One Hat Cyber Team

View File Name : CHAPT27.TEX

\section{$q$-Deformation of $U_q(so(3,{\Bbb C}))$}

$q$-Deformation of the orthogonal Lie algebra $so(3)$ was proposed by
Fairlie \cite{}. This nonstandard $q$-analog $U_q(so(3,{\Bbb C}))$ is
constructed  starting from $so(3,{\Bbb C})$
defined by generating elements $I_1$, $I_2$, $I_3$. Namely,
$U_q(so(3,{\Bbb C}))$ is the associative algebra generated by $I_1$,
$I_2$, $I_3$ satisfying the relations:
\begin{eqnarray}\label{soq3}
q^{1/2}I_1I_2-q^{-1/2}I_2I_1=I_3\nonumber\\
q^{1/2}I_2I_3-q^{-1/2}I_3I_2=I_1\\
q^{1/2}I_3I_1-q^{-1/2}I_1I_3=I_2\nonumber
\end{eqnarray}
Note that the Lie algebras $sl(2,{\Bbb C})$ and $so(3,{\Bbb C})$ are
isomorphic. However, the quantum algebra $U_q(sl(2,{\Bbb C})$ which is
a Hopf algebra, differs from $U_q(so(3,{\Bbb C}))$.

Let us describe $*$-structures  (involutions) over the algebra
$U_q(so(3,{\Bbb C}))$. It is clear that an involution
in an algebra with generators and relations is completely defined
by its values on the generators.
An involution $*$ may send linear combinations of generators to linear combinations of
generators. In this case $*$ is said to be an involution of the first order or a linear
involution. On the other hand, there might exist involutions
which map linear combinations of generators to the polynomials
in generators of the degree higher then one. We will
call such involutions nonlinear. If linear combinations of generators
are  mapped by
an involution to the polynomials of  the second degree  then we will call such
involutions quadratic.

Here we will consider, for the algebras $ U_q(so(3,{\Bbb C})) $,
all involutions of the first order and some quadratic involutions.

\begin{theorem} \label{th-inv1-eq}
\begin{itemize}
\item[$1)$] If $ q\in{\Bbb R} $, $|q|\ne 1$ then all involutions of the
first order  in the algebra $U_q(so(3,{\Bbb C})) $ are equivalent to
the following  involution:
\item[$a)$] $I_1^*=I_2$, $I_2^*$,
$I_3^*=\left\{
\begin{array}{ll}
I_3,&q>0\\
-I_3,&q<0
\end{array}\right.$.

\item[$2)$] If $ |q| = 1, q \neq \pm 1 $, then all involutions of the
first order  in the algebra $ U_q(so(3,{\Bbb C})) $ are equivalent to
the following
two inequivalent involutions:
\item[$a)$] $ I_1^* = I_1 $ , $ I_2^*=-I_2 $ , $ I_3^*=I_3 $ ,
\item[$b)$] $ I_1^* = -I_1 $ ,$ I_2^*=-I_2 $, $ I_3^*=-I_3 $ ,
\item[$3)$] If $q=-1$, then each involution of the first order
in the algebra $ U_q(so(3,{\Bbb C})) $ is either equivalent to 1a), or  2a),
or  2b)
\end{itemize}
\end{theorem}
\begin{proof}
By definition any involution of the first order in the algebra
$ U_q(so(3,{\Bbb C}))$
is defined, on the generators $I_1$, $I_2$ and $I_3$, by formulas of the form
\begin{eqnarray}
I_1^{*}=c_{11}I_1+c_{12}I_2+c_{13}I_3, \nonumber \\
I_2^{*}=c_{21}I_1+c_{22}I_2+c_{23}I_3, \label{invconst}\\
I_3^{*}=c_{31}I_1+c_{32}I_2+c_{33}I_3, \nonumber
\end{eqnarray}
where $C=[c_{jk}]$ is a complex $3\times3$ matrix.
The condition $(A^*)^*=A$ , $A\in U_q(so(3,{\Bbb C}))$, is satisfied if
and only if it
is satisfied for generators $I_1$, $I_2$ and $I_3$,
that is if and only if
\begin{equation}\label{invdoublmatr}
\overline{C}C=I,
\end{equation}
where $\overline{C}=[\overline{c_{jk}}]$ is the matrix obtained
from $C$ by the complex conjugation of all elements.
Applying  involution to the commutation relations (\ref{soq3})
and using the axioms for $*$ and
(\ref{invconst}) we get three noncommutative
polynomials in variables $I_1$, $I_2$ and $I_3$
which must be zero in $U_q(so(3,{\Bbb C}))$. Using the commutation
relations (\ref{soq3})
these three polynomials can be
rewritten in the degree graded lexicographically ordered form,
that is in the form where  $I_1$
does not appear after  $I_2$ and $I_3$
and $I_2$  does not appear after $I_3$, and the monomials of the higher
total degree appear first. These polynomials written  in the
lexicographically ordered form  will
be equal to zero if and only if
all their coefficients are zero, since it can be shown that the lexicographically
ordered monomials
form a basis in  $U_q(so(3,{\Bbb C}))$, that is $U_q(so(3,{\Bbb C}))$
is the PBW-type algebra.
So, we obtain an additional set of
equations for the constants $c_{jk}$. These equations and (\ref{invdoublmatr})
form a nonlinear system
of equations. Solving this system we  get all possible involutions of the first
order in $U_q(so(3,{\Bbb C}))$.
Finally, the obtained involutions can be classified up to isomorphism
of $*$-algebras.
The described calculations are lengthy and we leave them out.
\end{proof}

%Let $ |q|=1, q\neq 1 $ and $ q = e^{i\phi} $ for some
%$ \phi \in ( 0, 2\pi ) $.
It is possible to choose
hermitian  generators
in all the $*$-algebras.  Indeed,  the new generators
$a_1=(I_1+I_2)/2$,
$a_2=i(I_1-I_2)/2$,
$a_3=I_3$
are hermitian in $R_q^{1a}$. Moreover, it is easy to check that they
satisfy the relations:
\begin{eqnarray}
(q^{\frac{1}{2}}-q^{-\frac{1}{2}})(a_1^2+a_2^2)+i(q^{\frac{1}{2}}+
q^{-\frac{1}{2}})[a_1,a_2]=a_3,\nonumber\\
q^{\frac{1}{2}}a_1 a_3-q^{-\frac{1}{2}}a_3 a_1+i(q^{\frac{1}{2}}a_2
a_3-q^{-\frac{1}{2}}a_3 a_2)=a_1,\\
q^{\frac{1}{2}}a_3 a_1-q^{-\frac{1}{2}}a_1 a_3-i(q^{\frac{1}{2}}a_3
a_2-q^{-\frac{1}{2}}a_2 a_3)=a_2.\nonumber
\end{eqnarray}
In $R_q^{2a}$ the new generators
$ a_1=I_1$,
$ a_2=iI_2 $,
$ a_3=I_3 $
are hermitian and  satisfy
the following  relations:
\begin{eqnarray}
q^{\frac{1}{2}}a_1 a_2-q^{-\frac{1}{2}}a_2 a_1&=&ia_3, \nonumber \\
q^{\frac{1}{2}}a_2 a_3-q^{-\frac{1}{2}}a_3 a_2&=&ia_1, \label{rel:Mqa} \\
q^{\frac{1}{2}}a_3 a_1-q^{-\frac{1}{2}}a_1 a_3&=&-ia_2   \nonumber .
\end{eqnarray}
In $R_{q}^{2b}$ the new generators
$ a_1=iI_1$,
$ a_2= iI_2 $,
$ a_3=iI_3 $
are hermitian and satisfy the relations:
\begin{eqnarray}
q^{\frac{1}{2}}a_1 a_2-q^{-\frac{1}{2}}a_2 a_1&=&ia_3, \nonumber \\
q^{\frac{1}{2}}a_2 a_3-q^{-\frac{1}{2}}a_3 a_2&=&ia_1, \label{rel:Mqb} \\
q^{\frac{1}{2}}a_3 a_1-q^{-\frac{1}{2}}a_1 a_3&=&ia_2   \nonumber .
\end{eqnarray}

It turns out that the algebras
$U_q(so(3,{\Bbb C}))$ have  also some  quadratic involutions.

Let us substitute $I_3$ in the second and third relation in
(\ref{soq3})
by the left hand side of the first relation.
Then we will get the following two relations of the third order
for $I_1$ and $I_2$:
\begin{equation} \label{rel:Mq2}
\begin{tabular}{l}
$I_2^2 I_1 - (q+q^{-1}) I_2 I_1 I_2 + I_1 I_2^{2} = - I_1$ \\
$I_1^2 I_2 - (q+q^{-1}) I_1 I_2 I_1 + I_2 I_1^{2} = -I_2.$
\end{tabular}
\end{equation}

Let us find all involutions of the first degree in the algebra
with generators $I_1$ and $I_2$ and relations (\ref{rel:Mq2}).
The element $I_3$ is defined by relations (\ref{soq3}).
Applying the arguments given in the proof of Theorem~\ref{th-inv1-eq}
we get
the following statement
\begin{proposition}
Any  such involution is isomorphic either to involutions from
Theorem~\ref{th-inv1-eq} or to the following ones:
\begin{itemize} \item[$1)$] $q\in{\Bbb R}$, \hspace{0.6cm}
$I_1^*=I_1$, $I_2^*=I_2$, and $I_3^*=\left\{ \begin{array}{ll}
q^{\frac{1}{2}}I_2I_1-q^{-\frac{1}{2}}I_1I_2,&q>0,\\
-(q^{\frac{1}{2}}I_2I_1-q^{-\frac{1}{2}}I_1I_2),&q<0,
\end{array}\right.$

\item[$2)$] $q \in {\Bbb R}$, \hspace{0.6cm} $I_1^*=I_1$,
$I_2^*=-I_2$, and $I_3^* =\left\{
\begin{array}{ll}
-(q^{\frac{1}{2}}I_2I_1-q^{-\frac{1}{2}}I_1I_2),&q>0,\\
q^{\frac{1}{2}}I_2I_1-q^{-\frac{1}{2}}I_1I_2,&q<0,
\end{array}\right.$

\item[$3)$] $q \in {\Bbb R}$, \hspace{0.6cm} $I_1^*=-I_1$,
$I_2^*=-I_2$, and $I_3^* =\left\{
\begin{array}{ll}
(q^{\frac{1}{2}}I_2I_1-q^{-\frac{1}{2}}I_1I_2),&q>0,\\
-(q^{\frac{1}{2}}I_2I_1-q^{-\frac{1}{2}}I_1I_2),&q<0,
\end{array}\right.$

\item[$4)$] $|q|=1$, \hspace{0.6cm} $I_1^*=I_2$,
$I_2^*=I_1$, $I_3^* =
q^{-\frac{1}{2}}I_2I_1-q^{\frac{1}{2}}I_1I_2$,

\end{itemize}
\end{proposition}

We shall study $*$-representationsof the algebra
$U_q(so(3,{\Bbb C})$ by bounded operators in a Hilbert space $H$.
 Here we restrict ourselves by studing  representations of the
$*$-algebra defined by the involution $I_1^*=-I_1$, $I_2^*=-I_2$, which for
$q=1$ corresponds to the compact real form of $so(3)$. Henceforth
we denote it by $U_q(so(3))$.

Let us consider the following notations
$$[x]=\frac{x^q-x^{-q}}{q-q^{-1}},\quad d(m)=\frac{[m][m+1]}{[2m][2(m+1)]}.$$
Let $E_A(\cdot)$ denote the spectral measure of selfadjoint operator
$A$ and $({\Bbb F}(\cdot))_i$  the $i$-th coordinate of the function
${\Bbb F}:{\Bbb R}^n\rightarrow {\Bbb R}^n$.

Let $J_1$, $J_2$ be defined by $I_1=iJ_1$ and $I_2=iJ_2$. The new generators
$J_1$, $J_2$ of $U_q(so(3))$ are
 selfadjoint
 and satisfy the relations:
\begin{eqnarray}
J_2^2 J_1 - (q+q^{-1}) J_2 J_1 J_2 + J_1 J_2^{2} = J_1, \label{rel:soq3I}\\
J_1^2 J_2 - (q+q^{-1}) J_1 J_2 J_1 + J_2 J_1^{2} = J_2.\label{rel:soq3II}
\end{eqnarray}

{\bf Representations of $U_q(so(3))$, $q>0$}
\begin{theorem}\label{th1}
Any irreducible representation of $U_q(so(3))$, $q>0$, is finite-dimensional.
For any $n\geq 1$ irreducible representations in $H={\Bbb C}^n$ are unitarily
equivalent to the following  one:
$$ J_1e_k=[k-(n+1)/2]e_k,\qquad
 J_2e_k=\left\{
 \begin{array}{ll}
 \alpha_1e_2,& k=1,\\
 \alpha_ke_{k+1}+\alpha_{k-1}e_{k-1},& k\ne1, n,\\
 \alpha_{n-1}e_{n-1},&k=n,
 \end{array}\right.$$
where
$\alpha_k=(d(k-(n+1)/2)[k][n-k])^{1/2}$.

\end{theorem}
\begin{proof}
The proof of the theorem is based on the technigue of semilinear
relations developed in section 2.4 and the technique of dynamical
system.

Let $q=e^{\sigma}$, $\sigma\in{\Bbb  R}$ and $J_1$, $J_2$ be
selfadjoint operators in a Hilbert space $H$ satisfying relations
(\ref{rel:soq3I})--(\ref{rel:soq3II}).
Equation (\ref{rel:soq3I}) is linear with respect to $J_2$ with the
corresponding binary relation:
$$\Gamma=\{(t,s)\mid\Phi(t,s)\equiv t^2-(q+q^{-1})ts+s^2-1=0\}.$$
Let
$F_{1}(s)=sch\sigma+\sqrt{s^2sh^2\sigma+1}$,
$F_{2}(s)=sch\sigma-\sqrt{s^2sh^2\sigma+1}$ and
$sh\sigma=(q-q^{-1})/{2}$.
Then $\Phi(t,s)=(t-F_1(s))(t-F_2(s))$.
Consider the parametrization $s=\frac{sh\sigma
  x(s)}{sh\sigma}=[x(s)]$,
$x(s)\in{\Bbb R}$ which gives  $F_1(s)=[x(s)+1]$, $F_2(s)=[x(s)-1]$, and
$$\Gamma=\{([x+1],[x]), ([x-1],[x])\mid x\in{\Bbb R}\}.$$
Let $E_{J_1}(\cdot)$ be the resolution of the identity for the operator
$J_1$.
Then, by Theorem~\ref{cth2}, $J_1$, $J_2$ satisfy
(\ref{rel:soq3I})--(\ref{rel:soq3II}) if and only if
\begin{equation}\label{supp}
E_{J_1}(\Delta)J_2E_{J_1}(\Delta')=0,\quad \mbox{for any}\
\Delta,\Delta'\in{\frak B}({\Bbb R}),\
\Delta\times\Delta'\cap\Gamma=\emptyset
\end{equation}
For the operator $A_1$ defined by $\displaystyle J_1=\frac{sh\sigma A_1}{sh\sigma}$
condition (\ref{supp}) is equivalent to
$$E_{A_1}(\Delta)J_2E_{A_1}(\Delta')=0,\quad \mbox{for any}\
\Delta,\Delta'\in{\frak B}({\Bbb R}),\
\Delta\times\Delta'\cap\Gamma'=\emptyset,$$
where $\Gamma'=\{(s+1,s), (s-1,s)\mid s\in{\Bbb R}\}$.
It follows from Theorem~\ref{cth2} that the operators $A_1$, $J_2$
satisfy the following relation
$$A_1^2J_2-2A_1J_2A_1+J_2A_1^2=J_2.$$
Let $E_1=([A_1,J_2]+J_2)/2$, $E_1^*=([A_1,J_2]-J_2)/2$. Then one can
check that
\begin{equation}\label{dynrel}
\begin{tabular}{l}
$A_1E_1=E_1(A_1+I),\quad A_1E_1^*=E_1^*(A_1-I),$\\
$E_1^*E_1=F(E_1E_1^*,A_1),$
\end{tabular}
\end{equation}
where $F(x_1,x_2)=x_2\frac{ch((x_1-1)\sigma)}{ch((x_1+1)\sigma)}-
\frac{sh(x_1\sigma)}{2sh\sigma ch((x_1+1)\sigma)}$. Conversely, any
operators ($A_1$, $E_1$, $E_1^*$) satisfying (\ref{dynrel}) determine
a representation ($J_1$, $J_2=E_1+E_1^*$) of
(\ref{rel:soq3I})--(\ref{rel:soq3II}).

It is obvious that the pair ($J_1$, $J_2$) is irreducible if and only
if the family ($A_1$, $E_1$, $E_1^*$) is irreducible. Hence, istead of
representations of relations (\ref{rel:soq3I})--(\ref{rel:soq3II}) we can speak about the
representations of (\ref{dynrel}). To relation (\ref{dynrel}) there
correspond the dynamical system
 $(x_1,x_2)\rightarrow{\Bbb F}(x_1,x_2)\equiv (x_1+1,F(x_1+1,x_2)$ which has the measurable
section ${\Bbb R}^+\times [0,1)$. Therefore the joint spectral measure of
$A_1$, $E_1^*E_1$
is discrete if ($A_1$, $E_1$, $E_1^*$) is irreducible, and we can choose a basis consisting of its
eigenvectors. Then we have
\begin{eqnarray*}
&A_1e_{x_1,x_2}=x_1 e_{x_1,x_2},\quad E_1^*E_1e_{x_1,x_2}=x_2e_{x_1,x_2},\\
&E_1e_{x_1,x_2}=\sqrt{x_2}e_{{\Bbb F}(x_1,x_2)}, E_1^*e_{x_1,x_2}=
\sqrt{({\Bbb F}^{-1}(x_1,x_2))_2}e_{{\Bbb F}^{-1}(x_1,x_2)}
\end{eqnarray*}
where $(x_1,x_2)$ is taken from some orbit. The later can not hold for all
points of the orbit, since
$$\begin{array}{ll}
({\Bbb F}^{(k)}(x_1,x_2))_2=&
x_2\frac{ch(x\sigma)ch((x+1)\sigma)}{ch((x+k)\sigma)ch((x+k+1)\sigma)}-\\
&-\frac{sh(2x+k+1)\sigma)sh(k\sigma)}{4sh^2\sigma
  ch((x+k)\sigma)ch((x+k+1)\sigma)}\rightarrow-\frac{1}{4sh^2\sigma},
\end{array}$$
$k\rightarrow\pm\infty$, while $({\Bbb F}^{(k)}(x_1,x_2))_2$ are
    eigenvalues of the selfadjoint non-negative operator $E_1^*E_1$. Thus there
    exists the highest vector (vector $e_{x_1,x_2}$ with the lagest
    $x_1$) on which $E_1$ acts as zero and the lowest vector, on which
    $E_1^*$ is zero. Using this argument one can easily get the statement.
 \end{proof}
{\bf Representations of $U_q(so(3))$, $q<0$}
\begin{theorem}\label{th2}
Any irreducible representation of $U_q(so(3))$, $q>0$, is finite-dimensional.
For any $p\geq 1$ there exist four non-unitarily equivalent
irreducible representations of
dimension $2p$ and five irreducible representations of dimension
$2p-1$, which act as follows:

1. Four representations with any finite dimension, $\mbox{dim} H=n$
$$ J_1e_k=(-1)^{k-1}[k+(-1)^j/2]e_k,\quad
J_2e_k=\left\{
\begin{array}{ll}
\alpha_1e_2+(-1)^i[n][1/2]e_1,& k=1,\\
\alpha_ke_{k+1}+\alpha_{k-1}e_{k-1},& k\ne 1, n,\\
\alpha_{n-1}e_{n-1},&k=n,
\end{array}\right.$$
where
$\alpha_k=(d(k-1/2)[n-k][n+k])^{1/2}$, $i,j=0,1$;
\vskip0.5cm

2. One more representation for each odd dimension, $\mbox{dim} H=n=2p-1$
$$ J_1e_k=(-1)^{k-1}[k-(n+1)/2]e_k,\qquad
J_2e_k=\left\{
\begin{array}{ll}
\alpha_1e_2,& k=1,\\
\alpha_ke_{k+1}+\alpha_{k-1}e_{k-1},& k\ne 1, n,\\
\alpha_{n-1}e_{n-1},&k=n,
\end{array}\right.$$
$\alpha_k=(d(k-(n+1)/2)[k][n-k])^{1/2}$.
\end{theorem}
\begin{proof}
The proof essentially runs as that of Theorem~\ref{th1}.
Let $q=-e^{\sigma}$, $\sigma\in {\Bbb R}$ and $J_1$, $J_2$ be
selfadjoint operators satysfing (\ref{rel:soq3I})--(\ref{rel:soq3II}).
Let $\Gamma=\{\Phi(t,s)\equiv t^2-(q+q^{-1})ts+s^2=1\}$ be
characterisic binary relation corresponding to
(\ref{rel:soq3I}). Considering the same parametrization
$s=\frac{sh\sigma x(s)}{sh\sigma}\equiv [x(s)]$, $x(s)\in{\Bbb R}$, we
get $\Gamma=\{(-[x+1],[x]), (-[x-1],[x])\mid x\in{\Bbb R}\}$.
As before, $J_1$, $J_2$ satisfy (\ref{rel:soq3I}) if and only if
$J_2$ is concentrated on $\Gamma$ with respect to $J_1$, i.e.
$$
E_{J_1}(\Delta)J_2E_{J_1}(\Delta')=0,\quad \mbox{for any}\
\Delta,\Delta'\in{\frak B}({\Bbb R}),\
\Delta\times\Delta'\cap\Gamma=\emptyset
$$
which is equivalent to
$$
E_{A_1}(\Delta)J_2E_{A_1}(\Delta')=0,\quad \mbox{for any}\
\Delta,\Delta'\in{\frak B}({\Bbb R}),\
\Delta\times\Delta'\cap\Gamma'=\emptyset,
$$
where $J_1=\frac{sh\sigma A_1}{sh\sigma}$ and
$\Gamma'=\{(-(x+1),x), (-(x-1),x)\mid x\in{\Bbb R}\}=\{(t,s)\mid
t^2+2ts+s^2=1\}$. From this and Theorem~\ref{cth2} we conclude that
$A_1$, $J_2$ satisfy the relation
$$A_1^2J_2+2A_1J_2A_1+J_2A_1^2=J_2.$$
Let $E_1=-(\{A_1,J_2\}+J_2)/2$, $E_2=-(\{A_1,J_2\}-J_2)/2$. It is easy to
show that $E_1=E_1^*$, $E_2=E_2^*$, $J_2=E_1+E_2$ and
\begin{eqnarray}\label{dynrel3}
&A_1E_1=-E_1(A_1-I),\quad A_1E_1=-E_1(A_1+I),\\
&\displaystyle
E_1^2ch(\sigma(A_1+I))=E_2^2ch(\sigma(A_1-1))-\frac{sh\sigma
  A_1}{2sh\sigma}.
\label{dynrel4}
\end{eqnarray}
Conversely, any representation ($A_1$, $E_1$, $E_2$) of
(\ref{dynrel3})--(\ref{dynrel4})
defines a representation ($J_1=sh\sigma A_1/sh\sigma$, $J_2=E_1+E_2$)
of the algebra $U_q(so(3))$. Moreover, there is a one-to-one correspondence
between irreducible and unitary equivalent representations of both objects.

We can now proceed analogously to the proof of Theorem~\ref{} about
representations of graded $so(3)$.
By the same arguments, for any irreducible representation
($A_1$, $E_1$, $E_2$) in $H$ we can choose a basis consisting of
eigenvectors of the operator $A_1$. Then we have
$$A_1e_{\lambda}=\lambda e_{\lambda},\
E_1e_{\lambda}=a_1(\lambda)e_{1-\lambda},\
E_2e_{\lambda}=a_2(\lambda)e_{-1-\lambda},
$$
where $\lambda$ belongs to an  orbit
$\Omega=\{F_1^{(k)}F_2^{(m)}(\lambda),k,m\in {\Bbb N}\}$, where
$F_1(\lambda)=1-\lambda$, and $F_2(\lambda)=-1-\lambda$. The conditions for
$A_1$, $E_1$, $E_2$ to satisfy  relation  (\ref{dynrel4})
are the following:
\begin{eqnarray*}
&a_1(1-\lambda)=\overline{a_1}(\lambda),\quad
a_2(-1-\lambda)=\overline{a_2}(\lambda),\\
&|a_1(\lambda)|^2ch\sigma(\lambda+1)=|a_2(\lambda)|^2ch\sigma(\lambda-1)-
sh\sigma\lambda/2sh\sigma.
\end{eqnarray*}
As for the graded $so(3)$ we see that the last relation can not hold
for any point of the orbit $\Omega$ and there exist the highest
vector on which $E_2$ acts as zero and lowest vector on which the
operator $E_1$ is zero. This implies that the only orbits satysfying
these condition are those which contain $0$, and $\pm 1/2$.
Using these conditions one can easily get the statement.
\end{proof}
{\bf Representations of $U_q(so(3))$ for $q$ a root of unity}

Let $q=e^{i\sigma}$, $\sigma\in(-\pi,\pi)$. If $q$ is a root of unity,
then $\sigma=\frac{\pi k}{n}$, where $\frac{k}{n}$ is an irreducible
fraction. Let $ s=\left\{
\begin{array}{ll}
n,&k\quad\mbox{is even},\\
2n,&k\quad\mbox{is odd}.
\end{array}\right.$

\begin{theorem}
Let $\sigma=\pi\frac{k}{n}$, $\sigma\ne\pi l$.  Any irreducible
representations  of $U_q(so(3))$  is unitarily equivalent to one
of the following:

$1$. $H={\Bbb C}^s$,
$$ J_1e_m=[a+m]e_m,\quad
J_2e_m=\left\{
\begin{array}{ll}
\alpha_0e_1+e^{i\phi}\alpha_{s-1}e_{s-1},& m=0,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1}, &m\ne 0, s-1,\\
\alpha_{s-2}e_{s-2}+e^{-i\phi}\alpha_{s-1}e_{0}, &m=s-1,
\end{array}\right.$$
where $$\alpha_m=(\frac{b\cos a\sigma\cos(a+1)\sigma}{\cos(a+m+1)\sigma\cos (a+m)\sigma}-\frac{\sin m\sigma
\sin(2a+m+1)\sigma}{4\sin^2\sigma
\cos(a+m+1)\sigma\cos (a+m)\sigma})^{1/2},$$
the pair
$(a,b)$ belongs to the set
$\{(a,b)\in$ $M\times{\Bbb R}^+\mid \alpha_m>0, m=0,\ldots,s-1\}$,
$\phi\in [0,2\pi)$
and
$\sigma M=[-\pi/2,\pi/2]\setminus\{(\pi (2l+1)+m\sigma)/2\mid
l,m\in{\Bbb Z}\}$;

\vskip0.5cm

$2$.  $H={\Bbb C}^n$, $k$ is odd,
$$ J_1e_m=[a+m]e_m,\quad
J_2e_m=\left\{
\begin{array}{ll}
(-1)^i\lambda e_1+\alpha_1e_2,&  m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1}, & m\ne 1, n-1,
\\
\alpha_{n-1}e_{n-1}+(-1)^j\lambda e_n,& m=n,
\end{array}\right.$$
where
$$\alpha_m=
(\frac{\sin^2m\sigma}{4\sin^2\sigma \sin(m-1/2)\sigma\sin(m+1/2)\sigma}
-\lambda^2\frac{\sin^2(\sigma/2)}
{\sin(m-1/2)\sigma\sin(m+1/2)\sigma})^{1/2},$$
$a=-\frac{\pi}{2\sigma}-\frac{1}{2}$, $\lambda$ belongs to the set
$\{\lambda\in{\Bbb R}\mid$ $\alpha_m>0, m=1,\ldots,n-1\}$, $i,j=0,1$;
\vskip0.5cm

$3$. $H={\Bbb C}^{2n}$, $k$ is odd,
$$
J_1=\left(
\begin{array}{llll}
\lambda_1I_2&&&0\\
&\cdot&&\\
&&\ddots&\\
0&&&\lambda_nI_2
\end{array}\right),\quad
J_2=\left(
\begin{array}{llll}
Y_1&\alpha_1I_2&&\\
\alpha_1I_2&0&\ddots&\\
&\ddots&\ddots&\alpha_{n-1}I_2\\
&&\alpha_{n-1}I_2&Y_2
\end{array}\right),$$
where
$\displaystyle
\lambda_m=[a+m]$, $a=-\frac{\pi}{2\sigma}-\frac{1}{2}$,\quad
$Y_1=\left(
\begin{array}{ll}
\lambda&0\\
0&-\lambda
\end{array}\right)$, \quad $Y_2=\lambda\left(
\begin{array}{ll}
\cos\varphi&\sin\varphi\\
\sin\varphi&-\cos\varphi
\end{array}\right)$,

\noindent
$\alpha_m$ are the same as in $2$,
$\lambda$ belongs to the set
$\{\lambda\in{\Bbb R^+}\mid\alpha_m>0, m=1,\ldots,n-1\}$,
$\varphi\in(0,\pi)$;
\vskip0.5cm

$4$. $H={\Bbb C}^{n+1}$, $k$ is odd,
$$ J_1e_m=[a+m]e_m,\quad
J_2e_m=\left\{
\begin{array}{ll}
\alpha_1e_2,&m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1},&m\ne 1,n+1,\\
\alpha_{n}e_{n},&m=n+1,
\end{array}\right.$$
where $a=-\frac{\pi}{2\sigma}-1$,
$\alpha_1=\alpha_{n}=\frac{\sqrt{2}}{|q-q^{-1}|}$,
$\alpha_m=\frac{1}
{|q-q^{-1}|}$,  $m=2,\ldots, n-1$;
\vskip0.5cm
$5$. $H={\Bbb C}^{2n}$, $k$ is odd,

$$J_1=\left(
\begin{array}{llll}
\lambda_1J_1&&&0\\
&\lambda_2I_2&&\\
&&\ddots&\\
0&&&\lambda_{n+1}I_1\\
\end{array}\right),
J_2=\left(
\begin{array}{lllll}
0&X_1^*&&&\\
X_1&0&X_2^*&&\\
&X_2&\ddots&&\\
&&&\ddots & X_{n}^*\\
&&&X_{n}&0
\end{array}\right),$$
where $\lambda_m=[m-1-\frac{\pi}{2\sigma}]$,\quad
$X_1=\left(
\begin{array}{l}
\frac{\sqrt{2}}{|q-q^{-1}|}\cos\varphi\\
\frac{\sqrt{2}}{|q-q^{-1}|}\sin\varphi
\end{array}\right)$, \quad $X_{n}=(\frac{\sqrt{2}}{|q-q^{-1}|},0)$,

\vspace{0.1cm}
$X_m=\frac{1}{|q-q^{-1}|}I_2$,
$m=2,\ldots,n-1$, $\varphi\in (0,\pi/2)$;
\vskip0.5cm
$6$. $H={\Bbb C}^{(n+1)/2}$, $k$ is even,
$$ J_1e_m=(-1)^{k/2}[a+m]e_m,\quad
J_2e_m=\left\{
\begin{array}{ll}
\alpha_1e_2+\frac{(-1)^i}{q-q^{-1}}e_1,& m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1},& m\ne 1, (n+1)/2,\\
\alpha_{(n-1)/2}e_{(n-1)/2},&m=(n+1)/2,
\end{array}\right.$$
where $a=\frac{\pi}{2\sigma}-\frac{1}{2}$, $\alpha_m=\frac{1}{|q-q^{-1}|}$,
$m=1,\ldots, (n-3)/2$, $\alpha_{(n-1)/2}=\frac{\sqrt{2}}{|q-q^{-1}|}$,
$i=0,1$;
\vskip0.5cm
$7$. $H={\Bbb C}^n$, $k$ is even,

$J_1=\left(
\begin{array}{lllll}
\lambda_1I_2&&&0\\
&\ddots&&\\
&&\lambda_{(n-1)/2}I_2&\\
0&&&\lambda_{(n+1)/2}I_1
\end{array}\right)$,
$J_2=\left(
\begin{array}{lllll}
Y&X_1^*&&\\
X_1&0&&\\
&&\ddots & X_{(n-1)/2}^*\\
&&X_{(n-1)/2}&0
\end{array}\right)$,

\noindent
where
$\lambda_m=[\frac{\pi}{2\sigma}-\frac{1}{2}+m]$,\quad
$Y=\frac{1}{|q-q^{-1}|}\left(
\begin{array}{ll}
\cos\varphi&\sin\varphi\\
\sin\varphi&-\cos\varphi
\end{array}\right)$, $X_{(n-1)/2}=(\frac{\sqrt{2}}{|q-q^{-1}|},0)$,
$X_m=\frac{1}{|q-q^{-1}|}I_2$,
$m=1,\ldots,(n-3)/2$,
$\varphi\in (0,\pi)$;
\vskip0.5cm
$8$. $H={\Bbb C}^{p}$, $p<n$ if $k$ is odd, $p<(n+1)/2$ if $k$ is even,
$$ J_1e_m=(-1)^i[a+m]e_m,\quad
J_2e_m=\left\{
\begin{array}{ll}
\alpha_1e_2+(-1)^j\frac{\sin p\sigma}{2\sin(\sigma/2)\sin\sigma}e_1,&m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1},&m\ne 1,p,\\
\alpha_{p-1}e_{p-1},& m=p,
\end{array}\right.$$
where
$$\alpha_m=(\frac{\sin(m-l)\sigma\sin(l+m)
\sigma}{4\sin^2\sigma\sin(m-1/2)\sigma\sin(m+1/2)\sigma})^{1/2},$$
$a=\frac{\pi}{2\sigma}-\frac{1}{2}$, $p$ belongs to the set $\{p\in{\Bbb N}\mid
\alpha_m>0, 1\le m<p\}$, $i,j=0,1$;
\vskip0.5cm
$9$. $H={\Bbb C}^{p}$,
$$ J_1e_m=[a+m]e_m,\quad
J_2e_m=\left\{
\begin{array}{ll}
\alpha_1e_2,&m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1},&m\ne 1, p,\\
\alpha_{l-1}e_{l-1},&m=p,
\end{array}\right.$$
where $$\alpha_1=(-\frac{\sin(a+1)\sigma}{2\sin\sigma\cos(a+
2)\sigma)})^{1/2},\quad \alpha_m=(-\frac{\sin m\sigma\sin(2a+m+1)\sigma}
{4\sin^2\sigma\cos(a+m)\sigma\cos(a+m+1)\sigma})^{1/2},$$
$m\ne 1$,
the pair
$(a,p)$ belongs to the set $\{(a,p)\in{\Bbb R}\times{\Bbb N}\mid
\sigma a\ne\frac{\pi}{2}-l\sigma+\pi r$, $[p][2a+p+1]=0$,
$[a+m]\ne [a+n]$, $1\leq m<n\leq p$,
$\alpha_m>0, m=1,\ldots,p-1, r\in{\Bbb Z}\}$.
\vskip0.5cm
$10$. $H={\Bbb C}^1$, $J_1=0$, $J_2=0$.
\end{theorem}
\begin{proof}
Let $J_1=J_1^*$, $J_2=J_2^*$ be operators in a Hilbert space
satisfying (\ref{rel:soq3I})--(\ref{rel:soq3II}) and
$\Gamma=\{(t,s)\mid
\Phi(t,s)\equiv t^2-(q+q^{-1})ts+s^2-1\}$ the characteristic binary
relation corresponding to (\ref{rel:soq3I}). By Theorem~\ref{},
\begin{equation}\label{supp}
E_{J_1}(\Delta)J_2E_{J_1}(\Delta')=0,\quad \mbox{for any}\
\Delta,\Delta'\in{\frak B}({\Bbb R}),\
\Delta\times\Delta'\cap\Gamma=\emptyset,
\end{equation}
Let
denote by $S_0$ the set $\{s\in{\Bbb R}\mid (q-q^{-1})^2s^2+4<0\}$,
and by $S_1$ its complement.
Then $R\times S_0\cap \Gamma=\emptyset$ which implies
$J_2E_{J_1}(S_0)H=0$. Hence $H_0:=E_{J_1}(S_0)H$ is invariant with
respect to $J_1$, $J_2$ and any irreducible representation in $H_0$ is
one-dimensional given by
$$J_1=(\lambda),\quad J_2=(0),\quad \lambda\in S_0.$$
But, by (\ref{rel:soq3II}), $\lambda=0\notin S_0$.
Therefore the spectrum of $J_1$ belongs to $S_1$. Consider the
following parametrization of $S_1$: $\lambda=\sin x\sigma/\sin\sigma$,
$x\in {\Bbb R}$. Let $O(\{\lambda\})$ be the trajectory of the
point $\lambda$ with  respect to $\Gamma$, i.e., the minimal subset
$M\subset{\Bbb R}$ which contain $\lambda$ and satisfy the condition
$({\Bbb R}\setminus M)\times M\cap\Gamma=\emptyset$,
and $M\times({\Bbb R}\setminus M)\cap\Gamma=\emptyset$.
In our case we have $\displaystyle
O(\{\lambda\})=\{\frac{\sin(x+k)\lambda}{\sin\sigma}\mid k\in{\Bbb
  Z}\}$.

If ($J_1$, $J_2$) is irreducible, then the measure $E_{J_1}(\cdot)$ is
ergodic with respect to $\Gamma$, i.e., either $E_{J_1}(M)=0$ or
$E_{J_1}(M)=I$ for any set $M$ which is invariant with respect to
$\Gamma$. In fact, if it were not, we would conclude that $E_{J_1}(M)H$
or $E_{J_1}(M)H$ is a subspace invariant with respect to $J_1$, $J_2$.
Moreover, one can easily check that there esists a measurable section
of $(S_1,\Gamma)$, i.e.,
set which meets every trajectory only once.
It follows from this that any ergodic measure is concentrated on a
single  trajectory of
some points and hence the spectrum of the operator $J_1$ is discrete
and concentrated on a trajectory if only $(J_1,J_2)$ is an irreducible
pair.
Let $H_1=E_A(O(\{\frac{\sin (\pi/2)}{\sin\sigma}\}))H$,
$H_2=E_A(O(\{-\frac{\sin(\pi/2)}{\sin\sigma}\}))H$, if $k$ is even,
and $ H_1=E_A(O(\{\frac{\sin (\pi/2)}{\sin\sigma}\}))H$,
$H_2=E_A(O(\{\frac{\sin ((\pi-\sigma)/2)}{\sin\sigma}\}))H$,  if
$k$ is odd,  $H_3=(H_1\oplus H_2)^{\perp}$.

Any trajectory can be described geometrically in the following way: if
$t$, $s\in O(\{\lambda\})$ and $(t,s)\in\Gamma$ then we draw an edge
$\edge{t}{s}$ if $t\ne s$, and a loop $\ear{t}$ if $t=s$. Then we have
the following types of trajectories.

%I.
%$\begin{array}{ll}
%\cycle{\lambda_1}{\lambda_2}{\lambda_{s}},\
%\mbox{where}&
%\displaystyle\lambda_m=\frac{\sin((x+m)\sigma)}{\sin\sigma},\\
%&\displaystyle x\sigma\in [-\frac{\pi}{2},\frac{\pi}{2}]\setminus
%\{\frac{\pi(2l+1)+m\sigma}{2}\mid m,l\in{\Bbb Z}\},
%\end{array}$
%\vspace{0.1cm}

%II.
%$\earchear{\lambda_1}{\lambda_n}\qquad\qquad$, where
%$\displaystyle \lambda_m=-\frac{\cos(\sigma/2+m\sigma)}{\sin\sigma}$, $k$ is odd;
%\vspace{0.1cm}

%III.
%$\begin{array}{ll}
%\ch{\lambda_1}{\lambda_m}\qquad\qquad,\
%\mbox{where}& m\le n,\
%\displaystyle\lambda_i=\frac{\sin((x+i)\sigma)}{\sin\sigma},
%i=1,\ldots,m, \\
%&\lambda_i\ne\lambda_j, 1\le i<j\le m
%\end{array}$
%\vspace{0.1cm}

%IV.
%$\begin{array}{ll}
%\earch{\lambda_1}{\lambda_m}\qquad\qquad,\
%\mbox{where}& m\le n\  \mbox{and either}\
%\displaystyle\lambda_i=\frac{\cos(\sigma/2+i\sigma)}{\sin\sigma},\
%\mbox{or}\\
%&\displaystyle\lambda_i=-\frac{\cos(\sigma/2+i\sigma)}{\sin\sigma},
% \lambda_i\ne\lambda_j,1\le i\ne j\le m.
%\end{array}$

Any     trajectory     of    the    point    $t\notin
O(\{\frac{1}{\sin\sigma}\})\cup O(\{-\frac{1}{\sin\sigma}\})\cup
O(\{\frac{\cos(\sigma/2)}{\sin\sigma}\})\cup
O(\{-\frac{\cos(\sigma/2)}{sin\sigma}\})$ is a cycle of length  $s$,
i.e., the following graph:

$\begin{array}{ll}
\cycle{\lambda_1}{\lambda_2}{\lambda_{s}},\
\mbox{where}&
\lambda_m=\frac{\sin((x+m)\sigma)}{\sin\sigma},\\
& x\sigma\in [-\frac{\pi}{2},\frac{\pi}{2}]\setminus
\{\frac{\pi(2l+1)+m\sigma}{2}\mid m,l\in{\Bbb Z}\},
\end{array}$
\vspace{0.1cm}

The trajectories of the points  $\{\pm\frac{1}{\sin\sigma}\}$,
$\{\pm\frac{\cos(\sigma/2)}{\sin\sigma}\}$ are of the form:

\noindent
a) if $k=2(2p-1)$,
 $$\earch{\frac{\cos(\sigma/2)}{\sin\sigma}}{-\frac{1}{\sin\sigma}}
\qquad\qquad\qquad\earch{-\frac{\cos(\sigma/2)}{\sin\sigma}}
{\frac{1}{\sin\sigma}}$$
b) if $k=4p$,
 $$\earch{-\frac{\cos(\sigma/2)}{\sin\sigma}}{-\frac{1}{\sin\sigma}}
\qquad\qquad\qquad
\earch{\frac{\cos(\sigma/2)}{\sin\sigma}}{\frac{1}{\sin\sigma}}$$
c) if $k=2p-1$,
  $$\earchear{\frac{\cos(\sigma/2)}{\sin\sigma}}{-\frac{\cos(\sigma/2)}
{\sin\sigma}}\qquad\qquad\qquad
\ch{-\frac{1}{\sin\sigma}}{\frac{1}{\sin\sigma}}$$
\vspace{0.1cm}

2. In what follows we study  irreducuble representations  in  each
of the subspaces $H_i$, $i=1,2,3$.

2.1. If $J_1$, $J_2$ is an irreducible representation acting in $H_3$,  then
$\sigma(J_1)\subset O(\{\frac{\sin x\sigma}{\sin\sigma}\})
=\{\frac{\sin((x+k)\sigma)}{\sin\sigma}\mid 0\le k\le s-1\}$, moreover
$\{\pm\frac{1}{\sin\sigma}\}$, $\{\pm\frac{\cos\sigma/2}{\sin\sigma}\}
\notin O(\{\frac{\sin x\sigma}{\sin\sigma}\})$. Denote by $P_k$ the
projection onto the eigenspace of the operator $J_1$ corresponding to
$\lambda_k=\frac{\sin((x+k)\sigma)}{\sin\sigma}$.

It follows from
(\ref{supp}) that $J_2P_0H\subset P_1H\oplus P_{s-1}H$,
$J_2P_{s-1}H\subset P_1H\oplus P_{s-2}H$, and $J_2P_kH\subset
P_{k+1}H\oplus P_{k-1}H$ for $k\ne 0, s-1$. Thus the operator $J_2$ can
be  represented in the form $J_2=X+X^*$, where
$X=\sum_{k=0}^{s-2}P_{k+1}J_2P_k+P_0J_2P_{s-1}$. Moreover, the pair
($J_1$, $J_2$) is irreducible if and only if so is ($J_1$, $X$, $X^*$).
One can easily show that for $J_1$, $J_2$ to satisfy (\ref{rel:soq3II}) it
is necessary and sufficient that
$X$,  $X^*$ are additionally connected by the relations:
\begin{eqnarray}\label{ff2}
&\alpha_kX^*XP_k+\beta_kXX^*P_k=\gamma_kI,
\end{eqnarray}
where $\alpha_k=-2cos((x+k+1)\sigma)$,
$\beta_k=2cos((x+k-1)\sigma)$,
$\gamma_k=\frac{\sin((x+k)\sigma)}{\sin\sigma}$. Since
$\{\pm\frac{1}{\sin\sigma}\}$ does   not   belong   to  the  trajectory,
$\alpha_k\beta_k\ne 0$.

Let $0\le k\le s-1$ be the smallest number such that $\lambda_k\in\sigma(J_1)$.
Set $C_k=X^*XP_k$, and denote by $E_{C_k}(\cdot)$ the resolution of
the indentity for $C_k$. From (\ref{ff2}) it follows  that
$[A,X^s]=0$,  $[X^*,  X^s]=0$ which yields $X^s=cI$ if the collection
$J_1$,  $X$, $X^*$ is irreducible. From this we conclude that
$\oplus_{l=k}^{s-1}X^{l-k}E_{C_k}(\Delta)U$ is invariant with respect
to $J_1$,  $X$,  $X^*$ for any $\Delta\in{\frak B}({\Bbb  R})$ and
subspace $U$ such that $C_kU\subset U$.
 Hence, if ($J_1$, $X$, $X^*$) is irreducible then $\Delta$ is concentrated
in one point and we can choose a basis consisting of eigenvectors of
$J_1$, namely, $\{e_{\lambda_k},
Xe_{\lambda_k}/||Xe_{\lambda_k}||,\ldots,
X^{s-1}e_{\lambda_k}/||X^{s-k-1}e_{\lambda_k}||\}$, where
$e_{\lambda_k}$ is an eigenvector of $C_k$ which exists due to the last
arguments.
Let $X^*Xe_{\lambda_k}\equiv C_ke_{\lambda_k}=be_{\lambda_k}$,
$b\in{\Bbb R}$. Using (\ref{ff2}) one can get the action of the operators
$J_1$, $X$,
$X^*$ and $J_2$ on the basis. In particular, if
$\sigma(J_1)=\{\lambda_0, \lambda_1,\ldots,\lambda_{s-1}\}$ we will
have representations given by 1, otherwise, representations from
series 9.

2.2. If $J_1$,  $J_2$  is  an irreducible representation acting in $H_1$,
$H_2$, then $\sigma(J_1)$ is not necessary to be simple.  First let us
consider  the
case when the support of the operator $J_2$ with repspect to $J_1$ is
of the  form
\begin{equation}
\earchear{\lambda_1}{\lambda_n}\label{gr1}
\end{equation}
where $\lambda_1=-\frac{\cos(\sigma/2)}{\sin\sigma}$
$\lambda_n=\frac{\cos(\sigma/2)}{\sin\sigma}$, and $\sigma(J_1)=\{\lambda_1,\ldots,\lambda_n\}$.

Denote by   $P_k$   the   projection   onto  the  eigenspace  of  $J_1$
corresponding to the eigenvalue $\lambda_k=-\frac{\cos((2k-1)\sigma/2}
{\sin\sigma}$. As in the case 2.1 the operator $J_2$   can  be
represented  in  the  form
$J_2=X+X^*+Y$, where      $X=\sum_{k=1}^{n-1}P_{k+1}J_2P_k$,
$Y=P_1J_2P_1+P_nJ_2P_n$, and $Y\ne 0$,  $XP_k\ne 0$,  $k<n$. Moreover
($J_1$, $J_2$)  is irreducible if and only if the family
$J_1$,  $X$,  $X^*$, $Y$ is irreducible.
Clearly $J_1$,  $J_2$  satisfy the relation $Q(J_1,J_2)\equiv
J_2^2J_1-(q+q^{-1})J_2J_1J_2+J_1J_2^2-J_1=0$ if and only if
$P_sQ(J_1,J_2)P_k=0$ for any $1\le k,s\le n$.
>From this it follows that
\begin{eqnarray}
\begin{array}{ll}
\alpha_kX_k^*X_k+\beta_kX_{k-1}X_{k-1}^*=\gamma_kI,
\ k\ne 1,\  k\ne n,\\
\alpha_1X_1^*X_1+\beta_1Y^2P_1=\gamma_1I,\\
\beta_nX_{n-1}X_{n-1}^*+\alpha_nY^2P_n=\gamma_nI,\\
\end{array}\label{ff10}
\end{eqnarray}
where   $X_k=XP_k$, $\alpha_k=-2\sin((2k+1)\sigma/2)$,
$\beta_k=2\sin((2k-3)\sigma/2)$,
$\gamma_k=-\frac{\cos((2k-1)\sigma/2)}{\sin\sigma}$.

Set $D_0=(X^*)^{n-1}X^{n-1}P_1$. Then $P_1H\subset(\ker D_0)^{\perp}$,
because if it were not the case  we would conclude that
$W=\ker D_0\oplus X\ker D_0\oplus\ldots\oplus
X^{n-2}\ker  D_0$  is invariant with  respect to $J_1$, $J_2$, which
contradicts the fact that $\sigma(J_1)=\{\lambda_1,\ldots,\lambda_n\}$.

Let $X=U\sqrt{X^*X}$ be the polar decomposition of the operator $X$.
Put $Y_1=YP_1$,  $Y_2=(U^*)^{n-1}YU^{n-1}P_1$. Let  us prove that ($J_1$,
$J_2$) is irreducible if and only if  the pair ($Y_1$, $Y_2$) is irreducible.
In fact, if $\Xi$ is invariant with respect to $Y_1$, $Y_2$, then
$\Xi'=\Xi\oplus U\Xi\oplus \ldots U^{n-1}\Xi$ is invariant with respect to
$J_1$, $X$, $X^*$, $Y$, and hence with respect to $J_1$, $J_2$:
\begin{eqnarray*}
&XU^k\Xi=U\sqrt{X^*X}U^{k}\Xi=U^{k+1}({\bf F}^{(k)}(\frac{3\sigma-\pi}{2},
X^*X))_2\Xi\subset U^{k=1}\Xi,\\
&YU^{n-1}\Xi=U^{n-1}(U^*)^{n-1}YU^{n-1}\Xi\subset
U^{n-1}Y_1\Xi\subset U^{n-1}\Xi,
\end{eqnarray*}
where ${\bf F}(x,y)=(F_1(x,y), F_2(x,y))=(x+1,\frac{y\cos
x\sigma}{\cos((x+2)\sigma)}-\frac{sin((x+1)\sigma)}{2\sin\sigma
\cos((x+2)\sigma)})$ (here we use the fact that $UU^*$  is the
projection on $(\ker(A+\frac{\cos(\sigma/2)}{\sin\sigma}I))^{\perp}$).
Moreover, (\ref{ff10}) gives
\begin{eqnarray*}
\begin{array}{ll}
Q_2(Y_2)&=(U^*)^{n-1}XX^*U^{n-1}P_1=(U^*)^{n-2}X^*XU^{n-2}P_1=\\
&=({\bf F}^{(n-2)}(\frac{3\sigma-\pi}{2},X^*X))_2P_1=
=({\bf F}^{(m-2)}(\frac{3\sigma-\pi}{2}, Q_1(Y_1))=\\
&=Q_1(Y_1)+\alpha I,
\end{array}
\end{eqnarray*}
where
$Q_1(x)=(-\beta_1x^2+\gamma_1)/\alpha_1$,
$Q_2(x)=(-\alpha_n x^2+\gamma_n)/\beta_n$, and
$\alpha=({\bf F}^{(n-2)}(\frac{3\sigma-\pi}{2},x))_2-x=
\frac{\gamma_n}{\beta_n}-\frac{\gamma_1}{\alpha_1}$.
Hence $Y_1^2=Y_2^2$. For any irreducible pair ($Y_1$, $Y_2$) we have
$Y_1^2=Y_2^2=\lambda I$, $\lambda>0$. Denote by  $H_0$ the
corresponding representation space. By \cite{os}, $\mbox{dim} H_0\leq
2$. Let $\{e_k,k\in K\}$ be an orthonormal basis in $H_0$. Then $\{U^me_k\mid
m=1,\ldots, n-1, k\in K\}$ is an orthonormal basis in $H$ and the corresponding
operators $J_1$, $J_2$ acts as follows:
$$J_1=\left(
\begin{array}{llll}
\lambda_1I&&&\\
&\lambda_2I&&\\
&&\ddots&\\
&&&\lambda_nI
\end{array}\right), J_2=\left(
\begin{array}{llll}
Y_1&\mu_1(\lambda)&&\\
\mu_1(\lambda)I&0&\ddots&\\
&\ddots&\ddots&\mu_{n-1}(\lambda)I\\
&&\mu_{n-1}(\lambda)I&Y_2
\end{array}\right),$$
$ \begin{array}{ll}\mbox{where}\
\mu_m(\lambda)&=\sqrt{({\Bbb F}^{(m-1)}(\frac{\sigma-\pi}{2},
-\frac{\sin(\sigma/2)}{\sin
(3\sigma/2)}\lambda^2+\frac{\cos(\sigma/2)}{2\sin(3\sigma/2)
\sin\sigma}))_2}=\\
&=
\sqrt{\frac{\sin^2(m\sigma)-4\lambda^2\sin^2(\sigma/2)\sin^2\sigma}
{4\sin^2\sigma
\sin((2m-1)\sigma/2)\sin((2m+1)\sigma/2)}},\ m=1,\ldots,n-1,\end{array}$

\vspace{0.1cm}\noindent
$\lambda\in\{x\in{\Bbb R}\mid ({\Bbb F}^{(m-1)}(\frac{\sigma-\pi}{2},
-\frac{\sin(\sigma/2)}{\sin
(3\sigma/2)}x^2+\frac{\cos(\sigma/2)}{2\sin(3\sigma/2)
\sin\sigma}))_2>0, 1\le m\le n-1\}$.

By  \cite{os},  any
irreducible pair ($Y_1$,  $Y_2$) is unitarily equivalent  to
one of the following:

a) one-dimensional:     $Y_1=(\pm\lambda)$,    $Y_2=(\pm\lambda)$,
$\lambda> 0$;

b) two-dimensional: $Y_1=\left(
\begin{array}{ll}
\lambda&0\\
0&-\lambda
\end{array}\right)$, $Y_2=\lambda\left(
\begin{array}{ll}
\cos\varphi&\sin\varphi\\
\sin\varphi&-\cos\varphi
\end{array}\right)$, with $\lambda>0$, $\varphi\in(0,\pi)$.

Hence, all  irreducible representations $J_1$, $J_2$ have
dimensions $n$ or $2n$, moreover, two such  pairs ($J_1$,  $J_2$),
($J_1'$, $J_2'$) are unitarily equivalent if and only if
the corresponding  pairs  ($Y_1$,  $Y_2$),  ($Y_1'$, $Y_2'$) are
unitarily equivalent.
Thus we will get the representations 2 and 3 of the theorem.

If the support of $J_2$ is a  subgraph  $\Gamma_0$  of  (\ref{gr1}) and
$\Gamma_0$ does  not  coincide  with  (\ref{gr1}),  then  arguments
similar to  that  in  the  proof  of  Theorem~\ref{th3}  gives   that
$\sigma(J_1)$ is simple as soon as the pair $J_1$, $J_2$ is irreducible. Using
(\ref{ff10}), one can easily describe all non-unitarily equivalent
irreducible representations connected with the support $\Gamma_0$.

2.3. Let now the support of $J_2$ be the graph
\begin{equation}\label{gr2}
\ch{-\frac{1}{sin\sigma}}{\frac{1}{\sin\sigma}}
\end{equation}
or some its subgraph. Let $P_k$ be the projection onto  the
eigenspace of $J_1$ corresponding  to
$\lambda_k=-\frac{\cos (k-1)\sigma}{\sin\sigma}$, $X_k=P_{k+1}J_2P_k$,
$X=\sum_{k=1}^{n}X_k$. Using the same arguments we can conclude
that  $J_2=X+X^*$ and
the operators
$X_k$,  $X_k^*$  satisfy  the following relations:
\begin{eqnarray}
&&\alpha_kX_k^*X_k+\beta_kX_{k-1}X_{k-1}^*=\gamma_k I,\label{f30}
\end{eqnarray}
where $\alpha_k=-2\sin(k\sigma)$, $\beta_k=2\sin((k-2)\sigma)$,
$\gamma_k=-\frac{\cos (k-1)\sigma}{\sin\sigma}$, in particular,
$\alpha_{n}=0$, $\beta_2=0$, $\alpha_k\beta_m\ne 0$, $k\ne n$,
$m\ne 2$.
Moreover, the pair ($J_1$, $J_2$)
is irreducible if and only if the  triple ($J_1$, $X$, $X^*$) is irreducible.
If $\{\frac{1}{\sin\sigma}\}$ or $\{-\frac{1}{\sin\sigma}\}$ does  not
belong  to $\sigma(J_1)$ then $X$, $X^*$ satisfy the relation:
$$X^*X=F_1(XX^*,A)\quad \mbox{or}\quad XX^*=F_2(X^*X,A)$$
respectively, where
$$F_1(\mu,\lambda_k)=\left\{
\begin{array}{ll}
\frac{\gamma_k-\beta_k\mu}{\alpha_k},& k\ne n,\\
0,& \mbox{otherwise},
\end{array}\right.\quad
F_2(\mu,\lambda_k)=\left\{
\begin{array}{ll}
\frac{\gamma_k-\alpha_k\mu}{\beta_k},& k\ne 2,\\
0,& \mbox{otherwise}.
\end{array}\right.$$
Hence any irreducible representation can be realized in the space
$H=l_2(\Delta)$ by formulae:
$$ J_1e_{k}=\lambda_ke_{k},\quad Xe_k=\mu_ke_{k+1},$$
where $\Delta=\{\lambda_p,\lambda_{p+1},\ldots,\lambda_{m}\mid 0\le
p,m\le n\}$, and either $\frac{1}{\sin\sigma}\notin\Delta$ or
$-\frac{1}{\sin\sigma}\notin\Delta$, and
$\alpha_k\mu_k+\beta_k\mu_{k-1}=\gamma_k$ such that $\mu_k>0$ if
$\lambda_k,\lambda_{k+1}\in\Delta$ and $\mu_k=0$ if
$\lambda_k\notin\Delta$ or $\lambda_{k+1}\notin\Delta$.

If $\{\pm\frac{1}{\sin\sigma}\}\in\sigma(J_1)$, then the  problem  to
describe irreducible  representations ($J_1$, $J_2$) is reduced to
that of pairs of operators  satisfying  some  quadratic
relations. In order to show that, consider the following operators
in the subspace $P_2H$: $D_1=X_1X_1^*$,
$D_2=(X_{n}\ldots X_{2})^*X_{n}\ldots X_{2}$. From (\ref{f30}) we have
$X_1^*X_1=\frac{\gamma_1}{\alpha_1}I$, $X_{n}\ldots
X_2(X_{n}\ldots X_2)^*=\mu I$, where $\mu=\prod_{k=0}^{n-3}({\bf F}^{(k)}
(\frac{\gamma_n}{\beta_n},\lambda_{n}))_1\frac{\gamma_{n+1}}{\beta_{n+1}}$,
${\bf F}(x,\lambda_{k+1})=
(\frac{\gamma_k-\alpha_kx}{\beta_k},\lambda_{k})$. One can
check that $\frac{\gamma_1}{\alpha_1}=\frac{1}{2\sin^2\sigma}$.
Hence
$$D_1(D_1-\frac{1}{2sin^2\sigma}I)=0,\quad D_2(D_2-\mu I)=0.$$
Let $H_0$ be a subspace of $H$ which is invariant with respect to $D_1$,
$D_2$. Then
$$X_1^*H_0\oplus H_0\oplus X_2H_0\oplus X_3X_2H_0\oplus\ldots\oplus
X_{n}\ldots  X_2H_0$$
is  invariant  with respect to $J_1$,  $X$, $X^*$. In fact, using
(\ref{f30}) one can
easily prove that $X_k^*X_k\ldots X_2=\mu_kX_{k-1}\ldots X_2$, $1<k<n$
(here $\mu_k=1/4sin^2\sigma$) .
Moreover, $X_{n-1}\ldots X_2X_2^*\ldots X^*_{n-1}=\lambda I$,
where  $\lambda=\prod_{k=0}^{n-3}({\bf F}^{(k)}
(\frac{\gamma_{n}}{\beta_{n}},\lambda_{n}))_1$. Assuming that
$\lambda=0$, we will have that $X_3^*\ldots  X_{n-1}^*P_{n}H\oplus\ldots
P_{n}H\oplus X_{n}P_{n}H$ is invariant with respect to $J_1$,
$J_2$,  hence
$\sigma(J_1)\not\ni\{\frac{1}{\sin\sigma}\}$ if  the pair ($J_1$, $J_2$) is
irreducible, which contradicts the assumption. Thus $\lambda\ne 0$,
\begin{eqnarray*}
&&X_{n}^*X_{n}\ldots X_2H_0=\lambda^{-1}\lambda
X_{n}^*X_{n}\ldots
X_2H_0=\\
&&=\lambda^{-1}(X_{n-1}\ldots X_2X_2^*\ldots X_{n-1}^*)X_{n}^*
X_{n}\ldots X_2H_0\subset X_{n-1}\ldots  X_2H_0.
\end{eqnarray*}

By \cite{os}  any  irreducible  pair  ($D_1$,  $D_2$)  is  one  or
two-dimensional.
Let us describe the corresponding irreducible pairs $J_1$, $J_2$.
Denote by $U_i$ the
phase of the operator $X_i\ldots X_2$, $U_1^*$ the phase of $X_1^*$. If
$\mbox{dim} H_0\le 2$ and $e_1$, $e_2$ is an orthonormal basis in the
space  $H_0$ such that
$e_1\in(\ker D_2)^{\perp}$, then
$$D_1=\frac{1}{2\sin^2\sigma}\left(
\begin{array}{cc}
1+\cos\varphi&\sin\varphi\\
\sin\varphi&1-\cos\varphi
\end{array}\right),\qquad
D_2=\left(
\begin{array}{ll}
\mu&0\\
0&0
\end{array}\right),$$
$U_ie_1$, $U_ie_2$ is an orthonormal basis in $X_i\ldots X_2H_0$, $i\ne n$,
and the vectors $U_{n}e_1$,
$U_1(\cos(\varphi/2)e_1+\sin(\varphi/2)e_2)$ generate the spaces
$X_{n}\ldots X_2H_0$ and $X_1^*H_0$
respectively.
Note that $\cos(\varphi/2)e_1+\sin(\varphi/2)e_2 \in(\ker D_1)^{\perp}$.
To describe the action of the operator $J_2$ on the vectors of  this
base it is sufficient to know that for the operators $X_i$.

$X_{i+1}U_ie_k=(\prod_{l=2}^{i-1}\mu_l)^{-1/2}X_{i+1}\ldots X_2e_k=
\sqrt{\mu_i}U_{i+1}e_k=\frac{1}{2\sin\sigma}$, $i<n-1$;

$X_{n}U_{n}e_k=(\prod_{l=2}^{n-1}\mu_l)^{-1/2}X_{n}\ldots
X_2e_k=\left\{
\begin{array}{ll}
0,& k=2,\\
\frac{1}{\sqrt{2}\sin\sigma}U_{n}e_1,& k=1,
\end{array}\right.$

\noindent
$\begin{array}{ll}
X_1U_1^*(\cos(\varphi/2)e_1+\sin(\varphi/2)e_2)&=
\sqrt{2}\sin\sigma X_1X_1^*(\cos(\varphi/2)e_1+
\sin(\varphi/2)e_2)=\\
&=\frac{1}{\sqrt{2}\sin\sigma}(\cos(\varphi/2)e_1+\sin(\varphi/2)e_2).
\end{array}$

Thus we will get representations from series 5. If $\mbox{dim}
H_0=1$ then $D_1\ne 0$ and  $D_2\ne 0$, since otherwise
$\sigma(J_1)\ne\{\lambda_1,\ldots,\lambda_{n+1}\}$ if ($J_1$, $J_2$) is
irreducible.
Let ${\Bbb C}e_1=H_0$. Then using the same arguments one can show that
$\{U_1^*e_1,e_1,U_1e_1,\ldots U_{n}e_1\}$ is an orthogonal basis in
$H$ and the corresponding irreducible representation ($J_1$, $J_2$) is
given by formulae 6.

2.4. Suppose that the support of $J_2$ is
\begin{equation}\earch{\lambda_1}
{\lambda_{(n+1)/2}}\label{gr3}
\end{equation}
where $\lambda_1=\pm\frac{\cos\sigma/2}{\sin\sigma}$,
$\lambda_{(n+1)/2}=\frac{1}{\sin\sigma}$. Let us consider the  case
$\lambda_1=\frac{cos\sigma/2}{\sin\sigma}$ (for
$\lambda_1=-\frac{cos\sigma/2}{\sin\sigma}$ the proof is the same). The operator $J_2$  can
be represented in the  form  $J_2=X+X^*+Y$,  where
$X=\sum_{k=1}^{(n-1)/2}P_{k+1}J_2P_k$, $Y=P_1J_2P_1$, $P_k$ is the projection
on the eigenspace which corresponds to the eigenvalue $\lambda_k=
\frac{\cos(2k-1)\sigma/2}{\sin\sigma}$. Let $X_k=XP_k$. Then
\begin{eqnarray}
\begin{array}{ll}
\alpha_kX_k^*X_k+\beta_kX_{k-1}X_{k-1}^*=\gamma_kI, k\ne 1,\\
\alpha_1X_1^*X_1+\beta_1Y^2P_1=\gamma_1I,
\end{array}\label{ff31}
\end{eqnarray}
where $\alpha_k=-2\sin(\frac{(2k+1)\sigma}{2})$,
$\beta_k=2\sin(\frac{(2k-3)\sigma}{2})$,
$\gamma_k=-\frac{\cos((2k-1)\sigma/2)}{\sin\sigma}$,   in  particular,
$\alpha_k\ne 0$ for $k\ne (n-1)/2$, $\alpha_{(n-1)/2}=0$.
As above, the  problem of  describing  irreducible  representations
($J_1$, $J_2$) can be
reduced  to  that of
irreducible pairs  ($D_1$,  $D_2$)   satisfying   some   quadratic
relation. Here $D_1=Y$, $D_2=(X_{(n-1)/2}\ldots X_1)^*X_{(n-1)/2}\ldots X_1$
and $D_2(D_2-\mu I)=0$, $D_1^2=\frac{1}{4\sin^2\sigma} I$, $\mu=
\frac{1}{2^{2n-3}(\sin^2\sigma)^{n-1}}$,?????

If $H_0$  is  invariant  with respect to $D_1$,  $D_2$, then $H_0\oplus
XH_0\oplus\ldots\oplus X^{(n-1)/2}H_0$ is invariant with respect  to  $J_1$,
$X$, $X^*$. Moreover dimensions of the irreducible representations
are $2(\frac{n-1}{2})+1$ or $(n+1)/2$.  This  follows  by  the same method as in
the previous
case. We will get representations 6, 7.

If ($J_1$, $J_2$) is irreducible and $\sigma(J_1)$ is a proper subset
of $\{\lambda_1,\ldots,\lambda_{(n+1)/2}\}$ then, as before, we conclude that
$\sigma(J_1)$ is simple and the  irreducible representation is
realized in $l_2(\Delta)$, where
$\Delta=\{\lambda_p,\lambda_{p+1},\ldots,\lambda_m\}$, for some $1\le p,m\le
(n+1)/2$ and
either $\lambda_1\notin\Delta$ or $\lambda_{(n+1)/2}\notin\Delta$, by formulae
8 if $\lambda_1\in\Delta$ or 9 if $\lambda_1\notin\Delta$.
\end{proof}