Vector-Valued Polynomials and a Matrix Weight Function with B₂-Action. II
This is a sequel to [SIGMA 9 (2013), 007, 23 pages], in which there is a construction of a 2×2 positive-definite matrix function K(x) on R². The entries of K(x) are expressed in terms of hypergeometric functions. This matrix is used in the formula for a Gaussian inner product related to the standard...
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| Цитувати: | Vector-Valued Polynomials and a Matrix Weight Function with B₂-Action. II / C.F. Dunkl // Symmetry, Integrability and Geometry: Methods and Applications. — 2013. — Т. 9. — Бібліогр.: 1 назв. — англ. |
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irk-123456789-1492002019-02-20T01:26:03Z Vector-Valued Polynomials and a Matrix Weight Function with B₂-Action. II Dunkl, C.F. This is a sequel to [SIGMA 9 (2013), 007, 23 pages], in which there is a construction of a 2×2 positive-definite matrix function K(x) on R². The entries of K(x) are expressed in terms of hypergeometric functions. This matrix is used in the formula for a Gaussian inner product related to the standard module of the rational Cherednik algebra for the group W(B₂) (symmetry group of the square) associated to the (2-dimensional) reflection representation. The algebra has two parameters: k₀, k₁. In the previous paper K is determined up to a scalar, namely, the normalization constant. The conjecture stated there is proven in this note. An asymptotic formula for a sum of ₃F₂-type is derived and used for the proof. 2013 Article Vector-Valued Polynomials and a Matrix Weight Function with B₂-Action. II / C.F. Dunkl // Symmetry, Integrability and Geometry: Methods and Applications. — 2013. — Т. 9. — Бібліогр.: 1 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 33C52; 33C20 DOI: http://dx.doi.org/10.3842/SIGMA.2013.043 http://dspace.nbuv.gov.ua/handle/123456789/149200 en Symmetry, Integrability and Geometry: Methods and Applications Інститут математики НАН України |
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This is a sequel to [SIGMA 9 (2013), 007, 23 pages], in which there is a construction of a 2×2 positive-definite matrix function K(x) on R². The entries of K(x) are expressed in terms of hypergeometric functions. This matrix is used in the formula for a Gaussian inner product related to the standard module of the rational Cherednik algebra for the group W(B₂) (symmetry group of the square) associated to the (2-dimensional) reflection representation. The algebra has two parameters: k₀, k₁. In the previous paper K is determined up to a scalar, namely, the normalization constant. The conjecture stated there is proven in this note. An asymptotic formula for a sum of ₃F₂-type is derived and used for the proof. |
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Article |
| author |
Dunkl, C.F. |
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Dunkl, C.F. Vector-Valued Polynomials and a Matrix Weight Function with B₂-Action. II Symmetry, Integrability and Geometry: Methods and Applications |
| author_facet |
Dunkl, C.F. |
| author_sort |
Dunkl, C.F. |
| title |
Vector-Valued Polynomials and a Matrix Weight Function with B₂-Action. II |
| title_short |
Vector-Valued Polynomials and a Matrix Weight Function with B₂-Action. II |
| title_full |
Vector-Valued Polynomials and a Matrix Weight Function with B₂-Action. II |
| title_fullStr |
Vector-Valued Polynomials and a Matrix Weight Function with B₂-Action. II |
| title_full_unstemmed |
Vector-Valued Polynomials and a Matrix Weight Function with B₂-Action. II |
| title_sort |
vector-valued polynomials and a matrix weight function with b₂-action. ii |
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Інститут математики НАН України |
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2013 |
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http://dspace.nbuv.gov.ua/handle/123456789/149200 |
| citation_txt |
Vector-Valued Polynomials and a Matrix Weight Function with B₂-Action. II / C.F. Dunkl // Symmetry, Integrability and Geometry: Methods and Applications. — 2013. — Т. 9. — Бібліогр.: 1 назв. — англ. |
| series |
Symmetry, Integrability and Geometry: Methods and Applications |
| work_keys_str_mv |
AT dunklcf vectorvaluedpolynomialsandamatrixweightfunctionwithb2actionii |
| first_indexed |
2025-07-12T21:37:43Z |
| last_indexed |
2025-07-12T21:37:43Z |
| _version_ |
1837478727503577088 |
| fulltext |
Symmetry, Integrability and Geometry: Methods and Applications SIGMA 9 (2013), 043, 11 pages
Vector-Valued Polynomials
and a Matrix Weight Function with B2-Action. II
Charles F. DUNKL
Department of Mathematics, University of Virginia,
PO Box 400137, Charlottesville VA 22904-4137, USA
E-mail: cfd5z@virginia.edu
URL: http://people.virginia.edu/~cfd5z/home.html
Received February 15, 2013, in final form June 07, 2013; Published online June 12, 2013
http://dx.doi.org/10.3842/SIGMA.2013.043
Abstract. This is a sequel to [SIGMA 9 (2013), 007, 23 pages], in which there is a con-
struction of a 2 × 2 positive-definite matrix function K(x) on R2. The entries of K(x)
are expressed in terms of hypergeometric functions. This matrix is used in the formula for
a Gaussian inner product related to the standard module of the rational Cherednik alge-
bra for the group W (B2) (symmetry group of the square) associated to the (2-dimensional)
reflection representation. The algebra has two parameters: k0, k1. In the previous paper
K is determined up to a scalar, namely, the normalization constant. The conjecture stated
there is proven in this note. An asymptotic formula for a sum of 3F2-type is derived and
used for the proof.
Key words: matrix Gaussian weight function
2010 Mathematics Subject Classification: 33C52; 33C20
1 Introduction
This is a sequel to [1] and the definitions and notations from that paper are used here. Briefly,
we constructed a 2×2 positive-definite matrix function K(x) on R2 whose entries are expressed
in terms of hypergeometric functions. This matrix is used in the formula for a Gaussian inner
product related to the standard module of the rational Cherednik algebra for the group W (B2)
(symmetry group of the square) associated to the (2-dimensional) reflection representation. The
algebra has two parameters: k0, k1. In [1] K is determined up to the normalization constant,
henceforth denoted by c(k0, k1). The conjecture stated there is proven in this note.
Instead of trying to integrate K directly (a problem involving squares of hypergeometric func-
tions whose argument is x2
2/x
2
1) we compute a sequence of integrals in two ways: asymptotically
and exactly in terms of sums. Comparing the two answers will determine the value of c(k0, k1).
First the problem is reduced to a one-variable integral over the sector
{
(cos θ, sin θ) : 0 < θ < π
4
}
of the unit circle. With detailed information about the behavior of a function f(θ) near θ = 0
one can find an asymptotic value of
∫ π/4
0 θnf(θ)dθ. This part of the argument is described in
Section 2. The other part is produced by exploiting the relationship between the Laplacian and
integration over the circle. That is, the plan is to determine the result of applying appropriate
powers of the Laplacian to certain polynomials behaving like θn near θ = 0. This will be done
by establishing recurrence relations; their proofs are in Section 4. The main theorem and its
proof which combines the various ingredients are contained in Section 3.
Recall from [1, p. 18] that for 0 < x2 < x1 and u = x2
x1
L(u)11 = |u|k1
(
1− u2
)−k0 F
(
−k0,
1
2
− k0 + k1; k1 +
1
2
;u2
)
, (1)
mailto:cfd5z@virginia.edu
http://people.virginia.edu/~cfd5z/home.html
http://dx.doi.org/10.3842/SIGMA.2013.043
2 C.F. Dunkl
L(u)12 = − k0
k1 + 1
2
|u|k1
(
1− u2
)−k0 uF
(
1− k0,
1
2
− k0 + k1; k1 +
3
2
;u2
)
,
L(u)21 = − k0
1
2 − k1
|u|−k1
(
1− u2
)−k0 uF
(
1− k0,
1
2
− k0 − k1;
3
2
− k1;u2
)
,
L(u)22 = |u|−k1
(
1− u2
)−k0 F
(
−k0,
1
2
− k0 − k1;
1
2
− k1;u2
)
,
and [1, p. 20]
K(x) = L(u)T
[
d1 0
0 d2
]
L(u), (2)
d1 = c(k0, k1)
Γ
(
1
2 − k1
)2
cosπk0Γ
(
1
2 + k0 − k1
)
Γ
(
1
2 − k0 − k1
) ,
d2 = c(k0, k1)
Γ
(
1
2 + k1
)2
cosπk0Γ
(
1
2 + k0 + k1
)
Γ
(
1
2 − k0 + k1
) .
2 Integrals over the circle
First we reduce the integral to a sector of the unit circle by assuming homogeneity and an
invariance property. In general, the integral of a positively homogeneous function with respect to
Gaussian measure can be found by integrating over the sphere (circle). Furthermore the integral
over the sphere of a polynomial p homogeneous of degree 2n can be found by computing ∆np.
This method applies in the present situation, when replacing ∆ by ∆κ.
As before, elements of PV can be expressed as polynomials f1(x)t1 + f2(x)t2 or vectors
(f1(x), f2(x)), as needed. The Gaussian inner product is expressed as
〈f, g〉G =
∫
R2
f(x)K(x)g(x)T e−|x|
2/2dx,
and the normalization condition is equivalent to 〈(1, 0), (1, 0)〉G = 1. Recall from [1, p. 2] that
the fundamental bilinear form 〈·, ·〉τ on PV can be written as
〈f1(x)t1 + f2(x)t2, g1(x)t1 + g2(x)t2〉τ
= 〈t1, f1(D){g1(x)t1 + g2(x)t2}〉τ |x=0 + 〈t2, f2(D){g1(x)t1 + g2(x)t2}〉τ |x=0,
where f(D) and |x=0 denote f(D1,D2) and evaluation at x = 0, respectively. The abstract Gaus-
sian inner product is then defined as 〈f, g〉G = 〈e∆κ/2f, e∆κ/2g〉τ for f, g ∈ PV . The key property
of this inner product is 〈xif, g〉G = 〈f, xig〉G for i = 1, 2. Subsequently we can derive another
abstract inner product which in effect acts like the use of spherical polar coordinates in com-
puting integrals with respect to Gaussian measure. This is done in the following (S symbolizes
the sphere/circle):
Definition 1. For polynomials f ∈ PV,n, g ∈ PV,m let ` = m+n
2 and
〈f, g〉S :=
1
2``!
〈f, g〉G =
1
2``!
〈
e∆κ/2f, e∆κ/2g
〉
τ
, m ≡ n mod 2,
〈f, g〉S := 0, m− n ≡ 1 mod 2.
This is extended to all polynomials by linearity.
Next we relate the Gaussian integral of certain invariant polynomials to the abstract formula
for 〈·, ·〉S . Let xθ := (cos θ, sin θ), a generic point on the unit circle.
Matrix Weight Function 3
Lemma 1. Suppose f, g ∈ PV are relative invariants of the same type, that is, for some linear
character χ of W , (wf)(x) = f(xw)w−1 = χ(w)f(x) and (wg)(x) = χ(w)g(x) for each w ∈W ,
then ∫
R2
f(x)K(x)g(x)T e−|x|
2/2dx = 8
∫ ∞
0
e−r
2/2rdr
∫ π/4
0
f(rxθ)K(xθ)g(rxθ)
Tdθ.
Proof. Let C0 = {x : 0 < x2 < x1}, the fundamental chamber, then∫
R2
f(x)K(x)g(x)T e−|x|
2/2dx =
∑
w∈W
∫
C0
f(xw)K(xw)g(xw)T e−|x|
2/2dx
=
∑
w∈W
∫
C0
f(xw)w−1K(x)wg(xw)T e−|x|
2/2dx
=
∑
w∈W
χ(w)2
∫
C0
f(x)K(x)g(x)T e−|x|
2/2dx.
The statement follows from the fact χ(w)2 = 1 and the use of polar coordinates. Recall K is
positively homogeneous of degree zero. �
Proposition 1. If f, g ∈ PV are relative invariants of the same type and f ∈ PV,n, g ∈ PV,m
with m ≡ nmod 2 and ` = m+n
2 then
8
∫ π/4
0
f(xθ)K(xθ)g(xθ)
Tdθ =
1
2``!
〈f, g〉G = 〈f, g〉S .
If further n = 2q + 1 and m = 1 then
〈f, g〉S =
1
22q+1q!(q + 1)!
〈
∆q
κf, g
〉
τ
.
Proof. From Lemma 1 the factor relating 〈·, ·〉S to 〈·, ·〉G is
∫∞
0 e−r
2/2r2`+1dr = 2``!. Now
suppose m = 1 and ` = q + 1. By definition 〈f, g〉S = 〈e∆κ/2f,e∆κ/2g〉τ
2q+1(q+1)!
, then e∆κ/2g = g and the
degree-1 component of e∆κ/2f is 1
2qq!∆
q
κf (recall that 〈f, g〉τ = 0 when f and g have different
degrees of homogeneity). �
We will find exact expressions in the form of sums for ∆q
κf for certain polynomials in Section 4.
The idea underlying the asymptotic evaluation is this: suppose that g is continuous on [0, 1]
and satisfies |g(t) − g(0)| ≤ Ct for some constant, then
∫ 1
0 g(t)(1 − t)ndt = 1
n+1g(0) + O
(
1
n2
)
.
This formula can be adapted to the measure tαdt with α > −1. In the sequel we will use C, C ′
to denote constants whose values need not be explicit, as in the “big O” symbol.
Let φ := x2
1 − x2
2; then φ2 is W -invariant. Furthermore φ2np1,2 and φ2n+1p1,4 are all relative
invariants of the same type, that is, σ1f = σ+
12f = −f (recall p1,2 = −x2t1 + x1t2 and p1,4 =
−x2t1 − x1t2). We will evaluate 〈φ2np1,2, p1,2〉S and 〈φ2n+1p1,4, p1,2〉S . These polynomials peak
at θ = 0 and vanish at θ = π
4 . The following are used in the expressions for p1,4Kp
T
1,2 and
p1,2Kp
T
1,2. Set
h1(z) := F
(
−k0,
1
2
− k0 + k1;
3
2
+ k1; z
)
, (3)
h2(z) := F
(
−k0,−
1
2
− k0 − k1;
1
2
− k1; z
)
,
h3(z) := F
(
k0,
1
2
+ k0 + k1;
3
2
+ k1; z
)
,
4 C.F. Dunkl
h4(z) := F
(
k0,−
1
2
+ k0 − k1;
1
2
− k1; z
)
.
Each of these hypergeometric series satisfies the criterion for absolute convergence at z = 1 (for
real F (a, b; c; z) the condition is c − a − b > 0; here c − a − b = 1 ± 2k0), and so each satisfies
a bound of the form |h(z) − 1| ≤ Cz for 0 ≤ z ≤ 1. Recall the coordinate u = x2
x1
; on the unit
circle (in −π
2 < θ < π
2 )
xθ =
(
1√
1 + u2
,
u√
1 + u2
)
, φ =
1− u2
1 + u2
, dθ =
du
1 + u2
.
By use of the identities
F (a, b; c; z)− a
c
zF (a+ 1, b; c+ 1, z) = F (a, b− 1; c; z), (4)
F (a, b; c; z)− a
c
F (a+ 1, b; c+ 1; z) =
c− a
c
F (a, b; c+ 1; z),
we obtain for x = xθ, 0 < θ < π
4 and 0 < u < 1:
x2L11 − x1L12 =
uk1+1
(
1− u2
)−k0
(1 + u2)1/2
1 + 2k0 + 2k1
1 + 2k1
h1
(
u2
)
, (5)
x1L22 − x2L21 =
u−k1
(
1− u2
)−k0
(1 + u2)1/2
h2
(
u2
)
;
and
−x2L11 − x1L12 = −
uk1+1
(
1− u2
)k0
(1 + u2)1/2
1− 2k0 + 2k1
1 + 2k1
h3
(
u2
)
, (6)
−x2L21 − x1L22 = −
u−k1
(
1− u2
)k0
(1 + u2)1/2
h4
(
u2
)
.
The expressions for Lij (see (1)) appearing in (6) are first transformed with F (a, b; c; z) =
(1− z)c−a−bF (c− a, c− b; c; z) before using identities (4). These formulae will be used to obtain
asymptotic expressions for the integrals 〈φ2np1,2, p1,2〉S and 〈φ2n+1p1,4, p1,2〉S . The notation
a(n) ∼ b(n) means lim
n→∞
a(n)
b(n) = 1.
Lemma 2. Suppose α, γ > −1, and n = 2, 3, . . . then∫ 1
0
tα(1− t)n+γ(1 + t)β−ndt = (2n)−α−1Γ(α+ 1)
(
1 +O
(
1
n
))
.
Proof. The term
(
1−t
1+t
)n
is transformed to (1− v)n by the change of variable t = v
2−v . The
integral becomes
2−α−1
∫ 1
0
vα(1− v)n+γ
(
1− v
2
)−α−β−γ−2
dv = 2−α−1 Γ(α+ 1)Γ(n+ γ + 1)
Γ(n+ α+ γ + 2)
+R,
where R is bounded by 2−α−1C
∫ 1
0 v
α+1 (1− v)n+γ dv and
∣∣∣(1− v
2
)−α−β−γ−2 − 1
∣∣∣ ≤ Cv for 0 ≤
v ≤ 1. By Stirling’s formula Γ(n+γ+1)
Γ(n+α+γ+2) ∼ n
−α−1 and R ∼ C ′n−α−2 for some constant C ′. �
Matrix Weight Function 5
Corollary 1. Suppose g(t) is continuous and |g(t)− g(0)| ≤ Ct for 0 ≤ t ≤ 1 then∫ 1
0
tα(1− t)n+γ(1 + t)β−ng(t)dt = (2n)−α−1g(0)Γ(α+ 1)
(
1 +O
(
1
n
))
.
Proof. Break up the integrand as
tα(1− t)n+γ(1 + t)β−n{g(0) + (g(t)− g(0))};
by the Lemma the integral of the second part is bounded by C(2n)−α−2Γ(α+2)
(
1 +O
(
1
n
))
. �
Proposition 2. Suppose −1
2 < k0 ± k1 <
1
2 then
〈
φ2np1,2, p1,2
〉
S
= 8
∫ π/4
0
φ(xθ)
2np1,2(xθ)K(xθ)p1,2(xθ)
Tdθ
∼ 2πc(k0, k1)
cosπk0 cosπk1
22k1nk1−1/2Γ
(
1
2 + k1
)
Γ
(
1
2 + k0 + k1
)
Γ
(
1
2 − k0 + k1
) .
Proof. By definition
φ(xθ)
2np1,2(xθ)K(xθ)p1,2(xθ)
T
=
(
x2
1 − x2
2
)2n {
d1(x1L12 − x2L11)2 + d2(x1L22 − x2L21)2
}
.
Thus equation (5) implies
8
∫ π/4
0
φ(xθ)
2np1,2(xθ)K(xθ)p1,2(xθ)
Tdθ
= 8d1
(
1 + 2k0 + 2k1
1 + 2k1
)2 ∫ 1
0
(
1− u2
1 + u2
)2n (1− u2
)−2k0
(1 + u2)2 u2k1+2h1
(
u2
)2
du
+ 8d2
∫ 1
0
(
1− u2
1 + u2
)2n (1− u2
)−2k0
(1 + u2)2 u−2k1h2
(
u2
)2
du,
where h1 and h2 are from equation (3). The key fact is that hi(0) = 1 and
∣∣hi (u2
)
− 1
∣∣ ≤ Cu2
for 0 ≤ u ≤ 1 with some constant C, i = 1, 2. In each integral change the variable u = v1/2; the
first integral equals
1
2
(4n)−k1−3/2Γ
(
k1 +
3
2
)(
1 +O
(
1
n
))
(7)
and the second integral equals
1
2
(4n)k1−1/2Γ
(
1
2
− k1
)(
1 +O
(
1
n
))
. (8)
This is the dominant term in the sum because k1 − 1
2 > −k1 − 3
2 (that is, 2k1 + 1 > 0). Using
the value of d2 in equation (2) and the identity Γ
(
1
2 − k1
)
Γ
(
1
2 + k1
)
= π
cosπk1
we find
〈
φ2np1,2, p1,2
〉
S
∼ 4c(k0, k1)(4n)k1−1/2 Γ
(
1
2 − k1
)
Γ
(
1
2 + k1
)2
cosπk0Γ
(
1
2 + k0 + k1
)
Γ
(
1
2 − k0 + k1
)
=
2πc(k0, k1)
cosπk0 cosπk1
22k1nk1−1/2Γ
(
1
2 + k1
)
Γ
(
1
2 + k0 + k1
)
Γ
(
1
2 − k0 + k1
) . �
6 C.F. Dunkl
Proposition 3. Suppose −1
2 < k0 ± k1 <
1
2 then
〈
φ2n+1p1,4, p1,2
〉
S
= 8
∫ π/4
0
φ(xθ)
2n+1p1,4(xθ)K(xθ)p1,2(xθ)
Tdθ
∼ −2πc(k0, k1)
cosπk0 cosπk1
22k1nk1−1/2Γ
(
1
2 + k1
)
Γ
(
1
2 + k0 + k1
)
Γ
(
1
2 − k0 + k1
) .
Proof. By definition
φ(xθ)
2n+1p1,4(xθ)K(xθ)p1,2(xθ)
T = −
(
x2
1 − x2
2
)2n+1
× {d1(x2L11 + x1L12)(x2L11 − x1L12) + d2(x2L21 + x1L22)(x2L21 − x1L22)} .
Thus equations (5) and (6) imply
8
∫ π/4
0
φ(xθ)
2n+1p1,4(xθ)K(xθ)p1,2(xθ)
Tdθ = −8d1
(1− 2k0 + 2k1)(1 + 2k0 + 2k1)
(1 + 2k1)2
×
∫ 1
0
(
1− u2
1 + u2
)2n+1
u2k1+2
(1 + u2)2h1
(
u2
)
h3
(
u2
)
du
− 8d2
∫ 1
0
(
1− u2
1 + u2
)2n+1
u−2k1
(1 + u2)2h2
(
u2
)
h4
(
u2
)
du.
Arguing as in the previous proof, one obtains the same expressions (7) and (8) for the first and
second integrals respectively. �
3 Evaluation of the normalizing constant
The proof of the following appears in Section 4. (Recall (a)n :=
n∏
i=1
(a+ i− 1).)
Theorem 1. For arbitrary k0, k1 and n ≥ 0
〈
φ2np1,2, p1,2
〉
S
=
1
n!
(
1
2
)
n+1
n∑
j=0
(−n)2
j
j!
(−k1)j
(
1
2
+ k1 + k0
)
n+1−j
(
1
2
+ k1 − k0
)
n−j
,
〈
φ2n+1p1,4, p1,2
〉
S
= − 1
(n+ 1)!
(
1
2
)
n+1
×
n∑
j=0
(−n)j(−n−1)j
j!
(−k1)j
(
1
2
+ k1+ k0
)
n+1−j
(
1
2
+ k1− k0
)
n+1−j
.
Corollary 2. Suppose −k1 ± k0 /∈ 1
2 + N0 then
〈
φ2np1,2, p1,2
〉
S
=
(
1
2 + k1 + k0
)
n+1
(
1
2 + k1 − k0
)
n(
1
2
)
n+1
n!
× 3F2
(
−n,−n,−k1
−n− 1
2 − k1 − k0,−n+ 1
2 − k1 + k0
; 1
)
,
〈
φ2n+1p1,4, p1,2
〉
S
= −
(
1
2 + k1 + k0
)
n+1
(
1
2 + k1 − k0
)
n+1(
1
2
)
n+1
(n+ 1)!
× 3F2
(
−n,−n− 1,−k1
−n− 1
2 − k1 − k0,−n− 1
2 − k1 + k0
; 1
)
.
Matrix Weight Function 7
The next step is to compare the two hypergeometric series to 2F1
(
−n,−k1
−n−2k1
; 1
)
which equals
(1+k1)n
(1+2k1)n
(Chu–Vandermonde sum). The following lemma will be used with a = 1
2 + k1 + k0,
b = −1
2 + k1 − k0, and c = −k1.
Lemma 3. Suppose 0 < a < 1, −1 < b < 0 and c > −1 then
3F2
(
−n,−n− 1, c
−n− a,−n− b− 1
; 1
)
≤
(1 + a+ b+ c)n
(1 + a+ b)n
≤ 3F2
(
−n,−n, c
−n− a,−n− b
; 1
)
, c ≥ 0;
3F2
(
−n,−n, c
−n− a,−n− b
; 1
)
≤
(1 + a+ b+ c)n
(1 + a+ b)n
≤ 3F2
(
−n,−n− 1, c
−n− a,−n− b− 1
; 1
)
, c ≤ 0.
Proof. If c = 0 then each expression equals 1. The middle term equals 2F1
(
−n,c
−n−a−b ; 1
)
. For
0 ≤ i ≤ n set
si :=
(−n)2
i (c)i
i!(−n− a)i(−n− b)i
, ti :=
(−n)i(c)i
i!(−n− a− b)i
, ui :=
(−n)i(−n− 1)i(c)i
i!(−n− a)i(−n− b− 1)i
.
Note s0 = t0 = u0 = 1. From the relation (−n − d)i = (−1)i(n − i + 1 + d)i it follows that
sign(si) = sign(ti) = sign(ui) = sign((c)i) for each i with 1 ≤ i ≤ n (by hypothesis a + 1 > 2
and b+ 1 > 0). If c > 0 then sign((c)i) = 1 for all i and if −1 < c < 0 then sign((c)i) = −1 for
i ≥ 1. We find
si
ti
=
(−n)i(−n− a− b)i
(−n− a)i(−n− b)i
=
(n− i+ 1)(n− i+ 1 + a+ b)
(n− i+ 1 + a)(n− i+ 1 + b)
si−1
ti−1
, i ≥ 1,
and
m(m+ a+ b)
(m+ a)(m+ b)
= 1 +
−ab
(m+ a)(m+ b)
> 1, m ≥ 1,
because −ab > 0 and b > −1 (setting m = n − i + 1). This shows the sequence si
ti
is positive
and increasing. Also
ui
ti
=
(−n− 1)i(−n− a− b)i
(−n− a)i (−n− b− 1)i
=
(n− i+ 2) (n− i+ 1 + a+ b)
(n− i+ 1 + a)(n− i+ 2 + b)
ui−1
ti−1
, i ≥ 1,
(m+ 1)(m+ a+ b)
(m+ a)(m+ b+ 1)
= 1− b(a− 1)
(m+ a)(m+ b+ 1)
< 1, m ≥ 1,
because a < 1 and b < 0 (and a + b > −1). The sequence ui
ti
is positive and decreasing. If
c > 0 then si, ti, ui > 0 for 1 ≤ i ≤ n and si
ti
> 1 > ui
ti
implies si > ti > ui. This proves the
first inequalities. If −1 < c < 0 then si, ti, ui < 0 for 1 ≤ i ≤ n and thus si
ti
> 1 > ui
ti
implies
si < ti < ui. This proves the second inequalities. �
We will use a version of Stirling’s formula to exploit the lemma:
(a)n
(b)n
=
Γ (b)
Γ (a)
Γ (a+ n)
Γ (b+ n)
∼ Γ (b)
Γ (a)
na−b.
Theorem 2. Suppose −1
2 < k0 ± k1 <
1
2 then the normalizing constant
c(k0, k1) =
1
2π
cosπk0 cosπk1.
8 C.F. Dunkl
Proof. Denote the 3F2-sums in Corollary 2 by f1(n) and f2(n) respectively, then by Stirling’s
formula we obtain〈
φ2np1,2, p1,2
〉
S
∼
Γ
(
1
2
)
f1(n)
Γ
(
1
2 + k1 + k0
)
Γ
(
1
2 + k1 − k0
)n2k1− 1
2 ,
〈
φ2n+1p1,4, p1,2
〉
S
∼ −
Γ
(
1
2
)
f2(n)
Γ
(
1
2 + k1 + k0
)
Γ
(
1
2 + k1 − k0
)n2k1− 1
2 .
By Propositions 2 and 3 these imply for i = 1, 2
fi(n) ∼ 2πc(k0, k1)
cosπk0 cosπk1
× 22k1n−k1
Γ
(
1
2 + k1
)
Γ
(
1
2
) =
2πc(k0, k1)
cosπk0 cosπk1
n−k1
Γ(1 + 2k1)
Γ(1 + k1)
,
by the duplication formula. By Lemma 3 f1(n) ≤ (1+k1)n
(1+2k1)n
≤ f2(n) for 0 ≤ k1 <
1
2 , and the
reverse inequality holds for −1
2 < k1 < 0. The fact that (1+k1)n
(1+2k1)n
∼ Γ(1+2k1)
Γ(1+k1) n
−k1 completes the
proof. �
The weight function K is integrable if the inequalities −1
2 < k0, k1 <
1
2 are satisfied (and the
same value of c(k0, k1) applies). However K is not positive-definite and integrable unless −1
2 <
k0 ± k1 <
1
2 . It was shown in [1, p. 21] that detK = d1d2. By using the (now-known) value of
c(k0, k1) and the values of d1 and d2 (see (2)) we find detK = 1
4π2 cosπ (k0 + k1) cosπ (k0 − k1).
4 Formulae for
〈
φ2np1,2, p1,2
〉
S
and
〈
φ2n+1p1,4, p1,2
〉
S
The inner products are evaluated by computing ∆2n
κ
(
φ2np1,2
)
and ∆2n+1
κ
(
φ2n+1p1,4
)
by means
of recurrence relations. The start is a product formula for ∆κ (using ∂i to denote ∂
∂xi
):
Lemma 4. Suppose f(x) is a W -invariant polynomial and g (x, t) ∈ PV then
∆κ(fg)− f∆κ(g) = g∆f + 2〈∇f,∇g〉 (9)
+ 2k1
(
g(x,−t1, t2)
∂1f
x1
+ g(x, t1,−t2)
∂2f
x2
)
+ 2k0
(
g(x, t2, t1)
∂1f − ∂2f
x1 − x2
+ g(x,−t2,−t1)
∂1f + ∂2f
x1 + x2
)
.
Lemma 5. Suppose f ∈ PV,2n+1 then ∆n+1
κ |x|2f = 4(n+ 1)(n+ 2)∆n
κf .
Proof. If g ∈ PV,m then ∆κ|x|2g = 4(m + 1)g + |x|2∆κg by [1, p. 4, equation (4)]. Apply ∆κ
repeatedly to this expression, and by induction obtain ∆`
κ|x|2g = 4`(m− `+ 2)∆`−1
κ g+ |x|2∆`
κg
for ` = 1, 2, 3, . . .. Set g = f , m = 2n+ 1 and ` = n+ 1 then ∆n+1
κ f = 0. �
Recall that φ := x2
1 − x2
2; and φ2np1,2 and φ2n+1p1,4 are all relative invariants of the same
type as p1,2, that is, σ1f = σ+
12f = −f . Thus ∆2n
κ
(
φ2np1,2
)
and ∆2n+1
κ
(
φ2n+1p1,4
)
are both
scalar multiples of p1,2, because ∆κ commutes with the action of the group and p1,2 is the unique
degree-1 relative invariant of this type.
Proposition 4. For n = 0, 1, 2, 3, . . .
∆κφ
2np1,2 = −8n(1 + 2k1 + 2k0)φ2n−1p1,4 (10)
+ 8n(2n− 1− 2k0)|x|2φ2n−2p1,2,
∆κφ
2n+1p1,4 = −4(2n+ 1)(1 + 2k1 − 2k0)φ2np1,2 (11)
+ 8n(2n+ 1 + 2k0)|x|2φ2n−1p1,4.
Matrix Weight Function 9
Proof. We use Lemma 4 with f = φ2n and g = p1,2 or g = φp1,4. Simple computation shows
that
∆φ2n = 8n(2n− 1)|x|2φ2n−2, ∇φ2n = 4nφ2n−1 (x1,−x2) ,
∆κ (φp1,4) = −4(1 + 2k1 − 2k0)p1,2.
For g = p1,2 we find 2
〈
∇φ2n,∇g
〉
= −8nφ2n−1p1,4, the coefficient of k1 in (9) is −16nφ2n−1p1,4
and the coefficient of k0 is 32nφ2n−2x1x2 (x1t1 − x2t2). The first formula now follows from
x1x2 (x1t1 − x2t2) = −1
2
(
φp1,4 + |x|2p1,2
)
.
For g = φp1,4 we obtain 2
〈
∇φ2n,∇g
〉
= 8nφ2n−1
(
2|x|2p1,4 − φp1,2
)
. The coefficient of k1
in (9) is −16nφ2np1,2 and the coefficient of k0 is −32nφ2n−1x1x2 (x1t1 + x2t2). Similarly to the
previous case
x1x2 (x1t1 + x2t2) = −1
2
(
φp1,2 + |x|2p1,4
)
.
The proof of the second formula is completed by adding up the parts, including φ2n∆κφp1,4. �
We use this to set up a recurrence relation.
Definition 2. For n = 0, 1, 2, . . . the constants αn, βn implicitly depending on k0, k1 are defined
by 〈
φ2np1,2, p1,2
〉
S
= αn 〈p1,2, p1,2〉S = αn(1 + 2k1 + 2k0),〈
φ2n+1p1,4, p1,2
〉
S
= βn 〈p1,2, p1,2〉S = βn(1 + 2k1 + 2k0).
Also α′n := 24n(2n)!(2n+ 1)!αn and β′n := 24n+2(2n+ 1)!(2n+ 2)!βn.
Proposition 5. Suppose n = 0, 1, 2, . . . then ∆2n
κ
(
φ2np1,2
)
= α′np1,2, ∆2n+1
κ
(
φ2n+1p1,4
)
=
β′np1,2 and α′n, β′n satisfy the recurrence (with α′0 = 1, β′−1 = 0)
β′n = −4(2n+ 1)(1 + 2k1 − 2k0)α′n + 64n2(2n+ 1)(2n+ 1 + 2k0)β′n−1,
α′n = −8n(1 + 2k1 + 2k0)β′n−1 + 64n2(2n− 1)(2n− 1− 2k0)α′n−1, n ≥ 1.
Proof. By Proposition 1
24n+1(2n)!(2n+ 1)!
〈
φ2np1,2, p1,2
〉
S
=
〈
∆2n
κ
(
φ2np1,2
)
, p1,2
〉
τ
= αn〈p1,2, p1,2〉τ = 2αn〈p1,2, p1,2〉S .
Similarly
24n+3(2n+ 1)!(2n+ 2)!
〈
φ2n+1p1,4, p1,2
〉
S
=
〈
∆2n+1
κ
(
φ2n+1p1,4
)
, p1,2
〉
τ
= βn〈p1,2, p1,2〉τ = 2βn〈p1,2, p1,2〉S .
Apply ∆2n−1
κ to both sides of equation (10) to obtain
α′np1,2 = −8n(1 + 2k1 + 2k0)β′n−1p1,2 + 8n(2n− 1− 2k0)∆2n−1
κ
(
|x|2φ2n−2p1,2
)
.
By Lemma 5
∆2n−1
κ
(
|x|2φ2n−2p1,2
)
= 8n(2n− 1)∆2n−2
κ
(
φ2n−2p1,2
)
= 8n(2n− 1)α′n−1p1,2.
Apply ∆2n
κ to both sides of equation (11) to obtain
β′np1,2 = −4(2n+ 1)(1 + 2k1 − 2k0)α′np1,2 + 8n(2n+ 1 + 2k0)∆2n
κ
(
|x|2φ2n−1p1,4
)
,
and by the same lemma
∆2n
κ
(
|x|2φ2n−1p1,4
)
= 8n(2n+ 1)∆2n−1
κ
(
φ2n−1p1,4
)
= 8n(2n+ 1)β′n−1p1,2. �
10 C.F. Dunkl
By a simple computation with the change of scale for αn, βn we obtain the following:
Corollary 3. αn, βn satisfy the recurrence
βn = −1 + 2k1 − 2k0
2(n+ 1)
αn +
n(2n+ 1 + 2k0)
(n+ 1)(2n+ 1)
βn−1,
αn = −1 + 2k1 + 2k0
2n+ 1
βn−1 +
2n− 1− 2k0
2n+ 1
αn−1.
The following formulae arose from examining values of αn, βn for some small n, calcu-
lated by using the recurrence and symbolic computation. In the following we use relations
like (a)m (a+m) = (a)m+1 and 1
(m−1)! = m
m! .
Theorem 3. Suppose n = 0, 1, 2, . . . then
αn =
n∑
j=0
(−n)2
j
n!
(
3
2
)
n
j!
(−k1)j
(
3
2
+ k1 + k0
)
n−j
(
1
2
+ k1 − k0
)
n−j
,
βn = −
n∑
j=0
(−n)j (−1− n)j
(n+ 1)!
(
3
2
)
n
j!
(−k1)j
(
3
2
+ k1 + k0
)
n−j
(
1
2
+ k1 − k0
)
n+1−j
.
For brevity k+ := k1 +k0 and k− := k1−k0, as previously. We use induction on the sequence
α0 → β0 → α1 → β1 → α2 → · · · . The formulae are clearly valid for n = 0. Suppose they are
valid for some n− 1. Consider −1+2k1+2k0
2n+1 βn−1 + 2n−1−2k0
2n+1 αn−1; split up the j-term in αn−1 by
writing 1 = j−k1
n− 1
2
−k0
+
n− 1
2
+k1−k0−j
n− 1
2
−k0
, then
2n− 1− 2k0
2n+ 1
αn−1 =
1
(n− 1)!
(
3
2
)
n
n−1∑
j=0
1
j!
(1− n)2
j
(
3
2
+ k+
)
n−1−j
(12)
×
{
(−k1)j
(
1
2
+ k−
)
n−j
+ (−k1)j+1
(
1
2
+ k−
)
n−1−j
}
,
−1 + 2k1 + 2k0
2n+ 1
βn−1 (13)
=
1
2 + k+
n!
(
3
2
)
n
n−1∑
j=0
1
j!
(1− n)j(−n)j(−k1)j
(
3
2
+ k+
)
n−1−j
(
1
2
+ k−
)
n−j
.
The coefficient of (−k1)j in the sum of the first part of { } in (12) and (13) is
1
n!
(
3
2
)
n
j!
(1− n)j
(
3
2
+ k+
)
n−1−j
(
1
2
+ k−
)
n−j
[
n(1− n)j +
(
1
2
+ k+
)
(−n)j
]
=
1
n!
(
3
2
)
n
j!
(1− n)j
(
3
2
+ k+
)
n−1−j
(
1
2
+ k−
)
n−j
(−n)j
(
n− j +
1
2
+ k+
)
=
1
n!
(
3
2
)
n
j!
(1− n)j
(
3
2
+ k+
)
n−j
(
1
2
+ k−
)
n−j
(−n)j .
For j = 0 this establishes the validity of the j = 0 term in αn. For 1 ≤ j ≤ n replace j by j − 1
in the second part of { } in (12) and obtain
jn
n!
(
3
2
)
n
j!
(1− n)2
j−1
(
3
2
+ k+
)
n−j
(
1
2
+ k−
)
n−j
(−k1)j ;
Matrix Weight Function 11
adding all up leads to
1
n!
(
3
2
)
n
j!
(
3
2
+ k+
)
n−j
(
1
2
+ k−
)
n−j
{
(1− n)j(−n)j + nj(1− n)2
j−1
}
,
and the expression in { } evaluates to (−n)2
j . This proves the validity of the formula for αn.
To prove the formula for βn consider −1+2k1−2k0
2(n+1) αn + n(2n+1+2k0)
(n+1)(2n+1) βn−1 and as before split up
the j-term in βn−1 by writing 1 = j−k1
n+ 1
2
+k0
+
n+ 1
2
+k1+k0−j
n+ 1
2
+k0
. Then
−1 + 2k1 − 2k0
2(n+ 1)
αn = −
1
2 + k1 − k0
(n+ 1)!
(
3
2
)
n
n∑
j=0
1
j!
(−n)2
j (−k1)j
(
3
2
+ k+
)
n−j
(
1
2
+ k−
)
n−j
, (14)
n(2n+ 1 + 2k0)
(n+ 1)(2n+ 1)
βn−1 =
1
(n+ 1)!
(
3
2
)
n
n−1∑
j=0
1
j!
(−n)j+1(−n)j
(
1
2
+ k−
)
n−j
(15)
×
{
(−k1)j
(
3
2
+ k+
)
n−j
+ (−k1)j+1
(
3
2
+ k+
)
n−j−1
}
.
The coefficient of (−k1)j in the sum of (14) and the first part of { } in (15) is
−1
(n+ 1)!
(
3
2
)
n
j!
(−n)2
j
(
3
2
+ k+
)
n−j
(
1
2
+ k−
)
n−j
(
1
2
+ k− + n− j
)
=
−1
(n+ 1)!
(
3
2
)
n
j!
(−n)2
j
(
3
2
+ k+
)
n−j
(
1
2
+ k−
)
n+1−j
.
For j = 0 this establishes the validity of the j = 0 term in βn. For 1 ≤ j ≤ n replace j by j − 1
in the second part of { } in (15) and obtain
j
(n+ 1)!
(
3
2
)
n
j!
(−n)j(−n)j−1
(
1
2
+ k−
)
n−j+1
(
3
2
+ k+
)
n−j
;
adding all up leads to
−1
(n+ 1)!
(
3
2
)
n
j!
(−n)j
(
1
2
+ k−
)
n−j+1
(
3
2
+ k+
)
n−j
{(−n)j − j(−n)j−1}
and the expression in { } evaluates to (−1− n)j . This proves the validity of the formula for βn
and completes the induction.
This completes the proof of Theorem 1.
References
[1] Dunkl C.F., Vector-valued polynomials and a matrix weight function with B2-action, SIGMA 9 (2013), 007,
23 pages, arXiv:1210.1177.
http://dx.doi.org/10.3842/SIGMA.2013.007
http://arxiv.org/abs/1210.1177
1 Introduction
2 Integrals over the circle
3 Evaluation of the normalizing constant
4 Formulae for "426830A 2np1,2,p1,2"526930B S and "426830A 2n+1p1,4,p1,2"526930B S
References
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