A generalization of Kawanaka’s identity for Hall-Littlewood polynomials and applications

Masao Ishikawa^{1}, Frédéric Jouhet^{2} and Jiang Zeng^{2}

1 Department of Mathematics, Faculty of Education Tottori University, Tottori 680-8551, Japan,

               ishikawa@fed.tottori-u.ac.jp

2 Institut Camille Jordan, Université Claude Bernard (Lyon I) 43, bd du 11 Novembre 1918, 69622 Villeurbanne Cedex, France

             {jouhet,zeng}@math.univ-lyon1.fr

Abstract

An infinite summation formula of Hall-Littlewood polynomials due to Kawanaka is generalized to a finite summation formula, which implies, in particular, twelve more multiple q-identities of Rogers-Ramanujan type than those previously found by Stembridge and the last two authors.

Keywords: Symmetric functions, Hall-Littlewood polynomials, qseries, Rogers-Ramanujan type identities.

1 Introduction

Recently, starting from two infinite summation formulae for Hall-Littlewood polynomials, two of the present authors [7] have generalized a method due to Macdonald [9] to obtain new finite summation formulae for these polynomials. This approach permits them to extend Stembridge’s list of multiple qseries identities of Rogers-Ramanujan type [12]. Conversely these symmetric functions identities can be viewed as a generalization of Rogers-Ramanujan identities. In view of the numerous formulae of Rogers-Ramanujan type [11] one may speculate that there should be more such generalizations starting from other infinite summation formulae for Hall-Littlewood polynomials. However, as pointed out in [7], when one passes from an infinite summation to a finite summation, one may need to modify the coefficients normalizing Hall-Littlewood polynomials in order to obtain some useful formulae.

In this paper we take up on Kawanaka’s new infinite summation identities of Hall-Littlewood polynomials [8] and show that one of his formulae has a finite summation generalization.

We first need to recall some standard notations of q-series, which can be found in [5]. Set (x)_{0} := (x; q)_{0} = 1 and for n ≥ 1

(x)_{n} := (x; q)_{n} = ∏_{k=0}^{n}(1 − xq^{k}), (x)_{∞} : (x; q)_{∞} = ∏_{k=0}^{∞}(1 − xq^{k}).

For n ≥ 0 and r ≥ 1, set

(a_{1}, ⋯, a_{r}; q)_{n} = ∏_{i=1}^{r}(a_{i})_{n} (a_{1}, ⋯, a_{r}; q)_{∞} = ∏_{i=1}^{r}(a_{i})_{∞}.

The q-binomial identity [1] then reads as follows :

∑_{m≥0} \frac{(a)_{m}}{(q)_{m}} x^{m} = \frac{(ax)_{∞}}{(x)_{∞}} ,  (1)

which reduces to the finite q-binomial identity by substitution a → q^{−n} and x → q^{n}x:

(x)_{n} = ∑_{m≥0}(−1)^{m}q^{\binom{m}{2}}\left[{n \atop m}\right]x^{m}  (2)

and to the following identity of Euler when a = 0 :

\frac{1}{(x)_{∞}} = ∑_{m≥0} \frac{x^{m}}{(q)_{m}} .  (3)

Let n ≥ 1 be a fixed integer and S_{n} denote the group of permutations of the set {1, 2, …, n}. Let X = {x_{1}, …, x_{n}} be a set of indeterminates and q a parameter. For each partition λ = (λ_{1}, …, λ_{n}) of length ≤ n, if m_{i} := m_{i}(λ) is the multiplicity of part i in λ, then we also denote λ by (1^{m_{1}} 2^{m_{2}} …). Recall that the Hall-Littlewood polynomials P_{λ}(X, q) are defined by [9, p.208] :

P_{λ}(X, q) = ∏_{i≥1} \frac{(1 − q)^{m_{i}}}{(q)_{m_{i}}} ∑_{w∈S_{n}} w (x_{1}^{λ_{1}} … x_{n}^{λ_{n}} ∏_{i<j} \frac{x_{i} − qx_{j}}{x_{i} − x_{j}} ) .

Since ([9, p. 207])

∑_{w∈S_{n}} w (∏_{i<j} \frac{x_{i} − qx_{j}}{x_{i} − x_{j}} ) = \frac{(q)_{n}}{(1 − q)^{n}} ,

we see that the coefficient of x_{1}^{λ_{1}} … x_{n}^{λ_{n}} in P_{λ} is 1. Set

Φ(X) := ∏_{i} \frac{1 + qx_{i}}{1 − x_{i}} ∏_{j<k} \frac{1 − q^{2}x_{j}x_{k}}{1 − x_{j}x_{k}} .

Our starting point is the following result due to Kawanaka [8]:

∑_{λ} (∏_{i≥1}(−q)_{m_{i}}) P_{λ}(X, q^{2}) = Φ(X).  (4)

Since Kawanaka’s proof uses the representation theory of groups we shall give another proof using Pieri’s rule for Hall-Littlewood polynomials.

For each sequence ξ ∈ {±1}^{n}, set X^{ξ} := {x_{1}^{ξ_{1}}, …, x_{n}^{ξ_{n}}}. Our finite extension of Kawanaka’s formula then reads as follows:

Theorem 1. For k ≥ 1 the following identity holds

∑_{λ_{1}≤k} (∏_{i=1}^{k−1}(−q)_{m_{i}}) P_{λ}(X, q^{2}) = ∑_{ξ∈{±1}^{n}} Φ(X^{ξ}) ∏_{i} x_{i}^{k(1−ξ_{i})/2}.  (5)

Remark. In the case q = 0, the right-hand side of (5) can be written as a quotient of determinants and the formula reduces to a known identity of Schur functions [6].

For any partition λ it will be convenient to adopt the following notation :

(x)_{λ} := (x; q)_{λ} = (x)_{λ_{1}−λ_{2}}(x)_{λ_{2}−λ_{3}} ⋯

Note that this is not the standard notation for (x)_{λ} and corresponds to b_{λ′}(q) in [9, p.210].

We also introduce the following generalization of q-binomial coefficients

\left[{n \atop λ}\right] := \frac{(q)_{n}}{(q)_{n−λ_{1}}(q)_{λ}} ,

with the convention that \left[{n \atop λ}\right] = 0 if λ_{1} > n. If λ = (λ_{1}) we recover the classical q-binomial coefficient. Finally, for any partition λ we denote by l(λ) the length of λ, i.e., the number of its positive parts, and define n(λ) := ∑_{i} \binom{λ_{i}}{2} and n_{2}(λ) = ∑_{i} λ_{i}^{2}. When x_{i} = zq^{2i−2} for i ≥ 1 and then z is replaced by zq, formula (5) specializes to the following identity.

Corollary 1. For k ≥ 1 there holds

∑_{l(λ)≤k} (∏_{i=1}^{k−1}(−q)_{λ_{i}−λ_{i+1}}) z^{|λ|}q^{n_{2}(λ)} \left[{n \atop λ}\right]_{q^{2}}  (6)

= ∑_{r=0}^{n}(−1)^{r}z^{kr}q^{(k+1)r^{2}} \left[{n \atop r}\right]_{q^{2}} \frac{(−z)_{2n+1}}{(z^{2}q^{2r}; q^{2})_{n+1}} (1 − zq^{2r}).

Now, as in [7, 12], we can prove the following key q-identity which allows to produce identities of Rogers-Ramanujan type :

Theorem 2. For k ≥ 1,

∑_{l(λ)≤k} z^{|λ|}q^{2n(λ)} \frac{(a, b; q^{−1})_{λ_{1}}}{(−q)_{λ_{k}}(q)_{λ}}  (7)

= \frac{(−z/q)_{∞}}{(abz)_{∞}} ∑_{r≥0}(−1)^{r}z^{kr}q^{r+(2k+2)\binom{r}{2}} \frac{(a, b; q^{−1})_{r}}{(q^{2}; q^{2})_{r}} \frac{(azq^{r}, bzq^{r})_{∞}}{(z^{2}q^{2r−2}; q^{2})_{∞}} (1 − zq^{2r−1}).

This paper is organized as follows. In Section 2 we give a new proof of Kawanaka’s formula using Pieri’s rule for Hall-Littlewood polynomials since Kawanaka’s original proof uses the representation theory of groups. In section 3, we derive from Theorem 2 twelve multiple analogs of Rogers-Ramanujan type identities. In section 4 we give the proofs of Theorem 1 and Corollary 1, and some consequences, and defer the elementary proof, i.e., without using the Hall-Littlewood polynomials, of Theorem 2, Corollary 1 and other multiple q-series identities to section 5. To prove Theorems 1 and 2 we apply the generating function technique which was developped in [7, 9, 12].

2 Another proof of Kawanaka’s identity

Recall [9, p.230, Ex.1] the following summation of Hall-Littlewood polynomials :

∑_{μ} P_{μ}(X, q) = ∏_{i} \frac{1}{1 − x_{i}} ∏_{i<j} \frac{1 − qx_{i}x_{j}}{1 − x_{i}x_{j}} .

By replacing q by q^{2}, we get

∑_{μ} P_{μ}(X, q^{2}) = ∏_{i} \frac{1}{1 − x_{i}} ∏_{i<j} \frac{1 − q^{2}x_{i}x_{j}}{1 − x_{i}x_{j}} .  (8)

Note that

∑_{r≥0} e_{k}(X)q^{k} = ∏_{i}(1 + qx_{i}),  (9)

where e_{r}(X) stands for the r-th elementary symmetric function. Identities (8) and (9) imply

∑_{μ} ∑_{r} q^{r}P_{μ}(X, q^{2})e_{r}(X) = ∏_{i} \frac{1 + qx_{i}}{1 − x_{i}} ∏_{i<j} \frac{1 − q^{2}x_{i}x_{j}}{1 − x_{i}x_{j}} .

From [9, p.209, (2.8)], we have

P_{(1^{r})}(X, q) = e_{r}(X),

and this shows that

∑_{μ} ∑_{r} q^{r}P_{μ}(X, q^{2})P_{(1^{r})}(X, q^{2}) = ∏_{i} \frac{1 + qx_{i}}{1 − x_{i}} ∏_{i<j} \frac{1 − q^{2}x_{i}x_{j}}{1 − x_{i}x_{j}} .

Let f_{μν}^{λ}(q) be the coefficients defined by

P_{μ}(X, q)P_{ν}(X, q) = ∑_{λ} f_{μν}^{λ}(q)P_{λ}(X, q),

then, by [9, p.215 (3.2)] we have

f_{μ(1^{m})}^{λ}(q) = ∏_{i≥1} \left[{λ′_{i} − λ′_{i+1} \atop λ′_{i} − μ′_{i}}\right]_{q}

(and therefore f_{μ(1^{m})}^{λ}(q) = 0 unless λ \ μ is a vertical m-strip, or m-vs, which means λ ⊂ μ, |λ \ μ| = m and there is at most one cell in each row of the Ferrers diagram of λ \ μ). Thus we have

∑_{λ} ∑_{{μ \atop λ\μ vs}} q^{|λ−μ|} ∏_{i≥1} \left[{λ′_{i} − λ′_{i+1} \atop λ′_{i} − μ′_{i}}\right]_{q^{2}} P_{λ}(X, q^{2})

= ∏_{i} \frac{1 + qx_{i}}{1 − x_{i}} ∏_{i<j} \frac{1 − q^{2}x_{i}x_{j}}{1 − x_{i}x_{j}} .

Applying the identity (see [1], and [13] for a bijective proof) :

∑_{k=0}^{n} q^{k} \left[{n \atop k}\right]_{q^{2}} = ∏_{k=1}^{n}(1 + q^{k}),  (10)

we conclude that

∑_{{μ \atop λ\μ vs}} q^{|λ−μ|} ∏_{i≥1} \left[{λ′_{i} − λ′_{i+1} \atop λ′_{i} − μ′_{i}}\right]_{q^{2}} = ∏_{i≥1} ∏_{k=1}^{λ′_{i}−λ′_{i+1}}(1 + q^{k}),

which is precisely what we desired to prove. Remark. For a node v = (i, j) in the diagram of λ, the arm-length a(v) and the leg-length l(v) of λ at v are defined by a(v) = λ_{i} −j and l(v) = λ′_{j} −i respectively. Kawanaka [8, (5.2)] proved another identity for Hall-Littlewood polynomials :

∑_{λ} q^{o(λ)/2} ( ∏_{{v∈λ, a(v)=0 \atop l(v)even}}(1 − q^{l(v)+1})) P_{λ}(X, q) = ∏_{i≤j} \frac{1 − qx_{i}x_{j}}{1 − x_{i}x_{j}} ,  (11)

where the sum on the left is taken over all partitions λ such that m_{i}(λ) is even for odd i and

o(λ) = ∑_{i odd} m_{i}(λ).

It would be possible to prove this identity in the same manner as above.

There is a related identity about Hall-Littlewood polynomials in Macdonald’s book [9, p. 219] :

∑_{λ} q^{n(λ)} (∏_{j=1}^{l(λ)}(1 + q^{1−j}y)) P_{λ}(X, q) = ∏_{i≥1} \frac{1 + x_{i}y}{1 − x_{i}} .  (12)

3 Multiple identities of Rogers-Ramanujan type

We shall derive several identities of Rogers-Ramanujan type from Theorem 2. First we note that if z = q^{2} identity (7) reduces to

∑_{l(λ)≤k} q^{|λ|+n_{2}(λ)} \frac{(a, b; q^{−1})_{λ_{1}}}{(−q)_{λ_{k}}(q)_{λ}} = \frac{1}{(q, abq^{2})_{∞}}  (13)

× ∑_{r≥0}(−1)^{r}q^{(2k+1)r+(2k+2)\binom{r}{2}}(a, b; q^{−1})_{r}(aq^{r+2}, bq^{r+2})_{∞}(1 − q^{2r+1}),

and if z = q it becomes

∑_{l(λ)≤k} q^{n_{2}(λ)} \frac{(a, b; q^{−1})_{λ_{1}}}{(−q)_{λ_{k}}(q)_{λ}} = \frac{1}{(q, abq)_{∞}}  (14)

× ((aq, bq)_{∞} + 2 ∑_{r≥1}(−1)^{r}q^{(k+1)r^{2}}(a, b; q^{−1})_{r}(aq^{r+1}, bq^{r+1})_{∞}) .

We need the following two forms of Jacobi triple product identity [1, p.21] :

J(x, q) := (q, x, q/x)_{∞} = ∑_{r=0}^{∞}(−1)^{r}x^{r}q^{\binom{r}{2}}(1 − q^{2r+1}/x^{2r+1}) (15)
= 1 + ∑_{r=1}^{∞}(−1)^{r}x^{r}q^{\binom{r}{2}}(1 + q^{r}/x^{2r}). (16)

Theorem 3. For k ≥ 1, the following identities hold

∑_{l(λ)≤k} \frac{q^{|λ|+n_{2}(λ)}}{(−q)_{λ_{k}}(q)_{λ}} = \frac{(q^{2k+2}, q^{2k+1}, q; q^{2k+2})_{∞}}{(q)_{∞}} ,  (17)
∑_{l(λ)≤k} \frac{q^{|λ|+n_{2}(λ)−(λ_{1}^{2}+λ_{1})/2}(−q)_{λ_{1}}}{(−q)_{λ_{k}}(q)_{λ}} = \frac{(−q)_{∞}}{(q)_{∞}} (q^{2k+1}, q^{2k}, q; q^{2k+1})_{∞}, (18)
∑_{l(λ)≤k} \frac{q^{2|λ|+2n_{2}(λ)−λ_{1}^{2}}(−q; q^{2})_{λ_{1}}}{(−q^{2}; q^{2})_{λ_{k}}(q^{2}; q^{2})_{λ}} = \frac{(−q; q^{2})_{∞}}{(q^{2}; q^{2})_{∞}} (q^{4k+2}, q^{4k+1}, q; q^{4k+2})_{∞},(19)
∑_{l(λ)≤k} \frac{q^{2|λ|+2n_{2}(λ)−2λ_{1}^{2}−λ_{1}}(−q)_{2λ_{1}}}{(−q^{2}; q^{2})_{λ_{k}}(q^{2}; q^{2})_{λ}} = \frac{(−q)_{∞}}{(q)_{∞}} (q^{4k}, q^{4k−1}, q; q^{4k})_{∞},  (20)

∑_{l(λ)≤k} \frac{q^{n_{2}(λ)}}{(−q)_{λ_{k}}(q)_{λ}} = \frac{(q^{2k+2}, q^{k+1}, q^{k+1}; q^{2k+2})_{∞}}{(q)_{∞}} ,  (21)

∑_{l(λ)≤k} \frac{q^{n_{2}(λ)−(λ_{1}^{2}+λ_{1})/2}(−q)_{λ_{1}}}{(−q)_{λ_{k}}(q)_{λ}} = \frac{(−1)_{∞}}{(q)_{∞}} (q^{2k+1}, q^{k}, q^{k+1}; q^{2k+1})_{∞}, (22)

∑_{l(λ)≤k} \frac{q^{2n_{2}(λ)−λ_{1}^{2}}(−q; q^{2})_{λ_{1}}}{(−q^{2}; q^{2})_{λ_{k}}(q^{2}; q^{2})_{λ}} =

\frac{(−q; q^{2})_{∞}}{(q^{2}; q^{2})_{∞}} (q^{4k+2}, q^{2k+1}, q^{2k+1}; q^{4k+2})_{∞}. (23)

Proof. For identities (17)-(20), first set (a, b) = (0, 0), (−q^{−1}, 0), (−q^{−1/2}, 0) and (−q^{−1/2}, −q^{−1}) in (13), respectively, and then apply (15).

For identities (21)-(22), first set (a, b) = (0, 0), (−q^{−1/2}, 0) and (−q^{−1}, 0) in (14), respectively, and then apply (16).

Note that, for (19), (20) and (23), we need to replace q by q^{2} at last.

Theorem 4. For k ≥ 1, the following identities hold

∑_{l(λ)≤k} \frac{q^{|λ|+n_{2}(λ)−(λ_{1}^{2}+3λ_{1})/2}(−q)_{λ_{1}}(1 − q^{λ_{1}})}{(−q)_{λ_{k}}(q)_{λ}}

= \frac{(−q)_{∞}}{(q)_{∞}} (q^{2k+1}, q^{2k−1}, q^{2}; q^{2k+1})_{∞}, (24)
∑_{l(λ)≤k} \frac{q^{2|λ|+2n_{2}(λ)−λ_{1}^{2}−2λ_{1}}(−q; q^{2})_{λ_{1}}}{(−q^{2}; q^{2})_{λ_{k}}(q^{2}; q^{2})_{λ}}

= \frac{(−q; q^{2})_{∞}}{(q^{2}; q^{2})_{∞}} (q^{4k+2}, q^{4k−1}, q^{3}; q^{4k+2})_{∞},(25)
∑_{l(λ)≤k} \frac{q^{|λ|+n_{2}(λ)−λ_{1}}}{(−q)_{λ_{k}}(q)_{λ}} = \frac{(q^{2k+2}, q^{2k}, q^{2}; q^{2k+2})_{∞}}{(q)_{∞}} ,  (26)

∑_{l(λ)≤k} \frac{q^{|λ|+n_{2}(λ)−2λ_{1}}(1 − q^{2λ_{1}})}{(−q)_{λ_{k}}(q)_{λ}} = \frac{(q^{2k+2}, q^{2k−1}, q^{3}; q^{2k+2})_{∞}}{(q)_{∞}} ,  (27)

∑_{l(λ)≤k} \frac{q^{n_{2}(λ)−λ_{1}}}{(−q)_{λ_{k}}(q)_{λ}} = \frac{(−1)_{∞}}{(q^{2}; q^{2})_{∞}} (q^{2k+2}, q^{k}, q^{k+2}; q^{2k+2})_{∞}. (28)

Proof. For i ∈ {0, 1, 2}, denote by [b^{i}] the operation of extracting the coefficient of b^{i} in the corresponding identity. For (24)-(27), apply the following operations to (13) respectively : a = −q^{−1} and (1 − 1/q)[b], a = 0 and [b^{0}]+(1−1/q)[b], a = −q^{−1/2} and [b^{0}]+(1−1/q)[b], a = 0 and [b]+(1−1/q)[b^{2}], and then apply (15). Note that, for (25), we need to replace q by q^{2} at last.

For (28) apply the operations a = 0 and (1 − 1/q)[b] to (14) and then apply (16).

Remark. As speculated by the anonymous referee, all of the Rogers-Ramanujan type identities given in Theorems 3 and 4 are known.

For example, specializing equation (3.4) of Bressoud [4] (see also [3]) with k → k + 1 and r = 1, r = 2 and r = k + 1, respectively, we recover identities (17), (26)) and (21); while specializing equation (3.9) of Bressoud [4] with k → k + 1 and r = 1, r = 2 and r = k + 1, respectively, we recover (19), (25) and (23).

Since we derived all these identities in Theorems 3 and 4 from the two master identities (13) and (14), instead of identifying each identity individually, it suffices to identify the later two with known results in the literature. In 1984, by means of Bailey chains, Andrews proved a remarkable generalization of Bailey’s lemma [2, Thm. 1], which contains many multiple Rogers-Ramanujan type identities as special cases. In particular, identities (13) and (14) are limit cases of Andrews’ theorem. More precisely, to derive (13), set a = q, b_{k} = 1/a, c_{k} = 1/b in Andrews’ formula and let N, b_{1}, c_{1}, …, b_{k−1}, c_{k−1} → ∞, finally apply the Bailey pair E(3) of Slater’s paper [10]. To derive (14) we do the same thing except that we set a = 1 and apply the Bailey pair B(3) of Slater’s paper [10].

When k = 1, identities (17), (18), (20), (27) and (22) reduce directly to special cases of the q-binomial identity (1). For example, when k = 1 identity (20) reduces to

∑_{n=0}^{∞} \frac{q^{n}(−q; q^{2})_{n}}{(q^{2}; q^{2})_{n}} \frac{(−q^{2}; q^{2})_{∞}}{(q; q^{2})_{∞}} ,

which is the q-binomial identity (1) after substitutions q → q^{2}, a → −q and x → q. The other identities reduce to the following Rogers-Ramanujan type identities :

∑_{n=0}^{∞} \frac{q^{n^{2}+2n}(−q; q^{2})_{n}}{(q^{4}; q^{4})_{n}} = \frac{(−q; q^{2})_{∞}}{(q^{2}; q^{2})_{∞}} (q, q^{5}, q^{6}; q^{6})_{∞},  (29)

∑_{n=0}^{∞} \frac{q^{n^{2}}}{(q^{2}; q^{2})_{n}} = \frac{(q^{2}, q^{2}, q^{4}; q^{4})_{∞}}{(q)_{∞}} ,  (30)

∑_{n=0}^{∞} \frac{q^{n^{2}}(−q; q^{2})_{n}}{(q^{4}; q^{4})_{n}} = \frac{(−q; q^{2})_{∞}}{(q^{2}; q^{2})_{∞}} (q^{3}, q^{3}, q^{6}; q^{6})_{∞}.  (31)

Note that (30) is again a special case of the q-binomial identity (1) and (31) is (25) of Slater’s list [11].

4 Proof of Theorem 1 and consequences

4.1 Proof of Theorem 1

For any statement A it will be convenient to introduce the Boolean function χ(A), which is 1 if A is true and 0 if A is false. Consider the generating function

S(u) = ∑_{λ_{0},λ} ( ∏_{i=1}^{λ_{0}−1}(−q)_{m_{i}}) P_{λ}(X, q^{2}) u^{λ_{0}}

where the sum is over all partitions λ = (λ_{1}, …, λ_{n}) and the integers λ_{0} ≥ λ_{1}. Suppose λ = (μ_{1}^{r_{1}} μ_{2}^{r_{2}} … μ_{k}^{r_{k}}), where μ_{1} > μ_{2} > ⋯ > μ_{k} ≥ 0 and (r_{1}, …, r_{k}) is a composition of n.

Let S_{n}^{λ} be the set of permutations of S_{n} which fix λ. Each w ∈ S_{n}/S_{n}^{λ} corresponds to a surjective mapping f : X ⟶ {1, 2, …, k} such that |f^{−1}(i)| = r_{i}. For any subset Y of X, let p(Y) denote the product of the elements of Y (in particular, p(∅) = 1). We can rewrite Hall-Littlewood functions as follows :

P_{λ}(X, q^{2}) = ∑_{f} p(f^{−1}(1))^{μ_{1}} ⋯ p(f^{−1}(k))^{μ_{k}} ∏_{f(x_{i})<f(x_{j})} \frac{x_{i} − q^{2}x_{j}}{x_{i} − x_{j}} ,

summed over all surjective mappings f : X ⟶ {1, 2, …, k} such that |f^{−1}(i)| = r_{i}. Furthermore, each such f determines a filtration of X :

F : ∅ = F_{0} ⊊ F_{1} ⊊ ⋯ ⊊ F_{k} = X,  (32)

according to the rule x_{i} ∈ F_{l} ⟺ f(x_{i}) ≤ l for 1 ≤ l ≤ k. Conversely, such a filtration F = (F_{0}, F_{1}, …, F_{k}) determines a surjection f : X ⟶ {1, 2, …, k} uniquely. Thus we can write :

P_{λ}(X, q^{2}) = ∑_{F} π_{F} ∏_{1≤i≤k} p(F_{i} \ F_{i−1})^{μ_{i}},  (33)

summed over all the filtrations F such that |F_{i}| = r_{1} + r_{2} + ⋯ + r_{i} for 1 ≤ i ≤ k, and

π_{F} = ∏_{f(x_{i})<f(x_{j})} \frac{x_{i} − q^{2}x_{j}}{x_{i} − x_{j}} ,

where f is the function defined by F.

Now let ν_{i} = μ_{i}−μ_{i+1} if 1 ≤ i ≤ k−1 and ν_{k} = μ_{k}, thus ν_{i} > 0 if i < k and ν_{k} ≥ 0. Furthermore, let μ_{0} = λ_{0} and ν_{0} = μ_{0}−μ_{1} in the definition of S(u), so that ν_{0} ≥ 0 and μ_{0} = ν_{0} +ν_{1} +⋯+ν_{k}. Define c_{F} = ∏_{i=1}^{k}(−q)_{|F_{i}\F_{i−1}|} for any filtration F. Thus, since the lengths of columns of λ are |F_{j}| = r_{1} + ⋯ + r_{j} with multiplicities ν_{j} and r_{j} = m_{μ_{j}}(λ) for 1 ≤ j ≤ k, we have

∏_{i=1}^{λ_{0}−1}(−q)_{m_{i}} = c_{F} × (χ(ν_{k} = 0)(−q)_{|F_{k}\F_{k−1}|} + χ(ν_{k} ≠ 0))^{−1}
× (χ(ν_{0} = 0)(−q)_{|F_{1}|} + χ(ν_{0} ≠ 0))^{−1} .

Let F (X) be the set of filtrations of X. Summarizing we obtain

S(u) = ∑_{F∈F (X)} c_{F} π_{F} ∑_{ν_{1}>0}(u p(F_{1}))^{ν_{1}} ⋯ ∑_{ν_{k−1}>0}(u p(F_{k−1}))^{ν_{k−1}}

× ∑_{ν_{0}≥0} \frac{u^{ν_{0}}}{χ(ν_{0} = 0) (−q)_{|F_{1}|} + χ(ν_{0} ≠ 0)}

× ∑_{ν_{k}≥0} \frac{u^{ν_{k}} p(F_{k})^{ν_{k}}}{χ(ν_{k} = 0) (−q)_{|F_{k}\F_{k−1}|} + χ(ν_{k} ≠ 0)}.  (34)

For any filtration F of X set

A_{F}(X, u) = c_{F} ∏_{j} [\frac{p(F_{j})u}{1 − p(F_{j})u} + \frac{χ(F_{j} = X)}{(−q)_{|F_{j}\F_{j−1}|}} + \frac{χ(F_{j} = ∅)}{(−q)_{|F_{1}|}}] .

It follows from (34) that

S(u) = ∑_{F∈F (X)} π_{F}A_{F}(X, u).

Hence S(u) is a rational function of u with simple poles at 1/p(Y), where Y is a subset of X. We are now proceeding to compute the corresponding residue c(Y) at each pole u = 1/p(Y).

Let us start with c(∅). Writing λ_{0} = λ_{1} + k with k ≥ 0, we see that

S(u) = ∑_{λ} f_{λ}(q)P_{λ}(X, q^{2})u^{λ_{1}} ∑_{k≥0} \frac{u^{k}}{χ(k = 0)(−q)_{m_{λ_{1}}} + χ(k ≠ 0)}

= ∑_{λ} f_{λ}(q)P_{λ}(X, q^{2})u^{λ_{1}} (\frac{u}{1 − u} + \frac{1}{(−q)_{m_{λ_{1}}}} ) .

It follows from (4) that

c(∅) = [S(u)(1 − u)]_{u=1} = Φ(X).

For the computations of other residues, we need some more notations. For any Y ⊆ X, let Y′ = X \ Y and −Y = {x_{i}^{−1} : x_{i} ∈ Y }. Then

c(Y) = [∑_{F} π_{F}A_{F}(X, u)(1 − p(Y)u)]_{u=p(−Y)} .  (35)

If Y ∉ F, the corresponding summand is equal to 0. Thus we need only to consider the following filtrations F :

∅ = F_{0} ⊊ ⋯ ⊊ F_{t} = Y ⊊ ⋯ ⊊ F_{k} = X 1 ≤ t ≤ k.

We may then split F into two filtrations F_{1} and F_{2} :

F_{1} : ∅ ⊊ −(Y \ F_{t−1}) ⊊ ⋯ ⊊ −(Y \ F_{1}) ⊊ −Y,
F_{2} : ∅ ⊊ F_{t+1} \ Y ⊊ ⋯ ⊊ F_{k−1} \ Y ⊊ Y′.

Then, writing v = p(Y)u and c_{F} = c_{F_{1}} × c_{F_{2}}, we have

π_{F}(X) = π_{F_{1}}(−Y)π_{F_{2}}(Y′) ∏_{x_{i}∈Y,x_{j}∈Y′} \frac{1 − q^{2}x_{i}^{−1}x_{j}}{1 − x_{i}^{−1}x_{j}} ,

and A_{F}(X, u)(1 − p(Y)u) is equal to

A_{F_{1}}(−Y, v)A_{F_{2}}(Y′, v)(1 − v) (\frac{v}{1 − v} + \frac{χ(Y = X)}{(−q)_{|Y \F_{t−1}|}} )

× (\frac{v}{1 − v} + \frac{1}{(−q)_{|Y \F_{t−1}|}} )^{−1} (\frac{v}{1 − v} + \frac{1}{(−q)_{|F_{t+1}\Y |}} )^{−1} .

Thus when u = p(−Y), i.e., v = 1,

[π_{F}(X)A_{F}(X, u)(1 − p(Y)u)]_{u=p(−Y)} =
[π_{F_{1}}(−Y)A_{F_{1}}(−Y, v)(1 − v)π_{F_{2}}(Y′)A_{F_{2}}(Y′, v)(1 − v)]_{v=1}

× ∏_{x_{i}∈Y,x_{j}∈Y′} \frac{1 − q^{2}x_{i}^{−1}x_{j}}{1 − x_{i}^{−1}x_{j}} .

Using (35) and the result of c(∅), which can be written

[∑_{F} π_{F}A_{F}(X, u)(1 − u)]_{u=1} = Φ(X),

we get

c(Y) = Φ(−Y)Φ(Y′) ∏_{x_{i}∈Y,x_{j}∈Y′} \frac{1 − q^{2}x_{i}^{−1}x_{j}}{1 − x_{i}^{−1}x_{j}} .

Each subset Y of X can be encoded by a sequence ξ ∈ {±1}^{n} according to the rule : ξ_{i} = 1 if x_{i} ∉ Y and ξ_{i} = −1 if x_{i} ∈ Y. Hence

c(Y) = Φ(X^{ξ}).

Note also that

p(Y) = ∏_{i} x_{i}^{(1−ξ_{i})/2}, p(−Y) = ∏_{i} x_{i}^{(ξ_{i}−1)/2}.

Now, extracting the coefficients of u^{k} in the equation :

S(u) = ∑_{Y ⊆X} \frac{c(Y)}{1 − p(Y)u},

yields

∑_{λ_{1}≤k} (∏_{i=1}^{k−1}(−q)_{m_{i}}) P_{λ}(X, q^{2}) = ∑_{Y ⊆X} c(Y)p(Y)^{k}.

Finally, substituting the value of c(Y) in the above formula we obtain (5).

4.2 Proof of Corollary 1

Recall [9, p. 213] that if x_{i} = zq^{2i−2} (1 ≤ i ≤ n) then :

P_{λ′}(X, q^{2}) = z^{|λ|}q^{2n(λ)} \left[{n \atop λ}\right]_{q^{2}} .  (36)

Replacing each partition λ by its conjugate λ′ on the left-hand side of (5) yields the left-hand side of (6). Set

Ψ(X) = ∏_{i} \frac{1}{1 − x_{i}^{2}} ∏_{j<k} \frac{1 − q^{2}x_{j}x_{k}}{1 − x_{j}x_{k}} .

Then, for any ξ ∈ {±1}^{n} such that the number of ξ_{i} = −1 is r, 0 ≤ r ≤ n, we can write Φ(X^{ξ}) as follows:

Φ(X^{ξ})Ψ(X^{ξ}) ∏_{i} \frac{1 + qx_{i}^{ξ_{i}}}{1 − x_{i}^{ξ_{i}}} (1 − x_{i}^{2ξ_{i}}),  (37)

which is readily seen to equal 0 unless ξ ∈ {−1}^{r} ×{1}^{n−r}. Now, in the latter case, we have ∏_{i} x_{i}^{k(1−ξ_{i})/2} = z^{kr}q^{2k\binom{r}{2}},

∏_{i=1}^{n} \frac{1 + qx_{i}^{ξ_{i}}}{1 − x_{i}^{ξ_{i}}} (1 − x_{i}^{2ξ_{i}}) = \frac{(z^{2}; q^{4})_{n}}{z^{2r}q^{4\binom{r}{2}−r}} \frac{(−z/q; q^{2})_{r}}{(z; q^{2})_{r}} \frac{(−zq^{2r+1}; q^{2})_{n−r}}{(zq^{2r}; q^{2})_{n−r}} , (38)

and [12, p. 476] :

Ψ(X^{ξ}) = (−1)^{r}z^{2r}q^{6\binom{r}{2}} \left[{n \atop r}\right] \frac{1 − z^{2}q^{4r−2}}{(zq^{r−1})_{n+1}} .  (39)

Substituting these into the right-hand side of (5) we obtain the right-hand side of (6) after simple manipulations.

When n → +∞, since \left[{n \atop λ}\right] → \frac{1}{(q)_{λ}}, equation (6) reduces to :

∑_{l(λ)≤k} \frac{z^{|λ|}q^{2n(λ)}}{(−q)_{λ_{k}}(q)_{λ}} = (−z/q)_{∞} ∑_{r≥0} \frac{(−1)^{r}z^{kr}q^{r+(2k+2)\binom{r}{2}}}{(q^{2}; q^{2})_{r}(z^{2}q^{2r−2})_{∞}} (1 − zq^{2r−1}). (40)

Furthermore, as in Section 2, setting z = q^{2} and z = q in (40) yields (17) and (21), respectively.

5 Elementary approach and proof of Theorem 2

5.1 Preliminaries

We will need the following result, which corresponds to the case k → ∞ in (6), and can be proved in an elementary way :

Lemma 1. For n ≥ 0

∑_{λ} z^{|λ|}q^{2n(λ)}(−q)_{λ} \left[{n \atop λ}\right]_{q^{2}} = \frac{(−z)_{2n}}{(z^{2}; q^{2})_{n}} .  (41)

Proof. Recall the following identity, which is proved in [7] :

q^{\binom{m}{2}+n(μ)}\left[{n \atop m}\right]\left[{n \atop μ}\right] = ∑_{λ} q^{n(λ)}\left[{n \atop λ}\right] ∏_{i≥1} \left[{λ_{i} − λ_{i+1} \atop λ_{i} − μ_{i}}\right],  (42)

where the sum is over all partitions λ such that λ/μ is a horizontal m-strip, i.e., μ ⊆ λ, |λ/μ| = m and there is at most one cell in each column of the Ferrers diagram of λ/μ. We also need

∑_{λ} z^{|λ|}q^{n(λ)}\left[{n \atop λ}\right] = \frac{(−z)_{n}}{(z^{2})_{n}} ,  (43)

which can be found in [7, 12]. Using (43) with q replaced by q^{2} and (2), the right-hand side of (41) can be written

\frac{(−z; q^{2})_{n}}{(z^{2}; q^{2})_{n}} (−zq; q^{2})_{n} = ∑_{μ, m} z^{|μ|}q^{2n(μ)}\left[{n \atop μ}\right]_{q^{2}}z^{m}q^{2\binom{m}{2}+m}\left[{n \atop m}\right]_{q^{2}}

= ∑_{λ, m} z^{|λ|}q^{2n(λ)}\left[{n \atop λ}\right]_{q^{2}} ∏_{i≥1} ∑_{r_{i}≥0} q^{r_{i}} \left[{λ_{i} − λ_{i+1} \atop r_{i}}\right]_{q^{2}},

where the last equality follows from (42), setting r_{i} = λ_{i} − μ_{i} for i ≥ 1. Now we conclude by using (10).

Recall the following extension of the n → ∞ case of (41), which is Stembridge’s lemma 3.3 (b) in [12], and identity (60) in [7] :

∑_{λ} z^{|λ|}q^{2n(λ)} \frac{(a, b; q^{−1})_{λ_{1}}}{(q)_{λ}} = \frac{(az, bz)_{∞}}{(z, abz)_{∞}} .  (44)

Now, using (41), we are able to prove directly identity (6) in Corollary 1, and then using (44), to deduce an elementary proof of (7) in Theorem 2.

5.2 Elementary proof of Corollary 1

Consider the generating function of the left-hand side of (6) :

φ(u) = ∑_{k≥0} u^{k} ∑_{l(λ)≤k} \frac{(−q)_{λ}}{(−q)_{λ_{k}}} z^{|λ|}q^{2n(λ)} \left[{n \atop λ}\right]_{q^{2}}  (45)

= ∑_{λ} u^{l(λ)}z^{|λ|}q^{2n(λ)}(−q)_{λ} \left[{n \atop λ}\right]_{q^{2}} ∑_{k≥0} \frac{u^{k}}{(−q)_{λ_{k+l(λ)}}}

= ∑_{λ} u^{l(λ)}z^{|λ|}q^{2n(λ)}(−q)_{λ} \left[{n \atop λ}\right]_{q^{2}} (\frac{u}{1 − u} + \frac{1}{(−q)_{λ_{l(λ)}}} ) , (46)

where the last equality follows from the fact that λ_{k+l(λ)} = 0 unless k = 0. Now, each partition λ with parts bounded by n can be encoded by a pair of sequences ν = (ν_{0}, ν_{1}, ⋯, ν_{l}) and m = (m_{0}, ⋯, m_{l}) such that λ = (ν_{0}^{m_{0}}, …, ν_{l}^{m_{l}}), where n = ν_{0} > ν_{1} > ⋯ > ν_{l} > 0 and ν_{i} has multiplicity m_{i} ≥ 1 for 1 ≤ i ≤ l and ν_{0} = n has multiplicity m_{0} ≥ 0. Using the notation :

< α >= \frac{α}{1 − α}, u_{i} = z^{i}q^{i(i−1)} for i ≥ 0,

we can then rewrite (46) as follows :

φ(u) = ∑_{ν}(−q)_{ν} \left[{n \atop ν}\right]_{q^{2}} (< u > +\frac{1}{(−q)_{ν_{l}}} )

× ∑_{m} ((u_{n}u)^{m_{0}} + \frac{χ(m_{0} = 0)}{(−q)_{n−ν_{1}}}) ∏_{i=1}^{l}(u_{ν_{i}}u)^{m_{i}}

= ∑_{ν} \frac{(q^{2}; q^{2})_{n}}{(q)_{ν}} B_{ν},  (47)

where the sum is over all strict partitions ν = (ν_{0}, ν_{1}, …, ν_{l}) and

B_{ν} = (< u > +\frac{1}{(−q)_{ν_{l}}} ) (< u_{r}u > +\frac{1}{(−q)_{n−ν_{1}}} ) ∏_{i=1}^{l} < u_{ν_{i}}u > .

So φ(u) is a rational fraction with simple poles at u_{r}^{−1} for 0 ≤ r ≤ n. Let b_{r}(z, n) be the corresponding residue of φ(u) at u_{r}^{−1} for 0 ≤ r ≤ n. Then, it follows from (47) that

b_{r}(z, n) = ∑_{ν} \frac{(q^{2}; q^{2})_{n}}{(q)_{ν}} [B_{ν}(1 − u_{r}u)]_{u=u_{r}^{−1}} .  (48)

We shall first consider the cases where r = 0 or n. Using (46) and (41) we have

b_{0}(z, n) = [φ(u)(1 − u)]_{u=1} = \frac{(−z)_{2n}}{(z^{2}; q^{2})_{n}} .  (49)

Now, by (47) and(48) we have

b_{0}(z, n) = ∑_{ν} \frac{(q^{2}; q^{2})_{n}}{(q)_{ν}} (< u_{n} > +\frac{1}{(−q)_{n−ν_{1}}} ) ∏_{i=1}^{l} < u_{ν_{i}} >,  (50)

and

b_{n}(z, n) = ∑_{ν} \frac{(q^{2}; q^{2})_{n}}{(q)_{ν}} (< 1/u_{n} > +\frac{1}{(−q)_{ν_{l}}} ) ∏_{i=1}^{l} < u_{ν_{i}}/u_{n} >,  (51)

which, by setting μ_{i} = n − ν_{l+1−i} for 1 ≤ i ≤ l and μ_{0} = n, can be written as

b_{n}(z, n) = ∑_{μ} \frac{(q^{2}; q^{2})_{n}}{(q)_{μ}} (< 1/u_{n} > +\frac{1}{(−q)_{n−μ_{1}}} ) ∏_{i=1}^{l} < u_{n−μ_{i}}/u_{n} > . (52)

Comparing (52) with (50) we see that b_{n}(z, n) is equal to b_{0}(z, n) with z replaced by z^{−1}q^{−2n+2}. It follows from (49) that

b_{n}(z, n) = b_{0}(z^{−1}q^{−2n+2}, n) = (−1)^{n}q^{n^{2}} \frac{(−z/q)_{2n}}{(z^{2}q^{2n−2}; q^{2})_{n}} .  (53)

Consider now the case where 0 < r < n. Clearly, for each partition ν, the corresponding summand in (48) is not zero only if ν_{j} = r for some j, 0 ≤ j ≤ n. Furthermore, each such partition ν can be split into two strict partitions ρ = (ρ_{0}, ρ_{1}, …, ρ_{j−1}) and σ = (σ_{0}, …, σ_{l−j}) such that ρ_{i} = ν_{i} − r for 0 ≤ i ≤ j − 1 and σ_{s} = ν_{j+s} for 0 ≤ s ≤ l − j. So we can write (48) as follows :

b_{r}(z, n) = \left[{n \atop r}\right]_{q^{2}} ∑_{ρ} \frac{(q^{2}; q^{2})_{n−r}}{(q)_{ρ}} F_{ρ}(r) × ∑_{σ} \frac{(q^{2}; q^{2})_{r}}{(q)_{σ}} G_{σ}(r)

where for ρ = (ρ_{0}, ρ_{1}, …, ρ_{l}) with ρ_{0} = n − r,

F_{ρ}(r) = (< u_{n}/u_{r} > +\frac{1}{(−q)_{n−r−ρ_{1}}} ) ∏_{i=1}^{l(ρ)} < u_{ρ_{i}+r}/u_{r} >,

and for σ = (σ_{0}, …, σ_{l}) with σ_{0} = n,

G_{σ}(r) = (< 1/u_{r} > +\frac{1}{(−q)_{σ_{l}}} ) ∏_{i=1}^{l(σ)} < u_{σ_{i}}/u_{r} > .

Comparing with (50) and (52) and using (49) and (53) we obtain

b_{r}(z, n) = \left[{n \atop r}\right]_{q^{2}} b_{0}(zq^{2r}, n − r) b_{r}(z, r)
= (−1)^{r}q^{r+2\binom{n}{r}} \frac{(−z/q)_{2n+1}}{(z^{2}q^{2r−2}, q^{2})_{n+1}} (1 − zq^{4r−1}).

Finally, extracting the coefficients of u^{k} in the equation

φ(u) = ∑_{p=0}^{n} \frac{b_{r}(z, n)}{1 − u_{r}u},

and using the values for b_{r}(z, n) we obtain (6).

5.3 Proof of Theorem 2

Consider the generating function of the left-hand side of (7) :

φ_{ab}(u) := ∑_{k≥0} u^{k} ∑_{l(λ)≤k} z^{|λ|}q^{2n(λ)} \frac{(a, b; q^{−1})_{λ_{1}}}{(q)_{λ}(−q)_{λ_{k}}}
= ∑_{λ} ∑_{k≥0} u^{k+l(λ)}z^{|λ|}q^{2n(λ)} \frac{(a, b; q^{−1})_{λ_{1}}}{(q)_{λ}(−q)_{λ_{l(λ)+k}}} ,

where the sum is over all the partitions λ, and as before λ_{l(λ)+k} = 0 unless k = 0. Thus

φ_{ab}(u) = ∑_{λ} u^{l(λ)}z^{|λ|}q^{2n(λ)} \frac{(a, b; q^{−1})_{λ_{1}}}{(q)_{λ}} (\frac{u}{1 − u} + \frac{1}{(−q)_{λ_{l(λ)}}} ) .  (54)

As in the elementary proof of Corollary 1, we can replace any partition λ by a pair (ν, m), where ν is a strict partition consisting of distinct parts ν_{1}, ⋯, ν_{l} of λ, so that ν_{1} > ⋯ > ν_{l} > 0, and m = (m_{1}, …, m_{l}) is the sequence of multiplicities of ν_{i} for 1 ≤ i ≤ l. Therefore

φ_{ab}(u) = ∑_{ν, m} \frac{(a, b; q^{−1})_{ν_{1}}}{(q)_{ν}} (\frac{u}{1 − u} + \frac{1}{(−q)_{ν_{l}}} ) ∏_{i=1}^{l}(u_{ν_{i}}u)^{m_{i}}

= ∑_{ν} \frac{(a, b; q^{−1})_{ν_{1}}}{(q)_{ν}} (< u > +\frac{1}{(−q)_{ν_{l}}} ) ∏_{i=1}^{l} < u_{ν_{i}}u >, (55)

where the sum is over all the strict partitions ν. Each of the terms in this sum, as a rational function of u, has a finite set of simple poles, which may occur at the points u_{r}^{−1} for r ≥ 0. Therefore, each term is a linear combination of partial fractions. Moreover, the sum of their expansions converges coefficientwise. So φ_{ab} has an expansion

φ_{ab}(u) = ∑_{r≥0} \frac{c_{r}}{1 − uz^{r}q^{r(r−1)}} ,

where c_{r} denotes the formal sum of partial fraction coefficients contributed by the terms of (55). It remains to compute these residues c_{r} (r ≥ 0). By using (44) and (55), we get immediately

c_{0} = [φ_{ab}(u)(1 − u)]_{u=1} = \frac{(az, bz)_{∞}}{(z, abz)_{∞}} .

In view of (55), this yields the identity

∑_{ν} \frac{(a, b; q^{−1})_{ν_{1}}}{(q)_{ν}} ∏_{i=1}^{l} < u_{ν_{i}} >= \frac{(az, bz)_{∞}}{(z, abz)_{∞}} .  (56)

Clearly, a summand in (55) has a non zero contribution to c_{r} (r > 0) only if the corresponding partition ν has a part equal to r. For any partition ν such that ∃j | ν_{j} = r, set ρ_{i} := ν_{i} − r for 1 ≤ i < j and σ_{i} := ν_{i+j} for 0 ≤ i ≤ l − j, we then get two partitions ρ and σ, with σ_{i} bounded by r. Multiplying (55) by (1 − u_{r}u) and setting u = 1/u_{r} we obtain

c_{r} = ∑_{ρ} \frac{(a, b; q^{−1})_{ρ_{1}+r}}{(q)_{ρ}} ∏_{i=1}^{j−1} < u_{r+ρ_{i}}/u_{r} >

× ∑_{σ} \frac{1}{(q)_{σ}} (< 1/u_{r} > +\frac{1}{(−q)_{σ_{l−j}}} ) ∏_{i=1}^{l−j} < u_{σ_{i}}/u_{r} > .

In view of (51) the inner sum over σ is equal to b_{r}(z, r)/(q^{2}, q^{2})_{r}, and applying (53) we get

c_{r} = (−1)^{r}q^{r+2\binom{r}{2}} \frac{(−z/q)_{2r}}{(z^{2}q^{2r−2}, q^{2})_{r}} \frac{(a, b; q^{−1})_{r}}{(q^{2}; q^{2})_{r}}

× ∑_{ρ} \frac{(aq^{−r}, bq^{−r}; q^{−1})_{ρ_{1}}}{(q)_{ρ}} ∏_{i=1}^{j−1} < u_{r+ρ_{i}}/u_{r} > .

Now, the sum over ρ can be computed using (56) with a, b and z replaced by aq^{−r}, bq^{−r} and zq^{2r} respectively. After simplification, we obtain

c_{r} = (−1)^{r}q^{r+2\binom{r}{2}} \frac{(−z/q)_{∞}}{(z^{2}q^{2r−2}, q^{2})_{∞}} \frac{(a, b; q^{−1})_{r}}{(q^{2}; q^{2})_{r}} \frac{(azq^{r}, bzq^{r})_{∞}}{(abz)_{∞}} ,

which completes the proof.

Acknowledgement

The second and third authors are partially supported by EC’s IHRP Programme, within the Research Training Network “Algebraic Combinatorics in Europe”, grant HPRN-CT-2001-00272.

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