One Hat Cyber Team

Edit File: CHAPT25.TEX

\section{Operator relations and one-dimensional dynamical systems}
\label{sec_relations}

\subsection{Operator relations related to one-dimensional  dynamical
systems}

1. Consider an operator $X$, which satisfies, together with its adjoint
$X^*$, algebraic relation of the form
\begin{equation}
\label{xx}
XX^* = F(X^*X),
\end{equation}
where $F(\cdot)\colon \mathbb{R} \to \mathbb{R}$ is a measurable
with respect to Borel $\sigma$-algebra mapping.

If $F(\cdot)= P_n(\cdot)$ is a real polynomial, then the pair of
operators $X$, $X^*$ satisfying relation \eqref{xx} is a
representation of $*$-algebra $\mathcal{A}$ generated by elements
$x$, $x^*$, satisfying the relation $xx^* = P_n(x^*x)$. In some
non-polynomial cases
pairs of operators $X$, $X^*$ form representations of $C^*$-algebras
or other topological algebras, but we will not concentrate our
attention at the underlying algebraic objects, restricting ourselves
to the study of representations of the relation.

We have already considered examples of representations of the form
\eqref{xx} 
above. For the Hermitian $q$-plane $xx^* = q x^*x$, $q \in \mathbb{R}$, 
we have $F(\lambda) = q\lambda$, and for $q$-CCR $xx^* = qx^*x +1$, $q 
\in \mathbb{R}$, we have $F(\lambda) = q\lambda +1$.

Below we consider another examples of relations of the form \eqref{xx}.

\begin{example} Second order mappings. Consider the following relation
\[
xx^* = q x^*x(x^*x -I), \quad q \in \mathbb{R}^1.
\]
The corresponding polynomial is $F(\lambda) = q \lambda(\lambda -1)$. 
Structure of bounded representations of such relations hardly depends 
on the value of $q$ (see below).
\end{example}

\begin{example} Continued fractions.
Another interesting class of relations related to continuous fractions 
are the following relations
\[
xx^* = (a x^*x +c)(bx^*x +d)^{-1}, \quad a,b,c,d \in \mathbb{R}, 
b\ne 0, \ ad -bc \ne 0.
\]
Below, we will see how representations of the relation depend on values 
of the parameters.

In particular, two-parameter unit quantum disk algebra \cite{klles}
is generated by the relation
\begin{gather*}
qzz^* - z^*z = q-1 + \mu (1-zz^*)(1-z^*z), 
\\
 0\le \mu \le 1, \ 0\le 
q \le 1, \ (\mu, q) \ne (0,1),
\end{gather*}
which can be rewritten in the form \eqref{xx} with 
\[
F(\lambda) = \frac{(q+\mu) \lambda + 1 -q - \mu}{\mu \lambda + 1 - \mu}
\]
\end{example}

\begin{remark}
As we will see, in many examples unbounded operators $X$ arise as
natural representations of the relation. Therefore, it is necessary
to introduce an accurate operator sense of the relation for
unbounded operators. However, we will not 
discuss problems related to unbounded
operators here.
\end{remark}

2. Let $X$ be bounded. For a polar decomposition of the operator $X$
we have $X= UC$, where $C = C^* = (X^*X)^{1/2}$, $U$ is a partial
isometry, and $\ker U= \ker C = \ker X$, $U^*U$ is an orthogonal 
projection onto $(\ker C)^\perp$. Taking into
account relation \eqref{xx}, we get 
\begin{equation} \label{ucu}
UC^2U^* = F(C^2),
\end{equation} 
which implies
\begin{equation} \label{cu}
UC^2 = F(C^2) U, \quad C^2 U^* =U^* F(C^2).
\end{equation}
Notice that \eqref{ucu} implies $ \ker U^* = \ker X^* =\ker F(C^2)$ as well.

3. Now we establish some relations which follow from \eqref{xx}. First
notice that $X(X^*X) = F (X^*X) X$, and $(X^*X) X^* = X^* F(X^*X)$.
For any $k=1$, 2,\dots, we recursively get the relations $X(X^*X)^k
=(F(X^*X))^kX$, and $(X^*X)^k X^* = X^* (F(X^*X))^k$. From the two latter
relations we get $XP(X^*X) = P(X^*X)X$
for polynomials of $X^*X$, and in the case of bounded
operators, standard approximation procedure using the spectral
decomposition of the self-adjoint operator $X^*X$ yields
\begin{equation}\label{xx_fun}
X \phi(X^*X) = \phi(F(X^*X) X, \quad \phi(X^*X) X^* = X^*
\phi(F(X^*X))
\end{equation}
for any bounded measurable function $\phi(\cdot)$. In particular,
\eqref{xx_fun} implies that for all $n=1$, 2,~\dots,
\begin{equation} \label{xx_n}
X^n \phi(X^*X) = \phi(F^{ n}(X^*X) X^n, \quad \phi(X^*X) (X^*)^n
=(X^*)^n\phi(F^{ n}(X^*X)),
\end{equation}
where $F^{ n}(\cdot)$ is the $n$-th iteration of $F(\cdot)$.

We have the following statement.

\begin{proposition}\label{cu-props}
Let $X$ be a bounded operator satisfying \eqref{xx} with a bounded
measurable real function $F(\cdot)$, and let $X=UC$ be the polar
decomposition introduced above, such that $\ker C^2 = \ker U$, $\ker F(C^2) =
\ker U^*$.
Relation \eqref{xx} is
equivalent to any of the following:

a. $C^2 U^* = U^* F(C^2)$;

b. $X \phi(X^*X) = \phi(F(X^*X)) X$ for any bounded on the spectrum
of $X^*X$ measurable function $\phi(\cdot)$;

c. $\phi(C^2) U^* = U^* \phi(F(C^2))$  for any bounded on the
spectrum of $X^*X$ measurable function $\phi(\cdot)$;

d. $E_{C^2}(\Delta) U^* = U^* E_{C^2}(F^{-1}(\Delta))$, for all
measurable $\Delta\subset \mathbb{R}$;

\noindent
where $E_{C^2}(\cdot) $ is a resolution of the identity of the
operator $C^2$, and $F^{-1}(\Delta)$ denotes the full pre-image of
$\Delta$ (more precisely, full pre-image of $\Delta \cap 
F(\mathbb{R})$).
\end{proposition}

\begin{proof}
It was already shown that \eqref{xx} is equivalent to a), and that
\eqref{xx} implies b). To show that b) implies \eqref{xx} take
$\phi(\lambda) =\lambda$. Multiplying by $X^*$ from the right, we
have $XX^*XX^* = F(X^*X) XX^*$, which implies that on vectors
orthogonal to $\ker XX^*$ relation \eqref{xx} holds. On the other
hand, $\ker X^* = \ker U^* = \ker F(X^*X)$.

The proof that c) is equivalent to a) can be done the same way as b)
was derived. Relation d) is a partial case of c) with $\phi(\lambda) =
\chi_\Delta(\lambda)$ (notice that $E_{F(C^2)}(\Delta) =
E_{C^2}(F^{-1}(\Delta))$);
 using spectral
representation for measurable functions of the positive self-adjoint
operator $C^2$, $\phi(C^2) = \int \phi(\lambda\, dE_{C^2}(\lambda)$,
one can easily see that c) follows from d)
as well.
\end{proof}

4. Operators of the form \eqref{xx} form a subclass in the class of 
 centered
operators. Recall that a bounded operator $X$ is centered, if the
operators $X^l(X^*)^l$, $(X^*)^kX^k$, $k,l=1$, 2,\dots, form a
commuting family.

\begin{proposition}
 \label{xx-center}
A bounded operator $X$ satisfying \eqref{xx} is centered. If a pair
of operators, self-adjoint $C$, and partial isometry $U$ such that
$\ker U =  \ker C^2$, $\ker U^* = \ker F(C^*)$ satisfy
relation \eqref{cu}, then $U$ is a centered partial isometry.
\end{proposition}

\begin{proof}
Show that $X$ is centered. Write $A_k =X^k(X^*)^k$, $B_l = (X^*)^lX^l$,
$k,l\ge 1$. First consider the operators $A_k$. Applying \eqref{xx_n}, we
have
\begin{align*}
A_k &= X^k(X^*)^k= X^{k-1} (XX^*) (X^*)^{k-1}=
\\
&= X^{k-1}F(X^*X) (X^*)^{k-1} = F^{ n} (X^*X) X^{n-1}(X^*)^{n-1}
\\
& = F^{ n} (X^*X) X^{n-2}F(X^*X)(X^*)^{n-2}= \dotsc
\\
& = F^{ n} (X^*X) F^{ n-1} (X^*X)\dots F(X^*X).
\end{align*}
Since all the operators $A_k$, $k\ge 1$ are functions of the  single
operator $X^*X$, they commute with each other.

Now consider a pair $A_k$, $B_l$. Again applying \eqref{xx_n}, and
the obtained representation for $A_k$, we get
\begin{align*}
B_l A_k &= (X^*)^l X^l X^k (X^*)^k
\\
&= (X^*)^l X^l F^{ k} (X^*X)
F^{ k-1} (X^*X)\dots F(X^*X)
\\
&= (X^*)^l F^{ k+l} (X^*X)
F^{ k+l-1} (X^*X)\dots F^{l+1}(X^*X) X^l
\\
&= F^{ k} (X^*X)
F^{ k-1} (X^*X)\dots F(X^*X) (X^*)^l X^l= A_k B_l,
\end{align*}
and therefore, the operators $A_k$ and $B_l$, $k, l \ge 1$, also
commute with each other.

Now we show that the operators $B_l$, $l\ge 1$, commute with each
other, too. Indeed, for $k>l$ we have
\begin{align*}
B_k B_l &= (X^*)^k X^k (X^*)^l X^l =(X^*)^l (X^*)^{k-l} X^{k-l}  X^l
(X^*)^l X^l
\\
&= (X^*)^l B_{k-l} A_l X^l = (X^*)^l A_l B_{k-l} X^l
\\
&=
(X^*)^l X^l (X^*)^l (X^*)^{k-l} X^{k-l}  X^l = B_l B_k.
\end{align*}
Thus, the operator $X$ is centered.

It remains to prove the second statement of the proposition. Let
$\Delta  = \sigma(C^2) \setminus \{0\}$. Then $E_{C^2}(\Delta)$ is
the projection onto cokernel of $C^2$, and, since $\ker U = \ker C$,
we have $E_{C^2}(\Delta) = U^*U$. Then \eqref{cu} implies $UU^* =
UE(\Delta)U^* =  E_{C^2}(F^{-1}(\Delta))$. Similarly,
\begin{align*}
U^k (U^*)^k &= U^{k-1} U U^* (U^*)^{k-1}= U^{k-1} U U^* U U^*
(U^*)^{k-1}
\\
&= U^k E_{C^2}(\Delta) (U^*)^k = E_{C^2} (F^{-k}(\Delta)).
\end{align*}
Therefore, projections $U^k (U^*)^k$, $k\ge1$, commute with $U^*U$
and with each other. The rest of commuting relations can be obtained
the same way as it was done for $X$ in the proof of the first part of
the proposition.
\end{proof}

5. Now we will show that properties of representations of relation 
 \eqref{xx} depend on properties of the corresponding dynamical 
system $\lambda \mapsto F(\lambda)$, and study such dependency in 
details.

\begin{proposition}
Let the operators $C$ and $U$ satisfy relation \eqref{cu}. If 
$e_\lambda$ is an eigenvector  of the operator $C^2$, with eigenvalue 
$\lambda$, then $U^* e_\lambda$ is either zero or again an eigenvector of
$C^2$, 
with eigenvalue $F(\lambda)$.
\end{proposition}

\begin{proof}
Indeed, by \eqref{cu}, we have $C^2 U^* e_\lambda = U^* F(C^2) 
e_\lambda = F(\lambda) U^* e_\lambda$.
\end{proof}

In particular, if the operator $U$ is unitary, then $U^*e_\lambda$ is 
not zero, which implies that in this case $F(\cdot)$ maps the spectrum 
of $C^2$ into itself.

Similar fact holds for all points of the spectrum of the operator 
$C^2$.

\begin{proposition}
If the operator $X$ in \eqref{xx} is invertible (i.e., $U$ in 
\eqref{cu} is unitary), then $F(\cdot)$ maps the spectrum $\sigma(C^2)$ 
into itself.
\end{proposition}

\begin{proof}
Let $\lambda \in \sigma(C^2)$. Then there exists a sequence of unit 
vectors $e_n \in H$, $n\ge 1$, such that $\|C^2 e_n - \lambda e_n\| \to 
\infty$, $n \to \infty$. Then
\begin{gather*}
\|C^2 U^* e_n - F(\lambda) U^* e_n\| = \|U^*(F(C^2) e_n - F(\lambda) 
e_n)\| 
\\
= \|F(C^2) e_n - F(\lambda) e_n\| \to 0, \quad n\to \infty,
\end{gather*}
and $\|U^* e_n\|=1$, $n \ge 1$. This implies that $F(\lambda )\in 
\sigma(C^2)$. 
\end{proof}

Let a Borel  set $\Delta$ be such that $\Delta \subset F(\mathbb{R})$, and
$F^{-1}(\Delta) = \Delta$.
Proposition~\ref{cu-props} implies that we have $E_{C^2}(\Delta) U^* = U^* 
E_{C^2}(F^{-1}(\Delta)) = U^* E_{C^2}(\Delta)$, i.e., $E_{C^2}(\Delta)$ 
is a projection onto invariant subspace in $H$. Therefore, to study 
irreducible representations, we need to study the ``smallest'' 
invariant (in the indicated above sense) 
subsets, which would carry the spectrum of $C^2$. Below  we will 
use some facts about dynamical systems and their invariant 
sets, as well as properties of the corresponding spectral measures.

6. Recall some basic notions and facts about discrete dynamical systems. 
A discrete time one-dimensional dynamical system
\index{dynamical system}
is just a
(continuous or measurable) mapping ${\mathbb{R}^{1}}\ni\lambda
\mapsto F(\lambda)\in{\mathbb{R}^{1}}$.

First of all, we recall the notion of a trajectory or 
an orbit of a dynamical
system.
Traditionally, a trajectory or an orbit
\index{orbit}
\index{dynamical system, orbit of}
of the dynamical system
$F(\cdot):{\mathbb{R}^{1}}\mapsto{\mathbb{R}^{1}}$
is the set $\orb(\lambda)=\{\lambda,F(\lambda),F^{2}(\lambda),\dots\}=
\bigcup_{n=0}^{\infty} {F^{n}(\lambda)}$. Here
$F^{n}(\cdot)=F(F^{n-1}(\cdot))$, $n=1$, 2,~\dots and $F^{0}(\cdot)$
is the identity transformation. Since we are interested in the action 
of $F^{-1}$ as well (past), will mean that two points, $\lambda_1$ and
$\lambda_2$
belong to the same trajectory, if $\lambda_1 = F^m(\lambda_2)$
or $\lambda_2 = F^m(\lambda_1)$ 
for some $m$.
A point $\lambda_{0}\in{\mathbb{R}^{1}}$,
such that $F^{m}(\lambda_{0})=\lambda_{0}$ and $F^{n}(\lambda_{0})\neq
\lambda_{0}$
for $0<n<m$, is called a periodic point
\index{periodic point}
\index{dynamical system, periodic point of}
of period $m$.
The periodic points $\lambda_{0}, F(\lambda_{0}), \dots, F^{m-1}(\lambda_{0})$
form a cycle
\index{cycle}
\index{dynamical system, cycle of}
of period $m$.

Keeping in mind representations of relation \eqref{cu}, and the fact 
that those invariant sets should carry the spectral measure of the 
operator $C^2$, we need to 
consider a measurable
mapping of a measurable space. In this case, the mapping
$F(\cdot)$ gives rise to a mapping of a Borel measure on the
line by the formula
\[
d\rho(\lambda) \mapsto d\rho(F(\lambda)).
\]

A Borel measure $\rho(\cdot)$ is called quasi-invariant with
respect to $F(\cdot)$, if $\rho(F(\cdot))$ is absolutely
continuous with respect to $\rho(\cdot)$.

\begin{proposition}
If the operator $U$ in \eqref{cu} is unitary, and $F(\cdot)$ is 
one-to-one, then the spectral measure of $C^2$ is quasi-invariant with 
respect to $F(\cdot)$ and $F^{-1}(\cdot)$.
\end{proposition}

\begin{proof}
Indeed, the spectral measure can be taken in the form $\rho(\cdot) = 
(E_{C^2}(\cdot)\omega, \omega)$, where $\omega$ is a maximal spectral 
type vector. Then the statement follows from the equality 
$\rho(F^{-1}(\cdot)) = (E_{C^2}(F^{-1}(\cdot))\omega, \omega) = (U^* 
E_{C^2}(F^{-1}(\cdot)) 
\omega, U^*\omega) = (E_{C^2}(\cdot) U^* \omega, U^*\omega)$,
and from similar relation for $U$.
\end{proof}

We also need the notion of ergodicity of a measure. A Borel
measure $\rho(\cdot)$ is called ergodic with respect to action of
dynamical system $F(\cdot)$, if for any measurable
$F(\cdot)$-invariant set $\Delta\subset\mathbb{R}$ either
$\rho(\Delta)=0$ or $\rho(\mathbb{R}\setminus \Delta)=0$.

\begin{proposition}
If the pair $X$, $X^*$, satisfying \eqref{xx}, is irreducible, then the 
spectral measure of the operator $C^2$ is ergodic with respect to 
$F^{-1}(\cdot)$.
\end{proposition}

\begin{proof}
Indeed, otherwise there would exist a nontrivial invariant subset  
$\Delta\subset\sigma(C^2)$. As it was noticed above, in this case 
$E_{C^2}(\Delta)$ is a projection onto invariant subspace, which is  
nontrivial if and only if the spectral measure of $\Delta$ is not zero 
or one.
\end{proof}

The simplest invariant sets are orbits of the dynamical
system. Therefore, the simplest class of quasi-invariant
ergodic measures form atomic measures concentrated on an
orbit, i.e., the measures for which
\[
\rho(\lambda_k) = \rho_k>0, \quad \lambda_k = F^{ k}(\lambda), \
k \in \mathbb{Z},
\]
and zero elsewhere. The existence of non-atomic
quasi-invariant ergodic measures depends on topological
properties of the dynamical system. We have the following
fact.

\begin{proposition}
If a dynamical system $\lambda\mapsto F(\lambda)$ with one-to-one $F(\cdot)$ 
possesses a measurable
section, i.e., a measurable set which intersects any orbit
in a single point, then any ergodic measure is concentrated
on a single orbit of the dynamical system.
\end{proposition}

In the non-bijective case, similar statement holds (see Section~\ref{??}
below).

Below we study the correspondence between orbits and irreducible 
representations of the relation.

\subsection{Finite-dimensional representations}
 
1. Let us classify,  up to unitary equivalence,
the irreducible pairs $X$, $X^*$ of operators in a finite-dimensional
space $H$, obeying  relation \eqref{xx} (the irreducible
finite-dimensional representations of \eqref{xx}).

\begin{theorem} \label{fin_th}
Any cycle $O_\lambda=\{\lambda, F(\lambda), \dots , F^{n-1}(\lambda)\}$ of
period
$n$ such that $\lambda \ge0$, and $t>0$ for all other points $t
\in O_\lambda$,  
defines a family of $n$-dimensional irreducible representations
of \eqref{xx}:
\begin{equation}\label{one5}
X=\left(
\begin{array}{cccc}
0&\cdots&0&e^{i\phi}\lambda\\
F(\lambda)&0&&0\\
 &\ddots&\ddots&\vdots\\
0&&F^{n-1}(\lambda)&0\end{array}\right)
\end{equation}
where $\phi\in [0,2\pi)$, if $\lambda>0$, or single representation,
if $\lambda=0$.

These are all, up to unitary equivalence, different irreducible
representations of  \eqref{xx}.
\end{theorem}

\begin{proof}
Indeed, by a direct calculation one can check that representation
\eqref{one5} satisfy the necessary relation. Show that it is irreducible. Any bounded self-adjoint operator $T$ which commutes with $X$ and $X^*$, commutes with introduced above $C^2$ and $U$ as well. But in the selected basis $C^2$ is diagonal with distinct eigenvalues $\lambda$, \dots, $F^{n-1}(\lambda)$. Then $T$ is diagonal, and the commutation with $U$ implies $T= cI$.

We show that these are all irreducible representations of
\eqref{xx}.

Assume that $U$ is unitary. 
Since $\dim H < \infty$,
the operator $C^2$ can be diagonalized,
and have non-negative eigenvalues.
Take a unit eigenvector $e_t$ of $C^2$, $C^2 e_t = t e_t$. It follows from
\eqref{cu} that $U^*e_t=e_{F(t)}$ is again a unit eigenvector of $C^2$,
$C^2U^*e_t = F(t) e_{F(t)}$. Consider the sequence of
unit vectors $(U^*)^ke_t$, $k\ge 0$. Since there are only finite number of eigenvalues of $C^2$, we conclude that for some $n\ge1$
$F^n(t) = t$. Linear span $H_0$ of eigenspaces of $C^2$, corresponding to eigenvalues $t$, \dots, $F^{n-1}(t)$, is invariant with respect to $C^2$ and $U^*$.
 
We show that the space $H_0$ is invariant with respect to $U$,  too.
Assume that $Ue_t$ does not belong to $H_0$. Denote by $u$ the  projection
of $Ue_t$ on the orthogonal complement to $H_0$. From the relation  we
have that $u $ belongs to the space $H_1$ generated by eigenvectors 
corresponding to all eigenvalues $\lambda \in \Lambda =\{F(\lambda) = t, 
\lambda \ne
F^{n-1}(t)\}$.
For all $k>0$, sets $F^{-k}(\Lambda)$ are disjoint; we have $U^k H_1 \subset E_{C^2}(F^{-k}(\Lambda)) H_1$, and all these spaces are nonzero and orthogonal to each other. But this is impossible since $H$ is finite-dimensional. Therefore, $H_0$ is invariant with respect to $X$ and $X^*$, and due to the irreducibility, coincides with the whole $H$.

The operator $U^n$ commutes with $C^2$, since  $F^n(t)
=t$. It also obviously commutes with $U$ and 
$U^*$. Therefore, the irreducibility implies  $U^n
= \alpha I$, $|\alpha|=1$. In the basis $e_t$,  \dots,
$e_{F^{n-1}(t)}$ the operators act as needed.

Now consider the case of non-unitary $U$. Now $\ker C^2 =\ker U \ne
\{0\}$, and there
exists a unit vector $e_0$, for which $Ue_0=C^2 e_0=0$. Again, consider
vectors $(U^*)^k e_0$, $k=0$, 1,~\dots. Relation \eqref{cu} implies that
$e_k$ is either eigenvector of $C^2$ with eigenvalue $F^k(0)$, or zero
vector. Then, since
$\dim H<\infty$, we conclude that there exists $n$, for which $e_k=0$,
$k\ge n$, i.e., $e_{n-1} \in \ker U^*=\ker F(C^2)$, and $F^n(0)=0$, i.e.,
0
is a periodic point of period $n$, and the formula follows.
\end{proof}

2. The presented theorem reduces the problem of classification of
finite-dimensional irreducible representations $(C^2, U)$
to the description of the cycles of the dynamical system
${\mathbb{R}^{1}}\stackrel{F(\cdot)}{\mapsto}{\mathbb{R}^{1}}$.
So, let us look at how the facts about cycles of the
dynamical systems can be of use in the
context of representation theory.

Sharkovsky's theorem establishes the following order in the set of natural
numbers.

\begin{theorem}
Let $F\colon I\mapsto I$ be a continuous mapping of the closed
interval $I$ into itself. If the dynamical system possesses a
cycle of period $m$, then for any $m' \triangleleft m$,
there exists a cycle of period $m'$, where $\triangleleft$
denotes the following order on the set $\mathbb{N}$ of
natural numbers:
\begin{equation}
1 \triangleleft 2 \triangleleft \dots \triangleleft 2^{n}
\triangleleft \cdots \triangleleft
2^{2} \cdot 5 \triangleleft 2^{2} \cdot 3 \triangleleft \cdots
\triangleleft 2 \cdot 5 \triangleleft 2 \cdot 3
\triangleleft \cdot \triangleleft 9 \triangleleft 7 \triangleleft 5
\triangleleft 3.
\end{equation}
For any $m$, there exists a continuous mapping $\tilde F\colon I
\mapsto I$ such that the dynamical system has cycle of period $m$,
and does not have cycles of periods $m'$ for  $m \triangleleft m'$.
\end{theorem}

This theorem gives the following statement about dimensions of 
 irreducible representations of \eqref{xx}.

\begin{proposition}
Let $F\colon I\mapsto I$ be a continuous mapping of the interval $I$
into
itself. If there exist irreducible $m$-dimensional representations
of \eqref{xx} such that $\sigma(C^2) \subset I$, then
for any $m'\triangleleft m$,  there exists $m'$-dimensional
irreducible
representations of \eqref{xx}.

For any $m$, there exists a continuous mapping $\tilde F\colon I
\mapsto I$
such that  relation \eqref{xx} has  an $m$-dimensional irreducible
representation and does not have irreducible representations of a
dimension
$m'$ for  $m \triangleleft m'$.
\end{proposition}

\begin{corollary}
If   relation \eqref{xx} has irreducible
representations of the dimension
not equal to the power of $2$, than there are infinitely many
dimensions for
which \eqref{xx} has irreducible representations.
\end{corollary}

\begin{corollary}
An existence of a three-dimensional irreducible representation of
\eqref{xx}
implies that \eqref{xx} has irreducible representations of any
dimension
$n\in \mathbb{N}$.
\end{corollary}

\begin{example} {\em Second-order mapping. Finite-dimensional
representations.}
Consider the following relation
\[
XX^* = (X^*X -q)^2
\]
The corresponding dynamical system is generated by polynomial 
$P_q(\lambda)=(\lambda -q)^2$. 
According to the arguments 
above, all finite-dimensional representations are described in terms of 
cycles of this mapping. Let us look at how the value of $q$ affects the 
existence of cycles, and their order (see, e.g., \cite{smr})

For $q<-1/4$, there are no stationary points, and therefore, there are 
no cycles at all. For $q=1/4$, there exists a unique stationary point 
$\lambda=1/4$, and no other cycles. To this point there corresponds a 
circle of one-dimensional representations, $X= e^{i\phi}/4$, $\phi \in 
[0,2\pi)$.

For $-1/4 < q <0$ there are two stationary points, $x_{0,1} = (2q +1 
\pm \sqrt{4q +1})/2$, which give two one-dimensional families of 
irreducible representations. There are no other cycles, and no other 
irreducible representations.

As $q$ raises from 0 to $q^* \approx 1.4\dots$, cycles of order 2, 
$2^2$, \dots, $2^n$, and the corresponding families of irreducible 
representations of the corresponding dimensions arise.

For $q=q^*$, there exist cycles of any order $2^k$, $k\ge 1$, and no 
other cycles; any irreducible finite-dimensional representation has 
dimension $2^k$ for some $k\ge0$.

As $q>q^*$, raises, another cycles arise in the order described by the 
Sharkovsky theorem. For some $q$,  there are cycles of order 3, 
and therefore, cycles of all 
other orders. Therefore, for those $q$ 
there are irreducible representations of any 
dimension.
\end{example}

\begin{example}(Continuous fractions. Finite-dimensional representations)
Consider operator relations which arise from  the M\"obius
mapping
\begin{equation}\label{moebius-rel}
xx^* = (a x^*x +c)(bx^*x +d)^{-1}, \quad a,b,c,d \in \mathbb{R}, 
b\ne 0 >0, \ ad -bc \ne 0.
\end{equation}
According to Theorem~\ref{fin_th}, to describe finite-dimensional
representations of relation \eqref{moebius-rel}, one need to find cycles of
dynamical system generaed by the mapping
\begin{equation}\label{moebius-map}
F(z) = \frac{az +c}{bz +d}.
\end{equation}
To do it, we follow \cite{silv-wall}, see also \cite{jones-thron}. First
consider fixed points of the mapping. If $(d-a)^2 +4bc =0$, then there exists a
single fixed point $\xi_1=(a-d)/2b$;
otherwise, there are two fixed points, $\xi_1$,
$\xi_2$.

If there exists a single stationary point $\xi_1$, then $F$ is conjugate to the
shift in the complex plane, $F = \phi^{-1} \circ T \circ \phi$, where $\phi(z)=
1/(z-\xi_1)$, $T(w) = w+l$, $l = 2b/(a+d)$. We see that in this case $\xi_1$ is
a unique periodic point. Representations exist only if $\xi_1\ge0$;
for  $a=d$, $X=0$ is a single solution of
\eqref{moebius-rel}, for $\xi_1>0$ there exists a one-parameter family of
one-dimensional representations $X = \alpha \sqrt{\xi_1}$, $|\alpha| =1$.

If there are two stationary points, $\xi_1$, $\xi_2$, one can construct
one-dimensional representations quite similarly, if one or both these points
are non-negative. Also, write
\[
\phi(z) = \frac{z-\xi_1}{z-\xi_2};
\]
then $F = \phi^{-1} \circ T \circ \phi$, where $Tw = qw$ with $q =
(a-\xi_1b)/(a-\xi_2 b) = (d - \xi_2 b)/(d-\xi_1b)$. Now it is easy to see that
either $q$ is $n$-th root of identity for some $n$, and all points are periodic
with period $n$, or there are no periodic points at all. In the periodic case,
represetations corespond to orbits with all non-negative points, and these
representations are constructed according to Theorem~\ref{fin_th}.

Notice that in this example the rule about dimensions of representations
established by the Sharkovsky theorem does not hold: there can be only
one-dimensional and $n$-dimensional irreducible representations.
\end{example}

\subsection{Infinite-dimensional representations}

In order to describe general case, recall that according to 
Proposition~\ref{xx-center} the operator $U$ is a centered partial 
isometry. We will see that in the irreducible representation with 
non-unitary $U$, the pair $U$, $U^*$ is again irreducible; since all 
irreducible partial isometries can be easily described, this enables us 
to give a complete description of all irreducible representations in 
the non-unitary case.

In the unitary case, two classes of representations may arise: 
representations, in which $U$ acts as a shift in $l_2$ (in this case 
the spectrum of $C^2$ lies on a single orbit), and representations 
corresponding to non-trivial ergodic measures. The latter class of 
representations is too complicated to describe for the moment; however, 
non-trivial ergodic measures may arise only if the corresponding 
dynamical system does not have a measurable section.

1. We start with description of centered partial isometries.

\begin{theorem}\label{thiso}
Any irreducible centered partial isometry is
one of the following:

\begin{itemize}
\item[(i)]  one-dimensional unitary operator $U=\alpha $, $|\alpha
|=1;$

\item[(ii)]  unilateral shift operator $Ue_k=e_{k+1}$ on $l_2;$

\item[(iii)]  adjoint to the unilateral shift operator$;$

\item[(iv)]  finite-dimensional operator of the form $Ue_k=e_{k+1}$,
$k=1,\dots ,n-1$, $Ue_n=0$ on $\Bbb C^n$ for some $n=1,2,\dots $
\end{itemize}
\end{theorem}

\begin{proof}
We start with a simple fact.

\begin{proposition}
The operators $U^k(U^{*})^k$, $(U^{*})^lU^l$ $k,l=1,2,\dots $
are projections.
\end{proposition}

\begin{proof}
Since $UU^{*}U=U$ and the operators $U^{*}U$ and
$U^{k-1}(U^{*})^{k-1}$ commute we have by induction
\begin{align*}
U^k(U^*)^kU^k(U^*)^k&=
UU^{k-1}(U^*)^{k-1}U^*UU^{k-1}(U^*)^{k-1}U^*\\
&=UU^*UU^{k-1}(U^*)^{k-1}U^{k-1}(U^*)^{k-1}U^*\\
&=UU^{k-1}(U^*)^{k-1}U^*=U^{k}(U^*)^{k}.
\end{align*}
Similarly, since $U^{*}UU^{*}=U^{*}$ and $UU^{*}$ and
$(U^{*})^{k-1}U^{k-1}$ commute we get that $(U^{*})^kU^k$ is a
projection.
\end{proof}

Denote these projections by $P_k=(U^{*})^kU^k$,
$P_{-k}=U^k(U^{*})^k$, $ k=1,2,\dots $, $P_0=I$.

\begin{proposition}
For all $k\in \Bbb Z$ the following relations hold
\begin{equation}
\label{iso2}P_kU=UP_{k+1}.
\end{equation}
\end{proposition}

\begin{proof}
Indeed, for $k>0$ we have
\begin{align*}
P_{k}U &= (U^*)^{k}U^{k} U= (U^*)^{k}U^{k}UU^*U\\ &=
UU^*(U^*)^{k}U^{k}U\\ &= U(U^*)^{k+1}U^{k+1}=UP_{k+1},\\
P_{-k}U&=U^k(U^*)^k U\\ &= UU^{k-1}(U^*)^{k-1}U^*U\\ &=
UU^*UU^{k-1}(U^*)^{k-1}\\ &= UU^{k-1}(U^*)^{k-1}=UP_{-k+1}
\end{align*}
and
\begin{align*}
P_{-1}U&=(UU^*)U=UI=UP_0,\\ P_0U&=IU=U(U^*U)=UP_1.
\end{align*}
\end{proof}

The case when $U$ is unitary is trivial. Suppose that the
operator $U^*$ has a non-trivial kernel (the case of nontrivial
kernel of $U$ is similar). Let $f\in\ker U^*$. For every
$k=1,2,\dotsc$ consider the vector $(U^*)^kU^kf$. The following
situations can occur:

a) for all $k=1,2,\dotsc$ \ $(U^*)^kU^kf=f$. Then the vector
$f_0=f$ is a joint eigenvector of a commuting family $(P_k)$.

b) for some $k>0$ the following conditions hold:
\begin{align*}
(U^*)^lU^lf&=f,\qquad l=1,\dots,k-1, \\ (U^*)^kU^kf&\ne f.
\end{align*}
Put $f_0=f-(U^*)^kU^k f\ne0$. Then $(U^*)^kU^kf_0=0$, which
implies $U^kf_0=0$, \ $U^{k+1}f_0=0$ etc.\ and $f_0$ is a joint
eigenvector of a commuting family $(P_k)$.

In both the cases the relation \eqref{iso2} implies that
$f_0$, $Uf_0$, $U^2f_0$,\dots are orthogonal joint eigenspaces of
the family $(P_k)$ and can be chosen as the basis of the space.
The rest of the proof is obvious.
\end{proof}

2. To apply this theorem to the description of the irreducible
solutions of \eqref{xx}, we we need the following fact.

\begin{theorem}\label{thirr}
Let the pair $(X,X^*)$ satisfying \eqref{xx} be irreducible. If one
of the operators $X$, $X^*$ have a nonzero kernel, then the pair
$(U,U^*)$ is irreducible, i.e., any bounded operator commuting with
$U$ and $U^*$ is a multiple of the identity.
\end{theorem}

\begin{proof}
Put $C_0=I$, $C_k = (U^*)^kU^k$ for $k>0$, and $C_k = U^{|k|}(U^*)^{|k|}$
for $k<0$. 
Consider the commuting family $(C_k)_{k\in{\Bbb Z}}$. We show that
any bounded operator commuting with the operators $C_k$, $k\in{\Bbb
Z}$
commutes with $C^2$.

Denote $H_0=\ker U^*$. We claim that  for all $k\in {\Bbb Z}$
  $C_kH_0\subset H_0$.

Indeed,  for $k<0$ we have $C_k H_0=0$. For $k>0$ take an arbitrary
$f\in H_0$, then
\[
(U^*C_kf,U^*C_kf)=(UU^*C_kf,C_kf)=(C_kUU^*f,C_kf)=0.
\]
Thus $C_kf\in H_0$.

For all $l\ge0$ introduce a subspace $H_l=U^lH_0$. We now show
that
  for all $k\in {\Bbb Z}$  the operator $C_k$ maps the subspaces
$H_l$
  into themselves.

To show this, we use the fact that $f\in U^lH_0\iff (U^*)^lf\in H_0$.
For $k>0$ anf $f\in H_0$ we have $(U^*)^l C_k U^l f = C_{k+l}f\in
H_0$,
thus $C_kU^lf\in H_k$. The case of $k<0$ can be easily deduced from
the previous one.

Show that the subspaces $H_k$ are orthogonal to each other.

Indeed, for any $f_1,f_2\in H_0$, $k>l$
\[
(U^lf_1,U^kf_2)= (C_lf_1,U^{k-l}f_2)=((U^*)^{k-l}C_lf_1,f_2)=0,
\]
since $C_lf_1\in H_0$.

Now we show that the decomposition
\begin{equation}\label{aux2}
\textstyle
H=\bigoplus\limits_{k\ge0} H_k
\end{equation}
holds.

Indeed, it is obvious that $\tilde{H} =\bigoplus\limits_{k\ge0}
H_k$ is invariant with respect to the operator $U$. The invariance
under
the action of the operator $U^*$ follows from the fact that the
operators
$C_k$ preserve the subspaces $H_k$. To prove the invariance with
respect to
$C^2$ first observe that  $U^*C^2= F(C^2)U^*$  implies $C^2H_0\subset
H_0$.
Since $C^2U^k= U^kF^{ k}(C^2)$ and $C^2$ preserves $H_0$, the
operator
$C^2$ maps each $H_k$ into itself. The irreducibility implies
$\tilde{H}=H$.

Our purpose now is to show that  $H_0$ is an eigenspace of the
operator
$C^2$.

Let $C^2| _{H_0}=C^2_0\ne\lambda I$. Then there  exist a nontrivial
decomposition $H_0=H_0'\oplus H_0''$ into a direct sum of invariant
with respect to $C^2$ subspaces. Since $C_k$ commute with $C^2$, this
decomposition is also invariant with respect to $C_k$. Assume $H_k'=
U^kH_0'$, $H_k''=U^kH_0''$. By the commutativity of $C^2$ and $C_k$
we conclude that $C^2H_k'\subset H_k'$ and $\forall k$ \ $H_k'\perp
H_k''$.
Indeed, for $f_1\in H_0'$, $f_2\in H_0''$
\[
(U^kf_1,U^kf_2)=(f_1,C_kf_2)=0.
\]
Thus
$H=(\bigoplus\limits_{k\ge0}H_k')\oplus(\bigoplus\limits_{k\ge0}H_k''
)$
is a decomposition into the direct sum of invariant subspaces, which
contradicts the irreducibility.

As it is shown in \cite{romp}, all the subspaces $H_k$ are
eigenspaces
of the operator $C^2$, and \eqref{aux2} gives a decomposition of $H$
into the direct sum of eigenspaces.
Then $P_0=I-UU^*$ is a
projection
on $H_0$ and $P_k=U^kP_0(U^*)^k$ is a projection on $H_k$.

Let $X\in L(H)$ commutes with $U$ and $U^*$. Then  it commutes
with the projections $P_k$, $k\ge0$. But since $P_k$ are the
projections
on the eigenspaces of the operator $C^2$,  $X$ also commutes with $C^2$
and
thus $X=cI$.
\end{proof}

3. Now we apply the results on partial isometries to the description
of centered operators (note that the same result can be obtained
using \cite{vai}).
 Since we are interested in infinite-dimensional 
representations, only isometries, or co-isometries may arise in 
infinite-dimensional non-unitary case.

\begin{theorem}\label{thdeg}
All the irreducible representations of the relations \eqref{xx}
for which $\ker X\cup\ker X^*\ne\{0\}$ fall into one of the following
classes:
\begin{itemize}
\item[(i)] infinite-dimensional in $l_2$
$$
Xe_j=\lambda_je_{j+1},\qquad j=1,2,\dotsc,\quad \lambda_j>0 ;\quad 
\lambda _j = F^j(0);
$$

\item[(ii)] infinite dimensional in $l_2$
$$
Xe_1=0,\quad Xe_j=\lambda_je_{j-1},\qquad j=2,3,\dotsc,\quad 
\lambda_j>0; \quad \lambda_j \in F^{-j}(0).
$$
\end{itemize}
\end{theorem}

\begin{proof}
Indeed, the phase $U$ of the operator $X$
is a partial isometry (non-unitary), thus by Theorem~\ref{thirr}
the representation space for an irreducible $X$ is the same as
for irreducible $U$. Use Theorem~\ref{thiso} to represent the
operator $U$; the rest of the proof follows immediately from
\eqref{im}.
\end{proof}

4. Now, it remains to consider the case of unitary operator $U$. We 
distinguish between two cases: representations related to a single 
orbit, and representations related to a non-trivial ergodic 
quasi-invariant measure.

In there exists a measurable section of the dynamical system, then any 
ergodic measure is concentrated on a single orbit; in this case we will 
provide complete description of representations.

\begin{theorem}
Let the operator $X$ be invertible, and the dynamical system possesses 
a measurable section. Any 
infinite-dimensional irreducible representation of \eqref{xx} has the 
form
\[
Xe_k = \lambda_k e_{k+1}, \quad k \in \mathbb{Z},
\]
where $\lambda_k$, $k\in \mathbb{Z}$ is any sequence of positive 
numbers such that $F(\lambda_k^2) = \lambda_{k+1}^2$, $k \in \mathbb{Z}$.
Two such representations are unitarily equivalent if and only if the 
corresponding sequences coincide.
\end{theorem}

\begin{proof}
Let first the mapping $F(\cdot)$ be one-to-one on the spectrum of $C^2$.
In this case, any ergodic quasi-invariant measure is concentrated on a 
single orbit of the dynamical system, and is equivalent to atomic 
measure concentrated at points of the orbit. Choose any point $\lambda$ 
of the orbit, and take a unit eigenvector $e_0$, corresponding to 
$\lambda$. Write $e_k= U^ke_0$, $k\in \mathbb{Z}$. We claim that the 
space spanned by these vectors is invariant under $X$ and $X^*$. 
For $k<0$ we have
\begin{align*}
C^2 e_k &= C^2 U^{k} e_0 = C^2 {U^*}^{|k|} e_0 
\\
& = {U^*}^{|k|} F^{|k|}(C^2) e_0 = {U^*}^{|k|} F^{|k|}(\lambda) e_0
\\
&= F^{|k|}(\lambda) e_k.
\end{align*}
Also, since $F(\cdot)$ is one-to-one, in this case we 
have $C^2 U = U F^{-1}(C^2)$, which implies for each 
$k>0$:
\begin{align*}
C^2e_{k}& = C^2 {U} e_{k-1} = \dots 
\\
&=  C^2 U^k e_0 = U^k 
F^{-k}(C^2) e_0 = U^k F^{-k}(\lambda) e_0
\\
&= F^{-k}(\lambda) e_k.
\end{align*}
Therefore, all $e_k$ are eigenvalues of $C^2$, and hence, of $C$. In 
the chosen basis, the operator $X$ acts as stated in the theorem.

In a general case, consider a commuting family of self-adjoint 
operators, $C_k = U^k C^2 U^{-k}$. The relations are $C_k U^* = U^* 
C_{k+1}$. Now we have one-to-one action of $\mathbb{Z}$ on 
infinite-dimensional space $\mathbb{R}^\mathbb{Z}$. Using the same 
arguments, as above, one conclude that the joint spectrum of the
commutative 
family is concentrated on a single orbit, the representation space is 
generated by $\delta$-functions concentrated at points of the orbit, 
$U$ acts as a shift, and the formula follows.
\end{proof}

\begin{remark}
Notice that we do not assume that the dynamical system is one-to-one 
here. Looking closely at the spectrum of $C^2$, one sees that it can be 
just a chain of points, or a chain ``glued'' to a cycle or stationary 
point. In the case if the mapping of one-to-one on the spectrum, only 
chains may arise.
\end{remark}

If the dynamical system does not have a measurable section, 
representations listed in the theorem above, do not form, in general, 
the complete list of irreducible representations.

\begin{example} Consider quadratic mapping $\lambda \mapsto q 
\lambda(1-\lambda)^2$. For $q=4$, there exists a measurable subset
of the unit interval having full Lebesgue measure, on which the mapping 
is one-to-one. Moreover, the Lebesgue measure is ergodic, and 
therefore, operator $(X f)(\lambda) = \lambda \, (d (4 
\lambda(1-\lambda)^2)/d\lambda)^{1/2} f(4\lambda (1-\lambda)^2)$ in 
$L_2([0,1], d\lambda)$ gives an irreducible representation of the relation 
$XX^* = 4 X^*X(I-X^*X)$.
\end{example}

\begin{remark}
Notice that the description of nontrivial 
ergodic measures may be very complicated task; also, there can be lots 
of unitarily nonequivalent irreducible representations corresponding to 
the same nontrivial ergodic measure. 
\end{remark}

%\begin{example}(Continued fractions. Infinite-dimensional
%representations)
%\end{example}