# Triangular solutions to the reflection equation for
$U_{q}(\widehat{sl_{n}})$

Dmitry Kolyaskin and Vladimir V Mangazeev  Department of Theoretical &
Fundamental Physics, Research School of Physics, Australian National
University, Canberra, ACT 2601, Australia<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We study solutions of the reflection equation related to the quantum affine
algebra $U_{q}(\widehat{sl_{n}})$. First, we explain how to construct a family
of stochastic integrable vertex models with fixed boundary conditions. Then,
we construct upper- and lower-triangular solutions of the reflection equation
related to symmetric tensor representations of $U_{q}(\widehat{sl_{n}})$ with
arbitrary spin. We also prove the star-star relation for the Boltzmann weights
of the Ising-type model, conjectured by Bazhanov and Sergeev, and use it to
verify certain properties of the solutions obtained.

Keywords: integrable systems, stochastic processes, Yang-Baxter equation,
reflection equation, star-star relation

##  1 Introduction

Stochastic integrable vertex models play an important role in the analysis of
$1+1$-dimensional systems out-of-equilibrium [1, 2, 3], interacting particle
systems with exclusion [4, 5, 6, 7, 8], and KPZ universality [9, 10]. From one
side, they provide interesting examples of interacting Markovian systems; from
the other side, one can analyse their critical behavior and explicitly
calculate some physical quantities like correlation functions.

Corwin and Petrov in [11] (see also [12] for a review) showed that many
currently known integrable $1+1$-dim KPZ models come from suitable limits of
stochastic higher-spin vertex models. Originally, they studied these higher-
spin models on the line (see also [13, 14]). However, later, this was
generalized to half-line, half-quadrant processes with open and special fixed
boundary conditions (see, for example, [15, 12, 16, 17]).

Sklyanin [18] developed a general theory of quantum integrable systems with
boundaries based on the so-called reflection equation. The reflection equation
first appeared in a 2-dim factorised scattering theory on a half-line [19],
with the scattering matrix satisfying the Yang-Baxter equation [20, 21, 22,
23].

In this paper, we construct special triangular solutions of the reflection
equation for the higher-spin stochastic vertex models related to symmetric
tensor representations of $U_{q}(\widehat{sl_{n}})$. These solutions
generalise some of our results from [24].

It is important to note that we work with a stochastic $R$-matrix, which is a
twisted version [25] of the standard $U_{q}(\widehat{sl_{n}})$ $R$-matrix. As
was noticed in [26], this leads to a factorised $L$-operator, which may be why
all our solutions possess factorised matrix elements.

The paper is organised as follows. In Section 2 we review the theory of
boundary integrable systems and show how to construct stochastic versions of
the transfer-matrix and hamiltonian. In Section 3 we introduce required
notations, define the stochastic $U_{q}(\widehat{sl_{n}})$ $R$-matrix and
discuss its properties. We also introduce two higher-spin $L$-operators
required later in the text. In Section 4 we prove the star-star relation from
[27], which we use later. In Section 5 we construct right boundary solutions
for the reflection equation. In Section 6 we explain how to construct left
boundary solutions and prove their properties. In Section 7 we give the proof
of the reflection equation for a generalized model presented in [24]. In
Conclusion, we discuss the results obtained and unsolved problems. Finally, in
Appendices, we prove some technical results used in the text.

##  2 Reflection equation and commuting transfer-matrices

We start with the Yang-Baxter equation

$R_{12}\left(\frac{x}{y}\right)R_{13}\left(\frac{x}{z}\right)R_{23}\left(\frac{y}{z}\right)=R_{23}\left(\frac{y}{z}\right)R_{13}\left(\frac{x}{z}\right)R_{12}\left(\frac{x}{y}\right),$
(2.1)

whose solutions $R_{12}\left(x\right)\in\text{End}\left(V_{1}\otimes
V_{2}\right)$ act non-trivially in the tensor product of two vector spaces and
describe the behaviour of the system in the bulk. We assume that the
$R$-matrix satisfies the following properties:

  * •

$R^{t_{2}}_{12}\left(x\right)$ is non-degenerate, i.e. one can correctly
define the matrix

$\widetilde{R}_{12}\left(x\right)=\left(\left(R_{12}\left(x\right)^{t_{2}}\right)^{-1}\right)^{t_{2}}.$
(2.2)

  * •

Regularity condition

$R_{12}\left(1\right)=P_{12},$ (2.3)

where we assume that $V_{1}\cong V_{2}\equiv V$. Here $P$ is a permutation
operator, acting in a usual way $P(v\otimes w)=w\otimes v$, where $v,w\in V$.

  * •

Unitarity condition

$R_{12}(x)R_{21}\left(x^{-1}\right)=f(x)I\otimes I,$ (2.4)

where $f(x)$ is a scalar function satisfying $f(x)=f(x^{-1})$. In fact, one
can check that if $V_{1}\cong V_{2}$, then the last property directly follows
from the regularity condition (2.3) and the Yang-Baxter equation (2.1) (see,
for example, [28]).

  * •

Crossing unitarity condition

$M_{1}R_{12}^{t_{1}}(x)M_{1}^{-1}R_{21}^{t_{1}}\left(\left(rx\right)^{-1}\right)=g(x)I\otimes
I,$ (2.5)

where $r\in\mathbb{C}\setminus\\{0\\}$, and $M\in\text{End}(V)$ is a constant
matrix. Note that the crossing unitarity is a stronger condition than (2.2)
[29, 24], since the expression

$\widetilde{R}_{21}\left(x\right)=\dfrac{1}{g(1/(rx))}\left(M^{t_{1}}_{1}\right)^{-1}R_{12}\left(1/(rx)\right)M^{t_{1}}_{1}$
(2.6)

directly follows from (2.5). In particular, if we consider the $R$-matrix as
an intertwiner of finite-dimensional $U_{q}(\widehat{\mathfrak{g}})$ modules,
then it follows from the quantum group arguments that conditions of unitarity
(2.4) and crossing unitarity (2.5) are satisfied [30, 31]. In this instance,
$M=q^{-2\rho}$, $r=q^{h^{\vee}},$ where $\rho$ is a half-sum of positive
roots, and $h^{\vee}$ is a dual Coxeter number of the Lie algebra
$\mathfrak{g}.$ In what follows, we will focus on
$\mathfrak{g}=sl_{n}(\mathbb{C})$ case, and as we will see in the next
chapter, the matrix $M$ is diagonal and satisfies an additional property

$\left[M\otimes M,R_{12}(x)\right]=0.$ (2.7)

Now we come to describing the boundaries: for a given solution to the Yang-
Baxter equation (2.1), we define the so-called reflection equation

$R_{12}\left(\frac{x}{y}\right)K_{1}\left(x\right)R_{21}\left(xy\right)K_{2}\left(y\right)=K_{2}\left(y\right)R_{12}\left(xy\right)K_{1}\left(x\right)R_{21}\left(\frac{x}{y}\right)$
(2.8)

and the dual reflection equation

$\widetilde{K}_{2}\left(y\right)\widetilde{R}_{21}(xy)\widetilde{K}_{1}\left(x\right)R_{21}\left(\frac{y}{x}\right)=R_{12}\left(\frac{y}{x}\right)\widetilde{K}_{1}\left(x\right)\widetilde{R}_{12}(xy)\widetilde{K}_{2}\left(y\right),$
(2.9)

where the matrix $\widetilde{R}_{12}(x)$ is defined in (2.2). The solutions
$K_{i}\left(x\right),\widetilde{K}_{i}\left(x\right)$ act non-trivially on
$V_{i}$ $(i=1,2)$ and describe the right and left boundaries, respectively.
Knowing the solution to the reflection equation (2.8) and using the
commutativity property (2.7), the solution to the dual reflection equation
(2.9) can be expressed in the form

$\widetilde{K}_{1}(x)=M^{-1}K\left(\dfrac{1}{r^{1/2}x}\right).$ (2.10)

Following [18, 32], we can also consider another reflection equation related
to the inverse $R$-matrix $R_{21}\left(x^{-1}\right)\cong
R_{12}^{-1}\left(x\right)$:

$R_{21}\left(\frac{y}{x}\right)\bar{K}_{1}\left(x\right)R_{12}\left(\frac{1}{xy}\right)\bar{K}_{2}\left(y\right)=\bar{K}_{2}\left(y\right)R_{21}\left(\frac{1}{xy}\right)\bar{K}_{1}\left(x\right)R_{12}\left(\frac{y}{x}\right).$
(2.11)

It is stated that if $\bar{K}(x)$ is a solution to equation (2.11), then the
expression

$\widetilde{K}_{1}(x)=\operatorname{tr}_{0}\left(\bar{K}_{0}\left(\dfrac{1}{x}\right)\widetilde{R}_{01}\left(x^{2}\right)P_{01}\right)$
(2.12)

solves the dual reflection equation (2.9). The proof of this fact is quite
technical, and we provide it in A for the reader’s convenience. The mapping
(2.12) defines an isomorphism between the solutions to the dual reflection
equation (2.9) and the equation (2.11) (see [18] for this and other
isomorphisms): the inverse mapping has the form

$\bar{K}_{1}(x)=\operatorname{tr}_{0}\left(\widetilde{K}_{0}\left(\dfrac{1}{x}\right)R_{01}\left(\dfrac{1}{x^{2}}\right)P_{01}\right),$
(2.13)

which we also prove in A.

Substituting $y=1$ into the reflection equation (2.8), we get

$[R_{12}(x)K_{1}(x)R_{21}(x),K_{2}(1)]=0$ (2.14)

which should be true for any $x$. Apparently, the only reasonable solution to
(2.14) is $K(1)\sim I$. In this paper we shall always assume the normalization

$K(1)=\it{I}.$ (2.15)

Solutions to the reflection equation (2.8) and the dual reflection equation
(2.9) allow us to construct a double-row transfer matrix

$T(x)=\operatorname{tr}_{0}\left[\widetilde{K}_{0}\left(x\right)R_{01}\left(x\right)\dots
R_{0N}\left(x\right)K_{0}\left(x\right)R_{N0}\left(x\right)\dots
R_{10}\left(x\right)\right].$ (2.16)

Transfer-matrices (2.16) form a one-parameter family of mutually commuting
quantities:

$\left[T(x),T(y)\right]=0,$ (2.17)

and this property implies the integrability of the model. The proof of (2.17)
can be found in [18].

Taking into account (2.3) and (2.15), we obtain

$T(1)=\operatorname{tr}{\bigl{[}}\widetilde{K}\left(1\right){\bigr{]}}\,\boldsymbol{\it{I}}.$
(2.18)

Finally, we can calculate the Hamiltonian of the integrable system associated
with the transfer-matrix (2.16)

$H=-\dfrac{1}{4}\dfrac{\partial}{\partial x}\ln
T(x)\bigg{|}_{x=1}=B_{L}+\sum\limits_{k=1}^{N-1}H_{k,k+1}+B_{R},$ (2.19)

where

$B_{L}=\dfrac{1}{4}\dfrac{\bar{K}^{\prime}_{1}\left(1\right)}{\operatorname{tr}\widetilde{K}(1)},\quad
H_{k,k+1}=-\dfrac{1}{2}R^{\prime}_{k,k+1}(1)P_{k,k+1},\quad
B_{R}=-\dfrac{1}{4}{K}^{\prime}_{N}\left(1\right).$ (2.20)

The operators $B_{L}$ and $B_{R}$ act only in the first and the $N$-th
component of the Hilbert space. Let us notice that the operator $B_{L}$ at the
left boundary is proportional to the derivative of the matrix ${\bar{K}}(x)$
as can be easily seen from differentiating (2.13).

Now let us discuss under what conditions the Hamiltonian and the transfer-
matrix define stochastic processes.

We say that an $M\times M$ matrix $A_{ij}$ is (left) stochastic if it
satisfies the condition

$\sum_{i=1}^{M}A_{ij}=1\mbox{ for any }j\>\mbox{ or
}\>\langle\boldsymbol{1}|A=\langle\boldsymbol{1}|,\quad\langle\boldsymbol{1}|=\langle
1\ldots 1|,$ (2.21)

that is, $\langle\boldsymbol{1}|$ is a row vector whose components are all
equal to 1.

If the transfer-matrix $T(x)$ (2.16) is left stochastic,

$\langle\boldsymbol{1}|\otimes\ldots\otimes\langle\boldsymbol{1}|T(x)=\langle\boldsymbol{1}|\otimes\ldots\otimes\langle\boldsymbol{1}|,$
(2.22)

and matrix elements of $T(x)$ are positive then $T^{t}(x)$ defines a discrete-
time Markov process in a quantum space $V\otimes\ldots\otimes V$. The
corresponding Hamiltonian (2.19) satisfies the property

$\langle\boldsymbol{1}|\otimes\ldots\otimes\langle\boldsymbol{1}|H=0$ (2.23)

and the continuous-time Markov generator is defined by $M=-H^{t}$. The inverse
statement is harder to establish and it is not always true.

First, we call matrices $R_{12}(x)$, $K(x)$ and $\bar{K}(x)$ stochastic if
they are left stochastic

$\langle\boldsymbol{1}|\otimes\langle\boldsymbol{1}|R_{12}(x)=\langle\boldsymbol{1}|\otimes\langle\boldsymbol{1}|,\quad\langle\boldsymbol{1}|K(x)=\langle\boldsymbol{1}|,\quad\langle\boldsymbol{1}|\bar{K}(x)=\langle\boldsymbol{1}|.$
(2.24)

Differentiating (2.24) with respect to $x$ and using (2.19-2.20), we conclude
that

$\langle\boldsymbol{1}|\otimes\ldots\otimes\langle\boldsymbol{1}|H=0.$ (2.25)

Now let us remember that the Hamiltonian (2.19) commutes with the transfer-
matrix

$[T(x),H]=T(x)\,H-H\,T(x)=0.$ (2.26)

Applying $\langle\boldsymbol{1}|\otimes\ldots\otimes\langle\boldsymbol{1}|$
from the left, we obtain

$\langle\boldsymbol{1}|\otimes\ldots\otimes\langle\boldsymbol{1}|T(x)H=\langle\boldsymbol{1}|\otimes\ldots\otimes\langle\boldsymbol{1}|H\,T(x)=0.$
(2.27)

If the rank of the matrix $H$ is $\mbox{dim}(V)^{N}-1$, then
$\langle\boldsymbol{1}|\otimes\ldots\otimes\langle\boldsymbol{1}|$ is the only
left zero vector of $H$ and we conclude that

$\langle\boldsymbol{1}|\otimes\ldots\otimes\langle\boldsymbol{1}|T(x)=c(x)\langle\boldsymbol{1}|\otimes\ldots\otimes\langle\boldsymbol{1}|.$
(2.28)

Dividing $T(x)$ by a scalar factor $c(x)$, we obtain the condition (2.22). If
the rank of $H$ is smaller than $\mbox{dim}(V)^{N}-1$, we cannot say anything
about the stochasticity of $T(x)$. Indeed, in the case of periodic boundary
conditions the hamiltonian may have additional zero modes and the transfer
matrix is not stochastic in general. However, in all cases with fixed boundary
conditions, which we considered, the rank of the Hamiltonian $H$ was
$\mbox{dim}(V)^{N}-1$ and this implies (2.28).

Let us also notice that we don’t know how to show the stochasticity of the
transfer-matrix (2.16) directly from (2.24). The main difficulty is that
$T(x)$ is defined in terms of $\widetilde{K}(x)$ related to $\bar{K}(x)$ via
(2.12).

##  3 Stochastic $R$-matrix for $U_{q}(\widehat{sl_{n}})$

In this section we introduce the stochastic $R$-matrix related to the
symmetric tensor representations of the quantum affine algebra
$U_{q}(\widehat{sl_{n}})$ and consider its basic properties.

First, we start by introducing index notations. We denote by
$\boldsymbol{i}=\\{i_{1},\dots,i_{m}\\}$, the $m$-component set of non-
negative integers $i_{k}\in\mathbb{Z}_{+},$ $k=1,\dots m$. We also define some
operations on these sets:

$|\boldsymbol{a}|\stackrel{{\scriptstyle\text{def}}}{{=}}\sum_{k=1}^{m}a_{k},\qquad(\boldsymbol{a},\boldsymbol{b})\stackrel{{\scriptstyle\text{def}}}{{=}}\sum\limits_{k=1}^{m}a_{k}b_{k},\qquad
Q(\boldsymbol{a},\boldsymbol{b})\stackrel{{\scriptstyle\text{def}}}{{=}}\sum_{1\leq
l<k\leq m}a_{l}b_{k},$ (3.1)

where $\boldsymbol{a}$, $\boldsymbol{b}\in\mathbb{Z}_{+}^{m}$. The permutation
$\tau$ is defined as

$\tau\left\\{i_{1},\ldots,i_{m}\right\\}=\left\\{i_{m},i_{m-1},\ldots,i_{1}\right\\}.$
(3.2)

We will also need one more operation on indices: for a given $J\in\mathbb{N}$
and $\boldsymbol{i}\in\mathbb{Z}_{+}^{m}$, $|\boldsymbol{i}|\leq J$, we
introduce the mapping

$\sigma\left\\{i_{1},\ldots,i_{m}\right\\}=\left\\{i_{2},i_{3},\ldots,i_{m},J-|\boldsymbol{i}|\right\\}.$
(3.3)

For a positive integer $J\in\mathbb{N}$ and $n\geq 2$, we define a finite-
dimensional vector space $V_{J}^{(n)}$ with basis vectors having the form

$|\boldsymbol{i}\rangle=|i_{1},i_{2},\dots,i_{n-1}\rangle,\quad|\boldsymbol{i}|\leq
J,$ (3.4)

(do not confuse this notation with the row vectors from the previous section).
The dimension of the space $V_{J}^{(n)}$ is $\binom{J+n-1}{n-1}$. We will also
use an equivalent notation for the basis vectors of $V_{J}^{(n)}$:

$|\bar{\boldsymbol{i}}\rangle=|i_{1},i_{2},\dots,i_{n-1},i_{n}\rangle,\quad|\bar{\boldsymbol{i}}|=J,$
(3.5)

where we extended the vector $|\boldsymbol{i}\rangle$ by adding an additional
component $i_{n}=J-|\boldsymbol{i}|$. The introduced vector space naturally
inherits the structure of finite-dimensional symmetric tensor representations
of the quantum affine algebra $U_{q}(\widehat{sl_{n}})$ [14]. For further
purposes, it will be essential for us to consider a realization of the
symmetric tensor representation of $sl_{n}(\mathbb{C})$ Lie algebra on
$V_{J}^{(n)}$:

$e_{i}|\boldsymbol{\bar{\alpha}}\rangle=\alpha_{i+1}|\boldsymbol{\bar{\alpha}}+\boldsymbol{e}_{i}-\boldsymbol{e}_{i+1}\rangle,\quad
f_{i}|\boldsymbol{\bar{\alpha}}\rangle=\alpha_{i}|\boldsymbol{\bar{\alpha}}-\boldsymbol{e}_{i}+\boldsymbol{e}_{i+1}\rangle,\quad
h_{i}|\boldsymbol{\bar{\alpha}}\rangle=(\alpha_{i}-\alpha_{i+1})|\boldsymbol{\bar{\alpha}}\rangle,$
(3.6)

where $e_{i}$, $f_{i}$ and $h_{i}$ ($i=1,\dots n-1$) are the standard
$sl_{n}(\mathbb{C})$ generators,
$|\boldsymbol{\bar{\alpha}}\rangle=|\alpha_{1},\alpha_{2},\dots,\alpha_{n}\rangle$
as defined in (3.5) and $\boldsymbol{e}_{j}$ represents the vector with $1$ at
the $j$-th place and 0 otherwise. For what follows, we will need to find an
expression for the half-sum of positive roots $\rho$ of $sl_{n}(\mathbb{C})$,
and a simple analysis shows that it has the form

$2\rho|\boldsymbol{\bar{\alpha}}\rangle=\left(-2\sum\limits_{k=1}^{n-1}(n-k)\alpha_{k}+(n-1)J\right)|\boldsymbol{\bar{\alpha}}\rangle.$
(3.7)

Now we discuss the stochastic $R$-matrix related to symmetric tensor
representations, see [14, 26] for more details. For arbitrary positive
integers $I,J\in\mathbb{N}$ we introduce a linear operator
$S_{I,J}(x)\in\operatorname{End}(V^{(n)}_{I}\otimes V^{(n)}_{J})$, whose
action on the basis
$\left|\boldsymbol{i}^{\prime}\right\rangle\otimes\left|\boldsymbol{j}^{\prime}\right\rangle$
(in terms of (3.4)) has the form

$S_{I,J}(x)\left|\boldsymbol{i}^{\prime}\right\rangle\otimes\left|\boldsymbol{j}^{\prime}\right\rangle=\sum_{\boldsymbol{i},\boldsymbol{j}}\left[S_{I,J}(x)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}|\boldsymbol{i}\rangle\otimes|\boldsymbol{j}\rangle.$
(3.8)

The matrix elements can be written in a compact way as

$\displaystyle\left[S_{I,J}(x)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}=$
$\displaystyle=\delta_{\boldsymbol{i}+\boldsymbol{j},\boldsymbol{i}^{\prime}+\boldsymbol{j}^{\prime}}\sum_{\boldsymbol{m}+\boldsymbol{n}=\boldsymbol{i}+\boldsymbol{j}}\Phi_{q^{2}}\left(\boldsymbol{m}-\boldsymbol{j}\mid\boldsymbol{m};\frac{q^{J-I}}{x^{2}},\frac{q^{-I-J}}{x^{2}}\right)\Phi_{q^{2}}\left(\boldsymbol{n}\mid\boldsymbol{j}^{\prime};\frac{x^{2}}{q^{I+J}},q^{-2J}\right),$
(3.9)

where we defined the function

$\displaystyle\Phi_{q}(\boldsymbol{\gamma}\mid\boldsymbol{\beta};\lambda,\mu)$
$\displaystyle=q^{Q(\boldsymbol{\beta}-\boldsymbol{\gamma},\boldsymbol{\gamma})}\left(\frac{\mu}{\lambda}\right)^{|\boldsymbol{\gamma}|}\frac{(\lambda;q)_{|\boldsymbol{\gamma}|}\left(\frac{\mu}{\lambda};q\right)_{|\boldsymbol{\beta}|-|\boldsymbol{\gamma}|}}{(\mu;q)_{|\boldsymbol{\beta}|}}\prod_{s=1}^{n-1}\left[\begin{array}[]{l}\beta_{s}\\\
\gamma_{s}\end{array}\right]_{q}$ (3.10)

for $\boldsymbol{\beta}$, $\boldsymbol{\gamma}\in\mathbb{Z}_{+}^{n-1}$ and
$\lambda$, $\mu\in\mathbb{C}.$ Here we use the standard notation for
$q$-Pochhammer symbols

$(x;q)_{n}\stackrel{{\scriptstyle\text{def}}}{{=}}\begin{cases}\prod\limits_{k=0}^{n-1}\left(1-xq^{k}\right),&n>0,\\\
1,&n=0,\\\
\prod\limits_{k=0}^{-n-1}\left(1-xq^{n+k}\right)^{-1},&n<0,\end{cases}$ (3.11)

and $q$-binomial coefficients

$\left[\begin{array}[]{l}m\\\
k\end{array}\right]_{q}\stackrel{{\scriptstyle\text{def}}}{{=}}\frac{(q;q)_{m}}{(q;q)_{k}(q;q)_{m-k}}.$
(3.12)

The introduced $R$-matrix was constructed in [26] from the corresponding 3D
$R$-operators, where it was shown that it satisfies the Yang-Baxter equation
defined on $V^{(n)}_{I}\otimes V^{(n)}_{J}\otimes V^{(n)}_{K}$

$S_{I,J}\left(\frac{x}{y}\right)S_{I,K}\left(\frac{x}{z}\right)S_{J,K}\left(\frac{y}{z}\right)=S_{J,K}\left(\frac{y}{z}\right)S_{I,K}\left(\frac{x}{z}\right)S_{I,J}\left(\frac{x}{y}\right).$
(3.13)

We can also obtain different symmetries of the $R$-matrix (3) from the
symmetries of corresponding 3D $R$-matrix. Before we describe them, we define
a symmetric version of the $R$-matrix (3)

$\left[\bar{R}_{I,J}(x)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}=x^{|\boldsymbol{i}|-|\boldsymbol{i}^{\prime}|}q^{\left(\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}\right)-\left(\boldsymbol{i},\boldsymbol{j}\right)+J|\boldsymbol{i}|-I|\boldsymbol{j}^{\prime}|-Q(\boldsymbol{j},\boldsymbol{i})+Q(\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime})}\left[S_{I,J}(x)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}.$
(3.14)

One can check that in the case $n=2$ and $I=J=1$, the introduced $R$-matrix
corresponds to the symmetric 6-vertex model. It will be also convenient to
represent the function $\Phi$ from (3.10) in another form

$\Phi_{q^{2}}(\boldsymbol{\gamma}\mid\boldsymbol{\beta};\lambda^{2},\mu^{2})=\dfrac{\mu^{|\boldsymbol{\gamma}|}}{\lambda^{|\boldsymbol{\beta}|}}\dfrac{V_{\lambda}(\boldsymbol{\gamma},0)V_{\mu/\lambda}(\boldsymbol{\beta},\boldsymbol{\gamma})}{V_{\mu}(\boldsymbol{\beta},0)},$
(3.15)

where the function $V_{x}\left(\boldsymbol{a},\boldsymbol{b}\right)$ is
defined as

$V_{x}(\boldsymbol{a},\boldsymbol{b})=q^{2Q(\boldsymbol{a},\boldsymbol{b})-Q(\boldsymbol{a},\boldsymbol{a})-Q(\boldsymbol{b},\boldsymbol{b})}\left(\frac{q}{x}\right)^{|\boldsymbol{a}|-|\boldsymbol{b}|}\frac{\left(x^{2};q^{2}\right)_{|\boldsymbol{a}|-|\boldsymbol{b}|}}{\prod\limits_{i=1}^{n-1}\left(q^{2};q^{2}\right)_{a_{i}-b_{i}}}.$
(3.16)

This function was introduced in [27] in the formulation of Ising-type model
related to the $U_{q}(\widehat{sl_{n}})$ algebra, and the use of it will allow
us to write down a number of relations below in a more compact form.

Then we can express the symmetries, proven in [26] from the properties of the
3D $R$-matrix, in terms of the symmetric $R$-matrix (3.14) and the function
(3.16) as follows

$\left[\bar{R}_{I,J}(x)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}=\left[\bar{R}_{J,I}(x)\right]_{\tau\boldsymbol{j},\tau\boldsymbol{i}}^{\tau\boldsymbol{j}^{\prime},\tau\boldsymbol{i}^{\prime}},$
(3.17)
$\left[\bar{R}_{I,J}(x)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}=q^{\left[\boldsymbol{i^{\prime}},\boldsymbol{j^{\prime}}\right]-\left[\boldsymbol{i},\boldsymbol{j}\right]}\dfrac{V_{q^{-I}}(\boldsymbol{i},0)V_{q^{-J}}(\boldsymbol{j},0)}{V_{q^{-I}}(\boldsymbol{i}^{\prime},0)V_{q^{-J}}(\boldsymbol{j}^{\prime},0)}\left[\bar{R}_{I,J}(x)\right]_{\tau\boldsymbol{i}^{\prime},\tau\boldsymbol{j}^{\prime}}^{\tau\boldsymbol{i},\tau\boldsymbol{j}},$
(3.18)

where the indices are related by the permutation $\tau$ defined in (3.2). Here
we introduced an additional operation on indices

$\left[\boldsymbol{i},\boldsymbol{j}\right]=(\boldsymbol{i},\boldsymbol{j})-(I-|\boldsymbol{i}|)(J-|\boldsymbol{j}|).$
(3.19)

We also have another symmetry based on the transformation $\sigma$ from (3.3):

$\left[\bar{R}_{I,J}(x)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}=x^{(|\boldsymbol{i}|+|\sigma\boldsymbol{i}|)-(|\boldsymbol{i^{\prime}}|+|\sigma\boldsymbol{i^{\prime}}|)}\left[\bar{R}_{I,J}(x)\right]_{\sigma\boldsymbol{i},\sigma\boldsymbol{j}}^{\sigma\boldsymbol{i}^{\prime},\sigma\boldsymbol{j}^{\prime}}.$
(3.20)

Now let us recall the other basic properties of the stochastic $R$-matrix (3):

  * •

Stochastic condition

$\sum_{\boldsymbol{i},\boldsymbol{j}}\left[S_{I,J}(x)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}=1.$
(3.21)

This identity is based on the following summation rule:

$\sum_{\boldsymbol{\gamma}}\Phi_{q}(\boldsymbol{\gamma}\mid\boldsymbol{\beta};\lambda,\mu)=1,$
(3.22)

which will be important for us in describing stochastic processes. The proof
of this fact can be done by induction on $n$ and using the $q$-Vandermonde
summation formula from B (see [14] for details). As a consequence of the
summation formula (3.22), we directly obtain the stochastic condition (3.21).

  * •

Regularity

$S_{I,I}(1)=P_{12},$ (3.23)

where $P$ is a permutation operator, acting on $V^{(n)}_{I}\otimes
V^{(n)}_{I}$. This property can be easily checked from the definition of the
$R$-matrix (3).

  * •

Unitarity condition

$S_{I,J}(x)S_{J,I}\left(x^{-1}\right)=I\otimes I.$ (3.24)

  * •

Crossing unitarity condition

$M_{1}S_{12}^{t_{1}}(x)M_{1}^{-1}S_{21}^{t_{1}}\left(\left(q^{n}x\right)^{-1}\right)=g_{IJ}^{(n)}(x)I\otimes
I,$ (3.25)

where $M$ is the diagonal matrix

$M_{\boldsymbol{i}}^{\boldsymbol{j}}=q^{2\sum\limits_{k=1}^{n-1}(n-k)i_{k}}\delta_{\boldsymbol{i}}^{\boldsymbol{j}},$
(3.26)

and the scalar function in the right-hand side is

$g_{IJ}^{(n)}(x)=\dfrac{\left(x^{2}q^{2-I-J};q^{2}\right)_{I}\left(x^{2}q^{2n-I+J};q^{2}\right)_{I}}{\left(x^{2}q^{2-I+J};q^{2}\right)_{I}\left(x^{2}q^{2n-I-J};q^{2}\right)_{I}}.$
(3.27)

To verify that the scalar factor in the unitarity condition (3.24) is indeed
equal to $1$, one needs to apply the stochastic property (3.21). As mentioned
in the previous section, (3.25) follows from the general theory of quantum
groups. Using the formula (3.7) and the fact that the dual Coxeter number
$h^{\vee}=n$ for the Lie algebra $sl_{n}(\mathbb{C})$, we come to the
expression in the left-hand side of (3.25).

It remains to show that the function in the right-hand side of the crossing
unitarity relation is given by (3.27). It is enough to check the relation

$\sum\limits_{\boldsymbol{i^{\prime}},\boldsymbol{j^{\prime}}}q^{2\sum\limits_{k=1}^{n-1}(n-k)(i^{\prime\prime}_{k}-i^{\prime}_{k})}\left[S_{I,J}(x)\right]_{\boldsymbol{i^{\prime}},\boldsymbol{j}}^{\boldsymbol{i}^{\prime\prime},\boldsymbol{j}^{\prime}}\left[S_{J,I}\left((q^{n}x)^{-1}\right)\right]_{\boldsymbol{j^{\prime}},\boldsymbol{i}}^{\boldsymbol{j}^{\prime\prime},\boldsymbol{i}^{\prime}}=g_{IJ}^{(n)}(x)\delta_{\boldsymbol{i}}^{\boldsymbol{i^{\prime\prime}}}\delta_{\boldsymbol{j}}^{\boldsymbol{j^{\prime\prime}}}$
(3.28)

for any diagonal element, say, for
$\boldsymbol{i}=\boldsymbol{j}=\boldsymbol{i^{\prime\prime}}=\boldsymbol{j^{\prime\prime}}=0$
($\boldsymbol{i^{\prime}}$ and $\boldsymbol{j^{\prime}}$ are internal
summation indices). Note that the $R$-matrix contains an internal summation
(see definition (3)). However, we can remove internal sums by applying the
symmetry (3.17) to both factors in (3.28), since in this case the $R$-matrix
contains only one term. Then one can rewrite (3.28) using (3.15), and we come
to the formula

$\displaystyle\sum\limits_{\boldsymbol{j^{\prime}}}q^{-2\sum\limits_{k=1}^{n-1}kj^{\prime}_{k}+n|\boldsymbol{j^{\prime}}|}\frac{V_{q^{-J}}(\boldsymbol{j^{\prime}},0)V_{q^{-I}}(\boldsymbol{j^{\prime}},0)}{V_{\frac{q^{-\frac{I+J}{2}}}{x}}(\boldsymbol{j^{\prime}},0)V_{xq^{-\frac{I+J}{2}+n}}(\boldsymbol{j^{\prime}},0)}=g_{IJ}^{(n)}(x).$
(3.29)

This identity is a consequence of the equation (6.16), that will be considered
in the following sections. And this ends the proof of the crossing unitarity
relation.

In what follows, we will also need the expressions for $L$-operators, which we
consider as $n\times n$ matrices acting in the symmetric tensor representation
of $U_{q}(\widehat{sl_{n}})$. Namely, we introduce $(n-1)$-component vectors
$j_{0}=(0,\dots,0)$ (vector of zeros) and $e_{\alpha}=(0,\dots,0,1,0,\dots,0)$
with $1$ at the $\alpha$-th position ($\alpha=1,\dots,n-1$). Then we construct
$L$-operators acting on $V_{1}^{(n)}\otimes V_{J}^{(n)}$ and on
$V_{J}^{(n)}\otimes V_{1}^{(n)}$, respectively:

$\displaystyle\left[S_{1,J}(x)\right]_{e_{\alpha},\boldsymbol{j}}^{e_{\beta},\boldsymbol{l}}=$
(3.30)
$\displaystyle\begin{cases}\delta_{\boldsymbol{j},\boldsymbol{l}}q^{2|e_{\alpha}|(\sum\limits_{s=1}^{\alpha}j_{s}-J)}\frac{1-q^{J-1-2j_{\alpha}-2\delta_{\alpha,0}(J-|\boldsymbol{j}|)}x^{-2}}{1-q^{-1-J}x^{-2}}&\text{
if }\alpha=\beta,\\\
-\delta_{\boldsymbol{j}+e_{\alpha},\boldsymbol{l}+e_{\beta}}q^{2(\sum\limits_{s=1}^{\alpha-1}j_{s}-J-1)+\delta_{\beta,0}(J+1)}x^{-2\delta_{\beta,0}}\frac{1-q^{2+2j_{\alpha}}}{1-q^{-1-J}x^{-2}}&\text{
if }\alpha>\beta,\\\
-\delta_{\boldsymbol{j}+e_{\alpha},\boldsymbol{l}+e_{\beta}}q^{2\sum\limits_{s=1}^{\alpha-1}j_{s}-1-J-\delta_{\alpha,0}(J+1-2|\boldsymbol{j}|)}x^{-2|e_{\alpha}|}\frac{1-q^{2(1+j_{\alpha}+\delta_{\alpha,0}(J-|\boldsymbol{j}|))}}{1-q^{-1-J}x^{-2}}&\text{
if }\alpha<\beta,\end{cases}$
$\displaystyle\left[S_{J,1}(x)\right]_{\boldsymbol{j},e_{\alpha}}^{\boldsymbol{l},e_{\beta}}=$
(3.31)
$\displaystyle\begin{cases}\delta_{\boldsymbol{j},\boldsymbol{l}}q^{-2|\boldsymbol{j}|+2|e_{\alpha}|\sum\limits_{s=\alpha}^{n-1}j_{s}}\frac{1-q^{J-1-2j_{\alpha}-2\delta_{\alpha,0}(J-|\boldsymbol{j}|)}x^{-2}}{1-q^{-1-J}x^{-2}}&\text{
if }\alpha=\beta,\\\
-\delta_{\boldsymbol{j}+e_{\alpha},\boldsymbol{l}+e_{\beta}}q^{-1+J-\delta_{\beta,0}(1+J)-2\sum\limits_{s=1}^{\alpha}j_{s}}x^{-2|e_{\beta}|}\frac{1-q^{2+2j_{\alpha}}}{1-q^{-1-J}x^{-2}}&\text{
if }\alpha>\beta,\\\
-\delta_{\boldsymbol{j}+e_{\alpha},\boldsymbol{l}+e_{\beta}}q^{-2+\delta_{\alpha,0}(1-J)-2|e_{\alpha}|\sum\limits_{s=1}^{\alpha}j_{s}}x^{-2\delta_{\alpha,0}}\frac{1-q^{2+2j_{\alpha}+2\delta_{\alpha,0}(J-|\boldsymbol{j}|)}}{1-q^{-1-J}x^{-2}}&\text{
if }\alpha<\beta.\end{cases}$

These relations can be derived directly from the definition of the $R$-matrix
(3).

##  4 The star-star relation

Using any of the symmetries (3.17), (3.18), one can obtain the star-star
relation for $U_{q}(\widehat{sl_{n}})$, written in terms of the functions
(3.16):

$\displaystyle\frac{V_{y/y^{\prime}}(\boldsymbol{b},\boldsymbol{d})}{V_{y/y^{\prime}}(\boldsymbol{a},\boldsymbol{c})}\sum_{\boldsymbol{m}=\max(\boldsymbol{b},\boldsymbol{c})}^{\boldsymbol{a}}\frac{V_{x/y^{\prime}}(\boldsymbol{a},\boldsymbol{m})V_{y^{\prime}/x^{\prime}}(\boldsymbol{m},\boldsymbol{b})V_{y/x}(\boldsymbol{m},\boldsymbol{c})}{V_{y/x^{\prime}}(\boldsymbol{m},\boldsymbol{d})}=$
(4.1)
$\displaystyle=\frac{V_{x/x^{\prime}}(\boldsymbol{a},\boldsymbol{b})}{V_{x/x^{\prime}}(\boldsymbol{c},\boldsymbol{d})}\sum_{\boldsymbol{m}=\boldsymbol{d}}^{\min(\boldsymbol{b},\boldsymbol{c})}\frac{V_{x/y^{\prime}}(\boldsymbol{m},\boldsymbol{d})V_{y^{\prime}/x^{\prime}}(\boldsymbol{c},\boldsymbol{m})V_{y/x}(\boldsymbol{b},\boldsymbol{m})}{V_{y/x^{\prime}}(\boldsymbol{a},\boldsymbol{m})},$

which was presented by Bazhanov and Sergeev in [27] as a conjecture. Namely,
let us introduce the new matrix indices

$\boldsymbol{i}=\boldsymbol{b}-\boldsymbol{d},\quad\boldsymbol{j}=\boldsymbol{a}-\boldsymbol{b},\quad\boldsymbol{i}^{\prime}=\boldsymbol{a}-\boldsymbol{c},\quad\boldsymbol{j}^{\prime}=\boldsymbol{c}-\boldsymbol{d},$
(4.2)

which automatically imply the conservation law
$\boldsymbol{i}+\boldsymbol{j}=\boldsymbol{i}^{\prime}+\boldsymbol{j}^{\prime}$
coming from the Kronecker delta-symbol in the definition of the $R$-matrix
(3).

Now let us note the property of the function $\Phi$ with permuted indices

$\Phi_{q}(\tau\boldsymbol{\gamma}\mid\tau\boldsymbol{\beta};\lambda,\mu)=q^{Q(\boldsymbol{\gamma},\boldsymbol{\beta})-Q(\boldsymbol{\beta},\boldsymbol{\gamma})}\Phi_{q}(\boldsymbol{\gamma}\mid\boldsymbol{\beta};\lambda,\mu)$
(4.3)

and the property of the function $V$

$V_{x}(\boldsymbol{a},\boldsymbol{b})=q^{Q(\boldsymbol{a},\boldsymbol{b})-Q(\boldsymbol{b},\boldsymbol{a})}V_{x}(\boldsymbol{a}-\boldsymbol{b},0),$
(4.4)

which can be easily verified from the corresponding definitions (3.10),
(3.16).

If we take (3.17) or (3.18) and use (3.15) together with (4.3) and (4.4), one
can see that all pre-factors on both sides of (3.17) or (3.18) cancel out, and
we come to the star-star relation (4.1).

##  5 Stochastic right boundaries

In this section we construct special cases of the right integrable boundaries,
which are the upper-triangular solutions to the reflection equation and those
that are generated from the former by applying symmetries of the stochastic
$R$-matrix.

The reflection equation for the stochastic $R$-matrix (3) has the form

$S_{I,J}\left(\frac{x}{y}\right)K_{I}\left(x\right)S_{J,I}\left(xy\right)K_{J}\left(y\right)=K_{J}\left(y\right)S_{I,J}\left(xy\right)K_{I}\left(x\right)S_{J,I}\left(\frac{x}{y}\right).$
(5.1)

First, we show that for a solution $K_{J}^{(1)}(y)$ of (5.1), we can construct
another solution

$K_{J}^{(2)}(y)_{\boldsymbol{j}}^{\boldsymbol{l}}=\left(\frac{1}{y^{2}\mu
q^{J}}\right)^{-|\sigma\boldsymbol{j}|}\left(\frac{y^{2}}{\mu
q^{J}}\right)^{|\sigma\boldsymbol{l}|}K_{J}^{(1)}(y)_{\sigma\boldsymbol{j}}^{\sigma\boldsymbol{l}},$
(5.2)

where $\mu\in\mathbb{C}$ is an arbitrary parameter. To prove this statement,
we will write the reflection equation (5.1) for $K_{J}^{(2)}(y)$ in index
notation, but with both external and internal summation indices replaced by
$j\rightarrow\sigma^{-1}j$, defined in (3.3),

$\displaystyle\sum_{\begin{subarray}{c}\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime},\\\
\boldsymbol{i}^{\prime\prime},\boldsymbol{j}^{\prime\prime}\end{subarray}}\left[S_{I,J}\left(\frac{x}{y}\right)\right]_{\sigma^{-1}\boldsymbol{i},\sigma^{-1}\boldsymbol{j}}^{\sigma^{-1}\boldsymbol{i}^{\prime},\sigma^{-1}\boldsymbol{j}^{\prime}}\left[K_{I}^{(2)}(x)\right]_{\sigma^{-1}\boldsymbol{i}^{\prime}}^{\sigma^{-1}\boldsymbol{i}^{\prime\prime}}\left[S_{J,I}(xy)\right]_{\sigma^{-1}\boldsymbol{j}^{\prime},\sigma^{-1}\boldsymbol{i}^{\prime\prime}}^{\sigma^{-1}\boldsymbol{j}^{\prime\prime},\sigma^{-1}\boldsymbol{i}^{\prime\prime\prime}}\left[K_{J}^{(2)}(y)\right]_{\sigma^{-1}\boldsymbol{j}^{\prime\prime}}^{\sigma^{-1}\boldsymbol{j}^{\prime\prime\prime}}=$
(5.3)
$\displaystyle\sum_{\begin{subarray}{c}\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime},\\\
\boldsymbol{i}^{\prime\prime},\boldsymbol{j}^{\prime\prime}\end{subarray}}\left[K_{J}^{(2)}(y)\right]_{\sigma^{-1}\boldsymbol{j}}^{\sigma^{-1}\boldsymbol{j}^{\prime}}\left[S_{I,J}(xy)\right]_{\sigma^{-1}\boldsymbol{i},\sigma^{-1}\boldsymbol{j}^{\prime}}^{\sigma^{-1}\boldsymbol{i}^{\prime},\sigma^{-1}\boldsymbol{j}^{\prime\prime}}\left[K_{I}^{(2)}(x)\right]_{\sigma^{-1}\boldsymbol{i}^{\prime}}^{\sigma^{-1}\boldsymbol{i}^{\prime\prime}}\left[S_{J,I}\left(\frac{x}{y}\right)\right]_{\sigma^{-1}\boldsymbol{j}^{\prime\prime},\sigma^{-1}\boldsymbol{i}^{\prime\prime}}^{\sigma^{-1}\boldsymbol{j}^{\prime\prime\prime},\sigma^{-1}\boldsymbol{i}^{\prime\prime\prime}}$

Now let us note that the $R$-matrix (3) possesses another symmetry

$\left[S_{I,J}\left(x\right)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}=\left[S_{J,I}\left(x\right)\right]_{\tau\sigma\boldsymbol{j},\tau\sigma\boldsymbol{i}}^{\tau\sigma\boldsymbol{j}^{\prime},\tau\sigma\boldsymbol{i}^{\prime}},$
(5.4)

which follows from the relations (3.17) and (3.20).

Applying (5.4) and (3.17) to (5.3) and using the fact that

$\left[K^{(2)}_{J}(y)\right]_{\sigma^{-1}\boldsymbol{j}}^{\sigma^{-1}\boldsymbol{l}}=\left(\frac{1}{y^{2}\mu
q^{J}}\right)^{-|\boldsymbol{j}|}\left(\frac{y^{2}}{\mu
q^{J}}\right)^{|\boldsymbol{l}|}\left[K^{(1)}_{J}(y)\right]_{\boldsymbol{j}}^{\boldsymbol{l}},$
(5.5)

one can check that all pre-factors on both sides of (5.3) cancel, and we come
to the reflection equation for $K_{J}^{(1)}(y)$.

Using the symmetry (5.2) $k$ times, $k=1,\ldots,n-1$, we will obtain a set of
different solutions to the reflection equation (5.1). Moreover, we can adjust
the free parameter $\mu\in\mathbb{C}$ in (5.2) at each step to satisfy the
stochastic condition.

For example, the solutions $K^{(2,3)}(x)$ from [33] can be obtained from
$K^{(1)}(x)$ by using the symmetry (5.2) for $n=3$. However, the solution
$K^{(4)}$ can not be obtained in this way.

Now let us construct the upper-triangular boundary matrices. We will use the
same approach as described in [24] for the $U_{q}(\widehat{sl_{2}})$ case.
Namely, we will start from considering the reflection equation for $I=J=1$ and
show that it admits the solution

$K_{1}(y)_{\boldsymbol{j}}^{\boldsymbol{l}}=\Phi_{q^{2}}\left(\boldsymbol{j}|\boldsymbol{l};\dfrac{y^{2}}{\nu
q},\dfrac{1}{y^{2}\nu q}\right),$ (5.6)

which is a generalization for $U_{q}(\widehat{sl_{n}})$ with arbitrary $n$.
Here $\nu\in\mathbb{C}$ is a free parameter. Then, using this solution, we
will construct a system of linear recurrence equations for $I=1$ and arbitrary
$J.$ And, at the final step, we will show that an upper-triangular stochastic
solution to these equations has the form

$K_{J}(y)_{\boldsymbol{j}}^{\boldsymbol{l}}=\Phi_{q^{2}}\left(\boldsymbol{j}|\boldsymbol{l};\dfrac{y^{2}}{\nu
q^{J}},\dfrac{1}{y^{2}\nu q^{J}}\right),$ (5.7)

which satisfies the stochastic condition

$\sum\limits_{\boldsymbol{j}}K_{J}(y)_{\boldsymbol{j}}^{\boldsymbol{l}}=1,$
(5.8)

as can be seen from the formula (3.22).

It is convenient to consider the reflection equation (5.1) in a matrix form,
that is, to use the expressions for $L$-operators (3.30) and (3.31). In a
particular case $J=1$, we come to a simpler formula

$\left[S_{1,1}(x)\right]_{e_{\alpha},e_{\gamma}}^{e_{\beta},e_{\delta}}=\begin{cases}\delta_{\gamma,\delta}q^{2Q(e_{\gamma},e_{\alpha})-2(1-\delta_{\alpha,\gamma})|e_{\alpha}|}\frac{1-q^{-2\delta_{\alpha,\gamma}}x^{-2}}{1-q^{-2}x^{-2}}&\text{
if }\alpha=\beta,\\\
-\delta_{e_{\alpha}+e_{\gamma},e_{\beta}+e_{\delta}}q^{-2}x^{-2\delta_{\beta,0}}\frac{1-q^{2}}{1-q^{-2}x^{-2}}&\text{
if }\alpha>\beta,\\\
-\delta_{e_{\alpha}+e_{\gamma},e_{\beta}+e_{\delta}}q^{-2}x^{-2|e_{\alpha}|}\frac{1-q^{2}}{1-q^{-2}x^{-2}}&\text{
if }\alpha<\beta.\end{cases}$ (5.9)

The last expression allows us to analyze the structure of the reflection
equation for $I=J=1$. So we want to show that the right boundary (5.6)
satisfies the relation

$\displaystyle\sum\limits_{e_{\alpha^{\prime}},e_{\beta^{\prime}},e_{\alpha^{\prime\prime}},e_{\beta^{\prime\prime}}}\left[S_{1,1}\left(\dfrac{x}{y}\right)\right]_{e_{\alpha},e_{\beta}}^{e_{\alpha^{\prime}},e_{\beta^{\prime}}}\left[K_{1}\left(x\right)\right]_{e_{\alpha^{\prime}}}^{e_{\alpha^{\prime\prime}}}\left[S_{1,1}\left(xy\right)\right]_{e_{\beta^{\prime}},e_{\alpha^{\prime\prime}}}^{e_{\beta^{\prime\prime}},e_{\alpha^{\prime\prime\prime}}}\left[K_{1}\left(y\right)\right]_{e_{\beta^{\prime\prime}}}^{e_{\beta^{\prime\prime\prime}}}=$
(5.10)
$\displaystyle=\sum\limits_{e_{\alpha^{\prime}},e_{\beta^{\prime}},e_{\alpha^{\prime\prime}},e_{\beta^{\prime\prime}}}\left[K_{1}\left(y\right)\right]_{e_{\beta}}^{e_{\beta^{\prime}}}\left[S_{1,1}\left(xy\right)\right]_{e_{\alpha},e_{\beta^{\prime}}}^{e_{\alpha^{\prime}},e_{\beta^{\prime\prime}}}\left[K_{1}\left(x\right)\right]_{e_{\alpha^{\prime}}}^{e_{\alpha^{\prime\prime}}}\left[S_{1,1}\left(\dfrac{x}{y}\right)\right]_{e_{\beta^{\prime\prime}},e_{\alpha^{\prime\prime}}}^{e_{\beta^{\prime\prime\prime}},e_{\alpha^{\prime\prime\prime}}},$

where the summation goes over internal repeated indices, and the upper-
triangular matrix $K_{1}(x)$ has the form

$\left[K_{1}\left(x\right)\right]_{e_{\alpha}}^{e_{\beta}}=\begin{cases}1&\text{
if }\alpha=\beta=0,\\\ \frac{q\nu(1-x^{4})}{x^{2}(1-q\nu x^{2})}&\text{ if
}\alpha=0,\beta\neq 0,\\\ 0&\text{ if }\alpha\neq 0,\beta=0,\\\
\delta_{\alpha,\beta}\frac{x^{2}-q\nu}{x^{2}(1-q\nu x^{2})}&\text{ if
}\alpha\neq 0,\beta\neq 0.\end{cases}$ (5.11)

It is clear from the formulas (5.9) and (5.11) that the reflection equation
(5.10) has the same form for any $n$: indeed, it only depends on the external
indices $e_{\alpha}$, $e_{\beta}$, $e_{\alpha^{\prime\prime\prime}}$,
$e_{\beta^{\prime\prime\prime}}$ (more precisely, on the relative order of the
numbers $\alpha$, $\beta$, $\alpha^{\prime\prime\prime}$,
$\beta^{\prime\prime\prime}$), and each factor in the relation gives no more
than 2 nontrivial terms. This means that it is enough to check the correctness
of the equation for, say, $n=5$ (to consider all possibilities for index
permutations). And, indeed, the explicit calculation shows that the equation
is valid in this case, which proves (5.6).

Now, based on the solution for $I=1$, we can consider the reflection equation
for an arbitrary parameter $J$

$S_{1,J}\left(\frac{x}{y}\right)K_{1}\left(x\right)S_{J,1}\left(xy\right)K_{J}\left(y\right)=K_{J}\left(y\right)S_{1,J}\left(xy\right)K_{1}\left(x\right)S_{J,1}\left(\frac{x}{y}\right),$
(5.12)

where $K_{1}(x)$ is given by (5.6). Then, substituting (3.30), (3.31) in
(5.12) in matrix form and performing some cumbersome calculations, we obtain a
system of $n^{2}$ linear equations, that are polynomial in $x$ and $y$.
Decoupling with respect to $x$, we get a set of recurrence relations for
$K_{J}(y)_{\boldsymbol{j}}^{\boldsymbol{l}}.$ A careful analysis of these
equations shows that there are $n$ linearly independent ones:

$\displaystyle\nu\left(1-q^{2l_{1}}\right)\left[K_{J}(y)\right]_{\boldsymbol{j}}^{\boldsymbol{l}-e_{1}}$
$\displaystyle-q^{-J}y^{2}\left(q^{2j_{1}}-q^{2l_{1}}\right)\left[K_{J}(y)\right]_{\boldsymbol{j}}^{\boldsymbol{l}}-$
(5.13) $\displaystyle-\nu
y^{4}\left(1-q^{2j_{1}+2}\right)\left[K_{J}(y)\right]_{\boldsymbol{j}+e_{1}}^{\boldsymbol{l}}=0,$
$\displaystyle\left(q^{-2\sum\limits_{k=1}^{i}l_{k}}-q^{-2\sum\limits_{k=1}^{i}j_{k}}\right)\left[K_{J}(y)\right]_{\boldsymbol{j}}^{\boldsymbol{l}}-\sum\limits_{m=1}^{i}q^{-2\sum\limits_{k=1}^{m}l_{k}}\left(1-q^{2l_{m}}\right)\left[K_{J}(y)\right]_{\boldsymbol{j}}^{\boldsymbol{l}-e_{m}+e_{i+1}}$
(5.14)
$\displaystyle+q^{-2}\sum\limits_{m=1}^{i}q^{-2\sum\limits_{k=1}^{m}j_{k}}\left(1-q^{2+2j_{m}}\right)\left[K_{J}(y)\right]_{\boldsymbol{j}+e_{m}-e_{i+1}}^{\boldsymbol{l}}=0,\quad\quad
i=1,\dots,n-2,$
$\displaystyle\left(q^{-2j_{n-1}}-q^{-2l_{n-1}}\right)\left[K_{J}(y)\right]_{\boldsymbol{j}}^{\boldsymbol{l}}$
$\displaystyle-\left(1-q^{-2l_{n-1}}\right)\left[K_{J}(y)\right]_{\boldsymbol{j}}^{\boldsymbol{l}-e_{n-1}}+$
(5.15)
$\displaystyle+\left(1-q^{-2-2j_{n-1}}\right)\left[K_{J}(y)\right]_{\boldsymbol{j}+e_{n-1}}^{\boldsymbol{l}}=0,$

where we assume that
$\left[K_{J}(y)\right]_{\boldsymbol{j}}^{\boldsymbol{l}}=0$ unless
$j_{\alpha}\geq 0$, $l_{\alpha}\geq 0$, $|\boldsymbol{j}|\leq J$ and
$|\boldsymbol{l}|\leq J.$ And in the last step one can check that (5.7) solves
these recurrence equations. Thus we constructed the integrable upper-
triangular right boundary solution.

Applying (5.2) to (5.7) and choosing $\mu=\nu$, we obtain a stochastic lower-
triangular solution of the form

$\displaystyle K_{J}(y)_{\boldsymbol{j}}^{\boldsymbol{l}}$
$\displaystyle=\left(\frac{1}{y^{2}\nu
q^{J}}\right)^{-|\sigma\boldsymbol{j}|}\left(\frac{y^{2}}{\nu
q^{J}}\right)^{|\sigma\boldsymbol{l}|}\Phi_{q^{2}}\left(\sigma\boldsymbol{j}\mid\sigma\boldsymbol{l};\frac{y^{2}}{\nu
q^{J}},\frac{1}{y^{2}\nu q^{J}}\right)=$ (5.16)
$\displaystyle=\widehat{\Phi}_{q^{2}}\left(\sigma\boldsymbol{l}-\sigma\boldsymbol{j}\mid\sigma\boldsymbol{l};y^{-4},\frac{1}{y^{2}\nu
q^{J}}\right),$

where the operation on indices $\sigma$ is defined in (3.3), and we introduced
a new notation

$\widehat{\Phi}_{q}\left(\boldsymbol{j}\mid\boldsymbol{l};x,y\right)=q^{Q(\boldsymbol{j},\boldsymbol{l})-Q(\boldsymbol{l},\boldsymbol{j})}\Phi_{q}\left(\boldsymbol{j}\mid\boldsymbol{l};x,y\right).$
(5.17)

In the second line in (5.16) we applied the relation.

$\Phi_{q}(\boldsymbol{m}-\boldsymbol{j}\mid\boldsymbol{m},\mu/\lambda,\mu)=\Phi_{q}(\boldsymbol{j}\mid\boldsymbol{m},\lambda,\mu)q^{Q(\boldsymbol{j},\boldsymbol{m})-Q(\boldsymbol{m},\boldsymbol{j})}\mu^{-|\boldsymbol{j}|}\lambda^{|\boldsymbol{m}|}.$
(5.18)

The stochastic condition (5.8) for the lower-triangular solution follows from
the identity

$\displaystyle\sum\limits_{\boldsymbol{j}}x^{|\boldsymbol{l}|}y^{-|\boldsymbol{j}|}\Phi_{q}\left(\boldsymbol{j}\mid\boldsymbol{l};x,y\right)=\sum\limits_{\boldsymbol{j}}q^{Q(\boldsymbol{j},\boldsymbol{l})-Q(\boldsymbol{l},\boldsymbol{j})}\Phi_{q}\left(\boldsymbol{j}\mid\boldsymbol{l};\frac{y}{x},y\right)=$
(5.19)
$\displaystyle=\sum\limits_{\boldsymbol{j}}\Phi_{q}\left(\tau\boldsymbol{j}\mid\tau\boldsymbol{l};\frac{y}{x},y\right)=1,$

where we used the relations (5.18), (4.3) and the summation rule (3.22).

It is interesting to compare the obtained solutions (5.7) and (5.16) with
already known results. In [34] a full classification of the right boundaries
for the case $J=1$ was given. It is based on the classification with four
particles of two special types (four positive integers): slow ($s_{1}$,
$s_{2}$) and fast ($f_{1}$, $f_{2}$) ones, which satisfy the constraints

$1\leq s_{1}\leq s_{2}<f_{2}\leq f_{1}\leq n\quad\text{ and }\quad
f_{1}-f_{2}=s_{2}-s_{1}.$ (5.20)

Note that the authors of [34] use different index notations, starting from $1$
rather than $0$. Each solution depends on two real parameters $\alpha$,
$\gamma$ with the choice of two boundary matrices
$B^{0}\left(\alpha,\gamma|s_{1},s_{2},f_{2},f_{1}\right)$ and
$B\left(\alpha,\gamma|s_{1},s_{2},f_{2},f_{1}\right).$ Then one can check that
the solution (5.7) for $J=1$ corresponds to the case of the matrix
$B\left(\alpha,\gamma|s_{1},s_{2},f_{2},f_{1}\right)$ with the slow
$s_{1}=s_{2}=1$ and fast $f_{1}=f_{2}=n$ species and the parameters

$\alpha=0,\quad\gamma=\dfrac{1-q}{1-\nu q},$ (5.21)

and the lower-triangular solution (5.16) corresponds to the parameters

$\alpha=\dfrac{1-q}{1-\nu q},\quad\gamma=0.$ (5.22)

To illustrate the structure of the right boundary matrices (5.7), (5.16), we
listed them in D for a particular case $n=3$ and $J=2$.

##  6 Stochastic left boundaries

Now we will construct the left upper- and lower-triangular boundaries using
the results from the previous sections. The starting point for this
calculation is the dual reflection equation

$\widetilde{K}_{J}\left(y\right)\widetilde{S}_{J,I}(xy)\widetilde{K}_{I}\left(x\right)S_{J,I}\left(\frac{y}{x}\right)=S_{I,J}\left(\frac{y}{x}\right)\widetilde{K}_{I}\left(x\right)\widetilde{S}_{I,J}(xy)\widetilde{K}_{J}\left(y\right),$
(6.1)

where as usual

$\widetilde{S}_{I,J}\left(x\right)=\left(\left(S_{I,J}\left(x\right)^{t_{2}}\right)^{-1}\right)^{t_{2}}.$
(6.2)

First, let us notice that the matrix $M$ (3.26) satisfies the compatibility
condition (2.7):

$\left[M\otimes M,S_{I,J}(x)\right]=0.$ (6.3)

As discussed in Section 2, we can find the solution to (6.1) in terms of
$K_{I}(x)$ from (5.7)

$\widetilde{K}_{I}(x)=M^{-1}K_{I}\left(\dfrac{1}{q^{n/2}x}\right).$ (6.4)

However, we are also interested in solutions $\bar{K}_{I}(x)$ of the
reflection equation (2.11) which enter the definition of the hamiltonian
rates, see (2.20). The corresponding reflection equation takes the form:

$S_{J,I}\left(\frac{y}{x}\right)\bar{K}_{I}\left(x\right)S_{I,J}\left(\frac{1}{xy}\right)\bar{K}_{J}\left(y\right)=\bar{K}_{J}\left(y\right)S_{J,I}\left(\frac{1}{xy}\right)\bar{K}_{I}\left(x\right)S_{I,J}\left(\frac{y}{x}\right).$
(6.5)

First, let us discuss the symmetries of this equation. If
$\bar{K}_{J}^{(1)}(y)$ is a solution of (6.5), then

$\bar{K}_{J}^{(2)}(y)_{\boldsymbol{j}}^{\boldsymbol{l}}=\left(\frac{y^{2}}{\mu
q^{J}}\right)^{|\sigma\boldsymbol{j}|}\left(\frac{1}{y^{2}\mu
q^{J}}\right)^{-|\sigma\boldsymbol{l}|}\bar{K}_{J}^{(1)}(y)_{\sigma\boldsymbol{j}}^{\sigma\boldsymbol{l}}$
(6.6)

is also a solution. This can be shown by using the symmetry (5.4) of the
$R$-matrix and performing a calculation similar to (5.3).

Proceeding in a similar way, one can show that if $K_{J}(y)$ solves (5.1),
then

$\bar{K}_{J}(y)_{\boldsymbol{j}}^{\boldsymbol{l}}=K_{J}(y^{-1})_{\tau\sigma\boldsymbol{j}}^{\tau\sigma\boldsymbol{l}}$
(6.7)

solves (6.5). Let us emphasize that the formula (6.7) is a consequence of the
symmetry (5.4).

However, we also have a general construction for the matrix $\bar{K}(x)$ given
by (2.13) and (2.10). In the rest of this section we will show that it gives
the same result (6.7) up to the transformation (6.6).

Combining (2.13) with (6.4), we obtain

$\displaystyle\bar{K}_{I}(x)=\operatorname{tr}_{V_{J}}\left(M_{J}^{-1}K_{J}\left(\dfrac{x}{q^{\frac{n}{2}}}\right)S_{J,I}\left(\dfrac{1}{x^{2}}\right)P_{I,J}\right),$
(6.8)

where we assume that $I=J$ to correctly define the permutation operator acting
on two isomorphic spaces.

We start our analysis with the upper-triangular solutions (5.7). Substituting
(3.26) in (6.8) and replacing an arbitrary parameter
$\nu\rightarrow\dfrac{1}{\nu q^{n}}$, we obtain the following formula:

$\displaystyle\bar{K}_{J}(x)_{\boldsymbol{j}}^{\boldsymbol{j}^{\prime}}=\sum\limits_{\boldsymbol{i},\boldsymbol{i^{\prime}}}q^{-2\sum\limits_{k=1}^{n-1}(n-k)i_{k}}\left.K_{J}\left(\dfrac{x}{q^{n/2}}\right)_{\boldsymbol{i}}^{\boldsymbol{i}^{\prime}}\right|_{\nu\rightarrow\frac{1}{\nu
q^{n}}}\left[S_{J,J}\left(\dfrac{1}{x^{2}}\right)\right]_{\boldsymbol{i}^{\prime},\boldsymbol{j}}^{\boldsymbol{j}^{\prime},\boldsymbol{i}}.$
(6.9)

We claim that (6.9) can be transformed to the following form

$\displaystyle\bar{K}_{J}(y)_{\boldsymbol{j}}^{\boldsymbol{l}}$
$\displaystyle=\lambda_{J}^{(n)}(x)\left(\frac{y^{2}}{\nu
q^{J}}\right)^{-|\boldsymbol{j}|}\left(\frac{1}{y^{2}\nu
q^{J}}\right)^{|\boldsymbol{l}|}q^{2\left(Q(\boldsymbol{j},\boldsymbol{l})-Q(\boldsymbol{l},\boldsymbol{j})\right)}\Phi_{q^{2}}\left(\boldsymbol{j}|\boldsymbol{l};\dfrac{1}{y^{2}\nu
q^{J}},\dfrac{y^{2}}{\nu q^{J}}\right)=$ (6.10)
$\displaystyle=\lambda_{J}^{(n)}(x)\Phi_{q^{2}}\left(\boldsymbol{l}-\boldsymbol{j}|\boldsymbol{l};y^{4},\dfrac{y^{2}}{\nu
q^{J}}\right),$

where in the last line we used (5.18) and introduced the function

$\lambda_{J}^{(n)}(x)=\dfrac{\left(x^{-4}q^{2J+2};q^{2}\right)_{n-1}\left(\nu
x^{-2}q^{2-J};q^{2}\right)_{n-1}}{\left(x^{-4}q^{2};q^{2}\right)_{n-1}\left(\nu
x^{-2}q^{J+2};q^{2}\right)_{n-1}}.$ (6.11)

Combining the first line in (6.10) with the symmetry transformation (6.6), we
see that the formula (6.10) is equivalent to (6.7) up to a scalar factor
(6.11).

To prove the equivalence of (6.9) and (6.10), we want to apply certain
summation formulas. A simple analysis shows that, by using a symmetry
transformation

$\left[S_{J,J}\left(x\right)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}=x^{2\left(|\boldsymbol{i}^{\prime}|-|\boldsymbol{i}|\right)}q^{2J\left(|\boldsymbol{j}^{\prime}|-|\boldsymbol{i}|\right)}\left[S_{J,J}\left(x\right)\right]_{\tau\boldsymbol{j},\tau\boldsymbol{i}}^{\tau\boldsymbol{j}^{\prime},\tau\boldsymbol{i}^{\prime}}$
(6.12)

in (6.9), we obtain the terminating series, truncated in a natural way by $J$.

Rewriting all functions $\Phi$ in terms of $V$’s and canceling out the pre-
factors, we obtain that the equality of (6.9) and (6.10) is equivalent to the
proof of the relation

$\displaystyle\lambda_{J}^{(n)}(x)\frac{V_{\frac{q^{-J/2}}{\nu^{1/2}x}}(\boldsymbol{b},\boldsymbol{a})V_{x^{2}}(\boldsymbol{c},\boldsymbol{b})}{V_{\frac{xq^{-J/2}}{\nu^{1/2}}}(\boldsymbol{c},\boldsymbol{a})}=\frac{V_{\frac{q^{n}}{x^{2}}}(\boldsymbol{c},\boldsymbol{b})}{V_{q^{-J}}(\boldsymbol{c},\boldsymbol{a})}\sum\limits_{\boldsymbol{m}}V_{x^{2}}(\boldsymbol{b},\boldsymbol{m})V_{x^{2}}(\boldsymbol{c},\boldsymbol{m})V_{\frac{q^{-J}}{x^{2}}}(\boldsymbol{m},\boldsymbol{a})\times$
(6.13)
$\displaystyle\times\sum\limits_{\boldsymbol{d}}q^{-2\sum\limits_{k=1}^{n-1}k(d_{k}-c_{k})+n(|\boldsymbol{d}|-|\boldsymbol{c}|)}\frac{V_{q^{-J}}(\boldsymbol{d},\boldsymbol{b})V_{\nu^{1/2}xq^{-J/2}}(\boldsymbol{d},\boldsymbol{c})}{V_{\frac{\nu^{1/2}q^{-\frac{J}{2}+n}}{x}}(\boldsymbol{d},\boldsymbol{b})V_{x^{2}q^{-J}}(\boldsymbol{d},\boldsymbol{m})}.$

Here we introduced matrix indices

$\boldsymbol{i}=\boldsymbol{d}-\boldsymbol{c},\quad\boldsymbol{j}=\boldsymbol{b}-\boldsymbol{a},\quad\boldsymbol{i}^{\prime}=\boldsymbol{d}-\boldsymbol{b},\quad\boldsymbol{j}^{\prime}=\boldsymbol{c}-\boldsymbol{a}.$
(6.14)

In these notations, all indices satisfy the constraints

$0\leq a_{i}\leq m_{i}\leq b_{i}\leq c_{i}\leq d_{i},\quad|\boldsymbol{d}|\leq
J.$ (6.15)

Two summations in (6.13) can be done by using the summation formulas

$\displaystyle\sum\limits_{\boldsymbol{d}}q^{-2\sum\limits_{k=1}^{n-1}k(d_{k}-c_{k})+n(|\boldsymbol{d}|-|\boldsymbol{c}|)}$
$\displaystyle\frac{V_{q^{-J}}(\boldsymbol{d},\boldsymbol{b})V_{y}(\boldsymbol{d},\boldsymbol{c})}{V_{z}(\boldsymbol{d},\boldsymbol{b})V_{\frac{y}{z}q^{n-J}}(\boldsymbol{d},\boldsymbol{m})}=$
(6.16)
$\displaystyle=\mu_{J}^{(n)}(y,z)\frac{V_{q^{-J}}(\boldsymbol{c},\boldsymbol{b})V_{\frac{q^{n}}{z}}(\boldsymbol{b},\boldsymbol{m})}{V_{\frac{yq^{n}}{z}}(\boldsymbol{b},\boldsymbol{m})V_{\frac{z}{y}}(\boldsymbol{c},\boldsymbol{b})V_{\frac{q^{n-J}}{z}}(\boldsymbol{c},\boldsymbol{m})},$

where

$\mu_{J}^{(n)}(y,z)=\frac{\left(q^{2-2n}z^{2};q^{2}\right)_{n-1}\left(\frac{q^{2J-2n+2}z^{2}}{y^{2}};q^{2}\right)_{n-1}}{\left(q^{2J-2n+2}z^{2};q^{2}\right)_{n-1}\left(\frac{q^{2-2n}z^{2}}{y^{2}};q^{2}\right)_{n-1}},$
(6.17)

and

$\dfrac{V_{y/y^{\prime}}(\boldsymbol{c},\boldsymbol{b})}{V_{y/y^{\prime}}(\boldsymbol{c},\boldsymbol{a})}\sum\limits_{\boldsymbol{m}}\dfrac{V_{x/y^{\prime}}(\boldsymbol{m},\boldsymbol{a})V_{y^{\prime}/x^{\prime}}(\boldsymbol{b},\boldsymbol{m})V_{y/x}(\boldsymbol{c},\boldsymbol{m})}{V_{y/x^{\prime}}(\boldsymbol{c},\boldsymbol{m})}=\frac{V_{x/x^{\prime}}(\boldsymbol{b},\boldsymbol{a})V_{y/x}(\boldsymbol{c},\boldsymbol{b})}{V_{y/x^{\prime}}(\boldsymbol{c},\boldsymbol{a})}.$
(6.18)

The relation (6.16) is proven in C. The relation (6.18) easily follows from
the star-star relation (4.1). Indeed, let us substitute
$\boldsymbol{b}=\boldsymbol{a}$ in (4.1). Then the sum in the LHS disappears,
since $\boldsymbol{a}\geq\boldsymbol{m}$ and
$\boldsymbol{m}\geq\boldsymbol{b}=\boldsymbol{a}$ and (4.1) reduces to (6.18).

Since the solution to the reflection equation is defined up to an arbitrary
factor, we divide (6.10) by $\lambda_{J}^{(n)}(x)$ and obtain for the
integrable left upper-triangular boundary

$\bar{K}_{J}(y)_{\boldsymbol{j}}^{\boldsymbol{l}}=\Phi_{q^{2}}\left(\boldsymbol{l}-\boldsymbol{j}|\boldsymbol{l};y^{4},\dfrac{y^{2}}{\nu
q^{J}}\right),$ (6.19)

which satisfies the stochastic property due to (3.22).

Now let us construct the stochastic lower-triangular left boundary solutions.
They are obtained by applying the symmetry (6.6) with $\mu=\nu$ to the first
line of (6.10) and removing the scalar factor $\lambda_{J}^{(n)}(x)$. As a
result, we have

$\bar{K}_{J}(y)_{\boldsymbol{j}}^{\boldsymbol{l}}=\widehat{\Phi}_{q^{2}}\left(\sigma\boldsymbol{j}\mid\sigma\boldsymbol{l};\dfrac{1}{y^{2}\nu
q^{J}},\dfrac{y^{2}}{\nu q^{J}}\right),$ (6.20)

where $\widehat{\Phi}_{q}$ is defined in (5.17).

In D we listed left boundary matrices (6.19), (6.20) for a particular case
$n=3$ and $J=2$.

##  7 Another model

In this section we will not impose the requirement $I,J\in\mathbb{N}$, i.e the
$R$-matrix (3) acts in the tensor product of infinite dimensional modules with
basis vectors having the form

$|\boldsymbol{i}\rangle=|i_{1},i_{2},\dots,i_{n-1}\rangle,\quad
i_{k}\in\mathbb{Z}_{+}.$ (7.1)

The $R$-matrix (3) degenerates when we choose the spectral parameter as
$x=q^{(J-I)/2}$ [26, 14]:

$\left[S_{I,J}\left(q^{(J-I)/2}\right)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}=\delta_{\boldsymbol{i}+\boldsymbol{j},\boldsymbol{i}^{\prime}+\boldsymbol{j}^{\prime}}\Phi_{q^{2}}\left(\boldsymbol{i}\mid\boldsymbol{j}^{\prime};q^{-2I},q^{-2J}\right),$
(7.2)

which suggests to define a new stochastic non-difference type $R$-matrix

$\left[S_{12}(x,y)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i}^{\prime},\boldsymbol{j}^{\prime}}=\delta_{\boldsymbol{i}+\boldsymbol{j},\boldsymbol{i}^{\prime}+\boldsymbol{j}^{\prime}}\Phi_{q^{2}}\left(\boldsymbol{i}\mid\boldsymbol{j}^{\prime};x,y\right).$
(7.3)

Now we consider $x,y\in\mathbb{C}$ as spectral parameters and the Yang-Baxter
equation takes the form

$S_{12}\left(x,y\right)S_{13}\left(x,z\right)S_{23}\left(y,z\right)=S_{23}\left(y,z\right)S_{13}\left(x,z\right)S_{12}\left(x,y\right),$
(7.4)

which directly follows from (3.13). It is easy to see that the internal sums
in (7.4) contain only a finite number of terms, so there is no a problem with
convergency.

For a given solution to the Yang-Baxter equation (7.3), we define the
reflection equation of a non-difference type

$S_{12}(x,y)K_{1}(x,\bar{x})S_{21}(y,\bar{x})K_{2}(y,\bar{y})=K_{2}(y,\bar{y})S_{12}(x,\bar{y})K_{1}(x,\bar{x})S_{21}(\bar{y},\bar{x}).$
(7.5)

In [24] it was suggested that this equation admits an upper-triangular
stochastic solution of the form

$K(x,\bar{x})_{\boldsymbol{j}}^{\boldsymbol{l}}=\Phi_{q^{2}}({\boldsymbol{j}}\mid{\boldsymbol{l}};z^{2}x,z^{2}\bar{x}),$
(7.6)

where $z\in\mathbb{C}$ is an arbitrary parameter and there is no constraint
between spectral parameters $x$ and $\bar{x}$ (see the equation (5.10) from
[24]).

Here we give the proof of this fact suggested in [35], which is based on the
star-star relation (4.1) and an orthogonality property:

$\sum\limits_{\boldsymbol{m}}x^{|\boldsymbol{m}|-|\boldsymbol{b}|}y^{|\boldsymbol{m}|-|\boldsymbol{a}|}V_{x}\left(\boldsymbol{a},\boldsymbol{m}\right)V_{y}\left(\boldsymbol{m},\boldsymbol{b}\right)=V_{xy}\left(\boldsymbol{a},\boldsymbol{b}\right).$
(7.7)

The last equality directly follows from the relation (B.6) and the
$q$-Vandermonde summation formula (B.2). We start the proof by writing the
reflection equation (7.5) in index notation

$\displaystyle\sum\limits_{\boldsymbol{i^{\prime}},\boldsymbol{j^{\prime}},\boldsymbol{i^{\prime\prime}},\boldsymbol{j^{\prime\prime}}}\left[S_{12}\left(x^{2},y^{2}\right)\right]_{\boldsymbol{i},\boldsymbol{j}}^{\boldsymbol{i^{\prime}},\boldsymbol{j^{\prime}}}\left[K_{1}\left(x^{2},\bar{x}^{2}\right)\right]_{\boldsymbol{i^{\prime}}}^{\boldsymbol{i^{\prime\prime}}}\left[S_{21}\left(y^{2},\bar{x}^{2}\right)\right]_{\boldsymbol{j^{\prime}},\boldsymbol{i^{\prime\prime}}}^{\boldsymbol{j^{\prime\prime}},\boldsymbol{i^{\prime\prime\prime}}}\left[K_{2}\left(y^{2},\bar{y}^{2}\right)\right]_{\boldsymbol{j^{\prime\prime}}}^{\boldsymbol{j^{\prime\prime\prime}}}=$
(7.8)
$\displaystyle=\sum\limits_{\boldsymbol{i^{\prime}},\boldsymbol{j^{\prime}},\boldsymbol{i^{\prime\prime}},\boldsymbol{j^{\prime\prime}}}\left[K_{2}\left(y^{2},\bar{y}^{2}\right)\right]_{\boldsymbol{j}}^{\boldsymbol{j^{\prime}}}\left[S_{12}\left(x^{2},\bar{y}^{2}\right)\right]_{\boldsymbol{i},\boldsymbol{j^{\prime}}}^{\boldsymbol{i^{\prime}},\boldsymbol{j^{\prime\prime}}}\left[K_{1}\left(x^{2},\bar{x}^{2}\right)\right]_{\boldsymbol{i^{\prime}}}^{\boldsymbol{i^{\prime\prime}}}\left[S_{21}\left(\bar{y}^{2},\bar{x}^{2}\right)\right]_{\boldsymbol{j^{\prime\prime}},\boldsymbol{i^{\prime\prime}}}^{\boldsymbol{j^{\prime\prime\prime}},\boldsymbol{i^{\prime\prime\prime}}},$

where, as usual, the summation goes over repeated indices. Then we introduce
new external indices

$\boldsymbol{i}=\boldsymbol{b}-\boldsymbol{a},\quad\boldsymbol{j}=\boldsymbol{c}-\boldsymbol{b},\quad\boldsymbol{i^{\prime\prime\prime}}=\boldsymbol{e}-\boldsymbol{a},\quad\boldsymbol{j^{\prime\prime\prime}}=\boldsymbol{f}-\boldsymbol{e},$
(7.9)

summation indices for the left-hand side

$\boldsymbol{i^{\prime}}=\boldsymbol{c}-\boldsymbol{d}_{1},\quad\boldsymbol{j^{\prime}}=\boldsymbol{d}_{1}-\boldsymbol{a},\quad\boldsymbol{i^{\prime\prime}}=\boldsymbol{d}_{2}-\boldsymbol{d}_{1},\quad\boldsymbol{j^{\prime\prime}}=\boldsymbol{d}_{2}-\boldsymbol{e}$
(7.10)

and for the right-hand side

$\boldsymbol{i^{\prime}}=\boldsymbol{d}_{2}-\boldsymbol{d}_{1},\quad\boldsymbol{j^{\prime}}=\boldsymbol{d}_{2}-\boldsymbol{b},\quad\boldsymbol{i^{\prime\prime}}=\boldsymbol{f}-\boldsymbol{d}_{1},\quad\boldsymbol{j^{\prime\prime}}=\boldsymbol{d}_{1}-\boldsymbol{a}.$
(7.11)

After canceling out similar factor on both sides, the reflection equation in
terms of the functions $V_{x}\left(\boldsymbol{a},\boldsymbol{b}\right)$ takes
the form

$\displaystyle\sum\limits_{\boldsymbol{d}_{2}}\frac{\bar{x}^{|\boldsymbol{c}|}\bar{y}^{\left|\boldsymbol{d}_{2}\right|}}{x^{\left|\boldsymbol{d}_{2}\right|}y^{|\boldsymbol{f}|}}\frac{V_{yz}\left(\boldsymbol{d}_{2},\boldsymbol{e}\right)V_{\frac{\bar{x}}{x}}\left(\boldsymbol{d}_{2},\boldsymbol{c}\right)V_{\frac{\bar{y}}{y}}\left(\boldsymbol{f},\boldsymbol{d}_{2}\right)}{V_{z\bar{y}}(\boldsymbol{f},\boldsymbol{e})}\sum\limits_{\boldsymbol{d}_{1}}\frac{V_{\frac{y}{x}}\left(\boldsymbol{d}_{1},\boldsymbol{b}\right)V_{xz}\left(\boldsymbol{c},\boldsymbol{d}_{1}\right)V_{\frac{\bar{x}}{y}}\left(\boldsymbol{e},\boldsymbol{d}_{1}\right)}{V_{z\bar{x}}\left(\boldsymbol{d}_{2},\boldsymbol{d}_{1}\right)}=$
(7.12) $\displaystyle=$
$\displaystyle\sum\limits_{\boldsymbol{d}_{2}}\frac{\bar{x}^{\left|\boldsymbol{d}_{2}\right|}\bar{y}^{|\boldsymbol{c}|}}{x^{|\boldsymbol{f}|}y^{\left|\boldsymbol{d}_{2}\right|}}\frac{V_{yz}(\boldsymbol{c},\boldsymbol{b})V_{\frac{\bar{x}}{x}}\left(\boldsymbol{f},\boldsymbol{d}_{2}\right)V_{\frac{\bar{y}}{y}}\left(\boldsymbol{d}_{2},\boldsymbol{c}\right)}{V_{z\bar{y}}\left(\boldsymbol{d}_{2},\boldsymbol{b}\right)}\sum\limits_{\boldsymbol{d}_{1}}\frac{V_{\frac{\bar{y}}{x}}\left(\boldsymbol{d}_{1},\boldsymbol{b}\right)V_{xz}\left(\boldsymbol{d}_{2},\boldsymbol{d}_{1}\right)V_{\frac{\bar{x}}{\bar{y}}}\left(\boldsymbol{e},\boldsymbol{d}_{1}\right)}{V_{z\bar{x}}\left(\boldsymbol{f},\boldsymbol{d}_{1}\right)}.$

Then, sequentially applying the star-star relation (4.1) centered at
$\boldsymbol{d}_{1}$ and orthogonality formula (7.7) with respect to
$\boldsymbol{d}_{2}$ to both sides of the relation (7.12), we come to the
conclusion that the left and right sides coincide:

$\operatorname{LHS}=\operatorname{RHS}=\frac{V_{\frac{\bar{x}}{x}}(\boldsymbol{e},\boldsymbol{b})V_{yz}(\boldsymbol{c},\boldsymbol{b})}{V_{z\bar{y}}(\boldsymbol{f},\boldsymbol{e})}\sum\limits_{\boldsymbol{d}_{1}}\left(\frac{\bar{y}}{y}\right)^{\left|\boldsymbol{d}_{1}\right|}\frac{V_{xz}\left(\boldsymbol{d}_{1},\boldsymbol{e}\right)V_{\frac{\bar{x}}{y}}\left(\boldsymbol{d}_{1},\boldsymbol{c}\right)V_{\frac{\bar{y}}{x}}\left(\boldsymbol{f},\boldsymbol{d}_{1}\right)}{V_{z\bar{x}}\left(\boldsymbol{d}_{1},\boldsymbol{b}\right)},$
(7.13)

which ends the proof.

Note that $S_{12}^{t_{2}}\left(x,y\right)$ is not invertible, and we cannot
define $\widetilde{S}_{12}(x,y)$ and therefore the dual reflection equation.
However, the inverse $R$-matrix $S_{12}^{-1}\left(x,y\right)$ is well defined
and has the form

$S_{12}^{-1}\left(x,y\right)=S_{21}\left(y,x\right),$ (7.14)

which is a consequence of the orthogonality property (7.7). Therefore we can
try to define an analogue of the reflection equation (2.11) based on the
inverse $R$-matrix (7.14):

$S_{21}\left(y,x\right)\bar{K}_{1}\left(x,\bar{x}\right)S_{12}\left(\bar{x},y\right)\bar{K}_{2}\left(y,\bar{y}\right)=\bar{K}_{2}\left(y,\bar{y}\right)S_{21}\left(\bar{y},x\right)\bar{K}_{1}\left(x,\bar{x}\right)S_{12}\left(\bar{x},\bar{y}\right).$
(7.15)

From (7.7) one can establish that this equation admits a stochastic solution
of the form

$\bar{K}\left(x,\bar{x}\right)=\Phi_{q^{2}}\left({\boldsymbol{l}}-{\boldsymbol{j}}\mid{\boldsymbol{l}};\dfrac{\bar{x}}{x},\bar{x}\right),$
(7.16)

which, however, does not include an additional free parameter as in (7.6).

##  8 Conclusion

In this paper we constructed triangular solutions for the reflection equation
corresponding to symmetric tensor representations of
$U_{q}(\widehat{sl_{n}})$. It is quite remarkable that all constructed
solutions are built from the same function
$\Phi_{q}(\boldsymbol{\gamma}\mid\boldsymbol{\beta};\lambda,\mu)$ defined in
(3.10). As a result, both reflection and dual reflection equations (5.1),
(6.1) reduce to multiple sum identities among the $\Phi$-functions (or
$V$-functions (3.16)). It would be natural to expect that (5.1), (6.1) can be
proven by a repeated application of the star-star relation (4.1). However, we
failed to find a correct sequence of the star-star relations which would prove
these equations. This is the reason why we introduced the $L$-operators (3.30)
and (3.31) to prove that the solution (5.7) solves the reflection equation
(5.12). We are not aware of any other nontrivial identities among
$V$-functions except (4.1) and believe that such an algebraic proof may still
exist.

We expect that general stochastic solutions for the $U_{q}(\widehat{sl_{n}})$
algebra should have one extra parameter as demonstrated in [24] for
$U_{q}(\widehat{sl_{2}})$ and in [34] for $U_{q}(\widehat{sl_{n}})$ with
$J=1$. Since our triangular solutions are factorised, they may serve as
building blocks for general solutions similar to the expression of the
stochastic $R$-matrix (3). However, we have not yet found such a general
representation and aim to do that in the future.

##  9 Acknowledgements

We would like to thank V. Bazhanov and S. Sergeev for careful reading of the
manuscript and valuable comments at different stages of this project.

##  Appendix A Solution to the dual reflection equation

In this Appendix we show that (2.12) solves the dual reflection equation (2.9)
and prove the inverse mapping (2.13), see also [31] for similar calculations.

First, we notice two different forms of the Yang-Baxter equation

$R_{12}(x)\widetilde{R}_{23}(y)\widetilde{R}_{13}(xy)=\widetilde{R}_{13}(xy)\widetilde{R}_{23}(y)R_{12}(x)$
(A.1)
$\widetilde{R}_{12}(x)\widetilde{R}_{13}(xy){R}_{23}(y)={R}_{23}(y)\widetilde{R}_{13}(xy)\widetilde{R}_{12}(x)$
(A.2)

which can be proved using 2.1, 2.2 and 2.4.

Now we show that (2.12) solves the dual reflection equation (2.9). First, let
us calculate the product of the first three factors in the LHS of (2.9)

$\displaystyle\widetilde{K}_{2}\left(y\right)\widetilde{R}_{21}(xy)\widetilde{K}_{1}\left(x\right)=$
$\displaystyle\operatorname{tr}_{0}\left[\bar{K}_{0}\left(\frac{1}{y}\right)\widetilde{R}_{02}(y^{2})P_{02}\right]\widetilde{R}_{21}(xy)\operatorname{tr}_{0^{\prime}}\left[\bar{K}_{0^{\prime}}\left(\frac{1}{x}\right)\widetilde{R}_{0^{\prime}1}(x^{2})P_{0^{\prime}1}\right]$
(A.3)

Using the property $\tr_{i}A^{t_{i}}B^{t_{i}}=\tr_{i}AB$ and the definition
(2.2) , we transform (A.3) to

$\displaystyle\operatorname{tr}_{0}\left[\left(\bar{K}_{0}\left(\frac{1}{y}\right)P_{02}\right)^{t_{0}}\left(\widetilde{R}_{20}(y^{2})\right)^{t_{0}}\right]\widetilde{R}_{21}(xy)\operatorname{tr}_{0^{\prime}}\left[\bar{K}_{0^{\prime}}\left(\frac{1}{x}\right)\widetilde{R}_{0^{\prime}1}(x^{2})P_{0^{\prime}1}\right]=$
$\displaystyle\operatorname{tr}\left[\bar{K}_{0^{\prime}}\left(\frac{1}{x}\right)\left(\bar{K}_{0}\left(\frac{1}{y}\right)P_{02}\right)^{t_{0}}P_{0^{\prime}1}\left(\widetilde{R}_{20}(y^{2})\right)^{t_{0}}\widetilde{R}_{20^{\prime}}(xy)\widetilde{R}_{10^{\prime}}(x^{2})\right]=$
$\displaystyle\tr\left[\bar{K}_{0^{\prime}}\left(\frac{1}{x}\right)\left(\bar{K}_{0}\left(\frac{1}{y}\right)P_{02}\right)^{t_{0}}R_{0^{\prime}0}^{t_{0}}(xy)P_{0^{\prime}1}\widetilde{R}_{10}^{t_{0}}(xy)\widetilde{R}_{20}^{t_{0}}(y^{2})\widetilde{R}_{20^{\prime}}(xy)\widetilde{R}_{10^{\prime}}(x^{2})\right]=$
$\displaystyle\tr\left[\left(\bar{K}_{0^{\prime}}\left(\frac{1}{x}\right)R_{0^{\prime}0}(xy)\bar{K}_{0}\left(\frac{1}{y}\right)P_{02}P_{0^{\prime}1}\right)\left(\widetilde{R}_{20}(y^{2})\widetilde{R}_{10}(xy)\widetilde{R}_{20^{\prime}}(xy)\widetilde{R}_{10^{\prime}}(x^{2})\right)\right],$
(A.4)

where we defined
$\tr\stackrel{{\scriptstyle\text{def}}}{{=}}\tr_{0,0^{\prime}}$.

Multiplying (A.4) by $R_{21}\left({\displaystyle\frac{y}{x}}\right)$ from the
right and applying (A.1) twice, we bring the LHS of (2.9) to the form

$\displaystyle\text{LHS}=$
$\displaystyle\tr\left[\bar{K}_{0^{\prime}}\left(\frac{1}{x}\right)R_{0^{\prime}0}(xy)\bar{K}_{0}\left(\frac{1}{y}\right)P_{02}P_{0^{\prime}1}R_{21}\left(\frac{y}{x}\right)\widetilde{R}_{10}(xy)\widetilde{R}_{20}(y^{2})\widetilde{R}_{10^{\prime}}(x^{2})\widetilde{R}_{20^{\prime}}(xy)\right]=$
$\displaystyle\tr\left[R_{0^{\prime}0}\left(\dfrac{y}{x}\right)\bar{K}_{0}\left(\dfrac{1}{y}\right)R_{00^{\prime}}(xy)\bar{K}_{0^{\prime}}\left(\dfrac{1}{x}\right)P_{02}P_{0^{\prime}1}\widetilde{R}_{10}(xy)\widetilde{R}_{20}(y^{2})\widetilde{R}_{10^{\prime}}(x^{2})\widetilde{R}_{20^{\prime}}(xy)\right]=$
$\displaystyle\tr\left[\bar{K}_{0}\left(\dfrac{1}{y}\right)R_{00^{\prime}}(xy)\bar{K}_{0^{\prime}}\left(\dfrac{1}{x}\right)P_{02}P_{0^{\prime}1}\widetilde{R}_{10}(xy)\widetilde{R}_{20}(y^{2})\widetilde{R}_{10^{\prime}}(x^{2})\widetilde{R}_{20^{\prime}}(xy)R_{0^{\prime}0}\left(\dfrac{y}{x}\right)\right],$
(A.5)

where in the second line we used the reflection equation (2.11).

Repeating the above calculations for the RHS of the equation (2.9), we obtain

$\displaystyle\text{RHS}=$
$\displaystyle\tr\left[\bar{K}_{0}\left(\dfrac{1}{y}\right)R_{00^{\prime}}(xy)\bar{K}_{0^{\prime}}\left(\dfrac{1}{x}\right)P_{02}P_{0^{\prime}1}R_{0^{\prime}0}\left(\dfrac{y}{x}\right)\widetilde{R}_{10^{\prime}}(x^{2})\widetilde{R}_{10}(xy)\widetilde{R}_{20^{\prime}}(xy)\widetilde{R}_{20}(y^{2})\right].$
(A.6)

Applying (A.2) twice to (A.6), we bring it to (A.5) which completes the proof.

Now let us show that the inverse mapping from $\widetilde{K}(x)$ to
$\bar{K}(x)$ has the form (2.13)

$\bar{K}_{1}(x)=\operatorname{tr}_{0^{\prime}}\left[\widetilde{K}_{0^{\prime}}\left(\dfrac{1}{x}\right)R_{0^{\prime}1}\left(\dfrac{1}{x^{2}}\right)P_{0^{\prime}1}\right].$
(A.7)

The proof of this statement is based on the trace property

$\operatorname{tr}_{0^{\prime}}\left(P_{00^{\prime}}A_{10}B_{10^{\prime}}\right)=\left(A_{10}^{t_{0}}B_{10}^{t_{0}}\right)^{t_{0}},$
(A.8)

which is valid for any matrices $A$ and $B$. Then the following calculation
proves (A.7):

$\displaystyle\operatorname{tr}_{0^{\prime}}\left[\widetilde{K}_{0^{\prime}}\left(\dfrac{1}{x}\right)R_{0^{\prime}1}\left(\dfrac{1}{x^{2}}\right)P_{0^{\prime}1}\right]=\tr_{0,0^{\prime}}\left[\bar{K}_{0}\left(x\right)\widetilde{R}_{00^{\prime}}\left(\dfrac{1}{x^{2}}\right)P_{00^{\prime}}R_{0^{\prime}1}\left(\dfrac{1}{x^{2}}\right)P_{0^{\prime}1}\right]$
$\displaystyle=\operatorname{tr}_{0}\left[P_{01}\bar{K}_{1}\left(x\right)\operatorname{tr}_{0^{\prime}}\left\\{P_{00^{\prime}}\widetilde{R}_{10}\left(\dfrac{1}{x^{2}}\right)R_{10^{\prime}}\left(\dfrac{1}{x^{2}}\right)\right\\}\right]$
$\displaystyle=\operatorname{tr}_{0}\left[P_{01}\bar{K}_{1}\left(x\right)\left(\widetilde{R}_{10}^{t_{0}}\left(\dfrac{1}{x^{2}}\right)R_{10}^{t_{0}}\left(\dfrac{1}{x^{2}}\right)\right)^{t_{0}}\right]=\bar{K}_{1}(x).$
(A.9)

##  Appendix B Useful summation formulas

First, we list some well known summation formulas from [36].

The $q$-binomial theorem:

${}_{1}\phi_{0}\left(q^{-n};-;q,z\right)=\left(zq^{-n};q\right)_{n};$ (B.1)

the $q$-Vandermonde summation formula:

${}_{2}\phi_{1}\left(a,q^{-n};c;q,q\right)=\frac{(c/a;q)_{n}}{(c;q)_{n}}a^{n};$
(B.2)

the $q$-Pfaff-Saalschutz summation formula:

${}_{3}\phi_{2}\left[\begin{array}[]{c}a,b,q^{-n}\\\
c,abc^{-1}q^{1-n}\end{array};q,q\right]=\frac{(c/a,c/b;q)_{n}}{(c,c/ab;q)_{n}}.$
(B.3)

We also give two special $m$-dimensional extensions of the $q$-Pfaff-
Saalschutz sum:

$\displaystyle\sum_{k_{1},\ldots,k_{m}=0}q^{|\boldsymbol{k}|}\frac{(q^{-n},b;q)_{|\boldsymbol{k}|}}{\left(c,bc^{-1}q^{1-n}\prod\limits_{i=1}^{m}z_{i};q\right)_{|\boldsymbol{k}|}}\prod\limits_{j=1}^{m}\frac{\left(z_{j};q\right)_{k_{j}}}{(q;q)_{k_{j}}}z_{j}^{\sum\limits_{i<j}k_{i}}=\frac{\left(c\prod\limits_{i=1}^{m}z_{i}^{-1},\dfrac{c}{b};q\right)_{n}}{\left(c,\dfrac{c}{b}\prod\limits_{i=1}^{m}z_{i}^{-1};q\right)_{n}},$
(B.4)
$\sum_{k_{1},\ldots,k_{m}=0}q^{|\boldsymbol{k}|-\sum\limits_{i<j}k_{i}n_{j}}\frac{(a,b;q)_{|\boldsymbol{k}|}}{\left(c,abc^{-1}q^{1-|n|};q\right)_{|\boldsymbol{k}|}}\prod_{i=1}^{m}\frac{\left(q^{-n_{i}};q\right)_{k_{i}}}{(q;q)_{k_{i}}}=\frac{(c/a,c/b;q)_{|\boldsymbol{n}|}}{(c,c/ab;q)_{|\boldsymbol{n}|}}.$
(B.5)

Both relations can be proven as follows. First, we rewrite the summations in
the form
$\sum\limits_{k_{1},\ldots,k_{m}=0}\rightarrow\sum\limits_{l=0}\sum\limits_{|k|=l}$.
Then the result follows from (B.3) and the formula

$\sum_{|k|=l}\prod\limits_{j=1}^{m}z_{j}^{\sum\limits_{i<j}k_{i}}\prod_{i=1}^{m}\frac{\left(z_{i};q\right)_{k_{i}}}{(q;q)_{k_{i}}}=\frac{\left(\prod\limits_{i=1}^{m}z_{i};q\right)_{l}}{(q;q)_{l}},$
(B.6)

where the last equality can be easily proved by induction on $m$ and using the
$q$-Vandermonde summation formula (B.2).

##  Appendix C Proof of (6.16)

The proof of the (6.16) is based on (B.4). First, we shift the summation index
$\boldsymbol{d}\rightarrow\boldsymbol{d}+\boldsymbol{c}$ and use the
definition of the function $V$ (3.16) to get

$\displaystyle\sum\limits_{\boldsymbol{d}}q^{-2\sum\limits_{k=1}^{n-1}k(d_{k}-c_{k})+n(|\boldsymbol{d}|-|\boldsymbol{c}|)}\frac{V_{q^{-J}}(\boldsymbol{d},\boldsymbol{b})V_{y}(\boldsymbol{d},\boldsymbol{c})}{V_{z}(\boldsymbol{d},\boldsymbol{b})V_{\frac{y}{z}q^{n-J}}(\boldsymbol{d},\boldsymbol{m})}=$
(C.1)
$\displaystyle\frac{(q^{n-1}y)^{|\boldsymbol{c}|-|\boldsymbol{m}|}}{(q^{J}z)^{|\boldsymbol{b}|-|\boldsymbol{m}|}}\frac{q^{Q(\boldsymbol{c},\boldsymbol{c}-\boldsymbol{m})}}{q^{Q(\boldsymbol{c}-\boldsymbol{m},\boldsymbol{m})}}\frac{\prod\limits_{s=1}^{n-1}\left(q^{2};q^{2}\right)_{c_{s}-m_{s}}\left(q^{-2J};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{b}|}}{\left(\frac{q^{2n-2J}y^{2}}{z^{2}};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{m}|}\left(z^{2};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{b}|}}\times$
$\displaystyle\sum_{\boldsymbol{d}}\frac{\left(\frac{q^{2(|\boldsymbol{c}|-|\boldsymbol{b}|)}}{q^{2J}},y^{2};q^{2}\right)_{|\boldsymbol{d}|}q^{2\sum\limits_{k=1}^{n}(n-k)d_{k}+2Q(\boldsymbol{d},\boldsymbol{c}-\boldsymbol{m})}}{\left(\frac{q^{-2(J-n-(|\boldsymbol{c}|-|\boldsymbol{m}|))}y^{2}}{z^{2}},q^{2(|\boldsymbol{c}|-|\boldsymbol{b}|)}z^{2};q^{2}\right)_{|\boldsymbol{d}|}}\prod_{s=1}^{n-1}\frac{\left(q^{2\left(1+c_{s}-m_{s}\right)};q^{2}\right)_{d_{s}}}{\left(q^{2};q^{2}\right)_{d_{s}}}.$

After taking into account the relation

$q^{2\sum\limits_{k=1}^{n}(n-k)d_{k}+2Q(\boldsymbol{d},\boldsymbol{c}-\boldsymbol{m})}=q^{2|\boldsymbol{d}|}\prod_{j=1}^{n-1}q^{2\sum\limits_{i<j}\left(1+c_{j}-m_{j}\right)d_{i}},$
(C.2)

we can apply (B.4)

$\displaystyle\sum\limits_{\boldsymbol{d}}q^{-2\sum\limits_{k=1}^{n-1}k(d_{k}-c_{k})+n(|\boldsymbol{d}|-|\boldsymbol{c}|)}\frac{V_{q^{-J}}(\boldsymbol{d},\boldsymbol{b})V_{y}(\boldsymbol{d},\boldsymbol{c})}{V_{z}(\boldsymbol{d},\boldsymbol{b})V_{\frac{y}{z}q^{n-J}}(\boldsymbol{d},\boldsymbol{m})}=\frac{(q^{n-1}y)^{|\boldsymbol{c}|-|\boldsymbol{m}|}}{(q^{J}z)^{|\boldsymbol{b}|-|\boldsymbol{m}|}}\frac{q^{Q(\boldsymbol{c},\boldsymbol{c}-\boldsymbol{m})}}{q^{Q(\boldsymbol{c}-\boldsymbol{m},\boldsymbol{m})}}\times$
(C.3)
$\displaystyle\frac{\prod\limits_{s=1}^{n-1}\left(q^{2};q^{2}\right)_{c_{s}-m_{s}}\left(q^{-2J};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{b}|}}{\left(\frac{q^{2n-2J}y^{2}}{z^{2}};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{m}|}\left(z^{2};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{b}|}}\frac{\left(\frac{q^{2-2J}y^{2}}{z^{2}},\frac{q^{-2(J-n-|c|+|m|)}}{z^{2}};q^{2}\right)_{J-(|\boldsymbol{c}|-|\boldsymbol{b}|)}}{\left(\frac{q^{-2(J-n-|c|+|m|)}y^{2}}{z^{2}},\frac{q^{2-2J}}{z^{2}};q^{2}\right)_{J-(|\boldsymbol{c}|-|\boldsymbol{b}|)}}.$

Then successively applying the following identities:

$\left(aq^{-2J};q^{2}\right)_{J-(|\boldsymbol{c}|-|\boldsymbol{b}|)}=\dfrac{\left(q^{2}/a;q^{2}\right)_{J}}{\left(q^{2}/a;q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{b}|}}\left(-\dfrac{a}{q^{2}}\right)^{J-(|\boldsymbol{c}|-|\boldsymbol{b}|)}\frac{q^{(|\boldsymbol{c}|-|\boldsymbol{b}|)(|\boldsymbol{c}|-|\boldsymbol{b}|-1)}}{q^{J(J-1)}},$
(C.4)
$\left(aq^{-2(|\boldsymbol{c}|-|\boldsymbol{m}|)};q^{2}\right)_{J}=\dfrac{\left(a;q^{2}\right)_{J}\left(q^{2}/a;q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{m}|}}{\left(q^{2(1-J)}/a;q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{m}|}}q^{-2J(|\boldsymbol{c}|-|\boldsymbol{m}|)},$
(C.5)

we see, that the factors of the form $\left(a;q^{2}\right)_{J}$ are combined
in the function $\mu_{J}^{(n)}(y,z)$ from (6.17), and one obtains

$\displaystyle\text{LHS (\ref{first sum of
V})}=\mu_{J}^{(n)}(y,z)\frac{(q^{n-1}y)^{|\boldsymbol{c}|-|\boldsymbol{m}|}}{(q^{J}z)^{|\boldsymbol{b}|-|\boldsymbol{m}|}}\frac{q^{Q(\boldsymbol{c},\boldsymbol{c}-\boldsymbol{m})}}{q^{Q(\boldsymbol{c}-\boldsymbol{m},\boldsymbol{m})}}\dfrac{\prod\limits_{s=1}^{n-1}\left(q^{2};q^{2}\right)_{c_{s}-m_{s}}}{\left(\frac{q^{2n}y^{2}}{z^{2}};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{m}|}}\times$
(C.6)
$\displaystyle\dfrac{\left(q^{-2J};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{b}|}}{\left(\frac{z^{2}}{y^{2}};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{b}|}}\dfrac{\left(\frac{q^{2n}}{z^{2}};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{m}|}}{\left(\frac{q^{-2J+2n}}{z^{2}};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{m}|}}\dfrac{\left(\frac{q^{-2(-1+n+|\boldsymbol{c}|-|\boldsymbol{m}|)}z^{2}}{y^{2}};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{b}|}}{\left(q^{-2(-1+n+|\boldsymbol{c}|-|\boldsymbol{m}|)}z^{2};q^{2}\right)_{|\boldsymbol{c}|-|\boldsymbol{b}|}}.$

Substituting to the RHS of (6.16) the definition (3.16) of the function
$V_{x}(\boldsymbol{a},\boldsymbol{b})$, we come to the same expression (C.6).

##  Appendix D Some explicit matrices

Here we demonstrate the structure of some triangular boundary matrices. For
$J=1$ and arbitrary $n$ we refer to [33, 34]. We will focus on the simplest
case of the higher spin matrices for two particles, i.e when $J=2$ and $n=3$.
The basis is ordered as follows:

$|0,0\rangle,\quad|0,1\rangle,\quad|0,2\rangle,\quad|1,0\rangle,\quad|1,1\rangle,\quad|2,0\rangle.$
(D.1)

We also introduce the parameters

$\lambda=\frac{y^{2}}{\nu q^{2}},\quad\mu=\frac{1}{y^{2}\nu q^{2}},\quad
p=q^{2}.$ (D.2)

Then the upper- and lower-triangular matrices (5.7), (5.16), describing the
right boundary, have the form

$\displaystyle K^{(u)}_{2}(y)=$ (D.3)
$\displaystyle\left(\begin{array}[]{cccccc}1&\frac{\left(\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{1}}&\frac{\left(\frac{\mu}{\lambda};p\right)_{2}}{\left(\mu;p\right)_{2}}&\frac{\left(\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{1}}&\frac{\left(\frac{\mu}{\lambda};p\right)_{2}}{\left(\mu;p\right)_{2}}&\frac{\left(\frac{\mu}{\lambda};p\right)_{2}}{\left(\mu;p\right)_{2}}\\\
0&\frac{\mu}{\lambda}\frac{\left(\lambda;p\right)_{1}}{\left(\mu;p\right)_{1}}&\frac{\mu}{\lambda}\frac{\left(\lambda,\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{2}}{2\brack
1}_{p}&0&p\frac{\mu}{\lambda}\frac{\left(\lambda,\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{2}}&0\\\
0&0&\left(\frac{\mu}{\lambda}\right)^{2}\frac{\left(\lambda;p\right)_{2}}{\left(\mu;p\right)_{2}}&0&0&0\\\
0&0&0&\frac{\mu}{\lambda}\frac{\left(\lambda;p\right)_{1}}{\left(\mu;p\right)_{1}}&\frac{\mu}{\lambda}\frac{\left(\lambda,\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{2}}&\frac{\mu}{\lambda}\frac{\left(\lambda,\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{2}}{2\brack
1}_{p}\\\
0&0&0&0&\left(\frac{\mu}{\lambda}\right)^{2}\frac{\left(\lambda;p\right)_{2}}{\left(\mu;p\right)_{2}}&0\\\
0&0&0&0&0&\left(\frac{\mu}{\lambda}\right)^{2}\frac{\left(\lambda;p\right)_{2}}{\left(\mu;p\right)_{2}}\\\
\end{array}\right),$ $\displaystyle K^{(d)}_{2}(y)=$ (D.4)
$\displaystyle\left(\begin{array}[]{cccccc}\frac{\left(\lambda;p\right)_{2}}{\left(\mu;p\right)_{2}}&0&0&0&0&0\\\
0&\frac{\left(\lambda;p\right)_{2}}{\left(\mu;p\right)_{2}}&0&0&0&0\\\
0&0&\frac{\left(\lambda;p\right)_{2}}{\left(\mu;p\right)_{2}}&0&0&0\\\
\lambda\frac{\left(\lambda,\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{2}}{2\brack
1}_{p}&p\lambda\frac{\left(\lambda,\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{2}}&0&\frac{\left(\lambda;p\right)_{1}}{\left(\mu;p\right)_{1}}&0&0\\\
0&\lambda\frac{\left(\lambda,\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{2}}&\lambda\frac{\left(\lambda,\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{2}}{2\brack
1}_{p}&0&\frac{\left(\lambda;p\right)_{1}}{\left(\mu;p\right)_{1}}&0\\\
\lambda^{2}\frac{\left(\frac{\mu}{\lambda};p\right)_{2}}{\left(\mu;p\right)_{2}}&\lambda^{2}\frac{\left(\frac{\mu}{\lambda};p\right)_{2}}{\left(\mu;p\right)_{2}}&\lambda^{2}\frac{\left(\frac{\mu}{\lambda};p\right)_{2}}{\left(\mu;p\right)_{2}}&\lambda\frac{\left(\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{1}}&\lambda\frac{\left(\frac{\mu}{\lambda};p\right)_{1}}{\left(\mu;p\right)_{1}}&1\\\
\end{array}\right).$

The solutions (6.19), (6.20), related to the left boundary, have a similar
structure:

$\displaystyle\bar{K}^{(u)}_{2}(y)=$ (D.5)
$\displaystyle\left(\begin{array}[]{cccccc}1&\mu\frac{\left(\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{1}}&\mu^{2}\frac{\left(\frac{\lambda}{\mu};p\right)_{2}}{\left(\lambda;p\right)_{2}}&\mu\frac{\left(\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{1}}&\mu^{2}\frac{\left(\frac{\lambda}{\mu};p\right)_{2}}{\left(\lambda;p\right)_{2}}&\mu^{2}\frac{\left(\frac{\lambda}{\mu};p\right)_{2}}{\left(\lambda;p\right)_{2}}\\\
0&\frac{\left(\mu;p\right)_{1}}{\left(\lambda;p\right)_{1}}&\mu\frac{\left(\mu,\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{2}}{2\brack
1}_{p}&0&\mu\frac{\left(\mu,\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{2}}&0\\\
0&0&\frac{\left(\mu;p\right)_{2}}{\left(\lambda;p\right)_{2}}&0&0&0\\\
0&0&0&\frac{\left(\mu;p\right)_{1}}{\left(\lambda;p\right)_{1}}&\mu
p\frac{\left(\mu,\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{2}}&\mu\frac{\left(\mu,\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{2}}{2\brack
1}_{p}\\\
0&0&0&0&\frac{\left(\mu;p\right)_{2}}{\left(\lambda;p\right)_{2}}&0\\\
0&0&0&0&0&\frac{\left(\mu;p\right)_{2}}{\left(\lambda;p\right)_{2}}\\\
\end{array}\right),$ $\displaystyle\bar{K}^{(d)}_{2}(y)=$ (D.6)
$\displaystyle\left(\begin{array}[]{cccccc}\left(\frac{\lambda}{\mu}\right)^{2}\frac{\left(\mu;p\right)_{2}}{\left(\lambda;p\right)_{2}}&0&0&0&0&0\\\
0&\left(\frac{\lambda}{\mu}\right)^{2}\frac{\left(\mu;p\right)_{2}}{\left(\lambda;p\right)_{2}}&0&0&0&0\\\
0&0&\left(\frac{\lambda}{\mu}\right)^{2}\frac{\left(\mu;p\right)_{2}}{\left(\lambda;p\right)_{2}}&0&0&0\\\
\frac{\lambda}{\mu}\frac{\left(\mu,\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{2}}{2\brack
1}_{p}&\frac{\lambda}{\mu}\frac{\left(\mu,\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{2}}&0&\frac{\lambda}{\mu}\frac{\left(\mu;p\right)_{1}}{\left(\lambda;p\right)_{1}}&0&0\\\
0&p\frac{\lambda}{\mu}\frac{\left(\mu,\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{2}}&\frac{\lambda}{\mu}\frac{\left(\mu,\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{2}}{2\brack
1}_{p}&0&\frac{\lambda}{\mu}\frac{\left(\mu;p\right)_{1}}{\left(\lambda;p\right)_{1}}&0\\\
\frac{\left(\frac{\lambda}{\mu};p\right)_{2}}{\left(\lambda;p\right)_{2}}&\frac{\left(\frac{\lambda}{\mu};p\right)_{2}}{\left(\lambda;p\right)_{2}}&\frac{\left(\frac{\lambda}{\mu};p\right)_{2}}{\left(\lambda;p\right)_{2}}&\frac{\left(\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{1}}&\frac{\left(\frac{\lambda}{\mu};p\right)_{1}}{\left(\lambda;p\right)_{1}}&1\\\
\end{array}\right).$

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