Symmetry, Integrability and Geometry: Methods and Applications
SIGMA 10 (2014), 044, 23 pages
Vector Polynomials and a Matrix Weight
Associated to Dihedral Groups
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/
Received January 22, 2014, in final form April 10, 2014; Published online April 15, 2014
http://dx.doi.org/10.3842/SIGMA.2014.044
Abstract. The space of polynomials in two real variables with values in a 2-dimensional
irreducible module of a dihedral group is studied as a standard module for Dunkl operators.
The one-parameter case is considered (omitting the two-parameter case for even dihedral
groups). The matrix weight function for the Gaussian form is found explicitly by solving
a boundary value problem, and then computing the normalizing constant. An orthogonal
basis for the homogeneous harmonic polynomials is constructed. The coefficients of these
polynomials are found to be balanced terminating 4 F3 -series.
Key words: standard module; Gaussian weight
2010 Mathematics Subject Classification: 33C52; 20F55; 33C45
1
Introduction
For each irreducible two-dimensional representation of a dihedral reflection group there is a module of the algebra of operators on polynomials generated by multiplication and the Dunkl
operators. This algebra is called the rational Cherednik algebra of the group. The space of
vector-valued polynomials is equipped with a bilinear form which depends on one parameter
and is invariant under the group action. For a certain interval of parameter values the form can
be represented as an integral with respect to a positive-definite matrix weight function times
the Gaussian measure. In a previous paper this structure was analyzed for the group of type B2
where there are two free parameters. This paper concerns the Coxeter group of type I2 (m), the
full symmetry group of the regular m-gon, of order 2m. Even values for m would allow two
parameters but only the one parameter case is considered here.
The orthogonality properties of the Gaussian form are best analyzed by means of harmonic
homogeneous polynomials. These are studied in Section 2. An inductive approach is used
to produce the definitions. Closed forms are obtained for these polynomials in the complex
coordinate system. The formulae are explicit but do require double sums, that is, the coefficients
are given as terminating balanced 4 F3 -series. Section 3 contains the construction of an orthogonal
basis of harmonic polynomials and the structure constants with respect to the Gaussian form.
The results in Sections 2 and 3 are of algebraic flavor and hold for any value of κ. The sequel is
of analytic nature and the results hold for restricted values of κ. Section 4 sets up the differential
system for the weight function and then constructs the solution up to a normalizing constant.
Section 5 deals with the mechanics of integrating harmonic polynomials with respect to the
Gaussian weight. In Section 6 the normalizing constant is found by direct integration. The
survey [3] covering general information about Dunkl operators and related topics is accessible
online.
2
C.F. Dunkl
Fix m ≥ 3 and set ω := exp 2πi
m . The group I2 (m) (henceforth denoted by W ) contains m
reflections and m − 1 rotations. Although the group is real it is often useful to employ complex
coordinates for x = (x1 , x2 ) ∈ R2 , that is, z = x1 + ix2 , z = x1 − ix2 . Then the reflections
are expressed as zσj := zω j (0 ≤ j < m) and the rotations are z%j := zω j (1 ≤ j < m) The
(pairwise nonequivalent) 2-dimensional irreducible representations are given by
m−1
0 ω −j`
,
0 ≤ j < m.
τ` (σj ) = j`
,
1≤`≤
ω
0
2
Henceforth fix `. We consider the vector space Pm,` consisting of polynomials
f (z, z, t, t) = f1 (z, z)t + f2 (z, z)t
with the group action
σj f (z, z, t, t) = f2 zω j , zω −j ω −`j t + f1 zω j , zω −j ω `j t.
(1)
Fix a parameter κ; the Dunkl operators are defined by
m−1
Df (z, z, t, t) =
X f (z, z, ω `j t, ω −`j t) − f (zω j , zω −j , ω `j t, ω −`j t)
∂
f (z, z, t, t) + κ
,
∂z
z − zω j
(2)
j=0
m−1
X f (z, z, ω `j t, ω −`j t) − f (zω j , zω −j , ω `j t, ω −`j t)
∂
f (z, z, t, t) − κ
ωj .
Df (z, z, t, t) =
∂z
z − zω j
j=0
In these coordinates the Laplacian ∆κ = 4DD. The group covariant property D = σ0 Dσ 0 is
convenient for simplifying some proofs.
The rational Cherednik algebra associated with the data (I2 (m), κ) is the abstract algebra
formed from the product C[z, z]⊗C[D, D]⊗CI2 (m) (polynomials in the two sets of variables and
the group algebra of I2 (m)) with relations like D = σ0 Dσ 0 and z = σ0 zσ0 , and commutations
described in Proposition 1. The algebra acts on Pm,` by multiplication, and equations (2), (1)
for the respective three components. Thus there is a representation of the algebra as operators
on Pm,` , and so Pm,` is called the standard module with lowest degree component of isotype τ`
(that is, the representation τ` of I2 (m)).
The bilinear form on Pm,` is defined in the real coordinate system (t = s1 + is2 , D1 = D+D,
D2 = i(D−D)) by
hsj , sk i = δjk ,
hxj p(x, s), q(x, s)i = hp(x, s), Dj q(x, s)i,
hsj , q(x, s)i = hsj , q(0, s)i.
This definition is equivalent to
hp1 (x)s1 + p2 (x)s2 , q(x, s)i = hs1 , (p1 (D1 , D2 )q)(0, s)i + hs2 , (p2 (D1 , D2 )q)(0, s)i.
The form is real-bilinear and symmetric (see [2]). To transform the definition to the complex
coordinate system we impose the conditions
hcp, qi = chp, qi,
hp, cqi = chp, qi,
c ∈ C,
thus ht, ti = hs1P
+ is2 , s1 + is2 i = 2, ht, ti = 0,Pht, ti = 2. Now suppose p is given in complex
form p(z, z) =
ajk z j z k then set p∗ (z, z) =
ajk z k z j and replace xj by Dj and conjugate.
j,k
j,k
P
This leads to p∗ (D1 + iD2 , D1 − iD2 ) = p∗ (2D, 2D) (in more detail, if p =
ajk z j z k then
j,k
p∗ (2D, 2D)
=
P
j,k
ajk
k
2j+k Dj D ).
Vector Polynomials and a Matrix Weight Associated to Dihedral Groups
3
Then the form is given by
hp1 (z, z)t + p2 (z, z)t, q(z, z, t, t)i = ht, p∗1 (2D, 2D)qi|z=0 + ht, p∗2 (2D, 2D)qi|z=0 .
(3)
`
`
< κ < m
, as will
The form is defined for all κ. It is positive-definite provided that − m
be shown. This is related to the results of Etingof and Stoica [4, Proposition 4.3] on unitary
representations. The (abstract) Gaussian form is defined for f, g ∈ Pm,` by
hf, giG = e∆κ /2 f, e∆κ /2 g = e2DD f, e2DD g .
This has the important property that hpf, giG = hf, p∗ giG for any scalar polynomial p(z, z).
This motivates the attempt to express the pairing in integral form, specifically
Z
2
gKf ∗ e−|z| /2 dm2 (z),
(4)
hf, giG =
C
where f, g ∈ Pm,` are interpreted as vectors f1 (z, z)t + f2 (z, z)t 7→ (f1 , f2 ) and K is an integrable
Hermitian
2 × 2 matrix function.
We preview the formula for K on the fundamental chamber
π
z = reiθ : r > 0, 0 < θ < m
, in the real coordinate system, in terms of auxiliary parameters
and functions:
1
`
Set v := sin2 mθ
2 , δ := 2 − m ,
δ, −δ
κ/2
−κ/2
;v ,
f1 (κ, δ; v) := v (1 − v)
2 F1
1
2 +κ
δ
1 + δ, 1 − δ
f2 (κ, δ; v) := 1
v (κ+1)/2 (1 − v)(1−κ)/2 2 F1
;
v
,
3
2 +κ
2 +κ
2
Γ 12 + κ
,
H(κ, δ) :=
Γ 12 + κ + δ Γ 12 + κ − δ
and
f1 (κ, δ; v)
f2 (κ, δ; v)
L(θ) =
−f2 (−κ, δ; v) f1 (−κ, δ; v)
cos δmθ − sin δmθ
,
sin δmθ cos δmθ
then
cos πκ
0
T H(−κ, δ)
L(θ).
L(θ)
K(r, θ) =
0
H(κ, δ)
2π cos πδ
−1 K(z)τ (w) defines K on the other chambers z = reiθ : r > 0,
The equation K(zw)
=
τ
(w)
`
`
π
π
(j − 1) m
< θ < jm
, 2 ≤ j ≤ 2m.
∂
Usually we will write ∂u := ∂u
for a variable u.
2
Harmonic polynomials
In this section we construct a basis for the (vector-valued) harmonic (∆κ f = 0) homogeneous
polynomials. These are important here because for generic κ any polynomial in Pm,` has a unique
expansion as a sum of terms like (zz)n f (z, z, t, t) where f is homogeneous and ∆κ f = 0 (see [2,
(5), p. 4]); furthermore hf, giG = hf, gi for any harmonic polynomials f , g. Complex coordinates
are best suited for this study. To each monomial in z, z, t or t associate the degree and the
m-parity: (with a, b ∈ N0 )
1) the degree is given by deg(z a z b t) = deg(z a z b t) = a + b,
2) the m-parities of z a z b t, z a z b t are a − b + ` mod m, a − b − ` mod m, respectively.
If each monomial in a polynomial p has the same degree and m-parity then say that p has
this degree and m-parity (this implies p is homogeneous in z, z). The following is a product
rule.
4
C.F. Dunkl
Proposition 1. For f ∈ Pm,`
Dzf (z, z, t, t) = (zD + 1)f (z, z, t, t) + κ
m−1
X
f zω j , zω −j , ω `j t, ω −`j t ,
j=0
Dzf (z, z, t, t) = zDf (z, z, t, t) − κ
m−1
X
f zω j , zω −j , ω `j t, ω −`j t ω j .
j=0
Proof . Abbreviate gj = f (z, z, ω `j t, ω −`j t) and hj = f (zω j , zω −j , ω `j t, ω −`j t). Then
m−1
X zgj − zω j hj
∂
Dzf (z, z, t, t) = z
+ 1 f (z, z, t, t) + κ
∂z
z − zω j
j=0
m−1
X zgj − zhj
zhj − zω j hj
∂
= z
+ 1 f (z, z, t, t) + κ
+
∂z
z − zω j
z − zω j
j=0
= (zD + 1)f (z, z, t, t) + κ
m−1
X
hj .
j=0
The second formula is proved similarly.
With the aim of using Proposition 1 define
Tz f (z, z, t, t) :=
m−1
X
f zω j , zω −j , ω `j t, ω −`j t ,
j=0
Tz f (z, z, t, t) :=
m−1
X
f zω j , zω −j , ω `j t, ω −`j t ω j .
j=0
Lemma 1. Suppose that p ∈ Pm,` has m-parity r then Tz p = 0 for r 6= 0 mod m, Tz p = mσ0 p
for r = 0 mod m, Tz p = 0 for r 6= m − 1 mod m and Tz p = mσ0 p if r = m − 1 mod m.
Proof . Consider Tz z a z b t = z b z a t
m−1
P
ω (a−b+`)j ; the sum is zero if a−b+` 6= 0 mod m, otherwise
j=0
Tz z a z b t = mσ0 (z a z b t). The same argument applies to Tz z a z b t whose m-parity is a − b − `.
m−1
P (a−b+`+1)j
Similarly Tz z a z b t = z b z a t
ω
.
j=0
A polynomial p is called harmonic if DDp = 0. By using an inductive construction we do not
need to explicitly determine the action of D, D on general monomials.
Proposition 2. Suppose that p ∈ Pm,` has m-parity r with r 6= 0 mod m, and Dp = 0 then
Dzp = (zD + 1)p and Dzp = −κmσ0 p if r = m − 1 mod m, otherwise Dzp = 0.
Proof . By Proposition 1 Dzp = (zD + 1)p + κTz p and Tz p = 0 by the lemma. Also Dzp =
zDp − κTz p.
Notice that p = t or p = t with degree 0 and m-parities ` and m − ` respectively, satisfy the
hypotheses of the following.
Vector Polynomials and a Matrix Weight Associated to Dihedral Groups
5
Theorem 1. Suppose that p ∈ Pm,` has m-parity r with 1 ≤ r ≤ m − 1, deg p = n, zDp = np
and Dp = 0 then Dz s p = (n + s)z s−1 p for 0 ≤ s ≤ m − r, Dz s p = 0 for 0 ≤ s ≤ m − 1 − r
mκ
and Dz m−r p = −mκz m−1−r σ0 p. Furthermore q := z m−r p + m−r+n
z m−r σ0 p has m-parity 0 and
satisfies
(
2 )
mκ
Dq = (m − r + n) 1 −
z m−r−1 p,
Dq = 0,
m−r+n
Dzq = (m + n − r + 1)q,
Dzq = 0.
Proof . By Lemma 1 Tz z s p = 0 for 0 ≤ s < m − r. Arguing by induction suppose Dz s p =
(n + s)z s−1 p for some s satisfying 1 ≤ s < m − r then by Proposition 2 Dz s+1 p = (zD + 1)z s p =
((n + s) + 1)z s p. Similarly Tz z s p = 0 for 0 ≤ s ≤ m − 2 − r and Dz s p = 0 for 1 ≤ s ≤ m − 1 − r.
Also Dz m−r p = −mκσ0 (z m−r−1 p). Thus
Dq = Dz m−r p +
mκ
mκ
σ0 Dz m−r p = 0,
Dσ0 z m−r p = −mκσ0 z m−r−1 p +
m−r+n
m−r+n
and
mκ
Dσ 0 z m−r p
m−r+n
mκ
= (m − r + n)z m−r−1 p +
σ0 Dz m−r p
m−r+n
mκ
= (m − r + n)z m−r−1 p +
σ0 − mκσ0 z m−r−1 p
m − r + n)
(
2
mκ
z m−r−1 p.
= (m − r + n) 1 −
m−r+n
Dq = Dz m−r p +
The m-parity of q is zero and Dq = 0 thus Dzq = 0. By Proposition 1
(
2 )
mκ
z m−r p
Dzq = zDq + q + mκσ0 q = (m − r + n) 1 −
m−r+n
mκ
mκ
m−r
m−r
m−r
m−r
+z
p+
z
σ0 p + mκ z
σ0 p +
z
p
m−r+n
m−r+n
mκ
m−r
m−r
= (m − r + n + 1) z
p+
z
σ0 p .
m−r+n
This completes the proof.
Corollary 1. With the same hypotheses z s p, z s σ0 p for 0 ≤ s ≤ m − r, and zq, σ0 (zq) are
harmonic.
Proof . Each of these polynomials f satisfy Df = 0 or Df = 0 (recall Dσ 0 = σ0 D) with the
exception of z m−r p and σ0 (z m−r p). But DDz m−r p = (m − r + n)Dz m−r−1 p = 0.
Observe that D, D change the m-parity by −1, +1 respectively. The leading term of a homon
n
P
P
geneous polynomial
aj z n−j z j t+
bj z n−j z j t is defined to be (a0 z n +an z n )t+(b0 z n +bn z n )t.
j=0
j=0
There are 4 linearly independent harmonic polynomials of each degree ≥1. Recall σ0 p(z, z, t, t)
= p(z, z, t, t). By use of Theorem 1 we find there are two sequences of harmonic homogeneous
(1)
(2)
(1)
(2)
polynomials pn , σ0 pn : n ≥ 1 and pn , σ0 pn : n ≥ 1 with the properties:
(1)
(1)
(2)
(2)
1) the leading terms of pn , σ0 pn , pn , σ0 pn are z n t, z n t, z n t, z n t respectively,
6
C.F. Dunkl
(1)
(1)
2) p1 = zt, σ0 p1 = zt,
(1)
(1)
3) if n + ` 6= 0 mod m then pn+1 = zpn ,
(1)
(1)
4) if n + ` = 0 mod m then pn+1 = z pn +
(2)
(1) mκ
n σ0 pn ;
(2)
5) p1 = zt, σ0 p1 = zt,
(2)
(2)
6) if n − ` 6= 0 mod m then pn+1 = zpn ,
(2)
(2)
7) if n − ` = 0 mod m then pn+1 = z pn +
(2) mκ
n σ0 pn .
(1)
The consequence of these relations is that there exist polynomials Pn
m(n + 1) − ` + 1 and nm + ` + 1 respectively (n ≥ 0) such that
(1)
(1)
(1)
1) pr = z r t for 1 ≤ r ≤ m − ` and ps = z r Pn
(2)
(2)
(2)
2) pr = z r t for 1 ≤ r ≤ ` and ps = z r Pn
(2)
and Pn
of degrees
for s = m(n + 1) − ` + 1 + r and 0 ≤ r < m;
for s = nm + ` + 1 + r and 0 ≤ r < m.
The formulae imply recurrences:
(1)
= z m−`+1 t +
(2)
= z `+1 t +
1) P0
2) P0
mκ
m−` t
m−` zz
mκ
`
` zz t
(1)
(1)
and Pn+1 = z m Pn +
(2)
(2)
and Pn+1 = z m Pn +
mκ
m−1 σ P (1) ;
0 n
(n+1)m+(m−`) zz
mκ
m−1 σ P (2) .
0 n
(n+1)m+` zz
There is a common thread in these recurrences. Let λ be a parameter with λ > 0 and define
(1)
(2)
polynomials Qn (κ, λ; w, w), Qn (κ, λ; w, w) by
κ
(2)
(1)
Q0 (κ, λ; w, w) = ,
Q0 (κ, λ; w, w) = 1,
λ
κ
(1)
wQ(2)
Qn+1 (κ, λ; w, w) = wQ(1)
n (κ, λ; w, w) +
n (κ, λ; w, w),
λ+n+1
κ
(2)
(2)
Qn+1 (κ, λ; w, w) =
wQ(1)
n (κ, λ; w, w) + wQn (κ, λ; w, w).
λ+n+1
There is a closed form point evaluation:
Proposition 3. For λ > 0 and n ≥ 0
(λ + κ)n+1 + (λ − κ)n+1
,
2(λ)n+1
(λ + κ)n+1 − (λ − κ)n+1
.
Q(2)
n (κ, λ; 1, 1) =
2(λ)n+1
Q(1)
n (κ, λ; 1, 1) =
Proof . The statement is obviously true for n = 0. Arguing by induction suppose the statement
is valid for some n then
κ
(λ + κ)n+1 − (λ − κ)n+1
(λ + κ)n+1 + (λ − κ)n+1
(1)
Qn+1 (κ, λ; 1, 1) =
+
2(λ)n+1
λ+n+1
2(λ)n+1
(λ + κ)n+1
κ
(λ − κ)n+1
κ
=
1+
+
1−
2(λ)n+1
λ+n+1
2(λ)n+1
λ+n+1
(λ + κ)n+2 (λ − κ)n+2
=
+
,
2(λ)n+2
2(λ)n+2
and
(2)
Qn+1 (κ, λ; 1, 1)
κ
(λ + κ)n+1 + (λ − κ)n+1
(λ + κ)n+1 − (λ − κ)n+1
=
+
λ+n+1
2(λ)n+1
2(λ)n+1
(λ + κ)n+1
κ
(λ − κ)n+1
κ
=
1+
−
1−
.
2(λ)n+1
λ+n+1
2(λ)n+1
λ+n+1
This completes the proof.
Vector Polynomials and a Matrix Weight Associated to Dihedral Groups
7
Proposition 4. For n ≥ 0
m−` m m
m−` m m
m−` (2)
Pn(1) = z m−`+1 Q(1)
κ,
;
z
,
z
t
+
zz
Q
κ,
;
z
,
z
t,
n
n
m
m
` m m
` m m
(2)
`+1 (1)
` (2)
Pn = z Qn κ, ; z , z
t + zz Qn κ, ; z , z
t.
m
m
Proof . The statement is true for n = 0. Suppose that the first one holds for some n, then set
λ = m−`
m and
mκ
(1)
Pn+1 = z m Pn(1) +
zz m−1 σ0 Pn(1)
(n + 1)m + (m − `)
m m
m−` (2)
m m
= z m z m−`+1 Q(1)
κ,
λ;
z
,
z
t
+
zz
Q
κ,
κ;
z
,
z
t
n
n
κ
m m
m m
+
zz m−1 z m−`+1 Q(1)
t + z m−` zQ(2)
t
n κ, λ; z , z
n κ, λ; z , z
n+1+
λ
κ
m m
m m
= z m−`+1 t z m Q(1)
z m Q(2)
+
n κ, λ; z , z
n κ, λ; z , z
n+1+λ
κ
m m
m (1)
m m
κ,
κ;
z
,
z
+
z
Q
κ,
λ;
z
,
z
+ zz m−` t z m Q(2)
n
n
n+1+λ
(1)
(2)
= z m−`+1 Qn+1 κ, λ; z m , z m t + zz m−` Qn+1 κ, κ; z m , z m t;
this proves the first part by induction. The same argument applies with ` replaced by m − `
and t, t interchanged.
Proposition 5. For n ≥ 0
n
X
κ2 (n − j + 1)
1 − j, j − n, 1 − κ, 1 + κ
=w +
; 1 wn−j wj ,
4 F3
λ(λ + n)
2, λ + 1, −λ − n + 1
j=1
n
X κ
−j, j − n, −κ, +κ
;
1
wn−j wj .
Q(2)
(κ,
λ;
w,
F
w)
=
4 3
n
1, λ, −λ − n
λ+j
Q(1)
n (κ, λ; w, w)
n
j=0
(1)
Proof . Set Qn (κ, λ; w, w) =
n
P
j=0
recurrence is equivalent to a0,0 = 1, b0,0 =
an+1,j = an,j +
(2)
an,j wn−j wj and Qn (κ, λ; w, w) =
κ
bn,n+1−j ,
λ+n+1
κ
λ
n
P
bn,j wn−j wj , then the
j=0
and
bn+1,j = bn,j +
κ
an,n+1−j ,
λ+n+1
for 0 ≤ j ≤ n + 1 (with an,n+1 = 0 = bn,n+1 ). It is clear that an+1,0 = an,0 = 1 and bn+1,0 =
bn,0 = λκ . Now let 1 ≤ j ≤ n + 1. In the following sums the upper limit can be taken as n + 1.
κ
Proceeding by induction we verify that an+1,j − an,j = λ+n+1
bn,n+1−j using the formulae. The
left-hand side is
κ2 (n − j + 2)
1 − j, j − n − 1, 1 − κ, 1 + κ
;1
4 F3
λ(λ + n + 1)
2, λ + 1, −λ − n
κ2 (n − j + 1)
1 − j, j − n, 1 − κ, 1 + κ
−
;1
4 F3
λ(λ + n)
2, λ + 1, −λ − n + 1
n+1
κ2 X (1 − j)i (1 − κ)i (1 + κ)i (n − j + 2)(j − n − 1)i
(n − j + 1)(j − n)i
=
−
λ
(2)i (λ + 1)i i!
(λ + n + 1)(−λ − n)i
(λ + n)(−λ − n + 1)i
i=0
8
C.F. Dunkl
n+1
κ2 X (1 − j)i (1 − κ)i (1 + κ)i
(j − n − 2)i+1
(j − n − 1)i+1
=
−
λ
(i + 1)!(λ + 1)i i!
(−λ − n − 1)i+1
(−λ − n)i+1
i=0
n+1
κ2 X
(1 − j)i (1 − κ)i (1 + κ)i (j − n − 1)i (i + 1)(λ + j − 1)
(i + 1)!(λ + 1)i i!
(−λ − n − 1)i+2
i=0
κ2 (λ + j − 1)
1 − j, j − n − 1, 1 − κ, 1 + κ
=
;1 .
4 F3
λ(λ + n + 1)(λ + n)
1, λ + 1, −λ − n + 1
=
λ
The 4 F3 is balanced and we can apply the transformation (see [5, (16.4.14)]) to show
1 − j, j − n − 1, 1 − κ, 1 + κ
;1
4 F3
1, λ + 1, −λ − n + 1
(λ + 2 − j + n)j−1 (−λ + 2 − j)j−1
1 − j, j − n − 1, −κ, κ
=
;1
4 F3
(λ + 1)j−1 (−λ − n + 1)j−1
1, λ, −λ − n
λ+n+1−j
λ(λ + n)
bn,n−j+1 ,
=
(λ + j − 1)(λ + n + 1 − j)
κ
and this proves the first recurrence relation.
κ
Next verify bn+1,j − bn,j = λ+n+1
an,n+1−j . The left-hand side equals
n+1
κ X (−j)i (−κ)i (κ)i
(j − n − 1)i
(j − n)i
−
λ+j
i!i!(λ)i
(−λ − n − 1)i (−λ − n)i
i=0
n+1
j−n−1
j−n+i−1
κ X (−j)i (−κ)i (κ)i (j − n)i−1
−
=
λ+j
i!i!(λ)i (−λ − n)i−1
−λ − n − 1 −λ − n + i − 1
i=1
n+1
X
(−j)i (−κ)i (κ)i (j − n)i−1 (λ + j)i
i!i!(λ)i (−λ − n − 1)i+1
i=1
n+1
κ
κ2 j X (1 − j)s (j − n)s (1 − κ)s (1 + κ)s
=
λ + n + 1 λ(λ + n)
s!(2)s (λ + 1)s+1 (−λ − n + 1)s
s=0
κ
an,n+1−j ,
=
λ+n+1
=
κ
λ+j
where the index of summation is changed s = i − 1.
We consider the differentiation formulae. Set λ1 :=
we used δ :=
1
2
−
`
m;
m(n + λ1 ) + 1 and
thus δ =
(2)
deg Pn
1
2
− λ2 = λ1 −
1
2.
m−`
m ,
λ2 :=
For n ≥ 0 we have
`
m.
In the previous section
(1)
deg Pn
(2)
Dz r Pn(j) = (m(n + λj ) + r + 1)z r−1 Pn(j) ,
1 ≤ r ≤ m − 1,
κ
(j)
(j)
m−1 (j)
m−1
DPn = (m(n + λj ) + 1) z
Pn−1 +
z
σ0 Pn−1 ,
n + λj
(
2 )
κ
(j)
D2 Pn(j) = (m(n + λj ) + 1)m(n + λj ) 1 −
z m−2 Pn−1 ,
n + λj
Dz
m−1
Pn(j)
(1)
= mn + ` + 1 = m(n + λ2 ) + 1, then (using P−1 = z 1−` t and
P−1 = z 1+`−m t) for j = 1, 2
Dz r Pn(j) = 0,
= m(n + 1) − ` + 1 =
0 ≤ r ≤ m − 2,
= −mκz m−2 σ0 Pn(j) .
The other differentiations follow from D = σ0 Dσ0 .
Vector Polynomials and a Matrix Weight Associated to Dihedral Groups
9
(j)
Notice we have an evaluation formula for pn (1) (j = 1, 2): suppose n = m(k + λj ) + r with
1 ≤ r ≤ m then
1 (λ1 + κ)k+1
(t + t) +
2 (λ1 )k+1
1 (λ2 + κ)k+1
p(2)
(t + t) −
n (1) =
2 (λ2 )k+1
p(1)
n (1) =
3
1 (λ1 − κ)k+1
(t − t),
2 (λ1 )k+1
1 (λ2 − κ)k+1
(t − t).
2 (λ2 )k+1
The bilinear form on harmonic polynomials
Recall (formula (3)) that the form is given by
p1 (z, z)t + p2 (z, z)t, q(z, z, t, t) = t, p∗1 (2D, 2D)q |z=0 + t, p∗2 (2D, 2D)q |z=0 .
We will compute values for the harmonic polynomials. Suppose q is harmonic of degree n and
n
n
P
P
bj z n−j z j t with arbitrary coefficients {aj }, {bj }, then
p=
aj z n−j z j t +
j=0
j=0
*
2
−n
hp, qi =
t,
n
X
+
aj D
n−j
j
D q
*
+
t,
j=0
n
X
+
bj D
n−j
j
D q
j=0
n = t, a0 D + an D q + t, b0 Dn + bn D q ,
(5)
(1) (2) j
because Dn−j D q = 0 for 1 ≤ j ≤ n − 1. Thus pn , q = 2n t, Dn q and pn , q = 2n t, Dn q ;
n
n
(1)
(2)
this follows from the leading terms of pn , pn .
(1)
Define αn by Dn pn = αn t (we will show that there is no t component). If 1 ≤ n ≤ m − `
(1)
then pn = z n t and Dn z n t = n!t so αn = n!. Suppose n = m(k + λj ) + r + 1 for some k ≥ 0 and
(1)
(1)
(1)
(1)
(1)
1 ≤ r ≤ m − 1, then pn = z r Pk and Dpn = (m(n + λj ) + r + 1)z r−1 Pk = nz r−1 Pk . Repeat
(1)
(1)
this procedure to show Dr pn = n(n − 1) · · · (n − r + 1)Pk , that is, αn = (−1)r (−n)r αn−r .
(1)
(1)
Now suppose n = m(k + λj ) + 1 so that pn = Pk and
(
2 )
κ
(1)
2 (1)
D Pk = (m(k + λj ) + 1)m(k + λj ) 1 −
z m−2 Pk−1
k + λ1
(
2 )
κ
(1)
z m−2 Pk−1 ,
= n(n − 1) 1 −
k + λ1
(
2 )
κ
(1)
m (1)
D Pk = n(n − 1) 1 −
Dm−2 z m−2 Pk−1
k + λ1
(
2 )
κ
(1)
(n − 2) · · · (n − m + 1)Pk−1
= n(n − 1) 1 −
k + λ1
=
n!
(λ1 − κ + k)(λ1 + κ + k) (1)
Pk−1 .
(n − m)!
(λ1 + k)2
Induction on these results is used in the following
Proposition 6. For k = 0, 1, 2, . . . and n = m (k + λ1 ) + 1 = m (k + 1) − ` + 1
(1)
Dn Pk
(1)
= n!
(1) Pk , Pk
(λ1 − κ)k+1 (λ1 + κ)k+1
t,
(λ1 )2k+1
= 2n+1 n!
(λ1 − κ)k+1 (λ1 + κ)k+1
.
(λ1 )2k+1
(6)
10
C.F. Dunkl
Proof . Suppose k = 0 then
(1)
D2 P0
= (m − ` + 1)(m − `)
(λ1 − κ)(λ1 + κ) m−`−1
z
t
λ21
and Dm−`−1 z m−`−1 t = (m − ` − 1)!t, so the formula is valid for k = 0. Suppose the formula is
valid for some k ≥ 0 with n = m(k + 1) − ` + 1 then by (6)
(1)
Dm Pk+1 =
(1)
Dn+m Pk+1
(n + m)! (λ1 − κ + k + 1)(λ1 + κ + k + 1) (1)
Pk ,
n!
(λ1 + k + 1)2
(λ1 − κ)k+2 (λ1 + κ)k+2
t.
= (n + m)!
(λ1 )2k+2
(1)
This completes the induction. By Proposition 4 the coefficients of z n and z n in Pk are t and 0
(1) (1) (1) respectively. By equation (5) Pk , Pk
= 2n t, Dn Pk . The fact that ht, ti = 2 finishes the
proof.
Proposition 7. For k = 0, 1, 2, . . . and n = m(k + λ2 ) + 1 = mk + ` + 1
(2)
Dn Pk
(2)
= n!
(2) Pk , Pk
(λ2 − κ)k+1 (λ2 + κ)k+1
t,
(λ2 )2k+1
= 2n+1 n!
(λ2 − κ)k+1 (λ2 + κ)k+1
.
(λ2 )2k+1
Proof . The same argument used in the previous propositions works here.
The other basis polynomials can be handled as a consequence.
Proposition 8. Suppose j = 1, 2, k = 0, 1, 2, . . ., 1 ≤ r ≤ m − 1 and n = m (k + λj ) + r + 1
(j)
(j)
then pn = z r Pk and
(j) (j) (λj − κ)k+1 (λj + κ)k+1
.
pn , pn = 2n+1 n!
(λj )2k+1
(j)
(j)
(j)
n!
Proof . By Theorem 1 Dr pn = (n−r)!
Pk and deg Pk = n − r = m(k + λj ) + 1. The preceding
(j) (j) formulae for Pk , Pk
imply the result.
(1)
(2)
The trivial cases come from pn : 0 ≤ n ≤ m − ` and pn : 0 ≤ n ≤ ` . By Theorem 1
(1) (1) (1)
(2)
Dn pn = Dn z n t = n!t and pn , pn = 2n+1 n! for 0 ≤ n ≤ m − `. Similarly Dn pn = Dn z n t =
(2) (2) n!t and pn , pn = 2n+1 n! for 0 ≤ n ≤ `.
3.1
Orthogonal basis for the harmonics
(1)
(1) (2)
(2)
At each degree ≥ 1 we have the basis pn , σ0 pn , pn , σ0 pn : n ≥ 1 . These polynomials are
pairwise orthogonal with certain exceptions. The leading terms and relevant properties are (the
constants cn are not specified, and may vary from line to line):
(1)
(1)
(1)
1) pn : z n t, Dn pn = cn t, Dpn = 0 if n + ` 6= 0 mod m;
(1)
n
(1)
(1)
2) σ0 pn : z n t, D σ0 pn = cn t, Dσ0 pn = 0 if n + ` 6= 0 mod m;
(2)
(2)
(2)
3) pn : z n t, Dn pn = cn t, Dpn = 0 if n − ` 6= 0 mod m;
(2)
n
(2)
(2)
4) σ0 pn : z n t, D σ0 pn = cn t, Dσ0 pn = 0 if n − ` 6= 0 mod m.
(1)
(2)
(1) (2) If n ≥ 1 then span pn , σ0 pn ⊥ span pn , σ0 pn . This is clear from the above properties
(1) (2) when n ± ` 6= 0 mod m (note pn , pn
= 0 because ht, ti = 0). One example suffices to
Vector Polynomials and a Matrix Weight Associated to Dihedral Groups
11
(1)
(2) demonstrate the other cases: suppose n + ` = 0 mod m then n − ` 6= 0 mod m and pn , σ0 pn =
(2) 2n t, Dn σ0 pn = 0.
Consider the special cases n ± ` = 0 mod m. Recall
(j)
Dz m−1 Pk
(j)
= −mκσ0 z m−2 Pk .
(j)
Using the leading term of σ0 z m−1 Pk
we find that
(j)
(j) (j)
(j) σ0 z m−1 Pk , z m−1 Pk = −2mκ z m−2 Pk , z m−2 Pk 6= 0.
(j)
(j) (j)
(j) In this case there are two different natural orthogonal bases. One is pn + σ0 pn , pn − σ0 pn .
(1)
(1)
First suppose j = 1 and n = m(k + 2) − ` with k ≥ 0, then pn = z m−1 Pk
and
(1) (1) (λ1 − κ)k+1 (λ1 + κ)k+1
(1)
pn , pn = σ0 p(1)
= 2n+1 n!
,
n , σ0 pn
(λ1 )2k+1
n
n−1
(1) (1)
σ0 p(1)
= 2n t, D p(1)
= −2n mκ t, D
σ0 z m−2 Pk
n , pn
n
(λ1 − κ)k+1 (λ1 + κ)k+1
= −2n mκ(n − 1)!
ht, ti.
(λ1 )2k+1
These imply
(1)
(1) (1) (1)
(1)
(1)
pn + σ0 p(1)
= pn , pn − σ0 p(1)
= 0,
n , pn − σ0 pn
n , σ0 pn
(1)
(1) (1) (1) (1)
(1)
(1)
(1)
pn + σ0 pn , pn + σ0 pn = 2 pn , pn + 2 σ0 pn , σ0 pn
(λ1 − κ)k+2 (λ1 + κ)k+1
= 2n+2 n!
,
(λ1 )k+2 (λ1 )k+1
(1)
(j)
(1)
(1)
(1)
pn − σ0 p(1)
= 2 p(1)
− 2 σ0 p(1)
n , pn − σ0 pn
n , pn
n , σ0 pn
(λ1 − κ)k+1 (λ1 + κ)k+2
= 2n+2 n!
.
(λ1 )k+1 (λ1 )k+2
In these calculations we used
n − mκ
m(k + 1 − κ) + m − `
λ1 − κ + k + 1
n! − mκ(n − 1)! = n!
= n!
= n!
n
m(k + 1) + m − `
λ1 + k + 1
and
n! + mκ(n − 1)! = n!
λ1 + κ + k + 1
.
λ1 + k + 1
(1)
The result also applies when k = −1, that is, n = m − ` and pn = z m−` t.
(2)
(2)
Now suppose j = 2 and n = m(k + 1) + ` with k ≥ 0, then pn = z m−1 Pk and by a similar
argument
(λ2 − κ)k+2 (λ2 + κ)k+1
(2) (2)
(2)
p(2)
= 2n+2 n!
,
n + σ0 pn , pn + σ0 pn
(λ2 )k+2 (λ2 )k+1
(2)
(λ2 − κ)k+1 (λ2 + κ)k+2
(2)
(2)
pn − σ0 p(2)
= 2n+2 n!
.
n , pn − σ0 pn
(λ2 )k+1 (λ2 )k+2
Note n = m(k + 1 + λ2 ) and n ± mκ = m(λ2 ± κ + k + 1). The formula also applies to n = `
(2)
with k = −1 and pn = z ` t.
12
C.F. Dunkl
(j)
The other natural basis is pn +
(j)
form a linear combination fn =
Indeed
(j)
(j) (j)
(j)
mκ
(with pn = z m−1 Pk ); the idea is to
n σ0 pn , σ0 pn
(j)
(j)
(j) (j) pn + cσ0 pn so that Df = 0, which implies σ0 pn , fn = 0.
(j)
(j)
Dfn(j) = −mκσ0 z m−2 Pk + cnσ0 z m−2 Pk .
(j)
(j)
Another expression for fn is z1 Pk+1 , that is,
(2)
n = m(k + 1) − `,
κ, λ1 ; z m , z m t + z m−` Qk κ, λ1 ; z m , z m t,
m m
` (2)
m m
κ, λ2 ; z , z t + z Qn κ, λ2 ; z , z t,
n = mk + `.
(1)
fn(1) = z m−` Qk
fn(2) = z ` Q(1)
n
(j) (j) To compute fn , fn (by use of Dσ0 = σ0 D)
mκ m−2 (j) (j)
z
Pk
Dm−1 fn(j) = Dm−2 nz m−1 Pk − mκ
n
mκ 2
mκ 2 n!
(j)
m−2 m−1 (j)
=n 1−
1−
z
Pk =
D
Pk .
n
(n − m + 1)!
n
If j = 1 then n = m(k + 2) − ` and
1−
mκ 2
n
=
(k + 1 + λ1 − κ)(k + 1 + λ1 + κ)
,
(k + 1 + λ1 )2
and by the previous results (also valid for k = −1)
(1) (1) (λ1 − κ)k+2 (λ1 + κ)k+2
fn , fn = 2n+1 n!
.
(λ1 )2k+2
Similarly if j = 2 then n = m(k + 1) + ` and
(2) (2) (λ2 − κ)k+2 (λ2 + κ)k+2
.
fn , fn = 2n+1 n!
(λ2 )2k+2
`
Observe that − m
<κ<
4
`
m
= λ2 implies all the pairings hp, pi > 0 (by definition λ2 < λ1 ).
The Gaussian kernel
In the previous sections we showed that the forms h·, ·i and h·, ·iG are positive-definite on the
`
`
harmonic polynomials exactly when − m
<κ< m
. By use of general formulae ([2, (4), p. 4])
k
n
with γ(κ; τ ) = 0) for ∆κ ((zz) f (z, z, t, t)), where ∆κ f = 0 it follows that the forms are positivedefinite on Pm,` for the same values of κ. This together with the property hpf, giG = hf, p∗ giG
for any scalar polynomial p(z, z) suggests there should be an integral formulation of h·, ·iG .
The construction begins with suitable generalizations
of the integral for the scalar case and the
group I2 (m), namely |z m − z m |2κ exp − zz
.
2
In [2, Section 5] we derived conditions, holding for any finite reflection group, on the matrix
function K associated with the Gaussian inner product. The formula for W is stated in (4), and
the general conditions specialize to:
1) K(zw) = τ` (w)−1 K(z)τ` (w), w ∈ W ;
2) K satisfies a boundary
condition
at
the
walls
of
the
fundamental
chamber
C
:=
z = reiθ :
π
0 < θ < m , r > 0 ; see equation (9) below;
Vector Polynomials and a Matrix Weight Associated to Dihedral Groups
13
3) K = L∗ M L where M is a constant positive-definite matrix and L satisfies a differential
system.
The differential system (for a 2 × 2 matrix function L) is
∂z L(z, z) = κL(z, z)
m−1
X
j=0
∂z L(z, z) = κL(z, z)
m−1
X
j=0
1
τ` (σj ),
z − zω j
−ω j
τ` (σj ).
z − zω j
(As motivation for this formulation note that the analogous scalar weight function |z m −z m |2κ =
m−1
m−1
P
P −ωj
1
hh where h satisfies ∂z h = κh
and ∂z h = κh
.) It follows that
z−zω j
z−zω j
j=0
j=0
(z∂z + z∂z )L = 0,
thus L is positively homogeneous of degree 0 and depends only on eiθ for z = reiθ (r > 0,
−π < θ ≤ π). Further ∂θ = i(z∂z − z∂z ) and thus
∂θ L = iκL
m−1
X
j=0
z + zω j
τ` (σj ).
z − zω j
We use the following to compute the sums involving z − zω j .
Lemma 2. For n ∈ Z let r = n − 1 mod m and 0 ≤ r < m then (indeterminate w)
m−1
X
j=0
mwr
ω jn
=
,
w − ωj
wm − 1
m−1
X
j=0
ω jn
mz r z m−1−r
=
.
z − zω j
zm − zm
Proof . Assume |w| < 1 then
m−1
X
j=0
m−1
∞
m−1
X ω j(n−1)
X
X
ω jn
s
=
−
=
−
w
ω j(n−1−s) .
w − ωj
1 − wω −j
j=0
s=0
j=0
The inner sum equals m if n − 1 = s mod m, that is s = r + km for k ≥ 0, otherwise the sum
vanishes; thus
m−1
X
j=0
∞
X
ω jn
−mwr
r+km
=
−m
w
=
.
w − ωj
1 − wm
k=0
The identity is valid for all w ∈
/ {ω j }. Next
m−1
X
j=0
m−1
ω jn
mz r z m−1−r
1 X ω jn
m z r z −r
=
=
=
.
z
j
z − zω j
z
z z m z −m − 1
zm − zm
z −ω
j=0
Corollary 2. For 1 ≤ ` ≤ m − 1
m−1
X
j=0
z ` z m−`
z + zω j j`
ω
=
2m
,
z − zω j
zm − zm
m−1
X
j=0
z + zω j −j`
z m−` z `
ω
=
2m
.
z − zω j
zm − zm
14
C.F. Dunkl
Write L = L1 t + L2 t, then
∂θ L1 (θ) =
mκ iθ(m−2`)
e
L2 (θ),
sin mθ
As in [1, Theorem 4.3] let δ =
1
2
−
`
m,
∂θ L2 (θ) =
mκ −iθ(m−2`)
e
L1 (θ).
sin mθ
f1 = e−imδθ L1 , L
f2 = eimδθ L2 , then
L
e 2 (θ),
e 1 (θ) = −imδ L
e 1 (θ) + mκ L
∂θ L
sin mθ
mκ e
e 2 (θ).
L1 (θ) + imδ L
sin mθ
e 1 −L
e 2 s2 .
e 1 +L
e 2 s1 +i L
e 1 t+L
e2 t = L
Now changing {t, t} to real coordinates t = s1 +is2 we write L
e1 − L
e 2 and φ = mθ. The system is transformed to
e1 + L
e 2 , g2 = i L
Let g1 = L
∂φ g1 =
κ
g1 − δg2 ,
sin φ
∂φ g2 = δg1 −
Introduce the variable v = sin2
g1 =
v
1−v
φ
2
κ
g2 .
sin φ
and let
κ/2
h1 (v),
e 2 (θ) =
∂θ L
g2 =
v
1−v
(κ+1)/2
h2 (v).
Here we restrict θ to the fundamental chamber θ : 0 < θ <
∂
sin φ = 2(v(1 − v))1/2 and ∂φ
v = (v(1 − v))1/2 . Then
∂ v h1 = −
δ
h2 ,
1−v
∂v h2 =
π
m
so that 0 < v < 1 and
κ + 12
δ
h1 −
h2 ,
v
v(1 − v)
from which follows
v(1 −
v)∂v2 h1
1
+ κ + − v ∂v h1 + δ 2 h1 = 0.
2
This is the hypergeometric equation with solutions
δ, −δ
(1)
h1 = 2 F1 1
;v ,
2 +κ
!
1
− κ − δ, 12 − κ + δ
1
1
1
1 + δ, 1 − δ
(2)
−κ
−κ
+κ
2
; v = v 2 (1 − v) 2 2 F1
;v .
h1 = v 2 2 F1
3
3
2 −κ
2 −κ
By use of the differentiations
a, b
ab
a + 1, b + 1
∂v 2 F1
;v =
;v ,
2 F1
c
c
c+1
a, b
a, b
c
c−1
∂v v 2 F1
; v = cv
;v ,
2 F1
c+1
c
we find
(1)
h2
(2)
h2
=
1
2
δ
(1 − v) 2 F1
+κ
1 + δ, 1 − δ
;v ,
3
2 +κ
1
2
−κ
1
=−
(1 − v)v − 2 −κ 2 F1
δ
1
=−2
−κ
1
1
(1 − v)κ+ 2 v − 2 −κ
δ
1
2
− κ − δ, 12 − κ + δ
;v
1
2 −κ
δ, −δ
;v .
2 F1
1
2 −κ
!
Vector Polynomials and a Matrix Weight Associated to Dihedral Groups
15
All these solutions can be expressed in terms of the following
δ, −δ
κ/2
−κ/2
f1 (κ, δ; v) := v (1 − v)
;v ,
2 F1
1
2 +κ
1 + δ, 1 − δ
δ
(κ+1)/2
(1−κ)/2
f2 (κ, δ; v) := 1
v
(1 − v)
;v .
2 F1
3
2 +κ
2 +κ
(2)
Then (by replacing hj
(1)
g1
(2)
g1
=
=
v
1−v
κ/2
v
1−v
κ/2
by
−δ
1
−κ
2
(1)
h1
(2)
h1
(2)
hj
(7)
for j = 1, 2)
= f1 (κ, δ; v),
(1)
g2
= −f2 (−κ, δ; v),
(2)
g2
=
=
v
1−v
(κ+1)/2
v
1−v
(κ+1)/2
(1)
h2 = f2 (κ, δ; v),
(2)
h2 = f1 (−κ, δ; v).
The Wronskian of the system is f1 (κ, δ; v)f1 (−κ, δ; v) + f2 (κ, δ; v)f2 (−κ, δ; v) = 1. Returning
to L1 , L2 we find (for j = 1, 2)
(j)
e (j) = 1 eimδθ g (j) − ig (j) ,
L1 = eimδθ L
1
1
2
2
(j)
e (j) = 1 e−imδθ g (j) + ig (j) ,
L2 = e−imδθ L
2
1
2
2
and in the real coordinate system t = s1 + is2
(j)
(j)
(j)
(j)
Lj1 := L1 + L2 = cos mδθ g1 + sin mδθ g2 ,
(j)
(j) (j)
(j)
Lj2 := i L1 − L2 = − sin mδθ g1 + cos mδθ g2 .
So a fundamental solution in the real coordinates (for 0 < φ = mθ < π, recall v = sin2 φ2 ) is
L(φ) = Mf (v)Mδ (φ),
f1 (κ, δ; v)
f2 (κ, δ; v)
,
Mf (v) =
−f2 (−κ, δ; v) f1 (−κ, δ; v)
(8)
cos δφ − sin δφ
.
Mδ (φ) =
sin δφ cos δφ
(In the trivial case κ = 0 it follows from standard hypergeometric series formulae that f1 (0, δ; v)
= cos δφ and f2 (0, δ; v) = sin δφ, see Lemma 3.) As described in [2] the kernel for the Gaussian
2
2
inner product in the fundamental chamber C is given by K(φ)e−(x1 +x2 )/2 where
K(φ) = L(φ)T M L(φ)
and M is a constant positive-definite matrix determined by boundary conditions at the walls,
π
θ = 0 and θ = m
in this case. Suppose σ corresponds to one of the walls (that is, the hyperplane
fixed by σ) and x0 6= 0 is a boundary point of C with x0 σ = x0 then the condition is: for any
vectors ξ, η in the module for τ` (the underlying representation) if ξτ` (σ) = ξ and ητ` (σ) = −η
then
lim
x→x0 , x∈C
ξK(x)η T = 0,
equivalently
lim
x→x0 , x∈C
K(x) − τ` (σ)K(x)τ` (σ) = 0.
By the invariance property of K the condition is equivalent to
lim
K(x0 + x) − K(x0 + xσ) = 0,
x→0, x0 +x∈C
(9)
16
C.F. Dunkl
1 0
a weak type of continuity. Near φ = 0 consider τ` (σ0 ) =
and the boundary condition
0 −1
becomes
lim K(φ)12 = 0.
φ→0+
Let M0 =
a1 a2
and replace Mf (v) by (with nonzero constants c1 , c2 depending on δ and κ,
a2 a3
see (7))
2−κ φκ c1 φκ+1
L (φ) :=
,
c2 φ1−κ 2κ φ−κ
0
then lim Mδ (φ) = I and
φ→0
L0 (φ)T M0 L0 (φ)
12
= 2−κ c2 a1 φ1+2κ − 2κ c1 a3 φ1−2κ + a2 1 − c1 c2 φ2 .
The boundary condition immediately implies a2 = 0, then the positive-definite condition implies
a1 a3 6= 0 and thus 1 ± 2κ > 0. This bound (− 12 < κ < 21 ) also implies the integrability of K on
2
the circle (and with respect to e−|x| /2 dm2 (x)).
Next we consider the behavior of L(φ) near φ = π, that is, v near 1. We use (see [5, (15.8.4)])
a, b
Γ(c)Γ(c − a − b)
a, b
;v =
;1 − v
2 F1
2 F1
Γ(c − a)Γ(c − b)
c
1+a+b−c
c − a, c − b
c−a−b Γ(c)Γ(a + b − c)
+ (1 − v)
;1 − v .
2 F1
Γ(a)Γ(b)
1−a−b+c
Define
H(κ, δ) :=
Γ
1
2
2
Γ 12 + κ
.
+ κ + δ Γ 12 + κ − δ
The formula Γ(u)Γ(1 − u) =
Γ
1
2
π
sin πu
+ κ Γ − 12 − κ
=
Γ(δ)Γ(−δ)
1
2
implies
!
δ
sin πδ
.
+ κ cos πκ
Then
f1 (κ, δ; v) = H(κ, δ)f1 (−κ, δ; 1 − v) +
sin πδ
f2 (κ, δ; 1 − v),
cos πκ
sin πδ
f1 (κ, δ; 1 − v) − H(κ, δ)f2 (−κ, δ; 1 − v).
cos πκ
sin πδ 2
Note H(κ, δ)H(−κ, δ) + cos
= 1. Summarize the transformation rules as
πκ
f2 (κ, δ; v) =
(10)
sin πδ
f1 (κ, δ; v)
f2 (κ, δ; v)
H(κ, δ)
f1 (−κ, δ; 1 − v) −f2 (−κ, δ; 1 − v)
cos
πκ
=
,
sin πδ
−f2 (−κ, δ; v) f1 (−κ, δ; v)
f2 (κ, δ; 1 − v)
f1 (κ, δ; 1 − v)
− cos
πκ H(−κ, δ)
or set
sin πδ
H(κ, δ)
cos
πκ
,
=
sin πδ
− cos
πκ H(−κ, δ)
MH
0 1
Mσ =
,
1 0
Vector Polynomials and a Matrix Weight Associated to Dihedral Groups
17
then
Mf (v) = MH Mσ Mf (1 − v)Mσ ,
the same relation holds with v and 1 − v interchanged (indeed (MH Mσ )2 = I). The reflection
π
corresponding to the wall θ = m
is σ1 (zσ1 = zω so eiπ/m σ1 = eiπ/m ). Thus the boundary
condition is
lim K(φ) − τ` (σ1 )K(φ)τ` (σ1 ) = 0,
φ→π−
and (in the real coordinates)
"
#
2π`
cos 2π`
sin
m
m
τ` (σ1 ) =
.
2π`
sin 2π`
−
cos
m
m
Then
"
K(φ) − τ` (σ1 ) K(φ)τ` (σ1 ) = q(φ)
q(φ) = {K(φ)11 − K(φ)22 } sin
sin 2π`
m
− cos 2π`
m
− cos 2π`
m
− sin 2π`
m
#
,
2π`
2π`
− 2K (φ)12 cos
.
m
m
We want to determine M0 so that lim q(φ) = 0. We have
φ→π−
K(φ) = Mδ (φ)T Mf (v)T M0 Mf (v)Mδ (φ).
Let ψ = π − φ then the first-order approximations (as ψ → 0+ ) are
1
1 − v ∼ ψ2,
4
−κ
ψ
f1 (κ, δ; 1 − v) ∼
,
2
κ
ψ
f1 (−κ, δ; 1 − v) ∼
,
2
f2 (κ, δ; 1 − v) ∼
1
2
1−κ
ψ
,
2
1+κ
δ
ψ
,
−κ 2
δ
+κ
f2 (−κ, δ; 1 − v) ∼
1
2
and
κ


Mf (v) ∼ MH 
ψ
2
δ
1
+κ
2
1−κ
ψ
2
1+κ 
ψ
δ
− 1 −κ
2

2
.
−κ
ψ
2
We also find
Mδ (π − ψ) ∼ M1 + ψδM2 ,
"
#
π`
sin π`
−
cos
m
m
M1 =
,
π`
cos π`
sin
m
m
"
M2 =
cos π`
m
sin π`
m
− sin π`
m
cos π`
m
#
.
There is one more simplification: for an arbitrary M0 =
2π`
(J11 − J22 ) sin 2π`
m − 2J12 cos m = 2a2 . So
 κ

Mf (v)Mδ (π − ψ) ∼ MH 
ψ
2
δ
1
+κ
2
1−κ
ψ
2
a1 a2
a2 a3
if J = M1T M0 M1 then
1+κ 
ψ
δ
− 1 −κ
2

2
 (M1 + ψδM2 );
−κ
ψ
2
18
C.F. Dunkl
and discarding quantities tending to zero (as ψ → 0+ ) we need to consider the (1, 2)-entry of
−κ κ
−κ κ
2 ψ
0
2 ψ
0
0
T a1
MH
MH
,
0
2κ ψ −κ
0
2κ ψ −κ
0 a3
sin πδ
which equals (a1 H(κ, δ) − a3 H(−κ, δ)) cos
πκ . Thus with the normalization constant c(κ, δ)
0
T H(−κ, δ)
L(φ),
(11)
K(φ) = c(κ, δ)L(φ)
0
H(κ, δ)
for 0 < φ = mθ < π. The condition that the kernel be positive definite is H(κ, δ)H(−κ, δ) > 0,
and
`
`
+ κ sin π m
−κ
sin π m
cos π(κ + δ) cos π(κ − δ)
H(κ, δ)H(−κ, δ) =
=
,
cos2 πκ
cos2 πκ
`
<κ<
thus − m
c(κ, δ) =
5
`
m.
In Section 6 we show that
cos πκ
.
2π cos πδ
Integrals of harmonic polynomials
As in [2] define the circle-type pairing by
hf, giS =
1
2n n!
hf, giG =
1 2n n!
e∆κ /2 f, e∆κ /2 g
when deg f + deg g = 2n (n = 0, 1, . . .); the term “circle” refers to the polar coordinate system.
If f , g are harmonic then hf, giG = hf, gi. In the coordinate system {t, t} the matrix function
(denoted by K C for this basis) satisfies K C (z)∗ = K C (z) and K C (zw) = τ` (w)∗ K C (z)τ` (w) for
w ∈ W (recall τ` is unitary), and K C is positively homogeneous of degree 0. The Gaussian
pairing is
Z
2
hf, giG =
g(z)K C (z)f (z)∗ e−|z| /2 dm2 (z),
C
where dm2 (z) = dx1 dx2 (for z = x1 + ix2 ) and g(z) denotes (g1 (z), g2 (z)) with
g = g1 t + g2 t,
iθ the fundamental chamber is (r, θ) : r > 0,
similarly for
f
.
In
polar
coordinates
z
=
re
π
0<θ< m
. Then
Z π
X Z ∞
∗
m
−r2 /2
hf, giG =
e
rdr
g reiθ w K C eiθ w f reiθ w dθ
w∈W
=
w∈W
0
0
X Z
∞
e−r
2 /2
Z
rdr
0
π
m
∗
g reiθ w τ` (w)∗ K C eiθ τ` (w)f reiθ w dθ.
0
In particular, if f , g are homogeneous with deg f + deg g = 2n then
π
X Z m
∗
1
hf, giS = n hf, giG =
g eiθ w τ` (w)∗ K C eiθ τ` (w)f eiθ w dθ.
2 n!
0
(12)
w∈W
Now
r1 , r2 . That is, 0 ≤ r1 , r2 < m, f is a sum of monomials
a bsuppose f , g have m-parities
z z t : a − b + ` = r1 mod m and z a z b t : a − b − ` = r1 mod m and g has the same property
with r1 replaced by r2 . The generic rotation in W is
%j : f (z, z, t, t) 7→ f zω j , zω −j , tω `j , tω −`j ;
Vector Polynomials and a Matrix Weight Associated to Dihedral Groups
19
this maps the monomial z a z b t to ω j(a−b+`) z a z b t and z a z b t to ω j(a−b−`) z a z b t. By hypothesis
f (eiθ ρj )τ` (ρj )∗ = ω r1 j f (eiθ ). The reflection
σj : f (z, z, t, t) 7→ f zω j , zω −j , tω `j , tω −`j
maps the monomial z a z b t to ω j(a−b+`) z b z a t = and z a z b t to ω j(a−b−`) z b z a t. For a polynomial
p(z, z) let p∨ = p(z, z). So if f = f1 t + f2 t has m-parity r1 then σj f = ω r1 j (f1∨ t + f2∨ t). Break
up the sum in formula (12) into two parts, one for rotations (including the identity ρ0 ) and one
for the reflections we obtain
Z π
m−1
X
∗
m
hf, giS =
g eiθ K C eiθ f eiθ dθ
ω r2 j ω −r1 j
0
j=0
+
m−1
X
ω
r2 j
ω
−r1 j
Z
j=0
π
m
∗
gˆ eiθ K C eiθ f ˆ eiθ dθ,
0
where gˆ = (g2∨ , g1∨ ) and f ˆ = (f2∨ , f1∨ ). The sums vanish unless r1 = r2 in which case both
equal m.
Next we convert the matrix in (11) to the complex coordinate system. Recall
K(φ) = Mδ (φ)T Mf (v)T M0 Mf (v)Mδ (φ),
in the real coordinates, where
f1 (κ, δ; v)
f2 (κ, δ; v)
,
Mf (v) =
−f2 (−κ, δ; v) f1 (−κ, δ; v)
cos δφ − sin δφ
.
Mδ (φ) =
sin δφ cos δφ
To find the change-of-basis matrix observe g1 t + g2 t 7→ g1 (s1 + is2 ) + g2 (s1 − is2 ) = (g1 + g2 )s1 +
i(g1 − g2 )s2 . Let
1 i
,
B=
1 −i
then in the complex coordinates
K C = BMδ (φ)T Mf (v)T M0 Mf (v)Mδ (φ)B ∗ .
We find
e−iδφ 0
BMδ (φ) =
B,
0
eiδφ
`
and δφ = 12 − m
mθ. So
−iδφ
iδφ
e
0
0
C
T
∗ e
K =
BMf (v) M0 Mf (v)B
.
0 e−iδφ
0
eiδφ
T
Then using (8)
C
C
K11
= K22
= c(κ, δ)G1 (κ, δ, v),
G1 (κ, δ, v) := f1 (κ, δ; v)2 + f2 (κ, δ; v)2 H(−κ, δ) + f1 (−κ, δ; v)2 + f2 (−κ, δ; v)2 H(κ, δ),
C
C = e−2iδmθ c(κ, δ)G (κ, δ; v),
K12
= K21
(13)
2
2
2
G2 (κ, δ; v) := f1 (κ, δ; v) + if2 (κ, δ; v) H(−κ, δ) − f1 (−κ, δ; v) + i f2 (−κ, δ; v) H(κ, δ).
Recall v = sin2
φ
2
= 12 (1 − cos mθ) and e−2iδφ = e−iθ(m−2`) (= z m z 2` on the circle).
20
C.F. Dunkl
In the special case f = g = t we obtain
Z π
Z
m
C
C
2 = ht, tiS = m
K11
+ K22
dθ = 2m
0
π
m
C
K11
dθ.
(14)
0
In the “trivial” case κ = 0 the term f1 (0, δ; v)2 + f2 (0, δ; v)2 = 1 (see Lemma 3) and H(0, δ) =
(1)
1
cos πδ thus c(0, δ) = 2π cos
πδ . Consider the evaluation of hf, f iS where f = pn and n =
m(k + 1) − ` + r with k ≥ 0 and 1 ≤ r ≤ m; thus
m−` m m
m−` m m
(1)
(2)
t + z r z m−` Qk
f = z m−`+r Qk
κ,
;z ,z
κ,
;z ,z
t.
m
m
(1)
(2)
Because Qk and Qk have real coefficients we find f1∨ = f1 and f2∨ = f2 . The integrand is
(j)
(suppressing the arguments κ, m−`
m in Qk and omitting the factor c(κ, δ))
2 f1 f1 + f2 f2 G1 + 2e−2iδmθ f1 f2 G2 + 2e2iδmθ f2 f1 G2 ,
(1)
(2)
(1)
(2)
f1 f1 + f2 f2 = Qk z m , z m Qk z m , z m + Qk z m , z m Qk z m , z m ,
(2)
(1)
e−2iδmθ f1 f2 = z m z 2` z m−`+r z r z m−` Qk z m , z m Qk z m , z m
(2)
(1)
= z m Qk z m , z m Qk z m , z m ,
(2)
(1)
e2iδmθ f1 f2 = z m Qk z m , z m Qk z m , z m .
(Note zz = 1 on the circle.) The integrand is now expressed as a function of z m , that is, eiφ .
(2)
Similarly suppose f = pn with n = mk + ` + r so that
` m m
` m m
`+r (1)
(2)
r ` (2)
κ, ; z , z
t + z Qk
κ, ; z , z
t,
pn = z z Qk
m
m
(1)
(1)
(2)
(2)
e−2iδmθ f1 f2 = z m z 2` z r z ` z `+r Qk z m , z m Qk z m , z m = z m Qk z m , z m Qk z m , z m ,
(1)
(2)
e2iδmθ f1 f2 = z m Qk z m , z m Qk z m , z m .
6
The normalizing constant
In this section we find the normalizing constant for K by computing the integral in (14) and
using a method that separates the parameters δ and κ. To start we examine the special cases
f1 (0, δ; v) and f2 (0, δ; v).
Lemma 3. For −π < φ < π and any δ
2 φ
2 φ
f1 0, δ; sin
= cos δφ,
f2 0, δ; sin
= sin δφ.
2
2
Proof
. The first part is a standard formula (see [5, (15.4.12)]). Briefly, consider g(v) =
δ,−δ
2
2
2 F1
1/2 ; v ; it can be shown directly that ∂φ g(v) = −δ g(v), g(0) = 1 and g(1) = cos πδ;
p
note ∂φ = v(1 − v)∂v . Apply ∂φ to both sides of the first formula to obtain
δ(−δ) p
1 + δ, 1 − δ
v(1 − v) 2 F1
; v = −δ sin δφ;
1/2
3/2
the left hand side equals −δf2 (0, δ; v).
C + K C . Recall G (κ, δ; v) from (13).
Next we derive another expression for K11
1
22
Vector Polynomials and a Matrix Weight Associated to Dihedral Groups
Lemma 4. For − 12 < κ <
1
2
21
and δ > 0
G1 (κ, δ; v) = f1 (κ, δ; v)f1 (−κ, δ; 1 − v) + f1 (−κ, δ; v)f1 (κ, δ; 1 − v)
− f2 (κ, δ; v)f2 (−κ, δ; 1 − v) − f2 (−κ, δ; v)f2 (κ, δ; 1 − v).
Proof . From equation (10) we have (replacing κ by −κ and v by 1 − v)
sin πδ
f2 (−κ, δ; v),
cos πκ
sin πδ
H(−κ, δ)f1 (κ, δ; v)2 = f1 (κ, δ; v)f1 (−κ, δ; 1 − v) −
f1 (κ, δ; v)f2 (−κ, δ; v),
cos πκ
sin πδ
H(κ, δ)f1 (−κ, δ; v)2 = f1 (−κ, δ; v)f1 (κ, δ; 1 − v) −
f1 (−κ, δ; v)f2 (κ, δ; v),
cos πκ
H(−κ, δ)f1 (κ, δ; v) = f1 (−κ, δ; 1 − v) −
and
sin πδ
f1 (−κ, δ; v) − f2 (−κ, δ; 1 − v),
cos πκ
sin πδ
H(−κ, δ)f2 (κ, δ; v)2 =
f1 (−κ, δ; v)f2 (κ, δ; v) − f2 (κ, δ; v)f2 (−κ, δ; 1 − v),
cos πκ
sin πδ
H(κ, δ)f2 (−κ, δ; v)2 =
f1 (κ, δ; v)f2 (−κ, δ; v) − f2 (−κ, δ; v)f2 (κ, δ; 1 − v).
cos πκ
H(−κ, δ)f2 (κ, δ; v) =
Adding up the appropriate equations (the second two in each group) we get the desired result.
Note the cancellations of the terms f1 (κ, δ; v)f2 (−κ, δ; v) and f1 (−κ, δ; v)f2 (κ, δ; v).
Proposition 9. Suppose − 21 < κ <
Z
1
−1/2
G1 (κ, δ; v)(v(1 − v))
0
1
2
and 0 ≤ δ <
1
dv =
cos πκ
Z
1
1
2
then
G1 (0, δ; v)(v(1 − v))−1/2 dv.
0
Proof . By the symmetry v → 1 − v it suffices to integrate f1 (κ, δ; v)f1 (−κ, δ; 1 − v) and
f2 (κ, δ; v)f2 (−κ, δ; 1 − v). By definition
−κ
κ
f1 (κ, δ; v)f1 (−κ, δ; 1 − v) = v (1 − v)
∞
X
(δ)i (−δ)i (δ)j (−δ)j i
1
v (1 − v)j .
1
i!j! 2 + κ i 2 − κ j
i,j=0
Multiply by (v(1 − v))−1/2 and integrate over 0 < v < 1 to obtain
∞
X
(δ)i (−δ)i (δ)j (−δ)j
1
1
B i + + κ, j + − κ
2
2
i!j! 12 + κ i 12 − κ j
i,j=0
X
∞
(δ)i (−δ)i (δ)j (−δ)j
1
1
=Γ
+κ Γ
−κ
,
2
2
i!j!(i + j)!
i,j=0
because
1
1
1
1
1
1
1
B i + + κ, j + − κ =
Γ
+κ Γ
−κ
+κ
−κ .
2
2
Γ (i + j + 1)
2
2
2
i 2
j
The ratio to the same integral with κ replaced by 0 is
Γ 12 + κ Γ 12 − κ
1
.
=
1 2
cos
πκ
Γ 2
22
C.F. Dunkl
Similarly
f2 (κ, δ; v)f2 (−κ, δ; 1 − v) =
δ2
v 1+κ (1 − v)1−κ
1
1
+
κ
−
κ
2
2
∞
X
(1 + δ)i (1 − δ)i (1 + δ)j (1 − δ)j i
3
×
v (1 − v)j .
3
i!j!
+
κ
−
κ
2
i 2
j
i,j=0
Multiply by (v(1 − v))−1/2 and integrate over 0 < v < 1 to obtain
1
2
∞
X
(1 + δ)i (1 − δ)i (1 + δ)j (1 − δ)j
δ2
3
3
B i + + κ, j + − κ
2
2
i!j! 32 + κ i 23 − κ j
+ κ 12 − κ i,j=0
X
∞
(1 + δ)i (1 − δ)i (1 + δ)j (1 − δ)j
1
1
2
=Γ
+κ Γ
−κ δ
,
2
2
i!j!(i + j + 2)!
i,j=0
because
B i+
1
2
+ κ Γ 32 − κ 32 + κ i 32 − κ j
1
1
2 +κ
2 − κ Γ (i + j + 3)
3 + κ 3 − κ
1
1
2
i 2
j
=Γ
+κ Γ
−κ
.
2
2
(i + j + 2)!
Γ
+ κ, j + 32 − κ
1
=
+κ 2 −κ
3
2
3
2
The double series converges by Stirling’s formula and the comparison test. By the same argument
as for the first integral the proof is finished.
Lemma 5. Suppose 0 < δ <
Z
then
1
dv
π
1
f1 (0, δ; v)f1 (0, δ; 1 − v) p
= cos δπ +
sin πδ,
2
2δ
v(1 − v)
1
dv
π
1
f2 (0, δ; v)f2 (0, δ; 1 − v) p
= − cos δπ +
sin πδ.
2
2δ
v(1 − v)
0
Z
1
2
0
Proof . For v = sin2 φ2 one has dv =
Lemma 3 the two integrals equal
Z
p
v(1 − v)dφ. Also 1 − v = cos2
φ
2
= sin2
π−φ
2 ,
thus by
π
π
1
cos δπ +
sin δπ,
2
2δ
0
Z π
π
1
sin δπ,
sin δφ sin δ(π − φ)dφ = − cos δπ +
2
2δ
0
cos δφ cos δ(π − φ)dφ =
respectively.
Combining the lemma and the proposition we prove:
Proposition 10. Suppose − 12 < κ <
Z
0
1
1
2
and 0 ≤ δ <
G1 (κ, δ; v)(v(1 − v))−1/2 dv = 2π
cos δπ
.
cos κπ
1
2
then
Vector Polynomials and a Matrix Weight Associated to Dihedral Groups
23
To determine the normalizing constant: recall the requirement from (14)
Z
π
m
Z
π
C
K11
dθ
C
2 = ht, tiS = 2m
=2
K11
dφ
0
0
Z 1
cos δπ
G1 (κ, δ; v)(v(1 − v))−1/2 dv = 4πc(κ, δ)
= 2c(κ, δ)
,
cos
κπ
0
1 cos κπ
c(κ, δ) =
.
2π cos δπ
This implies
det K =
sin π
`
m
+ κ sin π
4π 2 sin2 π`
m
`
m
−κ

=
1 
1−
4π 2
sin πκ
sin π`
m
!2 

(and det K C = | det B|2 det K = 4 det K). The weight function K is actually integrable for
`
− 21 < κ < 12 but does not give a positive form when m
≤ |κ| < 12 .
References
[1] Dunkl C.F., Differential-difference operators and monodromy representations of Hecke algebras, Pacific J.
Math. 159 (1993), 271–298.
[2] Dunkl C.F., Vector-valued polynomials and a matrix weight function with B2 -action, SIGMA 9 (2013), 007,
23 pages, arXiv:1210.1177.
[3] Dunkl C.F., Dunkl operators and related special functions, arXiv:1210.3010.
[4] Etingof P., Stoica E., Unitary representations of rational Cherednik algebras, Represent. Theory 13 (2009),
349–370, arXiv:0901.4595.
[5] Olver F.W.J., Lozier D.W., Boisvert R.F., Clark C.W. (Editors), NIST handbook of mathematical functions,
U.S. Department of Commerce National Institute of Standards and Technology, Washington, DC, 2010.