One Hat Cyber Team

View File Name : chapt28.tex

\section{Many-dimensional dynamical systems}\label{sec:2.4}
\markright{2.4. Many-dimensional dynamical systems}

In this section we study representations of relations with several
generators using the many-dimensional dynamical system approach.

We start with some examples. Firstly, in Section~\ref{sec:2.4.1}
we consider the so-called ``direct
products'' of one-dimensional relations and show how to apply
results of Section~\ref{sec:2.1}  
to classify their irreducible representations. In
Section~\ref{sec:2.4.2}
we study the  more complicated ``triangular'' case and apply the
inductive algorithm for the description of irreducible representations to the
twisted CCR and twisted CAR algebras. Then (see Section~\ref{sec:2.4.3})
we consider families of
operators satisfying a general class of relations whose representations can
be described in terms of orbits of some many-dimensional dynamical
system. The results  obtained are illustrated with several examples. Namely,
we study irreducible representations of  a nonstandard quantum
sphere (see~\ref{sec:2.4.4}), Heisenberg relations for the quantum
$E_2$ group (see ~\ref{sec:2.4.5}), and a wide class
of Wick algebras containing 
$q_{ij}$-CCR, $\mu$-CCR  and other algebras (see~\ref{sec:2.4.6}).

Let us note that all examples presented below are Wick algebras with
a braided operator $T$ and additional relations that reduce these
algebras to the form, to which dynamical formalism can be applied.
For more details we address the reader to Section~\ref{sec:2.4.6}
where we
give all necessary definitions and facts concerning Wick algebras.

\subsection{``Direct products'' of one-dimensional dynamical systems}
\label{sec:2.4.1}
%\label{dirpr}

\textbf{1.} By the ``direct product'' of one-dimensional systems we
mean a dynamical system on $\mathbb{R}^d$ defined by the
family of mappings
${F}_i$,  $i=1$, \dots , $d$ with the property that the $i$-th
mapping  only changes the $i$-th coordinate, i.e.,
\[
{{F}}_i (\vec{\lambda})=(\lambda_1,\dots ,
\lambda_{i-1},f_i (\lambda_i),\lambda_{i+1},\dots
,\lambda_d)
,\qquad \vec{\lambda}\in\mathbb{R}^d.
\]

\noindent\textbf{2.} Let us give an example of a $*$-algebra such that its
irreducible representations can be classified  using the
direct product of one-dimensional systems.
Namely, we consider the direct product of $d$ one-dimensional $q$-CCR,
which is a particular case of the so-called $q_{ij}$-CCR algebra 
(see Section~\ref{sec:2.4.6}).
Consider the $*$-algebra:
\begin{align} 
\label{gather:dirpr}
\mathbb{C}\bigl<x_i, x_i^* & \mid x_i^*x_i = 1+q x_i x_i^*,\notag
\\*
&\quad x_i^* x_j = x_j x_i^*, \, i,j=1,..,d,\, 0<q<1 \bigr>.
\end{align}
The classification of irreducible representations is based on the
following proposition which shows that there are additional relations
between generators which hold automatically in any irreducible
representation by bounded operators.

\begin{proposition}
Let $\pi(\cdot)$ be a bounded representation of \eqref{gather:dirpr}. Then
$\pi(x_j x_i - x_i x_j)=0$.
\end{proposition}

\begin{proof}
A simple calculations show that $X_{ij}=\pi(x_ix_j-x_jx_i)$ satisfies the
relation $X_{ij}^*X_{ij}=q^2 X_{ij}X_{ij}^*$, which implies
that $X_{ij}=0$ (since the relation $x^* x = q x x^*$ does not have any
non-trivial bounded representation, see Section~\ref{sec:1.4.2}).
\end{proof}

Let us note that the elements $x_j x_i - x_i x_j$ generate a quadratic
Wick ideal (see Section~\ref{sec:2.4.6}). So, to study the irreducible
representations, one has to
consider a family of bounded operators $X_i$, $ X_i^*$, $i=1$, \dots,
$d$, satisfying the relations
\begin{gather}\label{align:dirpr}
X_i^* X_i  = 1+q X_i X_i^*,\quad
X_i^* X_j  = X_j X_i^*,\quad
X_i X_j  = X_j X_i.
\end{gather}
It is easy to see that this algebra is generated by $d$ commuting
algebras
\[
\mathfrak{A}_i =\mathbb{C}\bigl<X_i, X_i^* \mid X_i^* X_i=1+q X_i
X_i^*\bigr>.
\]

Let us now consider the polar decomposition $X_i^*=U_iC_i$. Using
relations~(\ref{align:dirpr}) one can rewrite the system in an
equivalent form,
\begin{gather}
\label{align:dsdp}
     C_i^2 U_i^*  = U_i^* (1+q C_i^2),\quad
     C_i^2 U_j^*  = U_j^* C_i^2 ,\nonumber
\\
     C_i C_j  = C_j C_i, \quad
     U_j U_i  = U_i U_j .
\end{gather}
As in the one-dimensional case, the operator $U_i^*$ determines the action
on the spectrum of $C_i^2$, $x_i\mapsto 1+q x_i$. The action of
$U_j^*$, $i\neq j$, on $\sigma(C_i^2)$ is identical, because of the relation
$C_i^2 U_j=U_j C_i^2$.

\medskip\noindent\textbf{3.}
  Now we are able to give a classification of bounded
representations up to  unitary equivalence using the
one-dimensional technique.

To classify bounded irreducible representations, one must describe
bounded orbits of the dynamical system on $\mathbb{R}_{+}$
determined by the mapping $f(\lambda)=1+ q \lambda$.
It has only two orbits,
\begin{enumerate}
\item
$\{f^{n}(0),\ n\ge 0\}$, the ``Fock'' orbit;
\item
the fixed point $\{{1}/{(1-q)}\}$.
\end{enumerate}
It is easy to see that the spectral projection $E_{C_1^2}(\mathcal O)$, where
$\mathcal O$ is an orbit  of the one-dimensional dynamical system 
$(f, \mathbb{R}_{+})$,
commutes with all operators of the
representation. Consequently, for an the irreducible representation, the spectrum of
the operator $C_1^2$ lies on a single orbit of the dynamical system.
Moreover, since $U_1^*$ is an isometry, the spectrum coincides with one of
the two orbits presented above.

Let us consider the case where $\sigma(C_1^2)=\{f^n(0),\, n\ge 0\}$.
We have:
\begin{gather*}
U_1^*=
\begin{pmatrix}
  0    &       &&        \\
  1 &  0     &   &     \\
    &   1   & 0 &       \\
      &       &\ddots&\ddots
 \end{pmatrix}
\otimes I,
\quad
C_1^2= 
\begin{pmatrix}
f(0)            &       & \\
  & f^2(0)            & \\
      &             &\ddots
\end{pmatrix}
\otimes I.
\end{gather*}
It is easy to deduce from the commutation relations that
\[
C_i^2 = 1\otimes \hat{C}_i^2 ,\quad U_i^*=1\otimes\hat{U}_i^*,
\qquad i\ge 2.
\]
Hence the family $\{\hat{U}_i , \hat{C}_i ,\, i\ge 2\}$ satisfies
relations~(\ref{align:dsdp}). Moreover, the family $\{\hat {C}_i ,
\hat{U}_i\}_{i=2}^d$ is irreducible if and only if the family $\{C_i , U_i\}_{i=1}^d$ is such.

Consider now the case where $\sigma (C_1^2)=\{{1}/({1-q})\}$. In
this case,
\[
C_1^2 = ({1-q})^{-1} I ,\quad U_1 = \lambda I ,\qquad |\lambda| = 1
\]
(since $U_1$ commutes with all operators of the representation).
Therefore,
in this case we deal with the ``direct product'' of $d-1$ copies of $q$-CCR algebras.
Let us introduce, for convenience, the operators on $l_2(\mathbb N)$
given by
\[
{S}=
\begin{pmatrix}
  0    &       &       & \\
  1 &  0     &       & \\
    &   1   & 0 &       \\
      &       &\ddots &\ddots
\end{pmatrix},
\quad
{C}= 
\begin{pmatrix}
f(0)            &       & \\
  & f^2(0)            & \\
             &      &\ddots
\end{pmatrix}.
\]

Combining the results obtained above and applying the induction in $d$, we can
formulate the following proposition.

\begin{proposition}
Fix a subset \/$\Phi =\{1\le i_1< i_2 <\cdots< i_k\le d\ ,
\ 1\le k\le d\}$. To each such subset, associate the following irreducible
representation\textup:
\begin{gather*}
C_i^2 =
\bigotimes_{j<i,j\not\in\Phi}I\otimes{C}\otimes
\bigotimes_{j>i,j\not\in\Phi}I ,\qquad i\not\in\Phi, 
\\
U_i^*=
\bigotimes_{j<i,j\not\in\Phi}I\otimes{S}\otimes
\bigotimes_{j>i,j\not\in\Phi}I ,\qquad i\not\in\Phi ,
\\
U_i = \lambda_i I ,\quad C_i^2 = ({1-q})^{-1}I ,\qquad i\in\Phi.
\end{gather*}
Then all irreducible bounded representations of the ``direct
 product'' of\/ $d$
copies of  the \/ $q$-CCR algebras can be obtained in such a way. Moreover,
representations are equivalent if and only if they correspond 
to the same $\Phi$.
\end{proposition}

\noindent\textbf{4.} It is obvious that one will classify
irreducible representations of the $*$-algebra
\[
\mathfrak{A}=\mathbb{C}\bigl<\mathfrak{A}_i \mid
[\mathfrak{A}_i,\mathfrak{A}_j]=0,\, i,j=1,\dots ,d\bigr>
\]
  in such a way. Here
\[
\mathfrak{A}_i=\mathbb{C}
\bigl<x_i, x_i^* \mid x_i^* x_i = f_i (x_i x_i^*)\bigr>
,\qquad i=1,\dots ,d,
\]
and $f_i(\cdot)\colon\mathbb{R}\mapsto\mathbb{R}$ is a one-to-one
measurable mapping such that the dynamical system $(f_i, \mathbb{R})$ has
a measurable section.

\subsection{``Triangular'' dynamical systems.}
%\label{triang}
\label{sec:2.4.2}

\textbf{1.} We call a dynamical system defined by a family of mappings
${F}_i$ triangular if
\[
{F}_i (\vec{\lambda})=(\lambda_1,\dots ,\lambda_{i-1},
 f_i(\lambda_1,\dots ,\lambda_i),q_{ii+1}\lambda_{i+1},\dots ,
q_{id}\lambda_d),
\]
$q_{ij}\in\mathbb{R}$, $j=i+1$, \dots , $d$,
where $f_i\colon\mathbb{R}^i\mapsto\mathbb{R}^i$ are measurable
functions.
Here we give examples of $*$-algebras connected
with ``triangular'' dynamical systems. These are  $\mu$-CCR
and $\mu$-CAR algebras connected with a well-known twisted CCR and
twisted CAR algebras constructed by 
Pusz and Woronowicz (see \cite{pusz_anti,pw,jorg}).

In the examples that follow, we show that, if the dynamical system
connected to
the operator family is
``triangular'', then the  description of classes of irreducible
representations can be obtained inductively applying only the one-dimensional
technique.

\medskip\noindent\textbf{2.}   $\mathbf{\mu}$-CCR algebra.
Here we study irreducible representations of the Wick algebra
connected
with the twisted CCR algebra of Pusz--Woronowicz.

Let us consider the following $*$-algebra:
\begin{align}\label{align:muccr}
\mathbb{C}\Bigl<x_i, x_i^* \mid x_i^* x_i & =  1+
\mu^2 x_i x_i^* - (1- \mu^2)
\sum_{j<i} x_j x_j^*, i=1,\dots, d,
\notag
\\
&x_i^* x_j  =  \mu x_j x_i^* ,\, \mu\in [0 , 1],\, i\neq j
\Bigr>
\end{align}

We show that, as in the previous example, the additional relations
$x_i x_j = \mu x_j x_i$,  $i>j$, hold automatically for  any
bounded representation of the $\mu$-CCR algebra.

\begin{proposition}\label{ccr-klein}
Let $\pi(\cdot)$ be an irreducible bounded representation 
of the $\mu$-CCR algebra. Then
$\pi (x_i x_j - \mu x_j x_i)=0$, $i>j$.
\end{proposition}

\begin{proof}
Denote $X_{ij}=\pi( x_j x_i - \mu x_i x_j)$, $ j>i$. Then it is
easy to obtain using the commutation rules (\ref{align:muccr}) that
     \begin{align*}
     X_{ij}^* X_{ij} & = \mu^6 X_{ij}X_{ij}^* -
\mu(1-\mu^2)\sum_{r<j}X_{ir}X_{ir}^*\\
     & - \mu^4(1-\mu^2)\sum_{j<s<i}X_{sj}X_{sj}^*+\mu^2(1-\mu^2)
\sum_{r<j<s<i}X_{sr}X_{sr}^*\\
     & - \mu^4(1-\mu^2)\sum_{r<j}X_{jr}X_{jr}^*
      + (1-\mu^2)^2(1+\mu^2)\sum_{r<s<j}X_{sr}X_{sr}^*.
     \end{align*}
This expression shows that $X_{ij}^*X_{ij}$ consists of the term $\mu^6
X_{ij}X_{ij}^*$ and the ones that have the index less than $(i,j)$ with respect to the
lexicographic ordering. Then the induction on the lexicographic ordering and the
fact that the relation $x^* x = q x x^*$ does not have non-trivial bounded
representations finishes the proof of proposition.
\end{proof}

This means that, as far as bounded representations 
are concerned, no information is
lost if the $\mu$-CCR relations are replaced with the complete twisted
CCR relations:
\begin{align*}
x_i^* x_i & =  1+ \mu^2 x_i x_i^* - (1- \mu^2)
\sum_{j<i} x_j x_j^*, \qquad i=1,\dots, d,
\\
x_i^* x_j & =  \mu x_j x_i^* ,\qquad i\neq j.
\\
x_j x_i & = \mu x_i x_j,\qquad i<j,\quad \mu\in [0 , 1].
\end{align*}
Let us now, as usual, consider the polar decomposition
$\pi(x_i^*) = U_i C_i$, $i=1$, \dots, $d$. Then the relations can be rewritten
in an equivalent form,
\begin{align}\label{gather:dsmucc}
     C_i^2 U_i^* &= U_i^*\Bigl(1+\mu^2 C_i^2 + \sum_{k<i} C_k^2\Bigr),
\qquad i=1, \dots, d,\notag
\\
     C_i^2 U_j^* &= U_j^*C_i^2 ,\qquad i<j ,\notag  %\ \ \ \ \ (*)  
\\
     C_i^2 U_j^* &= \mu^2 U_j^*C_i^2 ,\qquad i>j.
\end{align}
It follows from these relations that all\/ $U_i$ are co-isometries. It is also
easy to see  that the isometry $U_i^*$ acts non-trivially on the
spectrum of\/ $C_j^2$, $ i<j$, and the corresponding dynamical system
is triangular.

Let us classify the irreducible representations of these operator relations.
For an irreducible representation, the spectrum of $C_1^2$ coincides with
one of the two orbits:
\begin{enumerate}
\item
$\sigma (C_1^2)=\{ f^n(0)\ ,\ n>0\}$;
\item
$\sigma(C_1^2)=\{ {1}/({1-\mu^2 })\}$.
\end{enumerate}
Here $f(x)=1+\mu^2 x$, $ x\in\mathbb{R}$.
First, consider the case where $\sigma (C_1^2)=\{ f^n(0) ,\, n>0\} $.
In this case the operators $C_1^2$ and $U_1^*$ are
unitarily equivalent  to the following operators:
\begin{gather*}
C_1^2=
\begin{pmatrix}
f(0)     &       &        \\
  & f^2(0)     &        \\
            &      &\ddots
\end{pmatrix}
\otimes I,
\quad
U_1^*= 
\begin{pmatrix}
  0    &       &       & \\
  1 &  0     &       & \\
    &   1   & 0 &       \\
      &       &\ddots &\ddots
\end{pmatrix}
\otimes I,
\\
C_i^2=
\begin{pmatrix}
 1     &       &       & \\
  & \mu^2     &       & \\
     &     & \mu^4 &       \\
      &       &      &\ddots
\end{pmatrix}
\otimes \hat{C}_i^2,
\quad
U_i^*= 
I\otimes \hat{U}_i^*,\qquad i=2, \dots, d,
\end{gather*}
where the families $\{\hat{C}_i^2\}$, $\{\hat{U}_i\}$ satisfy the relations
\eqref{gather:dsmucc} with $d-1$ generator.

Consider now the case $\sigma(C_1^2)=\{ {1}/({1-\mu^2 })\}$.
Here $U_1$ is a unitary operator, and it is easy to deduce from
$C_i^2 U_i^* = \mu^2 U_i^* C_i^2$ that
$C_i^2 = 0$ for a bounded representation. This follows from the
fact that if\/ $U_1$ is unitary,
then $\sigma(C_i^2)$ is invariant under multiplication by
$\mu^2$, $\mu^{-2}$. Consequently, the spectrum of $C_i^2$,
$i=2$, \dots , $d$,
is bounded if and only if $\sigma(C_i^2)=\{0\}$ and, since $\ker U_i =
\ker C_i$, we have that $U_i = 0$. The operator $U_1$ then is in the center of
the representation, and $U_1^* = \lambda I$, $|\lambda| = 1$.

To obtain the final result, we combine these two cases.
\begin{proposition}
Fix a number $i$, $1\le i \le d$, and consider the following
representation of the $\mu$-CCR algebra\textup:
     \begin{align*}
     C_j^2  & = \bigotimes_{k=1}^{j-1}d(\mu)\otimes d(f)
\otimes\bigotimes_{k=j+1}^{i-1}I ,\qquad j<i, 
\\
     U_j^* & = \bigotimes_{k=1}^{j-1} I\otimes S
\otimes\bigotimes_{k=j+1}^{i-1}I ,\qquad j<i, 
\\
     C_i^2 & = \frac{1}{1-\mu^2}\bigotimes_{k=1}^{i-1}d(\mu),
\\
     U_i & = \lambda_i I, \quad
     C_j  = U_j = 0  ,\qquad j>i,
     \end{align*}
where
\[
d(\mu)= 
\begin{pmatrix}
 1     &       &       & \\
  & \mu^2     &       & \\
     &     & \mu^4 &       \\
      &       &      &\ddots
\end{pmatrix},
\quad
d(f)= 
\begin{pmatrix}
f(0)         &       & \\
  & f^2(0)            & \\
             &      &\ddots
\end{pmatrix},
\]
$S$ is the unilateral shift in $l_2(\mathbb{N})$, and $\lambda_i\in\mathbb{C}$,
$|\lambda_i|=1$.

Then all irreducible representations of the $\mu$-CCR algebra
coincide with the ones constructed for some $i$. Moreover, two representations are
equivalent if and only if they correspond to the same $i$ and $\lambda_i$.
\end{proposition}

\noindent\textbf{3.} $\mu$-CAR algebra.
 Now consider the anticommutative case.
By the $\mu$-CAR algebra we call the following $*$-algebra:
\begin{align*}
\mathbb{C}\Bigl<x_i, x_i^* \mid x_i^* x_i & =1-x_i x_i^*
- (1-\mu^2)\sum_{j<i}x_j x_j^*,\, i=1,\dots,d,
\\
&x_i^* x_j  = -\mu x_j x_i^*,\, i\ne j\Bigr>.
\end{align*}
\begin{remark}
Let us note that any representation $\pi(\cdot)$ 
of the $\mu$-CAR algebra is bounded, since
  $\pi(x_i^*x_i)\ge 0$ for any $i$, and, therefore,
\[
\mathbf{1}-\pi(x_i x_i^*)
- (1-\mu^2)\sum_{j<i}\pi(x_j x_j^*)\ge 0,
\]
which implies that $\Vert\pi(x_i x_i^*)\Vert\le 1$. Then $\pi(x_i)$ is bounded
for any $i=1$, \dots, $d$.
\end{remark}

To reduce the $\mu$-CAR algebra to a dynamical form, we have to find
additional relations for the generators $x_i$, $ x_j$ so that these
relations would be compatible with the
defining relations. For Wick algebras such relations are described by
Wick ideals, in particular, quadratic Wick ideals 
(see Section~\ref{sec:2.4.6}).
The largest quadratic Wick ideal of the $\mu$-CAR algebra is generated by the
following family of elements:
\[
x_i x_j +\mu x_j x_i,\quad 1\le j<i\le d,\quad\text{and} \quad
x_i^2,\quad i=1,\ldots ,d.
\]
According to the previous example, 
we have to show that the generators of this ideal are annihilated in
any irreducible representation of the $\mu$\nobreakdash-CAR algebra. However,
this is not true for the $\mu$-CAR algebra.
The largest quadratic ideal is very large. So, consider another
quadratic Wick ideal,
\[
\hat{\mathcal{I}}_2 =\bigl< x_i x_j +\mu x_j x_i,\, 1\le j<i\le d,
; x_i^2,\, i=1,\ldots ,d-1\bigr>.
\]

\begin{theorem}\label{car-klein}
For any irreducible representation $\pi(\cdot)$ of the 
$\mu$\nobreakdash-CAR algebra 
$\pi(\hat{\mathcal{I}}_{2})=\{0\}$ holds.
\end{theorem}

\begin{proof}
Let us denote  $X_i = \pi (x_i)$, $A=\pi(x_{1}^{2})$,
$B=\pi (x_{2}x_{1} +\mu x_{1}x_{2})$. Now we prove that $A=B=0$.
It is easy to see from the defining relations that
\begin{align}\label{align:muc}
A^{*}A=A A^{*},
\quad
A^{*}X_{k}=\mu^{2}X_{k}A^{*},\qquad k>1.
\end{align}
Since $A$ is a normal operator, we can use the Fuglede--Putnam
theorem and obtain the following relations:
\[
A X_{k}=\mu^{2}X_{k} A,\qquad k>1.
\]
(It is obvious that $A X_{1}=X_{1} A$).
Thus, in any irreducible representation, either $A=0$ or $\ker A =\{ 0\}$
(because $\ker A$ is an invariant subspace).
Let $\ker A =\{0\}$. Then we have:
\begin{gather}\label{gather:pos}
B^{*}B=\mu^{2} B B^{*} + (1-\mu^{4}) (1+\mu^{2}) A A^{*},\notag
\\
A^{*}B=\mu^{2} B A^{*}, \quad A B=\mu^{2}B A.
\end{gather}
It follows from the condition $\ker A=\{ 0 \}$ that $A A^{*}>0$. 
Equation~(\ref{gather:pos}) implies that $B^{*}B>0$.
Let us consider the polar decomposition
$ B^*  = W T$, $\ker W =\ker T$, $ T\ge 0$, $ T^2 = B B^*$,
and, since $B^* B>0$, we have that $W$ is a co-isometry.

Now we can rewrite  relations~(\ref{gather:pos}) in an equivalent
form:
\begin{gather}\label{align:wt}
T^2 W^* = W^* \bigl(\mu^2 T^2 + (1-\mu^4)(1+\mu^2)A A^*\bigr),\notag
\\
A W^*  =\mu^2 W^*A,
\quad
A T  = T A .
\end{gather}
These equalities and relations~(\ref{align:muc})
imply that the spectrum of $A$ coincides with the set
$\{ \lambda\mu^{2n}x_1,\, n\in\mathbb{N}\cup\{0\}\}$ for some
$\lambda\in \mathbb{C}$,
$|\lambda| = 1$, $x_1 > 0$. Since $W^*$ is an  isometry, the
eigenvalues of the operator $A$ have the same multiplicity, and
choosing the corresponding basis in the 
representation space we can write the operators
$A$ and $W$ in the following form:
\[
A =\begin{pmatrix}
\lambda x_1 I& &  \\
& \lambda x_1\mu^2 I  & \\
 & & \ddots
\end{pmatrix},
\quad
W = \begin{pmatrix}
0 & I & & \\
& 0 &I & \\
& & \ddots & \ddots 
\end{pmatrix} .
\]
The condition $T A =A T$ then gives that
\[
T =\begin{pmatrix}
T_0 & & & \\
& T_1 & & \\
& & T_2 & \\
& & & \ddots
\end{pmatrix}.
\]
Since  $\ker W = \ker T$, we have that $T_0 = 0$. Moreover,
it is easy to obtain from~(\ref{align:wt}) that
\[
T_{n}  = x_1 \,(1+\mu^{2})\, \mu^{n-1}(1-\mu^{2n})^{ {1}/{2}},
\qquad n\geq 1.
\]
Since  $X_{1}A=A X_{1}$, $X_{1}^{*}A=A X_{1}^{*}$, we get
\[
X_1=
\begin{pmatrix}
b_1 & & \\
& b_2 & \\
& & \ddots
\end{pmatrix}.
\]
Let us write the operator $B=TW^{*}$ in the matrix form,
\[
B= 
\begin{pmatrix}
0     &       &       & \\
T_{1} & 0     &       & \\
      & T_{2} & 0     & \\
      &       &\ddots &\ddots
\end{pmatrix}.
\]
The condition $X_{1}^{*}B=\mu B X_{1}^{*}$ expressed in terms of the
matrix coefficients takes the following form:
\[
b_{n+1}^{*} T_{n} =\mu \,T_{n} b_{n}^{*} .
\]
Let us note that $x_1\neq 0$, hence, $T_n\neq 0$, and we have
$b_{k}=\mu^{k-1} b_{1}$.
From the relation
$x_{1}^{*} x_{1}= 1-x_{1} x_{1}^{*} $
it follows that
\[
b_{1}^{*}b_{1}=1-b_{1}b_{1}^{*},
\quad
\mu^{2}b_{1}b_{1}^{*}=1-\mu^{2}b_{1}b_{1}^{*}.
\]
These equalities are compatible if and only if
$\mu^{2}=1$, which is impossible. Hence, $x_1 = 0$, $A=0$, and
 relation~(\ref{gather:pos}) yields that
$B^* B = \mu^2 B B^* $.
Since $B$ is a bounded operator,  $B=0$.

Let us denote $B_{k}=\pi (x_{k} x_{1} + \mu x_{1}x_{k})$, $2< k\le d$.
From the defining relations we obtain:
\[
B_{k}^*B_{k} = \mu^{2} B_{k} B_{k}^*
 + \mu^{2} (1-\mu^{2}) \sum_{1<i<k} B_{i} B_{i}^*
 + (1+\mu^{2}) (1-\mu^{4}) A A^*.
\]
It is easy to see that the induction in  $d$ gives that
$B_k = 0$, $k=3$, \dots, $d$.
Thus the operators $X_i$ can be decomposed in the tensor product,
 \[
 X_{1}= \begin{pmatrix}
 0 & 0 \\
 1 & 0
 \end{pmatrix}
 \otimes I,
 \quad
 X_{k}= \begin{pmatrix}
 1 & 0 \\
 0 & -\mu
 \end{pmatrix}
 \otimes\tilde{A}_{k},\qquad k>1,
 \]
where $\{\tilde{X}_{k},\ k>1\}$ satisfy the $\mu$-CAR relations with $d-1$
generator. Moreover, the set $\{X_i\}$ is irreducible if and only if
the set $\{\tilde{X}_k\}$ is irreducible,
\begin{gather*}
X_{j}^{2}=0,\Leftrightarrow \tilde{X}_{j}^{2}=0,\qquad j\ge 2,
\\*
X_{j}X_{i}+\mu X_{i}X_{j}=0\Leftrightarrow
\tilde{X}_j\tilde{X}_i +\mu\tilde{X}_i\tilde{X}_j = 0,
\qquad 2\le i<j\le d.
\end{gather*}
The proof is completed by induction in $d$.
\end{proof}


\subsection{Operator relations connected with many-dimen\-si\-onal dynamical
systems}
%\label{oprel}
\label{sec:2.4.3}

In this section, we consider families of operators satisfying a general
class
of relations  whose representations can be described in terms of orbits of some
dynamical system
acting on the spectrum of a commuting sub-family.

We consider representations of a family of operator relations
satisfied by the 
operators $X_j$, $j=1$, \dots, $n$, of the following form:
\begin{align}
X_j^*X_j & = F_j(X_1X_1^*, \dots, X_nX_n^*) ,\notag
\\
X_j^*X_k & = \mu_{jk}\, X_k X_j^*, \notag
\\
X_jX_k& = \lambda_{jk} \, X_k X_j, \label{rels-multi}
\end{align}
where $F_1(\cdot)$, \dots, $F_n(\cdot) \colon
\mathbb{R}^n \to \mathbb{R}$ are measurable mappings,
$\lambda_{jk}$, $\mu_{jk}>0$, $1\le j,k \le n$. Notice that the last two
relations in \eqref{rels-multi} imply that the operators $X_1X_1^*$,
\dots, $X_nX_n^*$ commute; this explains the meaning of the
functions used in the first relation.

\begin{remark}
Assuming that $X_1$, \dots, $X_n$ are, generally speaking, unbounded
closed densely defined operators which satisfy the relations
\eqref{rels-multi}, we need to take some care in writing relations
between unbounded operators; our approach is to rewrite them in a formally
equivalent form that would involve only bounded operators (this is similar
to the case of unbounded representations of a real Lie algebra, which can
be
described in terms of unitary representations of the corresponding Lie
group).
\end{remark}

Write the polar decompositions, $X_j = C_jU_j$, $j=1$, \dots, $n$,
where $C_j$ are non-negative, $U_j$ are partial isometries, and each $C_j$
is zero on vectors orthogonal to the range of $U_j$.

\begin{lemma}
Let the operators $X_j$, $j=1$,\dots, $n$, be bounded. Then the relations
\eqref{rels-multi} are equivalent to the following\textup:
\begin{align}
C_j^2 U_k& = q_{jk} U_kC_j^2,\qquad j\ne k, \notag
\\
C_j^2 U_j& = U_j F_j(C_1^2,\dots,C_n^2),\qquad
j=1,\dots n, \notag
\\
U_jU_k&=U_kU_j,\quad U_jU_k^*=U_k^*U_j,\qquad
j<k,\label{rels-cu}
\end{align}
where
\[
q_{jk} = \begin{cases}
 \mu_{jk}\lambda_{jk},& j<k,\\
\mu_{jk}\lambda_{jk}^{-1},& j>k.
\end{cases}
\]
Moreover, the operators $(U_j^*)^kU_j^k$, $U_l^m(U_l^*)^m$, $j$, $l=1$, \dots,
$n$;
$k$, $m=1$, $2$,~\dots, form a commuting
family
\textup(in particular, all $U_j$ are centered partial isometries\textup).
\end{lemma}

\begin{proof}
The proof is a rather straightforward calculation involving the fact that
$U_lU_l^*=\sign C_l$.
\end{proof}

According to \cite{ostur}, relations \eqref{rels-cu} can be rewritten in
the following form, involving only bounded operators. Introduce the mappings
of $\mathbb{R}^n$ into itself by
\begin{align}\notag
\bold F_l(\lambda_1,\dots,\lambda_n)
&=(q_{1l}\lambda_1,\dots,q_{l-1\,l}\lambda_{l-1},
\\*
&\qquad F_l\bigl(\lambda_1,\dots,x_n),q_{l+1\,l}\lambda_{l+1},
\dots,q_{nl}\lambda_n\bigr), \label{rels-ds}
\end{align}
$l=1$, \dots, $n$. Then the relations are equivalent to the following
\[
E(\Delta) U_l = U_l E(\mathbf{F}_l^{-1}(\Delta)),
\]
where $E(\cdot)$ is a joint resolution of the identity of the commuting
family $C_1^2$, \dots, $C_n^2$, $\Delta$ ranges over all measurable
subsets of $\mathbb{R}^n$, $l=1$, \dots, $n$. The latter relations
include only bounded operators, and will be used as a precise version of
the relations in the unbounded case.

According to \cite{ostur}, it makes sense to consider such relations
for which \allowbreak $\mathbf{F}_j(\mathbf{F}_k(\cdot))=
\mathbf{F}_k(\mathbf{F}_j(\cdot))$, $j\ne k$, which is equivalent to the
following equalities
\[
F_j(\mathbf{F}_k(\lambda_1,\dots,\lambda_n))= q_{jk}
F_j(\lambda_1,\dots,\lambda_n).
\]
In what follows, we are mostly interested in the case of the second order
relations, i.e., linear functions $F_j(\cdot)$, $j=1$, \dots, $n$. If
\[
F_j(\lambda_1, \dots, \lambda_n) = \sum_{l=1}^n
\phi_{jl}\lambda_l +\alpha_jI, \qquad
j=1,\dots,n,
\]
the conditions are
\begin{gather}
\phi_{jl}(q_{lk}-q_{jk})+\phi_{jk}\phi_{kl}=0,\qquad l\ne j,\ l\ne k, \notag
\\*
\phi_{jk}\phi_{kj}=0,\quad \phi_{jk}(\phi_{kk} - q_{jk})=0, \notag
\\*
\alpha_j(1-q_{jk}) +\alpha_k\phi_{jk}=0,
\label{ds-commute}
\end{gather}
for all $j$, $k=1$\dots, $n$, $j\ne k$.

In what follows, we will assume that the $n$-dimensional dynamical
system generated by the mappings $\mathbf{F}_1(\cdot)$, \dots,
$\mathbf{F}_n(\cdot)$ possesses a measurable section, a measurable set
which meets each orbit at a single point. In this case, for any
irreducible
representation of the relations, the spectral
measure of the commuting family $C_1^2$, \dots, $C_n^2$ is concentrated
on (a subset of) a single orbit, and we can classify all irreducible
representations up to unitary equivalence. In the case of more
complicated dynamical systems without a measurable section, non-trivial
ergodic measures can arise, which gives rise to a  much more complicated
structure of representations, including factor representations not of
type~I etc. Notice that the linear dynamical system of the form
\eqref{rels-ds} always possesses a measurable section.

We proceed with a  more detailed study of the irreducible collections $X_j$,
$j=1$, \dots, $n$, satisfying \eqref{rels-multi}, which correspond to an
orbit $\Omega$. Denote by $\Delta$ the support of the spectral measure
of the commuting family $C_j^2$, $j=1$, \dots, $n$. It is a general fact
that in the basis of eigenvectors of the commuting family, the operators
$X_j$ act as weighted shift operators \cite{ostur}, but we need to take
into account that $C_j\ge0$, and that $U_jU_j^*$ is a projection on the
co-kernel $(\ker C_j^2)^\perp$, $j=1$, \dots, $n$.

\begin{lemma} \label{lemma2}
For any $\mathbf{\lambda} = (\lambda_1, \dots, \lambda_n) \in \Delta$,
we have

\textup{i)} $\lambda_j\ge0$, $j=1$, \dots, $n$\textup;

\textup{ii)} either $\mathbf{F}_j(\mathbf{\lambda})\in \Delta$, or
$(\mathbf{F}_j(\mathbf{\lambda}))_j=0$\textup;

\textup{iii)} similarly, either $\mathbf{F}_j^{-1}(\mathbf{\lambda})
\in \Delta$, or $\mathbf{\lambda}_j=0$.
\end{lemma}

\begin{proof}
i) Indeed, since $C_j^2\ge 0$, we have $\lambda_j\ge0$, $j=1$, \dots, $n$.

ii) If $\mathbf{F}_j (\mathbf{\lambda}) \notin \Delta$, then $U_j
e_{\mathbf{\lambda}} =0$, where $e_{\mathbf{\lambda}}$ is the
basis eigenvector of
the commuting family corresponding to the joint eigenvalue
$\mathbf{\lambda}$.
Then we also have $U_j U_j^* e_{\mathbf{F}_j(\mathbf{\lambda})}=0$ and
\[
C_j^2e_{\mathbf{F}_j(\mathbf{\lambda})} =
(\mathbf{F}_j(\mathbf{\lambda}))_j\,
e_{\mathbf{F}_j(\mathbf{\lambda})} =0,
\]
which implies that $\mathbf{F}_j (\mathbf{\lambda}))_j =0$.

iii) Similarly, if $\mathbf{F}_j^{-1} (\mathbf{\lambda})\notin \Delta$,
then $U_j^*\, e_{\mathbf{\lambda}} =0$. Then $U_jU_j^*\,
e_{\mathbf{\lambda}}
=0$, and $C_j^2\, e_{\mathbf{\lambda}} = \lambda_j
e_{\mathbf{\lambda}}=0$, which gives
$\lambda_j=0$.
\end{proof}

\begin{corollary}
If for some $\mathbf{\lambda}=(\lambda_1,\dots,\lambda_n)
\in \Omega$, we have $\lambda_j>0$
and $(\mathbf{F}_j(\lambda))_j <0$, then $\mathbf{\lambda}\notin\Delta$.
This
condition implies that irreducible representations correspond only to the
orbits such that $\lambda_j >0$ implies
$(\mathbf{F}_j(\mathbf{\lambda}))_j\ge0$,
$(\bold{F}_j^{-1}(\mathbf{\lambda}))_j\ge0$. Notice also that \eqref{rels-ds}
implies that it follows from $\lambda_j>0$ that
$(\mathbf{F}_k(\mathbf{\lambda}))_j >0$ for
$k\ne j$.
\end{corollary}

Consider possible types of orbits and describe the corresponding irreducible
representations of \eqref{rels-multi}.

\begin{theorem}
Any irreducible representation can be realized in the space
$l_2(\Delta)$. For any\/ $l=1$, \dots, $n$  one of the
following cases holds\textup:

$a)$. The mapping\/ $\mathbf{F}_l(\cdot)$ possesses a stationary point
$\mathbf{\lambda}\in \Delta$ \textup(in this case all other points are also
stationary\textup). If $\lambda_l=0$, then $X_l=0$\textup;
otherwise, the operator $X_l$ has
the form
\[
X_l e_{\mathbf{\lambda}} = \beta_l \, \lambda_l \, e_{\mathbf{\lambda}},
\]
where $\beta_l$ is a parameter with the absolute value equal to one\textup;

$b)$. The mapping $\mathbf F_l(\cdot)$ does not have stationary points. In this case
the operator $X_l$ has the form
\begin{equation}
X_le_{\mathbf{\lambda}} = F_l(\mathbf{\lambda})\,
e_{\mathbf{F}_l(\mathbf{\lambda})}.
\end{equation}
The kernel of the operator $X_l$ is generated by vectors
$e_{\mathbf{\lambda}}$ such that $F_l(\mathbf{\lambda})=0$;
the kernel of $X_l^*$ is
generated by vectors $e_{\mathbf{\lambda}}$ for which $\lambda_l=0$.
\end{theorem}

\begin{proof}
The proof is essentially based on the following statements from
\cite{ostur}.

\begin{theorem}
Let the dynamical system on $\mathbb{R}^n$ generated by the mappings
$\mathbf{F}_l$, $l=1$, \dots, $n$, possess a measurable section. Then, for
each irreducible collection of operators $C_j$, $U_j$, $j=1$, \dots,
$n$,
satisfying \eqref{rels-cu}, the following holds.

i. There exists a unique orbit $\Omega$ of the dynamical system of full
spectral measure of the commuting collection $C_j$, $j=1$, \dots, $n$,
$E(\Omega)=1$\textup;

ii. If\/ $\ker U_l = \{0\}$, then the spectral measure is quasi-invariant
with respect to the mapping $\mathbf{F}_l(\cdot)$\textup; in the case of
unitary $U_l$, the measure is also quasi-invariant with respect to
$\mathbf{F}_l^{-1}(\cdot)$\textup;

iii. The joint spectrum of the commuting family $C_j$, $j=1$, \dots,
$n$, is simple.
\end{theorem}

\begin{theorem}
The irreducible collection $C_j$, $U_j$, $j=1$, \dots, $n$, satisfying
\eqref{rels-cu} acts in the space $l_2(\Delta)$, where  $\Delta \subset
\Omega$ is a subset of some orbit $\Omega$ \textup(for unitary $U_l$, $l=1$,
\dots, $n$, $\Delta=\Omega$\textup), by the following formulae
\[
C_l e_{\mathbf{\lambda}} = x_k e_{\mathbf{\lambda}},
\quad U_l e_{\mathbf{\lambda}} =
u_l(\mathbf{\lambda}) \, e_{\mathbf{F}_l(\mathbf{\lambda})},
\]
where $u_l(\mathbf{\lambda})$ are constants which determine the action of\/
$U_l$. The subset $\Delta$ satisfies the following ``boundary
conditions''\textup:
\begin{align}
u_l(\mathbf{\lambda}) & =0, \quad \forall \, \mathbf{\lambda}
\in \Delta \colon \mathbf{F}_l (\mathbf{\lambda}) \notin \Delta, \notag
\\
u_l(\mathbf{F}_l^{-1}(\mathbf{\lambda})) & =0, \quad \forall \,
\mathbf{\lambda} \in \Delta \colon \mathbf{F}_l^{-1}(\mathbf{\lambda})
\notin \Delta,
\end{align}
and is ``connected'' in the following sense\textup: 
$u_l(\mathbf{\lambda}) \ne 0$
for all $\mathbf{\lambda} \in \Delta$, such that
$\mathbf{F}_l(\mathbf{\lambda}) \in
\Delta$, $l=1$, \dots, $n$.
\end{theorem}


Let $\mathbf{\lambda}$ be a stationary point of the mapping
$\mathbf{F}_l(\cdot)$. If $\lambda_l =0$, then for all points
$\mathbf{\lambda} \in
\Delta$, the commutation of $\mathbf{F}_l(\cdot)$ and
$\mathbf{F}_k(\cdot)$ also implies that $\lambda_l =0$. Then $X_l=0$. If
$\lambda_l \ne 0$, then also $\lambda_l \ne 0$ for all
$\mathbf{\lambda} \in \Delta$. In this
case, the operator $U_l$ commutes with all the operators $X_j$,
$X_j^*$, and,
therefore, is a multiple of the identity.

In the case where the mapping $\mathbf{F}_l(\cdot)$ does not have
stationary points, the operator $U_l$ is unitarily equivalent to the shift
operator; taking into account that $X_lX_l^* =C_l^2$, we get the needed
formula for $X_l$.
\end{proof}


\subsection[Representations of the non-standard real
quantum
sphere]{Representations of the non-standard real quantum
sphere}
%\label{sub:qsphere}
\label{sec:2.4.4}

\textbf{1.} The algebra of functions on the non-standard three-dimensional
real quantum
sphere (see~\cite{noumi}) is an associative $*$-algebra generated by
the elements $x$, $y$,  $u$, $v$, $c$, and $d$ satisfying the
relations:
\begin{gather}
ux=qxu,\quad  vx=qxv,\quad yu =quy,\quad
yv=qvy,\notag
\\*
vu-uv = (q-q^{-1})\, d,\quad xy - q^{-1} uv =
yx -qvu =c+d,\notag 
\\*
dx=q^2 xd,\quad dv= q^2 vd ,\quad ud= q^2
du,\quad yd = q^2 dy,\label{sphere}
\end{gather}
with  $c$ lying in the center, and the 
involution defined by $x^* =y$, $u^*=-q^{-1}v$,
$c^*=c$, $d^*=d$. For the generators $x$, $u$,
$c$, $d$, the relations \eqref{sphere} have the form \eqref{rels-ds}, and
are equivalent to the following relations
\begin{gather}
ux=qxu,\quad u^*x = qxu^*, \notag
\\
u^*u = q^{-2}uu^* - (1-q^{-2})(xx^*-c),\quad
x^*x = q^2xx^* + (1-q^2)c,\notag
\\
d=xx^* +uu^*- c.\label{qsphere}
\end{gather}

\noindent\textbf{2.} The corresponding dynamical system on $\mathbb{R}^2$
is generated by the mappings
\begin{align*}
\mathbf{F}_1(\lambda_1, \lambda_2)& = (q^2 \lambda_1 + (1-q^2)c,
q^2\lambda_2),
\\
\mathbf{F}_2(\lambda_1, \lambda_2)& = (\lambda_1,q^{-2} \lambda_2
-(1-q^{-2}) (\lambda_1-c)),
\end{align*}
which satisfy, as one can see, the conditions \eqref{ds-commute}.

Any orbit of the dynamical system consists of the points
\begin{align*}
\mathbf{\lambda}^{(kl)}&=\mathbf{F}_1^k
(\mathbf{F}_2^l(\mathbf{\lambda})) 
\\
&=(q^{2k}\lambda_1 +(1-q^{2k})\, c, q^{2(k-l)} \lambda_2 -
q^{2k}(1-q^{-2l})(c- \lambda_1)),
\end{align*}
where $\mathbf{\lambda}=(\lambda_1, \lambda_2)$, $\mathbf{F}_l^k(\cdot)$
is the  $k$-th iterations of the mapping $\mathbf{F}_l(\cdot)$.

The mapping $\mathbf{F}_1(\cdot)$ has a single stationary point
$(c,0)$, Stationary points of the mapping $\mathbf{F}_2(\cdot)$ have the
coordinates $(\lambda, c-\lambda)$. There are no periodic points of
$\mathbf{F}_1(\cdot)$, $\mathbf{F}_2(\cdot)$,  apart from the
stationary ones.

\medskip\noindent\textbf{3.}
The following is a list of orbits, the corresponding sets $\Delta$,
and the corresponding irreducible representations.

1) A single stationary point $(c,0)$. For $c=0$, this orbit
corresponds to the trivial representation $X=U=0$, and for $c>0$, to the family
of one-dimensional irreducible representations $U=0$, $X=\alpha\,c$,
where $|\alpha|=1$. Therefore, the set of one-dimensional
representations is parametrized by points of the cone.

2) If $c>0$, then there exists a unique orbit that is invariant with
respect to the mapping $\bold F_2(\cdot)$ and satisfies the
conditions of Lemma~\ref{lemma2}. Namely, this orbit contains the point
$\bold \lambda=(0,c)$. The set $\Delta$ consists of the points
$\bold \lambda^{(k)}= ((1-q^{2k})c,q^{2k}c)$, and the irreducible
representation corresponding to this orbit is realized on the space
$l_2$ by the formulae:
\begin{align}
Xe_k& = ({(1-q^{2k})\, c})^{1/2} e_{k+1},\notag
\\*
Ue_k& = \alpha \,q^{k-1}\sqrt{c}\, e_k, \qquad
|\alpha|=1, \ k=1,2,\dotsc.\notag
\end{align}
The representations of this series are defined by the parameters
$c>0$, $\alpha\in S^1$.

3) If $c>0$, then the orbits that contain the points
$(c,y)$, $y>0$ lie in the first quadrant. They consist of the points
$(c, q^{2n})$, $n\in \mathbb Z$, and the set of these orbits is naturally
parametrized by points of real circle $S^1$. The corresponding
irreducible representations are realized on $l_2(\mathbb Z)$ by the
formulae:
\[
X e_k = \sqrt{c}\, e_{k+1},\quad
U e_k = \lambda \,q^{k}\, e_{k-1}, \qquad k \in \mathbb Z.
\]
The parameters $\lambda \in (q^2,1]$,  $c\ge0$,
determine the set of representations of this series.

4) Consider now the orbits containing the points
$(0, y)$, $y>c>0$. They are described
by $\lambda \in (c+q^2, c+1]\approx S^1$. The set $\Delta$ consists of
the points
$$
\bold x^{(k,l)}=\bigl((1-q^{2k})\, c, q^{2(k-l)}
\lambda+q^{2k}(1-q^{-2l})\, c\bigr),
\qquad k\ge0, \, l \in \mathbb Z.
$$
The irreducible representation corresponding to this orbit is realized
on $l_2(\mathbb N\times\mathbb Z)$ by the formulae:
\begin{align*}
Xe_{kl}&= ({(1-q^{2k})\, c})^{1/2}\,
e_{k+1,l},
\\
Ue_{kl}&=\bigl({q^{2(k-l-1)}\lambda
+q^{2k-2}(1-q^{-2l})\, c}\bigr)^{1/2}\, e_{k,l+1}, 
\\
&\qquad
k=1, 2, \dotsc;\ l\in \mathbb Z.
\end{align*}
For any $c\ge0$ and $\lambda \in (c+q^2,
c+1]$ there exists a unique representation of such a form.

5) The following series of representations that depend on a continuous
parameter corresponds (in the case where $c\ge0$) to the orbits containing
the points $(z,0)$, $z>c$. The set parameterizing these orbits can be
chosen to be the line segment $(c+q^2, c+1]\ni\lambda$. The set
$\Delta$ has the form $\Delta=\{\bold
\lambda^{(kl)}=(c-q^{2k}(c-\lambda), q^{2k}(1-q^{-2l})
(c-\lambda)), \ l\ge0, k\in \mathbb Z\}$, and the irreducible
representation that corresponds to $c,\lambda$
acts in
$l_2(\mathbb Z\times \mathbb N)$ by:
\begin{align}
X e_{kl}& =
({c-q^{2k+2}(c-\lambda)})^{1/2}\,e_{k+1,l},\notag
\\
Ue_{kl}&=q^k\,\bigl({(1-q^{-2l})(c-\lambda)}\bigr)^{1/2}\,e_{k,
 l+1},\notag\\
&\qquad k\in \Bbb Z, \ l=1,2,\dotsc.\notag
\end{align}
Unlike the previous series of orbits, in the case where $c<0$ the orbit
containing the point $(0,0)$ determines the following 
representation on
$l_2(\mathbb N\times \mathbb N)$:
\begin{align}
X e_{kl}&= ({(1-q^{-2k+2})\,c})^{1/2}\,e_{k-1,l},
\notag\\
U e_{kl}&= q^{-k}\,({(1-q^{-2l})\, c})^{1/2}\,
e_{k,l+1},\qquad k,l=1,2,\dotsc.
\notag
\end{align}

6) Finally, for $c>0$, consider the orbit containing $(0,0)$. The set
$\Delta$ has the form
$\Delta=\{\bold \lambda^{(kl)}=((1-q^{2k})\,
c,q^{2k}(1-q^{-2l})\, c), \
k\ge 0,\allowbreak\, l\le -1\}$. 
The representation acts on $l_2(\mathbb N\times \mathbb N)$:
\begin{align}
X e_{kl}&= ({(1-q^{2k})\, c})^{1/2}\, e_{k+1,
l},\notag\\
U e_{kl}&= q^{k-1}({(1-q^{2l})\,
c})^{1/2}\,e_{k,l-1},\notag\qquad k,l=1,\dotsc.
\end{align}
\begin{remark}
We note that representations described above include representations
by unbounded operators. Operators of representations in the cases 1),
2), and 6) are bounded.
\end{remark}

\noindent\textbf{4.}
It is easy to see that, for $c>0$,  using the following substitution
of variables,
\[
y_1 =((1-q^2)\,c)^{-{1}/{2}}x,\quad
y_2^* =((1-q^2)\,c)^{-{1}/{2}}u,
\]
the relations (\ref{qsphere})  can be reduced to the twisted CCR algebra of
Pusz--Woronowicz:
\begin{gather*}
y_1^* y_1  =1+q^2 y_1 y_1^*,\quad
y_2^* y_2  =1+q^2 y_2 y_2^* + (1-q^2) y_1 y_1^*,\\
y_1^* y_2  = q\, y_2 y_1^*,\quad
y_2 y_1  = q\,y_1 y_2.
\end{gather*}


\subsection[Heisenberg relations for the quantum $E(2)$\break
group]{Heisenberg relations for the quantum $E(2)$
group}
%\label{sub:Heis}
\label{sec:2.4.5}

\textbf{1.} In this section we study  $*$-representations
of the involutive
algebra $\cal A$ generated by the so-called Heisenberg relations
\cite{wor}. These relations connect generators of a quantum
deformation of $E(2)$ group and those of its dual.

First we recall the definition of
the Heisenberg relations for $E_q(2)$. Then we
prove some auxiliary assertions which allow us to rewrite the relations
in a more convenient  form for us. The relations which are obtained can be
considered in the framework of the general formalism developed above,
which provides a way of dealing  with unbounded representations. Finally,
a complete list of irreducible $*$\nobreakdash-representations of the Heisenberg
relations is given.

\medskip\noindent\textbf{2.}  The quantum deformation of the group of motions was
introduced and investigated in \cite{vakskor,woraff,wor}.
The algebra  of  ``functions on $E_q(2)$''  is
an algebra generated by the elements $v$ and $n$, $v$ being unitary
and $n$ being normal, which satisfy the relation
\begin{equation}\label{e2}
vn=qnv, \qquad q>0,
\end{equation}
subject to an additional condition: the spectrum of the operator $|n|$ lies in
$\{q^n ,\, n\in
\mathbb{Z}\}$. Although this condition is rather natural, we will not
assume it in our study of the representations; one can easily pick
representations satisfying this condition from a wider list of representations.

On the other hand, using the comultiplication in $E_q(2)$, the algebra
of ``continuous functions on $\hat E_q(2)$'', where
$\hat E_q(2)$ denotes the Pontryagin dual of $E_q(2)$, was
constructed
and investigated in \cite{wor}. This algebra is generated by
the elements $N$ and $b$ ($N$ is self-adjoint, $b$ is normal) with
the relation
\begin{equation}\label{coe2}
Nb=b\,(N+I).
\end{equation}


If one considers both  algebras represented on the same
Hilbert space, some natural relations between the generators
$v$, $n$ and the generators $N$, $b$  (the Heisenberg relations,
see~\cite{wor})
appear. These relations are:
\begin{gather}
vN=(N-I)\,v,\quad vb=q^{-1/2}bv, \quad nN=(N+I)\,n, \notag
\\
bn^*=q^{1/2}n^*b,\quad nb-q^{1/2}bn=
(1-q^2)\,q^{-\frac{N+I}{2}}v. \label{heis}
\end{gather}

\noindent\textbf{3.} In the sequel we consider a $*$-algebra $\cal A$ generated by
the elements $v$, $n$, $N$, and $b$ such that $v$ is unitary,
$N$ is self-adjoint, $n$ and $b$ are normal, and the generators
satisfy relations \eqref{e2}, \eqref{coe2}, \eqref{heis}.

Instead of $b$ and $N$, introduce the new generators, $d=bv^*$ and
$M=(1-q^2)\,q^{-({N+I})/{2}}$. Then the relations are:
\begin{gather*}
vn=qnv,\quad nn^*=n^*n,\quad Md=q^{1/2}dM, \quad d^*d =
q^{-1}dd^*,
\\
vM = q^{1/2} Mv,\quad vd= q^{-1/2}dv,\quad nM= q^{-1/2} Mn,
\\
nd^*=q^{-1/2}d^*n,\quad nd-q^{3/2}dn = M.
\end{gather*}

\noindent\textbf{4.} Considering the representations of the algebra $\cal A$,
we have to keep in
mind that the operator $M$ must be positive for $0<q<1$ and
negative for $q>1$. We will consider the case $0<q<1$ (the case
$q>1$ is quite similar).

\begin{lemma}
 Suppose we have a $*$-representation of the Heisenberg relations
\textup(generally speaking, by unbounded operators\textup) and  there
exists a vector $f\in H$ such that $f\in\ker n$ and
$$
ndf - q^{3/2}dnf = Mf
$$
\textup(it is supposed that the required operators are defined on $f$\textup).
Then $f=0$.
\end{lemma}

\begin{proof}
Indeed, since $n$ is normal, $f \in \ker n^*$.
But this implies that
\[
(Mf,f) = (ndf-q^{3/2}dn\,f,f) =     (ndf,f) = (df,n^*f) =0,
\]
which is impossible by  the positivity of $M$.
\end{proof}

Introduce the element
$$
y  = nMd + \frac{q^{-1/2}}{q^2-1}\, M^2.
$$

By the previous lemma we can suppose that, for ``good''
representations of the Heisenberg relations, the operators $n$ and
$M$ are
invertible and that, if we find $ y $, we will be able to
reconstruct $d$ as
$$
d= M^{-1} n^{-1}y +\frac{1}{1-q^2}\,n^{-1}M.
$$
Thus, replacing $d$ by $y$, we get the
following relations:
\begin{gather}
vn=qnv,\quad nn^*=n^*n,\quad y M = M y,  \quad [ y , y ^*]
=0,\notag
\\
vM = q^{1/2} Mv,\quad v y = q y v,\quad nM=
q^{-1/2} Mn, \notag
\\
n^* y =q y n^*,\quad n y =q y n. \label{ds}
\end{gather}

From now on, we deal with $*$-representations of the
algebra $\cal A$.

\begin{proposition}
There are no representations of \eqref{ds} by bounded operators.
\end{proposition}

\begin{proof}
Indeed, since $Mu = q^{-1/2} uM$ and $u$ is unitary, the spectrum of
$M$ is invariant under multiplication by $q^{-1/2}$. But since
$M>0$, the spectrum of $M$ does not contain zero and, thus, is
unbounded.
\end{proof}

\begin{remark}
As one can see, relations \eqref{ds} have the form \eqref{rels-cu}. This
fact enables to select the class of ``good'' unbounded
representations, which can be continued to representations of the
corresponding bounded operators (see Section~\ref{sec:2.4.3}), 
and to list irreducible representations from this
class.
\end{remark}
Introduce the following operators on the space $l_2(\mathbb{Z})$:
$$
Se_k = e_{k+1},\quad Te_k=ke_k,\quad Qe_k =
q^{k/2}e_k=e^{T/2}e_k \qquad k \in \mathbb{Z}.
$$

\begin{theorem}
All irreducible $*$-representations of the algebra $\cal A$,  up to
unitary equivalence, are:

\smallskip
$a)$ representations on $l_2(\mathbb{Z})\otimes l_2(\mathbb{Z})$\textup:
\begin{gather*}
 n =  \lambda \, S\otimes Q^2,
\quad
 v =  S^*\otimes S^*,
\quad
 N =  \alpha - T\otimes I,
\\
 b =  {q^{-\alpha/2-1}}{\lambda^{-1}} Q(S^*)^2\otimes
     Q^{-2}S^*;
\end{gather*}


$b)$ representations on $l_2(\mathbb{Z})\otimes
l_2(\mathbb{Z})\otimes l_2(\mathbb{Z})$\textup:
\begin{gather*}
n = \lambda\, S \otimes Q^2\otimes I,
\quad
v = S^*\otimes S^*\otimes S^*,
\quad
N = \alpha -T\otimes I\otimes I,
\\
b = \frac{\delta q^{(\alpha+1)/2}}{\lambda(1-q^2)} Q^{-1}(S^*)^2
     \otimes  Q^{-2} \otimes Q^2 S^*
\\
{} + {q^{-\alpha/2-1}}{\lambda^{-1}} Q(S^*)^2\otimes
     Q^{-2}S^*\otimes S^*,
\end{gather*}
where $\lambda$, $\delta \in (q,1]$, $\alpha \in [0,1)$.
\end{theorem}

\begin{proof} Applying the argument as above, we obtain that, up to
unitary equivalence,
$$
u_n= S \otimes I\otimes I,\quad
u_{ y } =  I\otimes S \otimes I,\quad
v = S^* \otimes S^* \otimes S^*
$$
for $\delta\ne0$ and, otherwise,
$$
u_n= S \otimes I,\quad
u_{ y } = 0,\quad
v = S^* \otimes S^*.
$$
Then, since
$$
b= M^{-1}n^{-1} y  v +({1-q^2})^{-1} n^{-1} M v.
$$
we get the necessary
expressions by using the expression for $M$.
\end{proof}

\subsection{Wick algebras related to dynamical systems}
%\label{sub:wick}
\label{sec:2.4.6}

\textbf{1.} In this section we consider some Wick algebras connected with
dynamical systems.

It was noted above that the relations \eqref{ds-commute} serve as some
consistency
conditions. They appeared in Section~\ref{sec:2.4.3} as conditions
for commutation of the mappings obtained from the defining relations.

In this section we look at this consistency from the point of view of
Wick algebras. We suppose that the generators $x_i^*$, $ x_j$,
$i$, $j=1$, \dots, $d$,
only satisfy  the relations
\begin{align*}
x_i^* x_i & = F_i (x_1x_1^*,\dots,x_d x_d^*),\qquad i=1,\dots,d,
\\
x_i^*x_j & = q_{ij}x_j x_i^*,\qquad i\neq j.
\end{align*}
Then additional relations for $x_i$, $x_j$,  $i\neq j$ define an ideal
$\mathcal{I}$ in the algebra generated by
$\{x_i^*, x_i,\, i=1,\dots,d\}$, and the consistency condition is
determined  as a special  property of the ideal $\mathcal{I}$.

\medskip\noindent\textbf{2.}
To do our reasoning more clearly let us present
some definitions and propositions.

 \begin{definition}
Let $I=\{1,\dots,d\}$ and $T_{ij}^{kl}\in \mathbb{C}$, $i$, $j$,
$k$, $l\in I$, be such that $T_{ij}^{kl} =
\bar{T}_{ji}^{lk}$. A Wick algebra with the coefficients
$\{T_{ij}^{kl} \}$ \textup(see \textup{\cite{jorg}}\textup) 
is a $*$-algebra $\mathcal{W}$ generated by
the elements $x_i$, $x_i^*$ and the defining relations
\[
x_i^* x_j = \delta_{ij}1+\sum_{k,l=1}^d T_{ij}^{kl} x_lx_k^*.
\]
\end{definition}

Denote by $\mathcal{H} = \langle e_{1},\ldots, e_d \rangle$ a
finite-dimensional space over $\mathbb{C}$, and by
$\mathcal{H}^*$ its formal dual.
$\mathcal{T}(\mathcal{H}, \mathcal{H}^*)$ will denote the
tensor algebra over $\mathcal{H}$, $\mathcal{H}^{*}$. Then
$\mathcal{W}$ can be canonically realized as
\[
\mathcal{T}(\mathcal{H},\mathcal{H}^*)\Big/\Bigl< e_i^*\otimes e_j -
\delta_{ij}1 - \sum T_{ij}^{kl} e_l\otimes e_k^* \Bigr>.
\]
In this realization, the subalgebra generated by
$\{x_i\}$ is identified with $\mathcal{T}(\mathcal{H})$.

It is obvious that any element of $\mathcal{W}$ can be uniquely
represented as a polynomial in the non-commuting variables
$a_i$, $a_i^*$, where in each monomial, the variables
$a_i$ are placed to the left of $a_j^*$. Such
monomials are called Wick ordered monomials, and they form a
basis in $\mathcal{W}$.

When studying the properties of $\mathcal{W}$, one can find
 the following operators useful:
\begin{align*}
T\colon & \mathcal{H} \otimes \mathcal{H}\to \mathcal{H}
 \otimes\mathcal{H}, \quad  T e_{k}\otimes e_{l} =
\sum_{i,j} T_{ik}^{lj}e_{i}\otimes e_{j},
\\
T_{i}\colon & \mathcal{H}^{\otimes n}\to\mathcal{H}^{\otimes n},
\quad
T_{i}=\underbrace{1\otimes \dots\otimes 1}_{i-1}{}\otimes T
\otimes{}\underbrace{1\otimes\cdots\otimes 1}_{n-i-1},
\\
R_{n}\colon & \mathcal{H}^{\otimes n}\to \mathcal{H}^{\otimes n},
\quad
R_{n}=1+T_1+T_1 T_2+\dots + T_1 T_2\dots T_{n-1}.
\end{align*}
These operators determine the Wick ordering on $\mathcal{W}$.

\begin{proposition}
Let $X\in\mathcal{H}^{\otimes n}$. Then
\begin{equation} \label{commutation_rule}
e_i^*\otimes X = \mu (e_i^*)\, R_n X + \mu (e_i^*)\sum_{k=1}^{d}
T_1 T_2 \dots T_n (X\otimes e_k)\, e_k^*,
\end{equation}
where $\mu (e_i^*)\colon\mathcal{T}(\mathcal{H})
 \to\mathcal{T}(\mathcal{H})$ is defined by
\[
\mu (e_i^*) 1=0,\quad \mu (e_i^*)\, e_{i_1}\otimes \cdots\otimes e_{i_n}=
\delta_{ii_1} e_{i_2}\otimes\cdots\otimes e_{i_n}.
\]
\end{proposition}

Let us note that the defining relations of a Wick algebra give a commutation
rule only for the generators $x_i^*$, $ x_j$, and do not induce any
relations for $x_i$, $x_j$. However such relations are very useful
in the study of representations of $\mathcal{W}$. It is obvious that
any relation determines a two-sided ideal. So, additional relations
for the generators $x_i$, $x_j$ can be described in the terms of special
ideals in $\mathcal{W}$.

\begin{definition}
A Wick ideal is a two-sided ideal $\mathcal I\subset
\mathcal{T}(\mathcal{H})$ such that
$\mathcal{T}(\mathcal{H}^{*})\mathcal I\subset
\mathcal I\mathcal{T}(\mathcal{H}^{*})$. If\/ $\mathcal I$ 
is generated by a set\/
$\mathcal I_{0}\subset\mathcal{H}^{\otimes{n}}$, then $\mathcal I$ is called
homogeneous Wick ideal of degree $n$.
\end{definition}

The following statement is a generalization of the fact
established in \cite{jorg} for $n=2$.

\begin{proposition}\label{wick-proj}
Let $P\colon\mathcal{H}^{\otimes {n}}
\to\mathcal{H}^{\otimes {n}}$ be a projection. Then
$\mathcal I_{n}= \langle P \mathcal{H}^{\otimes n} \rangle$ is a
Wick ideal if and only if

$1.$ $R_{n}P=0,$

$2.$ $[1\otimes (1-P)]\,T_{1}T_{2}\dots T_{n}[P\otimes 1]=0.$

Moreover, if\/ $T$ satisfies the braid condition $T_{1}T_{2}T_{1}=
T_{2}T_{1}T_{2}$ and $P$ is a projection to $\ker R_{n}$,
then condition $2$ is automatically fulfilled.
\end{proposition}
In the case where $d=2$, condition $1$ is called ``linear'',
condition $2$ is called ``quadratic''.

\medskip\noindent\textbf{3.} 
Let us consider the following class of Wick algebras:
 \[
x_{i}^{*} x_{i} = 1+\sum_{j=1}^{d} \lf x_{j}x_{j}^{*},\quad
x_{i}^{*}x_{j} = \lm\qu x_{j}x_{i}^{*}, \qquad i\ne j,
 \]
$0<\alpha_{ii}<1$, $q_{ij}=q_{ji}\in R_{+}$,
$\overline\lambda_{ij} =\lambda_{ji}$, $|\lm| = 1$.
We denote this class by $\ub(A,\Lambda,Q)$ with $ A=(\lf)$,
$\Lambda = (\lm)$, $Q = (\qu) $.

\begin{remark}
1. If we choose $\alpha_{ii}=\mu^2$, $\alpha_{ij}=\mu^2-1,\ j<i$,
$\alpha_{ij}=0,\ j>i$, $\lambda_{ij}=1$, and $q_{ij}=\mu$,
then the obtained algebra is the $\mu$-CCR algebra.

2. For $\alpha_{ii}=q_i$, $\alpha_{ij}=0,\ i\ne j$,  the algebra
becomes the
$q_{ij}$-CCR algebra defined by the following relations:
\[
x_i^*x_i  =1+q_i x_ix_i^*,\quad
x_i^* x_j = \lm\qu x_j x_i^*.
\]
\end{remark}

The purpose of this section is to describe algebras from this class
that have a quadratic ideal of maximal possible rank,
and to classify the $*$-representations of these algebras by bounded
operators.

\medskip\noindent\textbf{4.}
Let $\ub=\ub(A,\Lambda,Q) $. Then the operator
$T$ 
has the form:
    \begin{align*}
T e_{i}\otimes e_{i} &=\alpha_{ii}\,e_{i}\otimes e_{i}, \\
T \tpr &=\lf\,\tpr +\lambda_{ji}q_{ji}\,\otpr.
    \end{align*}
Then
     \begin{gather*}
 \hb\otimes\hb = \bigoplus_{i=1}^{d}\hb_{i}\oplus
\bigoplus_{i,j=1}^{d}\hb_{ij}, 
\\
 \hb_{i}=\langle e_{i}\otimes e_{i}\rangle,\quad \hb_{ij}=
\langle \tpr , \otpr \rangle .
     \end{gather*}
The ``linear condition''  means that $P$ must be a projection
to
 the
subspace generated by eigenvectors of\/ $T$ 
corresponding to the eigenvalue $-1$. Since
$\alpha_{ii}\not = -1$, the rank of $P$ is maximal if and
only if the equalities
   \begin{equation}
(\lf +1)(\alpha_{ji}+1)= q_{ij}q_{ji} \label{eq:max}
   \end{equation}
hold for all $i\not = j $, and
     \begin{displaymath}
P \,\hb\otimes\hb = \Bigl< \otpr - \frac{\lm \qu}{\lf+1}\,\tpr ,\, i<j \Bigr>.
     \end{displaymath}
 Denote the algebra $\ub(A,\Lambda,Q)$ for which
equations~\eqref{eq:max} hold by $\ub(A,\Lambda)$.
The ``quadratic condition'' takes the form:
   \begin{align}
\lf\alpha_{ji}&=0, \qquad i \ne j \nonumber,
\\
\lf(\lf+1-\alpha_{jj})&=0,\qquad  i \ne j,   \nonumber   
\\
\alpha_{ik}(\alpha_{kj}-\lf)+\lf\alpha_{jk}&=0,\qquad   i\ne j, \
 i\ne k ,\ j \ne k.\label{eqnarray:main}
   \end{align}
It is convenient to consider $\{\lf\}$ as a function
$\alpha\colon {I}\times{I}\to \mathbb R$
and denote it by $\lf=\alpha(i,j)$.

     \begin{remark}
If $\alpha$ is a solution of system \eqref{eqnarray:main}, then
for all $\pi\in S_{d}$, $\alp = \alpha(\pi(i),\pi(j))$ is also a solution,
and if $\hat x_{i}=x_{\pi_{i}}$, then the ``structure constants'' for
$\hat{x}_{i}$ are $\hat{\lambda}_{ij}=\alp$,
 $\hat{\lambda}_{ij}=\lambda^{\pi}(i,j)$.
Consequently, it will suffice to describe solutions of
 \eqref{eqnarray:main} up to the action of $S_{d}$.
     \end{remark}

   \begin{definition}
A solution $\alpha$ of \eqref{eqnarray:main}
is  canonical if\/ $\alpha(i,j) = 0$ for all
$i<j$.
   \end{definition}

  \begin{proposition}
Let $\alpha$ be an arbitrary solution of \eqref{eqnarray:main}. Then
exists $\pi\in S_{d}$ such that $\alpha^{\pi}$ is a canonical solution.
  \end{proposition}

We can suppose now that $\lf=0$, $\forall i<j$. Then
\eqref{eqnarray:main}
is reduced to the following:
   \begin{align*}
\lf(\alpha_{jk}-\alpha_{ik})&=0  , \quad  1\leq k<j<i\leq d,\\
\lf(1+\lf-\alpha_{j})&=0  , \quad  1\leq j<i\leq d,\\
\alpha_{j}&=\alpha_{jj},
   \end{align*}
where the second equation means only the fact that all non-zero
$\lf$ are equal to the same parameter $\alpha_{j}-1$ for fixed $j$
and $i>j$.

    \begin{definition}
A canonical solution is called decomposable if
\[
{I}= {I}_{1}\cup{I}_{2},\quad
{I}_{1}\cap{I}_{2}=\emptyset,
\]
and for all $i\in{I}_{1}$, $j\in{I}_{2}$,
$\lf=\alpha_{ji}=0$.
    \end{definition}

\begin{remark}  \label{remark:posm}
 If a canonical solution is decomposable, then there exists $\pi\in S_{d}$
such that $\alpha^{\pi}$ is decomposable and canonical, and
the set of indices has the form
\[
{I}_{1}=\{1,\dots ,m\}, \quad {I}_{2}=\{m+1,\ldots ,d\}.
\]
\end{remark}

 It is clear that if $\alpha_{21}=\dots=\alpha_{d1}=\alpha_{1}-1$,
then $\alpha$ is indecomposable.

  \begin{proposition}  \label{proposition:cal}
Let $\alpha$ be a canonical solution. Then it is indecomposable if and
only if $\alpha_{21}=\dots=\alpha_{d1}=\alpha_{1}-1$.
  \end{proposition}
Let $\alpha$ be a canonical solution, $A=(\lf)$. It follows from
Proposition~\ref{proposition:cal} and Remark~\ref{remark:posm},
that we can  suppose that, for any fixed $j$, all non-zero
$\lf$, $i>j$, are placed before all zeroes.
 Consider $\vec{k}=(k_{1},\ldots , k_{d-1})$, where
$i\leq k_{i}\leq d$ are natural numbers constructed
as follows: if, for a fixed $j$ and all $i>j$, $\lf=0$, then
$k_{j}=0$; otherwise $k_{j}$ is the greatest number $l$ such that
$\alpha_{lj}=\alpha_{j}-1$.
 The characteristic property of $\vec{k}$ is the following.

\begin{proposition}
If\/ $i>j$ and $i\leq k_{j}$, then $k_{i}\leq k_{j}$.
\end{proposition}

Conversely, let $\vec{k}$ be a vector with the characteristic
property, and $A=(\lf)$ be a matrix such that
$\alpha_{ii}=\alpha_{i}$, $\lf=0$, $i<j$.
Then,  $\lf=0$ $, \forall i>j$ if $k_j=j$;
otherwise  $\alpha_{lj}=\alpha_{j}-1$, $ j<l\leq k_{j}$, $\alpha_{lj}=0$,
$l>k_{j}$.
Then it is easy to verify that $A$ is a matrix of a canonical
solution. We will denote such a matrix by $A(\vec{k})$.
The following statement has been proved.

\begin{theorem}
Let $\alpha$ be a solution of system \eqref{eqnarray:main}. Then there exist
$\pi\in S_{d}$ and\/ $\vec{k}$ having the characteristic property
so that $(\lf^{\pi})=A(\vec{k})$. Conversely, for any $\vec{k}$
with characteristic property, $A=A(\vec{k})$ gives a solution.
\end{theorem}

\noindent\textbf{5.}
Now we describe irreducible representations of the  algebras obtained.
 Let $A=A(\vec{k})$, $\ub=\alg$. Then $\ub$ has the
largest quadratic ideal generated by
\[
X_{ij}=x_{j}x_{i} - \lm\qu\, x_{i}x_{j}, \qquad i<j.
\]

\begin{proposition}
Let $\pi(\cdot)$ be a bounded representation of\/  $\alg$. Then
$\pi(X_{ij})=0$.
\end{proposition}

\begin{remark}
The proof basically coincide with the proof of the
analogous fact for the twisted commutation relations
(see Section ~\ref{sec:2.4.2})
given by $\Lambda = 1$, $A=A(d,\dots , d)$, and
$\alpha_{j}=\mu^{2}$.
\end{remark}

This means that in order to describe irreducible representations of
$\ub$, it is necessary to  describe families of operators
$\{X_{i},\, i=1,\dots , d\}$ such that
\begin{align}
X_{i}^{*}X_{i}&=1+\alpha_{i}X_{i}X_{i}^{*}+ \sum_{j<i, k_j \ge i}
(\alpha_j -1)\,
X_{j}X_{j}^{*},
\nonumber
\\
X_{i}^{*}X_{j}&=\lm\qu X_{j}X_{i}^{*}, \qquad i<j, \notag
\\
X_{j}X_{i}&=\lm\qu X_{i}X_{j}, \qquad i<j, \nonumber 
\\
\qu^{2}&=\begin{cases} \alpha_i,&i<j,\, k_i \ge j,\\ 1,&
\text{otherwise}.\end{cases} \label{align:sys}
\end{align}
Let $X_{i}^{*}=U_{i}C_{i}$ be the polar decomposition. Then
system \eqref{align:sys} can be rewritten in an equivalent form:
\begin{gather*}
\mathbf{C}U_{i}^{*}=U_{i}^{*}\mathbf{F}_{i}(\mathbf{C}), \quad
\mathbf{C}=(C_{1}^{2},\ldots ,C_{d}^{2}),
\\
[C_{i},C_{j}]=0 , \quad U_{i}U_{j}=\overline{\lambda}_{ij}U_{j}U_{i} ,
\quad
U_{i}U_{j}^{*}=\lm U_{j}^{*}U_{i} , \qquad i<j,
\\
\mathbf{F}_{i}(x_{1},\dots,x_{d})
\\
=\bigl(x_{1},\dots ,x_{i-1},F_i(x_1, \dots, x_d),
q_{ii+1}^{2}x_{i+1},\dots ,q_{id}^{2}x_{d}\bigr),
\end{gather*}
where
\[
F_i(x_1, \dots, x_d) =1+\alpha_{i}x_{i}+\sum_{j<i,k_j \ge i}
(\alpha_j -1) \,x_j, \qquad i=1, \dots, d.
\]
Using the  technique of dynamical systems, we can reduce the problem of
describing irreducible representations of $\alg$
to an analogous problem for finite families of the unitary
operators $\{U_{i}\}$ that satisfy the relations
$U_{i}U_{j}=\lm U_{j}U_{i}$, $i<j$. First, introduce
some notations: let $D(\mu)$ denote an operator in $l_{2}(N)$ given by
\begin{align*}
D(\mu)e_{n}&=\mu^{n-1}e_{n}, \qquad n\in \mathbb{N},
\\
D(j,k_{i})&=
\begin{cases}
1,& j>k_{i},\\
D(\alpha_{j}),& j\le k_{i},
\end{cases}
\end{align*}
and let $S$ stand for the unilateral shift. 
Then $D(f_{j})\,e_{n}=f_{j}^{n-1}(0)\,e_n$,
where
$f_{j}(x)=1+\alpha_{j}x$, and $f^{n}(\cdot)$ denotes the $n$-th iteration
of $f$.

Let $1\leq i_{1}<\dots< i_{l}\leq d$ be natural numbers such
that
\[
k_{i_{j}}+1\leq i_{j+1},\qquad j=1,\dots ,l-1.
\]
Fix  such a family. Denote
$\Phi=\bigcup_{j=1}^{l}\{i_{j}+1,\ldots ,k_{i_{j}}\}$.
Construct the following irreducible representation for a
fixed family of $\{i_{1}\ldots ,i_{l}\}$:
\begin{align*}
C_{j}&=U_{j}=0, \qquad  \forall j\in\Phi,
\\
C_{j}^{2}&=\bigotimes_{i=1,i\notin\Phi}^{j-1} D(j,k_{i})
\otimes D(f_{j})\otimes I\otimes\cdots\otimes I , \qquad j\ne i_{k},
\\
U_{j}^{*}&=\bigotimes_{i=1,i\not\in\Phi}^{j-1} D(\lm)\otimes S
\otimes I\otimes\dots\otimes I, \qquad j\neq i_{k},
\\
U_{i_{k}}^{*}&=\bigotimes_{i<i_{k},i\not\in\Phi} D(\lambda_{ii_{k}})
\otimes\bigotimes_{i>i_{k},i\not\in\Phi} D(\overline{\lambda}_{ii_{k}})
\otimes \hat{U}_{i_{k}}^{*}, \qquad k=1,\dots ,l,
\\
C_{i_{k}}^{2}&=\frac{1}{1-\alpha_{i_{k}}}
\bigotimes_{i<i_{k},i\not\in\Phi} D(i,k_{i_{k}})\otimes I\otimes
\cdots \otimes I,
\quad k=1,\ldots ,l,
\end{align*}
where $\{\hat{U}_{i_{k}}\}$ is an irreducible family of unitary
operators
satisfying the relations
$\hat{U}_{i}\hat{U}_{j}=\lm\hat{U}_{j}\hat{U}_{j}$.


\begin{theorem}
All irreducible representations of the algebra $\mathcal U$ can be obtained in the way described
above\textup; moreover, two representations are unitarily equivalent if and only
if they correspond to the same family $\{i_{1},\dots ,i_{l}\}$, and
the corresponding unitary families are unitarily equivalent.
\end{theorem}

\begin{remark}
1. If at least one of $\lm$ is not a root of $1$, then
there exists a representation that is not of type one.

2. If all $\lm$ are roots of $1$, then the problem of classification
of families $\{U_{i}\}$ can be reduced to the case where $\lm^{q}=1$;
here
$q=p^{m}$ for some prime $p$. In this case, the families $\{U_{i}\}$ can be
described by a simple reduction algorithm.
\end{remark}

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