# Form factors of integrable higher-spin XXZ chains and the affine quantum-
group symmetry

Tetsuo Deguchi1111e-mail<EMAIL_ADDRESS>and Chihiro Matsui2,3
222e-mail<EMAIL_ADDRESS>
###### Abstract

We derive exactly scalar products and form factors for integrable higher-spin
XXZ chains through the algebraic Bethe-ansatz method. Here spin values are
arbitrary and different spins can be mixed. We show the affine quantum-group
symmetry, $U_{q}(\widehat{sl_{2}})$, for the monodromy matrix of the XXZ spin
chain, and then obtain the exact expressions. Furthermore, through the
quantum-group symmetry we explicitly derive the diagonalized forms of the $B$
and $C$ operators in the $F$-basis for the spin-1/2 XXZ spin chain, which was
conjectured in the algebraic Bethe-ansatz calculation of the XXZ correlation
functions. The results should be fundamental in studying form factors and
correlation functions systematically for various solvable models associated
with the integrable XXZ spin chains.

1 Department of Physics, Graduate School of Humanities and Sciences,
Ochanomizu University

2-1-1 Ohtsuka, Bunkyo-ku, Tokyo 112-8610, Japan

2 Department of Physics, Graduate School of Science, University of Tokyo

7-3-1 Hongo, Bunkyo-ku, Tokyo 113-0033, Japan

3 CREST, JST, 4-1-8 Honcho Kawaguchi, Saitama, 332-0012, Japan

##  1 Introduction

Correlation functions of the spin-1/2 XXZ spin chain have attracted much
attention in mathematical physics for more than a decade [1, 2, 3]. The
multiple-integral representations of XXZ correlation functions were first
derived in terms of the $q$-vertex operators [4]. Based on the algebraic Bethe
ansatz method, the determinant expressions [5] of the scalar products and the
norms of Bethe ansatz eigenstates were reconstructed in terms of the $F$-basis
[6], and then the XXZ correlation functions are derived under an external
magnetic field [7, 8]. The multiple-integral representations at zero
temperature have been extended into those at finite temperature [9].
Furthermore, dynamical structure factors have been evaluated by solving the
Bethe ansatz equations numerically [10, 11, 12, 13].

The Hamiltonian of the spin-1/2 XXZ chain under the periodic boundary
conditions is given by

${\cal H}_{\rm
XXZ}={\frac{1}{2}}\sum_{j=1}^{L}\left(\sigma_{j}^{X}\sigma_{j+1}^{X}+\sigma_{j}^{Y}\sigma_{j+1}^{Y}+\Delta\sigma_{j}^{Z}\sigma_{j+1}^{Z}\right)\,.$
(1.1)

Here $\sigma_{j}^{a}$ ($a=X,Y,Z$) are the Pauli matrices defined on the $j$th
site, and $\Delta$ the XXZ coupling. By $\Delta=(q+q^{-1})/2$ we define
parameter $q$, which plays a significant role in the paper. We note that $L$
denotes the number of the one-dimensional lattice sites. Solvable higher-spin
generalizations of the XXX and XXZ chains have been constructed by the fusion
method in several references [14, 15, 16, 17, 18, 19]. For instance, the
Hamiltonian of the solvable spin-$s$ XXX chain is given by the following [15]:

$H_{s}=J\sum_{j=1}^{L}Q_{2s}({\vec{S}}_{j}\cdot{\vec{S}}_{j+1})\,,$ (1.2)

where ${\vec{S}}_{j}$ are operators of spin $s$ acting on the $j$th site and
$Q_{2s}(x)$ is a polynomial of degree $2s$

$Q_{2s}(x)=\sum_{p=1}^{2s}\left(\sum_{k=1}^{p}{\frac{1}{k}}\right)\prod_{\ell=0,\ell\neq
j}^{2s}{\frac{x-x_{\ell}}{x_{j}-x_{\ell}}}\,.$ (1.3)

Here $x_{\ell}=[\ell(\ell+1)-2s(s+1)]/2$. At $T=0$ in the critical regime, it
is discussed that the low-excitation spectrum of the spin-$s$ XXX chain is
described in terms of the level-$k$ $SU(2)$ WZWN model where $k=2s$ [20].

In the present paper we derive exact expressions of scalar products and form
factors for the integrable higher-spin XXZ spin chains. Here different spins
can be mixed. We first show that the monodromy matrix of the XXZ spin chain
has the symmetry of the affine quantum group $U_{q}({\widehat{sl_{2}}})$.
Then, we derive the exact expressions taking advantage of the quantum group
symmetry. By a similarity transformation [21], we transform the symmetric
$R$-matrix into an asymmetric one, which is directly connected to the quantum
group $U_{q}(sl_{2})$. We derive projection operators from the asymmetric
$R$-matrices [19], and construct integrable higher-spin XXZ spin chains by the
fusion method similarly as the case of the XXX spin chain [14]. Here we make
an extensive use of the $q$-analogues of Young’s projection operators, which
play a central role in the $q$-analogue of the Schur-Weyl reciprocity of the
quantum group $U_{q}(sl_{2})$ [22, 23]. Hereafter, we call transformations on
the $R$-matrix gauge transformations. After we construct integrable higher-
spin models, applying the inverse transformation to them, we reduce the
asymmetric monodromy matrices into those of the symmetric $L$-operators
constructed from the symmetric $R$-matrices, and thus obtain scalar products
and form factors for the integrable higher-spin models in the standard
formulation.

Form factors and correlation functions have been discussed for integrable
higher-spin XXX models in previous researches [24, 25, 26, 27]. In the
approach of the so-called quantum inverse scattering problem for the
integrable $N$-state models, one has to construct the $N$-by-$N$ monodromy
matrix in order to express local operators in terms of the global operators.
However, it seems to be technically nontrivial to construct the $N$-by-$N$
monodromy matrix for the integrable higher-spin systems (see also [28]). By
the approach of the present paper, the calculational task is much reduced into
the minimal level. In fact, as a consequence of the $q$-analogue of the Schur-
Weyl duality, the exact expressions of scalar products and form factors are
derived from the formulas of the spin-1/2 XXZ chain by setting the
inhomogeneous parameters in the form of “complete $\ell$-strings”[29], and the
most general results are straightforwardly obtained.

The affine quantum-group invariance has another important consequence. In the
XXZ case we can explicitly prove the pseudo-diagonalized forms of the $B$ and
$C$ operators of the algebraic Bethe-ansatz method through the quantum-group
symmetry. Here we remark that the XXZ spin chain has no spin SU(2) symmetry.
In the pseudo-diagonal basis, the $B$ and $C$ operators are expressed as sums
of local spin operators $\sigma_{i}^{-}$ and $\sigma_{i}^{+}$ multiplied by
diagonal matrices, respectively, where each of the local spin operators
$\sigma_{i}^{\pm}$ are defined on one lattice-site, i.e. on the $i$th site.
The pseudo-diagonalized forms of the $B$ and $C$ operators were conjectured in
the algebraic Bethe-ansatz derivation of the XXZ correlation functions [6, 7,
8]. In fact, the $B$ operators create the Bethe states where the $C$ operators
are conjugate to them, and the pseudo-diagonalized forms play a central role
in the calculation of the scalar products and the norms of Bethe states. They
are also fundamental in the quantum inverse scattering problem, by which local
operators are expressed in terms of the global operators such as the $B$ and
$C$ operators. In the XXZ case, however, an explicit derivation of the pseudo-
diagonalized forms of the $B$ and $C$ operators has not been shown previously,
yet. Thus, the explicit derivation in the paper completes the algebraic Bethe-
ansatz formulation of the form factors and correlation functions of the
integrable XXZ spin chains. Here we remark that for the XXX case (i.e. the
isotropic case), the diagonalized forms have been shown by Maillet and Sanchez
de Santos in Ref. [6] by making an explicit use of the rotational
SU(2)-symmetry. However, the method for the XXX case does not hold for the XXZ
spin chain which has no SU(2) symmetry.

The derivation of the affine quantum-group symmetry of the monodromy matrix
should be not only theoretically interesting but also practically useful for
calculation. Here we remark that the infinite-dimensional symmetry,
$U_{q}({\widehat{sl_{2}}})$, was realized for the infinite XXZ spin chain with
the $q$-vertex operators [2, 4]. Thus, it should be an interesting open
problem how the affine quantum-group symmetry of the finite XXZ spin chain can
be related to that of the infinite XXZ spin chain. Furthermore, there are
several advantages in the present formulation of the affine quantum-group
symmetry. We derive the symmetry through gauge transformations. The
transformed asymmetric $R$-matrix is directly related to the quantum group
$U_{q}(sl_{2})$ so that we can systematically construct higher-spin
representations of the $R$-matrices through the $q$-analogue of the Young
symmetrizers. Here we should note that the gauge transformation connects the
$R$ matrix in the different gradings of $U_{q}(\widehat{sl_{2}})$. The
symmetric and asymmetric $R$-matrices are equivalent to that of the principal
and homogeneous gradings, respectively (see for instance §5.4 of [2]).
Moreover, we thus avoid technical difficulties appearing when we directly
derive the matrix representation of the universal $R$-matrix of the affine
quantum group, which is given by a product of infinite series of generators
[30]. Although one can construct matrix representations of the modified
universal $R$-matrix with its derivation $d$ dropped [30, 22](see A.2 of
[31]), it seems that the calculation is not quite straightforward when we
construct higher-spin representations.

The results of the present paper should be useful for calculating exact
expressions of correlation functions for various integrable models associated
with higher-spin XXZ chains. For instance, the $\tau_{2}$ model in the
$N$-dimensional nilpotent representation corresponds to the integrable
spin-$(N-1)/2$ XXZ spin chain with $q$ being a primitive $N$th root of unity
[32]. Here we remark that the $\tau_{2}$ model is closely related to the
$N$-state superintegrable chiral Potts model. Furthermore, there are several
possible physical applications, such as calculating form factors of quantum
impurity models through the result in the case of mixed spins. As an
illustrative example, we have calculated exact expressions for the emptiness
formation probability of the higher-spin XXZ spin chains, which we shall
discuss in a subsequent paper.

The content of the paper consists of the following: In section 2, we introduce
the symmetric $R$-matrix of the spin-1/2 XXZ spin chain, and define the
monodromy matrix [33]. We also define the action of the symmetric group on
products of $R$-matrices. In section 3 we derive the symmetry of the quantum
affine algebra for the monodromy matrix of the XXZ spin chain. We introduce
the asymmetric $R$-matrix and then derive it from the symmetric one by a gauge
transformation. We decompose the asymmetric $R$-matrix in terms of the
generators of the Temperley-Lieb algebra, and show the affine quantum-group
symmetry. We also derive it by a systematic method for expressing products of
$R$-matrices formulated in definition 2.21. In fact, all the fundamental
relations of the quantum inverse-scattering problem can be derived much more
simply without using the ${\hat{R}}$-matrix of Ref. [6], as shown in
Appendices A and B. In section 4 we construct the $R$ matrices of integrable
higher-spin XXZ spin chains with projection operators of $U_{q}(sl_{2})$ by
the fusion method. We also discuss the case of mixed spins. In section 5 we
formulate an explicit derivation of the pseudo-diagonalized forms of the
$B$-operators. We also show it for the $C$-operator in Appendix E. In section
6 we derive determinant expressions of scalar products for the higher-spin XXZ
spin chains. In section 7, for the higher-spin cases we show the method by
which we can express local operators in terms the global operators. We give
some useful formulas of the quantum inverse scattering problem for the higher
spin case. Finally, we derive some examples of form factors for the integrable
higher-spin XXZ spin chains.

##  2 $R$-matrices and $L$-operators

###  2.1 Symmetric $R$-matrix

We shall introduce the $R$-matrix for the XXZ spin chain [33]. We consider two
types of $R$-matrices, $R_{ab}(u)$ and $R_{ab}(\lambda,\mu)$. The $R$-matrix
with a single rapidity argument, $R_{ab}(u)$, acts on the tensor product of
two vector spaces $V_{a}$ and $V_{b}$, i.e. $R_{ab}(u)\in End(V_{a}\otimes
V_{b})$, where parameter $u$ is independent of $V_{a}$ or $V_{b}$. The
$R$-matrix with two rapidity arguments, $R_{ab}(\lambda,\mu)$, acts on the
tensor product of vector spaces with parameters, $V_{a}(\lambda)$ and
$V_{b}(\mu)$, i.e. $R_{ab}(\lambda,\mu)\in End(V_{a}(\lambda)\otimes
V_{b}(\mu))$.

Let us denote by $e^{a,b}$ such a matrix that has only one nonzero element
equal to 1 at entry $(a,b)$. We denote by $V$ the two-dimensional vector
space. We define the $R$-matrix acting on the tensor product $V\otimes V$ by

$R(u)=\sum_{a,b,c,d=1,2}R^{ab}_{cd}(u)e^{a,c}\otimes e^{b,d}\,.$ (2.1)

Here matrix elements $R^{ab}_{cd}(u)$ satisfy the charge conservation, i.e.
$R^{ab}_{cd}(u)=0$ unless $a+b=c+d$, and all the nonzero elements are given by
the following:

$\displaystyle R^{11}_{11}(u)$ $\displaystyle=$ $\displaystyle
R^{22}_{22}(u)=1\,,\quad R^{12}_{12}(u)=R^{21}_{21}(u)=b(u),$ $\displaystyle
R^{12}_{21}(u)$ $\displaystyle=$ $\displaystyle R^{21}_{12}(u)=c(u)\,,$ (2.2)

where functions $b(u)$ and $c(u)$ are given by

$b(u)={\frac{\sinh(u)}{\sinh(u+\eta)}}\,,\quad
c(u)={\frac{\sinh(\eta)}{\sinh(u+\eta)}}\,.$ (2.3)

Here, parameter $\eta$ is related to $q$ of $\Delta=(q+q^{-1})/2$ by
$q=\exp(\eta)$.

We now introduce operators acting on the $L$th power of tensor product of
vector spaces with parameters, $V(\lambda_{1})\otimes\cdots\otimes
V(\lambda_{L})$. We generalize the notation of (2.1). Let us take a pair of
integers $j$ and $k$ satisfying $1\leq j<k\leq L$. For a given set of matrix
elements $A^{a,\,b}_{c,\,d}(\lambda_{j},\lambda_{k})$ ($a,b,c,d=1,2$) we
define operators $A_{j,k}(\lambda_{j},\lambda_{k})$ and
$A_{k,j}(\lambda_{k},\lambda_{j})$ by

$\displaystyle A_{j,k}(\lambda_{j},\lambda_{k})$ $\displaystyle=$
$\displaystyle\sum_{a,b,\alpha,\beta=1,2}A^{a,\,\alpha}_{b,\,\beta}(\lambda_{j},\lambda_{k})I_{1}\otimes\cdots\otimes
I_{j-1}$ $\displaystyle\quad\otimes e^{a,b}_{j}\otimes
I_{j+1}\otimes\cdots\otimes I_{k-1}\otimes e_{k}^{\alpha,\beta}\otimes
I_{k+1}\otimes\cdots\otimes I_{L}\,,$ $\displaystyle
A_{k,j}(\lambda_{k},\lambda_{j})$ $\displaystyle=$
$\displaystyle\sum_{a,b,\alpha,\beta=1,2}A^{\alpha,\,a}_{\beta,\,b}(\lambda_{k},\lambda_{j})I_{1}\otimes\cdots\otimes
I_{j-1}$ (2.4) $\displaystyle\quad\otimes e^{a,b}_{j}\otimes
I_{j+1}\otimes\cdots\otimes I_{k-1}\otimes e_{k}^{\alpha,\beta}\otimes
I_{k+1}\otimes\cdots\otimes I_{L}\,.$

Here $I$ is the two-by-two unit matrix, and $I_{j}$ and $e_{j}^{a,b}$ act on
the $j$th vector space $V(\lambda_{j})$ of $V(\lambda_{1})\otimes\cdots\otimes
V(\lambda_{L})$. In terms of matrices we express operators $A_{j,k}$ and
$A_{k,j}$ for $j<k$ by

$A_{j,k}=\left(\begin{array}[]{cccc}A^{11}_{11}&A^{11}_{12}&A^{11}_{21}&A^{11}_{22}\\\
A^{12}_{11}&A^{12}_{12}&A^{12}_{21}&A^{12}_{22}\\\
A^{21}_{11}&A^{21}_{12}&A^{21}_{21}&A^{21}_{22}\\\
A^{22}_{11}&A^{22}_{12}&A^{22}_{21}&A^{22}_{22}\end{array}\right)_{[j,k]}\,,\quad
A_{k,j}=\left(\begin{array}[]{cccc}A^{11}_{11}&A^{11}_{21}&A^{11}_{12}&A^{11}_{22}\\\
A^{21}_{11}&A^{21}_{21}&A^{21}_{12}&A^{21}_{22}\\\
A^{12}_{11}&A^{12}_{21}&A^{12}_{12}&A^{12}_{22}\\\
A^{22}_{11}&A^{22}_{21}&A^{22}_{12}&A^{22}_{22}\end{array}\right)_{[j,k]}$
(2.5)

Here by the symbol $[j,k]$ we express that matrix element $A^{ab}_{cd}$
corresponds to $e_{j}^{a,c}\otimes e_{k}^{b,d}$ for $A_{j,k}$, and to
$e_{j}^{b,d}\otimes e_{k}^{a,c}$ for $A_{k,j}$.

Let us now introduce operators $R_{j,k}(\lambda_{j},\lambda_{k})$ and
$R_{k,j}(\lambda_{k},\lambda_{j})$ acting on the tensor product
$V(\lambda_{1})\otimes\cdots\otimes V(\lambda_{L})$. We define them by putting
$A^{ab}_{cd}(\lambda_{j},\lambda_{k})=R^{ab}_{cd}(\lambda_{j}-\lambda_{k})$ in
(2.4). Here the matrix elements $R^{ab}_{cd}(u)$ are given in (2.2). For
instance, setting $u=\lambda_{1}-\lambda_{2}$, we have explicitly

$R_{12}(\lambda_{1},\lambda_{2})=\left(\begin{array}[]{cccc}1&0&0&0\\\
0&b(u)&c(u)&0\\\ 0&c(u)&b(u)&0\\\ 0&0&0&1\end{array}\right)_{[1,2]}\,.$ (2.6)

The $R$-matrices satisfy the Yang-Baxter equations:

$R_{12}(\lambda_{1},\lambda_{2})R_{13}(\lambda_{1},\lambda_{3})R_{23}(\lambda_{2},\lambda_{3})=R_{23}(\lambda_{2},\lambda_{3})R_{13}(\lambda_{1},\lambda_{3})R_{12}(\lambda_{1},\lambda_{2})$
(2.7)

They also satisfy the inversion relations (unitarity conditions):

$R_{jk}(\lambda_{j},\lambda_{k})R_{kj}(\lambda_{k},\lambda_{j})=I^{\otimes
L}\qquad{\rm for}\,\,1\leq j,k\leq L\,.$ (2.8)

Here $I^{\otimes L}$ denotes the $L$th power of tensor product of $I$.

Hereafter we often abbreviate $R_{jk}(\lambda_{j},\lambda_{k})$ simply by
$R_{jk}$.

###  2.2 $L$-operators and the monodromy matrix

Let us introduce parameters $\xi_{1},\xi_{2},\ldots,\xi_{L}$, which we call
the inhomogeneous parameters. In the case of the monodromy matrix, we assume
that parameters $\lambda_{j}$ of the tensor product
$V(\lambda_{1})\otimes\cdots\otimes V(\lambda_{L})$ are given by the
inhomogeneous parameters, i.e. $\lambda_{j}=\xi_{j}$ for $j=1,2,\ldots,L$. Let
us denote by $0$ the suffix for the auxiliary space. We define $L$-operators
acting on the $m$th site for $m=1,2,\ldots,L$, by

$L_{m}(\lambda,\xi_{m})=R_{0m}(\lambda,\xi_{m})\,.$ (2.9)

We define the monodromy matrix acting on the $L$ lattice-sites in one
dimension by

$T_{0,12\cdots
L}(\lambda;\xi_{1},\ldots,\xi_{L})=L_{L}(\lambda,\xi_{L})L_{L-1}(\lambda,\xi_{L-1})\cdots
L_{2}(\lambda,\xi_{2})L_{1}(\lambda,\xi_{1})\,.$ (2.10)

We shall also denote it by $R_{0,12\cdots
L}(\lambda_{0};\xi_{1},\ldots,\xi_{L})$ in §2.4. Hereafter we often suppress
the symbols of inhomogeneous parameters and express the monodromy matrix
$T_{0,12\cdots L}(\lambda;\xi_{1},\ldots,\xi_{L})$ simply as $R_{0,12\cdots
L}(\lambda;\\{\xi_{j}\\})$ or $T_{0}(\lambda)$.

Let us consider two auxiliary spaces with suffices $a$ and $b$. We define
monodromy matrices $T_{a}(\lambda_{a})$ and $T_{b}(\lambda_{b})$ similarly as
(2.10) with $0$ replaced by $a$ and $b$, respectively. It is clear that they
satisfy the following Yang-Baxter equations.

$R_{ab}(\lambda_{a},\lambda_{b})T_{a}(\lambda_{a})T_{b}(\lambda_{b})=T_{b}(\lambda_{b})T_{a}(\lambda_{a})R_{ab}(\lambda_{a},\lambda_{b})\,.$
(2.11)

Let us introduce operator $A_{j}$ acting on the $j$th site by

$A_{j}=\sum_{a,b=1,2}A^{a}_{b}I_{0}\otimes\cdots\otimes I_{j-1}\otimes
e^{a,\,b}_{j}\otimes\cdots\otimes I_{L}$ (2.12)

We express it in terms of the matrix notation as follows:

$A_{j}=\left(\begin{array}[]{cc}A^{1}_{1}&A^{1}_{2}\\\
A^{2}_{1}&A^{2}_{2}\end{array}\right)_{[j]}$ (2.13)

We express the matrix elements of the monodromy matrix by

$T_{0,12\cdots
L}(u;\xi_{1},\ldots,\xi_{L})=\left(\begin{array}[]{cc}A_{12\cdots
L}(u;\xi_{1},\ldots,\xi_{L})&B_{12\cdots L}(u;\xi_{1},\ldots,\xi_{L})\\\
C_{12\cdots L}(u;\xi_{1},\ldots,\xi_{L})&D_{12\cdots
L}(u;\xi_{1},\ldots,\xi_{L})\end{array}\right)_{[0]}$ (2.14)

The transfer matrix, $t(u)$, is given by the trace of the monodromy matrix
with respect to the 0th space:

$\displaystyle t(u;\xi_{1},\ldots,\xi_{L})$ $\displaystyle=$
$\displaystyle{\rm{tr}_{0}}\left(T_{0,12\cdots
L}(u;\xi_{1},\ldots,\xi_{L})\right)$ (2.15) $\displaystyle=$ $\displaystyle
A_{12\cdots L}(u;\xi_{1},\ldots,\xi_{L})+D_{12\cdots
L}(u;\xi_{1},\ldots,\xi_{L})\,.$

Here we note that the transfer matrix $t(u)$ is nothing but the transfer
matrix of the six-vertex model defined on the two-dimensional square lattice
[34].

Hereafter, we shall often denote $B_{12\cdots L}(u;\xi_{1},\ldots,\xi_{L})$ by
$B(u;\\{\xi_{j}\\})$ or $B(u)$, briefly.

###  2.3 Products of $R$-matrices and the symmetric group

Let us consider the symmetric group ${\cal S}_{n}$ of $n$ integers,
$1,2,\ldots,n$. We denote by $\sigma$ an element of ${\cal S}_{n}$. Then
$\sigma$ maps $j$ to $\sigma(j)$ for $j=1,2,\ldots,n$

######  Definition 1.

Let $p$ be a sequence of $n$ integers, $1,2,\ldots,n$, and $\sigma$ an element
of the symmetric group ${\cal S}_{n}$. We define the action of $\sigma$ on $p$
by

$\sigma(p)=(p_{\sigma(1)},\ldots,p_{\sigma(n)})\,.$ (2.16)

Here we remark that $(\sigma_{A}\sigma_{B})\,p=\sigma_{B}(\sigma_{A}p)$ for
$\sigma_{A},\sigma_{B}\in{\cal S}_{n}$. We shall show it in Appendix A.

Let us recall that $R_{jk}$ denote $R_{jk}(\lambda_{j},\lambda_{k})$.

######  Definition 2.

Let $p=(p_{1},p_{2},\ldots,p_{n})$ be a sequence of $n$ integers,
$1,2,\ldots,n$. We define $R_{p_{1},\,p_{2}p_{3}\cdots p_{n}}$ and
$R_{p_{1}p_{2}\cdots p_{n-1},\,p_{n}}$ by

$\displaystyle R_{p_{1},\,p_{2}p_{3}\cdots p_{n}}$ $\displaystyle=$
$\displaystyle R_{p_{1}p_{n}}R_{p_{1}p_{n-1}}\cdots R_{p_{1}p_{2}}\,,$
$\displaystyle R_{p_{1}p_{2}\cdots p_{n-1},\,p_{n}}$ $\displaystyle=$
$\displaystyle R_{p_{1}p_{n}}R_{p_{2}p_{n}}\cdots R_{p_{n-1}p_{n}}\,.$ (2.17)

For $p=(1,2,\ldots,n)$ we have

$R_{1,23\cdots n}=R_{1n}R_{1n-1}\cdots R_{12}\,,\quad R_{12\cdots
n-1,n}=R_{1n}R_{2n}\cdots R_{n-1n}.$ (2.18)

We thus express the monodromy matrix as follows

$T_{0,12\cdots L}(\lambda_{0};\\{\xi_{k}\\})=R_{0,12\cdots
L}(\lambda_{0};\\{\xi_{k}\\})\,.$ (2.19)

Here we have assumed that $\lambda_{k}=\xi_{k}$ for $k=1,2,\ldots,n$.

Let us express by $s_{j}=(j\,\,j+1)$ such a permutation that maps $j$ to $j+1$
and $j+1$ to $j$ and does not change other integers.

######  Definition 3.

Let $p$ be a sequence of $n$ integers, $1,2,\ldots,n$. We define
$R^{s_{j}}_{p}$ by

$R^{s_{j}}_{p}=R_{p_{j},p_{j+1}}(\lambda_{p_{j}},\lambda_{p_{j+1}})\,.$ (2.20)

For the unit element $e$ of ${\cal S}_{n}$, we define $R^{e}_{p}$ by
$R^{e}_{p}=1$. For a given element $\sigma$ of ${\cal S}_{n}$, we define
$R^{\sigma}_{p}$ recursively by the following:

$R_{p}^{\sigma_{A}\sigma_{B}}=R_{\sigma_{A}(p)}^{\sigma_{B}}R_{p}^{\sigma_{A}}\,.$
(2.21)

We remark that every permutation $\sigma$ is expressed as a product of some
$s_{j}=(j\,j+1)$ with $j=1,2,\ldots,n-1$. We thus obtain $R_{p}^{\sigma}$ as a
product of $R_{p}^{s_{j}}$ for some $j$ s. For an illustration, let us
calculate $R^{(123)}_{(1,2,3)}$. Noting $(123)=(1\,2)(2\,3)$, we have

$\displaystyle R^{(12)(23)}_{(1,2,3)}$ $\displaystyle=$ $\displaystyle
R^{(23)}_{(2,1,3)}R^{(12)}_{(1,2,3)}$ (2.22) $\displaystyle=$ $\displaystyle
R_{13}R_{12}=R_{1,23}\,.$

Through the defining relations of the symmetric group ${\cal S}_{n}$ [35], we
can show that definition 2.21 is well defined. The proof is given in
proposition A.1 of Appendix A.

Let us denote by $\sigma_{c}$ such a cyclic permutation that maps $j$ to $j+1$
for $j=1,\ldots,n-1$ and $n$ to 1. We also express it as $\sigma_{c}=(12\cdots
n)$. Noting $(12\cdots n)=(1\,\,2)\cdots(n-1\,\,n)=s_{1}s_{2}\cdots s_{n-1}$,
we can show the following lemma

######  Lemma 4.

Let us denote by $p_{q}$ the sequence $p_{q}=(1,2,\ldots,n)$. For
$\sigma_{c}=(12\cdots n)$ we have

$R^{\sigma_{c}}_{p_{q}}=R_{1,2\cdots n}\,.$ (2.23)

The proof of lemma 2.23 is given in lemma A.6 of Appendix A.

###  2.4 ${\check{R}}$-matrices and permutation operators

Let us consider two-dimensional vector spaces $V_{a}$ and $V_{b}$. We define
permutation operator $\Pi_{ab}$ which maps elements of $V_{a}\otimes V_{b}$ to
those of $V_{b}\otimes V_{a}$ as follows.

$\Pi_{ab}\,v_{a}\otimes v_{b}=v_{b}\otimes v_{a}\,,\quad v_{a}\in
V_{a}\,,\quad v_{b}\in V_{b}\,.$ (2.24)

We define ${\check{R}}_{ab}(u)$ by

${\check{R}}_{ab}(u)=\Pi_{ab}R_{ab}(u)$ (2.25)

The operators ${\check{R}}_{ab}$ satisfy the Yang-Baxter equations

${\check{R}}_{12}(u){\check{R}}_{23}(u+v){\check{R}}_{12}(v)={\check{R}}_{23}(v){\check{R}}_{12}(u+v){\check{R}}_{23}(u)$
(2.26)

The operator ${\check{R}}_{ab}$ gives a linear map from $V_{a}\otimes V_{b}$
to $V_{b}\otimes V_{a}$. If $V_{a}$ and $V_{b}$ are equivalent, then we may
regard ${\check{R}}_{ab}$ as a map from $V_{a}^{\otimes 2}$ to $V_{a}^{\otimes
2}$.

We add 0 to the $n$ integers. For a given element $\sigma$ of the symmetric
group ${\cal S}_{n+1}$, we define $\Pi^{\sigma}$ acting on integers
$0,1,\ldots,n$, as follows. We first express $\sigma$ in terms of
$s_{j}=(j\,\,j+1)$ such as $\sigma=s_{j_{1}}s_{j_{2}}\cdots s_{j_{r}}$, and
then we define $\Pi^{\sigma}=\Pi^{(j_{1}\,j_{1}+1)\cdots(j_{r}\,j_{r}+1)}$ by

$\Pi^{\sigma}=\Pi_{j_{1},j_{1}+1}\Pi_{j_{2},j_{2}+1}\cdots\Pi_{j_{r},j_{r}+1}\,.$
(2.27)

######  Lemma 5.

We have the following relation between $R$-matrices and operators
${\check{R}}_{ab}$:

$R_{0,12\cdots n}=\Pi^{(01\cdots
n)}\,{\check{R}}_{n-1\,n}(\lambda_{0}-\xi_{n-1})\cdots{\check{R}}_{12}(\lambda_{0}-\xi_{2}){\check{R}}_{01}(\lambda_{0}-\xi_{1})$
(2.28)

##  3 The quantum group invariance

We shall show that the monodromy matrix, $T_{0,12\cdots
L}(\lambda;\xi_{1},\ldots,\xi_{L})$, has the symmetry of the affine quantum
group, $U_{q}(\widehat{sl_{2}})$.

###  3.1 Quantum group $U_{q}(sl_{2})$ and the asymmetric $R$-matrices

The quantum algebra $U_{q}(sl_{2})$ is an associative algebra over ${\bf C}$
generated by $X^{\pm},K^{\pm}$ with the following relations: [31]

$\displaystyle KK^{-1}$ $\displaystyle=$ $\displaystyle KK^{-1}=1\,,\quad
KX^{\pm}K^{-1}=q^{\pm 2}X^{\pm}\,,\quad\,,$ $\displaystyle{[}X^{+},X^{-}{]}$
$\displaystyle=$ $\displaystyle{\frac{K-K^{-1}}{q-q^{-1}}}\,.$ (3.1)

The algebra $U_{q}(sl_{2})$ is also a Hopf algebra over ${\bf C}$ with
comultiplication

$\displaystyle\Delta(X^{+})$ $\displaystyle=$ $\displaystyle X^{+}\otimes
1+K\otimes X^{+}\,,\quad\Delta(X^{-})=X^{-}\otimes K^{-1}+1\otimes X^{-}\,,$
$\displaystyle\Delta(K)$ $\displaystyle=$ $\displaystyle K\otimes K\,,$ (3.2)

and antipode: $S(K)=K^{-1}\,,S(X^{+})=-K^{-1}X^{+}\,,S(X^{-})=-X^{-}K$, and
coproduct: $\epsilon(X^{\pm})=0$ and $\epsilon(K)=1$.

In association with the quantum group, we define the $q$-integer of an integer
$n$ by $[n]_{q}=(q^{n}-q^{-n})/(q-q^{-1})$.

The universal $R$-matrix, ${\cal R}$, of $U_{q}(sl_{2})$ satisfies the
following relations:

${\cal R}\Delta(x)=\tau\circ\Delta(x){\cal R}\quad{\rm for\,all}\quad x\in
U_{q}(sl_{2})\,.$ (3.3)

Here $\tau$ denotes a permutation such that $\tau\,a\otimes b=b\otimes a$ for
$a,b\in U_{q}(sl_{2})$.

We now introduce some notation of a Hopf algebra. Let $x(1),\ldots,x(n)$ be
elements of Hopf algebra ${\cal A}$. For a given permutation $\sigma$ of
${\cal S}_{n}$, we define its action on the tensor product
$x(1)\otimes\cdots\otimes x(n)$ as follows:

$\sigma\circ\left(x(1)\otimes\cdots\otimes
x(n)\right)=x(\sigma^{-1}1)\otimes\cdots x(\sigma^{-1}n)\,.$ (3.4)

We note that ${\cal A}$ has coassociativity: $\left(\Delta\otimes
id\right)\Delta(x)=\left(id\otimes\Delta\right)\Delta(x)$ for any element $x$
of ${\cal A}$. We therefore denote it by $\Delta^{(2)}(x)$. We define
$\Delta^{(n)}(x)$ recursively by

$\Delta^{(n)}(x)=\left(\Delta^{(n-1)}\otimes id\right)\Delta(x)\quad{\rm
for}\,\,x\in{\cal A}.$ (3.5)

Let us now introduce the following asymmetric $R$-matrices:

$R^{\pm}(u)=\left(\begin{array}[]{cccc}1&0&0&0\\\ 0&b(u)&c^{\mp}(u)&0\\\
0&c^{\pm}(u)&b(u)&0\\\ 0&0&0&1\end{array}\right)\,,$ (3.6)

where $c^{\pm}(u)$ are defined by

$c^{\pm}(u)={\frac{e^{\pm u}\sinh(\eta)}{\sinh(u+\eta)}}\,.$ (3.7)

In the spin-1/2 representation of $U_{q}(sl_{2})$, we have the following
relations:

$R_{12}^{+}(u)\Delta(x)=\tau\circ\Delta(x)R^{+}_{12}(u)\quad{\rm for\,\,}\quad
x=X^{\pm},K\,.$ (3.8)

Here we remark that spectral parameter $u$ is arbitrary and independent of
$X^{\pm}$ or $K$. Similarly as in the symmetric case, we define the monodromy
matrix $R^{+}_{0,1\cdots n}$ by $R^{+}_{0,1\cdots n}=R^{+}_{0,n}\cdots
R^{+}_{0,1}$, and ${\check{R}}^{+}$ by
${\check{R}}^{+}_{12}(u)=\Pi_{12}R^{+}(u)$.

######  Lemma 6.

The monodromy matrix expressed in terms of ${\check{R}}$’s commutes with the
action of the quantum group $U_{q}(sl_{2})$:

${[}{\check{R}}^{+}_{L-1,L}(\lambda-\xi_{L})\cdots{\check{R}}^{+}_{1,2}(\lambda-\xi_{2}){\check{R}}^{+}_{0,1}(\lambda-\xi_{1}),\quad\Delta^{(L)}(x){]}=0\,,\quad{\rm
for}\,{\rm all}\,x\in U_{q}(sl_{2})\,.$ (3.9)

Here parameters $\lambda$, $\xi_{1},\ldots,\xi_{L}$ are independent of element
$x$ of $U_{q}(sl_{2})$.

We shall show lemma 6 and eqs. (3.8) through the Temperley-Lieb algebra in
§3.2.

Lemma 6 leads to the following symmetry relations of the monodromy matrix
$R^{+}_{0,12\cdots L}$ with respect to the quantum group $U_{q}(sl_{2})$:

######  Proposition 7.

Let $\sigma_{c}$ be a cyclic permutation: $\sigma_{c}=(01\cdots L)$. Then we
have

$R^{+}_{0,12\cdots
L}(\lambda;\xi_{1},\ldots,\xi_{L})\,\Delta^{(L)}(x)=\sigma_{c}\circ\Delta^{(L)}(x)\,R^{+}_{0,12\cdots
L}(\lambda;\xi_{1},\ldots,\xi_{L})\quad{\rm for\,\,all}\,\,x\in U_{q}(sl_{2})$
(3.10)

Here parameters $\lambda$, $\xi_{1},\ldots,\xi_{L}$ are independent of element
$x$ of $U_{q}(sl_{2})$.

###### Proof.

Making use of lemma 6 we show (3.10) from (2.28) and the following relation:

$\sigma_{c}\circ\Delta^{(L)}(x)=\Pi^{\sigma_{c}}\,\Delta^{(L)}(x)\,\left(\Pi^{\sigma_{c}}\right)^{-1}\,.$
(3.11)

∎

###  3.2 Derivation in terms of the Temperley-Lieb algebra

Let us define $U_{j}^{\pm}$ for $j=0,1,\ldots,L-1$, by

$U^{\pm}_{j}=\left(\begin{array}[]{cccc}0&0&0&0\\\ 0&q^{\mp}&-1&0\\\
0&-1&q^{\pm}&0\\\ 0&0&0&0\end{array}\right)_{[j,j+1]}\,.$ (3.12)

They satisfy the defining relations of the Temperley-Lieb algebra: [34]

$\displaystyle U^{\pm}_{j}U^{\pm}_{j+1}U^{\pm}_{j}$ $\displaystyle=$
$\displaystyle U^{\pm}_{j},$ $\displaystyle
U^{\pm}_{j+1}U^{\pm}_{j}U^{\pm}_{j+1}$ $\displaystyle=$ $\displaystyle
U^{\pm}_{j},\quad{\rm for}\,j=0,1,\ldots,L-2,$
$\displaystyle\left(U^{\pm}_{j}\right)^{2}$ $\displaystyle=$
$\displaystyle(q+q^{-1})\,U^{\pm}_{j}\quad{\rm for}\,j=0,1,\ldots,L-1,$
$\displaystyle U^{\pm}_{j}U^{\pm}_{k}$ $\displaystyle=$ $\displaystyle
U^{\pm}_{k}U^{\pm}_{j}\quad{\rm for}\,\,|j-k|>1\,.$ (3.13)

Let us now show commutation relations (3.10), making use of the Temperley-Lieb
algebra. The operator ${\check{R}}_{j,j+1}^{+}(u)$ is decomposed in terms of
the generators of the Temperley-Lieb algebra as follows [36].

${\check{R}}_{j,j+1}^{+}(u)=I-b(u)U_{j}^{+}\,.$ (3.14)

######  Lemma 8.

The monodromy matrix of the six-vertex model is expressed in terms of the
generators of the Temperley-Lieb algebra as follows.

$\displaystyle{\check{R}}^{+}_{L-1,L}(\lambda-\xi_{L})\cdots{\check{R}}^{+}_{1,2}(\lambda-\xi_{2}){\check{R}}^{+}_{0,1}(\lambda-\xi_{1})$
$\displaystyle=\sum_{k=0}^{L}(-1)^{k}\sum_{0\leq
i_{1}<\cdots<i_{k}<L}\,\left(\prod_{j=1}^{k}b(\lambda-\xi_{i_{j}})\right)U^{+}_{i_{k}}\cdots
U^{+}_{i_{2}}U^{+}_{i_{1}}\,.$ (3.15)

######  Lemma 9.

The generators $U^{+}_{j}$ commute with the generators of $U_{q}(sl_{2})$. For
$x=X^{\pm},K$ and for $j=0,1,\ldots,L-1$, we have in the tensor-product
representation

${\big{[}}U^{+}_{j},\Delta^{(L)}(x){\big{]}}=0\,.$ (3.16)

Proof of proposition 7.  From lemmas 8 and 3.16 we have lemma 6, which is
equivalent to proposition 7.

We now show that in the limit of taking $u$ to $-\infty$, ${\hat{R}}^{+}(u)$
is equivalent to the spin-1/2 matrix representation of the universal
$R$-matrix ${\cal R}$ of $U_{q}(sl(2))$. An explicit expression of ${\cal R}$
is given by

${\cal R}=q^{-{\frac{1}{2}}H\otimes
H}\exp_{q}\left(-(q-q^{-1})K^{-1}X^{+}\otimes X^{-}K\right)$ (3.17)

where $\exp_{q}x$ denotes the following series:

$\exp_{q}x=\sum_{n=0}^{\infty}{\frac{q^{-n(n-1)/2}}{[n]_{q}!}}x^{n}\,.$ (3.18)

Here $q$ is generic. We recall that $[n]_{q}$ denotes the $q$-integer of an
integer $n$: $[n]_{q}=(q^{n}-q^{-n})/(q-q^{-1})$. Putting $X^{+}=e^{1,2}$,
$X^{-}=e^{2,1}$ and $K={\rm diag}(q,q^{-1})$ in the series (3.17), we have the
following matrix representation.

$R_{{\frac{1}{2}},{\frac{1}{2}}}=\left(\begin{array}[]{cccc}q^{-1/2}&0&0&0\\\
0&q^{1/2}&-q^{1/2}(q-q^{-1})&0\\\ 0&0&q^{1/2}&0\\\
0&0&0&q^{-1/2}\end{array}\right)\,.$ (3.19)

Thus, for arbitrary $u$, we have the following:

$R^{+}_{12}(u)=c^{+}(u)\Pi_{12}+q^{-1/2}b(u)R_{{\frac{1}{2}},{\frac{1}{2}}}$
(3.20)

Therefore, we have ${R}^{+}(-\infty)=q^{1/2}R_{{\frac{1}{2}},{\frac{1}{2}}}$.

We remark that some relations equivalent to (3.16) have been shown in
association with the $sl(2)$ loop algebra symmetry of the XXZ spin chain at
roots of unity [37].

###  3.3 Gauge transformations

Let us introduce operators ${\Phi}_{j}$ with arbitrary parameters $\phi_{j}$
for $j=0,1,\ldots,L$ as follows:

${\Phi}_{j}=\left(\begin{array}[]{cc}1&0\\\
0&e^{\phi_{j}}\end{array}\right)_{[j]}=I^{\otimes(j)}\otimes\left(\begin{array}[]{cc}1&0\\\
0&e^{\phi_{j}}\end{array}\right)\otimes I^{\otimes(L-j)}.$ (3.21)

In terms of ${\chi}_{jk}={\Phi}_{j}{\Phi}_{k}$, we define a similarity
transformation on the $R$-matrix by

$R_{jk}^{\chi}={\chi}_{jk}R_{jk}{\chi}_{jk}^{-1}$ (3.22)

Explicitly, the following two matrix elements are transformed.

$\left(R_{jk}^{\chi}\right)^{21}_{12}=c(\lambda_{j},\lambda_{k})e^{\phi_{j}-\phi_{k}}\,,\quad\left(R_{jk}^{\chi}\right)^{12}_{21}=c(\lambda_{j},\lambda_{k})e^{-\phi_{j}+\phi_{k}}\,.$
(3.23)

We now put $\phi_{j}=\lambda_{j}$ in eq. (3.21) for $j=0,1,\ldots,L$. For
$j,k=0,1,\ldots,L$, we have

$R_{jk}^{\pm}(\lambda_{j},\lambda_{k})=\left(\chi_{jk}\right)^{\pm
1}\,R_{jk}(\lambda_{j},\lambda_{k})\,\left(\chi_{jk}\right)^{\mp 1}\,.$ (3.24)

Thus, the asymmetric $R$-matrices $R_{12}^{\pm}(\lambda_{1},\lambda_{2})$ are
derived from the symmetric one through the gauge transformation $\chi_{jk}$.

For the monodromy matrix, in terms of the inhomogeneous parameters,
$\xi_{1},\ldots,\xi_{L}$, we put $\lambda_{j}=\xi_{j}$ for $j=1,\ldots,L$. We
define ${\chi}_{012\cdots L}$ by ${\chi}_{012\cdots
L}=\Phi_{0}\Phi_{1}\cdots\Phi_{L}$. Then, the asymmetric monodromy matrices
are transformed into the symmetric one as follows.

$R_{0,12\cdots L}^{\pm}=\left(\chi_{012\cdots L}\right)^{\pm 1}\,R_{0,12\cdots
L}\left(\chi_{012\cdots L}\right)^{\mp 1}\,.$ (3.25)

We note that the asymmetric $R$-matrices ${\check{R}}^{\pm}_{j,j+1}(u)$ are
derived from the symmetric $R$-matrix through the gauge transformations, and
they are related to the Jones polynomial. [21]

###  3.4 Affine quantum group symmetry

The affine quantum algebra $U_{q}(\widehat{sl_{2}})$ is an associative algebra
over ${\bf C}$ generated by $X_{i}^{\pm},K_{i}^{\pm}$ for $i=0,1$ with the
following relations:

$\displaystyle K_{i}K_{i}^{-1}$ $\displaystyle=$ $\displaystyle
K^{-1}_{i}K_{i}=1\,,\quad K_{i}X_{i}^{\pm}K_{i}^{-1}=q^{\pm
2}X_{i}^{\pm}\,,\quad K_{i}X_{j}^{\pm}K_{i}^{-1}=q^{\mp
2}X_{j}^{\pm}\quad(i\neq j)\,,$ $\displaystyle{[}X_{i}^{+},X_{j}^{-}{]}$
$\displaystyle=$
$\displaystyle\delta_{i,j}\,{\frac{K_{i}-K_{i}^{-1}}{q-q^{-1}}}\,,$
$\displaystyle(X_{i}^{\pm})^{3}X_{j}^{\pm}$ $\displaystyle-$
$\displaystyle[3]_{q}\,(X_{i}^{\pm})^{2}X_{j}^{\pm}X_{i}^{\pm}+[3]_{q}\,X_{i}^{\pm}X_{j}^{\pm}(X_{i}^{\pm})^{2}-X_{j}^{\pm}(X_{i}^{\pm})^{3}=0\quad(i\neq
j)\,.$ (3.26)

The algebra $U_{q}(\widehat{sl_{2}})$ is also a Hopf algebra over ${\bf C}$
with comultiplication

$\displaystyle\Delta(X_{i}^{+})$ $\displaystyle=$ $\displaystyle
X_{i}^{+}\otimes 1+K_{i}\otimes
X_{i}^{+}\,,\quad\Delta(X_{i}^{-})=X_{i}^{-}\otimes K_{i}^{-1}+1\otimes
X_{i}^{-}\,,$ $\displaystyle\Delta(K_{i})$ $\displaystyle=$ $\displaystyle
K_{i}\otimes K_{i}\,,$ (3.27)

and antipode:
$S(K_{i})=K_{i}^{-1}\,,S(X_{i})=-K_{i}^{-1}X_{i}^{+}\,,S(X_{i}^{-})=-X_{i}^{-}K_{i}$.

We now introduce evaluation representations for $U_{q}(\widehat{sl_{2}})$
[22]. For a given complex number $a$ there is a homomorphism of algebras
$\varphi_{a}$: $U_{q}(\widehat{sl_{2}})\rightarrow U_{q}({sl_{2}})$ such that

$\displaystyle\varphi_{a}(X_{0}^{\pm})$ $\displaystyle=$
$\displaystyle\exp(\pm a)\,X^{\mp}\,,\quad\varphi_{a}(K_{0})=K^{-1}\,,$
$\displaystyle\varphi_{a}(X_{1}^{\pm})$ $\displaystyle=$ $\displaystyle
X^{\pm}\,,\quad\varphi_{a}(K_{1})=K\,.$ (3.28)

We denote by $(\pi,V)$ a representation of an algebra ${\cal A}$ such that
$\pi(x)$ give linear maps on vector space $V$ for $x\in{\cal A}$. For a given
finite-dimensional representation $(\pi_{V},V)$ of $U_{q}(sl_{2})$ we have a
finite-dimensional representation $(\pi_{V(a)},V(a))$ of
$U_{q}(\widehat{sl_{2}})$ through homomorphism $\varphi_{a}$, i.e.
$\pi_{V(a)}(x)=\pi_{V}(\varphi_{a}(x))$ for $x\in U_{q}(\widehat{sl_{2}})$. We
call $(\pi_{V(a)},V(a))$ or $V(a)$ the evaluation representation of $V$ and
nonzero parameter $a$ the evaluation parameter of $V(a)$. If $V$ is
$(2s+1)$-dimensional, then we also denote it by $V^{(2s)}(a)$. Hereafter we
express $2s$ by an integer $\ell$.

Similarly as (3.14), we have the following decomposition:

${\check{R}}_{j,j+1}^{-}(u)=I-b(u)U_{j}^{-}\,,\quad{\rm
for}\,\,j=0,1,\ldots,L-1.$ (3.29)

######  Lemma 10.

Generators $U^{-}_{j}$ commute with $X_{0}^{\pm}$ and $K_{0}$ of
$U_{q}(\widehat{sl_{2}})$ in the tensor-product representation
$V^{(1)}(a_{0})\otimes\cdots\otimes V^{(1)}(a_{L})$ with
$a_{0}=a_{1}=\cdots=a_{L}=a$. For $j=0,1,\ldots,L-1$, we have

${\big{[}}U^{-}_{j},\,\,\varphi_{a}^{\otimes(L+1)}\left(\Delta^{(L)}(x)\right){\big{]}}=0\,\quad\left(x=X_{0}^{\pm},\,K_{0}.\right)$
(3.30)

###### Proof.

Let us denote by $U^{-}$ the $4\times 4$ matrix given by the decomposition:
${\check{R}}(u)=I-b(u)U^{-}$. Through an explicit calculation we show that
$U^{-}$ and $\varphi_{a}\otimes\varphi_{b}\left(\Delta(X_{0}^{\pm})\right)$
commute if $a=b$:

${[}U^{-},\,\varphi_{a}\otimes\varphi_{a}\left(\Delta(X_{0}^{\pm})\right){]}=0\,.$
(3.31)

We derive (3.30) through (3.31). ∎

In the spin-1/2 representation of $U_{q}(sl_{2})$, we thus have the following
relations:

$R_{12}^{-}(u)\,\,\varphi_{a}^{\otimes 2}\Delta(x)=\varphi_{a}^{\otimes
2}\left(\tau\circ\Delta(x)\right)\,\,R^{-}_{12}(u)\quad{\rm for\,\,}\quad
x=X_{0}^{\pm},K_{0}\,.$ (3.32)

Here we note that $u$ is arbitrary and independent of $X_{0}^{\pm},K_{0}$.

Similarly as lemmas 8 and 3.16, we have from lemma 3.30 the quantum-group
symmetry of the monodromy matrix $R^{-}_{0,12\cdots L}$:

######  Proposition 11.

Let $\sigma_{c}$ be a cyclic permutation: $\sigma_{c}=(01\cdots L)$. In the
evaluation representation (3.28) we have, for $x=X_{0}^{\pm},K_{0}$, the
following:

$R^{-}_{0,12\cdots
L}(\lambda;\xi_{1},\ldots,\xi_{L})\,\varphi_{a}^{\otimes(L+1)}\left(\Delta^{(L)}(x)\right)=\varphi_{a}^{\otimes(L+1)}\left(\sigma_{c}\circ\Delta^{(L)}(x)\right)\,R^{-}_{0,12\cdots
L}(\lambda;\xi_{1},\ldots,\xi_{L})\,.$ (3.33)

Here parameters $\lambda$, $\xi_{1},\ldots,\xi_{L}$ are arbitrary and
independent of $x=X_{0}^{\pm},K_{0}$.

Let us now make a summary of the symmetry relations of $R^{+}_{12}$. Here we
recall that $R^{+}_{12}(\lambda_{1},\lambda_{2})\in End(V(\lambda_{1})\otimes
V(\lambda_{2}))$. For simplicity, we put $a=0$ in (3.33). Combining (3.8) and
(3.32) We have the following relations:

$\displaystyle R^{+}_{12}(\lambda_{1},\lambda_{2})\,\,\varphi_{0}^{\otimes
2}\left(\Delta(X_{1}^{\pm})\right)$ $\displaystyle=$
$\displaystyle\varphi_{0}^{\otimes
2}\left(\tau\circ\Delta(X_{1}^{\pm})\right)\,\,R^{+}_{12}(\lambda_{1},\lambda_{2})\,,$
$\displaystyle
R^{+}_{12}(\lambda_{1},\lambda_{2})\,(\chi_{12})^{2}\varphi_{0}^{\otimes
2}\left(\Delta(X_{0}^{\pm})\right)(\chi_{12})^{-2}$ $\displaystyle=$
$\displaystyle(\chi_{12})^{2}\varphi_{0}^{\otimes
2}\left(\tau\circ\Delta(X_{0}^{\pm})\right)(\chi_{12})^{-2}\,R^{+}_{12}(\lambda_{1},\lambda_{2})\,,$
$\displaystyle R^{+}_{12}(\lambda_{1},\lambda_{2})\,\varphi_{0}^{\otimes
2}\left(\Delta(K_{i}^{\pm})\right)$ $\displaystyle=$
$\displaystyle\varphi_{0}^{\otimes
2}\left(\tau\circ\Delta(K_{i}^{\pm})\right)\,R^{+}_{12}(\lambda_{1},\lambda_{2})\,{\rm
for}\,i=0,1\,.$

Let us now consider $\varphi_{a_{1}}\otimes\varphi_{a_{2}}$ with
$a_{j}=2\lambda_{j}$ for $j=1,2$. We have

$\varphi_{2\lambda_{1}}\otimes\varphi_{2\lambda_{2}}\left(\Delta(X_{0}^{\pm})\right)=(\chi_{12})^{2}\,\,\varphi_{0}\otimes\varphi_{0}\left(\Delta(X_{0}^{\pm})\right)\,\,(\chi_{12})^{-2}$
(3.35)

Thus, relations (LABEL:eq:affineCR) are now expressed as follows.

$R^{+}_{12}(\lambda_{1},\lambda_{2})\,\varphi_{2\lambda_{1}}\otimes\varphi_{2\lambda_{2}}\left(\Delta(x)\right)=\varphi_{2\lambda_{1}}\otimes\varphi_{2\lambda_{2}}\left(\tau\circ\Delta(x)\right)\,R^{+}_{12}(\lambda_{1},\lambda_{2})\,,\,\,{\rm
for}\,\,x=X_{0}^{\pm},X_{1}^{\pm},K_{0},K_{1}\,.$ (3.36)

In (3.36) all the parameters are now associated with the evaluation parameters
of the tensor product $V(2\lambda_{1})\otimes V(2\lambda_{2})$. Therefore, we
conclude that the asymmetric $R$-matrix $R^{+}_{12}(\lambda_{1},\lambda_{2})$
satisfies the affine quantum-group symmetry.

The fundamental commutation relations (3.10) and (3.33) are summarized as
follows.

######  Proposition 12.

Let $\sigma_{c}$ be a cyclic permutation: $\sigma_{c}=(01\cdots L)$. The
asymmetric $R$-matrix $R^{+}$ satisfies the commutation relations for the
affine-quantum group:

$R^{+}_{0,12\cdots L}(\lambda_{0})\left(\Delta^{(n)}(x)\right)_{01\cdots
L}=\left(\sigma_{c}\circ\Delta^{(L)}(x)\right)_{01\cdots L}R^{+}_{0,12\cdots
L}(\lambda_{0})\quad{\rm for\,\,all}\,\,x\in U_{q}({\widehat{sl_{2}}})\,.$
(3.37)

Here the symbol $(x)_{01\cdots n}$ denotes the matrix representation of $x$ in
the tensor product of evaluation representations, $V({2\lambda_{0}})\otimes
V({2\xi_{1}})\otimes\cdots\otimes V({2\xi_{L}})$.

###  3.5 Symmetry relations of $U_{q}(\widehat{sl_{2}})$ for all permutations

Let us generalize relations (3.37). Making an extensive use of relations
(3.36), we can show commutation relations for $R_{p}^{\sigma}$ for all
permutations $\sigma$. In fact, we can prove (3.37) also by the method for
showing proposition A.2. In Appendix A we shall show in proposition A.2 how we
generalize the symmetry relation of $R_{12}$ such as (3.36) into those of
$R_{p}^{\sigma}$ for any permutation $\sigma$.

We now formulate the symmetry relations in terms of the symmetric
$R$-matrices. Let us denote by ${\bar{\chi}}$ the inverse of the gauge
transformation $\chi$. We express by $\left(\Delta^{(n)}(x)\right)_{01\cdots
n}^{\bar{\chi}}$ the following:

$\left(\Delta^{(n)}(x)\right)_{01\cdots n}^{\bar{\chi}}=\left(\chi_{01\cdots
n}\right)^{-1}\,\varphi_{2\lambda_{0}}\otimes\varphi_{2\xi_{1}}\otimes\cdots\otimes\varphi_{2\xi_{n}}\left(\Delta^{(n)}(x)\right)\,\chi_{01\cdots
n}$ (3.38)

######  Proposition 13.

Let $p_{q}$ be an increasing sequence of $n+1$ integers:
$p_{q}=(0,1,2,\ldots,n)$, and $\sigma$ a permutation on $n+1$ integers,
$0,1,\ldots,n$. With the symmetric $R$-matrices, we have

$R^{\sigma}_{p_{q}}(\lambda_{0})\,\left(\Delta^{(n)}(x)\right)_{01\cdots
n}^{\bar{\chi}}=\left(\sigma\circ\Delta^{(n)}(x)\right)^{\bar{\chi}}_{01\cdots
n}\,R^{\sigma}_{p_{q}}(\lambda_{0})\quad{\rm for\,\,all}\,\,x\in
U_{q}({\widehat{sl_{2}}})\,.$ (3.39)

##  4 Projection operators and the fusion procedure

###  4.1 Projection operators

Let us recall that ${\check{R}}_{12}^{+}(u)$ has been defined by

${\check{R}}_{12}^{+}(u)=\Pi_{12}R_{12}^{+}(u)\,.$

We define operator $P_{12}^{+}$ by

$P_{12}^{+}={\check{R}}_{12}^{+}(\eta)$ (4.1)

Explicitly we have

$P_{12}^{+}=\left(\begin{array}[]{cccc}1&0&0&0\\\
0&{\frac{q}{[2]}}&{\frac{1}{[2]}}&0\\\
0&{\frac{1}{[2]}}&{\frac{q^{-1}}{[2]}}&0\\\
0&0&0&1\end{array}\right)_{[12]}\,.$ (4.2)

By making use of the matrix representation (4.2) it is easy to show that
operator $P_{12}^{+}$ is idempotent:

$\left(P_{12}^{+}\right)^{2}=P_{12}^{+}$ (4.3)

Thus, we may consider that the operator $P_{12}^{+}$ is a projection operator.
In fact, $P_{12}^{+}$ is nothing but the $q$-analogue of the Young operator
which projects out the spin-1 representation of $U_{q}(sl_{2})$ from of the
tensor product of two spin-1/2 representations. It should be noticed that in
the case of symmetric $R$-matrix, $R(\eta)^{2}$ is not equal to $R(\eta)$.

We now introduce projection operators for the spin-$s$ irreducible
representations of $U_{q}(sl_{2})$. Hereafter we set $\ell=2s$. We define
projection operator $P_{12\cdots\ell}^{(\ell)}$ acting on the tensor product
$V_{12\cdots\ell}$ of spin-1/2 representations $V$, i.e.
$V_{12\cdots\ell}=V^{\otimes\ell}$, by the following recursive relations:

$P^{(\ell)}_{12\cdots\ell}=P^{(\ell-1)}_{12\cdots\ell-1}{\check{R}}_{\ell-1,\ell}^{+}((\ell-1)\eta)P^{(\ell-1)}_{12\cdots\ell-1}$
(4.4)

Here $P_{12}^{(1)}=P_{12}^{+}$.

Making use of the Yang-Baxter equations (2.26) and through induction on $\ell$
we can show

$\left(P^{(\ell)}_{12\cdots\ell}\right)^{2}=P^{(\ell)}_{12\cdots\ell}$ (4.5)

Thus, operator $P^{(\ell)}_{12\cdots\ell}$ gives a projection operator.
Similarly, we define projection operators acting on
$V_{j\,j+1\cdots\,j+\ell-1}$ recursively by

$P^{(\ell)}_{j\,j+1\cdots j+\ell-1}=P^{(\ell-1)}_{j\,j+1\cdots
j+\ell-2}{\check{R}}_{j+\ell-2,\,j+\ell-1}^{+}((\ell-1)\eta)P^{(\ell-1)}_{j\,j+1\cdots
j+\ell-2}$ (4.6)

Hereafter we shall abbreviate $P^{(\ell)}_{j\,j+1\cdots j+\ell-1}$ by
$P^{(\ell)}_{j}$ .

From idempotency (4.5) and by the Yang-Baxter equations we can show the
following:

######  Lemma 14.

Suppose that inhomogeneous parameters
$\xi_{j},\xi_{j+1},\ldots,\xi_{j+\ell-1}$ are given by
$\xi_{j+i-1}=z-(i-1)\eta$ for $i=1,2,\ldots,\ell$ with a constant $z$. Then,
the monodromy matrix $R^{+}_{0,12\cdots L}$ satisfies the following property:

$P_{j}^{(\ell)}\,R^{+}_{0,12\cdots L}=P_{j}^{(\ell)}\,R^{+}_{0,12\cdots
L}\,P_{j}^{(\ell)}\,.$ (4.7)

Making use of (2.28) we can express projection operators in terms of
$R$-matrices.

$P^{(\ell)}_{1}=\left(\prod_{j=1}^{\ell}\Pi^{(j\,\,\ell-j+1)}\right)\,\,R^{+}_{\ell-1,\,\ell}\cdots
R^{+}_{2,\,3\cdots\ell}R^{+}_{1,\,2\cdots\ell}$ (4.8)

###  4.2 Fusion of monodromy matrices

####  4.2.1 The case of tensor product of spin-$s$ representations

We first consider the case of tensor product of spin-$s$ representations. We
set $L=N_{s}\ell$. We introduce a set of parameters, $\xi_{1}^{(\ell)}$,
$\xi_{2}^{(\ell)}$, …, $\xi_{L}^{(\ell)}$, as follows:

$\xi_{(k-1)\ell+j}^{(\ell)}={\zeta}_{k}-(j-1)\eta+(\ell-1)\eta/2\qquad\mbox{for}\quad
j=1,\ldots\ell,\,\,\mbox{and}\,\,k=1,\ldots,N_{s}.$ (4.9)

Let us set inhomogeneous parameters $\xi_{1},\xi_{2},\ldots,\xi_{L}$, by
$\xi_{j}=\xi_{j}^{(\ell)}$ for $j=1,2,\ldots,L$. We define the monodromy
matrix $T^{(\ell+)}_{0}(u;\zeta_{1},\ldots,\zeta_{N_{s}})$ acting on the
tensor product of spin-$s$ representations, $V^{(2s)}(\zeta_{1})\otimes
V^{(2s)}(\zeta_{N_{s}})$ by

$T^{(\ell+)}_{0}(\lambda_{0};\zeta_{1},\ldots,\zeta_{N_{s}})=\prod_{k=0}^{N_{s}-1}P^{(\ell)}_{\ell
k+1}\,\cdot\,R^{+}_{0,12\cdots
L}(\lambda_{0};\xi_{1}^{(\ell)},\ldots,\xi_{L}^{(\ell)})\,\cdot\,\prod_{k=0}^{N_{s}-1}P^{(\ell)}_{\ell
k+1}$ (4.10)

Making use of properties of projection operators (4.5) and (4.7), we can show
the Yang-Baxter equation [fusionXXX, 19]

$\displaystyle
R^{+}_{ab}(\lambda_{a}-\lambda_{b})T^{(\ell+)}_{a}(\lambda_{a};\zeta_{1},\ldots,\zeta_{N_{s}})T^{(\ell+)}_{b}(\lambda_{b};\zeta_{1},\ldots,\zeta_{N_{s}})$
(4.11) $\displaystyle=$ $\displaystyle
T^{(\ell+)}_{b}(\lambda_{b};\zeta_{1},\ldots,\zeta_{N_{s}})T^{(\ell+)}_{a}(\lambda_{a};\zeta_{1},\ldots,\zeta_{N_{s}})R^{+}_{ab}(\lambda_{a}-\lambda_{b})$

Through the inverse of the gauge transformation, we derive the symmetric
spin-$s$ monodromy matrix as follows:

$T^{(\ell)}_{0}(\lambda_{0};\zeta_{1},\ldots,\zeta_{N_{s}})=\prod_{k=0}^{N_{s}-1}\left(P^{(\ell)}_{\ell
k+1}\right)^{\bar{\chi}}\,\cdot\,R_{0,12\cdots
L}(\lambda_{0};\xi_{1}^{(\ell)},\ldots,\xi_{L}^{(\ell)})\,\cdot\,\prod_{k=0}^{N_{s}-1}\left(P^{(\ell)}_{\ell
k+1}\right)^{\bar{\chi}}$ (4.12)

where we have defined the transformed projectors by

$\left(P^{(\ell)}_{\ell k+1}\right)^{\bar{\chi}}=\left(\chi_{01\cdots
L}\right)^{-1}\,P^{(\ell)}_{\ell k+1}\,\left(\chi_{01\cdots L}\right)$ (4.13)

####  4.2.2 The case of mixed spins

Let us consider the tensor product of representations with different spins,
$s_{1},s_{2},\ldots,s_{r}$. Here we introduce $\ell_{j}$ by $\ell_{j}=2s_{j}$
for $j=1,2,\ldots,r$, and we assume that
$\ell_{1}+\ell_{2}+\cdots+\ell_{r}=L$. @ Let us introduce a set of parameters,
$\xi_{1}^{({\mbox{\boldmath$\ell$}})}$,
$\xi_{2}^{({\mbox{\boldmath$\ell$}})}$, …,
$\xi_{L}^{({\mbox{\boldmath$\ell$}})}$, as follows:

$\xi_{\ell_{1}+\cdots+\ell_{k-1}+j}^{({\mbox{\boldmath$\ell$}})}=\zeta_{k}-(j-1)\eta+(\ell_{k}-1)\eta/2\qquad\mbox{for}\quad
j=1,\ldots\ell_{k},\,\,\mbox{and}\,\,k=1,\ldots,r.$ (4.14)

We define the asymmetric monodromy matrix for the mixed spin case
$T^{({\mbox{\boldmath$\ell$}}+)}_{0}(u;\zeta_{1},\ldots,\zeta_{r})$ acting on
the tensor product representation
$V_{\ell_{1}}({\zeta_{1}})\otimes\cdots\otimes V_{\ell_{r}}({\zeta_{r}})$ by

$T^{({\mbox{\boldmath$\ell$}}+)}_{0}(\lambda_{0};\zeta_{1},\ldots,\zeta_{r})=\prod_{k=1}^{r}P^{(\ell_{k})}_{\ell_{1}+\cdots+\ell_{k-1}+1}\,\cdot\,R_{0,12\cdots
L}(\lambda_{0};\xi_{1}^{(\mbox{\boldmath$\ell$})},\ldots,\xi_{L}^{(\mbox{\boldmath$\ell$})})\,\cdot\,\prod_{k=1}^{r}P^{(\ell_{k})}_{\ell_{1}+\cdots+\ell_{k-1}+1}\,.$
(4.15)

It is easy to show that they satisfy the Yang-Baxter equations.

$\displaystyle
R^{+}_{ab}(\lambda_{a}-\lambda_{b})T^{({\mbox{\boldmath$\ell$}}+)}_{a}(\lambda_{a};\zeta_{1},\ldots,\zeta_{N_{s}})T^{({\mbox{\boldmath$\ell$}}+)}_{b}(\lambda_{b};\zeta_{1},\ldots,\zeta_{N_{s}})$
(4.16) $\displaystyle=$ $\displaystyle
T^{({\mbox{\boldmath$\ell$}}+)}_{b}(\lambda_{b};\zeta_{1},\ldots,\zeta_{N_{s}})T^{({\mbox{\boldmath$\ell$}}+)}_{a}(\lambda_{a};\zeta_{1},\ldots,\zeta_{N_{s}})R^{+}_{ab}(\lambda_{a}-\lambda_{b})$

We also define the symmetric monodromy matrix for the mixed spin case
$T^{({\mbox{\boldmath$\ell$}})}_{0}(u;\zeta_{1},\ldots,\zeta_{r})$ as follows.

$T^{({\mbox{\boldmath$\ell$}})}_{0}(\lambda_{0};\zeta_{1},\ldots,\zeta_{r})=\prod_{k=1}^{r}\left(P^{(\ell_{k})}_{\ell_{1}+\cdots+\ell_{k-1}+1}\right)^{\bar{\chi}}\,\cdot\,R_{0,12\cdots
L}(\lambda_{0};\xi_{1}^{(\mbox{\boldmath$\ell$})},\ldots,\xi_{L}^{(\mbox{\boldmath$\ell$})})\,\cdot\,\prod_{k=1}^{r}\left(P^{(\ell_{k})}_{\ell_{1}+\cdots+\ell_{k-1}+1}\right)^{\bar{\chi}}$
(4.17)

###  4.3 Higher-spin $L$-operators

We now define the basis vectors of the $(\ell+1)$-dimensional irreducible
representation of $U_{q}(sl_{2})$, $||\ell,n\rangle$ for $n=0,1,\ldots,\ell$
as follows. We define $||\ell,0\rangle$ by

$||\ell,0\rangle=|1\rangle_{1}\otimes|1\rangle_{2}\otimes\cdots|1\rangle_{\ell}$
(4.18)

Here $|\alpha\rangle_{j}$ for $\alpha=1,2$ denote the basis vectors of the
spin-1/2 representation defined on the $j$th position in the tensor product.
We define $||\ell,n\rangle$ for $n\geq 1$ by

$||\ell,n\rangle=\left(\Delta^{(\ell-1)}(X^{-})\right)^{n}||\ell,0\rangle\,{\frac{1}{[n]_{q}!}}\,.$
(4.19)

Then we have

$||\ell,n\rangle=\sum_{1\leq
i_{1}<\cdots<i_{n}\leq\ell}\sigma_{i_{1}}^{-}\cdots\sigma_{i_{n}}^{-}|0\rangle\,q^{i_{1}+i_{2}+\cdots+i_{n}-n\ell+n(n-1)/2}$
(4.20)

It is easy to show the following:

$P^{(\ell)}_{12\cdots\ell}||\ell,n\rangle=||\ell,n\rangle$ (4.21)

We define the conjugate vectors by the following conditions:

$\langle\ell,n||P^{(\ell)}_{12\cdots\ell}=\langle\ell,n||$ (4.22)

with the normalization condition: $\langle\ell,n||\,||\ell,n\rangle=1$. Let us
define the $q$-factorial, $[n]_{q}!$, by

$[n]_{q}!=[n]_{q}[n-1]_{q}\cdots[1]_{q}\,.$ (4.23)

For integers $m$ and $n$ satisfying $m\geq n$ we define the $q$-binomial
coefficients as follows

$\left[\begin{array}[]{c}m\\\
n\end{array}\right]_{q}={\frac{[m]_{q}!}{[m-n]_{q}![n]_{q}!}}$ (4.24)

Then we have the following expression of the conjugate vectors

$\langle\ell,n||=\left[\begin{array}[]{c}\ell\\\
n\end{array}\right]_{q}^{-1}\,q^{n(\ell-n)}\,\sum_{1\leq
i_{1}<\cdots<i_{n}\leq\ell}\langle
0|\sigma_{i_{1}}^{+}\cdots\sigma_{i_{n}}^{+}\,q^{i_{1}+\cdots+i_{n}-n\ell+n(n-1)/2}$
(4.25)

The projection operators are given explicitly as follows.

$P^{(\ell)}_{12\cdots\ell}=\sum_{n=0}^{\ell}||\ell,n\rangle\,\langle\ell,n||$
(4.26)

We define the $L$-operator of the spin-$\ell/2$ XXZ model by

$L^{(\ell+)}(\lambda_{0})=P^{(\ell)}_{1}\,R^{+}_{0,12\cdots\ell}\,P^{(\ell)}_{1}\,.$
(4.27)

and then by the inverse gauge transformation we have

$L^{(\ell)}(\lambda_{0})=\left(P^{(\ell)}_{1}\right)^{\bar{\chi}}\,R_{0,12\cdots\ell}\,\left(P^{(\ell)}_{1}\right)^{\bar{\chi}}\,.$
(4.28)

Let us define $|\ell,n\rangle$ and their conjugates $\langle\ell,n|$ by

$\displaystyle|\ell,n\rangle$ $\displaystyle=$ $\displaystyle
N(\ell,n)\,||\ell,n\rangle$ $\displaystyle\langle\ell,n|$ $\displaystyle=$
$\displaystyle\langle\ell,n||\,{\frac{1}{N(\ell,n)}}\,.$ (4.29)

The matrix elements of the $L$-operator are given by

$\langle\ell,a|\,L^{(\ell)}(\lambda)\,|\ell,b\rangle=\left(\begin{array}[]{cc}\langle\ell,a|\,L^{(\ell)}_{11}(\lambda)\,|\ell,b\rangle&\langle\ell,a|\,L^{(\ell)}_{12}(\lambda)\,|\ell,b\rangle\\\
\langle\ell,a|\,L^{(\ell)}_{21}(\lambda)\,|\ell,b\rangle&\langle\ell,a|\,L^{(\ell)}_{22}(\lambda)\,|\ell,b\rangle\end{array}\right)_{[0]}$
(4.30)

for $a,b=0,1,\ldots,\ell$. Choosing the normalization factors $N(\ell,n)$, we
can derive the following symmetric expression of the $L$-operator:

$L^{(\ell)}(\lambda)={\frac{1}{2\sinh(u+\ell\eta/2)}}\left(\begin{array}[]{cc}zK^{1/2}-z^{-1}K^{-1/2}&2\sinh\eta
X^{-}\\\ 2\sinh\eta X^{+}&zK^{-1/2}-z^{-1}K^{1/2}\end{array}\right)_{[0]}$
(4.31)

Here $X^{\pm}$ and $K$ are in the $(\ell+1)$-dimensional representation of
$U_{q}(sl_{2})$, and $u=\lambda-\xi_{1}+\ell\eta/2$ and $z=\exp u$. Explicitly
they are given by

$\displaystyle\langle\ell,a|\,X^{+}\,|\ell,b\rangle$ $\displaystyle=$
$\displaystyle\delta_{a,b-1}\,[\ell-a]_{q}\,,$
$\displaystyle\langle\ell,a|\,X^{-}\,|\ell,b\rangle$ $\displaystyle=$
$\displaystyle\delta_{a,b+1}\,[a]_{q}\,,$
$\displaystyle\langle\ell,a|\,K\,|\ell,b\rangle$ $\displaystyle=$
$\displaystyle\delta_{a,b}\,q^{\ell-2a}\quad{\rm
for}\,\,a,b=0,1,\ldots,\ell,.$ (4.32)

###  4.4 Algebraic Bethe-ansatz method for higher-spin cases

We now discuss the eigenvalues of the transfer matrix of an integrable higher-
spin XXZ spin chain constructed by the fusion method. We consider the case of
mixed spins, where we define the transfer matrix on the tensor product of
spin-$s_{j}$ representations for $j=1,2,\ldots,r$.

We define $A$, $B$, $C$, and $D$ operators of the algebraic Bethe ansatz for
higher-spin cases by the following matrix elements of the monodromy matrix:

$\left(\begin{array}[]{cc}A^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{0};\zeta_{1},\ldots,\zeta_{r})&B^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{0};\zeta_{1},\ldots,\zeta_{r})\\\
C^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{0};\zeta_{1},\ldots,\zeta_{r})&D^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{0};\zeta_{1},\ldots,\zeta_{r})\end{array}\right)=T^{({\mbox{\boldmath$\ell$}}+)}_{0}(\lambda_{0};\zeta_{1},\ldots,\zeta_{r})\,.$
(4.33)

In terms of projection operators we have

$\displaystyle B^{({\mbox{\boldmath$\ell$}}+)}_{1\cdots
r}(\lambda_{0};\zeta_{1},\ldots,\zeta_{r})=\prod_{k=1}^{r}P^{(\ell_{k})}_{\ell(k-1)+1}\,\cdot\
B^{+}_{12\cdots
L}(\lambda_{0};\xi_{1}^{({\mbox{\boldmath$\ell$}})},\ldots,\xi_{L}^{({\mbox{\boldmath$\ell$}})})\prod_{k=1}^{r}P^{(\ell_{k})}_{\ell(k-1)+1}\,.$
(4.34)

We define $A^{({\mbox{\boldmath$\ell$}})},B^{({\mbox{\boldmath$\ell$}})}$,
$C^{({\mbox{\boldmath$\ell$}})}$ and $D^{({\mbox{\boldmath$\ell$}})}$
similarly for the monodromy matrix
$T^{({\mbox{\boldmath$\ell$}})}_{0}(\lambda_{0};\zeta_{1},\ldots,\zeta_{r})$.

The operators $A^{({\mbox{\boldmath$\ell$}}+)}$s of the asymmetric monodromy
matrix $R^{+}_{0,12\cdots n}$ are related to the symmetric ones
$A^{({\mbox{\boldmath$\ell$}})}$ s as follows.

$\displaystyle R^{+}_{0,12\cdots
n}(\lambda_{0},\\{\zeta_{i}\\})=\left(\begin{array}[]{cc}A^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{0};\\{\zeta_{i}\\})&B^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{0};\\{\zeta_{i}\\})\\\
C^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{0};\\{\zeta_{i}\\})&D^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{0};\\{\zeta_{i}\\})\end{array}\right)$
(4.37) $\displaystyle=$ $\displaystyle\left(\begin{array}[]{cc}\chi_{12\cdots
L}A^{({\mbox{\boldmath$\ell$}})}(\lambda_{0};\\{\zeta_{i}\\})\chi_{12\cdots
L}^{-1}&e^{-\lambda_{0}}\,\chi_{12\cdots
L}B^{({\mbox{\boldmath$\ell$}})}(\lambda_{0};\\{\zeta_{i}\\})\chi_{12\cdots
L}^{-1}\\\ e^{\lambda_{0}}\,\chi_{12\cdots
L}C^{({\mbox{\boldmath$\ell$}})}(\lambda_{0};\\{\zeta_{i}\\})\chi_{12\cdots
L}^{-1}&\chi_{12\cdots
L}D^{({\mbox{\boldmath$\ell$}})}(\lambda_{0};\\{\zeta_{i}\\})\chi_{12\cdots
L}^{-1}\end{array}\right)$ (4.40)

It follows from the Yang-Baxter equations (4.16) that the $A,B,C,D$ operators
in the higher-spin case also satisfy the standard commutation relations.

$A^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{1})B^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{2})={\frac{1}{b(\lambda_{2}-\lambda_{1})}}B^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{2})A^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{1})-{\frac{c^{-}(\lambda_{2}-\lambda_{1})}{b(\lambda_{2}-\lambda_{1})}}B^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{1})A^{({\mbox{\boldmath$\ell$}}+)}(\lambda_{2})$
(4.41)

Through the inverse gauge transformation ${\bar{\chi}}$ we have

$A^{({\mbox{\boldmath$\ell$}})}(\lambda_{1})B^{({\mbox{\boldmath$\ell$}})}(\lambda_{2})={\frac{1}{b(\lambda_{2}-\lambda_{1})}}B^{({\mbox{\boldmath$\ell$}})}(\lambda_{2})A^{({\mbox{\boldmath$\ell$}})}(\lambda_{1})-{\frac{c(\lambda_{2}-\lambda_{1})}{b(\lambda_{2}-\lambda_{1})}}B^{({\mbox{\boldmath$\ell$}})}(\lambda_{1})A^{({\mbox{\boldmath$\ell$}})}(\lambda_{2})$
(4.42)

Therefore, we derive Bethe ansatz eigenvectors of the higher-spin transfer
matrix by the same method as the case of spin-1/2.

Let us denote by $|0\rangle$ the vacuum state where all spins are up. Noting

$\prod_{k=1}^{r}P^{(\ell_{k})}_{\ell(k-1)+1}|0\rangle=|0\rangle\,,$ (4.43)

it is easy to show the following relations:

$\displaystyle A^{({\mbox{\boldmath$\ell$}})}(\lambda)|0\rangle$
$\displaystyle=$ $\displaystyle
a^{({\mbox{\boldmath$\ell$}})}(\lambda;\\{\zeta_{k}\\})|0\rangle\,,$
$\displaystyle D^{({\mbox{\boldmath$\ell$}})}(\lambda)|0\rangle$
$\displaystyle=$ $\displaystyle
d^{({\mbox{\boldmath$\ell$}})}(\lambda;\\{\zeta_{k}\\})|0\rangle\,,$ (4.44)

where $a^{({\mbox{\boldmath$\ell$}})}(\lambda;\\{\zeta_{k}\\})$ and
$d^{({\mbox{\boldmath$\ell$}})}(\lambda;\\{\zeta_{k}\\})$ are given by

$\displaystyle a^{({\mbox{\boldmath$\ell$}})}(\lambda;\\{\zeta_{k}\\})$
$\displaystyle=$ $\displaystyle
a^{({\mbox{\boldmath$\ell$}})}(\lambda;\\{\xi_{j}^{({\mbox{\boldmath$\ell$}})}\\})=1\,,$
$\displaystyle d^{({\mbox{\boldmath$\ell$}})}(\lambda;\\{\zeta_{k}\\})$
$\displaystyle=$ $\displaystyle
d^{({\mbox{\boldmath$\ell$}})}(\lambda;\\{\xi_{j}^{({\mbox{\boldmath$\ell$}})}\\})=\prod_{j=1}^{L}b(\lambda-\xi_{j}^{({\mbox{\boldmath$\ell$}})})=\prod_{k=1}^{r}{\frac{\sinh(\lambda-\zeta_{k}-(\ell_{k}-1)\eta/2)}{\sinh(\lambda-\zeta_{k}+(\ell_{k}+1)\eta/2)}}$

Thus, the vector $B^{{(\mbox{\boldmath$\ell$})}}(\lambda_{1})\cdots
B^{(\mbox{\boldmath$\ell$})}(\lambda_{n})|0\rangle$ becomes an eigenvector of
the transfer matrix
$A^{({\mbox{\boldmath$\ell$}})}(\lambda)+D^{({\mbox{\boldmath$\ell$}})}(\lambda)$
with the following eigenvalue

$\tau^{({\mbox{\boldmath$\ell$}})}(\mu)=\prod_{j=1}^{n}{\frac{\sinh(\lambda_{j}-\mu+\eta)}{\sinh(\lambda_{j}-\mu)}}+\prod_{k=1}^{r}{\frac{\sinh(\mu-\zeta_{k}-(\ell_{k}-1)\eta/2)}{\sinh(\mu-\zeta_{k}+(\ell_{k}+1)\eta/2)}}\,\cdot\,\prod_{j=1}^{n}{\frac{\sinh(\mu-\lambda_{j}+\eta)}{\sinh(\mu-\lambda_{j})}}$
(4.46)

if rapidities $\tilde{\lambda}_{j}=\lambda_{j}+\eta/2$ satisfy the Bethe
ansatz equations

$\prod_{k=1}^{r}{\frac{\sinh(\tilde{\lambda}_{\alpha}-\zeta_{k}+\ell_{k}\eta/2)}{\sinh(\tilde{\lambda}_{\alpha}-\zeta_{k}-\ell_{k}\eta/2)}}=\prod_{\beta=1;\beta\neq\alpha}^{n}{\frac{\sinh(\tilde{\lambda}_{\alpha}-\tilde{\lambda}_{\beta}+\eta)}{\sinh(\tilde{\lambda}_{\alpha}-\tilde{\lambda}_{\beta}-\eta)}}$
(4.47)

##  5 Pseudo-diagonalization of the $B$ and $C$ operators

###  5.1 Diagonalizing the $A$ and $D$ operators

####  5.1.1 The $F$-basis

In order to formulate the derivation of the pseudo-diagonalized forms of $B$
and $C$ operators for the XXZ case, we briefly formulate some symbols and
review some useful formulas shown in Ref. [6] in SS5.1. First, we introduce
the $F$-basis .

######  Definition 15.

(Partial $F$ and total $F$) We define partial $F$ by

$\displaystyle F_{1,\,2\cdots n}$ $\displaystyle=$ $\displaystyle
e^{11}_{1}+e_{1}^{22}R_{1,\,2\cdots n}$ $\displaystyle F_{12\cdots n-1,\,n}$
$\displaystyle=$ $\displaystyle e^{22}_{n}+e_{n}^{11}R_{12\cdots n-1,\,n}$
(5.1)

We define total $F$ recursively with respect to $n$ by

$F_{12\cdots n}=F_{12\cdots n-1}F_{12\cdots n-1,\,n}$ (5.2)

######  Lemma 16.

(Cocycle conditions)

$\displaystyle F_{1,\,2}F_{12,\,3}$ $\displaystyle=$ $\displaystyle
F_{23}F_{1,\,23}$ $\displaystyle F_{1,\,2\cdots n-1}F_{12\cdots n-1\,n}$
$\displaystyle=$ $\displaystyle F_{2\cdots n-1,\,n}F_{1,\,2\cdots n}$ (5.3)

###### Proof.

Expressing the $F$-basis in terms of $R$-matrices through (5.1), we show that
the cocycle conditions of the $F$-basis are reduced to those of the
$R$-matrices, which are shown in Appendix B. ∎

From the cocycle conditions we have the following:

######  Lemma 17.

$F_{12\cdots n}=F_{2\cdots n}F_{1,2\cdots n}$ (5.4)

####  5.1.2 Basic properties of the $R$-matrix

Let us introduce some important properties of the $R$-matrix of the XXZ spin-
chain.

The $R$-matrix is invariant under the charge conjugation. For the symmetric
$R$-matrix, we define the charge conjugation operator ${\cal C}$ by

${\cal C}_{12\cdots n}=\sigma^{x}_{1}\cdots\sigma_{n}^{x}$ (5.5)

For a given operator $A\in End(V(\lambda_{1})\otimes\cdots\otimes
V(\lambda_{n}))$ we define ${\bar{A}}$ by ${\bar{A}}={\cal C}_{1\cdots
n}A{\cal C}_{1\cdots n}$. For instance, we define ${\bar{F}}_{0,1\cdots n}$ by

${\bar{F}}_{0,1\cdots n}={\cal C}_{01\cdots n}F_{0,1\cdots n}{\cal
C}_{01\cdots n}$ (5.6)

######  Proposition 18.

The charge conjugation operator ${\cal C}$ commutes with the monodromy matrix
of the symmetric $R$-matrix:

${[}{\cal C}_{01\cdots n},\,R_{0,1\cdots n}{]}=0$ (5.7)

We thus have ${\bar{A}}_{1\cdots n}(\lambda_{0})=D_{1\cdots n}(\lambda_{0})$
and ${\bar{B}}_{1\cdots n}(\lambda_{0})=C_{1\cdots n}(\lambda_{0})$.

######  Lemma 19 (Crossing symmetry).

The $R$-matrix has the crossing symmetry relation:

$\left(\gamma\otimes
I\right)R_{12}(\lambda_{1}-\eta,\lambda_{2})\left(\gamma\otimes
I\right)=b_{21}^{-1}R_{21}^{t_{1}}(\lambda_{2},\lambda_{1})$ (5.8)

where $b_{21}=b(\lambda_{2}-\lambda_{1})$ and $\gamma$ is given by

$\gamma=\sigma^{y}=\left(\begin{array}[]{cc}0&-i\\\ i&0\end{array}\right)$
(5.9)

Here the crossing symmetry is slightly different from [6].

######  Lemma 20.

(Crossing symmetry of the monodromy matrix)

$\gamma_{0}R_{0,1\cdots
n}(\lambda_{0}-\eta;\xi_{1},\ldots,\xi_{n})\gamma_{0}=\left(\prod_{i=1}^{n}b^{-1}(\xi_{i}-\lambda_{0})\right)R^{t_{0}}_{1\cdots
n,0}(\lambda_{0};\xi_{1},\ldots,\xi_{n})$ (5.10)

Let us introduce $\dagger$ operation. We shall use it when we pseudo-
diagonalize the $B$ operators.

######  Definition 21.

For $X_{1\cdots n}(\lambda_{1},\cdots,\lambda_{n})\in
End(V(\lambda_{1})\otimes\cdots\otimes V(\lambda_{n}))$ we define
$X^{\dagger}_{1\cdots n}$ by

$X^{\dagger}_{1\cdots n}(\lambda_{1},\ldots,\lambda_{n})=X^{t_{1}\cdots
t_{n}}_{1\cdots n}(-\lambda_{1},\cdots,-\lambda_{n})$ (5.11)

Here we note $(X^{\dagger})^{\dagger}=X$, and
$(XY)^{\dagger}=Y^{\dagger}X^{\dagger}$. It is easy to show
$R_{12}^{\dagger}=R_{21}$.

######  Lemma 22.

Under the $\dagger$ operation the monodromy matrix is given by the following:

$R^{\dagger}_{0,1\cdots n}=R_{0,1\cdots n}^{-1}=R_{1\cdots n,\,0}\\\ $ (5.12)

We define operators $A^{\dagger}_{1\cdots n}$, $B^{\dagger}_{1\cdots n}$,
$C^{\dagger}_{1\cdots n}$ and $D^{\dagger}_{1\cdots n}$ by

$R^{\dagger}_{0,1\cdots n}=\left(\begin{array}[]{cc}A^{\dagger}_{1\cdots
n}(\lambda_{0})&C^{\dagger}_{1\cdots n}(\lambda_{0})\\\ B^{\dagger}_{1\cdots
n}(\lambda_{0})&D^{\dagger}_{1\cdots n}(\lambda_{0})\end{array}\right)_{[0]}$
(5.13)

######  Proposition 23.

Under the $\dagger$ operation the monodromy matrix is given by

$\displaystyle R_{0,12\cdots
n}^{\dagger}=\left(\begin{array}[]{cc}A^{\dagger}_{1\cdots
n}(\lambda_{0})&C^{\dagger}_{1\cdots n}(\lambda_{0})\\\ B^{\dagger}_{1\cdots
n}(\lambda_{0})&D^{\dagger}_{1\cdots n}(\lambda_{0})\end{array}\right)_{[0]}$
(5.16)
$\displaystyle\quad=\left(\prod_{i=1}^{n}b(\xi_{i}-\lambda_{0})\right)\left(\begin{array}[]{cc}A_{1\cdots
n}(\lambda_{0}-\eta)&-B_{1\cdots n}(\lambda_{0}-\eta)\\\ -C_{1\cdots
n}(\lambda_{0}-\eta)&D_{1\cdots n}(\lambda_{0}-\eta)\end{array}\right)_{[0]}$
(5.19)

####  5.1.3 The diagonalized forms of operators $A$ and $D$

Let us give the diagonalized forms of the $A$ and $D$ operators [6].

The following criterion for the $F$-basis to be non-singular should be useful.

######  Proposition 24.

The determinants of the partial and total $F$ matrices are given by

${\rm det}F_{0,1\cdots n}=\prod_{j=1}^{n}b(\lambda_{0}-\xi_{j})\,,\quad{\rm
det}F_{1\cdots n}=\prod_{1\leq i<j\leq n}b(\xi_{i}-\xi_{j})$ (5.20)

Proposition 5.20 follows from lemma D.1 of Appendix D.

We can show the diagonalized forms of operators $A$ and $D$ as follows [6].

######  Proposition 25 (Diagonalization of $A$ and $D$).

$\displaystyle F_{1\cdots n}D_{1\cdots n}(\lambda_{0})F_{1\cdots n}^{-1}$
$\displaystyle=$
$\displaystyle\bigotimes_{i=1}^{n}\left(\begin{array}[]{cc}b_{0i}&0\\\
0&1\end{array}\right)_{[i]}$ (5.23) $\displaystyle{\bar{F}}_{1\cdots
n}A_{1\cdots n}(\lambda_{0}){\bar{F}}_{1\cdots n}^{-1}$ $\displaystyle=$
$\displaystyle\bigotimes_{i=1}^{n}\left(\begin{array}[]{cc}1&0\\\
0&b_{0i}\end{array}\right)_{[i]}\,,$ (5.26)

where $b_{0i}=b(\lambda_{0}-\xi_{i})$

######  Proposition 26 (Diagonalization of $A^{\dagger}$ and $D^{\dagger}$).

$\displaystyle F_{1\cdots n}A^{\dagger}_{1\cdots n}(\lambda_{0})F_{1\cdots
n}^{-1}$ $\displaystyle=$
$\displaystyle\bigotimes_{i=1}^{n}\left(\begin{array}[]{cc}1&0\\\
0&b_{i0}\end{array}\right)_{[i]}$ (5.29) $\displaystyle{\bar{F}}_{1\cdots
n}D^{\dagger}_{1\cdots n}(\lambda_{0}){\bar{F}}_{1\cdots n}^{-1}$
$\displaystyle=$
$\displaystyle\bigotimes_{i=1}^{n}\left(\begin{array}[]{cc}b_{i0}&0\\\
0&1\end{array}\right)_{[i]}$ (5.32)

where $b_{i0}=b(\xi_{i}-\lambda_{0})$

The derivation of the diagonalized forms and some useful formulas are briefly
reviewed in Appendix D.

For a given operator $X_{1\cdots n}(\lambda_{1},\cdots,\lambda_{n})\in
End(V(\lambda_{1})\otimes\cdots\otimes V(\lambda_{n}))$ we denote $FAF^{-1}$
by ${\widetilde{F}}$:

${\widetilde{X}}_{1\cdots n}=F_{12\cdots n}X_{12\cdots n}F_{12\cdots
n}^{-1}\,.$ (5.33)

For instance we have ${\widetilde{D}}_{1\cdots n}(\lambda_{0})={F}_{1\cdots
n}D_{1\cdots n}(\lambda_{0}){F}_{1\cdots n}^{-1}$.

###  5.2 Pseudo-diagonalization of the $B$ operator

Let us recall that the matrix elements of the monodromy matrix
$R^{+}_{0,1\cdots L}$ are related to the symmetric ones as follows.

$\displaystyle R^{+}_{0,12\cdots
L}(u;\xi_{1},\ldots,\xi_{L})=\left(\begin{array}[]{cc}A^{+}_{12\cdots
L}(u;\xi_{1},\ldots,\xi_{L})&B^{+}_{12\cdots L}(u;\xi_{1},\ldots,\xi_{L})\\\
C^{+}_{12\cdots L}(u;\xi_{1},\ldots,\xi_{L})&D^{+}_{12\cdots
L}(u;\xi_{1},\ldots,\xi_{L})\end{array}\right)_{[0]}$ (5.36) $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}\chi_{12\cdots L}A_{12\cdots
L}(u;\\{\xi_{j}\\})\left(\chi_{12\cdots
L}\right)^{-1}&e^{-\lambda_{0}}\chi_{12\cdots L}B_{12\cdots
L}(u;\\{\xi_{j}\\})\left(\chi_{12\cdots L}\right)^{-1}\\\
e^{\lambda_{0}}\chi_{12\cdots L}C_{12\cdots
L}(u;\\{\xi_{j}\\})\left(\chi_{12\cdots L}\right)^{-1}&\chi_{12\cdots
L}D_{12\cdots L}(u;\\{\xi_{j}\\})\left(\chi_{12\cdots
L}\right)^{-1}\end{array}\right)_{[0]}$ (5.39)

Then, from the quantum-group invariance (3.10) we have the following
commutation relations:

$\displaystyle B^{+}_{1\cdots n}(\lambda)$ $\displaystyle=$ $\displaystyle
D^{+}_{1\cdots
n}(\lambda)\Delta^{(n-1)}(X^{-})-q\Delta^{(n-1)}(X^{-})D^{+}_{1\cdots
n}(\lambda)$ (5.41) $\displaystyle C^{+}_{1\cdots n}(\lambda)$
$\displaystyle=$ $\displaystyle\Delta^{(n-1)}(X^{+})D^{+}_{1\cdots
n}(\lambda)-q^{-1}D^{+}_{1\cdots n}(\lambda)\Delta^{(n-1)}(X^{+})$ (5.42)

Here $X^{\pm}$ are generators of $U_{q}(sl_{2})$, and $\Delta^{(n-1)}(X^{-})$
denote the tensor-product representation of $\Delta^{(n-1)}(X^{-})$ acting on
the $n$ sites from the 1st to $n$th. We remark that more generally, we have
commutation relations (3.39) for the affine quantum group
$U_{q}({\widehat{sl_{2}}})$.

In this subsection we abbreviate the superscript $+$ for the asymmetric
monodromy matrix, for simplicity. In fact, the essential parts of formulas
such as the fundamental commutation relations are invariant under gauge
transformations if we express them in terms of the generators of the quantum
affine algebra $U_{q}(\widehat{sl_{2}})$ in the evaluation representation
(3.28). Here we remark that the matrix representation of the evaluation
representation of $U_{q}(\widehat{sl_{2}})$ can be changed through gauge
transformations.

Let us now introduce some symbols.

######  Definition 27.

We define operators ${\widehat{\delta}}_{jk}(\lambda_{j},\lambda_{k})$ for
$j,k$ satisfying $0\leq j<k\leq L$ by

${\widehat{\delta}}_{jk}(\lambda_{j},\lambda_{k})=\left(\begin{array}[]{cccc}1&0&0&0\\\
0&b^{-1}_{kj}&0&0\\\ 1&0&b^{-1}_{jk}&0\\\ 1&0&0&1\\\
\end{array}\right)_{[jk]}\,,$ (5.43)

where $b_{jk}=b(\lambda_{j}-\lambda_{k})$ and
$b_{kj}=b(\lambda_{k}-\lambda_{j})$. We define ${\widehat{\delta}}_{1\cdots
n}$ and ${\widehat{\delta}}_{0,1\cdots n}$ by

$\displaystyle{\widehat{\delta}}_{1\cdots n}$ $\displaystyle=$
$\displaystyle\prod_{1\leq j<k\leq
n}{\widehat{\delta}}_{jk}(\lambda_{j},\lambda_{k})$
$\displaystyle{\widehat{\delta}}_{0,1\cdots n}$ $\displaystyle=$
$\displaystyle{\widehat{\delta}}_{01\cdots n}{\widehat{\delta}}^{-1}_{1\cdots
n}=\prod_{j=1}^{n}{\widehat{\delta}}_{0j}(\lambda_{0},\lambda_{j})$ (5.44)

We define ${\widehat{\delta}}^{1\cdots n}_{i}$ by

${\widehat{\delta}}^{1\cdots n}_{i}=\widehat{\delta}_{i,i+1\cdots n1\cdots
i-1}=\prod_{j=1;j\neq i}^{n}{\widehat{\delta}}_{ij}$ (5.45)

Some useful formulas are given in Appendix D.

Let us denote $I^{\otimes m}\otimes\Delta^{(\ell-1)}(x)\otimes
I^{\otimes(n-\ell-m)}$ by $\Delta^{(\ell-1)}_{m+1\,m+2\,\cdots\,m+\ell}(x)$ or
$\Delta_{m+1\,m+2\,\cdots\,m+\ell}(x)$ for $x\in U_{q}(sl_{2})$ in the tensor-
product representation.

######  Lemma 28.

Let $X^{-}$ denote the generator of the quantum group $U_{q}(sl_{2})$ and
$X_{j}^{-}$ the spin-1/2 representation of $X^{-}$ acting on the $j$th site in
the tensor product representation $(V^{(1)})^{\otimes n}$. We have

$\widetilde{\Delta}_{1\cdots
n}(X^{-})=\left(X_{1}^{-}+e_{1}^{11}\widetilde{\Delta}_{2\cdots
n}(X^{-})\widetilde{A^{\dagger}}_{2\cdots
n}(\xi_{1})+e_{1}^{22}\widetilde{D}_{2\cdots
n}(\xi_{1})\widetilde{\Delta}_{2\cdots
n}(X^{-})\right)\widehat{\delta}_{1,2\cdots n}$ (5.46)

###### Proof.

Making use of (D.39) we show $F^{-1}_{1\cdots n}=F^{t_{1}\cdots
t_{n}}_{n\cdots 2,1}F_{2\cdots n}^{-1}\widehat{\delta}_{1,2\cdots n}$. We have

$\displaystyle\widetilde{\Delta}_{1\cdots n}(X^{-})$ $\displaystyle=$
$\displaystyle F_{2\cdots n}F_{1,2\cdots
n}{\Delta}^{(n-1)}(X^{-})F^{t_{1}\cdots t_{n}}_{n\cdots 2,1}F_{2\cdots
n}^{-1}\widehat{\delta}_{1,2\cdots n}$ (5.47)

Putting $F_{1,2\cdots n}=e_{1}^{11}+e_{1}^{22}R_{1,2\cdots n}$ and
$F^{t_{1}\cdots t_{n}}_{n\cdots 2,1}=e_{1}^{22}+R_{2\cdots n,1}e_{1}^{11}$, we
have

$\displaystyle F_{1,2\cdots n}{\Delta}^{(n-1)}(X^{-})F^{t_{1}\cdots
t_{n}}_{n\cdots 2,1}=\left(e_{1}^{11}+e_{1}^{22}R_{1,2\cdots
n}\right)\,{\Delta}^{(n-1)}(X^{-})\,\left(e_{1}^{22}+R_{2\cdots
n,1}e_{1}^{11}\right)$
$\displaystyle=e_{1}^{11}{\Delta}^{(n-1)}(X^{-})e_{1}^{22}+e_{1}^{11}{\Delta}^{(n-1)}(X^{-})R_{2\cdots
n,1}e_{1}^{11}$ $\displaystyle\quad+e_{1}^{22}R_{1,2\cdots
n}{\Delta}^{(n-1)}(X^{-})e_{1}^{22}+e_{1}^{22}R_{1,2\cdots
n}{\Delta}^{(n-1)}(X^{-})R_{2\cdots n,1}e_{1}^{11}$
$\displaystyle=0+e_{1}^{11}{\Delta}_{2\cdots n}(X^{-})R_{2\cdots
n,1}e_{1}^{11}+e_{1}^{22}R_{1,2\cdots n}{\Delta}_{2\cdots
n}(X^{-})e_{1}^{22}+X_{1}^{-}\,.$ (5.48)

Here we have made use of the following:

${\Delta}^{(n-1)}(X^{-})={\Delta}^{(n-2)}\Delta(X^{-})=X^{-}\otimes\Delta^{(n-2)}(K^{-1})+I\otimes\Delta^{(n-2)}(X^{-})\,.$
(5.49)

We thus have

$\widetilde{\Delta}_{1\cdots
n}(X^{-})=\left(X_{1}^{-}+e_{1}^{11}\widetilde{\Delta}_{2\cdots
n}(X^{-})\widetilde{R}_{2\cdots
n,1}e_{1}^{11}+e_{1}^{22}\widetilde{R}_{1,2\cdots
n}\widetilde{\Delta}_{2\cdots
n}(X^{-})e_{1}^{11}\right)\widehat{\delta}_{1,2\cdots n}\,.$ (5.50)

We obtain the case of $n$ from (5.12). ∎

######  Lemma 29.

In the tensor-product representation $(V^{(1)})^{\otimes n}$ we have

${\widetilde{\Delta}}_{1\cdots
n}(X^{-})=\sum_{i=1}^{n}X_{i}^{-}{\widehat{\delta}}^{1\cdots n}_{i}\,.$ (5.51)

###### Proof.

We show it by induction on $n$. The case of $n=1$ is trivial. Let us assume
the case of $n-1$. In eq. (5.46), the first term gives the following:
$X_{1}^{-}\widehat{\delta}_{1,2\cdots
n}=X_{1}^{-}\widehat{\delta}_{1}^{1\cdots n}$. Assuming (5.51) for
$\widehat{\Delta}_{2\cdots n}$ and putting it into the second term of (5.46),
we have

$\displaystyle e_{1}^{11}\widetilde{\Delta}_{2\cdots
n}(X^{-})\widetilde{A^{\dagger}}_{2\cdots
n}(\xi_{1})\widehat{\delta}_{1,2\cdots n}$ $\displaystyle=$ $\displaystyle
e_{1}^{11}\sum_{i=2}^{n}X_{i}^{-}\widehat{\delta}_{i}^{2\cdots
n}\bigotimes_{k=2}^{n}\left(\begin{array}[]{cc}1&0\\\
0&b_{k1}\end{array}\right)_{[k]}\widehat{\delta}_{1,2\cdots n}$
$\displaystyle=$
$\displaystyle\sum_{i=2}^{n}X_{i}^{-}\widehat{\delta}_{i}^{1\cdots
n}e_{1}^{11}$ $\displaystyle=$ $\displaystyle
e_{1}^{11}\sum_{i=2}^{n}X_{i}^{-}\widehat{\delta}_{i}^{1\cdots n}$ (5.55)

Here we have made use of (5.29). Similarly, we have

$e_{1}^{22}\widetilde{D}_{2\cdots n}(\xi_{1})\widetilde{\Delta}_{2\cdots
n}(X^{-})e_{1}^{22}\widehat{\delta}_{1,2\cdots
n}=e_{1}^{22}\sum_{i=2}^{n}X_{i}^{-}\widehat{\delta}_{i}^{1\cdots n}$ (5.56)

Thus, we have the case of $n$ as follows.

$\displaystyle{\widetilde{\Delta}}_{1\cdots n}(X^{-})$ $\displaystyle=$
$\displaystyle X_{1}^{-}{\widehat{\delta}}_{1}^{1\cdots
n}+(e_{1}^{11}+e_{1}^{22})\sum_{i=2}^{n}X_{1}^{-}{\widehat{\delta}}_{i}^{1\cdots
n}$ (5.57) $\displaystyle=$
$\displaystyle\sum_{i=1}^{n}X_{i}^{-}{\widehat{\delta}}_{i}^{1\cdots n}\,.$

∎

From the fundamental commutation relation (5.41) we have the following:

######  Lemma 30.

In the tensor product $(V^{(1)})^{\otimes n}$ we have

${\widetilde{B}}_{1\cdots
n}(\lambda)=\sum_{i=1}^{n}X_{i}^{-}({\widetilde{D}}_{1\cdots i-1,i+1\cdots
n}(\lambda)-q{\widetilde{D}}_{1\cdots
n}(\lambda))\widehat{\delta}_{i}^{1\cdots n}\,.$ (5.58)

###### Proof.

We transform the both sides of the fundamental commutation relation (5.41) by
$F_{1\cdots n}$, and put (5.23) and (5.51) into it, we have the following:

$\displaystyle{\widetilde{B}}_{1\cdots n}(\lambda)$ $\displaystyle=$
$\displaystyle{\widetilde{D}}_{1\cdots
n}(\lambda){\widetilde{\Delta}}_{1\cdots
n}(X^{-})-q{\widetilde{\Delta}_{1\cdots n}}(X^{-}){\widetilde{D}}_{1\cdots
n}(\lambda)$ (5.63) $\displaystyle=$
$\displaystyle\bigotimes_{j=1}^{n}\left(\begin{array}[]{cc}b_{0j}&0\\\
0&1\end{array}\right)_{[j]}\sum_{i=1}^{n}X_{i}^{-}{\widehat{\delta}}^{1\cdots
n}_{i}-q\sum_{i=1}^{n}X_{i}^{-}{\widehat{\delta}}^{1\cdots
n}_{i}\bigotimes_{j=1}^{n}\left(\begin{array}[]{cc}b_{0j}&0\\\
0&1\end{array}\right)_{[j]}$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{n}X_{i}^{-}({\widetilde{D}}_{1\cdots i-1,i+1\cdots
n}(\lambda)-q{\widetilde{D}}_{1\cdots
n}(\lambda))\widehat{\delta}_{i}^{1\cdots n}\,.$ (5.64)

∎

######  Proposition 31 (Pseudo-diagonalization of $B$ operator).

We have

${\widetilde{B}}_{1\cdots
n}(\lambda)=\sum_{i=1}^{n}c_{0i}^{-}\,X^{-}_{i}\bigotimes_{j=1;j\neq
i}^{n}\left(\begin{array}[]{cc}b_{0j}&0\\\
0&b_{ji}^{-1}\end{array}\right)_{[j]}\,,$ (5.65)

where $b_{0i}=b(\lambda_{0}-\xi_{i})$, $b_{ji}=b(\xi_{j}-\xi_{i})$ and
$c_{0i}^{-}=c^{-}(\lambda_{0}-\xi_{i})=\exp(-(\lambda_{0}-\xi_{i}))c(\lambda_{0}-\xi_{i})$.

###### Proof.

Let us denote $b_{0i}=\sinh(\lambda-\xi_{i})/\sinh(\lambda-\xi_{i}+\eta)$ and
$c_{0i}=\sinh(\eta)/\sinh(\lambda-\xi_{i}+\eta)$, by $b_{0i}$ and $c_{0i}$
respectively, . Putting $1-qb_{01}=c_{0i}^{-}$ in (5.58) we show

${\widetilde{B}}_{1\cdots
n}(\lambda)=\sum_{i=1}^{n}c_{0i}^{-}X_{i}^{-}{\widetilde{D}}_{1\cdots
i-1\,i+1\cdots n}{\widehat{\delta}}_{i}^{1\cdots n}\,.$ (5.66)

After some calculation, we have (5.65). ∎

Similarly, making use of lemmas E.1, E.2 and E.3, we can show the diagonalized
form of operator $C$.

######  Proposition 32.

Let $X_{i}^{+}$ denote the spin-1/2 representation of $X^{+}$ acting on the
$i$th site in the tensor product representation. We have

${\widetilde{C}}_{1\cdots
n}(\lambda_{0})=\sum_{i=1}^{n}c_{0i}^{+}\,X^{+}_{i}\bigotimes_{j=1;j\neq
i}^{n}\left(\begin{array}[]{cc}b_{0j}b_{ij}^{-1}&0\\\
0&1\end{array}\right)_{[j]}\,,$ (5.67)

where $b_{0i}=b(\lambda_{0}-\xi_{i})$, $b_{ji}=b(\xi_{j}-\xi_{i})$ and
$c_{0i}^{+}=c^{+}(\lambda_{0}-\xi_{i})$.

###  5.3 Pseudo-diagonalized forms of the symmetric $B$ and $C$ operators

Let us show the pseudo-diagonalized forms of the $B$ and $C$ operators of the
symmetric monodromy matrix $R_{0,1\cdots n}$. Here we recall that expressions
(5.65) and (5.67) are for $\widetilde{B}^{+}_{12\cdots n}(\lambda)$ and
$\widetilde{C}^{+}_{12\cdots n}(\lambda)$, respectively. They are matrix
elements of the asymmetric monodromy matrix $R^{+}_{0,1\cdots
n}=\chi_{01\cdots n}R_{0,1\cdots n}\chi^{-1}_{01\cdots n}$. We have the
following relations:

$\displaystyle B^{+}_{12\cdots n}(\lambda)$ $\displaystyle=$ $\displaystyle
e^{-\lambda}\chi_{01\cdots n}R_{0,1\cdots n}\chi^{-1}_{01\cdots n}$
$\displaystyle C^{+}_{12\cdots n}(\lambda)$ $\displaystyle=$ $\displaystyle
e^{\lambda}\chi_{01\cdots n}R_{0,1\cdots n}\chi^{-1}_{01\cdots n}$ (5.68)

Therefore, applying the inverse gauge transformation ${\bar{\chi}}$ to (5.65)
and (5.67), we obtain

${\widetilde{B}}_{1\cdots
n}(\lambda)=\sum_{i=1}^{n}c_{0i}\,\sigma^{-}_{i}\bigotimes_{j=1;j\neq
i}^{n}\left(\begin{array}[]{cc}b_{0j}&0\\\
0&b_{ji}^{-1}\end{array}\right)_{[j]}\,,$ (5.69)

and

${\widetilde{C}}_{1\cdots
n}(\lambda_{0})=\sum_{i=1}^{n}c_{0i}\,\sigma^{+}_{i}\bigotimes_{j=1;j\neq
i}^{n}\left(\begin{array}[]{cc}b_{0j}b_{ij}^{-1}&0\\\
0&1\end{array}\right)_{[j]}\,.$ (5.70)

Here we recall $c_{0i}=\sinh(\eta)/\sinh(\lambda-\xi_{i}+\eta)$.

We should remark that expressions (5.69) and (5.70) coincide with eq. (2.29)
and (2.30) of Ref. [7], respectively.

##  6 Scalar products formulas

###  6.1 Formula for higher-spin scalar products

Let us consider the case of tensor product of spin-$s$ representations. We
recall that $\ell=2s$ and $L=\ell N_{s}$. We introduce parameters
$\xi_{j}^{(\ell;\epsilon)}$ for $j=1,2,\ldots,L$, as follows:

$\xi_{(k-1)\ell+j}^{(\ell;\epsilon)}=\zeta_{k}-(j-1)\eta+\ell\eta/2+\epsilon
r_{j}\qquad j=1,\ldots\ell;k=1,\ldots,N_{s}\,.$ (6.1)

Here $r_{j}$ $(j=1,2,\ldots,\ell)$ are distinct and nonzero parameters, and
$\epsilon$ is an arbitrary small number. We also introduce the following
symbol:

$P^{(\ell)}_{1\cdots L}=\prod_{j=1}^{N_{s}}P^{(\ell)}_{(j-1)\ell+1}\,,\quad
P^{(\ell)\,{\bar{\chi}}}_{1\cdots
L}=\prod_{j=1}^{N_{s}}\left(P^{(\ell)}_{(j-1)\ell+1}\right)^{\bar{\chi}}$
(6.2)

Here we recall that $B$ operator acting on the tensor product of spin-$s$
representations, $\left(V^{(2s)}\right)^{\otimes N_{s}}$, is given by $B$
operator acting on the tensor product of spin-1/2 representations
$\left(V^{(1)}\right)^{\otimes L}$ with $L=N_{s}\ell$ and multiplied by the
projection operators:

$B^{(\ell)}_{1\cdots N_{s}}(u;\zeta_{1},\ldots,\zeta_{N_{s}})=P_{1\cdots
L}^{(\ell)}\,B_{1\cdots
L}^{(1)}(u;\xi_{1}^{(\ell)},\ldots,\xi_{L}^{(\ell)})\,P_{1\cdots
L}^{(\ell)}\,.$

We now define the scalar product for the spin-$\ell/2$ case as follows.

######  Definition 33.

Let $\\{\lambda_{\alpha}\\}$ ($\alpha=1,2,\ldots,n$) be a set of solutions of
the Bethe ansatz equations and $\\{\mu_{j}\\}$ ($j=1,2,\ldots,n$) be arbitrary
numbers. We define the scalar product
$S^{(\ell)}_{n}(\\{\mu_{j}\\},\\{\lambda_{\alpha}\\};\\{\zeta_{k}\\})$ by the
following:

$S^{(\ell)}_{n}(\\{\mu_{j}\\},\\{\lambda_{\alpha}\\};\\{\zeta_{k}\\})=\langle
0|\,C^{(\ell)}(\mu_{1})\cdots C^{(\ell)}(\mu_{n})B^{(\ell)}(\lambda_{1})\cdots
B^{(\ell)}(\lambda_{n})\,|0\rangle$ (6.3)

Here $C^{(\ell)}(\mu_{j})$ and $B^{(\ell)}(\lambda_{\alpha})$ abbreviate
$C^{(\ell)}_{1\cdots N_{s}}(\mu_{j};\\{\zeta_{j}\\})$ and $B^{(\ell)}_{1\cdots
N_{s}}(\lambda_{\alpha};\\{\zeta_{j}\\})$, respectively, and $\zeta_{k}$
denote the centers of $\ell_{k}$-strings of the inhomogeneous parameters
$\\{\xi_{k}^{\ell}\\}$.

We calculate the scalar product for the higher-spin XXZ chains by the formula
in the next proposition.

######  Proposition 34.

Let $\\{\lambda_{\alpha}\\}$ satisfy the Bethe ansatz equations for the
spin-$\ell/2$ case. The scalar product of the spin-$\ell/2$ XXZ spin chain is
reduced into that of the spin-1/2 XXZ spin chain as follows:

$S^{(\ell)}_{n}(\\{\mu_{j}\\},\\{\lambda_{\alpha}\\};\\{\zeta_{k}\\})=\lim_{\epsilon\rightarrow
0}\Bigg{[}S^{(1)}_{n}(\\{\mu_{j}\\},\\{\lambda_{\alpha}\\};\\{\xi^{(\ell;\epsilon)}_{k}\\})\Bigg{]}$
(6.4)

###### Proof.

We now calculate the scalar product making use of eq. (4.7) of lemma 4.7 as
follows.

$\displaystyle\langle 0|\,C^{(\ell)}_{1\cdots
N_{s}}(\mu_{1};\\{\zeta_{j}\\})\cdots C^{(\ell)}_{1\cdots
N_{s}}(\mu_{n};\\{\zeta_{j}\\})B^{(\ell)}_{1\cdots
N_{s}}(\lambda_{1};\\{\zeta_{j}\\})\cdots B^{(\ell)}_{1\cdots
N_{s}}(\lambda_{n};\\{\zeta_{j}\\})\,|0\rangle$ (6.5) $\displaystyle=$
$\displaystyle\langle 0|\left(P_{1\cdots
L}^{(\ell)\,{\bar{\chi}}}C^{(1)}_{1\cdots
L}(\mu_{1};\\{\xi_{j}^{(\ell)}\\})P_{1\cdots
L}^{(\ell)\,{\bar{\chi}}}\right)\,\cdots\,\left(P_{1\cdots
L}^{(\ell)\,{\bar{\chi}}}C^{(1)}_{1\cdots
L}(\mu_{n};\\{\xi_{j}^{(\ell)}\\})P_{1\cdots L}^{(\ell)\,{\bar{\chi}}}\right)$
$\displaystyle\quad\cdot\,\left(P_{1\cdots
L}^{(\ell)\,{\bar{\chi}}}B^{(1)}_{1\cdots
L}(\lambda_{1};\\{\xi_{j}^{(\ell)}\\})P_{1\cdots
L}^{(\ell)}\right)\,\cdots\left(P_{1\cdots
L}^{(\ell)\,{\bar{\chi}}}B^{(1)}_{1\cdots
L}(\lambda_{n};\\{\xi_{j}^{(\ell)}\\})P_{1\cdots
L}^{(\ell)}\right)\,|0\rangle$ $\displaystyle=$ $\displaystyle\langle
0|P_{1\cdots L}^{(\ell)\,{\bar{\chi}}}\,\times\,C^{(1)}_{1\cdots
L}(\mu_{1};\\{\xi_{j}^{(\ell)}\\})\,\cdots\,C^{(1)}_{1\cdots
L}(\mu_{n};\\{\xi_{j}^{(\ell)}\\})$ $\displaystyle\quad\cdot\,B^{(1)}_{1\cdots
L}(\lambda_{1};\\{\xi_{j}^{(\ell)}\\})\,\cdots B^{(1)}_{1\cdots
L}(\lambda_{n};\\{\xi_{j}^{(\ell)}\\})\,\times\,P_{1\cdots
L}^{(\ell)\,{\bar{\chi}}}\,|0\rangle$

Here we note that we have $\langle 0|P_{1\cdots L}^{(\ell)}=\langle 0|$ and
$P_{1\cdots L}^{(\ell)}|0\rangle=|0\rangle$. Moreover, we have $\langle
0|P_{1\cdots L}^{(\ell)\,{\bar{\chi}}}=\langle 0|$ and $P_{1\cdots
L}^{(\ell)\,{\bar{\chi}}}|0\rangle=|0\rangle$. We thus have

$\displaystyle\langle 0|\,C^{(\ell)}_{1\cdots
N_{s}}(\mu_{1};\\{\zeta_{j}\\})\cdots C^{(\ell)}_{1\cdots
N_{s}}(\mu_{n};\\{\zeta_{j}\\})B^{(\ell)}_{1\cdots
N_{s}}(\lambda_{1};\\{\zeta_{j}\\})\cdots B^{(\ell)}_{1\cdots
N_{s}}(\lambda_{n};\\{\zeta_{j}\\})\,|0\rangle$ $\displaystyle=$
$\displaystyle\langle 0|C^{(1)}_{1\cdots
L}(\mu_{1};\\{\xi_{j}^{(\ell)}\\})\,\cdots\,C^{(1)}_{1\cdots
L}(\mu_{n};\\{\xi_{j}^{(\ell)}\\})\,\cdot\,B^{(1)}_{1\cdots
L}(\lambda_{1};\\{\xi_{j}^{(\ell)}\\})\,\cdots B^{(1)}_{1\cdots
L}(\lambda_{n};\\{\xi_{j}^{(\ell)}\\})\,|0\rangle\,$

We evaluate the last line through the following limit of sending $\epsilon$ to
zero:

$\displaystyle\langle 0|C^{(1)}_{1\cdots
L}(\mu_{1};\\{\xi_{j}^{(\ell)}\\})\,\cdots\,C^{(1)}_{1\cdots
L}(\mu_{n};\\{\xi_{j}^{(\ell)}\\})\,\cdot\,B^{(1)}_{1\cdots
L}(\lambda_{1};\\{\xi_{j}^{(\ell)}\\})\,\cdots B^{(1)}_{1\cdots
L}(\lambda_{n};\\{\xi_{j}^{(\ell)}\\})\,|0\rangle\,$ (6.7) $\displaystyle=$
$\displaystyle\lim_{\epsilon\rightarrow 0}\langle 0|C^{(1)}_{1\cdots
L}(\mu_{1};\\{\xi_{j}^{(\ell;\epsilon)}\\})\cdots C^{(1)}_{1\cdots
L}(\mu_{n};\\{\xi_{j}^{(\ell;\epsilon)}\\})$ $\displaystyle\qquad\times
B^{(1)}_{1\cdots L}(\lambda_{1};\\{\xi_{j}^{(\ell;\epsilon)}\\})\,\cdots
B^{(1)}_{1\cdots L}(\lambda_{n};\\{\xi_{j}^{(\ell;\epsilon)}\\})\,|0\rangle$
$\displaystyle=$ $\displaystyle\lim_{\epsilon\rightarrow 0}\Bigg{[}\langle
0|{\widetilde{C}}^{(1)}_{1\cdots
L}(\mu_{1};\\{\xi_{j}^{(\ell;\epsilon)}\\})\cdots{\widetilde{C}}^{(1)}_{1\cdots
L}(\mu_{n};\\{\xi_{j}^{(\ell;\epsilon)}\\})$
$\displaystyle\qquad\quad\times\,{\widetilde{B}}^{(1)}_{1\cdots
L}(\lambda_{1};\\{\xi_{j}^{(\ell;\epsilon)}\\})\,\cdots\,{\widetilde{B}}^{(1)}_{1\cdots
L}(\lambda_{n};\\{\xi_{j}^{(\ell;\epsilon)}\\})\Bigg{]}\,|0\rangle\,$

∎

We evaluate the spin-1/2 scalar product taking the limit of sending $\epsilon$
to 0, so that we can make the determinant of $F_{L\cdots 21}$ being nonzero.
Here we remark that the operator $F_{L\cdots 21}$ appears in the pseudo-
diagonalization process of the $B$ and $C$ operators, as shown in Section 5,
and also that the determinant of $F_{L\cdots 21}$ vanishes at $\epsilon=0$,
when parameters $\xi_{j}$ are given by eq. (6.1). In fact, if we put some
inhomogeneous parameters $\xi_{j}$ in the form of a “complete $\ell$-string”
[29], that is, for some integers $\ell$, $m$ and a constant $z$, we have
$\xi_{m+j}=z-j\eta$ for $j=1,2,\ldots,\ell$, then the determinant of
$F_{L\cdots 21}$ vanishes. Here we also note that ${\rm det}F_{12\cdots L}\neq
0$ even at $\epsilon=0$.

Let us discuss the mixed spin case. We set the inhomogeneous parameters as
follows:

$\xi_{\ell_{1}+\cdots\ell_{k-1}+j}^{({\mbox{\boldmath$\ell$}};\epsilon)}=\zeta_{k}-(j-1)\eta+(\ell-1)\eta/2+\epsilon
r_{j}\qquad j=1,\ldots\ell;\,k=1,\ldots,r\,.$ (6.8)

Let us define $P^{({\mbox{\boldmath$\ell$}})}_{12\cdots L}$ and
$P^{({\mbox{\boldmath$\ell$}})\,{\bar{\chi}}}_{12\cdots L}$ by

$\displaystyle P^{({\mbox{\boldmath$\ell$}})}_{12\cdots L}$ $\displaystyle=$
$\displaystyle\prod_{k=1}^{r}P^{(\ell_{k})}_{\ell(k-1)+1}$ $\displaystyle
P^{({\mbox{\boldmath$\ell$}})\,{\bar{\chi}}}_{12\cdots L}$ $\displaystyle=$
$\displaystyle\prod_{k=1}^{r}\left(P^{(\ell_{k})}_{\ell(k-1)+1}\right)^{\bar{\chi}}$
(6.9)

It is easy to see the following:

$\displaystyle\langle 0|P^{({\mbox{\boldmath$\ell$}})}_{12\cdots L}$
$\displaystyle=$ $\displaystyle\langle 0|\,,\quad
P^{({\mbox{\boldmath$\ell$}})}_{12\cdots L}|0\rangle=|0\rangle\,,$
$\displaystyle\langle 0|P^{({\mbox{\boldmath$\ell$}})\,{\bar{\chi}}}_{12\cdots
L}$ $\displaystyle=$ $\displaystyle\langle 0|\,,\quad
P^{({\mbox{\boldmath$\ell$}})\,{\bar{\chi}}}_{12\cdots L}|0\rangle=|0\rangle$
(6.10)

We now define the scalar product for the mixed spin case as follows.

$S^{({\mbox{\boldmath$\ell$}})}_{n}(\\{\mu_{j}\\},\\{\lambda_{\alpha}\\};\\{\zeta_{k}\\})=\langle
0|\,C^{({\mbox{\boldmath$\ell$}})}(\mu_{1})\cdots
C^{({\mbox{\boldmath$\ell$}})}(\mu_{n})B^{({\mbox{\boldmath$\ell$}})}(\lambda_{1})\cdots
B^{({\mbox{\boldmath$\ell$}})}(\lambda_{n})\,|0\rangle$ (6.11)

Here $C^{({\mbox{\boldmath$\ell$}})}(\mu_{j})$ and
$B^{({\mbox{\boldmath$\ell$}})}(\lambda_{\alpha})$ abbreviate
$C^{({\mbox{\boldmath$\ell$}})}_{1\cdots r}(\mu_{j};\\{\zeta_{j}\\})$ and
$B^{({\mbox{\boldmath$\ell$}})}_{1\cdots
r}(\lambda_{\alpha};\\{\zeta_{j}\\})$, respectively.

######  Proposition 35.

Let $\\{\lambda_{\alpha}\\}$ satisfy the Bethe ansatz equations for the mixed
spin case. The scalar product of the mixed-spin XXZ spin chain is reduced into
that of the spin-1/2 XXZ spin chain as follows:

$S^{({\mbox{\boldmath$\ell$}})}_{n}(\\{\mu_{j}\\},\\{\lambda_{\alpha}\\};\\{\zeta_{k}\\})=\lim_{\epsilon\rightarrow
0}\Bigg{[}S^{(1)}_{n}(\\{\mu_{j}\\},\\{\lambda_{\alpha}\\};\\{\xi^{({\mbox{\boldmath$\ell$}};\epsilon)}_{k}\\})\Bigg{]}$
(6.12)

###  6.2 Determinant expressions of the scalar products

Let us review the result of the spin-1/2 case [7]. Suppose that
$\lambda_{\alpha}$ for $\alpha=1,2,\ldots,n$, are solutions of the Bethe
ansatz equations with in homogeneous parameters $\xi_{j}$ for
$j=1,2,\ldots,L$, the scalar product is defined by

$S_{n}(\\{\mu_{j}\\},\,\\{\lambda_{\alpha}\\};\\{\xi_{k}\\})=\langle
0|\prod_{j=1}^{n}C(\mu_{j};\\{\xi_{k}\\})\prod_{\alpha=1}^{n}B(\lambda_{\alpha};\\{\xi_{k}\\})|0\rangle$
(6.13)

Here $\mu_{j}$ for $j=1,2,\ldots,n$ are arbitrary. We note that
$S_{n}(\\{\mu_{j}\\},\,\\{\lambda_{\alpha}\\};\\{\xi_{k}\\})$ has been denoted
by $S_{n}^{(1)}(\\{\mu_{j}\\},\,\\{\lambda_{\alpha}\\};\\{\xi_{k}\\})$ in the
last subsection. Then, the exact expression of the scalar product has been
shown through the pseudo-diagonalized forms of the $B$ and $C$ operators as
follows [7]:

$\displaystyle
S_{n}(\\{\mu_{j}\\},\,\\{\lambda_{\alpha}\\};\\{\xi_{k}\\})=\langle
0|\prod_{j=1}^{n}\widetilde{C}(\mu_{j})\prod_{\alpha=1}^{n}\widetilde{B}(\lambda_{\alpha})|0\rangle$
(6.14) $\displaystyle=$
$\displaystyle{\frac{\prod_{\alpha=1}^{n}\prod_{j=1}^{n}\sinh(\mu_{j}-\lambda_{\alpha})}{\prod_{j>k}\sinh(\mu_{k}-\mu_{j})\prod_{\alpha<\beta}\sinh(\lambda_{\beta}-\lambda_{\alpha})}}\,{\rm
det}T(\\{\mu_{j}\\},\,\\{\lambda_{\alpha}\\};\\{\xi_{k}\\})$

Here the matrix elements $T_{ab}$ for $a,b=1,\ldots,n$, are given by

$T_{ab}={\frac{\partial}{\partial\lambda_{\alpha}}}\tau(\mu_{b},\\{\lambda_{k}\\};\\{\xi_{k}\\})$
(6.15)

where

$\displaystyle\tau(\mu,\\{\lambda_{k}\\};\\{\xi_{k}\\})$ $\displaystyle=$
$\displaystyle
a(\mu)\prod_{k=1}^{n}b^{-1}(\lambda_{k}-\mu)+d(\mu;\\{\xi_{j}\\};\\{\xi_{k}\\})\prod_{k=1}^{n}b^{-1}(\mu-\lambda_{k})$

and

$a(\mu)=1\,,\quad d(\mu;\\{\xi_{j}\\})=\prod_{j=1}^{L}b(\mu-\xi_{j})\,.$
(6.17)

Let us express the scalar product of the higher-spin case in terms of the
determinant of the matrix $T$. In the tensor product of spin-$\ell/2$
representations we have

$\displaystyle
S^{(\ell)}_{n}(\\{\mu_{j}\\},\,\\{\lambda_{\alpha}\\};\\{\zeta_{k}\\})$ (6.18)
$\displaystyle=$
$\displaystyle{\frac{\prod_{\alpha=1}^{n}\prod_{j=1}^{n}\sinh(\mu_{j}-\lambda_{\alpha})}{\prod_{j>k}\sinh(\mu_{k}-\mu_{j})\prod_{\alpha<\beta}\sinh(\lambda_{\beta}-\lambda_{\alpha})}}\,{\rm
det}T(\\{\mu_{j}\\},\,\\{\lambda_{\alpha}\\};\\{\xi_{k}^{(\ell)}\\})$

In the mixed-spin case we have

$\displaystyle
S^{({\mbox{\boldmath$\ell$}})}_{n}(\\{\mu_{j}\\},\,\\{\lambda_{\alpha}\\};\\{\zeta_{k}\\})$
(6.19) $\displaystyle=$
$\displaystyle{\frac{\prod_{\alpha=1}^{n}\prod_{j=1}^{n}\sinh(\mu_{j}-\lambda_{\alpha})}{\prod_{j>k}\sinh(\mu_{k}-\mu_{j})\prod_{\alpha<\beta}\sinh(\lambda_{\beta}-\lambda_{\alpha})}}\,{\rm
det}T(\\{\mu_{j}\\},\,\\{\lambda_{\alpha}\\};\\{\xi_{k}^{({\mbox{\boldmath$\ell$}})}\\})$

###  6.3 Norms of the Bethe states in the higher-spin case

For two sets of $n$ parameters, $\mu_{1},\ldots,\mu_{n}$ and
$\lambda_{1},\ldots,\lambda_{n}$, we define matrix elements $H_{a}b$ by

$H_{ab}(\\{\mu_{j}\\},\\{\lambda_{\alpha};\\{\xi_{k}\\})={\frac{\sinh\eta}{\sinh(\lambda_{a}-\mu_{b})}}\left({\frac{a(\mu_{b})}{d(\mu_{b})}}\prod_{k=1;\neq
a}^{n}\sinh(\lambda_{k}-\mu_{b}+\eta)-\prod_{k=1;\neq
a}^{n}\sinh(\lambda_{k}-\mu_{b}-\eta)\right)$ (6.20)

Let us assume that $\lambda_{1},\ldots,\lambda_{n}$ are solutions of the Bethe
ansatz equations. We have

${\rm det}H(\\{\lambda_{\alpha}\\},\\{\mu_{j}\\};\\{\xi_{k}\\})={\rm
det}T(\\{\lambda_{\alpha}\\},\\{\mu_{j}\\};\\{\xi_{k}\\})\prod_{\alpha=1}^{n}\prod_{j=1}^{n}\sinh(\mu_{j}-\lambda_{\alpha})\left(\prod_{j=1}^{n}d(\mu_{j})\right)^{-1}$
(6.21)

Let us now take the limit of sending $\mu_{j}$ to $\lambda_{j}$ for each $j$.
Then we have

$\lim_{\mu_{j}\rightarrow\lambda_{j}}detH(\\{\lambda_{\alpha}\\},\\{\mu_{j}\\};\\{\xi_{k}\\})=\sinh^{n}\eta\prod_{\beta=1}^{n}\prod_{m=1;m\neq\beta}^{n}\sinh(\lambda_{m}-\lambda_{\beta}-\eta)\cdot
det\Phi^{{}^{\prime}}(\\{\lambda_{\alpha}\\})$ (6.22)

where matrix elements $\Phi^{{}^{\prime}}_{ab}$ for $a,b=1,\ldots,n$, are
given by

$\Phi^{{}^{\prime}}_{ab}(\\{\lambda_{\alpha}\\};\\{\xi\\})=-{\frac{\partial}{\partial\lambda_{b}}}\left({\frac{a(\lambda_{a};\\{\xi_{j}\\})}{d(\lambda_{b};\\{\xi_{j}\\})}}\prod_{k=1;k\neq
a}^{n}{\frac{b(\lambda_{a}-\lambda_{k})}{b(\lambda_{k}-\lambda_{a})}}\right)$
(6.23)

Suppose that $\lambda_{\alpha}$ for $n=1,2,\ldots,n$ are solutions of the
Bethe ansatz equations. Gaudin’s formula for the square of he norm of the
Bethe state is given by

$\displaystyle{\mbox{\boldmath$N$}}_{n}(\\{\lambda_{\alpha}\\};\\{\xi_{j}\\})$
$\displaystyle=$ $\displaystyle\langle
0|\prod_{j=1}^{n}C(\lambda_{j})\prod_{j=1}^{n}B(\lambda_{k})|0\rangle$ (6.24)
$\displaystyle=$
$\displaystyle\sinh^{n}\eta\prod_{\alpha,\beta=1;\alpha\neq\beta}^{n}b^{-1}(\lambda_{\alpha}-\lambda_{\beta})\cdot{\rm
det}\Phi^{{}^{\prime}}({\lambda_{\alpha}};\\{\xi_{j}\\})$

Let us define the norm of the Bethe state for the mixed spin case of $\ell$ as
follows.

${\mbox{\boldmath$N$}}_{n}^{({\mbox{\boldmath$\ell$}})}(\\{\lambda_{\alpha}\\};\\{\xi^{({\mbox{\boldmath$\ell$}})}_{k}\\})=\langle
0|\prod_{j=1}^{n}C^{({\mbox{\boldmath$\ell$}})}(\lambda_{j})\prod_{j=1}^{n}B^{({\mbox{\boldmath$\ell$}})}(\lambda_{k})|0\rangle$
(6.25)

Then we have

${\mbox{\boldmath$N$}}_{n}^{({\mbox{\boldmath$\ell$}})}(\\{\lambda_{\alpha}\\};\\{\xi^{({\mbox{\boldmath$\ell$}})}_{k}\\})=\sinh^{n}\eta\prod_{\alpha,\beta=1;\alpha\neq\beta}^{n}b^{-1}(\lambda_{\alpha}-\lambda_{\beta})\cdot{\rm
det}\Phi^{{}^{\prime}}({\lambda_{\alpha}};\\{\xi_{j}^{\mbox{\boldmath$\ell$}}\\})$
(6.26)

##  7 Form factors and the inverse-scattering for the higher-spin case

###  7.1 Formulas of the quantum inverse scattering problem

Let us briefly review the derivation of the fundamental lemma of the quantum-
inverse scattering problem for the spin-1/2 XXZ spin chain [6]. It will thus
becomes clear how the pseudo-diagonalization of $B$ and $C$ are important.

######  Proposition 36.

Let us denote by $p_{q}$ sequence $p_{q}=(1,2,\ldots,n)$. Recall the notation
$\widetilde{R}_{0,p_{q}}=F_{12\cdots n}R_{0,12\cdots n}F^{-1}_{12\cdots n}$.
Then, $\widetilde{R}_{0,p_{q}}$ is invariant under any permutation. We have

$\widetilde{R}_{0,p_{q}}=\widetilde{R}_{0,\sigma(p_{q})}\,,\quad{\rm
for}\,\,\sigma\in{\cal S}_{n}\,.$ (7.1)

We shall show (7.1) in Appendix B.

The following lemma plays a central role in the quantum inverse-scattering
problem [7].

######  Lemma 37.

For arbitrary inhomogeneous parameters $\xi_{1},\xi_{2},\ldots,\xi_{L}$ we
have

$x_{i}=\prod_{\alpha=1}^{i-1}(A+D)(\xi_{\alpha})\,{\rm
tr}_{0}(x_{0}R_{0,1\cdots
L}(\xi_{i}))\,\prod_{\alpha=i+1}^{L}(A+D)(\xi_{\alpha})$ (7.2)

###### Proof.

For any operator $x_{0}$ defined on the auxiliary space, we have

$\displaystyle{\rm tr}_{0}(x_{0}R_{0,12\cdots L}(\lambda=\xi_{i}))$
$\displaystyle=$ $\displaystyle F_{12\cdots L}^{-1}{\rm
tr}_{0}(x_{0}\widetilde{R}_{0,12\cdots L}(\lambda=\xi_{i}))F_{12\cdots L}$
$\displaystyle=$ $\displaystyle F_{12\cdots L}^{-1}{\rm
tr}_{0}(x_{0}\widetilde{R}_{i,i+1\cdots L1\cdots
i-1}(\lambda=\xi_{i}))F_{12\cdots L}$ $\displaystyle=$ $\displaystyle
F_{12\cdots L}^{-1}F_{i\cdots L1\cdots i-1}\,{\rm tr}_{0}(x_{0}R_{i,i+1\cdots
L1\cdots i-1}(\lambda=\xi_{i}))\,F_{i\cdots L1\cdots i-1}^{-1}F_{12\cdots L}$
$\displaystyle=$ $\displaystyle\left(F_{i\cdots L1\cdots i-1}^{-1}F_{12\cdots
L}\right)^{-1}x_{i}(A(\xi_{i})+D(\xi_{i}))\,\left(F_{i\cdots L1\cdots
i-1}^{-1}F_{12\cdots L}\right)$

Here we have used (7.1), i.e. $\widetilde{R}_{0,12\cdots
L}=\widetilde{R}_{i,i+1\cdots L1\cdots i-1}$. From the expression of
$F^{-1}_{i\cdots L\,1\cdots i-1}F_{1\cdots L}$ we now have

${\rm tr}_{0}(x_{0}R_{0,1\cdots
L})=\prod_{\alpha=1}^{i-1}\left((A+D)(\xi_{\alpha})\right)^{-1}\cdot
x_{i}\cdot\prod_{\alpha=1}^{i}(A+D)(\xi_{\alpha})\,.$ (7.3)

∎

###  7.2 Quantum inverse-scattering problem for the higher-spin operators

Let us consider monodromy matrix $T^{+}_{0,1\cdots\ell N_{s}}$. Here we recall
$L=\ell N_{s}$. For simplicity, we shall suppress the superscript ‘$+$’ for
$A,B,C$ and $D$ operators through this subsection.

We recall the following: $\Delta^{(n-1)}(K)=K^{\otimes n}$ and

$\displaystyle\Delta^{(n-1)}(X^{+})$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{n}K^{\otimes(j-1)}\otimes X^{+}_{j}\otimes
I^{\otimes(n-j)}\,,$ $\displaystyle\Delta^{(n-1)}(X^{-})$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{n}I^{\otimes(j-1)}\otimes
X^{-}_{j}\otimes\left(K^{-1}\right)^{\otimes(n-j)}\,.$ (7.4)

It is useful to note that for $i=1,2,\ldots,\ell N_{s}$ we have

$K_{i}=\prod_{\alpha=1}^{i-1}(A+D)(\xi_{\alpha})(qA+q^{-1}D)(\xi_{i})\prod_{\alpha=i+1}^{\ell
N_{s}}(A+D)(\xi_{\alpha})\,.$ (7.5)

In the tensor product of $2N_{s}$ spin-1/2 representations,
$\left(V_{1}^{(1)}\otimes V_{2}^{(1)}\right)^{\otimes N_{s}}$, we have

$\displaystyle\Delta_{12}(X^{-})$ $\displaystyle=$
$\displaystyle\left(X^{-}_{1}\otimes K^{-1}_{2}+I_{1}\otimes
X^{-}_{2}\right)\otimes I^{\otimes 2(N_{s}-1)}$ (7.6) $\displaystyle=$
$\displaystyle\Bigg{\\{}\left(B(\xi_{1})\cdot\prod_{\alpha=2}^{2N_{s}}(A+D)(\xi_{\alpha})\right)\cdot\bigg{(}(A+D)(\xi_{1})(q^{-1}A+qD)(\xi_{2})\prod_{\alpha=3}^{2N_{s}}(A+D)(\xi_{\alpha})\bigg{)}$
$\displaystyle\quad+(A+D)(\xi_{1})\cdot
B(\xi_{2})\cdot\prod_{\alpha=3}^{2N_{s}}(A+D)(\xi_{\alpha})\Bigg{\\}}$

In the $\ell$th tensor product of spin-1/2 representations,
$V_{1}^{(1)}\otimes\cdots\otimes V_{\ell}^{(1)}$, we have

$P_{1\cdots\ell}^{(\ell)}\,\cdot\,\Delta^{(\ell-1)}(X^{\pm})\,\cdot\,P_{1\cdots\ell}^{(\ell)}=P_{1\cdots\ell}^{(\ell)}\,\cdot\,\Delta^{(\ell-1)}(X^{\pm})=\Delta^{(\ell-1)}(X^{\pm})\,\cdot\,P_{1\cdots\ell}^{(\ell)}$
(7.7)

In the tensor product of spin-$\ell/2$ representations,
$V_{1}^{(\ell)}\otimes\cdots\otimes V_{N_{s}}^{(\ell)}$, we have for
$i=1,2,\ldots,N_{s}$ the following relations:

$\displaystyle X_{i}^{-(\ell+)}=\Delta^{(\ell-1)}_{(i-1)\ell+1\,\cdots
i\ell}(X^{-})$
$\displaystyle\quad=\sum_{k=1}^{\ell}\prod_{\alpha=1}^{(i-1)\ell+k-1}(A+D)(\xi_{\alpha})\cdot
B(\xi_{(i-1)\ell+k})\cdot\prod_{\alpha=(i-1)\ell+k+1}^{\ell
N_{s}}(A+D)(\xi_{\alpha})$
$\displaystyle\quad\times\prod_{j=k+1}^{\ell}\prod_{\alpha=1}^{(i-1)\ell+j-1}(A+D)(\xi_{\alpha})\cdot(q^{-1}A+qD)(\xi_{(i-1)\ell+j})\cdot\prod_{\alpha=(i-1)\ell+j+1}^{\ell
N_{s}}(A+D)(\xi_{\alpha})$
$\displaystyle=\sum_{k=1}^{\ell}\prod_{\alpha=1}^{(i-1)\ell+k-1}(A+D)(\xi_{\alpha})\cdot
B(\xi_{(i-1)\ell+k})\cdot$
$\displaystyle\quad\times\prod_{j=k+1}^{\ell}(q^{-1}A+qD)(\xi_{(i-1)\ell+j})\cdot\prod_{\alpha=i\ell+1}^{\ell
N_{s}}(A+D)(\xi_{\alpha})\,.$ (7.8)

Here we have made use of the following:

$\prod_{\alpha=1}^{\ell N_{s}}(A+D)(\xi_{\alpha})=I^{\otimes\ell N_{s}}\,.$
(7.9)

Similarly, we can express $X_{i}^{+(\ell+)}$ and $K_{i}^{(\ell+)}$ as follows.

$\displaystyle X_{i}^{+(\ell+)}=\Delta^{(\ell-1)}_{(i-1)\ell+1\,\cdots
i\ell}(X^{+})$
$\displaystyle=\sum_{k=1}^{\ell}\prod_{\alpha=1}^{(i-1)\ell}(A+D)(\xi_{\alpha})\cdot\prod_{j=1}^{k-1}(qA+q^{-1}D)(\xi_{(i-1)\ell+j})\cdot
C(\xi_{(i-1)\ell+k})$ $\displaystyle\quad\prod_{\alpha=(i-1)\ell+k+1}^{\ell
N_{s}}(A+D)(\xi_{\alpha})\,.$ (7.10)

$\displaystyle K_{i}^{(\ell+)}=\Delta^{(\ell-1)}_{(i-1)\ell+1\,\cdots
i\ell}(K)$
$\displaystyle=\prod_{\alpha=1}^{(i-1)\ell}(A+D)(\xi_{\alpha})\,\prod_{j=1}^{\ell}(qA+q^{-1}D)(\xi_{(i-1)\ell+j})\,\prod_{\alpha=i\ell+1}^{\ell
N_{s}}(A+D)(\xi_{\alpha})\,.$ (7.11)

###  7.3 Useful formulas in the higher-spin case

Let us denote by $X^{\pm(\ell)}$ the matrix representations of generators
$X^{\pm}$ in the spin-$\ell/2$ representation of $U_{q}(sl_{2})$. Here we
recall that the matrix representations of $X^{\pm(\ell)}$ are obtained by
calculating the action of $\Delta^{(\ell-1)}(X^{\pm})$ on the basis
$\\{||\ell,n\rangle\\}$.

We explicitly calculate the actions of $\sigma_{1}^{-}=\sigma^{-}\otimes
I^{\otimes(\ell-1)}$ and
$\sigma_{\ell}^{+}=I^{\otimes(\ell-1)}\otimes\sigma^{+}$ on the basis
$\\{||\ell,n\rangle\\}$ in the spin-$\ell/2$ representation. Multiplying
projection operators to them, we obtain the following formulas:

$\displaystyle
P^{(\ell)}_{1\cdots\ell}\,\sigma_{1}^{-}\,P^{(\ell)}_{1\cdots\ell}$
$\displaystyle=$ $\displaystyle{\frac{1}{[\ell]_{q}}}\,X^{-(\ell+)}$
$\displaystyle
P^{(\ell)}_{1\cdots\ell}\,\sigma_{\ell}^{+}\,P^{(\ell)}_{1\cdots\ell}$
$\displaystyle=$ $\displaystyle{\frac{1}{[\ell]_{q}}}\,X^{-(\ell+)}$ (7.12)

Therefore, we have for $i=1,2,\ldots,N_{s}$, the following formulas:

$\displaystyle
P^{(\ell)}_{(i-1)\ell+1}\,X_{i}^{-(\ell+)}\,P^{(\ell)}_{(i-1)\ell+1}$
$\displaystyle={[}\ell{]}_{q}P^{(\ell)}_{(i-1)\ell+1}\,\prod_{\alpha=1}^{(i-1)\ell}(A^{+}+D^{+})(\xi_{\alpha})\cdot
B^{+}(\xi_{(i-1)\ell+1})\cdot\prod_{\alpha=(i-1)\ell+2}^{\ell
N_{s}}(A^{+}+D^{+})(\xi_{\alpha})\,P^{(\ell)}_{(i-1)\ell+1}$ (7.13)
$\displaystyle
P^{(\ell)}_{(i-1)\ell+1}\,X_{i}^{+(\ell+)}\,P^{(\ell)}_{(i-1)\ell+1}$
$\displaystyle=[\ell]_{q}P^{(\ell)}_{(i-1)\ell+1}\,\prod_{\alpha=1}^{i\ell-2}(A^{+}+D^{+})(\xi_{\alpha})\cdot
C^{+}(\xi_{i\ell-1})\cdot\prod_{\alpha=i\ell}^{\ell
N_{s}}(A^{+}+D^{+})(\xi_{\alpha})\,P^{(\ell)}_{(i-1)\ell+1}\,.$ (7.14)

Taking advantage of projection operators, we thus have shown that the
summation over $k$ arising from the $(\ell-1)$th comultiplication operation
can be calculated by a single term. This reduces the calculational task very
much.

In the derivation of (7.13), we first note

$\chi_{1\cdots L}\sigma_{(i-1)\ell+1}^{-}\chi_{1\cdots
L}^{-1}\exp(-\xi_{(i-1)\ell+1})\,,$ (7.15)

and then we show the following:

$\sigma_{(i-1)\ell+1}^{-}=\prod_{\alpha=1}^{(i-1)\ell}(A^{+}+D^{+})(\xi_{\alpha})\,B^{+}(\xi_{(i-1)\ell+1})\,\prod_{\alpha=(i-1)\ell+2}^{\ell
N_{s}}(A^{+}+D^{+})(\xi_{\alpha})\,.$ (7.16)

We shall show relations (7.12) in Appendix C.

We now introduce useful formulas expressing any given operator in the
spin-$\ell$ representation. Let us take two sets of integers
$i_{1},\ldots,i_{m}$ and $j_{1},\ldots,j_{n}$ satisfying $1\leq
i_{1}<\cdots<i_{m}\leq\ell$ and $1\leq j_{1}<\cdots<j_{n}\leq\ell$,
respectively. We can show the following:

$\displaystyle||\ell,m\rangle\langle\ell,n||$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{c}\ell\\\
m\end{array}\right]_{q}\,q^{m(m+1)/2-n(n-1)/2+n\ell-(i_{1}+\cdots+i_{m}+j_{1}+\cdots+j_{n})}$
(7.20)
$\displaystyle\times\,P_{1\cdots\ell}^{(\ell)}\left(\prod_{k=1}^{m}e_{i_{k}}^{21}\cdot\prod_{p=1;p\neq
i_{k},j_{q}}^{\ell}e_{p}^{22}\cdot\prod_{q=1}^{n}e_{j_{q}}^{12}\right)P_{1\cdots\ell}^{(\ell)}$

Setting $i_{1}=1,i_{2}=2$, …, $i_{m}=m$ and $j_{1}=1,j_{2}=2$, …, $j_{n}=n$,
we have for $m>n$

$||\ell,m\rangle\langle\ell,n||=\left[\begin{array}[]{c}\ell\\\
n\end{array}\right]_{q}\,q^{(\ell-n)}\,P_{1\cdots\ell}^{(\ell)}\left(\prod_{k=1}^{n}e_{k}^{22}\,\prod_{k=n+1}^{m}e_{k}^{21}\,\prod_{k=m+1}^{\ell}e_{k}^{11}\right)P_{1\cdots\ell}^{(\ell)}$
(7.21)

and for $m<n$

$||\ell,m\rangle\langle\ell,n||=\left[\begin{array}[]{c}\ell\\\
n\end{array}\right]_{q}\,q^{(\ell-n)}\,P_{1\cdots\ell}^{(\ell)}\left(\prod_{k=1}^{m}e_{k}^{22}\,\prod_{k=m+1}^{n}e_{k}^{12}\,\prod_{k=n+1}^{\ell}e_{k}^{11}\right)P_{1\cdots\ell}^{(\ell)}$
(7.22)

and for $m=n$

$||\ell,n\rangle\langle\ell,n||=\left[\begin{array}[]{c}\ell\\\
n\end{array}\right]_{q}\,q^{(\ell-n)}\,P_{1\cdots\ell}^{(\ell)}\left(\prod_{k=1}^{n}e_{k}^{22}\,\prod_{k=n+1}^{\ell}e_{k}^{11}\right)P_{1\cdots\ell}^{(\ell)}$
(7.23)

Let us denote by $E^{mn(\ell+)}$ the unit matrices acting on spin-$\ell$
representation $V^{(\ell)}$ for $m,n=0,1,\ldots,\ell$. We now define
$E_{i}^{mn(\ell+)}$ by the unit matrices acting on the $i$th component of the
tensor product $(V^{(\ell)})^{\otimes N_{s}}$. Explicitly, we have

$E_{i}^{mn\,(\ell+)}=(I^{(\ell)})^{\otimes(i-1)}\otimes
E^{mn}\otimes(I^{(\ell)})^{\otimes(N_{s}-i)}$ (7.24)

where $I^{(\ell)}$ denotes the $(\ell+1)\times(\ell+1)$ identity matrix. Then,
we derive the following formulas from (7.21), (7.22) and (7.23), respectively.
For $m>n$ we have

$\displaystyle E_{i}^{mn\,(\ell+)}$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{c}\ell\\\
m\end{array}\right]_{q}\,q^{n(\ell-n)}\,P^{(\ell)}_{1\cdots
L}\,\prod_{\alpha=1}^{(i-1)\ell}(A+D)(\xi_{\alpha})\prod_{k=1}^{n}D(\xi_{(i-1)\ell+k})\prod_{k=n+1}^{m}B(\xi_{(i-1)\ell+k})$
(7.28)
$\displaystyle\quad\times\,\prod_{k=m+1}^{\ell}A(\xi_{(i-1)\ell+k})\prod_{\alpha=i\ell+1}^{\ell
N_{s}}(A+D)(\xi_{\alpha})\,\,P^{(\ell)}_{1\cdots L}\,.$

For $m<n$ we have

$\displaystyle E_{i}^{mn\,(\ell+)}$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{c}\ell\\\
m\end{array}\right]_{q}\,q^{n(\ell-n)}\,P^{(\ell)}_{1\cdots
L}\,\prod_{\alpha=1}^{(i-1)\ell}(A+D)(\xi_{\alpha})\prod_{k=1}^{m}D(\xi_{(i-1)\ell+k})\prod_{k=m+1}^{n}C(\xi_{(i-1)\ell+k})$
(7.32)
$\displaystyle\quad\times\,\prod_{k=n+1}^{\ell}A(\xi_{(i-1)\ell+k})\prod_{\alpha=i\ell+1}^{\ell
N_{s}}(A+D)(\xi_{\alpha})\,\,P^{(\ell)}_{1\cdots L}\,.$

For $m=n$ we have

$\displaystyle E_{i}^{nn\,(\ell+)}$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{c}\ell\\\
n\end{array}\right]_{q}\,q^{n(\ell-n)}\,P^{(\ell)}_{1\cdots
L}\,\prod_{\alpha=1}^{(i-1)\ell}(A+D)(\xi_{\alpha})\prod_{k=1}^{n}D(\xi_{(i-1)\ell+k})$
(7.36)
$\displaystyle\quad\times\,\prod_{k=n+1}^{\ell}A(\xi_{(i-1)\ell+k})\prod_{\alpha=i\ell+1}^{\ell
N_{s}}(A+D)(\xi_{\alpha})\,\,P^{(\ell)}_{1\cdots L}\,.$

Let us now discuss the derivation of formula (7.20). It is easy to show the
following:

$\sigma_{i_{1}}^{-}\cdots\sigma_{i_{m}}^{-}|0\rangle\langle
0|\sigma_{j_{1}}^{+}\cdots\sigma_{j_{n}}^{+}=e_{i_{1}}^{21}\cdots
e_{i_{n}}^{21}\prod_{p=1;p\neq
i_{k},j_{q}}^{\ell}e_{p}^{11}e_{i_{1}}^{12}\cdots e_{i_{n}}^{12}$ (7.37)

Then, making use of expressions (4.20) and (4.25), we obtain (7.20).

###  7.4 Form factors for higher-spin operators

Making use of the fundamental lemma of the quantum inverse-scattering problem,
lemma 7.2, together with the useful formulas given in §7.3 such as (7.13) and
(7.14), and (7.28), (7.32) and (7.36), we can systematically calculate form
factors for the higher-spin cases. Here we note that the form factors
associated with generators $S^{\pm}$ of the spin $SU(2)$ have been derived for
the higher-spin XXX chains [27]. They are derived through the relations
corresponding to (7.8) and (7.10) in the limit of $q=1$.

For an illustration, let us calculate the following form factor:

$F_{n}^{-(\ell+)}(i;\,\\{\mu_{j}\\},\\{\lambda_{k}\\})=\langle
0|\prod_{j=1}^{n+1}C^{(\ell+)}(\mu_{j})\cdot
X_{i}^{-(\ell+)}\prod_{k=1}^{n}B^{(\ell+)}(\lambda_{k})|0\rangle\,,$ (7.38)

where $\\{\mu_{j}\\}$ and $\\{\lambda_{k}\\}$ are solutions of the Bethe
ansatz equations. Putting (7.13) into (7.38) and making use of the fact that
projector $P_{1\cdots L}^{(\ell)}$ commutes with the matrix elements of
$R_{0,1\cdots L}^{+}$, we have

$\displaystyle\langle 0|\prod_{j=1}^{n+1}C^{+}(\mu_{j})\cdot
X_{i}^{-(\ell+)}\prod_{k=1}^{n}B^{+}(\lambda_{k})|0\rangle$ (7.39)
$\displaystyle=$ $\displaystyle[\ell]_{q}\,\langle
0|\prod_{j=1}^{n+1}C^{+}(\mu_{j})\cdot
P_{(i-1)\ell+1}^{(\ell)}\prod_{\alpha=1}^{(i-1)\ell}(A^{+}+D^{+})(\xi_{\alpha})\cdot
B^{+}(\xi_{(i-1)\ell+1})\cdot$
$\displaystyle\quad\times\prod_{\alpha=(i-1)\ell+2}^{\ell
N_{s}}(A^{+}+D^{+})(\xi_{\alpha})P_{(i-1)\ell+1}^{(\ell)}\prod_{k=1}^{n}B^{+}(\lambda_{k})|0\rangle$
$\displaystyle=$
$\displaystyle[\ell]_{q}\,e^{\sum_{j=1}^{n+1}\mu_{j}-\sum_{k=1}^{n}\lambda_{k}-\xi_{(i-1)\ell+1}}{\frac{\phi_{(i-1)\ell}(\\{\mu_{j}\\}_{n+1})}{\phi_{(i-1)\ell+1}(\\{\lambda\\}_{n})}}\langle
0|\prod_{j=1}^{n+1}C(\mu_{j})\cdot
B(\xi_{(i-1)\ell+1})\prod_{k=1}^{n}B(\lambda_{k})|0\rangle$ $\displaystyle=$
$\displaystyle[\ell]_{q}\,e^{\sum_{j=1}^{n+1}\mu_{j}-\sum_{k=1}^{n}\lambda_{k}-\xi_{(i-1)\ell+1}}{\frac{\phi_{(i-1)\ell}(\\{\mu_{j}\\}_{n+1})}{\phi_{(i-1)\ell+1}(\\{\lambda\\}_{n})}}$
$\displaystyle\quad\times
S_{n+1}\left(\\{\mu_{j}\\}_{n+1},\\{\xi_{(i-1)\ell+1},\lambda_{1},\ldots,\lambda_{n}\\};\\{\xi_{k}^{(\ell)}\\}\right)$

Let us define the form factor in the symmetric case as follows.

$F_{n}^{-(\ell)}(i;\,\\{\mu_{j}\\},\\{\lambda_{k}\\})=\langle
0|\prod_{j=1}^{n+1}C(\mu_{j})\cdot\left(X_{i}^{-(\ell+)}\right)^{\bar{\chi}}\prod_{k=1}^{n}B(\lambda_{k})|0\rangle$
(7.40)

Here we recall that $\\{\mu_{j}\\}$ and $\\{\lambda_{k}\\}$ are solutions of
the Bethe ansatz equations. Then, we obtain the following expression.

$\displaystyle
F_{n}^{-(\ell)}(i;\,\\{\mu_{j}\\},\\{\lambda_{k}\\})=[\ell]_{q}\,{\frac{\phi_{(i-1)\ell}(\\{\mu_{j}\\}_{n+1})}{\phi_{(i-1)\ell+1}(\\{\lambda\\}_{n})}}$
$\displaystyle\qquad\times
S_{n+1}\left(\\{\mu_{j}\\}_{n+1},\\{\xi_{(i-1)\ell+1},\lambda_{1},\ldots,\lambda_{n}\\};\\{\xi_{k}^{(\ell)}\\}\right)\,.$
(7.41)

Let us define the form factor for $K$ by

$F^{K(\ell+)}_{n}(i,\,\\{\mu_{j}\\},\\{\lambda_{k}\\})=\langle
0|\prod_{j=1}^{n}C^{(\ell+)}(\mu_{j})\cdot\left(K_{i}^{(\ell+)}\right)^{\bar{\chi}}\prod_{k=1}^{n}B^{(\ell+)}(\lambda_{k})|0\rangle\,.$
(7.42)

Here we recall that $\\{\mu_{j}\\}$ and $\\{\lambda_{k}\\}$ are solutions of
the Bethe ansatz equations. Similarly, we can calculate as follows.

$\displaystyle
F^{K(\ell+)}_{n}(i,\,\\{\mu_{j}\\},\\{\lambda_{k}\\})=\exp\left({\sum_{j=1}^{n}\mu_{j}-\sum_{k=1}^{n}\lambda_{k}}\right){\frac{\phi_{(i-1)\ell}(\\{\mu_{j}\\}_{n})}{\phi_{i\ell}(\\{\lambda\\}_{n})}}$
$\displaystyle\quad\times\sum_{n=0}^{\ell}\left[\begin{array}[]{c}\ell\\\
n\end{array}\right]_{q}\,q^{\ell-2n+n(\ell-n)}\langle
0|\prod_{j=1}^{n}C(\mu_{j})\prod_{k=1}^{n}D(\xi_{(i-1)\ell+k})\prod_{k=n+1}^{\ell}A(\xi_{(i-1)\ell+k})\prod_{k=1}^{n}B(\lambda_{k})|0\rangle$
(7.45)

Through the commutation relation between $A$ and $B$ and that between $C$ and
$D$, the vacuum expectation of the product of $C$, $D$, $A$ and $B$ operators
can be expressed in terms of the sum of scalar products.

##  Appendix A Derivation of symmetry relations for $R_{p}^{\sigma}$

######  Lemma A.1.

Let $p$ be a sequence of $n$ integers, $1,2,\ldots,n$. For any
$\sigma_{A},\sigma_{B}\in{\cal S}_{n}$ we have

$(\sigma_{A}\sigma_{B})\,p=\sigma_{B}(\sigma_{A}p)\,.$ (A.1)

###### Proof.

Let us denote $p_{\sigma_{A}i}$ by $q_{i}$ for $i=1,2,\ldots,n$. We thus have

$\displaystyle\sigma_{B}(q_{1},\ldots,q_{n})$ $\displaystyle=$
$\displaystyle(q_{\sigma_{B}1},\ldots,q_{\sigma_{B}n})$ (A.2) $\displaystyle=$
$\displaystyle(p_{\sigma_{A}(\sigma_{B}1)},\ldots,p_{\sigma_{A}(\sigma_{B}n)})$
$\displaystyle=$
$\displaystyle(p_{\sigma_{A}\sigma_{B}\,1},\ldots,p_{\sigma_{A}\sigma_{B}\,n})$
$\displaystyle=$ $\displaystyle(\sigma_{A}\sigma_{B})\,p$

∎

######  Proposition A.1.

Definition 2.21 is well defined. That is, for
$R_{j,k}=R_{j,k}(\lambda_{j},\lambda_{k})$ the following relations hold:

$\displaystyle R_{p}^{(\sigma_{A}\sigma_{B})\sigma_{C}}$ $\displaystyle=$
$\displaystyle R_{p}^{\sigma_{A}(\sigma_{B}\sigma_{C})}$ (A.3) $\displaystyle
R_{p}^{s_{j}^{2}}$ $\displaystyle=$ $\displaystyle
R_{p}^{e}=1\quad{\mbox{f}or}\,j=1,2,\ldots,n\,,$ (A.4) $\displaystyle
R_{p}^{s_{i}s_{i+1}s_{i}}$ $\displaystyle=$ $\displaystyle
R_{p}^{s_{i+1}s_{i}s_{i+1}}\quad{\mbox{f}or}\,i=1,2,\ldots,n-1\,.$ (A.5)

###### Proof.

We recall that in terms of generators $s_{j}$ the defining relations of the
symmetric group ${\cal S}_{n}$ are given by
$s_{i}s_{i+1}s_{i}=s_{i+1}s_{i}s_{i+1}$ for $i=1,2,\ldots,n-1$ and
$s_{j}^{2}=1$ for $j=1,2,\ldots,n$ [35]. It thus follows that definition 2.21
is well defined if and only if conditions (A.3), (A.4) and (A.5) hold. We now
show them. Lemma A.1 leads to conditions (A.3). Conditions (A.4) are derived
from the inversion relations (unitarity conditions) (2.8). Finally we show
conditions (A.5) by the Yang-Baxter equations. ∎

######  Lemma A.2.

Let $p$ be a sequence of integers $1,2,\ldots,n$, and $R_{j,k}$ denote
$R_{j,k}(\lambda_{j},\lambda_{k})$ for $j,k=1,2,\ldots,n$. For
$\sigma_{c}=(12\cdots n)$ we have

$R^{\sigma_{c}}_{p}=R_{p_{1},p_{2}\cdots p_{n}}\,.$ (A.6)

###### Proof.

Noting $(12\cdots n)=(12)(23)\cdots(n-1\,n)=s_{1}s_{2}\cdots s_{n-1}$. we have

$\displaystyle R_{p}^{\sigma_{c}}$ $\displaystyle=$ $\displaystyle
R_{p}^{s_{1}s_{2}\cdots s_{n-1}}=R_{s_{1}p}^{s_{2}\cdots
s_{n-1}}R_{p}^{s_{1}}$ (A.7) $\displaystyle=$ $\displaystyle
R_{s_{2}(s_{1}p)}^{s_{3}\cdots
s_{n-1}}R_{s_{1}p}^{s_{2}}R_{p}^{s_{1}}=R_{(s_{1}s_{2})p}^{s_{3}\cdots
s_{n-1}}R_{s_{1}p}^{s_{2}}R_{p}^{s_{1}}$ $\displaystyle=$ $\displaystyle
R_{(s_{1}\cdots s_{n-2})p}^{s_{n-1}}\cdots
R_{(s_{1}s_{2})p}^{s_{3}}R_{s_{1}p}^{s_{2}}R_{p}^{s_{1}}$ $\displaystyle=$
$\displaystyle R_{(12\cdots n-1)p}^{s_{n-1}}\cdots
R_{(123)p}^{s_{3}}R_{(12)p}^{s_{2}}R_{p}^{s_{1}}$ $\displaystyle=$
$\displaystyle R_{p_{1},p_{n}}\cdots R_{p_{1},p_{3}}R_{p_{1},p_{2}}$
$\displaystyle=$ $\displaystyle R_{p_{1},p_{2}\cdots p_{n}}\,.$

∎

######  Proposition A.2.

Let ${\cal A}$ be a Hopf algebra and ${\cal R}$ is an element of ${\cal
A}\otimes{\cal A}$ such that ${\cal R}\Delta(x)=\tau\circ\Delta(x){\cal R}$
for all $x\in{\cal A}$. Suppose that ${\cal R}$ is given by ${\cal
R}=\sum_{a}r^{(a,1)}\otimes r^{(a,2)}$, where $r^{(a,1)}$, $r^{(a,2)}\in{\cal
A}$. We define $R_{j,k}$ by

$R_{j,k}=\sum_{a}id_{1}\otimes\cdots\otimes r_{j}^{(a,1)}\otimes\cdots\otimes
r_{k}^{(a,2)}\otimes\cdots\otimes id_{n}\quad\in{\cal A}^{\otimes n}\,.$ (A.8)

If $R_{j,k}$ satisfy the inversion relations and the Yang-Baxter equations:

$\displaystyle R_{12}R_{21}$ $\displaystyle=$ $\displaystyle id\,,$
$\displaystyle R_{12}R_{13}R_{23}$ $\displaystyle=$ $\displaystyle
R_{23}R_{13}R_{12}\,,$ (A.9)

then we have the following symmetry relations:

$R_{p_{q}}^{\sigma}\Delta^{(n-1)}(x)=\sigma\circ\Delta^{(n-1)}(x)R_{p_{q}}^{\sigma}\quad
x\in{\cal A}\,.$ (A.10)

Here $p_{q}$ denotes $p_{q}=(1,2,\ldots,n)$, a sequence of $n$ integers, and
we have defined $R_{p_{q}}^{\sigma}$ as in definition 2.21.

###### Proof.

Recall that any given permutation $\sigma$ is expressed as a product of
generators $s_{j}=(j\,j+1)$. We show symmetry relations (A.10) by induction on
the number of generators $s_{j}$ whose product gives permutation $\sigma$.
Suppose that (A.10) holds for $\sigma=\sigma_{A}$. We now show that (A.10)
holds for $\sigma=\sigma_{A}s_{j}$. Making use of eq. (2.21) in definition
2.21 we have

$R_{p_{q}}^{\sigma_{A}s_{j}}\Delta^{(n-1)}(x)=R_{\sigma_{A}(p_{q})}^{s_{j}}\,R_{p_{q}}^{\sigma_{A}}\Delta^{(n-1)}(x)=R_{\sigma_{A}(p_{q})}^{s_{j}}\,\sigma_{A}\circ\Delta^{(n-1)}(x)\,R_{p_{q}}^{\sigma_{A}}$
(A.11)

Expressing $\Delta^{(n-1)}(x)$ as $\Delta^{(n-1)}(x)=\sum x^{(1)}\otimes
x^{(2)}\otimes\cdots\otimes x^{(n)}$, we have

$\sigma_{A}\circ\Delta^{(n-1)}(x)=\sum x^{(\sigma_{A}^{-1}1)}\otimes
x^{(\sigma_{A}^{-1}2)}\otimes\cdots\otimes x^{(\sigma_{A}^{-1}n)}$ (A.12)

It now follows from definition 2.21 that we have
$R_{\sigma_{A}(p_{q})}^{s_{j}}=R_{\sigma_{A}j,\,\sigma_{A}(j+1)}$. Let us
denote $\sigma_{A}j$ and $\sigma_{A}(j+1)$ by $a$ and $b$, respectively. Then
we have $\sigma_{A}^{-1}a=j$ and $\sigma_{A}^{-1}b=j+1$. Assuming $a<b$ we
have

$\displaystyle
R_{\sigma_{A}j,\,\sigma_{A}(j+1)}\,\,\sigma_{A}\circ\Delta^{(n-1)}(x)$ (A.13)
$\displaystyle=$ $\displaystyle R_{a,b}\,\sum
x_{1}^{(\sigma_{A}^{-1}1)}\otimes
x_{2}^{(\sigma_{A}^{-1}2)}\otimes\cdots\otimes
x^{(\sigma_{A}^{-1}a)}_{a}\otimes\cdots\otimes
x^{(\sigma_{A}^{-1}b)}_{b}\otimes\cdots\otimes x^{(\sigma_{A}^{-1}n)}_{n}$
$\displaystyle=$ $\displaystyle\sum x^{(\sigma_{A}^{-1}1)}_{1}\otimes
x^{(\sigma_{A}^{-1}2)}_{2}\otimes\cdots\otimes
x^{(\sigma_{A}^{-1}b)}_{a}\otimes\cdots\otimes
x^{(\sigma_{A}^{-1}a)}_{b}\otimes\cdots\otimes
x^{(\sigma_{A}^{-1}n)}_{n}\,\,R_{a,b}$ $\displaystyle=$
$\displaystyle(\sigma_{A}s_{j})\circ\Delta^{(n-1)}(x)\,\,R_{\sigma_{A}j,\,\sigma_{A}(j+1)}$

Here we note that

$R_{a,b}\,x^{(\sigma_{A}^{-1}a)}_{a}\otimes
x^{(\sigma_{A}^{-1}b)}_{b}=x^{(\sigma_{A}^{-1}b)}_{a}\otimes
x^{(\sigma_{A}^{-1}a)}_{b}\,R_{a,b}$ (A.14)

since we have

$x^{(\sigma_{A}^{-1}a)}\otimes x^{(\sigma_{A}^{-1}b)}=x^{(j)}\otimes
x^{(j+1)}=\Delta({\bar{x}}^{(j)})$

where ${\bar{x}}^{(j)}$ are defined by

$\Delta^{(n-2)}(x)=\sum{\bar{x}}^{(1)}\otimes\cdots\otimes{\bar{x}}^{(n-1)}$

We therefore have

$\displaystyle
R_{\sigma_{A}(p_{q})}^{s_{j}}\,\sigma_{A}\circ\Delta^{(n-1)}(x)\,R_{p_{q}}^{\sigma_{A}}=(\sigma_{A}s_{j})\circ\Delta^{(n-1)}(x)\,R_{\sigma_{A}(p_{q})}^{s_{j}}\,R_{p_{q}}^{\sigma_{A}}$
$\displaystyle=(\sigma_{A}s_{j})\circ\Delta^{(n-1)}(x)\,R_{p_{q}}^{\sigma_{A}s_{j}}$
(A.15)

∎

Symmetry relations similar to (A.10) hold for products of monodromy matrices.
Let us consider $m$ auxiliary spaces with suffices $a(1),a(2),\ldots,a(m)$,
respectively. We denote the monodromy matrix
$T_{a(j)}(\lambda_{a(j)};\xi_{1},\ldots,\xi_{L})$ simply by $T_{a(j)}$. We
denote by $\Delta^{(m-1)}(T)$ the following operator:

$\Delta^{(m-1)}(T)=T_{a(1)}T_{a(2)}\cdots T_{a(m)}$ (A.16)

Let $\sigma$ an element of ${\cal S}_{m}$. We define the action of $\sigma$ on
$\Delta^{(m-1)}(T)$ by the following:

$\sigma\circ\Delta^{(m-1)}(T)=T_{a({\bar{\sigma}}1)}T_{a({\bar{\sigma}}2)}\cdots
T_{a({\bar{\sigma}}m)}$ (A.17)

Here ${\bar{\sigma}}$ denotes the inverse of $\sigma$:
${\bar{\sigma}}=\sigma^{-1}$. Then we have the following.

######  Proposition A.3.

Let $p_{q}$ be $p_{q}=(1,2,\ldots,m)$. For any $\sigma\in{\cal S}_{m}$ we have

$R^{\sigma}_{p_{q}}\Delta^{(m-1)}(T)=\sigma\circ\Delta^{(m-1)}(T)R^{\sigma}_{p_{q}}$
(A.18)

##  Appendix B Symmetric-group action on products of $R$-matrices and the
$F$-basis

######  Lemma B.1.

([6]) (i) Cocycle conditions hold for $n\leq L$.

$R_{2\cdots n-1,\,n}R_{1,\,2\cdots n}=R_{1,\,2\cdots n-1}R_{12\cdots n-1,\,n}$
(B.1)

(ii) The unitarity relations hold for $n\leq L$.

$R_{1,\,2\cdots n}R_{23\cdots n,\,1}=I^{\otimes L}$ (B.2)

###### Proof.

Cocycle conditions (B.1) are derived from the Yang-Baxter equations. ∎

Let us denote by the symbol $(p_{0},p_{1},p_{2},\ldots,p_{n})$ a sequence of
$n+1$ integers, $0,1,2,\ldots,n$, and we express it as $(p_{0},p)$ where $p$
denotes the subsequence $(p_{1},p_{2},\ldots,p_{n})$.

######  Lemma B.2.

Let $p$ be a sequence of $n$ integers, $1,2,\ldots,n$. For
$s_{j}=(j\,j+1)\in{\cal S}_{n}$ we have

$\displaystyle R_{p}^{s_{j}}R_{0,\,p}$ $\displaystyle=$ $\displaystyle
R_{0,\,s_{j}(p)}R_{p}^{s_{j}}$ $\displaystyle R_{p}^{s_{j}}R_{p,\,0}$
$\displaystyle=$ $\displaystyle R_{s_{j}(p),\,0}R_{p}^{s_{j}}$ (B.3)

###### Proof.

We first note $R_{p}^{s_{j}}=R_{p_{j},p_{j+1}}$. We have

$\displaystyle R^{s_{j}}_{p}R_{0,p}$ $\displaystyle=$ $\displaystyle
R_{p_{j},p_{j+1}}\,R_{0,p_{n}}\cdots R_{0,p_{1}}$ (B.4) $\displaystyle=$
$\displaystyle R_{0,p_{n}}\cdots R_{0,p_{j+2}}\cdot
R_{p_{j},p_{j+1}}R_{0,p_{j+1}}R_{0,p_{j}}\cdot R_{0,p_{j+2}}\cdots
R_{0,p_{1}}$

Applying the Yang-Baxter equations
$R_{p_{j},p_{j+1}}R_{0,p_{j+1}}R_{0,p_{j}}=R_{0,p_{j}}R_{0,p_{j+1}}R_{p_{j},p_{j+1}}$
we have

$\displaystyle R^{s_{j}}_{p}R_{0,p}$ $\displaystyle=$ $\displaystyle
R_{0,p_{n}}\cdots R_{0,p_{j+2}}\cdot R_{0,p_{j}}R_{0,p_{j+1}}\cdot
R_{0,p_{j-1}}\cdots R_{0,p_{1}}\cdot R_{p_{j},p_{j+1}}$ (B.5) $\displaystyle=$
$\displaystyle R_{0,s_{j}(p)}R_{p_{j},p_{j+1}}\,.$

∎

######  Proposition B.1.

Let $p$ be a sequence of $n$ integers, $1,2,\ldots,n$, and $\sigma$ a
permutation of the $n$ integers. We have

$\displaystyle R_{p}^{\sigma}R_{0,\,p}$ $\displaystyle=$ $\displaystyle
R_{0,\,\sigma(p)}R_{p}^{\sigma}$ $\displaystyle R_{p}^{\sigma}R_{p,\,0}$
$\displaystyle=$ $\displaystyle R_{\sigma(p),\,0}R_{p}^{\sigma}$ (B.6)

###### Proof.

Expressing permutation $\sigma$ as a product of generators $s_{j}$, and
applying (B.3) many times, we can show the symmetry relations. ∎

We define the action of $\sigma$ on the $F$-basis as follows.

$F_{\sigma(p)}=F_{p_{\sigma(1)}p_{\sigma(2)}\cdots p_{\sigma(n)}}$ (B.7)

######  Proposition B.2.

Let $p$ be a sequence of integers, $1,2,\ldots,n$. For $\sigma\in{\cal S}_{n}$
we have

$R^{\sigma}_{p}F_{0,p}=F_{0,\sigma(p)}R_{p}^{\sigma}$ (B.8)

We also have

$F_{p}=F_{\sigma(p)}R_{p}^{\sigma}$ (B.9)

###### Proof.

Expressing the $F$-basis in terms of the $R$-matrices we have

$\displaystyle R^{\sigma}_{p}F_{0,p}$ $\displaystyle=$ $\displaystyle
R^{\sigma}_{p}\left(e_{0}^{11}+e_{0}^{22}R^{\sigma}_{0,p}\right)$ (B.10)
$\displaystyle=$ $\displaystyle
e_{0}^{11}R^{\sigma}_{p}+e_{0}^{22}R^{\sigma}_{p}R^{\sigma}_{0,p}$
$\displaystyle=$ $\displaystyle
e_{0}^{11}R^{\sigma}_{p}+e_{0}^{22}R^{\sigma}_{0,\sigma(p)}R^{\sigma}_{p}$
$\displaystyle=$
$\displaystyle\left(e_{0}^{11}+e_{0}^{22}R^{\sigma}_{0,\sigma(p)}\right)R^{\sigma}_{p}$
$\displaystyle=$ $\displaystyle F_{0,\sigma(p)}R^{\sigma}_{p}$

We first show (B.9) with $\sigma=s_{j}$ for $j=1,2,\ldots,n-1$, and then we
derive it for all permutations $\sigma$. ∎

######  Lemma B.3.

The propagator $F^{-1}_{i\cdots L\,1\cdots i-1}F_{1\cdots L}$ is given by the
following:

$F^{-1}_{i\cdots L\,1\cdots i-1}F_{1\cdots
L}=\prod_{\alpha=1}^{i-1}\left(A_{1\cdots L}(\xi_{\alpha})+D_{1\cdots
L}(\xi_{\alpha})\right)$ (B.11)

###### Proof.

Let $\sigma_{c}$ be a cyclic permutation: $\sigma_{c}=(12\cdots L)$, and
$p_{q}$ the sequence $p_{q}=(1,2,\ldots,n)$. We have

$F_{i\cdots L\,1\cdots i-1}=F_{\sigma_{c}^{i-1}(p_{q})}=F_{1\cdots
L}R_{p_{q}}^{\sigma_{c}^{i-1}}$ (B.12)

and hence we have

$\displaystyle F^{-1}_{i\cdots L\,1\cdots i-1}F_{1\cdots L}$ $\displaystyle=$
$\displaystyle\left(F_{1\cdots
L}R_{p_{q}}^{\sigma_{c}^{i-1}}\right)^{-1}F_{1\cdots L}$ (B.13)
$\displaystyle=$ $\displaystyle R_{p_{q}}^{\sigma_{c}^{i-1}}\,F_{1\cdots
L}^{-1}F_{1\cdots L}$ $\displaystyle=$ $\displaystyle
R_{p_{q}}^{\sigma_{c}^{i-1}}$

We thus obtain

$\displaystyle R_{p_{q}}^{\sigma_{c}^{i-1}}$ $\displaystyle=$ $\displaystyle
R_{\sigma(p_{q})}^{\sigma_{c}^{i-2}}R_{p_{q}}^{\sigma_{c}}$ $\displaystyle=$
$\displaystyle R_{\sigma_{c}^{i-2}(p_{q})}^{\sigma_{c}}\cdots
R_{p_{q}}^{\sigma_{c}}$ $\displaystyle=$ $\displaystyle R_{i-1\cdots
L\,1\cdots i-2}^{\sigma_{c}}\cdots R_{2\cdots L1}^{\sigma_{c}}R_{12\cdots
L}^{\sigma_{c}}$ $\displaystyle=$ $\displaystyle R_{i-1,i\cdots L\,1\cdots
i-2}\cdots R_{2,3\cdots L1}R_{1,2\cdots L}$ $\displaystyle=$
$\displaystyle\prod_{\alpha=1}^{i-1}\left(A_{1\cdots
L}(\xi_{\alpha})+D_{1\cdots L}(\xi_{\alpha})\right)$

∎

##  Appendix C Formulas of the $q$-analogue

######  Lemma C.1.

For two integers $\ell$ and $n$ satisfying $0\leq n\leq\ell$ we have

$\sum_{1\leq
i_{1}<\cdots<i_{n}\leq\ell}q^{2i_{1}+\cdots+2i_{n}}=q^{n(\ell+1)}\,\left[\begin{array}[]{c}\ell\\\
n\end{array}\right]_{q}$ (C.1)

###### Proof.

We can show by induction on $\ell$ the $q$-binomial expansion as follows.

$\prod_{k=0}^{\ell-1}\left(1-zq^{2k}\right)=\sum_{n=0}^{\ell}(-z)^{n}q^{n(\ell-1)}\left[\begin{array}[]{c}\ell\\\
n\end{array}\right]_{q}$ (C.2)

It is now easy to show the following:

$\prod_{k=1}^{\ell}\left(1-zq^{2k}\right)=\sum_{n=0}^{\ell}(-z)^{n}\sum_{1\leq
i_{1}<\cdots<i_{n}\leq\ell}q^{2i_{1}+\cdots+2i_{n}}$ (C.3)

Comparing (C.3) with (C.2) we obtain formula (C.1). ∎

######  Lemma C.2.

The spin-$\ell$ matrix representations $X^{\pm(\ell+)}$ of the generators
$X^{\pm}$ of $U_{q}(sl_{2})$ are related to $\sigma_{\ell}^{+}$ and
$\sigma_{1}^{-}$, respectively, as follows.

$\displaystyle
P^{(\ell)}_{1\cdots\ell}\,\sigma_{1}^{-}\,P^{(\ell)}_{1\cdots\ell}$
$\displaystyle=$ $\displaystyle{\frac{1}{[\ell]_{q}}}\,X^{-(\ell+)}$ (C.4)
$\displaystyle
P^{(\ell)}_{1\cdots\ell}\,\sigma_{\ell}^{+}\,P^{(\ell)}_{1\cdots\ell}$
$\displaystyle=$ $\displaystyle{\frac{1}{[\ell]_{q}}}\,X^{+(\ell+)}$ (C.5)

###### Proof.

Expressing the projection operator $P^{(\ell)}_{1\cdots\ell}$ as

$P^{(\ell)}_{1\cdots\ell}=\sum_{n=0}^{\ell}||\ell,n\rangle\langle\ell,n||$

we have

$P^{(\ell)}_{1\cdots\ell}\sigma_{1}^{-}P^{(\ell)}_{1\cdots\ell}=\sum_{n=0}^{\ell-1}||\ell,n+1\rangle\langle\ell,n||\quad\langle\ell,n+1||\sigma_{1}^{-}||\ell,n\rangle$
(C.6)

Making use of (C.1) we have

$\langle\ell,n+1||\sigma_{1}^{-}||\ell,n\rangle=\frac{[n+1]}{[\ell]},$ (C.7)

and then we obtain (C.4). Similarly, we can show (C.5). ∎

##  Appendix D Formulas of the $F$-basis useful for the diagonalization.

Let us review some points of the diagonalization process of the $A$ and $D$
operators [6].

######  Lemma D.1.

Operators $A$ and $D$ are upper- and lower-triangular matrices, respectively.
Moreover, the eigenvalues of operators $A$ and $D$ are given by

$\displaystyle{\rm diag}\left(D_{12\cdots n}(\lambda_{0})\right)$
$\displaystyle=$
$\displaystyle\bigotimes_{i=1}^{n}\left(\begin{array}[]{cc}b_{0i}&0\\\
0&1\end{array}\right)_{[i]}\,,$ (D.3) $\displaystyle{\rm
diag}\left(A_{12\cdots n}(\lambda_{0})\right)$ $\displaystyle=$
$\displaystyle\bigotimes_{i=1}^{n}\left(\begin{array}[]{cc}1&0\\\
0&b_{0i}\end{array}\right)_{[i]}\,,$ (D.6)

where $b_{0i}=b(\lambda_{0}-\xi_{i})$.

###### Proof.

We can show it by induction on $n$. Noting $D_{12\cdots n}=C_{n}B_{1\cdots
n-1}+D_{n}D_{1\cdots n-1}$, we show ${\rm diag}(D_{1\cdots n}={\rm
diag}(D_{1\cdots n-1}(\lambda_{0})\otimes{\rm diag}(b_{0n},1)_{[n]}$. We can
show (D.6) similarly. ∎

######  Lemma D.2.

The partial $F_{0,1\cdots n}$ and ${\bar{F}}_{0,1\cdots n}$ are expressed as
follows.

$\displaystyle{F}_{0,1\cdots n}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}I_{1\cdots n}&0\\\ C_{1\cdots
n}(\lambda_{0})&D_{1\cdots n}(\lambda_{0})\end{array}\right)_{[0]}\,,$ (D.9)
$\displaystyle{\bar{F}}_{0,1\cdots n}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}A_{1\cdots n}(\lambda_{0})&B_{1\cdots
n}(\lambda_{0})\\\ 0&I_{1\cdots n}\end{array}\right)_{[0]}\,.$ (D.12)

###### Proof.

It is clear from definition 5.2 of the $F$-basis. ∎

Proof of propositions 25 and 26

For an illustration, we now derive the diagonalized form of the $D$ operator.
From $R_{0,1\cdots n}=R_{0,2\cdots n}R_{0,1}$ we have

$\displaystyle D_{12\cdots n}$ $\displaystyle=$ $\displaystyle C_{2\cdots
n}B_{1}+D_{2\cdots n}D_{1}$ (D.15) $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}b_{01}D_{2\cdots n}(\lambda_{0})&0\\\
c_{01}C_{2\cdots n}(\lambda_{0})&D_{2\cdots
n}(\lambda_{0})\end{array}\right)_{[0]}\,.$

We thus calculate

$\displaystyle F_{1\cdots n}D_{1\cdots n}(\lambda_{0})$ $\displaystyle=$
$\displaystyle F_{2\cdots n}\,F_{1,2\cdots n}\,D_{1\cdots n}(\lambda_{0})$
(D.20) $\displaystyle=$ $\displaystyle F_{2\cdots
n}\,\left(\begin{array}[]{cc}I_{2\cdots n}&0\\\ C_{2\cdots
n}(\lambda_{0})&D_{2\cdots
n}(\lambda_{0})\end{array}\right)_{[0]}\left(\begin{array}[]{cc}b_{01}D_{2\cdots
n}(\lambda_{0})&0\\\ c_{01}C_{2\cdots n}(\lambda_{0})&D_{2\cdots
n}(\lambda_{0})\end{array}\right)_{[0]}$ $\displaystyle=$ $\displaystyle{\rm
diag}(b_{01},1)_{[1]}\widetilde{D}_{2\cdots n}(\lambda_{0})F_{1\cdots n}$
(D.21)

Therefore, by induction we have the diagonalized form of operator $D$.
Similarly, we can diagonalize $A$, $A^{\dagger}$ ] and $D^{\dagger}$.

######  Lemma D.3.

The diagonalized form of $F_{0,1\cdots n}{\bar{F}}^{\dagger}_{0,1\cdots n}$ is
given by the following:

$F_{1\cdots n}\left(F_{0,1\cdots n}{\bar{F}}^{\dagger}_{0,1\cdots
n}\right)F_{1\cdots n}^{-1}={\hat{\delta}}^{-1}_{0,1\cdots n}$ (D.22)

###### Proof.

Making use of (5.12) we show

$\displaystyle F_{0,1\cdots n}{\bar{F}}_{0,1\cdots n}^{\dagger}=F_{0,1\cdots
n}{\cal C}_{01\cdots n}{F}_{0,1\cdots n}^{\dagger}{\cal C}_{01\cdots n}$
(D.31) $\displaystyle=$ $\displaystyle\left(\begin{array}[]{cc}1&0\\\
C_{1\cdots n}&D_{1\cdots
n}\end{array}\right)_{[0]}\left(\begin{array}[]{cc}0&{\cal C}_{1\cdots n}\\\
{\cal C}_{1\cdots
n}&0\end{array}\right)_{[0]}\left(\begin{array}[]{cc}1&C^{\dagger}_{1\cdots
n}\\\ 0&D_{1\cdots n}\end{array}\right)_{[0]}\left(\begin{array}[]{cc}0&{\cal
C}_{1\cdots n}\\\ {\cal C}_{1\cdots n}&0\end{array}\right)_{[0]}$
$\displaystyle=$ $\displaystyle\left(\begin{array}[]{cc}A^{\dagger}&0\\\
0&D_{1\cdots n}\end{array}\right)_{[0]}\,.$ (D.34)

Calculating $F_{12\cdots n}\cdot F_{0,1\cdots n}{\bar{F}}_{0,1\cdots
n}^{\dagger}\cdot F^{-1}_{12\cdots n}$ we have (D.22). ∎

######  Corollary D.1.

The inverse of the total $F$ is given as follows.

$F^{-1}_{1\cdots n}={\bar{F}}_{1\cdots n}^{\dagger}{\hat{\delta}}_{1\cdots n}$
(D.35)

###### Proof.

We show it by induction on $n$. Let us assume (D.35) for the case of $n$. We
have

${\bar{F}}^{\dagger}_{01\cdots n}\widehat{\delta}_{01\cdots
n}={\bar{F}}^{\dagger}_{01\cdots n}\left({\bar{F}}^{\dagger}_{1\cdots
n}\widehat{\delta}_{1\cdots n}\right)\widehat{\delta}_{0,1\cdots
n}={\bar{F}}^{\dagger}_{01\cdots n}{F}^{-1}_{1\cdots
n}\widehat{\delta}_{0,1\cdots n}$

From (D.22) we have ${\bar{F}}^{\dagger}_{01\cdots n}{F}^{-1}_{1\cdots
n}\widehat{\delta}_{0,1\cdots n}=F_{01\cdots n}^{-1}$, which corresponds to
(D.35) for the case of $n+1$. ∎

######  Lemma D.4.

The dagger of total $F$ is given by the charge conjugation of transposed total
$F$.

$F_{1\cdots n}^{\dagger}={\cal C}_{1\cdots n}F^{t_{1}\cdots t_{n}}_{n\cdots
21}{\cal C}_{1\cdots n}$ (D.36)

Or equivalently we have