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\section[Representations of two-dimensional Lie algebras, their
nonlinear transformations, and
semilinear relations]{Representations of two-dimensional Lie algebras, their
nonlinear transformations, and\break
semilinear rela\-ti\-ons}

\label{sec:1.3}

\markright{1.3. Lie
algebras  and semilinear relations}

\subsection{Representations of two-dimensional real Lie algebras and
their nonlinear transformations by bounded operators}\label{sec:1.3.1}

\textbf{1.}
Lie algebras $(IV_0)$, $(IV_1)$, $(IV_2)$, similarly to the relations
$(V_0)$, $(V_1)$, $(V_1')$, can be treated from a general point of
view.
Namely, all these relations are partial cases of the relation
\begin{equation}\label{deformedlie}
[A,B] = iP_2(A),
\end{equation}
where $P_2(A)$ is a real quadratic polynomial.
 
If $A= A^*$, $B=B^* \in L(H)$, then their study can be performed
by using the
following theorem.

\begin{theorem} \textup{(D. Kleinecke, F.V. Shirokov).}
If $P$ and $Q$ are
bounded operators, and $[P,[P,Q]]=0$, then the operator $[P, Q]$ is
quasi-nilpotent, i.e.,
\[
\lim_{n \to \infty} \sqrt[n]{\|[P, Q]^n\|} =0.
\]
\end{theorem}

\begin{proof}
The Kleinecke--Shirokov theorem follows, for example, from the 
formula $\ad_P^nQ^n= n!\, (\ad_PQ)^n$, where $\ad_PX = PX - XP$ is
a
bounded operator in $L(H)$, and $\|\ad _P\| \le 2 \|P\|$. Then
\par\vskip\abovedisplayskip\noindent
\hfill $\displaystyle\sqrt[n]{\|[P,Q]^n\|}
\le \frac 2{\sqrt[n]{n!}} \, \|P\|\cdot \|Q\|
\to
0, \qquad n \to \infty.$ \hfill
\end{proof}

In Section \ref{sec:1.3.3} below we will give many analogies  of this
theorem.

\medskip\noindent\textbf{2.}
Now we consider bounded self-adjoint operators
satisfying~\eqref{deformedlie}.

\begin{proposition}
Irreducible pairs, $A$, $B$ of bounded self-adjoint operators
which
satisfy the relation~\eqref{deformedlie} are one-dimensional, and they
are given by\textup:
$A=\lambda$, $B=\mu$, where $(\lambda, \mu) \in M = \{(\lambda, \mu)
\in
\mathbb{R}^2 \mid P_2(\lambda) =0\}$.

An arbitrary bounded pair has the form
\[
A = \int_M \lambda \, dE(\lambda, \mu), \quad B = \int_M \mu\,
dE(\lambda, \mu),
\]
where $E(\cdot, \cdot)$ is the resolution of the identity on $M$.
\end{proposition}

\begin{proof}
Indeed, \eqref{deformedlie} implies $[A, [A,B]]=0$, and by the
Kleinecke--Shirokov theorem, the operator $[A,B]$ is quasi-nilpotent.
But
since $[A,B]$ is skew-adjoint, it yields $[A,B]=0$. Then $P_2(A)=0$,
and
the statement follows from the spectral theorem for a pair of
commuting
self-adjoint operators.
\end{proof}

\begin{remark}
The  proposition  above implies that if the polynomial
$P_2(\cdot)$
has
no real roots, then there are no bounded self-adjoint pairs that
satisfy \eqref{deformedlie}. In particular, there are no bounded
pairs
that satisfy CCR (relation $(IV_1)$), or $[A,B] = i(A^2 +I)$
(relation
$(V_1)$).
\end{remark}

\begin{remark}
Relation $(V_2)$ $[A,B] = i(A^2 +B)$ can also be rewritten in the
equivalent form $[A, A^2 +B] = i(A^2 +B)$, which can be reduced by a
non-degenerate non-linear change of variables, $\tilde A = A^2 +B$,
$\tilde B = B$, to the form $[\tilde A, \tilde B] =i\tilde A$. Since
$\tilde A$ and $\tilde B$ are also bounded self-adjoint operators,
by the Kleinecke--Shirokov theorem, we have $[\tilde A, \tilde B]=0$,
and
$[A,B]=0$, which yields $A^2 =-B$.

Irreducible representations of relation$ (V_2)$ are one-dimensional,
$A=\lambda$, $B=\mu$, $(\lambda, \mu) \in M_{(V_2)} = \{
(\lambda, \mu) \in \mathbb{R}^2 \mid \lambda^2 = -\mu\}$.

An arbitrary pair of bounded self-adjoint operators satisfying
relation
$(V_2)$ has the form
\[
A = \int_{M_{(V_2)}} \lambda\, dE(\lambda, \mu), \quad A =
\int_{M_{(V_2)}}
\lambda\, dE(\lambda, \mu),
\]
where $E(\cdot, \cdot)$ is the resolution of the identity on
$M_{(V_2)}$.
\end{remark}

\noindent\textbf{3.}
If in the study of representations of relations given by the families
$(IV)$ and
$(V)$, we restrict ourselves to only the
finite-dimensional level, then
instead
of the Kleinecke--Shirokov
theorem, which was proved in 1956, we could apply
its
finite-dimensional variant, which is the following theorem proved much
earlier (1935). We formulate it in the simplest case.

\begin{theorem} \textup{(N. Jacobson).} If $\dim H < \infty$ and
$[P,[P,Q]]=0$,
then the operator $[P,Q]$ is nilpotent.
\end{theorem}

\begin{remark}
In the relations $(IV)$ and $(V)$ no new irreducible
representations arise
when passing from the finite-dimensional to the bounded case.


Of course, considering unbounded operators, for which the
Klei\-necke--Shirokov theorem fails, one obtains a much more consistent
representation theory of the relations $(IV)$ and $(V)$.
\end{remark}

\subsection{Pairs of operators connected by semilinear relations}
\label{sec:1.3.2}

In Sections \ref{sec:1.3.2}--\ref{sec:1.3.5} we study a wide class of
so-called semilinear relations which connect pairs of bounded
operators $A=A^*$, $B\in L(H)$.  They appear as a generalization
of the relations $(IV)$ and $(V)$ studied in  previous sections.
We prove many Kleinecke--Shirokov type theorems
(Section~\ref{sec:1.3.3}).  For general semilinear relations, we describe
properties of irreducible representations (Section~\ref{sec:1.3.4}). All
irreducible representations are classified up to unitary
equivalence for semilinear ${ F}_4$-relations (Section~\ref{sec:1.3.5}).

\medskip\noindent\textbf{1.}
 Consider bounded operators $A=A^*$,
$B\in L(H)$, which satisfy a relation of the form:
\begin{equation}\label{sr}
\sum_{i=1}^nf_i(A)Bg_i(A)=h(A),
\end{equation}
where $f_i(\cdot)$, $g_i(\cdot)$, $h(\cdot)$,  $i=1$, \dots, $n$, are
complex
bounded Borel mappings defined on ${\mathbb R}$ or a subset $D$,
$\sigma(A) \subset D$. Relation (\ref{sr}) is called semilinear
and the operators $A$, $B$ are called representation of (\ref{sr}).

If $f_i(\cdot)$, $g_i(\cdot)$,
$h(\cdot)$, $i=1$, \dots, $ n$, are polynomials, then the operators
$A$, $B$, $B^*$ define
a representation of the
$*$-algebra $\mathfrak A$ with three generators $a=a^*$, $b$, $b^*$
satisfying
the relation $\sum_{i=1}^nf_i(a)\,b\,g_i(a)=h(a)$.
This algebra is the quotient algebra of the free $*$-algebra with
three
generators
${\mathbb C}\langle a=a^*, b,b^*\rangle$
with respect to the two-sided $*$-ideal
generated by the element $\sum_{i=1}^nf_i(a)\,b\,g_i(a)-h(a)$.

Unless otherwise stated, we assume here $f_i$, $g_i$, $h$ to be
defined on the
whole ${\mathbb R}$. The general case can be easily derived from this
one.

\begin{remark} \rm
As before, if the operators $A=A^*$, $B$ are unbounded, then it is
necessary
to define the meaning of the operator equality (\ref{sr}).
We investigate the question of the 
``correct'' definition of relation (\ref{sr}) with unbounded
operators
for some special relations in \cite{sam_tur_sh}.
\end{remark}

\noindent\textbf{2.}
The study of bounded representations of (\ref{sr}) can be reduced to 
the study of operators satisfying 
the corresponding homogeneous relation:
\begin{equation}\label{hom}
\sum_{i=1}^nf_i(A)Bg_i(A)=0.
\end{equation}

If the function $\phi (t)=h(t)/\sum_{i=1}^nf_i(t)g_i(t)$ is bounded
on the
spectrum of the operator $A=A^*$, then the pair of operators $A$,
$B=\phi (A)$ is a particular solution of the inhomogeneous relation
(\ref{sr}),
and the general solution of (\ref{sr}), with the operator $A$ fixed,
consists
of all pairs
($A$, $B+\phi(A)$) where ($A$, $B$) satisfies (\ref{hom}). Let
$$
\tilde\phi(t)=
\begin{cases}
h(t)/\sum_{i=1}^nf_i(t)\,g_i(t), &\mbox{if}\
\sum_{i=1}^nf_i(t)\,g_i(t)\ne 0,\\
0,&\mbox{otherwise}.
\end{cases}
$$

\begin{proposition}
If \eqref{sr} has a \textup(bounded\textup)
solution $(A,B)$, then there exists
 $C>0$ such that
$$
|h(t)|\le C\Bigl|\sum_{i=1}^nf_i(t)g_i(t)\Bigr|, \qquad t\in \sigma(A).
$$
Any solution $A$, $B$ of
\eqref{sr} has the form $B=B'+\tilde\phi(A)$ where $(A, B')$ is
a
representation of \eqref{hom}.
\end{proposition}

\begin{proof}
Let $A$, $B$ satisfy \eqref{sr}, and  let
$\psi\colon L(H)\rightarrow {\mathfrak A}$ be a conditional expectation on
the $W^*$-algebra $\mathfrak A$ of all operators which commute with the
operator $A$.
Applying $\psi$ to both sides of equation \eqref{sr}, we obtain
$$
\sum_{i=1}^nf_i(A)\,\psi (B)\,g_i(A)=h(A).
$$
From this we conclude that $(A, \psi(B))$ is a solution of
\eqref{sr}
and
$$
\psi (B)\sum_{i=1}^nf_i(A)\,g_i(A)=h(A).
$$
Thus, $\psi(B)=\tilde\phi(A)$
on  the  image  of   the   operator
$\sum_{i=1}^n f_i(A)\,g_i(A)$.
Since $\psi(B)$ is a bounded operator, there exists $C>0$ such that
for every $t\in\sigma(A)$ and  $\sum_{i=1}^nf_i(t)\,g_i(t)\ne 0$ we
have
\[
|h(t)|\le C\Bigl|\sum_{i=1}^nf_i(t)g_i(t)\Bigr|.
\] 
On the other hand,
$\ker \sum_{i=1}^nf_i(A)\,g_i(A)\subset \ker h(A)$, which implies that
the
inequality holds for every $t\in\sigma(A)$, and
$(A, B-\tilde\phi(A))$ is a
representation of relation (\ref{hom}).
\end{proof}

It is easy to prove that the correspondence between irreducible
representations of (\ref{sr}) and (\ref{hom}) preserves unitary
equivalence.
So, in studying bounded representations,
we can restrict ourselves only to the homogeneous
relations (\ref{hom}).

\medskip\noindent\textbf{3.}
To the semilinear relation \eqref{hom}, we associate:
\begin{itemize}
\item[a)] the characteristic function:
$$
\Phi(t,s)=\sum_{i=1}^nf_i(t)\,g_i(s),\qquad (t,s\in {\mathbb R});
$$

\item[b)] the characteristic binary relation:
$$
\Gamma=\{(t,s)\mid \Phi(t,s)=0\}\subset {\mathbb R}^2;
$$

\item[c)] an oriented graph $({\mathbb R}, \Gamma)$, where the edge
\vrule width0pt height 0pt depth 8pt$\edgeright{t}{s}$,
$t$, $s\in {\mathbb R}$, belongs to the graph if and only
if $\Phi(t,s)=0$.
\end{itemize}

 If no confusion  arise, we will write simply $\Gamma$ for the
graph ($D$, $\Gamma$) and call it the graph of
relation (\ref{hom}).  Any subset $M\subset {\mathbb R}$ determines
a subgraph $\Gamma\restriction_M$ such that its vertices are points of $M$
and its edges are edges of $\Gamma$ which connect points of $M$.

If $\Phi(t,s)=0$ is equivalent to $\Phi(s,t)=0$, the
graph $\Gamma$ together with the edge $\edgeright{}{}$ also contains
the
edge $\edgeleft{}{}\!\!$. In this case, we will consider
the graph as  non-oriented.

In what follows $\Gamma_s$ denote the set
$\{(t,s)\mid \Phi(t,s)=0,\, \Phi(s,t)=0\}\subset {\mathbb R}^2$ or the
corresponding non-oriented graph.

Consider  some examples of semilinear relations.

1) Relations
$\ad_A(B)=[A,B]=AB-BA=0$ and $(\ad_A)^n(B)=[A,\dots[A,B]\dots]=0$
have
the characteristic functions of the form\allowbreak{}
$\Phi(t,s)=t-s$ and $\Phi(t,s)=(t-s)^n$ respectively, and define the
same
graph,  all connected components of which are the following:
\vrule width0pt height0pt depth8pt$\ear {\lambda}$, $\lambda\in{\mathbb R}$.

2) Characteristic  functions corresponding to the relations
\[
\ad_{A,-1}(B)=\{A,B\}=AB+BA=0
\]
and
$(\ad_{A,-1})^n(B)=\{A,\ldots\{A,B\}\ldots\}=0$ 
 are
$\Phi(t,s)=t+s$ and $\Phi(t,s)=(t+s)^n$, respectively. As
before, they define the same graph with connected components of
the form:  
$$ 
\ear {0},\quad \edge
{\lambda}{\smash{\hbox{\kern-5pt$\scriptstyle-\lambda$}}},\qquad
({\lambda>0}).  
$$

3) Let $ABA=\alpha B$, $\alpha\in{\mathbb R}$. Then $\Phi(t,s)=ts-\alpha$,
and
all connected components of the corresponding graph are of the form:

a) $\alpha\ne 0$
$$
\edge
{\alpha\lambda^{-1}}
{\lambda},\qquad({
\lambda^2\ne\alpha,0})\qquad \onepoint {0},\qquad
\ear{\smash{\pm\sqrt{\alpha}}}\ \
,\qquad ({\alpha>0}); 
$$

b) if $\alpha=0$, then any vertex $\lambda\in{\mathbb R}\setminus\{0\}$ is connected
with $0$ by the edge \vrule width0pt height0pt
depth8pt$\edge{\lambda}{0}$.

4) Any connected component of the graph corresponding to the relation
$$
A^2B-(q+q^{-1})ABA+BA^2=0,\qquad q\in{\mathbb
R}\setminus\{-1,0,1\},
$$
 is either an infinite chain:
$$
\qquad\chainl{\smash{q^{-1}t}}{t}{\smash{qt}} ,\qquad t\ne 0,
$$
or \vrule width0pt height0pt
depth8pt$\ear{0}.$


\subsection{Kleinecke--Shirokov type theorems} \label{sec:1.3.3}



We begin with the study of the structure of
operators $A=A^*$, $B$ satisfying \eqref{hom};
in particular, we will look at the
connection
between  the spectral properties of the operator $A$ and the structure
of the operator $B$.  It turns out that for a broad class of
semilinear relations the characteristic binary relation
corresponding to them completely defines solutions of \eqref{hom}.
This fact provides many  Kleinecke--Shirokov type
theorems.


\medskip\noindent\textbf{1.}
 We start by studying  finite-dimensional
representations of semilinear relations. Note that the same arguments
work
in the more general case where
the operator $A$ in the representation has discrete spectrum. So
let $\sigma (A)=\{\lambda_1,\ldots, \lambda_m\}$ and
$H_{\lambda_j}$ be eigenspaces of $A$ corresponding to
$\lambda_j$.  With respect to the decomposition
$H=H_{\lambda_1}\oplus \ldots \oplus H_{\lambda_m}$, the operator
$B$ can be written in the form of the block matrix:  $
B=(B_{ln})_{l,n=1}^m. $



\begin{proposition}\label{pr1} For
operators $A$, $B$ to define a representation of the relation
\eqref{hom}, it is  necessary and sufficient that the block
 matrix $B=(B_{sn})_{s,n=1}^m$ be supported 
by $\Gamma\restriction_{\sigma
(A)}$ \textup(i.e., $B_{sn}=0$ for any $(\lambda_s, \lambda_n)\notin
\Gamma$\textup).  
\end{proposition}



\begin{proof}
The statement follows immediately from the equality
\par\vskip\abovedisplayskip\hfill$\displaystyle
\Bigl(\sum _{i=1}f_i(A)\,B\,g_i(A)\Bigr)_{kj}=
\Phi (\lambda_k,\lambda_j)\,B_{kj},\qquad k,j=1,\ldots,m.
$\hfill
\end{proof}



\noindent\textbf{2.}
A pair $A=A^*$,  $B=B^*$, is a representation of relation \eqref{hom}
if and
only if
the block matrix $B=(B_{ij})_{i,j=1}^n$ is supported by
$$
\Gamma_s|_{\sigma(A)}=\{(t,s)\in\sigma(A)\times\sigma(A)\mid
\Phi(t,s)=\Phi(s,t)=0\}. 
$$ 
In fact, if $A=A^*$,
$B=B^*$ is a representation of (\ref{hom}), then $A$, $B$ also satisfy
the relation
$$
\sum_{i=1}^n\bar g_i(A)\,B\,\bar f_i(A)=0.
$$
Hence the
block matrix $B=(B_{ij})_{i,j=1}^m$ is supported by
$(\Gamma\cap\Gamma^*)|_{\sigma(A)}$, where
$\Gamma^*=\{(t,s)\in {\mathbb R}\times{\mathbb R} \mid \Phi(s,t)
=\sum_{i=1}^ng_i(t)\,f_i(s)=0\}$, which gives the required statement.



\medskip\noindent\textbf{3.}
 Now we will try to remove the condition in
Proposition~\ref{pr1} that
$\sigma(A)$ is discrete.
For this purpose, we will consider a more general situation.



Let $M$, $N$ be normal operators acting on Hilbert spaces $H_M$,
$H_N$,
respectively, and let $E_{M}(\cdot)$, $E_N(\cdot)$ be their spectral
measures.

\begin{definition}\label{def1}
We say that a subset ${ F}\subset{\mathbb C} \times{\mathbb C} $
$(M,N)$-supports an operator $B\colon H_N\rightarrow H_M$ if
$$
E_M(\alpha)\,B\,E_N(\beta)=0
$$
for any pair $(\alpha,\beta)$ of Borel sets such that
$(\alpha\times\beta)\cap{ F}=\emptyset$.
\end{definition}

It is not difficult to prove that there exists a smallest closed
set
${ F}$
supporting $B$ (take the complement to the union of all such open
sets $\alpha\times\beta$).
We will denote this set by $\supp_{M,N}(B)$. It is clear that
$\supp_{M,N}(B)\subset\sigma(M)\times\sigma(N)$; in particular,
it
is
a subset of ${\mathbb R} \times{\mathbb R} $ when $M$ and $N$ are
self-adjoint.
We shall also write $\supp_M(B)$ instead of
$\supp_{M,M}(B)$.



\begin{theorem} \label{th1-semi}
If $A$, $B$ is a representation of relation \eqref{hom}, then
$$
\supp_A(B)\subset \Gamma,
$$
where $\Gamma$ is the binary relation corresponding to \eqref{hom}.
\end{theorem}

\begin{proof}
Let $\alpha\times \beta$ do not intersect $\Gamma$. Denote by $\mu$
the
scalar spectral
measure of the operator $A$.
By Luzin's theorem, for every $\varepsilon>0$, there exist compact
sets $\alpha'\subset \alpha$ and $\beta'\subset \beta$ such that
$\mu(\alpha\setminus \alpha')<\varepsilon$, $\mu(\beta\setminus
\beta')<\varepsilon$,
and the functions $f_i'=f_i\restriction_{\alpha'}$,
$g_i'=g_irestriction_{\beta'}$ are continuous. Put $P=E_{A}(\alpha')$,
$Q=E_{A}(\beta')$, $B'=PBQ\colon QH\rightarrow PH$. The operator of
multiplication,
$$
\Delta\colon X\mapsto \sum_{i=1}^nf_i'(A)\,X\,g_i'(A),
$$
acts on the space  $L(QH, PH)$.
The spectrum of the operator $\Delta$ belongs to the set
$$
\Bigl\{\sum_{i=1}^nf_i(\lambda)\,g_i(\mu)\mid\lambda \in \alpha',\,\mu \in
\beta'\Bigr\}= \Phi(\alpha'\times \beta'), $$
see, for example \cite{137}.
By the condition of
the theorem, $\Phi(\alpha'\times \beta')$ does not contain zero,
hence the operator $\Delta$ is invertible. Since $\Delta(B')=0$, we
have that $B'=0$, i.e., $E_A(\alpha')\,BE_A(\beta')=0$. Letting
$\varepsilon \rightarrow 0$, we obtain  that
$E_A(\alpha)\,BE_A(\beta)=0$.
\end{proof}



In a similar way, more general results can be proved.

\begin{theorem} \label{th2-semi}
If $M$, $N$ are normal operators and
\begin{equation}\label{p20'}
\sum_{i=1}^nf_i(M)\,B\,g_i(N)=0,
\end{equation}
then
$$
\supp_{M,N}(B)\subset \Gamma,
$$
 where $\Gamma=\bigl\{(t,s)\in {\mathbb C} \times{\mathbb C} \mid
 \Phi(t,s)=\sum_{i=1}^n
f_i(t)\,g_i(s)=0\bigr\}$.
\end{theorem}


\medskip\noindent\textbf{4.}
For any $F\subset {\mathbb C} \times{\mathbb C} $ let us denote by
${\mathfrak
M}(F)$
 the set of all operators supported by $F$. If necessary, we will
write more explicitly: ${\mathfrak M}_A(F)$ or ${\mathfrak M}_{M,N}(F)$.
Similarly, we will denote by $\Delta$ (instead of
$\Delta_{M,N}$ or $\Delta_A$) the multiplication operator 
$$
X\mapsto\sum_{i=1}^nf_i(M)\,X\,g_i(N)
$$
on $L(H_N,H_M)$.



As analogue of Proposition~\ref{pr1}, it would be natural to
expect
that the equality
\begin{equation}\label{p20}
{\mathfrak M}(\Gamma)=\ker\Delta
\end{equation}
holds.
Theorem~\ref{th1-semi} shows that ${\mathfrak
M}(\Gamma)\supset\ker\Delta$.
We shall see that the inverse inclusion is true under some
restrictions on the smoothness of the functions $f_i$, $g_i$, and
it is not true in the general case.



Let us establish some additional results. Denote  by $\pr_1$, $\pr_2$
the
projections onto the components in $\sigma(M)\times\sigma(N)$. Let
$S=(\sigma(M)\times\sigma(N))\cap\Gamma$.

\begin{lemma}\label{l1}
Let $g_i\in\Lip\sigma(N)$, $k=1$, \dots, $n$. If
$\pr_1S=\sigma(M)$, then
$$
 \|\Delta\|\le 2\sum_{i=1}^n\|f_i\|\,
\|g_i\|_{\Lip}\diam\sigma(N) 
$$ 
where
$\|f_i\|=\sup\{f_i(t)\mid t\in\sigma(M)\}$.  
\end{lemma}

\begin{proof}
Set $C=2\sum_{i=1}^n\|f_i\|\, \|g_i\|_{\Lip}\diam\sigma(N)$ and
fix $s_0\in \sigma(N)$. Then 
$$
\Delta(B)=\sum_{i=1}^nf_i(M)\,B\,g_i(s_0)+\sum_{i=1}^nf_i(M)\,B\,(g_i(N)-g_i(
s_0)I).
$$
Since $\sigma(g_i(N)-g_i(s_0)I)=\{g_i(t)-g_i(s_0)\mid t\in
\sigma(N)\}$, we have that 
$$
 \|g_i(N)-g_i(s_0)I\|\le
\|g_i\|_{\Lip}\diam\sigma(N).  
$$ 
Thus,
$$
\Bigl\|\sum_{i=1}^nf_i(M)\,B\,(g_i(N)-g_i(s_0)I)\Bigr\|\le
\frac{1}{2}\,C\|B\|.
$$
Furthermore,
$\sigma\bigl(\sum_{i=1}^ng_i(s_0)\,f_i(M)\bigr)=\{\Phi(t,s_0)\mid t\in
\sigma(M)\}$.  By the condition of the theorem,  for any
$t\in\sigma(M)$ there exists $ s=s(t)\in \sigma(N) $ such that
$\Phi(t,s)=0$. Therefore,
\begin{gather*}
|\Phi(t,s_0)|=|\Phi(t,s_0)-\Phi(t,s(t))| 
\\
\Bigl|\sum_{i=1}^n(g_i(s_0)-g_i(s(t)))\,f_i(s)\Bigr|\le \frac{1}{2}\,C.
\end{gather*}
Hence, $\bigl\|\sum_{i=1}^ng_i(s_0)\,f_i(M)\bigr\|\le C/2$.
Thus,
$$
\Bigl\|\sum_{i=1}^nf_i(M)\,B\,g_i(s_0)\Bigr\|\le \frac{1}{2}\,C\|B\|,
$$ 
and
$\|\Delta (B)\|\le C\|B\|$.
\end{proof}

\begin{lemma}\label{l2}
Let $E(\cdot)$ be a projection-valued measure on a compact set $K$, let
$\alpha_1$,
\dots, $\alpha_N$ be Borel subsets of $K$ such that the intersection
of
any
$m+1$ subsets $\alpha_i$ is empty . Then
$$
\sum_{j=1}^N\|E(\alpha_j)\xi\|^2\le m\|\xi\|^2
$$
for any vector $\xi$.
\end{lemma}

\begin{proof}
Since $E(\alpha_j)=\int_K \chi_{\alpha_j}(t)\,dE(t)$, where
$\chi_{\alpha_j}$
is the characteristic function of $\alpha_j$, we have that
$$
\sum_{j=1}^NE(\alpha_j)=\int_K\Bigl(\sum_j\chi_{\alpha_j}(t)\Bigr)\,dE(t)\le
m\int_KdE(t)=mI.
$$
Therefore,
$\sum_{j=1}^N\|E(\alpha_j)\xi\|^2=\sum_{j=1}^N(E(\alpha_j)\xi,\xi)
\le m\,\|\xi\|^2$. \end{proof}

\begin{theorem} \label{cth2}
Let one of the following conditions hold\textup:
\begin{itemize}
\item[\textup{(a)}] $g_i$, $i=1,\ldots,n$, are polynomials\textup;

\item[\textup{(b)}] $g_i\in\Lip\sigma(N)$,  $i=1$,\dots, $n$, and the
Hausdorff dimension of $\sigma(M)$ is less then $2$.
\end{itemize}

 Then ${\mathfrak M}(\Gamma)=\ker\Delta$.
\end{theorem}
\begin{proof}  Let $B\in
{\mathfrak M}(\Gamma)$, then $\supp_{M,N}B\subset S$.  For any
$\beta\subset \sigma(N)$ we will denote the
set
$\pr_1(\pr_2^{-1}(\beta))$  by $\tilde{\beta}$. It is easy to show that
if $\beta$
is closed then so is $\tilde \beta$,
and that
$\pr_1(\pr_2^{-1}(\cup_{k=1}^{\infty}\beta_k))=\cup_{k=1}^{\infty}\tilde
{\beta_k}$, for any sequence $\{\beta_k\}$. Hence $\tilde{\beta}$
is a Borel set for any ${F}_{\sigma}$-set $\beta$ (here we
consider only such $\beta$).  Since
$(\sigma(M)\setminus\tilde\beta)\times \sigma(N)$ does not
intersect $S$, one has $(I-E_M(\tilde{\beta}))\,BE_N(\beta)=0$.

Let $\varepsilon>0$ and let
$\displaystyle\sigma(N)=\cup_{j=1}^K\alpha_j$, where
$\alpha_i\cap\alpha_j=\emptyset$,
$\diam\alpha_j<\varepsilon$, and $K=K(\varepsilon)$ is
the least possible for all such decompositions.
Then
\begin{equation}
\Delta(B)=\sum_{j=1}^K\Delta(B)E_N(\alpha_j)=
\sum_{j=1}^KE_M(\tilde\alpha_j)\,\Delta(B)E_N(\alpha_j).
\end{equation}
Set $\gamma=\bigl\{t\in \sigma(M)\colon\sum_{i=1}^nf_i(t)\,g_i(N)=0
\bigr\}$.
Then
\begin{equation}\label{p21}
E_M(\gamma)\Delta=0.
\end{equation}
In fact, for any $Y\in L(H_N, H_M)$, we have
$$
E_M(\gamma)\Delta(Y)=\sum_{i=1}^nE_M(\gamma)\,f_i(M)\,Yg_i(N),
$$
and
\begin{align*}
&(E_M(\gamma)\Delta(Y)\xi,
\nu)=\sum_{i=1}^n(Yg_i(N)\xi,
E_M(\gamma)\,(f_i(M))^*\nu)
\\
&\qquad{}=\sum_{i=1}^n\int_{\gamma}\bigl(Yg_i(N)\xi,
\overline{f_i(t)}\,dE_M(t)\nu\bigr)
\\ 
&\qquad=\int_{\gamma}\Bigl(\sum_{i=1}^nYg_i(N)\,f_i(t)
\xi,dE_M(t)\nu\Bigr)=0,
\end{align*}
for any $\xi\in
H_N$, $\nu\in H_M$.

Relation (\ref{p21}) implies
\[
\Delta (B)=\sum_{j=1}^KE_M(\hat{\alpha}_j)\,\Delta (B)\,E_N(\alpha_j),
\]
where $\hat{\alpha}_j=\tilde\alpha_j\setminus\gamma$. Hence,
\begin{gather}
|(\Delta(B)\xi, \nu)|=\Bigl|\sum_{j=1}^K\bigl(E_M(\hat{\alpha}_j)
\Delta (B)E_N(\alpha_j)\xi, E_M(\hat{\alpha}_j)\nu\bigr)\Bigr|
\notag 
\\
\le\Bigl(\sum_{j=1}^K\bigl\|E_M(\hat{\alpha}_j)\Delta
(B)E_N(\alpha_j)\xi\bigr\|^2\Bigr)^{1/2}
\Bigl(\sum_{j=1}^K\|E_M(\hat{\alpha}_j)\nu \|^2\Bigr)^{1/2}.\label{p22}
\end{gather}
Since $\hat{\alpha}_j\subset \pr_1((\hat\alpha_j\times\alpha_j)\cap
S)$, we can apply Lemma~\ref{l1} to the operators
$NE_N(\alpha_j)$, $ME_M(\hat{\alpha}_j)$. We obtain
\begin{gather*}
\|E_M(\hat{\alpha}_j)\Delta (B)E_N(\alpha_j)\|\le
C\|B\|\diam\alpha_j
\le C\|B\|\,\varepsilon,
\\[3pt]
\|E_M(\hat{\alpha}_j)\Delta (B)E_N(\alpha_j)\xi\|\le
C\|B\|\,\varepsilon \|E_N(\alpha_j)\xi\|,
\\
\sum_{j=1}^K\|E_M(\hat{\alpha}_j)\Delta (B)E_N(\alpha_j)\xi\|^2
\le C^2\|B\|^2\varepsilon^2\|\xi \|^2,
\end{gather*}
since the sets $\alpha_j$ are mutually disjoint.

First let  all $g_i(\cdot)$ be polynomials and let $m$ be the
greatest degree of the polynomials $g_i(\cdot)$.  Then, for any
$t\in\sigma(M)\setminus\gamma$, the equation $\Phi(t,s)=0$ does
not have more then $m$ roots. Thus, $t$ can not belong to more
than $m$ sets $\hat{\alpha}_j$,  because for any $t\in
\hat{\alpha}_j$ there exists $s(t)\in \alpha_j$ such that
$\Phi(t,s(t))=0$, and the sets $\alpha_j$ are disjoint.  Applying
Lemma~\ref{l2}, we can conclude that
$\sum_{j=1}^K\|E_M(\hat{\alpha}_j)\nu\|^2\le m\|\nu\|^2$.  Hence,
$|(\Delta (B)\xi, \nu)|\le C\|B\|\,\varepsilon \|\xi\|\,m^{1/2}\|\nu
\|^2$.
Letting $\varepsilon\rightarrow 0$, we obtain  $\Delta (B)=0$.

Now assume that the second condition holds. We can estimate the second
factor in the right-hand side of (\ref{p22}) as follows:  
$$
\Bigl(\sum_{j=1}^K\|E_M(\hat{\alpha}_j)\nu\|^2\Bigr)^{1/2}\le (K\|\nu
\|^2)^{1/2}=
K^{1/2}\|\nu \|.
$$
Thus $\|\Delta(B)\|\le C\|B\|\,\varepsilon K^{1/2}$
($K=K(\varepsilon)$).  Since $\dim
\sigma(M)<2$, $K(\varepsilon)=o(\varepsilon^{-2})$, this implies
$\Delta(B)=0$. \end{proof}
\begin{remark}\rm
The 
restriction on $\dim\sigma(M)$ in Theorem~\ref{cth2} can
be removed (see \cite{143}), but the proof  becomes much more
complicated, so we omit this generalization here.  
\end{remark}

\medskip\noindent\textbf{5.}
As a corollary from Theorem~\ref{cth2} we obtain the following
Kleinecke--Shirokov type theorems.

\begin{theorem} \label{c1}
If two polynomial semilinear relations have the same graph, then
their representations coincide.  \end{theorem}

\begin{example}
By Theorem~\ref{c1}, the semilinear relations $(\ad_{A,q})^nB=0$ and
$\ad_{A,q}B=AB-qBA=0$ have the same representations by the bounded
operators
$A=A^*$, $B$, since the characteristic functions $\Phi(t,s)=t-qs$ and
$\Phi(t,s)=(t-qs)^n$, corresponding to the first and the second
relations respectively, define the same graph.
In particular, for $q=1$ we obtain  that bounded operators
  $A=A^*$, $B$ satisfy the relation
  $(\ad_A)^nB=[A,\dots[A,B]\dots]=0$ if and only if
  $A$, $B$ commute, and for $q=-1$, a pair of bounded
  operators $A=A^*$, $B$ is a solution of the equation
  $(\ad_{A,-1})^nB=\{A,\ldots\{A,B\}\ldots\}=0$ iff $AB=-BA$.
  \end{example}


\noindent\textbf{6.}
As corollaries from Theorem~\ref{cth2} we will also give some
  results, where the form of the characteristic binary relation
 corresponding  to the  semilinear relation (\ref{hom}) is more
important then the form of the functions $f_i$, $g_i$.

\begin{theorem}\label{c2} Let $f$, $g$, be bounded Borel functions.
A pair of boun\-ded operators $A=A^*$, $B$ satisfies the equation $$
f(A)B=Bg(A) $$ if and only if $\supp_A(B)\subset\Gamma$,
where $\Gamma=\{(t,s)\mid f(t)=g(s)\}$.  
\end{theorem}

\begin{proof} Set $M=f(A)$, $N=g(A)$.  Since
$E_M(\alpha)=E_A(f^{-1}(\alpha))$ for any Borel set $\alpha$, the
condition $\supp_A(B)\subset\Gamma$ is equivalent to the
following one $$ \supp_{M,N}(B)\subset\{(t,s)\mid t=s\}.  $$
Theorem~\ref{cth2} gives the required statement. 
\end{proof}

In a similar way one shows that  Theorem~\ref{cth2} implies
the Fuglede--Putnam--Rozenblum theorem see, for  example~\cite{137}
and also Section~\ref{sec:1.4.3}.
In particular, if bounded operators $A=A^*$, $B$ satisfy  the
relation
$AB=qBA$, where $|q|=1$, then $AB=\bar qBA$, since
the sets $\{(t,s)\in{\mathbb R}\times{\mathbb R}\mid t=qs\}$  and
$\{(t,s)\in{\mathbb R}\times{\mathbb R}\mid t=\overline qs\}$ coincide.

\begin{remark}\rm
Sets of the form $\{(t,s)\mid f(t)=g(s)\}$, where $f$, $g$ are
bounded
Borel
 functions are called rectangular sets. For them, a much stronger
 result is true: an operator supported by a rectangular set $F$
 satisfies any
relation whose characteristic binary relation  contains $F$~\cite{143}.
The proof is complicated and we
restrict ourselves to a more particular result.
\end{remark}

\begin{theorem}\label{gr}
Let $F$ be the graph of a function, that is, $F=\{(t,s)\mid
s=\varphi(t)\}$, where
$\varphi$ is a bounded Borel function. If $\supp
_A(B)\subset
F$, then
the pair $A$, $B$ gives a representation of any relation whose
binary relation contains $F$.
\end{theorem}

\begin{proof} Let
$F\subset\Gamma=\bigl\{(t,s)\mid\sum_{i=1}^nf_i(t)g_i(s)=0\bigr\}$.
Since
$$
\supp_A(B)\subset \{(t,s)\mid s=\varphi(t)\}\subset
\{(t,s)\mid g_i(s)= g_i(\varphi(t))\},
$$
we have
$Bg_i(A)=g_i(\varphi(A))B$, $i=1$, \dots, $n$, by Theorem~\ref{c2}.
On
the
other hand,
$$
\sum_{i=1}^nf_i(t)\,g_i(\varphi(t))=0
$$
 for any $t\in \sigma(A)$,
hence $\sum_{i=1}^nf_i(A)\,g_i(\varphi(A))=0$. It follows that
\par\vskip\abovedisplayskip\hfill$\displaystyle
\sum_{i=1}^nf_i(A)Bg_i(A)=\sum_{i=1}^nf_i(A)(Bg_i(A)-g_i(\varphi(A))B)=0.
$\hfill
\end{proof}

\begin{corollary}\label{c4}
Let $A=A^*$, and let  there  exist a decomposition of $\sigma(A)$
into
Borel
sets $P_i$, $\sigma(A)=\cup_{i}P_i$, such that each $P_i\times
P_j\cap\Gamma$,
$i$, $j=1$, $2$,~\dots, is the graph of a function. Then $\ker
\Delta_A={\mathfrak M}_A(\Gamma)$.
\end{corollary}

\begin{proof}
Theorem~\ref{gr} implies $E_A(P_i)\Delta
(B)E_A(P_j)=0$ for any $i,j$ and hence  $\Delta (B)=0$.\end{proof}

\begin{remark}
Let us note that if\/
$\Gamma$ is such that  the set $\{\mu\in {\mathbb R}\mid
(\lambda,
\mu)\in \Gamma\}$ is
finite or countable for any $\lambda \in {\mathbb R}$, then  the
condition of
Corollary~\ref{c4} is satisfied.
\end{remark}

\medskip\noindent\textbf{7.}
As  was noticed, the converse of\/ Theorem~\ref{th2-semi} is
false in the general case. Here we construct an example which
simultaneously solves the Fuglede--Weiss problem. Namely, we show that
there exist a pair of bounded operators $A=A^*$, $B$ and
continuous functions $f_i$, $g_i$ such that
$$
\sum_{i=1}^mf_i(A)Bg_i(A)=0,\quad 
\text{but} \quad  \sum_{i=1}^m\bar{f}_i(A)B\,\bar{g}_i(A)\ne 0.
$$

\begin{example}\rm\label{ex}
Let $D({\mathbb R}^n)$ be the space of compactly supported infinitely
differentiable
functions on ${\mathbb R} ^n$, $D'({\mathbb R}^n)$ be its dual space,
$FL^1({\mathbb R} ^n)$  the Fourier algebra,
$PM({\mathbb R} ^n)$ the space dual to $FL^1({\mathbb R} ^n)$ (the space
of pseudo-measures), $F$  the Fourier transform, $\varphi*\psi$ the
convolution of two functions in $D({\mathbb R} ^n)$, $\tilde
\varphi(x)=\overline{\varphi(-x)}$.

Consider the following polynomial in six variables
$$
p(x_1,\ldots, x_6)=x_1^2+x_2^2+x_3^2-1+i(x_4^2+x_5^2+x_6^2-1).
$$

Let $s_i$, $r_i$, $i=1$, \dots, $m$, be polynomials satisfying 
the relation 
$$
p(x-y)=\sum_{i=1}^ms_i(x)\,r_i(y)
$$ 
for any
$x,y\in {\mathbb R} ^6$. Let $u,v\in D({\mathbb R} ^n)$ and $a_i=us_i$,
$b_i=vr_i$.
Obviously, $a_i$, $b_i\in D({\mathbb R}^6)$. Consider  the
operators $A_i=M_{a_i}$, $B_i=M_{b_i}$, $i=1$, \dots, $m$ in the space
$L^2({\mathbb R} ^6)$ (here $M_f$ is
the operator of multiplication by the function $f$). Since the
Fourier
transform
of a pseudo-measure $\Phi$ belongs to $L^{\infty}({\mathbb R} ^6)$,
the operator $T=F^{-1}M_{F\Phi}F$ is well defined in the
space $L^2({\mathbb R}^6)$.
Furthermore a direct computation shows that
\begin{equation}
\Bigl(\sum_{i=1}^mM_{b_i}TM_{a_i}\varphi,\psi\Bigr)=\langle
p\,\Phi,u\varphi*
\widetilde{\bar{v}\psi}\rangle
\end{equation}
for any $u$, $v$, $\varphi$, $\psi\in
D({\mathbb R}^6)$, where $\langle\cdot,\cdot\rangle$ is
the
pairing of
the spaces $D({\mathbb R}^6)$, $D'({\mathbb R}^6)$
(since $PM({\mathbb R} ^6)\subset D({\mathbb R} ^6)'$,
$p\,\Phi\in D'({\mathbb R}^6)$ for any
polynomial $p$
in six variables).
Since the set
$$
L=\{u\varphi*\tilde{\bar{v}\psi}\mid \varphi,
\psi,u,v\in D({\mathbb R}^6)\}
$$ is dense in $D({\mathbb R}^6)$,
the existence of a
pseudo-measure $\Phi$ such that $p\,\Phi=0$ and $\bar{p}\,\Phi\ne0$
would imply the existence of functions $u$, $v$ such that
\begin{align*}
\Bigl(\sum_{i=1}^mM^*_{b_i}TM^*_{a_i}\varphi, \psi\Bigr)&=
\langle\bar{p}\,\Phi,
\bar{u}\varphi*\widetilde{v\psi}\rangle\ne 0,
\\*[-8pt]
&\qquad\qquad \text{for
some $\varphi$, $\psi\in D({\mathbb R}^6)$}, 
\\
\Bigl(\sum_{i=1}^nM_{b_i}TM_{a_i}\varphi,\psi\Bigr)&=
\langle  p\,\Phi,u\varphi*\widetilde
{\bar
{v}\psi}\rangle=0,
\\*[-8pt]
&\qquad\qquad \text{for all $\varphi$, $\psi\in D({\mathbb R}^6).$}
\end{align*}
Hence, for the operator $A_i$, $B_i$, $T$ constructed relatively
to $\Phi$, $u$, $v$, we  obtain
\begin{equation}\label{prp}
\sum_{i=1}^mB_iTA_i=0 \quad \text{and} \quad
\sum_{i=1}^mB_i^*TA_i^*\ne 0.
\end{equation}
Since any finite
commutative family of normal operators can be realized as a family
of continuous functions of one self-adjoint operator, one can find
a self-adjoint operator $A$ and continuous functions $f_i$, $g_i$
such that $B_i=f_i(A)$, $A_i=g_i(A)$. Then (\ref{prp}) is
equivalent to the following 
$$ \sum_{i=1}^mf_i(A)\,T\,g_i(A)=0
\quad\text{and}\quad
\sum_{i=1}^m\bar{f}_i(A)\,T\,\bar{g}_i(A)\ne 0.
$$
Thus, it remains to prove the existence of a pseudo-measure $\Phi$
such
that $p\,\Phi=0$ and $\bar{p}\,\Phi\ne 0$. In order to do this, we
consider
the $\delta$\nobreakdash-mea\-sure
$\mu_0$ of the surface $S^2$ in
${\mathbb R} ^3$. Let $\mu=\mu_0\times\mu_0$. It is easy to show that
$$
|F\mu(\lambda)|=\Bigl|\int_{{\mathbb
R} ^6}e^{i(\lambda,x)}\,d\mu(x)\Bigr|=O(|\lambda
|^{-1}),
$$
as $|\lambda |\rightarrow +\infty$ in ${\mathbb R} ^6$.

 Since $F(\frac{\partial}{\partial x_i}f)=\lambda _iFf$,
the Fourier transform $F(\frac{\partial\mu}{\partial x_i})$ belongs
to
$L^{\infty}({\mathbb R} ^6)$, so that $\frac{\partial\mu}{\partial
x_i}\in
PM({\mathbb R} ^6)$. The pseudo-measure $\frac{\partial\mu}{\partial
x_i}$ has
a compact support. Thus $L\mu\in PM({\mathbb R} ^6)$ for any first-order
differential operator $L$ with smooth coefficients. Let
$$
L=(1+i)\,x_4\frac{\partial}{\partial
x_1}-(1-i)\,x_1\frac{\partial}{\partial
x_4},
\quad\text{and} \quad \Phi=L\mu.
$$
It is easy to see that $Lp=0$ and $L\bar{p}=4(1+i)\,x_1x_4$, which
implies that for
any function $\varphi$ in $D({\mathbb R}^6)$,
\begin{align*}
\langle
p\,
\Phi,\varphi\rangle
&=\langle L\mu,p\varphi\rangle= \langle  \mu,L(p\varphi)\rangle
=\langle\mu,pL(\varphi)\rangle=0,
 \\
\langle\bar{p}\,\Phi,\varphi\rangle&=\langle
\mu,L(\bar{p})\varphi\rangle + \langle\mu,
\bar{p}L
\varphi\rangle=
4(1+i)\langle x_1x_4\,\mu,\varphi\rangle.
\end{align*}
Thus, $p\,\Phi=0$ and $\bar p\,\Phi=4(1+i)\,x_1x_4\,\mu\ne 0.$
\end{example}

\begin{remark}
Instead of one operator $A=A^*$ one can consider a family of
commuting
self-adjoint operator operators  ${\mathbf A}=(A_k)_{k=1}^m$ which are
connected
with a bounded operator $B$ by a semilinear relation of the form
\begin{equation}\label{mul}
\sum_{i=1}^n f_i({\mathbf A})\,B\,g_i({\mathbf A})=0,
\end{equation}
where  $f_i({\mathbf A})$, $g_i({\mathbf A})$ are Borel bounded functions
of the family ${\mathbf A}$. As before, one can associate  the set 
$$
\Gamma=\{(t,s)\in {\mathbb R} ^m\times{\mathbb R}
^m \mid \sum_{i=1}^n f_i(t)g_i(s)=0\}
$$  to relation
\eqref{mul}. Support of the operator
$B$ ($\supp_{\mathbf A}B\subset \sigma ({\mathbf
A})\times\sigma({\mathbf A})$) with respect to the family ${\mathbf A}$
is defined in the same way as in Definition~\ref{def1} but with
the joint resolution of the identity $E_{\mathbf A}(\cdot)$ for the family
${\mathbf A}$ instead of $E_{M}(\cdot)$ and $E_{N}(\cdot)$.  If
${\mathfrak M}_{\mathbb A}(\Gamma)$ is the set of all bounded operators
whose support belongs to $\Gamma$ and $\ker\Delta_{\mathbb A}$ is the
operator of multiplication, $X\mapsto \sum_{i=1}^n f_i({\mathbf
A})\,B\, g_i({\mathbf A})$, then using similar
arguments to those given in Section~\ref{sec:1.3.3}, item~4,
one can prove that $\ker\Delta_{\mathbf
A}\subset{\mathfrak M}_{\mathbf A}(\Gamma)$. But the converse inclusion
is not true already for $m$ greater than 2 and polynomials $f_k$,
$g_k$. The corresponding example can be derived using similar
arguments to those given in Example~\ref{ex}.  
\end{remark}

\begin{remark}
As has been shown above, an operator $B$, $A$-supported by the set
$\Gamma=\{(t,s)\mid\sum_{i=1}^nf_i(t)\,g_i(s)=0\}$, might not  satisfy
the
corresponding
semilinear relation $\sum_{i=1}^nf_i(A)B\,g_i(A)=0$ if the functions
$f_i(\cdot)$,
$g_i(\cdot)$ are not smooth. But this is true  if the operator $B$ is
pseudo-integral. Let us recall the definition of
a pseudo-integral operator.

Let $\mu$ be the scalar spectral measure of a self-adjoint operator $A$,
let $H=\int_{\sigma(A)}^\oplus l_2(N(\lambda))\,d\mu(\lambda)$,
$N(\lambda)\in{\bf N} \cup\{\infty\})$, be the spectral resolution
of  $A$, and let $m$ be a bounded measure on ${\mathbb R} \times
{\mathbb R} $ such that the projections of $m$ on  both coordinates
are majorized by $\mu$. The measure $m$ is said to be  regular.
Since $l_2(N)$ for $N\in {\mathbb N}$ can be embedded into
$l_2(\infty)$, we can define
the bilinear form
$$
(\vec{x}, \vec{y})\mapsto \iint (\vec{x}(s),\vec{y}(t))\,dm(t,s),
$$
where $(\vec{x}(s),\vec{y}(t))$ is the scalar product in
$l_2(\infty)$.
If $m$ is regular,
then the bilinear form defines some bounded operator $T_m$.
The operator $T_m$ is called a ``pseudo-integral operator constructed
relatively
to $m$''.

It was shown by Arveson that the pseudo-integral operator $T_m$ is
supported by any pseudo-closed set on which the measure $m$ is
concentrated
(the
set $E\subset{\mathbb R} \times{\mathbb R} $ is called pseudo-closed if
its
complement is a union of countably many measurable sets of the form
$X\times Y\subset{\mathbb R} \times{\mathbb R} $).

Let $\Gamma$ be the characteristic binary relation corresponding to
the semilinear relation (\ref{hom}),
and $m$ a regular measure with $\supp m\subset\Gamma$. Then,
$A$, $T_m$ is a representation of relation (\ref{hom}). Indeed,
\begin{align*}
&\bigl(\sum_{i=1}^nf_i(A)\,T_m\,g_i(A)
\vec{x},\vec{y}\Bigr)=\sum_{i=1}^n\bigl(T_mg_i(A)
\vec{
x},
\overline{f_i}(A)\vec{y}\bigr)
\\
&\qquad{}=\sum_{i=1}^n\iint_{\Gamma}
g_i(s)\,f_i(t)\,(\vec{x}(s),\vec{y}(t))\,dm(t,s)
\\
&\qquad{}
=\iint_{\Gamma}\Phi(t,s)\,(\vec{x}(s),\vec{y}(t))\,dm(t,s)=0.
\end{align*}
\end{remark}

\subsection{Irreducible  representations of semilinear
relations} \label{sec:1.3.4}

Now we study irreducible  representations of semilinear relations,
i.e., an 
irreducible families of operators $A=A^*$, $B$, $B^*$ satisfying
$$
\sum_{i=1}^nf_i(A)\,B\,g_i(A)=0.
$$
It is clear that if $f_i$, $g_i$ are polynomials, then such a family of
operators defines an irreducible representation of the
$*$-algebra   generated by $a=a^*$, $b$, $b^*$ and the relation
$\sum_{i=1}^nf_i(a)\,b\,g_i(a)=0$.
In what follows we shall mean by a representation of the semilinear
relation
a triple $(A=A^*, B, B^*)$ satisfying the semilinear relation
instead
of the pair ($A=A^*, B$), if it does not lead to any confusion.

\medskip\noindent\textbf{1.}
We begin with  finite-dimensional re\-pre\-sen\-ta\-tions
and establish a connection between irreducible representations
of \eqref{hom} and the corresponding graph.

\begin{proposition}\label{connec}
If a family $(A, B, B^*)$  defines a
finite-dimensional irreducible representation of \eqref{hom}, then the
corresponding
graph $\Gamma\restriction_{\sigma(A)}$ is
connected. For every finite connected subgraph $(D,
\Gamma\restriction_{D})$,
there exists an irreducible representation $(A, B, B^*)$ with
$\sigma(A)=D$.
\end{proposition}

\begin{proof} In fact, if
$\Gamma\restriction_{\sigma(A)}=\Gamma_1\cup\Gamma_2$
is
the union of two connected subgraphs then, by Theorem~\ref{th1-semi},
the spectral subspaces $A$ corresponding to the vertices of\, $\Gamma_1$
and
$\Gamma_2$ are invariant with respect to $B$ and $B^*$. This shows that
$(A,
B, B^*)$ is reducible.

Let $\Gamma\restriction_{\sigma(A)}$ be connected. The family of
operators $A$, $B$, $B^*$ defined by
\begin{align*} 
A&= \begin{pmatrix} \lambda _1 &        &
0\\ & \ddots &  \\ 0      &        & \lambda _m \end{pmatrix},
\qquad
 \{\lambda_1,\dots,\lambda_m\}=D,
\\[-5pt]
&\qquad\qquad\qquad\qquad\qquad\qquad \lambda _i\ne \lambda _j,
\qquad i\ne j,\\[3pt]
B&=(b_{ij})_{i,j=1}^m,
\qquad
b_{ij}=
\begin{cases}
 0, & ( \lambda_i,\lambda_j)\notin \Gamma|_D,\\
 1, & ( \lambda_i,\lambda_j)\in \Gamma|_D,
\end{cases}
\end{align*}
is irreducible. Indeed, the relation $[C,A]=0$ implies that $C$ is
diagonal,
$C=\diag(c_1,\ldots,c_2)$. From $[C,B]=0$, it follows that
$c_kb_{kl}=b_{kl}c_l$. Since $\Gamma\restriction_{\sigma(A)}$
is connected, we
have
that
there exists a permutation $(l_1,\ldots,l_m)\in S_m$ such that
$(\lambda_{l_k},\lambda_{l_{k+1}})\in\Gamma\restriction_{\sigma(A)}$,
hence
$b_{l_k,l_{k+1}}=1$. This implies that $c_1=\ldots=c_m$, and
$C=c_1I$.
Moreover, we have $\sigma(A)=D$.
\end{proof}

We also give a reformulation of this statement for the symmetrical
case.
\begin{proposition}
If\/ $A=A^*$, $B=B^*$ is an irreducible finite-dimen\-sional
representation of
\eqref{hom},
then the graph $\Gamma_s\restriction_{\sigma(A)}$ is
connected. For every finite connected subgraph $(D,
\Gamma_s\restriction_{\sigma(A)})$,
there exists an irreducible pair $A=A^*$, $B=B^*$  satisfying
\eqref{hom}
such that $D=\sigma(A)$.
\end{proposition}
The proof is the same as the one given above, but with $\Gamma_s$
replacing
$\Gamma$. Note that the constructed operator $B$ is self-adjoint
because
the graph $\Gamma_s$ is symmetrical.


\medskip\noindent\textbf{2.}
In what follows, we investigate irreducible
representations of \eqref{hom} on a Hilbert space  $H$, where $H$
is not necessary finite-dimen\-sional. First, we prove an
analogue of the theorem on connectedness of the  graph
(Proposition~\ref{connec}). For this purpose, we recall some
definitions.

\begin{definition}
A subset $F\subset{\mathbb R}\times {\mathbb R}$ is  called
 marginally null
with
respect to  a measure $\mu$ \textup(or a class of measures equivalent to
$\mu$\textup)
if
$F\subset
(\alpha\times {\mathbb R})\cup({\mathbb R}\times \beta)$, where
$\mu(\alpha)=
\mu(\beta)=0$.
\end{definition}

Let, further,
\begin{align*}
\Gamma(M)&=\{y\mid\exists x\in M\colon (x,y)\in \Gamma\},
\\
\Gamma^{-1}(M)&=\{x\mid\exists y\in M\colon (x,y)\in \Gamma\},
\end{align*}
and let $M^{c}$ be the complement of the set $M$. In what
follows we shall assume that $\Gamma(M)$ and $\Gamma^{-1}(M)$
are Borel for any Borel set $M$. 

The concepts defined below
generalize those of quasi-invariance and ergodicity of a measure.
For graphs related to dynamical systems, we will also discuss these
notions in Section~\ref{sec:2.1.1}.

\begin{definition} We call a set $M\subset{\mathbb R}$ right-invariant
\textup(left-invariant\textup)
with respect to $\Gamma$ and a measure $\mu$, if
the set $M^{c}\times M\cap\Gamma$ \textup(respectively $M\times
M^{c}\cap\Gamma$\textup) is marginally null\textup;
$M$ is invariant if it is
left- and right-invariant. A measure $\mu$ is called \textup(left-,
right-\textup) quasi-invariant with respect to $\Gamma$ if the set
$\Gamma(M)\cup\Gamma^{-1}(M)$ \textup(respectively $\Gamma^{-1}(M)$,
$\Gamma(M)$\textup) is of non-zero measure for every Borel set $M$ such
that $\mu(M)\ne 0$, $\Gamma(M)$, $\Gamma^{-1}(M)\in {\mathfrak
B}({\mathbb R})$. A spectral type $\mu$ is called \textup(left-,
right\nobreakdash-\textup)
ergodic with respect to $\Gamma$ if any \textup(left-, right-\textup)
invariant
set is a $\mu$\nobreakdash-null set or has the $\mu$-null complement.
\end{definition}

Note that for any operator $A=A^*$ there exists the trivial
representation
$(A,0,0)$ of the relation (\ref{hom}). Therefore, it is natural to
consider
representations without trivial parts.

\begin{definition}
We call a representation $A$, $B$, $B^*$
$*$-complete if for any non-zero spectral subspace $W$ of the
operator $A$ one of the operators $B$ and $B^*$ is not equal to
zero on $W$.
\end{definition}

It  is easy to show that any irreducible representation of
dimension greater than one  is $*$-complete.

\begin{proposition}\label{pr4}
If a representation $(A, B,
B^*)$ of \eqref{hom} is $*$-comp\-lete, then the spectral measure
of the operator $A$ is quasi-invariant with respect to $\Gamma$.
\end{proposition}

\begin{proof}
If $E_A(M)\ne 0$ but $E_A(\Gamma^{-1}(M))=0$ and
$E_A(\Gamma(M))=0$, then 
$$
BE_A(M)= E_A(\Gamma^{-1}(M))BE_A(M)=0
$$
and 
$$
B^*E_A(M)=E_A(\Gamma(M))B^*E_A(M)=0,
$$
hence $B\restriction E_A(M)H=0$
and $B^*\restriction E_A(M)H=0$.
Thus, the representation $(A, B, B^*)$ is
not $*$-complete.
\end{proof}

One can  also show that if the spectral measure of $A$ is quasi-invariant,
then there exists a bounded operator $B$ such that the triple
$(A, B, B^*)$
is a $*$-complete representation of \eqref{hom}. But we omit the proof
here.

\begin{proposition}\label{therg}
If a representation $(A, B, B^*)$ of  \eqref{hom}
is irreducible, then the spectral measure of $A$ is ergodic with
respect to~$\Gamma$. 
\end{proposition}

\begin{proof} Let $(A, B, B^*)$ be an irreducible representation
of the
relation \eqref{hom}.
If the spectral type of $A$ is not ergodic, then there exists
$M\subset{\mathbb R}$ such that $E_A(M)\ne 0$, $E_A(M)\ne I$, and
the sets $\Gamma\cap(M^{c}\times M)$, $\Gamma\cap(M\times M^c)$ are
marginally null sets. Hence,
$\Gamma\cap
(M^{c}\times M)\subset(M_1\times{\mathbb R})\cup({\mathbb R}\times M_2)$,
where
$\mu(M_1)=\mu(M_2)=0$. From this it follows that $E_A(M^{c})BE_A(M)=
E_A(M^{c}\setminus M_1)BE_A(M\setminus M_2)=0$,
i.e., the spectral subspace  $E_A(M)H$  is invariant with respect
to $A$, $B$. In the same way one can show that $E_A(M)H$ is invariant
with respect to $A$, $B^*$.
This contradicts the irreducibility of $A$, $B$, $B^*$.
\end{proof}

Note that the converse statement is also true. 
Namely, for any ergodic measure $\mu$ 
there exists an irreducible  representation $A$, $B$, $B^*$ of relation
(\ref{hom}) such that $\mu$ is a spectral scalar measure of $A$.
However we will not discuss the proof here. 

\medskip\noindent\textbf{3.}
Let $(A, B)$ satisfy the following semilinear relation:
\begin{equation}\label{mk}
AB=BF(A),
\end{equation}
where $F(\cdot)$ is a bounded Borel mapping defined on ${\mathbb R}$.
It is easy to show that in this case the invariance of a set
$\Delta\in {\mathfrak B}({\mathbb R})$ with respect to  $\Gamma$ and a
measure $\mu$ means its invariance with respect to the mapping
$F(\cdot)$, i.e., $\mu(F(\Delta))=\mu(\Delta)$ and 
$\mu(F^{-1}(\Delta))=\mu(\Delta)$. A measure $\mu$ is quasi-invariant
with respect to $\Gamma$ if the measures $\mu(F(\cdot))$ and $\mu(F^{-1}(\cdot))$ are absolutely continuous with respect to
$\mu$. The
ergodicity of the spectral measure with respect to $\Gamma$ means
ergodicity of the measure $E_A(\cdot)$ with respect to
$F(\cdot)$ (i.e., for any $F(\cdot)$-invariant Borel set
$\Delta\subset{\mathbb R}$, either $E_A(\Delta)=0$ or
$E_A(\Delta)=I$).  Therefore, by Proposition~\ref{therg}, we have
that, if $(A, B, B^*)$ is an irreducible representation of
\eqref{mk}, then  $E_A(\cdot)$ is ergodic with respect to
$F(\cdot)$.

\medskip\noindent\textbf{4.}
In  the analysis of the representations of \eqref{mk}, the behavior
of
the dynamical system
$F\colon\mathbb{R} \to \mathbb{R}$ plays a central role.
The structure of representations of \eqref{mk} depends on the
structure of
the orbits
of the corresponding dynamical system (see Section~\ref{sec:2.1.1}
for details). Here we
give the corresponding
results for general semilinear relations.
We begin with the following definition.

\begin{definition}
A set $E$ is called\/ $\Gamma$-invariant if
$E\times E^c\cap\Gamma=\emptyset$ and $E^c\times
E\cap\Gamma=\emptyset$.
A minimal $\Gamma\restriction_M$-invariant set $O_M(E)$
containing the set $E$
is said to be a trajectory \textup(semi-trajectory\textup) of the set
$E\subset M$ with respect to $\Gamma\restriction_M$.
\end{definition}

The concept of a trajectory generalizes that of an orbit for
dynamical systems. The following result is an analogue of the theorem
on the
connectedness of the graph supporting an irreducible
finite-dimensional representation.

\begin{theorem} Let $M$ be a
compact set.

\textup(a\textup)
If there is $x\in M$ such that the trajectory $O_M(\{x\})$
 is dense in $M$, then $M$ is the spectrum
of an irreducible representation, i.e.,
there exists an irreducible representation
$(A, B, B^*)$ such that $\sigma(A)=M$.

\textup(b\textup) If the set $M$ is the spectrum of an irreducible
representation of \eqref{hom}, then the trajectory $O_M(G\cap M)$
is dense in $M$ for any open set $G$, where
$G\cap M\ne\emptyset$.
\end{theorem}

\begin{proof}
 Let $O_M(\{x\})$ be dense in $M$. Then there is a sequence
$\{\lambda_n\}_{n=1}^{\infty}\subset O_M(\{x\})$ which is dense in $M$.
Now let $\mu $ be a measure concentrated on
 $\{\lambda_n\}_{n=1}^{\infty}$,
and $\mu(\lambda_n)\ne 0$ for any $\lambda_n$. Then $\mu$ is ergodic.
In fact, let $S$ be invariant with respect to $\Gamma$, and
$\mu(S)\ne 0$,
$\mu(S^c)\ne 0$. Then
$S\cap\{\lambda_n\}_{n=1}^{\infty}\ne\emptyset$.
From this it follows that $S\cap O_M(\{x\})\ne \emptyset$. If
$x\in S$, then $O_M(\{x\})\subset S$, hence $\mu(S^c)=0$, which
contradicts the assumption.  Assume that $x\not\in S$, hence $x\in
S^c$. Since $S$ is invariant with respect to $\Gamma$, so is the
complement of $S$ and $O_M(\{x\})\cap S^c\ne\emptyset$.  From the
condition $x\in O_M(\{x\})\cap S^{c}$ we obtain $O_M(\{x\})\cap
S^c=O_M(\{x\})$.  Hence $O_M(\{x\})\subset S^c$, and $\mu(S)=0$,
which gives a contradiction.

\medskip\noindent\textbf{5.}
Let $G$ be an open set, $G\cap M\ne \emptyset$,
$\overline{O_M(G\cap
M)}\ne M$,
and $E_A(\cdot)$ be the spectral measure of the operator $A$ with
spectrum $M$. Then $E_A(G)\ne 0$ and there exists a compact set
$F\subset G\cap M$, $E_A(F)\ne 0$. The set $O_M(F)$ is a Borel
set.  Since $\overline{O_M(F)}\ne M$, $E_A(O_M(F))\ne I$. Thus,
$E_A(\cdot)$ is not ergodic with respect to $\Gamma\restriction_M$,
because
$O_M(F)$ is an invariant set.
\end{proof}

\begin{definition} A
set $\tau$ is said to be a measurable section  of $(D, \Gamma)$
if $\tau\subset D$ is a Borel set and every trajectory
$O_D(\{x\})$ with respect to $\Gamma\restriction_D$ intersects $\tau$ at
exactly one point.
\end{definition} 

\begin{proposition} Let $(D, \Gamma)$ have a measurable section. 
Then, for any  irreducible
representation of \eqref{hom} with $\sigma(A)\subset D$, there exists
a
unique trajectory
$O_D(\{x\})$ for which $E_A(O_D(\{x\}))=I$.
\end{proposition}

\begin{proof}
Let $\tau$ be a measurable section of $(D, \Gamma)$.
Assume that $\mu $ is not concentrated on any orbit $O_D(\{x\})$. We
prove that
there exists a partitions of $\tau$ into two sets
$\tau=\tau_1\cup\tau_2$, $\tau_1\cap\tau_2=\emptyset$, such that
$\mu(O_D(\tau_1))>0$, $\mu(O_D(\tau_2))>0$. Consider the trajectories
$T_i=O_D^i(\tau_i)$, $i=1$, $2$, for any two sets $\tau_i$ satisfying the
conditions
$\tau=\tau_1\cup\tau_2$, $\tau_1\cap\tau_2=\emptyset$. By the 
definition
of
a measurable section,
$T_1\cap T_2=\emptyset$ and $T_1\cup T_2=D$. Assuming that for any
such
decomposition $D=T_1\cup T_2$ one of the values $\mu(T_1)$ and
$\mu(T_2)$
 is equal to zero, we can find a decreasing sequence $\{T^k\}$ such
 that
 $E_A(T^k)=I$ for any $k$ and $\bigcap T^k=O_D(\{x\})$ for some
 $x\in\tau$.
 From the latter argument we can conclude that $\mu$ is concentrated on
 an
 orbit which contradicts the assumption.

 Let $\tau=\tau_1\cup\tau_2$ be the required decomposition, i.e.
 $\mu(O_D(\tau_i))>0$, $i=1,2$.  Since both sets $O_D(\tau_1)$ and
 $O_D(\tau_2)$
 are invariant with respect to $\Gamma|_D$, the existence of the
 decomposition implies a contradiction  to
ergodicity of the measure $\mu$. Thus, $\mu$ is concentrated
on the trajectory of some point $x$. 
\end{proof}

If there is no measurable section for the graph ($D$, $\Gamma$) of
the semilinear
relation \eqref{hom}, then the structure of representations with
bounded
operators
$(A, B, B^*)$ is
more complicated: there might exist
irreducible representations such that the spectrum of the operator
$A$ is not discrete.

\begin{remark}
The same theorems are valid for the
representation with $B=B^*$, but with $\Gamma_s$ instead of $\Gamma$.
\end{remark}

\subsection{Representations of semilinear ${F}_4$-relations}
\label{sec:1.3.5}

The problem of  describing  all irreducible
representations of semilinear relations up to unitary equivalence
might be very difficult.
The complexity of the description depends on the structure of the
corresponding graph.

\begin{theorem}
$1$. If all connected
components of the graph $\Gamma$ corresponding to a semilinear
relation
are  of the form:
$\onepoint {}$, $ \edgeright {}{}$,
then any irreducible representation $(A, B, B^*)$ of the
relation is one- or two-dimensional\textup:
\begin{itemize}
\item[\textup(i\textup)] $A=\lambda$, $B=0$, $\lambda\in{\mathbb
R}$\textup;

\item[\textup(ii\textup)] $A=\bigl(
\begin{smallmatrix}
\lambda_1&0\\
0&\lambda_2
\end{smallmatrix}\bigr)$,
$B=\bigl(
\begin{smallmatrix}
0&0\\
b&0
\end{smallmatrix}\bigr)$, where  $(\lambda_1, \lambda_2,
b)$ belongs to the set $K_1=\bigl\{(\lambda_1, \lambda_2, b)\in {\mathbb
R}^3\mid
\Phi(\lambda_1,\lambda_2)=0;\,\lambda_1\ne\lambda_2,
\,b>0\bigr\}$.
\end{itemize}

$2$. If all connected
components of the graph $\Gamma_s$
are of the form
$\onepoint {}$, $\ear{}$, $\edge {}{} $,
then any irreducible representation $(A, B=B^*)$ of the relation
is one- or two-dimensional\textup:
\begin{itemize}
\item[\textup(i\textup)] $A=\lambda$, $B=0$, $\lambda\in{\mathbb
R}$\textup;

\item[\textup(ii\textup)]
$A=\lambda$, $B=b$, where $(\lambda, b)$ belongs
to the set $K_2=\{(\lambda, b)\in {\mathbb R}^2\mid
\Phi(\lambda,\lambda)=0\}$\textup;

\item[\textup(iii\textup)] $A=\bigl(
\begin{smallmatrix}
\lambda_1&0\\
0&\lambda_2
\end{smallmatrix}\bigr)$,
$B=\bigr(
\begin{smallmatrix}
0&b\\
b&0
\end{smallmatrix}\bigr)$, where $(\lambda_1, \lambda_2,
b)$ belongs to  $K_3=\bigl\{(\lambda_1, \lambda_2, b)\in {\mathbb
R}^3\mid
\Phi(\lambda_1,\lambda_2)=\Phi(\lambda_2,\lambda_1)=0;\, \lambda_1\ne
\lambda_2,\,
b>0\bigr\}$.
\end{itemize}
\end{theorem}

Note that if $f_i$, $g_i$ are polynomials, then the algebra
${\mathbb C}\langle a,b\mid \sum_{i=1}^nf_i(a)\,b\,g_i(a)=0\rangle$
is an ${
F}_4$-algebra.

\begin{proof} Under the above condition we have
that $\Gamma$ is the graph of a bijective mapping
$\phi\colon N_1\rightarrow N_2$, where
$N_1\cap N_2=\emptyset$
(i.e., $\Gamma=\{(t,s)\in N_2\times N_1\mid t=\phi(s)\}$), and
$\Gamma_s$ is the
graph of a bijective mapping $\phi\colon N\rightarrow N$
(i.e., $\Gamma=\{(t,s)\in N^2\mid t=\phi(s)\}$).
Moreover, $\phi$ is measurable under the assumption that
$\Gamma(M)$ and $\Gamma^{-1}(M)$ are Borel sets for any Borel $M$.
Therefore, by
Corollary~\ref{cth2}, if $(A,B)\in L(H)$ is a representation of a
relation with  such a graph, then ($A$, $B$) satisfy the relation
$AB=B\phi(A)$, and the spectrum of the operator $A$ belongs to
$N_1\cup N_2$ in the first case, and to $N$ in the second one.
Clearly, the operators $A$, $BB^*$ commute.

Let us consider the first case. We have
$B^2=0$ and, therefore, $\ker B\ne 0$. Denote  the
subspace $ \ker B\cap\ker B^*$ by $H_1$. It is easy to show that $H_1$ is
invariant with respect to $A$, $B$, $B^*$, and  all
irreducible representations defined on $H_1$ are one-dimensional
and given by $(i)$.  Now assume that $\ker B\cap\ker
B^*=\emptyset$.  Set $H_0=\ker B\cap(\ker B^*)^{\perp}$.  Since
$H_0\subset E_A(\phi(N_1))H$, $B^*H_0\subset E_A(N_1)H$, and
$N_1\cap\phi(N_1)=\emptyset$, the subspaces $H_0$, $B^*H_0$ are
orthogonal. Moreover, $H_0$,
$B^*H_0$ are invariant with respect to $A$, $BB^*$, which imply that
given
$\Delta\in {\mathfrak
B}({\mathbb R}^2)$, the subspace $E(\Delta)H_0\oplus
B^*E(\Delta)H_0$ is invariant with respect to $A$, $B$, $B^*$,
where $E(\cdot)$  is the joint resolution of the identity for
the commuting pair of operators $A$, $BB^*$ restricted to $H_0$.
From this it follows that $\Delta$ is concentrated in one point if
$A$, $B$, $B^*$ is an irreducible family. Thus for such a
family of operators there exists a joint 
eigenvector $e\in H_0$ for $A$, $BB^*$, and $\{e,B^*e\}$ define an orthogonal basis
of the representation space. The corresponding irreducible
representation is given by $(ii)$.

In the case $A=A^*$, $B$ we have that the operators $A\phi(A)$,
$A+\phi(A)$
commute with $A$, $B$, and hence due to the irreducibility, they are
multiples of the identity, i.e., $A\phi(A)=a_1I$ and $A+\phi(A)=a_2I$.
Then $A^2-a_2A+a_1I=0$,
and so the spectrum of $A$ is $\sigma(A)=\{\lambda_1,\lambda_2\}$,
where
$\lambda_1$, $\lambda_2$ are the roots of the equation
$\lambda^2-a_2\lambda+a_1=0$. Hence the spectrum of $A$ is discrete
as soon as
($A$, $B$) is irreducible. In addition, $B^2$ commutes with $A$, $B$
and
is a multiple of the identity,
 $B^2=b^2I$. If $b\ne 0$ and $e_{\lambda_1}$ is an eigenvector of
 $A$, then
$e_{\lambda_1}$, $Be_{\lambda_1}$ define an invariant subspace;
moreover
$Be_{\lambda_1}$ is an eigenvector with the eigenvalue
$\phi(\lambda_1)=:\lambda_2$.
Therefore, by the irreducibility, the operators $A$, $B$ can be at most
two-dimensional. If $\phi(\lambda_1)\ne\lambda_1$ (i.e.,
$\Phi(\lambda_1,\lambda_1)\ne 0$),
then normalizing the orthogonal basis $e_{\lambda_1}$,
$Be_{\lambda_1}$ we get an orthogonal basis in which operators $A$,
$B$ are of the form $(iii)$.
For $\phi(\lambda)=\lambda$ one has that $A$, $B$ commute, and hence
we can
choose a joint eigenvector $e_{\lambda,b}$, which
define an invariant subspace. From this one can  conclude that the
corresponding irreducible representation is one dimensional and given
by $(i)$
or $(ii)$.
 \end{proof}

If the  graph $\Gamma$ corresponding to a semilinear relation
contains the subgraphs:
 $\ear {}$ or  $\twovec $ (and with any other orientation),
and the graph $\Gamma_s$ contains the subgraphs:
 $\earedge{}{} $, or  $\twoedge{}{}{} $,
then the problem of describing all irreducible representations $(A,
B,
B^*)$ and $(A, B=B^*)$ respectively becomes very complicated (the
corresponding $*$-algebra is wild). We refer the reader to
Sections~\ref{sec:3.1.1}
and
\ref{sec:3.1.2} for a precise
definition of  $*$-wild
algebras, and to Section~\ref{sec:3.1.4} for the 
proof of the above fact.



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