# Algebraic formulas and Geometric derivation
of Source Identities

Kohei Motegi and Ryo Ohkawa E-mail<EMAIL_ADDRESS>Faculty of Marine
Technology, Tokyo University of Marine Science and Technology,Etchujima 2-1-6,
Koto-Ku, Tokyo, 135-8533, Japan, E-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>Osaka Central Advanced Mathematical Institute, Osaka
Metropolitan University, 558-8585, Osaka, Japan, Research Institute for
Mathematical Sciences, Kyoto University, 606-8502, Kyoto, Japan.

###### Abstract

Source identities are fundamental identities between multivariable special
functions. We give a geometric derivation of rational and trigonometric source
identities. We also give a systematic derivation and extension of various
determinant representations for source functions which appeared in previous
literature as well as introducing the elliptic version of the determinants,
and obtain identities between determinants. We also show several
symmetrization formulas for the rational version.

##  1 Introduction

Identities between special functions play important roles not only in special
function theory itself but also various fields in mathematics and mathematical
physics such as geometric representation theory, supersymmetric gauge theory
and integrable systems. Typicial ones are transformation formulas for
hypergeometric functions which appear as wall-crossing phenomena in
supersymmetric gauge theory. For example, Kajihara’s transformation formulas
[1] for hypergeometric functions appeared in the context of supersymmetric
gauge theory [2, 3], and a geometric derivation of the rational version was
given by one of the authors and Yoshida [4]. The setting of the algebraic
variety used for the derivation was the handsaw quiver variety.

The derivation of transformation formulas for hypergeometric functions by
Kajihara [1] and an elliptic version by Kajihara-Noumi [5] (see also Rosengren
[6] for a different derivation) is based on specializations of more
fundamental identities called as source identities. The basic idea is that
taking multiple principal specializations of the original variables for the
source identities lead to transformation formulas for hypergeometric
functions.

For a positive integer $n$, we define $[1,\dots,n]:=\\{j\ |\ j=1,\dots,n\\}$
and denote the sum over all subsets of $[1,\dots,n]$ as
$\displaystyle\sum_{K\subset[1,\dots,n]}$. For a set of integers $K$, we
denote the number of elements of the set $K$ as $|K|$. We introduce the
$q$-infinite product $(u;q)_{\infty}:=\prod_{j=0}^{\infty}(1-uq^{j})$ and the
odd theta function

$\displaystyle\theta(u;p):=(u;p)_{\infty}(p/u;p)_{\infty}=\prod_{j=0}^{\infty}(1-up^{j})(1-p^{j+1}/u).$
(1.1)

Elliptic version of the source identity is

$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}q^{|K|(|K|-1)/2}\theta\Big{(}q^{|K|}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{\theta(qv_{j}/v_{i};p)}{\theta(v_{j}/v_{i};p)}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq
n\end{subarray}}\frac{\theta(u_{k}/v_{i};p)}{\theta(qu_{k}/v_{i};p)}$
$\displaystyle=$
$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}q^{|K|(|K|-1)/2}\theta\Big{(}q^{|K|}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{\theta(qu_{i}/u_{j};p)}{\theta(u_{i}/u_{j};p)}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq
n\end{subarray}}\frac{\theta(u_{i}/v_{k};p)}{\theta(qu_{i}/v_{k};p)},$ (1.2)

where $q,p,\Lambda,z,v_{1},\dots,v_{n},u_{1},\dots,u_{n}$ are complex
parameters. The elliptic identity (1.2) was derived in [5] by applying
difference operators on the Frobenius determinant formula which is an elliptic
version of the Cauchy determinant formula, and was used to derive
transformation formulas for elliptic hypergeometric functions by multiple
principal specializations.

For the trigonometric case, the source identity is

$\displaystyle\sum_{K\subset[1,\dots,m]}(-z)^{|K|}q^{|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{v_{i}-qv_{j}}{v_{i}-v_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq n\end{subarray}}\frac{v_{i}-u_{k}}{v_{i}-qu_{k}}$
$\displaystyle=$
$\displaystyle\frac{(z;q)_{\infty}}{(q^{m-n}z;q)_{\infty}}\sum_{K\subset[1,\dots,n]}(-q^{m-n}z)^{|K|}q^{|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{qu_{i}-u_{j}}{u_{i}-u_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m\end{subarray}}\frac{v_{k}-u_{i}}{v_{k}-qu_{i}},$ (1.3)

where $q,z,v_{1},\dots,v_{m},u_{1},\dots,u_{n}$ are complex parameters, and
the integers $m$ and $n$ are not necessarily the same. This source identity
was derived by Kirillov-Noumi [7] and Mimachi-Noumi [8], and was used in [8]
to construct integral representations for eigenfunctions of Macdonald
operators, and later used by Kajihara [1] to derive transformation formulas
for $q$-hypergeometric functions.

Rational version of the source identity is

$\displaystyle\sum_{K\subset[1,\dots,m]}(-z)^{|K|}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{v_{i}-v_{j}-c}{v_{i}-v_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq n\end{subarray}}\frac{v_{i}-u_{k}}{v_{i}-u_{k}-c}$
$\displaystyle=$
$\displaystyle(1-z)^{m-n}\sum_{K\subset[1,\dots,n]}(-z)^{|K|}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{u_{i}-u_{j}+c}{u_{i}-u_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m\end{subarray}}\frac{u_{i}-v_{k}}{u_{i}-v_{k}+c},$ (1.4)

where $c,z,v_{1},\dots,v_{m},u_{1},\dots,u_{n}$ are complex parameters.
Although not written in this form, one observes that the identity between
determinants in the paper by Gorsky-Zabrodin-Zotov [9] is equivalent to this
rational version of source identity. One can see the equivalence by using the
expansion of the determinants given in Belliard-Slavnov-Vallet [10]. In [9],
the identity was combined with the Bethe ansatz equations to derive duality
between classical and quantum integrable systems. The trigonometric version of
the identity between determinants is given in [11], which is equivalent to
(1.3).

One of the main purposes of this paper is to give a geometric understanding of
these identities. The summations of both sides in the source identities are
regarded as torus equivariant integrals over moduli of the following framed
quiver $Q\colon$

In [12] and [13], we derived wall-crossing formula for general setting of
framed quiver moduli. Applying this to the above framed quiver $Q$, we derive
the source identities.

Another purpose of this paper is to give various determinant forms of the
rational functions appearing in the source identities. As mentioned above,
rational/trigonometric source identities appear in the works in integrable
systems in disguise, i.e. in the form of identities between determinants.
Besides application to quantum-classical duality [9], one notes that special
cases $z=1$ and/or $m=n$ of the rational functions or determinants appear as
representations for (partial) domain wall boundary partition functions and
scalar products of the six-vertex models in the works of Izergin-Korepin [14,
15], Gromov-Sever-Vieira [16], Kostov [17, 18], Foda-Wheeler [19], Belliard-
Slavnov [20] and Minin-Pronko-Tarasov [21, 22]. Recently, $z$ generic and
$m\neq n$ case of the rational version of source functions appear in a more
generalized version of domain wall boundary partition functions by Belliard-
Pimenta-Slavnov [23]. We also remark this type of determinant form first
appeared in the context of the Bose gas in the work of Gaudin [24]. We give a
systematic derivation and extension of various forms of the determinants which
appeared in these literature. The idea is to start from the expressions of
source functions as rational functions with products of difference operators
applied, which appeared in Kajihara-Noumi [1, 5]. The systematic derivation
can be extended to an elliptic version by noting that (1.2) is originally
obtained by acting products of difference operators on an elliptic version of
Cauchy determinant in Kajihara-Noumi [5]. As a consequence of this systematic
derivation, we obtain identities between elliptic determinants.

This paper is organized as follows. We discuss the elliptic, trigonometric and
rational version of source functions and source identities in section 2, 3 and
4 respectively. We give a systematic derivation of various determinant forms
of source functions, and also a standard complex analytic proof of source
identities. For the trigonometric and rational case, we discuss a degeneration
from the elliptic and trigonometric version and also a geometric derivation of
the source identities. Additionaly, we prove a rational version of
symmetrization formulas of Lascoux for the rational case.

We summarize the notations which are frequently used in this paper. We denote
the set of variables $x_{1},\dots,x_{\ell}$ as $\overline{x}$. We frequently
use the sets $\overline{u}=\\{u_{1},\dots,u_{n}\\}$ and
$\overline{v}=\\{v_{1},\dots,v_{m}\\}$. We define additive and multiplicative
difference operators $T_{x}$ and $T_{q,x}$ acting on function $f(x)$ as
$T_{x}f(x):=f(x+c)$ and $T_{q,x}f(x):=f(qx)$ where $c$ and $q$ are complex
parameters. The inverse operators $T_{x}^{-1}$ and $T_{q,x}^{-1}$ act on
function $f(x)$ as $T_{x}^{-1}f(x):=f(x-c)$ and
$T_{q,x}^{-1}f(x):=f(q^{-1}x)$.

##  2 Elliptic version

In this section, we derive several determinant representations for elliptic
source functions and also give a complex analytic proof of the elliptic source
identity.

###  2.1 Determinant representations of elliptic source functions

We introduce two types of determinant representations for elliptic source
functions. We give an elliptic version of the determinants introduced in
Minin-Pronko-Tarasov [21, 22] and Belliard-Slavnov [20].

Recall the $q$-infinite product and the version of the odd theta function
which is used in this paper

$\displaystyle(u;q)_{\infty}=\prod_{j=0}^{\infty}(1-uq^{j}),$ (2.1)
$\displaystyle\theta(u;p)=(u;p)_{\infty}(p/u;p)_{\infty}=\prod_{j=0}^{\infty}(1-up^{j})(1-p^{j+1}/u).$
(2.2)

Let us introduce notations for the left and right hand side of (1.2):

$\displaystyle F^{ell(z)}_{n}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}q^{|K|(|K|-1)/2}\theta\Big{(}q^{|K|}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{\theta(qv_{j}/v_{i};p)}{\theta(v_{j}/v_{i};p)}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq
n\end{subarray}}\frac{\theta(u_{k}/v_{i};p)}{\theta(qu_{k}/v_{i};p)},$ (2.3)
$\displaystyle G^{ell(z)}_{n}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}q^{|K|(|K|-1)/2}\theta\Big{(}q^{|K|}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{\theta(qu_{i}/u_{j};p)}{\theta(u_{i}/u_{j};p)}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq
n\end{subarray}}\frac{\theta(u_{i}/v_{k};p)}{\theta(qu_{i}/v_{k};p)}.$ (2.4)

We call these functions as elliptic source functions in this paper.

The following elliptic analog of factorization of Vandermonde determinants is
derived by Rosengren-Schlosser [27]

$\displaystyle\det_{1\leq j,k\leq
n}\Big{(}\psi_{j}^{A_{n-1}}(u_{k};p,r)\Big{)}=\frac{(p;p)_{\infty}^{n}}{(p^{n};p^{n})_{\infty}^{n}}\theta\Bigg{(}r\prod_{\ell=1}^{n}u_{\ell};p\Bigg{)}\prod_{1\leq
i<j\leq n}u_{j}\theta(u_{i}/u_{j};p),$ (2.5)

where

$\displaystyle\psi_{j}^{A_{n-1}}(u;p,r):=u^{j-1}\theta(p^{j-1}(-1)^{n-1}ru^{n};p^{n}),\
\ \ j=1,\dots,n.$ (2.6)

See [28] for applications of the determinants to elliptic stochastic systems.

We also use the Frobenius determinant formula [29] which is an elliptic
analogue of Cauchy’s determinant formula

$\displaystyle\det_{1\leq i,j\leq n}\Bigg{(}\frac{\theta(\Lambda
u_{i}/v_{j};p)}{\theta(\Lambda;p)\theta(u_{i}/v_{j};p)}\Bigg{)}=\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}\prod_{1\leq
i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)v_{j}^{-1}\theta(v_{j}/v_{i};p)}{\theta(\Lambda;p)\prod_{1\leq
i,j\leq n}\theta(u_{i}/v_{j};p)}.$ (2.7)

The source functions can be rewritten using multiplicative difference
operators as

$\displaystyle F^{ell(z)}_{n}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\frac{\theta(\Lambda;p)\prod_{i,j=1}^{n}\theta(u_{i}/v_{j};p)}{\prod_{1\leq
i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)v_{j}^{-1}\theta(v_{j}/v_{i};p)}\prod_{j=1}^{n}(1-zT_{q,v_{j}}^{-1})\det_{1\leq
i,j\leq n}\Bigg{(}\frac{\theta(\Lambda
u_{i}/v_{j};p)}{\theta(\Lambda;p)\theta(u_{i}/v_{j};p)}\Bigg{)},$ (2.8)
$\displaystyle G^{ell(z)}_{n}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\frac{\theta(\Lambda;p)\prod_{i,j=1}^{n}\theta(u_{i}/v_{j};p)}{\prod_{1\leq
i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)v_{j}^{-1}\theta(v_{j}/v_{i};p)}\prod_{j=1}^{n}(1-zT_{q,u_{j}})\det_{1\leq
i,j\leq n}\Bigg{(}\frac{\theta(\Lambda
u_{i}/v_{j};p)}{\theta(\Lambda;p)\theta(u_{i}/v_{j};p)}\Bigg{)},$ (2.9)

which is essentially the same object used by Kajihara-Noumi [5] to derive
transformation formulas for elliptic hypergeometric functions from the source
identity
$F^{ell(z)}_{n}(\overline{u}|\overline{v})=G^{ell(z)}_{n}(\overline{u}|\overline{v})$.
Starting from these forms, it is easy to see
$F^{ell(z)}_{n}(\overline{u}|\overline{v})=G^{ell(z)}_{n}(\overline{u}|\overline{v})$
by inserting the difference operators into the determinants. Using the
factorization of the elliptic determinant (2.7), the right hand side of (2.8)
and (2.9) can be rewritten as

$\displaystyle F^{ell(z)}_{n}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\frac{\prod_{1\leq i,j\leq n}\theta(u_{i}/v_{j};p)}{\prod_{1\leq
i<j\leq
n}v_{j}^{-1}\theta(v_{j}/v_{i};p)}\prod_{j=1}^{n}(1-zT_{q,v_{j}}^{-1})\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}\prod_{1\leq
i<j\leq n}v_{j}^{-1}\theta(v_{j}/v_{i};p)}{\prod_{1\leq i,j\leq
n}\theta(u_{i}/v_{j};p)},$ (2.10) $\displaystyle
G^{ell(z)}_{n}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\frac{\prod_{1\leq i,j\leq n}\theta(u_{i}/v_{j};p)}{\prod_{1\leq
i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)}\prod_{j=1}^{n}(1-zT_{q,u_{j}})\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}\prod_{1\leq
i<j\leq n}u_{j}\theta(u_{i}/u_{j};p)}{\prod_{1\leq i,j\leq
n}\theta(u_{i}/v_{j};p)},$ (2.11)

and further expanding using

$\displaystyle\Big{(}\prod_{1\leq i<j\leq
n}v_{j}^{-1}\theta(v_{j}/v_{i};p)\Big{)}^{-1}\prod_{k\in
K}T_{q,v_{k}}^{-1}\Big{(}\prod_{1\leq i<j\leq
n}v_{j}^{-1}\theta(v_{j}/v_{i};p)\Big{)}=q^{|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{\theta(qv_{j}/v_{i};p)}{\theta(v_{j}/v_{i};p)},$ (2.12)
$\displaystyle\Big{(}\prod_{1\leq i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)\Big{)}^{-1}\prod_{k\in
K}T_{q,u_{k}}\Big{(}\prod_{1\leq i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)\Big{)}=q^{|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{\theta(qu_{i}/u_{j};p)}{\theta(u_{i}/u_{j};p)},$ (2.13)

gives (2.3) and (2.4) respectively, and this is how the elliptic source
identity (1.2) was derived in [5].

We derive an elliptic version of determinant representations whose
trigonometric and rational version which appeared in Minin-Pronko-Tarasov [21,
22].

######  Proposition 2.1.

We have the following determinant representations:

$\displaystyle
F^{ell(z)}_{n}(\overline{u}|\overline{v})=\frac{\theta(r\prod_{\ell=1}^{n}v_{\ell}^{-1};p)}{\det_{1\leq
i,j\leq
n}\Big{(}\sum_{k=1}^{n}p_{ik}\psi_{k}^{A_{n-1}}(v_{j}^{-1};p,r)\Big{)}}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(r\prod_{\ell=1}^{n}v_{\ell}^{-1};p)}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}p_{ik}\psi_{k}^{A_{n-1}}(v_{j}^{-1};p,r)$
$\displaystyle-z\Bigg{(}\frac{\theta\Big{(}q\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(qr\prod_{\ell=1}^{n}v_{\ell}^{-1};p)}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}p_{ik}\psi_{k}^{A_{n-1}}(qv_{j}^{-1};p,r)\prod_{\ell=1}^{n}\frac{\theta(u_{\ell}/v_{j};p)}{\theta(qu_{\ell}/v_{j};p)}\Bigg{)},$
(2.14) $\displaystyle
G^{ell(z)}_{n}(\overline{u}|\overline{v})=\frac{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)}{\det_{1\leq
i,j\leq n}\Big{(}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)\Big{)}}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)$
$\displaystyle-z\Bigg{(}\frac{\theta\Big{(}q\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(qr\prod_{\ell=1}^{n}u_{\ell};p)}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(qu_{j};p,r)\prod_{\ell=1}^{n}\frac{\theta(u_{j}/v_{\ell};p)}{\theta(qu_{j}/v_{\ell};p)}\Bigg{)}.$
(2.15)

Here, $\delta_{ij}$ is the Kronecker delta, and $r$, $p_{ij}$ and $q_{ij}$
$(i,j=1,\dots,n)$ are additional parameters such that $\det_{1\leq i,j\leq
n}(p_{ij})\not\equiv 0$, $\det_{1\leq i,j\leq n}(q_{ij})\not\equiv 0$.

###### Proof.

We show the case for $G^{ell(z)}_{n}(\overline{u}|\overline{v})$ (2.15) as
(2.14) can be derived in the same way. We start from the expression (2.11).
Using the elliptic analog of the factorization of Vandermonde determinants
(2.5), we can rewrite (2.11) as

$\displaystyle G^{ell(z)}_{n}(\overline{u}|\overline{v})=$
$\displaystyle\frac{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)\prod_{1\leq i,j\leq
n}\theta(u_{i}/v_{j};p)}{\det_{1\leq j,k\leq
n}\Big{(}\psi_{k}^{A_{n-1}}(u_{j};p,r)\Big{)}}$
$\displaystyle\times\prod_{j=1}^{n}(1-zT_{q,u_{j}})\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}\prod_{j=1}^{n}v_{j};p\Big{)}\det_{1\leq
j,k\leq
n}\Big{(}\psi_{k}^{A_{n-1}}(u_{j};p,r)\Big{)}}{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)\prod_{1\leq
i,j\leq n}\theta(u_{i}/v_{j};p)}.$ (2.16)

Multiplying the denominator and the numerator by some $\det_{1\leq i,k\leq
n}(q_{ik})\not\equiv 0$, where $q_{ik}$, $i,k=1,\dots,n$ are parameters
independent of $u_{i}$, $i=1,\dots,n$, we can rewrite (2.16) as

$\displaystyle G^{ell(z)}_{n}(\overline{u}|\overline{v})=$
$\displaystyle\frac{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)\prod_{1\leq i,j\leq
n}\theta(u_{i}/v_{j};p)}{\det_{1\leq i,k\leq n}(q_{ik})\det_{1\leq j,k\leq
n}\Big{(}\psi_{k}^{A_{n-1}}(u_{j};p,r)\Big{)}}\prod_{j=1}^{n}(1-zT_{q,u_{j}})$
$\displaystyle\times\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}\det_{1\leq
i,k\leq n}(q_{ik})\det_{1\leq j,k\leq
n}\Big{(}\psi_{k}^{A_{n-1}}(u_{j};p,r)\Big{)}}{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)\prod_{1\leq
i,j\leq n}\theta(u_{i}/v_{j};p)}$ $\displaystyle=$
$\displaystyle\frac{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)\prod_{1\leq i,j\leq
n}\theta(u_{i}/v_{j};p)}{\det_{1\leq i,j\leq
n}\Big{(}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)\Big{)}}\prod_{j=1}^{n}(1-zT_{q,u_{j}})$
$\displaystyle\times\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}\det_{1\leq
i,j\leq
n}\Big{(}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)\Big{)}}{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)\prod_{1\leq
i,j\leq n}\theta(u_{i}/v_{j};p)}$ $\displaystyle=$
$\displaystyle\frac{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)\prod_{1\leq i,j\leq
n}\theta(u_{i}/v_{j};p)}{\det_{1\leq i,j\leq
n}\Big{(}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)\Big{)}}\prod_{j=1}^{n}(1-zT_{q,u_{j}})$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)}\Bigg{)}^{\delta_{i1}}\frac{\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)}{\prod_{\ell=1}^{n}\theta(u_{j}/v_{\ell};p)}\Bigg{)}.$
(2.17)

Inserting the difference operators inside the determinant, we have

$\displaystyle
G^{ell(z)}_{n}(\overline{u}|\overline{v})=\frac{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)\prod_{1\leq
i,j\leq n}\theta(u_{i}/v_{j};p)}{\det_{1\leq i,j\leq
n}\Big{(}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)\Big{)}}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\prod_{j=1}^{n}(1-zT_{q,u_{j}})\Bigg{(}\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)}\Bigg{)}^{\delta_{i1}}\frac{\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)}{\prod_{\ell=1}^{n}\theta(u_{j}/v_{\ell};p)}\Bigg{)}\Bigg{)}$
$\displaystyle=$
$\displaystyle\frac{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)}{\det_{1\leq i,j\leq
n}\Big{(}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)\Big{)}}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)$
$\displaystyle-z\Bigg{(}\frac{\theta\Big{(}q\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(qr\prod_{\ell=1}^{n}u_{\ell};p)}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(qu_{j};p,r)\prod_{\ell=1}^{n}\frac{\theta(u_{j}/v_{\ell};p)}{\theta(qu_{j}/v_{\ell};p)}\Bigg{)}.$
(2.18)

∎

This form can be regarded as an elliptic analog of Minin-Pronko-Tarasov [21],
which will be explained in later sections.

We apply the same idea due to Kajihara-Noumi [5] now to determinant
representations instead of source functions. From (2.14), (2.15) and
$F^{ell(z)}_{n}(\overline{u}|\overline{v})=G^{ell(z)}_{n}(\overline{u}|\overline{v})$,
we have the following identity between determinants.

######  Theorem 2.2.

We have the following identity:

$\displaystyle\frac{\theta(r\prod_{\ell=1}^{n}v_{\ell}^{-1};p)}{\det_{1\leq
i,j\leq
n}\Big{(}\sum_{k=1}^{n}p_{ik}\psi_{k}^{A_{n-1}}(v_{j}^{-1};p,r)\Big{)}}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(r\prod_{\ell=1}^{n}v_{\ell}^{-1};p)}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}p_{ik}\psi_{k}^{A_{n-1}}(v_{j}^{-1};p,r)$
$\displaystyle-z\Bigg{(}\frac{\theta\Big{(}q\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(qr\prod_{\ell=1}^{n}v_{\ell}^{-1};p)}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}p_{ik}\psi_{k}^{A_{n-1}}(qv_{j}^{-1};p,r)\prod_{\ell=1}^{n}\frac{\theta(u_{\ell}/v_{j};p)}{\theta(qu_{\ell}/v_{j};p)}\Bigg{)}$
$\displaystyle=$
$\displaystyle\frac{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)}{\det_{1\leq i,j\leq
n}\Big{(}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)\Big{)}}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(r\prod_{\ell=1}^{n}u_{\ell};p)}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};p,r)$
$\displaystyle-z\Bigg{(}\frac{\theta\Big{(}q\Lambda\prod_{\ell=1}^{n}u_{\ell}/\prod_{\ell=1}^{n}v_{\ell};p\Big{)}}{\theta(qr\prod_{\ell=1}^{n}u_{\ell};p)}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(qu_{j};p,r)\prod_{\ell=1}^{n}\frac{\theta(u_{j}/v_{\ell};p)}{\theta(qu_{j}/v_{\ell};p)}\Bigg{)}.$
(2.19)

We next derive an elliptic analog of determinant representations which can be
regarded as an elliptic lift of the rational version introduced in Belliard-
Slavnov [20].

######  Proposition 2.3.

We have the following determinant representations:

$\displaystyle F_{n}^{ell(z)}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\frac{\theta(\Delta;p)}{\prod_{1\leq i<j\leq
n}v_{j}^{-1}\theta(v_{j}/v_{i};p)\eta_{j}\theta(\eta_{i}/\eta_{j};p)}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}}{\theta\Big{(}\Delta\prod_{i=1}^{n}\eta_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}}\Bigg{)}^{\delta_{i1}}\frac{\theta(\Delta\eta_{i}/v_{j};p)\prod_{k=1,k\neq
i}^{n}\theta(\eta_{k}/v_{j};p)}{\theta(\Delta;p)}$
$\displaystyle-z\Bigg{(}\frac{\theta\Big{(}q\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}}{\theta\Big{(}q\Delta\prod_{i=1}^{n}\eta_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}}\Bigg{)}^{\delta_{i1}}\frac{\theta(q\Delta\eta_{i}/v_{j};p)\prod_{k=1,k\neq
i}^{n}\theta(q\eta_{k}/v_{j};p)}{\theta(\Delta;p)}\prod_{k=1}^{n}\frac{\theta(u_{k}/v_{j};p)}{\theta(qu_{k}/v_{j};p)}\Bigg{)},$
(2.20) $\displaystyle G_{n}^{ell(z)}(\overline{u}|\overline{v})$
$\displaystyle=$ $\displaystyle\frac{\theta(\Delta;p)}{\prod_{1\leq i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)\eta_{j}^{-1}\theta(\eta_{j}/\eta_{i};p)}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}}{\theta\Big{(}\Delta\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}\eta_{j};p\Big{)}}\Bigg{)}^{\delta_{i1}}\frac{\theta(\Delta
u_{i}/\eta_{j};p)\prod_{k=1,k\neq
j}^{n}\theta(u_{i}/\eta_{k};p)}{\theta(\Delta;p)}$
$\displaystyle-z\Bigg{(}\frac{\theta\Big{(}q\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}}{\theta\Big{(}q\Delta\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}\eta_{j};p\Big{)}}\Bigg{)}^{\delta_{i1}}\frac{\theta(q\Delta
u_{i}/\eta_{j};p)\prod_{k=1,k\neq
j}^{n}\theta(qu_{i}/\eta_{k};p)}{\theta(\Delta;p)}\prod_{k=1}^{n}\frac{\theta(u_{i}/v_{k};p)}{\theta(qu_{i}/v_{k};p)}\Bigg{)}.$
(2.21)

Here, $\Delta$ and $\eta_{i}$ $(i=1,\dots,n)$ are additional parameters.

###### Proof.

We show the case for $G_{n}^{ell(z)}(\overline{u}|\overline{v})$ (2.21).
(2.20) can be proved in the same way. We start from the expression (2.11) and
first rewrite as follows

$\displaystyle G_{n}^{ell(z)}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\frac{\prod_{1\leq i,j\leq n}\theta(u_{i}/v_{j};p)}{\prod_{1\leq
i<j\leq n}u_{j}\theta(u_{i}/u_{j};p)\eta_{j}^{-1}\theta(\eta_{j}/\eta_{i};p)}$
$\displaystyle\times\prod_{j=1}^{n}(1-zT_{q,u_{j}})\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}\prod_{1\leq
i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)\eta_{j}^{-1}\theta(\eta_{j}/\eta_{i};p)}{\prod_{1\leq
i,j\leq n}\theta(u_{i}/v_{j};p)}$ $\displaystyle=$
$\displaystyle\frac{\prod_{1\leq i,j\leq n}\theta(u_{i}/v_{j};p)}{\prod_{1\leq
i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)\eta_{j}^{-1}\theta(\eta_{j}/\eta_{i};p)}\prod_{j=1}^{n}(1-zT_{q,u_{j}})$
$\displaystyle\times\prod_{1\leq i,j\leq
n}\frac{\theta(u_{i}/\eta_{j};p)}{\theta(u_{i}/v_{j};p)}\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}\prod_{1\leq
i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)\eta_{j}^{-1}\theta(\eta_{j}/\eta_{i};p)}{\prod_{1\leq
i,j\leq n}\theta(u_{i}/\eta_{j};p)}.$ (2.22)

Using the Frobenius determinant formula (2.7), we get

$\displaystyle G_{n}^{ell(z)}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\frac{\prod_{1\leq i,j\leq n}\theta(u_{i}/v_{j};p)}{\prod_{1\leq
i<j\leq n}u_{j}\theta(u_{i}/u_{j};p)\eta_{j}^{-1}\theta(\eta_{j}/\eta_{i};p)}$
$\displaystyle\times\prod_{j=1}^{n}(1-zT_{q,u_{j}})\prod_{1\leq i,j\leq
n}\frac{\theta(u_{i}/\eta_{j};p)}{\theta(u_{i}/v_{j};p)}\frac{\theta(\Delta;p)\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}}{\theta\Big{(}\Delta\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}\eta_{j};p\Big{)}}\det_{1\leq
i,j\leq n}\Bigg{(}\frac{\theta(\Delta
u_{i}/\eta_{j};p)}{\theta(\Delta;p)\theta(u_{i}/\eta_{j};p)}\Bigg{)}$
$\displaystyle=$ $\displaystyle\frac{\theta(\Delta;p)\prod_{1\leq i,j\leq
n}\theta(u_{i}/v_{j};p)}{\prod_{1\leq i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)\eta_{j}^{-1}\theta(\eta_{j}/\eta_{i};p)}$
$\displaystyle\times\prod_{j=1}^{n}(1-zT_{q,u_{j}})\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}}{\theta\Big{(}\Delta\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}\eta_{j};p\Big{)}}\Bigg{)}^{\delta_{i1}}\frac{\theta(\Delta
u_{i}/\eta_{j};p)\prod_{k=1,k\neq
j}^{n}\theta(u_{i}/\eta_{k};p)}{\theta(\Delta;p)\prod_{k=1}^{n}\theta(u_{i}/v_{k};p)}\Bigg{)}.$
(2.23)

Inserting the difference operators into the determinant, we have

$\displaystyle G_{n}^{ell(z)}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\frac{\theta(\Delta;p)\prod_{1\leq i,j\leq
n}\theta(u_{i}/v_{j};p)}{\prod_{1\leq i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)\eta_{j}^{-1}\theta(\eta_{j}/\eta_{i};p)}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}(1-zT_{q,u_{i}})\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}}{\theta\Big{(}\Delta\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}\eta_{j};p\Big{)}}\Bigg{)}^{\delta_{i1}}\frac{\theta(\Delta
u_{i}/\eta_{j};p)\prod_{k=1,k\neq
j}^{n}\theta(u_{i}/\eta_{k};p)}{\theta(\Delta;p)\prod_{k=1}^{n}\theta(u_{i}/v_{k};p)}\Bigg{)}$
$\displaystyle=$ $\displaystyle\frac{\theta(\Delta;p)}{\prod_{1\leq i<j\leq
n}u_{j}\theta(u_{i}/u_{j};p)\eta_{j}^{-1}\theta(\eta_{j}/\eta_{i};p)}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{\theta\Big{(}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}}{\theta\Big{(}\Delta\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}\eta_{j};p\Big{)}}\Bigg{)}^{\delta_{i1}}\frac{\theta(\Delta
u_{i}/\eta_{j};p)\prod_{k=1,k\neq
j}^{n}\theta(u_{i}/\eta_{k};p)}{\theta(\Delta;p)}$
$\displaystyle-z\Bigg{(}\frac{\theta\Big{(}q\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}}{\theta\Big{(}q\Delta\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}\eta_{j};p\Big{)}}\Bigg{)}^{\delta_{i1}}\frac{\theta(q\Delta
u_{i}/\eta_{j};p)\prod_{k=1,k\neq
j}^{n}\theta(qu_{i}/\eta_{k};p)}{\theta(\Delta;p)}\prod_{k=1}^{n}\frac{\theta(u_{i}/v_{k};p)}{\theta(qu_{i}/v_{k};p)}\Bigg{)}.$
(2.24)

∎

###  2.2 A complex anaytic proof of the elliptic source identity

Although it is easy to see the elliptic source identity follows from the
Frobenius determinant formula as first derived in [5], we give a direct
complex analytic proof. Some of the properties of elliptic source functions
derived in this section are elliptic version of the ones in [21, 22].

We prove the following version of the source identity

$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}q^{|K|(|K|-1)/2}\theta\Big{(}q^{|K|}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}$
$\displaystyle\times\prod_{\begin{subarray}{c}i\in K\\\ j\not\in
K\end{subarray}}\frac{\theta(qv_{j}/v_{i};p)}{\theta(v_{j}/v_{i};p)}\prod_{\begin{subarray}{c}j\not\in
K\\\ 1\leq k\leq
n\end{subarray}}\theta(qu_{k}/v_{j};p)\prod_{\begin{subarray}{c}i\in K\\\
1\leq k\leq n\end{subarray}}\theta(u_{k}/v_{i};p)$ $\displaystyle=$
$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}q^{|K|(|K|-1)/2}\theta\Big{(}q^{|K|}\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j};p\Big{)}$
$\displaystyle\times\prod_{\begin{subarray}{c}i\in K\\\ j\not\in
K\end{subarray}}\frac{\theta(qu_{i}/u_{j};p)}{\theta(u_{i}/u_{j};p)}\prod_{\begin{subarray}{c}j\not\in
K\\\ 1\leq k\leq
n\end{subarray}}\theta(qu_{j}/v_{k};p)\prod_{\begin{subarray}{c}i\in K\\\
1\leq k\leq n\end{subarray}}\theta(u_{i}/v_{k};p).$ (2.25)

Denote the left and right hand side of (2.25) as
$P_{n}^{ell(z)}(\overline{u}|\overline{v})$ and
$Q_{n}^{ell(z)}(\overline{u}|\overline{v})$ respectively. Note the relations
with $F_{n}^{ell(z)}(\overline{u}|\overline{v})$ and
$G_{n}^{ell(z)}(\overline{u}|\overline{v})$ are

$\displaystyle P_{n}^{ell(z)}(\overline{u}|\overline{v})$
$\displaystyle=\prod_{i=1}^{n}\prod_{k=1}^{n}\theta(qu_{k}/v_{i};p)F_{n}^{ell(z)}(\overline{u}|\overline{v}),$
(2.26) $\displaystyle Q_{n}^{ell(z)}(\overline{u}|\overline{v})$
$\displaystyle=\prod_{i=1}^{n}\prod_{k=1}^{n}\theta(qu_{k}/v_{i};p)G_{n}^{ell(z)}(\overline{u}|\overline{v}).$
(2.27)

The complex analytic proof is based on the following fact for elliptic
interpolation which can be seen for example in [30].

######  Proposition 2.4.

([30], Proposition 4.27) Let $t,y_{1},\dots,y_{n}\in\mathbb{C}^{*}$ be such
that $t\not\in p^{\mathbb{Z}}$, $y_{j}/y_{k}\not\in p^{\mathbb{Z}}\ (j\neq
k)$. Let $V$ be the space of functions that are analytic for $x\neq 0$ and
satisfy $f(px)=(-1)^{n}y_{1}\cdots y_{n}t^{-1}x^{-n}f(x)$. Then, any $f\in V$
is uniquely determined by the values $f(y_{1}),\dots,f(y_{n})$ as

$\displaystyle
f(x)=\sum_{j=1}^{n}f(y_{j})\frac{\theta(tx/y_{j};p)}{\theta(t;p)}\prod_{\begin{subarray}{c}k=1\\\
k\neq j\end{subarray}}^{n}\frac{\theta(x/y_{k};p)}{\theta(y_{j}/y_{k};p)}.$
(2.28)

Using $\theta(px;p)/\theta(x;p)=-1/x$, we can see the following holds.

######  Lemma 2.5.

We have

$\displaystyle P_{n}^{ell(z)}(\overline{u}|\overline{v})|_{v_{k}\to pv_{k}}$
$\displaystyle=(-p^{-1})^{n+1}q^{n}\Lambda
v_{k}^{-n-1}\prod_{i=1}^{n}u_{i}^{2}\prod_{\begin{subarray}{c}j=1\\\ j\neq
k\end{subarray}}^{n}v_{j}^{-1}P_{n}^{ell(z)}(\overline{u}|\overline{v}),$
(2.29) $\displaystyle Q_{n}^{ell(z)}(\overline{u}|\overline{v})|_{v_{k}\to
pv_{k}}$ $\displaystyle=(-p^{-1})^{n+1}q^{n}\Lambda
v_{k}^{-n-1}\prod_{i=1}^{n}u_{i}^{2}\prod_{\begin{subarray}{c}j=1\\\ j\neq
k\end{subarray}}^{n}v_{j}^{-1}Q_{n}^{ell(z)}(\overline{u}|\overline{v}),$
(2.30)

for $k=1,\dots,n$.

From Proposition 2.4 and Lemma 2.5, one notes that it is enough to check
$P_{n}^{ell(z)}(\overline{u}|\overline{v})=Q_{n}^{ell(z)}(\overline{u}|\overline{v})$
for $(n+1)^{n}$ distinct points in $(v_{1},\dots,v_{n})$. We can in fact check
the equality for the following
$(2n)\times(2n-1)\times\cdots\times(n+1)\geq(n+1)^{n}$ specializations
$v_{1}=q^{\epsilon_{1}}u_{p_{1}},\ v_{2}=q^{\epsilon_{2}}u_{p_{2}},\
\dots,v_{n}=q^{\epsilon_{n}}u_{p_{n}}$, $p_{1},\dots,p_{n}\in\\{1,\dots,n\\}$,
$\epsilon_{1},\dots,\epsilon_{n}\in\\{0,1\\}$ and $v_{i}\neq v_{j}$ for $i\neq
j$. First we check the following cases which correspond to $p_{i}=p_{j}$ for
some $i\neq j$.

######  Lemma 2.6.

If $v_{i}=u_{k}$, $v_{j}=qu_{k}$ for some $i,j,k\ (i\neq j)$, we have

$\displaystyle
P_{n}^{ell(z)}(\overline{u}|\overline{v})=Q_{n}^{ell(z)}(\overline{u}|\overline{v})=0.$
(2.31)

###### Proof.

We check the case $k=n$. From the expressions (2.25), one can check the
following relations hold when substituting $v_{n}=qu_{n}$ in
$P_{n}^{ell(z)}(\overline{u}|\overline{v})$ and
$Q_{n}^{ell(z)}(\overline{u}|\overline{v})$

$\displaystyle P_{n}^{ell(z)}(\overline{u}|\overline{v})|_{v_{n}=qu_{n}}$
$\displaystyle=$
$\displaystyle-z\prod_{j=1}^{n}\theta(u_{j}/qu_{n};p)\prod_{j=1}^{n-1}q\theta(u_{n}/v_{j};p)P_{n-1}^{ell(z)}(u_{1},\dots,u_{n-1}|v_{1},\dots,v_{n-1}|\Lambda),$
(2.32) $\displaystyle
Q_{n}^{ell(z)}(\overline{u}|\overline{v})|_{v_{n}=qu_{n}}$ $\displaystyle=$
$\displaystyle-z\prod_{j=1}^{n}\theta(u_{j}/qu_{n};p)\prod_{j=1}^{n-1}q\theta(u_{n}/v_{j};p)Q_{n-1}^{ell(z)}(u_{1},\dots,u_{n-1}|v_{1},\dots,v_{n-1}|\Lambda).$
(2.33)

From the factor $\prod_{j=1}^{n-1}q\theta(u_{n}/v_{j})$ in (2.32) and (2.33),
one notes that $P_{n}^{ell(z)}(\overline{u}|\overline{v})|_{v_{n}=qu_{n}}$ and
$Q_{n}^{ell(z)}(\overline{u}|\overline{v})|_{v_{n}=qu_{n}}$ vanish after
further substituting $v_{j}=u_{n}$ for some $j\ (1\leq j\leq n-1)$. The other
cases can be checked in the same way or follow by symmetry. ∎

We next assume $p_{i}\neq p_{j}$ for $i\neq j$. For $I=\\{1\leq
i_{1}<i_{2}<\cdots<i_{|I|}\leq n\\}$ and $J=\\{1\leq
j_{1}<j_{2}<\cdots<j_{|J|}\leq n\\}$ such that $I\cup J=[1,\dots,n]$, let us
denote by $\overline{v}=\\{\overline{u}_{I},q\overline{u}_{J}\\}$ the
following type of substitution

$\displaystyle
v_{q_{1}}=u_{i_{1}},\dots,v_{q_{|I|}}=u_{i_{|I|}},v_{r_{1}}=qu_{j_{1}},\dots,v_{r_{|J|}}=qu_{j_{|J|}},$
(2.34)

for some $\\{q_{1},\dots,q_{|I|},r_{1},\dots,r_{|J|}\\}=[1,\dots,n]$.

We have the following explicit evaluations.

######  Proposition 2.7.

We have

$\displaystyle
P_{n}^{ell(z)}(\overline{u}|\\{\overline{u}_{I},q\overline{u}_{J}\\})$
$\displaystyle=$
$\displaystyle(-z)^{|J|}q^{|J|(|J|-1)/2}\theta(\Lambda;p)\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}\theta(u_{i}/u_{j};p)\prod_{\begin{subarray}{c}i\in
I\\\ 1\leq j\leq
n\end{subarray}}\theta(qu_{j}/u_{i};p)\prod_{\begin{subarray}{c}i\in J\\\ j\in
J\end{subarray}}\theta(u_{i}/qu_{j};p),$ (2.35) $\displaystyle
Q_{n}^{ell(z)}(\overline{u}|\\{\overline{u}_{I},q\overline{u}_{J}\\})$
$\displaystyle=$
$\displaystyle(-z)^{|J|}q^{|J|(|J|-1)/2}\theta(\Lambda;p)\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}\theta(u_{i}/u_{j};p)\prod_{\begin{subarray}{c}i\in
I\\\ 1\leq j\leq
n\end{subarray}}\theta(qu_{j}/u_{i};p)\prod_{\begin{subarray}{c}i\in J\\\ j\in
J\end{subarray}}\theta(u_{i}/qu_{j};p).$ (2.36)

###### Proof.

After the substitution
$v_{q_{1}}=u_{i_{1}},\dots,v_{q_{|I|}}=u_{i_{|I|}},v_{r_{1}}=qu_{j_{1}},\dots,v_{r_{|J|}}=qu_{j_{|J|}}$,
one notes from the factors $\displaystyle\prod_{\begin{subarray}{c}j\not\in
K\\\ 1\leq k\leq n\end{subarray}}\theta(qu_{k}/v_{j};p)$ and
$\displaystyle\prod_{\begin{subarray}{c}i\in K\\\ 1\leq k\leq
n\end{subarray}}\theta(u_{k}/v_{i};p)$ that only the summand corresponding to
$K=\\{r_{1},\dots,r_{|J|}\\}$ survives and we get

$\displaystyle
P_{n}^{ell(z)}(\overline{u}|\\{\overline{u}_{I},q\overline{u}_{J}\\})$
$\displaystyle=$
$\displaystyle(-z)^{|J|}q^{|J|(|J|-1)/2}\theta(\Lambda;p)\prod_{\begin{subarray}{c}i\in
J\\\ j\in
I\end{subarray}}\frac{\theta(qu_{j}/qu_{i};p)}{\theta(u_{j}/qu_{i};p)}\prod_{\begin{subarray}{c}j\in
I\\\ 1\leq k\leq
n\end{subarray}}\theta(qu_{k}/u_{j};p)\prod_{\begin{subarray}{c}j\in J\\\
1\leq k\leq n\end{subarray}}\theta(u_{k}/qu_{j};p)$ $\displaystyle=$
$\displaystyle(-z)^{|J|}q^{|J|(|J|-1)/2}\theta(\Lambda;p)\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}\theta(u_{i}/u_{j};p)\prod_{\begin{subarray}{c}i\in
I\\\ 1\leq j\leq
n\end{subarray}}\theta(qu_{j}/u_{i};p)\prod_{\begin{subarray}{c}i\in J\\\ j\in
J\end{subarray}}\theta(u_{i}/qu_{j};p).$ (2.37)

The factorization of
$Q_{n}^{ell(z)}(\overline{u}|\\{\overline{u}_{I},q\overline{u}_{J}\\})$ after
the substitution can be checked in the same way. Substituting
$v_{q_{1}}=u_{i_{1}},\dots,v_{q_{|I|}}=u_{i_{|I|}},v_{r_{1}}=qu_{j_{1}},\dots,v_{r_{|J|}}=qu_{j_{|J|}}$,
we note from the factors $\displaystyle\prod_{\begin{subarray}{c}j\not\in K\\\
1\leq k\leq n\end{subarray}}\theta(qu_{j}/v_{k};p)$ and
$\displaystyle\prod_{\begin{subarray}{c}i\in K\\\ 1\leq k\leq
n\end{subarray}}\theta(u_{i}/v_{k};p)$ that only the summand corresponding to
$K=\\{r_{1},\dots,r_{|J|}\\}$ survives, and the factorized expression can be
rewritten as

$\displaystyle
Q_{n}^{ell(z)}(\overline{u}|\\{\overline{u}_{I},q\overline{u}_{J}\\})$
$\displaystyle=$
$\displaystyle(-z)^{|J|}q^{|J|(|J|-1)/2}\theta(\Lambda;p)\prod_{\begin{subarray}{c}i\in
J\\\ j\in
I\end{subarray}}\frac{\theta(qu_{i}/u_{j};p)}{\theta(u_{i}/u_{j};p)}$
$\displaystyle\times\prod_{\begin{subarray}{c}j\in J\\\ k\in
I\end{subarray}}\theta(u_{j}/u_{k};p)\prod_{\begin{subarray}{c}j\in J\\\ k\in
J\end{subarray}}\theta(u_{j}/qu_{k};p)\prod_{\begin{subarray}{c}i\in I\\\ k\in
I\end{subarray}}\theta(qu_{i}/u_{k};p)\prod_{\begin{subarray}{c}i\in I\\\ k\in
J\end{subarray}}\theta(u_{i}/u_{k};p)$ $\displaystyle=$
$\displaystyle(-z)^{|J|}q^{|J|(|J|-1)/2}\theta(\Lambda;p)\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}\theta(u_{i}/u_{j};p)\prod_{\begin{subarray}{c}i\in
I\\\ 1\leq j\leq
n\end{subarray}}\theta(qu_{j}/u_{i};p)\prod_{\begin{subarray}{c}i\in J\\\ j\in
J\end{subarray}}\theta(u_{i}/qu_{j};p).$ (2.38)

∎

From Lemma 2.6 and Proposition 2.7, we have checked (2.25) holds for enough
specializations, hence we conclude (2.25) itself holds.

##  3 Trigonometric version

In this section, we discuss trigonometric source identity as a degeneration
from elliptic source identity. We also give a geometric derivation and a
complex analytic proof as well as giving a systematic derivation of
determinant representations which can be regarded as a generalization of the
ones which appeared in previous literature.

###  3.1 From elliptic source identity to trigonometric source identity

We discuss the degeneration from elliptic source identity (1.2) to
trigonometric source identity (1.3).

We introduce $q$-integers, $q$-factorials and $q$-binomials
$\displaystyle[n]_{q}:=1+q+q^{2}+\cdots+q^{n-1}=\frac{1-q^{n}}{1-q}$,
$[n]_{q}!:=\prod_{j=1}^{n}[j]_{q}$, $\displaystyle\begin{bmatrix}n\\\
\ell\end{bmatrix}_{q}:=\frac{[n]_{q}!}{[\ell]_{q}![n-\ell]_{q}!}$ for $n,\ell$
nonnegative integers. To discuss degeneration, we also introduce notation for
the following rational function

$\displaystyle F_{n,m}^{trig(z)}(\overline{u}|\overline{v}|\Lambda)$
$\displaystyle:=$
$\displaystyle\sum_{K\subset[1,\dots,m]}(-z)^{|K|}q^{|K|(|K|-1)/2}(1-q^{|K|}\Lambda)\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{v_{i}-qv_{j}}{v_{i}-v_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq n\end{subarray}}\frac{v_{i}-u_{k}}{v_{i}-qu_{k}}.$ (3.1)

Also note $\theta(z;0)=1-z$. By taking the trigonometric limit $p\to 0$ of the
elliptic source identity (1.2), we get the following identity.

######  Theorem 3.1.

For $n\geq m$, we have

$\displaystyle\sum_{\ell=0}^{n-m}(-z)^{\ell}q^{\ell(\ell-1)/2}\displaystyle\begin{bmatrix}n-m\\\
\ell\end{bmatrix}_{q}F_{n,m}^{trig(q^{n-m}z)}(\overline{u}|\overline{v}|q^{\ell}\Lambda)$
$\displaystyle=$
$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}q^{|K|(|K|-1)/2}(1-q^{|K|}\Lambda)\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{qu_{i}-u_{j}}{u_{i}-u_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m\end{subarray}}\frac{v_{k}-u_{i}}{v_{k}-qu_{i}}.$ (3.2)

Note that when $\Lambda=0$, using the $q$-binomial theorem

$\displaystyle\sum_{\ell=0}^{n}z^{\ell}q^{\ell(\ell+1)/2}\begin{bmatrix}n\\\
\ell\end{bmatrix}_{q}=\prod_{j=1}^{n}(1+q^{j}z),$ (3.3)

(3.2) reduces to

$\displaystyle\prod_{j=1}^{n-m}(1-q^{j-1}z)F_{n,m}^{trig(q^{n-m}z)}(\overline{u}|\overline{v}|0)=\sum_{K\subset[1,\dots,n]}(-z)^{|K|}q^{|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{qu_{i}-u_{j}}{u_{i}-u_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m\end{subarray}}\frac{v_{k}-u_{i}}{v_{k}-qu_{i}}$ (3.4)

which becomes the trigonometric source identity (1.3) after redefining
$q^{n-m}z$ as $z$. Also note that equating both hand sides of the coefficients
of $\Lambda$ in (3.2), we get (3.4) with $z$ replaced by $qz$. This means
(3.2) is actually a combination of the trigonometric source identity (1.3).

###### Proof.

We prove by descending induction on $m$, starting from $m=n$. By taking the
limit $p\to 0$ of elliptic source identity (1.2) and redefining
$\Lambda\prod_{i=1}^{n}u_{i}/\prod_{j=1}^{n}v_{j}$ as $\Lambda$, we get the
case $m=n$ of (3.2), which corresponds to the initial case.

Suppose (3.2) holds, and derive (3.2) with $m$ replaced by $m-1$. We take the
limit $v_{m}\rightarrow\infty$ of both hand sides of (3.2). Dividing the sums
in $F_{n,m}^{trig(q^{n-m}z)}(\overline{u}|\overline{v}|q^{\ell}\Lambda)$ into
two parts according to whether $m\in K$ or $m\not\in K$, one can show

$\displaystyle\lim_{v_{m}\rightarrow\infty}F_{n,m}^{trig(q^{n-m}z)}(\overline{u}|\overline{v}|q^{\ell}\Lambda)=-q^{n-m}zF_{n,m-1}^{trig(q^{n-m+1}z)}(\overline{u}|\overline{v}|q^{\ell+1}\Lambda)+F_{n,m-1}^{trig(q^{n-m+1}z)}(\overline{u}|\overline{v}|q^{\ell}\Lambda).$
(3.5)

Then we find the limit $v_{m}\rightarrow\infty$ of the left hand side of (3.2)
can be rewritten in the following way

$\displaystyle\sum_{\ell=0}^{n-m}(-z)^{\ell}q^{\ell(\ell-1)/2}\displaystyle\begin{bmatrix}n-m\\\
\ell\end{bmatrix}_{q}\Bigg{(}-q^{n-m}zF_{n,m-1}^{trig(q^{n-m+1}z)}(\overline{u}|\overline{v}|q^{\ell+1}\Lambda)+F_{n,m-1}^{trig(q^{n-m+1}z)}(\overline{u}|\overline{v}|q^{\ell}\Lambda)\Bigg{)}$
$\displaystyle=$
$\displaystyle\sum_{\ell=0}^{n-m+1}(-z)^{\ell}q^{\ell(\ell-1)/2}\Bigg{(}q^{n-m-\ell+1}\displaystyle\begin{bmatrix}n-m\\\
\ell-1\end{bmatrix}_{q}+\displaystyle\begin{bmatrix}n-m\\\
\ell\end{bmatrix}_{q}\Bigg{)}F_{n,m-1}^{trig(q^{n-m+1}z)}(\overline{u}|\overline{v}|q^{\ell}\Lambda)$
$\displaystyle=$
$\displaystyle\sum_{\ell=0}^{n-m+1}(-z)^{\ell}q^{\ell(\ell-1)/2}\displaystyle\begin{bmatrix}n-m+1\\\
\ell\end{bmatrix}_{q}F_{n,m-1}^{trig(q^{n-m+1}z)}(\overline{u}|\overline{v}|q^{\ell}\Lambda).$
(3.6)

It is also easy to see the the limit $v_{m}\rightarrow\infty$ of the right
hand side of (3.2) is

$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}q^{|K|(|K|-1)/2}(1-q^{|K|}\Lambda)\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{qu_{i}-u_{j}}{u_{i}-u_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m-1\end{subarray}}\frac{v_{k}-u_{i}}{v_{k}-qu_{i}},$ (3.7)

hence we get

$\displaystyle\sum_{\ell=0}^{n-m+1}(-z)^{\ell}q^{\ell(\ell-1)/2}\displaystyle\begin{bmatrix}n-m+1\\\
\ell\end{bmatrix}_{q}F_{n,m-1}^{trig(q^{n-m+1}z)}(\overline{u}|\overline{v}|q^{\ell}\Lambda)$
$\displaystyle=$
$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}q^{|K|(|K|-1)/2}(1-q^{|K|}\Lambda)\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{qu_{i}-u_{j}}{u_{i}-u_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m-1\end{subarray}}\frac{v_{k}-u_{i}}{v_{k}-qu_{i}},$ (3.8)

which is nothing but (3.2) with $m$ replaced by $m-1$.

∎

###  3.2 Determinant representations of trigonometric source functions

We present determinant representations for trigonometric source functions. The
procedure of deriving determinant representations go pararell with the
elliptic case in most cases.

We introduce notations for the left and right hand side of (1.3)

$\displaystyle F_{n,m}^{trig(z)}(\overline{u}|\overline{v})$
$\displaystyle:=\sum_{K\subset[1,\dots,m]}(-z)^{|K|}q^{|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{v_{i}-qv_{j}}{v_{i}-v_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq n\end{subarray}}\frac{v_{i}-u_{k}}{v_{i}-qu_{k}},$ (3.9)
$\displaystyle G_{n,m}^{trig(z)}(\overline{u}|\overline{v})$
$\displaystyle:=\frac{1}{\prod_{j=1}^{n-m}(1-q^{-j}z)}\sum_{K\subset[1,\dots,n]}(-q^{m-n}z)^{|K|}q^{|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{qu_{i}-u_{j}}{u_{i}-u_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m\end{subarray}}\frac{v_{k}-u_{i}}{v_{k}-qu_{i}}.$ (3.10)

We call these rational functions as trigonometric source functions. First we
rewrite (3.9) and (3.10) in the following forms using multiplicative
difference operators

$\displaystyle F_{n,m}^{trig(z)}(\overline{u}|\overline{v})=$
$\displaystyle\sum_{K\subset[1,\dots,m]}(-zq^{m-n-1})^{|K|}q^{-|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{q^{-1}v_{i}-v_{j}}{v_{i}-v_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq n\end{subarray}}\frac{v_{i}-u_{k}}{q^{-1}v_{i}-u_{k}}$
$\displaystyle=$
$\displaystyle\frac{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})}{\prod_{1\leq
i<j\leq
m}(v_{j}-v_{i})}\prod_{j=1}^{m}(1-zq^{m-n-1}T_{q,v_{j}}^{-1})\frac{\prod_{1\leq
i<j\leq m}(v_{j}-v_{i})}{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})}.$ (3.11)
$\displaystyle G_{n,m}^{trig(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1}{\prod_{j=1}^{n-m}(1-q^{-j}z)}\frac{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})}{\prod_{1\leq
i<j\leq
n}(u_{j}-u_{i})}\prod_{j=1}^{n}(1-zq^{m-n}T_{q,u_{j}})\frac{\prod_{1\leq
i<j\leq n}(u_{j}-u_{i})}{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})}.$ (3.12)

Source functions in these forms appear in [1, 8] for example.

We use the $p=0$ degeneration of the factorization of the elliptic Vandermonde
determinants (2.5)

$\displaystyle\det_{1\leq j,k\leq
n}\Big{(}\psi_{j}^{A_{n-1}}(u_{k};0,r)\Big{)}=\Bigg{(}1-r\prod_{\ell=1}^{n}u_{\ell}\Bigg{)}\prod_{1\leq
i<j\leq n}(u_{j}-u_{i}),$ (3.13)

where

$\displaystyle\psi_{j}^{A_{n-1}}(u;0;r)=\begin{cases}1-(-1)^{n-1}ru^{n}&\text{if
$j=1$,}\\\ u^{j-1}&\text{if $j=2,\dots,n$.}\\\ \end{cases}$ (3.14)

The $p=0$ case of the Frobenius determinant formula (2.7) is

$\displaystyle\det_{1\leq i,j\leq n}\Bigg{(}\frac{v_{j}-\Lambda
u_{i}}{(1-\Lambda)(v_{j}-u_{i})}\Bigg{)}=\frac{\Big{(}\prod_{j=1}^{n}v_{j}-\Lambda\prod_{i=1}^{n}u_{i}\Big{)}\prod_{1\leq
i<j\leq n}(u_{j}-u_{i})(v_{i}-v_{j})}{(1-\Lambda)\prod_{1\leq i,j\leq
n}(v_{j}-u_{i})}.$ (3.15)

We can extend the determinant representations introduced by Minin-Pronko-
Tarasov [21, 22] to the case when $z$ is generic and $m$ and $n$ are not
necessarily the same.

######  Proposition 3.2.

We have the following determinant representations:

$\displaystyle F_{n,m}^{trig(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1-r\prod_{\ell=1}^{m}v_{\ell}}{\det_{1\leq i,j\leq
m}\Big{(}\sum_{k=1}^{m}p_{ik}\psi_{k}^{A_{m-1}}(v_{j};0,r)\Big{)}}$
$\displaystyle\times\det_{1\leq i,j\leq
m}\Bigg{(}\Bigg{(}\frac{1}{1-r\prod_{\ell=1}^{m}v_{\ell}}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{m}p_{ik}\psi_{k}^{A_{m-1}}(v_{j};0,r)$
$\displaystyle-q^{m-1}z\Bigg{(}\frac{1}{1-q^{-1}r\prod_{\ell=1}^{m}v_{\ell}}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{m}p_{ik}\psi_{k}^{A_{m-1}}(q^{-1}v_{j};0,r)\prod_{\ell=1}^{n}\frac{v_{j}-u_{\ell}}{v_{j}-qu_{\ell}}\Bigg{)},$
(3.16) $\displaystyle G_{n,m}^{(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1}{\prod_{j=1}^{n-m}(1-q^{-j}z)}\frac{1-r\prod_{\ell=1}^{n}u_{\ell}}{\det_{1\leq
i,j\leq n}\Big{(}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};0,r)\Big{)}}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{1}{1-r\prod_{\ell=1}^{n}u_{\ell}}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};0,r)$
$\displaystyle-q^{m-n}z\Bigg{(}\frac{1}{1-qr\prod_{\ell=1}^{n}u_{\ell}}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(qu_{j};0,r)\prod_{\ell=1}^{m}\frac{v_{\ell}-u_{j}}{v_{\ell}-qu_{j}}\Bigg{)}.$
(3.17)

Here, $r$, $p_{ij}$ $(i,j=1,\dots,m)$, $q_{ij}$ $(i,j=1,\dots,n)$ are
additional parameters such that $\det_{1\leq i,j\leq m}(p_{ij})\not\equiv 0$,
$\det_{1\leq i,j\leq n}(q_{ij})\not\equiv 0$.

The derivation is the same with Proposition 2.1, using (3.13) and (3.15)
instead.

The case $r=0$, $z=1$, $m=n$ of (3.16), (3.17) corresponds to the determinants
in Minin-Pronko-Tarasov [21]. We also remark that the parameter $r$ seems to
be related with the parameter $\alpha$ in their recent work [22].

Specializing to $r=0$ and $p_{ij}=q_{ij}=\delta_{ij}$, we get the following
type of determinant forms, which was first introduced by Kostov [17, 18] and
Foda-Wheeler [19] for the case $F_{n,m}^{trig(1)}(\overline{u}|\overline{v})$.

######  Corollary 3.3.

We have the following determinant representations:

$\displaystyle F_{n,m}^{trig(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1}{\prod_{1\leq i<j\leq m}(v_{j}-v_{i})}\det_{1\leq
i,j\leq
m}\Bigg{(}v_{j}^{i-1}-zq^{m-i}v_{j}^{i-1}\prod_{\ell=1}^{n}\frac{v_{j}-u_{\ell}}{v_{j}-qu_{\ell}}\Bigg{)},$
(3.18) $\displaystyle G_{n,m}^{trig(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1}{\prod_{j=1}^{n-m}(1-q^{-j}z)\prod_{1\leq i<j\leq
n}(u_{j}-u_{i})}$ $\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}u_{j}^{i-1}-zq^{m-n+i-1}u_{j}^{i-1}\prod_{\ell=1}^{m}\frac{v_{\ell}-u_{j}}{v_{\ell}-qu_{j}}\Bigg{)}.$
(3.19)

We call (3.18), (3.19) as scalar product type representation in this paper
since this kind of determinant representations was first obtained in [17, 18,
19] as a degeneration from the determinant formulas for scalar products of the
six vertex model by Slavnov [25].

On the other hand, another type of determinant representations below was also
obtained in Foda-Wheeler [19] for the case
$F_{n,m}^{trig(1)}(\overline{u}|\overline{v})$.

######  Proposition 3.4.

For $n\geq m$, we have the following determinant representations:

$\displaystyle F_{n,m}^{trig(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})}{\prod_{1\leq
i<j\leq m}(v_{j}-v_{i})\prod_{1\leq i<j\leq n}(u_{i}-u_{j})}\det_{1\leq
i,j\leq n}(Y),$ (3.20)

where $Y$ is an $n\times n$ matrix whose $(i,j)$-entry is given by

$\displaystyle
Y_{ij}=\left\\{\begin{array}[]{ll}\displaystyle\frac{1}{v_{i}-u_{j}}-zq^{m-n}\frac{1}{v_{i}-qu_{j}},&1\leq
i\leq m,\ \ \ 1\leq j\leq n\\\ u_{j}^{n-i},&m+1\leq i\leq n,\ \ \ 1\leq j\leq
n\end{array}\right..$ (3.23) $\displaystyle
G_{n,m}^{trig(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1}{\prod_{j=1}^{n-m}(1-q^{-j}z)}\frac{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})}{\prod_{1\leq
i<j\leq m}(v_{j}-v_{i})\prod_{1\leq i<j\leq n}(u_{i}-u_{j})}\det_{1\leq
i,j\leq n}(Z),$ (3.24)

where $Z$ is an $n\times n$ matrix whose $(i,j)$-entry is given by

$\displaystyle
Z_{ij}=\left\\{\begin{array}[]{ll}\displaystyle\frac{1}{v_{i}-u_{j}}-q^{m-n}z\frac{1}{v_{i}-qu_{j}},&1\leq
i\leq m,\ \ \ 1\leq j\leq n\\\ u_{j}^{n-i}-q^{m-i}zu_{j}^{n-i},&m+1\leq i\leq
n,\ \ \ 1\leq j\leq n\end{array}\right..$ (3.27)

We call (3.20), (3.24) as domain wall boundary type determinants as this type
of determinant representations was first obtained in [19] as a limit of the
Gaudin-Izergin-Korepin determinant for the domain wall boundary partition
functions of the six vertex model [24, 14, 15]. Here we give a systematic
derivation below.

###### Proof.

We use the following factorization of the Cauchy-Vandermonde determinants
which can be proved by induction on $n$

$\displaystyle\det_{1\leq i,j\leq n}(X)=\frac{\prod_{1\leq i<j\leq
m}(v_{j}-v_{i})\prod_{1\leq i<j\leq
n}(u_{i}-u_{j})}{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})},$ (3.28)

where $X$ is an $n\times n$ matrix whose $(i,j)$-entry is given by

$\displaystyle
X_{ij}=\left\\{\begin{array}[]{ll}\displaystyle\frac{1}{v_{i}-u_{j}},&1\leq
i\leq m,\ \ \ 1\leq j\leq n\\\ u_{j}^{n-i},&m+1\leq i\leq n,\ \ \ 1\leq j\leq
n\end{array}\right..$ (3.31)

Starting from (3.11) and using (3.28), we have

$\displaystyle
F_{n,m}^{trig(z)}(\overline{u}|\overline{v})=\frac{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})}{\prod_{1\leq
i<j\leq
m}(v_{j}-v_{i})}\prod_{i=1}^{m}(1-zq^{m-n-1}T_{q,v_{i}}^{-1})\frac{1}{\prod_{1\leq
i<j\leq n}(u_{i}-u_{j})}\det_{1\leq i,j\leq n}(X).$ (3.32)

Inserting $(1-zq^{m-n-1}T_{q,v_{i}}^{-1})$ in (3.32) into the $i$-th row of
$\det_{1\leq i,j\leq n}(X)$ for $i=1,\dots,m$ gives (3.20).

Starting from (3.12) and using (3.28), one gets

$\displaystyle G_{n,m}^{trig(z)}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\frac{1}{\prod_{j=1}^{n-m}(1-q^{-j}z)}\frac{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})}{\prod_{1\leq
i<j\leq
n}(u_{i}-u_{j})}\prod_{j=1}^{n}(1-zq^{m-n}T_{q,u_{j}})\frac{1}{\prod_{1\leq
i<j\leq m}(v_{j}-v_{i})}\det_{1\leq i,j\leq n}(X).$ (3.33)

Inserting $(1-zq^{m-n}T_{q,u_{j}})$ in (3.33) into the $j$-th column of
$\det_{1\leq i,j\leq n}(X)$ for $j=1,\dots,n$ gives (3.24).

∎

The following type of determinant representations is a one-parameter extension
of a trigonometric analog of the rational version by Belliard-Slavnov [20].

######  Proposition 3.5.

We have the following determinant representations:

$\displaystyle
F_{n,m}^{trig(z)}(\overline{u}|\overline{v})=\frac{1-\Delta}{\prod_{1\leq
i<j\leq m}(v_{j}-v_{i})(\eta_{i}-\eta_{j})}$ $\displaystyle\times\det_{1\leq
i,j\leq
m}\Bigg{(}\Bigg{(}\frac{1}{\prod_{\ell=1}^{m}v_{\ell}-\Delta\prod_{\ell=1}^{m}\eta_{\ell}}\Bigg{)}^{\delta_{i1}}\frac{(v_{i}-\Delta\eta_{j})\prod_{k=1,k\neq
j}^{m}(v_{i}-\eta_{k})}{1-\Delta}$
$\displaystyle-z\Bigg{(}\frac{1}{q^{-1}\prod_{\ell=1}^{m}v_{\ell}-\Delta\prod_{\ell=1}^{m}\eta_{\ell}}\Bigg{)}^{\delta_{i1}}\frac{(q^{-1}v_{i}-\Delta\eta_{j})\prod_{k=1,k\neq
j}^{m}(v_{i}-q\eta_{k})}{1-\Delta}\prod_{k=1}^{n}\frac{v_{i}-u_{k}}{v_{i}-qu_{k}}\Bigg{)},$
(3.34) $\displaystyle
G_{n,m}^{trig(z)}(\overline{u}|\overline{v})=\frac{1-\Delta}{\prod_{j=1}^{n-m}(1-q^{-j}z)}\frac{1}{\prod_{1\leq
i<j\leq n}(u_{j}-u_{i})(\eta_{i}-\eta_{j})}$ $\displaystyle\times\det_{1\leq
i,j\leq
n}\Bigg{(}\Bigg{(}\frac{1}{\prod_{\ell=1}^{n}u_{\ell}-\Delta\prod_{\ell=1}^{n}\eta_{\ell}}\Bigg{)}^{\delta_{i1}}\frac{(u_{i}-\Delta\eta_{j})\prod_{k=1,k\neq
j}^{n}(u_{i}-\eta_{k})}{1-\Delta}$
$\displaystyle-q^{m-n}z\Bigg{(}\frac{1}{q\prod_{\ell=1}^{n}u_{\ell}-\Delta\prod_{\ell=1}^{n}\eta_{\ell}}\Bigg{)}^{\delta_{i1}}\frac{(qu_{i}-\Delta\eta_{j})\prod_{k=1,k\neq
j}^{n}(qu_{i}-\eta_{k})}{1-\Delta}\prod_{k=1}^{m}\frac{u_{i}-v_{k}}{qu_{i}-v_{k}}\Bigg{)}.$
(3.35)

Here, $\Delta$, $\eta_{i}$ $(i=1,\dots,m)$ are additional parameters for
$F_{n,m}^{trig(z)}(\overline{u}|\overline{v})$, and $\Delta$, $\eta_{i}$
$(i=1,\dots,n)$ are additional parameters for
$G_{n,m}^{trig(z)}(\overline{u}|\overline{v})$.

The derivation of (3.34), (3.35) is the same with Proposition 2.3, using
(3.15) instead. Taking the limit $\Delta\to\infty$, we get the following forms
which are direct trigonometric analogue of the representations in Belliard-
Slavnov [20].

######  Corollary 3.6.

We have the following determinant representations:

$\displaystyle F_{n,m}^{trig(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1}{\prod_{1\leq i<j\leq
m}(v_{j}-v_{i})(\eta_{i}-\eta_{j})}$ $\displaystyle\times\det_{1\leq i,j\leq
m}\Bigg{(}\Bigg{(}\prod_{k=1,k\neq j}^{m}(v_{i}-\eta_{k})-z\prod_{k=1,k\neq
j}^{m}(v_{i}-q\eta_{k})\prod_{k=1}^{n}\frac{v_{i}-u_{k}}{v_{i}-qu_{k}}\Bigg{)},$
(3.36) $\displaystyle G_{n,m}^{trig(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1}{\prod_{j=1}^{n-m}(1-q^{-j}z)}\frac{1}{\prod_{1\leq
i<j\leq n}(u_{j}-u_{i})(\eta_{i}-\eta_{j})}$ $\displaystyle\times\det_{1\leq
i,j\leq n}\Bigg{(}\prod_{k=1,k\neq
j}^{n}(u_{i}-\eta_{k})-q^{m-n}z\prod_{k=1,k\neq
j}^{n}(qu_{i}-\eta_{k})\prod_{k=1}^{m}\frac{u_{i}-v_{k}}{qu_{i}-v_{k}}\Bigg{)}.$
(3.37)

###  3.3 A complex analytic proof of trigonometric source identity

We give a complex analytic proof of trigonometric source identity. See also
[8] for example for the original different proof of this identity. We prove
the following polynomial version

$\displaystyle\sum_{K\subset[1,\dots,m]}(-z)^{|K|}q^{|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{v_{i}-qv_{j}}{v_{i}-v_{j}}\prod_{\begin{subarray}{c}j\not\in
K\\\ 1\leq k\leq n\end{subarray}}(v_{j}-qu_{k})\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq n\end{subarray}}(v_{i}-u_{k})$ $\displaystyle=$
$\displaystyle\frac{1}{\prod_{j=1}^{n-m}(1-q^{-j}z)}$
$\displaystyle\times\sum_{K\subset[1,\dots,n]}(-q^{m-n}z)^{|K|}q^{|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{qu_{i}-u_{j}}{u_{i}-u_{j}}\prod_{\begin{subarray}{c}j\not\in
K\\\ 1\leq k\leq m\end{subarray}}(v_{k}-qu_{j})\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m\end{subarray}}(v_{k}-u_{i}).$ (3.38)

Let us denote the left hand side and the right hand side of (3.38) as
$P_{n,m}^{trig(z)}(\overline{u}|\overline{v})$ and
$Q_{n,m}^{trig(z)}(\overline{u}|\overline{v})$ respectively. We assume $m\leq
n$ for the proof.

Relations with $F_{n,m}^{trig(z)}(\overline{u}|\overline{v})$ and
$G_{n,m}^{trig(z)}(\overline{u}|\overline{v})$ are

$\displaystyle P_{n,m}^{trig(z)}(\overline{u}|\overline{v})$
$\displaystyle=\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-qu_{k})F_{n,m}^{trig(z)}(\overline{u}|\overline{v}),$
(3.39) $\displaystyle Q_{n,m}^{trig(z)}(\overline{u}|\overline{v})$
$\displaystyle=\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-qu_{k})G_{n,m}^{trig(z)}(\overline{u}|\overline{v}).$
(3.40)

We first show the following polynomial properties.

######  Lemma 3.7.

$P_{n,m}^{trig(z)}(\overline{u}|\overline{v})$ and
$Q_{n,m}^{trig(z)}(\overline{u}|\overline{v})$ are symmetric polynomials in
$v_{1},\dots,v_{m}$. The degree of $v_{i}\ (i=1,2,\dots,m)$ is at most $n$ for
both polynomials.

###### Proof.

From (3.18) and (3.19), we have the following determinant forms of
$P_{n,m}^{trig(z)}(\overline{u}|\overline{v})$ and
$Q_{n,m}^{trig(z)}(\overline{u}|\overline{v})$

$\displaystyle P_{n,m}^{trig(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1}{\prod_{1\leq i<j\leq m}(v_{j}-v_{i})}\det_{1\leq
i,j\leq
m}\Bigg{(}v_{i}^{j-1}\prod_{k=1}^{n}(v_{i}-qu_{k})-zq^{m-j}v_{i}^{j-1}\prod_{k=1}^{n}(v_{i}-u_{k})\Bigg{)},$
(3.41) $\displaystyle Q_{n,m}^{trig(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1}{\prod_{j=1}^{n-m}(1-q^{-j}z)\prod_{1\leq i<j\leq
n}(u_{j}-u_{i})}$ $\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}u_{i}^{j-1}\prod_{k=1}^{m}(v_{k}-qu_{i})-zq^{m-n+j-1}u_{i}^{j-1}\prod_{k=1}^{m}(v_{k}-u_{i})\Bigg{)},$
(3.42)

As for $P_{n,m}^{trig(z)}(\overline{u}|\overline{v})$, one can see from (3.41)
that $v_{i}=v_{j}$ coming from $\prod_{1\leq i<j\leq m}(v_{j}-v_{i})^{-1}$ is
an apparent pole since two rows of the determinant coincide in that case. As a
polynomial of $v_{i}$, one can see the contribution to the degree coming from
the determinant is at most $m+n-1$, and $\prod_{1\leq i<j\leq
m}(v_{j}-v_{i})^{-1}$ subtracts the degree by $m-1$, hence we find the degree
is at most $n$. Symmetry property is also easy to check. It is easier to check
the properties for $Q_{n,m}^{trig(z)}(\overline{u}|\overline{v})$ from (3.42).
∎

Due to Lemma 3.7, one notes that it is enough to check
$P_{n,m}^{trig(z)}(\overline{u}|\overline{v})=Q_{n,m}^{trig(z)}(\overline{u}|\overline{v})$
for $(n+1)^{m}$ distinct points in $(v_{1},\dots,v_{m})$. One can check the
equality for the following
$2n\times(2n-1)\times\cdots\times(2n-m+1)(\geq(n+1)^{m})$ distinct points:
$v_{1}=q^{\epsilon_{1}}u_{p_{1}},\ v_{2}=q^{\epsilon_{2}}u_{p_{2}},\
\dots,v_{m}=q^{\epsilon_{m}}u_{p_{m}}$, $p_{1},\dots,p_{m}\in\\{1,\dots,n\\}$,
$\epsilon_{1},\dots,\epsilon_{m}\in\\{0,1\\}$ and $v_{i}\neq v_{j}$ for $i\neq
j$. These specializations can be evaluated explicitly. It is easy to check the
following cases which correspond to $p_{i}=p_{j}$ for some $i\neq j$.

######  Lemma 3.8.

If $v_{i}=u_{k}$, $v_{j}=qu_{k}$ for some $i,j,k\ (i\neq j)$, we have

$\displaystyle
P_{n,m}^{trig(z)}(\overline{u}|\overline{v})=Q_{n,m}^{trig(z)}(\overline{u}|\overline{v})=0.$
(3.43)

###### Proof.

We check the case $k=n$. From the expressions (3.38), one can check the
following relations hold when substituting $v_{m}=qu_{n}$ in
$P_{n,m}^{trig(z)}(\overline{u}|\overline{v})$ and
$Q_{n,m}^{trig(z)}(\overline{u}|\overline{v})$

$\displaystyle P_{n,m}^{trig(z)}(\overline{u}|\overline{v})|_{v_{m}=qu_{n}}$
$\displaystyle=$
$\displaystyle(1-q)u_{n}z\prod_{j=1}^{n-1}(qu_{n}-u_{j})\prod_{j=1}^{m-1}q(v_{j}-u_{n})P_{n-1,m-1}^{trig(z)}(u_{1},\dots,u_{n-1}|v_{1},\dots,v_{m-1}),$
(3.44) $\displaystyle
Q_{n,m}^{trig(z)}(\overline{u}|\overline{v})|_{v_{m}=qu_{n}}$ $\displaystyle=$
$\displaystyle(1-q)u_{n}z\prod_{j=1}^{n-1}(qu_{n}-u_{j})\prod_{j=1}^{m-1}q(v_{j}-u_{n})Q_{n-1,m-1}^{trig(z)}(u_{1},\dots,u_{n-1}|v_{1},\dots,v_{m-1}).$
(3.45)

From the factor $\prod_{j=1}^{m-1}q(v_{j}-u_{n})$ in (3.44) and (3.45), one
notes that $P_{n,m}^{trig(z)}(\overline{u}|\overline{v})|_{v_{m}=qu_{n}}$ and
$Q_{n,m}^{trig(z)}(\overline{u}|\overline{v})|_{v_{m}=qu_{n}}$ vanish after
further substituting $v_{j}=u_{n}$ for some $j\ (1\leq j\leq m-1)$ gives 0.
The other cases can be checked in the same way or follow by symmetry. ∎

We next assume $p_{i}\neq p_{j}$ for $i\neq j$. For $I=\\{1\leq
i_{1}<i_{2}<\cdots<i_{|I|}\leq n\\}$ and $J=\\{1\leq
j_{1}<j_{2}<\cdots<j_{|J|}\leq n\\}$ such that $I\cup J$ is a set of $m$
distinct integers in $[1,\dots,n]$, let us denote by
$\overline{v}=\\{\overline{u}_{I},q\overline{u}_{J}\\}$ the following type of
substitution

$\displaystyle
v_{q_{1}}=u_{i_{1}},\dots,v_{q_{|I|}}=u_{i_{|I|}},v_{r_{1}}=qu_{j_{1}},\dots,v_{r_{|J|}}=qu_{j_{|J|}},$
(3.46)

for some $\\{q_{1},\dots,q_{|I|},r_{1},\dots,r_{|J|}\\}=[1,\dots,m]$. Note
$|I|+|J|=m$.

We have the following explicit evaluations.

######  Proposition 3.9.

We have

$\displaystyle
P_{n,m}^{trig(z)}(\overline{u}|\\{\overline{u}_{I},q\overline{u}_{J}\\})=(-z)^{|J|}q^{|I||J|+|J|(|J|-1)/2}\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}(u_{j}-u_{i})\prod_{\begin{subarray}{c}i\in I\\\
1\leq j\leq n\end{subarray}}(u_{i}-qu_{j})\prod_{\begin{subarray}{c}j\in J\\\
k\in[1,\dots,n]\backslash I\end{subarray}}(qu_{j}-u_{k}),$ (3.47)
$\displaystyle
Q_{n,m}^{trig(z)}(\overline{u}|\\{\overline{u}_{I},q\overline{u}_{J}\\})=(-z)^{|J|}q^{|I||J|+|J|(|J|-1)/2}\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}(u_{j}-u_{i})\prod_{\begin{subarray}{c}i\in I\\\
1\leq j\leq n\end{subarray}}(u_{i}-qu_{j})\prod_{\begin{subarray}{c}j\in J\\\
k\in[1,\dots,n]\backslash I\end{subarray}}(qu_{j}-u_{k}).$ (3.48)

###### Proof.

After the substitution
$v_{q_{1}}=u_{i_{1}},\dots,v_{q_{|I|}}=u_{i_{|I|}},v_{r_{1}}=qu_{j_{1}},\dots,v_{r_{|J|}}=qu_{j_{|J|}}$,
one notes from the factors $\displaystyle\prod_{\begin{subarray}{c}j\not\in
K\\\ 1\leq k\leq n\end{subarray}}(v_{j}-qu_{k})$ and
$\displaystyle\prod_{\begin{subarray}{c}i\in K\\\ 1\leq k\leq
n\end{subarray}}(v_{i}-u_{k})$ that only the summand corresponding to
$K=\\{r_{1},\dots,r_{|J|}\\}$ survives and we get

$\displaystyle
P_{n,m}^{trig(z)}(\overline{u}|\\{\overline{u}_{I},q\overline{u}_{J}\\})$
$\displaystyle=$
$\displaystyle(-z)^{|J|}q^{|J|(|J|-1)/2}\prod_{\begin{subarray}{c}i\in J\\\
j\in
I\end{subarray}}\frac{q(u_{i}-u_{j})}{qu_{i}-u_{j}}\prod_{\begin{subarray}{c}j\in
I\\\ 1\leq k\leq n\end{subarray}}(u_{j}-qu_{k})\prod_{\begin{subarray}{c}j\in
J\\\ 1\leq k\leq n\end{subarray}}(qu_{j}-u_{k})$ $\displaystyle=$
$\displaystyle(-z)^{|J|}q^{|I||J|+|J|(|J|-1)/2}\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}(u_{j}-u_{i})\prod_{\begin{subarray}{c}i\in I\\\
1\leq j\leq n\end{subarray}}(u_{i}-qu_{j})\prod_{\begin{subarray}{c}j\in J\\\
k\in[1,\dots,n]\backslash I\end{subarray}}(qu_{j}-u_{k}).$ (3.49)

Next, we check the case for
$Q_{n,m}^{trig(z)}(\overline{u}|\\{\overline{u}_{I},q\overline{u}_{J}\\})$.
Substituting
$v_{q_{1}}=u_{i_{1}},\dots,v_{q_{|I|}}=u_{i_{|I|}},v_{r_{1}}=qu_{j_{1}},\dots,v_{r_{|J|}}=qu_{j_{|J|}}$,
we note from the factor $\displaystyle\prod_{\begin{subarray}{c}j\in K\\\
1\leq k\leq m\end{subarray}}(v_{k}-u_{j})$ that summands labeled by $K$
satisfying $K\cap I\neq\phi$ vanish, and from the factor
$\displaystyle\prod_{\begin{subarray}{c}j\not\in K\\\ 1\leq k\leq
m\end{subarray}}(v_{k}-qu_{j})$ we note the summands vanish unless $J\subset
K$. Hence the sum can be restricted to $\displaystyle\sum_{J\subset
K\subset[1,\dots,n]\backslash I}$ and after simplifications, we find
$Q_{n,m}^{trig(z)}(\overline{u}|\\{\overline{u}_{I},q\overline{u}_{J}\\})$ can
be rewritten as

$\displaystyle
Q_{n,m}^{trig(z)}(\overline{u}|\\{\overline{u}_{I},q\overline{u}_{J}\\})$
$\displaystyle=$
$\displaystyle\frac{q^{|I||J|+|J|(|J|-1)/2}}{\prod_{j=1}^{n-m}(1-q^{-j}z)}\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}(u_{j}-u_{i})\prod_{\begin{subarray}{c}i\in I\\\
1\leq j\leq n\end{subarray}}(u_{i}-qu_{j})\prod_{\begin{subarray}{c}j\in J\\\
k\in[1,\dots,n]\backslash I\end{subarray}}(qu_{j}-u_{k})(-z)^{|J|}$
$\displaystyle\times\sum_{J\subset K\subset[1,\dots,n]\backslash
I}(-z)^{|K|-|J|}q^{-(|K|-|J|)(|K|-|J|+1)/2}\prod_{\begin{subarray}{c}i\in
K\backslash J\\\
j\in[1,\dots,n]\backslash\\{I,K\\}\end{subarray}}\frac{u_{i}-q^{-1}u_{j}}{u_{i}-u_{j}}.$
(3.50)

Finally, using Cauchy’s $q$-binomial identity (3.3) we note the following
factorization

$\displaystyle\sum_{J\subset K\subset[1,\dots,n]\backslash
I}(-z)^{|K|-|J|}q^{-(|K|-|J|)(|K|-|J|+1)/2}\prod_{\begin{subarray}{c}i\in
K\backslash J\\\
j\in[1,\dots,n]\backslash\\{I,K\\}\end{subarray}}\frac{u_{i}-q^{-1}u_{j}}{u_{i}-u_{j}}$
$\displaystyle=$
$\displaystyle\sum_{\ell=|J|}^{n-|I|}\sum_{\begin{subarray}{c}K:|K|=\ell\\\
J\subset K\subset[1,\dots,n]\backslash
I\end{subarray}}(-z)^{\ell-|J|}q^{-(\ell-|J|)(\ell-|J|+1)/2}\prod_{\begin{subarray}{c}i\in
K\backslash J\\\
j\in[1,\dots,n]\backslash\\{I,K\\}\end{subarray}}\frac{u_{i}-q^{-1}u_{j}}{u_{i}-u_{j}}$
$\displaystyle=$
$\displaystyle\sum_{\ell=|J|}^{n-|I|}(-z)^{\ell-|J|}q^{-(\ell-|J|)(\ell-|J|+1)/2}\begin{bmatrix}n-|I|-|J|\\\
\ell-|J|\end{bmatrix}_{q^{-1}}=\prod_{j=1}^{n-|I|-|J|}(1-q^{-j}z)=\prod_{j=1}^{n-m}(1-q^{-j}z).$
(3.51)

Note that we also used the following $q$-identity

$\displaystyle\sum_{K\subset[1,\dots,n]:|K|=\ell}\prod_{\begin{subarray}{c}i\in
K\\\ j\in[1,\dots,n]\backslash
K\end{subarray}}\frac{u_{i}-q^{-1}u_{j}}{u_{i}-u_{j}}=\begin{bmatrix}n\\\
\ell\end{bmatrix}_{q^{-1}},$ (3.52)

in the second equality of (3.51). (3.52) can be checked by noting that the
left hand side is a bounded function and the possible poles $u_{i}=u_{j}\
(i\neq j)$ are apparent singularities, hence using Liouville’s theorem and
taking the limit $u_{1}\to\infty,u_{2}\to\infty,\dots,u_{n}\to\infty$ in this
order, we get

$\displaystyle\sum_{K\subset[1,\dots,n]:|K|=\ell}\prod_{\begin{subarray}{c}i\in
K\\\ j\in[1,\dots,n]\backslash
K\end{subarray}}\frac{u_{i}-q^{-1}u_{j}}{u_{i}-u_{j}}=\sum_{K\subset[1,\dots,n]:|K|=\ell}q^{-\\#\\{(i,j)\
|\ i\in K,\ j\in[1,\dots,n]\backslash K,\ i>j\\}}.$ (3.53)

Hence it remains to show

$\displaystyle\sum_{K\subset[1,\dots,n]:|K|=\ell}q^{-\\#\\{(i,j)\ |\ i\in K,\
j\in[1,\dots,n]\backslash K,\ i>j\\}}=\displaystyle\begin{bmatrix}n\\\
\ell\end{bmatrix}_{q^{-1}},$ (3.54)

which can be proved by induction on $n$. Denote the left hand side and right
hand side of (3.54) as $f(n,\ell)$ and $g(n,\ell)$ respectively. It is easy to
show they satisfy the recursive relations

$\displaystyle f(n,\ell)$ $\displaystyle=f(n-1,\ell)+f(n-1,\ell-1)q^{\ell-n},$
(3.55) $\displaystyle g(n,\ell)$
$\displaystyle=f(n-1,\ell)+f(n-1,\ell-1)q^{\ell-n},$ (3.56)

and $f(n,0)=g(n,0)=1$ and $f(1,1)=g(1,1)=1$ which correspond to the initial
conditions.

Combining (3.50) and (3.51) gives

$\displaystyle
Q_{n,m}^{trig(z)}(\overline{u}|\\{\overline{u}_{I},\overline{u}_{J}+c\\})$
$\displaystyle=$
$\displaystyle(-z)^{|J|}q^{|I||J|+|J|(|J|-1)/2}\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}(u_{j}-u_{i})\prod_{\begin{subarray}{c}i\in I\\\
1\leq j\leq n\end{subarray}}(u_{i}-qu_{j})\prod_{\begin{subarray}{c}j\in J\\\
k\in[1,\dots,n]\backslash I\end{subarray}}(qu_{j}-u_{k}).$ (3.57)

∎

By showing Lemma 3.8 and Proposition 3.9, we checked (3.38) for enough
specializations of $(v_{1},\dots,v_{m})$ and hence the identity (3.38) itself
holds.

###  3.4 Geometric derivation of trigonometric source identity

We give a geometric derivation of the trigonometric source identity. We assume
$n\leq m$ and set $t:=q^{-1}$ and redefine $t^{-1}v_{i}$ $(i=1,\dots,m)$ and
$t^{m-1}z$ as $v_{i}$ and $z$. Then (1.3) is rewritten as

$\displaystyle\sum_{K\subset[1,\dots,m]}(-z)^{|K|}t^{|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{1-tv_{i}/v_{j}}{1-v_{i}/v_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq n\end{subarray}}\frac{1-tu_{k}/v_{i}}{1-u_{k}/v_{i}}$
$\displaystyle=$
$\displaystyle\prod_{j=1}^{m-n}(1-t^{m-j}z)\sum_{K\subset[1,\dots,n]}(-z)^{|K|}t^{|K|(|K|-1)/2}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{1-tu_{j}/u_{i}}{1-u_{j}/u_{i}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m\end{subarray}}\frac{1-tu_{i}/v_{k}}{1-u_{i}/v_{k}}.$ (3.58)

We show the version (3.58) of the trigonometric source identity. Applying the
$q$-binomial identity (3.3), we have

$\displaystyle\prod_{j=1}^{m-n}(1-t^{m-j}z)=\sum_{k=0}^{m-n}t^{nk}t^{k(k-1)/2}(-z)^{k}\displaystyle\begin{bmatrix}m-n\\\
k\end{bmatrix}_{t}.$ (3.59)

Using (3.59) and expanding in terms of $-z$ and taking the coefficients of
$(-z)^{\ell}$ of both hand sides of (3.58), we have

$\displaystyle
t^{\ell(\ell-1)/2}\sum_{\begin{subarray}{c}K\subset[1,\dots,m]\\\ \lvert
K\rvert=\ell\end{subarray}}\prod_{\begin{subarray}{c}i\in K\\\
j\in[1,\dots,m]\setminus
K\end{subarray}}\frac{1-tv_{i}/v_{j}}{1-v_{i}/v_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq n\end{subarray}}\frac{1-tu_{k}/v_{i}}{1-u_{k}/v_{i}}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\ell}t^{nk}t^{k(k-1)/2}\displaystyle\begin{bmatrix}m-n\\\
k\end{bmatrix}_{t}\times
t^{(\ell-k)(\ell-k-1)/2}\sum_{\begin{subarray}{c}K\subset[1,\dots,n]\\\ \lvert
K\rvert=\ell-k\end{subarray}}\prod_{\begin{subarray}{c}i\in K\\\
j\in[1,\ldots,n]\setminus
K\end{subarray}}\frac{1-tu_{j}/u_{i}}{1-u_{j}/u_{i}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k^{\prime}\leq
m\end{subarray}}\frac{1-tu_{i}/v_{k^{\prime}}}{1-u_{i}/v_{k^{\prime}}}.$
(3.60)

We give a geometric derivation of (3.60). For the description, it is useful to
introduce another version of $q$-integers, $q$-factorials and $q$-binomials
$\displaystyle(n)_{q}:={q^{n/2}-q^{-n/2}\over q^{1/2}-q^{-1/2}}$,
$(n)_{q}!:=\prod_{j=1}^{n}(j)_{q}$, for $n$ nonnegative integers. Note the
following relations $(n)_{q}=q^{(1-n)/2}[n]_{q}$,
$(n)_{q}!=q^{(n-n^{2})/4}[n]_{q}!$.

We apply the wall-crossing formula [13, Theorem 3.6] to the following framed
quiver $Q\colon$

We consider $Q$-representations on $V=\mathbb{C}^{\ell}$ with the framing
$W_{1}=\mathbb{C}^{n}\to V$ and the co-framing $V\to W_{2}=\mathbb{C}^{m}$.
Such $Q$-representations are parametrized by

$\mathbb{M}=\mathop{\mathrm{Hom}}\nolimits(W_{1},\mathbb{C}^{\ell})\oplus\mathop{\mathrm{Hom}}\nolimits(\mathbb{C}^{\ell},W_{2}).$

The general linear group $\operatorname{GL}(\mathbb{C}^{\ell})$ acts on
$\mathbb{M}$ by $g\cdot(z,w)=(gz,wg^{-1})$ for
$g\in\operatorname{GL}(\mathbb{C}^{\ell})$ and $(z,w)\in\mathbb{M}$, and
$\operatorname{GL}(\mathbb{C}^{\ell})$-orbits are regarded as isomorphism
classes of $Q$-representations.

We consider the character
$\operatorname{GL}(\mathbb{C}^{\ell})\to\mathbb{C}^{\ast},g\mapsto\text{det}(g)^{\pm
1}$. These define semistable loci

$\displaystyle\mathbb{M}^{+1}$ $\displaystyle=\\{(z,w)\in\mathbb{M}\mid
w\text{ is injective}\\}$ $\displaystyle\mathbb{M}^{-1}$
$\displaystyle=\\{(z,w)\in\mathbb{M}\mid z\text{ is surjective}\\}.$

The framed quiver moduli $M^{\pm}_{Q}(\ell)$ of $\pm 1$-semistable
$Q$-representations on $\mathbb{C}^{\ell}$ are given by
$M^{\pm}_{Q}(\ell)=\mathbb{M}^{\pm 1}/\text{GL}(\mathbb{C}^{\ell})$. We
consider the diagonal torus $\mathbb{T}=(\mathbb{C}^{\ast})^{n+m}$ of
$\text{GL}(W_{1})\times\text{GL}(W_{2})$, and $\mathbb{T}$-action on
$M^{\pm}_{Q}(\ell)$ induced by the natural
$\text{GL}(W_{1})\times\text{GL}(W_{2})$-action on $\mathbb{M}$ by the
conjugation. Concretely, we have $(u,v)\cdot(z,w)=(zu^{-1},vw)$ for
$(u,v)\in\mathbb{T}$ and $(z,w)\in\mathbb{M}$ where we
$u\in(\mathbb{C}^{\ast})^{n}$ and $v\in(\mathbb{C}^{\ast})^{m}$. We take
$\mathbb{T}$-equivariant $K$-theoretic integrals

$\int^{[K]}_{M^{\pm}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(TM^{\pm}_{Q}(\ell))=\int^{[K]}_{M^{\pm\zeta}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(TM^{\pm}_{Q}(\ell))\cdot\mathop{\mathrm{Td}}\nolimits_{M^{\pm}_{Q}(\ell)}$

of the $K$-theoretic Euler class

$\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(TM^{\pm}_{Q}(\ell))=\operatorname{ch}(\wedge_{-t}T^{\ast}M^{\pm}_{Q}(\ell))$

where
$\mathop{\mathrm{Td}}\nolimits_{M^{\pm}_{Q}(\ell)}={\mathop{\mathrm{Eu}}\nolimits(TM^{\pm}_{Q}(\ell))\over\mathop{\mathrm{Eu}}\nolimits_{K}(TM^{\pm}_{Q}(\ell))}$
is the Todd class with
$\mathop{\mathrm{Eu}}\nolimits_{K}(TM^{\pm}_{Q}(\ell))=\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(TM^{\pm}_{Q}(\ell))|_{t=1}$.
We see that
$\int^{[K]}_{M^{\zeta^{+}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(TM^{\zeta^{+}}_{Q}(\ell))$
is equal to

$\displaystyle\sum_{\begin{subarray}{c}K\subset[1,\dots,m]\\\ \lvert
K\rvert=\ell\end{subarray}}\prod_{\begin{subarray}{c}i\in K\\\
j\in[1,\dots,m]\setminus
K\end{subarray}}\frac{1-tv_{i}/v_{j}}{1-v_{i}/v_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq n\end{subarray}}\frac{1-tu_{k}/v_{i}}{1-u_{k}/v_{i}},$

and
$\int^{[K]}_{M^{\zeta^{-}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{-}}_{Q}(\ell))$
is equal to

$\displaystyle\sum_{\begin{subarray}{c}K\subset[1,\dots,n]\\\ \lvert
K\rvert=\ell\end{subarray}}\prod_{\begin{subarray}{c}i\in K\\\
j\in[1,\ldots,n]\setminus
K\end{subarray}}\frac{1-tu_{j}/u_{i}}{1-u_{j}/u_{i}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m\end{subarray}}\frac{1-tu_{i}/v_{k}}{1-u_{i}/v_{k}}.$

By [13, Theorem 3.6], we have

$\displaystyle\int^{[K]}_{M^{\zeta^{+}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{+}}_{Q}(\ell))-\int^{[K]}_{M^{\zeta^{-}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{-}}_{Q}(\ell))$
$\displaystyle=\sum_{k=1}^{\ell}\sum_{{\boldsymbol{\mathfrak{I}}\in\operatornamewithlimits{Dec}(\ell)\atop|\boldsymbol{\mathfrak{I}}|=k}}{[\ell-k]_{t}!\over[\ell]_{t}!}\prod_{i=1}^{j}{[d_{i}-1]_{t}!\over
t-1}\gamma^{K}_{d_{i}}(t)t^{-(\ell-i)d_{i}}(t^{s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i})+md_{i}}-t^{s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})+nd_{i}})$
$\displaystyle\times\int^{[K]}_{M^{\zeta^{-}}_{Q}(\ell-k)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{-}}_{Q}(\ell-k)).$

where $\operatornamewithlimits{Dec}(\ell)$ is the set of collections
$\boldsymbol{\mathfrak{I}}=(\mathfrak{I}_{1},\ldots,\mathfrak{I}_{j})$
satisfying
$\mathfrak{I}_{1}\sqcup\cdots\sqcup\mathfrak{I}_{j}\subset[1,\dots,\ell]$ and

$\min(\mathfrak{I}_{1})>\cdots>\min(\mathfrak{I}_{j}).$

Here we set $d_{i}=|\mathfrak{I}_{i}|$, and
$|\boldsymbol{\mathfrak{I}}|=d_{1}+\cdots+d_{j}$,

$s^{K}(\mathfrak{I}_{1},\mathfrak{I}_{2})=\left|\\{(i,j)\in\mathfrak{I}_{1}\times\mathfrak{I}_{2}\mid
i<j\\}\right|.$

We have

$\displaystyle\gamma^{K}_{d}(t)=\int^{[K]}_{M_{Q^{\sharp}}(d)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M_{Q^{\sharp}}(d))=\begin{cases}1&d=1\\\
0&d\neq 1\end{cases}$ (3.61)

for the following framed quiver $Q^{\sharp}\colon$

Hence we have

$\displaystyle\int^{[K]}_{M^{\zeta^{+}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{+}}_{Q}(\ell))-\int^{[K]}_{M^{\zeta^{-}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{-}}_{Q}(\ell))$
$\displaystyle=\sum_{k=1}^{\ell}\sum_{\begin{subarray}{c}\boldsymbol{\mathfrak{I}}\in\operatornamewithlimits{Dec}(\ell)\\\
|\boldsymbol{\mathfrak{I}}|=k\\\
|\mathfrak{I}_{1}|=\cdots=|\mathfrak{I}_{k}|=1\end{subarray}}\frac{[\ell-k]_{t}!}{[\ell]_{t}!}\prod_{i=1}^{k}\frac{t^{s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i})+m}-t^{s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})+n}}{t-1}\cdot
t^{-\ell+i}\int^{[K]}_{M^{\zeta^{-}}_{Q}(\ell-k)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{-}}_{Q}(\ell-k)).$

Since
$\sum_{i=1}^{k}i-\ell=\frac{k(k+1)-2kl}{2}=\frac{(\ell-k)(\ell-k-1)-\ell(\ell-1)}{2}$,
this is equivalent to

$\displaystyle
t^{\ell(\ell-1)/2}\int^{[K]}_{M^{\zeta^{+}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{+}}_{Q}(\ell))-t^{\ell(\ell-1)/2}\int^{[K]}_{M^{\zeta^{-}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{-}}_{Q}(\ell))$
$\displaystyle=\sum_{k=1}^{\ell}\sum_{\boldsymbol{\mathfrak{I}}\in\operatornamewithlimits{Dec}(\ell)\atop{|\boldsymbol{\mathfrak{I}}|=k\atop|\mathfrak{I}_{1}|=\cdots=|\mathfrak{I}_{k}|=1}}{[\ell-k]_{t}!\over[\ell]_{t}!}\prod_{i=1}^{k}\frac{t^{s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i})+m}-t^{s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})+n}}{t-1}$
$\displaystyle\times
t^{\frac{(\ell-k)(\ell-k-1)}{2}}\int^{[K]}_{M^{\zeta^{-}}_{Q}(\ell-k)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{-}}_{Q}(\ell-k)).$
(3.62)

We now transform (3.62) to (3.60). To this end, we investigate the sum in the
right hand side of (3.62). Using
$s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})+s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i})=\ell-i$,
we note the factors in (3.62) can be rewritten as

$\displaystyle\frac{[\ell-k]_{t}!}{[\ell]_{t}!}\prod_{i=1}^{k}\frac{t^{s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i})+m}-t^{s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})+n}}{t-1}$
$\displaystyle=$ $\displaystyle
t^{(\ell-k)^{2}/4-(\ell-k)/4-\ell^{2}/4+l/4}\frac{(\ell-k)_{t}!}{(\ell)_{t}!}$
$\displaystyle\times\prod_{i=1}^{k}t^{(m+n-1+s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})+s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i}))/2}(m-n-s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})+s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i}))_{t}$
$\displaystyle=$ $\displaystyle t^{(k^{2}+k)/4-\ell
k/2}{(\ell-k)_{t}!\over(\ell)_{t}!}\prod_{i=1}^{k}t^{(m+n-1+\ell-i)/2}(m-n-s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})+s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i}))_{t}$
$\displaystyle=$ $\displaystyle
t^{(m+n-1)k/2}\frac{(\ell-k)_{t}!}{(\ell)_{t}!}\prod_{i=1}^{k}(m-n-s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})+s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i}))_{t}.$
(3.63)

The right hand side of (3.62) can be rewritten as

$\displaystyle\sum_{k=1}^{\ell}t^{(m+n-1)k/2}{(\ell-k)_{t}!\over(\ell)_{t}!}\sum_{\boldsymbol{\mathfrak{I}}\in\operatornamewithlimits{Dec}(\ell)\atop{|\boldsymbol{\mathfrak{I}}|=k\atop|\mathfrak{I}_{1}|=\cdots=|\mathfrak{I}_{k}|=1}}\prod_{i=1}^{k}(m-n-s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})+s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i}))_{t}$
$\displaystyle\times
t^{\frac{(\ell-k)(\ell-k-1)}{2}}\int^{[K]}_{M^{\zeta^{-}}_{Q}(\ell-k)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{-}}_{Q}(\ell-k)).$
(3.64)

Writing the subsets as $\mathfrak{I}_{i}=\\{h_{i}\\}$ $(\ell\geq
h_{1}>\cdots>h_{k}\geq 1)$, we have

$\displaystyle s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})$
$\displaystyle=s^{K}([\ell]\backslash\\{h_{1},\dots,h_{i}\\},\\{h_{i}\\})=h_{i}-1,$
(3.65) $\displaystyle s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i})$
$\displaystyle=s^{K}(\\{h_{i}\\},[\ell]\backslash\\{h_{1},\dots,h_{i}\\})=\ell-
h_{i}-i+1,$ (3.66)

and hence the second sum in (3.64) can be simplified using (3.70) in the the
proposition below as

$\displaystyle\sum_{\begin{subarray}{c}\boldsymbol{\mathfrak{I}}\in\operatornamewithlimits{Dec}(\ell)\\\
|\boldsymbol{\mathfrak{I}}|=k\\\
|\mathfrak{I}_{1}|=\cdots=|\mathfrak{I}_{k}|=1\end{subarray}}\prod_{i=1}^{k}(m-n-s^{K}(\boldsymbol{\mathfrak{I}}_{>i},\mathfrak{I}_{i})+s^{K}(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i}))_{t}$
$\displaystyle=$ $\displaystyle\sum_{\ell\geq h_{1}>\cdots>h_{k}\geq
1}\prod_{i=1}^{k}(\ell-2h_{i}-i+2+m-n)_{t}=\frac{(m-n)_{t}!(\ell)_{t}!}{(k)_{t}!(m-n-k)_{t}!(\ell-k)_{t}!}.$
(3.67)

Therefore, (3.64) can be simplified as

$\displaystyle\sum_{k=1}^{\ell}t^{(m+n-1)k/2}{(m-n)_{t}!\over(k)_{t}!(m-n-k)_{t}!}\times
t^{{(\ell-k)(\ell-k-1)\over
2}}\int^{[K]}_{M^{\zeta^{-}}_{Q}(\ell-k)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{-}}_{Q}(\ell-k))$
$\displaystyle=$
$\displaystyle\sum_{k=1}^{\ell}t^{nk}t^{k(k-1)/2}\displaystyle\begin{bmatrix}m-n\\\
k\end{bmatrix}_{t}\times
t^{\frac{(\ell-k)(\ell-k-1)}{2}}\int^{[K]}_{M^{\zeta^{-}}_{Q}(\ell-k)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{-}}_{Q}(\ell-k)).$
(3.68)

The right hand side of (3.62) is simplified as (3.68), and moving the second
term in the left hand side to the right hand side, (3.62) can be rewritten as

$\displaystyle
t^{\ell(\ell-1)/2}\int^{[K]}_{M^{\zeta^{+}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{+}}_{Q}(\ell))$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\ell}t^{nk}t^{k(k-1)/2}\displaystyle\begin{bmatrix}m-n\\\
k\end{bmatrix}_{t}\times
t^{\frac{(\ell-k)(\ell-k-1)}{2}}\int^{[K]}_{M^{\zeta^{-}}_{Q}(\ell-k)}\mathop{\mathrm{Eu}}\nolimits_{K}^{t}(M^{\zeta^{-}}_{Q}(\ell-k)).$
(3.69)

This is nothing but the identity (3.60).

Finally we show the following proposition used in (3.67).

######  Proposition 3.10.

We have

$\displaystyle\sum_{\ell\geq h_{1}>\cdots>h_{k}\geq
1}\prod_{i=1}^{k}(\ell-2h_{i}-i+2+m-n)_{t}=\frac{(m-n)_{t}!(\ell)_{t}!}{(k)_{t}!(m-n-k)_{t}!(\ell-k)_{t}!}.$
(3.70)

###### Proof.

We show by induction on $\ell$. For $\ell=k$, it is easy to see that both hand
sides of (3.70) are $(m-n)_{t}(m-n-1)_{t}\cdots(m-n-k+1)_{t}$. Next we assume
that (3.70) hold for $\ell\geq k$ and show (3.70) for $\ell+1$. We divide the
summation in left hand side of (3.70) as

$\sum_{\ell+1\geq h_{1}>\cdots>h_{k}\geq 1}\
=\sum_{\ell+1=h_{1}>h_{2}>\cdots>h_{k}\geq 1}\ +\sum_{\ell\geq
h_{1}>h_{2}>\cdots>h_{k}\geq 1}\ ,$

and apply the induction hypothesis to each sum. The first sum becomes

$\displaystyle\sum_{\ell+1=h_{1}>h_{2}>\cdots>h_{k}\geq
1}\prod_{i=1}^{k}(\ell-2h_{i}-i+3+m-n)_{t}$ $\displaystyle=$
$\displaystyle(-\ell+m-n)_{t}\sum_{\ell\geq h_{2}>\cdots>h_{k}\geq
1}\prod_{i=2}^{k}(\ell-2h_{i}-i+3+m-n)_{t}$ $\displaystyle=$
$\displaystyle(-\ell+m-n)_{t}\sum_{\ell\geq h_{1}>\cdots>h_{k-1}\geq
1}\prod_{i=1}^{k-1}(\ell-2h_{i}-i+2+m-n)_{t}$ $\displaystyle=$
$\displaystyle(-\ell+m-n)_{t}\frac{(m-n)_{t}!(\ell)_{t}!}{(k-1)_{t}!(m-n-k+1)_{t}!(\ell-k+1)_{t}!}.$
(3.71)

The second sum can be rewritten as

$\displaystyle\sum_{\ell\geq h_{1}>h_{2}>\cdots>h_{k}\geq
1}\prod_{i=1}^{k}(\ell-2h_{i}-i+3+m-n)_{t}=\frac{(m-n+1)_{t}!(\ell)_{t}!}{(k)_{t}!(m-n+1-k)_{t}!(\ell-k)_{t}!}.$
(3.72)

Combining (3.71) and (3.72) and using

$\displaystyle(x-u)_{t}(y-v)_{t}-(x-v)_{t}(y-u)_{t}=(x-y)_{t}(u-v)_{t},$
(3.73)

for $x=m-n,y=k-1,u=\ell,v=-1$, we get

$\displaystyle\sum_{\ell+1\geq h_{1}>\cdots>h_{k}\geq
1}\prod_{i=1}^{k}(\ell-2h_{i}-i+3+m-n)_{t}$ $\displaystyle=$
$\displaystyle(-\ell+m-n)_{t}\frac{(m-n)_{t}!(\ell)_{t}!}{(k-1)_{t}!(m-n-k+1)_{t}!(\ell-k+1)_{t}!}+\frac{(m-n+1)_{t}!(\ell)_{t}!}{(k)_{t}!(m-n+1-k)_{t}!(\ell-k)_{t}!}$
$\displaystyle=$
$\displaystyle\frac{(m-n)_{t}!(\ell)_{t}!}{(k)_{t}!(m-n-k+1)_{t}!(\ell-k+1)_{t}!}((-\ell+m-n)_{t}(k)_{t}+(m-n+1)_{t}(\ell-k+1)_{t})$
$\displaystyle=$
$\displaystyle\frac{(m-n)_{t}!(\ell+1)_{t}!}{(k)_{t}!(m-n-k)_{t}!(\ell-k+1)_{t}!},$
(3.74)

which completes the proof. ∎

##  4 Rational version

In this section, we discuss the rational version of the source identity. In
addition to presenting the results which are rational version of the ones in
previous section for the trigonometric case, we also prove a rational version
of symmetrization formulas by Lascoux.

###  4.1 Reduction from trigonometric to rational source identity

Here we briefly discuss the degeneration from trigonometric to rational source
identity.

We start from the $m=n$ case of the trigonometric source identity (1.3).
Replacing $zq^{-1/2}$ by $z$, $m=n$ case of (1.3) can be rewritten as

$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{q^{-1/2}v_{i}^{1/2}v_{j}^{-1/2}-q^{1/2}v_{j}^{1/2}v_{i}^{-1/2}}{v_{i}^{1/2}v_{j}^{-1/2}-v_{j}^{1/2}v_{i}^{-1/2}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq
n\end{subarray}}\frac{v_{i}^{1/2}u_{k}^{-1/2}-u_{k}^{1/2}v_{i}^{-1/2}}{q^{-1/2}v_{i}^{1/2}u_{k}^{-1/2}-q^{1/2}u_{k}^{1/2}v_{i}^{-1/2}}$
$\displaystyle=$
$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{q^{1/2}u_{i}^{1/2}u_{j}^{-1/2}-q^{-1/2}u_{j}^{1/2}u_{i}^{-1/2}}{u_{i}^{1/2}u_{j}^{-1/2}-u_{j}^{1/2}u_{i}^{-1/2}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq
m\end{subarray}}\frac{v_{k}^{1/2}u_{i}^{-1/2}-u_{i}^{1/2}v_{k}^{-1/2}}{q^{-1/2}v_{k}^{1/2}u_{i}^{-1/2}-q^{1/2}u_{i}^{1/2}v_{k}^{-1/2}}.$
(4.1)

Introduce $x_{i}$, $y_{j}$, $c$ through $u_{i}=e^{x_{i}}$, $v_{j}=e^{y_{j}}$,
$q^{-1/2}=e^{-c}$. Then (4.1) is rewritten using trigonometric functions as

$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{\mathrm{sinh}(y_{i}-y_{j}-c)}{\mathrm{sinh}(y_{i}-y_{j})}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq
n\end{subarray}}\frac{\mathrm{sinh}(y_{i}-x_{k})}{\mathrm{sinh}(y_{i}-x_{k}-c)}$
$\displaystyle=$
$\displaystyle\sum_{K\subset[1,\dots,n]}(-z)^{|K|}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{\mathrm{sinh}(x_{i}-x_{j}+c)}{\mathrm{sinh}(x_{i}-x_{j})}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq
m\end{subarray}}\frac{\mathrm{sinh}(x_{i}-y_{k})}{\mathrm{sinh}(x_{i}-y_{k}+c)}.$
(4.2)

Taking the rational limit $\sinh w\to w$ $(w\to 0)$ and redefining $x_{i}$ and
$y_{j}$ as $u_{i}$ and $v_{j}$ gives the $m=n$ case of the rational source
identity (1.3). Next, taking the limiting procedure in the same way as
described in the proof in Theorem 3.1, we get (1.3) for $n>m$.

###  4.2 Determinant representations of rational source functions

The systematic derivation of determinant forms of rational functions appearing
in the source identity (1.4) can be applied to the rational case as well. We
introduce notations for both hand sides of (1.4)

$\displaystyle F_{n,m}^{(z)}(\overline{u}|\overline{v})$
$\displaystyle:=\sum_{K\subset[1,\dots,m]}(-z)^{|K|}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{v_{i}-v_{j}-c}{v_{i}-v_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq n\end{subarray}}\frac{v_{i}-u_{k}}{v_{i}-u_{k}-c},$ (4.3)
$\displaystyle G_{n,m}^{(z)}(\overline{u}|\overline{v})$
$\displaystyle:=(1-z)^{m-n}\sum_{K\subset[1,\dots,n]}(-z)^{|K|}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{u_{i}-u_{j}+c}{u_{i}-u_{j}}\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m\end{subarray}}\frac{u_{i}-v_{k}}{u_{i}-v_{k}+c}.$ (4.4)

We call these functions as rational source functions. We note the form (4.3)
for $F_{n,m}^{(1)}(\overline{u}|\overline{v})$ appeared in the context of
partition functions first in Gromov-Sever-Vieira [16]. See also [19]. For the
case of generic $z$, $F_{n,m}^{(z)}(\overline{u}|\overline{v})$ and
$G_{n,m}^{(z)}(\overline{u}|\overline{v})$ appear as representations of the
most generalized domain wall boundary partition functions in Belliard-Pimenta-
Slavnov [23].

To derive determinant forms, we rewrite rational source functions in the
following forms using additive difference operators

$\displaystyle F_{n,m}^{(z)}(\overline{u}|\overline{v})$
$\displaystyle=\frac{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})}{\prod_{1\leq
i<j\leq m}(v_{j}-v_{i})}\prod_{j=1}^{m}(1-zT_{v_{j}}^{-1})\frac{\prod_{1\leq
i<j\leq m}(v_{j}-v_{i})}{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})},$ (4.5)
$\displaystyle G_{n,m}^{(z)}(\overline{u}|\overline{v})$
$\displaystyle=(1-z)^{m-n}\frac{\prod_{i=1}^{n}\prod_{k=1}^{m}(u_{i}-v_{k})}{\prod_{1\leq
i<j\leq n}(u_{j}-u_{i})}\prod_{j=1}^{n}(1-zT_{u_{j}})\frac{\prod_{1\leq
i<j\leq n}(u_{j}-u_{i})}{\prod_{i=1}^{n}\prod_{k=1}^{m}(u_{i}-v_{k})}.$ (4.6)

In the same way with deriving Proposition 3.2 starting from (4.5) and (4.6)
instead, we get the following Minin-Pronko-Tarasov [21, 22] type determinant
represenations.

######  Proposition 4.1.

We have the following determinant representations:

$\displaystyle F_{n,m}^{(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1-r\prod_{\ell=1}^{m}v_{\ell}}{\det_{1\leq i,j\leq
m}\Big{(}\sum_{k=1}^{m}p_{ik}\psi_{k}^{A_{m-1}}(v_{j};0,r)\Big{)}}$
$\displaystyle\times\det_{1\leq i,j\leq
m}\Bigg{(}\Bigg{(}\frac{1}{1-r\prod_{\ell=1}^{m}v_{\ell}}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{m}p_{ik}\psi_{k}^{A_{m-1}}(v_{j};0,r)$
$\displaystyle-z\Bigg{(}\frac{1}{1-r(v_{j}-c)\prod_{\ell=1,\ell\neq
j}^{m}v_{\ell}}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{m}p_{ik}\psi_{k}^{A_{m-1}}(v_{j}-c;0,r)\prod_{\ell=1}^{n}\frac{v_{j}-u_{\ell}}{v_{j}-u_{\ell}-c}\Bigg{)},$
(4.7) $\displaystyle G_{n,m}^{(z)}(\overline{u}|\overline{v})=$
$\displaystyle(1-z)^{m-n}\frac{1-r\prod_{\ell=1}^{n}u_{\ell}}{\det_{1\leq
i,j\leq n}\Big{(}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};0,r)\Big{)}}$
$\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\Bigg{(}\frac{1}{1-r\prod_{\ell=1}^{n}u_{\ell}}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j};0,r)$
$\displaystyle-z\Bigg{(}\frac{1}{1-r(u_{j}+c)\prod_{\ell=1,\ell\neq
j}^{n}u_{\ell}}\Bigg{)}^{\delta_{i1}}\sum_{k=1}^{n}q_{ik}\psi_{k}^{A_{n-1}}(u_{j}+c;0,r)\prod_{\ell=1}^{m}\frac{u_{j}-v_{\ell}}{u_{j}-v_{\ell}+c}\Bigg{)}.$
(4.8)

Here, $r$, $p_{ij}$ $(i,j=1,\dots,m)$, $q_{ij}$ $(i,j=1,\dots,n)$ are
additional parameters such that $\det_{1\leq i,j\leq m}(p_{ij})\not\equiv 0$,
$\det_{1\leq i,j\leq n}(q_{ij})\not\equiv 0$.

The case $r=0$, $p_{ij}=q_{ij}=\delta_{ij}$ of (4.7), (4.8) are the scalar
product type representations first considered by Kostov [17, 18].

######  Corollary 4.2.

We have the following determinant representations:

$\displaystyle F_{n,m}^{(z)}(\overline{u}|\overline{v})$
$\displaystyle=\prod_{1\leq i<j\leq m}\frac{1}{v_{j}-v_{i}}\det_{1\leq i,j\leq
m}\Bigg{(}v_{j}^{i-1}-z(v_{j}-c)^{i-1}\prod_{\ell=1}^{n}\frac{v_{j}-u_{\ell}}{v_{j}-u_{\ell}-c}\Bigg{)},$
(4.9) $\displaystyle G_{n,m}^{(z)}(\overline{u}|\overline{v})$
$\displaystyle=(1-z)^{m-n}\prod_{1\leq i<j\leq
n}\frac{1}{u_{j}-u_{i}}\det_{1\leq i,j\leq
n}\Bigg{(}u_{j}^{i-1}-z(u_{j}+c)^{i-1}\prod_{\ell=1}^{m}\frac{u_{j}-v_{\ell}}{u_{j}-v_{\ell}+c}\Bigg{)}.$
(4.10)

The following domain wall boundary type determinant representations first
considered by Foda-Wheeler [19] can also be derived in the same way with
deriving Proposition 3.4.

######  Proposition 4.3.

For $n\geq m$, we have the following determinant representations:

$\displaystyle F_{n,m}^{(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})}{\prod_{1\leq
i<j\leq m}(v_{j}-v_{i})\prod_{1\leq i<j\leq n}(u_{i}-u_{j})}\det_{1\leq
i,j\leq n}(Y),$ (4.11)

where $Y$ is an $n\times n$ matrix whose $(i,j)$-entry is given by

$\displaystyle
Y_{ij}=\left\\{\begin{array}[]{ll}\displaystyle\frac{1}{v_{i}-u_{j}}-z\frac{1}{v_{i}-u_{j}-c},&1\leq
i\leq m,\ \ \ 1\leq j\leq n\\\ u_{j}^{n-i},&m+1\leq i\leq n,\ \ \ 1\leq j\leq
n\end{array}\right..$ (4.14)

$\displaystyle G_{n,m}^{(z)}(\overline{u}|\overline{v})=$
$\displaystyle(1-z)^{m-n}\frac{\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k})}{\prod_{1\leq
i<j\leq m}(v_{j}-v_{i})\prod_{1\leq i<j\leq n}(u_{i}-u_{j})}\det_{1\leq
i,j\leq n}(Z),$ (4.15)

where $Z$ is an $n\times n$ matrix whose $(i,j)$-entry is given by

$\displaystyle
Z_{ij}=\left\\{\begin{array}[]{ll}\displaystyle\frac{1}{v_{i}-u_{j}}-z\frac{1}{v_{i}-u_{j}-c},&1\leq
i\leq m,\ \ \ 1\leq j\leq n\\\ u_{j}^{n-i}-z(u_{j}+c)^{n-i},&m+1\leq i\leq n,\
\ \ 1\leq j\leq n\end{array}\right..$ (4.18)

Finally, the following is a one-parameter extension of the determinant
representations by Belliard-Slavnov [20]. These representations can be derived
in the same way with Proposition 3.5.

######  Proposition 4.4.

We have the following determinant representations:

$\displaystyle
F_{n,m}^{(z)}(\overline{u}|\overline{v})=\frac{1-\Delta}{\prod_{1\leq i<j\leq
m}(v_{j}-v_{i})(\eta_{i}-\eta_{j})}$ $\displaystyle\times\det_{1\leq i,j\leq
m}\Bigg{(}\Bigg{(}\frac{1}{\prod_{\ell=1}^{m}v_{\ell}-\Delta\prod_{\ell=1}^{m}\eta_{\ell}}\Bigg{)}^{\delta_{i1}}\frac{(v_{i}-\Delta\eta_{j})\prod_{k=1,k\neq
j}^{m}(v_{i}-\eta_{k})}{1-\Delta}$
$\displaystyle-z\Bigg{(}\frac{1}{(v_{i}-c)\prod_{\ell=1,\ell\neq
i}^{m}v_{\ell}-\Delta\prod_{\ell=1}^{m}\eta_{\ell}}\Bigg{)}^{\delta_{i1}}\frac{(v_{i}-\Delta\eta_{j}-c)\prod_{k=1,k\neq
j}^{m}(v_{i}-\eta_{k}-c)}{1-\Delta}\prod_{k=1}^{n}\frac{v_{i}-u_{k}}{v_{i}-u_{k}-c}\Bigg{)},$
(4.19) $\displaystyle
G_{n,m}^{(z)}(\overline{u}|\overline{v})=(1-z)^{m-n}\frac{1-\Delta}{\prod_{1\leq
i<j\leq n}(u_{j}-u_{i})(\eta_{i}-\eta_{j})}$ $\displaystyle\times\det_{1\leq
i,j\leq
n}\Bigg{(}\Bigg{(}\frac{1}{\prod_{\ell=1}^{n}u_{\ell}-\Delta\prod_{\ell=1}^{n}\eta_{\ell}}\Bigg{)}^{\delta_{i1}}\frac{(u_{i}-\Delta\eta_{j})\prod_{k=1,k\neq
j}^{n}(u_{i}-\eta_{k})}{1-\Delta}$
$\displaystyle-z\Bigg{(}\frac{1}{(u_{i}+c)\prod_{\ell=1,\ell\neq
i}^{n}u_{\ell}-\Delta\prod_{\ell=1}^{n}\eta_{\ell}}\Bigg{)}^{\delta_{i1}}\frac{(u_{i}-\Delta\eta_{j}+c)\prod_{k=1,k\neq
j}^{n}(u_{i}-\eta_{k}+c)}{1-\Delta}\prod_{k=1}^{m}\frac{u_{i}-v_{k}}{u_{i}-v_{k}+c}\Bigg{)}.$
(4.20)

Taking the limit $\Delta\to\infty$ gives the determinants in [20].

######  Corollary 4.5.

We have the following determinant representations:

$\displaystyle F_{n,m}^{(z)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{1}{\prod_{1\leq i<j\leq
m}(v_{j}-v_{i})(\eta_{i}-\eta_{j})}$ $\displaystyle\times\det_{1\leq i,j\leq
m}\Bigg{(}\prod_{k=1,k\neq j}^{m}(v_{i}-\eta_{k})-z\prod_{k=1,k\neq
j}^{m}(v_{i}-\eta_{k}-c)\prod_{k=1}^{n}\frac{v_{i}-u_{k}}{v_{i}-u_{k}-c}\Bigg{)},$
(4.21) $\displaystyle G_{n,m}^{(z)}(\overline{u}|\overline{v})=$
$\displaystyle(1-z)^{m-n}\frac{1}{\prod_{1\leq i<j\leq
n}(u_{j}-u_{i})(\eta_{i}-\eta_{j})}$ $\displaystyle\times\det_{1\leq i,j\leq
n}\Bigg{(}\prod_{k=1,k\neq j}^{n}(u_{i}-\eta_{k})-z\prod_{k=1,k\neq
j}^{n}(u_{i}-\eta_{k}+c)\prod_{k=1}^{m}\frac{u_{i}-v_{k}}{u_{i}-v_{k}+c}\Bigg{)}.$
(4.22)

###  4.3 A complex analytic proof of rational source identity

We give a complex analytic proof of the rational source identity which goes
parallel with the trigonometric case. A different proof showing as an identity
between determinants can be found in [9]. We prove the following polynomial
version

$\displaystyle\sum_{K\subset[1,\dots,m]}(-z)^{|K|}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{v_{i}-v_{j}-c}{v_{i}-v_{j}}\prod_{\begin{subarray}{c}j\not\in
K\\\ 1\leq k\leq n\end{subarray}}(v_{j}-u_{k}-c)\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq n\end{subarray}}(v_{i}-u_{k})$ $\displaystyle=$
$\displaystyle(1-z)^{m-n}\sum_{K\subset[1,\dots,n]}(-z)^{|K|}\prod_{\begin{subarray}{c}i\in
K\\\ j\not\in
K\end{subarray}}\frac{u_{i}-u_{j}+c}{u_{i}-u_{j}}\prod_{\begin{subarray}{c}j\not\in
K\\\ 1\leq k\leq m\end{subarray}}(v_{k}-u_{j}-c)\prod_{\begin{subarray}{c}i\in
K\\\ 1\leq k\leq m\end{subarray}}(v_{k}-u_{i}).$ (4.23)

Let us denote the left hand side and the right hand side of (4.23) as
$P_{n,m}^{(z)}(\overline{u}|\overline{v})$ and
$Q_{n,m}^{(z)}(\overline{u}|\overline{v})$ respectively. Relations with
$F_{n,m}^{(z)}(\overline{u}|\overline{v})$ and
$G_{n,m}^{(z)}(\overline{u}|\overline{v})$ are

$\displaystyle P_{n,m}^{(z)}(\overline{u}|\overline{v})$
$\displaystyle=\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k}-c)F_{n,m}^{(z)}(\overline{u}|\overline{v}),$
(4.24) $\displaystyle Q_{n,m}^{(z)}(\overline{u}|\overline{v})$
$\displaystyle=\prod_{i=1}^{m}\prod_{k=1}^{n}(v_{i}-u_{k}-c)G_{n,m}^{(z)}(\overline{u}|\overline{v}).$
(4.25)

We assume $m\leq n$ for the proof. First, we find the following properties
hold.

######  Lemma 4.6.

$P_{n,m}^{(z)}(\overline{u}|\overline{v})$ and
$Q_{n,m}^{(z)}(\overline{u}|\overline{v})$ are symmetric polynomials in
$v_{1},\dots,v_{m}$. The degree of $v_{i}\ (i=1,2,\dots,m)$ is at most $n$ for
both polynomials.

###### Proof.

From the determinant forms of $F_{n,m}^{(z)}(\overline{u}|\overline{v})$ (4.9)
and $G_{n,m}^{(z)}(\overline{u}|\overline{v})$ (4.10), we have the following
determinant forms of $P_{n,m}^{(z)}(\overline{u}|\overline{v})$ and
$Q_{n,m}^{(z)}(\overline{u}|\overline{v})$

$\displaystyle P_{n,m}^{(z)}(\overline{u}|\overline{v})$ $\displaystyle=$
$\displaystyle\prod_{1\leq i<j\leq m}\frac{1}{v_{j}-v_{i}}\det_{1\leq i,j\leq
m}\Bigg{(}(v_{i}+c)^{j-1}\prod_{k=1}^{n}(v_{i}-u_{k}-c)-zv_{i}^{j-1}\prod_{k=1}^{n}(v_{i}-u_{k})\Bigg{)},$
(4.26) $\displaystyle Q_{n,m}^{(z)}(\overline{u}|\overline{v})$
$\displaystyle=$ $\displaystyle(1-z)^{m-n}\prod_{1\leq i<j\leq
n}\frac{1}{u_{j}-u_{i}}\det_{1\leq i,j\leq
n}\Bigg{(}u_{i}^{j-1}\prod_{k=1}^{m}(v_{k}-u_{i}-c)-z(u_{i}+c)^{j-1}\prod_{k=1}^{m}(v_{k}-u_{i})\Bigg{)}.$
(4.27)

Lemma 4.6 can be checked in the same way with the trigonometric case using
these determinant forms. ∎

Due to Lemma 4.6, one notes that it is enough to check
$P_{n,m}^{(z)}(\overline{u}|\overline{v})=Q_{n,m}^{(z)}(\overline{u}|\overline{v})$
for $(n+1)^{m}$ distinct points in $(v_{1},\dots,v_{m})$, and we check the
equality for the following
$2n\times(2n-1)\times\cdots\times(2n-m+1)(\geq(n+1)^{m})$ distinct points
$v_{1}=u_{p_{1}}+\epsilon_{1},\ v_{2}=u_{p_{2}}+\epsilon_{2},\
\dots,v_{m}=u_{p_{m}}+\epsilon_{m}$, $p_{1},\dots,p_{m}\in\\{1,\dots,n\\}$,
$\epsilon_{1},\dots,\epsilon_{m}\in\\{0,c\\}$ and $v_{i}\neq v_{j}$ for $i\neq
j$. It is easy to check the following cases which correspond to $p_{i}=p_{j}$
for some $i\neq j$.

######  Lemma 4.7.

If $v_{i}=u_{k}$, $v_{j}=u_{k}+c$ for some $i,j,k\ (i\neq j)$, we have

$\displaystyle
P_{n,m}^{(z)}(\overline{u}|\overline{v})=Q_{n,m}^{(z)}(\overline{u}|\overline{v})=0.$
(4.28)

###### Proof.

We check the case $k=n$ as the other cases can be checked in the same way. It
is easy to see from the expressions in (4.23) that substituting
$v_{m}=u_{n}+c$ in $P_{n,m}^{(z)}(\overline{u}|\overline{v})$ and
$Q_{n,m}^{(z)}(\overline{u}|\overline{v})$, we have

$\displaystyle P_{n,m}^{(z)}(\overline{u}|\overline{v})|_{v_{m}=u_{n}+c}$
$\displaystyle=-cz\prod_{j=1}^{n-1}(u_{n}-u_{j}+c)\prod_{j=1}^{m-1}(v_{j}-u_{n})P_{n-1,m-1}^{(z)}(u_{1},\dots,u_{n-1}|v_{1},\dots,v_{m-1}),$
(4.29) $\displaystyle
Q_{n,m}^{(z)}(\overline{u}|\overline{v})|_{v_{m}=u_{n}+c}$
$\displaystyle=-cz\prod_{j=1}^{n-1}(u_{n}-u_{j}+c)\prod_{j=1}^{m-1}(v_{j}-u_{n})Q_{n-1,m-1}^{(z)}(u_{1},\dots,u_{n-1}|v_{1},\dots,v_{m-1}).$
(4.30)

Since there is an overall factor $\prod_{j=1}^{m-1}(v_{j}-u_{n})$ in (4.29)
and (4.30), further substituting $v_{j}=u_{n}$ for some $j\ (1\leq j\leq
m-1)$, both $P_{n,m}^{(z)}(\overline{u}|\overline{v})|_{v_{m}=u_{n}+c}$ and
$Q_{n,m}^{(z)}(\overline{u}|\overline{v})|_{v_{m}=u_{n}+c}$ vanish. Since both
$P_{n,m}^{(z)}(\overline{u}|\overline{v})$ and
$Q_{n,m}^{(z)}(\overline{u}|\overline{v})$ are symmetric polynomials in $v$
variables, the other cases follow by symmetry. ∎

We next assume $p_{i}\neq p_{j}$ for $i\neq j$. For $I=\\{1\leq
i_{1}<i_{2}<\cdots<i_{|I|}\leq n\\}$ and $J=\\{1\leq
j_{1}<j_{2}<\cdots<j_{|J|}\leq n\\}$ such that $I\cup J$ is a set of $m$
distinct integers in $[1,\dots,n]$, let us denote by
$\overline{v}=\\{\overline{u}_{I},\overline{u}_{J}+c\\}$ the following type of
substitution

$\displaystyle
v_{q_{1}}=u_{i_{1}},\dots,v_{q_{|I|}}=u_{i_{|I|}},v_{r_{1}}=u_{j_{1}}+c,\dots,v_{r_{|J|}}=u_{j_{|J|}}+c,$
(4.31)

for some $\\{q_{1},\dots,q_{|I|},r_{1},\dots,r_{|J|}\\}=[1,\dots,m]$. Note
$|I|+|J|=m$.

We have the following explicit evaluations.

######  Proposition 4.8.

We have

$\displaystyle
P_{n,m}^{(z)}(\overline{u}|\\{\overline{u}_{I},\overline{u}_{J}+c\\})=(-z)^{|J|}\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}(u_{j}-u_{i})\prod_{\begin{subarray}{c}i\in I\\\
1\leq j\leq n\end{subarray}}(u_{i}-u_{j}-c)\prod_{\begin{subarray}{c}j\in J\\\
k\in[1,\dots,n]\backslash I\end{subarray}}(u_{j}-u_{k}+c),$ (4.32)
$\displaystyle
Q_{n,m}^{(z)}(\overline{u}|\\{\overline{u}_{I},\overline{u}_{J}+c\\})=(-z)^{|J|}\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}(u_{j}-u_{i})\prod_{\begin{subarray}{c}i\in I\\\
1\leq j\leq n\end{subarray}}(u_{i}-u_{j}-c)\prod_{\begin{subarray}{c}j\in J\\\
k\in[1,\dots,n]\backslash I\end{subarray}}(u_{j}-u_{k}+c).$ (4.33)

###### Proof.

It is easy to check the case for
$P_{n,m}^{(z)}(\overline{u}|\\{\overline{u}_{I},\overline{u}_{J}+c\\})$. After
substituting
$v_{q_{1}}=u_{i_{1}},\dots,v_{q_{|I|}}=u_{i_{|I|}},v_{r_{1}}=u_{j_{1}}+c,\dots,v_{r_{|J|}}=u_{j_{|J|}}+c$,
one notes from the factors $\displaystyle\prod_{\begin{subarray}{c}j\not\in
K\\\ 1\leq k\leq n\end{subarray}}(v_{j}-u_{k}-c)$ and
$\displaystyle\prod_{\begin{subarray}{c}i\in K\\\ 1\leq k\leq
n\end{subarray}}(v_{i}-u_{k})$ that only the summand corresponding to
$K=\\{r_{1},\dots,r_{|J|}\\}$ survives, and we get

$\displaystyle
P_{n,m}^{(z)}(\overline{u}|\\{\overline{u}_{I},\overline{u}_{J}+c\\})$
$\displaystyle=$ $\displaystyle(-z)^{|J|}\prod_{\begin{subarray}{c}i\in J\\\
j\in
I\end{subarray}}\frac{u_{i}-u_{j}}{u_{i}-u_{j}+c}\prod_{\begin{subarray}{c}j\in
I\\\ 1\leq k\leq n\end{subarray}}(u_{j}-u_{k}-c)\prod_{\begin{subarray}{c}j\in
J\\\ 1\leq k\leq n\end{subarray}}(u_{j}-u_{k}+c)$ $\displaystyle=$
$\displaystyle(-z)^{|J|}\prod_{\begin{subarray}{c}i\in I\\\ j\in
J\end{subarray}}(u_{j}-u_{i})\prod_{\begin{subarray}{c}i\in I\\\ 1\leq j\leq
n\end{subarray}}(u_{i}-u_{j}-c)\prod_{\begin{subarray}{c}j\in J\\\
k\in[1,\dots,n]\backslash I\end{subarray}}(u_{j}-u_{k}+c).$ (4.34)

Next, we check the case for
$Q_{n,m}^{(z)}(\overline{u}|\\{\overline{u}_{I},\overline{u}_{J}+c\\})$.
Substituting
$v_{q_{1}}=u_{i_{1}},\dots,v_{q_{|I|}}=u_{i_{|I|}},v_{r_{1}}=u_{j_{1}}+c,\dots,v_{r_{|J|}}=u_{j_{|J|}}+c$,
we note from the factor $\displaystyle\prod_{\begin{subarray}{c}i\in K\\\
1\leq k\leq m\end{subarray}}(v_{k}-u_{i})$ that summands labeled by $K$
satisfying $K\cap I\neq\phi$ vanish. We also note from the factor
$\displaystyle\prod_{\begin{subarray}{c}j\not\in K\\\ 1\leq k\leq
m\end{subarray}}(v_{k}-u_{j}-c)$ that summands vanish unless $J\subset K$.
Therefore we can restrict the sum to $\displaystyle\sum_{J\subset
K\subset[1,\dots,n]\backslash I}$ and after simplifications, we can write
$Q_{n,m}^{(z)}(\overline{u}|\\{\overline{u}_{I},\overline{u}_{J}+c\\})$ as

$\displaystyle
Q_{n,m}^{(z)}(\overline{u}|\\{\overline{u}_{I},\overline{u}_{J}+c\\})$
$\displaystyle=$ $\displaystyle\prod_{\begin{subarray}{c}i\in I\\\ j\in
J\end{subarray}}(u_{j}-u_{i})\prod_{\begin{subarray}{c}i\in I\\\ 1\leq j\leq
n\end{subarray}}(u_{i}-u_{j}-c)\prod_{\begin{subarray}{c}j\in J\\\
k\in[1,\dots,n]\backslash
I\end{subarray}}(u_{j}-u_{k}+c)(-z)^{|J|}(1-z)^{m-n}$
$\displaystyle\times\sum_{J\subset K\subset[1,\dots,n]\backslash
I}(-z)^{|K|-|J|}\prod_{\begin{subarray}{c}i\in K\backslash J\\\
j\in[1,\dots,n]\backslash\\{I,K\\}\end{subarray}}\frac{u_{i}-u_{j}+c}{u_{i}-u_{j}}.$
(4.35)

Next, we note the factorization of the sum in (4.35)

$\displaystyle\sum_{J\subset K\subset[1,\dots,n]\backslash
I}(-z)^{|K|-|J|}\prod_{\begin{subarray}{c}i\in K\backslash J\\\
j\in[1,\dots,n]\backslash\\{I,K\\}\end{subarray}}\frac{u_{i}-u_{j}+c}{u_{i}-u_{j}}$
$\displaystyle=$
$\displaystyle\sum_{\ell=|J|}^{n-|I|}\sum_{\begin{subarray}{c}K:|K|=\ell\\\
J\subset K\subset[1,\dots,n]\backslash
I\end{subarray}}(-z)^{\ell-|J|}\prod_{\begin{subarray}{c}i\in K\backslash J\\\
j\in[1,\dots,n]\backslash\\{I,K\\}\end{subarray}}\frac{u_{i}-u_{j}+c}{u_{i}-u_{j}}$
$\displaystyle=$
$\displaystyle\sum_{\ell=|J|}^{n-|I|}(-z)^{\ell-|J|}\dbinom{n-|I|-|J|}{\ell-|J|}=(1-z)^{n-|I|-|J|}=(1-z)^{n-m}.$
(4.36)

Here we used the well-known identity

$\displaystyle\sum_{K\subset[1,\dots,n]:|K|=\ell}\prod_{\begin{subarray}{c}i\in
K\\\ j\in[1,\dots,n]\backslash
K\end{subarray}}\frac{u_{i}-u_{j}+c}{u_{i}-u_{j}}=\dbinom{n}{\ell}.$ (4.37)

Combining (4.35) and (4.36) gives

$\displaystyle
Q_{n,m}^{(z)}(\overline{u}|\\{\overline{u}_{I},\overline{u}_{J}+c\\})=(-z)^{|J|}\prod_{\begin{subarray}{c}i\in
I\\\ j\in J\end{subarray}}(u_{j}-u_{i})\prod_{\begin{subarray}{c}i\in I\\\
1\leq j\leq n\end{subarray}}(u_{i}-u_{j}-c)\prod_{\begin{subarray}{c}j\in J\\\
k\in[1,\dots,n]\backslash I\end{subarray}}(u_{j}-u_{k}+c).$ (4.38)

∎

From Lemma 4.7 and Proposition 4.8, we checked the equality for enough
specializations of $(v_{1},\dots,v_{m})$ and hence, (4.23) follows.

###  4.4 Rational version of Lascoux’s symmetrization formulas

We prove symmetrization formulas which are rational version of Lascoux’s
symmetrization formulas [26], Theorem 3 and Theorem 4, which special cases of
rational source functions appear. We introduce notations for divided
difference operators and symmetrization for functions of $u$-variables

$\displaystyle\partial_{k}f(u_{1},\dots,u_{n})$ $\displaystyle:=$
$\displaystyle\frac{f(u_{1},\dots,u_{k},u_{k+1},\dots,u_{n})-f(u_{1},\dots,u_{k+1},u_{k},\dots,u_{n})}{u_{k}-u_{k+1}},\
\ \ k=1,\dots,n-1,$ (4.39)
$\displaystyle\mathrm{Sym}_{c}(f(u_{1},\dots,u_{n})):=\sum_{w\in
S_{n}}w\cdot\Bigg{(}\Delta(12\dots n)f(u_{1},\dots,u_{n})\Bigg{)},$ (4.40)

where $\displaystyle\Delta(k_{1}k_{2}\cdots k_{n}):=\prod_{1\leq i<j\leq
n}\frac{u_{k_{i}}-u_{k_{j}}-c}{u_{k_{i}}-u_{k_{j}}}$. We also define $\theta$
to be shifting the index of variables $\theta u_{k}:=u_{k+1}$ with
$u_{k+n}:=u_{k}+c$.

The following is a rational version of [26], Theorem 3.

######  Theorem 4.9.

We have

$\displaystyle-c\
\mathrm{Sym}_{c}\Bigg{(}(1-\theta)^{n-1}\prod_{j=2}^{n}\prod_{k=1}^{n}(u_{j}-v_{k})f(u_{1})\Bigg{)}$
$\displaystyle=$
$\displaystyle(n-1)!P_{n,n}^{(z=1)}(\overline{u}|\overline{v})\partial_{n-1}\cdots\partial_{1}f(u_{1}).$
(4.41)

###### Proof.

This can be proved in the same way with the proof of Theorem 3 in [26]. One
can show that both hand sides of (4.41) are symmetric in $u_{1},\dots,u_{n}$.
The symmetry of the $u$ variables of the right hand side can be easily checked
from the form (4.26). Since both hand sides are symmetric in
$u_{1},\dots,u_{n}$, it is enough to check the equality between coefficients
of $f(u_{1})$ of both hand sides. Using
$\displaystyle(1-\theta)^{n-1}=\sum_{\ell=1}^{n}\binom{n-1}{\ell-1}(-1)^{\ell-1}\theta^{\ell-1}$,
we can write the left hand side of (4.41) as

$\displaystyle\sum_{\ell=1}^{n}c(-1)^{\ell}\binom{n-1}{\ell-1}\sum_{w\in
S_{n}}w\cdot\Bigg{(}\Delta(12\cdots n)$
$\displaystyle\times\prod_{j=2}^{n+1-\ell}\prod_{k=1}^{n}(u_{j+\ell-1}-v_{k})\prod_{j=n+2-\ell}^{n}\prod_{k=1}^{n}(u_{j+\ell-1-n}-v_{k}+c)f(u_{\ell})\Bigg{)}$
$\displaystyle=$
$\displaystyle\sum_{\ell=1}^{n}c(-1)^{\ell}\binom{n-1}{\ell-1}\sum_{w\in
S_{n}}w\cdot\Bigg{(}\Delta(\ell,2,3,\cdots,\ell-1,1,\ell+1,\cdots,n)$
$\displaystyle\times\prod_{j=\ell+1}^{n}\prod_{k=1}^{n}(u_{j}-v_{k})\prod_{j=2}^{\ell}\prod_{k=1}^{n}(u_{j}-v_{k}+c)f(u_{1})\Bigg{)}.$
(4.42)

On the other hand, the coefficient of $f(u_{1})$ of the right hand side of
(4.41) is

$\displaystyle\frac{(n-1)!}{\prod_{j=2}^{n}(u_{1}-u_{j})}P_{n,n}^{(z=1)}(\overline{u}|\overline{v}),$
(4.43)

hence what we need to show is the following identity

$\displaystyle\sum_{\ell=1}^{n}c(-1)^{\ell}\binom{n-1}{\ell-1}\sum_{w\in
S_{n}:w(1)=1}w\cdot\Bigg{(}\Delta(\ell,2,3,\cdots,\ell-1,1,\ell+1,\cdots,n)$
$\displaystyle\times\prod_{j=\ell+1}^{n}\prod_{k=1}^{n}(u_{j}-v_{k})\prod_{j=2}^{\ell}\prod_{k=1}^{n}(u_{j}-v_{k}+c)\Bigg{)}=\frac{(n-1)!}{\prod_{j=2}^{n}(u_{1}-u_{j})}P_{n,n}^{(z=1)}(\overline{u}|\overline{v}).$
(4.44)

We prove this identity by viewing both hand sides as polynomials in
$v_{1},\dots,v_{n}$. One can easily see that the degree of $v_{j},j=1,\dots,n$
of the left hand side of (4.44) is at most $n-1$. From the determinant
representation of $F_{n,n}^{(z)}(\overline{u}|\overline{v})$ (4.11), we have

$\displaystyle P_{n,n}^{(1)}(\overline{u}|\overline{v})=$
$\displaystyle\frac{\prod_{i=1}^{n}\prod_{k=1}^{n}(v_{i}-u_{k})(v_{i}-u_{k}-c)}{\prod_{1\leq
i<j\leq n}(v_{j}-v_{i})\prod_{1\leq i<j\leq n}(u_{i}-u_{j})}\det_{1\leq
j,k\leq n}\Bigg{(}\frac{1}{v_{j}-u_{k}}-\frac{1}{v_{j}-u_{k}-c}\Bigg{)},$
(4.45)

which is nothing but the Gaudin-Izergin-Korepin determinant [14, 15, 24]. We
can check that the degree of $v_{j},j=1,\dots,n$ of this determinant and hence
the right hand side of (4.44) is also at most $n-1$.

Since the degree of $v_{j}$ $(j=1,\dots,n)$ is at most $n-1$, it is enough to
check the equality (4.44) for $n^{n}$ distinct points in
$(v_{1},\dots,v_{n})$. We check the following
$(2n-1)\times(2n-2)\times\cdots\times n$ $(\geq n^{n})$ cases: specializing
$v_{1},\dots,v_{n}$ to subsets of
$\\{u_{1},u_{2},u_{2}+c,\dots,u_{n},u_{n}+c\\}$.

It is easy to see from the expression that the left hand side of (4.44)
vanishes if $v_{i}=u_{k}$, $v_{j}=u_{k}+c$ for some $i,j,k\ (i\neq j)$ and
$2\leq k\leq n$. The right hand side also vanishes due to Lemma 4.7. We next
consider the other cases which we denote by
$\overline{u}=\\{\overline{v},\overline{v}+c\\}$ in the previous subsection.
By symmetry, it is enough to take $I={1,2,\dots,|I|}$ and $J={|I|+1,\dots,n}$,
i.e., consider the substitution
$v_{1}=u_{1},\dots,v_{|I|}=u_{|I|},v_{|I|+1}=u_{|I|+1}+c,\dots,v_{n}=u_{n}+c$.
One notes that the summands in the left hand side which survive after this
substitution are those which satisfy $\ell=|I|$ and
$\\{w(2),\dots,w(|I|)\\}=\\{2,\dots,|I|\\}$,
$\\{w(|I|+1),\dots,w(n)\\}=\\{|I|+1,\dots,n\\}$ in addition to $w(1)=1$. The
left hand side reduces to

$\displaystyle c(-1)^{|I|}\binom{n-1}{|I|-1}\sum_{\begin{subarray}{c}w\in
S_{n}:w(1)=1\\\ \\{w(2),\dots,w(|I|)\\}=\\{2,\dots,|I|\\}\\\
\\{w(|I|+1),\dots,w(n)\\}=\\{|I|+1,\dots,n\\}\end{subarray}}w\cdot\Bigg{(}\Delta(|I|,2,3,\cdots,|I|-1,1,|I|+1,\cdots,n)\Bigg{)}$
$\displaystyle\times\Bigg{(}\prod_{j=|I|+1}^{n}\prod_{k=1}^{n}(u_{j}-v_{k})\prod_{j=2}^{|I|}\prod_{k=1}^{n}(u_{j}-v_{k}+c)\Bigg{)}\Bigg{|}_{v_{1}=u_{1},\dots,v_{|I|}=u_{|I|},v_{|I|+1}=u_{|I|+1}+c,\dots,v_{n}=u_{n}+c}.$
(4.46)

Using the well-known identity

$\displaystyle\sum_{w\in S_{n}}\Delta(k_{1}k_{2}\dots k_{n})=n!,$ (4.47)

one finds

$\displaystyle\sum_{\begin{subarray}{c}w\in S_{n}:w(1)=1\\\
\\{w(2),\dots,w(|I|)\\}=\\{2,\dots,|I|\\}\\\
\\{w(|I|+1),\dots,w(n)\\}=\\{|I|+1,\dots,n\\}\end{subarray}}w\cdot\Bigg{(}\Delta(|I|,2,3,\cdots,|I|-1,1,|I|+1,\cdots,n)\Bigg{)}$
$\displaystyle=$
$\displaystyle(|I|-1)!(n-|I|)!\prod_{j=2}^{|I|}\prod_{k=|I|+1}^{n}\frac{u_{j}-u_{k}-c}{u_{j}-u_{k}}\prod_{j=2}^{|I|}\frac{u_{j}-u_{1}-c}{u_{j}-u_{1}}\prod_{k=|I|+1}^{n}\frac{u_{1}-u_{k}-c}{u_{1}-u_{k}}.$
(4.48)

We can also easily see

$\displaystyle\Bigg{(}\prod_{j=|I|+1}^{n}\prod_{k=1}^{n}(u_{j}-v_{k})\prod_{j=2}^{|I|}\prod_{k=1}^{n}(u_{j}-v_{k}+c)\Bigg{)}\Bigg{|}_{v_{1}=u_{1},\dots,v_{|I|}=u_{|I|},v_{|I|+1}=u_{|I|+1}+c,\dots,v_{n}=u_{n}+c}$
$\displaystyle=$
$\displaystyle\prod_{j=|I|+1}^{n}\prod_{k=1}^{|I|}(u_{j}-u_{k})\prod_{j=|I|+1}^{n}\prod_{k=|I|+1}^{n}(u_{j}-u_{k}-c)\prod_{j=2}^{|I|}\prod_{k=1}^{|I|}(u_{j}-u_{k}+c)\prod_{j=2}^{|I|}\prod_{k=|I|+1}^{n}(u_{j}-u_{k}).$
(4.49)

Inserting (4.48) and (4.49) into (4.46) and simplifying and rearranging, we
have the following expression for (4.46) which is the specialization of the
left hand side of (4.44)

$\displaystyle\sum_{\ell=1}^{n}c(-1)^{\ell}\binom{n-1}{\ell-1}\sum_{w\in
S_{n}:w(1)=1}w\cdot\Bigg{(}\Delta(\ell,2,3,\cdots,\ell-1,1,\ell+1,\cdots,n)$
$\displaystyle\times\prod_{j=\ell+1}^{n}\prod_{k=1}^{n}(u_{j}-v_{k})\prod_{j=2}^{\ell}\prod_{k=1}^{n}(u_{j}-v_{k}+c)\Bigg{)}$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\prod_{j=2}^{n}(u_{1}-u_{j})}(-1)^{n-|I|}\prod_{j=|I|+1}^{n}\prod_{k=1}^{|I|}(u_{j}-u_{k})\prod_{j=1}^{|I|}\prod_{k=1}^{n}(u_{j}-u_{k}-c)\prod_{j=|I|+1}^{n}\prod_{k=|I|+1}^{n}(u_{j}-u_{k}+c).$
(4.50)

On the other hand, from the case $m=n$, $I={1,2,\dots,|I|}$ and
$J={|I|+1,\dots,n}$ of Proposition 4.8, we find the specialization of the
right hand side of (4.44) is

$\displaystyle\frac{(n-1)!}{\prod_{j=2}^{n}(u_{1}-u_{j})}P_{n,n}^{(z=1)}(\overline{u}|\overline{v})\Bigg{|}_{v_{1}=u_{1},\dots,v_{|I|}=u_{|I|},v_{|I|+1}=u_{|I|+1}+c,\dots,v_{n}=u_{n}+c}$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\prod_{j=2}^{n}(u_{1}-u_{j})}(-1)^{n-|I|}\prod_{j=|I|+1}^{n}\prod_{k=1}^{|I|}(u_{j}-u_{k})\prod_{j=1}^{|I|}\prod_{k=1}^{n}(u_{j}-u_{k}-c)\prod_{j=|I|+1}^{n}\prod_{k=|I|+1}^{n}(u_{j}-u_{k}+c),$
(4.51)

which is exactly the same with the specialization of the left hand side
(4.50).

Since we checked the equality for enough specializations of
$(v_{1},\dots,v_{n})$, we conclude the identity (4.44) and hence (4.41) holds.

∎

The following identity is the rational version of [26], Theorem 4.

######  Theorem 4.10.

We have

$\displaystyle\mathrm{Sym}_{c}\Bigg{(}(1-\tau)^{n}\prod_{j=1}^{n}\prod_{k=1}^{n}\frac{u_{j}-v_{k}-c}{u_{j}-v_{k}}\Bigg{)}=\frac{n!}{\prod_{j=1}^{n}\prod_{k=1}^{n}(u_{j}-v_{k})}P_{n,n}^{(z=1)}(\overline{u}|\overline{v}+c).$
(4.52)

Here $\tau$ acts on variables and functions as $\tau u_{j}:=u_{j+1}$,
$j=1,\dots,n-1$ and $\tau\left(\frac{u_{n}-v_{k}-c}{u_{n}-v_{k}}\right):=1$.

###### Proof.

This can be shown in the same way with the proof of Theorem 4 in [26]. First,
by direct computation, we find the following relation

$\displaystyle(1-\tau)^{n}\prod_{j=1}^{n}\prod_{k=1}^{n}\frac{u_{j}-v_{k}-c}{u_{j}-v_{k}}$
$\displaystyle=$
$\displaystyle\prod_{j=1}^{n}\prod_{k=1}^{n}\frac{1}{u_{j}-v_{k}}\sum_{j=1}^{n}(-1)^{j-1}\binom{n-1}{j-1}\prod_{\ell=1}^{j-1}\prod_{k=1}^{n}(u_{\ell}-v_{k})\prod_{\ell=j+1}^{n}\prod_{k=1}^{n}(u_{\ell}-v_{k}-c)f(u_{j}),$
(4.53)

where

$\displaystyle
f(u_{j})=\prod_{k=1}^{n}(u_{j}-v_{k}-c)-\prod_{k=1}^{n}(u_{j}-v_{k}).$ (4.54)

We also note the following equality holds

$\displaystyle\sum_{j=1}^{n}(-1)^{j-1}\binom{n-1}{j-1}\prod_{\ell=1}^{j-1}\prod_{k=1}^{n}(u_{\ell}-v_{k})\prod_{\ell=j+1}^{n}\prod_{k=1}^{n}(u_{\ell}-v_{k}-c)f(u_{j})$
$\displaystyle=$
$\displaystyle(1-\theta)^{n-1}\prod_{j=2}^{n}\prod_{k=1}^{n}(u_{j}-v_{k}-c)f(u_{1}).$
(4.55)

Combining (4.53) and (4.55), we get

$\displaystyle(1-\tau)^{n}\prod_{j=1}^{n}\prod_{k=1}^{n}\frac{u_{j}-v_{k}-c}{u_{j}-v_{k}}=\prod_{j=1}^{n}\prod_{k=1}^{n}\frac{1}{u_{j}-v_{k}}(1-\theta)^{n-1}\prod_{j=2}^{n}\prod_{k=1}^{n}(u_{j}-v_{k}-c)f(u_{1}),$
(4.56)

and further combining with (4.41), we have

$\displaystyle-c\
\mathrm{Sym}_{c}\Bigg{(}(1-\tau)^{n}\prod_{j=1}^{n}\prod_{k=1}^{n}\frac{u_{j}-v_{k}-c}{u_{j}-v_{k}}\Bigg{)}$
$\displaystyle=$ $\displaystyle-c\
\mathrm{Sym}_{c}\Bigg{(}\prod_{j=1}^{n}\prod_{k=1}^{n}\frac{1}{u_{j}-v_{k}}(1-\theta)^{n-1}\prod_{j=2}^{n}\prod_{k=1}^{n}(u_{j}-v_{k}-c)f(u_{1})\Bigg{)}$
$\displaystyle=$
$\displaystyle\prod_{j=1}^{n}\prod_{k=1}^{n}\frac{1}{u_{j}-v_{k}}\times(-c)\
\mathrm{Sym}_{c}\Bigg{(}(1-\theta)^{n-1}\prod_{j=2}^{n}\prod_{k=1}^{n}(u_{j}-v_{k}-c)f(u_{1})\Bigg{)}$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\prod_{j=1}^{n}\prod_{k=1}^{n}(u_{j}-v_{k})}P_{n,n}^{(z=1)}(\overline{u}|\overline{v}+c)\partial_{n-1}\cdots\partial_{1}f(u_{1}).$
(4.57)

Finally, we note the explicit form of
$\partial_{n-1}\cdots\partial_{1}f(u_{1})$ where $f(u_{1})$ is the $j=1$ case
of (4.54) can be obtained from the coefficient of $u_{1}^{n-1}$ of $f(u_{1})$
since lower degree terms vanish after applying the divided difference
operators. We find $\partial_{n-1}\cdots\partial_{1}f(u_{1})=-nc$ and
inserting into (4.57) gives (4.52). ∎

###  4.5 Geometric derivation of rational source identity

The rational version (1.4) of the source identity is also obtained from the
cohomological version of the wall-crossing formula. We follow the similar
arguments as in §3.4 taking the usual Euler class

$\mathop{\mathrm{Eu}}\nolimits^{\theta}(M^{\zeta^{\pm}}_{Q}(\ell))=\mathop{\mathrm{Eu}}\nolimits(TM^{\zeta^{\pm}}_{Q}(\ell)\otimes\mathbb{C}_{e^{\theta}})$

instead of $\mathop{\mathrm{Eu}}\nolimits^{K}(M^{\zeta^{\pm}}_{Q}(\ell))$.

We see that
$\int_{M^{\zeta^{+}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits^{\theta}(M^{\zeta^{+}}_{Q}(\ell))$
is equal to

$\displaystyle\sum_{K\subset[n+1,\dots,n+m]\atop\lvert
K\rvert=\ell}\prod_{i\in K\atop j\in[n+1,\dots,n+m]\setminus
K}{a_{j}-a_{i}+\theta\over a_{j}-a_{i}}\prod_{i\in K\atop 1\leq k\leq
n}{a_{i}-a_{k}+\theta\over a_{i}-a_{k}}$ $\displaystyle=$
$\displaystyle\sum_{K\subset[1,\dots,m]\atop\lvert K\rvert=\ell}\prod_{i\in
K\atop j\not\in K}{v_{i}-v_{j}-c\over v_{i}-v_{j}}\prod_{i\in K\atop 1\leq
k\leq n}{v_{i}-u_{k}\over v_{i}-u_{k}-c},$

and
$\int_{M^{\zeta^{-}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits^{\theta}(M^{\zeta^{-}}_{Q}(\ell))$
is equal to

$\displaystyle\sum_{K\subset[1,\dots,n]\atop\lvert K\rvert=\ell}\prod_{i\in
K\atop j\in[1,\ldots,n]\setminus K}{a_{i}-a_{j}+\theta\over
a_{i}-a_{j}}\prod_{i\in K\atop n+1\leq k\leq n+m}{a_{k}-a_{i}+\theta\over
a_{k}-a_{i}}$ $\displaystyle=$
$\displaystyle\sum_{K\subset[1,\dots,n]\atop\lvert K\rvert=\ell}\prod_{i\in
K\atop j\not\in K}{u_{i}-u_{j}+c\over u_{i}-u_{j}}\prod_{i\in K\atop 1\leq
k\leq m}{u_{i}-v_{k}\over u_{i}-v_{k}+c},$

where we put $\theta=-c$, and $a_{j}=\begin{cases}-u_{j}-c&j=1,\ldots,n\\\
-v_{j}&j=n+1,\ldots,n+m\end{cases}$. These are coefficients of $z^{\ell}$ in
generating series appearing in both sides of (1.4).
Wall-crossing formula [12, Theorem 7.2] induces

$\displaystyle\int_{M^{\zeta^{+}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits^{\theta}(M^{\zeta^{+}}_{Q}(\ell))-\int_{M^{\zeta^{-}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits^{\theta}(M^{\zeta^{-}}_{Q}(\ell))$
$\displaystyle=\sum_{k=1}^{\ell}\sum_{\boldsymbol{\mathfrak{I}}\in\operatornamewithlimits{Dec}(\ell)\atop|\boldsymbol{\mathfrak{I}}|=k}{(\ell-k)!\over\ell!}\prod_{i=1}^{j}(d_{i}-1)!\gamma_{d_{i}}(\theta)(s(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i})-(n-m)d_{i})\int_{M^{\zeta^{-}}_{Q}(\ell-k)}\mathop{\mathrm{Eu}}\nolimits^{\theta}(M^{\zeta^{-}}_{Q}(\ell)).$

Here we set

$s(\mathfrak{I}_{1},\mathfrak{I}_{2})=\left|\\{(i,j)\in\mathfrak{I}_{1}\times\mathfrak{I}_{2}\mid
i<j\\}\right|-\left|\\{(i,j)\in\mathfrak{I}_{1}\times\mathfrak{I}_{2}\mid
i>j\\}\right|,$

and we have

$\displaystyle\gamma_{d}(\theta)=\int_{M_{Q^{\sharp}}(d)}\mathop{\mathrm{Eu}}\nolimits^{\theta}(TM^{\zeta^{-}}_{Q^{\sharp}}(d))=\begin{cases}1&d=1\\\
0&d\neq 1.\end{cases}$ (4.58)

By (4.58), we can re-write wall-crossing formula as

$\displaystyle\int_{M^{\zeta^{+}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits^{\theta}((M^{\zeta^{+}}_{Q}(\ell)))-\int_{M^{\zeta^{-}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits^{\theta}(M^{\zeta^{-}}_{Q}(\ell))$
$\displaystyle=\sum_{k=1}^{\ell}\sum_{\boldsymbol{\mathfrak{I}}\in\operatornamewithlimits{Dec}(\ell)\atop{|\boldsymbol{\mathfrak{I}}|=k\atop|\mathfrak{I}_{1}|=\cdots=|\mathfrak{I}_{k}|=1}}{(\ell-k)!\over\ell!}\prod_{i=1}^{j}(s(\mathfrak{I}_{i},\boldsymbol{\mathfrak{I}}_{>i})+(m-n))\int_{M^{\zeta^{-}}_{Q}(\ell-k)}\mathop{\mathrm{Eu}}\nolimits^{\theta}((M^{\zeta^{-}}_{Q}(\ell))).$

Since (1.4) is equivalent to

$\int_{M^{\zeta^{+}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits^{\theta}((M^{\zeta^{+}}_{Q}(\ell)))-\int_{M^{\zeta^{-}}_{Q}(\ell)}\mathop{\mathrm{Eu}}\nolimits^{\theta}((M^{\zeta^{-}}_{Q}(\ell)))\\\
=\sum_{k=1}^{\ell}\binom{m-n}{k}\int_{M^{\zeta^{-}}_{Q}(\ell-k)}\mathop{\mathrm{Eu}}\nolimits^{\theta}((M^{\zeta^{-}}_{Q}(\ell))),$

we must show that

$\displaystyle\sum_{{\ell\geq h_{1}>\cdots>h_{k}\geq
1}}{(\ell-k)!\over\ell!}\prod_{i=1}^{k}(s(\\{h_{i}\\},[1,\ldots,\ell]\setminus\\{h_{1},\ldots,h_{i}\\})+(m-n))=\binom{m-n}{k}.$
(4.59)

This follows from taking the limit $t\to 1$ of (3.70).

## Acknowledgement

The authors thank Vitaly Tarasov for useful discussions and sending their
preprint. K.M. is supported by Grant-in-Aid for Scientific Research 21K03176,
20K03793, JSPS. R.O. is supported by Grant-in-Aid for Scientific Research
21K03180, JSPS.

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