# A $q$-Morris constant term identity for the Lie algebra $A_{n}$ and its
symmetric function generalizations

Yue Zhou  School of Mathematics and Statistics, HNP-LAMA, Central South
University, Changsha 410083, P.R. China<EMAIL_ADDRESS>
(Date: March 12, 2023)

###### Abstract.

It is well-known that the Selberg integral is equivalent to the Morris
constant term identity. In 2009 Warnaar obtained the Selberg integral for the
Lie algebra $A_{n}$. In this paper, from the point view of constant term
identities, we obtain a $q$-Morris constant term identity of type $A_{n}$ and
its several symmetric function generalizations. The type $A_{n}$ $q$-Morris
identity looks as if a constant term version of Warnaar’s $A_{n}$ Selberg
integral to some extend.

Keywords: constant term identity; $q$-Morris identity; symmetric function;
Selberg integral; Warnaar’s $A_{n}$ Selberg integral.

###### 2010 Mathematics Subject Classification:

05A30, 33D70, 05E05

##  1\. Introduction

In 1944, Atle Selberg [13] gave the following remarkable multiple integral:

(1.1)
$\int_{0}^{1}\cdots\int_{0}^{1}\prod_{i=1}^{k}z_{i}^{\alpha-1}(1-z_{i})^{\beta-1}\prod_{1\leq
i<j\leq k}|z_{i}-z_{j}|^{2\gamma}\mathrm{d}z_{1}\cdots\mathrm{d}z_{k}\\\
=\prod_{j=0}^{k-1}\frac{\Gamma(\alpha+j\gamma)\Gamma(\beta+j\gamma)\Gamma\big{(}1+(j+1)\gamma\big{)}}{\Gamma\big{(}\alpha+\beta+(k+j-1)\gamma\big{)}\Gamma(1+\gamma)},$

where $\alpha,\beta,\gamma$ are complex parameters such that

$\mathrm{Re}(\alpha)>0,\quad\mathrm{Re}(\beta)>0,\quad\mathrm{Re}(\gamma)>-\min\\{1/n,\mathrm{Re}(\alpha)/(n-1),\mathrm{Re}(\beta)/(n-1)\\}.$

When $k=1$, the above Selberg integral reduces to the Euler beta integral.

It is well-known that the Selberg integral is equivalent to the Morris
constant term identity [11]

(1.2)
$\operatorname*{CT}_{z_{0},\dots,z_{k}}\prod_{i=1}^{k}(1-z_{0}/z_{i})^{a}(1-z_{i}/z_{0})^{b}\prod_{1\leq
i\neq j\leq
k}(1-z_{i}/z_{j})^{c}=\prod_{i=0}^{k-1}\frac{(a+b+ic)!\big{(}(i+1)c\big{)}!}{(a+ic)!(b+ic)!c!}$

for nonnegative integers $a,b,c$, where
$\operatorname*{CT}\limits_{z_{i}}L(z_{i})$ denotes taking the constant term
of the Laurent polynomial (series) $L(z_{i})$. Note that we can set $z_{0}=1$
in (1.2) since we take the constant term of a homogeneous Laurent polynomial.
In his Ph.D. thesis [11], Morris also conjectured the following $q$-analogue
constant term identity

(1.3)
$\operatorname*{CT}_{z_{0},\dots,z_{k}}\prod_{i=1}^{k}(z_{0}/z_{i})_{a}(qz_{i}/z_{0})_{b}\prod_{1\leq
i<j\leq
k}(z_{i}/z_{j})_{c}(qz_{j}/z_{i})_{c}=\prod_{i=0}^{k-1}\frac{(q)_{a+b+ic}(q)_{(i+1)c}}{(q)_{a+ic}(q)_{b+ic}(q)_{c}},$

where $(y)_{c}=(y;q)_{c}:=(1-y)(1-yq)\cdots(1-yq^{c-1})$ is the $q$-factorial
for a positive integer $c$ and $(y)_{0}:=1$. In 1988, Habsieger [6] and Kadell
[8] independently proved Askey’s conjectured $q$-analogue of the Selberg
integral [3]. Expressing their $q$-analogue integral as a constant term
identity they thus proved Morris’ $q$-constant term conjecture (1.3). Now the
constant term identity (1.3) is called the Habsieger–Kadell $q$-Morris
identity.

For nonnegative integers $k_{1},k_{2},\dots,k_{n}$, we write

$z^{(s)}=(z_{1}^{(s)},\dots,z_{k_{s}}^{(s)})$

and $z=(z^{(1)},\dots,z^{(n)})$. We also make use of the Vandermonde-type
products for alphabets $x=(x_{1},\dots,x_{l})$ and $y=(y_{1},\dots,y_{m})$,
which are given by

$\Delta(x)=\prod_{1\leq i<j\leq
l}(x_{i}-x_{j}),\qquad\Delta(x,y)=\prod_{i=1}^{l}\prod_{j=1}^{m}(x_{i}-y_{j}).$

In 2009, based on [14], Warnaar [15] obtained a Selberg integral for the Lie
algebra $A_{n}$, see Theorem 1.1 below. Then the original Selberg integral
(1.1) corresponds to the $A_{1}$ case, and Tarasov and Varchenko’s result [14]
is the $A_{2}$ case.

######  Theorem 1.1.

For a positive integer $n$ let $k_{1}\geq k_{2}\geq\dots\geq k_{n}\geq 0$ be
integers and $k_{0}=k_{n+1}=0$. Let
$\alpha,\beta_{1},\dots,\beta_{n},\gamma\in\mathbb{C}$ be such that

$\mathrm{Re}(\alpha)>0,\quad\mathrm{Re}(\beta_{1})>0,\dots,\mathrm{Re}(\beta_{n})>0,\quad-\min\Big{\\{}\frac{\mathrm{Re}(\alpha)}{k_{1}-1},\frac{1}{k_{1}}\Big{\\}}<\mathrm{Re}(\gamma)<\frac{1}{k_{1}}$

and

$-\frac{\mathrm{Re}(\beta_{s})}{k_{s}-k_{s+1}-1}<\mathrm{Re}(\gamma)<\frac{\beta_{r}+\dots+\beta_{s}}{s-r}$

for $1\leq r\leq s\leq n$. Then

(1.4)
$\displaystyle\int_{C_{\gamma}^{k_{n},\dots,k_{1}}[0,1]}\prod_{s=1}^{n}\Big{(}|\Delta(z^{(s)})|^{2\gamma}\prod_{i=1}^{k_{s}}(z_{i}^{(s)})^{\alpha_{s}-1}(1-z_{i}^{(s)})^{\beta_{s}-1}\Big{)}\prod_{s=1}^{n-1}|\Delta(z^{(s+1)},z^{(s)})|^{-\gamma}\mathrm{d}z$
$\displaystyle\qquad=\prod_{1\leq r\leq s\leq
n}\prod_{i=1}^{k_{s}-k_{s+1}}\frac{\Gamma\big{(}\beta_{r}+\dots+\beta_{s}+(i+r-s-1)\gamma\big{)}}{\Gamma\big{(}\alpha_{r}+\beta_{r}+\dots+\beta_{s}+(i+r-s+k_{r}-k_{r-1}-2)\gamma\big{)}}$
$\displaystyle\qquad\quad\times\prod_{s=1}^{n}\prod_{i=1}^{k_{s}}\frac{\Gamma\big{(}\alpha_{s}+(i-k_{s-1}-1)\gamma\big{)}\Gamma(i\gamma)}{\Gamma(\gamma)},$

where $\alpha_{1}=\alpha$, $\alpha_{2}=\dots=\alpha_{n}=1$ and
$\mathrm{d}z=\mathrm{d}z^{(1)}\dots\mathrm{d}z^{(n)}$.

We omit the explicit expression for the integration domain
$C_{\gamma}^{k_{n},\dots,k_{1}}[0,1]$ due to its complexity. The reader who is
interested can refer [15, (4.9)]. Recently, Albion, Rains and Warnaar [1]
generalized the above integral to an AFLT type by adding a product of two Jack
polynomials.

For nonnegative integers $a,b,k_{1},k_{2}$ and a positive integer $c$, let

(1.5)
$L_{k_{1},k_{2}}(a,b,c;z^{(1)},z^{(2)})=\frac{\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{b}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}.$

Denote

(1.6)
$L_{k_{1},k_{2}}(a,b,c):=\operatorname*{CT}_{z^{(1)}}L_{k_{1},k_{2}}(a,b,c;z^{(1)},z^{(2)}),$

where we suppress the variables $z^{(2)}$ in the left-hand side. When taking
constant terms, we always explain $(1-dz_{i}^{(s)}/z_{j}^{(s+1)})^{-1}$ as

$\sum_{i=0}^{\infty}(dz_{i}^{(s)}/z_{j}^{(s+1)})^{i}$

for $d\in\mathbb{C}(q)$ and $s=1,\dots,n$ throughout this paper. That is, we
assume all terms of the form $|dz_{i}^{(s)}/z_{j}^{(s+1)}|<1$. In (1.6), we
need the explanation of $(1-dz_{i}^{(1)}/z_{j}^{(2)})^{-1}$ to obtain the
constant term. By convention, we define the product of zero term as 1. Then

$L_{k_{1},0}(a,b,c)=\operatorname*{CT}_{z^{(1)}}\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{b}\prod_{1\leq
i<j\leq k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}$

equals

(1.7)
$M_{k_{1}}(a,b,c)=\prod_{i=0}^{k_{1}-1}\frac{(q)_{a+b+ic}(q)_{(i+1)c}}{(q)_{a+ic}(q)_{b+ic}(q)_{c}}$

by the $q$-Morris identity (1.3).

For $n$ a positive integer, let $k_{1},\dots,k_{n+1}$, $a_{1},\dots,a_{n}$ and
$b_{1},\dots,b_{n}$ be nonnegative integers such that $k_{1}\geq
k_{2}\geq\cdots\geq k_{n+1}$, $a_{1}=a,a_{2}=\cdots=a_{n}=0$, and denote
$\sigma_{s}=b_{1}+\cdots+b_{s}$ for $s=1,\dots,n$ and $\sigma_{0}=0$. In this
paper, we find that

######  Theorem 1.2.

If $a+\sigma_{s}+s\geq sc$ for $s=1,\dots,n$ and $k_{n+1}=0$, then

(1.8)
$\operatorname*{CT}_{z}\prod_{s=1}^{n}\frac{\prod_{i=1}^{k_{s}}(1/z_{i}^{(s)})_{a_{s}}(qz_{i}^{(s)})_{b_{s}}\prod_{1\leq
i<j\leq
k_{s}}(z_{i}^{(s)}/z_{j}^{(s)})_{c}(qz_{j}^{(s)}/z_{i}^{(s)})_{c}}{\prod_{i=1}^{k_{s}}(z_{i}^{(s)})^{b_{s}+1-c}\prod_{j=1}^{k_{s+1}}\prod_{i=1}^{k_{s}}(q^{b_{s}+1-c}z_{i}^{(s)}/z_{j}^{(s+1)})_{c}}\\\
\quad=(-1)^{\sum_{s=1}^{n}k_{s}(b_{s}+1-c)}q^{\sum_{s=1}^{n}k_{s}\binom{b_{s}+2-c}{2}}\prod_{s=1}^{n}M_{k_{s}}\big{(}a+\sigma_{s}+s(1-c),c-1,c\big{)}.$

If readers compare the similarity of (1.8) and (1.4) with the equivalence
between the Morris identity (1.2) and the Selberg integral (1.1), they may
find that the constant term identity (1.8) looks as if a constant term version
of Warnaar’a $A_{n}$ Selberg integral (1.4). But we do not know how to connect
them so far. Note that in this paper, we obtain more general results than
Theorem 1.2 by adding symmetric functions, see Theorem 6.4 and Theorem 6.5
below.

The structure of this paper is as follows. In the next section, we introduce
some basic notation and results of symmetric functions and plethystic
notation. In Section 3, we introduce the main tool of this paper — a splitting
formula. Using the formula, we find a family of vanishing constant terms. In
Section 4, we determine the expression of $L_{k_{1},k_{2}}(a,b,c)$. In Section
5, we study the $b=c-1$ case of $L_{k_{1},k_{2}}(a,b,c)$. In the last section,
we obtain several results equivalent to those in Section 5.

##  2\. Symmetric functions and plethystic notation

In this section, we introduce some basic results of symmetric functions and
plethystic notation.

A partition is a sequence $\lambda={(\lambda_{1},\lambda_{2},\dots)}$ of
nonnegative integers such that ${\lambda_{1}\geq\lambda_{2}\geq\cdots}$ and
only finitely many $\lambda_{i}$ are positive. The length of a partition
$\lambda$, denoted $\ell(\lambda)$ is defined to be the number of nonzero
$\lambda_{i}$. We adopt the convention of not displaying the tail of zeros of
a partition. We say that $|\lambda|=\lambda_{1}+\lambda_{2}+\cdots$ is the
size of the partition $\lambda$. The most commonly used partial order for
partitions should be the dominance order, denoted $\leq$. If $\lambda$ and
$\mu$ are partitions of the same size, then $\lambda\leq\mu$ if
$\lambda_{1}+\cdots+\lambda_{i}\leq\mu_{1}+\cdots+\mu_{i}$ for all $i\geq 1$.

For a field $\mathbb{F}$, denote $\Lambda_{\mathbb{F}}$ by the ring of
symmetric functions with coefficients in $\mathbb{F}$. In the following, we
introduce four bases of $\Lambda$ (with $\mathbb{F}=\mathbb{Q}$ and
$\mathbb{F}=\mathbb{Q}(q,t)$).

Let $X=\\{x_{1},x_{2},\dots\\}$ be an alphabet (a set of countably many
variables). For $r$ a positive integer, let $p_{r}$ be the power sum symmetric
function in $X$, defined by

$p_{r}=\sum_{i\geq 1}x_{i}^{r}.$

In addition, we set $p_{0}=1$. For a partition
$\lambda=(\lambda_{1},\lambda_{2},\dots)$, let

$p_{\lambda}=p_{\lambda_{1}}p_{\lambda_{2}}\cdots.$

The $p_{r}$ are algebraically independent over $\mathbb{Q}$, and the
$p_{\lambda}$ form a basis of $\Lambda_{\mathbb{Q}}$ [10]. That is,

$\Lambda_{\mathbb{Q}}=\mathbb{Q}[p_{1},p_{2},\dots].$

The elementary symmetric function is defined by

$e_{r}=\sum_{1\leq i_{1}<\dots<i_{r}}x_{i_{1}}\cdots x_{i_{r}},\quad
e_{0}=1,\quad\mbox{and}\quad
e_{\lambda}=e_{\lambda_{1}}e_{\lambda_{2}}\cdots.$

The $e_{\lambda}$ form a basis of $\Lambda_{\mathbb{Q}}$ [10].

The complete symmetric function $h_{r}(X)$ can be defined in terms of its
generating function as

(2.1) $\sum_{r\geq 0}h_{r}(X)y^{r}=\prod_{i\geq 0}\frac{1}{1-yx_{i}}.$

Like $p_{\lambda}$ and $e_{\lambda}$, define

$h_{\lambda}=h_{\lambda_{1}}h_{\lambda_{2}}\cdots.$

The $h_{\lambda}$ also form a basis of $\Lambda_{\mathbb{Q}}$ [10]. (The
$e_{\lambda}$ and the $h_{\lambda}$ are in fact bases of
$\Lambda_{\mathbb{Z}}$, but we do not need these in this paper.)

Let $F=\mathbb{Q}(q,t)$ be the field of rational functions in $q$ and $t$ with
coefficients in $\mathbb{Q}$ and
$(y)_{\infty}=(y;q)_{\infty}:=(1-y)(1-yq)\cdots$ be the infinity
$q$-factorial. Define the modified complete symmetric function
$g_{r}(X;q,t)\in\Lambda_{F}(X)$ by its generating function

$\sum_{r\geq 0}g_{r}(X;q,t)y^{r}=\prod_{i\geq
1}\frac{(tx_{i}y;q)_{\infty}}{(x_{i}y;q)_{\infty}},$

and for a partition $\lambda=(\lambda_{1},\lambda_{2},\dots)$ define

$g_{\lambda}=g_{\lambda}(q,t)=g_{\lambda}(X;q,t):=\prod_{i\geq
1}g_{\lambda_{i}}(X;q,t).$

By [10, Chapter VI, (2.12)], the $g_{\lambda}$ form a basis of $\Lambda_{F}$.

Plethystic or $\lambda$-ring notation is a device to facilitate computations
in the ring of symmetric functions. We briefly introduce plethystic notation.
For more details, see [7, 9, 12].

For an alphabet $X=\\{x_{1},x_{2},\dots\\}$ and a field $\mathbb{F}$, we
additively write $X:=x_{1}+x_{2}+\cdots$, and use plethystic brackets to
indicate this additive notation:

$f(X)=f(x_{1},x_{2},\dots)=f[x_{1}+x_{2}+\cdots]=f[X],\quad\text{for
$f\in\Lambda_{\mathbb{F}}$.}$

We introduce a consistent arithmetic on alphabets in terms of the basis of
power sums. In particular, a power sum whose argument is the sum, difference
or Cartesian product of two alphabets $X$ and $Y$ is defined as

(2.2a) $\displaystyle p_{r}[X+Y]$ $\displaystyle=p_{r}[X]+p_{r}[Y],$ (2.2b)
$\displaystyle p_{r}[X-Y]$ $\displaystyle=p_{r}[X]-p_{r}[Y],$ (2.2c)
$\displaystyle p_{r}[XY]$ $\displaystyle=p_{r}[X]p_{r}[Y].$

In general we cannot give meaning to division by an arbitrary alphabet and
only division by $1-t$ (the difference of two one-letter alphabets with
“letters” $1$ and $t$ respectively) is meaningful. In particular

(2.3) $p_{r}\Big{[}\frac{X}{1-t}\Big{]}=\frac{p_{r}[X]}{1-t^{r}}.$

Note that the alphabet $1/(1-t)$ may be interpreted as the infinite alphabet
$1+t+t^{2}+\cdots$. Indeed, by (2.2a) and (2.2c)

$p_{r}[X(1+t+t^{2}+\cdots)]=p_{r}[X]\sum_{k=0}^{\infty}p_{r}[t^{k}]=p_{r}[X]\sum_{k=0}^{\infty}t^{kr}=\frac{p_{r}[X]}{1-t^{r}}.$

If $f$ is a homogeneous symmetric function of degree $k$ then

(2.4) $f[aX]=a^{k}f[X]$

for a single-letter alphabet $a$. Note that by (2.2a), $f[2X]=f[X+X]=2f[X]$.
Moreover, we can extend this to

(2.5) $f[rX]=rf[X]$

for $r\in\mathbb{F}$. Note that this leads to some notational ambiguities, and
whenever not clear from the context we will indicate if a symbol such as $a$
or $r$ represents a letter or a binomial element111In [9, p. 32] Lascoux
refers to $r\in\mathbb{F}$ as a binomial element.. Throughout this paper, we
take $q$ and $q^{-1}$ as single-letter alphabets.

We need the next simple result. One can find a proof in [7, Theorem 1.27].

######  Proposition 2.1.

Let $X$ and $Y$ be two alphabets. For $r$ a nonnegative integer,

(2.6) $\displaystyle h_{r}[X+Y]$
$\displaystyle=\sum_{i=0}^{r}h_{i}[X]h_{r-i}[Y],$ (2.7) $\displaystyle
h_{r}[-X]$ $\displaystyle=(-1)^{r}e_{r}[X].$

Using plethystic notation, we can write $g_{r}$ as

(2.8) $g_{r}(X;q,t)=h_{r}\Big{[}\frac{1-t}{1-q}X\Big{]}.$

By (2.6) and (2.8), it is easy to obtain the next result.

######  Proposition 2.2.

Let $X$ and $Y$ be two alphabets. For $r$ a nonnegative integer,

(2.9) $g_{r}[X+Y]=\sum_{i=0}^{r}g_{i}[X]g_{r-i}[Y].$

###### Proof.

By (2.8),

$g_{r}[X+Y]=h_{r}\Big{[}\frac{1-t}{1-q}(X+Y)\Big{]}=\sum_{i=0}^{r}h_{r-i}\Big{[}\frac{1-t}{1-q}X\Big{]}h_{i}\Big{[}\frac{1-t}{1-q}Y\Big{]}.$

Here the last equality holds by (2.6). Using (2.8) again, we obtain (2.9). ∎

##  3\. A splitting formula

In this section, we give a splitting formula for a rational function
$F_{n}(y,w)$ defined in (3.2) below. Using the splitting formula, we find a
family of vanishing constant terms.

Before presenting the splitting formula, we need the next simple result.

######  Lemma 3.1.

Let $i$ and $j$ be positive integers. Then, for $t$ an integer such that
$0\leq t\leq j$,

(3.1a)
$\frac{(1/y)_{i}(qy)_{j}}{(q^{-t}/y)_{i}}=q^{it}(q^{1-i}y)_{t}(q^{t+1}y)_{j-t},$
and for $-1\leq t\leq j-1$, (3.1b)
$\frac{(y)_{j}(q/y)_{i}}{(q^{-t}/y)_{i}}=q^{i(t+1)}(q^{-i}y)_{t+1}(q^{t+1}y)_{j-t-1}.$

Note that the $t=j$ case of (3.1a) (taking $y\mapsto y/q$) is the standard
fact in [5, Equation (I.13)].

###### Proof.

For $0\leq t\leq j$,

$\frac{(1/y)_{i}(qy)_{j}}{(q^{-t}/y)_{i}}=\frac{(q^{i-t}/y)_{t}(qy)_{j}}{(q^{-t}/y)_{t}}=\frac{(-1/y)^{t}q^{it-\binom{t+1}{2}}(q^{1-i}y)_{t}(qy)_{j}}{(-1/y)^{t}q^{-\binom{t+1}{2}}(qy)_{t}}=q^{it}(q^{1-i}y)_{t}(q^{t+1}y)_{j-t}.$

Taking $y\mapsto y/q$ and $t\mapsto t+1$ in (3.1a) yields (3.1b) for $-1\leq
t\leq j-1$. ∎

For positive integers $n,c$, let

(3.2) $F_{n}(y,w)=\frac{\prod_{1\leq i<j\leq
n}(y_{i}/y_{j})_{c}(qy_{j}/y_{i})_{c}}{\prod_{l=1}^{n}(y_{l}/w)_{c}},$

where $w$ is a parameter such that all terms of the form $q^{u}y_{i}/w$ in
(3.2) satisfy $|q^{u}y_{i}/w|<1$. Hence,

$\frac{1}{1-q^{u}y_{i}/w}=\sum_{j\geq 0}(q^{u}y_{i}/w)^{j}.$

The rational function $F_{n}(y,w)$ admits the following partial fraction
expansion. We refer it as the splitting formula for $F_{n}(y,w)$.

######  Proposition 3.2.

Let $F_{n}(y,w)$ be defined as in (3.2). Then

(3.3)
$F_{n}(y,w)=\sum_{i=1}^{n}\sum_{j=0}^{c-1}\frac{A_{ij}}{1-q^{j}y_{i}/w},$

where

(3.4)
$A_{ij}=\frac{q^{(n-1)jc+(n-i)c}}{(q^{-j})_{j}(q)_{c-j-1}}\prod_{l=1}^{i-1}\big{(}q^{1-c}y_{i}/y_{l}\big{)}_{j}\big{(}q^{j+1}y_{i}/y_{l}\big{)}_{c-j}\prod_{l=i+1}^{n}\big{(}q^{-c}y_{i}/y_{l}\big{)}_{j+1}\big{(}q^{j+1}y_{i}/y_{l}\big{)}_{c-j-1}\\\
\times\prod_{\begin{subarray}{c}1\leq u<v\leq n\\\ u,v\neq
i\end{subarray}}(y_{u}/y_{v})_{c}(qy_{v}/y_{u})_{c}.$

Note that the $A_{ij}$ are polynomials in $y_{i}$. Proposition 3.2 is the
$\beta_{i}=c$ case of [4, Theorem 3.3] and the $n_{0}=0$ case of [16,
Proposition 3.5]. For completeness, we give a proof below.

###### Proof.

By partial fraction decomposition of $F_{n}(y,w)$ with respect to $1/w$, we
can rewrite $F_{n}(y,w)$ as (3.3) and

(3.5) $A_{ij}=F_{n}(y,w)(1-q^{j}y_{i}/w)|_{w=q^{j}y_{i}}\quad\text{for
$i=1,\dots,n$ and $j=0,\dots,c-1$.}$

Carrying out the substitution $w=q^{j}y_{i}$ in $F_{n}(y,w)(1-q^{j}y_{i}/w)$
for $i=1,\dots,n$ yields

(3.6)
$A_{ij}=\frac{1}{(q^{-j})_{j}(q)_{c-j-1}}\prod_{l=1}^{i-1}\frac{(y_{l}/y_{i})_{c}(qy_{i}/y_{l})_{c}}{(q^{-j}y_{l}/y_{i})_{c}}\prod_{l=i+1}^{n}\frac{(y_{i}/y_{l})_{c}(qy_{l}/y_{i})_{c}}{(q^{-j}y_{l}/y_{i})_{c}}\\\
\times\prod_{\begin{subarray}{c}1\leq u<v\leq n\\\ v,u\neq
i\end{subarray}}(y_{u}/y_{v})_{c}(qy_{v}/y_{u})_{c}.$

Using (3.1) with $(i,j,t,y)\mapsto(c,c,j,y_{i}/y_{l})$, we have

(3.7a)
$\frac{(y_{l}/y_{i})_{c}(qy_{i}/y_{l})_{c}}{(q^{-j}y_{l}/y_{i})_{c}}=q^{jc}\big{(}q^{1-c}y_{i}/y_{l}\big{)}_{j}\big{(}q^{j+1}y_{i}/y_{l}\big{)}_{c-j}\quad\text{for
$j=0,\dots,c$},$ and (3.7b)
$\frac{(y_{i}/y_{l})_{c}(qy_{l}/y_{i})_{c}}{(q^{-j}y_{l}/y_{i})_{c}}=q^{(j+1)c}\big{(}q^{-c}y_{i}/y_{l}\big{)}_{j+1}\big{(}q^{j+1}y_{i}/y_{l}\big{)}_{c-j-1}\quad\text{for
$j=-1,\dots,c-1$}.$

Substituting (3.7) into (3.6) gives (3.4). ∎

By Proposition 3.2, we find that a family of constant terms vanish.

######  Lemma 3.3.

Let $k_{1},k_{2},c$ be integers such that $c>0$ and $k_{1}>k_{2}\geq 0$. For
integers $t_{1},t_{2},\dots,t_{k_{1}}$, set $\\{t_{l}\leq 0\mid
l=1,\dots,k_{1}\\}=\\{b_{1},\dots,b_{m}\\}$ and $\\{t_{l}>0\mid
l=1,\dots,k_{1}\\}=\\{d_{1},\dots,d_{p}\\}$ (assume $d_{1}\leq
d_{2}\leq\cdots\leq d_{p}$). If
$\sum_{l=1}^{m}b_{l}+\sum_{l=1}^{p-k_{2}}d_{l}>0$ then

$C:=\operatorname*{CT}_{z^{(1)}}\frac{\prod_{1\leq i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{u=1}^{k_{1}}(z^{(1)}_{u})^{t_{u}}\prod_{i=1}^{k_{1}}\prod_{j=1}^{k_{2}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}=0.$

Note that if $p\leq k_{2}$ then we set the sum $\sum_{j=1}^{p-k_{2}}d_{j}=0$
by convention. The condition $\sum_{l=1}^{m}b_{l}+\sum_{l=1}^{p-k_{2}}d_{l}>0$
can not hold for $k_{2}\geq k_{1}$. Because in this case
$\sum_{l=1}^{p-k_{2}}d_{l}=0$ by $k_{2}\geq k_{1}\geq p$ and all the
$b_{l}\leq 0$.

###### Proof.

For $k_{2}=0$

$C=\operatorname*{CT}_{z^{(1)}}\prod_{i=1}^{k_{1}}(z^{(1)}_{i})^{-t_{i}}\prod_{1\leq
i<j\leq k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}.$

The above constant term vanishes since the Laurent polynomial inside the
operator $\operatorname*{CT}\limits_{z^{(1)}}$ is not homogeneous by
$\sum_{i=1}^{m}b_{i}+\sum_{j=1}^{p}d_{j}=\sum_{i=1}^{k_{1}}t_{i}>0$. Hence, we
can assume $k_{2}>0$ in the following of the proof. We proceed the proof by
induction on $k_{1}$. It is clear that the lemma holds for $k_{1}=1$ (This
forces $k_{2}=0$ by $k_{2}<k_{1}$). Assume the lemma holds for $k_{1}\mapsto
k_{1}-1$.

Using Proposition 3.2 with $(n,y,w)\mapsto(k_{1},z^{(1)},z_{1}^{(2)})$, we can
write

(3.8)
$C=\operatorname*{CT}_{z^{(1)}}\sum_{i=1}^{k_{1}}\sum_{j=0}^{c-1}\frac{A_{ij}}{(1-q^{j}z^{(1)}_{i}/z^{(2)}_{1})\prod_{u=1}^{k_{1}}(z^{(1)}_{u})^{t_{u}}\prod_{u=1}^{k_{1}}\prod_{v=2}^{k_{2}}(z_{u}^{(1)}/z_{v}^{(2)})_{c}},$

where

$A_{ij}=\frac{q^{(k_{1}-1)jc+(k_{1}-i)c}}{(q^{-j})_{j}(q)_{c-j-1}}\prod_{l=1}^{i-1}\big{(}q^{1-c}z^{(1)}_{i}/z^{(1)}_{l}\big{)}_{j}\big{(}q^{j+1}z^{(1)}_{i}/z^{(1)}_{l}\big{)}_{c-j}\\\
\times\prod_{l=i+1}^{k_{1}}\big{(}q^{-c}z^{(1)}_{i}/z^{(1)}_{l}\big{)}_{j+1}\big{(}q^{j+1}z^{(1)}_{i}/z^{(1)}_{l}\big{)}_{c-j-1}\prod_{\begin{subarray}{c}1\leq
u<v\leq k_{1}\\\ u,v\neq
i\end{subarray}}(z^{(1)}_{u}/z^{(1)}_{v})_{c}(qz^{(1)}_{v}/z^{(1)}_{u})_{c}.$

Denote the summand in (3.8) by $Q_{ij}$. We prove that
$\operatorname*{CT}\limits_{z^{(1)}}Q_{ij}=0$ for all the $i,j$. Note that we
view $Q_{ij}$ as a Laurent series in $z^{(1)}_{i}$. That is

$Q_{ij}=\frac{A_{ij}}{\prod_{u=1}^{k_{1}}(z^{(1)}_{u})^{t_{u}}\prod_{\begin{subarray}{c}u=1\\\
u\neq
i\end{subarray}}^{k_{1}}\prod_{v=2}^{k_{2}}(z_{u}^{(1)}/z_{v}^{(2)})_{c}}\sum_{l\geq
0}(q^{j}z_{i}^{(1)}/z_{1}^{(2)})^{l}\prod_{v=2}^{k_{2}}\prod_{m=0}^{c-1}\sum_{l\geq
0}(q^{m}z_{i}^{(1)}/z_{v}^{(2)})^{l}.$

For a fixed integer $i$, if $t_{i}<0$ then $Q_{ij}$ is in fact a power series
in $z^{(1)}_{i}$ with no constant term. It follows that
$\operatorname*{CT}\limits_{z^{(1)}}Q_{ij}=\operatorname*{CT}\limits_{z^{(1)}_{i}}Q_{ij}=0$.
If $t_{i}\geq 0$ then by the expression for $A_{ij}$ and taking the constant
term of $Q_{ij}$ with respect to $z^{(1)}_{i}$, we can write
$\operatorname*{CT}\limits_{z^{(1)}_{i}}Q_{ij}$ as a finite sum of the form

$R:=\frac{\prod_{\begin{subarray}{c}1\leq u<v\leq k_{1}\\\ u,v\neq
i\end{subarray}}(z_{u}^{(1)}/z_{v}^{(1)})_{c}(qz_{v}^{(1)}/z_{u}^{(1)})_{c}}{r\cdot\prod_{\begin{subarray}{c}u=1\\\
u\neq
i\end{subarray}}^{k_{1}}(z^{(1)}_{u})^{t_{u}+s_{u}}\prod_{\begin{subarray}{c}u=1\\\
u\neq
i\end{subarray}}^{k_{1}}\prod_{v=2}^{k_{2}}(z_{u}^{(1)}/z_{v}^{(2)})_{c}},$

where $r\in\mathbb{Q}(q)[z^{(2)}]\setminus\\{0\\}$ and all the $s_{u}$ are
nonnegative integers. Take $t^{\prime}_{u}=t_{u}+s_{u}$ and denote
$\\{t^{\prime}_{u}\leq 0\mid
u\in\\{1,\dots,k_{1}\\}\setminus\\{i\\}\\}=\\{b^{\prime}_{1},\dots,b^{\prime}_{m^{\prime}}\\}$
and $\\{t^{\prime}_{u}>0\mid
u\in\\{1,\dots,k_{1}\\}\setminus\\{i\\}\\}=\\{d^{\prime}_{1},\dots,d^{\prime}_{p^{\prime}}\\}$
(assume $d^{\prime}_{1}\leq\cdots\leq d^{\prime}_{p^{\prime}}$). We can see
that $\sum_{l=1}^{m^{\prime}}b^{\prime}_{l}\geq\sum_{l=1}^{m}b_{l}$,
$p^{\prime}\geq p-1$ and $\sum_{l=1}^{i}d^{\prime}_{l}\geq\sum_{l=1}^{i}d_{l}$
for any $1\leq i\leq p-1$. To see the last inequality holds, we can imagine
two positive integer sequences: $d^{\prime}_{1}\leq\cdots\leq
d^{\prime}_{p^{\prime}}$ is obtained from $d_{1}\leq\cdots\leq d_{p}$ by
deleting one element and adding a nonnegative integer to each element. Then

$\sum_{l=1}^{p^{\prime}-k_{2}+1}d^{\prime}_{l}+\sum_{l=1}^{m^{\prime}}b^{\prime}_{l}\geq\sum_{l=1}^{p-k_{2}}d^{\prime}_{l}+\sum_{l=1}^{m}b_{l}\geq\sum_{l=1}^{p-k_{2}}d_{l}+\sum_{l=1}^{m}b_{l}>0.$

By the induction hypothesis we conclude that all the form of
$\operatorname*{CT}\limits_{z^{(1)}}R=0$. Then
$\operatorname*{CT}\limits_{z^{(1)}}Q_{ij}$ also vanishes for $t_{i}\geq 0$.
Since $C$ is a finite sum of the $\operatorname*{CT}\limits_{z^{(1)}}Q_{ij}$,
we have $C=0$. ∎

By Lemma 3.3, we can get the next two corollaries.

######  Corollary 3.4.

Let $k_{1},k_{2},a,c$ be nonnegative integers and
$t_{1},\dots,t_{k_{1}}\in\mathbb{Z}$ such that $k_{1}>k_{2}$, $c>0$,
$t_{i}\leq a$ for $i=1,\dots,k_{1}$, and $\sum_{i=1}^{k_{1}}t_{i}>k_{2}a$.
Then

$\operatorname*{CT}_{z^{(1)}}\frac{\prod_{1\leq i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{i=1}^{k_{1}}(z^{(1)}_{i})^{t_{i}}\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}=0.$

###### Proof.

Let $\\{t_{i}\leq 0\mid i=1,\dots,k_{1}\\}=\\{b_{1},\dots,b_{m}\\}$ and
$\\{t_{i}>0\mid i=1,\dots,k_{1}\\}=\\{d_{1},\dots,d_{p}\\}$ (assume $d_{1}\leq
d_{2}\leq\cdots\leq d_{p}$). By $t_{i}\leq a$, we have

$\sum_{i=1}^{m}b_{i}+\sum_{i=1}^{p-k_{2}}d_{i}\geq\sum_{i=1}^{m}b_{i}+\sum_{i=1}^{p}d_{i}-k_{2}a=\sum_{i=1}^{k_{1}}t_{i}-k_{2}a>0.$

The last inequality holds by the condition $\sum_{i=1}^{k_{1}}t_{i}>k_{2}a$ in
the corollary. Hence, the corollary holds by Lemma 3.3. ∎

######  Corollary 3.5.

Let $k_{1},k_{2}$ be nonnegative integers such that $k_{1}>k_{2}$. Let
$c,t_{1},\dots,t_{k_{1}}$ be positive integers. Then

$\operatorname*{CT}_{z^{(1)}}\frac{\prod_{1\leq i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{i=1}^{k_{1}}(z^{(1)}_{i})^{t_{i}}\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}=0.$

###### Proof.

Let $B:=\\{t_{i}\leq 0\mid i=1,\dots,k_{1}\\}=\\{b_{1},\dots,b_{m}\\}$ and
$D:=\\{t_{i}>0\mid i=1,\dots,k_{1}\\}=\\{d_{1},\dots,d_{p}\\}$. Then
$B=\emptyset$ and $D=\\{t_{1},t_{2},\dots,t_{k_{1}}\\}$. Together with
$k_{1}>k_{2}$ and all the $t_{i}\geq 1$, we have

$\sum_{i=1}^{m}b_{i}+\sum_{i=1}^{p-k_{2}}d_{i}=\sum_{i=1}^{k_{1}-k_{2}}t_{i}>0.$

By Lemma 3.3, the corollary follows. ∎

##  4\. The expression for $L_{k_{1},k_{2}}(a,b,c)$

Recall the definitions of $L_{k_{1},k_{2}}(a,b,c;z^{(1)},z^{(2)})$ and
$L_{k_{1},k_{2}}(a,b,c)$ in (1.5) and (1.6) respectively. In
$L_{k_{1},k_{2}}(a,b,c;z^{(1)},z^{(2)})$, by expanding the product
$\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{b}$ directly,
we find that the degree and the lower degree of $z_{0}$ should be $k_{1}a$ at
most and $-k_{1}b$ at least respectively. But under a certain condition, we
find that after taking the constant term of
$L_{k_{1},k_{2}}(a,b,c;z^{(1)},z^{(2)})$ with respect to $z^{(1)}$ (not w.r.t.
$z_{0}$), the terms of $z_{0}^{l}$ for $l<0$ and $l>k_{2}a$ vanish. More
precisely, we will show that $L_{k_{1},k_{2}}(a,b,c)$ is of the form

(4.1)
$M_{k_{1}}(a,b,c)\prod_{j=1}^{k_{2}}(q^{c-1-b}z_{0}/z^{(2)}_{j})_{a+b+1-c}\times(1+C_{1}z_{0}+\cdots+C_{k_{2}(c-1-b)}z_{0}^{k_{2}(c-1-b)})$

for $k_{1}\geq k_{2}$ and $b+1\leq c\leq a+b+1$ in this section.

###  4.1. The negative powers

In this subsection, we show that all the terms of $L_{k_{1},k_{2}}(a,b,c)$
with negative powers of $z_{0}$ vanish.

For a Laurent series (polynomial) $f$, denote by $[z^{l}]f$ the coefficient of
$z^{l}$ in $f$.

######  Lemma 4.1.

Let $L_{k_{1},k_{2}}(a,b,c)$ be defined as in (1.6). For a negative integer
$l$,

$[z_{0}^{l}]L_{k_{1},k_{2}}(a,b,c)=0.$

###### Proof.

Let

(4.2) $F_{k_{1},k_{2}}(c):=\frac{\prod_{1\leq i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}.$

For an integer $l$ (not only negative),

$\displaystyle[z_{0}^{l}]L_{k_{1},k_{2}}(a,b,c)$
$\displaystyle=\operatorname*{CT}_{z_{0},z^{(1)}}z_{0}^{-l}L_{k_{1},k_{2}}(a,b,c)$
$\displaystyle=\operatorname*{CT}_{z_{0},z^{(1)}}z_{0}^{-l}\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{b}\times
F_{k_{1},k_{2}}(c)$ (4.3)
$\displaystyle=\operatorname*{CT}_{z_{0},z^{(1)}}z_{0}^{-l}\prod_{i=1}^{k_{1}}(-z_{0}/z_{i}^{(1)})^{a}q^{\binom{a}{2}}(q^{1-a}z_{i}^{(1)}/z_{0})_{a+b}\times
F_{k_{1},k_{2}}(c).$

By the well-known $q$-binomial theorem [2, Theorem 3.3]

$(y)_{n}=\sum_{t\geq 0}q^{\binom{t}{2}}\genfrac{[}{]}{0.0pt}{}{n}{t}(-y)^{t},$

we can expand (4.3) as

$\displaystyle[z_{0}^{l}]L_{k_{1},k_{2}}(a,b,c)$
$\displaystyle=\operatorname*{CT}_{z_{0},z^{(1)}}z_{0}^{-l}\prod_{i=1}^{k_{1}}(-z_{0}/z_{i}^{(1)})^{a}q^{\binom{a}{2}}\sum_{t_{i}\geq
0}q^{\binom{t_{i}}{2}}\genfrac{[}{]}{0.0pt}{}{a+b}{t_{i}}(-q^{1-a}z_{i}^{(1)}/z_{0})^{t_{i}}\times
F_{k_{1},k_{2}}(c)$
$\displaystyle=\operatorname*{CT}_{z_{0},z^{(1)}}\sum_{t_{1},\dots,t_{k_{1}}\geq
0}z_{0}^{k_{1}a-|t|-l}\prod_{i=1}^{k_{1}}(-z^{(1)}_{i})^{t_{i}-a}q^{\binom{a}{2}+\binom{t_{i}}{2}+(1-a)t_{i}}\genfrac{[}{]}{0.0pt}{}{a+b}{t_{i}}\times
F_{k_{1},k_{2}}(c).$

Here $|t|:=\sum_{i=1}^{k_{1}}t_{i}$ and
$\genfrac{[}{]}{0.0pt}{}{n}{t}=(q^{n-t+1})_{t}/(q)_{t}$ is the $q$-binomial
coefficient for nonnegative integers $n$ and $t$. Taking the constant term
with respect to $z_{0}$, we have

(4.4)
$[z_{0}^{l}]L_{k_{1},k_{2}}(a,b,c)=\operatorname*{CT}_{z^{(1)}}\sum_{\begin{subarray}{c}t_{1},\dots,t_{k_{1}}\geq
0\\\
|t|=k_{1}a-l\end{subarray}}\prod_{i=1}^{k_{1}}(-z^{(1)}_{i})^{t_{i}-a}q^{\binom{a}{2}+\binom{t_{i}}{2}+(1-a)t_{i}}\genfrac{[}{]}{0.0pt}{}{a+b}{t_{i}}\times
F_{k_{1},k_{2}}(c).$

Since $F_{k_{1},k_{2}}(c)$ only contributes nonnegative powers of the
$z^{(1)}_{i}$, the summand in (4.4) is not homogeneous in $z^{(1)}$ for $l<0$.
It follows that the constant term of each summand in (4.4) with respect to
$z^{(1)}$ equals zero, and so is the sum. ∎

###  4.2. The positive powers

As the statement in the beginning of this section, the degree of $z_{0}$ in
$L_{k_{1},k_{2}}(a,b,c)$ should be $k_{1}a$ at most. But we find that
$[z_{0}^{l}]L_{k_{1},k_{2}}(a,b,c)=0$ if $k_{1}\geq k_{2}$ and $l>k_{2}a$ in
this subsection. That is the content of the next lemma.

######  Lemma 4.2.

Let $L_{k_{1},k_{2}}(a,b,c)$ be defined as in (1.6). If $k_{1}\geq k_{2}$ and
the integer $l>k_{2}a$, then

$[z_{0}^{l}]L_{k_{1},k_{2}}(a,b,c)=0.$

###### Proof.

It is clear that the lemma holds for $k_{2}=k_{1}$ by the definition of
$L_{k_{1},k_{2}}(a,b,c)$. In the following of the proof, we assume
$k_{1}>k_{2}$. Changing $t_{i}\mapsto a-t_{i}$ for each $i$ in (4.4), we have

(4.5)
$[z_{0}^{l}]L_{k_{1},k_{2}}(a,b,c)=\sum_{\begin{subarray}{c}t_{1},\dots,t_{k_{1}}\leq
a\\\
|t|=l\end{subarray}}\prod_{i=1}^{k_{1}}(-1)^{-t_{i}}q^{\binom{a}{2}+\binom{a-t_{i}}{2}+(1-a)(a-t_{i})}\genfrac{[}{]}{0.0pt}{}{a+b}{a-t_{i}}\\\
\times\operatorname*{CT}_{z^{(1)}}\frac{\prod_{1\leq i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{i=1}^{k_{1}}(z^{(1)}_{i})^{t_{i}}\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}},$

where $|t|=\sum_{i=1}^{k_{1}}t_{i}$. All the constant term in the sum of (4.5)
vanish by Corollary 3.4 if $l=|t|=\sum_{i=1}^{k_{1}}t_{i}>k_{2}a$. Then the
lemma follows. ∎

In the following of this subsection, we show that
$\prod_{l=1}^{k_{2}}(q^{c-1-b}z_{0}/z_{l}^{(2)})_{a+b+1-c}$ is a factor of
$L_{k_{1},k_{2}}(a,b,c)$. The next lemma implies this.

######  Lemma 4.3.

If $b+1\leq c\leq a+b+1$ and $k_{1}\geq k_{2}$, then

$L_{k_{1},k_{2}}(a,b,c)=0$

by taking $z_{0}=z_{l}^{(2)}q^{m}$ for $l=1,\dots,k_{2}$ and
$m=1-a,\dots,b+1-c$.

###### Proof.

Carrying out the substitution $z_{0}=z_{l}^{(2)}q^{m}$ in
$L_{k_{1},k_{2}}(a,b,c)$ yields

$\displaystyle L:$
$\displaystyle=L_{k_{1},k_{2}}(a,b,c)\big{|}_{z_{0}=z_{l}^{(2)}q^{m}}$
$\displaystyle=\operatorname*{CT}_{z^{(1)}}\frac{\prod_{i=1}^{k_{1}}(q^{m}z_{l}^{(2)}/z_{i}^{(1)})_{a}(q^{1-m}z_{i}^{(1)}/z_{l}^{(2)})_{b}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}$
$\displaystyle=\operatorname*{CT}_{z^{(1)}}\frac{\prod_{i=1}^{k_{1}}(-z_{l}^{(2)}/z_{i}^{(1)})^{a}q^{ma+\binom{a}{2}}(q^{1-a-m}z_{i}^{(1)}/z_{l}^{(2)})_{a+b}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{l}^{(2)})_{c}\prod_{\begin{subarray}{c}j=1\\\
j\neq
l\end{subarray}}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}.$

Since $1-a-m\leq 1-a-(-a+1)=0$ and $1-a-m+(a+b-1)=b-m\geq b-(b+1-c)=c-1$,
$(z_{i}^{(1)}/z_{l}^{(2)})_{c}$ is in fact a factor of
$(q^{1-a-m}z_{i}^{(1)}/z_{l}^{(2)})_{a+b}$ for $i=1,\dots,k_{1}$. Hence,

(4.6)
$L=\operatorname*{CT}_{z^{(1)}}\prod_{i=1}^{k_{1}}(-z_{l}^{(2)}/z_{i}^{(1)})^{a}q^{ma+\binom{a}{2}}(q^{1-a-m}z_{i}^{(1)}/z_{l}^{(2)})_{a+m-1}(q^{c}z_{i}^{(1)}/z_{l}^{(2)})_{b-m-c+1}\\\
\times\frac{\prod_{1\leq i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{\begin{subarray}{c}j=1\\\
j\neq
l\end{subarray}}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}.$

From $b+1\leq c$, we have $a+m-1+(b-m-c+1)=a+b-c\leq a-1$. Then, by expanding
$(q^{1-a-m}z_{i}^{(1)}/z_{l}^{(2)})_{a+m-1}(q^{c}z_{i}^{(1)}/z_{l}^{(2)})_{b-m-c+1}$
in the right-hand side of (4.6) we know that $L$ is a finite sum of the form

(4.7)
$d\cdot(z_{l}^{(2)})^{\sum_{i=1}^{k_{1}}t_{i}}\operatorname*{CT}_{z^{(1)}}\frac{\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{i=1}^{k_{1}}(z_{i}^{(1)})^{t_{i}}\prod_{\begin{subarray}{c}j=1\\\
j\neq
l\end{subarray}}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}},$

where $d\in\mathbb{Q}(q)$ and $t_{i}\geq 1$ for $i=1,\dots,k_{1}$. By
Corollary 3.5, every constant term of the form (4.7) vanishes. Thus, the lemma
follows. ∎

To show that $L_{k_{1},k_{2}}(a,b,c)$ is of the form (4.1), the last step is
to prove

$[z_{0}^{0}]L_{k_{1},k_{2}}(a,b,c)=M_{k_{1}}(a,b,c).$

By the expression for $L_{k_{1},k_{2}}(a,b,c)$ in (1.6),

$[z_{0}^{0}]L_{k_{1},k_{2}}(a,b,c)=\operatorname*{CT}_{z_{0},z^{(1)}}\frac{\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{b}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}.$

Only those terms that are homogeneous in $z_{0}$ and the $z_{i}^{(1)}$ with
(total) degree 0 contribute to the constant term. Thus, only the first term
(i.e., 1) of the expansion of the denominator
$\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})^{-1}_{c}$
contributes to $[z_{0}^{0}]L_{k_{1},k_{2}}(a,b,c)$. Then, by the $q$-Morris
identity (1.3)

$[z_{0}^{0}]L_{k_{1},k_{2}}(a,b,c)=\operatorname*{CT}_{z_{0},z^{(1)}}\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{b}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}=M_{k_{1}}(a,b,c).$

By Lemmas 4.1–4.3 and the fact that
$[z_{0}^{0}]L_{k_{1},k_{2}}(a,b,c)=M_{k_{1}}(a,b,c)$, it is straightforward to
obtain that $L_{k_{1},k_{2}}(a,b,c)$ is of the form (4.1) for $b+1\leq c\leq
a+b+1$ and $k_{1}\geq k_{2}$.

##  5\. The $b=c-1$ case

In this section, we are concerned about the $b=c-1$ case of (4.1).

Taking $b=c-1$ in (4.1) we obtain an explicit constant term identity.

######  Lemma 5.1.

For $k_{1}\geq k_{2}$,

(5.1)
$L_{k_{1},k_{2}}(a,c-1,c)=\operatorname*{CT}_{z^{(1)}}\frac{\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}\\\
=M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}.$

An immediate consequence of Lemma 5.1 is the next result.

######  Proposition 5.2.

For $k_{1}\geq k_{2}$, let $\lambda$ be a partition such that
$\ell(\lambda)\leq k_{1}-k_{2}$. Then

(5.2)
$\displaystyle\operatorname*{CT}_{z^{(1)}}\frac{h_{\lambda}\big{[}\frac{1-q^{c}}{1-q}Z^{(1)}\big{]}\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{z_{0}^{|\lambda|}\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}$
$\displaystyle\quad=(-1)^{|\lambda|}\prod_{i=1}^{\ell(\lambda)}q^{\binom{\lambda_{i}}{2}}\genfrac{[}{]}{0.0pt}{}{a}{\lambda_{i}}\times
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}$
$\displaystyle\quad=h_{\lambda}\Big{[}\frac{q^{a}-1}{1-q}\Big{]}\times
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a},$

where $Z^{(1)}=z_{1}^{(1)}+\cdots+z_{k_{1}}^{(1)}$.

###### Proof.

Denote by $L$ the constant term in (5.2). By the generating function for the
complete symmetric functions in (2.1),

$L=\operatorname*{CT}_{z^{(1)},z_{k_{2}+1}^{(2)},\dots,z_{k_{2}+\ell(\lambda)}^{(2)}}\prod_{j=1}^{\ell(\lambda)}(z_{k_{2}+j}^{(2)})^{\lambda_{j}}\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}\\\
\times\frac{\prod_{1\leq i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{z_{0}^{|\lambda|}\prod_{j=1}^{k_{2}+\ell(\lambda)}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}.$

Since $k_{2}+\ell(\lambda)\leq k_{1}$, we can apply Lemma 5.1 with
$k_{2}\mapsto k_{2}+\ell(\lambda)$ and obtain

$\displaystyle L$
$\displaystyle=\operatorname*{CT}_{z_{k_{2}+1}^{(2)},\dots,z_{k_{2}+\ell(\lambda)}^{(2)}}z_{0}^{-|\lambda|}\prod_{j=1}^{\ell(\lambda)}(z_{k_{2}+j}^{(2)})^{\lambda_{j}}M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}+\ell(\lambda)}(z_{0}/z^{(2)}_{j})_{a}$
$\displaystyle=M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}\operatorname*{CT}_{z_{k_{2}+1}^{(2)},\dots,z_{k_{2}+\ell(\lambda)}^{(2)}}z_{0}^{-|\lambda|}\prod_{j=1}^{\ell(\lambda)}(z_{k_{2}+j}^{(2)})^{\lambda_{j}}(z_{0}/z^{(2)}_{k_{2}+j})_{a}.$

Using the well-known $q$-binomial theorem [2, Theorem 3.3]

$(y)_{n}=\sum_{t\geq 0}q^{\binom{t}{2}}\genfrac{[}{]}{0.0pt}{}{n}{t}(-y)^{t},$

we have

$\displaystyle L$
$\displaystyle=M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}\operatorname*{CT}_{z_{k_{2}+1}^{(2)},\dots,z_{k_{2}+\ell(\lambda)}^{(2)}}z_{0}^{-|\lambda|}\prod_{j=1}^{\ell(\lambda)}(z_{k_{2}+j}^{(2)})^{\lambda_{j}}\sum_{t_{j}\geq
0}q^{\binom{t_{j}}{2}}\genfrac{[}{]}{0.0pt}{}{a}{t_{j}}(-z_{0}/z^{(2)}_{k_{2}+j})^{t_{j}}$
$\displaystyle=M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}\times(-1)^{|\lambda|}\prod_{i=1}^{\ell(\lambda)}q^{\binom{\lambda_{i}}{2}}\genfrac{[}{]}{0.0pt}{}{a}{\lambda_{i}}.$

The right-most equality of (5.2) follows from (5.3) below. ∎

We complete the proof of Proposition 5.2 by the next result.

######  Lemma 5.3.

For $a$ and $r$ nonnegative integers,

(5.3)
$h_{r}\Big{[}\frac{q^{a}-1}{1-q}\Big{]}=(-1)^{r}q^{\binom{r}{2}}\genfrac{[}{]}{0.0pt}{}{a}{r}.$

###### Proof.

By (2.7),

$h_{r}\Big{[}\frac{q^{a}-1}{1-q}\Big{]}=(-1)^{r}e_{r}\Big{[}\frac{1-q^{a}}{1-q}\Big{]}=(-1)^{r}e_{r}(1,q,\dots,q^{a-1})=(-1)^{r}q^{\binom{r}{2}}\genfrac{[}{]}{0.0pt}{}{a}{r}.$

Here the last equality holds by [10, Page 26, Example 3]. ∎

By (2.8), we can rewrite (5.2) as

(5.4)
$\operatorname*{CT}_{z^{(1)}}\frac{g_{\lambda}(z^{(1)};q,q^{c})\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{z_{0}^{|\lambda|}\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}\\\
=g_{\lambda}\Big{(}\Big{[}\frac{q^{a}-1}{1-q^{c}}\Big{]};q,q^{c}\Big{)}\times
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}$

for $k_{1}\geq k_{2}$ and $\ell(\lambda)\leq k_{1}-k_{2}$. Then, we obtain a
few further results.

######  Proposition 5.4.

For $F=\mathbb{Q}(q,t)$ and partitions $\lambda^{(1)},\dots,\lambda^{(m)}$,
let $f_{\lambda^{(1)}},\dots,f_{\lambda^{(m)}}\in\Lambda_{F}$ be homogeneous
symmetric functions of degrees $|\lambda^{(1)}|,\dots,|\lambda^{(m)}|$
respectively such that
$f_{\lambda^{(i)}}=\sum_{\ell(\mu)\leq\ell(\lambda^{(i)})}d_{\mu}g_{\mu}$ for
$i=1,\dots,m$. Here $d_{\mu}\in F$. Then, for $k_{1}\geq k_{2}$,
$\sum_{i=1}^{m}\ell(\lambda^{(i)})\leq k_{1}-k_{2}$ and any alphabets $X_{i}$
independent of $z^{(1)}$,

(5.5)
$\operatorname*{CT}_{z^{(1)}}\prod_{i=1}^{m}f_{\lambda^{(i)}}\big{(}\big{[}Z^{(1)}+X_{i}\big{]};q,q^{c}\big{)}\frac{\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{z_{0}^{\sum_{i=1}^{m}|\lambda^{(i)}|}\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}\\\
=\prod_{i=1}^{m}f_{\lambda^{(i)}}\Big{(}\Big{[}\frac{q^{a}-1}{1-q^{c}}+\frac{X_{i}}{z_{0}}\Big{]};q,q^{c}\Big{)}\times
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}.$

###### Proof.

Since the $g_{\nu}$ form a basis of $\Lambda_{F}$ and
$f_{\lambda^{(i)}}=\sum_{\ell(\mu)\leq\ell(\lambda^{(i)})}d_{\mu}g_{\mu}$, we
can write $\prod_{i=1}^{m}f_{\lambda^{(i)}}=\sum d_{\nu}g_{\nu}$, where the
sum ranges over all partitions $\nu$ such that
$\ell(\nu)\leq\sum_{i=1}^{m}\ell(\lambda^{(i)})\leq k_{1}-k_{2}$. Then, by the
linearity it suffices to prove that

(5.6)
$L:=\operatorname*{CT}_{z^{(1)}}z_{0}^{-|\nu|}\prod_{i=1}^{\ell(\nu)}g_{\nu_{i}}\big{[}Z^{(1)}+X_{i}\big{]}\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}\times
F_{k_{1},k_{2}}(c)\\\
=\prod_{i=1}^{\ell(\nu)}g_{\nu_{i}}\Big{[}\frac{q^{a}-1}{1-q^{c}}+\frac{X_{i}}{z_{0}}\Big{]}\times
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}$

for a partition $\nu$ such that $\ell(\nu)\leq k_{1}-k_{2}$. Here and in the
following of the proof $g_{\nu}=g_{\nu}(q,q^{c})$, and $F_{k_{1},k_{2}}(c)$
and $M_{k_{1}}(a,b,c)$ are defined in (4.2) and (1.7) respectively. By (2.9),
we can write $L$ as

$\displaystyle L$
$\displaystyle=\operatorname*{CT}_{z^{(1)}}z_{0}^{-|\nu|}\prod_{i=1}^{\ell(\nu)}\sum_{t_{i}=0}^{\nu_{i}}g_{t_{i}}[X_{i}]g_{\nu_{i}-t_{i}}[Z^{(1)}]\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}\times
F_{k_{1},k_{2}}(c)$ $\displaystyle=\sum_{\begin{subarray}{c}0\leq
t_{i}\leq\nu_{i}\\\
i=1,\dots,\ell(\nu)\end{subarray}}z_{0}^{-|t|}\prod_{i=1}^{\ell(\nu)}g_{t_{i}}[X_{i}]\operatorname*{CT}_{z^{(1)}}z_{0}^{-|\nu|+|t|}\prod_{i=1}^{\ell(\nu)}g_{\nu_{i}-t_{i}}[Z^{(1)}]\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}\times
F_{k_{1},k_{2}}(c),$

where $|t|=\sum_{i=1}^{\ell(\nu)}t_{i}$. By Proposition 5.2 or its deformation
(5.4),

$\displaystyle L$ $\displaystyle=\sum_{\begin{subarray}{c}0\leq
t_{i}\leq\nu_{i}\\\
i=1,\dots,\ell(\nu)\end{subarray}}z_{0}^{-|t|}\prod_{i=1}^{\ell(\nu)}g_{t_{i}}[X_{i}]g_{\nu_{i}-t_{i}}\Big{[}\frac{q^{a}-1}{1-q^{c}}\Big{]}\times
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}$
$\displaystyle=\prod_{i=1}^{\ell(\nu)}\sum_{t_{i}=0}^{\nu_{i}}z_{0}^{-t_{i}}g_{t_{i}}[X_{i}]g_{\nu_{i}-t_{i}}\Big{[}\frac{q^{a}-1}{1-q^{c}}\Big{]}\times
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}.$

Using (2.4) and (2.9) gives

$L=\prod_{i=1}^{\ell(\nu)}g_{\nu_{i}}\Big{[}\frac{q^{a}-1}{1-q^{c}}+\frac{X_{i}}{z_{0}}\Big{]}\times
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}.\qed$

Before giving the next result, we introduce a determinant transformation
formula between the $p_{n}$ and the $h_{n}$.

######  Lemma 5.5.

_[10, Page 28]_ For a nonnegative integer $n$,

(5.7) $(-1)^{n-1}p_{n}=\left|\begin{array}[]{ccccc}h_{1}&1&0&\cdots&0\\\
2h_{2}&h_{1}&1&\cdots&0\\\ \vdots&\vdots&\vdots&\vdots&\vdots\\\
nh_{n}&h_{n-1}&h_{n-2}&\cdots&h_{1}\end{array}\right|.$

We are concerned about constant term identities for symmetric functions in
$\Lambda_{\mathbb{Q}(q,t)}$ in Theorem 5.4. The next result is a similar
result about symmetric functions in $\Lambda_{\mathbb{Q}}$.

######  Proposition 5.6.

For partitions $\lambda^{(1)},\dots,\lambda^{(m)}$, let
$f_{\lambda^{(1)}},\dots,f_{\lambda^{(m)}}\in\Lambda_{\mathbb{Q}}$ be
homogeneous symmetric functions of degrees
$|\lambda^{(1)}|,\dots,|\lambda^{(m)}|$ respectively. Let $X_{i}$ and $Y_{i}$
be alphabets independent of $z^{(1)}$ for $i=1,\dots,m$. Then, for $k_{1}\geq
k_{2}$ and $\sum_{i=1}^{m}|\lambda^{(i)}|\leq k_{1}-k_{2}$

(5.8)
$\operatorname*{CT}_{z^{(1)}}\prod_{i=1}^{m}f_{\lambda^{(i)}}\big{[}Y_{i}Z^{(1)}+X_{i}\big{]}\frac{\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{z_{0}^{\sum_{i=1}^{m}|\lambda^{(i)}|}\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}\\\
=\prod_{i=1}^{m}f_{\lambda^{(i)}}\Big{[}\frac{q^{a}-1}{1-q^{c}}Y_{i}+\frac{X_{i}}{z_{0}}\Big{]}\times
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}.$

###### Proof.

It suffices to prove that (5.8) holds for power sum symmetric functions by the
linearity since the $p_{\lambda}$ form a basis of $\Lambda_{\mathbb{Q}}$. That
is

(5.9)
$L:=\operatorname*{CT}_{z^{(1)}}z_{0}^{-|\lambda|}\prod_{i=1}^{l}p_{\lambda_{i}}\big{[}Y_{i}Z^{(1)}+X_{i}\big{]}\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}\times
F_{k_{1},k_{2}}(c)\\\
=\prod_{i=1}^{l}p_{\lambda_{i}}\Big{[}\frac{q^{a}-1}{1-q^{c}}Y_{i}+\frac{X_{i}}{z_{0}}\Big{]}\times
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a},$

where $|\lambda|\leq k_{1}-k_{2}$, $l=\ell(\lambda)$, the $X_{i}$ and the
$Y_{i}$ are independent of $z^{(1)}$ for $i=1,\dots,l$, and
$F_{k_{1},k_{2}}(c)$ is defined as in (4.2). By (2.2a) and (2.2c),

$L=\operatorname*{CT}_{z^{(1)}}z_{0}^{-|\lambda|}\prod_{i=1}^{l}\Big{(}p_{\lambda_{i}}\Big{[}\frac{1-q}{1-q^{c}}Y_{i}\Big{]}p_{\lambda_{i}}\Big{[}\frac{1-q^{c}}{1-q}Z^{(1)}\Big{]}+p_{\lambda_{i}}[X_{i}]\Big{)}\\\
\times\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}F_{k_{1},k_{2}}(c).$

Expanding the first product yields

$\displaystyle L$
$\displaystyle=\operatorname*{CT}_{z^{(1)}}z_{0}^{-|\lambda|}\sum_{(t_{1},\dots,t_{l})\in\mathbb{Z}_{2}^{l}}\prod_{i=1}^{l}\Big{(}p_{\lambda_{i}}\Big{[}\frac{1-q}{1-q^{c}}Y_{i}\Big{]}p_{\lambda_{i}}\Big{[}\frac{1-q^{c}}{1-q}Z^{(1)}\Big{]}\Big{)}^{t_{i}}(p_{\lambda_{i}}[X_{i}])^{1-t_{i}}$
$\displaystyle\qquad\times\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}F_{k_{1},k_{2}}(c)$
$\displaystyle=\sum_{(t_{1},\dots,t_{l})\in\mathbb{Z}_{2}^{l}}z_{0}^{-\sum_{i=1}^{l}\lambda_{i}(1-t_{i})}\prod_{i=1}^{l}\Big{(}p_{\lambda_{i}}\Big{[}\frac{1-q}{1-q^{c}}Y_{i}\Big{]}\Big{)}^{t_{i}}(p_{\lambda_{i}}[X_{i}])^{1-t_{i}}$
$\displaystyle\qquad\times\operatorname*{CT}_{z^{(1)}}z_{0}^{-\sum_{i=1}^{l}\lambda_{i}t_{i}}\prod_{i=1}^{l}\Big{(}p_{\lambda_{i}}\Big{[}\frac{1-q^{c}}{1-q}Z^{(1)}\Big{]}\Big{)}^{t_{i}}\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{c-1}\times
F_{k_{1},k_{2}}(c).$

Write

(5.10) $\prod_{i=1}^{l}\big{(}p_{\lambda_{i}}\big{)}^{t_{i}}=\sum
d_{\nu}h_{\nu}$

using (5.7). Since $|\nu|=\sum_{i=1}^{l}\lambda_{i}t_{i}\leq|\lambda|\leq
k_{1}-k_{2}$, we have $\ell(\nu)\leq k_{1}-k_{2}$ for each $\nu$ in (5.10).
Thus, we can apply Proposition 5.2 and then (5.10) to obtain

$\displaystyle L$
$\displaystyle=\sum_{(t_{1},\dots,t_{l})\in\mathbb{Z}_{2}^{l}}z_{0}^{-\sum_{i=1}^{l}\lambda_{i}(1-t_{i})}\prod_{i=1}^{l}\Big{(}p_{\lambda_{i}}\Big{[}\frac{1-q}{1-q^{c}}Y_{i}\Big{]}\Big{)}^{t_{i}}(p_{\lambda_{i}}[X_{i}])^{1-t_{i}}$
$\displaystyle\qquad\times\prod_{i=1}^{l}\Big{(}p_{\lambda_{i}}\Big{[}\frac{q^{a}-1}{1-q}\Big{]}\Big{)}^{t_{i}}\cdot
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}$
$\displaystyle=\sum_{(t_{1},\dots,t_{l})\in\mathbb{Z}_{2}^{l}}\prod_{i=1}^{l}\Big{(}p_{\lambda_{i}}\Big{[}\frac{q^{a}-1}{1-q^{c}}Y_{i}\Big{]}\Big{)}^{t_{i}}\Big{(}p_{\lambda_{i}}\Big{[}\frac{X_{i}}{z_{0}}\Big{]}\Big{)}^{1-t_{i}}\cdot
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}$
$\displaystyle=\prod_{i=1}^{l}\Big{(}p_{\lambda_{i}}\Big{[}\frac{q^{a}-1}{1-q^{c}}Y_{i}\Big{]}+p_{\lambda_{i}}\Big{[}\frac{X_{i}}{z_{0}}\Big{]}\Big{)}\cdot
M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}.$

By (2.2a),

$L=\prod_{i=1}^{l}p_{\lambda_{i}}\Big{[}\frac{q^{a}-1}{1-q^{c}}Y_{i}+\frac{X_{i}}{z_{0}}\Big{]}M_{k_{1}}(a,c-1,c)\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a}.$

Hence, we complete the proof of (5.9) and the proposition follows. ∎

For a positive integer $n$, let $m_{1},\dots,m_{n}$ be nonnegative integers.
For a fixed $s\in\\{1,\dots,n\\}$, let $\lambda^{(s,j)}$ be partitions for
$j=1,\dots,m_{s}$. Recall that $z^{(s)}=(z^{(s)}_{1},\dots,z^{(s)}_{k_{s}})$
for $s=1,\dots,n$ and $z=(z^{(1)},z^{(2)},\dots,z^{(n)})$. By iterating (5.5)
with $z_{0}=1$, we obtain the next more general result.

######  Theorem 5.7.

Let $k_{1},\dots,k_{n+1}$ and $a_{1},\dots,a_{n}$ be nonnegative integers such
that $k_{1}\geq k_{2}\geq\cdots\geq k_{n+1}$ and
$a_{1}=a,a_{2}=\cdots=a_{n}=0$. Let
$f_{\lambda^{(s,j)}}\in\Lambda_{\mathbb{Q}(q,t)}$ be homogeneous symmetric
functions of degrees $|\lambda^{(s,j)}|$ such that
$f_{\lambda^{(s,j)}}=\sum_{\ell(\mu)\leq\ell(\lambda^{(s,j)})}d_{\mu}g_{\mu}$,
where $d_{\mu}\in\mathbb{Q}(q,t)$. If
$\sum_{j=1}^{m_{s}}\ell(\lambda^{(s,j)})\leq k_{s}-k_{s+1}$ for every $s$ and
all the alphabets $X^{(s)}_{j}$ are independent of $z$, then

(5.11)
$\displaystyle\operatorname*{CT}_{z}\prod_{s=1}^{n}\frac{\prod_{i=1}^{k_{s}}(1/z_{i}^{(s)})_{a_{s}}(qz_{i}^{(s)})_{c-1}\prod_{1\leq
i<j\leq
k_{s}}(z_{i}^{(s)}/z_{j}^{(s)})_{c}(qz_{j}^{(s)}/z_{i}^{(s)})_{c}}{\prod_{j=1}^{k_{s+1}}\prod_{i=1}^{k_{s}}(z_{i}^{(s)}/z_{j}^{(s+1)})_{c}}$
$\displaystyle\quad\times\prod_{s=1}^{n}\prod_{j=1}^{m_{s}}f_{\lambda^{(s,j)}}\big{(}\big{[}Z^{(s)}+X^{(s)}_{j}\big{]};q,q^{c}\big{)}$
$\displaystyle\qquad=\prod_{s=1}^{n}\prod_{j=1}^{m_{s}}f_{\lambda^{(s,j)}}\Big{(}\Big{[}\frac{q^{a}-1}{1-q^{c}}+X^{(s)}_{j}\Big{]};q,q^{c}\Big{)}\prod_{s=1}^{n}M_{k_{s}}(a,c-1,c)\prod_{j=1}^{k_{n+1}}(1/z^{(n+1)}_{j})_{a}.$

Similar to Theorem 5.7, we can also iterate (5.8) with $z_{0}=1$ and obtain
the next result.

######  Theorem 5.8.

Let the $k_{i}$, $m_{i}$ and $a_{i}$ be the same as in Theorem 5.7. Let
$f_{\lambda^{(s,j)}}\in\Lambda_{\mathbb{Q}}$ be homogeneous symmetric
functions of degrees $|\lambda^{(s,j)}|$ such that
$\sum_{j=1}^{m_{s}}|\lambda^{(s,j)}|\leq k_{s}-k_{s+1}$ for every $s$. If the
$X^{(s)}_{j}$ and the $Y^{(s)}_{j}$ are alphabets independent of $z$, then

(5.12)
$\operatorname*{CT}_{z}\prod_{s=1}^{n}\prod_{j=1}^{m_{s}}f_{\lambda^{(s,j)}}\big{[}A^{(s,j)}\big{]}\prod_{s=1}^{n}\frac{\prod_{i=1}^{k_{s}}(1/z_{i}^{(s)})_{a_{s}}(qz_{i}^{(s)})_{c-1}\prod_{1\leq
i<j\leq
k_{s}}(z_{i}^{(s)}/z_{j}^{(s)})_{c}(qz_{j}^{(s)}/z_{i}^{(s)})_{c}}{\prod_{j=1}^{k_{s+1}}\prod_{i=1}^{k_{s}}(z_{i}^{(s)}/z_{j}^{(s+1)})_{c}}\\\
=\prod_{s=1}^{n}\prod_{j=1}^{m_{s}}f_{\lambda^{(s,j)}}\Big{[}\frac{q^{a}-1}{1-q^{c}}Y^{(s)}_{j}+X^{(s)}_{j}\Big{]}\prod_{s=1}^{n}M_{k_{s}}(a,c-1,c)\prod_{j=1}^{k_{n+1}}(1/z^{(n+1)}_{j})_{a},$

where $A^{(s,j)}=(z^{(s)}_{1}+\cdots+z^{(s)}_{k_{s}})Y^{(s)}_{j}+X^{(s)}_{j}$.

##  6\. Equivalent constant term identities

In this section, we give several constant term identities equivalent to those
in Section 5. All these equivalence are built on the next lemma.

######  Lemma 6.1.

For $k_{1}\geq k_{2}$ and $a+b+1\geq c$,

(6.1)
$\operatorname*{CT}_{z^{(1)}}\frac{z_{0}^{(b+1-c)k_{1}}\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{b}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{i=1}^{k_{1}}(z_{i}^{(1)})^{b+1-c}\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(q^{b+1-c}z_{i}^{(1)}/z_{j}^{(2)})_{c}}\\\
=(-1)^{k_{1}(b+1-c)}q^{k_{1}\binom{b+2-c}{2}}\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a+b+1-c}\cdot
M_{k_{1}}(a+b+1-c,c-1,c).$

Taking $b=c-1$, Lemma 6.1 reduces to Lemma 5.1. Hence, Lemma 6.1 is seemingly
more general. But the two lemmas are in fact equivalent. We will show the
equivalence in the following proof of Lemma 6.1.

###### Proof.

We prove the lemma by transforming (5.1) to (6.1).

We can write the left-hand side of (5.1) as

$\operatorname*{CT}_{z^{(1)}}\prod_{i=1}^{k_{1}}(-z_{0}/z_{i}^{(1)})^{a}q^{\binom{a}{2}}(q^{1-a}z_{i}^{(1)}/z_{0})_{a+c-1}\frac{\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}.$

Using the above and taking $a\mapsto a+b+1-c$, (5.1) becomes

$\operatorname*{CT}_{z^{(1)}}\prod_{i=1}^{k_{1}}(-z_{0}/z_{i}^{(1)})^{a+b+1-c}q^{\binom{a+b+1-c}{2}}(q^{c-a-b}z_{i}^{(1)}/z_{0})_{a+b}\frac{\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)}/z_{j}^{(2)})_{c}}\\\
=\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a+b+1-c}\cdot
M_{k_{1}}(a+b+1-c,c-1,c).$

Note that the substitution $a\mapsto a+b+1-c$ is valid by $a+b+1\geq c$. Take
$(z_{0},z^{(2)}_{j})\mapsto(z_{0}q^{c-b-1},z^{(2)}_{j}q^{c-b-1})$ in the
above. This substitution does not change the constant term. Thus

(6.2)
$\operatorname*{CT}_{z^{(1)}}\prod_{i=1}^{k_{1}}(-q^{c-b-1}z_{0}/z_{i}^{(1)})^{a+b+1-c}q^{\binom{a+b+1-c}{2}}(q^{1-a}z_{i}^{(1)}/z_{0})_{a+b}\frac{\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(q^{b+1-c}z_{i}^{(1)}/z_{j}^{(2)})_{c}}\\\
=\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a+b+1-c}\cdot
M_{k_{1}}(a+b+1-c,c-1,c).$

We can rewrite (6.2) as

(6.3)
$\displaystyle\operatorname*{CT}_{z^{(1)}}\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})^{b+1-c}q^{\binom{a}{2}}(-z_{0}/z_{i}^{(1)})^{a}(q^{1-a}z_{i}^{(1)}/z_{0})_{a+b}\frac{\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(q^{b+1-c}z_{i}^{(1)}/z_{j}^{(2)})_{c}}$
$\displaystyle\quad=(-1)^{k_{1}(b+1-c)}q^{k_{1}\big{(}\binom{a}{2}+(b+1-c)(a+b+1-c)-\binom{a+b+1-c}{2}\big{)}}$
$\displaystyle\qquad\times\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a+b+1-c}\cdot
M_{k_{1}}(a+b+1-c,c-1,c).$

Substituting

$q^{\binom{a}{2}}(-z_{0}/z_{i}^{(1)})^{a}(q^{1-a}z_{i}^{(1)}/z_{0})_{a+b}=(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{b}$

and

$\binom{a}{2}+(b+1-c)(a+b+1-c)-\binom{a+b+1-c}{2}=\binom{b+2-c}{2}$

into (6.3) yields (6.1). ∎

By Lemma 6.1 we obtain the next result. The proof is similar to the proof of
Proposition 5.2 using Lemma 5.1. Hence, we omit the proof.

######  Proposition 6.2.

For $a+b+1\geq c$, $k_{1}\geq k_{2}$ and a partition $\lambda$ such that
$\ell(\lambda)\leq k_{1}-k_{2}$,

(6.4)
$\operatorname*{CT}_{z^{(1)}}\frac{z_{0}^{(b+1-c)k_{1}-|\lambda|}h_{\lambda}\big{[}\frac{1-q^{c}}{1-q}Z^{(1)}\big{]}\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{b}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{i=1}^{k_{1}}(z_{i}^{(1)})^{b+1-c}\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(q^{b+1-c}z_{i}^{(1)}/z_{j}^{(2)})_{c}}\\\
\quad=(-1)^{k_{1}(b+1-c)}q^{k_{1}\binom{b+2-c}{2}}h_{\lambda}\Big{[}\frac{q^{a}-q^{c-b-1}}{1-q}\Big{]}\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a+b+1-c}\cdot
M_{k_{1}}(a+b+1-c,c-1,c),$

where $Z^{(1)}=z_{1}^{(1)}+\cdots+z_{k_{1}}^{(1)}$.

The identity (5.2) in Proposition 5.2 is the $b=c-1$ case of (6.4). However,
it is not hard to transform (5.2) into (6.4) along the same line as the proof
of Lemma 6.1. Hence, the two identities are equivalent.

By (2.8), we can rewrite (6.4) as

(6.5)
$\displaystyle\operatorname*{CT}_{z^{(1)}}\frac{g_{\lambda}(z^{(1)};q,q^{c})\prod_{i=1}^{k_{1}}(z_{0}/z_{i}^{(1)})_{a}(qz_{i}^{(1)}/z_{0})_{b}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{z_{0}^{|\lambda|-(b+1-c)k_{1}}\prod_{i=1}^{k_{1}}(z_{i}^{(1)})^{b+1-c}\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(q^{b+1-c}z_{i}^{(1)}/z_{j}^{(2)})_{c}}$
$\displaystyle\quad=(-1)^{k_{1}(b+1-c)}q^{k_{1}\binom{b+2-c}{2}}g_{\lambda}\Big{(}\Big{[}\frac{q^{a}-q^{c-b-1}}{1-q^{c}}\Big{]};q,q^{c}\Big{)}$
$\displaystyle\qquad\times\prod_{j=1}^{k_{2}}(z_{0}/z^{(2)}_{j})_{a+b+1-c}\cdot
M_{k_{1}}(a+b+1-c,c-1,c)$

for $k_{1}\geq k_{2}$, $\ell(\lambda)\leq k_{1}-k_{2}$ and $a+b+1\geq c$.
Using (6.5) with $z_{0}=1$, we obtain the next result equivalent to
Proposition 5.4. The proof is similar to the proof of Proposition 5.4.

######  Proposition 6.3.

For $F=\mathbb{Q}(q,t)$ and partitions $\lambda^{(1)},\dots,\lambda^{(m)}$,
let $f_{\lambda^{(1)}},\dots,f_{\lambda^{(m)}}\in\Lambda_{F}$ be homogeneous
symmetric functions of degrees $|\lambda^{(1)}|,\dots,|\lambda^{(m)}|$
respectively such that
$f_{\lambda^{(i)}}=\sum_{\ell(\mu)\leq\ell(\lambda^{(i)})}d_{\mu}g_{\mu}$ for
$i=1,\dots,m$. Here $d_{\mu}\in F$. Then, for $a+b+1\geq c$, $k_{1}\geq
k_{2}$, $\sum_{i=1}^{m}\ell(\lambda^{(i)})\leq k_{1}-k_{2}$ and any alphabets
$X_{i}$ independent of $z^{(1)}$

(6.6)
$\displaystyle\operatorname*{CT}_{z^{(1)}}\frac{\prod_{i=1}^{m}f_{\lambda^{(i)}}\big{(}\big{[}Z^{(1)}+X_{i}\big{]};q,q^{c})\prod_{i=1}^{k_{1}}(1/z_{i}^{(1)})_{a}(qz_{i}^{(1)})_{b}\prod_{1\leq
i<j\leq
k_{1}}(z_{i}^{(1)}/z_{j}^{(1)})_{c}(qz_{j}^{(1)}/z_{i}^{(1)})_{c}}{\prod_{i=1}^{k_{1}}(z_{i}^{(1)})^{b+1-c}\prod_{j=1}^{k_{2}}\prod_{i=1}^{k_{1}}(q^{b+1-c}z_{i}^{(1)}/z_{j}^{(2)})_{c}}$
$\displaystyle\quad=(-1)^{k_{1}(b+1-c)}q^{k_{1}\binom{b+2-c}{2}}\prod_{i=1}^{m}f_{\lambda^{(i)}}\Big{(}\Big{[}\frac{q^{a}-q^{c-b-1}}{1-q^{c}}+X_{i}\Big{]};q,q^{c}\Big{)}$
$\displaystyle\qquad\times\prod_{j=1}^{k_{2}}(1/z^{(2)}_{j})_{a+b+1-c}\cdot
M_{k_{1}}(a+b+1-c,c-1,c).$

For $n$ a positive integer, let $k_{1},\dots,k_{n+1}$, $m_{1},\dots,m_{n}$,
$a_{1},\dots,a_{n}$ and $b_{1},\dots,b_{n}$ be nonnegative integers such that
$k_{1}\geq k_{2}\geq\cdots\geq k_{n+1}$ and $a_{1}=a,a_{2}=\cdots=a_{n}=0$.
For $1\leq s\leq n$, $1\leq j\leq m_{s}$ and $\lambda^{(s,j)}$ partitions, let
$f_{\lambda^{(s,j)}}\in\Lambda_{F}=\Lambda_{\mathbb{Q}(q,t)}$ be homogeneous
symmetric functions of degrees $|\lambda^{(s,j)}|$. For $s=1,\dots,n$, denote
$\sigma_{s}=b_{1}+\cdots+b_{s}$, $\sigma_{0}=0$, and
$Z^{(s)}=z^{(s)}_{1}+\cdots+z^{(s)}_{k_{s}}$. By iterating (6.6) we obtain
Theorem 6.4 below, which is equivalent to Theorem 5.7.

######  Theorem 6.4.

Suppose
$f_{\lambda^{(s,j)}}=\sum_{\ell(\mu)\leq\ell(\lambda^{(s,j)})}d_{\mu}g_{\mu}$.
If $a+\sigma_{s}+s\geq sc$, $\sum_{j=1}^{m_{s}}\ell(\lambda^{(s,j)})\leq
k_{s}-k_{s+1}$ for every $s$, and all the alphabets $X^{(s)}_{j}$ are
independent of $z$, then

(6.7)
$\displaystyle\operatorname*{CT}_{z}\prod_{s=1}^{n}\frac{\prod_{i=1}^{k_{s}}(1/z_{i}^{(s)})_{a_{s}}(qz_{i}^{(s)})_{b_{s}}\prod_{1\leq
i<j\leq
k_{s}}(z_{i}^{(s)}/z_{j}^{(s)})_{c}(qz_{j}^{(s)}/z_{i}^{(s)})_{c}}{\prod_{i=1}^{k_{s}}(z_{i}^{(s)})^{b_{s}+1-c}\prod_{j=1}^{k_{s+1}}\prod_{i=1}^{k_{s}}(q^{b_{s}+1-c}z_{i}^{(s)}/z_{j}^{(s+1)})_{c}}$
$\displaystyle\qquad\times\prod_{s=1}^{n}\prod_{j=1}^{m_{s}}f_{\lambda^{(s,j)}}\big{(}\big{[}Z^{(s)}+X^{(s)}_{j}\big{]};q,q^{c}\big{)}$
$\displaystyle\quad=(-1)^{\sum_{s=1}^{n}k_{s}(b_{s}+1-c)}q^{\sum_{s=1}^{n}k_{s}\binom{b_{s}+2-c}{2}}\prod_{s=1}^{n}M_{k_{s}}\big{(}a+\sigma_{s}+s(1-c),c-1,c\big{)}$
$\displaystyle\qquad\times\prod_{j=1}^{k_{n+1}}(1/z^{(n+1)}_{j})_{a+\sigma_{n}+n(1-c)}\prod_{s=1}^{n}\prod_{j=1}^{m_{s}}f_{\lambda^{(s,j)}}\Big{(}\Big{[}\frac{q^{a+\sigma_{s-1}+(s-1)(1-c)}-q^{c-b_{s}-1}}{1-q^{c}}+X^{(s)}_{j}\Big{]};q,q^{c}\Big{)}.$

Along the same line to obtain Theorem 5.8, we can obtain its equivalent result
in the next theorem.

######  Theorem 6.5.

For $s=1,\dots,n$, if $f_{\lambda^{(s,j)}}\in\Lambda_{\mathbb{Q}}$ such that
$\sum_{j=1}^{m_{s}}|\lambda^{(s,j)}|\leq k_{s}-k_{s+1}$, all the alphabets
$X^{(s)}_{j}$ and $Y^{(s)}_{j}$ are independent of $z$, and
$a+\sigma_{s}+s\geq sc$, then

(6.8)
$\displaystyle\operatorname*{CT}_{z}\prod_{s=1}^{n}\prod_{j=1}^{m_{s}}f_{\lambda^{(s,j)}}\big{[}Z^{(s)}Y^{(s)}_{j}+X^{(s)}_{j}\big{]}\prod_{s=1}^{n}\frac{\prod_{i=1}^{k_{s}}(1/z_{i}^{(s)})_{a_{s}}(qz_{i}^{(s)})_{b_{s}}\prod_{1\leq
i<j\leq
k_{s}}(z_{i}^{(s)}/z_{j}^{(s)})_{c}(qz_{j}^{(s)}/z_{i}^{(s)})_{c}}{\prod_{i=1}^{k_{s}}(z_{i}^{(s)})^{b_{s}+1-c}\prod_{j=1}^{k_{s+1}}\prod_{i=1}^{k_{s}}(q^{b_{s}+1-c}z_{i}^{(s)}/z_{j}^{(s+1)})_{c}}$
$\displaystyle\quad=(-1)^{\sum_{s=1}^{n}k_{s}(b_{s}+1-c)}q^{\sum_{s=1}^{n}k_{s}\binom{b_{s}+2-c}{2}}\prod_{s=1}^{n}\prod_{j=1}^{m_{s}}f_{\lambda^{(s,j)}}\Big{[}\frac{q^{a+\sigma_{s-1}+(s-1)(1-c)}-q^{c-b_{s}-1}}{1-q^{c}}Y^{(s)}_{j}+X^{(s)}_{j}\Big{]}$
$\displaystyle\qquad\times\prod_{s=1}^{n}M_{k_{s}}\big{(}a+\sigma_{s}+s(1-c),c-1,c\big{)}\prod_{j=1}^{k_{n+1}}(1/z^{(n+1)}_{j})_{a+\sigma_{n}+n(1-c)}.$

### Acknowledgements

This work was supported by the National Natural Science Foundation of China
(No. 12171487).

## References

  * [1] S. P. Albion, E. M. Rains and S. O. Warnaar, _AFLT-type Selberg integrals_ , Commun. Math. Phys. 388 (2021), 735–791.
  * [2] G. E. Andrews, _The Theory of Partitions_ , Addison-Wesley, Reading, Mass.,1976, Encyclopedia of Mathematics and its Applications, Vol. 2.
  * [3] R. Askey, _Some basic hypergeometric extensions of integrals of Selberg and Andrews_ , SIAM J. Math. Anal. 11 (1980), 938–951.
  * [4] T. W. Cai, _Macdonald symmetric functions of rectangular shapes_ , J. Combin. Theory Ser. A 128 (2014), 162–179.
  * [5] G. Gasper and M. Rahman, _Basic Hypergeometric Series_ , Encyclopedia of Mathematics and its Applications, Vol. 35, second edition, Cambridge University Press, Cambridge, 2004.
  * [6] L. Habsieger, _Une $q$-intégrale de Selberg et Askey_, SIAM J. Math. Anal. 19 (1988) 1475–1489.
  * [7] J. Haglund, _The $q,t$-Catalan Numbers and the Space of Diagonal Harmonics: With an appendix on the combinatorics of Macdonald polynomials_, Univ. Lecture Ser., vol. 41, Amer. Math. Soc., Providence, RI, (2008).
  * [8] K. W. J. Kadell, _A proof of Askey’s conjectured $q$-analogue of Selberg’s integral and a conjecture of Morris_, SIAM J. Math. Anal. 19 (1988), 969–986.
  * [9] A. Lascoux, _Symmetric Functions and Combinatorial Operators on Polynomials_ , CBMS Reg. Conf. Ser. Math., Vol. 99, Amer. Math. Soc., Providence, RI, 2003.
  * [10] I. G. Macdonald, _Symmetric Functions and Hall Polynomials_ , 2nd edition, The Clarendon Press, Oxford University Press, 1995.
  * [11] W. G. Morris, _Constant Term Identities for Finite and Affine Root System: Conjectures and Theorems_ , Ph.D. thesis, Univ. Wisconsin–Madison, 1982.
  * [12] E. M. Rains and S. O. Warnaar, _Bounded Littlewood identities_ , Mem. Amer. Math. Soc., 270 (2021), No 1317, vii+115pp.
  * [13] A. Selberg, _Bemerkninger om et multipelt integral_ , Norsk Mat. Tidsskr. 26 (1944), 71–78.
  * [14] V. Tarasov and A. Varchenko, _Selberg-type integrals associated with $sl_{3}$_, Lett. Math. Phys., 65 (2003), 173–185.
  * [15] S. O. Warnaar, _A Selberg integral for the Lie algebra $A_{n}$_, Acta Math. 203 (2009), 269–304.
  * [16] G. Xin and Y. Zhou, Symmetric function generalizations of the $q$-Baker–Forrester ex-conjecture and Selberg-type integrals, arXiv: 2210.13469.