Séminaire Lotharingien de Combinatoire 66 (2012), Article B66b

VECTOR VALUED MACDONALD POLYNOMIALS

C. F. DUNKL^{†} AND J.-G. LUQUE^{‡}

Abstract. This paper defines and investigates nonsymmetric Macdonald polynomials with values in an irreducible module of the Hecke algebra of type A_{N−1}. These polynomials appear as simultaneous eigenfunctions of Cherednik operators. Several objects and properties are analyzed, such as the canonical bilinear form which pairs polynomials with those arising from reciprocals of the original parameters, and the symmetrization of the Macdonald polynomials. The main tool of the study is the Yang–Baxter graph. We show that these Macdonald polynomials can be easily computed following this graph. We give also an interpretation of the symmetrization and the bilinear forms applied to the Macdonald polynomials in terms of the Yang–Baxter graph.

1. Introduction

For each partition λ of N there is an irreducible module of the Hecke algebra of type A_{N−1} whose basis is labeled by standard tableaux of shape λ. This paper defines and analyzes nonsymmetric Macdonald polynomials with values in such modules. The double affine Hecke algebra generated by multiplication by coordinate functions, q-type Dunkl operators, the Hecke algebra and a q-shift acts on these polynomials. They appear as simultaneous eigenfunctions of the associated Cherednik operators. There is a canonical bilinear form which pairs these polynomials with those arising from the reciprocals of the original parameters. The Macdonald polynomials and their reciprocalparameter versions constitute a biorthogonal set of the form. The values of the form are found explicitly.

There are symmetric Macdonald polynomials in this structure. They are labeled by column-strict tableaux of shape λ (non-decreasing entries in each row, strictly increasing in each column). Formulae for these polynomials in terms of nonsymmetric Macdonald polynomials are derived and the values of the bilinear form are obtained in this case. There are analogous results for antisymmetric Macdonald polynomials, which are labeled by row-strict tableaux. There is a hook-length type formula for the bilinear form evaluated at the minimal symmetric polynomial associated with λ.

2010 Mathematics Subject Classification. 05E10, 05E05, 05E15, 20C08,20C30, 20C40, 33D80.

†Dept. of Mathematics, University of Virginia, Charlottesville VA 22904-4137, U.S.

email: cfd5z@virginia.edu ‡Université de Rouen, Laboratoire d’Informatique, du Traitement de l’Information et des Systèmes (LITIS - EA 4108), Avenue de l’Université - BP 8 76801 Saint-Étienne-du-Rouvray Cedex, Fr.

email: Jean-Gabriel.Luque@univ-rouen.fr This paper is partially supported by the ANR project PhysComb (ANR-08-BLAN-0243-2).

In the study of one-variable orthogonal polynomials the very simple graph 0 → 1 → 2 → ⋯ symbolizes the Gram–Schmidt process used to produce the polynomials. In the present multi-variable setting, the Yang–Baxter graph displays how each Macdonald polynomial is produced. Each arrow corresponds to either an adjacent transposition or an affine step (u_{1}, …, u_{N}) → (u_{2}, …, u_{N}, u_{1} + 1). This idea is developed in Section 4.

In Section 2 we give the basic definitions of the Hecke algebra, its modules, and the machinery necessary to describe the leading terms of Macdonald polynomials. Section 3 begins with the simplest two-dimensional module associated to the partition (2, 1) of N = 3. We describe how the basic operations arise in this situation and thus motivate our general definitions. The rest of the section gives the definitions and proves the fundamental relations, notably the braid relations, for the vector-valued situation. A key part is played by the triangularity property of the Cherednik operators with respect to a natural partial order on monomials.

Section 4 contains the description of the simultaneous eigenfunctions, the spectral vectors, the transformation rules for the action of the generators of the Hecke algebra on the polynomials, and the Yang–Baxter graph.

Section 5 concerns the connected components of the Yang–Baxter graph modified by the removal of the affine edges. Here we find the conditions under which the component contains a unique symmetric or antisymmetric polynomial.

The bilinear form is defined and evaluated in Section 6. The method of evaluation relies on relatively simple calculations of the effects of a single arrow in the Yang–Baxter graph. The minimal symmetric polynomials are studied in this section. The hook-length formula for the bilinear form gives some information about aspherical modules of the double affine Hecke algebra, a topic to be pursued in future work.

The paper concludes with a symbol index and a list of basic relations for quick reference.

2. Double affine Hecke algebra

2.1. Definitions and basic properties. Consider the elements T_{i} and w verifying the following four relations:

(1) (T_{i} + t_{1})(T_{i} + t_{2}) = 0,

(2) T_{i}T_{i+1}T_{i} = T_{i+1}T_{i}T_{i+1},

(3) T_{i}T_{j} = T_{j}T_{i} for |i − j| > 1,

(4) T_{i}w = wT_{i−1}. These operators act on the right on ℂ(t_{1}, t_{2}, q)[x_{1}, …, x_{N}] by

(1) T_{i} := \overline{π}_{i}(t_{1} + t_{2}) − t_{2}s_{i},

(2) w := τ_{1}s_{1} ⋯ s_{N−1}, where \overline{π}_{i} = ∂_{i}x_{i+1}, ∂_{i} is the divided difference defined by

∂_{i} = (1 − s_{i})\frac{1}{x_{i} − x_{i+1}} ,

s_{i} denotes the transposition (i, i + 1) and

f(x_{1}, …, x_{N})τ_{i} = f(x_{1}, …, x_{i−1}, qx_{i}, x_{i+1}, …, x_{N}).

Note that the parameter t_{1} could be omitted, since dividing each T_{i} by t_{1} we obtain

\frac{1}{t_{1}} T_{i}^{t_{1},t_{2},q} = T_{i}^{1, \frac{t_{2}}{t_{1}} ,q}.

For simplicity, we will use the parameters t_{1} = 1 and t_{2} = −s. Then the quadratic relation is (T_{i} + 1)(T_{i} − s) = 0 and T_{i} := \overline{π}_{i}(1 − s) + ss_{i}. Note that these operators have interesting commutation properties with respect to the multiplications by x_{i}:

(2.1)  x_{i}T_{i} − T_{i}x_{i+1} − (1 − s)x_{i+1} = 0
(2.2)  x_{i+1}T_{i} − T_{i}x_{i} + (1 − s)x_{i+1} = 0.

The double affine Hecke algebra is defined as

H_{N}(q, s) := ℂ(s, q)[T_{1}, …, T_{N−1}, w^{±1}, x_{1}^{±1}, …, x_{N}^{±1}].

The double affine Hecke algebra admits a maximal commutative subalgebra generated by the Cherednik elements:

ξ_{i} := s^{i−N}T_{i−1}^{−1} ⋯ T_{1}^{−1}wT_{N−1} ⋯ T_{i}.

The (nonsymmetric) Macdonald polynomials are the simultaneous eigenfunctions of the Cherednik operators. This implies that one can compute them using the Yang–Baxter graphs: the spectral vector of 1 is

ζ = [(\frac{1}{s})^{i−1}]_{1≤i≤N} .

The nonaffine edges act by s_{i} on the spectral vector and by T_{i} − _{\frac{1−s}{\frac{ζ[i+1]}{ζ[i]} −1}} on the polyno-

mials. The affine edges act by w on the spectral vector and by Φ_{q} := T_{1}^{−1} ⋯ T_{N−1}^{−1}x_{N} on the polynomial. Note that there exists a shifted version. All of that is contained in the papers [10, 1].

From [1], we define a (q, s)-version of the Dunkl operator:

(1) D_{N} := (1 − s^{N−1}ξ_{N})x_{N}^{−1},

(2) D_{i} := \frac{1}{s}T_{i}D_{i+1}T_{i}. These operators generalize the Dunkl operator for the double affine Hecke algebra. For instance, one has

D_{i+1}T_{i} = −sT_{i}^{−1}D_{i}, −T_{i}D_{i+1} + (1 − s)D_{i} + D_{i}T_{i} = 0,

−D_{i+1}T_{i}^{−1} − (1 − \frac{1}{s})D_{i+1} + T_{i}^{−1}D_{i} = 0, (2.3)

[D_{i}, T_{j}] = 0 when |i − j| > 1.

The (q, s)-Dunkl operators have also interesting commutation properties with respect to the operator w, namely

(2.4)  D_{i+1}w = wD_{i}, 1 ≤ i ≤ N − 1,
(2.5)  qD_{1}w = wD_{N}

Note also that the operators D_{i} commute with each other:

[D_{i}, D_{j}] = 0, 1 ≤ i, j ≤ N. (2.6)

2.2. Modules for the Hecke algebra.

Definition 2.1. A tableau of shape λ is a filling with integers which is weakly increasing in each row and in each column. In the sequel row-strict means (strictly) increasing in each row and column-strict means (strictly) increasing in each column.

A reverse standard tableau (RST) is obtained by filling the shape λ with integers 1, …, N subject to the condition that entries along rows and columns are strictly decreasing. We denote by Tab_{λ}, the set of the RST with shape λ.

Let 𝕋 be an RST. We define the vector of contents of τ as the vector CT_{𝕋} with the property that CT_{𝕋}[i] is the content of i in 𝕋. (The coordinates of the cell are (ROW_{𝕋}[i], COL_{𝕋}[i]), row and column; CT_{𝕋}[i] = COL_{𝕋}[i] − ROW_{𝕋}[i].)

Example 2.2.

CT 2 5 4 6 3 1 = [2, −2, 1, 0, −1, 0].

As in [3, 4] (see also [9]), let us introduce the pairwise commuting Murphy elements

L_{N} := 0,
L_{i} := T_{i} + \frac{1}{s}T_{i}L_{i+1}T_{i}, 1 ≤ i < N.

Let V_{λ} be the vector space spanned by (independent) {𝕋 : 𝕋 ∈ Tab (λ)}. The action of H_{N} (q, s) on V_{λ} satisfies

𝕋L_{i} = s\frac{1 − s^{CT_{𝕋}[i]}}{1 − s} 𝕋, 1 ≤ i ≤ N.

These equations determine {𝕋} up to scalar multiplication. There is a modification of the Murphy elements which is actually more useful for our applications.

Definition 2.3. For 1 ≤ i ≤ N let ϕ_{i} := s^{i−N}T_{i}T_{i+1} ⋯ T_{N−1}T_{N−1} ⋯ T_{i}, or, equivalently, ϕ_{N} = 1 and ϕ_{i} = \frac{1}{s}T_{i}ϕ_{i+1}T_{i} for 1 ≤ i < N.

Proposition 2.4. We have ϕ_{i} = 1 + \frac{s−1}{s}L_{i} for 1 ≤ i ≤ N, and if 𝕋 ∈ Tab (λ) then 𝕋ϕ_{i} = s^{CT_{𝕋}[i]}𝕋.

Proof. Use downward induction. The statement is true for i = N. Suppose the statement is true for ϕ_{i+1}, then

ϕ_{i} = \frac{1}{s}T_{i} (1 + \frac{s − 1}{s}L_{i+1}) T_{i}

= \frac{1}{s} (T_{i}^{2} + \frac{s − 1}{s}T_{i}L_{i+1}T_{i}) = \frac{1}{s} ((s − 1) T_{i} + s + \frac{s − 1}{s}T_{i}L_{i+1}T_{i})

= 1 + \frac{s − 1}{s}L_{i}.

Thus 𝕋ϕ_{i} = (1 + \frac{s − 1}{s} \frac{s (1 − s^{CT_{𝕋}[i]})}{1 − s}) 𝕋 = s^{CT_{𝕋}[i]}𝕋. □

There is an important commutation relation.

Lemma 2.5. Suppose 1 ≤ i, j ≤ N − 1 and i ≠ j, j + 1. Then T_{j}^{−1}ϕ_{i}T_{j} = ϕ_{i}.

Proof. If j < i − 1 the result follows from T_{k}T_{j} = T_{j}T_{k} for |i − j| ≥ 2. If j > i then (note that T_{j}^{−1}T_{j−1}T_{j} = T_{j−1}T_{j}T_{j−1}^{−1})

s^{N−i}T_{j}^{−1}ϕ_{i}T_{j} = T_{j}^{−1}T_{i} ⋯ T_{N−1}T_{N−1} ⋯ T_{i}T_{j}
= T_{i} ⋯ T_{j−2}T_{j}^{−1}T_{j−1}T_{j} ⋯ T_{j}T_{j−1}T_{j} ⋯ T_{i}
= T_{i} ⋯ T_{j−2}T_{j−1}T_{j}T_{j−1}^{−1}T_{j+1} ⋯ T_{j+1}T_{j}T_{j−1}T_{j} ⋯ T_{i}
= T_{i} ⋯ T_{j−2}T_{j−1}T_{j}T_{j+1} ⋯ T_{j+1}T_{j−1}^{−1}T_{j−1}T_{j}T_{j−1} ⋯ T_{i}
= s^{N−i}ϕ_{i}.
□

We describe the action of T_{i} on 𝕋. There are two special cases:

if ROW_{𝕋}[i]) = ROW_{𝕋}[i + 1] then 𝕋T_{i} = s𝕋,
if COL_{𝕋}[i] = COL_{𝕋}[i + 1] then 𝕋T_{i} = −𝕋.

Otherwise, if we denote by 𝕋^{(i,j)} the tableau 𝕋 where the entries i and j have been interchanged, the tableaux 𝕋^{(i,i+1)} is an RST. If ROW_{𝕋}[i] < ROW_{𝕋}[i + 1] (implying COL_{𝕋}[i] > COL_{𝕋}[i + 1]) then

(2.7)  𝕋T_{i} = 𝕋^{(i,i+1)} − \frac{1 − s}{1 − s^{COL_{𝕋}[i+1]−COL_{𝕋}[i]}} 𝕋.

Note that this is a formula for 𝕋^{(i,i+1)}. If ROW_{𝕋}[i] > ROW_{𝕋}[i+1] (implying COL_{𝕋}[i] < COL_{𝕋}[i + 1]) then set m := CT_{𝕋}[i + 1] − CT_{𝕋}[i] (which is > 0 by the hypothesis). It follows that

(2.8)  𝕋T_{i} = \frac{s − 1}{1 − s^{m}} 𝕋 + \frac{s (1 − s^{m+1}) (1 − s^{m−1})}{(1 − s^{m})^{2}}𝕋^{(i,i+1)}.

Formally this gives also the special cases m = 1 when COL_{𝕋}[i] = COL_{𝕋}[i + 1] and m = −1 when ROW_{𝕋}[i] = ROW_{𝕋}[i + 1].

2.3. Hecke elements associated to a multi-index. Denote S := T_{1} ⋯ T_{N−1} and θ := s_{1} ⋯ s_{N−1} . Observe that if i > 1

(2.9)  T_{i}S = ST_{i−1} and s_{i}θ = θs_{i−1}.

For each multi-index u = [u_{1}, …, u_{N}] we define

(2.10)  T_{u} = \begin{cases}1 & if u = [0, …, 0], \\ T_{[u_{N} −1,u_{1},...,u_{N−1}]}S & if u_{N} > 0, \\ T_{[u_{1},...,u_{i−1},0,u_{i},0,...,0]}T_{i} & if u_{i} > 0. \end{cases}

Example 2.6. Let u = [0, 1, 0, 2] then T_{u} = ST_{3}T_{2}ST_{3}S:

\begin{matrix}[0, 0, 0, 0] \\ [0, 0, 0, 1] \\ S \\ [0, 0, 1, 0] \\ T_{3} \\ [0, 1, 0, 0] \\ T_{2} \\ [1, 0, 0, 1] \\ S \\ [1, 0, 1, 0] \\ T_{3} \\ [0, 1, 0, 2] \\ S \end{matrix}

Since we use only braid relations and commutations, if u[j] > u[j + 1] one has

(2.11)  T_{u} = T_{us_{j}}T_{j}.

Hence, the vector T_{u} can be obtained by any product of the type A_{1} ⋯ A_{k} where A_{i} ∈ {S} ∪ {T_{i} : i = 1, …, N − 1} are such that

(1) We obtain u from [0, …, 0] by applying a_{1} ⋯ a_{k} where a_{i} = s_{j} if A_{i} = T_{j} and a_{i} = θ if A_{i} = S.

(2) If a_{i} = s_{j} then u′ := u · a_{1} ⋯ a_{i−1} verifies u′[j] < u′[j + 1].

Example 2.7. One has

T_{[0102]} = ST_{3}T_{2}ST_{3}S
= ST_{3}T_{2}T_{1}T_{2}T_{3}T_{3}T_{1}T_{2}T_{3}
= ST_{3}T_{1}T_{2}T_{1}T_{3}T_{3}T_{1}T_{2}T_{3}
= ST_{3}T_{1}T_{2}T_{3}T_{1}T_{1}T_{3}T_{2}T_{3}
= ST_{3}T_{1}T_{2}T_{3}T_{1}T_{1}T_{2}T_{3}T_{2}
= ST_{3}ST_{1}ST_{2}

Graphically this is

\begin{matrix}[0, 0, 0, 0] \\ [0, 0, 0, 1] \\ S \\ [0, 0, 1, 0] \\ T_{3} \\ [0, 1, 0, 1] \\ S \\ [0, 1, 0, 0] \\ T_{2} \\ [1, 0, 0, 1] \\ S \\ T_{1} \\ [1, 0, 1, 0] \\ T_{3} \\ [0, 1, 0, 2] \\ S \\ [1, 0, 0, 2] \\ S \\ T_{2} \end{matrix}

Remark 2.8. The construction of T_{u} can be illustrated in terms of braids. The generators T_{i} and S are interpreted as

T_{i} = S = u_{N} ⋮u_{i+1} u_{i} ⋮u_{1} u_{N} ⋮u_{i} u_{i+1} ⋮u_{1} u_{N} u_{N−1} ⋮u_{2} u_{1} u_{1} + 1 u_{N} ⋮u_{3} u_{2}

For instance for u = [0, 1, 0, 2] one obtains the braid:

\begin{matrix}0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \\ 1 \\ 2 u[4] \\ 0 u[3] \\ 1 u[2] \\ 0 u[1] \\ S \\ T_{3}T_{2} \\ S \\ T_{3} \\ S \end{matrix}

We introduce the creation operator

ℭ_{i} := (ST_{N−1} ⋯ T_{i})^{i}

This operator has the property that, if v = [v[1], …, v[N]] is a partition, then

T_{v}ℭ_{i} = T_{[v[1]+1,...,v[i]+1,v[i+1],...,v[N]]}

is the partition obtained from v by adding 1 to the i first entries. As a consequence, the element associated to a partition is a product of creation operators

T_{[v_{1},...,v_{N}]} = ℭ_{1}^{v_{1}−v_{2}} ⋯ ℭ_{N−1}^{v_{N−1}−v_{N}} ℭ_{N}^{v_{N}} .

Example 2.9. Consider the computation of T_{[2,1,0]} in the following figure.

\begin{matrix}[0, 0, 0] \\ [0, 0, 1] \\ S \\ [0, 1, 0] \\ T_{2} \\ [1, 0, 0] \\ T_{1} \\ ℭ_{1} \\ [0, 0, 2] \\ S \\ [0, 2, 0] \\ T_{2} \\ [2, 0, 1] \\ S \\ [2, 1, 0] \\ T_{2} \\ ℭ_{2} \end{matrix}

8 C. F. DUNKL AND J.-G. LUQUE

Setting ϕ̃_{i} := s^{N−i}ϕ_{i} = T_{i} ⋯ T_{N−1}T_{N−1} ⋯ T_{i}, one obtains the following factorization. Proposition 2.10. For each i, we have

ℭ_{i} = ϕ̃_{1} ⋯ ϕ̃_{i}.

In order to provide a proof, we need the following auxiliary result. Lemma 2.11. Let i − k > 1. Then we have

(T_{i−k} ⋯ T_{i}) (ST_{N−1} ⋯ T_{i}) = (ST_{N−1} ⋯ T_{i+1}) (T_{i−k−1} ⋯ T_{i}) .

Proof. By eq. (2.9), one has

T_{i} (ST_{N−1} ⋯ T_{i}) = ST_{i−1} (T_{N−1} ⋯ T_{i})
= (ST_{N−1} ⋯ T_{i+1}) (T_{i−1}T_{i}) .

Hence, using eq. (2.9) iteratively, one obtains

(T_{i−k} ⋯ T_{i}) (ST_{N−1} ⋯ T_{i}) = (T_{i−k} ⋯ T_{i−1}) (ST_{N−1} ⋯ T_{i+1}) (T_{i−1}T_{i})
= S (T_{i−k−1} ⋯ T_{i−2}) (T_{N−1} ⋯ T_{i+1}) (T_{i−1}T_{i})
= (ST_{N−1} ⋯ T_{i+1}) (T_{i−k−1} ⋯ T_{i}),

as expected. □ Proof of Proposition 2.10. Applying Lemma 2.11 iteratively, one has

ϕ̃_{1}ϕ̃_{2} ⋯ ϕ̃_{i} = (ST_{N−1} ⋯ T_{i}) (T_{i−1} ⋯ T_{2}) (ST_{N−1} ⋯ T_{2}) ϕ̃_{3} ⋯ ϕ̃_{i}
= (ST_{N−1} ⋯ T_{i})^{2} (T_{i−2}T_{i−1}) ⋯ (T_{1}T_{2}) ϕ̃_{3} ⋯ ϕ̃_{i}
= (ST_{N−1} ⋯ T_{i})^{2} (T_{i−2}T_{i−1}) ⋯ (T_{2}T_{3}) ST_{N−1} ⋯ T_{3}ϕ̃_{4} ⋯ ϕ̃_{i}
= (ST_{N−1} ⋯ T_{i})^{3} (T_{i−3}T_{i−2}T_{i−1}) ⋯ (T_{1}T_{2}T_{3}) ϕ̃_{4} ⋯ ϕ̃_{i}
= (ST_{N−1} ⋯ T_{i})^{4} (T_{i−4}T_{i−3}T_{i−2}T_{i−1}) ⋯ (T_{1}T_{2}T_{3}T_{4}) ϕ̃_{5} ⋯ ϕ̃_{i}
= ⋯
= (ST_{N−1} ⋯ T_{N−i})^{i}.
□

As a consequence, if 𝕋 is an RST and v is a partition, one has

(2.12)  𝕋T_{v} = s^{∗}𝕋,

where ∗ denotes an integer which depends only on v and 𝕋. 2.4. Rank function. There is a unique element of H_{N} (q, s) associated to each σ ∈ 𝔖_{N}. The length of σ ∈ 𝔖_{N} is

ℓ (σ) := # {(i, j) : 1 ≤ i < j ≤ N, i · σ > j · σ} .

There is a shortest expression σ = s_{i_{1}} ⋯ s_{i_{ℓ(σ)}} and a unique element T̃_{σ} ∈ H_{N} (q, s) defined by

(2.13)  T̃_{σ} = T_{i_{1}} ⋯ T_{i_{ℓ(σ)}}.

For any s_{i}, ℓ (s_{i}σ) = ℓ (σ) ± 1; if ℓ (s_{i}σ) = ℓ (σ) + 1 then T̃_{s_{i}σ} = T_{i}T̃_{σ}, and if ℓ (s_{i}σ) = ℓ (σ) − 1 then T̃_{s_{i}σ} = T_{i}^{−1}T̃_{σ}. Similarly, if ℓ (σs_{i}) = ℓ (σ) + 1 then T̃_{σs_{i}} = T̃_{σ}T_{i}, and if ℓ (σs_{i}) = ℓ (σ) − 1 then T̃_{σs_{i}} = T̃_{σ}T_{i}^{−1}. The following fact will be used in the analysis of the raising operator for polynomials.

Proposition 2.12. Let σ ∈ 𝔖_{N}. Then T̃_{σ}^{−1}T̃_{θ}T̃_{θ^{−1}σ} = s^{N−(1·σ)}ϕ_{1·σ}.

Proof. Use induction on ℓ (σ). The statement is true for ℓ (σ) = 0, σ = 1, because T̃_{θ}T̃_{θ^{−1}} = T_{1} ⋯ T_{N−1}T_{N−1} ⋯ T_{1} = s^{N−1}ϕ_{1}. Suppose the statement is true for all σ′ with ℓ (σ′) ≤ n and ℓ (σ) = n + 1. For some k one has ℓ (σs_{k}) = ℓ (σ) − 1. Set σ′ := σs_{k} and i := 1 · σ′. Then we have T̃_{σ} = T̃_{σ′}T_{k}. If ℓ (θ^{−1}σ′s_{k}) = ℓ (θ^{−1}σ′) − 1 then T̃_{θ^{−1}σ} = T̃_{θ^{−1}σ′}T_{k}^{−1} and

T̃_{σ}^{−1}T̃_{θ}T̃_{θ^{−1}σ} = T_{k}^{−1}T̃_{σ′}^{−1}T̃_{θ}T̃_{θ^{−1}σ′}T_{k}^{−1}
= s^{N−i}T_{k}^{−1}ϕ_{i}T_{k}^{−1},

by the inductive hypothesis. If ℓ (θ^{−1}w′s_{k}) = ℓ (θ^{−1}σ′) + 1 then T̃_{θ^{−1}σ} = T̃_{θ^{−1}σ′}T_{k} and T̃_{σ}^{−1}T̃_{θ}T̃_{θ^{−1}σ} = s^{N−i}T_{k}^{−1}ϕ_{i}T_{k} by a similar argument.

Let i_{1} = k · σ^{′−1} and i_{2} = (k + 1) · σ^{′−1}. By hypothesis, we have i_{1} < i_{2}. Let j_{1} = k ·(θ^{−1}σ′)^{−1} = i_{1} ·θ and j_{2} = (k + 1)·(θ^{−1}σ′)^{−1} = i_{2} ·θ. Then ℓ (θ^{−1}σ′s_{k}) = ℓ (θ^{−1}σ′)+1 if and only if j_{1} < j_{2}. (Note that j · θ = j − 1 if j > 1 and 1 · θ = N.) Since i_{2} > i_{1} ≥ 1, it follows that j_{2} = i_{2} −1. If i_{1} = 1 then j_{1} = N > j_{2}, and so ℓ (θ^{−1}σ′s_{k}) = ℓ (θ^{−1}σ′)−1, k = 1 · σ′ = i. This implies 1 · σ = i + 1 and T̃_{σ}^{−1}T̃_{θ}T̃_{θ^{−1}σ} = s^{N−i}T_{i}^{−1}ϕ_{i}T_{i}^{−1} = s^{N−i−1}ϕ_{i+1}. If i_{1} > 1 then j_{1} = i_{1}−1 < j_{2} and ℓ (θ^{−1}σ′s_{k}) = ℓ (θ^{−1}σ′)+1. In this case 1·σ′ ≠ k, k+1, and so s^{N−i}T_{k}^{−1}ϕ_{i}T_{k} = s^{N−i}ϕ_{i}, by Lemma 2.5; also 1 · σ = 1 · σ′ = i; and this completes the induction. □

Consider the rank function of a multi-index v = [v[1], …, v[N]] as an element of 𝔖_{N},

r_{v} [i] := # {j : 1 ≤ j ≤ i, v [j] ≥ v [i]} + # {j : i < j ≤ N, v [j] > v [i]} .

Example 2.13. (1) If v = [4, 2, 2, 3, 2, 1, 4, 4] then r_{v} = [1, 5, 6, 4, 7, 8, 2, 3]. (2) If v is a (decreasing) partition then r_{v} = id.

The length of r_{v} is

ℓ (r_{v}) := #inv(v),

with inv(v) := {(i, j) : 1 ≤ i < j ≤ N, v [i] < v [j]} being the number of inversions in v (note that for i < j we have r_{v} [i] > r_{v}[j] if and only if v [i] < v[j]). There is a shortest expression r_{v} = s_{i_{1}} ⋯ s_{i_{ℓ(rv)}} and an element R_{v} ∈ H_{N} (q, s) defined by

R_{v} := T_{i_{ℓ(r{α})}}^{−1} ⋯ T_{i_{1}}^{−1} = T̃_{r_{v}}^{−1}.

We have the following auxiliary result.

Lemma 2.14. (1) If v[i] > v[i + 1] then R_{vs_{i}} = R_{v}T_{i}^{−1}.

(2) If v[i] < v[i + 1] then R_{vs_{i}} = R_{v}T_{i}.

(3) If v[i] = v[i + 1] then R_{v}T_{i} = T_{r_{v}[i]}R_{v}.

Proof. (1) If v [i] > v [i + 1] then r_{vs_{i}} = s_{i}r_{v}and #inv (vs_{i}) = #inv (v) + 1, so

R_{vs_{i}} = R_{v}T_{i}^{−1}.

(2) Similarly if v [i] < v [i + 1] then R_{vs_{i}} = R_{v}T_{i}.

(3) If v [i] = v [i + 1] and k = r_{v}[i] then s_{i}r_{v} = r_{v}s_{k} and ℓ (s_{i}r_{v}) = ℓ (r_{v}) + 1 (one extra inverted pair (k + 1, k)); thus T̃_{s_{i}r_{v}} = T_{i}T̃_{r_{v}} and T̃_{r_{v}s_{k}} = T̃_{r_{v}}T_{k}. Hence, R_{v}T_{i} = T_{k}R_{v}.

□

We compare the elements T_{v} and R_{v} in terms of T_{v}R_{v}^{−1}. We need to consider three cases:

(1) If T_{[0,...,0]} = I then r_{v} = I = T_{[0,...,0]}.

(2) In the case T_{[v_{1},v_{2},...,v_{i−1},v_{i},0,...]} = T_{[v_{1},v_{2},...,v_{i−1},0,v_{i},0,...]}T_{i} (v_{i} ≥ 1, i < N), we see that #inv (v · s_{i}) = #inv (v) + 1, hence r_{v·s_{i}} = s_{i}r_{v} (see Lemma 2.14 (1)) and T̃_{r_{v·s_{i}}} = T_{i}T̃_{r_{v}}, R_{v·s_{i}} = R_{v}T_{i}^{−1}. So we have

(2.14)  T_{v·s_{i}}R_{v·s_{i}}^{−1} = T_{v}R_{v}^{−1}.

(3) If T_{vΨ} = T_{v}S (with vΨ := (v_{2}, v_{3}, …, v_{N}, v_{1} + 1)), then we have

r_{vΨ} = s_{N−1}s_{N−2} ⋯ s_{1}r_{v} = θ^{−1}r_{v},

where θ = s_{1}s_{2} ⋯ s_{N−1}. By Proposition 2.12 (let k = r_{v} [1]), we obtain

T̃_{r_{v}}^{−1}T̃_{θ}T̃_{θ^{−1}r_{v}} = s^{N−k}ϕ_{k},
s^{−N+k}ϕ_{k}^{−1}R_{v}S = R_{vΨ},

and thus

(2.15)  T_{vΨ}R_{vΨ}^{−1} = s^{N−k}T_{v}SS^{−1}R_{v}^{−1}ϕ_{k} = s^{N−k}T_{v}R_{v}^{−1}ϕ_{k}.

As a consequence, we are able to derive the following result.

Proposition 2.15. T_{v}R_{v}^{−1} is in the commutative algebra generated by

{ϕ_{i} : 1 ≤ i ≤ N}

for each v, and it acts by scalar multiplication (by powers of s) on each 𝕋 (recall that 𝕋ϕ_{i} = s^{CT (i,𝕋)}𝕋, 1 ≤ i ≤ N). Furthermore we have

T_{v} = ∏_{i=1}^{N} (s^{N−i}ϕ_{i})^{v_{i}^{+}} R_{v}.

Proof. By eq. (2.14), we see that, if the formula is true for v with v_{j} = 0 for j > i and v_{i} ≥ 1, then it is also true for v · s_{i} (note that (v · s_{i})^{+} = v^{+}). Using induction, suppose the formula is true for all v with |v| ≤ n, for some n ≥ 0 (the case n = 0 is trivially satisfied). Let |v| = n + 1. Using the case 2 step as often as necessary, assume that v_{N} ≥ 1. Thus v = uψ with |u| = n, and r_{v} = θ^{−1}r_{u}; in particular, let k = r_{v} [N] = r_{u} [1]. Then v^{+} = (u_{1}^{+}, …, u_{k}^{+} + 1, …, u_{N}^{+}) (u has exactly k − 1 entries > u_{1}, and thus v has exactly k entries ≥ v_{N} = u_{1} + 1, including v_{N}; hence v_{k}^{+} = v_{N} = u_{1} + 1 = u_{k}^{+} + 1). By eq. (2.15) and the inductive hypothesis, we obtain

T_{v}R_{v}^{−1} = (s^{N−k}ϕ_{k}) T_{u}R_{u}^{−1} = (s^{N−k}ϕ_{k}) ∏_{i=1}^{N} (s^{N−i}ϕ_{i})^{u_{i}^{+}} ,

and this proves the claim. □

In particular, if v is a partition then T_{v} = ∏_{i=1}^{N} (s^{N−i}ϕ_{i})^{v_{i}}.

3. Vector valued polynomials

3.1. First Examples. To motivate our definitions we consider the simplest two-dimen-sional situation: N = 3, isotype λ = (2, 1). A basis for the representation of {T_{1}, T_{2}} is

f_{1} = sx_{1} − \frac{1}{s + 1} (x_{2} + x_{3}) ,
f_{2} = x_{2} − \frac{1}{s}x_{3}.

Then f_{1}T_{2} = sf_{1}, f_{2}T_{2} = −f_{2} and

f_{1}T_{1} = −\frac{1}{s + 1}f_{1} + \frac{s (1 + s + s^{2})}{(1 + s)^{2}}f_{2},
f_{2}T_{1} = f_{1} + \frac{s^{2}}{1 + s}f_{2}.

We aim to set up a Macdonald-type structure in {p_{1} (x) f_{1} + p_{2} (x) f_{2}}. Firstly define operators Ti^{′} acting on pairs [p_{1}, p_{2}] so that

[p_{1}, p_{2}] Ti^{′} · [f_{1}, f_{2}] = (p_{1}f_{1} + p_{2}f_{2}) T_{i}, i = 1, 2,

where [a_{1}, a_{2}] · [b_{1}, b_{2}] := a_{1}b_{1} + a_{2}b_{2}. Indeed, we have

[p_{1}, p_{2}] T1^{′} = [p_{1}T_{1} − \frac{1 + s + s^{2}}{1 + s} p_{1}s_{1} + p_{2}s_{1}, p_{2}T_{1} − \frac{s}{1 + s}p_{2}s_{1} + \frac{s (1 + s + s^{2})}{(1 + s)^{2}}p_{1}s_{1}] ,
[p_{1}, p_{2}] T2^{′} = [p_{1}T_{2}, p_{2}T_{2} − (s + 1) p_{2}s_{2}] .

The inverses are obtained from the quadratic relation: T_{i}^{′−1} = \frac{1}{s} (Ti^{′} + 1 − s).

Secondly, we need a definition of w (to be generalized in the sequel). The relation wT_{1} = T_{2}w must be satisfied. The braid relation gives a solution T_{2} (T_{1}T_{2}) = (T_{1}T_{2}) T_{1}. Using w′ = T_{1}T_{2}, let

f_{1}w′ = −\frac{s}{1 + s}f_{1} − \frac{s (1 + s + s^{2})}{(1 + s)^{2}}f_{2},
f_{2}w′ = sf_{1} − \frac{s^{2}}{1 + s}f_{2}.

Then w′T_{1} = T_{2}w′, acting on span {f_{1}, f_{2}}. Now define

[p_{1}, p_{2}] w = [−\frac{s}{1 + s}p_{1}w + sp_{2}w, −\frac{s (1 + s + s^{2})}{(1 + s)^{2}}p_{1}w − \frac{s^{2}}{1 + s}p_{2}w] .

Set

ξ_{1} = s^{−2}wT2^{′}T1^{′},
ξ_{2} = s^{−1}T_{1}^{′−1}wT2^{′},
ξ_{3} = T_{2}^{′−1}T_{1}^{′−1}w.

These operators commute. Here are the degree 1 simultaneous eigenfunctions:

[− (1 + s) x_{3}, sx_{3}] ,

[x_{3}, \frac{1 + s + s^{2}}{1 + s} x_{3}] ,

[(s + 1) x_{2} + \frac{q (1 − s^{2})}{1 − qs}x_{3}, x_{2} − \frac{sq (1 − s)}{1 − qs}x_{3}] ,

[x_{2} − \frac{q (1 − s)}{s (q − s)}x_{3}, −\frac{1 + s + s^{2}}{s (1 + s)} \{x_{2} + \frac{q (1 − s)}{q − s}x_{3}\}] .

[\frac{q (1 − s)}{1 − q^{2}s} \{sx_{2} − x_{3}\} , x_{1} + \frac{sq (1 − s)}{(1 + s) (1 − q^{2}s)} \{x_{2} + x_{3}\}] ,

[x_{1} + \frac{qs (1 − s)}{(1 + s) (q − s^{2})} \{x_{2} + x_{3}\} , −\frac{q (1 + s + s^{2}) (1 − s)}{(1 + s)^{2} (q − s^{2})} \{x_{2} − sx_{3}\}] .

To generalize this setup to an arbitrary irreducible module V_{λ} (basis corresponding to Tab_{λ}), we need to define w; a necessary condition is that there be an intertwining operator S on V so that ST_{i} = T_{i+1}S for 1 ≤ i < N. The correct definition is S = T_{1}T_{2} ⋯ T_{N−1}. Indeed,

ST_{i} = T_{1} ⋯ T_{i−1}T_{i}T_{i+1}T_{i}T_{i+2} ⋯ T_{N−1}
= T_{1} ⋯ T_{i−1}T_{i+1}T_{i}T_{i+1}T_{i+2} ⋯ T_{N−1}
= T_{i+1}S.

Definition 3.1. The space of vector valued polynomials for the isotype λ (a partition of N) will be denoted by M_{λ} := ℂ[x_{1}, …, x_{N}] ⊗ V_{λ}.

The elements of M_{λ} are linear combinations of x^{v}𝕋, where x^{v} := x_{1}^{v[1]} ⋯ x_{N}^{v[N]}. We denote by ‘normal symbols’ (s_{i}, T_{i}, w, ξ_{i}, etc.) the operators acting only on tableaux. The operators acting only on letters will be denoted with superscript ^{x} (s_{i}^{x}, T_{i}^{x}, w^{x}, ξ_{i}^{x}, etc.). The operators acting on both letters and tableaux will be denoted by bold symbols (s_{i}, T_{i}, w, ξ_{i}, etc.).

3.2. Action of the double Hecke algebra on vectors. Denote δ_{i}^{x} := T_{i}^{x} − s · s_{i}^{x} = ∂_{i}^{x}x_{i+1}(1 − s) and T_{i} := δ_{i}^{x} + s_{i}^{x}T_{i}. We have the following fact.

Lemma 3.2. The operator T_{i} satisfies the quadratic relation:

(3.1)  (T_{i} + 1)(T_{i} − s) = 0.

Proof. From

∂_{i}^{x}x_{i+1}∂_{i}^{x} = ∂_{i}^{x}∂_{i}^{x}x_{i+1} + ∂_{i}^{x}s_{i}^{x}(x_{i+1}∂_{i}^{x}) = −∂_{i}^{x},

we deduce

(3.2)  δ_{i}^{x}^{2} − (1 − s)^{2}∂_{i}^{x}x_{i+1} = −(1 − s)δ_{i}^{x}.

Moreover, from

∂_{i}^{x}x_{i+1}s_{i}^{x} + s_{i}^{x}∂_{i}^{x}x_{i+1} = ∂_{i}^{x}(x_{i} − x_{i+1}) = 1 − s_{i}^{x},

one obtains

(3.3)  δ_{i}^{x}s_{i}^{x} + s_{i}^{x}δ_{i}^{x} = (1 − s)(1 − s_{i}^{x}).

Now, expanding (T_{i} + 1)(T_{i} − s), from eqs. (3.2) and (3.3) we observe that

(T_{i} + 1)(T_{i} − s) = (δ_{i}^{x}^{2} + (1 − s)δ_{i}^{x}) + (δ_{i}^{x}s_{i}^{x} + s_{i}^{x}δ_{i}^{x} + (1 − s)(s_{i}^{x} − 1))T_{i}
= 0,
□

We find also commutation relations.

Lemma 3.3. If |i − j| > 1, we have

(3.4)  T_{i}T_{j} = T_{j}T_{i}.

Proof. First we expand

(3.5)  T_{i}T_{j} = δ_{i}^{x}δ_{j}^{x} + δ_{i}^{x}s_{j}^{x}T_{j} + s_{i}^{x}δ_{j}^{x}T_{i} + s_{i}^{x}s_{j}^{x}T_{i}T_{j}.

But since |i − j| > 1, one has straightforwardly s_{i}^{x}s_{j}^{x} = s_{j}^{x}s_{i}^{x}, T_{i}T_{j} = T_{j}T_{i}, δ_{i}^{x}s_{j}^{x} = s_{j}^{x}δ_{i}^{x} and δ_{i}^{x}δ_{j}^{x} = δ_{j}^{x}δ_{i}^{x}. Using these relations in eq. (3.5), we find the result. □

To prove the braid relations, we need the following preliminary results.

Lemma 3.4. (1) s_{i}^{x}s_{i+1}^{x}s_{i}^{x}T_{i}T_{i+1}T_{i} = s_{i+1}^{x}s_{i}^{x}s_{i+1}^{x}T_{i+1}T_{i}T_{i+1}, (2) δ_{i}^{x}δ_{i+1}^{x}δ_{i}^{x} = δ_{i+1}^{x}δ_{i}^{x}δ_{i+1}^{x},

(3) δ_{i+1}^{x}s_{i}δ_{i+1}^{x} = s_{i}^{x}δ_{i+1}^{x}δ_{i}^{x} + δ_{i}^{x}δ_{i+1}^{x}s_{i}^{x} + (s − 1)s_{i}^{x}δ_{i+1}^{x}s_{i}^{x},

(4) δ_{i}^{x}s_{i+1}^{x}δ_{i}^{x} = s_{i+1}^{x}δ_{i}^{x}δ_{i+1}^{x} + δ_{i+1}^{x}δ_{i}^{x}s_{i+1}^{x} + (s − 1)s_{i+1}^{x}δ_{i}^{x}s_{i+1}^{x},

(5) δ_{i}^{x}s_{i+1}^{x}s_{i}^{x} = s_{i+1}^{x}s_{i}^{x}δ_{i+1}^{x},

(6) δ_{i+1}^{x}s_{i}^{x}s_{i+1}^{x} = s_{i}^{x}s_{i+1}^{x}δ_{i}^{x},

(7) s_{i}^{x}δ_{i+1}^{x}s_{i}^{x} = s_{i+1}^{x}δ_{i}^{x}s_{i+1}^{x}.

Proof. The first identity is trivial, but the others need to be proved. The simplest way to check these formulae is the direct verification of the action on a monomial x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}. For instance, the second equality follows from

x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}δ_{i}^{x}δ_{i+1}^{x}δ_{i}^{x} = (1 − s)^{3} \frac{x_{i+2}^{2}x_{i+1}}{V (x_{i}, x_{i+1}, x_{i+2})} det ( \begin{matrix}x_{i}^{a} & x_{i}^{b} & x_{i}^{c} \\ x_{i+1}^{a} & x_{i+1}^{b} & x_{i+1}^{c} \\ x_{i+2}^{a} & x_{i+2}^{b} & x_{i+2}^{c} \end{matrix} )
= x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}δ_{i+1}^{x}δ_{i}^{x}δ_{i+1}^{x},

where V (x_{1}, x_{2}, x_{3}) := ∏_{0<i<j<4}(x_{i} − x_{j}) denotes the Vandermonde determinant. The other identities can be verified in a similar fashion (for simplicity we omit the superscript x on δ and s):

14 C. F. DUNKL AND J.-G. LUQUE

(3)

x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}δ_{i}s_{i+1}δ_{i}

= (1 − s)^{2} \frac{x_{i+2}x_{i+1}}{V (x_{i}, x_{i+1}, x_{i+1})} [x_{i+2}^{a} (x_{i+2}(x_{i}^{b}x_{i+1}^{c} − x_{i}^{c}x_{i+1}^{b}) − (x_{i}^{b}x_{i+1}^{c+1} − x_{i}^{c+1}x_{i+1}^{b}))

−x_{i+2}^{b} (x_{i+2}(x_{i}^{a}x_{i+1}^{c} − x_{i}^{c}x_{i+1}^{a}) − (x_{i}^{a}x_{i+1}^{c+1} − x_{i}^{c+1}x_{i+1}^{a}))]

= x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}(s_{i+1}δ_{i}δ_{i+1} + δ_{i+1}δ_{i}s_{i+1} + (s − 1)s_{i+1}δ_{i}s_{i+1}). (4)

x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}δ_{i+1}s_{i}δ_{i+1} = (1 − s)^{2} \frac{x_{i+1}}{V (x_{i}, x_{i+1}, x_{i+1})}
× [x_{i}^{b} (x_{i}(x_{i+1}^{a}x_{i+2}^{c+1} − x_{i+1}^{c+1}x_{i+2}^{a}) − (x_{i+1}^{a+1}x_{i+2}^{c+1} − x_{i+1}^{c+1}x_{i+2}^{a+2}))
−x_{i}^{c} (x_{i}(x_{i+1}^{a}x_{i+2}^{b+1} − x_{i+1}^{b+1}x_{i+2}^{a}) − (x_{i+1}^{a+1}x_{i+2}^{b+1} − x_{i+1}^{b+1}x_{i+2}^{a+1}))]
= x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}(s_{i}δ_{i+1}δ_{i} + δ_{i}δ_{i+1}s_{i} + (s − 1)s_{i}δ_{i+1}s_{i}).

(5)

x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}(s_{i+1}s_{i}δ_{i+1}) = (1 − s)x_{i+1}x_{i+2}^{a} \frac{x_{i}^{b}x_{i+1}^{c} − x_{i}^{c}x_{i+1}^{b}}{x_{i} − x_{i+1}}
= x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}(δ_{i}s_{i+1}s_{i}).

(6)

x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}(s_{i}s_{i+1}δ_{i}) = (1 − s)x_{i+2}x_{i}^{c} \frac{x_{i+1}^{a}x_{i+2}^{b} − x_{i+1}^{b}x_{i+2}^{a}}{x_{i+1} − x_{i+2}}
= x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}(δ_{i+1}s_{i}s_{i+1}).

(7)

x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}(s_{i}δ_{i+1}s_{i}) = (1 − s)x_{i+2}x_{i+1}^{b} \frac{x_{i}^{a}x_{i+2}^{c} − x_{i}^{c}x_{i+2}^{a}}{x_{i} − x_{i+2}}
= x_{i}^{a}x_{i+1}^{b}x_{i+2}^{c}(s_{i+1}δ_{i}s_{i+1}).
□

Next we show that the operators {T_{i}} satisfy the braid relations. Proposition 3.5. For each i < N − 1, one has

(3.6)  T_{i}T_{i+1}T_{i} = T_{i+1}T_{i}T_{i+1}.

Proof. Expanding the braid T_{i}T_{i+1}T_{i}, we obtain

T_{i}T_{i+1}T_{i} = δ_{i}^{x}δ_{i+1}^{x}δ_{i}^{x} + δ_{i}^{x}s_{i+1}^{x}δ_{i}^{x}T_{i+1}

+ (s_{i}^{x}δ_{i+1}^{x}δ_{i}^{x} + δ_{i}^{x}δ_{i+1}^{x}s_{i}^{x}) T_{i} + s_{i}^{x}s_{i+1}^{x}s_{i}^{x}T_{i+1}T_{i} + s_{i}^{x}s_{i+1}^{x}δ_{i}^{x}T_{i}T_{i+1} + s_{i}^{x}δ_{i+1}^{x}s_{i}^{x}T_{i}^{2} + s_{i}^{x}s_{i+1}^{x}s_{i}^{x}T_{i}T_{i+1}T_{i}.

Using the fact that T_{i}^{2} = (s − 1)T_{i} + s, we obtain

T_{i}T_{i+1}T_{i} = δ_{i}^{x}δ_{i+1}^{x}δ_{i}^{x} + δ_{i}^{x}s_{i+1}^{x}δ_{i}^{x}T_{i+1}
+ (s_{i}^{x}δ_{i+1}^{x}δ_{i}^{x} + δ_{i}^{x}δ_{i+1}^{x}s_{i}^{x} + (s − 1)s_{i}^{x}δ_{i+1}^{x}s_{i}^{x}) T_{i} + s_{i}^{x}sx_{i+1}s_{i}^{x}T_{i+1}T_{i}
+ s_{i}^{x}s_{i+1}^{x}δ_{i}^{x}T_{i}T_{i+1} + ss_{i}^{x}δ_{i+1}^{x}s_{i}^{x} + s_{i}^{x}s_{i+1}^{x}s_{i}^{x}T_{i}T_{i+1}T_{i}.

Now applying Lemma 3.4, we arrive at the desired result. □

Next we examine the relation between the generators T_{i} and the multiplication by an indeterminate x_{i}. There are three identities satisfied by these operations.

Proposition 3.6. (1) x_{i}T_{i} − T_{i}x_{i+1} − (1 − s)x_{i+1} = 0; (2) x_{i+1}T_{i} − T_{i}x_{i} + (1 − s)x_{i+1} = 0;

(3) x_{i}T_{j} = T_{j}x_{i} when |i − j| > 1.

Proof. (1) One has

x_{i}δ_{i}^{x} = (1 − s)x_{i}∂_{i}^{x}x_{i+1}
= (1 − s)∂_{i}^{x}x_{i+1}^{2} + (1 − s)x_{i+1}
= δ_{i}^{x}x_{i+1} + (1 − s)x_{i+1}.

Hence, we obtain

x_{i}T_{i} = [δ_{i}^{x} + s_{i}^{x}T_{i}] x_{i+1} + (1 − s)x_{i+1}
= T_{i}x_{i+1} + (1 − s)x_{i+1},

as expected.

(2) The second equality is proved in the same way, by observing that

x_{i+1}δ_{i}^{x} = (1 − s)x_{i+1}∂_{i}^{x}x_{i+1}
= (1 − s)∂_{i}^{x}x_{i+1}x_{i} − (1 − s)x_{i+1}
= δ_{i}^{x}x_{i} − (1 − s)x_{i+1}.

(3) The third equality is straightforward.

□

Now, we examine the affine action and set

w = τ_{1}^{x}θ^{x}S,

where θ^{x} = s_{1}^{x} ⋯ s_{N−1}^{x} and S = T_{1} ⋯ T_{N−1}. When i < N − 1, one has

wT_{i} = (τ_{1}^{x}θ^{x}S)(δ_{i}^{x} + s_{i}^{x}T_{i}).

But since i < N − 1, one has

τ_{1}^{x}θ^{x}∂_{i}^{x}x_{i+1} = τ_{1}^{x}∂_{i+1}^{x}x_{i+2}θ^{x},

and i + 1 > 1 implies τ_{1}^{x}∂_{i+1}^{x}x_{i+2} = ∂_{i+1}^{x}x_{i+2}τ_{1}^{x}. Hence,

τ_{1}^{x}θ^{x}δ_{i}^{x} = δ_{i+1}^{x}τ_{1}^{x}θ^{x}.

One easily obtains τ_{1}^{x}θ^{x}s_{i}^{x} = s_{i+1}^{x}τ_{1}^{x}θ^{x} and ST_{i} = T_{i+1}S. From this, we deduce the following commutation relation.

16 C. F. DUNKL AND J.-G. LUQUE

Lemma 3.7. For each i, we have wT_{i} = T_{i+1}w.

From Lemmas 3.2, 3.3, 3.7, Propositions 3.5 and 3.6, we obtain the following result. Theorem 3.8. The algebra ℂ(s, q)[x_{1}^{±1}, …, x_{N}^{±1}, T_{1}, …, T_{N−1}, w^{±1}] is isomorphic to H_{N}(s, q). More precisely, the morphism sends T_{i} to T_{i}, w to w, and x_{i} to x_{i}. 3.3. Cherednik and Dunkl operators. Definition 3.9. In this context, the (vector valued) Cherednik operators are defined by

ξ_{i} = s^{i−N}T_{i−1}^{−1} ⋯ T_{1}^{−1}wT_{N−1} ⋯ T_{i},

where

T_{i}^{−1} = \frac{1}{s}(T_{i} + (1 − s)) = \frac{1}{s}((1 − s)(∂_{i}^{x}x_{i+1} + 1) + s_{i}^{x}T_{i}).

It follows immediately that

(3.7)  [ξ_{i}, ξ_{j}] = 0,

since, by Theorem 3.8, the operators ξ_{i} are the image of the Cherednik operators ξ_{i}. Furthermore, the tableaux are simultaneous eigenfunctions of the Cherednik elements, and the associated spectral vectors can be expressed in terms of contents. Proposition 3.10. For each tableau 𝕋, one has

𝕋ξ_{i} = s^{CT_{𝕋}[i]}𝕋.

Proof. Since, (1) 𝕋T_{i} = 𝕋T_{i},

(2) 𝕋T_{i}^{−1} = 𝕋T_{i}^{−1},

(3) 𝕋w = 𝕋S, one has 𝕋ξ_{i} = 𝕋ϕ_{i}. Hence, the result follows from Proposition 2.4. □

In the aim to define the Dunkl–Cherednik operators, we set F_{N} = 1 − ξ_{N}. Proposition 3.11. The operator F_{N} is divisible by x_{N}, that is, for each P ∈ ℂ[x_{1}, …, x_{N}] ⊗ V_{λ}, P F_{N} = x_{N}Q with Q ∈ ℂ[x_{1}, …, x_{N}] ⊗ V_{λ}. Proof. We prove the result by induction on N. Suppose first that N = 2. Then our operator is

F_{2} = 1 − \frac{1}{s}((1 − s)(∂_{1}^{x}x_{2} + 1) + s_{1}^{x}T_{1})(τ_{1}^{x}s_{1}^{x}T_{1})

= 1 − \frac{1}{s}((1 − s)(∂_{1}^{x}x_{2} + 1)τ_{1}^{x}s_{1}^{x} + s_{1}^{x}τ_{1}^{x}s_{1}^{x}T_{1}^{2}).

From T_{1}^{2} = (s − 1)T_{1} + s and s_{1}τ_{1}s_{1} = τ_{2}, one obtains

F_{2} = 1 − \frac{1}{s} ((1 − s)(∂_{1}^{x}x_{2} + 1 − s_{1})τ_{1}^{x}s_{1}^{x}T_{1} + sτ_{2}^{x}) .

Note that

∂_{1}^{x}x_{2} + 1 − s_{1}^{x} = ∂_{1}^{x}x_{1}

implies

F_{2} = \frac{s − 1}{s}q∂_{1}^{x}τ_{1}^{x}s_{1}^{x}T_{1}x_{2} + 1 − τ_{2}^{x}.

But, for any polynomial P, one has

P (x_{1})x_{2}^{b}(1 − τ_{2}^{x}) = \begin{cases}0 & if b = 0, \\ P (x_{1})x_{2}^{b}(1 − q^{b}) & if b > 0. \end{cases}

This proves the result for N = 2.

Now suppose that N > 2. Then we have

F_{N} = 1 − T_{N−1}^{−1} ⋯ T_{1}^{−1}(τ_{1}^{x}s_{1}^{x} ⋯ s_{N−1}^{x}T_{1} ⋯ T_{N−1}).

Similarly to the case N = 2, one obtains

F_{N} = 1 − \frac{1}{s}T_{N−1}^{−1} ⋯ T_{2}^{−1} ((sq∂_{1}^{x}τ_{1}^{x}s_{1}^{x} ⋯ s_{N−1}^{x}T_{1} ⋯ T_{N−1})x_{N−1}
+sτ_{2}^{x}s_{2}^{x} ⋯ s_{N−1}^{x}T_{2} ⋯ T_{N−1}) .

So it suffices to prove that the operator 1−T_{N−1}^{−1} ⋯ T_{2}^{−1}s_{2}^{x} ⋯ s_{N−1}^{x}T_{2} ⋯ T_{N−1} is divisible by x_{N}. Observing that

1 − T_{N−1}^{−1} ⋯ T_{2}^{−1}(s_{2}^{x} ⋯ s_{N−1}^{x}T_{2} ⋯ T_{N−1}) = θ^{−1}F_{N−1}θ,

the result follows by induction. □

Definition 3.12. The vector valued Dunkl operators are defined by D_{N} := F_{N}x_{N}^{−1} and D_{i} := \frac{1}{s}T_{i}D_{i+1}T_{i}.

As for the Cherednik operators, Theorem 3.8 implies that the classical relations hold. For instance, one has

[D_{i}, D_{j}] = 0,

and the relations with respect to the generators T_{i} are

(3.8) D_{i+1}T_{i} = −sT_{i}^{−1}D_{i}, −T_{i}D_{i+1} + (1 − s)D_{i} + D_{i}T_{i} = 0,

−D_{i+1}T_{i}^{−1} − (1 − \frac{1}{s})D_{i+1} + T_{i}^{−1}D_{i} = 0 (3.9)

[D_{i}, T_{j}] = 0 if |i − j| > 1.

Note that identities (1) and (2) of Proposition 3.6 are equivalent to x_{i}T_{i} = sx_{i+1}T_{i}^{−1} or sx_{i+1} = T_{i}x_{i}T_{i} (these are dual to the D_{i} relations D_{i} = (1/s)T_{i}D_{i+1}T_{i} ).

3.4. Triangularity of the Cherednik operators. Let v be a vector. In the sequel we will denote by v^{+} (respectively v^{R}) the unique decreasing (respectively increasing) partition whose entries are obtained by permuting those of v.

Let \overline{π}_{i}^{x} = ∂_{i}^{x}x_{i+1} = \frac{1}{1−s}δ_{i}^{x}, π_{i}^{x} = ∂_{i}^{x}x_{i+1} + 1, and, more generally, π_{ij}^{x} = ∂_{ij}^{x}x_{j} + 1.

Observe that, if i < j, then one has

(3.10)  x^{v}π_{ij}^{x} = ∑_{v′⊴v}(∗)x^{v′},

where (∗) denotes a coefficient and ⊴ is the dominance order on vectors defined by

v ⊴ v′ if and only if \begin{cases}v^{+} ≺ v^{′+} & if v^{+} ≠ v^{′+}, \\ v ≺ v′ & if v^{+} = v^{′+}. \end{cases}

Here, ≺ denotes the (classical) dominance order on partitions given by

v ≺ v′ if and only if, for each i, v[1] + ⋯ + v[i] ≤ v′[1] + ⋯ + v′[i].

Indeed, it suffices to understand the computation of x_{1}^{a}x_{2}^{b}π_{1}. We have three cases to consider:

(1) if a < b, then

x_{1}^{a}x_{2}^{b}π_{1}^{x} = − ∑_{i=1}^{b−a−1} x_{i}^{a+i}x_{2}^{b−i}.

In this case, one has x_{1}^{a}x_{2}^{b}π_{1}^{x} = ∑_{v^{′+}≺[b,a]}(∗)x^{v′}.

(2) if a = b, then

x_{1}^{a}x_{2}^{b}π_{1}^{x} = x_{1}^{a}x_{2}^{b}.

(3) if a > b, then

x_{1}^{a}x_{2}^{b}π_{1}^{x} = ∑_{i=0}^{a−b} x_{i}^{a−i}x_{2}^{b+i},

and the leading term in this expression is x^{[a,b]}. Similarly,

(3.11)  x^{v}\overline{π}_{ij}^{x} = ∑_{v′⊴v}(∗)x^{v′}.

With these notations, write

T_{i} = (∗)\overline{π}_{i}^{x} + (∗)s_{i}^{x}T_{i}

and

T_{i}^{−1} = (∗)π_{i}^{x} + (∗)s_{i}^{x}T_{i}.

Here, (∗) denotes a certain coefficient (we need not know it to follow the computation). Observe that, for each j, we have

T_{1}^{−1}s_{1}^{x} ⋯ s_{j−1}^{x} = [(∗)π_{1}^{x} + (∗)s_{1}^{x}T_{1}]s_{1}^{x} ⋯ s_{j−1}^{x} = [(∗)π_{1}^{x} + (∗)T_{1}]s_{2}^{x} ⋯ s_{j−1}^{x}

since π_{1}^{x}s_{1}^{x} = π_{1}^{x}. But π_{1}^{x}s_{2}^{x} ⋯ s_{j−1}^{x} = s_{2}^{x} ⋯ s_{j−1}^{x}π_{1,j}^{x}, and hence

T_{1}^{−1}s_{1}^{x} ⋯ s_{j−1}^{x} = s_{2}^{x} ⋯ s_{j−1}^{x}[(∗)\overline{π}_{1j}^{x} + (∗)T_{1}].ya

Iterating the process, one finds

(3.12)  T_{j−1}^{−1} ⋯ T_{1}^{−1}s_{1}^{x} ⋯ s_{j−1}^{x} = [(∗)\overline{π}_{j−1j}^{x} + (∗)T_{j−1}] ⋯ [(∗)\overline{π}_{1j}^{x} + (∗)T_{1}].

One has also

s_{j}^{x} ⋯ s_{N−1}^{x}T_{N−1} = s_{j}^{x} ⋯ s_{N−1}^{x}[(∗)\overline{π}_{N−1}^{x} + (∗)s_{N−1}^{x}T_{N−1}];

but s_{N−1}^{x}∂_{N−1}^{x} = −∂_{N−1}^{x}, and hence

s_{j}^{x} ⋯ s_{N−1}^{x}T_{N−1} = s_{j}^{x} ⋯ s_{N−2}^{x}[(∗)\overline{π}_{N−1}^{x} + (∗)T_{N−1}].

Since s_{j}^{x} ⋯ s_{N−2}^{x}\overline{π}_{N−1}^{x} = \overline{π}_{j,N}^{x}s_{j}^{x} ⋯ s_{N−2}^{x}, one obtains

s_{j}^{x} ⋯ s_{N−1}^{x}T_{N−1} = [(∗)\overline{π}_{j,N}^{x} + (∗)T_{N−1}]s_{j}^{x} ⋯ s_{N−2}^{x}.

Iterating this process, one finds

(3.13)  s_{j}^{x} ⋯ s_{N−1}^{x}T_{N−1} ⋯ T_{j} = [(∗)\overline{π}_{j,N}^{x} + (∗)T_{N−1}] ⋯ [(∗)\overline{π}_{j,j+1}^{x} + (∗)T_{j}].

Now, with these notations, the Cherednik operator reads

ξ_{j} = [(∗)π_{j−1}^{x} + (∗)s_{j−1}^{x}T_{j−1}] ⋯ [(∗)π_{1}^{x} + (∗)s_{1}^{x}T_{1}] τ_{1}^{x}s_{1}^{x} ⋯ s_{N−1}^{x}S

[(∗)\overline{π}_{N−1}^{x} + (∗)s_{N−1}^{x}T_{N−1}] ⋯ [(∗)\overline{π}_{j}^{x} + (∗)s_{j}^{x}T_{j}] .

Now apply eq. (3.12) and (3.13) to obtain

ξ_{i} = (∗)[(∗)π_{j−1,N}^{x} + (∗)T_{j−1}] ⋯ [(∗)π_{1,N}^{x} + (∗)T_{1}](τ_{j}^{x}S)

[(∗)\overline{π}_{j,N−1}^{x} + (∗)T_{N−1}] ⋯ [(∗)\overline{π}_{j,j+1}^{x} + (∗)T_{j}] ,

where x_{i}τ_{j}^{x} = x_{i} if i ≠ j and x_{j}τ_{j}^{x} = qx_{j}.

From (3.10) and (3.11), we obtain

(3.14)  𝕋x^{v}ξ_{i} = 𝕋 [x^{v}H_{v} + ∑_{v′◁v} x^{v′}H_{v′}] ,

with H_{u} ∈ H_{N}(q, s) (we apply an algebraic combination of π^{x} and \overline{π}^{x} to x^{v}, and the operator τ_{j}^{x} does not change the exponents). Finally, we arrive at the following theorem.

Theorem 3.13. We have

x^{v}𝕋ξ_{j} = x^{v}(𝕋 · H_{v}) + ∑_{v′◁v} x^{v′}(𝕋 · H_{v′}),

where H_{u} ∈ H_{N}(q, s).

Proof. Eq. 3.14 gives

x^{v}𝕋ξ_{j} = 𝕋x^{v}ξ_{j}

= x^{v}(𝕋H_{v}) + ∑_{v′◁v} x^{v′}(𝕋 · H_{v′}).

□

4. Eigenfunctions of Cherednik operators

4.1. Yang–Baxter graph. As in [6], we construct a Yang–Baxter-type graph with vertices labeled by 4-tuples (𝕋, ζ, v, σ), where 𝕋 is an RST, ζ is a vector of length N (ζ will be called the spectral vector), v ∈ ℕ^{N}, and σ ∈ 𝔖_{N}. First, consider an RST of shape λ and write a vertex labeled by the 4-tuple (𝕋, CT_{𝕋}^{s}, 0^{N}, [1, …, N]), where CT_{𝕋}^{s}[i] = s^{CT_{𝕋}[i]}. Now, we consider the action of the elementary transposition of 𝔖_{N} on the 4-tuple given by

(𝕋, ζ, v, σ)s_{i} := \begin{cases}(𝕋, ζs_{i}, vs_{i}, σs_{i}) & if v[i + 1] ≠ v[i], \\ (𝕋^{(σ[i],σ[i+1])}, ζs_{i}, v, σ) & if v[i] = v[i + 1] \\ & and 𝕋^{(σ[i],σ[i+1])} ∈ Tab_{λ}, \\ (𝕋, ζ, v, σ) & otherwise, \end{cases}

where 𝕋^{(i,j)} denotes the filling obtained by permuting the values i and j in 𝕋. Consider also the affine action given by

(𝕋, ζ, v, σ)Ψ := (𝕋, [ζ[2], …, ζ[N], qζ[1]], [v[2], …, v[N], v[1] + 1], [σ_{2}, …, σ_{N}, σ_{1}]).

In the sequel we will use the notation vΨ^{q} = [v_{2}, …, v_{N}, qv_{1}].

Example 4.1.

(1) ({31 \atop 542}, [s, 1, q^{2}, qs^{2}, qs^{−1}], [0, 0, 2, 1, 1], [45123]) s_{2}

= ({31 \atop 542}, [s, q^{2}, 1, qs^{2}, qs^{−1}], [0, 2, 0, 1, 1], [41523]) .
(2) ({31 \atop 542}, [s, 1, q^{2}, qs^{2}, qs^{−1}], [0, 0, 2, 1, 1], [45123]) s_{4}
= ({21 \atop 543}, [s, 1, q^{2}, qs^{−1}, qs^{2}], [0, 0, 2, 1, 1], [45123]) .
(3) ({31 \atop 542}, [s, 1, q^{2}, q^{2}s^{2}, qs^{−1}], [0, 0, 2, 1, 1], [45123]) s_{1}
= ({31 \atop 542}, [s, 1, q^{2}, qs^{2}, qs^{−1}], [0, 0, 2, 1, 1], [45123]) .
(4) ({31 \atop 542}, [s, 1, q^{2}, qs^{2}, qs^{−1}], [0, 0, 2, 1, 1], [45123]) Ψ
= ({31 \atop 542}, [1, q^{2}, qs^{2}, qs^{−1}, qs], [0, 2, 1, 1, 1], [51234]) .

Definition 4.2. If λ is a partition, denote by 𝕋_{λ} the tableau obtained by filling the shape λ from bottom to top and left to right by the integers {1, …, N} in decreasing order.

The graph G_{λ}^{q,s} is the infinite directed graph constructed from the 4-tuple

(𝕋_{λ}, CT_{𝕋_{λ}}^{s}, [0^{N}], [1, 2, …, N]),

called the root, adding vertices and edges according to the following rules:

(1) We add an arrow labeled by s_{i} from the vertex (𝕋, ζ, v, σ) to (𝕋′, ζ′, v′, σ′) if (𝕋, ζ, v, σ)s_{i} = (𝕋′, ζ′, v′, σ′) and v[i] < v[i + 1] or v[i] = v[i + 1], and τ is obtained from τ′ by interchanging the position of two integers k < ℓ such that k is in the south-east of ℓ (i.e., CT_{𝕋}(k) ≥ CT_{𝕋}(ℓ) + 2).

(2) We add an arrow labeled by Ψ from the vertex (𝕋, ζ, v, σ) to (𝕋′, ζ′, v′, σ′) if (𝕋, ζ, v, σ)Ψ = (𝕋′, ζ′, v′, σ′).

(3) We add an arrow s_{i} from the vertex (τ, ζ, v, σ) to ∅ if (𝕋, ζ, v, σ)s_{i} = (𝕋, ζ, v, σ). An arrow of the form

(𝕋, ζ, v, σ) (𝕋, ζ′, v′, σ′) s_{i} or Ψ

will be called a step. The other arrows will be called jumps, and in particular an arrow

\begin{matrix}(𝕋, ζ, v, σ) \\ ∅ \\ s_{i} \end{matrix}

will be called a fall; the other jumps will be called correct jumps.

As usual, a path is a finite sequence of consecutive arrows in G_{λ} starting from the root, and it is denoted by the vector of the labels of its arrows. Two paths 𝔓_{1} = (a_{1}, …, a_{k}) and 𝔓_{2} = (b_{1}, …, b_{ℓ}) are said to be equivalent (denoted by 𝔓_{1} ≡ 𝔓_{2}) if they lead to the same vertex.

We note that, when v[i] = v[i + 1], part (1) of Definition 4.2 is equivalent to the following statement: 𝕋′ is obtained from 𝕋 by interchanging σ_{v}[i] and σ_{v}[i+1] = σ_{v}[i]+1, where σ_{v}[i] is to the south-east of σ_{v}[i] + 1, that is, CT_{𝕋}[σ_{v}[i]] − CT_{𝕋}[σ_{v}[i] + 1] ≥ 2.

Example 4.3. The arrow below is a correct jump,

{{31 \atop 542} ,[s,1,q^{2},qs^{2},qs^{−1}] \atop [0,0,2,1,1],[45123]} {{21 \atop 543} ,[s,1,q^{2},qs^{−1},qs^{2}] \atop [0,0,2,1,1],[45123]} s_{4} ,

whilst

{{31 \atop 542} ,[q,1,q^{2},qs^{2},qs^{−1}] \atop [0,0,2,1,1],[45123]} {{31 \atop 542} ,[s,q^{2},1,qs^{−1},qs^{2}] \atop [0,2,0,1,1],[41523]} s_{2}

is a step. The arrows

{{31 \atop 542} ,[s,1,q^{2},qs^{2},qs^{−1}] \atop [0,0,2,1,1],[45123]} {{21 \atop 543} ,[s,1,q^{2},qs^{−1},qs^{2}] \atop [0,0,2,1,1],[45123]} s_{4}

and

{{31 \atop 542} ,[s,1,q^{2},qs^{2},qs^{−1}] \atop [0,0,2,1,1],[45123]} {{31 \atop 542} ,[s,q^{2},1,qs^{−1},qs^{2}] \atop [0,2,0,1,1],[41523]} s_{2}

are not allowed.

The graph G_{λ}^{q,s} is very similar to the Yang–Baxter graph G_{λ} described in [6]: only the spectral vectors change. Indeed, these are the same graphs, but with different labels: the spectral vector of G_{λ}^{q,s} is obtained from G_{λ} by sending aα + b to q^{a}s^{b}. Hence, many properties still apply. An example is given in the next proposition.

Proposition 4.4. All the paths joining two given vertices in G_{λ} have the same length.

For a given 4-tuple (𝕋, ζ, v, σ), the values of ζ and σ are determined by those of 𝕋 and v, as provided by the following proposition.

Proposition 4.5. If (𝕋, ζ, v, σ) is a vertex in G_{λ}, then σ = r_{v} and ζ[i] = q^{v[i]}s^{CT_{𝕋}[σ[i]]}.

We will set ζ_{v,𝕋} := ζ.

Example 4.6. Consider the RST

τ = 3 7 4 1 8 6 5 2

and the vector v = [6, 2, 4, 2, 2, 3, 1, 4]. One has r_{v} = [1, 5, 2, 6, 7, 4, 8, 3] and CT_{𝕋} = [1, 3, −2, 0, 2, 1, −1, 0]. Consequently,

ζ_{v,τ} = [q^{6}s, q^{2}s^{2}, q^{4}s^{3}, q^{2}s^{1}, q^{2}s^{−1}, q^{3}, q, q^{4}s^{−2}].

Hence, the 4-tuple

( \begin{matrix}3 \\ 7 & 4 & 1 \\ 8 & 6 & 5 & 2 \end{matrix} , [q^{6}, q^{2}s^{2}, q^{4}s^{3}, q^{2}s^{1}, q^{2}s^{−1}, q^{3}, q, q^{4}s^{−2}], [6, 2, 4, 2, 2, 3, 1, 4], [1, 5, 2, 6, 7, 4, 8, 3])

labels a vertex of G_{431}^{q,s}.

As a consequence, we obtain the following result.

Corollary 4.7. Let (𝕋, v) be a pair consisting of 𝕋 ∈ Tab(λ) (where λ is a partition of N) and a multi-index v ∈ ℕ^{N}. Then there exists a unique vertex in G_{λ}^{q,s} labeled by a 4-tuple of the form (𝕋, ζ, v, σ).

We will write 𝔙_{𝕋,ζ,v,σ} := (𝕋, v) .

Conversely, all the information can be retrieved from the spectral vector ζ — the exponents of q give v, the rank function of v gives σ, and the exponent of s in the spectral vector gives the content vector which does uniquely determine the RST 𝕋.

For simplicity, when needed, we will label the vertices by pairs (𝕋, v) or by the associated spectral vector ζ_{v,𝕋}.

Example 4.8. In Figure 1, the first several vertices are labeled by pairs (𝕋, v) of the graph G_{21}^{q,s}, while, in Figure 2, the vertices are labeled by spectral vectors.

Definition 4.9. We define the subgraph G_{𝕋}^{q,s} as the graph obtained from G_{λ}^{q,s} by erasing all the vertices labeled by RST’s other than 𝕋 and the associated arrows. Such a graph is connected.

The graph G_{λ}^{q,s} is the union of the graphs G_{𝕋}^{q,s} connected by jumps. Furthermore, if G_{𝕋}^{q,s} and G_{𝕋′}^{q,s} are connected by a succession of jumps, then there is no step from G_{𝕋′}^{q,s} to G_{𝕋}^{q,s}. Since the graphs G_{𝕋}^{q,s} are connected graphs, we infer the following result.

Proposition 4.10. Each vertex (𝕋, v) is obtained from (𝕋, 0^{N}) by a sequence of steps.

Example 4.11. In Figures 1 and 2, the graph G_{21}^{q,s} is constituted by the two graphs Gq,s_{{1 \atop 32}} and Gq,s_{{2 \atop 31}} connected by jumps (in blue).

4.2. Macdonald polynomials from scratch. Following [1], we define the operator

Φ = T_{1}^{−1} ⋯ T_{N−1}^{−1}x_{N},

which satisfies

Φξ_{j} = ξ_{j+1}Φ, 1 ≤ j < N,
Φξ_{N} = qξ_{1}Φ.

The operator Φ is injective (its kernel is {0}).

Let λ be a partition and G_{λ}^{q,s} be the associated graph. We construct the set of polynomials (P_{𝔓})_{𝔓 path in G_{λ}} using the following recurrence rules:

(1) P_{[]} := (𝕋_{λ}).

{2 \atop 31}, [000] {1 \atop 32}, [000] s_{1} {1 \atop 32}, [001] {1 \atop 32}, [010] {1 \atop 32}, [100] {2 \atop 31}, [001] {2 \atop 31}, [010] {2 \atop 31}, [100] s_{2} s_{1} s_{2} s_{1} Ψ Ψ {1 \atop 32}, [011] {2 \atop 31}, [011] {1 \atop 32}, [101] {2 \atop 31}, [101] {1 \atop 32}, [002] {2 \atop 31}, [002] Ψ Ψ Ψ Ψ Ψ Ψ s_{2} s_{1} s_{1} Ψ {1 \atop 32}, [110] {2 \atop 31}, [110] s_{2} s_{2} s_{1} {1 \atop 32}, [020] {1 \atop 32}, [200] {2 \atop 31}, [020] {2 \atop 31}, [200] s_{2} s_{1} s_{2} s_{1}

Figure 1. The first vertices labeled by pairs (𝕋, v) of the graph G_{21}^{q,s} where we omit to write the vertex ∅ and the associated arrows.

(2) If 𝔓 = [a_{1}, …, a_{k−1}, s_{i}], then

P_{𝔓} := P_{[a_{1},...,a_{k−1}]} (T_{i} + \frac{1 − s}{1 − \frac{ζ[i+1]}{ζ[i]}} ) ,

[s, s^{−1}, 1] [s^{−1}, s, 1] s_{1} [s, 1, q^{−1}] [s, qs^{−1}, 1] [qs^{−1}, s, 1] [s^{−1}, 1, qs] [s^{−1}, qs, 1] [qs, s^{−1}, 1] s_{2} s_{1} s_{2} s_{1} Ψ Ψ [1, qs^{−1}, qs] [1, qs, qs^{−1}] [qs^{−1}, 1, qs] [qs, 1, qs^{−1}] [s, 1, q^{2}s^{−1}] [s^{−1}, 1, q^{2}s] Ψ Ψ Ψ Ψ Ψ Ψ s_{2} s_{1} s_{1} Ψ [qs^{−1}, qs, 1] [qs, qs^{−1}, 1] s_{2} s_{2} s_{1} [s, q^{2}s^{−1}, 1] [q^{2}s^{−1}, s, 1] [s^{−1}, q^{2}s, 1] [q^{2}s, s^{−1}, 1] s_{2} s_{1} s_{2} s_{1}

Figure 2. The first vertices labeled by spectral vector of the graph G_{21}^{q,s}. where the vector ζ is defined by

(𝕋_{λ}, CT_{𝕋_{λ}}^{s}, 0^{N}, [1, 2, …, N])a_{1} ⋯ a_{k−1} = (𝕋, ζ, v, σ).

(3) If 𝔓 = [a_{1}, …, a_{k−1}, Ψ], then

P_{𝔓} = P_{[a_{1},...,a_{k−1}]}Φ.

One has the following theorem.

Theorem 4.12. Let 𝔓 = [a_{0}, …, a_{k}] be a path in G_{λ}^{q,s} from the root to (𝕋, ζ, v, σ) with no fall. The polynomial P_{𝔓} is a simultaneous eigenfunction of the operators ξ_{i}. Furthermore, the eigenvalues of ξ_{i} associated to P_{𝔓} are equal to ζ[i].

Consequently, P_{𝔓} does not depend on the path, but only on the end point (𝕋, ζ, v, σ), and it will be denoted by P_{v,𝕋} or alternatively by P_{ζ}. The family (P_{v,𝕋})_{v,𝕋} forms a basis of M_{λ} of simultaneous eigenfunctions of the Cherednik operators.

Proof. We prove the result by induction on the length k. If k = 0 then the result follows from Proposition 3.10.

Suppose now that k > 0 and let

(𝕋′, ζ′, v′, r_{v′}) = (𝕋_{λ}, CT_{𝕋_{λ}}^{q,s}, 0^{N}, [1, …, N])a_{1} ⋯ a_{k−1}.

By induction, P_{[a_{1},...,a_{k−1}]} is a simultaneous eigenfunction of the operators ξ_{i} such that the associated vector of eigenvalues is given by

P_{[a_{1},...,a_{k−1}]}ξ_{i} = ζ′[i]P_{[a_{1},...,a_{k−1}]}.

The argument depends on the value of the last operator a_{k}.

(1) If a_{k} = Ψ is an affine arrow, then 𝕋 = 𝕋′, ζ = [ζ′[2], …, ζ′[N], qζ′[1]], v = v′Ψ,

r_{v} = r_{v′}[2, …, N, 1] and P_{𝔓} = J_{[a_{1},...,a_{k−1}]}Φ. If i ≠ N, then

P_{𝔓}ξ_{i} = P_{[a_{1},...,a_{k−1}]}Φξ_{i}
= P_{[a_{1},...,a_{k−1}]}ξ_{i+1}Φ
= ζ′[i + 1]P_{𝔓}
= ζ[i]P_{𝔓}.

If i = N, then

P_{𝔓}ξ_{N} = P_{[a_{1},...,a_{k−1}]}Φξ_{N}
= P_{[a_{1},...,a_{k−1}]}qξ_{1}Φ
= (ζ′[1]q)P_{𝔓}
= ζ[N]P_{𝔓}.

(2) Suppose now that a_{k} = s_{i} is a non-affine arrow. Then we have ζ = ζ′s_{i}, v = v′s_{i}, and

P_{𝔓} = P_{[a_{1},...,a_{k−1}]} (T_{i} + \frac{1 − s}{1 − \frac{ζ′[i+1]}{ζ′[i]}} ) .

If j ≠ i, i + 1, then

P_{𝔓}ξ_{j} = P_{[a_{1},...,a_{k−1}]} (T_{i} + \frac{1 − s}{1 − \frac{ζ′[i+1]}{ζ′[i]}} ) ξ_{j}

= P_{[a_{1},...,a_{k−1}]}ξ_{j} (T_{i} + \frac{1 − s}{1 − \frac{ζ′[i+1]}{ζ′[i]}} )

= ζ′[j]P_{𝔓} = ζ[j]P_{𝔓}.

If j = i, then

P_{𝔓}ξ_{i} = P_{[a_{1},...,a_{k−1}]} (T_{i} + \frac{1 − s}{1 − \frac{ζ′[i+1]}{ζ′[i]}} ) ξ_{i}

= P_{[a_{1},...,a_{k−1}]} (ξ_{i+1}T_{i} + (1 − s) (−1 + \frac{1}{1 − \frac{ζ′[i+1]}{ζ′[i]}} ) ξ_{i})

= P_{[a_{1},...,a_{k−1}]} (ζ′[i + 1]T_{i} + (1 − s) (−1 + \frac{1}{1 − \frac{ζ′[i+1]}{ζ′[i]}} ) ζ′[i])

= ζ′[i + 1]P_{[a_{1},...,a_{k−1}]} (T_{i} + \frac{1 − s}{1 − \frac{ζ′[i+1]}{ζ′[i]}} )

= ζ[i]P_{𝔓}.

If j = i + 1, then

P_{𝔓}ξ_{i+1} = P_{[a_{1},...,a_{k−1}]} (𝕋_{i} + \frac{1 − s}{1 − \frac{ζ′[i+1]}{ζ′[i]}} ) ξ_{i+1}

= P_{[a_{1},...,a_{k−1}]} (sξ_{i}T_{i}^{−1} + ξ_{i+1} \frac{1 − s}{1 − \frac{ζ′[i+1]}{ζ′[i]}} )

= P_{[a_{1},...,a_{k−1}]} (ζ′[i]T_{i} + ζ′[i](1 − s) + ζ′[i + 1]\frac{1 − s}{1 − \frac{ζ′[i+1]}{ζ′[i]}} )

= ζ′[i]P_{[a_{1},...,a_{k−1}]} (T_{i} + \frac{1 − s}{1 − \frac{ζ′[i+1]}{ζ′[i]}} )

= ζ[i + 1]P_{𝔓}.

□

Example 4.13. Figure 3 illustrates how to obtain the first values of the polynomial P_{ζ} for isotype (2, 1).

Besides Φ = T_{1}^{−1} ⋯ T_{N−1}^{−1}x_{N}, there is another raising operator, namely Φ′ := wx_{N}.

P_{[s,s^{−1},1]} P_{[s^{−1},s,1]} 𝕋_{1} + \frac{1−s}{1−s^{2}} P_{[s,1,qs^{−1}]} P_{[s,qs^{−1},1]} P_{[qs^{−1},s,1]} P_{[s^{−1},1,qs]} P_{[s^{−1},qs,1]} P_{[qs,s^{−1},1]} 𝕋_{2} + \frac{1−s}{1−qs^{−1}} 𝕋_{1} + \frac{1−s}{1−qs^{−2}} 𝕋_{2} + \frac{1−s}{1−qs} 𝕋_{1} + \frac{1−s}{1−qs^{2}} Φ Φ P_{[1,qs^{−1},qs]} P_{[1,qs,qs^{−1}]} P_{[qs^{−1},1,qs]} P_{[qs,1,qs^{−1}]} P_{[s,1,q^{2}s^{−1}]} P_{[s^{−1},1,q^{2}s]} Φ Φ Φ Φ Φ Φ 𝕋_{2} + \frac{1−s}{1−s^{−2}} 𝕋_{1} + \frac{1−s}{1−qs^{−1}} 𝕋_{1} + \frac{1−s}{1−qs} Φ P_{[qs^{−1},qs,1]} P_{[qs,qs^{−1},1]} 𝕋_{2} + \frac{1−s}{1−qs} 𝕋_{2} + \frac{1−s}{1−qs^{−1}} 𝕋_{1} + \frac{1−s}{1−s^{−2}} P_{[s,q^{2}s^{−1},1]} P_{[q^{2}s^{−1},s,1]} P_{[s^{−1},q^{2}s,1]} P_{[q^{2}s,s^{−1},1]} 𝕋_{2} + \frac{1−s}{1−q^{2}s} 𝕋_{1} + \frac{1−s}{1−q^{2}s^{2}} 𝕋_{2} + \frac{1−s}{1−q^{2}s^{−1}} 𝕋_{2} + \frac{1−s}{1−q^{2}s^{−2}}

Figure 3. The first Macdonald polynomials for isotype (21).

Proposition 4.14. We have

Φ′ = s^{N−1}ξ_{1}Φ,

and, if v ∈ ℕ_{0}^{N}, 𝕋 ∈ Tab_{λ}, then P_{v,T} Φ′ = s^{N−1+CT_{𝕋}[r_{v}[1]]}q^{v[1]}P_{v,T} Φ.

Proof. From ξ_{1} = s^{1−N}wT_{N−1} ⋯ T_{1}, it follows that

ξ_{1}T_{1}^{−1}T_{2}^{−1} ⋯ T_{N−1}^{−1} = s^{1−N}w,
ξ_{1}Φ = s^{1−N}Φ′.

Moreover, we have P_{v,𝕋}ξ_{1} = q^{v[1]}s^{N−1+CT_{𝕋}[r_{v}[1]]}P_{v,𝕋}. □

Note that it is easier to compute P Φ′ for a polynomial P.

4.3. Leading terms. We will write x^{v,𝕋} := x^{v}𝕋R_{v}. By abuse of language, x^{v,𝕋} will be referred to as a monomial. Note that the space M_{λ} is spanned by the set of polynomials

M_{λ} := {x^{v,𝕋} : v ∈ ℕ^{N}, 𝕋 ∈ Tab_{λ}},

which can be naturally endowed with the order ◁ defined by

x^{v,𝕋} ◁ x^{v′,𝕋′} if and only if v ◁ v′.

Theorem 4.15. The leading term (up to constant multiple) of P_{v,𝕋} is x^{v,𝕋}.

Proof. Theorem 3.13 shows that the leading term of P_{v,𝕋} is x^{v}𝕋H_{v} for some H_{v} ∈ H_{N}(q, s) (because the eigenvalues determine q^{v[i]}).

Use induction on #inv(v) = # {(i, j) : 1 ≤ i < j ≤ N, v [i] < v [j]}. The claim is true for partitions v, that is, in the case where #inv(v) = 0. Suppose the claim is true for all u with #inv(u) ≤ k and #inv(v) = k + 1. There is some i for which v [i] < v [i + 1]. By Theorem 4.12, we know that p := P_{v,𝕋} (T_{i} + \frac{(1−s)ζ[i]}{ζ[i]−ζ[i+1]}) is a ξ-eigenvector with eigenvalues [ζ[1], …, ζ[i+1], ζ[i], …], where ζ[j] = ζ_{v,𝕋}[j]. The list of eigenvalues implies that the leading term of p is x^{v·s_{i}}𝕋′ for some 𝕋′ ∈ V_{λ}. In fact, pξ_{j} = q^{vs_{i}[j]}s^{CT_{𝕋}[r_{vs_{i}}]}p for all j, and so the inductive hypothesis (#inv(vs_{i}) = #inv(v) − 1) implies that p is a scalar multiple of P_{vs_{i},𝕋} and has leading term x^{vs_{i}}𝕋T_{r_{vs_{i}}}. The only appearance of x^{vs_{i}} in p comes from x^{v}𝕋H_{v}T_{i} (by dominance, x^{vs_{i}} does not appear in P_{v,𝕋}).

However, if v [i] < v [i + 1] and 𝕋 ∈ V_{λ}, then

(4.1)  x^{v}𝕋T_{i} = x^{v}δ_{i}𝕋 + x^{vs_{i}} (𝕋T_{i})

= −(1 − s)x^{v}𝕋 + x^{vs_{i}} (𝕋T_{i}) +_{∑_{v′◁v} ; ℙ_{v′}∈V_{λ}}x^{v′}ℙ_{v′}.

Hence, by (4.1), we have

x^{v}𝕋H_{v}T_{i} = −(1 − s)x^{v}𝕋H_{v} + x^{vs_{i}} (𝕋H_{v}T_{i}) +_{∑_{v′◁v} ; ℙ_{v′}∈V_{λ}}x^{v′}ℙ_{v′}.

Thus 𝕋H_{v}T_{i} = 𝕋R_{r_{vs_{i}}} and

𝕋H_{v} = 𝕋R_{r_{vs_{i}}}T_{i}^{−1} = 𝕋R_{r_{v}},

by Lemma 2.14. This completes the proof of the theorem. □

The last theorem has the following consequence.

Corollary 4.16. Let 𝔓 = [a_{1}, …, a_{k}] such that a_{k} is a fall. Then we have P_{𝔓} = 0.

Proof. Without loss of generality, we can suppose that [a_{1}, …, a_{k−1}] is a path without fall. By Theorem 4.12, there exists a pair (v, 𝕋) such that P_{v,𝕋} = P_{[a_{1},...,a_{k−1}]}. On the other hand, by Theorem 4.15, one has

P_{v,𝕋} = x^{v,𝕋} +_{∑_{v′◁v} ; ℙ_{v′}∈V_{λ}}x^{v′}ℙ_{v′}.

Since a_{k} is a fall, one has

P_{𝔓} = x^{v}ℙ +_{∑_{v′◁v} ; ℙ_{v′}∈V_{λ}}x^{v′}ℙ_{v′},

with ℙ ∈ V_{λ}. Since P_{𝔓} is a simultaneous eigenfunction of the Cherednik operators, it is proportional to P_{v,𝕋}. Noting that the associated eigenvectors are uniquely determined, one obtains P_{𝔓} = 0. □

4.4. Action of T_{i}. We have more formulae than those exhibited in the proof of Theorem 4.12. Examples are given by the following theorem.

Proposition 4.17. Suppose that v ∈ ℕ_{0}^{N}, 𝕋 ∈ Tab_{λ}, v [i] = v [i + 1] for some i, and k := r_{v}[i], m := CT_{𝕋}[k + 1] − CT_{𝕋}[k]. Then we have:

(1) if CT_{𝕋}[k + 1] = CT_{𝕋}[k] − 1 then P_{v,𝕋}T_{i} = sP_{v,𝕋};

(2) if CT_{𝕋}[k + 1] = CT_{𝕋}[k] + 1 then P_{v,𝕋}T_{i} = −P_{v,𝕋};

(3) if CT_{𝕋}[k + 1] ≤ CT_{𝕋}[k] − 2 then P_{v,𝕋}T_{i} = P_{v,𝕋^{(k,k+1)}} − \frac{1−s}{1−s^{m}} P_{v,𝕋};

(4) if CT_{𝕋}[k + 1] ≥ CT_{𝕋}[k] + 2 then P_{v,𝕋}T_{i} = \frac{s(1−s^{m+1})(1−s^{m−1})}{(1−s^{m})^{2}}P_{v,𝕋^{(k,k+1)}} − \frac{1−s}{1−s^{m}} P_{v,𝕋}.

We introduce a partial order which will be used to compare eigenvalues, that is, the spectral vectors.

Definition 4.18. For integers n_{1}, m_{1}, n_{2}, m_{2} define

q^{n_{1}}s^{m_{1}} ≻ q^{n_{2}}s^{m_{2}} if and only if n_{1} > n_{2} or n_{1} = n_{2}, m_{1} ≤ m_{2} − 2;
q^{n_{1}}s^{m_{1}} ≁ q^{n_{2}}s^{m_{2}} if and only if n_{1} = n_{2}, |m_{1} − m_{2}| = 1.

We will also write q^{n_{1}}s^{m_{1}} > q^{n_{2}}s^{m_{2}} if n_{1} > n_{2}.

This formulation is used to unify the various recursion relations. Note that, if ζ = ζ_{v,𝕋} is a spectral vector, then we have necessarily ζ[i] ≠ ζ[i + 1] for each i. Indeed, either v[i] <> v[i + 1] or v[i] = v[i + 1], and the contents are different (since an RST can not have adjacent entries on a diagonal).

Here is a unified transformation formula. Theorem 4.12 is implicitly used.

Proposition 4.19. Suppose that v ∈ ℕ_{0}^{N}, 𝕋 ∈ Tab_{λ}, and 1 ≤ i < N. Then

(4.2) P_{ζ} (T_{i} + \frac{(1 − s) ζ[i]}{ζ[i] − ζ[i + 1]}) = \{ \begin{matrix}P_{ζs_{i}} if ζ[i + 1] ≻ ζ[i], \\ \begin{matrix}(ζ[i]−sζ[i+1]) & (sζ[i]−ζ[i+1]) \\ (ζ[i]−ζ[i+1]) & ^{2} \end{matrix}P_{ζs_{i}} if ζ[i] ≻ ζ[i + 1], \\ 0 if ζ[i] ≁ ζ[i + 1]. \end{matrix}

and

(4.3)  P_{ζ}Φ = P_{ζΨ^{q}}.

This proposition shows that we can easily use the spectral vector ζ instead of the pair (v, 𝕋) for labeling the Macdonald polynomials (assuming that ζ = ζ_{v,𝕋} for a given vector v and a given tableau 𝕋).

Indeed, we showed that, if ζ is a spectral vector and ζ[i] ≻ ζ[i+1] or ζ[i] ≺ ζ[i+1], then ζs_{i} is also a spectral vector. Such an action is called a permissible transposition. If ζ[i] ≁ ζ[i + 1] then ζ · s_{i} is not a spectral vector. We use some of the ideas developed by [14], see Theorem 5.8, p. 22 there. Let μ be a decreasing partition. Suppose that μ [i] = μ [j], i < j, and CT_{𝕋}[i]) = CT_{𝕋}[j] = a. Then {a − 1, a + 1} ⊂ {CT_{𝕋}[i + 1], …, CT_{𝕋}[j − 1]}. That is, there exists k with i < k < j such that CT_{𝕋}[k] = a + 1, and μ [k] = μ [i] (because of the partition property). Thus the spectral vector ζ contains a substring (q^{μ[i]}s^{a}, q^{μ[i]}s^{a+1}, q^{μ[i]}s^{a}) (preserving the order from ζ; it is impossible to move q^{μ[i]}s^{a} past q^{μ[i]}s^{a+1} with a permissible transposition), and adjacent entries of a spectral vector can not be equal.

One description of permissible permutations is as the set of permutations of ζ in which each pair (ζ[i], ζ[j]) with ζ[i] ≁ ζ[j] maintains its order, that is, if i < j and ζ[i] ≁ ζ[j] and (ζ[i · σ])_{i=1}^{N} is a spectral vector then i · σ < j · σ. The structure of permissible permutations is analyzed in Section 5.1.

For example, take λ = (3, 2) , μ = (1, 1, 1, 1, 0),

𝕋 = 2 1 5 4 3 ,
ζ = (q, qs^{−1}, qs^{2}, qs, 1) ,
ζ[1] ≁ ζ[2] ≻ ζ[3] ≁ ζ[4] ≻ ζ[5].

However, also ζ[1] ≁ ζ[4], so the order of the pairs (ζ[1], ζ[2]) , (ζ[3], ζ[4]) , (ζ[1], ζ[4]) must be preserved in the permissible permutations (of which there are 25). Observe that ζ is a maximal element, in the sense that only ≻ and ≁ occur in the comparisons of adjacent elements. Clearly there must be a minimal element (if ζ[i] ≻ ζ[i + 1], then apply s_{i} to ζ). In the example this is

ζ = (1, qs^{2}, q, qs, qs^{−1})
= ζ_{(0,1,1,1,1),𝕋_{1}},
𝕋_{1} = 4 2 5 3 1 .

To finish this discussion, we show that the maximal and minimal elements are unique. By the definition of ≻, we need only consider the possible arrangements of ζ[i], ζ[i + 1], …, ζ[j] where μ [i − 1] > μ [i] = ⋯ = μ [j] > μ [j + 1] (or i = 1, or j = N and μ [N] > 0). Let

inv (μ, 𝕋) = {(i, j) : μ [i] = μ [j] , i < j, ζ_{μ,𝕋}[i] ≺ ζ_{μ,𝕋}[j]} .

We showed that there is a unique RST 𝕋_{0}, where (ζ_{μ,𝕋_{0}}[i])_{i=1}^{N} is a permissible permutation of ζ and #inv (μ, 𝕋_{0}) = 0. By a similar argument there is a unique RST 𝕋_{1} which maximizes inv (μ, 𝕋). The minimum spectral vector is (ζ_{μ^{R},𝕋_{1}}[i])_{i=1}^{N}, where μ^{R} [i] = μ [N + 1 − i], 1 ≤ i ≤ N.

According to the previous remark, we will use the following notations.

Definition 4.20. If ζ = ζ_{v,𝕋}, we write

inv_{◁}(ζ) := {(i, j) : 1 ≤ i < j ≤ N, ζ[i] ◁ ζ[j]},

for ◁ ∈ {<, >, ≺, ≻}. If ζ = ζ_{v,𝕋}, then we write ζ^{+} = ζ_{v^{+},𝕋}. Note that ζ^{+}[1] ≥ ζ^{+}[2] ≥ ⋯ ≥ ζ^{+}[N]. We set

inv(ζ) := inv_{<}(ζ) = inv(v).

The action of the symmetric group 𝔖_{N} on the spectral vector is defined by

(4.4) ζs_{i} = \begin{cases}[ζ[1], …, ζ[i − 1], ζ[i + 1], ζ[i], ζ[i + 2], …, ζ[N]] & if ζ[i] ≺ ζ[i + 1], \\ & or ζ[i] ≻ ζ[i + 1], \\ ζ & otherwise. \end{cases}

Say ζ′ ≺ ζ if and only if there exists a sequence of elementary transpositions (s_{i_{1}}, …, s_{i_{k}}) such that

ζ_{0} = ζ, ζ_{1} = ζ_{0}s_{i_{1}}, …, ζ_{k} = ζs_{i_{1}} ⋯ s_{i_{k}} = ζ′

and, for each j < k, ζ_{j}[i_{j+1}] ≺ ζ_{j}[i_{j+1} + 1].

5. Stable subspaces

5.1. Connected components. We denote by H_{λ}^{q,s} the graph obtained from G_{λ}^{q,s} by removing the affine edges, all the falls, and the vertex ∅.

Recall that v^{+} is the unique decreasing partition obtained by permuting the entries of v.

Definition 5.1. Let v ∈ ℕ^{N} and 𝕋 ∈ Tab_{λ} (where λ a partition). We define the filling T (𝕋, v) as the one obtained from 𝕋 by replacing i by v^{+}[i] for each i..

As in [6], the following fact holds true.

Proposition 5.2. Two 4-tuples (𝕋, ζ, v, σ) and (𝕋′, ζ′, v′, σ′) are in the same connected component of H_{λ}^{q,s} if and only if T (𝕋, v) = T (𝕋, v′).

This shows that the connected components of H_{λ}^{q,s} are indexed by the T (𝕋, μ), where μ is a partition.

Definition 5.3. We denote by H_{T}^{q,s} the connected component associated with T in H_{λ}^{q,s}. The component H_{T}^{q,s} will be said to be 1-compatible if T is a column-strict tableau. The component H_{T}^{q,s} will be said to be (−1)-compatible if T is a row-strict tableau.

Note that each connected component has a unique minimal element (i.e., an element without antecedent) called its root and denoted by

root(T) := (𝕋_{root(T)}, ζ_{root(T)}, v_{root(T)}, r_{root(T)}),

and a unique maximal element called its sink and denoted by

sink(T) := (𝕋_{sink(T)}, ζ_{sink(T)}, v_{sink(T)}, r_{sink(T)}).

.

With the notations of the previous section, we have v_{sink(T)} = v^{+} and 𝕋_{sink(T)} = 𝕋_{0} for any pair (v, 𝕋) ∈ T. In the same way, v_{root(T)} = v_{sink(T)}^{R} and 𝕋_{root(T)} = 𝕋_{1}.

Example 5.4. Let μ = [2, 1, 1, 0, 0] and λ = [3, 2]. There are four connected components with vertices labeled by permutations of μ in H_{λ}^{q,s}. The possible values of T (𝕋, μ) are

{12 \atop 001}, {02 \atop 011}, {01 \atop 012}, and {11 \atop 002}.

The 1-compatible components are Hq,s_{{12 \atop 001}} and Hq,s_{{11 \atop 002}} , while there is only one (−1)-compa-tible component Hq,s_{{01 \atop 012}} . The component Hq,s_{{02 \atop 011}} is neither 1-compatible nor (−1)-compa-tible.

The component Hq,s_{{12 \atop 001}} contains vertices of Gq,s_{{31 \atop 542}} and Gq,s_{{21 \atop 543}} connected by jumps. In Figure 4 we have drawn the components Hq,s_{{11 \atop 002}} and Hq,s_{{01 \atop 012}} .

[s^{−1}, 1, q, qs, q^{2}s^{2}] [s^{−1,1,q,q^{2}s^{2},qs}] [s^{−1}, q, 1, qs, q^{2}s^{2}] {01 \atop 012} [s^{−1}, 1, q^{2}s^{2}, q, qs] {[s^{−1},q,1,q^{2}s^{2},qs] {42 \atop 531} \atop [01021]} {[q,s^{−1},1,qs,q^{2}s^{2}] {42 \atop 531} \atop [10012]} s_{4} s_{2} s_{3} s_{2} s_{4} s_{1} [s, 1, qs^{2}, qs^{−1}, q^{2}] [s, 1, qs^{−1}, qs^{2}, q^{2}] [s, 1, qs^{2}, q^{2}, qs^{−1}] [s, qs^{2}, 1, qs^{−1}, q^{2}] [s, 1, q^{2}, qs^{2}, qs^{−1}] [s, qs^{2}, 1, q^{2}, qs^{−1}] [qs^{2}, s, 1, qs^{−1}, q^{2}] {12 \atop 001} [s, 1, qs^{−1}, q^{2}, qs^{2}] [s, qs^{−1}, 1, qs^{2}, q^{2}] [s, 1, q^{2}, qs^{−1}, qs^{2}] [s, qs^{−1}, 1, q^{2}, qs^{2}] [qs^{−1}, s, 1, qs^{2}, q^{2}] s_{4} s_{2} s_{3} s_{2} s_{4} s_{1} s_{4} s_{2} s_{3} s_{2} s_{4} s_{1} s_{3} s_{4} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Figure 4. Two connected components of H_{32}^{q,s}

Example 5.5. Consider the tableau T = {01 \atop 00} . The graph H_{T}^{q,s} is given by:

\begin{matrix}[s, s^{−1}, 1, q] \\ [s, s^{−1}, q, 1] \\ [s, q, s^{−1}, 1] \\ [q, s, s^{−1}, 1] \\ [s^{−1}, s, 1, q] \\ [s^{−1}, s, q, 1] \\ [s^{−1}, q, s, 1] \\ [q, s^{−1}, s, 1] \\ s_{1} \\ s_{1} \\ s_{2} \\ s_{3} \\ s_{2} \\ s_{1} \\ s_{3} \\ s_{2} \\ s_{1} \end{matrix}

The sink is indicated by a red disk and the root by a green disk.

By abuse of language, we write ζ ∈ T to mean that ζ appears in a vertex of the connected component H_{T}^{q,s}.

Definition 5.6. In the same way, we define std_{0}T of T to be the reverse standard tableau of shape λ obtained by the following process:

(1) Denote by |T |_{i} the number of occurrences of i in T.

(2) Read the tableau T from the left to the right and the bottom to the top and replace occurrences of i in the order of their appearance by the numbers N − |T |_{0} − ⋯ − |T |_{i−1}, N − |T |_{0} − ⋯ − |T |_{i−1} − 1, … N − |T |_{0} − ⋯ − |T |_{i}.

Let T be a filling of shape λ. Then std_{1}T of T is defined to be the reverse standard tableau of shape λ obtained by the following process:

(1) Denote by |T |_{i} the number of occurrences of i in T.

(2) Read the tableau T from the bottom to the top and the left to the right and replace occurrences of i in the order of their appearance by the numbers N − |T |_{0} − ⋯ − |T |_{i−1}, N − |T |_{0} − ⋯ − |T |_{i−1} − 1, … N − |T |_{0} − ⋯ − |T |_{i}.

Example 5.7. To construct std_{0} \binom{0 1}{0 0 2} we first write:

\begin{matrix}0 & 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & . & . \\ . & . & . & 1 & . \\ . & . & . & . & 2 \end{matrix}

and we renumber entries in increasing order from the bottom to the top and the right to the left:

\begin{matrix}0 & 0 & 0 & 1 & 2 \\ 5 & 4 & 3 & . & . \\ . & . & . & 2 & . \\ . & . & . & . & 1 \end{matrix}

We obtain std_{0} \binom{0 1}{0 0 2} = 4 2 5 3 1 .

Pictorially, we construct std_{1} \binom{0 1}{0 0 2} writing:

\begin{matrix}0 & 0 & 2 & 0 & 1 \\ 0 & 0 & . & 0 & . \\ . & . & . & . & 1 \\ . & . & 2 & . & . \end{matrix}

and renumbering entries in increasing order from the bottom to the top and the right to the left:

\begin{matrix}0 & 0 & 2 & 0 & 1 \\ 5 & 4 & . & 3 & . \\ . & . & . & . & 2 \\ . & . & 1 & . & . \end{matrix}

This gives std_{1} \binom{0 1}{0 0 2} = 3 2 5 4 1 .

Alternatively, one has

std_{0}(T)[i, j] := #{(k, l) : T [k, l] > T [i, j]} + #{(k, l) : k > i, T [k, l] = T [i, j]} + #{(i, l) : l ≥ j, T [i, l] = T [i, j]}

and

std_{1}(T)[i, j] := #{(k, l) : T [k, l] > T [i, j]} + #{(k, l) : l > j, T [k, l] = T [i, j]} + #{(k, j) : k ≥ i, T [k, j] = T [i, j]}.

We can characterize the root and the sink of a connected component.

Lemma 5.8. One has:

(1) 𝕋_{root(T)} = std_{0}T and 𝕋_{sink(T)} = std_{1}T,

(2) v_{root(T)} = v^{R} and v_{sink(T)} = v^{+}.

Proof. First observe that T (std_{0}(T), v) = T (std_{1}(T), v) = T by construction. So we have (v^{R}, std_{0}(T)), (v^{+}, std_{1}(T)) ∈ H_{T}^{q,s}.

Since, v^{R} is an increasing partition, each arrow

(𝕋, u) (std_{0}(T), v^{R}) s_{i}

is a jump (i.e., u = v^{R}). Let [i, j] be a cell of std_{0}(T) and k = std_{0}(T)[i, j]. Let [i′, j′] be the cell such that k + 1 = std_{0}(T)[i′, j′]. From the definition of std_{0}(T), we have either T [i, j] ≠ T [i′, j′] or j = j′ or i < i′ and j > j′ (that is, CT_{std_{0}(T)}[k] < CT_{std_{0}(T)}[k+1]−1). Hence, such a row does not exist and (std_{0}(T), v^{R}) has no antecedent in H_{T}^{q,s}. This is equivalent to std_{0}(T) = 𝕋_{root(T)}.

In an equivalent way, we find that there is no arrow in H_{T}^{q,s} of the form

(std_{1}(T), v^{+}) (𝕋, u) s_{i}

Consequently, we have std_{1}(T) = 𝕋_{sink(T)}. □

Example 5.9. We write Example 5.5 in terms of tableaux:

[0001]^{{31 \atop 42}} ; [0010]^{{31 \atop 42}} ; [0100]^{{31 \atop 42}} ; [1000]^{{31 \atop 42}} ; [0001]^{{21 \atop 43}} ; [0010]^{{21 \atop 43}} ; [0100]^{{21 \atop 43}} ; [1000]^{{21 \atop 43}} ; s_{1} ; s_{1} ; s_{2} ; s_{3} ; s_{2} ; s_{1} ; s_{3} ; s_{2} ; s_{1}

We observe that std_{0} \binom{01}{00} = {31 \atop 42} = 𝕋_{root}\binom{01}{00} and std_{1} \binom{01}{00} = {21 \atop 43} = 𝕋_{sink}\binom{01}{00}.

Remark 5.10. As a consequence, we obtain that, if m_{i} denotes the number of occurrences of i among the entries of T, then

r_{root(T)} = […, m_{0} + ⋯ + m_{i} + 1, …, m_{0} + ⋯ + m_{i+1} + 1, …, m_{0} + 1, …, m_{0} + m_{1}, 1, …, m_{0}],

and r_{sink(T)} = [1, …, N].

The notion of (±1)-compatibility is easily detectable from the root and the sink.

Lemma 5.11. If H_{T}^{q,s} is 1-compatible then, for each i, i and i + 1 are not in the same column of 𝕋_{root(T)}. If H_{T}^{q,s} is (−1)-compatible then, for each i, i and i + 1 are not in the same row of 𝕋_{sink(T)}.

Proof. From Lemma 5.8, we have 𝕋_{root(T)} = std_{0}(T) and 𝕋_{sink(T)} = std_{1}(T). But if k and k + 1 are in the same column of std_{0}(T), where we suppose that std_{0}(T)[i, j] = k, then std_{0}(T)[i, j + 1] = k + 1, and the only possibility is that T [i, j] = T [i, j + 1], which contradicts the fact that T is a column-strict tableau. Similarly, if k and k + 1 are in the same row of std_{1}(T), then T [i, j] = T [i + 1, j] for some (i, j), which contradicts the fact that T is a row-strict tableau. □

Now we have all the material for an interpretation of the (±1)-compatibility in terms of spectral vectors.

Proposition 5.12. If H_{T}^{q,s} is 1-compatible then, for each i, ζ_{root(T)}[i] ≁ ζ_{root(T)}[i + 1] implies ζ_{root(T)}[i] = sζ_{root(T)}[i+1]. If H_{T}^{q,s} is (−1)-compatible then, for each i, ζ_{sink(T)}[i] ≁ ζ_{sink(T)}[i + 1] implies ζ_{sink(T)}[i] = s^{−1}ζ_{sink(T)}[i + 1].

Proof. This is just the translation of Lemma 5.11 in terms of spectral vectors. □

5.2. Invariant subspaces. The Yang–Baxter graph and the previous section imply that we can characterize the irreducible subspaces U of polynomials invariant under H_{N}(s) and {ξ_{i} : 1 ≤ i ≤ N}, that is, UT_{i}, Uξ_{i} ⊂ U.

Definition 5.13. Let T be a tableau with increasing row and column entries. We denote by M_{T} the space generated by the polynomials P_{ζ} with ζ ∈ T.

Example 5.14. For instance, M 0 1 0 0 is spanned by

{P_{[s,s^{−1},1,q]}, P_{[s,s^{−1},q,1]}, P_{[s,q,s^{−1},1]}, P_{[q,s,s^{−1},1]}P_{[s^{−1},s,1,q]}, P_{[s^{−1},s,q,1]}, P_{[s^{−1},q,s,1]}, P_{[q,s^{−1},s,1]}}.

The spaces M_{T} are the irreducible invariant subspaces.

Proposition 5.15. We have M_{T} T_{i}, M_{T} ξ_{i} ⊂ M_{T}. Furthermore, if U is a proper subspace of M_{T}, then UT_{i} ⊄ U or Uξ_{i} ⊄ U.

Proof. Let U be a subspace of M_{T} T_{i} such that UT_{i}, Uξ_{i} ⊂ U. The operators ξ_{i} being simultaneously diagonalizable, U is spanned by a set of polynomials {P_{ζ_{1}}, …, P_{ζ_{k}}} with k ∈ ℕ and ζ_{i} ∈ 𝕋. But from the Yang–Baxter construction, if there exists ζ ∈ T such that P_{ζ} ∈ U, then, for each ζ ∈ T, P_{ζ} ∈ U. So U is not a proper subspace. □

In the rest of the section, we investigate the dimension of the spaces M_{T}. The dimension of such a space equals the number of permutations of the vector of the entries of T multiplied by the number of tableaux 𝕋 appearing in H_{T}. The first number is easy to obtain, but for the second we need some additional considerations.

Suppose that μ, λ are partitions with μ ⊂ λ (μ[i] ≤ λ[i] for all i), |μ| = k, |λ| = n, then the set {(i, j) : 1 ≤ i ≤ ℓ (λ) , μ [i] < j ≤ λ [i]} is the skew diagram λ\μ. The basic step in determining the dimension of a connected component is to find the number (denoted dim (λ\μ)) of RST’s of shape λ\μ, that is, the number of ways the numbers (n − k) , (n − k − 1) , …, 1 can be entered in λ\μ so that the entries decrease in each row and in each column. Equivalently, we ask for the number of standard Young tableaux of shape λ\μ (see [12] or [15] for the definition). There is a classical formula for this number (see [15, Cor. 7.16.3]). If det (a_{ij}) denotes the determinant of the matrix (a_{ij})_{i,j=1}^{p}, where p ≥ ℓ (λ) (the formula is independent of p), the formula says that

dim (λ\μ) = (n − k)! det [\frac{1}{(λ [i] − μ [j] − i + j)!}] ,

Here, \frac{1}{p!} = 0 for p = −1, −2, … (as in the zeros of \frac{1}{Γ(p+1)}).

Now consider a tableau T. Let M denote the maximum entry (also of any v in this component) and let

μ_{m} = {(i, j) ∈ T : T (i, j) ≤ m} , 0 ≤ m ≤ M.

Then each μ_{m} is the Ferrers diagram of a partition, μ_{m} ⊂ μ_{m+1} (possibly μ_{m} = μ_{m+1} for some m and dim (μ_{m+1}\μ_{m}) = 1 trivially), and v^{+} [j] = m if j is an entry in μ_{m}\μ_{m−1}. The number of RST’s in the connected component of T is

dim (μ_{0}\ [0]) ∏_{m=1}^{M} dim (μ_{m}\μ_{m−1}) ,

and the number of permutations of v^{+} is N!/ (|μ_{0}|! ∏_{m=1}^{M} (|μ_{m}| − |μ_{m−1}|)!); the dimension of the component is

N! det [\frac{1}{(μ_{0} [i] − i + j)!}] ∏_{m=1}^{M} det [\frac{1}{(μ_{m} [i] − μ_{m−1} [j] − i + j)!}] .

This product can be restricted to the values of m for which μ_{m−1} ≠ μ_{m}, that is, the set of entries of v^{+}.

Example 5.16. (1) Consider again the tableau T = 0 1 0 0 . Then μ_{0} = [2, 1] and

μ_{1} = [2, 2]. Hence,

dim (μ_{0}\ [0]) = 3! det [ 1 1^{\frac{1}{2} \frac{1}{6}} ] = 2,
dim (μ_{1}\μ_{0}) = 1! det [ \begin{matrix}1 & \frac{1}{2} \\ 0 & 1 \end{matrix} ] = 1.

Consequently, the number of tableaux 𝕋 in T equals 2. The tableaux are 2 1 4 3

and 3 1 4 2 . So the dimension of M_{T} is \frac{4!}{3!1!} × 2 = 8.

(2) Consider the bigger example given by the tableau T = \begin{matrix}1 & 2 \\ 0 & 0 & 1 \end{matrix} (see Figure 4).

Here μ_{0} = [2], μ_{1} = [3, 1], and μ_{2} = [3, 2]. We compute

dim (μ_{0}\ [0]) = 2! det [ 0 1^{\frac{1}{2} \frac{1}{6}} ] = 1,
dim (μ_{1}\μ_{0}) = 2! det [ \begin{matrix}1 & \frac{1}{24} \\ 0 & 1 \end{matrix} ] = 2,
dim (μ_{2}\μ_{1}) = 1! det [ \begin{matrix}1 & \frac{1}{6} \\ 0 & 1 \end{matrix} ] = 1.

There are two tableaux; the graph decomposes into two parts when we remove the jump edges. The dimension of M_{T} is \frac{5!}{2!2!1!} × 2 = 60.

(3) Consider T = \begin{matrix}0 & 1 \\ 0 & 1 & 2 \end{matrix} (see Figure 4). One has μ_{0} = [1, 1], μ_{1} = [2, 2], and

μ_{2} = [3, 2]. Hence, we have only one tableau in the connected component. Graphically, there is no jump (blue arrow) in the connected component H_{T}^{q,s}. The dimension of M_{T} is 30.

5.3. Symmetrizer/Antisymmetrizer. We define the operator

S_{N} := ∑_{σ∈𝔖_{N}} T̃_{σ},

where T̃_{σ} = T_{i_{1}} ⋯ T_{i_{k}} if there is a shortest expression σ = s_{i_{1}} ⋯ s_{i_{k}}. The operator S_{N} is an s-deformation of the classical symmetrizer in the following sense.

38 C. F. DUNKL AND J.-G. LUQUE

Proposition 5.17. For each i, one has

S_{N}T_{i} = sS_{N}.

Proof. It suffices to split the sum as

(5.1)  S_{N}T_{i} = _{∑_{σ∈𝔖_{N}} ; ℓ(σs_{i})>ℓ(σ)}T̃_{σ}T_{i} +_{∑_{σ∈𝔖_{N}} ; ℓ(σs_{i})<ℓ(σ)}T̃_{σ}T_{i}.

We use the quadratic relation to write the second sum as

∑_{σ∈𝔖_{N}} ; ℓ(σs_{i})<ℓ(σ)T̃_{σ}T_{i} = (s − 1)_{∑_{σ∈𝔖_{N}} ; ℓ(σs_{i})<ℓ(σ)}T̃_{σs_{i}}T_{i} + s_{∑_{σ∈𝔖_{N}} ; ℓ(σs_{i})<ℓ(σ)}T̃_{σs_{i}}.

But

∑_{σ∈𝔖_{N}} ; ℓ(σs_{i})<ℓ(σ)T̃_{σs_{i}}T_{i} = _{∑_{σ∈𝔖_{N}} ; ℓ(σs_{i})>ℓ(σ)}T̃_{σ}

Hence,

∑_{σ∈𝔖_{N}} ; ℓ(σs_{i})<ℓ(σ)T̃_{σ}T_{i} = (s − 1)_{∑_{σ∈𝔖_{N}} ; ℓ(σs_{i})>ℓ(σ)}T̃_{σ} + s_{∑_{σ∈𝔖_{N}} ; ℓ(σs_{i})<ℓ(σ)}T̃_{σs_{i}}.

Substituting this in (5.1), we obtain the result. □

As a consequence, we obtain the following result. Corollary 5.18. The operator S_{N} satisfies

S_{N}^{2} = ϕ_{N}(s)S_{N},

where ϕ_{N}(s) := ∏_{j=2}^{N} \frac{1−s^{j}}{1−s} is the Poincaré polynomial of 𝔖_{N}. Proof. From Proposition 5.17, one obtains

S_{N}^{2} = S_{N} ∑_{σ∈𝔖_{N}} T̃_{u} = ∑_{σ∈𝔖_{n}} s^{ℓ(σ)}S_{N} = ϕ_{N}(s)S_{N}.

□

Alternatively, we define

S′_{N} = _{∑_{σ∈𝔖_{N}} ; ℓ(σ)=k,σ=s_{i_{1}}···s_{i_{k}}}T_{i_{1}}^{−1} ⋯ T_{i_{k}}^{−1}.

This operator satisfies

(5.2)  S′_{N}T_{i} = sS′_{N}

and

(5.3)  S′_{N}^{2} = ϕ_{N} (\frac{1}{s}) S′_{N}.

The action of the symmetrizer on leading terms has some nice properties.

Lemma 5.19. Let v and 𝕋 be such that COL_{𝕋}[r_{v}[i]] = COL_{𝕋}[r_{v}[i]+1] and v[i] = v[i+1] for some i. Then

x^{v,𝕋}S_{N} = 0.

Proof. We have

x^{v,𝕋}T_{i} = x^{v}δ_{i}^{x}𝕋 + x^{v}s_{i}𝕋R_{v}T_{i}.

But v[i] = v[i + 1] implies x^{v}δ_{i}^{x} = 0 and, since COL_{𝕋}[r_{v}[i]] = COL_{𝕋}[r_{v}[i] + 1], we have 𝕋T_{r_{v}[i]} = −𝕋. Hence

(5.4)  x^{v,𝕋}T_{i} = x^{v}𝕋T_{r_{v}[i]}R_{v} = −x^{v,𝕋}.

Now we split the sum x^{v,𝕋}S_{N} into two sums,

x^{v,𝕋}S_{N} = x^{v,𝕋} ∑_{ℓ(s_{i}σ)<ℓ(σ)} T̃_{σ} + x^{v,𝕋} ∑_{ℓ(s_{i}σ)>ℓ(σ)} T̃_{σ}

= x^{v,𝕋} ∑_{ℓ(s_{i}σ)<ℓ(σ)} T_{i}T̃_{s_{i}σ} + x^{v,𝕋} ∑_{ℓ(s_{i}σ)>ℓ(σ)} T̃_{σ}.

From eq. (5.4), one obtains

x^{v,𝕋}S_{N} = −x^{v,𝕋} ∑_{ℓ(s_{i}σ)<ℓ(σ)} T̃_{s_{i}σ} + x^{v,𝕋} ∑_{ℓ(s_{i}σ)>ℓ(σ)} T̃_{σ}

= −x^{v,𝕋} ∑_{ℓ(s_{i}σ)>ℓ(σ)} T̃_{σ} + x^{v,𝕋} ∑_{ℓ(s_{i}σ)>ℓ(σ)} T̃_{σ}

= 0.

□

In the same way, we define

A_{N} = ∑_{σ∈𝔖_{N}}(−s)^{ℓ(σ)}\overline{T}_{σ},

where \overline{T}_{σ} = T_{i_{1}}^{−1} ⋯ T_{i_{k}}^{−1} given there is a shortest expression σ = s_{i_{1}} ⋯ s_{i_{k}}. The operator A_{N} satisfies the following relation.

Proposition 5.20. For each i, we have

A_{N}T_{i} = −A_{N}.

Proof. The proof is very close to the proof of Proposition 5.17 and is left to the reader.

□

The operator A_{N} satisfies the following analog of Corollary 5.18.

Corollary 5.21. We have

A_{N}^{2} = ϕ_{N}(s)A_{N}.

Lemma 5.22. Let v and 𝕋 be such that ROW_{𝕋}[r_{v}[i]] = ROW_{𝕋}[r_{v}[i] + 1] and v[i] = v[i + 1] for some i. Then

x^{v,𝕋}A_{N} = 0.

Lemma 5.23. Let v = [v[1] < ⋯ < v[N]] and 𝕋 be such that, for each i, v[i] = v[i + 1] implies COL_{𝕋}[r_{v}[i]] = COL_{𝕋}[r_{v}[i] + 1]. Then the coefficient of x^{v,𝕋} in x^{v,𝕋}A_{N} equals ∏_{i} s^{m_{i}}ϕ_{m_{i}}(s), where m_{i} denotes the number of parts i in v.

5.4. Symmetric/Antisymmetric polynomials. For ζ = ζ_{v,𝕋} and ζs_{i} = ζ_{v′,𝕋′}, we set

𝔰_{ζ}^{i} := P_{ζs_{i}} + \frac{s− \frac{ζ[i+1]}{ζ[i]}}{1− \frac{ζ[i+1]}{ζ[i]}} P_{ζ} and 𝔞_{ζ}^{i} := P_{ζs_{i}} − \frac{1−s \frac{ζ[i+1]}{ζ[i]}}{1− \frac{ζ[i+1]}{ζ[i]}} P_{ζ}.

Lemma 5.24. If ζ[i + 1] ≻ ζ[i], then we have

𝔰_{ζ}^{i}T_{i} = s𝔰_{ζ}^{i} and 𝔞_{ζ}^{i}T_{i} = −𝔞_{ζ}^{i}.

Proof. We prove only the result for 𝔰_{ζ}^{i} since the proof is very similar for 𝔞_{ζ}^{i}. Recall that Proposition 4.19 yields

P_{ζ}T_{i} = P_{ζs_{i}} − (1 − s)\frac{ζ[i]}{ζ[i] − ζ[i + 1]}P_{ζ}

and

P_{ζs_{i}}T_{i} = \frac{(ζ[i + 1] − sζ[i])(sζ[i + 1] − ζ[i])}{(ζ[i + 1] − ζ[i])^{2}}P_{ζ} − (1 − s)\frac{ζ[i + 1]}{ζ[i + 1] − ζ[i]}P_{ζs_{i}}.

Hence,

𝔰_{ζ}^{i}T_{i} = (\frac{(ζ[i + 1] − sζ[i])(sζ[i + 1] − ζ[i])}{(ζ[i + 1] − ζ[i])^{2}} − \frac{(1 − s)ζ[i](s − \frac{ζ[i+1]}{ζ[i]})}{(ζ[i] − ζ[i + 1])(1 − \frac{ζ[i+1]}{ζ[i]})}) P_{ζ}

+ (\frac{s − \frac{ζ[i+1]}{ζ[i]}}{1 − \frac{ζ[i+1]}{ζ[i]}} − (1 − s)\frac{ζ[i + 1]}{ζ[i + 1] − ζ[i]}) P_{ζs_{i}}

= sP_{ζs_{i}} + s (\frac{s − \frac{ζ[i+1]}{ζ[i]}}{1 − \frac{ζ[i+1]}{ζ[i]}} )

= s𝔰_{ζ}^{i}.

□

Let 𝔣 = ∑_{ζ∈T} b_{ζ}P_{ζ} ∈ M_{T} be a symmetric polynomial, i.e., 𝔣T_{i} = s𝔣 for each i.

Lemma 5.25. If ζ[i + 1] ≻ ζ[i], then \frac{b_{ζ}}{b_{ζs_{i}}} = \frac{sζ[i]−ζ[i+1]}{ζ[i]−ζ[i+1]}.

Proof. Since 𝔣T_{i} = s𝔣, we have

(b_{ζ}P_{ζ} + b_{ζs_{i}}P_{ζs_{i}}) T_{i} = s (b_{ζ}P_{ζ} + b_{ζs_{i}}P_{ζs_{i}}) .

Then b_{ζ}P_{ζ} + b_{ζs_{i}}P_{ζs_{i}} is proportional to 𝔰_{ζ}^{i}. This finishes the proof of the lemma. □

Since each vertex of T is connected to sink(T) by a series of edges

\begin{matrix}ζ \\ ζs_{i} \\ s_{i} \end{matrix}

the polynomial 𝔣 is unique up to a global multiplicative coefficient, and b_{ζ} ≠ 0 for all ζ if 𝔣 ≠ 0.

If T [i, j] = T [i, j + 1] for some (i, j) then ζ_{root(T)}[k] = q^{n}s^{m} ≁ ζ_{root(T)}[k + 1] = q^{n}s^{m+1} for some k. Indeed, T [i, j] = T [i, j + 1] implies v_{root(T)}[k] = v_{root(T)}[k + 1], hence r_{v_{root(T)}}[k] + 1 = r_{v_{root(T)}}[k + 1]. It follows that m = CT_{𝕋_{root(T)}}[ℓ] and m + 1 = CT_{𝕋_{root(T)}}[ℓ + 1] for some ℓ.

Example 5.26. If T = 0 1 0 0 , we have

root(T) = ( \begin{matrix}3 & 1 \\ 4 & 2 \end{matrix} , [s, s^{−1}, 1, q], [0, 0, 0, 1], [2, 3, 4, 1]) .

We have T [1, 1] = T [1, 2] = 0. The corresponding cells in the tableau 𝕋_{root(T)} are 𝕋_{root(T)}[1, 1] = 4 and 𝕋_{root(T)}[1, 2] = 3. So ℓ = 3, k = 2, and m = −1 = CT_{𝕋_{root(T)}}[3] = CT_{𝕋_{root(T)}}[4] − 1.

From 𝔣T_{k} = s𝔣, one deduces b_{ζ_{root(T)}} = s(s−1)^{−1}ζ_{root(T)}[k]− \frac{ζ_{root(T)}[k+1]}{ζ_{root(T)}[k]}b_{ζ_{root(T)}}. Finally,

\frac{ζ_{root(T)}[k]}{ζ_{root(T)}[k]−ζ_{root(T)}[k+1]} = \frac{1}{1−s} implies b_{ζ_{root(T)}} = 0 and 𝔣 = 0.

In the other cases, the coefficients b_{ζ} are not zero, and they can be computed via the recurrence given in Lemma 5.25. More precisely, setting b_{ζ_{root(T)}} = 1, and b_{ζs_{i}} = \frac{ζ[i]−ζ[i+1]}{sζ[i]−ζ[i+1]}b_{ζ} if ζ[i + 1] ≻ ζ[i], we define the polynomial

𝔐_{T} = ∑_{ζ∈T} b_{ζ}P_{ζ},

which is the unique generator of the subspace of symmetric polynomials of M_{T}. So, one arrives at the following result.

Theorem 5.27. The subspace of M_{T} of symmetric polynomials is (1) a 1-dimension space generated by 𝔐_{T} if T is a strict-column tableau;

(2) a 0-dimension space in the other cases.

Example 5.28. Consider the graph Hq,s_{{11 \atop 00}} (see Figure 5). The polynomial

𝔐_{{11 \atop 00}} = P_{[s,1,q,qs^{−1}]} + \frac{1 − q}{s − q} P_{[s,q,1,qs^{−1}]} + \frac{(1 − q)}{(s^{2} − q)}P_{[q,s,1,qs^{−1}]} + \frac{(1 − q)}{(s^{2} − q)}P_{[s,q,qs^{−1},1]}

+ \frac{(1 − q)(s − q)}{(s^{2} − q)^{2}}P_{[q,s,qs^{−1}]} + \frac{(1 − q)(s − q)}{(s^{2} − q)(s^{3} − q)}P_{[q,qs^{−1},s,1]}

is symmetric.

In the same way, define b_{ζ_{root(T)}}^{a} = 1, and b_{ζs_{i}}^{a} = −\frac{ζ[i]−ζ[i+1]}{ζ[i]−sζ[i+1]}b_{ζ}^{a} if ζ[i + 1] ≻ ζ[i], and the polynomial

𝔐_{T}^{a} = ∑_{ζ∈T} b_{ζ}^{a}P_{ζ}.

We then have the following result.

[s, 1, q, qs^{−1}] [s, q, 1, qs^{−1}] [s, q, qs^{−1}, 1] [q, s, 1, qs^{−1}] [q, s, qs^{−1}, 1] [q, qs^{−1}, s, 1] {s_{2} \atop × \frac{1−q}{s−q}} {s_{1} \atop × \frac{s−q}{s2−q}} {s_{3} \atop × \frac{s−q}{s2−q}} {s_{1} \atop × \frac{s−q}{s2−q}} {s_{3} \atop × \frac{s−q}{s2−q}} {s_{2} \atop × \frac{s2−q}{s3−q}}

Figure 5. The graph Hq,s_{{11 \atop 00}}

Theorem 5.29. The subspace of M_{T}^{a} of antisymmetric polynomials is

(1) a 1-dimension space generated by 𝔐_{T}^{a} if T is a strict-row tableau;

(2) a 0-dimension space in the other cases.

5.5. The group of permutations leaving T invariant. Let T be a filling of shape λ with increasing rows and strictly increasing columns.

To each i we associate the pair COORD_{T} [i] = (COL_{std_{1}(T)}[i], ROW_{std_{1}(T)}[i]). An elementary transposition s_{i} acts on T by permuting the cells COORD_{T} [i] and COORD_{T} [i+ 1].

For a tableau 𝕋, we denote by 𝔖_{𝕋} the maximal subgroup of 𝔖_{N} leaving invariant the sets of entries of each line.

Example 5.30. For instance, consider the tableau 𝕋 = 3 2 5 4 1 . We have 𝔖_{𝕋} =

𝔖_{{1,4,5}} × 𝔖_{{2,3}}.

We denote also by 𝔖_{T} the maximal subgroup of 𝔖_{std_{1}(T)} leaving T invariant.

Example 5.31. Let T = 1 1 0 0 1 . Then we have std_{1}(T) = 3 2 5 4 1 and

𝔖_{T} = 𝔖_{{2,3}} × 𝔖_{{4,5}} × 𝔖_{{1}} ⊂ 𝔖_{std_{1}(T)} = 𝔖_{{1,4,5}} × 𝔖_{{2,3}}.

Let 𝔖_{r}(T) be the subgroup of 𝔖_{N} leaving the partition v_{sink(T)} invariant.

Example 5.32. Again, with T = 1 1 0 0 1 , we have v_{sink(T)} = [1, 1, 1, 0, 0] and

𝔖_{r}(T) = 𝔖_{{1,2,3}} × 𝔖_{{4,5}}.

Observe that 𝔖_{T} = 𝔖_{std_{1}(T)} ∩ 𝔖_{r}(T). This implies that for each σ ∈ 𝔖_{T} one has (v_{sink(T)}, std_{1}(T))σ = (v_{sink(T)}, std_{1}(T)).

Remark 5.33. In terms of spectral vectors, we have ζ_{sink(T)}σ = ζ_{sink(T)} (here we use the action defined in eq. (4.4)). The property of T to have only strictly increasing columns can also be interpreted in terms of spectral vectors. Indeed, for each i, we have

(5.5) ζ_{sink(T)}[i] ≻ ζ_{sink(T)}[i + 1] or ζ_{sink(T)}[i] = q^{n}s^{m+1} ≁ ζ_{sink(T)}[i + 1] = q^{n}s^{m}.

Example 5.34. Consider the tableau T = 1 1 0 0 1 . We compute ζ_{sink(T)} from the vector

v_{sink(T)} = [1, 1, 1, 0, 0] and the tableau std_{1}(T) = 3 2 5 4 1 . Here, r_{sink(T)} = [1, 2, 3, 4, 5],

hence ζ_{sink(T)} = [s^{2}q, q, s^{−1}q, s, 1]. Observe that ζ_{sink(T)}[1] ≻ ζ_{sink(T)}[2], ζ_{sink(T)}[2] ≁ ζ_{sink(T)}[3], with \frac{ζ_{sink(T)}[2]}{ζ_{sink(T)}[3]} = s, ζ_{sink(T)}[3] ≻ ζ_{sink(T)}[4] and ζ_{sink(T)}[4] ≁ ζ_{sink(T)}[5], with

\frac{ζ_{sink(T)}[4]}{ζ_{sink(T)}[5]} = s.

Let σ_{T} be the minimal permutation such that ζ_{rootT} σ_{T} = ζ_{sink(T)}. It can be used to characterize the group 𝔖_{T}.

Lemma 5.35. The group 𝔖_{T} is the subgroup of 𝔖_{N} consisting of the permutations σ such that ℓ(σ_{T} σ) = ℓ(σ_{T}) + ℓ(σ).

Furthermore, we will use the following result.

Lemma 5.36. For each permutation σ such that ζ_{root(T)} ⪯ ζ_{root(T)}σ one has:

P_{ζ_{root(T)}} T̃_{σ} = P_{ζ_{root(T)}σ} + ∑_{ζ′≺ζ_{root(T)}σ}(∗)P_{ζ′}.

Proof. We prove the result by induction on the length of σ. If σ = id, then the result is obvious. Now suppose that σ = σ′s_{j}, with ℓ(σ) = ℓ(σ′)+1, and so ζ_{root(T)}σ ≺ ζ_{root(T)}σs_{j}. Then T̃_{σ} = T̃_{σ′}T_{j}. Furthermore, using the induction hypothesis, we obtain

(5.6)  P_{ζ_{root(T)}} T̃_{σ} = P_{ζ_{root(T)}σ′}T_{j} + ∑_{ζ′≺ζ_{root(T)}σ}(∗)P_{ζ′}T_{j}.

44 C. F. DUNKL AND J.-G. LUQUE

But P_{ζ_{root(T)}σ′}T_{j} = P_{ζ_{root(T)}σs_{j}} + (∗)P_{ζ_{root(T)}σ}. Furthermore, since ζ′ ≺ ζ_{root(T)}σ′, we have ζ′s_{j} ≺ ζ_{root(T)}σ. But

P_{ζ′}T_{j} = (∗)P_{ζ′s_{j}} + (∗)P_{ζ′}.

Hence, by substituting this in (5.6), we arrive at the claimed result. □

We are now in the position to deduce the following auxiliary result. Lemma 5.37. Denote by β_{T}^{σ} the coefficient of P_{ζ_{sink(T)}} in P_{ζ_{root(T)}} T̃_{σ}. Then we have:

(1) If σ_{T}^{−1}σ ∉ 𝔖_{T}, then β_{T}^{σ} = 0.

(2) If σ_{T}^{−1}σ ∈ 𝔖_{T}, then β_{T}^{σ} = s^{ℓ(σ)−ℓ(σ_{T})}. Proof. Part (1) is a direct consequence of Lemma 5.36. To show Part (2), we first use Lemma 5.36 and write P_{ζ_{root(T)}} T̃_{σ_{T}} = P_{ζ_{sink(T)}} + ∑_{ζ≺ζ_{sink(T)}}(∗)P_{ζ}. Now set τ := σ_{T}^{−1}σ ∈ 𝔖_{T} and observe that, for each element τ′ ∈ 𝔖_{T}, ζτ′ = ζ_{sink(T)} implies ζ = ζ_{sink(T)}. Hence, the coefficient of ζ_{sink(T)} in ∑_{ζ≺ζ_{sink(T)}}(∗)P_{ζ} T̃_{τ} is 0. It follows that β_{T}^{σ} equals the coefficient of ζ_{sink(T)} in P_{ζ_{sink(T)}} T̃_{τ}. But 𝔖_{T} is generated by transpositions s_{i} such that ζ_{sink(T)}[i] = qs^{m+1} ≁ ζ_{sink(T)}[i + 1] = q^{n}s^{m} (see eq. (5.5)). This implies P_{ζ_{sink(T)}}s_{i} = sP_{ζ_{sink(T)}}. Hence, P_{ζ_{sink(T)}}τ = s^{ℓ(τ)}P_{ζ_{sink(T)}}. Since, from Lemma 5.35, ℓ(τ) = ℓ(σ) − ℓ(σ_{T}), we obtain the desired result. □ Proposition 5.38. The coefficient of P_{ζ_{sink(T)}} in P_{ζ_{root(T)}}S_{N} equals the Poincaré polynomial ϕ_{T} (s) of 𝔖_{T}. Proof. We write

P_{ζ_{root(T)}}S_{N} = P_{ζ_{root(T)}} ∑_{σ∈𝔖_{T}} T̃_{σ_{T}} T̃_{σ} + P_{ζ_{root(T)}} ∑_{ℓ(σ_{T} σ)<ℓ(σ_{T})+ℓ(σ)} T̃_{σ}.

By Lemma 5.37, the coefficient of P_{ζ_{sink(T)}} in

P_{ζ_{root(T)}} ∑_{ℓ(σ_{T} σ)<ℓ(σ_{T})+ℓ(σ)} T̃_{σ} = P_{ζ_{root(T)}} ∑_{σ_{T}^{−1}σ∉𝔖_{T}} T̃_{σ}

is 0. Furthermore, Lemma 5.37 implies

P_{ζ_{root(T)}} ∑_{σ∈𝔖_{T}} T̃_{σ_{T}} T̃_{σ} = P_{ζ_{sink(T)}} ∑_{σ∈𝔖_{T}} T̃_{σ} + ∑_{ζ≺ζ_{sink(T)}}(∗)P_{ζ} T̃_{σ}.

But, since ζ ≠ ζ_{sink(T)}, the coefficient of P_{ζ_{sink(T)}} in P_{ζ} T̃_{σ} is 0. Hence, the coefficient of P_{ζ_{sink(T)}} in P_{ζ_{root(T)}}S_{N} equals the coefficient of P_{ζ_{sink(T)}} in P_{ζ_{sink(T)}} ∑_{σ∈𝔖_{T}} T̃_{σ}. The result now follows from Lemma 5.37. □

The polynomial 𝔐_{T} is proportional to any P_{ζ}S_{N} for ζ ∈ T. In fact, we can compute the proportion factor. Theorem 5.39. We have

𝔐_{T} = \frac{b_{ζ_{sink(T)}}}{ϕ_{T} (s)} P_{ζ_{root(T)}}S_{N}

.

Proof. It suffices to compare the coefficient of P_{ζ_{sink(T)}} in 𝔐_{T} (given by Theorem 5.27) and in P_{ζ}S_{N} (given by Proposition 5.38). □

Example 5.40. Consider the tableau T = 1 0 0 1 . Here, ζ_{root(T)} = [s, 1, qs^{2}, qs^{−1}] and

ζ_{sink(T)} = [qs^{−1}, qs^{2}, s, 1]. The images of ζ_{root(T)} by

𝔖_{4} = {[1, 2, 3, 4], [1, 2, 4, 3], [1, 3, 2, 4], [1, 3, 4, 2], [1, 4, 2, 3], [1, 4, 3, 2],
[2, 1, 3, 4], [2, 1, 4, 3], [2, 3, 1, 4], [2, 3, 4, 1], [2, 4, 1, 3], [2, 4, 3, 1],
[3, 1, 2, 4], [3, 1, 4, 2], [3, 2, 1, 4], [3, 2, 4, 1], [3, 4, 2, 3], [3, 4, 2, 1]
[4, 1, 2, 3], [4, 1, 3, 2], [4, 2, 1, 3], [4, 2, 3, 1], [4, 3, 1, 2], [4, 3, 2, 1]}

are

\begin{matrix}[s, 1, qs^{2}, qs^{−1}], & [s, 1, qs^{−1}, qs^{2}], & [s, qs^{−1}, 1, qs^{2}], & [s, qs^{−1}, qs^{2}, 1], \\ [s, qs^{2}, 1, qs^{−1}], & [s, qs^{2}, qs^{−1}, 1], & [s, 1, qs^{2}, qs^{−1}], & [s, 1, qs^{−1}, qs^{2}], \\ [s, qs^{−1}, 1, qs^{2}], & [s, qs^{−1}, qs^{2}, 1], & [s, qs^{2}, 1, qs^{−1}], & [s, qs^{2}, qs^{−1}, 1], \\ [qs^{2}, s, 1, qs^{−1}], & [qs^{2}, s, qs^{−1}, 1], & [qs^{2}, s, 1, qs^{−1}], & [qs^{2}, s, qs^{−1}, 1], \\ [qs^{2}, qs^{−1}, s, 1], & [qs^{2}, qs^{−1}, s, 1], & [qs^{−1}, s, 1, qs^{2}], & [qs^{−1}, s, qs^{2}, 1], \\ [qs^{−1}, s, 1, qs^{2}], & [qs^{−1}, s, qs^{2}, 1], & [qs^{−1}, qs^{2}, s, 1], & [qs^{−1}, qs^{2}, s, 1], \end{matrix}

respectively. Only two permutations give ζ_{sink(T)}: [4, 3, 1, 2] and [4, 3, 2, 1]. Indeed, one computes σ_{T} by choosing a maximal path in the Yang–Baxter graph: σ_{T} = s_{2}s_{3}s_{1}s_{2}s_{1} = [4, 3, 1, 2]. The group 𝔖_{T} is the order-two group 𝔖_{T} = 𝔖_{{3,4}}. We see that acting by T_{3} on P_{[qs^{−1},qs^{2},s,1]} gives sP_{[qs^{−1},qs^{2},s,1]}. Hence,

P_{[qs^{−1},qs^{2},s,1]}(1 + T_{3}) = (1 + s)P_{[qs^{−1},qs^{2},s,1]} = ϕ_{T} (s)P_{[qs^{−1},qs^{2},s,1]}.

Note that ϕ_{T} (s) is the product of the ϕ_{λ}(s) for each row λ = [a_{1}^{m_{1}}, …, a_{k}^{m_{k}}] of T, where ϕ_{λ}(s) = ∏_{i} ϕ_{m_{1}}(s).

In the same way, we prove a similar formula for antisymmetric polynomials.

Theorem 5.41. We have

𝔐_{T}^{a} = \frac{b_{ζ_{sink(T)}}^{a}}{ϕ\overline{T} (s)} P_{ζ_{root(T)}}A_{N},

where \overline{T} denotes the conjugate of T (that is, the tableau obtained by exchanging rows and columns).

Proof. Similarly to Lemma 5.37, we denote by \overline{β}_{T}^{σ} the coefficient of P_{ζ_{sink(T)}} in P_{ζ_{root(T)}} \overline{T}_{σ}. We then have the following:

(1) If σ−1 ; \overline{T}σ ∉ 𝔖\overline{T}, then \overline{β}_{T}^{σ} = 0.

(2) If σ−1 ; \overline{T}σ ∈ 𝔖\overline{T}, then \overline{β}_{T}^{σ} = (−1)^{ℓ(σ)−ℓ(σT̃)}.

Using these properties, we prove as in Proposition 5.38 that the coefficient of P_{ζ_{sink(T)}} in P_{ζ_{root(T)}}A_{N} equals the Poincaré polynomial ϕ\overline{T} (s). The result follows. □

Example 5.42. Consider the tableau T = 1 0 0 1 . Here, ζ_{root(T)} = [s^{−1}, 1, qs, qs^{−2}] and

ζ_{sink(T)} = [qs^{−2}, qs, s^{−1}, 1]. The images of ζ_{root(T)} by 𝔖_{4} are:

\begin{matrix}[s^{−1}, 1, sq, qs^{−2}], & [s^{−1}, 1, qs^{−2}, sq], & [s^{−1}, sq, 1, qs^{−2}], & [s^{−1}, sq, qs^{−2}, 1], \\ [s^{−1}, qs^{−2}, 1, sq], & [s^{−1}, qs^{−2}, sq, 1], & [s^{−1}, 1, sq, qs^{−2}], & [s^{−1}, 1, qs^{−2}, sq], \\ [s^{−1}, sq, 1, qs^{−2}], & [s^{−1}, sq, qs^{−2}, 1], & [s^{−1}, qs^{−2}, 1, sq], & [s^{−1}, qs^{−2}, sq, 1], \\ [sq, s^{−1}, 1, qs^{−2}], & [sq, s^{−1}, qs^{−2}, 1], & [sq, s^{−1}, 1, qs^{−2}], & [sq, s^{−1}, qs^{−2}, 1], \\ [sq, qs^{−2}, s^{−1}, 1], & [sq, qs^{−2}, s^{−1}, 1], & [qs^{−2}, s^{−1}, 1, sq], & [qs^{−2}, s^{−1}, sq, 1], \\ [qs^{−2}, s^{−1}, 1, sq], & [qs^{−2}, s^{−1}, sq, 1], & [qs^{−2}, sq, s^{−1}, 1], & [qs^{−2}, sq, s^{−1}, 1]. \end{matrix}

Only two permutations give ζ_{sink(T)}: [4, 3, 1, 2] and [4, 3, 2, 1]. These permutations gen-

erate 𝔖\overline{T} with \overline{T} = 1 0 0 1 .

5.6. Minimal symmetric/antisymmetric polynomials. We have seen that for a given isotype λ the symmetric polynomials are indexed by column-strict tableaux T of shape λ. There exists only one tableau filling of λ such that the sum of its entries is minimal. This tableau is obtained by filling the first row with 0’s, the second with 1’s, etc. Let

T_{λ} := \begin{matrix}m − 1 & … & m − 1 \\ ⋮ & & ⋮ \\ 1 & … & … & 1 \\ 0 & … & … & … & 0 \end{matrix}

if λ = [λ_{1}, …, λ_{m}], with λ_{1} ≥ ⋯ ≥ λ_{m}, and the number of i’s among the entries of T_{λ} equals λ_{i}.

Example 5.43. Let λ = [5, 3, 2, 2, 1], then

T_{λ} = 4 3 3 2 2 1 1 1 0 0 0 0 0 .

Corollary 5.44. The space of the minimal symmetric polynomials for isotype λ is spanned by 𝔐_{T_{λ}}, and similarly the space of minimal antisymmetric polynomials is spanned by 𝔐_{\overline{T} \overline{λ}}^{a}, where \overline{λ} denotes the conjugate partition of λ.

Example 5.45. Consider the isotype λ = [5, 3, 2, 2, 1]. Then \overline{λ} = [5, 4, 2, 1, 1] and

T\overline{λ} = 4 3 2 2 1 1 1 1 0 0 0 0 0 .

Hence, the space of minimal antisymmetric polynomials for isotype λ is spanned by

𝔐a ; 0 0 1 0 1 0 1 2 0 1 2 3 4 .

6. Bilinear form

6.1. Bilinear form on the space V_{λ}. To define a pairing for V_{λ}, introduce the dual Hecke algebra H_{N} (q^{−1}, s^{−1}); we use ∗ to indicate objects associated with H_{N} (q^{−1,}s^{−1}), e.g., T_{i}^{∗}, (c_{0} + c_{1}s)^{∗} = c_{0} + \frac{c_{1}}{s}. Recall that, when acting on V_{λ}, T_{i} = T_{i}. There is a bilinear form on V_{λ}^{∗} × V_{λ}, (u^{∗}, v) ↦ ⟨u^{∗}, v⟩ ∈ ℚ (s, q), such that ⟨u^{∗}T_{i}^{∗}, vT_{i}⟩ = ⟨u^{∗}, v⟩ for 1 ≤ i < N, and such that 𝕋_{1}, 𝕋_{2} ∈ Tab_{λ}, 𝕋_{1} ≠ 𝕋_{2} implies ⟨𝕋_{1}^{∗}, 𝕋_{2}⟩ = 0; the latter property follows from the eigenvalues of L_{i}, since ⟨u^{∗}ϕ_{i}^{∗}, vϕ_{i}⟩ = ⟨u^{∗}, v⟩. We establish a formula for ⟨𝕋^{∗}, 𝕋⟩ after the following recurrence relation.

Lemma 6.1. If 𝕋 ∈ Tab_{λ} and m := CT_{𝕋}[i] − CT_{𝕋}[i + 1]) ≥ 2 then 𝕋^{(i,i+1)} ∈ Tab_{λ} and

⟨(𝕋^{(i,i+1)})^{∗}, 𝕋^{(i,i+1)}⟩ = \begin{matrix}(1 − s^{m−1}) & (1 − s^{m+1}) \\ (1 − s^{m}) & 2 \end{matrix} ⟨𝕋^{∗}, 𝕋⟩ .

Proof. The equation 𝕋T_{i} = 𝕋^{(i,i+1)} − \frac{1−s}{1−s^{−m}} 𝕋 implies

⟨𝕋^{∗}, 𝕋⟩ = ⟨𝕋^{(i,i+1)}^{∗}, 𝕋^{(i,i+1)}⟩ + \begin{matrix}(1 − s^{−1}) & (1 − s) \\ (1 − s^{m}) & (1 − s^{−m}) \end{matrix} ⟨𝕋^{∗}, 𝕋⟩ ,

thus

⟨𝕋^{(i,i+1)}^{∗}, 𝕋^{(i,i+1)}⟩ = (1 − \frac{s^{m−1} (1 − s)^{2}}{(1 − s^{m})^{2}} ) ⟨𝕋^{∗}, 𝕋⟩ .

□

Definition 6.2. For 𝕋 ∈ Tab_{λ} let

ν (𝕋) := _{∏_{1≤i<j≤N} ; CT_{𝕋}[i]−CT_{𝕋}[j]≤−2} (1 − s^{CT_{𝕋}[j]−CT_{𝕋}[i]−1}) (1 − s^{CT_{𝕋}[j]−CT_{𝕋}[i]+1}) ; (1 − s^{CT_{𝕋}[j]−CT_{𝕋}[i]})^{2}.

Proposition 6.3. The bilinear form defined by ⟨𝕋_{1}^{∗}, 𝕋_{2}⟩ = 0 for 𝕋_{1} ≠ 𝕋_{2} and ⟨𝕋^{∗}, 𝕋⟩ = ν (𝕋) (for 𝕋, 𝕋_{1}, 𝕋_{2}) and extended by linearity satisfies ⟨P^{∗}T_{i}^{∗}, QT_{i}⟩ = ⟨P^{∗}, Q⟩ for all P^{∗}, Q, i.

Proof. It suffices to show that ⟨𝕋^{∗}T_{i}^{∗}, 𝕋T_{i}⟩ = ⟨𝕋^{∗}, 𝕋⟩ for all 𝕋. If 𝕋T_{i} = s𝕋 then 𝕋^{∗}T_{i}^{∗} = s^{−1}𝕋^{∗} and ⟨𝕋^{∗}T_{i}^{∗}, 𝕋T_{i}⟩ = s^{−1}s ⟨𝕋^{∗}, 𝕋⟩. The case 𝕋T_{i} = −𝕋 is treated similarly. Otherwise, consider the pair (𝕋, 𝕋^{(i,i+1)}) with CT_{𝕋}[i]−CT_{𝕋}[i+1] ≥ 2. There is only one factor in ν (𝕋^{(i,i+1)}) different from ν (𝕋), the one corresponding to j = i + 1. The proof then follows from Lemma 6.1 and CT_{𝕋^{(i,i+1)}}[i] = CT_{𝕋}[i+1], CT_{𝕋^{(i,i+1)}}[i+1] = CT_{𝕋}[i]. □

Any other bilinear form satisfying ⟨P^{∗}T_{i}^{∗}, QT_{i}⟩ = ⟨P^{∗}, Q⟩ is a constant multiple of the above form.

6.2. Bilinear form on the space M_{λ}. Consider the bilinear form ⟨ , ⟩ defined by

(6.1)  ⟨𝕋_{1}^{∗}, 𝕋_{2}⟩ = δ_{𝕋_{2},𝕋_{2}}ν(𝕋_{1})

and

(6.2)  ⟨P x_{i}, Q⟩ = ⟨P, QD_{i}⟩.

This form has the following property.

Proposition 6.4. We have

⟨P (T_{i}^{∗})^{±1}, Q⟩ = ⟨P, QT_{i}^{∓1}⟩.

Proof. We proceed by induction on the degree of the polynomials. The start of the induction is given by the inner product on the tableaux. Using the induction hypothesis, we have from eq. (3.8) and Proposition 3.6

(6.3)  ⟨P x_{i}T_{i}^{∗}, Q⟩ = ⟨P x_{i}, QT_{i}^{−1}⟩,
(6.4)  ⟨P x_{i+1}(T_{i}^{∗})^{−1}, Q⟩ = ⟨P x_{i+1}, QT_{i}⟩,

and (6.5)

⟨P x_{j}(T_{i}^{∗})^{±1}, Q⟩ = ⟨P x_{j}, QT_{i}^{∓1}⟩ = ⟨P, QT_{i}^{∓1}D_{j}⟩ = ⟨P, QD_{j}T_{i}^{∓1}⟩ when |i − j| > 1.

Indeed, one has

⟨P x_{i+1}(T_{i}^{∗})^{−1}, Q⟩ = \frac{1}{s}⟨P T_{i}^{∗}^{−1}x_{i+1}, Q⟩ = \frac{1}{s}⟨P T_{i}^{∗}^{−1}, QD_{i+1}⟩ = \frac{1}{s}⟨P, QD_{i+1}T_{i}⟩,

using the induction hypothesis. Hence,

⟨P x_{i+1}(T_{i}^{∗})^{−1}, Q⟩ = ⟨P, QD_{i+1}T_{i}⟩ = ⟨P, QT_{i}^{−1}D_{i}⟩ = ⟨P x_{i}, QT_{i}^{−1}⟩

which gives (6.3). The proofs of (6.4) and (6.5) are similar.

Now, by Proposition 3.6, one has

⟨P x_{i+1}T_{i}^{∗}, Q⟩ = ⟨P T_{i}^{∗}x_{i} − (1 − \frac{1}{s})P x_{i+1}, Q⟩ = ⟨P, Q(D_{i}T_{i}^{−1} − (1 − \frac{1}{s})D_{i+1})⟩

by induction hypothesis. Hence, by (3.8), one obtains

(6.6)  ⟨P x_{i+1}T_{i}^{∗}, Q⟩ = ⟨P, QT_{i}^{−1}D_{i+1}⟩ = ⟨P x_{i+1}, QT_{i}^{−1}⟩.

Similarly, one has

(6.7)  ⟨P x_{i}T_{i}^{∗}^{−1}, Q⟩ = ⟨P x_{i}, QT_{i}⟩.

Eqs. (6.3), (6.4),(6.5), (6.6) and (6.7) give the result. □

Now one has also the following equalities involving the operator w:

D_{i+1}w^{−1} = w^{−1}D_{i}, x_{i+1}w = wx_{i+1}, i ≠ 1,

and

D_{N}w^{−1} = qw^{−1}D_{1}, x_{1}w = qwx_{N}.

This entails the following identity.

VECTOR VALUED MACDONALD POLYNOMIALS 49

Proposition 6.5. We have

⟨P w^{∗}^{±1}, Q⟩ = ⟨P, Qw^{∓1}⟩.

From Proposition 6.5 and 6.5 one deduces the following theorem. Theorem 6.6. We have

(1) ⟨P ξ_{i}^{∗}, Q⟩ = ⟨P, Qξ_{i}^{−1}⟩,

(2) ⟨P_{ζ}^{∗}, P_{ζ′}⟩ = (∗)δ_{ζ,ζ′}, where (∗) denotes a certain coefficient which remains to be computed. 6.3. Computation of ⟨P_{ζ}^{∗}, P_{ζ}⟩. First we establish some recurrences. Proposition 6.7. Let ζ = ζ_{v,𝕋} for some 𝕋 ∈ Tab_{λ} and v ∈ ℕ^{N}. Suppose that ζ[i+1) ≻ ζ[i] for some i. Then

⟨P_{ζs_{i}}^{∗}, P_{ζs_{i}}⟩ = \frac{(1 − s\frac{ζ[i+1]}{ζ[i]}) (s − \frac{ζ[i+1]}{ζ[i]})}{s (1 − \frac{ζ[i+1]}{ζ[i]})^{2}} ⟨P_{ζ}^{∗}, P_{ζ}⟩ .

Proof. From eq. (4.2), we infer P_{ζ}T_{i} = −_{\frac{1−s}{1− \frac{ζ[i+1]}{ζ[i]}}} P_{ζ} + P_{ζs_{i}}. Thus

⟨P_{ζ}^{∗}, P_{ζ}⟩ = ⟨P_{ζ}^{∗}T_{i}^{∗}, P_{ζ}T_{i}⟩

= (\frac{1 − s}{1 − \frac{ζ[i+1]}{ζ[i]}} ) (\frac{1 − s}{1 − \frac{ζ[i+1]}{ζ[i]}} )^{∗} ⟨P_{ζ}^{∗}, P_{ζ}⟩ + ⟨P_{ζs_{i}}^{∗}, P_{ζs_{i}}⟩ .

Hence

⟨P_{ζs_{i}}^{∗}, P_{ζs_{i}}⟩ = (1 − \begin{matrix}(1 − s) & (1 − s^{−1}) \\ (1 − \frac{ζ[i+1]}{ζ[i]}) & (1 − \frac{ζ[i]}{ζ[i+1]}) \end{matrix}) ⟨P_{ζ}^{∗}, P_{ζ}⟩

= \frac{(1 − s\frac{ζ[i+1]}{ζ[i]}) (s − \frac{ζ[i+1]}{ζ[i]})}{s (1 − \frac{ζ[i+1]}{ζ[i]})^{2}} ⟨P_{ζ}^{∗}, P_{ζ}⟩ .

□

Definition 6.8. We define

E_{a}(ζ) = ∏_{(i,j)∈inv(ζ)} \frac{1 − s^{a} \frac{ζ[j]}{ζ[i]}}{1 − \frac{ζ[j]}{ζ[i]}}

and

E(ζ) = E_{1}(ζ)E_{−1}(ζ).

Proposition 6.9. Let ζ = ζ_{v,𝕋} for some v ∈ ℕ^{N} and 𝕋 ∈ Tab_{λ}. One has ⟨P_{ζ}^{∗}, P_{ζ}⟩ = E (ζ)^{−1} ⟨P_{ζ^{+}}^{∗}, P_{ζ^{+}}⟩ .

Proof. we argue by induction on #inv (ζ). The statement is trivially true for #inv (ζ) = 0, that is, if ζ = ζ^{+}. Suppose the statement is true for all ζ′ = ζ_{v′,𝕋′} with #inv (ζ′) ≤ n and #inv (ζ) = n + 1. Thus ζ [i] < ζ [i + 1] for some i < N. By Proposition 6.7, we have

⟨P_{ζ}^{∗}, P_{ζ}⟩ = \begin{matrix}(1 − \frac{ζ[i+1]}{ζ[i]}) & 2 \\ (1 − s\frac{ζ[i+1]}{ζ[i]}) & (1 − s^{−1} \frac{ζ[i+1]}{ζ[i]}) \end{matrix} ⟨P_{ζs_{i}}^{∗}, P_{ζs_{i}}⟩ ,

thus

\frac{⟨P_{ζ}^{∗}, P_{ζ}⟩}{⟨P_{ζs_{i}}^{∗}, P_{ζ·s_{i}}⟩} = \frac{E (ζ · s_{i})}{E (ζ)}.

This completes the induction since #inv (ζs_{i}) = #inv (ζ) − 1. □

Alternatively, the computation of ⟨P_{ζ}^{∗}, P_{ζ}⟩ can be related to the root or the sink of the connected component of ζ.

Proposition 6.10. Let ζ = ζ_{v,𝕋} for some v and 𝕋. Let H_{T}^{q,s} be the connected component of ζ. We define the values

S(ζ) = ∏_{(i,j)∈inv_{≺}(ζ)} \frac{(1 − s\frac{ζ[j]}{ζ[i]})(1 − s^{−1} \frac{ζ[j]}{ζ[i]})}{(1 − \frac{ζ[j]}{ζ[i]})^{2}}

and

R(ζ) = ∏_{(i,j)∈inv_{≻}(ζ)} \frac{(1 − s\frac{ζ[j]}{ζ[i]})(1 − s^{−1} \frac{ζ[j]}{ζ[i]})}{(1 − \frac{ζ[j]}{ζ[i]})^{2}}.

Then one has (1) ⟨P_{ζ}^{∗}, P_{ζ}⟩ = S(ζ)^{−1} ⟨P_{ζ_{sink(T)}}^{∗}, P_{ζ_{sink(T)}}⟩,

(2) ⟨P_{ζ}^{∗}, P_{ζ}⟩ = R(ζ) ⟨P_{ζ_{root(T)}}^{∗}, P_{ζ_{root(T)}}⟩.

Proof. The proof works as in Proposition 6.9, using induction on ♯inv_{◁}(ζ) (◁ ∈ {≺, ≻ }), since there is a unique maximal (respectively minimal) element in the connected component: the sink (respectively the root). These elements are connected to ζ by a sequence of steps or jumps. □

There holds as well the following identity.

Proposition 6.11. Let ζ = ζ_{v,𝕋} for some v ∈ ℕ^{N} and 𝕋 ∈ Tab_{λ}. Then one has

⟨P_{ζΨ^{q}}^{∗}, P_{ζΨ^{q}}⟩ = (1 − qζ[1])⟨P_{ζ}^{∗}, P_{ζ}⟩.

Proof. From Proposition 4.19 one has

⟨P_{ζΨ^{q}}^{∗}, P_{ζΨ^{q}}⟩ = ⟨P_{ζ}^{∗}Φ^{∗}, P_{ζ}Φ⟩ = ⟨P_{ζ}^{∗} (T_{1}^{−1} ⋯ T_{N−1}^{−1})^{∗} x_{N}, P_{ζ}T_{1}^{−1} ⋯ T_{N−1}^{−1}x_{N}⟩.

But Proposition 6.4 implies

⟨P_{ζ}^{∗} (T_{1}^{−1} ⋯ T_{N−1}^{−1})^{∗} x_{N}, P_{ζ}T_{1}^{−1} ⋯ T_{N−1}^{−1}x_{N}⟩

= ⟨P_{ζ}^{∗}, P_{ζ}T_{1}^{−1} ⋯ T_{N−1}^{−1}x_{N}D_{N}T_{N−1} ⋯ T_{1}⟩, and, by D_{N} = (1 − ξ_{N})x_{N}^{−1}, we obtain

⟨P_{ζΨ^{q}}^{∗}, P_{ζΨ^{q}}⟩ = ⟨P_{ζ}^{∗}, P_{ζ}⟩ − ⟨P_{ζ}^{∗}, P_{ζΨ^{q}}ξ_{N}x_{N}^{−1}T_{N−1} ⋯ T_{1}⟩
= ⟨P_{ζ}^{∗}, P_{ζ}⟩ − (ζΨ^{q})[N]⟨P_{ζ}^{∗}, P_{ζΨ^{q}}Φ^{−1}⟩.

Using again Proposition 4.19, we find

⟨P_{ζΨ^{q}}^{∗}, P_{ζΨ^{q}}⟩ = (1 − (ζΨ^{q})[N])⟨P_{ζ}^{∗}, P_{ζ}⟩.

Since (ζΨ^{q})[N] = qζ[1], we arrive at the desired result. □

Definition 6.12. We let χ(i, j) = 1 if j < i and χ(i, j) = 0 if j ≥ i.

Let ρ(a, b) = \frac{(a−s^{−1}b)(a−sb)}{(a−b)^{2}} and

△(ζ) := ∏_{j=1}^{N}_{∏_{ζ[i]≻ζ[j]q^{k}} ; k≥χ(i,j)}ρ(ζ[j]q^{k}, ζ[i]).

Let □(q^{n}s^{m}) = (q; qs^{m})_{n}, where (a; q)_{n} = (1 − a)(1 − qa) ⋯ (1 − q^{n−1}a), and □(ζ) = ∏_{i=1}^{N} □(ζ[i]).

Example 6.13. Let ζ := [q^{2}s^{−1}, qs^{2}, qs, q] be the spectral vector associated with v =

[2, 1, 1, 1] and 𝕋 = 1 4 3 2 . Then we have

□(ζ) = □(q^{2}s^{−1})□(qs^{2})□(qs)□(q) = (1 − qs^{−1})(1 − q^{2}s^{−1})(1 − qs^{2})(1 − qs)(1 − q).

With the aim of computing △(ζ), we list the triples (i, j, k) such that ζ[i] ≻ ζ[j]q^{k}. Here we find 6 triples:

(1, 2, 0), (1, 2, 1), (1, 3, 0), (1, 3, 1), (1, 4, 0), (4, 2, 0).

Note that (1, 4, 1) does not occur in the list since q^{2}s^{−1} ≁ q^{2}. Furthermore, there is no factor corresponding to (4, 2, 0) in △(ζ) because χ(4, 2) = 1. Hence, △(ζ) is a product of 5 factors:

△(ζ) = ρ(qs^{2}, q^{2}s^{−1})ρ(q^{2}s^{2}, q^{2}s^{−1})ρ(qs, q^{2}s^{−1})ρ(q^{2}s, q^{2}s^{−1})ρ(q, q^{2}s^{−1})

= \begin{matrix}(q − s^{4}) & (s^{2} + 1) & (−1 + q) \\ (−s^{3} + q) & (s^{2} + 1 + s) & (q − s) \end{matrix}

With these notations, one has the following auxiliary result.

Lemma 6.14. (1) If ζ = ζ_{0^{N} ,𝕋} then △(ζ) = ν(𝕋) and □(ζ) = 1.

(2) If ζ = ζ_{v,𝕋} with ζ[ℓ + 1] ≻ ζ[ℓ] then △(ζs_{ℓ}) = ρ(ζ[ℓ], ζ[ℓ + 1])△(ζ) and □(ζs_{ℓ}) = □(ζ).

(3) If ζ = ζ_{v,𝕋} then △(ζΨ^{q}) = △(ζ) and □(ζΨ^{q}) = (1 − qζ[1])□(ζ).

Proof. (1) First note that, if ζ = ζ_{0^{N} ,𝕋}, then there is no occurrence of q in ζ, so we have □(ζ) = 1. Moreover, we have

ν(𝕋) = _{∏_{1≤i<j≤N} ; CT_{𝕋}[i]−CT_{𝕋}[j]≤−2}ρ(ζ[i], ζ[j])

= ∏_{j=1}^{N} ∏_{ζ[i]≻ζ[j]} ρ(ζ[j], ζ[i]).

(2) Obviously we have □(ζs_{ℓ}) = □(ζ). Furthermore,

\frac{△(ζs_{ℓ})}{△(ζ)} = \frac{_{∏_{ζs_{ℓ}[ℓ]≻ζs_{ℓ}[ℓ+1]q^{k}} ; k≥χ(ℓ,ℓ+1)}ρ(ζs_{ℓ}[ℓ + 1]q^{k}, ζs_{ℓ}[ℓ])}{_{∏_{ζ[ℓ+1]≻ζ[ℓ]q^{k}} ; k≥χ(ℓ+1,ℓ)}ρ(ζ[ℓ]q^{k}, ζ[ℓ + 1])}

= \frac{_{∏_{ζ[ℓ+1]≻ζ[ℓ]q^{k}} ; k≥0}ρ(ζ[ℓ]q^{k}, ζ[ℓ + 1])}{_{∏_{ζ[ℓ+1]≻ζ[ℓ]q^{k}} ; k≥1}ρ(ζ[ℓ]q^{k}, ζ[ℓ + 1])}

= ρ(ζ[ℓ], ζ[ℓ + 1]).

This proves the result.

(3) One has □(ζΨ^{q}) = (1 − (ζΨ^{q})[N])□(ζ) = (1 − qζ[1])□(ζ). Furthermore,

\frac{△(ζΨ^{q})}{△(ζ)} = ∏_{i=1}^{N−1} [\frac{_{∏_{(ζΨ^{q})[i]≻ζΨ^{q}[N]q^{k}} ; k≥0}ρ((ζΨ^{q})[N]q^{k}, (ζΨ^{q})[i])}{_{∏_{ζ[i+1]≻ζ[1]q^{k}} ; k≥1}ρ(ζ[1]q^{k}, ζ[i + 1])}

×\frac{_{∏_{(ζΨ^{q})[N]≻ζΨ^{q}[i]q^{k}} ; k≥1}ρ((ζΨ^{q})[i]q^{k}, (ζΨ^{q})[N])}{_{∏_{ζ[1]≻ζ[i+1]q^{k}} ; k≥0}ρ(ζ[i + 1]q^{k}, ζ[i])}] .

But (ζΨ^{q})[N] = qζ[1] and (ζΨ^{q})[i] = ζ[i + 1]. Hence, (ζΨ^{q})[i] ≻ (ζΨ^{q})[N]q^{k} for k ≥ 0 implies ζ[i + 1] ≥ ζ[1]q^{k+1}. In the same way (ζΨ^{q})[N] ≻ ζ[i]q^{k} for k ≥ 1

VECTOR VALUED MACDONALD POLYNOMIALS 53

implies ζ[1] ≻ ζ[i + 1]q^{k−1}. Hence, the quotient simplifies to

\frac{△(ζΨ^{q})}{△(ζ)} = 1,

as expected.

□

We deduce the following result. Theorem 6.15. Let ζ = ζ_{v,𝕋}). Then the value of the pairing ⟨P_{ζ}^{∗}, P_{ζ}⟩ is

⟨P_{ζ}^{∗}, P_{ζ}⟩ = □(ζ)△(ζ).

Proof. Comparing the statement of Lemma 6.14 to Propositions 6.7, 6.11, and 6.3, we see that ⟨P_{ζ}^{∗}, P_{ζ}⟩ and □(ζ)△(ζ) satisfy the same recurrence rules and have the same values when ζ = ζ_{0^{N} ,𝕋}. □ 6.4. Computation of ⟨𝔐_{T}^{∗} , 𝔐_{T} ⟩. First observe that

(6.8)  ⟨P, QS_{N}⟩ = ⟨P S′_{N}^{∗}, Q⟩.

We use Theorem 5.39 to write

⟨𝔐_{T}^{∗} , 𝔐_{T} ⟩ = \frac{b_{ζ_{sink(T)}}}{ϕ_{T} (s)} ⟨𝔐_{T}^{∗} , P_{ζ_{root(T)}}S_{N}⟩.

Hence, by eq. (6.8), we have

⟨𝔐_{T}^{∗} , 𝔐_{T} ⟩ = \frac{b_{ζ_{sink(T)}}}{ϕ_{T} (s)} ⟨𝔐_{T}^{∗} S′_{N}^{∗}, P_{ζ_{root(T)}}⟩.

Since 𝔐_{T} is symmetric, eq. (5.3) gives

⟨𝔐_{T}^{∗} , 𝔐_{T} ⟩ = b_{ζ_{sink(T)}} \frac{ϕ_{N}(s)}{ϕ_{T} (s)}⟨𝔐_{T}^{∗} , P_{ζ_{root(T)}}⟩.

Hence,

⟨𝔐_{T}^{∗} , 𝔐_{T} ⟩ = \frac{ϕ_{N}(s)}{ϕ_{T} (s)}b_{ζ_{sink(T)}}b_{ζ_{root(T)}}^{∗}⟨P_{ζ_{root(T)}}^{∗}, P_{ζ_{root(T)}}⟩.

Using the normalization described in Section 5.4, we have b_{ζ_{root(T)}} = 1. Theorem 6.16. We have

⟨𝔐_{T}^{∗} , 𝔐_{T} ⟩ = \frac{ϕ_{N}(s)}{ϕ_{T} (s)}b_{ζ_{sink(T)}}⟨P_{ζ_{root(T)}}^{∗}, P_{ζ_{root(T)}}⟩.

In the same way, for antisymmetric polynomials we obtain the following result. Theorem 6.17. We have

⟨𝔐_{T}^{a}^{∗}, 𝔐_{T}^{a} ⟩ = \frac{ϕ_{N}(s)}{ϕ\overline{T} (s)}b_{ζ_{sink(T)}}^{a}⟨P_{ζ_{root(T)}}^{∗}, P_{ζ_{root(T)}}⟩.

Proof. The proof works as in the symmetric case, but it uses the operator A′_{N} with the property that

⟨P, QA_{N}⟩ = ⟨P A′_{N}^{∗}, Q⟩.

This operator is the antisymmetrizer

A′_{N} = ∑_{σ∈𝔖_{N}}(−s)^{ℓ(T)}T̃_{σ}

satisfying

A′_{N}^{2} = ϕ_{N}(\frac{1}{s})A′_{N}.

Hence, by a similar reasoning we arrive at the claimed result. □

6.5. Hook-length type formula for minimal polynomials. The topic of this section is simpler formulae for ⟨𝔐_{T_{λ}}^{∗}, 𝔐_{T_{λ}}⟩ for a decreasing partition λ in the situation where the entries of T are constant in each row. The formulae are then specialized to the minimal symmetric/antisymmetric polynomials. In this case they are expressions in terms of hook-lengths.

First consider a partition μ where μ = [μ_{1}^{λ[m]}, …, μ_{m}^{λ[1]}] with μ_{1} > ⋯ > μ_{m}. Let

𝕋 = \begin{matrix}λ[m] & … & 1 \\ λ[m − 1] + λ[m] & … & … & λ[m] + 1 \\ ⋮ & & & ⋮ \\ λ[1] + ⋯ + λ[m] & … & … & … & λ[2] + ⋯ + λ[m] + 1 \end{matrix}

be the RST obtained by filling the shape λ with 1, …, N(= λ[1] + ⋯ + λ[N]) row by row and

T = \begin{matrix}μ_{1} & … & μ_{1} \\ ⋮ & & ⋮ \\ μ_{m−1} & … & … & μ_{m−1} \\ μ_{m} & … & … & … & μ_{m} \end{matrix}

be the column strict tableau obtained by filling the shape λ with the entries of μ row by row. Then μ = v_{sink(T)} and 𝕋 = 𝕋_{sink(T)}. Hence,

(6.9) ζ_{sink(T)} = [q^{μ_{1}}s^{λ[m]−m}, …, q^{μ_{1}}s^{1−m}, q^{μ_{2}}s^{1−m+λ[m−1]}, …, q^{μ_{2}}s^{2−m}, …,

q^{μ_{m}}s^{−1+λ[1]}, …, q^{μ_{m}}].

Example 6.18. Let λ = [3, 3, 2] and μ = [3, 3, 2, 2, 2, 1, 1, 1]. We construct

𝕋 = 2 1 5 4 3 8 7 6

and

T = 3 3 2 2 2 1 1 1 .

Here, ζ_{sink(T)} = [q^{3}s^{−1}, q^{3}s^{−2}, q^{2}s, q^{2}, q^{2}s^{−1}, qs^{2}, qs, q].

VECTOR VALUED MACDONALD POLYNOMIALS 55

We have

⟨P_{ζ_{root(T)}}^{∗}, P_{ζ_{root(T)}}^{∗}⟩ = S(ζ_{root(T)})^{−1}⟨P_{ζ_{sink(T)}}^{∗}, P_{ζ_{sink(T)}}⟩
= S(ζ_{root(T)})^{−1}△(ζ_{sink(T)})□(ζ_{sink(T)}),

where (6.10) ζ_{root(T)} = [q^{μ_{m}}s^{−1+λ[1]}, …, q^{μ_{m}}, …, q^{μ_{2}}s^{1−m+λ[m−1]}, …, q^{μ_{2}}s^{2−m},

q^{μ_{1}}s^{λ[m]−m}, …, q^{μ_{1}}s^{1−m}].

By telescoping, we find

(6.11)  S(ζ_{root(T)}) = ∏_{1≤i<j≤m} \frac{(1 − q^{μ_{j}−μ_{i}}s^{j−i−λ[m−i+1]})(1 − q^{μ_{j}−μ_{i}}s^{j−i+λ[m−j+1]})}{(1 − q^{μ_{j}−μ_{i}}s^{j−i})(1 − q^{μ_{j}−μ_{i}}s^{j−i+λ[m−j+1]−λ[m−i+1]})}.

First we compute △(ζ_{sink(T)}), and following eq. (6.9) we write

△(ζ_{sink(T)}) = ⟨𝕋^{∗}, 𝕋⟩♢,

with

♢ := ∏_{1≤i<j≤m} ∏_{k=0}^{μ_{i}−μ_{j}−1} ∏_{a=1}^{λ[m−i+1]} ∏_{b=1}^{λ[m−j+1]} \frac{(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+b−a−1})(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+b−a+1})}{(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+b−a})^{2}}.

Indeed, ⟨P_{ζ_{sink(T)}}^{∗}, P_{ζ_{sink(T)}}⟩ splits into two factors: the first factor ⟨𝕋^{∗}, 𝕋⟩ does not depend on q, all the factors of the second factor ♢□(ζ_{sink(T)}) involve q. By telescoping, we have

∏_{b=1}^{λ[m−j+1]} \frac{(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+b−a−1})(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+b−a+1})}{(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+b−a})^{2}} (6.12)

= \frac{(1 − q^{μ_{j}−μ_{i}+k}s^{j−i−a})(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+λ[m−j+1]−a+1})}{(1 − q^{μ_{j}−μ_{i}+k}s^{j−i−a+1})(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+λ[m−j+1]−a})},

∏_{a=1}^{λ[m−i+1]} \frac{1 − q^{μ_{j}−μ_{i}+k}s^{j−i−a}}{1 − q^{μ_{j}−μ_{i}+k}s^{j−i−a+1}} = \frac{1 − q^{μ_{j}−μ_{i}}s^{j−i−λ[m−i+1]}}{1 − q^{μ_{j}−μ_{i}}s^{j−i}} , (6.13)

and

(6.14) ∏_{a=1}^{λ[m−i+1]} \frac{1 − q^{μ_{j}−μ_{i}+k}s^{j−i+λ[m−j+1]−a+1}}{1 − q^{μ_{j}−μ_{i}+k}s^{j−i+λ[m−j+1]−a}} = \frac{1 − q^{μ_{j}−μ_{i}+k}s^{j−i+λ[m−j+1]}}{1 − q^{μ_{j}−μ_{i}+k}s^{j−i+λ[m−j+1]−λ[m−i+1]}} .

So, equations (6.12), (6.13), and (6.14) give

♢ = ∏_{1≤i<j≤m} ∏_{k=0}^{μ_{i}−μ_{j}−1} \frac{(1 − q^{μ_{j}−μ_{i}+k}s^{j−i−λ[m−i+1]})(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+λ[m−j+1]})}{(1 − q^{μ_{j}−μ_{i}+k}s^{j−i})(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+λ[m−j+1]−λ[m−i+1]})}.

Note that, by eq. (6.11), we have

\frac{♢}{S(ζ_{root(T)})} = ∏_{1≤i<j≤m} ∏_{k=1}^{μ_{i}−μ_{j}−1} \frac{(1 − q^{μ_{j}−μ_{i}+k}s^{j−i−λ[m−i+1]})(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+λ[m−j+1]})}{(1 − q^{μ_{j}−μ_{i}+k}s^{j−i})(1 − q^{μ_{j}−μ_{i}+k}s^{j−i+λ[m−j+1]−λ[m−i+1]})}

= ∏_{1≤i<j≤m} \frac{(qs^{i−j+λ[m−i+1]}; q)_{μ_{i}−μ_{j}−1}(qs^{i−j−λ[m−j+1]}; q)_{μ_{i}−μ_{j}−1}}{(qs^{i−j+λ[m−i+1]−λ[m−j+1]}, q)_{μ_{i}−μ_{j}−1}(qs^{i−j}, q)_{μ_{i}−μ_{j}−1}} .

Furthermore,

□(ζ_{sink(T)}) = ∏_{i=1}^{m} ∏_{j=1}^{λ[m−i+1]}(qs^{j−m+i−1}; q)_{μ_{i}}.

Hence,

(6.15) ⟨P_{ζ_{root(T)}}^{∗}, P_{ζ_{root(T)}}⟩ = ⟨𝕋^{∗}, 𝕋⟩ ∏_{i=1}^{m} ∏_{j=1}^{λ[m−i+1]}(qs^{j−m+i−1}; q)_{μ_{i}}

× ∏_{1≤i<j≤m} \frac{(qs^{i−j+λ[m−i+1]}; q)_{μ_{i}−μ_{j}−1}(qs^{i−j−λ[m−j+1]}; q)_{μ_{i}−μ_{j}−1}}{(qs^{i−j+λ[m−i+1]−λ[m−j+1]}, q)_{μ_{i}−μ_{j}−1}(qs^{i−j}, q)_{μ_{i}−μ_{j}−1}} .

We find also

b_{ζ_{sink(T)}} = ∏_{1≤i<j≤m} ∏_{a=1}^{λ[m−i+1]} \frac{1 − q^{μ_{j}−μ_{i}}s^{j−i+1−a}}{1 − q^{μ_{j}−μ_{i}}s^{λ[m−j+1]−i+j+1−a}} .

Now we specialize μ = m − i. The tableau T then becomes

T = \begin{matrix}m − 1 & … & m − 1 \\ ⋮ & & ⋮ \\ 1 & … & … & 1 \\ 0 & … & … & … & 0 \end{matrix} .

For convenience, consider the normalization 𝔐̃_{T} := b_{ζ_{sink(T)}}^{−1}𝔐_{T} .

Furthermore, we set ∇_{λ} := \frac{ϕ_{T} (s)}{ϕ_{N} (s)} \frac{⟨𝔐̃_{T}^{∗} ,𝔐̃_{T} ⟩}{⟨𝕋^{∗},𝕋⟩}. So, we have

∇_{λ} = (b_{sink(T)}^{−1})^{∗} \frac{⟨P_{ζ_{root(T)}}^{∗}, P_{ζ_{root(T)}}}{⟨𝕋^{∗}, 𝕋⟩} ⟩.

From equality (6.15), we obtain

(6.16) ∇_{λ} = ∏_{i=1}^{m} ∏_{j=1}^{λ[m−i+1]}(qs^{j−m+i−1}; q)_{i−1}

× ∏_{1≤i<j≤m} \frac{(qs^{i−j+λ[m−i+1]}; q)_{j−i−1}(qs^{i−j−λ[m−j+1]}; q)_{j−i−1}}{(qs^{i−j+λ[m−i+1]−λ[m−j+1]}, q)_{j−i−1}(qs^{i−j}, q)_{j−i−1}}

× ∏_{a=1}^{λ[m−i+1]} \frac{1 − q^{j−i}s^{i−j+a−λ[m−j+1]−1}}{1 − q^{j−i}s^{i−j+a−1}} .

Note that this formula remains valid when λ[m] = 0:

∇_{[λ[1],...,λ[m−1],0]} := ∇_{[λ[1],...,λ[m−1]]}.

Let λ′ = [λ[1], λ[2], …, λ[m − 1], λ[m] − 1] be the partition obtained from λ by subtracting 1 from its last part. We denote by T′ and 𝕋′ the associated tableaux. Example 6.19. For instance, if λ = [6, 3, 2] then

T = 2 2 1 1 1 0 0 0 0 0 0 and 𝕋 = 2 1 5 4 3 11 10 9 8 7 6 .

In this case λ′ = [6, 3, 1] and

T = 2 1 1 1 0 0 0 0 0 0 and 𝕋 = 1 4 3 2 10 9 8 7 6 5

One has

(6.17)  \frac{∇_{λ}}{∇_{λ′}} = (qs^{λ[m]−m}; q)_{m−1}

× ∏_{j=2}^{m} [\frac{(qs^{1−j+λ[m]}; q)_{j−2}(qs^{λ[m]−λ[m−j+1]−j}; q)_{j−2}}{(qs^{λ[m]−j}; q)_{j−2}(qs^{1−j+λ[m]−λ[m−j+1]}; q)_{j−2}}

× \frac{(1 − q^{j−1}s^{λ[m]−λ[m−j+1]−j})}{(1 − q^{j−1}s^{λ[m]−j})}]

= (qs^{λ[m]−m}; q)_{m−1}

× ∏_{j=2}^{m} [\frac{(qs^{λ[m]−j+1}; q)_{j−2}(qs^{λ[m]−λ[m−j+1]−j}; q)_{j−1}}{(qs^{λ[m]−j}; q)_{j−1}(qs^{1−j+λ[m]−λ[m−j+1]}; q)_{j−2}} ] .

Observing that

∏_{j=2}^{m} \frac{(qs^{λ[m]−j+1}; q)_{j−2}}{(qs^{λ[m]−j}; q)_{j−1}} = \frac{1}{(qs^{λ[m]−m}; q)_{m−1}} ,

58 C. F. DUNKL AND J.-G. LUQUE

we see that eq. (6.17) gives

(6.18)  \frac{∇_{λ}}{∇_{λ′}} = ∏_{j=2}^{m} \frac{(qs^{λ[m]−λ[m−j+1]−j}; q)_{j−1}}{(qs^{1−j+λ[m]−λ[m−j+1]}; q)_{j−2}}

= ∏_{i=1}^{m−1} \frac{(qs^{λ[m]−λ[i]+i−m−1}; q)_{m−i}}{(qs^{λ[m]−λ[i]+i−m}; q)_{m−i−1}} .

As usual, we define the arm, leg, and hook lengths of a node (x, y) ∈ λ by

� _{λ}[x, y] = λ[y] − x, � _{λ}[x, y] = \overline{λ}[x] − y and � _{λ}[x, y] = � _{λ}[x, y] + � _{λ}[x, y] + 1, where \overline{λ} is the conjugate of λ. Remark 6.20. Note that we use French notation for Ferrers diagrams. For instance, the Ferrers diagram λ = [4, 2, 1] is

3 □
2 □ □
1 □ □ □ □
y/x 1 2 3 4

The coordinates of the node × in the diagram

\begin{matrix}□ \\ □ & □ \\ □ & × & □ & □ \end{matrix}

are [2, 1]. We have

� _{λ}[2, 1] = λ[2] − 2 = 2, � _{λ}[2, 1] = \overline{λ}[1] − 1 = 1 and � _{λ}[2, 1] = 4.

Graphically, the values of � _{λ}[2, 1] (respectively of � _{λ}[2, 1], or of � _{λ}[2, 1]) are obtained by counting the numbers of � (respectively of � , or of symbols {� , � , ×}) in the following diagram:

\begin{matrix}□ \\ □ & � \\ □ & × & � & � \end{matrix}

Let

H_{λ} := ∏_{y=1}^{ℓ(λ)−1} ∏_{x=1}^{λ[i]}(qs^{−� _{λ}[x,y]}; q)_{� _{λ}[x,y]}.

The changes from H_{λ′} to H_{λ} come from the node {(λ[m], y) : 1 ≤ i ≤ m − 1}; each hook-length and each leg-length increases by 1, thus

(6.19)  \frac{H_{λ}}{H_{λ′}} = ∏_{i=1}^{m−1} \frac{(s^{λ[m]−λ[i]+i−m−1}; q)_{m−i}}{(qs^{λ[m]−λ[i]+i−m}; q)_{m−i−1}} .

Hence,

(6.20)  \frac{∇_{λ}}{∇_{λ′}} = \frac{H_{λ}}{H_{λ′}} .

Using eq. (6.20) we obtain

(6.21)  H_{λ} = ∇_{λ}.

It remains to compute ⟨𝕋^{∗}, 𝕋⟩. We start from

⟨𝕋^{∗}, 𝕋⟩ = _{∏_{1≤i<j≤N} ; CT_{𝕋}[i]−CT_{𝕋}[j]≤−2} \frac{(1 − s^{CT_{𝕋}[i]−CT_{𝕋}[j]−1})(1 − s^{CT_{𝕋}[i]−CT_{𝕋}[j]+1})}{(1 − s^{CT_{𝕋}[i]−CT_{𝕋}[j]})^{2}},

and we analyze this product in terms of nodes:

(6.22) ⟨𝕋^{∗}, 𝕋⟩ = ∏_{(x,y)∈λ}_{∏_{1≤t≤\overline{λ}[x]−y,1≤z≤λ[y]} ; (x−y−t)−(z−t)≤−2} \frac{(1 − s^{(x−y−t)−(z−t)+1})(1 − s^{(x−y−t)−(z−t)+1})}{(1 − s^{(x−y−t)−(z−t)})^{2}}.

Indeed, consider the set I_{λ} of the pairs [(x, y), (z, t)] of nodes verifying 𝕋[x, y] < 𝕋[z, t] and (x − y) ≤ z − t − 2. This set splits into N disjoint (possibly empty) sets:

E_{(x,y)} := \{[(x, y + t), (z, y)] : 1 ≤ t ≤ \overline{λ}[x] − y, 1 ≤ z ≤ λ[y], (x − y − t) − (z − t) ≤ −2}.

Example 6.21. Consider the partition λ = [3, 2]. In this case,

𝕋 = 2 1 5 4 3 with contents \begin{matrix}−1 & 0 \\ 0 & 1 & 2 \end{matrix} .

Consequently,

I_{λ} = {[(2, 2), (3, 1)], [(1, 2), (3, 1)], [(1, 2), (2, 1)]},
E_{(1,1)} = {[(1, 2), (2, 1)], [(1, 2), (3, 1)]},
E_{(2,1)} = {[(2, 2), (3, 1)]}

and

E_{(3,1)} = E_{(1,2)} = E_{(2,2)} = ∅.

Hence,

⟨𝕋^{∗}, 𝕋⟩ = ∏_{[(x,y),(z,t)]∈I_{λ}} \frac{(1 − s^{x−y−z+t−1})(1 − s^{x−y−z+t+1})}{(1 − s^{x−y−z+t})^{2}}

= ∏_{(x,y)∈λ} ∏_{[(z_{1},t_{1}),(z_{2},t_{2})]∈E_{(x,y)}} \frac{(1 − s^{z_{1}−t_{1}+t_{2}−z_{2}−1})(1 − s^{z_{1}−t_{1}+t_{2}−z_{2}+1})}{(1 − s^{z_{1}−t_{1}+t_{2}−z_{2}})^{2}},

and we arrive at (6.22).

Let us compute the products

E_{(x,y)} := ∏_{[(z_{1},t_{1}),(z_{2},t_{2})]∈E_{(x,y)}} \frac{(1 − s^{z_{1}−t_{1}+t_{2}−z_{2}−1})(1 − s^{z_{1}−t_{1}+t_{2}−z_{2}+1})}{(1 − s^{z_{1}−t_{1}+t_{2}−z_{2}})^{2}}.

Observe that, if [(x, y + t), (z, y)] ∈ E_{(x,y)}, then t and z have bounds 1 ≤ z ≤ λ[y] and 1 ≤ t ≤ \overline{λ}[x] − y, z + t − x − 2 ≥ 0. Hence,

E_{(x,y)} := ∏_{t=1}^{\overline{λ}[x]−y} ∏_{z=max\{1,x+2−t\}}^{λ[y]} \frac{(1 − s^{x−t−z+1})(1 − s^{x−t−z−1})}{(1 − s^{x−t−z})^{2}}.

By telescoping, we find

(6.23)  E_{(x,y)} = ∏_{t=1}^{\overline{λ}[x]−y} \frac{(1 − s^{max\{1,x+2−t\}−x+t−1})(1 − s^{λ[y]−x+t+1})}{(1 − s^{max\{1,x+2−t\}−x+t})(1 − s^{λ[y]−x+t})}.

We also find

(6.24)  ∏_{t=1}^{\overline{λ}[x]−y} \frac{(1 − s^{λ[y]−x+t+1})}{(1 − s^{λ[y]−x+t})} = \frac{1 − s^{λ[y]−x+\overline{λ}[x]−y+1}}{1 − s^{λ[y]−x+1}} = \frac{1 − s^{� _{λ}[x,y]}}{1 − s^{� _{λ}[x,y]}} .

But, if \overline{λ}[x] − y ≤ x, then max\{1, x + 2 − t\} = x + 2 − t, for 1 ≤ t ≤ \overline{λ}[x] − y, and

(6.25)  ∏_{t=1}^{\overline{λ}[x]−y} \frac{(1 − s^{max\{1,x+2−t\}−x+t−1})}{(1 − s^{max\{1,y+2−t\}−x+t})} = (\frac{1 − s}{1 − s^{2}} )^{\overline{λ}[x]−y} .

If \overline{λ}[x] − y > x, then we use telescoping to show that

(6.26)  ∏_{t=1}^{\overline{λ}[x]−y} \frac{(1 − s^{max\{1,x+2−t\}−x+t−1})}{(1 − s^{max\{1,y+2−t\}−x+t})} = (\frac{1 − s}{1 − s^{2}} )^{x} \frac{1 − s}{1 − s^{\overline{λ}[x]−a−b+1}} .

Eqs. (6.25) and (6.26) give (6.27)

∏_{t=1}^{\overline{λ}[x]−y} \frac{(1 − s^{max\{1,x+2−t\}−x+t−1})}{(1 − s^{max\{1,y+2−t\}−x+t})} = (\frac{1 − s}{1 − s^{2}} )^{min\{x,� _{λ}[x,y]\}} \frac{1 − s}{1 − s^{max\{1,� _{λ}[x,y]−x+1\}}} .

Hence, by (6.24) and (6.27), we obtain (6.28)

E_{(x,y)} = (\frac{1 − s}{1 − s^{2}} )^{min\{x,� _{λ}[x,y]\}} \frac{(1 − s) (1 − s^{� _{λ}[x,y]})}{(1 − s^{max\{1,� _{λ}[x,y]−x+1\}}) (1 − s^{� _{λ}[x,y]})}.

Finally, eqs. (6.21), (6.22), and (6.28) lead to the following result.

Theorem 6.22. We have

⟨𝔐̃_{T}^{∗} , 𝔐̃_{T} ⟩ = ∏_{(x,y)∈λ} [(\frac{1 − s}{1 − s^{2}} )^{min\{x,� _{λ}[x,y]\}}

× \frac{(1 − s)(−s)^{� _{λ}[x,y]} (s^{−� _{λ}[x,y]}; q)_{� _{λ}[x,y]+1}}{(1 − s^{max\{1,� _{λ}[x,y]−x+1\}}) (1 − s^{� _{λ}[x,y]})}] .

For a rational expression f (s) let ιf (s) = f (s^{−1}). Here are some immediate consequences:

ιν (𝕋) = ν (𝕋) ,
CT\overline{𝕋} (i) = −CT_{𝕋} (i) , 1 ≤ i ≤ N,
ζ_{v,\overline{𝕋}} = q^{v[i]}s^{CT\overline{𝕋}(i)} = q^{v[i]}s^{−CT_{𝕋}(i)} = ιζ_{v,𝕋}.

If 𝕋_{1}, 𝕋_{2} ∈ Tab_{λ} then

\frac{ν (𝕋_{1})}{ν (𝕋_{2})} = \frac{ν (\overline{𝕋_{2}})}{ν (\overline{𝕋_{1}})}.

If ρ_{q} (m, n) = \begin{matrix}(qs^{n−1}; q) & m & (qs^{n+1}; q) & m \\ (qs^{n}; q) & {2 \atop m} \end{matrix} then ιρ_{q} (m, n) = ρ_{q} (m, −n). Using this in the

formula for ⟨P_{v,𝕋}^{∗}, P_{v,𝕋}⟩, we obtain

ι (\frac{⟨P_{v,𝕋}^{∗}, P_{v,𝕋}⟩}{ν (𝕋)}) = \frac{⟨P_{v,\overline{𝕋}}^{∗}, P_{v,\overline{𝕋}}⟩}{ν (\overline{𝕋})}.

Now suppose that λ is a partition of N, and 𝕋, T are the tableaux corresponding to the minimal antisymmetric polynomial.

Example 6.23. If λ = (3, 2), then

T = 0 1 0 1 2 , 𝕋 = 4 2 5 3 1 .

As for symmetric polynomials, we set

𝔐̃_{T}^{a} = (b_{ζ}^{a})^{−1} 𝔐_{T}^{a} .

Our formulae show that

ι (\frac{⟨𝔐̃_{T}^{a∗}, 𝔐̃_{T}^{a} ⟩}{ν (𝕋)}) = \frac{⟨𝔐̃∗ ; \overline{T} , 𝔐̃\overline{T} ⟩}{ν (\overline{𝕋})}

= \frac{ϕ_{N} (s)}{∏_{i=1}^{ℓ(\overline{λ})} ϕ_{\overline{λ}[i]} (s)} ∏_{(i,j)∈\overline{λ}} (qs^{−� \overline{λ}[i,j]}; q)_{� \overline{λ}[i,j]}

= \frac{ϕ_{N} (s)}{∏_{i=1}^{ℓ(\overline{λ})} ϕ_{\overline{λ}[i]} (s)} ∏_{(j,i)∈λ} (qs^{−� _{λ}[j,i]}; q)_{� _{λ}[j,i]} .

This leads to the following theorem.

Theorem 6.24. We have

⟨𝔐̃_{T}^{a∗}, 𝔐̃_{T}^{a} ⟩ = ν (𝕋) \frac{ϕ_{N} (s^{−1})}{∏_{i=1}^{λ[1]} ϕ_{λ′[i]} (s^{−1})} ∏_{(i,j)∈λ} (qs^{� _{λ}[i,j]}; q)_{� _{λ}[i,j]} .

Example 6.25. For λ = (3, 2), we have

⟨𝔐̃_{T}^{a∗}, 𝔐̃_{T}^{a} ⟩ = \frac{ϕ_{5} (s^{−1})}{ϕ_{2} (s^{−1})^{2}} (qs^{4}; q)_{2} (qs^{3}; q)_{1} (qs^{2}; q)_{1}

= s^{−8} \frac{ϕ_{5} (s)}{ϕ_{2} (s)^{2}} (1 − qs^{4}) (1 − q^{2}s^{4}) (1 − qs^{3}) (1 − qs^{2}) .

Note that ν(𝕋) does not always equal 1. For instance, we have

ν ( \begin{matrix}6 \\ 7 \\ 8 & 4 & 2 \\ 9 & 5 & 3 & 1 \end{matrix} ) = \frac{1 + s^{2}}{(1 + s)^{2}} .

7. Conclusion

Throughout this paper, we have constructed and analyzed a Macdonald-type structure for vector valued polynomials, that is, polynomials whose coefficients belong to an irreducible module of the Hecke algebra. The “classical” Macdonald polynomials are recovered for the trivial representation and then correspond to the shapes λ = (n), n ∈ ℕ. Thanks to the Yang–Baxter graph we have found algorithms and some explicit formulae for computing the Macdonald polynomials, their (anti)symmetrizations, their scalar products, etc., and we have given graphical interpretations of these properties.

We remark that almost everything works as for vector valued Jack polynomials [6], and that the Jack polynomials are recovered as a limit case of Macdonald polynomials, as expected (setting q = s^{α} and sending s to 1).

It remains to consider some constructions that could illuminate this theory. For instance, the shifted Macdonald polynomials could be defined by slightly changing the raising operators. For the trivial representation, shifted Macdonald polynomials are easier to manipulate than the homogeneous ones since they can be defined by vanishing properties [10, 11]. We have seen in [6], that this is no longer the case for shifted vector valued Jack polynomials for a generic irreducible module. But this research is not yet complete, and we speculate that the vanishing properties arise when considering some polynomial representations of the Hecke algebra.

Comparing the results in [5] and [8], we find similarities between the concepts of singular non-symmetric Macdonald polynomials and highest weight symmetric Macdonald polynomials. We hope that this similarity extends to vector valued polynomials. In this context, minimal symmetric polynomials should play a special role and, perhaps, provide applications to the study of the fractional quantum Hall effect. The fractional quantum Hall effect is a state of matter with elusive physical properties whose theoretical study was pioneered by Laughlin based on wave functions describing the many-body state of the interacting electrons. Some of these wave functions (called Read–Rezayi states; see [13]) are multivariate symmetric polynomials with special vanishing properties, and it was shown, combining minimality of the polynomials for the vanishing properties and results of [7], that they are Jack polynomials for a specialization of the parameter α (see e.g. [2]). It would be interesting to know if we can identify other relevant wave functions from vector valued Jack or Macdonald polynomials.

References

[1] T. H. Baker and P. J. Forrester, A q-analogue of the type A Dunkl operator and integral kernel, Int. Math. Res. Notices 14 (1997), 667–686.

[2] B. A. Bernevig and F. D. M. Haldane, Fractional quantum Hall states and Jack polynomials, Phys. Rev. Lett. 100, 246802 (2008), 4 pages.

[3] R. Dipper and G. James, Representations of Hecke algebras of general linear groups, Proc. London Math. Soc. (3) 52 (1986), 2–52.

[4] R. Dipper and G. James, Blocks and idempotents of Hecke algebras of general linear groups, Proc. London Math. Soc. (3) 54 (1987), 57–82.

[5] C. F. Dunkl, Singular polynomials and modules for the symmetric groups, Int. Math. Res. Not. 39 (2005), 2409–2436.

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[7] B. Feigin, M. Jimbo, T. Miwa and E. Mukhin, A differential ideal of symmetric polynomials spanned by Jack polynomials at β = −\frac{r−1}{k+1}, Int. Math. Res. Notices 23 (2002), 1223–1237.

[8] T. Jolicoeur and J.-G. Luque, Highest weight Macdonald and Jack Polynomials, J. Phys. A: Math. Theor. 44 (2011), 055204, 21 pages.

[9] J. Katriel, B. Abdesselam and A. Chakrabarti, The character table of the Hecke algebra H_{n}(q) in terms of traces of products of Murphy operators, arχiv:q-alg/9502018.

[10] A. Lascoux,Yang–Baxter graphs, Jack and Macdonald polynomials, Ann. Combin. 5 (2001), 397– [11] A. Lascoux, Schubert and Macdonald polynomials, a parallel preprint, available at

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[13] N. Read and E. H. Rezahi, Beyond paired quantum Hall states: parafermions and incompressible states in the first excited Landau level, Phys.Rev. B 59 (1999) 8084–8092.

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Index

� _{λ}[x, y] arm, 58

□(q^{n}s^{m}) = (q, qs^{m})_{n}, 51 □(ζ), 51 b_{ζ}, 40

χ(i, j) = 1 if j < i and χ(i, j) = 0 if j ≥ i, 51 ℭ_{i} a creation operator, 7 COL_{𝕋} the vector of columns of 𝕋, 4 CT_{𝕋} the vector of contents of 𝕋, 4

δ_{i}^{x} := T_{i}^{x} − s · s_{i}^{x}, 12 D_{i} a Dunkl operator, 3 ♢, 55 ∂_{i} divided difference, 2

E_{a}(ζ), 49 E(ζ), 49

F_{N} = 1 − ξ_{N}, 16

G_{λ} Yang–Baxter graph associated to λ, 20 G_{𝕋} a subgraph of G_{λ}, 22

H_{N}(q, s) double affine Hecke algebra, 3 H_{λ} the graph obtained from G_{λ} by removing

the affine edges, all the falls, and the vertex ∅, 31

� _{λ}[x, y] hook length, 58 H_{T} a connected component of H_{λ}, 31

\overline{λ} conjugate of λ, 46 ⟨ , ⟩ bilinear form, 48

� _{λ}[x, y] leg, 58 ◁ : x^{v,𝕋} ◁ x^{v′,𝕋′} if and only if v ◁ v′ , 28 L_{i} Murphy elements, 4

M_{λ} the space of vector valued polynomials for

the isotype λ , 12 𝔐_{T} symmetric Macdonald polynomial, 41 𝔐_{T}^{a} antisymmetric Macdonald polynomial, 42

∇_{λ}, 56 ν (𝕋), 47

Φ_{q} := T_{1}^{−1} ⋯ T_{N−1}^{−1}x_{N}, 3 ϕ_{i} := s^{i−N}T_{i}T_{i+1} ⋯ T_{N−1}T_{N−1} ⋯ T_{i}, 4 Φ = T_{1}^{−1} ⋯ T_{N−1}^{−1}x_{N}, 22 ϕ̃_{i} = s^{N−i}ϕ_{i}, 8 ϕ_{T} (s) Poincaré polynomial of 𝔖_{T}, 44 π_{i} isobaric divided difference, 2 \overline{π}_{i}, 2 P_{𝔓} Macdonald polynomial associated to a

path in G_{λ}, 23 ≺ dominance order on partitions, 18 P_{v,𝕋} Macdonald polynomial associated to a

pair (v, 𝕋), 25 P_{ζ} Macdonald polynomial associated to the

spectral vector ζ, 25

(a; q) = (1 − a)(1 − qa) ⋯ (1 − q^{n−1}a), 51

r_{v} rank function of v, 9

R(ζ), 50

ρ(a, b) = \frac{(a−s^{−1}b)(a−sb)}{(a−b)^{2}}, 51 root(T) root of H_{T}, 31 ROW_{𝕋} the vector of rows of 𝕋, 4 R_{v} a Hecke element associated to a

multi-index v, 9

S := T_{1} ⋯ T_{N−1}, 5

S(ζ), 50 s_{i}, 12 sink(T) sink of H_{T}, 31 s_{i}^{x}, 12 𝔖_{T}, 43 𝔖_{𝕋}, 42

𝕋 an RST, 4

τ_{i} : f(x_{1}, …, x_{N}) →

f(x_{1}, …, x_{i−1}, qx_{i}, x_{i+1}, …, x_{N}), 2 θ := s_{1} ⋯ s_{N−1}, 5 T_{i} generator of the Hecke algebra, 2 T_{i}, 12 T_{i}^{x}, 12 △(ζ), 51 T̃_{σ} a Hecke element associated to a

permutation, 8 T (𝕋, v) the filling obtained by replacing each i

by v^{+}[i] in 𝕋, 31 T_{u} a Hecke element associated to a

multi-index u, 5

⊴ dominance order on vectors, 18

V_{λ} the vector space of tableaux, 4

v^{+} the unique decreasing partition whose entries are obtained by permuting those of v, 17

v^{R} the unique increasing partition whose entries are obtained by permuting those of v, 17 𝔙_{𝕋,ζ,v,σ} := (𝕋, v), 22

w affine operation in the double affine Hecke

algebra, 2 w_{i}, 12 w^{x}, 12

ξ_{i} a Cherednik element, 3 ξ_{i}, 12 ξ_{i}^{x}, 12 x^{v,𝕋} := x^{v}𝕋R_{v}, 28

ζ_{v,𝕋} spectral vector associated to (v, 𝕋), 22

Appendix A. Some useful formulae for the affine double Hecke algebra

A.1. Hecke algebra of type A_{N−1}. The generators of H_{N} (s) are T_{1}, T_{2}, …, T_{N−1} with s^{n} ≠ 1 for 1 ≤ n ≤ N. The generators satisfy the relations:

(T_{i} − s) (T_{i} + 1) = 0, T_{i}^{2} = (s − 1) T_{i} + s,
T_{i}^{−1} = \frac{1}{s} (T_{i} − s + 1) ,
T_{i}T_{i+1}T_{i} = T_{i+1}T_{i}T_{i+1}, 1 ≤ i < N,
T_{i}T_{j} = T_{j}T_{i}, |i − j| > 1.

Let S = T_{1}T_{2} ⋯ T_{N−1}. Then T_{i}S = ST_{i−1} for 1 < i ≤ N − 1 and T_{j}S^{N} = S^{N}T_{j} for 1 ≤ j < N. Indeed,

T_{i}S = T_{1} ⋯ T_{i−2}T_{i}T_{i−1}T_{i}T_{i+1} ⋯ T_{N−1}
= T_{1} ⋯ T_{i−2}T_{i−1}T_{i}T_{i−1}T_{i+1} ⋯ T_{N−1}
= ST_{i−1},

and

T_{j}S^{N} = S^{j−1}T_{1}S^{N−j+1}
= S^{j−1} ((s − 1) T_{1} + s) (T_{2} ⋯ T_{N−1}S) S^{N−j−1}
= (s − 1) S^{N} + sS^{j−1} (ST_{1} ⋯ T_{N−2}) S^{N−j−1},
S^{N}T_{j} = S^{j+1}T_{N−1}S^{N−j−1}
= S^{j}T_{1}T_{2} ⋯ T_{N−2} ((s − 1) T_{N−1} + s) S^{N−j−1}
= (s − 1) S^{N} + sS^{j}T_{1} ⋯ T_{N−2}S^{N−j−1}.

A consequence of the above derivation is

T_{1}S^{2} = S^{2}T_{N−1}.

The Murphy elements are ϕ_{i} = s^{i−N}T_{i}T_{i+1} ⋯ T_{N−1}T_{N−1} ⋯ T_{i}. Let ϕi^{′} = s^{N−i}ϕ_{i} and

S_{i} = T_{i}T_{i+1} ⋯ T_{N−1} for 1 ≤ i < N. Then ϕi^{′}ϕi+1^{′} ⋯ ϕN−1^{′} = S_{i}^{N+1−i}. Indeed, for

i = N − 1, both sides equal T_{N−1}^{2}. Note that S_{i}T_{j} = T_{j+1}S_{i} for i ≤ j < N. Now suppose the statement is true for some i > 1. We compute

S_{i−1}^{N+1−i} = S_{i−1}^{N−i}T_{i−1}S_{i} = T_{N−1}S_{i−1}^{N−i}S_{i} = T_{N−1}S_{i−1}^{N−i−1}T_{i−1}S_{i}^{2}
= T_{N−1}T_{N−2}S_{i−1}^{N−i−1}S_{i}^{2} = T_{N−1}T_{N−2}S_{i−1}^{N−i−2}T_{i−1}S_{i}^{3}
= ⋯ = T_{N−1}T_{N−2} ⋯ T_{i−1}S_{i}^{N−i+1},

multiply both sides on the left by S_{i−1} = T_{i−1} ⋯ T_{N−1}, and use the inductive hypothesis, to obtain

S_{i−1}^{N+2−i} = T_{i−1} ⋯ T_{N−1}T_{N−1} ⋯ T_{i−1}S_{i}^{N+1−i}

= ϕi−1^{′}S_{i}^{N+1−i} = ϕi−1^{′}ϕi^{′} ⋯ ϕN−1^{′}.

Thus, S^{N} = s^{N(N−1)/2}ϕ_{1}ϕ_{2} ⋯ ϕ_{N−1}.

Adjoin an invertible operator w with relations

wT_{i} = T_{i+1}w, 1 ≤ i < N − 1,
w^{2}T_{N−1} = T_{1}w^{2},
w^{N}T_{i} = T_{i}w^{N}, 1 ≤ i < N.

A.2. Action on polynomials. Let P = 𝕂 [x_{1}, …, x_{N}], where 𝕂 is an extension field of ℚ (s, q). On P there is a representation of H_{N} (s),

p (x) T_{i} = (1 − s) \frac{p (x) − p (xs_{i})}{x_{i} − x_{i+1}} + sp (xs_{i}) , 1 ≤ i < N,

where xs_{i} = (x_{1}, …, x_{i+1}, x_{i}, …) (s_{i} is the transposition (i, i + 1)),

p (x) w = p (qx_{N}, x_{1}, x_{2}, …, x_{N−1}) .

Denote the multiplication operator p (x) ↦ x_{i}p (x) by x_{i}, 1 ≤ i ≤ N. Then we have

x_{i}T_{j} = T_{j}x_{i}, j ≠ i, i − 1,
x_{i}T_{i} = sT_{i}^{−1}x_{i+1}, x_{i} = sT_{i}^{−1}x_{i+1}T_{i}^{−1},
x_{i+1}w = wx_{i}, 1 ≤ i < N,
x_{1}w = qwx_{N}.

A.3. q-Dunkl operators. There are pairwise commuting operators D_{1}, …, D_{N} (dual to the multiplication operators) with relations

D_{i}T_{j} = T_{j}D_{i}, j ≠ i, i − 1,
sT_{i}^{−1}D_{i} = D_{i+1}T_{i}, D_{i} = \frac{1}{s}T_{i}D_{i+1}T_{i},
D_{i+1}w = wD_{i}, 1 ≤ i < N,
qD_{1}w = wD_{N}.

They act on polynomials by

p (x) D_{N} = (p (x) − s^{N−1}p (x) T_{N−1}^{−1}T_{N−2}^{−1} ⋯ T_{1}^{−1}w) x_{N}^{−1},
D_{i} = \frac{1}{s}T_{i}D_{i+1}T_{i} = w^{−1}D_{i+1}w, 1 ≤ i < N.

The Cherednik operators satisfy

ξ_{N} = s^{1−N} (1 − D_{N}x_{N}) ,
ξ_{i} = \frac{1}{s}T_{i}ξ_{i+1}T_{i}, 1 ≤ i < N.