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\section{Some classes of $*$-algebras with 3 and 4 generators}
\label{sec:2.2}
\markright{2.2. Algebras with 3 and 4 generators}

The method developed in Section \ref{sec:2.1} can be
carried over to other classes of operator relations. The idea
behind this is to consider some commutative family of self-adjoint
operators, and to see how other operators act on their (generalized)
eigenvectors.

\subsection{Representations of graded $so(3)$ and four-tup\-les of
projections  satisfying a linear relation}\label{sec:2.2.1}

In this section, we study irreducible representations of a graded analogue
of the Lie
algebra $so(3)$, and show how they are related to four-tuples of
projections satisfying a linear relation of a special type.

Consider a triple of bounded self-adjoint operators $A$, $B$, $C$
satisfying the following relations
\begin{equation}\label{graded-so3}
\{A,B\} = C, \quad \{B,C\} = A, \quad \{C, A\} = B,
\end{equation}
where $\{A, B\} = AB+BA$ denotes the anticommutator of the operators $A$
and $B$. As was mentioned above, such triples of self-adjoint
operators can be considered as representations of the graded $so(3)$
algebra. We describe all such irreducible families up to unitary
equivalence.

On the other hand, consider a four-tuple of orthogonal projections,
$P_1$, $P_2$, $P_3$, $P_2$ that are connected by a linear relation of
the form
\[
\alpha\,(P_1 + P_2 + P_3 + P_4) =I.
\]
It turns out that the unitary description of such collections of
projections is closely related to representations of \eqref{graded-so3}.

\begin{theorem}\label{th:gleb}
Any irreducible family of self-adjoint operators  satisfying
\eqref{graded-so3} is finite-dimensional. For any $l\ge1$ there exist
four irreducible representations of dimension $2l$, and
five irreducible representations of dimension $2l-1$, which act as
follows.

\textup 1.
Four representations with any finite dimension $\dim H = n$\textup:
\begin{align*}
A e_k &= \alpha_1\,(-1)^{k+1}\Bigl(k - \frac12\Bigr) \, e_k, \qquad k=1,
\dots, n,
\\
B e_1& =  \frac{\alpha_2 n}2\, e_1 + \frac{\alpha_1}2 \,\sqrt{n^2
-1}\, e_2,
\\
B e_k& = \frac{\alpha_1\, (-1)^{k}}2\, \sqrt{n^2 - (k-1)^2}\, e_{k-1} 
\\
&\quad{}+  \frac{\alpha_1 (-1)^{k+1}}2\, \sqrt{n^2 - (k)^2}\, e_{k+1},
 \qquad k=2, \dots, n-1,
\\
B e_n & = \frac{\alpha_1 (-1)^{n-1}}2\, \sqrt{2n-1}\, e_{n-1},
\\
C e_1 & =  \frac{\alpha_2 n}2\, e_1 + \frac12 \,\sqrt{n^2 -1}\, e_2,
\\
C e_k& = \frac12\, \sqrt{n^2 - (k-1)^2}\, e_{k-1} 
\\
& \quad{}+ \frac12\, \sqrt{n^2 - (k)^2}\, e_{k+1},
 \qquad k=2, \dots, n-1,
\\
C e_n & = \frac12\, \sqrt{2n-1}\, e_{n-1},
\qquad \text{where $\alpha_1$, $\alpha_2 = \pm1$};
\end{align*}

\textup2.
An additional representation for each odd value of the dimension, $\dim H = n =
2l-1$\textup:
\begin{align*}
Ae_k & = (-1)^{k+1}\,\frac{n+1-2k}2\, e_k, \qquad k=1, \dots, n,
\\
B e_1 & = \frac12\,\sqrt{n-1}\, e_2,
\\
Be_k & = \frac{(-1)^{k}}2 \, \sqrt{(k-1)(n-k+1)} \, e_{k-1}
\\
&\quad{} + \frac{(-1)^{k+1}}2 \, \sqrt{k(n-k)} \, e_{k+1},
\qquad k=2, \dots, n-1,
\\
Be_n & = \frac12 \,\sqrt{n-1} \, e_{n-1},
\\
C e_1 & = \frac12\,\sqrt{n-1}\, e_2,
\\
Ce_k & = \frac12 \, \sqrt{(k-1)(n-k+1)} \, e_{k-1} 
\\
&\quad{} +\frac12 \, \sqrt{k(n-k)} \, e_{k+1},
\qquad k=2, \dots, n-1,
\\
Ce_n & = \frac12 \,\sqrt{n-1} \, e_{n-1}.
\end{align*}
\end{theorem}

\begin{proof}
The proof of the theorem is essentially based on the same  ideas
as the proof for
non-graded $so(3)$. Introduce the operators $E_0 = A$, $E_1 = B+C$, $E_2 =
C-B$. It is easy to check that the relations are equivalent to
\begin{equation}\label{graded-su2}
\{E_0, E_1\} = E_1, \quad \{E_0, E_2\} = -E_2, \quad E_1^2 - E_2^2  =
2E_0,
\end{equation}
and that the operator
\[
\Delta = A^2 + B^2 + C^2 = E_0^2 + E_1^2 - E_0 = E_0^2 + E_2^2 + E_0
\]
commutes with $A$, $B$, $C$. This implies, in particular, that any
irreducible representation is bounded, since $A^2$, $B^2$, $C^2$ are
positive
operators, and their sum is a multiple of the identity.

One can easily see from relations \eqref{graded-su2} that $E_1$ maps
an eigenvector $e_\lambda$ of $E_0$ to an
eigenvector $e_{1-\lambda}$, and
$E_2$ maps $e_\lambda$ into $e_{-1-\lambda}$. Then we have a dynamical
system on $\mathbb{R}$ generated by two flips with respect to the points
$1/2$, and $-1/2$. For an irreducible representation, the spectral measure
of $E_0$ must be ergodic with respect to the action of this dynamical
system, since, otherwise, any ergodic component generates an invariant
subspace. The dynamical system possesses a measurable section, i.e., a
set
that meets every orbit only once. For such a set one can take $[-1/2,
1/2]$. Then any ergodic measure is concentrated at a single orbit of
some point.

Thus, the spectral measure of $E_0$ is discrete, and we can choose a
basis consisting of its eigenvectors. Then we have:
\[
E_0 e_\lambda = \lambda e_\lambda, \quad E_1 e_\lambda = a_1(\lambda) \,
e_{1-\lambda}, \quad E_2 e_\lambda = a_2(\lambda) \, e_{-1-\lambda},
\]
where $\lambda$ are taken from some orbit. It remains to find a condition
on $a_1$, $a_2$ so that the relation $E_1^2 - E_2^2 = 2E_0$ would hold, and
to check whether the these conditions can be satisfied, and the representation
is irreducible (the ergodicity is a necessary condition, but not
sufficient
in general).

One can easily check that necessary and sufficient conditions for
$a_1$, $a_2$ to form a representation are the following:
\[
a_1(1-\lambda) = \overline{ a_1(\lambda)}, \quad a_2(-1-\lambda) =
\overline{
a_2(\lambda)}, \quad a_1^2(\lambda) - a_2^2(\lambda) = 2\lambda
\]
for almost all $\lambda$ taken with respect to the spectral measure of $E_0$.
In particular, these relations imply that both $a_1^2$ and $a_2^2$ are
uniquely determined by the value of $a_1$ at a single point of a non-zero
spectral measure. Actually, there exists $\phi>0$ such that
\[
a_1^2(\lambda) = -(\lambda -1/2)^2 +\phi, \quad a_2^2(\lambda) =
-(\lambda +1/2)^2 + \phi
\]
on the spectrum of $E_0$.

But the latter relations cannot hold for all points of the orbit, since
in the right-hand sides we have functions decreasing to $-\infty$, while
the left-hand sides must be non-negative. To avoid the contradiction, we
need to demand that, on the highest vector (the vector
$e_\lambda$ with the
largest $\lambda$), the operator $E_2$ acts as zero, and on the lowest
vector, the operator $E_1$ is zero. This means that zeroes of  both the
functions $a_1$ and $a_2$ must belong to the same orbit.

One can easily check that these conditions can be satisfied only for
three orbits, corresponding to the points $0$, and $\pm1/2$. In each
case, only a discrete number of values of $\phi$ are admissible, one for
each dimension. Also, the representation corresponding to the orbit that
contains zero has an odd dimension.

To complete the proof, it remains to notice that for any orbit containing
zero there corresponds a unique irreducible representation, while the
orbits of $\pm1/2$ carry two representations each, depending on the sign
of $a_1(1/2)$ or $a_2(-1/2)$. Finding admissible values for $\phi$ and
restoring the operators $A$, $B$, $C$ from $E_0$, $E_1$, $E_2$ is a
routine calculation.
\end{proof}

Now we consider another problem that appeared to be related to the one just
considered.

The problem is related to general non-orthogonal resolutions of the
identity. Let $A_1$, \dots, $A_n$ be positive self-adjoint operators in
a finite-dimensional space $H$ such that
\[
\sum_{k=1}^n A_k =I.
\]
Writing the spectral decomposition for each of the operators $A_k$, we see
that the latter sum can be rewritten as
\begin{equation}\label{proj-gen}
\sum_{l=1}^m \alpha_l P_l =I,\qquad 0<\alpha_l \le 1,
\end{equation}
where $P_l$ are orthogonal projections. If all $\alpha_l=1$, the
projections commute, and we get an ordinary resolution of the identity.

The complexity of the description problem for families of projections
that form a non-orthogonal resolution of the identity depends on the
number of the projections in \eqref{proj-gen}.

Two projections satisfying \eqref{proj-gen} are orthogonal; therefore,
$\alpha_1=\alpha_2 =1$.

A triple of projections satisfying \eqref{proj-gen} has only one- or
two-dimen\-sional irreducible representations.

The problem of the unitary
description of five or more projections satisfying
\eqref{proj-gen}, is very complicated (``wild'', see
Section~\ref{sec:3.1.3}
below).

Consider four orthogonal projections, $P_1$, $P_2$, $P_3$, $P_4$ that
satisfy a special case of the relation~\eqref{proj-gen} with all
$\alpha_l=\alpha$,
\begin{equation}\label{proj-4}
\alpha(P_1+P_2+P_3+P_4 )= I.
\end{equation}
We study for which $\alpha$ solutions exist, and give their
unitary description.

\begin{theorem}
Solutions of \eqref{proj-4} exist for the following values of 
$\alpha$\textup:

\textup1.
$\alpha=\frac12$. There are six one-dimensional, and a continuous
family of two-dimensional representations\textup;

\textup2. $\alpha =\frac12 -\frac1{4k}$, $k\in \mathbb{N}$. There is one
irreducible representation with $\dim H =2k-1$\textup;

\textup3. $\alpha =\frac12 +\frac1{4k}$, $k\in \mathbb{N}$. There is one
irreducible representation with $\dim H =2k+1$\textup;

\textup4. 
$\alpha =\frac12 -\frac1{4k+2}$, $k\in \mathbb{N}$. There are four
irreducible representations with $\dim H =k$\textup;

\textup5. $\alpha =\frac12 +\frac1{4k-2}$,  $k\in
\mathbb{N}$. There are four
irreducible representations with $\dim H =k$.
\end{theorem}

\begin{proof}
Introduce self-adjoint unitary operators
$R_i = 2P_i -I$, $i=1$, \dots, $4$.
Then
\[
\sum_{i=1}^4 R_i = \frac{2-4\alpha}\alpha\, I =2hI,
\]
where we put $h=(1-2\alpha)/\alpha$.

First consider the case $\alpha=1/2$ ($h=0$). Introduce the elements
\[
X_1=(R_1+R_4)/2,\quad X_2=(R_2+R_4)/2,\quad X_3=(R_3+R_4)/2.
\]
Then
\begin{align*}
R_1& = X_1 - X_2 -X_3, &R_3& = -X_1 - X_2 +X_3,
\\
R_2& = -X_1 + X_2 -X_3, &R_4& = X_1 + X_2 +X_3,
\end{align*}
and the relation holds if and only if $X_1$, $X_2$, $X_3$ are pairwise
anti-commuting self-adjoint operators such that
\[
\Delta=X_1^2+X_2^2 + X_3^2 =I.
\]
Then an irreducible representation is either one-dimensional,
one of $X_i=\pm1$, and the others are zeros, or two-dimensional,
\begin{gather*}
X_1 = a \begin{pmatrix}1&0\\0&-1\end{pmatrix}, \quad
X_2= b\begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad
X_3= c\begin{pmatrix}0&i\\-i&0\end{pmatrix},
\\
a^2 + b^2 + c^2=1,
\end{gather*}
and either $a>0$, $b>0$, $c \in \mathbb R$, or $a=0$, $b>0$, $c>0$, or $a>0$, $b=0$, $c>0$. 
These representations form the first family in the statement of the
theorem. The formulas for $P_k$ are
\begin{align*}
P_1&=\frac12
\begin{pmatrix}1+a&-b-ic\\-b+ic&1-a\end{pmatrix}, &
P_2&=\frac12 \begin{pmatrix}1-a& b-ic\\
b+ic&1+a\end{pmatrix},
\\
P_3&=\frac12
\begin{pmatrix}1-a&-b+ic\\-b-ic&1+a\end{pmatrix}, &
P_4&=\frac12 \begin{pmatrix}1+a&
b+ic\\-b-ic&1-a\end{pmatrix}.
\end{align*}

For $\alpha\ne1/2$ ($h\ne0$),
introduce the elements $X_1$, $X_2$, $X_3$ such that
\begin{align*}
X_1& = \frac1{2h}\,(R_2+R_3 -hI),
\\
X_2& = \frac1{2h}\,(R_1+R_3 -hI),
\\
X_3& = \frac1{2h}\,(R_1+R_2 -hI).
\end{align*}
Then
\begin{align*}
R_1&=h\,(-X_1+X_2 +X_3 +1/2),
\\
R_2&=h\,(X_1-X_2 +X_3 +1/2),
\\
R_3&=h\,(X_1+X_2 -X_3 +1/2),
\\
R_4&=h\,(-X_1-X_2 -X_3 +1/2),
\end{align*}
and the relation \eqref{proj-4} is equivalent to the following set of
relations
\begin{gather*}
\{X_1, X_2\} = X_3, \quad \{X_2, X_3\} = X_1, \quad \{X_3, X_1\} = X_2,
\\*
h^2\, (\Delta +1/4) =1,
\end{gather*}
where, as above, $\Delta = X_1^2 + X_2^2 + X_3^2$. Now, we can easily
describe all irreducible representations of \eqref{proj-4} in terms of
the irreducible representations of \eqref{graded-so3} with one extra
restriction $h^2\, (\Delta +1/4) =1$.

i. Representations of odd dimension, corresponding to the orbit
containing zero. Let $\dim H=n=2k+1$, then $\Delta =\frac{n^2-1}4I$,
which implies that $h=\pm 2/n$ and $\alpha =\frac12 -\frac1{2(n+1)}$, or
$\alpha =\frac12 +\frac1{2(n-1)}$, which give cases 2 and 3 of the
theorem. The projections are three-diagonal matrices that can easily be
restored from the corresponding representation of the graded $so(3)$.

ii. Representations with an arbitrary dimension $n$,  corresponding
to the orbits
containing $1/2$, or $-1/2$. Now $\Delta = (n^2 -1/4)\, I$, which
implies $h=\pm 1/n$, and $\alpha=\frac12 -\frac1{2(2n+1)}$, or
$\alpha=\frac12 -\frac1{2(2n+1)}$. This gives cases 4 and 5.
\end{proof}

\begin{remark}
One can consider the following relation
\begin{equation}\label{rel:four-center}
P_1 + P_2 +P_3 +P_4 = Z,
\end{equation}
where $P_i$ are orthogonal projections, and $Z$ commutes with them.
Since, for an irreducible representation,  the
central element $Z$ is scalar,
$Z= \alpha I$, we can apply the latter theorem to the description of
irreducible representations of \eqref{rel:four-center}. Indeed, the set of
irreducible representations of \eqref{rel:four-center} consists of
representations of \eqref{proj-4} with all $\alpha$, i.e., all
representations of the graded $so(3)$ algebra \eqref{graded-so3}, and all
representations of triples of anti-commuting self-adjoint operators with
the sum of squares equal to the identity.
\end{remark}

\subsection{Representations of a class of quadratic algebras with three
generators}\label{sec:2.2.2}

Consider an algebra with three generators $X$, $Y$, $Z$ and the
relations
\begin{align*}
XY - q YX &= \mu Y,
\\
ZX - qXZ &=\mu Z,
\\
\alpha YZ - \beta ZY &= P(X),
\end{align*}
where $q$, $\mu$,  $\alpha$, $\beta \in \mathbb{C}$, and $P(\cdot)$ is a
quadratic polynomial in $X$.

In what follows, we assume that $\mu$, $q\in \mathbb{R}$, $q \ne \pm1$, and
the algebra is equipped with an involution defined on the generators
by $X^* = X$, $Y^* = Z$. Notice that the ideal generated by the
relations is a $*$-ideal. With such an involution, the introduced family
of relations includes the Lie algebra $su(2)$  (for $q=1$),
 and many of its deformations
which arise in recent papers in physics.

For $q \ne 1$, one can rewrite the relation with $\mu=0$;
just replace $X$ with $X+\lambda I$ for an appropriate value of
$\lambda$. In what follows, we assume that $\mu=0$.

Introducing self-adjoint generators, $A = X$, $B = \frac12 (Y +Z)$, $C
= \frac12(Y-Z)$, the relations will take the form
\begin{align*}
(1+q) [A,B] &= -i \,(1-q)\{A,C\},
\\
(1+q) [A,C] &=i \,(1-q)\{A, B\},
\\
-i\,(\alpha - \beta) [B,C] &+ (\alpha + \beta) (B^2 - C^2) = \textstyle
\frac12 P(A).
\end{align*}

Now we turn to the study of irreducible representations of the introduced
algebras, i.e., the operators $X$, $Y$ that satisfy the relations
\begin{equation}
XY = qYX,
\quad\alpha \,YY^* + \beta \,Y^*Y = P(X).\label{three-gen}
\end{equation}

Using the arguments that are quite similar to those used in the previous section, one
sees that $Y$ maps eigenspaces of the self-adjoint operator $X$ into
eigenspaces, and the corresponding dynamical system is $\lambda \mapsto
q\lambda$; however, we need not assume that $X$ is positive, and we have
no restrictions on the kernels of the operators. The dynamical system
$\lambda \mapsto q\lambda$ possesses a measurable section, therefore, the
spectral measure of $X$ in any irreducible representation is concentrated
on a single orbit.

Introduce the set
\[
M = \begin{cases} [1,|q|) \cup (-|q|, -1] \cup \{0\},& |q| >1,
\\ (|q|,1 ] \cup [-1, -|q|) \cap \{0\} ,& |q|<1.
\end{cases}
\]
$M$ is a measurable section of the dynamical system.

If the orbit consists of the zero point, we have $X=0$, and we have the
relation $\alpha YY^* +\beta Y^*Y = P(0) I$ which is either the $q$-plane,
or the $q$-CCR relation considered in Section~\ref{sec:1.4.2}. In what
follows, we study the representations corresponding to non-zero orbits.

\begin{theorem}
Unitary representations corresponding to the orbit $O_\lambda$,
$\lambda \in M \setminus \{0\}$, may be described in terms of the following\textup:

\begin{itemize}
\item[\textup(a\textup)]  infinite sequences $y_k>0$, $k \in \mathbb{Z}$,
for which
\begin{equation} \label{rel:recurse}
\alpha\, y_k^2 +\beta\, y_{k+1}^2 = P(q^k \lambda)
\end{equation}
holds for all $k \in \mathbb{Z}$ \textup(as
a rule, these representations form a
continuous family indexed by $\lambda$\textup)\textup;

\item[\textup(b\textup)] infinite sequences $y_k$, $k \ge l$, with $l$
fixed, such that $y_l =0$, and $y_k >0$, $k >l$, and \eqref{rel:recurse}
holds for all $k \ge l$ \textup(representations with  the lowest
weight\textup)\textup;

\item[\textup(c\textup)] infinite sequences $y_k$, $k \le l$, with $l$
fixed, such that for $y_l =0 $, and $y_k >0$, $k <l$, and
\eqref{rel:recurse} holds for all $k \le l$ \textup(representations with
highest weight\textup)\textup;

\item[\textup(d\textup)] finite sequences $y_k$, $l \le k \le m$, with $l$
and $m$ fixed, such that $y_l = y_m =0$, and $y_k >0$ for $l<k<m$, and
\eqref{rel:recurse} holds for all $l \le k \le m$
\textup(finite-dimensional representations\textup).
\end{itemize}
In series \textup(a\textup), the representations are unbounded.
In series \textup(b\textup) and \textup(c\textup), the
representations  may be bounded or unbounded. Besides the mentioned representations,
there can still be one-dimensional representations $Y=0$, $P(X)=0$, and
the representations corresponding to the zero orbit.
\end{theorem}

\begin{proof}
Using the same arguments as in the previous section, we
see that the spectrum of $X$ lies in $\{q^k\lambda, k \in
\mathbb{Z}\}$, where $\lambda\ne0$ is an initial point from $M$,
and the space $H$
is a direct sum of its eigenspaces $H_k$ corresponding to the eigenvalues
$\lambda_k = q^k \lambda$. The operator $Y$ maps $H_{k-1}$ into $H_k$;
write $Y_k\colon H_{k-1}\to H_k$ for the corresponding restrictions. From
the second relation in \eqref{three-gen}, we have $\alpha \,Y_k
Y^*_k + \beta\,
Y_{k+1}^* Y_{k+1} = P(q^k\lambda )\,I$.

We  will show that all $H_k$ are either zero or one-dimensional. Indeed,
$Y_kY_k^*$ and $Y_{k+1}^*Y_{k+1}$ are commuting self-adjoint operators in
$H_k$. Take an invariant subspace $H_k^0 \subset H_k$, then the image of
$H_k^0$
under the action of $Y$, $Y^*$ and $X$ is invariant in $H$. Thus,
$H_k$ does not have  proper
subspaces. Thus, the operators $Y_kY_k^*$ and $Y_{k+1}^*Y_{k+1}$ are
scalar, and we obtain the relations $\alpha\,|y_k|^2 + \beta \,|y_{k+1}|^2
=
P(q^k\lambda)$ for all $k$. Passing to a unitarily equivalent
representation, we can assume that $y_k \ge0$. To complete the proof one
can notice that the subspaces $\bigoplus _{k \ge l}H_k$ and $\bigoplus_{k
\le l} H_k$ are invariant if and only if $y_l =0$.
\end{proof}

\begin{remark}
Consider a generalization of the given class of quadratic algebras.
Replace the second relation in \eqref{three-gen} with a general
second-order
relation connecting $X$, $Y$, $Y^*$,
\begin{gather*}
a_{11} X^2 + a_{22} Y^2 + a_{33} (Y^*)^2 + a_{12} XY+ a_{21} YX
\\
{}+ a_{13} XY^* + a_{31} Y^*  + a_{23}  YY^* + a_{32} Y^* Y
\\
{}+ a_1 X + a_2 Y + a_3 Y^* + aI =0.
\end{gather*}
One can show that, in any representation, the latter relation is equivalent
to the following system:
\begin{gather*}
a_{11} X^2 + a_1 X +  aI+ a_{23}  YY^* + a_{32} Y^* Y
 =0,
\\
a_{12} XY+ a_{21} YX + a_2 Y =0,
\quad
a_{13} XY^* + a_{31} Y^* X  + a_3 Y^* =0,
\\
a_{22} Y^2 =0,
\quad
a_{22} Y^2 =0.
\end{gather*}
\end{remark}

\subsection{Operator relations connected with a dynamical system on a plane}
\label{sec:2.2.3}

Let $A=A^*$, $X$, $X^*$ be bounded operators that satisfy
relations of the form:
\begin{align}
AX&=XF_1(A),\notag\\
X^*X&=F_2(A,XX^*),\label{d2}
\end{align}
where $F_1(\cdot)\colon{\mathbb R}\rightarrow{\mathbb R}$,
$F_2(\cdot,\cdot)\colon{\mathbb R}^2\rightarrow{\mathbb R}$ are measurable
mappings.
It follows from the first relation in (\ref{d2}) that the operators $A$,
$XX^*$ commute and, hence, $F_2(A,XX^*)$ is well defined.

\begin{remark}
Instead of  assuming that the operator $A$ is self-adjoint, one can
consider the case of unitary or normal operator as well. In this case,
we consider the mapping $F_1\colon \mathbb{T} \to \mathbb{T}$,
and $F_2 \colon
\mathbb{T}
\times \mathbb{R} \to \mathbb{R}$ for the unitary operator $A$, and $F_1\colon
\mathbb{C} \to \mathbb{C}$, and $F_2 \colon
\mathbb{C} \times \mathbb{R} \to \mathbb{R}$ in the normal case. The
mapping $F$ introduced below, defines a dynamical system on
$\mathbb{T}\times \mathbb{R}$ or $\mathbb{C} \times \mathbb{R}$,
respectively. All statements about representations of relations
\eqref{d2} also hold true in these cases (of course, now, the
spectrum of $A$ belongs to the circle or complex plane).
\end{remark}

As before, let us consider  the polar decomposition  of the
operator $X^*=UC$, where $C=C^*=(XX^*)^{1/2}$, $U$ is a partial
isometry such that $\ker U=\ker C$. Using the relations
(\ref{d2}) one can obtain
\begin{equation}
AU^*=U^*F_1(A),\quad C^2U^*=U^*F_2(A,C^2),\label{d4}
\end{equation}
and $U$ is a centered partial isometry with $\ker U^* = \ker F_2(A,
C^2)$.
Conversely, any  triple of the operators $A=A^*$, $C\geq 0$, and
a centered partial isometry $U$ satisfying
(\ref{d4}) and such that $\ker U=\ker C$,
and $\ker U^* = \ker F_2(A,C^2)$, define a
representation $A$, $X=CU^*$ of the relations (\ref{d2}).

Let $F=(F_1,F_2)\colon{\mathbb R}^2\rightarrow{\mathbb R}^2$. For $k\in
\mathbb N$,
we will  denote by $F^{ k}(\cdot)$ the $k$-th iteration of $F(\cdot)$
and, for $\lambda\in{\mathbb R}$, $n=1$, $2$, by $F_n^{ k}(\lambda)$
the $n$-th coordinate of $F^{ k}(\lambda)$.

Analogously, the relations (\ref{d2}) correspond
to a two-dimensional dynamical system, $F(\cdot):{\mathbb
R}^2\to{\mathbb R}^2$. The possibility of classifying  all
irreducible representations of the relations depends on the
properties of the dynamical system.

\begin{proposition}\label{bprop}
Let \textup($A=A^*, X$\textup)
be a representation of \eqref{d2}
on a space $H$. Then $H$ can be decomposed into orthogonal
subspaces $H_1$ and $H_2$, invariant with respect to $A$, $X$,
$X^*$ such that the phase $U$ of $X$ is unitary in $H_1$ and\/ $\ker
U\cup\ker U^*\ne \{0\}$ in $H_2$.
\end{proposition}

Similarly to the case of relation (\ref{xx}),
irreducible representations of (\ref{d2}) in $H_2$ can
be completely described.  There is a correspondence between
irreducible representations and orbits of the dynamical
system going through a point with zero second coordinate.
Moreover, since $C^2\geq 0$, the spectral measure of the pair
($A$, $C^2$) is concentrated on that part of the orbit where the
second coordinates are non-negative.  Namely, we have the
following description of irreducible representations.

\begin{proposition} \label{three-fock}
Any irreducible representation \textup($A, X$\textup) of
\eqref{d2} such that $\ker X\cup\ker X^*\ne \{0\}$ is
unitarily equivalent to one of the following\textup:

\textup{(i)}. $H={\mathbb C}^n$, $n\in {\mathbb N}$,
\begin{align*}
A&=
\begin{pmatrix}
\lambda&&&\smash{\text{\lower 5pt \hbox{\Large $0$}}}\\
&F_1(\lambda,0)&&\\
&&\ddots&\\
&\smash{\text{\Large $0$}}&&F^{ (n-1)}_1(\lambda,0)
\end{pmatrix},
\\
X&=
\begin{pmatrix}
0&&\smash{\text{\lower 5pt \hbox to 0pt {\quad\Large $0$\hss}}}&\\
F_2(\lambda,0)&\ddots&&\\
&\ddots&0&\\
\smash{\text{\Large $0$}}&&F^{ (n-1)}_2(\lambda,0)&0
\end{pmatrix},
\end{align*}
where $\lambda$ belongs to the set 
$$
\sigma_n=\{\lambda\in{\mathbb
R}\mid F_2^{ k}(\lambda,0)>0,\, k=1,\dots,n-1,\, F_2^{
n}(\lambda,0)=0\};
$$

\textup{(ii)}. $H=l_2({\mathbb N})$,
$$
Ae_k=F_1^{ (k-1)}(\lambda,0)\,e_k,\quad Xe_k=F_2^{
k}(\lambda,0)\,e_{k+1},
$$
where $\lambda$ belongs to the set
$\sigma_{\infty}=\{\lambda\in{\mathbb R}\mid F_2^{
k}(\lambda,0)>0,\, k\in {\mathbb N}\}$\textup;

\textup{(iii)}. $H=l_2({\mathbb N})$,
$$Ae_k=\lambda_ke_k,\quad
Xe_k=\mu_{k-1}e_{k-1},$$ where
$\lambda_k=F_1(\lambda_{k+1})$,
$\mu_k=F_2(\lambda_{k+1},\mu_{k+1})$, $\mu_1=0$, and $\mu_k>0$,
$k=2$, \dots .
\end{proposition}

\begin{remark} Note that not all finite-dimensional representations are
necessarily related to cycles of the dynamical system (see
examples below in this section).
\end{remark}

The
 possibility of the description of irreducible representations in
 $H_1$ depends on topological properties of the
 two-dimensional dynamical system.

Below, we assume that  the mapping $F(\cdot)$ is one-to-one.

\begin{proposition} \label{three-uni}
If the
 dynamical system $F(\cdot)\colon{\mathbb R}^2\rightarrow{\mathbb R}^2$ has a
 measurable section, then any irreducible representation is
unitarily equivalent to one of the following\textup:

\textup{(i)}. $H={\mathbb C}^n$, $n\in {\mathbb N}$,
\begin{align*}
A&=
\begin{pmatrix}
\lambda&&&\smash{\text{\lower 5pt \hbox{\Large $0$}}}\\
&F_1(\lambda,\mu)&&\\
&&\ddots&\\
\smash{\text{\hbox to 0pt{\quad\Large
$0$\hss}}}&&&F^{ (n-1)}_1(\lambda,\mu)
\end{pmatrix},
\\
X&=\begin{pmatrix}
0&&&e^{i\varphi}\mu\\
F_2(\lambda,\mu)&\ddots&&\\
&\ddots&0&\\
\smash{\text{\hbox{\Large $0$}}}&&F^{ (n-1)}_2(\lambda,\mu)&0
\end{pmatrix},
\end{align*}
where $\lambda$, $\mu$ belong to the set
$$
\sigma_n=\{\lambda\in{\mathbb R}\mid F_2^{ k}(\lambda,\mu)>0,\,
k=1,\ldots,n-1,\, F^{\circ n}(\lambda,\mu)=(\lambda,\mu)\},
$$
$\varphi\in[0,2\pi)$\textup;

\textup{(ii)}. $H=l_2({\mathbb Z})$,
$$Ae_k=\lambda_ke_k,\quad
Xe_k=\mu_{k-1}e_{k-1},$$ where
$\lambda_k=F_1(\lambda_{k+1})$,
$\mu_k=F_2(\lambda_{k+1},\mu_{k+1})$,  and $\mu_k>0$,
$k\in {\mathbb Z}$.
\end{proposition}

\begin{remark}
 If there exists an ergodic quasi-invariant measure
which is not concentrated on a single orbit, then one can construct factor representations of the relation which are
not of type I, provided that all
second coordinates of the points of the orbit are positive.
\end{remark}

Below in this sections we consider two examples of relations from this
class: representations of real forms of Witten's first deformation, and
representations of the Sklyanin algebra in the degenerate case (they
correspond to representations of the quantum $sl_2$ group).

\subsection{Representation of real forms of Witten's first
deformation}\label{sec:2.2.4}

\textbf{1.}    Studying  the Jones polynomials, their generalizations
    and their connections with ``vertex models'' in two-dimensional
 statistical mechanics, Witten (see \cite{103})
introduced Hopf algebra deformations
of the universal enveloping algebra of $su(2)$. There is a  family
of associative
algebras that depend on a real parameter $p$. These algebras are
defined
by the generators $E_{0}$, $E_{+}$, $E_{-}$ and the 
following relations:
\begin{gather}
 p\,E_{0}E_{+}- p^{-1}\,E_{+}E_{0}=E_{+}, \notag%\label{witt1}
\\{}
 [E_{+},E_{-}]=E_{0} -(p - p^{-1} ) \,E_{0}^{2}, \notag %\label{witt2}
\\*
 p\,E_{-}E_{0}- {p^{-1}}\,E_{0}E_{-}=E_{-}.      \label{witt3}
\end{gather}

In item~2 we introduce a class of  $*$-algebra
structures in Witten's first deformation. In items~3 an 4 we give
a description
of irreducible representations of these $*$-algebras on a Hilbert space.
We  use the method of ``dynamical relations'' developed above in this section.
Notice, that some unbounded representations arise in a natural way here;
however, for unbounded operators we restrict  ourselves
with a  description
of
a certain class of representations, while in the bounded case, we give a
complete unitary description of irreducible representations.

\medskip\noindent\textbf{2.}
Real forms. Witten's
first deformation is a family of associative algebras
$A_{p}$ given by  generators and relations \eqref{witt3},
  $p \in (0,1)$ is a parameter.  $A_{1} =su(2)$;  $A_{0}$ has the
  following
 relations  $E_{+}E_{0}=E_{0}E_{+}=E_{0}^{2}=0$.
We consider involutions in $A_{p}$, which are obtained from involutions
in the free algebra, and reduce the linear subspace generated by the
relations \eqref{witt3}.

We will say that involutions are equivalent if the corresponding real forms
are isomorphic.

\begin {lemma}
There are  two inequivalent involutions in Witten's first
deformation\textup:
 \begin{gather}
  E_{0}^{*}=E_{0},\quad E_{+}^{*}=E_{-}  ,           \label{witt4}
   \\
  E_{0}^{*}=E_{0}, \quad E_{+}^{*}=-E_{-} .    \label{witt5}
 \end{gather}
\end{lemma}

\noindent\textbf{3.} 
  Relations \eqref{witt3}
 with the involution given by \eqref{witt4} or \eqref{witt5} have the form
\eqref{d2}.
The corresponding dynamical system on $\mathbb{R}^{2}$ is generated
by
\[
F(x,y)=(p^{-1}(1+p^{-1}x), q(qy-x+(p-p^{-1})x^{2})),
\]
were $q=1$ $(q=-1)$ for the first (second) real form.

It is a difficult problem to find positive orbits of a two-dimensional
nonlinear dynamical system. We have avoided these difficulties by using
the Casimir element,
\[
C_{p}=p^{-1}E_{+}E_{-}+p\,E_{-}E_{+}+E_{0}^{2}.
\]
For any irreducible representation $\pi$, we have $\pi(C_{p})=
\mu I$, were $ \mu $
is a complex number.

We will  work with the following system:
\begin{equation}
E_{0}E_{+}=E_{+}f(E_{0}),%\label{witt6}
\quad
 E_{+}^{*}E_{+}=G_{ \nu}(E_{0}) , \label{witt7}
\end{equation}
where $\nu=\mu p$,
\begin{gather*}
f(X)=p^{-1}(1+p^{-1}x),
\\
 G_{ \nu}(y)=\frac{q}{1+p^{2}}\,(-y-p^{-1}y^{2}+ \nu I).
\end{gather*}

\begin {lemma}
      For any irreducible representation $ \pi $ of the real form $A_{p}$
there is a unique  $\nu$ \textup($ \nu\ge 0$ for the first real
form\textup),
such that $ \pi $ is a representation of  \eqref{witt7}.

For an arbitrary  $ \nu $ \textup($\nu
\ge 0$ for the first real form\textup),
      every irreducible representation
of \eqref{witt7} with $\dim\pi>1$
is a representation of~$A_{p}$.
\end {lemma}

The dynamical system corresponding to relations
\eqref{witt7}
is, actually,
one-dimensional, linear, and depends on one real parameter.

Every irreducible representation of (\ref{witt7}) is
determined by
the subset $   \Delta \subset \mathbb{R}^{2}$,
\[
\Delta =\{(\lambda_{k},\mu_{k}),j<k<J\},
\]
where $ \lambda_{k+1}=f(\lambda_{k})$, $\mu_{k+1}=G_{\nu}(\lambda_{k})$,
 $\mu_{k}\ge 0$, $\mu_{k}=0 $ for   $k=j+1 > -\infty$ and $k=J-1
<+\infty$;
$j$, $J$ are integer or infinities; $l_{2}(\Delta )$ is a Hilbert space
with an
orthonormal base $\{{\base }\colon(\lambda_{k},\mu_{k})\in
\Delta\}$,
\begin{align*}
 T(E_{0})\,\base &=\lambda_{k}\base,
\\
 T(E_{+})\,\base&=\mu_{k+1}^{1/2}\basep ,\qquad j<k+1<J,
\\
 T(E_{-})\,\base&=\mu_{k}^{1/2}\basem ,\qquad j<k-1,
\\
 T(E_{+})\,\baseJ& =0,\quad T(E_{-})\,\basej =0.
\end{align*}

\medskip\noindent\textbf{4.}
 {Classification of representations.}
\begin {theorem}
    Every irreducible representation of the first real form is
bounded.

\textup{1.}  For every non-negative integer $m$ there is a representation
of dimension $m+1$
  with 
\begin{align*}
\nu&=\frac{p}4\,\Bigl(\Bigl(\frac{(1-p^{2m})(1+p^{2})}
{(1+p^{2m})(1-p^{2})}\Bigr)^{2}-1\Bigr),
\\
\Delta_{\nu}&= \{f(k,x_{1}),-1<k \le m+1 \};
\end{align*}

\textup{2.} There is a family of one-dimensional representations\textup:
$ E_{0}={p}{(p^{2}-1)^{-1}} $,
$ E_{+}=\lambda$,
$ E_{-}= \bar \lambda $\textup, where $ \lambda $ is a complex number,
$ \nu={p^{3}}{(1-p^{2})^{-2}}$\textup;

\textup{3.}
 For every $ \nu \in [{p^{3}}{(1-p^{2})^{-2}};+\infty) $ there is a
representation with the highest
weight, $\Delta_{\nu}= \{f(k,x_{2}),k<1 \} $.

\textup{4.}
 For every $\nu \in [{p^{3}}{(1-p^{2})^{-2}};+\infty)$ there is a
representation with the lowest weight,
$\Delta_{\nu}= \{f(k,x_{1}),k<1 \}$.
\end{theorem}

In the theorem we used the notation $ f(k,x)= \frac{1}{p^{2k}}
(x+ \frac{p^{2k}-1}{p^{2}-1}p)$;  $ g_{\nu}(x)=-x-p^{-1}x^{2}+
\nu$,
 $ x_{1}<x_{2}$ are roots of the equation $ g_{\nu}(x)=0$.

\begin {theorem}
  There are bounded and unbounded irreducible representation of the second
real form in an infinite-dimensional Hilbert space, 
which are given, except for the one-dimensional
 representation, by the following\textup:
\begin{itemize}
\item[$1.$]  bounded representations with the highest
weight such that
\[
 \nu \in [-{p}/{4};{p^{3}}{(1-p^{2})^{-2}}),
\quad\Delta_{\nu}=\{ f(k,x_{1}),k<1 \};
\]

\item[$2.$]  unbounded representations\textup:
\begin{itemize}

\item[$(a)$]  a family with the  highest weight such that
\[
\nu \in (-{p}/{4} ;\, 0),
 \quad\Delta_{\nu}=\{ f(k,x_{2}),k<1 \};
\]
a family with the highest weight such that
\[
 \nu \in ( {p^{3}}{(1-p^{2})^{-2}} ;+ \infty),\quad
    \Delta_{\nu}=\{f(k,x_{1}),k>-1\};
\]
    \item[$(b)$] a family with the lowest weight such that
\[
 \nu \in [{p}/{4} ;\,0),\quad \ddd \{ f(k,x_{2}),k>-1 \};
\]

a family with the lowest weight such that
\[
\nu \in ({p^{3}}{(1-p^{2})^{-2}} ;+\infty),\quad
 \ddd \{f(k,x_{1}),k>-1\};
\]
    \item[$(c)$]  families without  highest and lowest weights\textup:

a family indexed by $(\lambda,\nu)$, where
\begin{gather*}
 \lambda \in (-p\ ;-{p}/{2})
\cup (- \frac{p}{2}\ ; 0],\quad \nu \in (-\infty ;
\lambda(\lambda+p)p^{-1}),
 \\
 \Delta_{( \lambda, \nu)}= \{f(k, \lambda),k \in \mathbb{Z}  \};
\end{gather*}

a  family indexed by $(\lambda,\nu)$, where
\begin{gather*}
(\lambda,\nu) \in[-p^{-2} ;-p-1) \times (- \infty ;\p),
\\
 \Delta_{( \lambda, \nu)}= \{f(k, \lambda),\,k \in \mathbb{Z} \}.
\end{gather*}
\end{itemize}
\end{itemize}
\end {theorem}

\subsection{Representations of the Sklyanin algebra and $U_q(sl(2))$}
\label{sec:2.2.5}

In this subsection we apply the technique developed to a study of
representations of two algebras that arise naturally in
physical models.

\medskip\noindent\textbf{1.}
The Sklyanin algebra $\mathcal{F}$ was introduced in \cite{sklyan} as
an
algebra over $\mathbb{C}$ generated by elements $S_0$, $S_1$, $S_2$,
$S_3$ that satisfy the relations:
\begin{gather}
[S_0, S_3] =0, \quad [S_0, S_1] = i J \{S_2, S_3\}, \quad [S_0, S_2] =
-i J\{S_3, S_1\}, \notag
\\{}
[S_1, S_2] = 2i S_0 S_3, \quad [S_2, S_3] = i \{S_0, S_1\}, \quad [S_3,
S_1] = i \{S_0, S_2\}, \label{rel_sklyanin}
\end{gather}
where $J$ is a real parameter. The center of $\mathcal{F}$ is generated
by two elements, $\Delta_1 = S_0^2 - J S_3^2$ and $\Delta_2 = S_0^2 +
S_1^2 + S_2^2 + S_3^2$.

\medskip\noindent\textbf{2.}
The quantum algebra $U_q(sl(2))$ \cite{kul_re,drinf,jimbo} is a
complex algebra generated by elements  $k$, $k^{-1}$, $X$, $Y$
satisfying the relations
\begin{gather*}
kk^{-1} = k^{-1} k =1, \quad
k X = q Xk,\quad
kY = q^{-1} Yk,
\\*{}
[X,Y] = \frac{k^2 - k^{-2}}{q-q^{-1}},
\end{gather*}
where $q$ is a complex parameter, $q \ne -1$, $0$, $1$. The center is
generated by $C = XY + {(q^{-1} k^2 + q k^{-2})}/{(q-q^{-1})^2}$.

\begin{remark}
The transposition $k \leftrightarrow k^{-1}$ defines an isomorphism between
$U_q(sl(2))$ and $U_{q^{-1}}(sl(2))$.
\end{remark}

\noindent\textbf{3.}
The two algebras introduced are closely related. Indeed, introduce
the correspondence
\[
q(J) = \begin{cases} \frac{1+\sqrt{J}}{1-\sqrt{J}}, & \text{if $J>0$
and $J\ne
1$,} \\
\frac{1+ i \sqrt{-J}}{1- i \sqrt {-J}}, & \text{if $J<0$.} \end{cases}
\]
It establishes a one-to-one mapping between the sets $\{ J \in
\mathbb{R} \colon J \ne 0, 1\}$, and $\{ q \in \mathbb{R} \colon |q|
>1\} \cup \{ q \in \mathbb{C} \colon |q| =1,\, \im q >0\}$. Also,
introduce the elements
\begin{align*}
A_1 &= \begin{cases} S_0 + \sqrt{J}\,
S_3, & J>0, \\ S_0 + i \,\sqrt{-J}\, S_3,
& J<0, \end{cases}
\\
A_2 &= \begin{cases} S_0 - \sqrt{J}\,
S_3, & J>0, \\ S_0 - i\, \sqrt{-J} \,S_3,
& J<0, \end{cases}
\\
X &= \begin{cases} \frac12 \,\sqrt{1-J}\,(S_1 + i S_2), & J\le 1, \\
i\frac12 \, \sqrt{J-1}\,(S_1+ i S_2),
& J>1, \end{cases}
\\
Y &= \begin{cases} \frac12\, \sqrt{1-J}\,(S_1 - i S_2), & J\le 1, \\
i\frac12 \, \sqrt{J-1}\,(S_1- i S_2).
& J>1, \end{cases}
\end{align*}

\begin{proposition}
For $J \ne 0$, $1$, there is an algebraic isomorphism between
$\mathcal{F}/(\Delta_1 -I)$, and $U_{q(J)}(sl(2))$.
\end{proposition}

\begin{proof}
The relations \eqref{rel_sklyanin} are equivalent to the following
\begin{gather*}
A_1 X = q X A_1, \quad A_1 Y = q^{-1} Y A_1,
\\
A_2 Y = q Y A_2, \quad A_2 X = q^{-1} XA_2,
\\{}
[A_1, A_2] = 0,
\quad [X, Y] = \frac1{q-q^{-1}}(A_1^2 - A_2^2),
\end{gather*}
where $q = q(J)$. Set $k = A_1$, $k^{-1}= A_2$, and notice that $A_1A_2
= \Delta_1$.
\end{proof}

\noindent\textbf{4.}
Now we describe the real forms of $\mathcal{F}$. We will assume that
involution preserves natural grading related to the chosen generators
$S_i$, $i=0$, 1, 2, 3.

\begin{proposition}
There are three real forms of $\mathcal{F}$, namely\textup:

$\mathcal{F}_1:$ $S_i^* = S_i$, $i=0$, $1$, $2$, $3;$

$\mathcal{F}_2:$ $S_0^* = S_0$, $S_3^* = S_3$, $S_1^* = S_1$, $S_2^* =
S_2;$

$\mathcal{F}_3:$ $S_0^* = S_0$, $S_3^* = -S_3$, $S_1^* = S_1$, $S_2^* =
- S_2$, if $J \le 1$, and $S_1^* = - S_1$, $S_2 ^* = S_2$, if $J >1$.

For $J<0$, the $*$-algebras $\mathcal{F}_1$ and $\mathcal{F}_2$
are $*$-isomorphic.
\end{proposition}

\begin{proof} The proof is a routine calculation. For $J < 0 $, the
isomorphism $j\colon \mathcal{F}_1 \to \mathcal{F}_2$ is given by
$j(S_0 ) = -\sqrt{-J} S_3$, $j(S_3) = S_o/ \sqrt{-J}$, $j(S_1) = iS_2$,
$j(S_2) = - S_1$.
\end{proof}

The corresponding real forms of $U_q(sl(2))$ are the following:

$su_q(2)$ : $k^* = k$, $X^* = Y$ for $q>0$; $X^* = -Y$ for $q<0$, $k^*
= k^{-1}$ for $q \in \mathbb{T}$;

$su_q(1,1)$ : $k^* = k$, $X^* = -Y$ for $q>0$; $X^* = Y$ for $q<0$, $k^*
= k^{-1}$ for $q \in \mathbb{T}$;

$sl_q(2, \mathbb{R})$ : $X^* = X$, $Y^* = Y$, $k^* = k^{-1}$, if $q \in
\mathbb{R}$; $k^* = k^{-1}$ for $q \in \mathbb{R}$.

For $q \in \mathbb{T}$, the $*$-algebra $su_q(1,1)$ is isomorphic to
$su_q(2)$.

\begin{remark}
In \cite{masuda_etal} $*$-Hopf algebra structures on $U_q(sl(2))$ are
studied. In our list, the involution is compatible with the Hopf
algebra structure in the cases of $su_q(2)$, $su_q(1,1)$ with $q \in
\mathbb{R}$, and $sl_q(2, \mathbb{R})$ with $q \in \mathbb{T}$.
\end{remark}

\noindent\textbf{5.}
Now we go to the study of $*$-representations of real forms of
the Sklyanin algebra and the corresponding real forms of $U_q(sl(2))$.

First notice that the relations in $\mathcal{F}$
are homogeneous, therefore, for any $\lambda \in
\mathbb{R}$, $\lambda \ne 0$, the operators $(S_i)$ form representations of
$\mathcal{F}$ if and only if so do the operators $(\lambda S_i)$. The same
holds for $U_q(sl(2))$, and we agree to identify representations
$(\lambda S_i)$
with different $\lambda \ne 0$ of $\mathcal{F}$, and $(k, X,Y)$ and
$(-k, X,Y)$ of $U_q(sl(2))$, as well as the unitarily equivalent ones.

\begin{remark}
In the case $J=0$, each irreducible representation of $\mathcal{F}$ is
either one-dimensional with $S_0 =  0$, or is such that $S_0 = \frac12 I$,
and $S_1$, $S_2$, $S_3$ form an irreducible $*$-representation of
$sl(2)$.
\end{remark}

Below, we will study irreducible
$*$-representations of the Sklyanin algebra, and of the deformed
enveloping algebra $U_q(sl(2))$. We will see that in the case
$\mathcal{F}_1$, and $su_q(2)$ with $J<0$, and $|q|   =1$ which is 
not a root of
1, the description problem includes such a problem for the irrational rotation
algebra, and is, therefore, rather complicated.

Let us begin with $J>0$; in this case, $q\in\mathbb{R}$, $|
q|>1$, $A_1$,
$A_2$, $ k$ are self-adjoint, and $X^*=Y$ for $J\le 1$
$(q>1)$, and $X^*=-Y$ for $J>1$ ($q< -1)$.

\begin{theorem}
If\/ $J=1$, then $\mathcal{F}_1$ has the following irreducible
representations\textup:
\begin{itemize}
\item[$1.$] one-dimensional\textup: $A_1=\pm 1$,
$A_2=1$, $ X=0$; $ A_1=A_2=0$,
$X\in\mathbb{T}$\textup;

\item[$2.$] $n$-dimensional $(n=2,3,\ldots)$:
\begin{align*}
 A_1 f_1&=0,\quad\dots,\quad A_1f_{n-1}=0,\quad  
A_1f_n=\pm f_n,
\\
 A_2 f_1&=0,\quad \dots,\quad A_2f_{n-1}=0,\quad A_2f_n=0,
\\
 Xf_1&=f_2,\quad\dots,\quad Xf_{n-1}=f_n,\quad Xf_n=0;
\end{align*}

\item[$3.$] infinite-dimensional\textup:
\begin{itemize}
\item [$(i)$] $A_1=0$, $A_2f_1=f_1$, $A_2f_n=0$, $ n=2$,
$3$, \dots\textup;
$Xf_n=f_{n+1}$, $ n=1$, $2$,\dots\textup;
\item [$(ii)$]  $A_2=0$, $A_1f_1=f_1$, $A_1f_n=0$, $ n=2$, $3$,
\dots\textup;
$Xf_1=0$, $Xf_n=f_{n-1}$, $ n=2$, $3 $, \dots .
\end{itemize}
\end{itemize}
\end{theorem}

\begin{proof}
In this case we have the following relations:
\begin{gather*}
[A_1,A_2]=0,\quad XA_1=A_1X^*=0,\\
[X,X^*]  =A_1^2-A_2^2,\quad A_2X=X^*A_2=0
\end{gather*}
which imply that $A_1$,  $ A_2$, $ XX^*$, $ X^*X$ are pairwise commuting
operators and
\[
\mathcal{H}=\mathcal{H}_0\oplus\mathcal{H}_1
\oplus\mathcal{H}_2\oplus\mathcal{H}_3,
\]
where:

$A_1=A_2=0$, $ XX^*=X^*X>0$ in $\mathcal{H}_0$,

$A_1=X^*=0$, $ X^*X=A_2^2>0$ in $\mathcal{H}_1$,

$A_2=X=0$, $ XX^*=A_1^2>0$ in $\mathcal{H}_2$,

$X=X^*=0$, $ A_1^2=A_2^2>0$ in $\mathcal{H}_3$.

The subspace $\mathcal{H}_3$ is invariant with respect to $A_1$, $ A_2$,
$X$, $ X^*$, thus irreducible representations are one-dimensional there.

Consider now its orthogonal complement. Note that
$X\mathcal{H}_3^{\perp}\subset\mathcal{H}_0\oplus\mathcal{H}_2$,
$X^*\mathcal{H}_3^{\perp}\subset\mathcal{H}_0\oplus\mathcal{H}_1$. Set
$\mathcal{H}_1^{(1)}=\{f\in\mathcal{H}_1 \mid Xf\in\mathcal{H}_2\}$,
$\mathcal{H}_1^{(n+1)}=\{f\in\mathcal{H}_1 \mid
Xf,\ldots,X^nf\in\mathcal{H}_0,\, X^{n+1}f\in\mathcal{H}_2\}$,
$n\in\mathbb{N}$; $\mathcal{H}_1^{(\infty)}=\{f\in\mathcal{H}_1 \mid
X^nf\in\mathcal{H}_0,\, n\in\mathbb{N}\}$,
$\mathcal{H}_2^{(\infty)}=\{f\in\mathcal{H}_2 \mid
(X^*)^nf\in\mathcal{H}_0,\, n\in\mathbb{N}\}$. Then:
\begin{itemize}

\item [(i)] the subspace
$\mathcal{H}^{(n)}=\oplus_{j=0}^nX^j\mathcal{H}_1^{(n)}$ is invariant
and the irreducibility implies that $\dim X^n\mathcal{H}_1^{(n)}=1$,
$j=1$, $2$, \dots; therefore $\dim \mathcal{H}^{(n)}=n+1$;

\item [(ii)] the subspace
$\tilde{\mathcal{H}}_1=
\oplus_{n=0}^{\infty}X^n\mathcal{H}_1^{(\infty)}$ is invariant and the
irreducibility implies that $\dim X^n\mathcal{H}_1^{(\infty)}=1$,
$n\in\mathbb{N}$;

\item [(iii)] the subspace
$\tilde{\mathcal{H}}_2=
\oplus_{n=0}^{\infty}X^n\mathcal{H}_2^{(\infty)}$ is invariant and
in the irreducible case, $\dim X^n\mathcal{H}_2^{(\infty)}=1$,
$n\in\mathbb{N}$.
\end{itemize}
On the invariant subspace
$\tilde{\mathcal{H}}_0=\{f\in\mathcal{H}_0 \mid X^n
f\in\mathcal{H}_0,\, (X^*)^n f\in\mathcal{H}_0;\, n\in\mathbb{N}\}$, the
operator $X$ is normal, and irreducible representations are
one-dimensional.

Finally we note that for any representation, we have the following
decomposition:
\[
\mathcal{H}=\bigoplus_{n=1}^{\infty}\mathcal{H}^{(n)}\oplus
\tilde{\mathcal{H}}_0\oplus\tilde{\mathcal{H}}_1
\oplus\tilde{\mathcal{H}}_2\oplus\mathcal{H}_3,
\]
which completes the proof.
\end{proof}

\begin{theorem}
For $q\in\mathbb{R}$, $| q|>1$, the $*$-algebra $su_{q}(2)$ has the
following irreducible representations \textup($n\in\mathbb{N}$\textup{):}
\begin{gather*}
k f_j  = | q| ^{{(1-n)}/{2}}q^{j-1} f_j,\quad
X f_j  = (| [j]_q\, [n-j]_q|)^{{1}/{2}}f_{j+1},
\\
j=1,2,\ldots,n
\end{gather*}
\textup(we
use a standard notation $[m]_q=(q^m - q^{-m})/(q-q^{-1})$\textup).
\end{theorem}

\begin{proof}
Apply Propositions~\ref{three-fock} and \ref{three-uni}
to the self-adjoint operator $A=k$ and the map
$F(l,\lambda)=(ql,\lambda-{(l_1^2-l_2^{-2})}/{| q-q^{-1}|})$. We
see that
\[
F^n(l,\lambda)=\lambda-[n]_q\,
\frac{l^2q^{n-1}-l^{-2}q^{1-n}}{| q-q^{-1}|}\rightarrow
-\infty,\qquad n\rightarrow\pm\infty.
\]
It means that only the finite-dimensional case is realized.
\end{proof}

Now we consider the case where $J>1$.

\begin{proposition}
For $J>0$, $ J\neq 1$, the $*$-algebra $\mathcal{F}_1$ has the following
irreducible representations\textup:
\begin{itemize}
\item[$1.$] one-dimensional\textup: $A_1=A_2=0,\ X\in\mathbb{T}$\textup;

\item[$2.$] $n$-dimensional \textup($n\in\mathbb{N}$\textup{):}
$A_1=k$, $ A_2=\pm k^{-1}$\textup;
$k$, $ X$, $ Y$ generate an irreducible representations of
$su_q(2)$\textup;

\item[$3.$] infinite-dimensional\textup:
\begin{itemize}
\item [$(i)$]
$A_1 f_n = q^{-n}f_n$, $A_2=0$,
\[
 Xf_n =
\biggl({\frac{[n]_q \,q^{1-n}}{| q-q^{-1}|}}\biggr)^{1/2} f_{n-1},\qquad
n=0,1,2,\dots;
\]

\item [$(ii)$]
$A_1=0$, $A_2f_n=q^{-n}f_n$,
\[
Xf_n =
\biggl({\frac{[n+1]_q\,
q^{1-n}}{| q-q^{-1}|}}\biggr)^{1/2}f_{n+1},\qquad n=0,1,2.\dots.
\]
\end{itemize}
\end{itemize}
\end{proposition}

\begin{proof}
Apply Propositions~\ref{three-fock} and \ref{three-uni}
to $A=A_1+iA_2$ and the mapping
\[
F\colon (l_1,l_2,\lambda)\mapsto
\bigl(ql_1,q^{-1}l_2,\,\lambda-{(l_1^2-l_2^2)}/{|
q-q^{-1}|}\bigr).
\]
The orbits
with $l_1l_2\neq 0$ give the case 2); the orbits with $l_1=0$ or
$l_2=0$ give the cases 1) and 3).
\end{proof}

Now consider the case $J< 0$; now $q=e^{i\eta}\in\mathbb{T}$,
$0<\eta<\pi$, and $A_1^*=A_2,\ k^*=k^{-1}$, $X^*=Y$; also note
that $[m]_q={\sin(m\eta)}/{\sin\eta}$.

Decompose ${\eta}/{\pi}$ into a continued fraction (which is
finite if $q$ is a root of $1$, i.e., if
${\eta}/{\pi}\in\mathbb{Q}$, and infinite otherwise):
\[
\frac{\eta}{\pi}=0+\frac{1}{a_1+\displaystyle\frac{1}{a_2+\dots}}.
\]
The denominators of convergents of the continued fraction are:
\[
l_0=1,\quad l_1=a_1,\quad l_p=l_{p-2}+a_p l_{p_1},\qquad p\ge 2.
\]
Then define an increasing sequence of integers (finite if\/ $q$ is a
root of 1):
\begin{align*}
r_m^{(1)} & = m,\qquad m=1,2,\dots,a_1,
\\
r_m^{(p)} & = l_{p-2}+m\, l_{p-1},\qquad m=1,2,\ldots,a_p;\quad p\ge 2.
\end{align*}
For example, when $\eta={\pi}/{N}$, these numbers are
$1$, $2$, \dots, $N$; when $\eta ={\pi(\sqrt{5}-1)}/{2}$, they are
the Fibonacci sequence $1$, $2$, $3$, $5$,~\dots .

\begin{theorem}
For\/ $q\in\mathbb{T}$ such that\/ $q$ is not a root of $1$, the
$*$-algebra
$su_q(2)$ has the following irreducible representations\textup:
\begin{itemize}
\item[$1.$] finite-dimensional with dimension $n=r_m^{(p)}$,
$p\in\mathbb{N}$\textup:
\begin{align}
k f_j & =i\exp \bigl(i^{\,c}({(1-n)}/{2}+j-1)\eta)\,f_j ,\nonumber
\\
X f_j & = \sqrt{(-1)^c\, [j]q\,[n-j]q}\,f_{j+1}, \label{vos}
\end{align}
where $c=0$, $1$ is such that the expression under the square root is
positive\textup;

\item[$2.$] two degenerate infinite-dimensional\textup:
for $j=1$, $2$, \dots,
\begin{align*}
k f_j & =i\exp (i({\eta}/{2}+(j-1)\eta))\,f_j,
\\
X f_j & = [j]q\, f_{j+1} ;
\end{align*}
and
\begin{align*}
k f_j & =i\exp (-i ({\eta}/{2}+(j-1)\eta))\,f_j\\
X f_j & = [j-1]q \,f_{j-1} ;
\end{align*}

\item[$3.$]
non-degenerate infinite-dimensional, which are representations
of the irrational rotation algebra $\mathcal{A}_{\eta}$.
\end{itemize}
\end{theorem}

\begin{proof}
Now we apply Propositions~\ref{three-fock} and \ref{three-uni}
to the unitary $A=k$ and
the map $F\colon (e^{i\phi},\lambda)\mapsto
(e^{i(\phi+\eta)},\lambda-{\sin 2\phi}/{\sin \eta})$.

If $\lambda_0=\lambda_n=0$, then it is easy to see that
$$
\lambda_j=c\, \sin (n-j)\eta\,\sin j\eta/\sin^2 \eta,
$$ 
where $c=\pm
1$. The requirement $\sign (\sin ((n-j)\eta)\sin j\eta)=\const$ is
equivalent to $[{(n-j)\eta}/{\pi}]+[{j\eta}/{\pi}]=\const$. The
numbers satisfying this condition are $n=r_m^{(p)}$.

Let $X^*=U\sqrt{B}$ and $U$ be unitary. If the central
element $C=cI$, then $c>{1}/{2}$ and the operators
$B=cI+ ({e^{-i\eta}k^2+e^{i\eta}k^{-2}})/{4}$, $k$, $U$ form a 
representation of the irrational rotation algebra
$\mathcal{A}_\eta$: $U k=e^{(i\eta)}kU$.
\end{proof}

\begin{theorem}
Let $q\in\mathbb{T}$ be a root of\, $1$, $\eta={\pi M}/{N}$.
Then $su_q(2)$ has the following irreducible representations \textup(all
finite-dimensional\textup{):}
\begin{itemize}
\item[$1.$] ordinary\textup:
$n=r_m^{(p)}$, $1\le n\le N$\textup; $k$ and $X$ are as in
~\eqref{vos}\textup;

\item[$2$.] continuous series\textup:
$n=N$ for $M$ even, and $n=2N$ for $M$ odd\textup:
\begin{align*}
k f_j & =  \exp (\phi+(j-1)\eta)\,f_j,
\\
Xf_j & = \biggl({h-\frac{\sin(2\phi+(j-1)\eta)}{\sin
\eta}\,[j]_q}\biggr)^{1/2} f_{j+1},
\\
&\qquad j=1,\dots,n-1,
\\
X f_n & = E^{i\mu}\sqrt{h}f_1, 
\end{align*}
where
$\phi\in\tau=[{(\pi+\eta)}/{2}-{\pi}/{2n},
{(\pi+\eta)}/{2}+{\pi}/{2N})$, $e^{i\mu}\in\mathbb{T}$.
\end{itemize}
\end{theorem}

\begin{proof}
The mapping $F(\cdot)$
is defined for all $q\in\mathbb{T}$ in the same way. The
difference from the previous case is that each orbit now is cyclic
with the period $n=\min(l\colon e^{il\eta}=1)$.

Note that the dynamical system in this situation has a Borel
section $\tau\times\mathbb{R}_{+}$. It means that each irreducible
representation with unitary $U$ corresponds to a single orbit.
\end{proof}

\begin{theorem}
For $J<0$, the $*$-algebra $\mathcal{F}_1$ has the following irreducible
representations\textup:
\begin{itemize}
\item[$1.$] one-dimensional\textup: $A_1=A_2=0$, $ X\in\mathbb{T}$\textup;

\item[$2.$] $A_1=k$, $A_2=k^{-1}$\textup;
$k,\ X,\ Y$ generate an irreducible
representation of $su_q(2)$.
\end{itemize}
\end{theorem}

\begin{proof}
Recall that $A_1^*=A_2$ and $A_1^*A_1=A_1A_1^*$ commute with $X$ and
$X^*$, so the irreducible representation with degenerate $A_1$ can only be
trivial.
\end{proof}

\noindent\textbf{6.}
Now consider representations $\mathcal{F}_2$. Recall that in this
case $J>0$,
$q\in\mathbb{R}$, $| q| >1$, $A_1$, $A_2$,
$k$ are self-adjoint, and $X^*=-Y$ if\/ $J\le 1$ $(q>1)$, $X^*=Y$ if\/
$J>1$
$(q<-1)$.

\begin{proposition}
For $J=1$, $\mathcal{F}_2$ has only one-dimensional irreducible
representations.
\end{proposition}

\begin{proof}
As in the case of $\mathcal{F}_1$, we have that $A_1$, $A_2$, $ X^*X$,
$XX^*$ are commuting self-adjoint operators, and $X^*X-XX^*=A_1^2-A_2^2$.
Hence $\mathcal{H}=\mathcal{H}_0\oplus\mathcal{H}_1\oplus\mathcal{H}_2
\oplus\mathcal{H}_3$, where all $\mathcal{H}_j$ are invariant and:

$A_1=A_2=0$, $X$ is normal in $\mathcal{H}_0$;

$A_1=X^*=0$, $A_2^2=-X^*X\Rightarrow A_2=X=0$ in $\mathcal{H}_1$;

$A_2=X=0$, $A_1^2=-XX^*\Rightarrow A_1=X^*=0$ in $\mathcal{H}_2$;

$X=X^*=0$, $A_1^2=A_2^2$ in $\mathcal{H}_3$.
\end{proof}

\begin{theorem} \label{pat}
For $q\in\mathbb{R}$, $|q|>1$, the $*$-algebra $su_q(1,1)$ has the
following irreducible representations\textup:
\begin{itemize}
\item[$1.$] one-dimensional\textup: $k=1$, $X=0$\textup;

\item[$2.$] with the lowest weight\textup:
\begin{align*}
k f_j & = l\, q^{j-1}f_j,
\\
X f_j & =
\biggl({[j]_q\,\frac{l^2q^{j-1}-l^{-2}q^{1-j}} {|
q-q^{-1}|}}\biggr)^{1/2}f_{j+1},
\end{align*}
where $l>1$, $j=1$, $2$, \dots\textup;

\item[$3.$] with the highest weight\textup:
\begin{align*}
k f_j & = l\, q^{-j}f_j,
\\
X f_j & =
\biggl({[j]_q\,\frac{l^{-2}q^j-l^{2}q^{-j}}{| q-q^{-1}|}}
\biggr)^{1/2}f_{j-1},
\end{align*}
where $0<l<|q|$, $ j=1$, $2$, \dots\textup;

\item[$4.$] non-degenerate\textup:
\begin{align*}
k f_j & = l\, q^{j}f_j,
\\
X f_j & =
\biggl( {h+[j]_q\frac{l^2q^{j-1}-l^{-2}q^{1-j}}
{|q-q^{-1}|}}\biggr)^{1/2}f_{j-1},
\end{align*}
where $1\le l<|q|$, $ h>0$,  $j\in\mathbb{Z}$.
\end{itemize}
\end{theorem}

\begin{proof}
Apply Propositions~\ref{three-fock} and \ref{three-uni}
to the self-adjoint $A=k$ and the
mapping $F(l,\lambda)=\bigl(ql,\lambda+({l^2-l^{-2}})/| q-q^{-1}|\bigr)$.
It is easy to show that, save for the trivial representation, we have only 
infinite-dimensional representations with degenerate $X$ or $X^*$.

The dynamical system defined by $F(\cdot)$ has a Borel section. So in the
case of unitary $U$ (where $X^*=U\sqrt{B}$), each irreducible
representation corresponds to a single orbit.
\end{proof}

\begin{remark}
1. It is easy to see that an infinite-dimensional representation has a
classical limit only if $l^2\in q^{{\mathbb{Z}}/{2}}$, and so the
spectrum of $k$ belongs to $q^{{\mathbb{Z}}/{2}}$.

2. Note that all representations presented above, except for
one-dimensional, are unbounded. Since we don't give an accurate
definition of representation of $su_q(1,1)$ by unbounded operators,
we don't claim that the presented list of representations includes all
irreducible representations by unbounded operators.
\end{remark}

\begin{proposition}
For $J>1$, $J\neq 0$, the $*$-algebra $\mathcal{F}_2$ has the following
irreducible representations\textup:
\begin{itemize}
\item[$1.$] one-dimensional\textup: $A_1=A_2=0$, $ X\in\mathbb{T}$\textup;

\item[$2.$] infinite-dimensional inherited from $su_q(1,1)$\textup:
$A_1=k$,
$A_2=\pm k^{-1}$\textup;
$k$, $X$  are as in Theorem~\textup{\ref{pat}}\textup;

\item[$3.$] with the highest weight\textup: for $j=1$, $2$, \dots,
\begin{gather*}
A_1  =0,\quad A_2f_j=q^jf_j,
\\
Xf_j =\biggl({[j-1]_q\,\frac{q^j}{|q-q^{-1}|}}\biggr)^{1/2}f_{j-1};
\end{gather*}

\item[$4.$] with the lowest weight\textup: for $j=1,2,\ldots$,
\begin{gather*}
A_2  =0,\quad A_1 f_j=q^jf_j,\\
Xf_j =\biggl({[j]_q\,\frac{q^{j+1}}{|q-q^{-1}|}}\biggr)^{1/2}f_{j+1};
\end{gather*}

\item[$5.$] non-degenerate\textup: for $j\in\mathbb{Z}$,
\begin{gather*}
A_1  =0,\quad A_2f_j=q^{-j}f_j,
\\*
Xf_j=\biggl(h-[j]_q\,\frac{q^{-j-1}}{|q-q^{-1}|}\biggr)^{1/2}f_{j+1},
\end{gather*}
where $h>{|q|^{-1}(q-q^{-1})^{-2}}$\textup;
\begin{align*}
A_2 & =0,\quad A_1 f_j=q^{j-1}f_j,\\
Xf_j &=\biggl({h+[j]_q\,\frac{q^{j-1}}{|q-q^{-1}|}}\biggr)^{1/2}f_{j+1},
\end{align*}
where $h>({q^2-1})^{-1}$.
\end{itemize}
\end{proposition}

\begin{proof}
Apply Propositions~\ref{three-fock} and \ref{three-uni}
to $A=A_1+iA_2$ and the mapping
$F\colon (l_1,l_2,\lambda)\mapsto
(ql_1,\,q^{-1}l_2,\,\lambda+({l_1^2-l_2^2})/{|q-q^{-1}|})$. Then
the orbits with $l_1l_2\neq 0$ give the case 2), the orbits with
$l_1=0$ or $l_2=0$ give the cases 1), 3), 4), 5).

Once again, for the non-degenerate case with unitary $U$, we use the
existence of a Borel section for this dynamical system.
\end{proof}

\begin{remark}
Representations corresponding to the cases 2)--5) are unbounded.
\end{remark}

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