MacMahon KZ equation for Ding-Iohara-Miki algebra

Panupong<EMAIL_ADDRESS>and
Hiroaki<EMAIL_ADDRESS>

aGraduate School of Mathematics, Nagoya University, Nagoya, 464-8602, Japan
bKMI, Nagoya University, Nagoya, 464-8602, Japan

> Abstract. We derive a generalized Knizhnik-Zamolodchikov equation for the
> correlation function of the intertwiners of the vector and the MacMahon
> representations of Ding-Iohara-Miki algebra. These intertwiners are cousins
> of the refined topological vertex which is regarded as the intertwining
> operator of the Fock representation. The shift of the spectral parameter of
> the intertwiners is generated by the operator which is constructed from the
> universal $R$ matrix. The solutions to the generalized KZ equation are
> factorized into the ratio of two point functions which are identified with
> generalizations of the Nekrasov factor for supersymmetric quiver gauge
> theories.

##  1\. Introduction

Quantum Knizhnik-Zamolodchikov (q-KZ) equation is originally introduced for
the correlation function of the intertwiners of the quantum affine algebra
[27] [46] [31] [19]. In [9] we have shown that one can develop a parallel
story for Ding-Iohara-Miki (DIM) algebra [18] [34], which is regarded as the
quantum toroidal algebra of $\mathfrak{gl}_{1}$. The trivalent intertwining
operator of the Fock representations of DIM algebra agrees with what is called
the refined topological vertex [6] [7] [29] [2]. Consequently, the correlation
function of the intertwiners gives a building block of the Nekrasov partition
functions of the five dimensional supersymmetric quiver gauge theories in
accord with the AGT correspondence.

Our main interest in the generalized q-KZ equation comes from the AGT
correspondence and its generalizations [5] [35] [48] [14], which tell us that
the conformal blocks of two dimensional conformal field theories and the
instanton partition functions, and hence the low-energy effective actions of
the supersymmetric gauge theories are related. This correspondence allows us
to study problems on one side from the perspective of the other side. For
example, the modular properties of the six dimensional Seiberg-Witten theory
with adjoint hypermultiplet can be explained by using the elliptic Knizhnik-
Zamolodchikov equation of DIM algebra [10].

In the paper [9], the authors gave a general method of deriving the
generalized KZ equation for the Fock intertwiners, and wrote out explicit
solutions to the equation. The general solutions are written in terms of a
product of the propagators of the Fock intertwiners (the refined topological
vertex). Moreover, as expected from the AGT correspondence, the solution
relates to the Nekrasov function which gives the instanton partition function.
We know from the papers [20] [22] that the MacMahon modules are, in a sense, a
generalization of the Fock modules. More precisely, the MacMahon modules can
be constructed as the inductive limit of an inductive system whose objects are
Fock modules. Moreover, we are able to express the MacMahon intertwiner in
terms of the Fock intertwiners [11]. Thus, it is natural to ask whether we can
generalize the story in [9] to derive the generalized KZ equation for the
MacMahon representation of DIM algebra. In this paper we will show this is
indeed the case.

Let $\Psi_{\lambda}(u)$ be the intertwiner of the Fock representation of DIM
algebra, where $\lambda$ is a Young diagram which labels a basis of the Fock
representation and $u$ is the spectral parameter. The $(q,t)$-KZ equation
derived in [9] implies that the correlation function is expressed as a product
of two point functions like the Wick theorem for free fields. One can check
that the inverse of the two point function agrees with the Nekrasov factor
which is a building block of five dimensional quiver gauge theories with the
equivariant parameters $(q_{1},q_{2})=(e^{\epsilon_{1}},e^{\epsilon_{2}})$,
[38] [37] [42] [26]. Namely, if we define

$N_{\lambda\mu}(u,v)=\langle\Psi_{\lambda}(u)\Psi_{\mu}(v)\rangle^{-1},$ (1.1)

then we can show that

$\displaystyle N_{\lambda\mu}(u,v)$
$\displaystyle=\prod_{\genfrac{}{}{0.0pt}{}{\square\in\lambda}{\blacksquare\in\mu}}\frac{\left(1-q_{1}\frac{\chi_{\square}(u)}{\chi_{\blacksquare}(v)}\right)\left(1-q_{2}\frac{\chi_{\square}(u)}{\chi_{\blacksquare}(v)}\right)}{\left(1-\frac{\chi_{\square}(u)}{\chi_{\blacksquare}(v)}\right)\left(1-q_{1}q_{2}\frac{\chi_{\square}(u)}{\chi_{\blacksquare}(v)}\right)}\cdot\prod_{\square\in\lambda}\left(1-q_{1}q_{2}\frac{\chi_{\square}(u)}{v}\right)\cdot\prod_{\blacksquare\in\mu}\left(1-\frac{u}{\chi_{\blacksquare}(v)}\right)$
$\displaystyle=\prod_{\square\in\lambda}\left(1-q_{1}^{-\ell_{\mu}(\square)}q_{2}^{a_{\lambda}(\square)+1}\frac{u}{v}\right)\prod_{\blacksquare\in\mu}\left(1-q_{1}^{\ell_{\lambda}(\blacksquare)+1}q_{2}^{-a_{\mu}(\blacksquare)}\frac{u}{v}\right),$
(1.2)

where for $\square=(i,j)\in\lambda$

$\chi_{\square}(u):=u\cdot q_{1}^{i-1}q_{2}^{j-1},$ (1.3)

and

$a_{\mu}(\square)=\mu_{i}-j,\qquad\ell_{\mu}(\square)=\mu^{\vee}_{j}-i.$ (1.4)

###  1.1. Strategy of deriving q-KZ equation

Let $\Psi_{\alpha}(z)$ be the component of the DIM intertwiner in general.
Namely $\alpha$ stands for 1d, 2d and 3d Young diagrams for the vector, Fock
and MacMahon representations, respectively. Our first task is to construct the
shift operator which generates the shift of the spectral parameter;

$p^{z\partial_{z}}\Psi_{\alpha}(z)=\mathcal{T}_{\alpha}^{-}(\mathfrak{q}^{-3/2}z)\cdot\Psi_{\alpha}(z)\cdot\mathcal{T}_{\alpha}^{+}(\mathfrak{q}^{-1/2}z).$
(1.5)

This is a quantum version of the Sugawara construction
$L_{-1}\sim\sum_{n}:J_{n}J_{-n-1}:$ for the classical KZ equation. One of the
technical problems in deriving the generalized KZ equation is the construction
of such a shift operator. In [9] and [10], the shift operator is identified
with the composition of the intertwiner and the dual intertwiner by tuning
their spectral parameters appropriately. Remarkably the shift parameter is
fixed to be $p=\mathfrak{q}^{-2}=q_{1}q_{2}$. It turns out that up to the
normalization, the same shift operator is obtained in terms of the universal
$R$ matrix $\mathcal{R}_{0}$ of DIM algebra (see section 2.4);

$\mathcal{T}^{+}=[\rho^{V}\otimes\rho^{H}](\mathcal{R}_{0}),\qquad\mathcal{T}^{-}=[\rho^{H}\otimes\rho^{V}](\mathcal{R}_{0}),$
(1.6)

where $\rho^{V}$ and $\rho^{H}$ are the vertical and the horizontal
representations, respectively (see section 3).

The second ingredient is the commutation relations of the intertwiners and the
shift operator;

$\mathcal{T}_{\alpha}^{\pm}(z)\Psi_{\beta}(w)=\mathcal{R}_{\beta\alpha}\left(\mathfrak{q}^{n}\frac{z}{w}\right)\Psi_{\beta}(w)\mathcal{T}_{\alpha}^{\pm}(z),$
(1.7)

where $\mathcal{R}_{\beta\alpha}$ is the (diagonal) $R$ matrix of DIM algebra
and the power $n$ of $\mathfrak{q}$ depends on $\mathcal{T}_{\alpha}^{\pm}$.
We also have a similar commutation relation for the dual intertwiner
$\Psi_{\beta}^{*}(w)$, where we should change the power $n$ appropriately. The
relation (1.7) replaces the commutation (OPE) relations of the vertex
operators with the current algebra in the classical case. The $R$ matrix
appearing in (1.7) can be identified with

$\mathcal{R}=[\rho^{V_{1}}\otimes\rho^{V_{2}}](\mathcal{R}_{0}),$ (1.8)

which implies that $\mathcal{R}_{\beta\alpha}$ appears in the commutation
relation of the intertwiners [8] [11]. If we define the shift operator by the
composition of $\Psi$ and $\Psi^{*}$, (1.7) follows from this property. On the
other hand, the commutation relation (1.7) also comes from the fact that both
the shift operator and the $R$ matrix are derived from the same object; the
universal $R$ matrix.

After all these are prepared, the shift of the spectral parameter of the
intertwiners is achieved by an insertion of a pair of the shift operators
$\mathcal{T}_{\alpha}^{\pm}(z)$. Then we can use the commutation relations to
move them to the left or the right most position of the strip of the
intertwiners, where they act on the vacuum state. In this precess a product of
$R$ matrices is produced. Since the shift operators
$\mathcal{T}_{\alpha}^{\pm}(z)$ have only the positive or the negative modes
of free bosons, the vacuum is their eigenstate and consequently we obtain a
difference equation for the correlation function of the intertwiners, which we
identify with a generalized KZ equation for DIM algebra.

###  1.2. Solutions to the generalized KZ equation

For simplicity, let us assume the correlation function does not involve the
dual intetwiners111For the case with the dual intertwiners, see the main text
for detail.. In this case our generalized KZ equation for DIM algebra
canonically takes the following form

$\langle\Psi_{\alpha_{1}}(z_{1})\cdots\Psi_{\alpha_{k}}(pz_{k})\cdots\Psi_{\alpha_{n}}(z_{n})\rangle=A_{k}\cdot\langle\Psi_{\alpha_{1}}(z_{1})\cdots\Psi_{\alpha_{k}}(z_{k})\cdots\Psi_{\alpha_{n}}(z_{n})\rangle,$
(1.9)

where

$A_{k}=\prod_{i<k}R_{\alpha_{i}\alpha_{k}}(p^{-1}z_{i}/z_{k})^{-1}\cdot\prod_{k<j}R_{\alpha_{k}\alpha_{j}}(z_{k}/z_{j}).$
(1.10)

As mentioned before the shift parameter is $p=\mathfrak{q}^{-2}=q_{1}q_{2}$,
which is fixed by the choice of the (horizontal) Fock space. This is
schematically the same form as the q-KZ equation for the quantum affine
algebra [27] [46] [31] [19]. Note that, contrary to the case of the quantum
affine algebra, the $R$ matrices appearing in Eq.(1.9) is diagonal with
respect to the label $\alpha$ of basis of the vertical (or evaluation)
representation and their ordering does not matter. Due to such an abelian
nature of the $R$ matrix, the solutions to our KZ equation are factorized into
the ratio of fundamental building blocks which are given by two point
functions. We find the two point functions of MacMahon KZ equation can be
regarded as a generalization of the Nakrasov factor (1.2).

###  1.3. Organization of material

The paper is organized as follows; Sections 2 – 4 are preliminaries. We start
by reviewing the definition of DIM algebra and some of its properties in
section 2. We introduce the universal $R$ matrix of DIM algebra. Then, in
section 3, we concern with the representations of DIM algebra. The coproduct
of DIM algebra is crucial for defining the representation by the tensor
product. We show that we can construct the vertical Fock representation from
the vector representations which are the simplest vertical representation of
DIM algebra. By applying a similar method, we can construct the so-called
MacMahon representation from the Fock representation. We also introduce the
horizontal Fock representation by using the deformed Heisenberg algebra. In
section 4 we give explicit expressions of the trivalent intertwiners, where
vertical representations are vector, Fock, and MacMahon representations,
respectively. We also discuss the dual intertwiners.

Before embarking on the task of derivation of the MacMahon KZ equation, we
discuss the vector KZ equation in section 5, since basic ideas are well-
illustrated in a simplified setting. Following the method of the paper [9], we
construct the shift operator $\mathcal{T}$ which plays an important role in
the derivation. We also show an alternative way to construct the shift
operator $\mathcal{T}$ from the universal $R$-matrix, which is useful in our
construction of the shift operator in MacMahon case.

Sections 6 and 7 are the main part of the paper. In section 6 we use the
method discussed in section 5 to construct the shift operator and provide a
derivation of the MacMahon KZ equation. Finally, in section 7, we solve the
MacMahon KZ equation and show that the solution can be regarded as a
generalized Nekrasov function. Some of the techinical computation are
presented in Appendix.

###  1.4. Definitions and useful formulas

Before ending the introduction, let us introduce the theta function and also a
well-known lemma.

The theta function $\theta_{p}(z)$ is defined by

$\displaystyle\theta_{p}(z):=(p;p)_{\infty}(z;p)_{\infty}(pz^{-1};p)_{\infty}=(1-z)\prod_{k=1}^{\infty}(1-p^{k})(1-p^{k}z)(1-p^{k}z^{-1}),$
(1.11)

where we use the infinite product;

$\big{(}x;p\big{)}_{\infty}:=\prod\limits_{j\geq
0}(1-p^{j}x)=\exp\left(-\sum_{k=1}^{\infty}\frac{x^{k}}{k(1-p^{k})}\right).$
(1.12)

It is easy to check

$\theta_{p}(p^{n}z)=(-z)^{n}p^{-\frac{1}{2}n(n-1)}\theta_{p}(z).$ (1.13)

Finally, we state a particular case of the Campbell-Baker-Hausdorff (CBH)
formula.

######  Lemma 1.1.

If $[A,B]$ is central, i.e. $[[A,B],A]=[[A,B],B]=0$, then

$e^{A}e^{B}=e^{B}e^{A}e^{[A,B]}.$ (1.14)

This formula is repeatedly used for computing the commutation relations of
intertwining operators or the vertex operators.

##  2\. Ding-Iohara-Miki algebra

In this section we provide a quick review of the definition of Ding-Iohara-
Miki (DIM) algebra, which is the quantum toroidal algebra of
$\mathfrak{gl}_{1}$. The material in this and the next section is based on the
papers [20] [11].

###  2.1. Definition of DIM algebra

The intriguing triality of DIM algebra becomes manifest by using the
parameters $(q_{1},q_{2},q_{3})$ with $q_{1}q_{2}q_{3}=1$. We assume they are
generic in the sense that for any $a,b,c\in\mathbb{Z}$,

$\displaystyle q^{a}_{1}q^{b}_{2}q^{c}_{3}=1\implies a=b=c.$ (2.1)

We may parametrize $(q_{1},q_{2},q_{3})$ by
$\mathfrak{q},\mathfrak{d}\in\mathbb{C}$ as follows;

$\displaystyle q_{1}=\frac{\mathfrak{d}}{\mathfrak{q}},\quad
q_{2}=\frac{1}{\mathfrak{d}\mathfrak{q}},\quad q_{3}=\mathfrak{q}^{2}.$ (2.2)

But note that this parametrization breaks the triality of DIM algebra. We
define the DIM algebra
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$ to be
the associative algebra with the generators
$E_{k},F_{k},K_{0}^{\pm},H_{r}\,\,(k\in\mathbb{Z},r\in\mathbb{Z}\backslash\\{0\\})$
and $C$. Introducing the generating functions (currents)222 Compared with
[11], the normalization of $H_{\pm r}$ is changed by the factor
$(\mathfrak{q}-\mathfrak{q}^{-1})$.;

$\displaystyle E(z)=\sum_{k\in\mathbb{Z}}E_{k}z^{-k},\quad
F(z)=\sum_{k\in\mathbb{Z}}F_{k}z^{-k},\quad
K^{\pm}(z)=K_{0}^{\pm}\exp\left(\pm\sum_{r=1}^{\infty}H_{\pm r}z^{\mp
r}\right),$ (2.3)

and the structure function

$\displaystyle g(z,w)$ $\displaystyle:=(z-q_{1}w)(z-q_{2}w)(z-q_{3}w),$ (2.4)

we can write the following defining relations :

$\displaystyle C$ $\displaystyle\text{ is central},$ (2.5) $\displaystyle
K_{0}^{+}K_{0}^{-}$ $\displaystyle=1=K_{0}^{-}K_{0}^{+},$ (2.6) $\displaystyle
K^{\pm}(z)K^{\pm}(w)$ $\displaystyle=K^{\pm}(w)K^{\pm}(z),$ (2.7)
$\displaystyle\frac{g(C^{-1}z,w)}{g(Cz,w)}~{}K^{-}(z)K^{+}(w)$
$\displaystyle=\frac{g(w,C^{-1}z)}{g(w,Cz)}~{}K^{+}(w)K^{-}(z),$ (2.8)
$\displaystyle g(z,w)~{}K^{\pm}(C^{(1\mp 1)/2}z)E(w)$
$\displaystyle+g(w,z)~{}E(w)K^{\pm}(C^{(1\mp 1)/2}z)=0,$ (2.9) $\displaystyle
g(w,z)~{}K^{\pm}(C^{(1\pm 1)/2}z)F(w)$
$\displaystyle+g(z,w)~{}F(w)K^{\pm}(C^{(1\pm 1)/2}z)=0,$ (2.10)
$\displaystyle\left[E(z),F(w)\right]=\tilde{g}$
$\displaystyle\left(\delta(C\frac{w}{z})~{}K^{+}(z)-\delta(C\frac{z}{w})~{}K^{-}(w)\right),$
(2.11) $\displaystyle g(z,w)~{}E(z)E(w)$ $\displaystyle+g(w,z)~{}E(w)E(z)=0,$
(2.12) $\displaystyle g(w,z)~{}F(z)F(w)$ $\displaystyle+g(z,w)~{}F(w)F(z)=0,$
(2.13)

where the multiplicative delta function is defined by

$\delta(z)=\sum_{n\in\mathbb{Z}}z^{n}.$ (2.14)

It is convenient to employ the notation

$\displaystyle\kappa_{n}$
$\displaystyle:=\prod_{i=1}^{3}(q_{i}^{\frac{n}{2}}-q_{i}^{-\frac{n}{2}})=\prod_{i=1}^{3}(q_{i}^{n}-1)=\prod_{i=1}^{3}(1-q_{i}^{-n})=\sum_{i=1}^{3}(q_{i}^{n}-q_{i}^{-n}),$
(2.15)

which satisfies $\kappa_{-n}=-\kappa_{n}$. We choose the normalization of
(2.11) as $\tilde{g}=\kappa_{1}^{-1}$.

######  Remark 2.1.

One should bear in mind that actually there are the Serre’s relations in the
defining relations. Since we do not use the Serre’s relations in this paper,
we do not write it out.

By investigating the defining relations we see that $K_{0}^{\pm}$ is
central333 In [23] $K_{0}^{-}$ is denoted by $C^{\perp}$.. Thus we conclude
that the DIM algebra has two-dimensional center $(C,K_{0}^{\pm})$. Note that
$K_{0}^{+}$ is the inverse of $K_{0}^{-}$. We also see that (2.8) implies

$\displaystyle[H_{r},H_{s}]$
$\displaystyle=\delta_{r+s,0}\frac{\kappa_{r}}{r}(C^{r}-C^{-r}).$ (2.16)

###  2.2. Coproduct

The DIM algebra has a coproduct
$\Delta:U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})\rightarrow
U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})\otimes
U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$ defined
by

$\displaystyle\Delta(E(z))$ $\displaystyle=E(z)\otimes 1+K^{-}(C_{1}z)\otimes
E(C_{1}z),$ (2.17) $\displaystyle\Delta(F(z))$ $\displaystyle=F(C_{2}z)\otimes
K^{+}(C_{2}z)+1\otimes F(z),$ (2.18) $\displaystyle\Delta(K^{+}(z))$
$\displaystyle=K^{+}(z)\otimes K^{+}(C_{1}^{-1}z),$ (2.19)
$\displaystyle\Delta(K^{-}(z))$ $\displaystyle=K^{-}(C_{2}^{-1}z)\otimes
K^{-}(z),$ (2.20) $\displaystyle\Delta(C)$ $\displaystyle=C\otimes C,$ (2.21)

where $C_{1}=C\otimes 1$ and $C_{2}=1\otimes C$.

We will see that the coproduct structure plays an important role in defining
the tensor product representation of DIM. However, we would like to give a
remark here that $\Delta$ is not a coproduct in the strict sense. The reason
is that it contains an infinite summation of elements which is not defined in
general. Hence, when we use $\Delta$ for defining the action of tensor product
representation, we have to check every time that the action is well-defined.

###  2.3. Grading operators

The DIM algebra has a bi-grading defined by two grading operators
$d_{1},d_{2}$ which satisfy

$\displaystyle[d_{1},E(z)]=-E(z),\qquad[d_{1},F(z)]=F(z),\qquad[d_{1},H(z)]=0,$
(2.22)
$\displaystyle[d_{2},E(z)]=z\partial_{z}E(z),\qquad[d_{2},F(z)]=z\partial_{z}F(z),\qquad[d_{2},H(z)]=z\partial_{z}H(z).$
(2.23)

In [23] the degrees with respect to $d_{1}$ and $d_{2}$ are called principal
degree and homogeneous degree, respectively. Hence, the degree of the vertical
spectral parameter $z$ counts the $d_{2}$-grading and the generators
$E_{k},F_{k}$ and $H_{r}$ have gradings $(-1,k),(1,k)$ and $(0,r)$,
respectively. The generators with higher $d_{1}$-grading are given by multiple
commutators of $E_{k}$ and $F_{k}$. Later we will introduce the horizontal
spectral parameter $u$ which counts the $d_{1}$-grading. The
$SL(2,\mathbb{Z})$ automorphism of the DIM algebra acts on this bi-grading
[34]. The grading operators are important in the expression of the universal
$R$-matrix which is the main tool for deriving the generalized Knizhnik-
Zamolodchikov equation for MacMahon intertwiner.

###  2.4. Universal $R$ matrix

The quantum toroidal algebra allows a quantum (Drinfeld) double construction
[17]. Consequently it has a quasi-triangular structure, which implies the
existence of a universal $R$ matrix. According to [23] (see also [39]), the
universal $R$-matrix $\mathcal{R}$ of DIM algebra factorizes as follows;

$\mathcal{R}=\mathfrak{q}^{c^{\perp}\otimes d^{\perp}+d^{\perp}\otimes
c^{\perp}}\mathcal{R}_{+}\mathcal{R}_{0}\mathcal{R}_{-},$ (2.24)

where $\mathfrak{q}^{c^{\perp}}=K_{0}^{-}$ and $d^{\perp}=d_{1}$ (the grading
operator for the principal degree). What is most relevant in the present paper
is the Cartan factor of $\mathcal{R}$444The definition of $\kappa_{n}$ in this
paper is $-\kappa_{n}$ in [23].;

$\displaystyle\mathfrak{q}^{c^{\perp}\otimes d^{\perp}+d^{\perp}\otimes
c^{\perp}}\cdot\mathcal{R}_{0}=(K_{0}^{-}\otimes\mathfrak{q}^{d_{1}})(\mathfrak{q}^{d_{1}}\otimes
K_{0}^{-})\exp\big{\\{}\sum_{n=1}^{\infty}n\kappa_{n}h_{-n}\otimes
h_{n}\big{\\}},$ (2.25)

where $h_{\pm n}$ is defined via $\kappa_{n}h_{\pm n}=\pm H_{\pm n}$. Note
that the universal $R$ matrix we use in the present paper is $P\mathcal{R}$
with $P(a\otimes b)=b\otimes a$ in [23].

It is known that DIM algebra acts on the equivariant $K$-theory ;
$\displaystyle{\oplus_{n=0}^{\infty}}K_{G}(\mathrm{Hilb}^{n}(\mathbb{C}^{2}))$
of the Hilbert schemes of $n$ points on $\mathbb{C}^{2}$. Hence, we can also
define the $R$-matrix by using the ideas coming from geometry [33]. It is
interesting to find that the $R$-matrix
$[\rho^{V_{1}}\otimes\rho^{V_{2}}](\mathcal{R}_{0})$ featured in our
generalized KZ equation (1.9) coincides with the infinite slope $R$-matrix
$R_{\infty}$ which is ubiquitous in the Khoroshkin-Tolstoy factorization of
the slope $s$ $R$-matrix introduced in [43]. As noticed in [43] $R_{\infty}$
corresponds to multiplication by a class of normal bundles in $K$-theory and
is diagonal in the fixed point basis of the torus action. It is an intriguing
challenge to work out a possible link of the quantum difference equation (1.9)
to those in [43].

##  3\. Representations of DIM algebra

In this section we concern with the representation of DIM algebra. Note that
in this paper we use the word representation and module interchangeably. We
start this section with the definition of level.

######  Definition 3.1 (Level).

Let $V$ be a representation of DIM algebra
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$. We say
that the representation $V$ is of level
$(\gamma_{1},\gamma_{2})\in\mathbb{C}^{2}$ if $C$ and $K_{0}^{-}$ act as
constant multiplications by $\gamma_{1}$ and $\gamma_{2}$, respectively.

We will call a representation with $\gamma_{1}=1$ vertical representation.
Note that the condition $\gamma_{1}=1$ is kept intact under taking the tensor
product. By (2.16) $H_{r}$ are mutually commuting for the vertical
representation. Hence the vertical representation allows a basis which
simultaneously diagonalizes the Cartan modes $H_{r}$. There are three natural
vertical representations of DIM algebra; vector, Fock and MacMahon
representations. A basis which diagonalizes $H_{r}$ is labeled by 1d, 2d and
3d Young diagrams.

###  3.1. Vector representations

We start with the vector representation which is considered as the simplest
vertical representation. Though DIM algebra is completely symmetric in
parameters $(q_{1},q_{2},q_{3})$, the symmetry is broken at the level of
representation in general. In order to define the vector representation we
have to choose one of three parameters as prefered. Accordingly there are
three kinds of vector representations $\rho^{V^{(k)}}~{}(k=1,2,3)$ [51]. For a
parameter $v\in\mathbb{C}$ we consider a vector space $V(v)$ over $\mathbb{C}$
with a basis $\\{[v]_{i}\big{|}\,i\in\mathbb{Z}\\}$. The vector representation
$V^{(k)}$ is defined as follows :

$\displaystyle K^{+}(z)[v]_{i}$
$\displaystyle=\tilde{\psi}_{k}(q_{k}^{i}v/z)[v]_{i},$ (3.1) $\displaystyle
K^{-}(z)[v]_{i}$ $\displaystyle=\tilde{\psi}_{k}(q_{k}^{-i-1}z/v)[v]_{i},$
(3.2) $\displaystyle E(z)[v]_{i}$
$\displaystyle=(1-q_{k})^{-1}~{}\delta(q_{k}^{i+1}v/z)[v]_{i+1},$ (3.3)
$\displaystyle F(z)[v]_{i+1}$
$\displaystyle=(1-q_{k}^{-1})^{-1}~{}\delta(q_{k}^{i+1}v/z)[v]_{i}.$ (3.4)

Here

$\displaystyle\tilde{\psi}_{1}(z)=\frac{(1-q_{2}^{-1}z)(1-q_{3}^{-1}z)}{(1-z)(1-q_{1}z)},$
(3.5)

and $\tilde{\psi}_{2}(z)$ and $\tilde{\psi}_{3}(z)$ are defined by the cyclic
permutation of $(q_{1},q_{2},q_{3})$. The original vector representation in
[20] is $\rho^{V^{(1)}}$. In the following we will choose the same one and
simply denote it by $\rho^{V}$. If we introduce
$\displaystyle\psi(z)=\mathfrak{q}\frac{1-q^{-1}_{3}z}{1-z}$, then we can
express the $\tilde{\psi}_{1}(z)$ as555 Note that we can exchange $q_{2}$ and
$q_{3}$.

$\displaystyle\tilde{\psi}_{1}(z)=\psi(z)\psi(q^{-1}_{2}z)^{-1}.$ (3.6)

It is $\psi(z)$ rather than $\tilde{\psi}_{1}(z)$ which plays a main role
later, for example see Eqs.(3.22) and (3.23) in section 3.3. It is
straightforward to show that $V(v)$ with the above action really forms an
irreducible representation with level $(1,1)$.

######  Remark 3.2.

We give two remarks here.

  1. (i)

In the equations (3.1) – (3.4), $\tilde{\psi}$ and $\delta$ are formal power
series.

  2. (ii)

The vector representation is not a highest-weight representation.

###  3.2. Tensor product representation of the vector representation

Now perform the tensor product

$\displaystyle V(v_{1})\otimes V(v_{2})\otimes\cdots\otimes V(v_{n}).$ (3.7)

So we obtain a representation of
$\underbrace{U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})\otimes\cdots\otimes
U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})}_{\text{
n times}}$. By using $\Delta^{n-1}$,

$\displaystyle\Delta^{n-1}(K^{\pm}(z))$
$\displaystyle=\overbrace{K^{\pm}(z)\otimes\cdots\otimes K^{\pm}(z)}^{n},$
(3.8) $\displaystyle\Delta^{n-1}(E(z))$
$\displaystyle=\sum_{k=1}^{n}\overbrace{K^{-}(z)\otimes\cdots
K^{-}(z)}^{k-1}\otimes E(z)\otimes\overbrace{1\otimes\cdots\otimes 1}^{n-k},$
(3.9) $\displaystyle\Delta^{n-1}(F(z))$
$\displaystyle=\sum_{k=1}^{n}\overbrace{1\otimes\cdots\otimes 1}^{k-1}\otimes
F(z)\otimes\overbrace{K^{+}(z)\otimes\cdots\otimes K^{+}(z)}^{n-k}.$ (3.10)

we obtain a representation of
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$.

However, there are two issues here. First, we see from Eq.(3.9) that the
action of $E(z)$ results in a product of the formal power series
$\tilde{\psi}$ and $\delta$ which is not defined in general, since it contains
an infinite summation of elements in we obtain a representation of
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$. The
solution to this issue is to perform a regularization. More precisely, in the
tensor product representation we treat $\tilde{\psi}$ to be a function instead
of formal power series.

Second, after regularization, $\tilde{\psi}$ is a function. From Eq.(3.5), we
see that it contains poles. If we choose $v_{1},v_{2},\dots,v_{n}$ not
carefully, then it might hit the poles generated from $\tilde{\psi}$. This
means that the parameters $v_{1},v_{2},\dots,v_{n}$ can not be chosen
arbitrarily. Then, it is natural to ask what is a condition for
$v_{1},v_{2},\dots,v_{n}$ to assure that the action defined by $\Delta^{n-1}$
on $V(v_{1})\otimes V(v_{2})\otimes\cdots\otimes V(v_{n})$ does not hit the
poles. An answer is given in the following lemma which we take from the paper
[20].

######  Lemma 3.3.

If the parameters $v_{1},v_{2},\dots,v_{n}\in\mathbb{C}$ satisfy the condition
that for any $1\leq i<j\leq n$,

$\displaystyle\frac{v_{j}}{v_{i}}\neq q^{k}_{1}\quad\forall k\in\mathbb{Z},$

then $V(v_{1})\otimes V(v_{2})\otimes\cdots\otimes V(v_{n})$ is a well-defined
representation of DIM algebra.

As a consequence of Lemma 3.3, we get that the representation $V^{n}(v)$
defined by

$\displaystyle V^{n}(v)=V(v)\otimes V(q_{2}v)\otimes\cdots\otimes
V(q^{n-1}_{2}v)$ (3.11)

is well-defined. For each
$\lambda=(\lambda_{1},\dots,\lambda_{n})\in\mathbb{Z}^{n}$, we define the
state

$\displaystyle|\lambda\rangle:=[v]_{\lambda_{1}-1}\otimes[q_{2}v]_{\lambda_{2}-1}\otimes\cdots\otimes[q^{n-1}_{2}v]_{\lambda_{n}-1}.$
(3.12)

It is easy to see that
$\\{|\lambda\rangle\big{|}\,\lambda=(\lambda_{1},\dots,\lambda_{n})\in\mathbb{Z}^{n}\\}$
forms a basis of $V^{n}(v)$.

Note that the representation $V^{n}(v)$ is reducible since it contains a
nonzero proper DIM-submodule $W^{n}(v)$ which is defined by

$\displaystyle
W^{n}(v)=\operatorname{span}_{\mathbb{C}}\\{|\lambda\rangle\big{|}\,\lambda\in\mathcal{P}^{n}\\}.$
(3.13)

Here

$\displaystyle\mathcal{P}^{n}=\\{\lambda=(\lambda_{1},\cdots,\lambda_{n})\in\mathbb{Z}^{N}\big{|}\lambda_{1}\geq\cdots\geq\lambda_{n}\\}.$
(3.14)

###  3.3. Fock representation

Once we have constructed the $n$-th tensor product representation $V^{n}(v)$
for any $n\in\mathbb{Z}^{\geq 1}$, it is natural to think about the
representation of DIM algebra which collects all of the
$V^{n}(v)\,\,(n\in\mathbb{Z}^{\geq 1})$ together under an identification

$\displaystyle|(\lambda_{1},\dots,\lambda_{n})\rangle\sim|(\lambda_{1},\dots,\lambda_{n},0)\rangle\sim|(\lambda_{1},\dots,\lambda_{n},0,0)\rangle\sim\cdots$
(3.15)

and so on. This recalls us the notion of inductive limit. Of course, to
perform the inductive limit we need to have the inductive system in our hand
first.

Here we construct the inductive system. First we define $W^{n,+}(v)$ to be the
subspace

$\displaystyle\operatorname{span}_{\mathbb{C}}\\{|\lambda\rangle\big{|}\,\lambda\in\mathcal{P}^{n,+}\\}\subseteq
V^{n}(v)$ (3.16)

where
$\mathcal{P}^{n,+}=\\{\lambda=(\lambda_{1},\cdots,\lambda_{n})\in\mathbb{Z}^{N}\big{|}\lambda_{1}\geq\cdots\geq\lambda_{n}\geq
0\\}$. Then we construct the inductive system of vector spaces.

$\displaystyle
W^{1,+}(v)\xrightarrow{\tau_{1}}W^{2,+}(v)\xrightarrow{\tau_{2}}W^{3,+}(v)\cdots$
(3.17)

where $\tau_{n}:W^{n,+}(v)\rightarrow W^{n+1,+}(v)$ sends
$|(\lambda_{1},\dots,\lambda_{n})\rangle$ to
$|(\lambda_{1},\dots,\lambda_{n},0)\rangle$. Then, we take the inductive limit
of the above inductive system.

$\displaystyle\mathcal{F}(v)=\mathop{\lim_{\longrightarrow}}W^{n,+}(v).$
(3.18)

Being an inductive limit, the vector space $\mathcal{F}(v)$ is spanned by
$\\{|\lambda\rangle\big{|}\,\lambda\in\mathcal{P}^{+}\\}$ where

$\displaystyle\mathcal{P}^{+}=\\{\lambda=(\lambda_{1},\lambda_{2},\dots)\big{|}\,\lambda_{i}\geq\lambda_{i+1},\lambda_{i}\in\mathbb{Z},\lambda_{i}=0\text{
for sufficiently large }i\\}.$ (3.19)

The next task is to endow the structure of
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$-module
on $\mathcal{F}(v)$. From Eq.(3.18), we know that $\mathcal{F}(v)$ is the
disjoint union of $W^{n,+}(v)\,\,(n\in\mathbb{Z}^{\geq 1})$ modulo the
identification $\sim$ in (3.15). Accordingly, it is natural to use the
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$-module
structure on each $W^{n,+}$ to construct the
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$-module
structure on $\mathcal{F}(v)$. More precisely, if a partition
$\lambda=(\lambda_{1},\lambda_{2},\dots,\lambda_{l},0,0,\dots)$, then we
regard $|\lambda\rangle$ as an element of $V^{l}(v)$ and use the
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$-module
structure of $V^{l}(v)$.

Unfortunately, the above action is not compatible with the inductive system
(3.17). That is, if we denote the representation $V^{l}$ by
$\rho^{l}:U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})\rightarrow\mathop{\rm
End}V^{l}$ and if
$\lambda=(\lambda_{1},\lambda_{2},\dots,\lambda_{l},0,0,\dots)$, then we find
that

$\displaystyle\rho^{l}\big{(}K^{\pm}(z)\big{)}|(\lambda_{1},\lambda_{2},\dots,\lambda_{l})\rangle\neq\rho^{l+1}\big{(}K^{\pm}(z)\big{)}|(\lambda_{1},\lambda_{2},\dots,\lambda_{l},0)\rangle,$
(3.20)

and

$\displaystyle\rho^{l}\big{(}F(z)\big{)}|(\lambda_{1},\lambda_{2},\dots,\lambda_{l})\rangle\neq\rho^{l+1}\big{(}F(z)\big{)}|(\lambda_{1},\lambda_{2},\dots,\lambda_{l},0)\rangle.$
(3.21)

To make it compatible with the inductive system, we need to define the action
$\bar{\rho}:U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})\rightarrow\operatorname{End}\big{(}\mathcal{F}(v)\big{)}$
as follows : for each partition
$\lambda=(\lambda_{1},\lambda_{2},\dots,\lambda_{l},0,0,\dots)$ we determine

$\displaystyle\bar{\rho}\big{(}K^{\pm}(z)\big{)}|\lambda\rangle=\beta^{\pm}_{l+1}((v/z)^{\pm})\cdot\rho^{l+1}\big{(}K^{\pm}(z)\big{)}|\lambda\rangle,$
$\displaystyle\bar{\rho}\big{(}E(z)\big{)}|\lambda\rangle=\rho^{l+1}\big{(}E(z)\big{)}|\lambda\rangle,$
$\displaystyle\bar{\rho}\big{(}F(z)\big{)}|\lambda\rangle=\beta^{+}_{l+1}(v/z)\rho^{l+1}\big{(}F(z)\big{)}|\lambda\rangle,$
(3.22)

where

$\displaystyle\beta^{+}_{l}(v/z)=\psi(q^{-1}_{1}q^{l-1}_{2}v/z)^{-1}\quad\quad\beta^{-}_{l}(z/v)=\psi(q^{-l}_{2}z/v).$
(3.23)

This action is well-defined and compatible with the structure of inductive
system. This is assured by the following theorem.

######  Theorem 3.4.

Let $\lambda=(\lambda_{1},\lambda_{2},\dots,\lambda_{l},0,0,\dots)$ be a
partition. Then, for any $k\in\mathbb{Z}^{\geq 1}$

$\displaystyle\bar{\rho}\big{(}K^{\pm}(z)\big{)}|\lambda\rangle=\beta^{\pm}_{l+k}((v/z)^{\pm})\cdot\rho^{l+k}\big{(}K^{\pm}(z)\big{)}|\lambda\rangle,$
$\displaystyle\bar{\rho}\big{(}E(z)\big{)}|\lambda\rangle=\rho^{l+k}\big{(}E(z)\big{)}|\lambda\rangle,$
$\displaystyle\bar{\rho}\big{(}F(z)\big{)}|\lambda\rangle=\beta^{+}_{l+k}(v/z)\rho^{l+k}\big{(}F(z)\big{)}|\lambda\rangle.$
(3.24)

Thus, we have equipped a
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$-module
structure to $\mathcal{F}(v)$. We call
$\big{(}\bar{\rho}:U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})\rightarrow\mathcal{F}(v),\mathcal{F}(v)\big{)}$
the Fock representation of DIM algebra. The Fock representation is irreducible
representation of level $(1,\mathfrak{q})$ [20] [11].

###  3.4. MacMahon representation

Up to now we have constructed the Fock representation from the vector
representations. Now we construct the MacMahon representation from the Fock
representations by a similar process used in section 3.3.

Analogous to Eq.(3.11), we define

$\displaystyle\mathcal{F}^{n}(v)=\mathcal{F}(v)\otimes\mathcal{F}(q_{3}v)\otimes\cdots\otimes\mathcal{F}(q_{3}^{n-1}v).$
(3.25)

It is clear that the subset

$\displaystyle\\{|\Lambda\rangle:=|\Lambda^{(1)}\rangle\otimes\cdots\otimes|\Lambda^{(n)}\rangle\big{|}\,\Lambda^{(1)},\dots,\Lambda^{(n)}\in\mathcal{P}^{+}\\}$
(3.26)

forms a basis of $\mathcal{F}^{n}(v)$. Note that sometimes we write
$|\Lambda^{(1)},\dots,\Lambda^{(n)}\rangle$ for
$|\Lambda^{(1)}\rangle\otimes\cdots\otimes|\Lambda^{(n)}\rangle$.

Next we construct the subspace $\mathcal{M}^{n}(v)$ of $\mathcal{F}^{n}(v)$
spanned by plane partitions, i.e. the
$|\Lambda^{(1)},\dots,\Lambda^{(n)}\rangle$ which satisfy the condition

$\displaystyle\Lambda^{(k)}_{i}\geq\Lambda^{(k+1)}_{i}\quad\quad\forall i,k.$
(3.27)

This step is analogous to Eq.(3.16). Then, we collect these subspaces
$\mathcal{M}^{n}(v)$ together by running $n$ over $\mathbb{Z}^{\geq 1}$, and
then form an inductive system as (3.17).

Now we would like to endow a structure of
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$-module
to the vector space $\varinjlim_{n}\mathcal{M}^{n}(v)$ which is the inductive
limit of the above-mentioned inductive system. As usual we first try to use
the action as in the representation $\mathcal{F}^{n}(v)$. Again the problem
arises : the action is not compatible with the structure of the inductive
system. So we need a modification.

To make it compatible with the inductive system, we need to define the action
$\bar{\varrho}:U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})\rightarrow\operatorname{End}\big{(}\varinjlim_{n}\mathcal{M}^{n}(v)\big{)}$
as follows : for each 3d partition
$\Lambda=\big{(}\Lambda^{(1)},\dots,\Lambda^{(l)},0,0,\dots\big{)}$ we
determine

$\displaystyle\bar{\varrho}\big{(}K^{\pm}(z)\big{)}|\Lambda\rangle=\gamma^{\pm}_{l+1}((v/z)^{\pm})\cdot\varrho^{l+1}\big{(}K^{\pm}(z)\big{)}|\Lambda\rangle,$
$\displaystyle\bar{\varrho}\big{(}E(z)\big{)}|\Lambda\rangle=\varrho^{l+1}\big{(}E(z)\big{)}|\Lambda\rangle,$
$\displaystyle\bar{\varrho}\big{(}F(z)\big{)}|\Lambda\rangle=\gamma^{+}_{l+1}(v/z)\cdot\varrho^{l+1}\big{(}F(z)\big{)}|\Lambda\rangle,$
(3.28)

where

$\displaystyle\gamma^{+}_{l}(v/z)=\frac{K^{-1/2}(1-Kv/z)}{\mathfrak{q}^{-l}(1-q_{3}^{l}v/z)},\quad\quad\gamma^{-}_{l}(z/v)=\frac{K^{1/2}\big{(}1-\frac{z}{Kv}\big{)}}{\mathfrak{q}^{l}\big{(}1-\frac{z}{q^{l}_{3}v}\big{)}}.$
(3.29)

Here $K$ is an arbitrary parameter which arises by the prescription of making
the inductive system consistent [20]. The appearance of this continuous
parameter $K$ is one of the most intriguing aspects of the MacMahon
representation.

It is straightforward to check that the action $\bar{\varrho}$ is well-defined
and compatible with the structure of the inductive system. Thus we have
equipped the
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$-module
structure to $\varinjlim_{n}\mathcal{M}^{n}(v)$. We call
$\big{(}\bar{\varrho}:U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})\rightarrow\mathop{\rm
End}(\varinjlim_{n}\mathcal{M}^{n}(v)),\varinjlim_{n}\mathcal{M}^{n}(v)\big{)}$
the MacMahon module. It is easy to see that the MacMahon module has level
$(1,K^{1/2})$. Because of this, from now on we denote it by
$\mathcal{M}(K;v)$. As was shown in [22] [15], when
$K=q_{1}^{a}q_{2}^{b}q_{3}^{c},~{}a,b,c\in\mathbb{Z}_{\geq 0}$, the MacMahon
representation is reducible. We can reduce the representation space to that
spanned by plane partitons with a pit at $(a+1,b+1,c+1)$. In particular for
$K=q_{3}$ the pit is at $(1,1,2)$ and the plane partition has only the first
layer. Thus the representation is reduced to the Fock representation.

In the cases of the vector and the Fock representations, one of the parameters
$(q_{1},q_{2},q_{3})$ of the DIM algebra plays a distinguished role.
Consequently there are three kinds of the vector and the Fock
representations666The conventional choice, which we follow in this paper, is
to choose $q_{1}$ for the vertical representation and $q_{3}$ for the Fock
representation. For the existence of the intertwiner we have to choose the
parameters of the vector and the Fock representations differently.. On the
other hand in the MacMahon representation three parameters are treated as an
equal footing and the triality of DIM algebra is manifest.

###  3.5. Horizontal Fock representation

Up to now we only discussed about vertical representations. In this subsection
we construct a horizontal representation of DIM algebra with $C\neq 1$. As in
the case of the vector and the vertical Fock representations, there are three
kinds of horizontal Fock representations, for which $C=q_{k}^{\frac{1}{2}}$.
In the following we fix $k$ and write $\mathfrak{q}=q_{k}^{\frac{1}{2}}$. The
conventional horizontal representation corresponds to the choice $k=3$. When
$C=\mathfrak{q}$, from Eq.(2.16) we obtain the Heisenberg algebra

$\displaystyle[H_{r},H_{s}]$
$\displaystyle=\delta_{r+s,0}\frac{\kappa_{r}}{r}(\mathfrak{q}^{r}-\mathfrak{q}^{-r}).$
(3.30)

There is a well-known representation of Heisenberg algebra whose
representation space $\mathcal{F}$ is the Fock space of a (deformed) free
boson with a creation operators $a_{-r}\,\,(r>0)$ acting on the vacuum state
$|0\rangle$. The vacuum state is annihilated by the annihilation operator
$a_{r}\,\,(r>0)$. Namely, we define the horizontal Fock representation
by777Since the only difference of $H_{r}$ and $a_{r}$ is the normalization, we
will use them interchangeably in this paper.

$\displaystyle\rho_{H}^{(\mathfrak{q},1)}(H_{r}):=\frac{\kappa_{r}}{r}a_{r},\qquad[a_{r},a_{s}]=\delta_{r+s,0}\frac{r}{\kappa_{r}}(\mathfrak{q}^{r}-\mathfrak{q}^{-r}).$
(3.31)

Now we try to endow $\mathcal{F}$ with the
$U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})$-module
structure. We construct the action
$\rho_{H}^{(\mathfrak{q},1)}:U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})\rightarrow\mathop{\rm
End}\mathcal{F}$ by

$\displaystyle\rho_{H}^{(\mathfrak{q},1)}\big{(}E(z)\big{)}=V^{-}(\mathfrak{q}^{-1/2}z)V^{+}(\mathfrak{q}^{1/2}z),$
$\displaystyle\rho_{H}^{(\mathfrak{q},1)}\big{(}F(z)\big{)}=V^{-}(\mathfrak{q}^{1/2}z)^{-1}V^{+}(\mathfrak{q}^{-1/2}z)^{-1},$
$\displaystyle\rho_{H}^{(\mathfrak{q},1)}\big{(}K^{\pm}(\mathfrak{q}^{1/2}z)\big{)}=V^{\pm}(\mathfrak{q}^{\pm
1}z)V^{\pm}(\mathfrak{q}^{\mp 1}z)^{-1},$ (3.32)

where

$\displaystyle
V^{\pm}(z)=\exp\bigg{(}\mp\sum_{r=1}^{\infty}\frac{1}{r}\frac{\kappa_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}a_{\pm
r}z^{\mp r}\bigg{)}.$ (3.33)

It is straightforward to show that
$\big{(}\rho_{H}^{(\mathfrak{q},1)},\mathcal{F}\big{)}$ is a representation of
DIM algebra of level $(\mathfrak{q},1)$.

Now for a given $\gamma_{2}\in\mathbb{C}\backslash\\{0\\}$ we try to construct
a horizontal representation of DIM algebra of level
$(\mathfrak{q},\gamma_{2})$ by generalizing the above formula. While the
representation space is the same as above, say $\mathcal{F}$, the action is
modified to be
$\rho_{H}^{(\mathfrak{q},\gamma_{2})}:U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1})\rightarrow\mathop{\rm
End}\mathcal{F}$ by multiplying the zero mode factors
$\textbf{e}(z),\textbf{f}(z),\textbf{k}^{\pm}(z)$;

$\displaystyle\rho_{H}^{(\mathfrak{q},\gamma_{2})}\big{(}E(z)\big{)}=\rho_{H}^{(\mathfrak{q},1)}\big{(}E(z)\big{)}\textbf{e}(z),$
$\displaystyle\rho_{H}^{(\mathfrak{q},\gamma_{2})}\big{(}F(z)\big{)}=\rho_{H}^{(\mathfrak{q},1)}\big{(}F(z)\big{)}\textbf{f}(z),$
$\displaystyle\rho_{H}^{(\mathfrak{q},\gamma_{2})}\big{(}K^{\pm}(\mathfrak{q}^{1/2}z)\big{)}=\rho_{H}^{(\mathfrak{q},1)}\big{(}K^{\pm}(\mathfrak{q}^{1/2}z)\big{)}\textbf{k}^{\pm}(z)$
(3.34)

where $\textbf{e}(z),\textbf{f}(z),\textbf{k}^{\pm}(z)$ should satisfy the
condition

$\displaystyle\textbf{e}(z)\textbf{f}(\mathfrak{q}^{\mp
1}z)=\textbf{k}^{\pm}(\mathfrak{q}^{\mp 1/2}z),$ (3.35)

and

$\displaystyle\textbf{k}^{\pm}(z)=\textbf{k}^{\pm}(0)=\gamma_{2}^{\mp 1}.$
(3.36)

It is straightforward to check that
$\big{(}\rho_{H}^{(\mathfrak{q},\gamma_{2})},\mathcal{F}\big{)}$ is a
representation of DIM algebra of level $(\mathfrak{q},\gamma_{2})$.

When $\gamma_{2}=\mathfrak{q}^{N}$, by introducing the spectral parameter
$u\in\mathbb{C}$ of the horizontal representation, we can solve the conditions
(3.35) and (3.36) as follows;

$\displaystyle\textbf{e}(z)=\bigg{(}\frac{\mathfrak{q}}{z}\bigg{)}^{N}u,\quad\quad\textbf{f}(z)=\bigg{(}\frac{\mathfrak{q}}{z}\bigg{)}^{-N}u^{-1},\quad\quad\textbf{k}^{\pm}(z)=\mathfrak{q}^{\mp
N},$ (3.37)

which was originally employed in [2]. We denote the representation with this
choice of $\textbf{e}(z),\textbf{f}(z)$ and $\textbf{k}^{\pm}(z)$ by
$\mathcal{F}^{(\mathfrak{q},\mathfrak{q}^{N})}_{u}$, which is a representation
of DIM algebra of level $(\mathfrak{q},\mathfrak{q}^{N})$. Recall that the
MacMahon representation has a continuous parameter $K$ for the second level
$\gamma_{2}$. Hence we can no longer use (3.37) when the MacMahon
representation is involved. This is the reason why we need mode general zero
mode algebra of $\textbf{e}(z),\textbf{f}(z),\textbf{k}^{\pm}(z)$.

##  4\. Trivalent intertwiners

In this section, we give a quick review on the trivalent intertwiners and the
dual intertwiners. A trivalent intertwiner
$\Psi:\mathcal{V}\otimes\mathcal{H}\rightarrow\mathcal{H}^{\prime}$, where
$\mathcal{V}$ is a vertical representation and
$\mathcal{H},\mathcal{H}^{\prime}$ are horizontal representations, is
determined by the intertwining relation

$\displaystyle a\Psi=\Psi\Delta(a)\quad\quad\forall a\in
U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1}).$ (4.1)

Taking a basis $\\{\alpha\\}$ of $\mathcal{V}$ which diagonalizes $H_{r}$, we
define the $\alpha$-component of the intertwiner $\Psi_{\alpha}(\bullet)$ by

$\displaystyle\Psi_{\alpha}(\bullet)=\Psi(\alpha\otimes\bullet):\mathcal{H}\rightarrow\mathcal{H}^{\prime}.$
(4.2)

By using $\Psi_{\alpha}(\bullet)$ we can express the intertwining relation
(4.1) as

$\displaystyle K^{+}(z)\Psi_{\alpha}$
$\displaystyle=(\alpha|K^{+}(z)|\alpha)~{}\Psi_{\alpha}K^{+}(z),$ (4.3)
$\displaystyle K^{-}(\mathfrak{q}z)\Psi_{\alpha}$
$\displaystyle=(\alpha|K^{-}(z)|\alpha)~{}\Psi_{\alpha}K^{-}(\mathfrak{q}z),$
(4.4) $\displaystyle E(z)\Psi_{\alpha}$
$\displaystyle=\sum_{\beta}(\beta|E(z)|\alpha)~{}\Psi_{\beta}+(\alpha|K^{-}(z)|\alpha)~{}\Psi_{\alpha}E(z),$
(4.5) $\displaystyle F(z)\Psi_{\alpha}$
$\displaystyle=\sum_{\beta}(\beta|F(\mathfrak{q}z)|\alpha)~{}\Psi_{\beta}K^{+}(\mathfrak{q}z)+\Psi_{\alpha}F(z).$
(4.6)

When $\mathcal{V}$ is the vertical vector/Fock/MacMahon modules, we call the
intertwiner
$\Psi:\mathcal{V}\otimes\mathcal{H}\rightarrow\mathcal{H}^{\prime}$
vector/Fock/MacMahon intertwiner, respectively. In the following we will
summarize explicit expressions of these intertwiners which were derived in
[11]. We also determine the dual intertwiners (see (4.28) – (4.31) for the
definition) which were not given in [11].

###  4.1. Vector Intertwiner

In terms of the basis $\\{[v]_{n}\big{|}n\in\mathbb{Z}\\}$ of $V(v)$ (see
section 3.1), we define the components of the intertwiner by

$\displaystyle\mathbb{I}_{n}(v)(\bullet)=\mathbb{I}([v]_{n-1}\otimes\bullet)\colon\mathcal{H}\to\mathcal{H}^{\prime}.$
(4.7)

A solution to the intertwining relations is

$\displaystyle\mathbb{I}_{n}(v)$
$\displaystyle=z_{n}\tilde{\mathbb{I}}_{n}(v),\quad\tilde{\mathbb{I}}_{n}(v)=\tilde{\mathbb{I}}_{0}(q_{1}^{n}v),\quad
n\in\mathbb{Z},$ (4.8) $\displaystyle\tilde{\mathbb{I}}_{0}(v)$
$\displaystyle=\exp\left(-\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{\mathfrak{q}^{-r/2}}{1-q_{1}^{r}}v^{r}\right)\exp\left(\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{\mathfrak{q}^{-r/2}}{1-q_{1}^{-r}}v^{-r}\right).$
(4.9)

Here

$\displaystyle z_{0}(v)=1,\quad
z_{n}(v)=q_{2}^{-n}\prod_{j=1}^{n}\mathbf{e}(q_{1}^{j-1}v)\quad(n>0),\quad
z_{n}(v)=q_{2}^{-n}\prod_{j=n}^{-1}\mathbf{e}(q_{1}^{j}v)^{-1}\quad(n<0).$
(4.10)

For the existence of the intertwiner the zero modes
$\textbf{e}(z),\textbf{f}(z),\textbf{k}^{\pm}(z)$ of $\mathcal{H}$ and
$\textbf{e}^{\prime}(z),\textbf{f}^{\prime}(z),\textbf{k}^{\pm\prime}(z)$ of
$\mathcal{H}^{\prime}$ have to be related by888Recall that
$\textbf{k}^{\pm\prime}(z)=\gamma^{\mp 1}_{2}$.  [21];

$\displaystyle\gamma^{\prime}_{2}=\gamma_{2},\quad\quad\textbf{e}^{\prime}(z)=q^{-1}_{2}\textbf{e}(z),\quad\quad\textbf{f}^{\prime}(z)=q_{2}\textbf{f}(z).$
(4.11)

From now on we are going to write $\gamma$ for $\gamma_{2}$ for the sake of
convenience.

###  4.2. Fock intertwiner

Similarly for the set of partitions $\\{|\lambda\rangle\\}$ which forms a
basis of $\mathcal{F}(v)$, the $\lambda$-component of the Fock intertwiner is
defined by

$\displaystyle\Phi_{\lambda}(v)=\Phi\big{(}|\lambda\rangle\otimes\bullet\big{)}:\mathcal{H}\rightarrow\mathcal{H}^{\prime}.$
(4.12)

A solution to the intertwining relations is

$\displaystyle\Phi_{\lambda}(v)=z_{\lambda}\mathcal{G}_{\lambda}^{-1}\exp\left(\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\mathfrak{q}^{-r/2}v^{r}\left(\sum_{(i,j)\in\lambda}x_{i,j}^{r}-\frac{1}{(1-q_{1}^{r})(1-q_{2}^{r})}\right)\right)$
$\displaystyle\times\exp\left(-\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\mathfrak{q}^{-r/2}v^{-r}\left(\sum_{(i,j)\in\lambda}x_{i,j}^{-r}-\frac{1}{(1-q_{1}^{-r})(1-q_{2}^{-r})}\right)\right),$
(4.13)

where $x_{i,j}=q_{1}^{j-1}q_{2}^{i-1}$. The zero mode factor is

$\displaystyle
z_{\lambda}(v)=\prod_{i=1}^{\ell(\lambda)}\prod_{j=1}^{\lambda_{i}}\left(-\mathfrak{q}q_{2}^{i-1}x_{i,j}^{-1}\right)\mathbf{e}(x_{i,j}v)=q_{2}^{n(\lambda)}(-\mathfrak{q})^{|\lambda|}\prod_{(i,j)\in\lambda}x_{i,j}^{-1}\mathbf{e}(x_{i,j}v),$
(4.14)

and the normalization factor is

$\displaystyle\mathcal{G}_{\lambda}$
$\displaystyle=\prod_{\square\in\lambda}\left(1-q_{1}^{-a_{\lambda}(\square)}q_{2}^{l_{\lambda}(\square)+1}\right).$
(4.15)

The Fock intertwiner exists, if and only if the following relations between
the zero modes of $\mathcal{H}$ and $\mathcal{H}^{\prime}$ are satisfied [2];

$\displaystyle\gamma^{\prime}=\mathfrak{q}\gamma,\quad\quad\mathbf{e}^{\prime}(z)=(-\mathfrak{q}v/z)\mathbf{e}(z),\quad\quad\mathbf{f}^{\prime}(z)=(-\mathfrak{q}v/z)^{-1}\mathbf{f}(z).$
(4.16)

Compared with (3.37), this means that the level and the spectral parameter are
shifted by $\mathfrak{q}^{N}\to\mathfrak{q}^{N+1}$ and $u\to-uv$.

###  4.3. MacMahon intertwiner

Now let’s consider the MacMahon intertwiner whose vertical representation is
$\mathcal{M}(K;v)$. We know that the set of plane partitions
$\\{|\Lambda\rangle\\}$ forms a basis of $\mathcal{M}(K;v)$. We define the
$\Lambda$-component of the MacMahon intertwiner by

$\displaystyle\Xi_{\Lambda}(K;v)=\Xi\big{(}|\Lambda\rangle\otimes\bullet\big{)}:\mathcal{H}\rightarrow\mathcal{H}^{\prime}.$
(4.17)

A solution to the intertwining relations is

$\displaystyle\Xi_{\Lambda}(K;v)=z_{\Lambda}(K;v)\mathcal{M}^{[n]}(K)\tilde{\Phi}^{[n]}_{\Lambda}(v)\Gamma_{n}(K;v),\qquad
n>h(\Lambda),$ (4.18)

where

$\displaystyle\Gamma_{n}(K;v)=\exp\left(\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{q_{3}^{nr}-K^{r}}{\kappa_{r}}\mathfrak{q}^{-r/2}v^{r}\right)\exp\left(\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{q_{3}^{-nr}-K^{-r}}{\kappa_{r}}\mathfrak{q}^{-r/2}v^{-r}\right)$
(4.19)

is the vacuum contribution which survives for $\Lambda=\varnothing$. The zero
mode factor is

$\displaystyle
z_{\Lambda}(K;v)=\prod_{k=1}^{h(\Lambda)}\prod_{(i,j)\in\Lambda^{(k)}}\frac{K^{1/2}}{\mathfrak{q}^{k-1}}\frac{\theta_{q_{3}}(q_{3}^{k-1}/x_{ijk})}{\theta_{q_{3}}(K/x_{ijk})}\mathbf{e}(x_{ijk}v),$
(4.20)

where $x_{ijk}=q_{1}^{i}q_{2}^{j}q_{3}^{k}$. Note that for later convenience
we have exchanged $q_{1}$ and $q_{2}$ in the original formula in [11]. This is
equivalent to taking the transpose of the Young diagram $\lambda$ which labels
the basis of the Fock representation. Or we may exchange the role of $q_{1}$
and $q_{2}$ in the construction of the Fock representation in section 3.3.
Namely we begin with the vector representation $\rho^{V^{(2)}}$ and choose
$q_{1}$ as the shift parameter in the tensor product. Furthermore,

$\displaystyle\tilde{\Phi}^{[n]}_{\Lambda}(v)$
$\displaystyle=\tilde{\Phi}_{\Lambda^{(1)}}(v)\circ\cdots\circ\tilde{\Phi}_{\Lambda^{(n)}}(q_{3}^{n-1}v)$
(4.21)

is a composition of the Fock intertwiners $\tilde{\Phi}$ without the zero mode
factor. Finally the factor $\mathcal{M}^{[n]}(K)$ is defined via

$\displaystyle:\tilde{\Phi}^{[n]}_{\varnothing}(v)\Gamma_{n}(K;v):=\mathcal{M}^{[n]}(K)\tilde{\Phi}^{[n]}_{\varnothing}(v)\Gamma_{n}(K;v).$
(4.22)

The notation $n>h(\Lambda)$ in Eq.(4.18) means that we can choose any integer
$n$ which is greater than $h(\Lambda)$. All of them give the same
$\Xi_{\Lambda}(K;v)$. It can be shown that

$\displaystyle\Xi_{\Lambda}(K;v)=z_{\Lambda}\mathcal{G}_{\Lambda}^{-1}$
$\displaystyle\exp\left(\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\mathfrak{q}^{-r/2}v^{r}\left(\sum_{(i,j,k)\in\Lambda}x_{ijk}^{r}+\frac{1-K^{r}}{\kappa_{r}}\right)\right)$
$\displaystyle\quad\times\exp\left(-\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\mathfrak{q}^{-r/2}v^{-r}\left(\sum_{(i,j,k)\in\Lambda}x_{ijk}^{-r}-\frac{1-K^{-r}}{\kappa_{r}}\right)\right),$
(4.23)

where the factor $\mathcal{G}_{\Lambda}$ is defined by

$\displaystyle:\tilde{\Phi}^{[n]}_{\Lambda}(v)\Gamma_{n}(K;v):=\mathcal{G}_{\Lambda}\mathcal{M}^{[n]}(K)\tilde{\Phi}^{[n]}_{\Lambda}(v)\Gamma_{n}(K;v).$
(4.24)

As motivated from the Fock intertwiners, we can construct the MacMahon
intertwiner from the Fock intertwiners by composing them together with the
correction intertwiner [11]. From this method of construction, we obtain the
explicit form of $\mathcal{G}_{\Lambda}$. Nevertheless, in this paper, the
explicit form is not relevant and we do not state it here. We obtain the
following relations between the zero modes of $\mathcal{H}$ and
$\mathcal{H}^{\prime}$

$\displaystyle\gamma^{\prime}=K^{1/2}\gamma,\quad\mathbf{e}^{\prime}(z)=K^{1/2}\frac{\theta_{q_{3}}(v/z)}{\theta_{q_{3}}(Kv/z)}\mathbf{e}(z),\quad\mathbf{f}^{\prime}(z)=\frac{\theta_{q_{3}}(\mathfrak{q}Kv/z)}{\theta_{q_{3}}(\mathfrak{q}v/z)}\mathbf{f}(z),$
(4.25)

Note that when $K=q_{3}$, (4.25) reduces to (4.16).

###  4.4. Dual intertwiners

Now we turn to the construction of the dual intertwiner
$\Psi^{*}:\mathcal{H}^{\prime}\rightarrow\mathcal{H}\otimes\mathcal{V}$ ,
which is determined from the intertwining relation :

$\displaystyle\Delta(a)\Psi^{*}=\Psi^{*}a,\quad\quad\forall a\in
U_{\mathfrak{q},\mathfrak{d}}(\widehat{\widehat{\mathfrak{gl}}}_{1}).$ (4.26)

Taking the same basis $\\{\alpha\\}$ of $\mathcal{V}$ used to define the
components of $\Psi$, we can define the $\alpha$-component of the dual
intertwiner $\Psi^{*}_{\alpha}:\mathcal{H}^{\prime}\rightarrow\mathcal{H}$ by

$\displaystyle\Psi^{*}(\bullet)=\sum_{\alpha}\Psi^{*}_{\alpha}(\bullet)\otimes|\alpha\rangle,\quad\quad\bullet\in\mathcal{H}^{\prime}.$
(4.27)

We can express the intertwining relation (4.26) in terms of the
$\alpha$-component as

$\displaystyle\Psi^{*}_{\alpha}K^{+}(\mathfrak{q}z)$
$\displaystyle=(\alpha|K^{+}(z)|\alpha)K^{+}(\mathfrak{q}z)\Psi^{*}_{\alpha},$
(4.28) $\displaystyle\Psi^{*}_{\alpha}K^{-}(z)$
$\displaystyle=(\alpha|K^{-}(z)|\alpha)K^{-}(z)\Psi^{*}_{\alpha},$ (4.29)
$\displaystyle\Psi^{*}_{\alpha}E(z)$
$\displaystyle=K^{-}(\mathfrak{q}z)\sum_{\beta}(\alpha|E(\mathfrak{q}z)|\beta)\Psi^{*}_{\beta}+E(z)\Psi^{*}_{\alpha},$
(4.30) $\displaystyle\Psi^{*}_{\alpha}F(z)$
$\displaystyle=\sum_{\beta}(\alpha|F(z)|\beta)\Psi^{*}_{\beta}+(\alpha|K^{+}(z)|\alpha)F(z)\Psi^{*}_{\alpha}.$
(4.31)

When $\mathcal{V}$ is the vertical vector/Fock/MacMahon modules, we call the
dual intertwiner
$\Psi^{*}:\mathcal{H}^{\prime}\rightarrow\mathcal{H}\otimes\mathcal{V}$ the
vector/Fock/MacMahon dual intertwiner, respectively. In the following we give
their explicit expressions in each case.

We know that $\big{\\{}[v]_{n}\big{|}~{}n\in\mathbb{Z}\big{\\}}$ forms a basis
of $V(v)$. Then, according to (4.27), the $n$-component
$\mathbb{I}^{*}_{n}(v):\mathcal{H}^{\prime}\rightarrow\mathcal{H}$ of the dual
intertwiner satisfies

$\displaystyle\mathbb{I}^{*}(v)=\sum_{n\in\mathbb{Z}}\mathbb{I}^{*}_{n}(v)\otimes[v]_{n-1}.$
(4.32)

Thus, the intertwining relations become

$\displaystyle\mathbb{I}^{*}_{n}(v)K^{+}(\mathfrak{q}z)$
$\displaystyle=\tilde{\psi}(q^{n-1}_{1}v/z)K^{+}(\mathfrak{q}z)\mathbb{I}^{*}_{n}(v),$
(4.33) $\displaystyle\mathbb{I}^{*}_{n}(v)K^{-}(z)$
$\displaystyle=\tilde{\psi}(q^{-n}_{1}z/v)K^{-}(z)\mathbb{I}^{*}_{n}(v),$
(4.34) $\displaystyle\mathbb{I}^{*}_{n}(v)E(z)$
$\displaystyle=(1-q_{1})^{-1}\delta(\mathfrak{q}^{-1}q^{n-1}_{1}v/z)K^{-}(\mathfrak{q}z)\mathbb{I}^{*}_{n-1}(v)+E(z)\mathbb{I}^{*}_{n}(v),$
(4.35) $\displaystyle\mathbb{I}^{*}_{n}(v)F(z)$
$\displaystyle=(1-q^{-1}_{1})^{-1}\delta(q^{n}_{1}v/z)\mathbb{I}^{*}_{n+1}(v)+\tilde{\psi}(q^{n-1}_{1}v/z)F(z)\mathbb{I}^{*}_{n}(v).$
(4.36)

A solution to the intertwining relations is

$\displaystyle\mathbb{I}^{*}_{n}(v)=z^{*}_{n}\exp\bigg{(}\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{\mathfrak{q}^{r/2}(q^{n}_{1}v)^{r}}{1-q^{r}_{1}}\bigg{)}\exp\bigg{(}-\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{\mathfrak{q}^{r/2}(q^{n}_{1}v)^{-r}}{1-q^{-r}_{1}}\bigg{)},$
(4.37)

where

$\displaystyle
z^{*}_{n}=\begin{cases}q^{n}_{2}\prod_{j=1}^{n}\textbf{f}(q^{j-1}_{1}v)\quad;~{}n>0,\\\
q^{n}_{2}\prod_{j=n}^{-1}\textbf{f}(q^{j}_{1}v)^{-1}\quad;~{}n<0.\end{cases}$
(4.38)

As the condition for the existence of the dual intertwiner, we get the
following relations [21];

$\displaystyle\gamma^{\prime}=\gamma,\quad\quad\textbf{e}^{\prime}(z)=q^{-1}_{2}\textbf{e}(z),\quad\quad\textbf{f}^{\prime}(z)=q_{2}\textbf{f}(z).$
(4.39)

We define the $\lambda$-component of the dual Fock intertwiner
$\Phi^{*}_{\lambda}(v)$ by

$\displaystyle\Phi^{*}(v)=\sum_{\lambda}\Phi^{*}_{\lambda}(v)\otimes|\lambda\rangle.$
(4.40)

A solution to the dual intertwining relations is

$\displaystyle\Phi^{*}_{\lambda}(v)=z^{*}_{\lambda}(v)\mathcal{G}^{[n]*}B^{*}_{n}(v)\tilde{\mathbb{I}}^{[n]*}_{\lambda}(v),~{}~{}n>\ell(\lambda),$
(4.41)

where

$\displaystyle
B^{*}_{n}(v)=\exp\bigg{(}\sum_{r=1}^{\infty}\frac{1}{k_{r}}(\mathfrak{q}^{3/2}q^{n}_{2}v)^{r}H_{-r}\bigg{)}\exp\bigg{(}-\sum_{r=1}^{\infty}\frac{1}{\kappa_{r}}(\mathfrak{q}^{1/2}q^{n}_{2}v)^{-r}H_{r}\bigg{)},$
(4.42)

$\tilde{\mathbb{I}}^{[n]*}_{\lambda}:=\tilde{\mathbb{I}}^{*}_{\lambda_{n}}(q^{n-1}_{2}v)\circ\cdots\circ\tilde{\mathbb{I}}^{*}_{\lambda_{1}}(v)$,
and the normalization factor $\mathcal{G}^{[n]*}$ is defined by
$:B^{*}_{n}(v)\tilde{\mathbb{I}}^{[n]*}_{\emptyset}(v):=\mathcal{G}^{[n]*}B^{*}_{n}(v)\tilde{\mathbb{I}}^{[n]*}_{\emptyset}(v)$.
The factor $z^{*}_{\lambda}(v)$ is

$\displaystyle
z^{*}_{\lambda}(v)=\mathfrak{q}^{|\lambda|}\prod_{i=1}^{l(\lambda)}\prod_{j=1}^{\lambda_{i}}\big{(}-\mathfrak{q}q^{i-1}_{2}x^{-1}_{i,j}\big{)}^{-1}\textbf{f}(x_{ij}v).$
(4.43)

It can also be shown that

$\displaystyle\Phi^{*}_{\lambda}(v)=z^{*}_{\lambda}\mathcal{G}^{*\,-1}_{\lambda}\exp\bigg{[}\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\mathfrak{q}^{r/2}v^{r}\Big{(}\frac{q^{lr}_{2}}{(1-q^{r}_{1})(1-q^{r}_{2})}+\sum_{k=1}^{l}\frac{(q^{\lambda_{k}}_{1}q^{k-1}_{2})^{r}}{1-q^{r}_{1}}\Big{)}\bigg{]}$
$\displaystyle\times\exp\bigg{[}\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\mathfrak{q}^{r/2}v^{-r}\Big{(}\sum_{k=1}^{l}\frac{(q^{\lambda_{k}-1}_{1}q^{k-1}_{2})^{-r}}{1-q^{r}_{1}}-\frac{q^{-rl}_{2}}{(1-q^{-r}_{1})(1-q^{-r}_{2})}\Big{)}\bigg{]}.$
(4.44)

Here $\mathcal{G}^{*}_{\lambda}$ is defined via the equation

$\displaystyle:B^{*}_{n}(v)\tilde{\mathbb{I}}^{[n]*}_{\lambda}(v):=\mathcal{G}^{*}_{\lambda}\mathcal{G}^{[n]*}B^{*}_{n}(v)\tilde{\mathbb{I}}^{[n]*}_{\lambda}(v).$
(4.45)

Similarly we can define the $\Lambda$-component of the MacMahon intertwiner
$\Xi^{*}_{\Lambda}(K,v)$ by

$\displaystyle\Xi^{*}(K;v)=\sum_{\Lambda}\Xi^{*}_{\Lambda}(K;v)\otimes|\Lambda\rangle.$
(4.46)

A solution to the intertwining relations is

$\displaystyle\Xi_{\Lambda}^{*}(K;v)=z_{\Lambda}^{*}(K;v)\mathcal{M}^{[n]*}(K)\tilde{\Phi}^{[n]*}_{\Lambda}(v)\Gamma^{*}_{n}(K;v),$
(4.47)

where

$\displaystyle\Gamma^{*}_{n}(K;v)=\exp\bigg{(}-\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{q^{nr}_{3}-K^{r}}{\kappa_{r}}\mathfrak{q}^{r/2}v^{r}\bigg{)}\exp\bigg{(}-\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{q^{-nr}_{3}-K^{-r}}{\kappa_{r}}\mathfrak{q}^{r/2}v^{-r}\bigg{)},$
(4.48) $\displaystyle
z^{*}_{\Lambda}(K;v)=\prod_{k=1}^{h(\Lambda)}\prod_{(i,j)\in\Lambda^{(k)}}\Bigg{(}\frac{K^{1/2}}{\mathfrak{q}^{k}}\frac{\theta_{q_{3}}(q^{k}_{3}x^{-1}_{ijk})}{\theta_{q_{3}}(Kx^{-1}_{ijk})}\Bigg{)}^{-1}\textbf{f}(x_{ijk}v),$
(4.49)

and

$\displaystyle\tilde{\Phi}^{[n]*}_{\Lambda}(v)=\tilde{\Phi}^{*}_{\Lambda^{(1)}}(v)\circ\cdots\circ\tilde{\Phi}^{*}_{\Lambda^{(n)}}(q^{n-1}_{3}v).$
(4.50)

The factor $\mathcal{M}^{[n]*}(K)$ is defined by the equation

$\displaystyle:\tilde{\Phi}^{[n]*}_{\emptyset}(v)\Gamma^{*}_{n}(K;v):=\mathcal{M}^{[n]*}(K)\tilde{\Phi}^{[n]*}_{\emptyset}(v)\Gamma^{*}_{n}(K;v).$
(4.51)

As in the case of the dual Fock intertwiner, we can determine an explicit form
of $\Xi^{*}_{\Lambda}(K;v)$ like Eq.(4.44), which we will compute in section
6.

##  5\. Generalized KZ equation for the vector intertwiners

In this section we derive the generalized KZ equation for the vector
intertwiners as a warm-up exercise before attacking the more complicated case
of the MacMahon intertwiners. The first step is to construct the shift
operator $\mathcal{T}$.

###  5.1. Construction of the shift operator

In [9] the shift operator is constructed as the composition of the intertwiner
and the dual intertwiner with appropriate spectral parameter. We first recall
the relation (4.11) on the level of the vector intertwiner discussed in the
previous section. Comparing the choice of the zero modes (3.37) with (4.11)
saying there is no change of the level of $\mathcal{H}$ and
$\mathcal{H}^{\prime}$, we see $u^{\prime}=q^{-1}_{2}u$. We define the
operator $\mathcal{T}^{m}_{n}$ by the composition of the intertwiner and the
dual intertwiner;

$\displaystyle\mathcal{T}^{m}_{n}(N,u|v,w):=\mathbb{I}^{*}_{n}(w)\mathbb{I}_{m}(v)\quad:\quad\mathcal{F}_{u}^{(\mathfrak{q},\mathfrak{q}^{N})}\longrightarrow\mathcal{F}_{u}^{(\mathfrak{q},\mathfrak{q}^{N})}.$
(5.1)

Pictorially, it is represented as Fig.1.

$w$$(0,0)$$(1,N)$$(1,N)$$\mathbb{I}^{*}_{n}(w)$$q_{2}^{-1}u$$u$$\circ$$v$$(0,0)$$(1,N)$$(1,N)$$\mathbb{I}_{m}(v)$$u$$q_{2}^{-1}u$
Figure 1. Composition $\circ$ of the vector intertwiner and the dual vector
intertwiner. Contrary to the Fock intertwiner, the level $(1,N)$ of the
horizontal Fock space is kept intact and the shift of the spectral parameter
$u$ is independent of the spectral parameters $v,w$ of the vertical
representation.

Using the commutation relation (3.30), we see that

$\displaystyle\mathcal{T}^{m}_{n}(N,u|v,w)=z^{*}_{n}$
$\displaystyle\big{(}u,N\big{|}w\big{)}z_{m}\big{(}u,N\big{|}v\big{)}$
$\displaystyle\cdot\exp\Big{(}-\sum_{r=1}^{\infty}\frac{q_{1}^{-r(n-1)}w^{-r}(q^{m}_{1}v)^{r}}{r}\frac{\mathfrak{q}^{r}(1-q^{r}_{2})}{1-q^{r}_{1}}\Big{)}:\tilde{\mathbb{I}}^{*}_{n}(w)\tilde{\mathbb{I}}_{m}(v):~{},$
(5.2)

where $:\quad:$ denotes the normal ordering. The next step is to determine the
value of the shift parameter $p$ such that we can express $\mathbb{I}_{n}(pv)$
and $\mathbb{I}_{n}^{*}(pv)$ by the action of $\mathcal{T}$ operator. Consider
the quantity $\mathbb{I}_{n}(pv)$. From the explicit expression of the vector
intertwiner explained in section 4.1, we see that

$\displaystyle\mathbb{I}_{n}(pv)=$
$\displaystyle\frac{z_{n}(u,N|pv)}{z_{n}(u,N|v)}\cdot\exp\Big{(}-\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{(\mathfrak{q}^{-1/2}q^{n}_{1}v)^{r}}{1-q^{r}_{1}}(p^{r}-1)\Big{)}$
$\displaystyle\cdot\mathbb{I}_{n}(v)\cdot\exp\Big{(}\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{(\mathfrak{q}^{1/2}q^{n-1}_{1}v)^{-r}}{1-q^{r}_{1}}(1-p^{-r})\Big{)}.$
(5.3)

To express the exponential factors in (5.3) by $\mathcal{T}$ operator, let us
consider Eq.(5.2) in the case $m=n$.

$\displaystyle\mathcal{T}^{n}_{n}(N,u|v,w)=$ $\displaystyle
z^{*}_{n}\big{(}u,N\big{|}w\big{)}z_{n}\big{(}u,N\big{|}v\big{)}\exp\Big{(}-\sum_{r=1}^{\infty}\frac{q_{1}^{r}w^{-r}v^{r}}{r}\frac{\mathfrak{q}^{r}(1-q^{r}_{2})}{1-q^{r}_{1}}\Big{)}$
$\displaystyle\cdot\exp\Big{(}\sum_{r=1}^{\infty}\frac{H_{-r}}{(\mathfrak{q}^{r}-\mathfrak{q}^{-r})(1-q^{r}_{1})}\big{(}(\mathfrak{q}^{1/2}q^{n}_{1}w)^{r}-(\mathfrak{q}^{-1/2}q^{n}_{1}v)^{r}\big{)}\Big{)}$
$\displaystyle\cdot\exp\Big{(}\sum_{r=1}^{\infty}\frac{H_{r}}{(\mathfrak{q}^{r}-\mathfrak{q}^{-r})(1-q^{r}_{1})}\big{(}(\mathfrak{q}^{-1/2}q^{n-1}_{1}w)^{-r}-(\mathfrak{q}^{1/2}q^{n-1}_{1}v)^{-r}\big{)}\Big{)}.$
(5.4)

Thus, if $w=\mathfrak{q}v$, the positive modes disappear

$\displaystyle\mathcal{T}^{n}_{n}(N,u|v,\mathfrak{q}v)=$
$\displaystyle\exp\Big{(}-\sum_{r=1}^{\infty}\frac{q^{r}_{1}}{r}\frac{1-q_{2}^{r}}{1-q^{r}_{1}}\Big{)}\cdot\exp\Big{(}\sum_{r=1}^{\infty}\frac{H_{-r}\mathfrak{q}^{r/2}}{1-q^{r}_{1}}(q^{n}_{1}v)^{r}\Big{)}.$
(5.5)

where we have used $z^{*}_{n}\big{(}u,N\big{|}\mathfrak{q}^{\pm
1}v\big{)}z_{n}\big{(}u,N\big{|}v\big{)}=1$ for $\gamma=1$. On the other hand,
if $w=\mathfrak{q}^{-1}v$, the negative modes disappear

$\displaystyle\mathcal{T}^{n}_{n}(N,u|v,\mathfrak{q}^{-1}v)=$
$\displaystyle\exp\Big{(}-\sum_{r=1}^{\infty}\frac{q^{r}_{1}\mathfrak{q}^{2r}(1-q^{r}_{2})}{r(1-q^{r}_{1})}\Big{)}\cdot\exp\Big{(}\sum_{r=1}^{\infty}\frac{H_{r}\mathfrak{q}^{r/2}}{1-q^{r}_{1}}(q^{n-1}_{1}v)^{-r}\Big{)}.$
(5.6)

By tuning the spectral parameters of $\mathbb{I}^{*}_{n}(w)$ and
$\mathbb{I}_{m}(v)$, we define999See section 5.5 for the reason why we define
them in this way.

$\displaystyle\mathcal{T}^{+}_{n}(N,u|v)$
$\displaystyle:=\mathcal{T}^{n}_{n}(N,u|\mathfrak{q}^{1/2}v,\mathfrak{q}^{-1/2}v)^{-1},$
$\displaystyle\mathcal{T}^{-}_{n}(N,u|v)$
$\displaystyle:=\mathcal{T}^{n}_{n}(N,u|\mathfrak{q}^{-1/2}v,\mathfrak{q}^{1/2}v).$
(5.7)

Thus,

$\displaystyle\mathbb{I}_{n}(\mathfrak{q}^{-2}v)=$
$\displaystyle\frac{z_{n}(u,N|\mathfrak{q}^{-2}v)}{z_{n}(u,N|v)}\cdot\exp\bigg{(}\sum_{r=1}^{\infty}\frac{q^{r}_{1}}{r}\frac{1-q^{r}_{2}}{1-q^{r}_{1}}(1-\mathfrak{q}^{2r})\bigg{)}$
$\displaystyle\cdot\mathcal{T}^{-}_{n}(N,q_{2}^{-1}u|\mathfrak{q}^{-3/2}v)\cdot\mathbb{I}_{n}(v)\cdot\mathcal{T}^{+}_{n}(N,u|\mathfrak{q}^{-1/2}v).$
(5.8)

It is remarkable that the shift parameter is fixed to be
$p=\mathfrak{q}^{-2}$. In the above tuning the spectral parameters of the
intertwiners and the dual intertwiners are shifted by $\mathfrak{q}^{\pm 1}$.
This reminds us of exactly the same shift of the spectral parameters in the
Higgsed DIM network [47] [51] [24] [25]101010Upon the completion of the
present work, we noticed [52], which shows the universal $R$ matrix of DIM
algebra plays a significant role in the Higgsed network calculus. In
particular it is interesting that the shift operator in the present paper is
physically identified with the crossing of branes in [52]. . It seems that
this coincidence is related to the fact the shift operator (5.7) is obtained
from the universal $R$ matrix as we show later in section 5.5.

Similarly from the explicit expression of the dual vector intertwiner
explained in section 4.4, we see that

$\displaystyle\mathbb{I}^{*}_{n}(pv)=$
$\displaystyle\frac{z^{*}_{n}(u,N|pv)}{z^{*}_{n}(u,N|v)}\cdot\exp\bigg{[}\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{\mathfrak{q}^{r/2}(q^{n}_{1}v)^{r}}{1-q^{r}_{1}}(p^{r}-1)\bigg{]}$
$\displaystyle\cdot\mathbb{I}^{*}_{n}(v)\cdot\exp\bigg{[}-\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{\mathfrak{q}^{r/2}(q^{n}_{1}v)^{-r}}{1-q^{-r}_{1}}(p^{-r}-1)\bigg{]}.$

From Eqs.(5.5) and (5.6), we get that

$\displaystyle\mathbb{I}^{*}_{n}(\mathfrak{q}^{-2}v)=$
$\displaystyle\frac{z^{*}_{n}(u,N|\mathfrak{q}^{-2}v)}{z^{*}_{n}(u,N|v)}\cdot\exp\bigg{[}\sum_{r=1}^{\infty}\frac{q^{r}_{1}(1-q^{r}_{2})}{r(1-q^{r}_{1})}(\mathfrak{q}^{2r}-1)\bigg{]}$
$\displaystyle\mathcal{T}^{-}_{n}(N,q_{2}u|\mathfrak{q}^{-1/2}v)^{-1}\cdot\mathbb{I}^{*}_{n}(v)\cdot\mathcal{T}^{+}_{n}(N,u|\mathfrak{q}^{-3/2}v)^{-1}.$
(5.9)

###  5.2. Commutation of vector intertwiner and the shift operators

From the previous subsection, we see that $\mathcal{T}$ operators are
expressed as the composition of the intertwiner and the dual vector
intertwiner. Thus, to calculate the commutation of the vector intertwiner and
$\mathcal{T}$ operators, we first calculate the commutation between the vector
intertwiners and the dual vector intertwiners themselves.

$w$$(0,0)$$(1,N)$$(1,N)$$\mathbb{I}^{*}_{n}(w)$$q_{2}u$$u$$\circ$$v$$(0,0)$$(1,N)$$(1,N)$$\mathbb{I}_{m}(v)$$u$$q_{2}u$
Figure 2. Composition $\circ$ of the dual vector intertwiner and the vector
intertwiner. Contrary to the Fock intertwiner, the level $(1,N)$ of the
horizontal Fock space is kept intact and the shift of the spectral parameter
$u$ is independent of the spectral parameters $v,w$ of the vertical
representation.

The result of the composition $\mathbb{I}^{*}_{n}(w)\mathbb{I}_{m}(v)$ of
Fig.1 has already given in eq.(5.2). On the other hand the composition
$\mathbb{I}_{m}(v)\mathbb{I}^{*}_{n}(w)$ of Fig.2 is

$\displaystyle\mathbb{I}_{m}(v)\mathbb{I}^{*}_{n}(w)=z_{m}$
$\displaystyle\big{(}q_{2}u,N\big{|}v\big{)}z^{*}_{n}(q_{2}u,N\big{|}w)$
$\displaystyle\cdot\exp\bigg{(}-\sum_{r=1}^{\infty}\frac{1}{r}(\mathfrak{q}^{1/2}q^{m-1}_{1}v)^{-r}(\mathfrak{q}^{1/2}q^{n}_{1}w)^{r}\mathfrak{q}^{r}\frac{1-q^{r}_{2}}{1-q^{r}_{1}}\bigg{)}:\tilde{\mathbb{I}}_{m}(v)\tilde{\mathbb{I}}^{*}_{n}(w):~{}.$
(5.10)

Since

$\displaystyle
z_{m}\big{(}q_{2}u,N\big{|}v\big{)}=q_{2}^{m}z_{m}\big{(}u,N\big{|}v\big{)},\qquad
z^{*}_{n}\big{(}q_{2}u,N\big{|}w\big{)}=q_{2}^{-n}z^{*}_{n}(u,N\big{|}w),$
(5.11)

we get that

$\displaystyle\mathbb{I}_{m}(v)\mathbb{I}^{*}_{n}(w)=q_{2}^{m-n}\exp\bigg{(}\sum_{r=1}^{\infty}\frac{q^{r}_{1}\mathfrak{q}^{r}}{r}\frac{1-q^{r}_{2}}{1-q^{r}_{1}}\big{[}q^{-rn}_{1}q^{mr}_{1}(v/w)^{r}-q^{-mr}_{1}q^{nr}_{1}(w/v)^{r}\big{]}\bigg{)}\mathbb{I}^{*}_{n}(w)\mathbb{I}_{m}(v).$
(5.12)

$v^{\prime}$$(0,0)$$(1,N)$$(1,N)$$\mathbb{I}_{m}(v^{\prime})$$q_{2}^{-1}u$$q_{2}^{-2}u$$\circ$$v$$(0,0)$$(1,N)$$(1,N)$$\mathbb{I}_{n}(v)$$u$$q_{2}^{-1}u$
Figure 3. Composition $\circ$ of the vector intertwiner and the vector
intertwiner. Contrary to the Fock intertwiner, the level $(1,N)$ of the
horizontal Fock space is kept intact and the shift of the spectral parameter
$u$ is independent of the spectral parameters $v,v^{\prime}$ of the vertical
representation.

Next let us calculate $\mathbb{I}_{m}(v^{\prime})\mathbb{I}_{n}(v)$ of Fig.3.
The result is

$\displaystyle\mathbb{I}_{m}(v^{\prime})\mathbb{I}_{n}(v)=z_{m}$
$\displaystyle\big{(}q_{2}^{-1}u,N\big{|}v^{\prime}\big{)}z_{n}(u,N\big{|}v)$
$\displaystyle\cdot\exp\bigg{(}\sum_{r=1}^{\infty}(q^{m-1}_{1}v^{\prime})^{-r}(q^{n}_{1}v)^{r}\frac{1}{r}\cdot\frac{1-q^{r}_{2}}{1-q^{r}_{1}}\bigg{)}:\tilde{\mathbb{I}}_{m}(v^{\prime})\tilde{\mathbb{I}}_{n}(v):~{}.$
(5.13)

The formula for $\mathbb{I}_{n}(v)\mathbb{I}_{m}(v^{\prime})$ is obtained by
interchanging the parameters $v\leftrightarrow v^{\prime}$, and
$n\leftrightarrow m$. Hence, by (5.11), we see that

$\displaystyle\mathbb{I}_{n}(v)\mathbb{I}_{m}(v^{\prime})=q_{2}^{-n+m}\exp\bigg{(}\sum_{r=1}^{\infty}\frac{q^{r}_{1}}{r}\frac{1-q^{r}_{2}}{1-q^{r}_{1}}\big{[}\frac{(q^{m}_{1}v^{\prime})^{r}}{(q^{n}_{1}v)^{r}}-\frac{(q^{n}_{1}v)^{r}}{(q^{m}_{1}v^{\prime})^{r}}\big{]}\bigg{)}\mathbb{I}_{m}(v^{\prime})\mathbb{I}_{n}(v).$
(5.14)

A similar computation gives

$\displaystyle\mathbb{I}^{*}_{m}(w^{\prime})\mathbb{I}^{*}_{n}(w)=q_{2}^{n-m}\exp\bigg{(}\sum_{r=1}^{\infty}\frac{\mathfrak{q}^{2r}q^{r}_{1}}{r}\frac{1-q^{r}_{2}}{1-q^{r}_{1}}\big{[}\frac{(q^{n}_{1}w)^{r}}{(q^{m}_{1}w^{\prime})^{r}}-\frac{(q^{m}_{1}w^{\prime})^{r}}{(q^{n}_{1}w)^{r}}\big{]}\bigg{)}\mathbb{I}^{*}_{n}(w)\mathbb{I}^{*}_{m}(w^{\prime}).$
(5.15)

By introducing the function

$\displaystyle\Upsilon\big{(}\alpha\big{|}x\big{)}=\exp\bigg{(}\sum_{r=1}^{\infty}\frac{\alpha^{r}(1-q^{r}_{2})}{r(1-q^{r}_{1})}(x^{r}-x^{-r})\bigg{)},$
(5.16)

we can write the equations (5.12), (5.14) and (5.15) more compactly;

$\displaystyle\mathbb{I}_{m}(v)\mathbb{I}^{*}_{n}(w)$
$\displaystyle=q_{2}^{m-n}\Upsilon\big{(}q_{1}\mathfrak{q}~{}\big{|}~{}\frac{v}{w}q^{m-n}_{1}\big{)}\mathbb{I}^{*}_{n}(w)\mathbb{I}_{m}(v),$
(5.17) $\displaystyle\mathbb{I}_{n}(v)\mathbb{I}_{m}(v^{\prime})$
$\displaystyle=q_{2}^{-n+m}\Upsilon\big{(}q_{1}~{}\big{|}~{}q^{m-n}_{1}\frac{v^{\prime}}{v}\big{)}\mathbb{I}_{m}(v^{\prime})\mathbb{I}_{n}(v),$
(5.18) $\displaystyle\mathbb{I}^{*}_{m}(w^{\prime})\mathbb{I}^{*}_{n}(w)$
$\displaystyle=q_{2}^{n-m}\Upsilon\big{(}\mathfrak{q}^{2}q_{1}~{}\big{|}~{}q_{1}^{n-m}\frac{w}{w^{\prime}}\big{)}\mathbb{I}^{*}_{n}(w)\mathbb{I}^{*}_{m}(w^{\prime}).$
(5.19)

Now we are ready to calculate the commutation relation between $\mathcal{T}$
operator and the vector intertwiner $\mathbb{I}_{n}(v)$. Consider the
composition
$\mathcal{T}^{m}_{n}\big{(}N,q^{-1}u\big{|}v,w\big{)}\mathbb{I}_{k}(z)$
corresponding to Fig.4 below.

$v$$(0,0)$$\mathbb{I}^{*}_{n}(w)$$w$$(0,0)$$q_{2}^{-2}u$$(1,N)$$(1,N)$$(1,N)$$\mathbb{I}_{m}(v)$$q_{2}^{-1}u$$q_{2}^{-1}u$$\circ$$z$$(0,0)$$(1,N)$$(1,N)$$\mathbb{I}_{k}(z)$$u$$q_{2}^{-1}u$
Figure 4. Composition $\circ$ of the vector intertwiner and the shift
operator.

Then, by using (5.17) and (5.18), we obtain

$\displaystyle\mathcal{T}^{m}_{n}\big{(}N,q^{-1}_{2}u\big{|}v,w\big{)}\mathbb{I}_{k}(z)$
$\displaystyle=\mathbb{I}^{*}_{n}(w)\mathbb{I}_{m}(v)\mathbb{I}_{k}(z)=\mathbb{I}^{*}_{n}(w)\bigg{(}q_{2}^{-m+k}\Upsilon\big{(}q_{1}\big{|}q^{k-m}_{1}\frac{z}{v}\big{)}\bigg{)}\mathbb{I}_{k}(z)\mathbb{I}_{m}(v)$
$\displaystyle=q_{2}^{-m+k}\Upsilon\big{(}q_{1}\big{|}q_{1}^{k-m}\frac{z}{v}\big{)}q_{2}^{n-k}\Upsilon\big{(}q_{1}\mathfrak{q}\big{|}q^{k-n}_{1}\frac{z}{w}\big{)}^{-1}\mathbb{I}_{k}(z)\mathbb{I}^{*}_{n}(w)\mathbb{I}_{m}(v)$
$\displaystyle=q_{2}^{n-m}\Upsilon\big{(}q_{1}\big{|}q_{1}^{k-m}\frac{z}{v}\big{)}\Upsilon\big{(}q_{1}\mathfrak{q}\big{|}q^{n-k}_{1}\frac{w}{z}\big{)}\mathbb{I}_{k}(z)\mathcal{T}^{m}_{n}\big{(}N,u\big{|}v,w\big{)}.$
(5.20)

Here we have used the property
$\Upsilon\big{(}\alpha\big{|}x\big{)}^{-1}=\Upsilon\big{(}\alpha\big{|}\frac{1}{x}\big{)}$
which follows directly from the definition (5.16). Note that, during the
calculation, we do not need to be careful too much about the horizontal
spectral parameters, since we have dealt with this issue when we calculated
the commutation of the vector intertwiner and dual vector intertwiner. The
only thing we need to be careful is the initial point,
$\mathcal{T}^{m}_{n}\big{(}N,q_{2}^{-1}u\big{|}v,w\big{)}\mathbb{I}_{k}(z)$,
and the final point,
$\mathbb{I}_{k}(z)\mathcal{T}^{m}_{n}\big{(}N,u\big{|}v,w\big{)}$.

Now consider the composition
$\mathcal{T}^{m}_{n}\big{(}N,q_{2}u\big{|}v,w\big{)}\mathbb{I}^{*}_{k}(z)$
corresponding to Fig.5 below.

$v$$(0,0)$$\mathbb{I}^{*}_{n}(w)$$w$$(0,0)$$u$$(1,N)$$(1,N)$$(1,N)$$\mathbb{I}_{m}(v)$$q_{2}u$$q_{2}u$$\circ$$z$$(0,0)$$(1,N)$$(1,N)$$\mathbb{I}^{*}_{k}(z)$$u$$q_{2}u$
Figure 5. Composition $\circ$ of the dual vector intertwiner and the shift
operator.

We can compute

$\displaystyle\mathcal{T}^{m}_{n}\big{(}N,q_{2}u\big{|}v,w\big{)}\mathbb{I}^{*}_{k}(z)$
$\displaystyle=\mathbb{I}^{*}_{n}(w)\mathbb{I}_{m}(v)\mathbb{I}^{*}_{k}(z)=q_{2}^{m-k}\Upsilon\big{(}q_{1}\mathfrak{q}\big{|}\frac{v}{z}q^{m-k}_{1}\big{)}\mathbb{I}^{*}_{n}(w)\mathbb{I}^{*}_{k}(z)\mathbb{I}_{m}(v)$
$\displaystyle=q_{2}^{m-k}\Upsilon\big{(}q_{1}\mathfrak{q}\big{|}\frac{v}{z}q^{m-k}_{1}\big{)}q_{2}^{k-n}\Upsilon\big{(}\mathfrak{q}^{2}q_{1}\big{|}q^{k-n}_{1}\frac{z}{w}\big{)}\mathbb{I}^{*}_{k}(z)\mathbb{I}^{*}_{n}(w)\mathbb{I}_{m}(v)$
$\displaystyle=q_{2}^{m-n}\Upsilon\big{(}q_{1}\mathfrak{q}\big{|}\frac{v}{z}q^{m-k}_{1}\big{)}\Upsilon\big{(}\mathfrak{q}^{2}q_{1}\big{|}q^{k-n}_{1}\frac{z}{w}\big{)}\mathbb{I}^{*}_{k}(z)\mathcal{T}^{m}_{n}\big{(}N,u\big{|}v,w\big{)}.$
(5.21)

From Eqs. (5.20), (5.21) and the definition (5.7) we see that

$\displaystyle\mathcal{T}^{+}_{n}(N,u\big{|}v)\mathbb{I}_{k}(z)$
$\displaystyle=\frac{\Upsilon(q_{1}\big{|}q^{n-k}_{1}\mathfrak{q}^{1/2}\frac{v}{z})}{\Upsilon(q_{1}\mathfrak{q}\big{|}q^{n-k}_{1}\mathfrak{q}^{-1/2}\frac{v}{z})}\mathbb{I}_{k}(z)\mathcal{T}^{+}_{n}(N,q_{2}u\big{|}v),$
(5.22) $\displaystyle\mathcal{T}^{-}_{n}(N,u\big{|}v)\mathbb{I}_{k}(z)$
$\displaystyle=\frac{\Upsilon(q_{1}\big{|}q^{k-n}_{1}\mathfrak{q}^{1/2}\frac{z}{v})}{\Upsilon(q_{1}\mathfrak{q}\big{|}q^{k-n}_{1}\mathfrak{q}^{-1/2}\frac{z}{v})}\mathbb{I}_{k}(z)\mathcal{T}^{-}_{n}(N,q_{2}u\big{|}v),$
(5.23) $\displaystyle\mathcal{T}^{+}_{n}(N,u\big{|}v)\mathbb{I}^{*}_{k}(z)$
$\displaystyle=\frac{\Upsilon(q_{1}\mathfrak{q}\big{|}q^{k-n}_{1}\mathfrak{q}^{-1/2}\frac{z}{v})}{\Upsilon(q_{1}\mathfrak{q}^{2}\big{|}q^{k-n}_{1}\mathfrak{q}^{1/2}\frac{z}{v})}\mathbb{I}^{*}_{k}(z)\mathcal{T}^{+}_{n}(N,q_{2}^{-1}u\big{|}v),$
(5.24) $\displaystyle\mathcal{T}^{-}_{n}(N,u\big{|}v)\mathbb{I}^{*}_{k}(z)$
$\displaystyle=\frac{\Upsilon(q_{1}\mathfrak{q}\big{|}q^{n-k}_{1}\mathfrak{q}^{-1/2}\frac{v}{z})}{\Upsilon(q_{1}\mathfrak{q}^{2}\big{|}q^{n-k}_{1}\mathfrak{q}^{1/2}\frac{v}{z})}\mathbb{I}^{*}_{k}(z)\mathcal{T}^{-}_{n}(N,q^{-1}_{2}u\big{|}v).$
(5.25)

The function $\Upsilon(\alpha\big{|}x)$ involves both the positive and the
negative powers of the parameter $x$. However the combinations appearing in
Eqs.(5.22) – (5.25) have only the positive or the negative powers. Hence, we
expect that the proportional factors in these equations can be written in
terms of the $R$-matrix of the vector representation. We will see this is in
fact the case in the next subsection.

###  5.3. $R$-matrix of the vector representation

We can define the $R$-matrix of the vector representation of the DIM algebra
in terms of the universal $R$-matrix $\mathcal{R}_{0}$ (2.25);

$\displaystyle\bigg{[}\big{(}\rho^{V}_{v_{1}}\otimes\rho^{V}_{v_{2}}\big{)}\big{(}\mathcal{R}_{0}\big{)}\bigg{]}\bigg{[}[v_{1}]_{m}\otimes[v_{2}]_{l}\bigg{]}=R_{ml}\big{(}\frac{v_{1}}{v_{2}}\big{)}\bigg{[}[v_{1}]_{m}\otimes[v_{2}]_{l}\bigg{]}.$
(5.26)

Thus, to determine the $R_{ml}(\frac{v_{1}}{v_{2}})$ we have to calculate the
following quantity :

$\displaystyle\exp\bigg{(}\sum_{r=1}^{\infty}r\kappa_{r}\rho^{V}_{v_{1}}(h_{-r})\otimes\rho^{V}_{v_{2}}(h_{r})\bigg{)}\bigg{[}[v_{1}]_{m}\otimes[v_{2}]_{l}\bigg{]}.$
(5.27)

From Eqs.(2.3), (3.1) and (3.2), we see that

$\rho^{V}_{v}(H_{\pm r})[v]_{i}=\pm\frac{(q^{i}_{1}v)^{\pm r}}{r}q^{(1\pm
1)r/2}_{1}(q^{\pm r}_{2}-1)(q^{\pm r}_{3}-1)[v]_{i}.$ (5.28)

Hence,

$\rho^{V}_{v}(h_{\pm r})[v]_{i}=\pm\frac{1}{\kappa_{r}}\rho^{V}_{v}(H_{\pm
r})[v]_{i}=\frac{(q^{i}_{1}v)^{\pm r}}{r\kappa_{r}}q^{(1\pm 1)r/2}_{1}(q^{\pm
r}_{2}-1)(q^{\pm r}_{3}-1)[v]_{i}.$ (5.29)

Accordingly, we obtain

$\displaystyle R_{ml}\Big{(}\frac{v_{1}}{v_{2}}\Big{)}$
$\displaystyle=\exp\bigg{[}-\sum_{r=1}^{\infty}\frac{1}{r}\Big{(}q^{m-l}_{1}\frac{v_{1}}{v_{2}}\Big{)}^{-r}\frac{q^{r}_{1}(1-q^{r}_{2})(1-q^{r}_{3})}{1-q^{r}_{1}}\bigg{]}$
$\displaystyle=\frac{\bigg{(}q^{l-m+1}_{1}\frac{v_{2}}{v_{1}};q_{1}\bigg{)}_{\infty}\bigg{(}q^{l-m}_{1}\frac{v_{2}}{v_{1}};q_{1}\bigg{)}_{\infty}}{\bigg{(}\frac{q^{l-m}_{1}}{q_{2}}\frac{v_{2}}{v_{1}};q_{1}\bigg{)}_{\infty}\bigg{(}\frac{q^{l-m}_{1}}{q_{3}}\frac{v_{2}}{v_{1}};q_{1}\bigg{)}_{\infty}},$
(5.30)

where we have used the formula (1.12).

By direct calculation, it is easy to see that the following relation

$\displaystyle
R_{nl}(x)=\frac{\Upsilon\big{(}q_{1}\mathfrak{q}^{m}\big{|}q^{n-l}_{1}\mathfrak{q}^{m}x\big{)}}{\Upsilon\big{(}q_{1}\mathfrak{q}^{m+1}\big{|}q^{n-l}_{1}\mathfrak{q}^{m-1}x\big{)}}$
(5.31)

holds for any $m\in\mathbb{Z}$. Combining (5.31) with
$\Upsilon(\alpha\big{|}x)^{-1}=\Upsilon(\alpha\big{|}x^{-1})$, we can write
Eqs.(5.22) – (5.25) in terms of the $R$ matrix;

$\displaystyle\mathcal{T}^{+}_{n}(N,u\big{|}v)\mathbb{I}_{k}(z)$
$\displaystyle=R_{nk}(\mathfrak{q}^{1/2}\frac{v}{z})\mathbb{I}_{k}(z)\mathcal{T}^{+}_{n}(N,q_{2}u\big{|}v),$
(5.32) $\displaystyle\mathcal{T}^{-}_{n}(N,u\big{|}v)\mathbb{I}_{k}(z)$
$\displaystyle=R_{kn}(\mathfrak{q}^{1/2}\frac{z}{v})\mathbb{I}_{k}(z)\mathcal{T}^{-}_{n}(N,q_{2}u\big{|}v),$
(5.33) $\displaystyle\mathcal{T}^{+}_{n}(N,u\big{|}v)\mathbb{I}^{*}_{k}(z)$
$\displaystyle=\frac{1}{R_{nk}(\mathfrak{q}^{-1/2}\frac{v}{z})}\mathbb{I}^{*}_{k}(z)\mathcal{T}^{+}_{n}(N,q_{2}^{-1}u\big{|}v),$
(5.34) $\displaystyle\mathcal{T}^{-}_{n}(N,u\big{|}v)\mathbb{I}^{*}_{k}(z)$
$\displaystyle=\frac{1}{R_{kn}(\mathfrak{q}^{-1/2}\frac{z}{v})}\mathbb{I}^{*}_{k}(z)\mathcal{T}^{-}_{n}(N,q^{-1}_{2}u\big{|}v).$
(5.35)

###  5.4. KZ equation for the vector intertwiners

We define the correlation function of vector intertwiners and its duals by

$\displaystyle\mathcal{G}_{\lambda_{1},\cdots,\lambda_{n}}^{\mu_{1},\cdots,\mu_{m}}\big{(}$
$\displaystyle u_{n}\big{|}~{}{w_{1},\cdots,w_{m},z_{1},\cdots,z_{n}}\big{)}$
$\displaystyle:=$
$\displaystyle\langle\emptyset|\mathbb{I}^{*}_{\mu_{1}}\big{(}q_{2}^{-(n-m)}u_{n}\big{|}w_{1}\big{)}\cdots\mathbb{I}^{*}_{\mu_{m-1}}\big{(}q_{2}^{-(n-2)}u_{n}\big{|}w_{m-1}\big{)}\mathbb{I}^{*}_{\mu_{m}}\big{(}q_{2}^{-(n-1)}u_{n}\big{|}w_{m}\big{)}$
$\displaystyle\cdot~{}\mathbb{I}_{\lambda_{1}}\big{(}q_{2}^{-(n-1)}u_{n}\big{|}z_{1}\big{)}\cdots\mathbb{I}_{\lambda_{k}}\big{(}q_{2}^{-(n-k)}u_{n}\big{|}z_{k}\big{)}\cdots\mathbb{I}_{\lambda_{n}}\big{(}u_{n}\big{|}z_{n}\big{)}|\emptyset\rangle.$
(5.36)

Consider the shift of the spectral parameter of the vector intertwiner;

$\displaystyle(\mathfrak{q}^{-2})^{z_{k}\frac{\partial}{\partial z_{k}}}$
$\displaystyle\mathcal{G}_{\lambda_{1},\cdots,\lambda_{n}}^{\mu_{1},\cdots,\mu_{m}}\big{(}u_{n}\big{|}~{}{w_{1},\cdots,w_{m},z_{1},\cdots,z_{n}}\big{)}=$
$\displaystyle\langle\emptyset|$
$\displaystyle\mathbb{I}^{*}_{\mu_{1}}\big{(}q_{2}^{-(n-m)}u_{n}\big{|}w_{1}\big{)}\cdots\mathbb{I}^{*}_{\mu_{m-1}}\big{(}q_{2}^{-(n-2)}u_{n}\big{|}w_{m-1}\big{)}\mathbb{I}^{*}_{\mu_{m}}\big{(}q_{2}^{-(n-1)}u_{n}\big{|}w_{m}\big{)}$
$\displaystyle\cdot~{}\mathbb{I}_{\lambda_{1}}\big{(}q_{2}^{-(n-1)}u_{n}\big{|}z_{1}\big{)}\cdots\mathbb{I}_{\lambda_{k}}\big{(}q_{2}^{-(n-k)}u_{n}\big{|}\mathfrak{q}^{-2}z_{k}\big{)}\cdots\mathbb{I}_{\lambda_{n}}\big{(}u_{n}\big{|}z_{n}\big{)}|\emptyset\rangle.$
(5.37)

Here we explicitly write the information of incoming horizontal for each
vector intertwiner. The only independent horizontal spectral parameter is of
the incoming horizontal of $\mathbb{I}_{\lambda_{n}}$. The other parameters
are determined automatically by the property of vector intertwiners. From
(5.8) we know that

$\displaystyle\mathbb{I}_{\lambda_{k}}\big{(}$ $\displaystyle
q_{2}^{-(n-k)}u_{n}\big{|}\mathfrak{q}^{-2}z_{k}\big{)}=\exp\bigg{(}\sum_{r=1}^{\infty}\frac{q^{r}_{1}}{r}\frac{1-q^{r}_{2}}{1-q^{r}_{1}}(1-\mathfrak{q}^{2r})\bigg{)}$
$\displaystyle\cdot~{}\mathcal{T}^{-}_{\lambda_{k}}\big{(}N,q^{-(n-k)}_{2}u_{n}\big{|}\mathfrak{q}^{-3/2}z_{k}\big{)}\cdot\mathbb{I}_{\lambda_{k}}\big{(}q_{2}^{-(n-k)}u_{n}\big{|}z_{k}\big{)}\cdot\mathcal{T}^{+}_{\lambda_{k}}\big{(}N,q^{-(n-k)}_{2}u_{n}\big{|}\mathfrak{q}^{-1/2}z_{k}\big{)}.$
(5.38)

Hence, we can rewrite
$\mathbb{I}_{\lambda_{k}}\big{(}\mathfrak{q}^{-(n-k)}u_{n}\big{|}\mathfrak{q}^{-2}z_{k}\big{)}$
in (5.37) by using (5.38), which implies

$\displaystyle(\mathfrak{q}^{-2})^{z_{k}\frac{\partial}{\partial z_{k}}}$
$\displaystyle\mathcal{G}_{\lambda_{1},\cdots,\lambda_{n}}^{\mu_{1},\cdots,\mu_{m}}\big{(}u_{n}\big{|}~{}{w_{1},\cdots,w_{m},z_{1},\cdots,z_{n}}\big{)}$
$\displaystyle=\langle\emptyset|$
$\displaystyle\mathbb{I}^{*}_{\mu_{1}}\big{(}q_{2}^{-(n-m)}u_{n}\big{|}w_{1}\big{)}\cdots\mathbb{I}^{*}_{\mu_{m-1}}\big{(}q_{2}^{-(n-2)}u_{n}\big{|}w_{m-1}\big{)}\mathbb{I}^{*}_{\mu_{m}}\big{(}q_{2}^{-(n-1)}u_{n}\big{|}w_{m}\big{)}$
$\displaystyle\cdot~{}\mathbb{I}_{\lambda_{1}}\big{(}q_{2}^{-(n-1)}u_{n}\big{|}z_{1}\big{)}\cdots\mathbb{I}_{\lambda_{k-1}}\big{(}q_{2}^{-(n-k+1)}u_{n}\big{|}z_{k-1}\big{)}$
$\displaystyle\cdot~{}\mathcal{T}^{-}_{\lambda_{k}}\big{(}N,q^{-(n-k+1)}_{2}u_{n}\big{|}\mathfrak{q}^{-3/2}z_{k}\big{)}\cdot\mathbb{I}_{\lambda_{k}}\big{(}q_{2}^{-(n-k)}u_{n}\big{|}z_{k}\big{)}\cdot\mathcal{T}^{+}_{\lambda_{k}}\big{(}N,q^{-(n-k)}_{2}u_{n}\big{|}\mathfrak{q}^{-1/2}z_{k}\big{)}$
$\displaystyle\cdot~{}\mathbb{I}_{\lambda_{k+1}}\big{(}q_{2}^{-(n-k-1)}u_{n}\big{|}z_{k+1}\big{)}\cdots\mathbb{I}_{\lambda_{n}}\big{(}u_{n}\big{|}z_{n}\big{)}|\emptyset\rangle$
$\displaystyle\cdot\exp\bigg{(}\sum_{r=1}^{\infty}\frac{q^{r}_{1}}{r}\frac{1-q^{r}_{2}}{1-q^{r}_{1}}(1-\mathfrak{q}^{2r})\bigg{)}.$
(5.39)

Now by using (5.32), (5.33) and (5.35), we can transport
$\mathcal{T}^{-}_{\lambda_{k}}$ and $\mathcal{T}^{+}_{\lambda_{k}}$ in (5.39)
until they hit the left and the right vacua. Then, using (5.5) and (5.6), we
calculate
$\langle\emptyset|\mathcal{T}^{-}_{\lambda_{k}}\big{(}N,q^{m-n}_{2}u_{n}\big{|}\mathfrak{q}^{-3/2}z_{k}\big{)}$
and
$\mathcal{T}^{+}_{\lambda_{k}}\big{(}N,u_{n}\big{|}\mathfrak{q}^{-1/2}z_{k}\big{)}|\emptyset\rangle$.
The result cancels the exponential factor in (5.39). Hence, we finally obtain

$\displaystyle(\mathfrak{q}^{-2})^{z_{k}\frac{\partial}{\partial
z_{k}}}\mathcal{G}_{\lambda_{1},\cdots,\lambda_{n}}^{\mu_{1},\cdots,\mu_{m}}\big{(}u_{n}\big{|}~{}{w_{1},\cdots,w_{m},z_{1},\cdots,z_{n}}\big{)}=A_{k}\cdot\mathcal{G}_{\lambda_{1},\cdots,\lambda_{n}}^{\mu_{1},\cdots,\mu_{m}}\big{(}u_{n}\big{|}~{}{w_{1},\cdots,w_{m},z_{1},\cdots,z_{n}}\big{)},$
$\displaystyle
A_{k}:=\prod_{j=1}^{m}\Big{[}R_{\mu_{j}\lambda_{k}}\Big{(}\frac{\mathfrak{q}w_{j}}{z_{k}}\Big{)}\Big{]}\prod_{j=1}^{k-1}\Big{[}R_{\lambda_{j}\lambda_{k}}\Big{(}\mathfrak{q}^{2}\frac{z_{j}}{z_{k}}\Big{)}^{-1}\Big{]}\prod_{j=k+1}^{n}\Big{[}R_{\lambda_{k}\lambda_{j}}\Big{(}\frac{z_{k}}{z_{j}}\Big{)}\Big{]}.$
(5.40)

Now we consider the shift of the spectral parameter of the dual vector
intertwiner;

$\displaystyle(\mathfrak{q}^{-2})^{w_{j}\frac{\partial}{\partial w_{j}}}$
$\displaystyle\mathcal{G}_{\lambda_{1},\cdots,\lambda_{n}}^{\mu_{1},\cdots,\mu_{m}}\big{(}u_{n}\big{|}~{}{w_{1},\cdots,w_{m},z_{1},\cdots,z_{n}}\big{)}$
$\displaystyle=$
$\displaystyle\langle\emptyset|\mathbb{I}^{*}_{\mu_{1}}\big{(}q^{-(n-m)}_{2}u_{n}\big{|}w_{1}\big{)}\cdots\mathbb{I}^{*}_{\mu_{j-1}}\big{(}q_{2}^{-(n-m+j-2)}u_{n}\big{|}w_{j-1}\big{)}$
$\displaystyle\cdot~{}\mathbb{I}^{*}_{\mu_{j}}\big{(}q_{2}^{-(n-m+j-1)}u_{n}\big{|}\mathfrak{q}^{-2}w_{j}\big{)}\mathbb{I}^{*}_{\mu_{j+1}}\big{(}q_{2}^{-(n-m+j)}u_{n}\big{|}w_{j+1}\big{)}\cdots\mathbb{I}^{*}_{\mu_{m}}\big{(}q_{2}^{-(n-1)}u_{n}\big{|}w_{m}\big{)}$
$\displaystyle\cdot~{}\mathbb{I}_{\lambda_{1}}\big{(}q_{2}^{-(n-1)}u_{n}\big{|}z_{1}\big{)}\cdots\mathbb{I}_{\lambda_{n}}\big{(}u_{n}\big{|}z_{n}\big{)}|\emptyset\rangle.$
(5.41)

We can rewrite

$\displaystyle\mathbb{I}^{*}_{\mu_{j}}$
$\displaystyle\big{(}q_{2}^{-(n-m+j-1)}u_{n}\big{|}\mathfrak{q}^{-2}w_{j}\big{)}=\exp\bigg{[}\sum_{r=1}^{\infty}\frac{q^{r}_{1}(1-q^{r}_{2})}{r(1-q^{r}_{1})}(\mathfrak{q}^{2r}-1)\bigg{]}$
$\displaystyle\cdot~{}\mathcal{T}^{-}_{\mu_{j}}\big{(}N,q^{-(n-m+j-1)}_{2}u_{n}\big{|}\mathfrak{q}^{-1/2}w_{j}\big{)}^{-1}\cdot\mathbb{I}^{*}_{\mu_{j}}(w_{j})\cdot\mathcal{T}^{+}_{\mu_{j}}\big{(}N,q^{-(n-m+j)}_{2}u_{n}\big{|}\mathfrak{q}^{-3/2}w_{j}\big{)}^{-1}.$
(5.42)

Now we move the operator $\mathcal{T}^{+}_{\mu_{j}}$ to the rightmost and
$\mathcal{T}^{-}_{\mu_{j}}$ to the leftmost as above. By using (5.5) and
(5.6), we can calculate
$\langle\emptyset|\mathcal{T}^{-}_{\mu_{j}}\big{(}N,q^{-(n-m)}_{2}u_{n}\big{|}\mathfrak{q}^{-1/2}w_{j}\big{)}^{-1}$
and
$\mathcal{T}^{+}_{\mu_{j}}\big{(}N,u_{n}\big{|}\mathfrak{q}^{-3/2}w_{j}\big{)}^{-1}|\emptyset\rangle$.
The result of this is the cancellation of the exponential factor in (5.42).
The final result is

$\displaystyle(\mathfrak{q}^{-2})^{w_{j}\frac{\partial}{\partial
w_{j}}}\mathcal{G}_{\lambda_{1},\cdots,\lambda_{n}}^{\mu_{1},\cdots,\mu_{m}}\big{(}u_{n}\big{|}~{}{w_{1},\cdots,w_{m},z_{1},\cdots,z_{n}}\big{)}=A_{j}^{*}\cdot\mathcal{G}_{\lambda_{1},\cdots,\lambda_{n}}^{\mu_{1},\cdots,\mu_{m}}\big{(}u_{n}\big{|}~{}{w_{1},\cdots,w_{m},z_{1},\cdots,z_{n}}\big{)},$
$\displaystyle
A_{j}^{*}:=\prod_{l=1}^{j-1}\Big{[}R_{\mu_{l}\mu_{j}}\Big{(}\frac{w_{l}}{w_{j}}\Big{)}^{-1}\Big{]}\prod_{l=j+1}^{m}\Big{[}R_{\mu_{j}\mu_{l}}\Big{(}\mathfrak{q}^{-2}\frac{w_{j}}{w_{l}}\Big{)}\Big{]}\prod_{k=1}^{n}\Big{[}R_{\mu_{j}\lambda_{k}}\Big{(}\mathfrak{q}^{-1}\frac{w_{j}}{z_{k}}\Big{)}^{-1}\Big{]}.$
(5.43)

We call the system of Eqs.(5.40) and (5.43) generalized KZ equation for the
vector representation.

###  5.5. Shift operator from the universal $R$-matrix

Before proceeding to the generalized KZ equation for the MacMahon
representation, in this section we provide an alternative method to obtain the
shift operator. This method is based on the use of the universal $R$-matrix of
DIM algebra $\mathcal{R}_{0}$ (2.25). Consider the quantity
$\bigg{\\{}[\rho^{V}_{v}\otimes\rho^{H}](\mathcal{R}_{0})\bigg{\\}}[v]_{n-1}\otimes|\bullet\rangle$.
Here $\rho^{V}_{v}$ is the vector representation with the spectral parameter
$v$ described in section 3.1, and $\rho^{H}$ is the horizontal representation
described in section 3.5. Up to a proportional factor, we see that

$\displaystyle\bigg{\\{}[\rho^{V}_{v}\otimes\rho^{H}](\mathcal{R}_{0})\bigg{\\}}[v]_{n-1}\otimes|\bullet\rangle\sim\bigg{\\{}\exp\big{[}\sum_{r=1}^{\infty}r\kappa_{r}\rho^{V}(h_{-r})\otimes\rho^{H}(h_{r})\big{]}\bigg{\\}}\bigg{(}[v]_{n-1}\otimes|\bullet\rangle\bigg{)}.$
(5.44)

Eq.(5.29) tells us that

$\displaystyle\bigg{\\{}[\rho^{V}_{v}\otimes\rho^{H}]($
$\displaystyle\mathcal{R}_{0})\bigg{\\}}[v]_{n-1}\otimes|\bullet\rangle$
$\displaystyle\sim\bigg{\\{}\exp\big{[}\sum_{r=1}^{\infty}\frac{q^{-nr}_{1}(1-q^{-r}_{3})(1-q^{-r}_{2})}{v^{r}}\otimes\frac{\rho^{H}(H_{r})}{\kappa_{r}}\big{]}\bigg{\\}}\bigg{(}[v]_{n-1}\otimes|\bullet\rangle\bigg{)}$
$\displaystyle=\displaystyle\bigg{\\{}\exp\big{[}-\sum_{r=1}^{\infty}\frac{q^{-(n-1)r}_{1}}{v^{r}}\frac{H_{r}}{1-q^{r}_{1}}\big{]}\bigg{\\}}\bigg{(}[v]_{n-1}\otimes|\bullet\rangle\bigg{)}.$
(5.45)

Note that here we suppress the tensor product notation and $\rho^{H}$. The
context will tell us either $H_{r}$ plays a role of generator of the DIM
algebra or the operator on a representation. If we compare this operator with
Eq.(5.6), we see

$[\rho^{V}_{v}\otimes\rho^{H}](\mathcal{R}_{0})\sim\mathcal{T}^{n}_{n}(N,u|\mathfrak{q}^{1/2}v,\mathfrak{q}^{-1/2}v)^{-1}=\mathcal{T}^{+}_{n}\big{(}N,u\big{|}v\big{)}.$
(5.46)

Now consider the quantity
$\bigg{\\{}[\rho^{V}_{v}\otimes\rho^{H}](P\mathcal{R}_{0})\bigg{\\}}$. Here
the operator $P$ switches the position between the elements in tensor product,
i.e. $P(h_{-r}\otimes h_{r})=h_{r}\otimes h_{-r}$. This time Eq.(5.29) implies
that

$\displaystyle\bigg{\\{}\exp\big{[}\sum_{r=1}^{\infty}r\kappa_{r}\rho^{V}(h_{r})\otimes$
$\displaystyle\rho^{H}(h_{-r})\big{]}\bigg{\\}}\bigg{(}[v]_{n-1}\otimes|\bullet\rangle\bigg{)}$
$\displaystyle=\displaystyle\bigg{\\{}\exp\big{[}\sum_{r=1}^{\infty}(q^{n}_{1}v)^{r}\frac{1}{1-q^{r}_{1}}H_{-r}\big{]}\bigg{\\}}\bigg{(}[v]_{n-1}\otimes|\bullet\rangle\bigg{)}.$
(5.47)

If we compare this operator with Eq.(5.5), we see that

$[\rho^{V}_{v}\otimes\rho^{H}](P\mathcal{R}_{0})\sim\mathcal{T}^{n}_{n}(N,u|\mathfrak{q}^{-1/2}v,\mathfrak{q}^{1/2}v)=\mathcal{T}^{-}_{n}\big{(}N,u\big{|}v\big{)}.$
(5.48)

##  6\. Generalized KZ equation for MacMahon intertwiners

###  6.1. MacMahon shift operator

In the last section we learn that the formulas (5.46) and (5.48) provide us
the shift operators $\mathcal{T}^{\pm}$ from the universal $R$ matrix.
Replacing the vector representation with the MacMahon representation, we thus
consider the quantity

$\displaystyle\Bigg{\\{}\bigg{[}$
$\displaystyle\rho^{M}_{v,K_{1}}\otimes\rho^{H}\bigg{]}\big{(}\mathcal{R}_{0}\big{)}\Bigg{\\}}\big{(}|\Pi,v,K_{1}\rangle\otimes|\bullet\rangle\big{)}$
$\displaystyle=\Bigg{\\{}(K_{0}^{-}\otimes\mathfrak{q}^{d_{1}})(\mathfrak{q}^{d_{1}}\otimes
K_{0}^{-})\exp\big{[}\sum_{n=1}^{\infty}n\kappa_{n}\rho^{M}_{v,K_{1}}(h_{-n})\otimes\rho^{H}(h_{n})\big{]}\Bigg{\\}}\big{(}|\Pi,v,K_{1}\rangle\otimes|\bullet\rangle\big{)},$
(6.1)

where $\rho^{M}_{v,K_{1}}$ represents the vertical MacMahon module with the
spectral parameter $v$ and level $K_{1}$, and $H$ denotes the horizontal Fock
module with the spectral parameter $u$. To calculate
$\bigg{[}\rho^{M}_{v,K_{1}}(h_{-n})\otimes\rho^{H}(h_{n})\bigg{]}\Big{(}|\Pi,v,K_{1}\rangle\otimes|\bullet\rangle\Big{)}$
we need to know
$\bigg{[}\rho^{M}_{v,K_{1}}(h_{-n})\bigg{]}\Big{(}|\Pi,v,K_{1}\rangle\Big{)}$,
which has been calculated in [11]. We systematically state the result as the
following lemma111111Recall the exchange of $q_{1}$ and $q_{2}$ in the
formulas of [11]. We have corrected typos in [11].;

######  Lemma 6.1.

For any $n(\neq 0)\in\mathbb{Z}$

$\displaystyle\rho^{M}_{v,K}(h_{n})|\Pi,v,K\rangle=\frac{v^{n}}{n}\bigg{[}\frac{1-K^{n}}{\kappa_{n}}+\sum_{(i,j,k)\in\Pi}q^{ni}_{1}q^{nj}_{2}q^{nk}_{3}\bigg{]}|\Pi,v,K\rangle.$

Thus, we obtain

$\displaystyle\Bigg{\\{}$
$\displaystyle\bigg{[}\rho^{M}_{v,K_{1}}\otimes\rho^{H}\bigg{]}\big{(}\mathcal{R}_{0}\big{)}\Bigg{\\}}\big{(}|\Pi,v,K_{1}\rangle\otimes|\bullet\rangle\big{)}$
$\displaystyle=\alpha\,\Bigg{\\{}$
$\displaystyle\exp\bigg{[}-\sum_{n=1}^{\infty}n\kappa_{n}\bigg{(}\frac{v^{-n}}{n}\bigg{[}\frac{1-K^{-n}}{\kappa_{-n}}+\sum_{(i,j,k)\in\Pi}q^{-ni}_{1}q^{-nj}_{2}q^{-nk}_{3}\bigg{]}\otimes\frac{H_{n}}{\kappa_{n}}\bigg{)}\bigg{]}\Bigg{\\}}\big{(}|\Pi,v,K_{1}\rangle\otimes|\bullet\rangle\big{)}.$
(6.2)

where we have used the fact that $d_{1}$ commutes with $H_{r}$ and $\alpha$ is
the multiplicative constant in front of the action of
$(K_{0}^{-}\otimes\mathfrak{q}^{d_{1}})(\mathfrak{q}^{d_{1}}\otimes
K_{0}^{-})$ on the state $|\Pi,v,K_{1}\rangle\otimes|\bullet\rangle$.
Thus121212The notation $\mathcal{T}^{+}_{\Pi}(N\big{|}v,K)$ resembles the
notation of $\mathcal{T}$ operator of the vector module. However, it is just a
matter of notation.,

$\displaystyle\mathcal{T}^{+}_{\Pi}(v,K)\sim\exp\bigg{(}\sum_{n=1}^{\infty}\bigg{[}\frac{1-K^{-n}}{\kappa_{n}}-\sum_{(i,j,k)\in\Pi}q^{-ni}_{1}q^{-nj}_{2}q^{-nk}_{3}\bigg{]}v^{-n}H_{n}\bigg{)},$
(6.3)

where we have suppressed the tensor product.

Similarly,

$\displaystyle\Bigg{\\{}\bigg{[}\rho^{M}_{v,K_{1}}\otimes\rho^{H}\bigg{]}\big{(}\mathcal{P}\mathcal{R}_{0}\big{)}\Bigg{\\}}\big{(}|\Pi,v,K_{1}\rangle\otimes|\bullet\rangle\big{)}$
$\displaystyle=\Bigg{\\{}(K_{0}^{-}\otimes\mathfrak{q}^{d_{1}})(\mathfrak{q}^{d_{1}}\otimes
K_{0}^{-})\exp\bigg{[}\sum_{n=1}^{\infty}\bigg{(}n\kappa_{n}\frac{v^{n}}{n}\bigg{[}\frac{1-K^{n}}{\kappa_{n}}+\sum_{(i,j,k)\in\Pi}q^{ni}_{1}q^{nj}_{2}q^{nk}_{3}\bigg{]}$
$\displaystyle\quad\quad\otimes(-1)\frac{H_{-n}}{\kappa_{n}}\bigg{)}\bigg{]}\Bigg{\\}}\big{(}|\Pi,v,K_{1}\rangle\otimes|\bullet\rangle\big{)}.$
(6.4)

Note that we have applied Lemma 6.1 as before. Thus,

$\displaystyle\Bigg{\\{}\bigg{[}\rho^{M}_{v,K_{1}}\otimes\rho^{H}\bigg{]}\big{(}\mathcal{P}\mathcal{R}_{0}\big{)}\Bigg{\\}}\big{(}|\Pi,v,K_{1}\rangle\otimes|\bullet\rangle\big{)}$
$\displaystyle=\alpha\,\,\Bigg{\\{}\exp\bigg{[}-\sum_{n=1}^{\infty}\bigg{(}n\kappa_{n}\frac{v^{n}}{n}\bigg{[}\frac{1-K^{n}}{\kappa_{n}}+\sum_{(i,j,k)\in\Pi}q^{ni}_{1}q^{nj}_{2}q^{nk}_{3}\bigg{]}\otimes\frac{H_{-n}}{\kappa_{n}}\bigg{)}\bigg{]}\Bigg{\\}}\big{(}|\Pi,v,K_{1}\rangle\otimes|\bullet\rangle\big{)}.$
(6.5)

Suppressing the tensor product, we are able to conclude that

$\displaystyle\mathcal{T}^{-}_{\Pi}(v,K)\sim\exp\bigg{(}-\sum_{n=1}^{\infty}\bigg{[}\frac{1-K^{n}}{\kappa_{n}}+\sum_{(i,j,k)\in\Pi}q^{ni}_{1}q^{nj}_{2}q^{nk}_{3}\bigg{]}v^{n}H_{-n}\bigg{)}.$
(6.6)

###  6.2. Effect of the shift operator

As in the case of the vector and the Fock intertwiners we expect that we are
able to write the shift of the spectral parameter in term of the shift
operators $\mathcal{T}^{+}$ and $\mathcal{T}^{-}$. Schematically,

$\displaystyle\Xi(pv)=\mathcal{T}^{-}\cdot\Xi(v)\cdot\mathcal{T}^{+}.$ (6.7)

However, the value of $p$ in (6.7) is not arbitrary but only certain
appropriate values are allowed. We first determine these appropriate values of
$p$.

Recall that the MacMahon intertwiner is given by (see (4.18))

$\displaystyle\Xi_{\Lambda}(K;v)=z_{\Lambda}(K;v)\mathcal{M}^{[n]}(K)\tilde{\Phi}^{[n]}_{\Lambda}(v)\Gamma_{n}(K;v)$
(6.8)

where $\Gamma_{n}(K;v)$ and $z_{\Lambda}(K;v)$ are given by (4.19) and (4.20).
Note also that (see (4.21))

$\displaystyle\tilde{\Phi}^{[n]}_{\Lambda}(v)$
$\displaystyle=\tilde{\Phi}_{\Lambda^{(1)}}(v)\circ\cdots\circ\tilde{\Phi}_{\Lambda^{(n)}}(q_{3}^{n-1}v),$
(6.9)

where the tilde means the intertwiner without the zero mode factor. In the
following, introducing the horizontal spectral parameter $u$, we use the zero
mode factor

$\displaystyle\textbf{e}(z)=K^{1/2}\frac{\theta_{q_{3}}(z^{-1})}{\theta_{q_{3}}(Kz^{-1})}u.$
(6.10)

so that we can write

$\displaystyle
z_{\Lambda}(K;v)=\prod_{k=1}^{h(\Lambda)}\prod_{(i,j)\in\Lambda^{(k)}}\frac{K}{\mathfrak{q}^{k-1}}\frac{\theta_{q_{3}}(q_{3}^{k-1}/x_{ijk})}{\theta_{q_{3}}(K/x_{ijk})}\frac{\theta_{q_{3}}(1/(x_{ijk}v))}{\theta_{q_{3}}(K/(x_{ijk}v))}u.$
(6.11)

Let us calculate $\Xi_{\Lambda}(K;pv)$. Firstly, we see that

$\displaystyle\Gamma_{n}(K;pv)=$
$\displaystyle\exp\bigg{(}\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{q^{nr}_{3}-K^{r}}{\kappa_{r}}(\mathfrak{q}^{-1/2}v)^{r}(p^{r}-1)\bigg{)}$
$\displaystyle\cdot\Gamma_{n}(K;v)\exp\bigg{(}\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{q^{-nr}_{3}-K^{-r}}{\kappa_{r}}(\mathfrak{q}^{1/2}v)^{-r}(p^{-r}-1)\bigg{)},$
(6.12)

and

$\displaystyle z_{\Lambda}(K;pv)$
$\displaystyle=\prod_{k=1}^{h(\Lambda)}\prod_{(i,j)\in\Lambda^{(k)}}\frac{K}{\mathfrak{q}^{k-1}}\frac{\theta_{q_{3}}(q_{3}^{k-1}/x_{ijk})}{\theta_{q_{3}}(K/x_{ijk})}\frac{\theta_{q_{3}}(1/(x_{ijk}pv))}{\theta_{q_{3}}(K/(x_{ijk}pv))}u$
$\displaystyle=z_{\Lambda}(K;v)\prod_{k=1}^{h(\Lambda)}\prod_{(i,j)\in\Lambda^{(k)}}\frac{\theta_{q_{3}}(1/(x_{ijk}pv))}{\theta_{q_{3}}(K/(x_{ijk}pv))}\frac{\theta_{q_{3}}(K/(x_{ijk}v))}{\theta_{q_{3}}(1/(x_{ijk}v))}.$
(6.13)

To calculate $\tilde{\Phi}^{[n]}_{\Lambda}(pv)$ we need to know an explicit
expression of $\tilde{\Phi}^{[n]}_{\Lambda}(v)$ first. It is convenient to
introduce a notation

$\chi_{\pm r}^{(k)}(q_{1},q_{2}):=\sum_{(i,j)\in\Lambda^{(k)}}x^{\pm
r}_{ij}-\frac{1}{(1-q^{\pm r}_{1})(1-q^{\pm r}_{2})},$ (6.14)

Using this notation, we state its expression systematically as the following
lemma;

######  Lemma 6.2.

$\displaystyle\tilde{\Phi}^{[n]}_{\Lambda}(v)=c(\Lambda)\,\exp\big{[}\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\sum_{k=1}^{n}(\mathfrak{q}^{-1/2}q^{k-1}_{3}v)^{r}\chi_{r}^{(k)}(q_{1},q_{2})\big{]}$
$\displaystyle\exp\big{[}-\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\sum_{k=1}^{n}(\mathfrak{q}^{1/2}q^{k-1}_{3}v)^{-r}\chi_{-r}^{(k)}(q_{1},q_{2})\big{]},$

where $c(\Lambda)$ is a constant which may depend on $\Lambda$.

###### Proof.

By definition,

$\displaystyle\tilde{\Phi}^{[n]}_{\Lambda}(v)$
$\displaystyle=\tilde{\Phi}_{\Lambda^{(1)}}(v)\circ\cdots\circ\tilde{\Phi}_{\Lambda^{(n)}}(q_{3}^{n-1}v)$
$\displaystyle=\mathcal{G}^{-1}_{\Lambda^{(1)}}\mathcal{G}^{-1}_{\Lambda^{(2)}}\cdots\mathcal{G}^{-1}_{\Lambda^{(n)}}\big{(}\text{exponential
factor}\big{)}$ (6.15)

where (exponential factor) in the above equation is equal to

$\displaystyle\prod_{k=1}^{n}\Bigg{[}\exp\big{[}\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}(\mathfrak{q}^{-1/2}q^{k-1}_{3}v)^{r}\big{(}\sum_{(i,j)\in\Lambda^{(k)}}x^{r}_{ij}-\frac{1}{(1-q^{r}_{1})(1-q^{r}_{2})}\big{)}\big{]}$
$\displaystyle\exp\big{[}-\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}(\mathfrak{q}^{1/2}q^{k-1}_{3}v)^{-r}\big{(}\sum_{(i,j)\in\Lambda^{(k)}}x_{ij}^{-r}-\frac{1}{(1-q^{-r}_{1})(1-q^{-r}_{2})}\big{)}\big{]}\Bigg{]}.$
(6.16)

Applying the formula (1.14) to Eq.(6.16), we obtain the desired result. ∎

######  Remark 6.3.

Recall that the aim of this subsection is to calculate the appropriate value
of $p$ by comparing with the exponential factor appearing in (6.3) and (6.6).
For this purpose, only the exponential factor of $H_{\pm r}$ is instrumental,
and we do not pay attention to the form of the proportional factor $c$. For
the sake of convenience we introduce the notation $\sim$ to mean equal up to a
proportional factor.

By Lemma 6.2

$\displaystyle\tilde{\Phi}^{[n]}_{\Lambda}$
$\displaystyle(pv)=\exp\big{[}\sum_{r=1}^{\infty}\frac{(p^{r}-1)H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\sum_{k=1}^{n}(\mathfrak{q}^{-1/2}q^{k-1}_{3}v)^{r}\chi_{r}^{(k)}(q_{1},q_{2})\big{]}$
$\displaystyle\cdot\tilde{\Phi}^{[n]}_{\Lambda}(v)\exp\big{[}-\sum_{r=1}^{\infty}\frac{(p^{-r}-1)H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\sum_{k=1}^{n}(\mathfrak{q}^{1/2}q^{k-1}_{3}v)^{-r}\chi_{-r}^{(k)}(q_{1},q_{2})\big{]}.$
(6.17)

Merging (6.12) and (6.17), we see that

$\Xi_{\Lambda}(K;pv)\sim\widetilde{\mathcal{T}}^{-}_{\Lambda}(p)\cdot\,\Xi_{\Lambda}(K;v)\cdot\,\widetilde{\mathcal{T}}^{+}_{\Lambda}(p),$
(6.18)

where $\widetilde{\mathcal{T}}^{-}_{\Lambda}(p)$ and
$\widetilde{\mathcal{T}}^{+}_{\Lambda}(p)$ are

$\exp\bigg{(}\sum_{r=1}^{\infty}\frac{(p^{r}-1)H_{-r}}{(\mathfrak{q}^{r}-\mathfrak{q}^{-r})}(\mathfrak{q}^{-1/2}v)^{r}\Big{\\{}\frac{q^{nr}_{3}-K^{r}}{\kappa_{r}}+\sum_{k=1}^{n}q^{r(k-1)}_{3}\chi_{r}^{(k)}(q_{1},q_{2})\Big{\\}}\bigg{)},$

and

$\exp\bigg{(}\sum_{r=1}^{\infty}\frac{(p^{-r}-1)H_{r}}{(\mathfrak{q}^{r}-\mathfrak{q}^{-r})}(\mathfrak{q}^{1/2}v)^{-r}\Big{\\{}\frac{q^{-nr}_{3}-K^{-r}}{\kappa_{r}}-\sum_{k=1}^{n}q^{-r(k-1)}_{3}\chi_{-r}^{(k)}(q_{1},q_{2})\Big{\\}}\bigg{)},$

respectively. Again note that when we switch the $\exp(H_{-r})$ factor of
$\Gamma_{n}(K;pv)$ and the $\exp(-H_{r})$ factor of
$\tilde{\Phi}^{[n]}_{\Lambda}(pv)$ the additional proportional factor appears
from the formula (1.14). However, under the sign $\sim$ we are able to pay no
attention to it.

Now compare the $\exp(H_{-r})$ factor of (6.18) with those of (6.6). We would
like to find a value of $p$ which renders them to be equal up to a shift of
spectral parameter $v$. That is, we require that

$\widetilde{\mathcal{T}}^{-}_{\Lambda}(p)=\exp\bigg{(}-\sum_{n=1}^{\infty}\bigg{[}\frac{1-K^{n}}{\kappa_{n}}+\sum_{(i,j,k)\in\Lambda}q^{ni}_{1}q^{nj}_{2}q^{nk}_{3}\bigg{]}\frac{v^{n}}{\lambda^{n}}H_{-n}\bigg{)}.$
(6.19)

for some constant $\lambda$. We find that the above equality holds if and only
if

$\displaystyle\frac{p^{r}-1}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}$
$\displaystyle\bigg{\\{}(\mathfrak{q}^{-1/2}v)^{r}\big{(}\frac{q^{nr}_{3}-K^{r}}{\kappa_{r}}\big{)}+\sum_{k=1}^{n}(\mathfrak{q}^{-1/2}q^{k-1}_{3}v)^{r}\chi_{r}^{(k)}(q_{1},q_{2})\bigg{\\}}$
$\displaystyle=-\frac{v^{r}}{\lambda^{r}}\bigg{[}\frac{1-K^{n}}{\kappa_{n}}+\sum_{(i,j,k)\in\Lambda}q^{ni}_{1}q^{nj}_{2}q^{nk}_{3}\bigg{]}.$
(6.20)

Recall that we have exchanged $q_{1}$ and $q_{2}$ in the Fock representation;
$x_{ij}:=q^{i-1}_{1}q^{j-1}_{2}$, hence
$\displaystyle{\sum_{k=1}^{n}}q^{(k-1)r}_{3}\displaystyle{\sum_{(i,j)\in\Lambda^{(k)}}}x^{r}_{ij}=\sum_{(i,j,k)\in\Lambda}x^{r}_{ijk}$.
We see that for each $r\in\mathbb{Z}^{\geq 1}$, we should have

$\displaystyle
p^{r}=\frac{\lambda^{r}-\mathfrak{q}^{3r/2}+\mathfrak{q}^{-r/2}}{\lambda^{r}}.$
(6.21)

To make the value of $p$ independent of $r$ we have to choose
$\lambda=\mathfrak{q}^{3/2}$, and hence we get that $p=\mathfrak{q}^{-2}$.
Thus we see that the shift parameter is fixed. Thus, we have confirmed that

$\widetilde{\mathcal{T}}^{-}_{\Lambda}(\mathfrak{q}^{-2})=\mathcal{T}^{-}_{\Lambda}(\mathfrak{q}^{-3/2}v,K).$
(6.22)

We expect that the value $p=\mathfrak{q}^{-2}$ also works for the
$\exp(H_{r})$ as well. More precisely, when $p=\mathfrak{q}^{-2}$ we should
check that there exists $\lambda\in\mathbb{C}$ such that

$\widetilde{\mathcal{T}}^{+}_{\Lambda}(p)=\mathcal{T}^{+}_{\Lambda}(\frac{v}{\lambda},K),$
(6.23)

which means

$\displaystyle\frac{(p^{-r}-1)H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}$
$\displaystyle(\mathfrak{q}^{1/2}v)^{-r}\bigg{\\{}\frac{q^{-nr}_{3}-K^{-r}}{\kappa_{r}}-\sum_{k=1}^{n}(q^{k-1}_{3})^{-r}\chi_{-r}^{(k)}(q_{1},q_{2})\bigg{\\}}\bigg{)}$
$\displaystyle=\kappa_{r}\bigg{(}\frac{v^{-r}}{\lambda^{-r}}\bigg{[}\frac{1-K^{-r}}{\kappa_{r}}-\sum_{(i,j,k)\in\Lambda}q^{-ri}_{1}q^{-rj}_{2}q^{-rk}_{3}\bigg{]}\frac{H_{n}}{\kappa_{n}}\bigg{)}.$
(6.24)

By using the summation formula for geometric series, the above equality
becomes

$\displaystyle(p^{-r}-1)=\lambda^{r}(\mathfrak{q}^{3r/2}-\mathfrak{q}^{-r/2}).$
(6.25)

If $p=\mathfrak{q}^{-2}$, we obtain $\lambda=\mathfrak{q}^{1/2}$. Hence,

$\widetilde{\mathcal{T}}^{+}_{\Lambda}(\mathfrak{q}^{-2})=\mathcal{T}^{+}_{\Lambda}(\mathfrak{q}^{-1/2}v,K)$
(6.26)

is satisfied. In summary we get the relation

$\displaystyle\Xi_{\Lambda}(K;\mathfrak{q}^{-2}v)=K^{|\Lambda|}~{}\mathcal{T}^{-}_{\Lambda}(\mathfrak{q}^{-3/2}v,K)\cdot\Xi_{\Lambda}(K;v)\cdot\mathcal{T}^{+}_{\Lambda}(\mathfrak{q}^{-1/2}v,K),$
(6.27)

where we fix the proportional factor $K^{|\Lambda|}$ in appendix.

###  6.3. Case of the dual intertwiner

Recall that the dual MacMahon intertwiner is;

$\displaystyle\Xi_{\Lambda}^{*}(K;v)=z_{\Lambda}^{*}(K;v)\mathcal{M}^{[n]*}(K)\tilde{\Phi}^{[n]*}_{\Lambda}(v)\Gamma^{*}_{n}(K;v),$
(6.28)

where $\Gamma^{*}_{n}(K;v)$ and $z^{*}_{\Lambda}(K;v)$ are given by (4.48) and
(4.49). We also have (see (4.50));

$\displaystyle\tilde{\Phi}^{[n]*}_{\Lambda}(v)=\tilde{\Phi}^{*}_{\Lambda^{(1)}}(v)\circ\cdots\circ\tilde{\Phi}^{*}_{\Lambda^{(n)}}(q^{n-1}_{3}v).$
(6.29)

We will use the zero mode

$\displaystyle\textbf{f}(z)=\frac{\theta_{q_{3}}(\mathfrak{q}Kz^{-1})}{\theta_{q_{3}}(\mathfrak{q}z^{-1})}u^{-1},$
(6.30)

so that we can write

$\displaystyle
z^{*}_{\Lambda}(K;v)=\prod_{k=1}^{h(\Lambda)}\prod_{(i,j)\in\Lambda^{(k)}}\Bigg{(}\frac{K^{1/2}}{\mathfrak{q}^{k}}\frac{\theta_{q_{3}}(q^{k}_{3}x^{-1}_{ijk})}{\theta_{q_{3}}(Kx^{-1}_{ijk})}\Bigg{)}^{-1}\frac{\theta_{q_{3}}(\mathfrak{q}K\cdot(x_{ijk}v)^{-1})}{\theta_{q_{3}}(\mathfrak{q}\cdot(x_{ijk}v)^{-1})}u^{-1}.$
(6.31)

In order to calculate $\Xi^{*}_{\Lambda}(K;pv)$ we need to calculate
$z_{\Lambda}^{*}(K;pv),\tilde{\Phi}^{[n]*}_{\Lambda}(pv)$ and
$\Gamma^{*}_{n}(K;pv)$. From the above equations we see that

$\displaystyle\Gamma^{*}_{n}(K;pv)=$
$\displaystyle\exp\bigg{(}-\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{q^{nr}_{3}-K^{r}}{\kappa_{r}}(\mathfrak{q}^{1/2}v)^{r}(p^{r}-1)\bigg{)}$
$\displaystyle\cdot\Gamma^{*}_{n}(K;v)\exp\bigg{(}-\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}\frac{q^{-nr}_{3}-K^{-r}}{\kappa_{r}}(\mathfrak{q}^{-1/2}v)^{-r}(p^{-r}-1)\bigg{)},$
(6.32)

and

$\displaystyle
z^{*}_{\Lambda}(K;pv)=z^{*}_{\Lambda}(K;v)\prod_{k=1}^{h(\Lambda)}\prod_{(i,j)\in\Lambda^{(k)}}\bigg{[}\frac{\theta_{q_{3}}(\mathfrak{q}\cdot(x_{ijk}v)^{-1})}{\theta_{q_{3}}(\mathfrak{q}K\cdot(x_{ijk}v)^{-1})}\frac{\theta_{q_{3}}(\mathfrak{q}K\cdot(x_{ijk}pv)^{-1})}{\theta_{q_{3}}(\mathfrak{q}\cdot(x_{ijk}pv)^{-1})}\bigg{]}.$
(6.33)

Using the notation (6.14), we can state an explicit expression of
$\tilde{\Phi}^{[n]*}_{\Lambda}(v)$ as the following lemma;

######  Lemma 6.4.

$\displaystyle\tilde{\Phi}^{[n]*}_{\Lambda}(v)=\tilde{c}(\Lambda)$
$\displaystyle\exp\bigg{[}-\sum_{m=1}^{n}\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}(\mathfrak{q}^{1/2}q^{m-1}_{3}v)^{r}\chi_{r}^{(m)}(q_{1},q_{2})\bigg{]}$
$\displaystyle\cdot\exp\bigg{[}\sum_{m=1}^{n}\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}(\mathfrak{q}^{-1/2}q^{m-1}_{3}v)^{-r}\chi_{-r}^{(m)}(q_{1},q_{2})\bigg{]}.$

where $\tilde{c}(\Lambda)$ is a constant which may depend on $\Lambda$.

###### Proof.

By definition

$\displaystyle\tilde{\Phi}^{[n]*}_{\Lambda}(v)$
$\displaystyle=\tilde{\Phi}^{*}_{\Lambda^{(1)}}(v)\circ\cdots\circ\tilde{\Phi}^{*}_{\Lambda^{(n)}}(q^{n-1}_{3}v)$
$\displaystyle=\tilde{\mathcal{G}}^{-1}_{\Lambda^{(1)}}\cdots\tilde{\mathcal{G}}^{-1}_{\Lambda^{(n)}}\big{(}\text{exponential
factor}\big{)}$ (6.34)

where $\big{(}\text{exponential factor}\big{)}$ is

$\displaystyle\prod_{m=1}^{n}\Bigg{[}\exp\big{[}\sum_{r=1}^{\infty}\frac{H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}(\mathfrak{q}^{1/2}q^{m-1}_{3}v)^{r}\bigg{(}\frac{1}{(1-q^{r}_{1})(1-q^{r}_{2})}-\sum_{(i,j)\in\Lambda^{(m)}}x^{r}_{ij}\bigg{)}\big{]}$
$\displaystyle\cdot\exp\big{[}\sum_{r=1}^{\infty}\frac{H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}(\mathfrak{q}^{-1/2}q^{m-1}_{3}v)^{-r}\bigg{(}\sum_{(i,j)\in\Lambda^{(m)}}x^{-r}_{ij}-\frac{1}{(1-q^{-r}_{1})(1-q^{-r}_{2})}\bigg{)}\big{]}\Bigg{]}.$
(6.35)

Applying (1.14) to Eq.(6.35), we obtain the desired result. ∎

Now Lemma 6.4 implies that

$\displaystyle\tilde{\Phi}^{[n]*}_{\Lambda}$
$\displaystyle(pv)=\exp\bigg{[}-\sum_{m=1}^{n}\sum_{r=1}^{\infty}\frac{(p^{r}-1)H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}(\mathfrak{q}^{1/2}q^{m-1}_{3}v)^{r}\chi_{r}^{(m)}(q_{1},q_{2})\bigg{]}$
$\displaystyle\cdot\tilde{\Phi}^{[n]*}_{\Lambda}(v)\cdot\exp\bigg{[}\sum_{m=1}^{n}\sum_{r=1}^{\infty}\frac{(p^{-r}-1)H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}(\mathfrak{q}^{-1/2}q^{m-1}_{3}v)^{-r}\chi_{-r}^{(m)}(q_{1},q_{2})\bigg{]}.$
(6.36)

Now we merge the results (6.32) and (6.36) together. Then, we see that

$\displaystyle\Xi^{*}_{\Lambda}(K;pv)\sim$
$\displaystyle\exp\bigg{[}-\sum_{r=1}^{\infty}\frac{(p^{r}-1)H_{-r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}(\mathfrak{q}^{1/2}v)^{r}\big{\\{}\sum_{m=1}^{n}q^{(m-1)r}_{3}\chi_{r}^{(m)}(q_{1},q_{2})+\frac{q^{nr}_{3}-K^{r}}{\kappa_{r}}\big{\\}}\bigg{]}\cdot\Xi^{*}_{\Lambda}(K;v)$
$\displaystyle\cdot\exp\bigg{[}\sum_{r=1}^{\infty}\frac{(p^{-r}-1)H_{r}}{\mathfrak{q}^{r}-\mathfrak{q}^{-r}}(\mathfrak{q}^{-1/2}v)^{-r}\big{\\{}\sum_{m=1}^{n}q^{-(m-1)r}_{3}\chi_{-r}^{(m)}(q_{1},q_{2})-\frac{q^{-nr}_{3}-K^{-r}}{\kappa_{r}}\big{\\}}\bigg{]}.$
(6.37)

It is straightforward to see that if we take $p=\mathfrak{q}^{-2}$,

$\displaystyle\Xi^{*}_{\Lambda}(K;\mathfrak{q}^{-2}v)=K^{-|\Lambda|}~{}\mathcal{T}^{-}_{\Lambda}\big{(}\mathfrak{q}^{-1/2}v,K\big{)}^{-1}\cdot\Xi^{*}_{\Lambda}(K;v)\cdot\mathcal{T}^{+}_{\Lambda}\big{(}\mathfrak{q}^{-3/2}v,K\big{)}^{-1},$
(6.38)

where we explicitly determine the proportional factor $K^{-|\Lambda|}$ in
appendix.

###  6.4. MacMahon KZ equation

All the commutation relations between the MacMahon intertwiners and the shift
operators are obtained by using equation (1.14). It turns out that we can
write the commutation relations compactly by introducing

$\tilde{R}^{K_{1},K_{2}}_{\Pi,\Lambda}\big{(}\frac{u}{v}\big{)}:=\exp\bigg{[}-\sum_{r=1}^{\infty}\frac{\kappa_{r}}{r}\Big{(}\frac{u}{v}\Big{)}^{-r}\Big{(}\sum_{(i,j,k)\in\Lambda}x^{r}_{ijk}+\frac{1-K^{r}_{2}}{\kappa_{r}}\Big{)}\Big{(}\sum_{(i,j,k)\in\Pi}x^{-r}_{ijk}-\frac{1-K^{-r}_{1}}{\kappa_{r}}\Big{)}\bigg{]},$
(6.39)

The commutation relations are

$\displaystyle\mathcal{T}^{+}_{\Lambda}(v,K)\Xi_{\Lambda^{\prime}}(K^{\prime},v^{\prime})$
$\displaystyle=\tilde{R}^{K,K^{\prime}}_{\Lambda,\Lambda^{\prime}}\big{(}\frac{v}{\mathfrak{q}^{-1/2}v^{\prime}}\big{)}\cdot\Xi_{\Lambda^{\prime}}(K^{\prime},v^{\prime})\mathcal{T}^{+}_{\Lambda}(v,K),$
(6.40)
$\displaystyle\mathcal{T}^{+}_{\Lambda}(v,K)\Xi^{*}_{\Lambda^{\prime}}(K^{\prime},v^{\prime})$
$\displaystyle=\tilde{R}^{K,K^{\prime}}_{\Lambda,\Lambda^{\prime}}\big{(}\frac{v}{\mathfrak{q}^{1/2}v^{\prime}}\big{)}^{-1}\cdot\Xi^{*}_{\Lambda^{\prime}}(K^{\prime},v^{\prime})\mathcal{T}^{+}_{\Lambda}(v,K),$
(6.41)
$\displaystyle\mathcal{T}^{-}_{\Lambda}(v,K)\Xi_{\Lambda^{\prime}}(K^{\prime},v^{\prime})$
$\displaystyle=\tilde{R}^{K^{\prime},K}_{\Lambda^{\prime},\Lambda}\big{(}\frac{v^{\prime}}{\mathfrak{q}^{-1/2}v}\big{)}\cdot\Xi_{\Lambda^{\prime}}(K^{\prime},v^{\prime})\mathcal{T}^{-}_{\Lambda}(v,K),$
(6.42)
$\displaystyle\mathcal{T}^{-}_{\Lambda}(v,K)\Xi^{*}_{\Lambda^{\prime}}(K^{\prime},v^{\prime})$
$\displaystyle=\tilde{R}^{K^{\prime},K}_{\Lambda^{\prime},\Lambda}\big{(}\frac{v^{\prime}}{\mathfrak{q}^{1/2}v}\big{)}^{-1}\cdot\Xi^{*}_{\Lambda^{\prime}}(K^{\prime},v^{\prime})\mathcal{T}^{-}_{\Lambda}(v,K).$
(6.43)

They should be compared with the corresponding relations (5.32)–(5.35) for the
intertwiners of the vector representation.

We can determine the relation between
$\tilde{R}^{K_{1},K_{2}}_{\Pi,\Lambda}\big{(}\frac{u}{v}\big{)}$ and the
MacMahon R-matrix $R^{K_{1},K_{2}}_{\Pi,\Lambda}\big{(}\frac{u}{v}\big{)}$
which appears in the paper [11]. From (6.39) we get that

$\tilde{R}^{K_{1},K_{2}}_{\Pi,\Lambda}\big{(}\frac{u}{v}\big{)}=K^{\frac{-|\Lambda|}{2}}_{1}K^{-\frac{|\Pi|}{2}}_{2}R^{K_{1},K_{2}}_{\Pi,\Lambda}\big{(}\frac{u}{v}\big{)}\exp\bigg{[}\sum_{r=1}^{\infty}\frac{1}{r}\Big{(}\frac{u}{v}\Big{)}^{-r}\frac{(1-K^{r}_{2})(1-K^{-r}_{1})}{\kappa_{r}}\bigg{]}.$
(6.44)

Note that the factor

$\exp\bigg{[}\sum_{r=1}^{\infty}\frac{1}{r}\Big{(}\frac{u}{v}\Big{)}^{-r}\frac{(1-K^{r}_{2})(1-K^{-r}_{1})}{\kappa_{r}}\bigg{]}$
(6.45)

is the vacuum contribution to $\tilde{R}^{K_{1},K_{2}}_{\Pi,\Lambda}$ or the
normalization factor to make $R^{K_{1},K_{2}}_{\varnothing,\varnothing}=1$.

Now we are ready to derive the generalized KZ equation for MacMahon
intertwiner. We define the correlation function by

$\displaystyle
G^{\Omega^{1}\cdots\Omega^{m}}_{\Lambda^{1}\cdots\Lambda^{n}}\Big{(}^{K^{\prime}_{1},\dots,K^{\prime}_{m},K_{1}\dots,K_{n}}_{w_{1},\dots,w_{m},v_{1},\dots,v_{n}}\Big{)}:=\langle\emptyset|\Xi^{*}_{\Omega^{1}}(K^{\prime}_{1};w_{1})\cdots\Xi^{*}_{\Omega^{m}}(K^{\prime}_{m};w_{m})\Xi_{\Lambda^{1}}(K_{1};v_{1})\cdots\Xi_{\Lambda^{n}}(K_{n};v_{n})|\emptyset\rangle.$
(6.46)

Let us first consider the shift of the spectral parameter $v_{i}$;

$\displaystyle(\mathfrak{q}^{-2})^{v_{i}\frac{\partial}{\partial
v_{i}}}G^{\Omega^{1}\cdots\Omega^{m}}_{\Lambda^{1}\cdots\Lambda^{n}}\Big{(}^{K^{\prime}_{1},\dots,K^{\prime}_{m},K_{1}\dots,K_{n}}_{w_{1},\dots,w_{m},v_{1},\dots,v_{n}}\Big{)}$
$\displaystyle=\langle\emptyset|\Xi^{*}_{\Omega^{1}}(K^{\prime}_{1};w_{1})\cdots\Xi^{*}_{\Omega^{m}}(K^{\prime}_{m};w_{m})\Xi_{\Lambda^{1}}(K_{1};v_{1})\cdots\Xi_{\Lambda^{i-1}}(K_{i-1};v_{i-1})\Xi_{\Lambda^{i}}(K_{i};\mathfrak{q}^{-2}v_{i})$
$\displaystyle\Xi_{\Lambda^{i+1}}(K_{i+1};v_{i+1})\cdots\Xi_{\Lambda^{n}}(K_{n};v_{n})|\emptyset\rangle.$
(6.47)

From (6.27), the above equation becomes

$\displaystyle{K_{i}}^{|\Lambda_{i}|}\langle\emptyset|\Xi^{*}_{\Omega^{1}}(K^{\prime}_{1};w_{1})\cdots\Xi^{*}_{\Omega^{m}}(K^{\prime}_{m};w_{m})\Xi_{\Lambda^{1}}(K_{1};v_{1})\cdots\Xi_{\Lambda^{i-1}}(K_{i-1};v_{i-1})$
$\displaystyle\cdot\mathcal{T}^{-}_{\Lambda^{i}}\big{(}\mathfrak{q}^{-3/2}v_{i},K_{i}\big{)}\cdot\Xi_{\Lambda^{i}}(K_{i};v_{i})\cdot\mathcal{T}^{+}_{\Lambda^{i}}\big{(}\mathfrak{q}^{-1/2}v_{i},K_{i}\big{)}\Xi_{\Lambda^{i+1}}(K_{i+1};v_{i+1})\cdots\Xi_{\Lambda^{n}}(K_{n};v_{n})|\emptyset\rangle~{}.$
(6.48)

Next we move the operator
$\mathcal{T}^{+}_{\Lambda^{i}}\big{(}v_{i},K_{i}\big{)}^{-1}$ to the right by
using the result in the section 6.4, which implies

$\displaystyle\mathcal{T}^{+}_{\Lambda^{i}}\big{(}\mathfrak{q}^{-1/2}v_{i},K_{i}\big{)}\cdot\Xi_{\Lambda^{i+1}}(K_{i+1};v_{i+1})\cdots\Xi_{\Lambda^{n}}(K_{n};v_{n})|\emptyset\rangle$
$\displaystyle=\bigg{(}\prod_{l=i+1}^{n}\tilde{R}^{K_{i},K_{l}}_{\Lambda^{i},\Lambda^{l}}\big{(}\frac{v_{i}}{v_{l}}\big{)}\bigg{)}\Xi_{\Lambda^{i+1}}(K_{i+1};v_{i+1})\cdots\Xi_{\Lambda^{n}}(K_{n};v_{n})|\emptyset\rangle.$
(6.49)

Thus, at this step we see that

$\displaystyle(\mathfrak{q}^{-2})^{v_{i}\frac{\partial}{\partial
v_{i}}}G^{\Omega^{1}\cdots\Omega^{m}}_{\Lambda^{1}\cdots\Lambda^{n}}\Big{(}^{K^{\prime}_{1},\dots,K^{\prime}_{m},K_{1}\dots,K_{n}}_{w_{1},\dots,w_{m},v_{1},\dots,v_{n}}\Big{)}={K_{i}}^{|\Lambda_{i}|}\bigg{(}\prod_{l=i+1}^{n}\tilde{R}^{K_{i},K_{l}}_{\Lambda^{i},\Lambda^{l}}\big{(}\frac{v_{i}}{v_{l}}\big{)}\bigg{)}$
$\displaystyle\langle\emptyset|\Xi^{*}_{\Omega^{1}}(K^{\prime}_{1};w_{1})\cdots\Xi^{*}_{\Omega^{m}}(K^{\prime}_{m};w_{m})\Xi_{\Lambda^{1}}(K_{1};v_{1})\cdots\Xi_{\Lambda^{i-1}}(K_{i-1};v_{i-1})$
$\displaystyle\cdot\mathcal{T}^{-}_{\Lambda^{i}}\big{(}\mathfrak{q}^{-3/2}v_{i},K_{i}\big{)}\Xi_{\Lambda^{i}}(K_{i};v_{i})\Xi_{\Lambda^{i+1}}(K_{i+1};v_{i+1})\cdots\Xi_{\Lambda^{n}}(K_{n};v_{n})|\emptyset\rangle.$
(6.50)

We then move the operator
$\mathcal{T}^{-}_{\Lambda^{i}}\big{(}\mathfrak{q}^{-3/2}v_{i},K_{i}\big{)}$ to
the left. In the same manner as above, we obtain

$\displaystyle(\mathfrak{q}^{-2})^{v_{i}\frac{\partial}{\partial
v_{i}}}G^{\Omega^{1}\cdots\Omega^{m}}_{\Lambda^{1}\cdots\Lambda^{n}}\Big{(}^{K^{\prime}_{1},\dots,K^{\prime}_{m},K_{1}\dots,K_{n}}_{w_{1},\dots,w_{m},v_{1},\dots,v_{n}}\Big{)}=\mathcal{A}_{i}\cdot
G^{\Omega^{1}\cdots\Omega^{m}}_{\Lambda^{1}\cdots\Lambda^{n}}\Big{(}^{K^{\prime}_{1},\dots,K^{\prime}_{m},K_{1}\dots,K_{n}}_{w_{1},\dots,w_{m},v_{1},\dots,v_{n}}\Big{)},$
$\displaystyle\mathcal{A}_{i}:=K_{i}^{|\Lambda^{i}|}\bigg{(}\prod_{l=i+1}^{n}\tilde{R}^{K_{i},K_{l}}_{\Lambda^{i},\Lambda^{l}}\big{(}\frac{v_{i}}{v_{l}}\big{)}\bigg{)}\bigg{(}\prod_{k=1}^{i-1}\tilde{R}^{K_{k},K_{i}}_{\Lambda^{k},\Lambda^{i}}\big{(}\mathfrak{q}^{2}\frac{v_{k}}{v_{i}}\big{)}^{-1}\bigg{)}\bigg{(}\prod_{s=1}^{m}\tilde{R}^{K^{\prime}_{s},K_{i}}_{\Omega^{s},\Lambda^{i}}\big{(}\mathfrak{q}\frac{w_{s}}{v_{i}}\big{)}\bigg{)}.$
(6.51)

Next let us consider the shift of the spectral parameter $w_{i}$;

$\displaystyle(\mathfrak{q}^{-2})^{w_{i}\frac{\partial}{\partial
w_{i}}}G^{\Omega^{1}\cdots\Omega^{m}}_{\Lambda^{1}\cdots\Lambda^{n}}\Big{(}^{K^{\prime}_{1},\dots,K^{\prime}_{m},K_{1}\dots,K_{n}}_{w_{1},\dots,w_{m},v_{1},\dots,v_{n}}\Big{)}$
$\displaystyle=\langle\emptyset|\Xi^{*}_{\Omega^{1}}(K^{\prime}_{1};w_{1})\cdots\Xi^{*}_{\Omega^{i-1}}(K^{\prime}_{i-1};w_{i-1})\Xi^{*}_{\Omega^{i}}(K^{\prime}_{i};\mathfrak{q}^{-2}w_{i})\Xi^{*}_{\Omega^{i+1}}(K^{\prime}_{i+1};w_{i+1})\cdots\Xi^{*}_{\Omega^{m}}(K^{\prime}_{m};w_{m})$
$\displaystyle\cdot\Xi_{\Lambda^{1}}(K_{1};v_{1})\cdots\Xi_{\Lambda^{n}}(K_{n};v_{n})|\emptyset\rangle.$
(6.52)

From (6.38) we get that (6.52) is equal to

$\displaystyle(K^{\prime}_{i})^{-|\Omega^{i}|}\langle\emptyset|\Xi^{*}_{\Omega^{1}}(K^{\prime}_{1};w_{1})\cdots\Xi^{*}_{\Omega^{i-1}}(K^{\prime}_{i-1};w_{i-1})\mathcal{T}^{-}_{\Omega^{i}}(\mathfrak{q}^{-1/2}w_{i},K^{\prime}_{i})^{-1}\cdot\Xi^{*}_{\Omega^{i}}(K^{\prime}_{i};w_{i})\mathcal{T}^{+}_{\Omega^{i}}(\mathfrak{q}^{-3/2}w_{i},K^{\prime}_{i})^{-1}$
$\displaystyle\cdot\Xi^{*}_{\Omega^{i+1}}(K^{\prime}_{i+1};w_{i+1})\cdots\Xi^{*}_{\Omega^{m}}(K^{\prime}_{m};w_{m})\Xi_{\Lambda^{1}}(K_{1};v_{1})\cdots\Xi_{\Lambda^{n}}(K_{n};v_{n})|\emptyset\rangle$
(6.53)

As above we can move $\mathcal{T}^{+}_{\Omega^{i}}$ to the right and
$\mathcal{T}^{-}_{\Omega^{i}}$ to the left by using the result in section 6.4.
We see that (6.53) is equal to

$\displaystyle(K^{\prime}_{i})^{-|\Omega^{i}|}\Big{(}\prod_{k=i+1}^{m}\tilde{R}^{K^{\prime}_{i},K^{\prime}_{k}}_{\Omega^{i},\Omega^{k}}\big{(}\mathfrak{q}^{-2}\frac{w_{i}}{w_{k}}\big{)}\Big{)}\Big{(}\prod_{l=1}^{n}\tilde{R}^{K^{\prime}_{i},K_{l}}_{\Omega^{i},\Lambda^{l}}\big{(}\mathfrak{q}^{-1}\frac{w_{i}}{v_{l}}\big{)}^{-1}\Big{)}\Big{(}\prod_{s=1}^{i-1}\tilde{R}^{K^{\prime}_{s},K^{\prime}_{i}}_{\Omega^{s},\Omega^{i}}\big{(}\frac{w_{s}}{w_{i}}\big{)}^{-1}\Big{)}$
$\displaystyle\langle\emptyset|\Xi^{*}_{\Omega^{1}}(K^{\prime}_{1};w_{1})\cdots\Xi^{*}_{\Omega^{i-1}}(K^{\prime}_{i-1};w_{i-1})\Xi^{*}_{\Omega^{i}}(K^{\prime}_{i};w_{i})\cdots\Xi^{*}_{\Omega^{m}}(K^{\prime}_{m};w_{m})\Xi_{\Lambda^{1}}(K_{1};v_{1})\cdots\Xi_{\Lambda^{n}}(K_{n};v_{n})|\emptyset\rangle.$
(6.54)

Hence, we obtain

$\displaystyle(\mathfrak{q}^{-2})^{w_{i}\frac{\partial}{\partial
w_{i}}}G^{\Omega^{1}\cdots\Omega^{m}}_{\Lambda^{1}\cdots\Lambda^{n}}\Big{(}^{K^{\prime}_{1},\dots,K^{\prime}_{m},K_{1}\dots,K_{n}}_{w_{1},\dots,w_{m},v_{1},\dots,v_{n}}\Big{)}=\mathcal{A}_{i}^{*}\cdot
G^{\Omega^{1}\cdots\Omega^{m}}_{\Lambda^{1}\cdots\Lambda^{n}}\Big{(}^{K^{\prime}_{1},\dots,K^{\prime}_{m},K_{1}\dots,K_{n}}_{w_{1},\dots,w_{m},v_{1},\dots,v_{n}}\Big{)},$
$\displaystyle\mathcal{A}_{i}^{*}:=(K^{\prime}_{i})^{-|\Omega^{i}|}\Big{(}\prod_{k=i+1}^{m}\tilde{R}^{K^{\prime}_{i},K^{\prime}_{k}}_{\Omega^{i},\Omega^{k}}\big{(}\mathfrak{q}^{-2}\frac{w_{i}}{w_{k}}\big{)}\Big{)}\Big{(}\prod_{l=1}^{n}\tilde{R}^{K^{\prime}_{i},K_{l}}_{\Omega^{i},\Lambda^{l}}\big{(}\mathfrak{q}^{-1}\frac{w_{i}}{v_{l}}\big{)}^{-1}\Big{)}\Big{(}\prod_{s=1}^{i-1}\tilde{R}^{K^{\prime}_{s},K^{\prime}_{i}}_{\Omega^{s},\Omega^{i}}\big{(}\frac{w_{s}}{w_{i}}\big{)}^{-1}\Big{)}.$
(6.55)

The system of Eqs.(6.51) and (6.55) is the generalized KZ equation for
MacMahon intertwiners.

##  7\. Solutions to the generalized KZ equation

In this section we discuss solutions to the generalized KZ equation derived in
the last section. Since the generalized KZ equation is a system of linear
difference equations for the vertical spectral parameters of the intertwiners,
solutions are given up to a multiplicative constant or a
$\mathfrak{q}^{2}=q_{3}$ periodic function. In particular we will drop the
normalization factors which are independent of the spectral parameters131313
This does not mean that the normalization factor is unimportant. For example,
the normalization factor of the Fock intertwiner is closely related to the
theory of Macdonald function.. Due to the abelian nature of the $R$ matrices
(recall they are diagonal), the solutions factorize into the product or more
precisely the ration of the fundamental solutions, namely two point functions,
which reminds us of the Wick theorem for free field correlators.

###  7.1. Correlation function of the vector intertwiners

Since both the intertwiner and the dual intertwiner keep the level of the
horizontal Fock representation, we may assume that it is zero without loss of
generality. Under this assumption $z_{n}(v)$ and $z_{n}^{*}(v)$ are
independent of the spectral parameter (see (3.37)) and hence for simplicity we
will drop these factors in the following formulas; As mentioned above we
expect that the general correlation function
$\mathcal{G}_{\lambda_{1},\cdots,\lambda_{n}}^{\mu_{1},\cdots,\mu_{m}}\big{(}u_{n}\big{|}~{}{w_{1},\cdots,w_{m},z_{1},\cdots,z_{m}}\big{)}$
can be expressed in terms of the two-point function. Thus, let us begin with
the two-point correlator

$A_{nm}(v,w):=\langle
0\big{|}\mathbb{I}_{n}(v)\mathbb{I}_{m}(w)\big{|}0\rangle=\exp\bigg{[}\sum_{r=1}^{\infty}\Big{(}\frac{q^{n}_{1}v}{q^{m}_{1}w}\Big{)}^{-r}\frac{q_{1}^{r}}{r}\frac{1-q^{r}_{2}}{1-q^{r}_{1}}\bigg{]}.$
(7.1)

Thus, we can confirm the generalized KZ equation for two-point function;

$\displaystyle\frac{A_{nm}(\mathfrak{q}^{-2}v,w)}{A_{nm}(v,w)}=R_{nm}\Big{(}v/w\Big{)}.$
(7.2)

By using (7.2) one can show that

$\displaystyle\mathcal{G}_{\lambda_{1},\cdots,\lambda_{n}}^{\mu_{1},\cdots,\mu_{m}}\big{(}u_{n}\big{|}~{}{w_{1},\cdots,w_{m},z_{1},\cdots,z_{m}}\big{)}\frac{\prod\limits_{\begin{subarray}{c}i,j=1\\\
i<j\end{subarray}}^{n}A_{\lambda_{i}\lambda_{j}}\Big{(}\frac{z_{i}}{z_{j}}\Big{)}\prod\limits_{\begin{subarray}{c}i,l=1\\\
l<i\end{subarray}}^{m}A_{\mu_{l}\mu_{i}}\Big{(}\frac{w_{l}}{\mathfrak{q}^{2}w_{i}}\Big{)}}{\prod_{j=1}^{m}\prod_{l=1}^{n}A_{\mu_{j}\lambda_{l}}\Big{(}\frac{\mathfrak{q}w_{j}}{\mathfrak{q}^{2}z_{l}}\Big{)}}$
(7.3)

solves the generalized KZ equation. We should emphasize again that the formula
is valid up to a multiplicative constant.

The AGT correspondence motivates us to compare $A_{nm}(v,w)$ with the Nekrasov
factor (1.2). Under the normalization

$\displaystyle\tilde{A}_{nm}(v,w):=\frac{A_{nm}(v,w)}{A_{00}(v,w)}=\exp\bigg{[}\sum_{r=1}^{\infty}(q^{(m-n)r}_{1}-1)\Big{(}\frac{v}{w}\Big{)}^{-r}\frac{q^{r}_{1}}{r}\frac{1-q^{r}_{2}}{1-q^{r}_{1}}\bigg{]}.$
(7.4)

by the relation (1.12), we see

$\displaystyle\tilde{A}_{nm}(v,w)=\frac{\Big{(}\frac{wq_{1}}{v};q_{1}\Big{)}_{\infty}\Big{(}\frac{wq^{m-n+1}_{1}q_{2}}{v};q_{1}\Big{)}_{\infty}}{\Big{(}\frac{wq_{1}q_{2}}{v};q_{1}\Big{)}_{\infty}\Big{(}\frac{wq^{m-n+1}_{1}}{v};q_{1}\Big{)}_{\infty}}=\begin{cases}\frac{N_{\emptyset,m-n}(w/q_{2},v)}{N_{\emptyset,m-n}(w,v)};~{}m\geq
n\\\ \frac{N_{n-m,\emptyset}(q_{2}w,v)}{N_{n-m,\emptyset}(w,v)};~{}n\geq
m\end{cases}$ (7.5)

where the non-negative integer in the subscript stands for the Young diagram
with a single row. The insertion of the screening operator is required to
obtain more general solution. In particular with the contour integral
associated with the spectral parameter of the screening operators the
correlation function (7.3) provides building block of the partition function
of three dimensional quiver gauge theories, or the $K$ theoretic lift of the
vortex counting function [49] [3] [4] [13] [16] [1] [41] [12] [44] [32]. Quite
recently an elliptic lift of the correlation function of the vector
intertwiners is discussed in [28] based on the Higgsed network calculus and
the elliptic DIM algebra [45] [30] [40].

###  7.2. Correlation function of the MacMahon intertwiners

We also expect that the general correlation function
$G^{\Omega^{1}\cdots\Omega^{m}}_{\Lambda^{1}\cdots\Lambda^{n}}\Big{(}^{K^{\prime}_{1},\dots,K^{\prime}_{m},K_{1}\dots,K_{n}}_{w_{1},\dots,w_{m},v_{1},\dots,v_{n}}\Big{)}$
can be expressed in terms of the two-point correlators. We first provide an
explicit expression of
$\langle\emptyset|\Xi_{\Lambda}(K_{1};v)\Xi_{\Lambda^{\prime}}(K_{2};w)|\emptyset\rangle$.
Up to the normalization factor such as $\mathcal{G}^{-1}_{\Lambda}$ which is
independent of the spectral parameter, it is

$\displaystyle A^{K_{1}K_{2}}_{\Lambda\Lambda^{\prime}}(v,w):$
$\displaystyle=\langle\emptyset|\Xi_{\Lambda}(K_{1};v)\Xi_{\Lambda^{\prime}}(K_{2};w)|\emptyset\rangle$
$\displaystyle=z_{\Lambda}(K_{1};v)z_{\Lambda^{\prime}}(K_{2};w)E^{K_{1}K_{2}}_{\Lambda\Lambda^{\prime}}(v,w),$
(7.6)

where

$\displaystyle E^{K_{1}K_{2}}_{\Lambda\Lambda^{\prime}}$ $\displaystyle(v,w)$
$\displaystyle=\exp$
$\displaystyle\bigg{[}-\sum_{r=1}^{\infty}\Big{(}\frac{w}{v}\Big{)}^{r}\bigg{(}\sum_{(i,j,k)\in\Lambda^{\prime}}x^{r}_{ijk}+\frac{1-K^{r}_{2}}{\kappa_{r}}\bigg{)}\bigg{(}\sum_{(i,j,k)\in\Lambda}x^{-r}_{ijk}-\frac{1-K^{-r}_{1}}{\kappa_{r}}\bigg{)}\frac{(1-q^{r}_{1})(1-q^{r}_{2})}{r}\bigg{]}.$
(7.7)

But from (6.39) and $\kappa_{r}=(q^{r}_{1}-1)(q^{r}_{2}-1)(q^{r}_{3}-1)$ we
obtain

$\displaystyle\tilde{R}^{K_{1},K_{2}}_{\Pi,\Lambda}\big{(}\frac{v}{u}\big{)}$
$\displaystyle=\frac{\exp\bigg{(}\displaystyle{\sum_{r=1}^{\infty}}\frac{(1-q^{r}_{1})(1-q^{r}_{2})}{r}\big{(}\frac{u}{v}\big{)}^{r}\bigg{(}\sum_{(i,j,k)\in\Lambda}x^{r}_{ijk}+\frac{1-K^{r}_{2}}{\kappa_{r}}\bigg{)}\bigg{(}\sum_{(i,j,k)\in\Pi}x^{-r}_{ijk}-\frac{1-K^{-r}_{1}}{\kappa_{r}}\bigg{)}\bigg{)}}{\exp\bigg{(}\displaystyle{\sum_{r=1}^{\infty}}\frac{(1-q^{r}_{1})(1-q^{r}_{2})}{r}\big{(}\frac{q_{3}u}{v}\big{)}^{r}\bigg{(}\sum_{(i,j,k)\in\Lambda}x^{r}_{ijk}+\frac{1-K^{r}_{2}}{\kappa_{r}}\bigg{)}\bigg{(}\sum_{(i,j,k)\in\Pi}x^{-r}_{ijk}-\frac{1-K^{-r}_{1}}{\kappa_{r}}\bigg{)}\bigg{)}}$
$\displaystyle=\frac{E^{K_{1}K_{2}}_{\Pi\Lambda}(\mathfrak{q}^{-2}v,u)}{E^{K_{1}K_{2}}_{\Pi\Lambda}(v,u)},$
(7.8)

which is nothing but the generalized KZ equation for two point function up to
a factor coming from the zero modes. Hence, we see that
$G^{\Omega^{1}\cdots\Omega^{m}}_{\Lambda^{1}\cdots\Lambda^{n}}\Big{(}^{K^{\prime}_{1},\dots,K^{\prime}_{m},K_{1}\dots,K_{n}}_{w_{1},\dots,w_{m},v_{1},\dots,v_{n}}\Big{)}$
has to be proportional to

$\displaystyle\frac{\prod\limits_{\begin{subarray}{c}j,l=1\\\
j<l\end{subarray}}^{n}E^{K_{j},K_{l}}_{\Lambda^{j},\Lambda^{l}}(v_{j},v_{l})\prod\limits_{\begin{subarray}{c}l,j=1\\\
j<l\end{subarray}}^{m}E^{K^{\prime}_{j},K^{\prime}_{l}}_{\Omega^{j},\Omega^{l}}(w_{j},\mathfrak{q}^{2}w_{l})}{\prod_{l=1}^{n}\prod_{s=1}^{m}E^{K^{\prime}_{s},K_{l}}_{\Omega^{s},\Lambda^{l}}(w_{s},\mathfrak{q}v_{l})}.$
(7.9)

More precisely, we get that up to a multiplicative constant

$\displaystyle
G^{\Omega^{1}\cdots\Omega^{m}}_{\Lambda^{1}\cdots\Lambda^{n}}\Big{(}^{K^{\prime}_{1},\dots,K^{\prime}_{m},K_{1}\dots,K_{n}}_{w_{1},\dots,w_{m},v_{1},\dots,v_{n}}\Big{)}$
$\displaystyle=$
$\displaystyle\bigg{(}\prod_{i=1}^{n}z_{\Lambda^{i}}(K_{i};v_{i})\bigg{)}\bigg{(}\prod_{i=1}^{m}z^{*}_{\Omega^{i}}(K^{\prime}_{i};w_{i})\bigg{)}\frac{\prod\limits_{\begin{subarray}{c}j,l=1\\\
j<l\end{subarray}}^{n}E^{K_{j},K_{l}}_{\Lambda^{j},\Lambda^{l}}(v_{j},v_{l})\prod\limits_{\begin{subarray}{c}l,j=1\\\
j<l\end{subarray}}^{m}E^{K^{\prime}_{j},K^{\prime}_{l}}_{\Omega^{j},\Omega^{l}}(w_{j},\mathfrak{q}^{2}w_{l})}{\prod_{l=1}^{n}\prod_{s=1}^{m}E^{K^{\prime}_{s},K_{l}}_{\Omega^{s},\Lambda^{l}}(w_{s},\mathfrak{q}v_{l})},$
$\displaystyle=\frac{\prod_{l=1}^{n}z_{\Lambda^{l}}(K_{l};v_{l})z_{\Lambda^{l}}(K_{l};\mathfrak{q}v_{l})\prod_{s=1}^{m}z^{*}_{\Omega^{s}}(K^{\prime}_{s};w_{s})z_{\Omega^{s}}(K^{\prime}_{s};w_{s})}{\prod\limits_{\begin{subarray}{c}j,l=1\\\
j<l\end{subarray}}^{n}z_{\Lambda^{j}}(K_{j};v_{j})z_{\Lambda^{l}}(K_{l};v_{l})\cdot\prod\limits_{\begin{subarray}{c}l,j=1\\\
j<l\end{subarray}}^{m}z_{\Omega^{j}}(K^{\prime}_{j};w_{j})z_{\Omega^{l}}(K^{\prime}_{l};\mathfrak{q}^{2}w_{l})}$
$\displaystyle\qquad\cdot\frac{\prod\limits_{\begin{subarray}{c}j,l=1\\\
j<l\end{subarray}}^{n}A^{K_{j},K_{l}}_{\Lambda^{j},\Lambda^{l}}(v_{j},v_{l})\prod\limits_{\begin{subarray}{c}l,j=1\\\
j<l\end{subarray}}^{m}A^{K^{\prime}_{j},K^{\prime}_{l}}_{\Omega^{j},\Omega^{l}}(w_{j},\mathfrak{q}^{2}w_{l})}{\prod_{l=1}^{n}\prod_{s=1}^{m}A^{K^{\prime}_{s},K_{l}}_{\Omega^{s},\Lambda^{l}}(w_{s},\mathfrak{q}v_{l})}.$
(7.10)

In view of the AGT correspondence, it is quite curious to see if the solutions
to the MacMahon KZ-equation give some generalization of the Nekrasov function.
Let us analyze the two-points correlator
$A^{K_{1}K_{2}}_{\Lambda\Lambda^{\prime}}(v,w)$ (see (7.6)). Defining the
normalized two-point function by

$\tilde{A}^{K_{1}K_{2}}_{\Lambda\Lambda^{\prime}}(v,w):=\frac{A^{K_{1}K_{2}}_{\Lambda\Lambda^{\prime}}(v,w)}{A^{K_{1}K_{2}}_{\emptyset\emptyset}(v,w)},$
(7.11)

we see that

$\displaystyle\log$
$\displaystyle\tilde{A}^{K_{1}K_{2}}_{\Lambda\Lambda^{\prime}}(v,w)=\log\bigg{(}z_{\Lambda}(K_{1};v)z_{\Lambda^{\prime}}(K_{2};w)\bigg{)}+\mathrm{(I)}+\mathrm{(II)}+\mathrm{(III)},$
(7.12)

where

$\displaystyle\mathrm{(I)}$
$\displaystyle=-\sum_{r=1}^{\infty}\Big{(}\frac{w}{v}\Big{)}^{r}\frac{(1-q^{r}_{1})(1-q^{r}_{2})}{r}\bigg{(}\sum_{(i^{\prime},j^{\prime},k^{\prime})\in\Lambda^{\prime}}x^{r}_{i^{\prime}j^{\prime}k^{\prime}}\sum_{(i,j,k)\in\Lambda}x^{-r}_{ijk}\bigg{)}$
$\displaystyle=\sum_{(i^{\prime},j^{\prime},k^{\prime})\in\Lambda^{\prime}}\sum_{(i,j,k)\in\Lambda}$