n−m+1 f0
2m−2n−1 (z) ,
f1
2n+1 (z) =
(
κ0 + 1
2
)
n
(m− n− 1)!(
κ0 + 1
2
)
m−n−1
n!
(zz)2n−m+1 f1
2m−2n−1 (z) .
Proof. Similarly to the even case we have
f0
2n+1 (z) = r2n+1 cos θ
(
κ0 + 1
2
)
n+1
n! 2F1
(
−n, n−m + 1
κ0 + 1
2
;
1− cos 2θ
2
)
,
f0
2m−2n−1 (z) = r2m−2n−1 cos θ
(
κ0 + 1
2
)
m−n
(m− n− 1)! 2F1
(
− (m− n− 1) ,−n
κ0 + 1
2
;
1− cos 2θ
2
)
,
and
f1
2n+1 (z) = ir2n+1 sin θ
(
κ1 + n + 1
2
) (
κ0 + 3
2
)
n
n! 2F1
(
−n, n−m + 1
κ0 + 3
2
;
1− cos 2θ
2
)
,
f1
2m−2n−1 (z) = ir2m−2n−1 sin θ
(
κ1 + m− n− 1
2
) (
κ0 + 3
2
)
m−n−1
(m− n− 1)!
× 2F1
(
− (m− n− 1) ,−n
κ0 + 3
2
;
1− cos 2θ
2
)
.
Thus
f1
2n+1 (z)
f1
2m−2n−1 (z)
= r4n−2m+2
(m− n− 1)!
(
κ0 + 1
2
)
n+1
(
−m− κ0 + n + 1
2
)
n!
(
κ0 + 1
2
)
m−n
(
−κ0 − n− 1
2
)
= r4n−2m+2
(m− n− 1)!
(
κ0 + 1
2
)
n
n!
(
κ0 + 1
2
)
m−n−1
. �
Proposition 9. Suppose − (κ0 + κ1) = m ∈ N and 0 ≤ n < m then
f2n (z) =
(
κ0 + 1
2
)
n
(m− n− 1)!(
κ0 + 1
2
)
m−n
n!
(zz)2n−m zf2m−2n−1 (z) .
Proof. We use the expressions from Proposition 6. First we show for 0 ≤ j ≤ min (n, m− n)
that
En−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
=
(
κ0 + 1
2
)
m−n
(n− j)!(
κ0 + 1
2
)
n
(m− n− j)!
Em−n−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
.
16 C.F. Dunkl
Indeed by Proposition 3(
κ0 +
1
2
)
j
En−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
=
(
κ0 + 1
2
)
n
(n− j)! 3F2
(
j − n, n−m + j, j + 1
2
κ0 + j + 1
2 , 2j + 1
; 1
)
,
and (
κ0 +
1
2
)
j
Em−n−j
(
κ0, κ1; j +
1
2
, j +
1
2
)
=
(
κ0 + 1
2
)
m−n
(m− n− j)! 3F2
(
n−m + j, j − n, j + 1
2
κ0 + j + 1
2 , 2j + 1
; 1
)
.
Let
g1 (z) :=
n!(
κ0 + 1
2
)
n
f2n (z) , g2 (z) :=
(m− n− 1)!(
κ0 + 1
2
)
m−n
(zz)2n−m zf2m−2n−1 (z) ,
and for j ≥ 0 let
bj :=
1
22jj! 3F2
(
n−m + j, j − n, j + 1
2
κ0 + j + 1
2 , 2j + 1
; 1
)
.
Then
g1 (z) =
n∑
j=−n
(n−m + 1)|j|−1
n!
(n− |j|)!
b|j| (n−m + j) zn+jzn−j ,
g2 (z) =
m−n∑
j=n−m
(−n)|j|
(m− n− 1)!
(m− n− |j|)!
b|j| (m− n + j) zn+jzn−j .
Thus c (g1;n, n) = b0 = c (g2;n, n). Suppose |j| ≥ 1, then
c (g1;n + j, n− j) = (n−m + 1)|j|−1 (−n)|j| (−1)j (n−m + j) b|j|
for |j| ≤ n, and the equation remains valid if n < |j| ≤ m−n because (−n)|j| = 0 for |j| > n. Also
c (g2;n + j, n− j) = (−n)|j| (n + 1−m)|j|−1 (−1)j−1 (m− n− j) b|j|, and the equation remains
valid if m−n < |j| ≤ n (that is n−m+ |j|−1 ≥ 0). Thus c (g1;n + j, n− j) = c (g2;n + j, n− j)
for |j| ≤ max (n, m− n) and g1 = g2. �
Note that if k = 0, 1, 2, . . . then (−k)j = 0 for j > k. Recall the structural constants for the
Poisson kernel Pn from equations (2.5)–(2.10). These are rational functions of κ0, κ1 defined
for all (κ0, κ1) /∈ K0.
Proposition 10. Suppose − (κ0 + κ1) = m ∈ N, 1 ≤ n ≤ m− 1, and n 6= m
2 then
λ0
2m−2n
λ0
2n
= −
((
κ0 + 1
2
)
n
(m− n)!(
κ0 + 1
2
)
m−n
n!
)2
,
λ1
2m−2n
λ1
2n
= −
((
κ0 + 1
2
)
n
(m− n− 1)!(
κ0 + 1
2
)
m−n
(n− 1)!
)2
,
λ0
2m = −
(
m!(
κ0 + 1
2
)
m
)2
, λ1
2m = 0.
Proof. Recall λ0
2n =
n!(κ0+κ1+1)n−1(κ0+κ1+2n)
(κ0+ 1
2)n
(κ1+ 1
2)n
(for n ∈ N). Also
(
κ1 + 1
2
)
n
=
(
−m− κ0 + 1
2
)
n
=
(−1)n (κ0 + 1
2 + m− n
)
n
= (−1)n (κ0+ 1
2)m
(κ0+ 1
2)m−n
, and similarly
(
κ1 + 1
2
)
m−n
= (−1)m−n (κ0+ 1
2)m
(κ0+ 1
2)n
.
Thus
λ0
2m−2n
λ0
2n
= (−1)m
( (
κ0 + 1
2
)
n(
κ0 + 1
2
)
m−n
)2
(m− n)! (1−m)m−n−1 (m− 2n)
n! (1−m)n−1 (−m + 2n)
.
Polynomials Associated with Dihedral Groups 17
But
(1−m)m−n−1
(1−m)n−1
= (−1)m (m−1)!(m−n)!
(m−1)!n! (note (−k)j = (−1)j k!
(k−j)! for k ∈ N0). Next λ1
2n =
(n−1)!(κ0+κ1+1)n(κ0+κ1+2n)
(κ0+ 1
2)n
(κ1+ 1
2)n
. Similarly we find
λ1
2m−2n
λ1
2n
= (−1)m
( (
κ0 + 1
2
)
n(
κ0 + 1
2
)
m−n
)2
(m− n− 1)! (1−m)m−n (m− 2n)
(n− 1)! (1−m)n (−m + 2n)
,
and
(1−m)m−n
(1−m)n
= (−1)m (m−1)!(m−n−1)!
(m−1)!(n−1)! . The special case λ0
2m follows from setting n = 0 in the
first formula. The term (κ0 + κ1 + 1)m shows λ1
2m = 0. �
The following two propositions are proven by similar calculations.
Proposition 11. Suppose − (κ0 + κ1) = m ∈ N, 0 ≤ n ≤ m− 1, and n 6= m−1
2 then
λ0
2m−2n−1
λ0
2n+1
= −
((
κ0 + 1
2
)
n+1
(m− n− 1)!(
κ0 + 1
2
)
m−n
n!
)2
,
λ1
2m−2n−1
λ1
2n+1
= −
((
κ0 + 1
2
)
n
(m− n− 1)!(
κ0 + 1
2
)
m−n−1
n!
)2
.
Proposition 12. Suppose − (κ0 + κ1) = m ∈ N, 0 ≤ n ≤ m− 1 then
λ2m−2n−1
λ2n
= −
((
κ0 + 1
2
)
n
(m− n− 1)!(
κ0 + 1
2
)
m−n
n!
)2
.
Proposition 13. Suppose − (κ0 + κ1) = m ∈ N then λ0
m = 0 = λ1
m. If n > 2m then λ0
n = 0 =
λ1
n. If n ≥ 2m then λn = 0.
Proof. Since both λ0
n and λ1
n contain the factor (κ0 + κ1 + n) for n even or odd, it follows that
λ0
m = 0 = λ1
m. The term (κ0 + κ1 + 1)j vanishes for j > m, and j = k − 1 for λ0
2k, j = k for
each of λ2k, λ2k+1, λ1
2k, λ0
2k+1, λ1
2k+1. �
Theorem 3. Suppose − (κ0 + κ1) = m ∈ N then PN (z, w) = 0 for N > 2sm and PN (z, w) +
(zzww)N−sm P2sm−N (z, w) = 0 for 0 ≤ N ≤ 2sm.
Proof. If N = sk > 2sm then Psk (z, w) = λ0
kf
0
k (zs) f0
k (ws) + λ1
kf
1
k (zs) f1
k (ws) and λ0
k =
λ1
k = 0 by Proposition 13. If N = sk + t with 1 ≤ t < s and N > 2sm then Psk+t (z, w) =
λk
(
ztwtfk (zs) fk (ws) + ztwtfk (zs) fk (ws)
)
, k ≥ 2sm and λk = 0. If N = sm then λ0
m = λ1
m =
0. Suppose N = 2ns and 0 < N < 2sm (so 0 < n < m), then
P2ns (z, w) = λ0
2n
((
κ0 + 1
2
)
n
(m− n)!(
κ0 + 1
2
)
m−n
n!
)2
(zzww)(2n−m)s f0
2m−2n (zs) f0
2m−2n (ws)
+ λ1
2n
((
κ0 + 1
2
)
n
(m− n− 1)!(
κ0 + 1
2
)
m−n
(n− 1)!
)2
(zzww)(2n−m)s f1
2m−2n (zs) f1
2m−2n (ws) ,
thus by Propositions 7 and 10
(zzww)(m−2n)s P2ns (z, w) + P2ms−2ns (z, w)
= λ0
2n
((
κ0 + 1
2
)
n
(m− n)!(
κ0 + 1
2
)
m−n
n!
)2
+
λ0
2m−2n
λ0
2n
f0
2m−2n (zs) f0
2m−2n (ws)
18 C.F. Dunkl
+ λ1
2n
((
κ0 + 1
2
)
n
(m− n− 1)!(
κ0 + 1
2
)
m−n
(n− 1)!
)2
+
λ1
2m−2n
λ1
2n
f1
2m−2n (zs) f1
2m−2n (ws) = 0.
For the special case N = 2sm we have P2sm (z, w) = λ0
2m
((κ0+ 1
2)m
m!
)2
(zzww)ms = − (zzww)ms P0
because λ1
2m = 0, P0 = 1 and λ0
2m = −
(
m!
(κ0+ 1
2)m
)2
. Similarly by use of Propositions 8 and 11
we show the result holds for N = (2n + 1) s for 0 ≤ n < m and 2n+1 6= m. Suppose N = sk + t
with 1 ≤ t < s and 0 < N < 2sm, then 2sm−N = s (2m− k − 1)+(s− t). One of k, 2m−k−1
is even so assume k = 2n with 0 ≤ n < m (otherwise replace N by 2sm−N and t by s− t). By
Propositions 9 and 12
λ2nztwtf2n (zs) f2n (ws) +λ2m−2n−1 (zzww)(2n−m)s+t zs−tws−tf2m−2n−1 (zs) f2m−2n−1 (ws)
= λ2n (zw)(2n−m)s+t (zw)(2n−m+1)s f2m−2n−1 (zs) f2m−2n−1 (ws)
×
((
κ0 + 1
2
)
n
(m− n− 1)!(
κ0 + 1
2
)
m−n
n!
)2
+
λ2m−2n−1
λ2n
= 0.
Add this equation to its complex conjugate to show
P2ns+t (z, w) + (zzww)(2n−m)s+t P(2m−2n)s−t (z, w) = 0. �
Theorem 4. For n, m∈N equation (3.1) for Kn(z, w) has a removable singularity at κ0+κ1=−m.
Proof. Consider the series
Kn (z, w) = 2−n
bn/2c∑
j=0
1
j! (sκ0 + sκ1 + 1)n−j
(zzww)j Pn−2j (z, w) .
The possible poles occur at n− j ≥ sm (that is, (1− sm)n−j = 0) and the multiplicities do not
exceed 1. Thus there are no poles if n < sm. If n − 2j > 2sm then Pn−2j (z, w) is divisible
by (κ0 + κ1 + m), by Theorem 3, and the singularity is removable. It remains to consider
the case n − 2j ≤ 2sm and n − j ≥ sm. Suppose j = j0 satisfies these inequalities and let
j1 = n − j0 − sm. Then j1 ≥ 0 and n − 2j1 = 2sm − n + 2j0 ≥ 0, hence j = j1 appears in the
sum. But 2sm − (n− 2j0) = n − 2j1 so Theorem 3 applies. We can assume j1 ≤ j0. Consider
the following subset of the sum for Kn (z, w):
(zzww)j0 Pn−2j0 (z, w)
j0! (sκ0 + sκ1 + 1)n−j0
+
(zzww)j1 Pn−2j1 (z, w)
j1! (sκ0 + sκ1 + 1)n−j1
=
(zzww)j0
j0! (sκ0 + sκ1 + 1)n−j0
Cn,j0 .
with
Cn,j0 = Pn−2j0 (z, w) +
j0! (zzww)j1−j0 Pn−2j1 (z, w)
j1! (sκ0 + sκ1 + 1 + n− j0)j0−j1
.
The expression Cn,j0 has no pole at κ0 + κ1 = −m since 1 − sm + n − j0 ≥ 1. Indeed
(1− sm + n− j0)j0−j1
= (j1 + 1)j0−j1
= j0!/j1!. In the special case n − 2j0 = ms, and
j0 = j1 = (n− sm) /2 we replace Cn,j0 by Pn−2j0 (z, w). By Theorem 3 Cn,j0 = 0 when
κ0 + κ1 = −m, thus Cn,j0 is divisible by (κ0 + κ1 + m) s. The sum of the two terms (j = j0 and
j = j1) has a removable singularity there. �
Polynomials Associated with Dihedral Groups 19
The expressions for V
(
zazb
)
are derived from the series (3.1) for Kn (z, w) thus the result
about singularities at κ0 + κ1 = −m being removable by grouping the expansion into certain
pairs applies. Note that in the above proof the paired terms are Pn−2j0 and Pn−2j1 with j0+j1 =
n− sm. To analyze V
(
zazb
)
it suffices to identify the pairs. For the case a ≡ b mod s and a ≥ b
let a = us + r, b = vs + r and 0 ≤ r < s. The paired indices in the sum from Theorem 2 consist
of {(k, k′) : 0 ≤ k < k′ ≤ v, k + k′ = u + v −m} . Indeed for k, k′ with 0 ≤ k, k′ ≤ v define j
by a + b − 2j = (u + v − 2k) s so that j = ks + r and similarly set j′ := k′s + r. The pairing
condition j + j′ = a + b− sm is equivalent to k + k′ = u + v −m. Thus k, k′ are paired exactly
when k + k′ = u + v −m and 0 ≤ k, k′ ≤ v.
For the case a− b ≡ t mod s, and with a = us + r + t > b = vs + r, 0 ≤ r < s, 1 ≤ t < s the
pairing in the formula from Theorem 1 combines terms from the first sum with corresponding
terms in the second. For the first sum suppose 0 ≤ k ≤ v and j := ks + r so that a + b− 2j =
t + (u + v − 2k) s. For the second sum let 1 −
⌊
r+t
s
⌋
≤ k′ ≤ v and let j′ := (k′ − 1) s + r + t
so that a + b− 2j′ = (s− t) + (u + v − 2k′ + 1) s. The pairing condition j + j′ = a + b− sm is
equivalent to k + k′ = u + v + 1 − m. To remove the singularities at κ0 + κ1 = −m combine
the term in the first sum of index k with the term in the second of index k′ for all pairs (k, k′)
satisfying k + k′ = u + v + 1−m, 0 ≤ k ≤ v, 1−
⌊
r+t
s
⌋
≤ k′ ≤ v.
References
[1] Berenstein A., Burman Y., Quasiharmonic polynomials for Coxeter groups and representations of Cherednik
algebras, math.RT/0505173.
[2] Dunkl C., Differential-difference operators associated to reflection groups, Trans. Amer. Math. Soc. 311
(1989), 167–183.
[3] Dunkl C., Poisson and Cauchy kernels for orthogonal polynomials with dihedral symmetry, J. Math. Anal.
Appl. 143 (1989), 459–470.
[4] Dunkl C., Operators commuting with Coxeter group actions on polynomials, in Invariant Theory and
Tableaux, Editor D. Stanton, Springer, Berlin – Heidelberg – New York, 1990, 107–117.
[5] Dunkl C., Integral kernels with reflection group invariance, Can. J. Math. 43 (1991), 1213–1227.
[6] Dunkl C., de Jeu M., Opdam E., Singular polynomials for finite reflection groups, Trans. Amer. Math. Soc.
346 (1994), 237–256.
[7] Dunkl C., Xu Y., Orthogonal polynomials of several variables, Encycl. of Math. and its Applications, Vol. 81,
Cambridge University Press, Cambridge, 2001.
[8] Rösler M., Positivity of Dunkl’s intertwining operator, Duke Math. J. 98 (1999), 445–463, q-alg/9710029.
[9] Scalas F., Poisson integrals associated to Dunkl operators for dihedral groups, Proc. Amer. Math. Soc., 133
(2005), 1713–1720.
[10] Xu Y., Intertwining operator and h-harmonics associated with reflection groups, Can. J. Math. 50 (1998),
193–208.
http://arxiv.org/abs/math.RT/0505173
http://arxiv.org/abs/q-alg/9710029
1 Introduction
2 The Poisson kernel
3 Expressions for coefficients
4 The intertwining operator
5 Singular values
References
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