# Reflection Matrices for the $(n+1)(2n+1)$ Vertex Models: Diagonal Solutions

A. Lima- Santos

###### Abstract

We have find the diagonal $K$ matrix solutions of the reflection equations for
a class of vertex models. These models have $(n+1)(2n+1)$ vertices and are
defined as two set of (n + 1) $R$ matrices, solutions of the equations of
Yang-Baxter equations. For a given value of n we find
n!-$\frac{1}{2}(n-2)(n-1)$ $K$ diagonal matrices.

Universidade Federal de São Carlos, Departamento de Física, Caixa Postal 676,
CEP 13569-905, São Carlos, Brasil

##  1 Introduction

In a previous work [2] we broke up the $A_{n-1}^{(1)}$ symmetry of the
solution of the Yang-Baxter equations (YBE), the $R$-matrix [1]. The symmetry
break releases many parameters of the model. These parameters are the values
of the derivative of the entries of the matrix $R(u)$, in a particular value
of the parameter spectral u, $\>u=0$ .

For $n>1$ the R (u) has $(n+1)(2n+1)$ entries,after setting some parameters we
can see that, $n(n+3)/2+3$ free parameters remain in the two sets of n + 1
matrices $R$.

With so many parameters at our disposal, we can manipulate the solutions until
obtaining possible desired results. This freedom can be used to open up many
questions about integrality.

In this article after presenting the solutions of (YBE), with a difference of
two spectral parameters, we are going to treat the boundary Yang-Baxter
equations (BYBE ) or reflection equations (RE).

The RE are a natural extension of the YBE with an addition of one element, the
so-called reflection matrix or K-matrix. In the literature on integrable
models, we will find many works on this subject [3, 4, 5]. We will use a well-
known approach to calculations [9, 10] to find the diagonal solutions,. In the
appendix A we also present an essay of calculations for non-diagonal
solutions.

Solutions of (RE) are important in the study of critical phenomena for the
calculation of free energies and critical exponents [14]

##  2 The Model

The Yang-Baxter equation(YBE) can be seen as a matrix relation defined in
$End(V\otimes V\otimes V)$, where $V$ is an $n$\- dimensional complex vector
space. In a particular and usual case, it reads:

$R_{12}(u)R_{13}(u+v)R_{23}(v)-R_{23}(v)R_{13}(u+v)R_{12}(u)$ (2.1)

where the arguments u and v, called spectral parameters, have values in C. The
solution of the YBE is an R matrix defined in End (V $\otimes$ V ). The
indexed matrices $R_{ij}$ appearing in (1) are defined in End (V $\otimes$ V
$\otimes$ V ) through the formulas

$R_{12}=R\otimes I,\qquad R_{23}=I\otimes R,\qquad R_{13}=P_{23}R_{12}P_{23}$
(2.2)

where $I\in End(V$) is the identity matrix, $P\in End(V\otimes V)$ is the
permutation matrix (defined by the relation $P(A\otimes B)P=B\otimes A$ for
$A,B\in End(V))$ and $P_{12}=P\otimes I,P_{23}=I\otimes P$.

From the quantum group invariant representation for non-exceptional affine Lie
algebra $A_{n-1}^{(1)}$ [1],we consider the following generalization for a
R-matrix solution

$R(u)=\sum_{i=1}^{n+1}a_{ii}(u)\mathrm{e}_{ii}\otimes\mathrm{e}_{ii}+\sum_{i\neq
j}^{n+1}b_{ij}(u)\mathrm{e}_{ii}\otimes\mathrm{e}_{jj}+\sum_{i\neq
j}^{n+1}c_{ij}(u)\mathrm{\mathrm{e}}_{ij}\otimes\mathrm{e}_{ji}$ (2.3)

where $e_{ij}$ are the Weyl matrices $(e_{ij})_{ab}=\delta_{ia}\delta_{jb},$
acting for a $n+1$ dimensional vector space $V$ at the site $n$ . The R-matrix
elements $a_{ii}(u)$, $b_{ij}(u)$ and $c_{ij}(u)$ are fixed by the YBE.
Breaking of the original symmetry releases several parameters.

At this point we observe that (2.3) no longer has $A_{n-1}^{(1)}$ symmetry.
But using the fact that there are many free parameters the particular choice

$a_{ij}(u)=a(u),\quad b_{ij}(u)=b(u),\;c_{ij}(u)=\mathbf{c}\qquad(i<j),\quad
c_{ij}(u)=\mathbf{c}\exp(u)$ (2.4)

where $\mathbf{c}$ is a constant, we are recovering the $A_{n-1}^{(1)}$
symmetry[1]. Thus, we understand that the result presented in this work with
the choice (2.4) reproduces all those solved with the $A_{n-1}^{(1)}$
symmetry.

For each value of $n$>1, the R matrix has $n+1$ diagonal entries $a_{ii}(u)$,
that are determined by recurrence relative to $a_{11}(u)$

$a_{kk}(u)=a_{11}(u)+(\alpha_{kk}-\alpha_{11})F(u),\qquad k=2,...,n+1$ (2.5)

where

$\alpha_{jj}=\left(\frac{d}{du}a_{jj}(u)\right)_{u=0}\quad and\quad
F(u)=\frac{e^{\kappa_{1}u}-e^{\kappa_{2}u}}{-\kappa_{2}+\kappa_{1}}$ (2.6)

where

$\kappa_{1}=\mu_{13}+\mu_{21}-\mu_{23}$ (2.7)

and

$\kappa_{2}=-\mu_{13}+\mu_{12}+\mu_{23}.$ (2.8)

The remaining $n(n+1)$diagonal entries

$b_{ij}(u)=\beta_{ij}F(u),\qquad i\neq j=\\{1,...,n+1\\}$ (2.9)

where

$\beta_{ij}=\left(\frac{d}{du}b_{ij}(u)\right)_{u=0}$ (2.10)

with the constraints

$\beta_{ji}\beta_{ij}=\beta_{21}\beta_{12}=(\kappa_{1}-\alpha_{11})(\kappa_{2}-\alpha_{11})$
(2.11)

The number of fixed $\beta_{ij}$ is $\frac{1}{2}$ $n-1)(n+2)+1$.

The $n(n+1)$ off-diagonal matrix elements

$c_{ij}(u)=e^{\mu_{ij}u}$ (2.12)

All $\mu_{ij}$ of the $c_{ij}(u)$ below od main diagonal are fixed by the
relation

$\mu_{ji}=\kappa_{1}+\kappa_{2}-\mu_{ij},\qquad j>i$ (2.13)

The number of fixed $\mu_{ji}$ is$\frac{1}{2}$ $(n-1)(n+2)$and some $\mu_{ij}$
above the main diagonal can be fixed by two relation

$\mu_{ij}=\kappa_{1}-\mu_{1i}+\mu_{1j}\mathrm{\quad
and}\quad\mu_{ij}=\kappa_{2}-\mu_{1i}+\mu_{1j},\quad j<i$ (2.14)

The number of fixed parameters $\mu_{ij}$ is $(n+1)(n-2)/2$. It means that we
have two sets of solutions. With these relations the Yang-Baxter equation and
its derivatives are solved by two sets of $2^{n}$ solutions of the following
$n$ equations

$(\alpha_{kk}-\alpha_{11})(\alpha_{kk}+\alpha_{11}-\kappa_{1}-\kappa_{2})=0$
(2.15)

Therefore we have two differentes values for the $a_{kk}(u)$ and $n$
parameters $\alpha_{ii}$ are fixed. It mean that our R matrix solutions have
$n(n+3)/2+3$ free parameters

$a_{kk}(u)=A(u)=\frac{(\kappa_{1}-\alpha_{11})\exp(\kappa_{2}u)-(\kappa_{2}-\alpha_{11})\exp(\kappa_{1}u)}{\kappa_{1}-\kappa_{2}}$
(2.16)

For $\alpha_{ii}=\alpha_{11}$ and

$a_{kk}(u)=B(u)=\frac{(\kappa_{1}-\alpha_{11})\exp(\kappa_{1}u)-(\kappa_{2}-\alpha_{11})\exp(\kappa_{2}u)}{\kappa_{1}-\kappa_{2}}$
(2.17)

For $\alpha_{kk}=\kappa_{1}+\kappa_{2}-\alpha_{11}$.

As presented and discussed in [2], the $2^{n}$ solutions have been reduced for
$n+1$ different $R$ matrices

##  3 The Reflection Equations

In the current literature we can find several works on the solutions of the
reflection equations with $A_{n-1}$symmetry, this is with (2.4). Among them we
cite [6, 7, 8, 11, 12, 13]

While the YBE reading of the conservation of the linear moment , the RE
reading of the conservation of energy.

Given a solution R of YBE, we can build a monodromy matrix $T(u)$, whose trace
defines a row to row transfer matrix $\tau(u).$

The existence of a K matrix allows us to reflect T(u) on a wall. The reflected
$T(-u)^{-1}$ which in turn will be reflected on a second wall by the matrix
$K^{+}$, dual of K. Thus we have the Sklyanin’s transfer matrix [3]

$t(u)=Tr_{a}(K^{{}^{+}}(u)T(u)K(u)T^{-1}(-u))$ (3.1)

where $Tr_{a}$means trce in the auxiliary space.

In our case there is an isomorphism [4]

$K^{+}(u)=K(-u-\rho)^{t}M$ (3.2)

where $\rho$ is the crossing parameter,$M$ is the crossing matrix and $t$
means tranposition.

The translational invariance in the vertical (temporal) direction of the
monodromy on the wall with $K(u)$ is given by the RE [3, 4]:

$RE=R(u-v)K_{1}(u)PR(u+v)PK_{2}(v)=K_{2}(v)-K_{2}(v)R(u+v)K_{1}(u)PR(u-v)P$
(3.3)

where $K_{1}=K\otimes I$ , $K_{2}=I\otimes K$ and $P=R(0)$ .

This functional equation can be properly transformed into an algebraic
equation, through its derivatives with respect to the parameter v at point
$v=0$.

$DRE=-B_{2}R(u)K_{1}(u)PR(u)P+R(u)K_{1}(u)PR(u)PB_{2}-2D(u)K_{1}(u)PR(u)P+2R(u)K_{1}(u)PR(u)P$
(3.4)

Similarly we have another equation for the derivatives with respect to $u$,
but unlike YBE, we just need to use (3.4)

The diagonal matrices have the form.

To obtain (18) we have used the definitions

$D(u)=\left[\frac{\partial R(u\pm v)}{\partial v}\right]_{v=0},\qquad\qquad
B=\left[\frac{dK(v)}{dv}\right]_{v=0}$ (3.5)

and the regular condition $K(0)=I$.

From (2.3) we get the derivative of $R(u)$:

$D(u)=\sum_{i=1}^{n+1}a_{ii}(u)^{\prime}\mathrm{e}_{ii}\otimes\mathrm{e}_{ii}+\sum_{i\neq
j}^{n+1}b_{ij}(u)^{\prime}\mathrm{e}_{ii}\otimes\mathrm{e}_{jj}+\sum_{i\neq
j}^{n+1}c_{ij}(u)^{\prime}\mathrm{\mathrm{e}}_{ij}\otimes\mathrm{e}_{ji}$
(3.6)

For a given $n$ the K-matrix is given by and its derivative at the point $v=0$
are

$K(u)=\sum_{i=1}^{n+1}k_{ii}(u)\,\mathrm{e}(i,i)\qquad
B=\sum_{i=1}^{n+1}\theta_{ii}(u)\,\mathrm{e}(i,i)$ (3.7)

where $\mathrm{e}(i,j)$ are $n+1$ by $n+1$ matrices with entries
$[e(i,j)]_{ab}=\delta_{ai}\delta_{bj}$.

Taking into account the differential equation (3.4) with (3.7) and the date of
the n+1 solutions of the correponding Yang-Baxter equations discussed in [2].

###  3.1 Case n=2

In this case

$K(u)=\left(\begin{array}[]{ccc}k_{11}(u)&0&0\\\ 0&k_{22}(u)&0\\\
0&0&k_{33}(u)\end{array}\right)\qquad
B=\left(\begin{array}[]{ccc}\theta_{{}_{11}}&0&0\\\ 0&\theta_{22}&0\\\
0&0&\theta_{33}\end{array}\right)$ (3.8)

This completes the data for (18)

From the equation $\textrm{DRE}[$2,4] we find $k_{22}(u)$

$k_{22}(u)=\frac{[(2\mu_{21}+\theta_{11}-\theta_{22})F(u)-2F^{\prime}(u)]c_{21}(u)k_{11}(u)}{[(2\mu_{12}+\theta_{22}-\theta_{11})F(u)-2F^{\prime}(u)]c_{12}(u)}$
(3.9)

From the equation $\textrm{DRE}[3,7]$we get $k_{33}(u)$

$k_{33}(u)=\frac{[(2\mu_{31}+\theta_{11}-\theta_{33})F(u)-2F^{\prime}(u)]c_{31}(u)k_{11}(u)}{[(2\mu_{13}+\theta_{33}-\theta_{11})F(u)-2F^{\prime}(u)]c_{13}(u)}$
(3.10)

where $c_{ij}$, $F$ and $F^{\prime}$ are given by (2.12) and (2.6)

Now, from $\textrm{DRE}[6,8]$ and $\textrm{DRE}[8,6]$ we find two constraints

$eq_{1}=(-\theta_{33}+\theta_{11}+2\mu_{21}-2\mu_{23})(-\theta_{33}+\theta_{22}+2\mu_{12}-2\mu_{13})$
(3.11)
$eq_{2}=(-\theta_{11}+\theta_{22}+2\mu_{13}-2\mu_{23})(-\theta_{33}+\theta_{11}+2\mu_{21}-2\mu_{23})$
(3.12)

which provides two solutions:

$\displaystyle\theta_{22}=$ $\displaystyle\theta_{11}-2\mu_{13}+2\mu_{23}$
(3.13) $\displaystyle\theta_{33}=$
$\displaystyle\theta_{11}-2\mu_{23}+2\mu_{21}$ $\displaystyle\theta_{33}=$
$\displaystyle\theta_{22}+2\mu_{12}-2\mu_{13}$ (3.14)

Note that in the solutions we have one value for $\theta_{22}$ and two values
for $\theta_{33}.$

Being more explicit

$K_{1}=\left(\begin{array}[]{ccc}k_{11}&0&0\\\
0&k_{11}\mathrm{e}^{{}^{(-2u(\mu_{13}-\mu_{23})}}&0\\\
0&0&k_{11}\mathrm{e}^{(2u(\mu_{21}-\mu_{23})}\end{array}\right)$ (3.15)

and

$K_{2}=\left(\begin{array}[]{ccc}k_{11}&0&0\\\
0&k_{11}\mathrm{e}^{(-2u(\mu_{13}-\mu_{23})}&0\\\
0&0&k_{11}\mathrm{e^{(2u(\mu_{12}-2\mu_{13}+\mu_{23})}}\end{array}\right)$
(3.16)

where $k_{11}$=$k_{11}(u)$ is an arbitrary function with $k_{11}(0)=1$ and
$k^{\prime}_{11}(0)=\theta_{11}.$

With an appropriate choice of free parameters, for example, if we fix
$\mu_{23}$ as $\mu_{23}=\mu_{13}+\frac{1}{2}(\mu_{{}_{21}}-\mu_{12})$we will
get $K_{2}=K_{1}$, giving us a better understanding of the free parameters.

###  3.2 Case n=3

Now we have

$K(u)=\left(\begin{array}[]{cccc}k_{11}(u)\\\ &k_{22}(u)\\\ &&k_{33}(u)\\\
&&&k_{44}(u)\end{array}\right),\quad
B=\left(\begin{array}[]{cccc}\theta_{11}\\\ &\theta_{22}\\\ &&\theta_{33}\\\
&&&\theta_{44}\end{array}\right)$ (3.17)

From $\text{$DRE[2,5]$},$$DRE[3,9]$ and $DRE[4,13]$ we obtain $k_{22}$,
$k_{33}$ and $k_{44}$, respectively

$k_{22}(u)=\frac{[(2\mu_{21}+\theta_{11}-\theta_{22})F(u)-2F^{\prime}(u)]c_{21}(u)k_{11}(u)}{[(2\mu_{12}+\theta_{22}-\theta_{11})F(u)-2F^{\prime}(u)]c_{12}(u)},$
(3.18)
$k_{33}(u)=\frac{[(2\mu_{31}+\theta_{11}-\theta_{33})F(u)-2F^{\prime}(u)]c_{31}(u)k_{11}(u)}{[(2\mu_{13}+\theta_{33}-\theta_{11})F(u)-2F^{\prime}(u)]c_{13}(u)}.$
(3.19)

and

$k_{44}(u)=\frac{[(2\mu_{41}+\theta_{11}-\theta_{44})F(u)-2F^{\prime}(u)]c_{41}(u)k_{11}(u)}{[(2\mu_{14}+\theta_{44}-\theta_{11})F(u)-2F^{\prime}(u)]c_{14}(u)}$
(3.20)

Using these amplitudes the remaining
equations,$\textrm{DRE}[7,10],$$\textrm{DRE}[8,14]$ and $\textrm{DRE}[12,15]$,
leave us with three constraints , respectively

$eq_{1}=(2\mu_{21}-2\mu_{23}+\theta_{11}-\theta_{33})(2\mu_{13}-2\mu_{23}-\theta_{11}+\theta_{22})(2\mu_{12}-2\mu_{13}+\theta_{22}-\theta_{33})$
$eq_{2}=(2\mu_{13}-2\mu_{23}-\theta_{11}+\theta_{22})(2\mu_{13}-2\mu_{14}-\theta_{44}+\theta_{22})(2\mu_{13}-2\mu_{23}+2\mu_{21}-2\mu_{14}+\theta_{11})$
$eq_{3}=(2\mu_{14}-2\mu_{13}+\theta_{44}-\theta_{33})(2\mu_{12}-4\mu_{13}+2\mu_{23}+\theta_{11}-\theta_{33})(2\mu_{14}-2\mu_{13}+2\mu_{23}-2\mu_{21}+\theta_{44}-\theta_{11})$
(3.21)

For the solutions of this set of equation, we have find one $\theta_{22},$ two
$\theta_{33}$and three $\theta_{44}$ , therefore, six solutions:

$\displaystyle\theta_{22}=$ $\displaystyle\theta_{11}-2\mu_{13}+2\mu_{23}$
(3.22) $\displaystyle\theta_{33}=$
$\displaystyle\theta_{11}+2\mu_{21}-2\mu_{23}$ $\displaystyle\theta_{33}=$
$\displaystyle\theta_{22}+2\mu_{12}-2\mu_{13}$ (3.23) $\displaystyle*$
$\displaystyle\theta_{44}=$
$\displaystyle\theta_{11}+2\mu_{13}+2\mu_{21}-2\mu_{23}-2\mu_{14}$
$\displaystyle\theta_{44}=$ $\displaystyle\theta_{22}+2\mu_{12}-2\mu_{14}$
$\displaystyle\theta_{44}=$ $\displaystyle\theta_{33}+2\mu_{13}-2\mu_{14}$
(3.24)

Now let’s look at the parameters $\theta_{ii}$ for the case $n=3$ are the same
as the case $n=2$ plus more three new parameters $\theta_{44}$ . It means that
each of the $3$ by $3$ matrices of the case $n=2$ are contained in three 4 by
4 matrices for the case $n=3$. Due to the free parameters, coming from the $R$
matrix.

The new three values for $k_{44}(u)$ are

$\displaystyle k_{44_{1}}(u)=$ $\displaystyle
e^{2u(\mu_{13}-\mu_{14}+\mu_{21}-\mu_{23})}k_{11}(u)$ $\displaystyle
k_{44_{2}}(u)=$ $\displaystyle
e^{2u(\mu_{12}-\mu_{14}-\mu_{13}-+\mu_{23})}k_{11}(u)$ $\displaystyle
k_{44_{3}}(u)=$
$\displaystyle\mathrm{e^{2u(\mu_{13}-\mu_{14}+\mu_{21}-\mu_{23})}k_{11}(u)}$
(3.25)

for $K_{1}$ and for $K_{2}$.

Note that while we know that $k_{44_{1}}=k_{44_{3}}$ we will take this into
account only later

This construction will now be extended to the other vertex models.

###  3.3 Case $n>3$

Now we will present a generalization for these results for a given value of
$n$

Since there are two sets of $n+1$ matrices R different from each other by the
values of $\mu_{ij}$ with $i<j$ (2.14). Let’s choose the set with
$a_{11}(u)=A(u)$ 2.16 and $\mu_{ij}=\kappa_{2}-\mu_{1i}+\mu_{ij}$.

From the $(n+1)^{4}$entries of (18) we will only need $n+1$ in order to find
the $k_{ii}$matrix entries, namely

$\mathrm{DRE}[a,(i-1)(n+1)+1],\Longrightarrow k_{ii}(u),\qquad i=2,3,...,n+1$
(3.26)

After that, we can use the function

$V(i,j,u)=((2\mu_{ij}-\theta_{jj}+\theta_{ii})F(u)-2F^{\prime}(u))c_{ij}(u)$
(3.27)

to write all the diagonal elements of the reflection matrices for all values
of $n$ in a general form

$k_{ii}(u)=\frac{V(i,1,u)}{V(1,i,u)}k_{11}(u),\qquad\qquad a=2,3,...,n+1$
(3.28)

where $k_{11}(u)$ is an arbitrary function such that $k_{11}(0)=1$ and
$\left[\frac{d}{du}k_{11}(u)\right]_{u=0}=\theta_{11}$.

From these data we see that they do not depend on $a_{11}(u)$ and therefore
are also valid for $a_{11}(u)=B(u)$ . It means the solutions for the $n+1$
matrices of the set considered. Moreover, everything is valid for the second
set as long as $\kappa_{2}$ is exchanged for $\kappa_{1}$or taking
$\mu_{23}=\mu_{13}+\frac{1}{2}(\mu_{21}-\mu_{12})$. Considerably increasing
the number of solutions.

Thus, and we are left with the task to find the parameters
$\theta_{\alpha\alpha}$ depend on $\theta_{11}$ and $\mu_{ij}$ the free
parameters of the $n+1$ $R$-matrices, solutions of YBE.

This task takes a lot of time,but we have the reward of being able to get all
solutions with the help of the following expression

$\theta_{ii}=f_{ijk}=\theta_{jj}+2\mu_{ik}-2\mu_{jk}\qquad\qquad(i>j)$ (3.29)

The index $k$ is special, it has to be different from $i$ and $j$, it means
that there are $n-1$ possible values to be assigned to $k$.

It is clear that for high values of $n$ it is difficult to determine the value
of $k$. We managed to get rid of this difficulty:

As already described for the first two values of n, for a given value of $n$,
we only need to calculate the $n$ $\theta_{n+1,n+1}$because all others$(n-1)!$
$\theta_{ii}$ have already been calculated. The inclusion of these $n$ new
values, defines the $n!$ diagonal solutions of the our vertex model. This is
done by choosing the smallest among the possible values of $k$.

Thus the two solutions for $n=2$ are

$\begin{array}[]{cccc}1:&\theta_{22}=f_{213}&,&\theta_{33}=f_{312}\\\
2:&\theta_{22}=f_{213}&,&\theta_{33}=f_{321}\end{array}$ (3.30)

and the six solution for $n=3$ are

$\begin{array}[]{ccccccc}1:&&\theta_{22}=f_{213}&&\theta_{33}=f_{312}&&\theta_{44}=f_{412}\\\
2:&&\theta_{22}=f_{213}&&\theta_{33}=f_{312}&&\theta_{44}=f_{421}\\\
3:&&\theta_{22}=f_{213}&&\theta_{33}=f_{312}&&\theta_{44}=f_{431}\\\ \\\
4:&&\theta_{22}=f_{213}&&\theta_{33}=f_{321}&&\theta_{44}=f_{412}\\\
5:&&\theta_{22}=f_{213}&&\theta_{33}=f_{321}&&\theta_{44}=f_{421}\\\
\mathbf{6}:&&\theta_{22}=\mathbf{f}_{213}&&\theta_{33}=\mathbf{f}_{321}&&\theta_{44}=\mathbf{f}_{431}\end{array}$
(3.31)

Bold identifies the repeated solution.So, we found five different solutions
from each other

To better understand this approach, we will also present the solutions for
case $n=4$

The $n+1$ diagonal entries of the $K$ -matrix

$\begin{array}[]{ccc}k_{11}(u)&=&1\,k_{11}(u)\\\
k_{22}(u)&=&\frac{V(2,1,u)}{V(1,2,u)}\,k_{11}(u)\\\
k_{33}(u)&=&\frac{V(3,1,u)}{V(1,3,u)}\,k_{11}(u)\\\
k_{44}(u)&=&\frac{V(4,1,u)}{V(1,4,u)}\,k_{11}(u)\\\
k_{55}(u)&=&\frac{V(5,1,u)}{V(1,5,u)}\,k_{11}(u)\end{array}$ (3.32)

where the $V(i,1,u)$,$\qquad i=2,3,4,5$ are given by (3.27)

The next step is to find the parameters $\theta_{ii}$ contained in 3.32, that
are already given by (3.29).

After finding the $\theta_{ii}$ we have to identify which and how many are
repeated. In terms of $\mathbf{f}_{ijk}$ they are those with $i=n+1$,
$j=3,4,..,n$ and $k$ obeying the rule the smallest possible values.

For a given $n$ we have

$S(n)=\frac{1}{2}(n-2)(n-1)$ (3.33)

It means we’ve found

$T(n)=n!-S(n)$ (3.34)

different solutions.

Agreeing with what we have previously presented: $T(2)=2,\qquad T(3)=5$. It is
at least interesting to explicitly find $T(4)=$ $21$ solutions for the case
$n=4$ case.

##  4 Conclusion

One of the advantages of diagonal solutions is that the spectrum of their
Hamiltonian with diagonal boundaries are directly solved by the Bethe ansätze
[15], see also [16].

The matrices $K$ diagonal solutions of (RE), presented in this work have a
peculiar behavior: For a given value of $n$ its $n+1$ by $n+1$ diagonal is the
$n$ by $n$ digonal of $n-1$ case plus the element $k_{n+1\,n+1}$$(u)$ . Thus,
as the approach used is valid for $n=1$, $k_{11}(u)$ and $k_{22}(u)$ define
the diagonal solution of the six vertex- model.

This is also expected for non-diagonal solutions, which we hope to presented
in the future. The calculations of non-diagonal $K$ matrices are more
extensive and require a lot of time and their generalization is not as simple
as in the diagonal case, see for example the case $n=2$ as presented in [17]

##  Appendix A An essay on non-diagonal solutions

Going back to the reflection equation (3.4) but now with

$K(u)=\sum_{i,j=1}^{n+1}k_{ij}(u)\,\mathrm{e}(i,j)\qquad,K(0)=I,\quad
B=\sum_{i,j=1}^{n+1}\theta_{ij}(u)\,\mathrm{e}(i,j)$ (A.1)

The equations $DRE[i,i]=0$ are all resolved by relations

$\theta_{ij}k_{ji}(u)=\theta_{ji}k_{ij}(u),\quad(i\neq j)$ (A.2)

Now we choose a particular $k_{ij}(u)$ to be different from zero, wth
$\theta_{ij}\neq 0$ and write all remaining elements in terms of the
$k_{ij}(u)$ and from the remaining equations we will find a very strong
restriction for the matrix elements: for a particular $k_{ij}(u)\neq 0$ the
elements different from zero are $k_{ii}(u),k_{jj}(u)$ and $k_{ji}(u)$ .

$S=\\{k_{ij}(u)\neq 0,\quad k_{ji}(u)\neq 0,\quad k_{ii}(u)\neq 0,\quad
k_{jj}(u)\neq 0\\}$ (A.3)

these four elements different from zero define a block within the matrix
$K(u)$

$B_{ab}=\left[\begin{array}[]{ccc}k_{aa}&\ldots&\mathbf{k}_{ab}\\\
\vdots&\ddots&\vdots\\\ k_{ba}&\ldots&k_{bb}\end{array}\right]$ (A.4)

where $a<b$ . The number of blocks $B_{ab}$ within K define the types of
solutions

We note that all entries of are diagonal and also non-zero due to the normal
condition.

We identify the$K(u)$ solution with one block $B_{ab}$ by $K_{ab}^{[I]}$(u)
and with two blocks by $K_{ab}^{[II]}(u)$

The solutions with one block have the form

$K_{ab}^{[I]}(u)=k_{ab}(u)\mathrm{e}(a,b)+k_{ba}\mathrm{e}(b,a)+\sum_{i}^{n+1}k_{ii}\mathrm{e}(i,i),\qquad,a<b,\\{(a=1,..,n),\qquad(b=2,...,n+1)\\}$
(A.5)

we count $\frac{n}{2}(n+1)$ $K_{ab}^{[I]}$ matrices each with $n+3$ non-null
elements.

For six-vertex model we have one solution

$K_{12}^{[I]}=\left(\begin{array}[]{cc}k_{11}&\mathbf{k}_{12}\\\
k_{21}&k_{22}\end{array}\right),$ (A.6)

for fifteen-vertex model we have three solutions

$K_{12}^{[I]}=\left(\begin{array}[]{ccc}k_{11}&\mathbf{k}_{12}&0\\\
k_{21}&k_{22}&0\\\
0&0&k_{33}\end{array}\right),\;K_{13}^{[I]}=\left(\begin{array}[]{ccc}k_{11}&0&\mathbf{k}_{13}\\\
0&k_{22}&0\\\
k_{31}&0&k_{33}\end{array}\right),\;K_{23}^{[I]}=\left(\begin{array}[]{ccc}k_{11}&0&0\\\
k_{21}&k_{22}&\mathbf{k}_{23}\\\ 0&k_{32}&k_{33}\end{array}\right)$ (A.7)

We also found a second type of solution containing two blocks $B_{ab}$ and
$B_{cd}$ defining a second type of solution $K_{ab}^{[II]}$. The existence of
blocks implies the following condition for non-diagonal entries

$k_{ij}(u)k_{ji}(u)=k_{rs}(u)k_{rs}(u)$ (A.8)

since

$i+j=r+s,\mathrm{\qquad mod\>}(n+1)$ (A.9)

As $i<j$ and $r<s$ the choice $r=i-1,s=j+1$ is enough to solve (A.9) and we
find the Following form for $K_{ab}^{[II]}$ :

$K_{ab}^{]II]}(u)=k_{ab}(u)\mathrm{\,e}(a,b)+k_{ba}(u)\mathrm{\,e}(b,a)+k_{a+b,n+1}(u)\mathrm{e}(a+b,n+1)+k_{n+1,a+b}(u)\mathrm{\,e}(n+1,a+b)$
$+\sum_{i=1}^{n+1}k_{ii}(u)\,\mathrm{e}(i,i)$ (A.10)

for $a+b\leq n$.

For $a+b\geq n+1$ the solution has the form

$K_{ab}^{]II]}(u)=k_{ab}(u)\mathrm{\,e}(a,b)+k_{ba}(u)\mathrm{\,e}(b,a)+k_{a+1,b-1}(u)\mathrm{e}(a+1,b-1)+k_{b-1,a+1}(u)\mathrm{\,e}(b-1,a+1)$
$+\sum_{i=1}^{n+1}k_{tt}(u)\,\mathrm{e}(i,i)$ (A.11)

The $28$-vertex model has six type-I solutions and two type-II solutions

$K_{12}^{[II]}u)=\left(\begin{array}[]{cccc}k_{11}&\mathbf{k}_{12}&0&0\\\
k_{21}&k_{22}&0&0\\\ 0&0&k_{33}&\mathbf{k}_{34}\\\
0&0&k_{43}&k_{44}\end{array}\right),K_{14}^{[II]}=\left(\begin{array}[]{cccc}k_{11}&0&0&\mathbf{k}_{14}\\\
0&k_{22}&\mathbf{k}_{23}&0\\\ 0&k_{32}&k_{33}&0\\\
k_{41}&0&0&k_{44}\end{array}\right)$ (A.12)

We found $\frac{1}{2}(n+1)(n-2)$ of these type-II solutions.

In reference [12] the non-diagonal trigonometric solutions for the
$A_{n-1}^{(1)}$ models were classified and in reference [13] the rational case
is addressed.

We will present the non-diagonal solutions elsewhere.

Let’s conclude this essay with some additional information:

  * •

Our $n+1$ $R$ matrices have the same shape as the $R$ matrix of algebra
$A_{n-1}^{(1)}$, hence our K matrices, but with different entries.

  * •

In the calculations of the $\\{K_{ab}^{(I]}(u),K_{ab}^{(II]}(u)\\}$ we need to
bind some free parameters of $R$.

  * •

For a given value of $n$, the solution $K_{ab}^{[I]}(u)$ contains the
$K_{ab}^{[I]}(u)$ of the previous values of $n$. In the limit $n=1$ ,
$K_{ab}^{[I]}(u)$ is the solution for the six-vertex. Its well-known solution
is reproduced in the reduction (2.4).

  * •

Even with so many free parameters it was not possible to find a solution with
all non-zero entries.

  * •

We still have nothing to write about the possibilities of finding solutions
type -III, as well as the possibilities to consider other types of blocks

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