One Hat Cyber Team

View File Name : chapt27.tex

\section{Representations of $q$-deformed $U(so(3,{\mathbb
C}))$}\label{sec:2.3}
\subsection{Real forms of $U_q(so(3, {\mathbb C}))$}\label{sec:2.3.1}
\markright{2.3. Representations of $q$-deforemd $U(so(3, \mathbb R))$}

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

Let us describe involutions on the algebra
$U_q(so(3,{\mathbb C}))$. 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 it is said to be an involution of the first order.
 On the other hand, there might exist involutions
which map linear combinations of generators to  polynomials
of generators of the degree higher then one. 
If linear combinations of generators
are  mapped by
the involution to polynomials of  the second degree  then we will call such
involutions quadratic.

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

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

\item[$2).$] If\/ $ |q| = 1$, $q \neq \pm 1 $, then any involution of the
first order  in the algebra $ U_q(so(3,{\mathbb C})) $ is equivalent to
one of the following
two inequivalent involutions\textup:
\begin{itemize}
\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 $.
\end{itemize}

\item[$3).$] If\/ $q=-1$, then each involution of the first order
in the algebra $ U_q(so(3,{\mathbb C})) $ is 
 equivalent either to \textup{1)},
or  $2a)$, or  $2b)$.
\end{itemize}
\end{proposition}

\begin{proof}
The proof consists of trivial but rather lengthy calculations, and 
we leave them out.
\end{proof}

 The algebras
$U_q(so(3,{\mathbb C}))$  also have some  quadratic involutions. Since the element
$I_3$ is not independent and determined by $I_1$ and $I_2$, the algebra 
$U_q(so(3,{\mathbb C}))$  is generated by $I_1$ and $I_2$.
If we substitute $I_3$ with the left hand side of the first relation 
into the second and third relation in
(\ref{soq3}), we will get the following two 
relations of the third order
for $I_1$ and $I_2$:
\begin{align} 
I_2^2 I_1 - (q+q^{-1}) I_2 I_1 I_2 + I_1 I_2^{2} &= - I_1, \notag
\\
I_1^2 I_2 - (q+q^{-1}) I_1 I_2 I_1 + I_2 I_1^{2} &= -I_2.\label{rel:Mq2}
\end{align}

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 the relations (\ref{soq3}).
We have
the following statement.

\begin{proposition}
Any  such involution is isomorphic either to one of the involutions from
Proposition~\textup{\ref{th-inv1-eq}}, or to one of the following ones\textup:

$1)$ $q\in{\mathbb R}$, 
\[
I_1^*=I_1,\quad I_2^*=I_2,\quad I_3^*=\begin{cases}
q^{{1}/{2}}I_2I_1-q^{-{1}/{2}}I_1I_2,&q>0,\\
-(q^{{1}/{2}}I_2I_1-q^{-{1}/{2}}I_1I_2),&q<0,
\end{cases}
\]

$2)$ $q \in {\mathbb R}$, 
\[
I_1^*=I_1,\quad
I_2^*=-I_2, \quad I_3^* =\begin{cases}
-(q^{{1}/{2}}I_2I_1-q^{-{1}/{2}}I_1I_2),&q>0,\\
q^{{1}/{2}}I_2I_1-q^{-{1}/{2}}I_1I_2,&q<0,
\end{cases}
\]

$3)$ $q \in {\mathbb R}$, 
\[
I_1^*=-I_1,\quad I_2^*=-I_2, I_3^* =\begin{cases}
(q^{{1}/{2}}I_2I_1-q^{-{1}/{2}}I_1I_2),&q>0,\\
-(q^{{1}/{2}}I_2I_1-q^{-{1}/{2}}I_1I_2),&q<0,
\end{cases}
\]

$4)$ $|q|=1$, 
\[
I_1^*=I_2, \quad
I_2^*=I_1, \quad I_3^* =
q^{-{1}/{2}}I_2I_1-q^{{1}/{2}}I_1I_2.
\]
\end{proposition}

\subsection{Representations of $U_q(so(3,\mathbb C))$}\label{sec:2.3.2}

We will study bounded $*$-representations of the algebra
$U_q(so(3,{\mathbb C}))$ on a Hilbert space $H$.
We restrict ourselves to a study of   representations of the
$*$-algebra defined by the involution $I_1^*=-I_1$, $I_2^*=-I_2$ 
(for $q\in{\mathbb R}$ and $|q|=1$, $q\ne\pm 1$), which, for
$q=1$, corresponds to the compact real form of $so(3)$. Henceforth
we denote it by $U_q(so(3))$.

We assume the following notations 
$$
[x]_q=\frac{q^x-q^{-x}}{q-q^{-1}},\quad 
d_q(m)=\frac{1}{(q^m+q^{-m})(q^{m+1}+q^{-(m+1)})}.
$$
Let $E_A(\cdot)$ denote the spectral measure of the self-adjoint operator
$A$ and $({\mathbf F}(\cdot))_i$  the $i$-th coordinate of a function
${\mathbf F}\colon{\mathbb R}^n\rightarrow {\mathbb R}^n$.

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
 self-adjoint 
 and satisfy the relations:
\begin{align} 
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{align}

\noindent\textbf{1.} 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 $H={\mathbb C}^n$ are unitarily
equivalent to the following  one\textup:
\begin{align*}
J_1e_k&=[k-(n+1)/2]_qe_k,
\\
 J_2e_k&= \begin{cases}
 \alpha_1e_2,& k=1,\\
 \alpha_ke_{k+1}+\alpha_{k-1}e_{k-1},& 2 \le k \le  n-1 ,\\
 \alpha_{n-1}e_{n-1},&k=n,
 \end{cases}
\end{align*}
where
$\alpha_k=(d_q(k-(n+1)/2)[k]_q[n-k]_q)^{1/2}$.

\end{theorem}
\begin{proof}
The proof of the theorem is based on the technique of semilinear
relations developed in Section~\ref{sec:1.3} and the technique of dynamical
systems.

Let $q=e^{\sigma}$, $\sigma\in{\mathbb  R}$, and $J_1$, $J_2$ be
self-adjoint operators in a Hilbert space $H$ satisfying relations
\eqref{rel:soq3I}--\eqref{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
\begin{align*}
F_{1}(s)&=s\cosh \sigma+\sqrt{s^2\sinh^2\sigma+1},
\\
F_{2}(s)&=s\cosh\sigma-\sqrt{s^2\sinh^2\sigma+1},
\end{align*}
 and
$\sinh\sigma=(q-q^{-1})/{2}$.
Then $\Phi(t,s)=(t-F_1(s))(t-F_2(s))$.
Consider the parameterization $s=\sinh\sigma
  x(s)(\sinh\sigma)^{-1} =[x(s)]_q$,
$x(s)\in{\mathbb R}$,
which gives  $F_1(s)=[x(s)+1]_q$, $F_2(s)=[x(s)-1]_q$, and
$$
\Gamma=\{([x+1]_q,[x]_q), ([x-1]_q,[x]_q)\mid x\in{\mathbb R}\}.
$$
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
\eqref{rel:soq3I}  if and only if
\begin{equation}\label{supp}
E_{J_1}(\Delta)J_2E_{J_1}(\Delta')=0,
\end{equation}
for any $\Delta$, $\Delta'\in{\mathfrak B}({\mathbb R})$, 
$\Delta\times\Delta'\cap\Gamma=\emptyset$. For
the operator $A_1$ defined by $
J_1={\sinh\sigma A_1}/{\sinh\sigma}$,
condition (\ref{supp}) is equivalent to the following one:
$$
E_{A_1}(\Delta)J_2E_{A_1}(\Delta')=0,\quad \mbox{for any}\
\Delta,\Delta'\in{\mathfrak B}({\mathbb R}),\
\Delta\times\Delta'\cap\Gamma'=\emptyset,
$$
where $\Gamma'=\{(s+1,s), (s-1,s)\mid s\in{\mathbb R}\}$.
It follows from Theorem~\ref{cth2} that the operators $A_1$, $J_2$
satisfy the relation
$$
A_1^2J_2-2A_1J_2A_1+J_2A_1^2=J_2.
$$
Let $E_1=([A_1,J_2]+J_2)/2$, $E_1^*=([A_1,J_2]-J_2)/2$. Then one can
check that
\begin{gather}
A_1E_1=E_1(A_1+I),\quad A_1E_1^*=E_1^*(A_1-I),\notag
\\
E_1^*E_1=F(A_1,E_1E_1^*),\label{dynrel}
\end{gather}
where 
\[
F(x_1,x_2)=x_2\,\frac{\cosh((x_1-1)\sigma)}{\cosh((x_1+1)\sigma)}-
\frac{\sinh(x_1\sigma)}{2\sinh\sigma \cosh((x_1+1)\sigma)}.
\]
 Conversely, any
operators $A_1$, $E_1$, $E_1^*$ satisfying \eqref{dynrel} determine
a representation $J_1$, $J_2=E_1+E_1^*$ of
\eqref{rel:soq3I}--\eqref{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 and there is a one-to one 
correspondence between classes of unitary equivalent representations 
of these families. Hence, instead of
representations of relations (\ref{rel:soq3I})--(\ref{rel:soq3II}) we can
talk about the
representations of (\ref{dynrel}). To relation (\ref{dynrel}) there
corresponds the dynamical system
 $(x_1,x_2)\mapsto{\mathbf
F}(x_1,x_2)\equiv (x_1+1,F(x_1+1,x_2)$ which has the  measurable
section $[0,1)\times{\mathbb R}$. Therefore the joint spectral measure of
$A_1$, $E_1^*E_1$,
is discrete and concentrated on an orbit if ($A_1$, $E_1$, $E_1^*$) is 
irreducible, and we can choose a basis consisting of its
eigenvectors. Then we have
\begin{gather*}
A_1e_{x_1,x_2}=x_1\, e_{x_1,x_2},\quad E_1^*E_1e_{x_1,x_2}=x_2\,e_{x_1,x_2},
\\
E_1e_{x_1,x_2}=x_2^{1/2} e_{{\mathbf F}(x_1,x_2)}, 
\quad
E_1^*e_{x_1,x_2}=
({\mathbf F}^{-1}(x_1,x_2))_2^{1/2}e_{{\mathbf F}^{-1}(x_1,x_2)}
\end{gather*}
where $(x_1,x_2)$ is taken from some orbit. The later cannot hold for all
points of the orbit, since
\begin{align*}
({\mathbf F}^{(k)}(x_1,x_2))_2&=
x_2\,\frac{\cosh(x_1\sigma)\cosh((x_1+1)\sigma)}
{\cosh((x_1+k)\sigma)\cosh((x_1+k+1)\sigma)}
\\
&\quad{}-\frac{\sinh(2x_1+k+1)\sigma)\sinh(k\sigma)}{4\sinh^2\sigma
  \cosh((x_1+k)\sigma)\cosh((x_1+k+1)\sigma)}
\\&
\to -\frac{1}{4\sinh^2\sigma},\qquad k\rightarrow\pm\infty,
\end{align*}
 while $({\mathbf F}^{(k)}(x_1,x_2))_2$ are
    eigenvalues of the self-adjoint
non-negative operator $E_1^*E_1$. Thus there
    exists the highest vector (vector $e_{x_1,x_2}$ with the largest
    $x_1$) 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}

\noindent\textbf{2.} 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 unitarily non-equivalent
irreducible representations of
dimension $2p$ and five irreducible representations of dimension
$2p-1$, which act as follows\textup:

\textup1. Four representations with any finite dimension, $\dim H=n$,
\begin{align*}
 J_1e_k &=(-1)^{k-1}\,[k+(-1)^j/2]_{-q}\,e_k,
\\*
J_2e_k& =\begin{cases}
\alpha_1\,e_2+(-1)^i\,[n]_{-q}[1/2]_{-q}\,e_1,& k=1,\\
\alpha_k\, e_{k+1}+\alpha_{k-1}\,e_{k-1},& 2\le k \le  n-1,\\
\alpha_{n-1}\,e_{n-1},&k=n,
\end{cases}
\end{align*}
where
$\alpha_k=(d_{-q}(k-1/2)\,[n-k]_{-q}\,[n+k]_{-q})^{1/2}$, $i$, $j=0$, $1$.
\vskip0.5cm

\textup{2.}
One more representation for each odd dimension, $\dim H=n=2p-1$,
\begin{align*}
 J_1e_k& =(-1)^{k-1}\,[k-(n+1)/2]_{-q}\,e_k,
\\
J_2e_k&=\begin{cases}
\alpha_1\,e_2,& k=1,\\
\alpha_k\,e_{k+1}+\alpha_{k-1}\,e_{k-1},& 2 \le k\le n-1,\\
\alpha_{n-1}\,e_{n-1},&k=n,
\end{cases}
\end{align*}
where $\alpha_k=(d_{-q}(k-(n+1)/2)\,[k]_{-q}\,[n-k]_{-q})^{1/2}$.
\end{theorem}

\begin{proof}
The proof essentially goes in the same way  as that of Theorem~\ref{th1}.
Let $q=-e^{\sigma}$, $\sigma\in {\mathbb R}$, and $J_1$, $J_2$ be
self-adjoint operators satisfying \eqref{rel:soq3I}--\eqref{rel:soq3II}.
Let $\Gamma=\{\Phi(t,s)\equiv t^2-(q+q^{-1})ts+s^2=1\}$ be the 
the characteristic binary relation corresponding to
(\ref{rel:soq3I}). Considering the same parameterization
$s={\sinh\sigma
x(s)}/{\sinh\sigma}\equiv [x(s)]_{-q}$, $x(s)\in{\mathbb R}$, we
get $\Gamma=\{(-[x+1]_{-q},[x]_{-q}), 
(-[x-1]_{-q},[x]_{-q})\mid x\in{\mathbb R}\}$.
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.,
$$
E_{J_1}(\Delta)J_2E_{J_1}(\Delta')=0,\quad \mbox{for any}\
\Delta,\Delta'\in{\mathfrak B}({\mathbb R}),\
\Delta\times\Delta'\cap\Gamma=\emptyset,
$$
which is equivalent to
$$
E_{A_1}(\Delta)J_2E_{A_1}(\Delta')=0,\quad \mbox{for any}\
\Delta,\Delta'\in{\mathfrak B}({\mathbb R}),\
\Delta\times\Delta'\cap\Gamma'=\emptyset,
$$
where $J_1={\sinh\sigma A_1}/{\sinh\sigma}$ and
$\Gamma'=\{(-(x+1),x), (-(x-1),x)\mid x\in{\mathbb R}\}=\{(t,s)\mid
t^2+2ts+s^2=1\}$. 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{gather}%\label{dynrel3}
A_1E_1=-E_1(A_1-I),\quad A_1E_1=-E_1(A_1+I),\notag
\\
E_1^2\cosh(\sigma(A_1+I))=E_2^2\cosh(\sigma(A_1-1))-\frac{\sinh\sigma
  A_1}{2\sinh\sigma}.
\label{dynrel4}
\end{gather}
Conversely, any representation $A_1$, $E_1$, $E_2$ of
\eqref{dynrel4}
defines a representation $J_1=\sinh\sigma A_1/\sinh\sigma$, $J_2=E_1+E_2$
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 as in the proof of Theorem~\ref{th:gleb} on
representations of graded $so(3)$.
By the same arguments, we can choose a basis consisting of
eigenvectors of the operator $A_1$ for any irreducible representation
$A_1$, $E_1$, $E_2$ in $H$. Then we have
$$
A_1e_{\lambda}=\lambda\, e_{\lambda},\quad 
E_1e_{\lambda}=a_1(\lambda)\,e_{1-\lambda},\quad 
E_2e_{\lambda}=a_2(\lambda)\,e_{-1-\lambda},
$$
where $\lambda$ belongs to the  orbit
$\Omega=\{F_1^{(k)}F_2^{(m)}(\lambda),k,m\in {\mathbb Z}_+\}$ and
$F_1(\lambda)=1-\lambda$, $F_2(\lambda)=-1-\lambda$. The conditions for
$A_1$, $E_1$, $E_2$ to satisfy  relation  \eqref{dynrel4}
are the following:
\begin{gather*}
a_1(1-\lambda)=\overline{a_1(\lambda)},\quad
a_2(-1-\lambda)=\overline{a_2(\lambda)},
\\
|a_1(\lambda)|^2\cosh\sigma(\lambda+1)=|a_2(\lambda)|^2\cosh\sigma(\lambda-1)-
\sinh\sigma\lambda/2\sinh\sigma.
\end{gather*}
Similarly to the case of 
 the graded $so(3)$, we see that the last relation cannot 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 satisfying
these condition are those which contain $0$ and $\pm 1/2$.
Using these conditions one can easily get the statement.
\end{proof}

\noindent\textbf{3.}
Representations of $U_q(so(3))$, $q$ is a root of unity.

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

In what follows, we denote by $I_1$ and $I_2$ the identity
$1\times 1$ respectively $2\times 2$ matrices. 

\begin{theorem}\label{th:soq3root}
Let $\sigma=\pi k/{n}$, $\sigma\ne\pi l$.  Any irreducible
representations  of $U_q(so(3))$  is unitarily equivalent to one
of the following\textup:

$1$. $H={\mathbb C}^s$, $ J_1e_m=[a+m]_q\,e_m$,
\[
J_2e_m=\begin{cases}
\alpha_0\,e_1+e^{i\phi}\alpha_{s-1}\,e_{s-1},& m=0,\\
\alpha_m\,e_{m+1}+\alpha_{m-1}\,e_{m-1}, &1\le m \le  s-2,\\
\alpha_{s-2}\,e_{s-2}+e^{-i\phi}\alpha_{s-1}\,e_{0}, &m=s-1,
\end{cases}
\]
where 
\begin{align*}
\alpha_m&=\biggl(\frac{b\cos
a\sigma\cos(a+1)\sigma}{\cos(a+m+1)\sigma\cos (a+m)\sigma}
\\
&\qquad{}-\frac{\sin m\sigma
\sin(2a+m+1)\sigma}{4\sin^2\sigma
\cos(a+m+1)\sigma\cos (a+m)\sigma}\biggr)^{1/2},
\end{align*}
the pair 
$(a,b)$ belongs to the set 
$\{(a,b)\in$ $M\times{\mathbb R}^+\mid \alpha_m>0,\, m=0,\dots,s-1\}$,
$\phi\in [0,2\pi)$,
and\/
$\sigma M=[-\pi/2,\pi/2]\setminus\{(\pi (2l+1)+m\sigma)/2\mid
l,m\in{\mathbb Z}\}$\textup;

\smallskip
$2$.  $H={\mathbb C}^n$, $k$ is odd, $ J_1e_m=[a+m]_qe_m$,
\[
J_2e_m=\begin{cases}
(-1)^i\lambda \,e_1+\alpha_1\,e_2,&  m=1,\\
\alpha_m\,e_{m+1}+\alpha_{m-1}\,e_{m-1}, & 2\le m \le n-1,\\
\alpha_{n-1}\,e_{n-1}+(-1)^j\lambda\, e_n,& m=n,
\end{cases}
\]
where
\begin{align*}
\alpha_m&= \biggl(\frac{\sin^2m\sigma}{4\sin^2\sigma
\sin(m-1/2)\sigma\sin(m+1/2)\sigma}
\\
&\qquad{}-\lambda^2\,\frac{\sin^2(\sigma/2)}
{\sin(m-1/2)\sigma\sin(m+1/2)\sigma}\biggr)^{1/2},
\end{align*}
$a=-\pi/(2\sigma)-1/2$, $\lambda$ belongs to the set 
$\{\lambda\in{\mathbb R}^+\mid$ $\alpha_m>0,\, m=1,\ldots,n-1\}$, $i$, $j=0$,
$1$\textup;

\smallskip
$3$. $H={\mathbb C}^{2n}$, $k$ is odd, 
\begin{gather*}
J_1=
\begin{pmatrix}
\lambda_1I_2&&0\\
&\ddots&\\
0&&\lambda_nI_2
\end{pmatrix},
\\
J_2=\begin{pmatrix}
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{pmatrix},
\end{gather*}
where
$\lambda_m=[a+m]_q$, $a=-\pi/(2\sigma)-1/2$,
\[
Y_1=\begin{pmatrix}
\lambda&0\\
0&-\lambda
\end{pmatrix}, \quad Y_2=\lambda
\begin{pmatrix}
\cos\varphi&\sin\varphi\\
\sin\varphi&-\cos\varphi
\end{pmatrix},
\] 
$\alpha_m$ is the same as in $2$,
$\lambda$ belongs to the set 
$\{\lambda\in{\mathbb R^+}\mid\alpha_m>0,\, m=1,\ldots,n-1\}$,
$\varphi\in(0,\pi)$\textup;

\smallskip
$4$. $H={\mathbb C}^{n+1}$, $k$ is odd, $J_1e_m=[a+m]_qe_m$,
\[
J_2e_m=\begin{cases}
\alpha_1e_2,&m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1},&2\le m\le n,\\
\alpha_{n}e_{n},&m=n+1,
\end{cases}
\]
where $a=-\pi/(2\sigma)-1$,
$\alpha_1=\alpha_{n}=\sqrt{2}\, |q-q^{-1}|^{-1}\!$,
$\alpha_m=
{|q-q^{-1}|}^{-1}\!$,  $m=2$, \dots, $n-1$\textup;

\smallskip
$5$. $H={\mathbb C}^{2n}$, $k$ is odd,
\begin{gather*}
J_1=\begin{pmatrix}
\lambda_1I_1&&0\\
&\ddots&\\
0&&\lambda_{n+1}I_1\\
\end{pmatrix},
\quad
J_2=\begin{pmatrix}
0&X_1^*&&\\
X_1&0&\ddots&\\
&\ddots&\ddots & X_{n}^*\\
&&X_{n}&0
\end{pmatrix},
\end{gather*}
where $\lambda_m=[m-1-\pi/(2\sigma]_q$,
\[
X_1=\begin{pmatrix}
\sqrt{2}\,|q-q^{-1}|^{-1}\cos\varphi\\
\sqrt{2}\,|q-q^{-1}|^{-1}\sin\varphi
\end{pmatrix}, \quad X_{n}=(\sqrt{2}\,|q-q^{-1}|^{-1},0),
\]
$X_m=|q-q^{-1}|^{-1}I_2$,
$m=2$, \dots, $n-1$, $\varphi\in (0,\pi/2)$\textup;

\smallskip
$6$. $H={\mathbb C}^{(n+1)/2}$, $k$ is even, $ J_1e_m=(-1)^j[a+m]_q\,e_m$,
\[
J_2e_m=\begin{cases}
\alpha_1\,e_2+(-1)^i|q-q^{-1}|^{-1}e_1,& m=1,\\
\alpha_m\,e_{m+1}+\alpha_{m-1}\,e_{m-1},& 2\le m \le  (n-1)/2,\\
\alpha_{(n-1)/2}\,e_{(n-1)/2},&m=(n+1)/2,
\end{cases}
\]
where $a=\pi/({2\sigma})-{1}/{2}$, $\alpha_m={|q-q^{-1}|}^{-1}$, 
$m=1$, \dots, $(n-3)/2$, $\alpha_{(n-1)/2}=\sqrt{2}\,|q-q^{-1}|^{-1}$, 
$i=0$, $1$, $j=0$, $1$\textup;

\smallskip
$7$. $H={\mathbb C}^n$, $k$ is even,
\begin{gather*}
J_1=\begin{pmatrix}
\lambda_1I_2&&0\\
&\ddots&\\
0&&\lambda_{p+1}I_1
\end{pmatrix},
\quad
J_2=\begin{pmatrix}
Y&X_1^*&&\\
X_1&0&\ddots&\\
&\ddots&\ddots & X_{p}^*\\
&&X_{p}&0
\end{pmatrix},
\end{gather*}
where $p=(n-1)/2$,
$\lambda_m=(-1)^j[\pi/({2\sigma})-1/2+m]_q$,
\[
Y=|q-q^{-1}|^{-1}
\begin{pmatrix}
\cos\varphi&\sin\varphi\\
\sin\varphi&-\cos\varphi
\end{pmatrix}, \quad
X_{p}=(\sqrt{2}\,|q-q^{-1}|^{-1},0),
\]
$X_m=|q-q^{-1}|^{-1}I_2$,
$m=1$, \dots, $p-1$, 
$\varphi\in (0,\pi)$, $j=0$, $1$\textup;

\smallskip
$8$. $H={\mathbb C}^{p}$, $p<n$ if $k$ is odd, $p<(n+1)/2$ if $k$ is even,
\begin{align*}
J_1e_m&=(-1)^i\,[a+m]_q\,e_m,
\\
J_2e_m&=\begin{cases}
\alpha_1\,e_2+(-1)^j\frac{\sin p\sigma}{2\sin(\sigma/2)\sin\sigma}\,e_1,&m=1,\\
\alpha_m\,e_{m+1}+\alpha_{m-1}\,e_{m-1},&2 \le m\le p-1,\\
\alpha_{p-1}\,e_{p-1},& m=p,
\end{cases}
\end{align*}
where 
$$
\alpha_m=\biggl(\frac{\sin(m-l)\sigma\sin(m+l)
\sigma}{4\sin^2\sigma\sin(m-1/2)\sigma\sin(m+1/2)\sigma} 
\biggr)^{1/2},
$$
$a={\pi}/({2\sigma})-{1}/{2}$,
$p\in \{p\in{\mathbb Z}_+\mid
\alpha_m>0,\, 1\le m<p\}$, $i$, $j=0$, $1$\textup;

\smallskip
$9$. $H={\mathbb C}^{p}$, $ J_1e_m=[a+m]_q\,e_m$,
\[
J_2e_m=\begin{cases}
\alpha_1e_2,&m=1,\\
\alpha_me_{m+1}+\alpha_{m-1}e_{m-1},&2\le m\le p-1,\\
\alpha_{l-1}e_{l-1},&m=p,
\end{cases}
\]
where 
\begin{align*}
\alpha_1&=\biggl(-\frac{\sin(a+1)\sigma}{2\sin\sigma\cos(a+
2)\sigma)}\biggr)^{1/2},
\\
\alpha_m&=\biggl(-\frac{\sin m\sigma\sin(2a+m+1)\sigma}
{4\sin^2\sigma\cos(a+m)\sigma\cos(a+m+1)\sigma}\biggr)^{1/2},
\end{align*}
$m\ne 1$,
the pair
$(a,p)$ belongs to the set $\{(a,p)\in{\mathbb R}\times{\mathbb Z}_+\mid
\sigma a\ne\frac{\pi}{2}-l\sigma+\pi r$, $[p]_q[2a+p+1]_q=0$,
$[a+m]_q\ne [a+n]_q$, $1\leq m<n\leq p$,
$\alpha_m>0, m=1,\ldots,p-1, r\in{\mathbb Z}\}$.

\smallskip
$10$. $H={\mathbb C}^1$, $J_1=0$, $J_2=0$.
\end{theorem}

\begin{proof}
Let $J_1=J_1^*$, $J_2=J_2^*$ be operators on 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{cth2},
\begin{equation}\label{supp1}
E_{J_1}(\Delta)J_2E_{J_1}(\Delta')=0,
\end{equation}
for any $\Delta$, $\Delta'\in{\mathfrak B}({\mathbb R})$,
$\Delta\times\Delta'\cap\Gamma=\emptyset$.
Denote by $S_0$ the set $\{s\in{\mathbb R}\mid (q-q^{-1})^2s^2+4<0\}$,
and by $S_1$ its complement.
Then $R\times S_0\cap \Gamma=\emptyset$, which implies that
$J_2E_{J_1}(S_0)H=0$. 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 and given by
$$J_1=(\lambda),\quad J_2=(0),\qquad \lambda\in S_0.$$
But, by (\ref{rel:soq3II}), $\lambda=0$ which does not belong to $S_0$.
Therefore, the spectrum of $J_1$ belongs to $S_1$. Consider the
following parameterization of $S_1$: $\lambda=\sin x\sigma/\sin\sigma$,
$x\in {\mathbb R}$. Let $\mathcal O(\{\lambda\})$ be the trajectory of the
point $\lambda$ with  respect to $\Gamma$, i.e., the minimal subset
$M\subset{\mathbb R}$ which contains $\lambda$ and satisfies the condition
$({\mathbb R}\setminus M)\times M\cap\Gamma=\emptyset$,
and $M\times({\mathbb R}\setminus M)\cap\Gamma=\emptyset$.
In our case,  
$\mathcal O(\{\lambda\})=\{\sin(x+k)\sigma/\sin\sigma\mid k\in{\mathbb Z}\}$.

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 the case, 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 exists a measurable section
of $(S_1,\Gamma)$, i.e., a
set which meets every trajectory only once.
Thus any ergodic measure is concentrated on a
single  trajectory of
some point, and, hence the spectrum of the operator $J_1$ is discrete
and concentrated on a trajectory if $(J_1,J_2)$ is an irreducible
pair.
Let $H_1=E_A(\mathcal O(\{\frac{\sin (\pi/2)}{\sin\sigma}\}))H$,
$H_2=E_A(\mathcal O(\{-\frac{\sin(\pi/2)}{\sin\sigma}\}))H$, if $k$ is even,
and $ H_1=E_A(\mathcal O(\{\frac{\sin (\pi/2)}{\sin\sigma}\}))H$,
$H_2=E_A(\mathcal O(\{\frac{\sin ((\pi-\sigma)/2)}{\sin\sigma}\}))H$,  if
$k$ is odd,  $H_3=(H_1\oplus H_2)^{\perp}$.

Any trajectory can be described geometrically in the following way: if\/
$t$, $s\in \mathcal O(\{\lambda\})$ 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.

Any     trajectory     of    the    point    $t\notin
\mathcal O(\{1/{\sin\sigma}\})\cup \mathcal O(\{-{1}/{\sin\sigma}\})\cup
\mathcal O(\{{\cos(\sigma/2)}/{\sin\sigma}\})\cup
\mathcal O(\{-{\cos(\sigma/2)}/{\sin\sigma}\})$ is a cycle of length  $s$,
i.e., the following graph:
\vrule width0pt depth 9pt$\!\cycle{\lambda_1}{\lambda_2}{\lambda_{s}}$,
where
$\lambda_m={\sin((x+m)\sigma)}/{\sin\sigma}$,
$x\sigma\in [-\pi/2,\pi/2]\setminus
\{(\pi(2l+1)+m\sigma)/2\mid m,l\in{\mathbb Z}\}$,


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

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 $k=2p-1$,
  $$
\earchear{\frac{\cos(\sigma/2)}{\sin\sigma}}{-\frac{\cos(\sigma/2)}
{\sin\sigma}}\qquad\qquad\qquad
\ch{-\frac{1}{\sin\sigma}}{\frac{1}{\sin\sigma}}
$$

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

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

It follows from condition
(\ref{supp1})  that $J_2P_0H_3\subset P_1H_3\oplus P_{s-1}H_3$,
$J_2P_{s-1}H_3\subset P_1H_3\oplus P_{s-2}H_3$, and $J_2P_kH_3\subset
P_{k+1}H_3\oplus P_{k-1}H_3$ 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 the triple ($J_1,X,X^*$) is
irreducible.
One can easily show that  $J_1$, $J_2$ satisfy (\ref{rel:soq3II}) iff
$X$,  $X^*$ are additionally connected by the relations:
\begin{equation}\label{ff2}
\alpha_kX^*XP_k+\beta_kXX^*P_k=\gamma_kI,
\end{equation}
where $\alpha_k=-2\cos((x+k+1)\sigma)$,
$\beta_k=2\cos((x+k-1)\sigma)$,
$\gamma_k={\sin((x+k)\sigma)}/{\sin\sigma}$. Since
$\{\pm{1}/{\sin\sigma}\}$ 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 the resolution of
the identity for $C_k$  by $E_{C_k}(\cdot)$.  
From \eqref{ff2} it follows  that
$[A,X^s]=0$,  $[X^*,  X^s]=0$, which yields $X^s=cI$ if
($J_1, X, X^*$) is irreducible. From this we conclude that
$\oplus_{l=k}^{s-1}X^{l-k}E_{C_k}(\Delta)\,W$ is invariant with respect
to $J_1$,  $X$,  $X^*$ for any $\Delta\in{\mathfrak B}({\mathbb  R})$ and
a subspace $W$ such that $C_kW\subset W$.
 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, $\{e_{\lambda_k},
Xe_{\lambda_k}/||Xe_{\lambda_k}||,\ldots,
X^{s-k-1}e_{\lambda_k}/||X^{s-k-1}e_{\lambda_k}||\}$, where
$e_{\lambda_k}$ is an eigenvector of $C_k$ which exists due to the last
arguments.
Let $X^*Xe_{\lambda_k}\equiv C_ke_{\lambda_k}=b\,e_{\lambda_k}$,
$b\in{\mathbb R}_+$. Using (\ref{ff2}) one can get the action of the operators
$J_1$, $X$,
$X^*$, and $J_2$ on the basis. In particular, if
$\sigma(J_1)=\{\lambda_0, \lambda_1,\dots,\lambda_{s-1}\}$, we will
have representations of series~1, otherwise, representations from
series 9.


2. If $J_1$,  $J_2$  is  an irreducible representation acting in $H_1$,
$H_2$, then it is not necessary for $\sigma(J_1)$ to be simple.  First let us
consider  the
case where $J_1$ is concentrated on the trajectory of the form
\begin{equation}
\earchear{\raise5pt\hbox{\kern5pt$\scriptstyle\lambda_1$}}{\raise 5pt
\hbox{\kern5pt$\scriptstyle\lambda_n$}}\label{gr1}
\end{equation}
where $\lambda_1=-\frac{\cos(\sigma/2)}{\sin\sigma}$,
$\lambda_n=\frac{\cos(\sigma/2)}{\sin\sigma}$,  and
$\sigma(J_1)=\{\lambda_1,\dots,\lambda_n\}$.

Denote  the   projection   onto  the  eigenspace  of  $J_1$
corresponding to the eigenvalue $\lambda_k=-{\cos((2k-1)\sigma/2}/
{\sin\sigma}$ by   $P_k$. As before,  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 $Y\ne 0$,  $XP_k\ne 0$, for  $k<n$. 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 we have:
\begin{align}
\alpha_kX_k^*X_k+\beta_kX_{k-1}X_{k-1}^*&=\gamma_kI,
\qquad k\ne 1,\  k\ne n,\notag
\\
\alpha_1X_1^*X_1+\beta_1Y^2P_1&=\gamma_1I,\notag
\\
\beta_nX_{n-1}X_{n-1}^*+\alpha_nY^2P_n&=\gamma_nI,
\label{ff10}
\end{align}
where   $X_k=XP_k$, $\alpha_k=-2\sin((2k+1)\sigma/2)$,
$\beta_k=2\sin((2k-3)\sigma/2)$,
$\gamma_k=-{\cos((2k-1)\sigma/2)}/{\sin\sigma}$.

Set $D_0=(X^*)^{n-1}X^{n-1}P_1$. Then $P_1H\subset(\ker D_0)^{\perp}$,
because if this 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 $W$ is invariant with respect to $Y_1$, $Y_2$, then
$W'=W\oplus UW\oplus \ldots U^{n-1}W$ is invariant with respect to
$J_1$, $X$, $X^*$, $Y$, and hence with respect to $J_1$, $J_2$:
\begin{align*}
XU^kW&=U\sqrt{X^*X}U^{k}W
\\
&=U^{k+1}
\bigl({\mathbf F}^{(k)}\bigl((\sigma-\pi)/{2},
X^*X)\bigr)_2\bigr)^{1/2} W\subset U^{k+1}W,
\\
YU^{n-1}W&=U^{n-1}(U^*)^{n-1}YU^{n-1}W\subset
U^{n-1}Y_1W\subset U^{n-1}W,
\end{align*}
where 
\begin{align*}
\mathbf F(x,y)&=(F_1(x,y), F_2(x,y))
\\
&=\bigg(x+1,\frac{y\cos
x\sigma}{\cos((x+2)\sigma)}-\frac{\sin((x+1)\sigma)}{2\sin\sigma
\cos((x+2)\sigma)}\biggr).
\end{align*}
 Here we use the fact that $UU^*$  is the
projection on the
orthogonal complement to
$\ker(A+(\cos(\sigma/2)/\sin\sigma)I)$.
Moreover, (\ref{ff10}) gives
\begin{align*}
Q_2(Y_2)&=(U^*)^{n-1}XX^*U^{n-1}P_1=(U^*)^{n-2}X^*XU^{n-2}P_1
\\
&=\bigl({\mathbf F}^{(n-2)}((\sigma-\pi)/{2},X^*X)\bigr)_2P_1
\\
&= \bigl({\mathbf F}^{(n-2)}((\sigma-\pi)/{2}, Q_1(Y_1)\bigr)
=Q_1(Y_1)+\alpha I,
\end{align*}
where
$Q_1(x)=(-\beta_1x^2+\gamma_1)/\alpha_1$,
$Q_2(x)=(-\alpha_n x^2+\gamma_n)/\beta_n$, and
$\alpha=({\mathbf F}^{(n-2)}((\sigma-\pi)/{2},x))_2-x=
\gamma_n/\beta_n-\gamma_1/\alpha_1$.
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 the
corresponding representation space by  $H_0$. Then $\dim H_0\leq
2$. 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 $J_1$, $J_2$ act as follows:
\begin{gather*}
J_1=\begin{pmatrix}
\lambda_1I&&\\
&\ddots&\\
&&\lambda_nI
\end{pmatrix},
\\
J_2=
\begin{pmatrix}
Y_1&\mu_1(\lambda)I&&\\
\mu_1(\lambda)I&0&\ddots&\\
&\ddots&\ddots&\mu_{n-1}(\lambda)I\\
&&\mu_{n-1}(\lambda)I&Y_2
\end{pmatrix},
\end{gather*}
where
\begin{align*}
\mu_m(\lambda)&=\biggl(\mathbf F^{m-1}\biggl(\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}\biggr)\biggr)_2^{1/2}
\\
&=\biggl(\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)}\biggr)^{1/2},
\\
&\qquad\qquad  m=1,\dots,n-1,
\\
\lambda&\in \Bigl\{x\in{\mathbb
R}\Bigm | \Bigl({\mathbf F}^{(m-1)}\Bigl(\frac{\sigma-\pi}{2},
-\frac{\sin(\sigma/2)}{\sin
(3\sigma/2)}x^2
\\
&\qquad\qquad\qquad{}+\frac{\cos(\sigma/2)}{2\sin(3\sigma/2)
\sin\sigma}\Bigr)\Bigr)_2>0,\, 1\le m\le n-1\biggr\}.
\end{align*}

Any
irreducible pair ($Y_1, Y_2$) is unitarily equivalent  to
one of the following:

a) one-dimensional: $Y_1=(-1)^i\lambda$,   $Y_2=(-1)^j\lambda$,
$\lambda> 0$, $i,j=0,1$;

b) two-dimensional: 
\[
Y_1=\begin{pmatrix}
\lambda&0\\
0&-\lambda
\end{pmatrix},\quad Y_2=\lambda\begin{pmatrix}
\cos\varphi&\sin\varphi\\
\sin\varphi&-\cos\varphi
\end{pmatrix},
\]
 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 of series 2 and 3 of the theorem.


If $\Gamma_0=\Gamma\restriction_{\sigma(J_1)}$ is  a proper  subgraph  
 of  (\ref{gr1}),  then  arguments
similar to  those  in 1)  give   that
$\sigma(J_1)$ is simple if the pair $J_1$, $J_2$ is irreducible. Using
(\ref{ff10}), one can easily describe all unitarily non-equivalent
irreducible representations connected with the support $\Gamma_0$.

3. Let now $\Gamma\restriction_{\sigma(J_1)}$ be the following graph
\begin{equation}\label{gr2}
\ch{-\frac{1}{\sin\sigma}\ \ }{\frac{1}{\sin\sigma}}
\end{equation}
or one of its subgraphs. Let $P_k$ be the projection onto  the
eigenspace of $J_1$ corresponding  to
$\lambda_k=-\cos ((k-1)\sigma)/\sin\sigma$, $k=1$, \dots, $n+1$,
 $X_k=P_{k+1}J_2P_k$,
$X=\sum_{k=1}^{n}X_k$. Using the same arguments we can conclude
that  $J_2=X+X^*$, and that
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 $\alpha_k=-2\sin(k\sigma)$, $\beta_k=2\sin((k-2)\sigma)$,
$\gamma_k=-\cos ((k-1)\sigma)/\sin\sigma$; 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 $\{{1}/{\sin\sigma}\}$ or $\{-{1}/{\sin\sigma}\}$ 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
\begin{align*}
F_1(\mu,\lambda_k)&=\begin{cases}
\frac{\gamma_k-\beta_k\mu}{\alpha_k},& k\ne n,\\
0,& \mbox{otherwise},
\end{cases}
\\
F_2(\mu,\lambda_k)&=\begin{cases}
\frac{\gamma_k-\alpha_k\mu}{\beta_k},& k\ne 2,\\
0,& \mbox{otherwise}.
\end{cases}
\end{align*}
Hence, any irreducible representation can be realized in the space
$H=l_2(\Delta)$ by the 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 ${1}/{\sin\sigma}\notin\Delta$ or
$-{1}/{\sin\sigma}\notin\Delta$, 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 $\{\pm{1}/{\sin\sigma}\}\in\sigma(J_1)$, then the  problem  of
describing 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 $\mu=\prod_{k=0}^{n-3}({\mathbf F}^{k}
(\lambda_{n},\frac{\gamma_n}{\beta_n}))_2\frac{\gamma_{n+1}}{\beta_{n+1}}$, 
${\mathbf F}(\lambda_{k+1},x)=
(\lambda_{k},(\gamma_k-\alpha_kx)/{\beta_k})$. One can
check that ${\gamma_1}/{\alpha_1}={1}/({2\sin^2\sigma})$.
Hence
$$
D_1(D_1-{1}/({2\sin^2\sigma})I)=0,\quad D_2(D_2-\mu I)=0.
$$
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 $\mu_k=1/4\sin^2\sigma$) .
Moreover, 
$$
X_{n-1}\dots X_2X_2^*\dots X^*_{n-1}=\lambda I,
$$
where  $\lambda=\prod_{k=0}^{n-3}({\mathbf F}^{(k)}
(\lambda_{n},{\gamma_{n}}/{\beta_{n}}))_2$. 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
$\sigma(J_1)\not\ni\{{1}/{\sin\sigma}\}$ if  the pair ($J_1$, $J_2$) is
irreducible, which contradicts the assumption. Thus $\lambda\ne 0$,
\begin{gather*}
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{gather*}


Any  irreducible  pair  ($D_1,  D_2$)  is  one-  or
two-dimensional.
Let us describe the corresponding irreducible pairs $J_1$, $J_2$.
Denote the
phase of the operator $X_i\ldots X_2$ by $U_i$,
the phase of $X_1^*$ by $U_1^*$.
If
$\dim H_0\le 2$ and $e_1$, $e_2$ is an orthonormal basis in the
space  $H_0$ such that
$e_1\in(\ker D_2)^{\perp}$, then
$$
D_1=\frac{1}{2\sin^2\sigma}
\begin{pmatrix}
1+\cos\varphi&\sin\varphi\\
\sin\varphi&1-\cos\varphi
\end{pmatrix},\quad
D_2=\begin{pmatrix}
\mu&0\\
0&0
\end{pmatrix},
$$
$U_ie_1$, $U_ie_2$ is an orthonormal basis in $X_i\ldots X_2H_0$, 
$2\leq i< n$, and 
the vectors $U_{n}e_1$ and 
$U_1(\cos(\varphi/2)e_1+\sin(\varphi/2)e_2)$ 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 this
basis, it is sufficient to know it for the operators $X_i$. We have 
\begin{align*}
X_{i+1}U_ie_k&=\Bigl(\prod_{l=2}^{i-1}\mu_l\Bigr)^{-1/2}X_{i+1}\dots X_2e_k=
\sqrt{\mu_i}\,U_{i+1}e_k
\\*
&=({2\sin\sigma})^{-1}U_{i+1}e_k,\qquad i<n-1;
\\
X_{n}U_{n-1}e_k&=\Bigl(\prod_{l=2}^{n-1}\mu_l\Bigr)^{-1/2}X_{n}\dots
X_2\,e_k
\\*
&=\begin{cases}
0,& k=2,\\
({\sqrt{2}\sin\sigma})^{1/2}U_{n}e_1,& k=1,
\end{cases}
\\
&\hskip-4emX_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)=
\\
&=(\sqrt{2}\sin\sigma)^{-1}(\cos(\varphi/2)\,e_1+\sin(\varphi/2)\,e_2).
\end{align*}

Thus we will get representations from series 5. If $\dim
H_0=1$, then $D_1\ne 0$ and  $D_2\ne 0$, since otherwise
$\sigma(J_1)\ne\{\lambda_1,\ldots,\lambda_{n+1}\}$.
If ${\mathbb C}e_1=H_0$, 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 ($J_1, J_2$) 
form series 6.

4. Suppose that the support of $J_2$ is
\begin{equation}\earch{\raise5pt\hbox{\ $\scriptstyle\lambda_1$}}
{\raise5pt\hbox{$\scriptstyle\quad\quad\lambda_{(n+1)/2}$}}\label{gr3}
\end{equation}
where $\lambda_1=\pm({\cos\sigma/2})/{\sin\sigma}$,
$\lambda_{(n+1)/2}={1}/{\sin\sigma}$. Let us consider the  case
$\lambda_1=({\cos\sigma/2})/{\sin\sigma}$ (for
$\lambda_1=-({\cos\sigma/2})/{\sin\sigma}$
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 $\lambda_k=
({\cos((2k-1)\sigma/2)})/{\sin\sigma}$. Let $X_k=XP_k$. Then
\begin{align}
\alpha_kX_k^*X_k+\beta_kX_{k-1}X_{k-1}^*&=\gamma_kI, \qquad k\ne 1,\notag
\\
\alpha_1X_1^*X_1+\beta_1Y^2P_1&=\gamma_1I,
%\label{ff31}
\label{ss}
\end{align}
where $\alpha_k=-2\sin((2k+1)\sigma/2)$,
$\beta_k=2\sin((2k-3)\sigma/2)$,
$\gamma_k=-\cos((2k-1)\sigma/2)/\sin\sigma$;  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
($J_1, J_2$) 
reduces  to  that of
irreducible pairs  ($D_1, D_2$)   satisfying   some   quadratic
relation. Here 
\[
D_1=Y,\quad D_2=(X_{(n-1)/2}\ldots X_1)^*X_{(n-1)/2}\ldots X_1,
\]
and $D_2(D_2-\mu I)=0$, $D_1^2={1}/({4\sin^2\sigma}) I$, $\mu$ is defined
by 
\[
X_{(n-1)/2}\dots
X_1(X_{(n-1)/2}\dots
X_1)^*=\mu I,
\] which can be easily obtained from (\ref{ss}).


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, the  dimensions of the irreducible representations
are $(n+1)/2$.  This  follows  by  the same method as 
in the previous
case and we  get representations 6, 7.

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

\noindent\textbf{4.}
Representations of\/ $U_q(so(3))$, $q$ is not a root of unity.
Let $q=e^{i\sigma}$, where $\sigma\notin\pi{\mathbb Q}$.
Analysis similar to that as in the proof of Theorem~\ref{th:soq3root} shows that
if the pair ($J_1, J_2$) defines a non-trivial irreducible representation, 
then $\sigma(J_1)\in S_1=[{-1}/{\sin\sigma},{1}/{\sin\sigma}]$. 
Moreover,
the operator $J_2$ is concentrated on 
$\Gamma=\{(t,s)\mid t^2-(q+q^{-1})\,ts+s^2=0\}$. Any trajectory of
a point $\lambda=\sin x\sigma/\sin\sigma\in S_1$ with respect to
$\Gamma$ is of the form
$ \mathcal O(\{\lambda\})=\{\sin(x+k)\sigma/\sin\sigma\mid
k\in{\mathbb Z}\}$ which is 
clearly dense in $S_1$, and there is no measurable section of ($S_1, \Gamma$).
In this case there exist  irreducible representations which are not 
concentrated on a trajectory. The description of such representations is
problematic. 
The description of irreducible representations related to trajectories can be 
obtained  using
the same technique (see \cite{100} for the concrete formulae).


%%% Local Variables: 
%%% mode: latex
%%% TeX-master: "the"
%%% TeX-master: "the"
%%% End: