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\begin{document}

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\title[%
 On operator relations]{%
 On operator relations,
 centered operators, and nonbijective dynamical systems}
\author{Vasyl'  Ostrovs'ky\u\i{}}
\address{%
 Institute of Mathematics\\
 National Academy of
 Sciences of Ukraine\\
 Tereshchenkivs'ka str.\ 3\\
 252601 Kyiv 4\\ 
 Ukraine}
\email{ostrov@imat.gluk.apc.org}
\date{\today}
\thanks{This research was not supported by any foundation}
%\keywords{<???>}
\subjclass{Primary \ \ \ , Secondary \ \ \ \ \ }

\begin{abstract}
 We consider pairs of operators $(A,B)$, with self-adjoint
 $A$ and centered $B$ satisfying relation $AB = BF(A)$,
 where $F(\cdot)\colon \mathbb{R} \to \mathbb{R}$ is a measurable 
 non one-to-one mapping. The case of completely non-unitary
 centered partial isometry $B$ is investigated in details.
 It is shown that the problem of unitary classification of
 such pairs is wild, but under the additional condition
 that $\ker B$ is an eigenvalue of $A$ (which is satisfied,
 in particular, when
 one studies relation of the form $XX^* = F(X^*X)$),  
 the complete unitary classification is provided. 
\end{abstract}

\maketitle

% End of top matter ---------------------------------------

\section*{Introduction}
In papers \cite{bos,romp,vaisam1} etc., pairs of (unbounded)
operators $A$, $B$, satisfying algebraic relation of the
form
\begin{equation}\label{abbf}
AB = BF(A),
\end{equation}
were studied, where $A$ is self-adjoint and $B$ is some
bounded or closed unbounded operator, $F(\cdot)\colon
\mathbb{R} \to \mathbb{R}$ is a one-to-one measurable
mapping. To avoid ambiguity considering algebraic relations
for unbounded operators, in \cite{bos, romp} the relation is
rewritten in terms of bounded functions of the operators. If
no additional conditions are imposed on $B$, then the
unitary classification problem for such pairs is wild, and
only a commutative model can be constructed
\cite{bos,berkon,romp}. In previous papers structure
problems were investigated for relation~\eqref{abbf} with
self-adjoint \cite{fa}, unitary or normal
\cite{vaisam1,romp}, centered partially isometric \cite{non}
operator $B$.

In the present paper, we show that the condition of being
centered is a natural one for the operator $B$, which
together with $A$ satisfies relation \eqref{abbf}. If the
dynamical system generated by $F(\cdot)$ is simple and acts
freely, then in the irreducible case $B$ is a weighted shift
operator, and $A$ is a multiplication operator.

If the dynamical system does not have a measurable section
or has cycles (or stationary points), then the problem of
unitary description of pairs $A$, $B$, satisfying
\eqref{abbf} is complicated: there exist pairs which
generate a factor which is not of type I. Note, however,
that in the case of one-to-one $F(\cdot)$ the relation is
not $p$-wild in the sense of \cite{pirsam}, i.e., the
description problem for pairs satisfying \eqref{abbf} does
not contain the same problem for arbitrary pairs (without
any relations).

We establish some properties of (bounded and unbounded)
centered operators, which enable one to get an information
about the structure of pairs $A$, $B$, satisfying
\eqref{abbf} with centered $B$. In particular, for
one-to-one $F(\cdot)$, the fact that $B$ or $B^*$ has a
nonzero kernel implies that the structure can be easily
described independently on properties of the dynamical
system.

In Section~2 we consider the case of non one-to-one
dynamical systems. If $\ker B = \{0\}$, the situation is
similar to the one-to-one case. But otherwise, unlike for
bijective case, the situation is much more complicated. In
fact, if there exist a point $\lambda$ such that
$F^{-1}(\lambda)$ contains three or more points, then the
classification problem for pairs, satisfying \eqref{abbf} is
$p$-wild even if $B$ is a centered partial isometry.
However, if one assume that $\ker B$ is invariant under $A$,
a complete description of all such irreducible pairs is
obtained. The mentioned condition is always satisfied for
relations of the form $XX^* = F(X^*X)$, which form a wide
class of relations arising in physical models (see
\cite{greek} etc.).

\section{One-to-one dynamical systems}

The main idea of the method which was used in \cite{romp,vaisam1} to
calculate all irreducible representations of the pair $A$,
$B$, satisfying \eqref{abbf} is based on the fact that under
some conditions imposed on the dynamical system (existence
of a measurable section) and for operator $B$ from some
class (self-adjoint, unitary, normal etc.), in any
irreducible representation the operator $A$ is a
multiplication operator and $B$ is a (bilateral or
unilateral) weighted shift operator. Below, we will show
that the widest natural class of such operators is the class
of centered operators introduced in \cite{mormu}. First, we
recall the definition of centered operator and list some of
their properties.

\subsection{Centered operators}
In this subsection, we study some properties of centered
operators, which are used below in the study of operator
relations.

In \cite{mormu}, a bounded operator $B$ was called centered
if the operators $B^k(B^*)^k$, $(B^*)^kB^k$ form a commuting
family i.e., for all $k,j\in \mathbb{N}$
\begin{equation}\label{cent}
[B^k(B^*)^k,B^j(B^*)^j]= [B^k(B^*)^k,(B^*)^jB^j]=
[(B^*)^kB^k,(B^*)^jB^j]=0.
\end{equation}

Let $B$ be a closed operator on a complex Hilbert space $H$.
Suppose that for each $k$ the operator $B^k$ is densely
defined. Then one can consider a family of self-adjoint
operators 
$$
A_k=B^k(B^*)^k,\quad B_k=(B^*)^kB^k.
$$
Similarly to the bounded case we say that the closed
unbounded operator $B$ is centered if any pair of operators
from the family $(A_k,B_j)$, $k,j\in \mathbb{N}$, commute in
the sense of commutation of their resolutions of the
identity.

In this section we rewrite the relations \eqref{cent} in the
form which enables one to investigate them using the
dynamical systems formalism developed in
\cite{bos,romp,vai}. We show that the irreducible (or
factor) representations of the relations fall into the two
cases: the case of $\ker B\cup\ker B^*=\{0\}$ and the
degenerated cases, and study them separately.

Let $B=UC$, $C= \sqrt{B^*B}$, $\ker U =\ker B$, be a polar
decomposition of the operator $B$. We also denote by
$E_{\mathbf A,\mathbf B}(\cdot,\cdot)$ the joint resolution
of the identity of the commuting self-adjoint family
$(A_k,B_k)_{k\in\mathbb{N}}$. The following proposition was
announced in \cite{ostur2}.

\begin{proposition} 
Relations \eqref{cent} are equivalent to the following
relations
\begin{equation}\label{im}
E_{\mathbf A,\mathbf B}(\Delta)U=UE(F^{-1}(\Delta)),\qquad
\Delta \in\mathfrak{B}
(\mathbb{R}^{\mathbb{N}}\times\mathbb{R}^{\mathbb{N}}),
\end{equation}
where the mapping $F^{-1}(\cdot)$ is defined as follows:
\begin{multline*}
F^{-1}(\mathbf x,\mathbf y)
=F^{-1}((x_1,x_2,\ldots),(y_1,y_2,\ldots))
\\
=\begin{cases}
((x_2/x_1,x_3/x_1,\ldots),(x_1,y_1x_1,y_2x_1,\ldots)),&
 x_1\ne0,\\
((0,0,\ldots),(0,0,\ldots)),&x_1=0.
\end{cases}
\end{multline*}
The phase $U$ of the operator $B$ is a centered operator.
\end{proposition}

\begin{proof}
Since the operators $A_k$ and $B_k$ form a commuting family,
there exists a dense in $H$ set $\mathcal{D}$ consisting of
entire vectors for these operators. Notice that
$\mathcal{D}\subset D(B^k)\cap D((B^*)^k)$ for any $k$.
Thus, for any $f,g\in \mathcal{D}$, $k\in \mathbb{N}$,
\begin{align*}
(Cf,U^*A_kg)&= (UCf, A_kg)=(Bf,A_kg) =(Bf, B^k(B^*)^kg)
\\
  &=(B^*Bf,A_{k-1}B^*g)=(B_1f, A_{k-1}B^*g)
\\
  &=(UCA_{k-1}B_1f,g)=(UA_{k-1}B_1Cf,g),
\end{align*}
which implies $(Cf,U^*A_kg)=(Cf,B_1A_{k-1} U^* g)$. Denote
by $P$ the projection on $(\ker C)^\perp$, then, since $\ker
C=\ker U$ and $UP=U$ we have $A_kUf=UA_{k-1}B_1f$. Similarly
one can obtain the equalities $A_1Uf=UB_1f$ and
$B_kA_1Uf=UB_{k+1}f$.

Applying the results of \cite{romp} to $A_k$, $A_{k-1}$ and
$B_1$ one can show that for any Borel set $\Delta\subset
\mathbb{R}^{\mathbb{N}}$ the following relation hold:
\begin{equation}\label{f1}
E_{B_1,\mathbf A}(\mathbb{R}\times\Delta)U=
UE_{B_1,\mathbf A}(F_1^{-1}(\mathbb{R}\times\Delta)),
\end{equation}
where 
$$
F_1^{-1}(y_1,x_1,x_2,x_3,\ldots)=
(x_1,x_2/x_1,x_3/x_1,\ldots).
$$
Notice that $F_1^{-1}(\cdot)$ can be defined
for $x_1=0$ as well. Indeed, since $E_{B_1,\mathbf
A}(\{0\}\times\mathbb{R} \times\mathbb{R}\times\cdots )$ is
the projection on $\ker B_1=\ker U$, the expression in the
right-hand side of \eqref{f1} is zero. Thus, for convenience,
we can put 
$$ 
F_1^{-1}(y_1,0,x_2,x_3,\ldots)=(0,0,0,\ldots).
$$

In a similar manner, from the equation $B_kA_1Uf=UB_{k+1}f$
one can derive the equality
\begin{equation}\label{f2}
E_{A_1,\mathbf B}(\mathbb{R}\times \Delta)U^*=
U^*E_{A_1,\mathbf B}(F_2^{-1}(\mathbb{R}\times \Delta))
\end{equation}
with
$$
F_2^{-1}(x_1,y_1,y_2,y_3,\ldots)=
(y_1,y_2/y_1,y_3/y_1,\ldots).
$$
Passing to the adjoint operators in \eqref{f2} and combining
it with \eqref{f1} we get the needed relations \eqref{im}.

The fact that $U$ is centered follows directly from
\eqref{im}.

The proof that a collection of non-negative self-adjoint
operators $(A_k,B_k)$ and a centered $U$ (being a partial
isometry) satisfying relations \eqref{im} generate a
centered operator, is a direct calculation.
\end{proof}

\begin{corollary}
For an unbounded centered operator $B$ there exists a dense
domain $\mathcal{D}$ on which operators $B^k$, $(B^*)^k$,
$k\in \mathbb{N}$, are defined, and which is invariant under
these operators.
\end{corollary}

\begin{corollary}
Let $B$ be a centered \textup{(}in general setting\textup{,}
closed unbounded\textup{)} operator on a Hilbert space
$H$\textup{.} The space $H$ can be decomposed into a direct
sum of the two invariant with respect to $B$\textup{,} $B^*$
subspaces\textup{,} $H=H_0\oplus H_1$ such that in $H_1$ \
$\ker B\cup \ker B^*=\{0\}$ and $H_0$ is generated by $\ker
B\cup \ker B^*$ (see \cite{mormu})\textup{.} 
\end{corollary}

The degenerated representations can be completely described up to
the unitary equivalence. The structure of 
representations in $H$ is more complex.

\begin{example}
Consider an operator $B$ satisfying the relation 
$$
BB^*=F(B^*B),
$$
where $F(\cdot)\colon \mathbb{R}\to \mathbb{R}$ is a
measurable one-to-one function (such operators are always
centered). If the dynamical system $x\to F(x)$ possesses
non-atomic ergodic measure, then there exist a
representation of this relation which generates a factor
which is not of type I \cite{romp,vai}.
\end{example}

To finish this subsection, we give the list of irreducible
centered partial isometries, essentially obtained in
\cite{mormu} (see also the bibliography therein).

Let $U$ be a centered partial isometry, i.e., 
\begin{equation}\label{iso1}
UU^*UU^*=UU^*,\qquad U^*UU^*U=U^*U.
\end{equation}
If the operator $U$ is unitary, then it is obviously centered
and satisfy \eqref{iso1}. The irreducible unitary operators are 
one-dimensional.
If the operator $U$ is an isometry. or co-isometry (adjoint to
isometry), but is not unitary, there exists a unique irreducible
solution of \eqref{iso1} which is the unilateral shift operator on $l_2$
\cite{halm2}.

\begin{proposition}\label{thiso}
Any centered irreducible representation of \eqref{iso1} 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 $\mathbb{C}^n$ for some $n=1,2,\dots $
\end{itemize}
\end{proposition}

\subsection{Relations with centered operator}
The following theorem shows that the class  of centered
operators is the widest natural one for relations
\eqref{abbf}.

\begin{theorem}
If the operator $B$ is
centered, and the dynamical system $\lambda \mapsto
F(\lambda)$ is one-to-one and simple and $F(\cdot)$ acts freely 
then the spectrum of
the operator $A$ in an irreducible representation is
simple. The pair is unitarily equivalent to one of pairs of the form
%
\begin{equation}\label{reps}
\begin{split}
Ae_k& =\lambda_k e_k, \quad Be_k = b_ke_{k+1}\qquad k \in
\mathbb{Z}, \quad(H=l_2(\mathbb{Z})),
\nonumber
\\
Ae_k& =\lambda_k e_k, \quad Be_k =b_k e_{k+1}\qquad k \ge1
,\quad(H=l_2(\mathbb{N})),
\nonumber
\\
Ae_k& =\lambda_k e_k, \quad Be_0 =0, \quad Be_k =
b_ke_{k+1}\qquad k \le1, \quad(H=l_2(\mathbb{Z}_-)),
\nonumber
\\
Ae_k& =\lambda_k e_k, \quad Be_n=0,\quad Be_k =
b_ke_{k+1}\qquad 1\le k \le n-1,\quad(H= \mathbb{C}^n),
\end{split}
\end{equation}
%
where for the last pair $n\ge1$ is some number, and
$(\lambda_k)$ is any sequence of real numbers such that for all
$k$, $F(\lambda_k) = \lambda_{k+1}$\textup{;} $b_k$ are arbitrary
positive numbers. 
\end{theorem}

\begin{proof}
First, it follows from \eqref{abbf} that in the irreducible
representation the spectral
measure of $A$ is ergodic with respect to $F(\cdot)$. Since
the dynamical system is simple, the spectral measure is
concentrated on a single orbit $O_\lambda$. This imply the
decomposition
\begin{equation}\label{decomp}
H= \bigoplus_{x \in O_\lambda } H_x
\end{equation}
into a direct sum of the eigenspaces of $A$, $A|_{H_x} =xI$.

To complete the proof, we show that  dimension
of each $H_x$ is 0 or 1. Let $H_x\ne \{0\}$. Since the
dynamical system acts 
freely, the von~Neumann algebra of operators in $H_x$ is
generated by the spectral projections of operators
$X^k(X^*)^k$, $(X^*)^kX^k$, $k=1$, \dots. But these
operators form a commuting family. Irreducibility implies
now $\dim H_x=1$ (see also Lemma~2 in \cite{ostur}).

Formulas \eqref{reps} follow now from decomposition
\eqref{decomp}. Note that if some of $b_k$ is zero, then
the space can be split into a direct sum of invariant
subspaces which would contradict the irreducibility.
Passing to a unitarily equivalent representation, one can
make  $b_k>0$ for all $k$.

The fact that formulas \eqref{reps} give an irreducible
representation of the relation with centered $B$ is a
straightforward calculation.  
\end{proof}

Note that self-adjoint, unitary, normal, isometric and
operators satisfying relation $XX^* =
f(X^*X)$ all are examples of centered operators. When
considering such particular classes of operators $B$, one
should make some assumptions on the coefficients $b_k$ in
\eqref{reps}. For example, if one assume that $U$ is
unitary operator, then $b_k=1$ for all $k$ and only
representations in $l_2(\mathbb{Z})$ arise.

\subsection{Decomposition for pairs with centered partial isometry}

To apply theorem on classification  of centered partial isometries
to the description of the irreducible
solutions of \eqref{abbf}, we prove an auxiliary statement which
may be of independent use.

Consider a commuting family $\mathbf A$ of self-adjoint operators 
and a closed operator $B$ satisfying for all $k$ the relations
\begin{equation}\label{aux1}
A_kB =BF_k(\mathbf A).
\end{equation}
Here $F_k(\cdot)$ are measurable one-to-one mappings ${\mathbb{R}}\to{\mathbb{R}}$;
the operator sense of the relations for unbounded operators is provided
in \cite{romp}. We also assume that the operator $B$ is centered.

\begin{theorem}\label{thirr}
Let the pair $(A,B)$ satisfying \eqref{aux1} be irreducible\textup{.} If one 
of the operators $B$\textup{,} $B^*$ have a nonzero kernel\textup{,} then the pair
$(B,B^*)$ is irreducible\textup{,} i.e.\textup{,} any bounded operator commuting with
$B$ and $B^*$ is a multiple of the identity\textup{.}
\end{theorem}

\begin{proof}
Consider the commuting family $(C_k)_{k\in{\mathbb{Z}}}$ (the operator
$C_0$ is not yet defined, but we can put $C_0=I$). We show that
any bounded operator commuting with the operators $C_k$, $k\in{\mathbb{Z}}$
commutes with $A$.

We divide the proof into several steps.

1. Denote $H_0=\ker B^*$. We claim that {\it for all $k\in {\mathbb{Z}}$, 
  $C_kH_0\subset H_0$.}

Indeed,  for $k<0$ we have $C_k H_0=0$. For $k>0$ take an arbitrary 
$f\in H_0$, then
$$
(B^*C_kf,B^*C_kf)=(BB^*C_kf,C_kf)=(C_kBB^*f,C_kf)=0.
$$
Thus $C_kf\in H_0$.

2. For all $l\ge0$ introduce a subspace $H_l=B^lH_0$. We now show that
  for any $k\in {\mathbb{Z}}$, {\it the operator $C_k$ maps the subspaces $H_l$ 
  into themselves}.

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

3. Show that the {\it subspaces $H_k$ are orthogonal to each other}.

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

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

Indeed, it is obvious that $\tilde{H} =\bigoplus_{k\ge0}
H_k$ is invariant with respect to the operator $B$. The invariance under
the action of the operator $B^*$ follows from the fact that the operators
$C_k$ preserve the subspaces $H_k$. To prove the invariance with respect to
$A$ first observe that  $B^*A= F(A)B^*$  implies $AH_0\subset H_0$.
Since $AB^k= B^kF^{\circ k}(A)$ and $A$ preserves $H_0$, the operator
$A$ maps each $H_k$ into itself. The irreducibility implies $\tilde{H}=H$.

5. Our purpose now is to show that {\it $H_0$ is an eigenspace of the operator
$A$.}

Let $A| _{H_0}=A_0\ne\lambda I$. Then there  exist a nontrivial
decomposition $H_0=H_0'\oplus H_0''$ into a direct sum of invariant
with respect to $A_0$ subspaces. Since $C_k$ commute with $A$, this 
decomposition is also invariant with respect to $C_k$. Assume $H_k'=
B^kH_0'$, $H_k''=B^kH_0''$. By the commutativity of $A$ and $C_k$
we conclude that $AH_k'\subset H_k'$ and $\forall k$ \ $H_k'\perp H_k''$.
Indeed, for $f_1\in H_0'$, $f_2\in H_0''$ 
$$
(B^kf_1,B^kf_2)=(f_1,C_kf_2)=0.
$$
Thus $H=(\bigoplus_{k\ge0}H_k')\oplus(\bigoplus_{k\ge0}H_k'')$
is a decomposition into the direct sum of invariant subspaces, which
contradicts the irreducibility.

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

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

The proven proposition shows that in the case of degenerate
$B$, all representations can be described independently of
the structure of the dynamical system by the formulas \eqref{reps} (see also
\cite{vai}).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Non one-to-one dynamical systems}

\subsection{Unitary $B$}
Case of unitary operator $U$ can be reduced to the case of
a one-to-one dynamical system, see, e.g., \cite{vaisam1}.
Here, we only formulate the condition of simplicity of the
dynamical system given in \cite{vaisam1}. This condition will be used in this paper
whenever we refer to simple (non one-to-one) dynamical
system.

\begin{proposition}
The dynamical system $\tilde F\colon \mathbb{R}^\mathbb{Z} \to
\mathbb{R}^\mathbb{Z}$ is simple if and only if
the dynamical system $F\colon \mathbb{R}\to \mathbb{R}$ for non
one-to-one $F$ possesses the following property\textup{:}

there exists a partition
$\mathbb{R} = \bigcup_{n \in \mathbb{N}} \Delta_n$, $\Delta_k \in
\mathfrak{B}(\mathbb{R})$ such that

\textup{1)} for any $k$ there exists $j$ such that
$F(\Delta_k)=\Delta_j$ and $F(\cdot)$ is one-to-one on
$\Delta_k$\textup{; }

\textup{2)} if for some $n$\textup{,} $k$ the mapping $F^n(\cdot)$ maps
$\Delta_k$ into itself\textup{,} then $F^n(\cdot)$ is identity on
$\Delta_k$. 
\end{proposition}

\subsection{Case of nonzero kernel}
In the case of non one-to-one dynamical system and
nonunitary centered partial isometry $B$, the situation
differs from one with unitary $B$ or one-to-one $F(\cdot)$.
We start with a simple example.

\begin{example}
Let the mapping $F(\cdot)$ be such that for three distinct
points $\lambda_1$, $\lambda_2$, $\lambda_3$, we have
$F(\lambda_1)=\lambda_3$ and $F(\lambda_2)= \lambda_3$.
Then the pair
$$
A = \left( \begin{array}{ccc} \lambda_1& 0&0\\ 0&\lambda_2
&0\\ 0&0& \lambda_3 \end{array}\right), \quad U = \left(
\begin{array}{ccc} 0 & 0& 0\\ 0&0&0\\ 1/2 &
1/2&0\end{array}\right) ,
$$
gives an irreducible representation of the relation $AU =
UF(A) $. 
\end{example}

This example shows that unlike in the previous cases, there
can be more than one eigenvalue of $A$ which are mapped by
$B$ into another eigenvalue. This phenomenon has the
following consequence.

\begin{theorem}
Let the mapping $F(\cdot)$ has a point $\lambda$ such that
set $F^{-1}(\lambda)$ contains at least three points. Then
the relation $AB= BF(A)$ with self-adjoint $A$ and centered
partial isometry $B$ is wild\textup{,} i.e.\textup{,} the problem of unitary
classification of its representations contains the same
problem for arbitrary pairs of self-adjoint operators.
\end{theorem}

\begin{proof}
Let $\lambda_1$, $\lambda_2$, $\lambda_3$ be three distinct
points such that $F(\lambda_i)=F(\lambda_2)=
F(\lambda_3)=\mu$. We assume that $\lambda_1\ne \mu$, and
$\lambda_2 \ne \mu$. Let $P_1$, $P_2$, $P_3$ be arbitrary 
projection operators in some Hilbert space $\mathfrak{H}$ such
that $P_1\perp P_2$. Put $\mathfrak{H}'$ to be a Hilbert space
isomorphic to a range of $P_3$.  In the space $H=\mathfrak{H}\oplus \mathfrak
{H}'$ introduce the operators
$$
A=\lambda_1 P_1+\lambda_2 P_2 + \lambda_3 (I-P_1-P_2) + \mu
P_0,\quad B = UP_3,
$$
where $P_1$, $P_2$, $P_3$ are extended to
operators in $H$ putting them on $\mathfrak{H}'$ to be zero,
$P_0$ is a projection onto $\mathfrak{H}'$ and $U$ is a
(canonical) isometry between $\range(P_3)$ and $\mathfrak
{H}'$.

For such operators $A$ and $B$ the following holds:

a) $A$ is self-adjoint, $B$ is a centered partial isometry.
Indeed, $BB^*= P_0$, $B^*B = P_3$ and $B^2 = (B^*)^2 =0$,
and since $P_3\perp P_0$, $B$ is centered;

b) $AB = BF(A)$; it follows directly from the definition of
the operators;

c) The pair $A$, $B$ in $H$ is irreducible if and only if the
triple of projection operators $P_1$, $P_2$, $P_3$
($P_1\perp P_2$) in $\mathfrak{H}$ is irreducible. Indeed, let a
projection $P$ in $\mathfrak{H}$ commutes with $P_1$, $P_2$,
$P_3$. Put $\tilde P = P + UPU^*$, where, as before, $P$ is
extended to a projection in $H$. Then, taking into account
that $P$ has its range in $\mathfrak{H}$ and $UPU^*$ has its
range in $\mathfrak{H}'$, we have $P_1UPU^* = P_2UPU^* =0$ and
\begin{align*}
A\tilde P& = (\lambda_1 P_1 + \lambda_2 P_2 + \lambda_3 (I -
P_1 + P_2) + \mu P_0 )(P + UPU^*)
\\
& =(P + UPU^*) (\lambda_1
P_1 + \lambda_2 P_2 + \lambda_3 (I - 
P_1 + P_2) + \mu P_0 ) \\
&= \tilde P A,
\end{align*}
and since the range of $P_3$ is in $\mathfrak{H}$, and $U^*U=P_3$
\begin{align*}
B\tilde P& = UP_3(P +UPU^*) = UP_3 P= UP P_3
\\
&=UPU^*UP_3 = (UPU^*+P)UP_3 = \tilde P B.
\end{align*}
Conversely, let $\tilde P$ be a projection in $H$ commuting
with $A$ and $B$. Since $\tilde P$ commutes with $A$, it
also commutes with $P_1$ and $P_2$ being spectral
projections of $A$. Moreover, since $P$ commutes with $B$,
it also commutes with $B^*B=P_3$. Thus, the restriction of $P$ to $\mathfrak
{H}$ commutes with the restrictions to $\mathfrak{H}$ of $P_1$,
$P_2$, and $P_3$.

d) Two pairs $A$, $B$, and $A'$, $B'$ in $H$ and $H'$ are
unitarily equivalent if and only if the corresponding
triples $P_1$, $P_2$, $P_3$, and $P_1'$, $P_2'$, $P_3'$,
($P_1\perp P_2$, $P_1'\perp P_2'$) are unitarily
equivalent. One can check it similarly to the previous
paragraphs.

The conditions listed above mean that we constructed an exact
and strict
functor from the category of triples of projections, two of
which are orthogonal, into the category of representations
of the relation $AB = BF(A)$ \cite{krupir}. But since the
former problem is wild \cite{krusam}, the same holds for
the latter one.  
\end{proof}

Now, we give an additional condition, which makes the
classification problem ``tame''.

\begin{theorem}\label{good}
Let $A$\textup{,} $B$\textup{,} be an irreducible pair\textup{,} satisfying
relation~\eqref{abbf}\textup{,} where $B$ is a \textup{(}non-unitary\textup{)}
centered partial isometry with $\ker B\ne \{0\}$ being
eigenspace of $A$. Then the
problem of description of all irreducible representations of
this relation is tame. Any irreducible representation has
the form \eqref{reps}.
\end{theorem}

\begin{proof}
We consider only irreducible case. The general case can be
reduced to it easily.

Let $A\restriction \ker B = \lambda I$. It follows from the
relation that
\[
E_A(\Delta)B = BE_A(F^{-1}(\Delta), \qquad \forall \Delta \in
\mathfrak{B}(\mathbb{R}).
\]
This implies, in turn, that $E_A(F^{-1}\lambda)$ is a
projection on $B^*\ker B$. We calim that tha latter space is an
eigenspace of  $A$. Indeed, let $F^{-1}(\lambda)$ contains more
than one point of spectrum of the operator $A$, i.e.,
$F^{-1}(\lambda)$ can be decomposed into union of two
measurable sets, $F^{-1}(\lambda)= \Delta_1 \cup \Delta_2$,
such that each of them has nonzero spectral measure. Choose a
vector $f\ne0$ such that $E_A(\Delta_1) f = f$. If vector
$f'= U^*Uf$ is equal to $f$ then the space spanned into this
vector and its images under action of the operators is
invariant, and by the irreducibility we get $E_A(\Delta_2) =0$,
which contradicts the assumption. On the other hand, since
\[
B f' = B B^*B f =Bf,
\]
we conclude that $f-f' \in \ker B$, which is also impossible.
Therefore, $B^*\ker B$ is again an eigenspace of $A$. Repeating
this procedure, we conclude that for each $n\ge1$ subspace
$(B^*)^n$ is an eigenspace of $A$, and their direct sum is the
whole representation space.

Applying results of the previous section, we get that for
$B$ being (nonunitary) partial isometry the spectral
measure of $A$ is atomic independently of a topological
srtucture of the d.s.\ and $F(\cdot)$ is one-to-one on the
support of the spectral measure.  
\end{proof}

The assumption of the theorem is satisfied,
in particular, if $\ker B= \ker A$. This
case covers an important class of relations of the form
$XX^* = F(X^*X)$ (see Example~\ref{ed} below).

\begin{example}\label{ed}
\textit{Relation $XX^* = F(X^*X)$}. Consider operator relation
of the form $XX^* = F(X^*X)$ with closed densely defined
operator $X$ and a non-bijective $F$. The operator $X$
is centered, and for its polar
decomposition $X= BC$, $C= \sqrt{X^*X}$ we have (see \cite{vai}) $C^2B^* =
B^*F(C^2)$. Moreover, $\ker B = \ker C$, and thus,
\[
\ker B^* =\ker X^*=\ker XX^* = \ker F(C^2).
\]
Since the operator $C^2$ commutes with $F(C^2)$, it preserves
$\ker B^*$. Therefore, the condition of Theorem~\ref{good} is
satisfied. This enables one to get a complete list of irreducible
representations of the relation, for which the operator $X$ has
non-unitary phase. Notice that the unitary part may have much
more complicated structure \cite{vai}.
\end{example}

The author thanks Prof. Yu.~S.~Samoilenko for kind attention
and helpfull discussions.

%\bibliography{ref}
%\bibliographystyle{amsplain}

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\end{document}