Continued Fractions for Rogers-Szegö polynomials

Qing-Hu Hou Center for Combinatorics, LPMC Nankai University, Tianjin 300071, P. R. China Email: contact@eyou.com

Alain Lascoux CNRS, Institut Gaspard Monge, Université de Marne-la-Vallée 77454 Marne-la-Vallée Cedex, France and Center for Combinatorics, LPMC Nankai University, Tianjin 300071, P.R. China Alain.Lascoux@univ-mlv.fr

Yan-Ping Mu Center for Combinatorics, LPMC Nankai University, Tianjin 300071, P. R. China Email: myphb@eyou.com

Abstract

We evaluate different Hankel determinants of Rogers-Szegö polynomials, and deduce from it continued fraction expansions for the generating function of RS polynomials. We also give an explicit expression of the orthogonal polynomials associated to moments equal to RS polynomials, and a decomposition of the Hankel form with RS polynomials as coefficients.

1 Definitions and Notations

Let (a)_{n} = (a; q)_{n} = (1−a)(1−aq) ⋯ (1−aq^{n−1}). The q-binomial coefficient, or the Gauss polynomial, is

\left[{n \atop k}\right] = \frac{(q; q)_{n}}{(q; q)_{k}(q; q)_{n−k}} = \frac{(q^{n} − 1)(q^{n−1} − 1) ⋯ (q^{n−k+1} − 1)}{(q^{k} − 1)(q^{k−1} − 1) ⋯ (q − 1)},

which is equal to the number of k-dimensional vector subspaces of the ndimensional vector space V_{n}(q) over the finite field GF (q).

The Rogers-Szegö polynomials are defined by

h_{n}(t, q) = ∑_{k=0}^{n} \left[{n \atop k}\right]t^{k} = (q)_{n} ∑ \frac{t^{k}}{(q)_{k}(q)_{n−k}} .  (1.1)

They are some variants of q-Hermite polynomials which play an important role in the theory of basic hypergeometric series [1]. For t = 1, Rogers-Szegö polynomials specialize to the Galois numbers of Goldman and Rota [5, p. 77] (cf. also [7]):

G_{n} = ∑_{k=0}^{n} \left[{n \atop k}\right].

Let us introduce some notations related to symmetric functions (cf. [8]). Given two sets of indeterminates A, B (we say alphabets), the complete symmetric functions S^{i}(A − B), i ∈ ℤ, of the formal difference A − B are defined by the following generating function (using another variable z) :

σ_{z}(A − B) = ∑_{i≥0} S^{i}(A − B)z^{i} = \frac{∏_{b∈B}(1 − bz)}{∏_{a∈A}(1 − az)}.  (1.2)

More generally, for any polynomial P = ∑_{(c,u)} c u, c ∈ ℂ, u monomial, and any extra variable q, we define S^{i}(P) and S^{i}(P/(1−q)) by the generating functions

σ_{z}( ∑_{(c,u)} cu) = ∑_{i} z^{i}S^{i}(P) = ∏_{(c,u)}(1 − zu)^{−c}  (1.3)

σ_{z} (\frac{P}{1 − q} ) = ∑_{i} z^{i}S^{i} (\frac{P}{1 − q} ) = ∏_{(c,u)} ∏_{j=0}^{∞}(1 − zq^{j}u)^{−c} (1.4)

With these notations, a variable x is identified with the alphabet {x} of order 1, and (1 − q)^{−1} with the infinite alphabet {1, q, q^{2}, …}. Any formal series ∑_{n≥0} z^{n}S^{n}, with S^{0} = 1, can be formally factorized, i.e. identified with σ_{z}(A) for some alphabet A such that S^{n} = S^{n}(A), n ∈ ℕ.

Our basic ring R is the ring of rational functions in two variables q, t and some alphabets A, B, . . ..

For any element θ ∈ R, any n ∈ ℕ, any I = (i_{1}, …, i_{n}) ∈ ℕ^{n}, the Schur function S_{I}(θ) is equal to the determinant [9] :

S_{I}(θ) = |S^{i_{s}+s−r}(θ)|_{1≤r,s≤n}.

2 Recursion for Rogers-Szegö Polynomials

Taking coefficients of powers of z in (1.2), one gets

S^{n}(A ± B) = ∑_{k=0}^{n} S^{k}(A) S^{n−k}(±B).  (2.1)

From (2.1), we see that the Rogers-Szegö Polynomial h_{n}(t, q) can be written as:

(q)_{n}S^{n} (\frac{1 + t}{1 − q} ) .

Lemma 2.1 Let B, C be two alphabets of order less than or equal to m. Then S_{n} := S^{n} ( \frac{B−C}{1−q} ), n ∈ ℕ, satisfies the following recurrence:

∑_{r=0}^{m} (S^{r}(−B) − S^{r}(−C)q^{n−r}) S_{n−r} = 0.  (2.2)

Proof. Writing −B + \frac{B−C}{1−q} = −C + q \frac{B−C}{1−q}, then from (2.1), together with the fact that S^{n}(−B) = 0 = S^{n}(−C), for n > m, we get that

∑ S^{r}(−B)S^{n−r} (\frac{B − C}{1 − q} ) = ∑ S^{r}(−C)q^{n−r}S^{n−r} (\frac{B − C}{1 − q} ) .

□

Lemma 2.1 implies the following recursion :

Corollary 2.2 Let h_{n}(t, q) be the Rogers-Szegö polynomial. Then

h_{n}(t, q) = (1 + t)h_{n−1}(t, q) − t(q^{n−1} − 1)h_{n−2}(t, q).  (2.3)

Proof. Taking B = {1, t}, C = ∅ in Lemma 2.1, we have

(1 − q^{n})S^{n} (\frac{1 + t}{1 − q} ) − (1 + t)S^{n−1} (\frac{1 + t}{1 − q} ) + tS^{n−2} (\frac{1 + t}{1 − q} ) = 0. (2.4)

Substituting S^{n} (\frac{1+t}{1−q} ) = h_{n}(t, q)/(q)_{n} in Equation (2.4), we derive the recursion immediately. □

Remark. Goldman and Rota [5] proved the recursion

G_{n+1} = 2G_{n} + (q^{n} − 1)G_{n−1} , G_{0} = 1 , G_{1} = 2 ,

using vector spaces over finite fields. Nijenhuis, Solow and Wilf [10] gave a bijective proof for the same recursion. K. Hikami and B. Basu-Mallick [6] gave also a recursion for generalized Rogers-Szegö polynomials.

3 Determinants of Rogers-Szegö Polynomials

Theorem 3.1 Let 𝔸 be the alphabet defined by S^{i}(𝔸) = h_{i}(t, q), i ∈ ℕ. We have

S_{n^{n}}(𝔸) = t^{\binom{n}{2}}q^{\binom{n+1}{3}}(q)_{0}(q)_{1} ⋯ (q)_{n−1}h_{n}(t, q^{−1}),  (3.1)
S_{n^{n},m}(𝔸) = (q)_{0}(q)_{1} ⋯ (q)_{n}t^{\binom{n+1}{2}}q^{\binom{n+1}{3}}\left[{m \atop n}\right]h_{m−n}(t, q). (3.2)

In particular,

S_{n^{n+1}}(𝔸) = (q)_{0}(q)_{1} ⋯ (q)_{n}t^{\binom{n+1}{2}}q^{\binom{n+1}{3}}.  (3.3)

Proof. Denote X = \frac{1+t}{1−q}. By definition,

S_{n^{n}}(𝔸) = |h_{n+j−i}(t, q)|_{1≤i,j≤n}
= |(q)_{n+j−i}S^{n+j−i}(X)|_{1≤i,j≤n}
= (q)_{1} ⋯ (q)_{n} · |(q^{n−i+2})_{j−1}S^{n+j−i}(X)|_{1≤i,j≤n}. (3.4)

We can write recursion (2.4) as

(q^{k})_{m}S^{k+m−1}(X)

= (1 + t)(q^{k})_{m−1}S^{k+m−2}(X) − t(1 − q^{m−1})(q^{k})_{m−2}S^{k+m−3}(X)
−tq^{m−1}(q^{k−1})_{m−1}S^{k+m−3}(X).  (3.5)

Applying identity (3.5) on columns n, n − 1, …, 2 of the determinant in (3.4) successively, it becomes (−t)^{n−1}q^{n−2}q^{n−3} ⋯ q^{0}

× | \begin{matrix}S^{n}(X) & S^{n−1}(X) & (q^{n})_{1}S^{n}(X) & ⋯ & (q^{n})_{n−2}S^{2n−3}(X) \\ S^{n−1}(X) & S^{n−2}(X) & (q^{n−1})_{1}S^{n−1}(X) & ⋯ & (q^{n−1})_{n−2}S^{2n−4}(X) \\ ⋯ & ⋯ & ⋯ & ⋯ \\ S^{2}(X) & S^{1}(X) & (q^{2})_{1}S^{2}(X) & ⋯ & (q^{2})_{n−1}S^{n−1}(X) \\ S^{1}(X) & S^{0}(X) & (q)_{1}S^{1}(X) & ⋯ & (q)_{n−2}S^{n−2}(X) \end{matrix} | .

Repeating the same transformation, we will finally reach

(−t)^{\binom{n}{2}}q^{\binom{n}{3}} | \begin{matrix}S^{n}(X) & S^{n−1}(X) & ⋯ & S^{1}(X) \\ S^{n−1}(X) & S^{n−2}(X) & ⋯ & S^{0}(X) \\ ⋯ & ⋯ & ⋯ \\ S^{2}(X) & S^{1}(X) & ⋯ & 0 \\ S^{1}(X) & S^{0}(X) & ⋯ & 0 \end{matrix} | = t^{\binom{n}{2}}q^{\binom{n}{3}}S_{1^{n}}(X).

Noting that

S_{1^{n}}(X) = (−1)^{n}S^{n} (−\frac{1 + t}{1 − q} ) = \frac{q^{\binom{n}{2}}}{(q)_{n}} h_{n}(t, q^{−1}),

we obtain

S_{n^{n}}(𝔸) = t^{\binom{n}{2}}q^{\binom{n+1}{3}}(q)_{1} ⋯ (q)_{n−1}h_{n}(t, q^{−1}).

For m ≥ 0, define

W_{n,m} = \frac{S_{n^{n},m}(𝔸)}{(q)_{0}(q)_{1} ⋯ (q)_{n}}

= | \begin{matrix}S^{n}(X) & (q^{n+1})_{1}S^{n+1}(X) & ⋯ & (q^{n+1})_{n−1}S^{2n−1}(X) & (q^{n+1})_{m}S^{n+m}(X) \\ S^{n−1}(X) & (q^{n})_{1}S^{n}(X) & ⋯ & (q^{n})_{n−1}S^{2n−2}(X) & (q^{n})_{m}S^{n+m−1}(X) \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ S^{1}(X) & (q^{2})_{1}S^{2}(X) & ⋯ & (q^{2})_{n−1}S^{n}(X) & (q^{2})_{m}S^{m+1}(X) \\ S^{0}(X) & (q)_{1}S^{1}(X) & ⋯ & (q)_{n−1}S^{n−1}(X) & (q)_{m}S^{m}(X) \end{matrix} |.

Applying identity (3.5) on the last column of W_{n,m}, we transform it into

W_{n,m} = (1 + t)W_{n,m−1} − t(1 − q^{m−1})W_{n,m−2}

+ | \begin{matrix}S^{n}(X) & ⋯ & (q^{n+1})_{n−1}S^{2n−1}(X) & (q^{n})_{m−1}S^{n+m−2}(X) \\ S^{n−1}(X) & ⋯ & (q^{n})_{n−1}S^{2n−2}(X) & (q^{n−1})_{m−1}S^{n+m−3}(X) \\ ⋯ & ⋯ & ⋯ & ⋯ \\ S^{1}(X) & ⋯ & (q^{2})_{n−1}S^{n}(X) & (q)_{m−1}S^{m−1}(X) \\ S^{0}(X) & ⋯ & (q)_{n−1}S^{n−1}(X) & 0 \end{matrix} |

After successive such transformations, one finally obtains

W_{n,m} = (1+t)W_{n,m−1}−t(1−q^{m−1})W_{n,m−2}+t^{n}q^{m−1}q^{\binom{n−1}{2}}W_{n−1,m−1}. (3.6)

Noting that

t^{\binom{n+1}{2}}q^{\binom{n+1}{3}}\left[{m \atop n}\right]h_{m−n}(t, q)

satisfies the recursion (3.6) and coincides with W_{n,m} for (0, m), (n, 0) and (n, 1), we get (3.2). □

4 Continued Fractions for Rogers-Szegö Polynomial

There are three main types of continued fraction expressions for formal series. They are all obtained by evaluating some Hankel determinants. We refer to Stieltjes [11], Wall [12] and Brezinski [4]. The determinants given by Theorem 3.1 allow us to write the following three continued fractions.

Theorem 4.1 The generating function of Rogers-Szegö polynomials h_{n}(t, q) is equal to each of the three following continued fractions:

∑_{n=0}^{∞} h_{n}(t, q)z^{n} = \frac{1}{1 − \frac{zγ_{1}}{1 − \frac{zγ_{2}}{1 − \frac{zγ_{3}}{⋯}}}}

= \frac{1}{1 − ζ_{0}z + \frac{β_{1}z^{2}}{1 − ζ_{1}z + \frac{β_{2}z^{2}}{1 − ζ_{2}z + \frac{β_{3}z^{2}}{⋯}}}} ,

∑_{n=0}^{∞} h_{n}(t, q)z^{−n−1} = \frac{1}{z + \frac{1}{−\frac{1}{α_{1}} + \frac{1}{\frac{z}{α_{2}} + \frac{1}{−\frac{1}{α_{3}} + ⋯}}}} ,

where

α_{m} = \{ \begin{matrix}(−1)^{n−1} \frac{q^{\binom{n}{2}}h_{n}(t, q^{−1})h_{n−1}(t, q^{−1})}{t^{n−1}(q)_{n−1}}, & m = 2n − 1, \\ (−1)^{n} \frac{t^{n}(q)_{n}}{q^{\binom{n}{2}}h_{n}^{2}(t, q^{−1})}, & m = 2n, \end{matrix}

γ_{m} = \{ \begin{matrix}\frac{q^{n−1}h_{n}(t, q^{−1})}{h_{n−1}(t, q^{−1})}, & m = 2n − 1, \\ −t(1 − q^{n})\frac{h_{n−1}(t, q^{−1})}{h_{n}(t, q^{−1})}, & m = 2n, \end{matrix}

β_{n} = q^{n−1}t(1 − q^{n}),
ζ_{n} = q^{n}(1 + t).

Remark. In our notations, Stieltjes’ parameters, for the series ∑ z^{n}S^{n}(𝔸), are [8]

α_{m} = \{ \begin{matrix}(−1)^{n−1} \frac{S_{n^{n}}(𝔸)S_{(n−1)^{n−1}}(𝔸)}{S_{(n−1)^{n}}(𝔸)^{2}}, & m = 2n − 1, \\ (−1)^{n} \frac{S_{n^{n+1}}(𝔸)S_{(n−1)^{n}}(𝔸)}{S_{n^{n}}(𝔸)^{2}}, & m = 2n, \end{matrix}

β_{n} = \frac{S_{n^{n+1}}(𝔸)S_{(n−2)^{n−1}}(𝔸)}{S_{(n−1)^{n}}(𝔸)^{2}} = −γ_{2n}γ_{2n−1},
γ_{n} = α_{n}α_{n−1},
ζ_{n} = \frac{S_{n^{n},n+1}(𝔸)}{S_{n^{n+1}}(𝔸)} − \frac{S_{(n−1)^{(n−1)},n}(𝔸)}{S_{(n−1)^{n}}(𝔸)}.

5 Hankel Forms

To any alphabet 𝔸 is associated a Hankel form

H(𝔸; x) = ∑_{i,j=0}^{∞} S_{i+j}(𝔸) x_{i}x_{j} .

The problem of decomposing Hankel forms into sums of squares has also been solved by Stieltjes [3]. He found the coefficients, that we denote by ⋆, such that

H(𝔸; x) = (x_{0} + ⋆x_{1} + ⋯)^{2} + ⋆(x_{1} + ⋆x_{2} + ⋯)^{2} + ⋆(x_{2} + ⋆x_{3} + ⋯)^{2} + · · · .

Theorem 5.1 (Stieltjes) Let 𝔸 be a generic alphabet. With 𝔸^{0} = 𝔸, define the alphabets 𝔸^{1}, 𝔸^{2}, … by the generating series

σ_{z}(𝔸^{k}) = S_{k^{k+1}}(𝔸)^{−1} ∑_{j=0}^{∞} S_{k^{k}, k+j}(𝔸)z^{j} .

Then the form K(𝔸; x, y) = ∑_{i,j=0}^{∞} S_{i+j}(𝔸)x^{i}y^{j} decomposes as

K(A; x, y) = σ_{1}((x + y)𝔸^{0}) − xyS_{11}(𝔸) σ_{1}((x + y)𝔸^{1}) + ⋯

+ ⋯ + (−xy)^{n} \frac{S_{n^{n+1}}(𝔸)}{S_{(n−1)^{n}}(𝔸)}σ_{1}((x + y)𝔸^{n}) + · · · .

Substituting Equation (3.2) into the above identity, we instantly derive that

Corollary 5.2 For Rogers-Szegö polynomials h_{n}(t, q), we have

∑_{i,j=0}^{∞} h_{i+j}(t, q)x^{i}y^{j} = ∑_{n=0}^{∞}(−t)^{n}(q)_{n}q^{\binom{n}{2}}

× (∑_{u=0}^{∞} h_{u}(t, q)\left[{n + u \atop u}\right]x^{n+u}) (∑_{v=0}^{∞} h_{v}(t, q)\left[{n + v \atop v}\right]y^{n+v})

Comparing the coefficients of t^{m}s^{n} of both sides, we get the following formula due to Askey and Ismail [2]:

h_{m+n}(t, q) = ∑_{k=0}^{min\{m,n\}} \left[{n \atop k}\right]\left[{m \atop k}\right](−t)^{k}(q)_{k}q^{\binom{k}{2}}h_{m−k}(t, q)h_{n−k}(t, q).

6 Orthogonal Polynomials

Let ∫ be a linear functional on polynomials in x such that there exists a family of orthogonal polynomials {P_{n}(x)}:

∫ P_{m}(x)P_{n}(x) = 0 if m ≠ n , ∫ P_{n}(x)P_{n}(x) = c_{n} ≠ 0 .

c_{n} is called the normalization constant.

Identifying the moments ∫ x^{n} to the complete functions S^{n}(A) of some alphabet A, we can rewrite the classical expressions of orthogonal polynomials in term of symmetric functions:

P_{n}(x) = S_{n^{n}}(A − x), c_{n} = (−1)^{n}S_{(n−1)^{n}}(A) S_{n^{n+1}}(A), n ≥ 0. (6.1)

Theorem 6.1 Let ∫_{𝔸} be the linear functional with moments equal to the Rogers-Szegö polynomials: ∫_{𝔸} x^{n} = h_{n}(t, q). Then

P_{n}(x) = ∑_{k=0}^{n}(−1)^{k}\left[{n \atop k}\right]q^{\binom{k}{2}}h_{k}(t, q^{−1})x^{n−k} = (q)_{n}S^{n} (\frac{x − 1 − t}{1 − q}) ,

are its associated orthogonal polynomials, with normalization constant

c_{n} = t^{n}q^{\binom{n}{2}}(q − 1)(q^{2} − 1) ⋯ (q^{n} − 1).

Proof. Start with

S_{n^{n}}(𝔸 − x) = | \begin{matrix}S^{n}(𝔸) & S^{n+1}(𝔸) & ⋯ & S^{2n−1}(𝔸) & x^{n} \\ S^{n−1}(𝔸) & S^{n}(𝔸) & ⋯ & S^{2n−2}(𝔸) & x^{n−1} \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ S^{1}(𝔸) & S^{2}(𝔸) & ⋯ & S^{n}(𝔸) & x^{1} \\ S^{0}(𝔸) & S^{1}(𝔸) & ⋯ & S^{n−1}(𝔸) & 1 \end{matrix} | ,

and write X = \frac{1+t}{1−q}. The same kind of transformation as for Theorem 3.1 allows us to write the preceding determinant as

(q)_{0}(q)_{1} ⋯ (q)_{n} | \begin{matrix}S^{n}(X) & ⋯ & (q^{n+1})_{n−1}S^{2n−1}(X) & x^{n}/(q)_{n} \\ S^{n−1}(X) & ⋯ & (q^{n})_{n−1}S^{2n−2}(X) & x^{n−1}/(q)_{n−1} \\ ⋯ & ⋯ & ⋯ & ⋯ \\ S^{1}(X) & ⋯ & (q^{2})_{n−1}S^{n}(X) & x^{1}/(q)_{1} \\ S^{0}(X) & ⋯ & (q)_{n−1}S^{n−1}(X) & 1 \end{matrix} |

= (q)_{0}(q)_{1} ⋯ (q)_{n}(−t)^{\binom{n}{2}}q^{\binom{n}{3}}
| \begin{matrix}S^{n}(X) & S^{n−1}(X) & ⋯ & S^{1}(X) & x^{n}/(q)_{n} \\ S^{n−1}(X) & S^{n−2}(X) & ⋯ & S^{0}(X) & x^{n−1}/(q)_{n−1} \\ ⋯ & ⋯ & ⋯ & ⋯ & ⋯ \\ S^{1}(X) & S^{0}(X) & ⋯ & 0 & x^{1}/(q)_{1} \\ S^{0}(X) & 0 & ⋯ & 0 & 1 \end{matrix} |
= (q)_{1} ⋯ (q)_{n}t^{\binom{n}{2}}q^{\binom{n}{3}} ∑_{k=0}^{n}(−1)^{n−k} \frac{x^{n−k}}{(q)_{n−k}} S_{1^{k}}(X)
= (−1)^{n}(q)_{1} ⋯ (q)_{n−1}t^{\binom{n}{2}}q^{\binom{n}{3}} ∑_{k=0}^{n}(−1)^{k}\left[{n \atop k}\right]q^{\binom{k}{2}}h_{k}(t, q^{−1})x^{n−k},

After normalization, we finally get the polynomials

P_{n}(x) = ∑_{k=0}^{n}(−1)^{k}\left[{n \atop k}\right]q^{\binom{k}{2}}h_{k}(t, q^{−1})x^{n−k} = (q)_{n}S^{n} (\frac{x − 1 − t}{1 − q}) ,

with normalization constant

c_{n} = t^{n}q^{\binom{n}{2}}(q − 1)(q^{2} − 1) ⋯ (q^{n} − 1).

□

Acknowledgments. This work was done under the auspices of the National 973 Project on Mathematical Mechanization and the National Science Foundation of China.

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