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\begin{document}

In this section we will study $*$-representationsof the algebra 
$U_q(so(3,{\Bbb C})$ by bounded operators in a Hilbert space $H$.

Let us recall that the algebra $U_q(so(3,{\Bbb C})$ is defined as the
complex 
algebra with unit element generated by the elements $I_1$, $I_2$, $I_3$ 
satisfying the relations
\begin{eqnarray}
{}[I_1,I_2]_q=q^{1/2}I_1I_2-q^{-1/2}I_2I_1=I_3\label{rel1}\\
{}[I_2,I_3]_q=q^{1/2}I_2I_3-q^{-1/2}I_3I_2=I_1\label{rel2}\\
{}[I_3,I_1]_q=q^{1/2}I_3I_1-q^{-1/2}I_1I_3=I_2\label{rel3}
\end{eqnarray}
Since, by (\ref{rel1}), the element $I_3$ is determined by the elements
$I_1$,
$I_2$, the algebra $U_q(so(3,{\Bbb C}))$ is generated by $I_1$, $I_2$ and
the 
relations
\begin{eqnarray} 
I_2^2 I_1 - (q+q^{-1}) I_2 I_1 I_2 + I_1 I_2^{2} = - I_1
\label{rel:soq3I}\\  
I_1^2 I_2 - (q+q^{-1}) I_1 I_2 I_1 + I_2 I_1^{2} = -I_2.
\label{rel:soq3II}
\end{eqnarray}
which we get substituting the expresion for $I_3$.
Proposition~\ref{} defines involutions over the algebra $U_q(so(3,{\Bbb
C})$ 
(real forms). 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 $so(3)$.

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 $({\bf F}(\cdot))_i$  the $i$-th coordinate of the function
${\bf F}:{\Bbb R}^n\rightarrow {\Bbb R}^n$.

{\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:
$$ I_1e_k=[k-(n+1)/2]e_k,\qquad
 I_2e_k=\left\{
 \begin{array}{ll}
 \alpha_1e_2,& k=1,\\
 \alpha_ke_{k+1}+\alpha_{k-1}e_{k-1},& k\ne1, n,\\
 \alpha_{n-1}e_{n-1},&k=n,
 \end{array}\right.$$
where
$\alpha_k=(d(k-(n+1)/2)[k][n-k])^{1/2}$, $k=1,\ldots,n-1$.

\end{theorem} 
\begin{proof}
The proof of the theorem is based on the technigue of semilinear
relations developed in \cite{}. 

Let $q=e^{\sigma}$, $\sigma\in{\Bbb  R}$ and $I_1$, $I_2$ 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 $I_2$ 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 $E_{I_1}(\cdot)$ be the resolution of the identity for the operator
$I_1$.
Then, by Theorem~\ref{}, $I_1$, $I_2$ satisfy 
(\ref{rel:soq3I})--(\ref{rels0q3II}) if and only if
\begin{equation}\label{supp}
E_{I_1}(\Delta)I_2E_{I_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 $\displaystyle I_1=\frac{sh\sigma
A_1}{sh\sigma}$
condition (\ref{supp}) is equivalent to
$$E_{A_1}(\Delta)I_2E_{A_1}(\Delta')=0,\quad \mbox{for any}\
\Delta,\Delta'\in{\frak B}({\Bbb R}),\ 
\Delta\times\Delta'\cap\Gamma'=\emptyset,$$
where $\Gamma'=\{(s+1,s), (s-1,s)\mid s\in{\Bbb R}\}$.
It follows from Theorem~\ref{} that the operators $A_1$, $I_2$ satysfy
the following relation
$$A_1^2I_2-2A_1I_2A_1+I_2A_1^2=I_2.$$
Let $E_1=([A_1,I_2]+I_2)/2$, $E_1^*=([A_1,I_2]-I_2)/2$. 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 ($I_1$, $I_2=E_1+E_1^*$) of  
(\ref{rel:soq3I})--(\ref{rel:s0q3II}).

It is obvious that the pair ($I_1$, $I_2$) 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 ${\Bbb R}^+\times [0,1)$. Hence 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, $\mbox{dim} H=n$
$$ I_1e_k=(-1)^{k-1}[k+(-1)^j/2]e_k,\quad
I_2e_k=\left\{
\begin{array}{ll}
\alpha_1e_2+(-1)^i[n][1/2]e_1,& k=1,\\
\alpha_ke_{k+1}+\alpha_{k-1}e_{k-1},& k\ne 1, n,\\
\alpha_{n-1}e_{n-1},&k=n,
\end{array}\right.$$
where
$\alpha_k=(d(k-1/2)[n-k][n+k])^{1/2}$, $k=1,\ldots,n-1$, $i,j=0,1$;
\vskip0.5cm

2. One more representation for each odd dimension, $\mbox{dim} H=n=2p-1$
$$ I_1e_k=(-1)^{k-1}[k-(n+1)/2]e_k,\qquad
I_2e_k=\left\{
\begin{array}{ll}
\alpha_1e_2,& k=1,\\
\alpha_ke_{k+1}+\alpha_{k-1}e_{k-1},& k\ne 1, n,\\
\alpha_{n-1}e_{n-1},&k=n,
\end{array}\right.$$
$\alpha_k=(d(k-(n+1)/2)[k][n-k])^{1/2}$, $k=1,\ldots,n-1$.
\end{theorem}
\begin{proof}
The proof essentially runs as that of Theorem~\ref{th1}.
Let $q=-e^{\sigma}$, $\sigma\in {\Bbb R}$ and $I_1$, $I_2$ 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, $I_1$, $I_2$ satisfy (\ref{rel:soq3I}) if and only if
$I_2$ is concentrated on $\Gamma$ with respect to $I_1$, i.e.
$$ 
E_{I_1}(\Delta)I_2E_{I_1}(\Delta')=0,\quad \mbox{for any}\
\Delta,\Delta'\in{\frak B}({\Bbb R}),\ 
\Delta\times\Delta'\cap\Gamma=\emptyset
$$
which is equivalent to
$$ 
E_{A_1}(\Delta)I_2E_{A_1}(\Delta')=0,\quad \mbox{for any}\
\Delta,\Delta'\in{\frak B}({\Bbb R}),\ 
\Delta\times\Delta'\cap\Gamma'=\emptyset,
$$
where $I_1=\frac{sh\sigma A_1}{sh\sigma}$ 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{} we conclude that
$A_1$, $I_2$ satisfy the relation
$$A_1^2I_2+2A_1I_2A_1+I_2A_1^2=I_2.$$
Let $E_1=-(\{A_1,I_2\}+I_2)/2$, $E_2=-(\{A_1,I_2\}-I_2)/2$. It is easy to
show that $E_1=E_1^*$, $E_2=E_2^*$, $I_2=E_1+E_2$ 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 ($I_1=sh\sigma A_1/sh\sigma$, $I_2=E_1+E_2$)
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
$F_1(\lambda)=1-\lambda$, $F_2(\lambda)=-1-\lambda$. 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:

$1$. $H={\Bbb C}^s$,
$$ I_1e_m=[a+m]e_m,\quad
I_2e_m=\left\{
\begin{array}{ll}
\alpha_0e_1+e^{i\phi}\alpha_{s-1}e_{s-1},& m=0,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1}, &m\ne 0, s-1,\\
\alpha_{s-2}e_{s-2}+e^{-i\phi}\alpha_{s-1}e_{0}, &m=s-1,
\end{array}\right.$$
where $\alpha_m=(d(a+m+1)(d(a)^{-1}b-[m][2a+m+1]))^{1/2}$,

\vspace{0.1cm}
$(a,b)\in\{(a,b)\in$ $M\times{\Bbb R}^+\mid \alpha_m>0, m=0,\ldots,s-1\}$,
$\phi\in [0,2\pi)$,

\vspace{0.1cm}
$\sigma M=[-\pi/2,\pi/2]\setminus\{\frac{\pi (2l+1)+m\sigma}{2}\mid
l,m\in{\Bbb Z}\}$;

\vskip0.5cm

$2$.  $H={\Bbb C}^n$, $k$ is odd,
$$ I_1e_m=[a+m]e_m,\quad
I_2e_m=\left\{
\begin{array}{ll}
(-1)^i\lambda e_1+\alpha_1e_2,&  m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1}, & m\ne 1, n-1,
\\
\alpha_{n-1}e_{n-1}+(-1)^j\lambda e_n,& m=n,
\end{array}\right.$$
where $a=-\frac{\pi}{2\sigma}-\frac{1}{2}$, $\alpha_m=
(d(a+m)[m]^2-\lambda\frac{[1/2]}{[m-1/2][m+1/2]})^{1/2}$,
$m=1,\ldots,n-1$,

\vspace{0.1cm}
$\lambda\in\{\lambda\in{\Bbb R}\mid$ $\alpha_m>0, m=1,\ldots,n-1\}$,
$i,j=0,1$;
\vskip0.5cm

$3$. $H={\Bbb C}^{2n}$, $k$ is odd,
$$
I_1=\left(
\begin{array}{llll}
\lambda_1I_2&&&0\\
&\cdot&&\\
&&\ddots&\\
0&&&\lambda_nI_2
\end{array}\right),\quad
I_2=\left(
\begin{array}{llll}
Y_1&\alpha_1I_2&&\\
\alpha_1I_2&0&\ddots&\\
&\ddots&\ddots&\alpha_{n-1}I_2\\
&&\alpha_{n-1}I_2&Y_2
\end{array}\right),$$
where
$\displaystyle
\lambda_m=[a+m]$,\quad
$Y_1=\left(
\begin{array}{ll}
\lambda&0\\
0&-\lambda
\end{array}\right)$, \quad $Y_2=\lambda\left(
\begin{array}{ll}
\cos\varphi&\sin\varphi\\
\sin\varphi&-\cos\varphi
\end{array}\right)$,

\vspace{0.1cm}
$a$, $\alpha_m$ are the same as in $2$,
$\lambda\in\{\lambda\in{\Bbb R^+}\mid\alpha_m>0, m=1,\ldots,n-1\}$,

\vspace{0.1cm}
$\varphi\in[0,2\pi)$;
\vskip0.5cm

$4$. $H={\Bbb C}^n$, $k$ is odd,
$$ I_1e_m=[a+m]e_m,\quad
I_2e_m=\left\{
\begin{array}{ll}
\alpha_1e_2,&m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1},&m\ne 1,n,\\
\alpha_{n-1}e_{n-1},&m=n,
\end{array}\right.$$
where $a=-\frac{\pi}{2\sigma}-1$, 
$\alpha_1=\alpha_{n-1}=\frac{\sqrt{2}}{|q-q^{-1}|}$,
$\alpha_m=\frac{1}
{|q-q^{-1}|}$,  $m=2,\ldots, n-1$;
\vskip0.5cm
$5$. $H={\Bbb C}^{2(n-1)}$, $k$ is odd,

$$I_1=\left(
\begin{array}{llll}
\lambda_1I_1&&&0\\
&\lambda_2I_2&&\\
&&\ddots&\\
0&&&\lambda_nI_1\\
\end{array}\right),
I_2=\left(
\begin{array}{lllll}
0&X_1^*&&&\\
X_1&0&X_2^*&&\\
&X_2&\ddots&&\\
&&&\ddots & X_{n-1}^*\\
&&&X_{n-1}&0
\end{array}\right),$$
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 $X_{n-1}=(\frac{1}{|q-q^{-1}|},0)$,

\vspace{0.1cm}
$X_m=\frac{1}{|q-q^{-1}|}I_2$,
$m=2,\ldots,n-2$, $\varphi\in [0,\pi)$;
\vskip0.5cm
$6$. $H={\Bbb C}^{n}$, $k$ is even,
$$ I_1e_m=(-1)^{k/2}[a+m]e_m,\quad
I_2e_m=\left\{
\begin{array}{ll}
\alpha_1e_2+\frac{(-1)^i}{q-q^{-1}}e_1,& m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1},& m\ne 1, n,\\
\alpha_{n-1}e_{n-1},&m=n,
\end{array}\right.$$
where $a=\frac{\pi}{2\sigma}+\frac{1}{2}$,
$\alpha_m=\frac{1}{|q-q^{-1}|}$, $i=0,1$;
\vskip0.5cm
$7$. $H={\Bbb C}^{2n-1}$, $k$ is even,

$I_1=\left(
\begin{array}{lllll}
\lambda_1I_2&&&0\\
&\ddots&&\\
&&\lambda_{n-1}I_2&\\
0&&&\lambda_nI_1
\end{array}\right)$,
$I_2=\left(
\begin{array}{lllll}
Y&X_1^*&&\\
X_1&0&&\\
&&\ddots & X_{n-1}^*\\
&&X_{n-1}&0
\end{array}\right)$,

\noindent
where
$\lambda_m=[\frac{\pi}{2\sigma}+\frac{1}{2}+m]$,\quad
$Y=\frac{1}{|q-q^{-1}|}\left(
\begin{array}{ll}
\cos\varphi&\sin\varphi\\
\sin\varphi&-\cos\varphi
\end{array}\right)$, 
$X_m=\frac{1}{|q-q^{-1}|}I_2$,
$m=1,\ldots,n-2$, $X_{n-1}=(\frac{\sqrt{2}}{|q-q^{-1}|},0)$, 
$\varphi\in [0,2\pi)$;
\vskip0.5cm
$8$. $H={\Bbb C}^{\bf p}$
$$ I_1e_m=(-1)^i[a+m]e_m,\quad
I_2e_m=\left\{
\begin{array}{ll}
\alpha_1e_2+(-1)^j[p][a+1]e_1,&m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1},&m\ne 1,p,\\
\alpha_{p-1}e_{p-1},& m=p,
\end{array}\right.$$
where $a=\frac{\pi}{2\sigma}-\frac{1}{2}$, 
$\alpha_m=(d(x+m)[m-p][m+p])^{1/2}$,

\vspace{0.1cm}
$p\in\{p\in{\Bbb N}\mid
\alpha_m>0, 1\le m<p\}$, $i,j=0,1$;
\vskip0.5cm
$9$. $H={\Bbb C}^{\bf p}$,
$$ I_1e_m=[a+m]e_m,\quad
I_2e_m=\left\{
\begin{array}{ll}
\alpha_1e_2,&m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1},&m\ne 1, p,\\
\alpha_{l-1}e_{l-1},&m=p,
\end{array}\right.$$
where $\alpha_1=(-[a+1](q^{a+2}+q^{-a-2})^{-1})^{1/2}$,\quad

\vspace{0.1cm}
 $\alpha_m=(-d(a+m)[m][2a+m+1])^{1/2}$,
$m\ne 1$,

\vspace{0.1cm}
$(a,p)\in\{(a,p)\in{\Bbb R}\times{\Bbb N}\mid
\sigma a\ne\frac{\pi}{2}-l\sigma+\pi r$, $[p][2a+p+1]=0$, 

\vspace{0.1cm}
$[(m-l)/2](q^{a+(m+p)/2}+q^{-(a+(m+p)/2)})\ne 0,$

\vspace{0.1cm}
$\alpha_m>0, m=1,\ldots,p-1, r\in{\Bbb Z}\}$.
\vskip0.5cm
$10$. $H={\Bbb C}^1$, $I_1=\lambda$, $I_2=0$, where 
$\lambda^2<-\frac{4}{(q-q^{-1})^2}$.
\end{theorem}
\begin{proof}
Let $I_1=I_1^*$, $I_2=I_2^*$ 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_{I_1}(\Delta)I_2E_{I_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
$I_2E_{I_1}(S_0)H=0$. Hence $H_0:=E_{I_1}(S_0)H$ is invariant with
respect to $I_1$, $I_2$ and any irreducible representation in $H_0$ is
one-dimensional given by 
$$I_1=(\lambda),\quad I_2=(0),\quad \lambda\in S_0.$$

Assume now that the spectrum of $I_1$ 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 ($I_1$, $I_2$) is irreducible, then the measure $E_{I_1}(\cdot)$ is
ergodic with respect to $\Gamma$, i.e., either $E_{I_1}(M)=0$ or
$E_{I_1}(M)=I$ for any set $M$ which is invariant with respect to
$\Gamma$. In fact, if it were not, we would conclude that $E_{I_1}(M)H$
or $E_{I_1}(M)H$ is a subspace invariant with respect to $I_1$, $I_2$.
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 $I_1$ is discrete
and concentrated on a trajectory if only $(I_1,I_2)$ 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 grafically 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 $I_1$, $I_2$ is an irreducible representation acting in $H_3$,
then
$\sigma(I_1)\subset O(\{\frac{\sin x\sigma}{\sin\sigma}\})
=\{\frac{\sin((x+k)\sigma)}{\sin\sigma}\mid 0\le k\le s-1\}$, 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 $I_1$ corresponding to
$\lambda_k=\frac{\sin((x+k)\sigma)}{\sin\sigma}$. 

It follows from
(\ref{supp}) that $I_2P_0H\subset P_1H\oplus P_{s-1}H$,
$I_2P_{s-1}H\subset P_1H\oplus P_{s-2}H$, and $I_2P_kH\subset
P_{k+1}H\oplus P_{k-1}H$ for $k\ne 0, s-1$. Thus the operator $I_2$ can
be  represented in the form $I_2=X+X^*$, where
$X=\sum_{k=0}^{s-2}P_{k+1}I_2P_k+P_0I_2P_{s-1}$. Moreover, the pair
($I_1$, $I_2$) is irreducible if and only if so is ($I_1$, $X$, $X^*$).
One can easily show that for $I_1$, $I_2$ 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 least number such that
$\lambda_k\in\sigma(I_1)$.
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
$I_1$,  $X$, $X^*$ is irreducible. From this we conclude that 
$\oplus_{l=k}^{s-1}X^lE_{C_k}(\Delta)U$ is invariant with respect
to $I_1$,  $X$,  $X^*$ for any $\Delta\in{\frak B}({\Bbb  R})$ and
subspace $U$ such that $C_kU\subset U$. 
 Hence, if ($I_1$, $X$, $X^*$) is irreducible then $\Delta$ is
 concentrated
in one point and we can choose a basis consisting of eigenvectors of
$I_1$, namely, $\{e_{\lambda_k},
Xe_{\lambda_k}/||Xe_{\lambda_k}||,\ldots,
X^{s-1}e_{\lambda_k}/||X^{s-1}e_{\lambda_k}||\}$, 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
$I_1$, $X$,
$X^*$ and $I_2$ on the basis. In particular, if
$\sigma(I_1)=\{\lambda_0, \lambda_1,\ldots,\lambda_{s-1}\}$. we will
have representations given by 1, otherwise, representations from
series 9.


2.2. If $I_1$,  $I_2$  is  an irreducible representation acting in $H_1$,
$H_2$, then $\sigma(I_1)$ is not necessary to be simple.  First let us
consider  the
case when the support of the operator $I_2$ with repspect to $I_1$ 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
$\sigma(I_1)=\{\lambda_1,\ldots,\lambda_n\}$.

Denote by   $P_k$   the   projection   onto  the  eigenspace  of  $I_1$
corresponding to the eigenvalue $\lambda_k=-\frac{\cos((2k-1)\sigma/2}
{\sin\sigma}$. As in the case 2.1 the operator $I_2$   can  be
represented  in  the  form
$I_2=X+X^*+Y$, where      $X=\sum_{k=1}^{n-1}P_{k+1}I_2P_k$,
$Y=P_1I_2P_1+P_nI_2P_n$, and $Y\ne 0$,  $XP_k\ne 0$,  $k<n$. Moreover
($I_1$, $I_2$)  is irreducible if and only if the family 
$I_1$,  $X$,  $X^*$, $Y$ is irreducible.
Clearly $I_1$,  $I_2$  satisfy the relation $Q(I_1,I_2)\equiv
I_2^2I_1-(q+q^{-1})I_2I_1I_2+I_1I_2^2-I_1=0$ if and only if
$P_sQ(I_1,I_2)P_k=0$ 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 $I_1$, $I_2$, which
contradicts the fact that $\sigma(I_1)=\{\lambda_1,\ldots,\lambda_n\}$.

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 ($I_1$,
$I_2$) 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
$I_1$, $X$, $X^*$, $Y$, and hence with respect to $I_1$, $I_2$:
\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 $\{e_k,k\in K\}$, $\mbox{dim}K\le 2$, 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 $I_1$, $I_2$ acts as follows:
$$I_1=\left(
\begin{array}{llll}
\lambda_1I&&&\\
&\lambda_2I&&\\
&&\ddots&\\
&&&\lambda_nI
\end{array}\right), I_2=\left(
\begin{array}{llll}
Y_1&\mu_1(\lambda)&&\\
\mu_1(\lambda)I&0&\ddots&\\
&\ddots&\ddots&\mu_{n-1}(\lambda)I\\
&&\mu_{n-1}(\lambda)I&Y_2
\end{array}\right),$$
$ \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 $\lambda>0$, $\varphi\in[0,2\pi)$.

Hence, all  irreducible representations $I_1$, $I_2$ have
dimensions $n$ or $2n$, moreover, two such  pairs ($I_1$,  $I_2$),
($I_1'$, $I_2'$) 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 from the statement.


If the support of $I_2$ is a  subgraph  $\Gamma_0$  of  (\ref{gr1})
($\Gamma_0$ does  not  coincide  with  (\ref{gr1})),  then  arguments
similar to  that  in  the  proof  of  Theorem~\ref{th3}  gives   that
$\sigma(I_1)$ is simple as soon as the pair $I_1$, $I_2$ 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 $I_2$ 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 $I_1$ corresponding  to
$\lambda_k=-\frac{\cos (k-1)\sigma}{\sin\sigma}$, $X_k=P_{k+1}I_2P_k$,
$X=\sum_{k=1}^{n-1}X_k$. Using the same arguments we can conclude
that  $I_2=X+X^*$ 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,
$\alpha_{n-1}=0$, $\beta_2=0$, $\alpha_k\beta_m\ne 0$, $k\ne n-1$,
$m\ne 2$.
Moreover, the pair ($I_1$, $I_2$)
is irreducible if and only if the  triple ($I_1$, $X$, $X^*$) is
irreducible. 
If $\{\frac{1}{\sin\sigma}\}$ or $\{-\frac{1}{\sin\sigma}\}$ does  not
belong  to $\sigma(I_1)$ then $X$, $X^*$ satisfy the relation:
$$X^*X=F_1(XX^*,A)\quad \mbox{or}\quad XX^*=F_2(X^*X,A)$$
respectively, where 
$$F_1(\mu,\lambda_k)=\left\{
\begin{array}{ll}
\frac{\gamma_k-\beta_k\mu}{\alpha_k},& k\ne n-1,\\
0,& \mbox{otherwise},
\end{array}\right.\quad
F_2(\mu,\lambda_k)=\left\{
\begin{array}{ll}
\frac{\gamma_k-\alpha_k\mu}{\beta_k},& k\ne 2,\\
0,& \mbox{otherwise}.
\end{array}\right.$$
Hence any irreducible representation can be realized in the space
$H=l_2(\Delta)$ by formulae:
$$ I_1e_{k}=\lambda_ke_{k},\quad Xe_k=\mu_ke_{k+1},$$
where $\Delta=\{\lambda_p,\lambda_{p+1},\ldots,\lambda_{m}\mid 0\le
p.m\le n-1\}$, 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 $\{\pm\frac{1}{\sin\sigma}\}\in\sigma(I_1)$, then the  problem  to
describe irreducible  representations ($I_1$, $I_2$) is reduced to
that of pairs of operators  satisfying  some  quadratic
relations. In order to show that, consider the following operators  
in the subspace $P_2H$: $D_1=X_1X_1^*$,
$D_2=(X_{n-1}\ldots X_{2})^*X_{n-1}\ldots X_{2}$. From (\ref{f30}) we have
$X_1^*X_1=\frac{\gamma_1}{\alpha_1}I$, $X_{n-1}\ldots
X_2(X_{n-1}\ldots X_2)^*=\mu I$, where $\mu=\prod_{k=0}^{n-3}({\bf
F}^{(k)}
(\frac{\gamma_n}{\beta_n},\lambda_{n-1}))_1$, ${\bf F}(x,\lambda_k)=
(\frac{\gamma_k-\alpha_kx}{\beta_k},\lambda_{k-1})$. One can 
check that $\frac{\gamma_1}{\alpha_1}=\frac{1}{2\sin^2\sigma}$, 
$\mu=\frac{1}{2^{2n-5}(\sin^2\sigma)^{n-2}}$.
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
$$X_1^*H_0\oplus H_0\oplus X_2H_0\oplus X_3X_2H_0\oplus\ldots\oplus
X_{n-1}\ldots  X_2H_0$$
is  invariant  with respect to $I_1$,  $X$, $X^*$. In fact, using
(\ref{f30}) one can
easily prove that $X_k^*X_k\ldots X_2=\mu_kX_{k-1}\ldots X_2$, $1<k<n-1$
(here $\mu_k=1/4sin^2\sigma$) .
Moreover, $X_{n-2}\ldots X_2X_2^*\ldots X^*_{n-2}=\lambda I$,
where  $\lambda=\prod_{k=0}^{n-4}({\bf F}^{(k)}
(\frac{\gamma_{n-1}}{\beta_{n-1}},\lambda_{n-1}))_1$. Assuming that
$\lambda=0$, we will have that $X_3^*\ldots  X_{n-2}^*P_{n-1}H\oplus\ldots
P_{n-1}H\oplus X_{n-1}P_{n-1}H$ is invariant with respect to $I_1$,
$I_2$,  hence
$\sigma(I_1)\not\ni\{\frac{1}{\sin\sigma}\}$ if  the pair ($I_1$, $I_2$)
is
irreducible, which contradicts the assumption. Thus $\lambda\ne 0$,
\begin{eqnarray*}
&&X_{n-1}^*X_{n-1}\ldots X_2H_0=\lambda^{-1}\lambda
X_{n-1}^*X_{n-1}\ldots
X_2H_0=\\
&&=\lambda^{-1}(X_{n-2}\ldots X_2X_2^*\ldots X_{n-2}^*)X_{n-1}^*
X_{n-1}\ldots X_2H_0\subset X_{n-2}\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 $I_1$, $I_2$. 
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 $X_i\ldots X_2H_0$, $i\ne
n-1$,
and the vectors $U_{n-1}e_1$,
$U_1(\cos(\varphi/2)e_1+\sin(\varphi/2)e_2)$ generate the spaces
$X_{n-1}\ldots X_2H_0$ 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 $I_2$ on the vectors of  this
base it is sufficient to know that for the operators $X_i$.

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

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

\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
$\sigma(I_1)\ne\{\lambda_1,\ldots,\lambda_n\}$ if ($I_1$, $I_2$) is
irreducible.
Let ${\Bbb C}e_1=H_0$. Then using the same arguments one can show that
$\{U_1^*e_1,e_1,U_1e_1,\ldots U_{n-1}e_1\}$ is an orthogonal basis in
$H$ and the corresponding irreducible representation ($I_1$, $I_2$) is
given by formulae 6. 

2.4. Suppose that the support of $I_2$ is
\begin{equation}\earch{\lambda_1}
{\lambda_n}\label{gr3}
\end{equation}
where $\lambda_1=\pm\frac{\cos\sigma/2}{\sin\sigma}$,
$\lambda_n=\frac{1}{\sin\sigma}$. 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 $I_2$  can
be represented in the  form  $I_2=X+X^*+Y$,  where
$X=\sum_{k=1}^{n-1}P_{k+1}I_2P_k$, $Y=P_1I_2P_1$, $P_k$ 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 $\alpha_k=-2\sin(\frac{(2k+1)\sigma}{2})$,
$\beta_k=2\sin(\frac{(2k-3)\sigma}{2})$,
$\gamma_k=-f(\frac{\cos((2k-1)\sigma/2)}{\sin\sigma}$,   in  particular,
$\alpha_k\ne 0$ for $k\ne n-1$, $\alpha_{n-1}=0$. 
As above, the  problem of  describing  irreducible  representations
($I_1$, $I_2$) can be 
reduced  to  that of
irreducible pairs  ($D_1$,  $D_2$)   satisfying   some   quadratic
relation. Here $D_1=Y$, $D_2=(X_{n-1}\ldots X_1)^*X_{n-1}\ldots X_1$
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 $H_0\oplus
XH_0\oplus\ldots\oplus X^{n-1}H_0$ is invariant with respect  to  $I_1$,
$X$, $X^*$. Moreover dimensions of the irreducible representations
are $2(n-1)+1$ or $n$.  This  follows  by  the same method as in the
previous
case. We will get representations 6, 7.

If ($I_1$, $I_2$) is irreducible and $\sigma(I_1)$ is a proper subset
of $\{\lambda_1,\ldots,\lambda_n\}$ then, as before, we conclude that
$\sigma(I_1)$ is simple and the  irreducible representation is
realized in $l_2(\Delta)$, where
$\Delta=\{\lambda_p,\lambda_{p+1},\ldots,\lambda_m\}$, for some $1\le
p,m\le
n$ and 
either $\lambda_1\notin\Delta$ or $\lambda_n\notin\Delta$, by formulae
8 if $\lambda_1\in\Delta$ or 9 if $\lambda_1\notin\Delta$.

\end{proof}




\end{document}