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\section{On representations of some nuclear algebras}
\subsection{Commutative models}\label{sec:2.5.1}
\markright{2.5. Representations of some nuclear algebras}

In previous sections, considering representations of
operator relations, we have seen that the complete
unitary description of irreducible representations
is possible only if the underlying dynamical system has a
measurable section. Now we will look at
what can be said about the relation in the case where the dynamical system
does not necessarily
have a measurable section. Of course, one should not expect a complete
unitary description; however, we will see that in a certain sense, a uniform
realization (commutative model) of
all representations as operators of multiplication and weighted
shifts can be constructed.

Instead of considering the relation of the form $XX^* = F(X^*X)$ or its
multi-dimensional version, we will consider a family
of commuting self-adjoint or
normal operators $\mathbf{A} = (A_k)$, and a family $\mathbf{B} = (B_j)$
which satisfy more general relations
\[
A_k B_j = B_j F_{kj}(\mathbf{A})
\]
(we allow families $\mathbf{A}$ and $\mathbf{B}$ to have different
cardinality, and no
conditions on the joint spectrum of $\mathbf{A}$ or
conditions of the form $\ker A_k = \ker B_k$ are assumed).

We start with the case of a unitary operator $U$ and a self-adjoint
operator $A$ satisfying a relation of the form $AU = UF(A)$, where
the  mapping $F(\cdot)$ is
measurable and one-to-one on the spectrum of $A$.

\begin{theorem}
Let a self-adjoint operator $A$ and a unitary operator $U$ satisfy the
relation $AU = UF(A)$ with a measurable mapping $F(\cdot)$ which is
one-to-one on the  spectrum
of $A$. The space $H$ can be uniquely decomposed
into a direct sum, $H = H_1 \oplus H_2 \oplus\dots \oplus H_\infty$, of
invariant subspaces\textup; each $H_m$ is unitarily equivalent to $\oplus _{k=1}
^ m L_2 (\mathbb{R}^1, d\mu_m)$ in such a way that $\mu_m(\cdot)$ is
a probability $F$-quasi-invariant measure, $m=1$, \dots,~$\infty$,
and the operators in $H_m$ act as follows\textup:
\begin{align}
(Af)(\lambda) & = \lambda \, f(\lambda),\notag
\\
(Uf)(\lambda) & = u_m(\lambda)\,\rho_m(\lambda)^{1/2}\,
f(F^{-1}(\lambda)),
\notag
\\
(U^*f)(\lambda) & = u_m^*(F(\lambda))\, \rho_m(F(\lambda))^{-1/2}\,
f(F(\lambda)),\label{model:uni}
\end{align}
where $f(\cdot)$ is a vector function
in $\oplus _{k=1}^m L_2(\mathbb{R}^1,
d\mu_m)$, the functions $\rho_m(\lambda)
= d\mu_m(F^{-1}(\lambda))/d\mu(\lambda)$
are the Radon--Nikodym derivatives, $u_m(\cdot)$ are measurable mappings
taking values in unitary operators on $\mathbb{C}^m$.
\end{theorem}

\begin{proof}
The decomposition of the space $H$, and formula for $A$
follow from the decomposition of the self-adjoint operator $A$ with
respect to multiplicity of its spectrum. We need to show that the subspaces
$H_m$ are invariant with respect to $U$, $U^*$.

We shall prove that each measure $\mu_m$ is quasi-invariant. Let $h_1$,
\dots, $h_m$ be an orthonormal basis in $\mathbb{C}^m$. For any measurable
$\Delta\subset \mathbb{R}$, consider functions $f_k(\lambda) =
\chi_\Delta(\lambda) h_k\subset H_m$, ($\chi_\Delta(\cdot)$ is the
characteristic function of $\Delta$) which are obviously orthogonal in
$H$. From the relation that connects $A$ and $U$, we have that 
\begin{align*}
U^*f_k(\lambda)& = U^*
E_A(\Delta) f_k(\lambda) 
\\
&= E(F^{-1}(\Delta)) U^* f_k(\lambda) =
\chi_{F^{-1}(\Delta)}(\lambda) U^* f_k(\lambda),
\end{align*}
 and all $U^*
f_k(\cdot)$, $k=1$, \dots, $m$
have supports in $F^{-1}(\Delta)$. Since the operator $U$ is unitary, the
latter functions again are orthonormal; this implies that for almost all
$\lambda$ with respect to the spectral measure of $A$, the vectors
$(U^*f_k)(\lambda)$, $k=1$, \dots, $m$, are orthogonal, and therefore, the
spectral multiplicity of $A$ at points of $F(\Delta)$ is not less than
at
points of $\Delta$. Applying the same arguments to the operator $U$
instead of $U^*$, we conclude that the spectral multiplicity
of $A$ is invariant with respect to $F(\cdot)$. This
implies that $H_m$ is an invariant subspace, and that $\mu_k$ is
quasi-invariant.

In the subspace $H_m$, introduce the unitary operator
\[
(V_kf)(\lambda) = \bigl(d\mu(F(\lambda))/d\mu(\lambda)\bigr)^{1/2} 
f(F(\lambda)).
\]
Then the operator $\tilde U=UV$ commutes with $A$, and, therefore,
$(\tilde U f)(\lambda) = u_m(\lambda)\, f(\lambda)$. From $U = \tilde U
V^*$,
we have the representation required.
\end{proof}

Instead of decomposing $H$ into a direct sum  with respect to the
multiplicity of the spectrum of $A$, one can decompose $H$ into a direct
integral of generalized eigenspaces of $A$. Then the theorem above can be
reformulated as follows.

\begin{theorem}
Let $A$ and $U$ be as in the previous theorem. Then $H$ can be decomposed
into a direct integral,
\[
H= \int_{\mathbb{R}^1}^\oplus H_\lambda\, d\mu(\lambda),
\]
where the Borel probability measure $\mu$ is $F(\cdot)$-quasi-invariant,
and $N(\lambda) = \dim H_\lambda$ is invariant with respect to $F(\cdot)$,
so that the operators take the form
\begin{align}
(Af)(\lambda) & =\lambda \, f(\lambda), \notag
\\
(Uf)(\lambda) & = u(\lambda) \, \rho(\lambda)^{1/2} \, f(F^{-1}(\lambda)),
\notag
\\
(U^*f)(\lambda) &= u^* (F(\lambda)) \, \rho (F(\lambda))^{-1/2} \,
f(F(\lambda)) ,\label{model:uni2}
\end{align}
where $\rho(\lambda) = d\mu(F^{-1}(\lambda))/ d\mu(\lambda)$ is the
Radon--Nikodym derivative, $u(\cdot)$ is a unitary measurable
operator-valued function.
\end{theorem}


The following statement gives conditions of irreducibility and unitary
equivalence of representations in the form provided by the previous
theorem.

\begin{proposition}
If representation \eqref{model:uni2} is irreducible, then the measure
$\mu(\cdot)$ is ergodic, and the dimension function $N(\cdot)$ is
constant $\mu$-a.e.
If the measure $\mu(\cdot)$ is ergodic, and
$\dim H_\lambda=1$ $\mu$-a.e.,  \textup(i.e.,
$H = L_2(\mathbb{R}, d\mu)$\textup),
then the representation is irreducible.

Two representations, corresponding to measure $\mu$, multiplicity
function $N(\cdot)$, and multiplier $u(\cdot)$, and corresponding to
the triple $\tilde \mu$, $\tilde N(\cdot)$, and $\tilde u(\cdot)$, are
unitarily equivalent if and only if the measures $\mu$ and $\tilde \mu$
are equivalent \textup(have the same zero sets\textup),
multiplicity functions
coincide $\mu$-a.e., and the multipliers are equivalent in the following
sense\textup: there exists a measurable unitary operator-valued function
$v(\cdot)$ such that
\[
\tilde u(\lambda) = v(\lambda)\, u(\lambda)\,v^*(F^{-1}(\lambda)),
\qquad \text{for $\mu$-almost all $\lambda$.}
\]
\end{proposition}

\begin{proof}
If the measure is not ergodic, its support can be decomposed into the union
of two invariants subsets of positive measure. Such a decomposition gives
rise to a decomposition of $H$ into a direct sum of invariant subspaces.
Since $N(\cdot)$ is $F(\cdot)$-invariant, it is  constant $\mu$-a.e.

Let $\mu$ be ergodic and $N(\lambda) = 1$ for $\mu$-almost all $\lambda$.
Any bounded self-adjoint operator $C$ that commutes with $A$ is the
operator of multiplication by a bounded measurable function $c(\cdot)$. If
$C$ commutes with $U$, then this function is invariant, and by the 
ergodicity, is constant $\mu$-a.e., i.e., the pair $A$, $U$ is irreducible.

Now consider two pairs $A$, $U$ on $H$ corresponding to a triple $\mu$,
$N(\cdot)$, $u(\cdot)$, and $\tilde A$, $\tilde U$ on
$\tilde H$ corresponding
to $\tilde \mu$, $\tilde  N(\cdot)$, $\tilde u(\cdot)$. If these pairs are
unitarily equivalent, then the spectral measures of the self-adjoint
operators $A$ and $\tilde A$ are equivalent, and $N(\cdot) =\tilde
N(\cdot)$ $\mu$-a.e.; now we can naturally identify $H$ and $\tilde H$.
A unitary operator $V$ on
$H$ that intertwine these pairs commutes with $A$, and
therefore, is a multiplication by a measurable unitary
operator-valued function $v(\cdot)$. It directly follows from
\eqref{model:uni2} that the relation $\tilde U = VUV^*$ is equivalent to
$\tilde u(\lambda) = v(\lambda)\, u(\lambda)\, v^*(F^{-1}(\lambda))$.
\end{proof}

\begin{remark}
One can easily see that the same statement holds true for a finite
or countable commuting
family
of self-adjoint or normal operators $\mathbf{A} = (A_k)_{k \in X}$
which are
related with a unitary operator $U$ by the relations $A_k U =
UF_k(\mathbf{A})$, $k \in X$.
Indeed, we actually use the properties of the joint
resolution of the identity of this family, $E_\mathbf{A}(\cdot)$. To prove
the theorem, we use the relation $E_\mathbf{A}(\Delta) U = U
E_A(\mathbf F^{-1}(\mathbf{A}))$, where $\mathbf{F}\colon
\mathbb{R}^X \to \mathbb{R}^X$ is $\mathbf{F}(\cdot) =
(F_k(\cdot))_{k\in X}$. The only difference is that the dynamical
system acts on $\mathbb{R}^X$, and the spectral measure $\mu$ should
be considered on this space.

Moreover, since the decomposition of $H$ is constructed from the
commutative family $\mathbf{A}$, the
commutative model can be constructed for a finite or countable
family of unitary operators $\mathbf{U}=(U_j)_{j \in Y}$ which satisfy
the relations $A_k B_j = B_j F_{kj}(\mathbf{A})$, $k \in X$, $j \in Y$.

To extend this theorem to the case of an arbitrary (uncountable) number
of operators, one
should require the existence of a nuclear rigging for the operators
\cite{bos,umz88}.
\end{remark}

Now consider
a family of commuting bounded self-adjoint (or normal) operators
$(A_k)$, where $k \in X$ ranges over a finite or countable set, and a
family of bounded operators $(B_j)$,  $j
\in  Y$, which is connected with the
operators $A_k$ by the following relations
\begin{equation}\label{models-relations}
A_k B_j = B_j F_{kj}(\mathbf{A}),
\end{equation}
where $F_{kj}$ is a measurable function
of the commuting family
$\mathbf{A} = (A_k)$, $k \in X$, $j\in  Y$, such that $\mathbf{F}_j(\cdot)
= (F_{kj}(\cdot))_{k \in X} \colon \mathbb{R}^X \to \mathbb{R}^X$ is
an injective on the joint spectrum of $\mathbf{A}$ measurable mapping.

\begin{theorem}\label{model-common}
Let families  $(A_k)$, $A_k= A_k^*$, $k \in X$, $(B_j)$, $j \in  Y$,
 satisfy relations \eqref{models-relations} on a Hilbert space $H$. Then
$H$ can be decomposed into the direct integral
\[
H= \int_{\mathbb{R}^X}^\oplus H_\lambda \, d\mu(\lambda), \qquad \lambda=
(\lambda_k)_{k\in X},
\]
and the operators act as follows\textup:
\begin{align}
(A_k f)(\lambda) & = \lambda_k \, f(\lambda),\notag
\\
(B_j f)(\lambda) & = b_j(\lambda)\, \chi_{\Delta_0^j}(\lambda) \,
\rho_j(\lambda)^{1/2} \, f(\mathbf{F}_j^{-1}(\lambda)), \notag
\\
(B_j^* f)(\lambda) & =
b_j^*(\mathbf{F}_j(\lambda))\,\chi_{\mathbf{F}_j^{-1}
(\Delta_0^j)}(\lambda)\,
\rho_j(\mathbf{F}_j(\lambda))^{-1/2}\, f(\mathbf{F}_j(\lambda)).
\end{align}
Here $\Delta_0^j$
is the set defined by the property that for any measurable
$\delta \subset \Delta_0^j$, we have $B_j
E_{\mathbf{A}}(\delta) \ne 0$\textup; the measure
$\chi_{\Delta_0^j}(\lambda)
\, d\mu(\mathbf{F}^{-1}_j(\lambda))$ is absolutely continuous with
respect to $d\mu(\lambda)$, $\rho_j(\cdot)$ is the corresponding
Radon--Nikodym derivative, $b_j(\cdot)$ is a measurable operator-valued
function, $b_j(\lambda) \ne 0$ for $\mu$-almost all $\lambda \in
\Delta_0^j$.
\end{theorem}

\begin{proof}
The proof of the theorem in a general situation is based on the theory of
decomposition with respect to generalized eigenvectors of families of self-adjoint
operators, and uses techniques beyond the scope of the book; we consider
only the case where the commutative family $\mathbf{A}$ has a simple joint
spectrum in order to illustrate the idea of the proof.

In this case, the representation space is $L_2(\mathbb{R}^X, d\mu)$, where
$\mu$ is the spectral measure of the commutative family, which acts as
the operators of
multiplication, $(A_k f)(\lambda) = \lambda_k \, f(\lambda)$, $k
\in X$. Take the vector $e(\lambda)\equiv 1$; then for all characteristic functions of
measurable subsets of\/ $\mathbb{R}^X$, we have
\begin{align*}
(U_j \chi_\delta)(\lambda)& = (U_jE_{\mathbf{A}}(\delta)\, e)(\lambda) 
\\
&=
E_{\mathbf{A}}(\mathbf{F}_j(\delta)) (U_je)(\lambda) = e_j(\lambda)\,
\chi_\delta(\mathbf{F}^{-1}(\lambda)),
\end{align*}
where we write $e_j(\lambda) = (U_j e)(\lambda)$. Now we show that the
measure $\chi_{\Delta_0}(\lambda)\, d\mu(F^{-1}(\lambda))$ is absolutely
continuous with respect to $d\mu(\lambda)$. Indeed, for any measurable
$\delta$ we
have $E_{\mathbf{A}}(\delta) U_j = U_j E_{\mathbf{A}}
(\mathbf{F} _j^{-1}(\delta))$,
and since $U_j^*U_j =P_j$ commutes with
$E_{\mathbf{A}}(\mathbf{F}_j^{-1}(\delta))$, we see that $(P_jf)(\lambda) =
\chi_{\Delta^j_0}(\lambda)$. Now we have
\[
P_j
E_{\mathbf{A}}(\mathbf{F}^{-1}(\delta)) =
U_j^* E_{\mathbf{A}}(\delta)
U_j,
\]
which gives the  absolute continuity required, since $\mu(\cdot) =
(E_{\mathbf{A}}(\cdot)e,e)$. Notice that $\supp e_j(\cdot)
= \Delta^j_0$; then we can rewrite it as $e_j(\lambda) = u_j(\lambda)\,
\rho_j(\lambda)^{1/2}$, which proves the formula for characteristic
functions of measurable sets, and therefore, for the whole $H$.
\end{proof}

\subsection{Centered operators}\label{sec:2.5.2}

Recall that
a bounded operator $T$ is called centered
if the operators $T^k(T^*)^k$, $(T^*)^kT^k$ form a commutative family, i.e.,
for all $k,j\in \mathbb N$
\begin{gather}
[T^k(T^*)^k,T^j(T^*)^j]=  [T^k(T^*)^k,(T^*)^jT^j] \notag
\\=
[(T^*)^kT^k,(T^*)^jT^j]=0.\label{cent}
\end{gather}

In this section we study bounded centered
 operators. We rewrite  relations \eqref{cent} in the form
which enables one  to construct
a commutative model for centered operators,
and show that the problem
of the unitary classification of centered operators is
not wild in the sense discussed below in Chapter~3.
Similarly to Section~\ref{sec:2.1.1},
we describe centered operators
such that $\ker T\ne\{0\}$ or $\ker T^*\ne \{0\}$.
We also describe, up to the unitary equivalence, all finite-dimensional
irreducible centered operators.

\medskip\noindent\textbf{1.}
We will rewrite relations \eqref{cent} in a form
that will enable us to investigate the relations using the formalism 
of dynamical systems
developed above in this chapter. We will show that irreducible
(or factor) representations of the relations fall into two cases:
the case where $\ker T\cup\ker T^*=\{0\}$ and
degenerate cases (similarly to the Wold decomposition of isometries),
which are studied separately.

Let $T=UC$ be the polar decomposition of the operator $T$.
We also introduce the commuting self-adjoint operators  $A_k =
T^k(T^*)^k$, $B_k = (T^*)^k T^k$, $k \ge1$, and
denote the joint
resolution of the identity
of the commutative self-adjoint family $(A_k,B_k)_{k\in\mathbb N}$ 
 by $E_{\mathbf A,\mathbf B}(\cdot,\cdot)$.

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

The phase $U$ of the operator $T$ is a centered operator.
\end{proposition}

\begin{proof}
For any $k\in \mathbb N$
\begin{align*}
A_k UC& = T^k (T^*)^k T = T A_{k-1}B_1 = UC A_{k-1} B_1
\\
&= U A_{k-1} B_1 C.
\end{align*}
Denote by $P$ the
projection on $(\ker C)^\perp$. Then, since $\ker C=\ker U$ and
$UP=U$,
we have $A_kU=UA_{k-1}B_1$. Similarly one can obtain the equalities
$A_1U=UB_1$ and $B_kA_1U=UB_{k+1}$.

Applying the similar arguments as in Section~\ref{sec:2.2.2} 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 holds:
\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
$$
F_1^{-1}(y_1,x_1,x_2,x_3,\ldots)=
(x_1,x_2/x_1,x_3/x_1,\dotsc).
$$
Note that the right-hand side of \eqref{f1} can be defined
even for $x_1=0$. Indeed, since $E_{B_1,\mathbf A}(\{0\}\times\mathbb R
\times\mathbb R\times\dotsb )$ is a projection on $\ker B_1=\ker U$,
the whole expression is zero. Thus, for convenience, we can set
$$
F_1^{-1}(y_1,0,x_2,x_3,\dotsc)=(0,0,0,\dotsc).
$$

In a similar manner, one
can easily derive  from the equation $B_kA_1U=UB_{k+1}$ 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
$$
F_2^{-1}(x_1,y_1,y_2,y_3,\ldots)=
(y_1,y_2/y_1,y_3/y_1,\dotsc).
$$
Passing to the adjoint operators in \eqref{f2} and combining it with
\eqref{f1} we get relations \eqref{im}.

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

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

It is a standard argument similar to the one in Section~\ref{sec:2.2.2}
that allows to prove the following 
decomposition of centered operators similar to
the Wold decomposition of isometries.

\begin{proposition}
Let $T$ be a centered  operator in
a Hilbert space $H$. The space $H$ can be decomposed into the direct
sum of the two invariant with respect to $T$, $T^*$ subspaces,
$H=H_0\oplus H_1$ such that\/ $\ker T\cup \ker T^*=\{0\}$ in $H_1$, and
$H_0$ is generated by $\ker T\cup \ker T^*$.
\end{proposition}

The degenerate representations can be completely described up to
a unitary equivalence (see below). The structure of
representations in $H_1$ is more complicated; however, realization of
centered operators as ``operator-valued weighted shifts'' follows from
Theorem~\ref{model-common} if applied to relations~\eqref{im}.

\begin{theorem}\label{thmodel}
 Let
$T$ be a centered  operator with zero kernel and dense image.
 Then
it can be realized in the space $L_2(\mathbb R_+^\infty\times
\mathbb R_+^\infty,
\mathcal H, d\mu)$ of vector-valued square-integrable functions
having their values in a certain Hilbert space $\mathcal H$ by the formula
\begin{equation}\label{model}
 (Tf)(\mathbf
x,\mathbf y) =x_1^{1/2} u(\mathbf x,\mathbf y)(d\mu(F(\mathbf x,\mathbf y))/
d\mu(\mathbf x,\mathbf y))^{1/2}f)F(\mathbf x,\mathbf y),
\end{equation}
where $F(\cdot,\cdot)$ is introduced above, $\mu(\cdot,\cdot)$ is a
$F(\cdot,\cdot)$-quasi-invariant probability Borel measure and
$u(\cdot,\cdot)$ is a
unitary measurable operator-valued  function.

Conversely, any collection $\mathcal H$, $\mu(\cdot,\cdot)$, 
and $u(\cdot,\cdot)$
which has the mentioned properties generates a centered
operator  by the formula above.
\end{theorem}

Representations with non-zero kernel can be completely classified.

\begin{theorem}\label{thdeg}
All irreducible representations of  \eqref{cent}
for which $\ker T\cup\ker T^*\ne\{0\}$ fall into the following
classes\textup:
\begin{itemize}
\item[$(i)$] finite-dimensional representations in $\mathbb C^n$
\begin{align*}
Te_j&=\lambda_je_{j+1},\qquad j=1,\dotsc, n-1,\ \lambda_j>0,\\
Te_n&=0;
\end{align*}

\item[$(ii)$] infinite-dimensional in $l_2$
$$
Te_j=\lambda_je_{j+1},\qquad j=1,2,\dotsc,\ \lambda_j>0;
$$

\item[$(iii)$] infinite-dimensional in $l_2$
$$
Te_1=0,\quad Te_j=\lambda_je_{j-1},\qquad j=2,3,\dotsc,\, \lambda_j>0.
$$
\end{itemize}
\end{theorem}

\begin{proof}
Indeed, the phase $U$ of the operator $T$
is a centered partial isometry (non-unitary); in quite the same way as in
Theorem~\ref{thirr-x} we get that
the representation space for an irreducible $T$ is the same as
for an irreducible $U$. Use Theorem~\ref{thiso-x} to represent the
operator $U$; the rest of the proof follows immediately from \eqref{im}.
\end{proof}

\noindent\textbf{3.}
Finally, we give a complete list of finite-dimensional
irreducible centered operators. We have already seen that there
is a family of finite-dimensional representations (Theorem~\ref{thdeg}).
In fact, there are some irreducible finite-dimensional representations
with a non-degenerate $T$.

\begin{theorem}
All irreducible finite-dimensional representations of \eqref{cent}
are either the finite-dimensional representations described by
Theorem~\textup{\ref{thdeg}} or  representations of the form\textup:
\begin{align*}
Te_j&=\lambda_je_{j+1},\qquad j=1,\dotsc,n-1,\\
Te_n&=\alpha\lambda_ne_1,\qquad \lambda_j>0, \
\lambda_j\ne\lambda_k \text{ \ for \ } j\ne k,\quad |\alpha|=1.
\end{align*}
\end{theorem}

\begin{proof}
Indeed, it is easy to see that representation \eqref{model} is
finite-dimensional only if the measure $\mu(\cdot)$ is concentrated on a
periodic orbit of $F(\cdot)$. Irreducibility in this case implies that
the spectrum of the operator $B_1$ is simple. Passing to a unitarily
equivalent realization we get the necessary formulae.
\end{proof}

\begin{corollary}
Any factor generated by a centered operator is hyperfinite.
\end{corollary}

\begin{proof}
Indeed, \eqref{im} imply that the corresponding
von~Neumann algebra is either of type $I$, if the representation is
degenerate, or a crossed product of a commutative algebra
by the group $\mathbb Z$. By \cite{dye} we get the assertion.
\end{proof}

\begin{corollary}
Since for a pair of self-adjoint operators there
exists a non-hyperfinite factor representations, the
description problem for centered operators by the previous
statement is not wild in the sense discussed in Section~\ref{sec:3.1}.
\end{corollary}



\subsection{Representations of Cuntz algebras}

We consider representations of the Cuntz algebras $\mathcal{O}_n$.
Recall
that the Cuntz algebra is generated by $n$ isometries,
$S_1$, \dots, $S_n$, which satisfy the following relations
\begin{equation}
S_i^*S_i = I, \qquad i=1,\dots,n, \quad \sum_{i=1}^n S_iS_i^*
 =I.
\end{equation}
Notice that the relations imply that $S_i^*S_j=0$ for $i\ne j$.

Our goal is to construct a commutative model for representations
of the Cuntz algebra and to show how it can be used to study
representations of the Cuntz algebra. The main statement here is
the
following theorem.

\begin{theorem}
For any representation of the Cuntz algebra $\mathcal{O}_n$
the following form for $S_1$, \dots,  $S_n$ holds\textup:
\begin{align}
H &= \int_{\mathbb{Z}_n^\infty}^\oplus H_x\, d\mu(x), \notag
\\
(S_i f)(x_1, x_2,\dots) & = \delta_i(x_1)\, U_i(x_2, x_3, \dots)
\notag
\\*
&\quad{}\times\biggl(\frac{d (\delta_i(x_1) \otimes \mu(x_2, x_3,
\dots)}{d\mu(x_1, x_2, \dots)}\biggr)^{1/2}\, f(x_2, x_3, \dots),
\notag
\\
(S_i^*f)(x_1, x_2, \dots)& = U_i^*(x_1, x_2, \dots) \notag
\\*
&\quad{}\times
\biggl(\frac{d\mu(i, x_1, x_2, \dots)}{d\mu(x_1, x_2, \dots)}
\biggr)^{1/2} \, f(i, x_1, x_2, \dots), 
\label{cuntz_model}
\end{align}
Here, $\mu(\cdot)$ is a probability measure defined on the cylinder
$\sigma$-algebra, quasi-invariant with respect to the transformations
$\mu(x_1,
x_2,\dots)\mapsto \delta_i(x_1)\otimes \mu(x_2, x_3,\dots)$, $i=1$,
$2$,~\dots\textup; $H_x$ is a measurable field of Hilbert spaces such
that
$d(x) = \dim H_x$ is invariant $\mu$-a.e. with respect to
the transformations
$d(x_1, x_2, \dots) \mapsto d(i, x_1, x_2, \dots)$\textup; $U_1(x)$,
\dots, $U_n(x)$ are measurable unitary operator-valued functions.

The ergodicity of the spectral measure $\mu$ is a necessary
condition
of the irreducibility of the representation\textup; in the case of a simple
joint
spectrum \textup(if  $H_x$ are one-dimensional $\mu$-a.e.\textup), 
the ergodicity
is
also sufficient for the irreducibility.

Two representations of the form \eqref{cuntz_model} are unitary
equivalent if and only if\textup:

i. the spectral measures $\mu$ and $\tilde \mu$ are equivalent\textup;

ii. the multiplicity functions $d(x) = \dim H_x$ and $\tilde d(x) =
\dim
\tilde H_x$ coincide $\mu$-a.e.\textup;

iii. the collections of unitary operator functions $(U_i(x))$ and
$(\tilde U_i(x))$ are equivalent in the following
sense\textup: there exists a measurable unitary operator-valued function
$V(x)$ such that
\begin{multline*}
V^*(x_1, x_2,\dots)\,U_i^*(x_1, x_2, \dots)\, V(1,x_1,x_2,\dots) 
\\
=\tilde U_i^*(x_1,x_2,\dots), \qquad i=1,\dots, n.
\end{multline*}

Conversely, a family consisting of a quasi-invariant measure
$\mu(\cdot)$, a dimension function $d(\cdot)$, and a collection of
unitary operator-valued functions $(U_i(\cdot))$,
possessing the listed properties, determines a representation of the
Cuntz algebra $\mathcal{O}_n$.
\end{theorem}

\begin{proof}
First we introduce some notations. For any multi-index
$\alpha=(\alpha_1, \dots, \alpha_s)$, $\alpha_j \in \mathbb{Z}_n$,
$s\ge0$, write
$S_\alpha= S_{\alpha_1}\dots S_{\alpha_s}$, $P_{\alpha}
= S_{\alpha}S_{\alpha}^*$. The set of all finite multi-indices
$\alpha$ will be denoted by $\Gamma$. The following statement
is obtained by an easy direct calculation.

\begin{proposition}
The operators $P_\alpha$,
where $\alpha$ ranges over $\Gamma$, form a commuting family of
projections. Also,
\begin{gather}
S_i P_{\alpha_1\dots \alpha_s} = P_{i\alpha_1\dots \alpha_s} S_i,
\quad
S_i^*P_{\alpha_1\dots \alpha_s} =
\delta_{i\alpha_1}P_{\alpha_2\dots \alpha_s}S_i^*,\notag
\\
\label{cuntz_ds}
i,\alpha_1,\dots, \alpha_s=1,\dots,n, \qquad
s=1, 2,\dots\,. 
\end{gather}
\end{proposition}

According to Theorem~\ref{model-common}, relations \eqref{cuntz_ds} imply that the
operators $S_1$, \dots, $S_n$ act as weighted operator-valued
shifts on the joint spectrum of the commuting family
$(P_\alpha)_{\alpha \in \Gamma}$ in the space of Fourier images
relative to this commuting
family. Our further task is to describe the joint spectrum of the
 commuting family of projections mentioned above, and to
study the corresponding dynamical system on it.

According to the spectral theorem for an infinite commuting family of
bounded self-adjoint operators (see,
for example, \cite{book} etc.), for any $\alpha \in \Gamma$ we have
\[
P_\alpha = \int_{\{0,1\}^\Gamma} \lambda(\alpha) \,
dE(\lambda(\cdot)),
\]
where $\{0,1\}^\Gamma \ni \lambda(\cdot)$ is a set of measurable
functions on
$\Gamma$ taking values 0 or 1, and $E(\cdot)$ is the joint resolution
of the identity for the commuting family of projections defined on
the cylinder $\sigma$-algebra in $\{0,1\}^\Gamma$.

For any $(i_1, \dots, i_k) \in \Gamma$, we have
\begin{align*}
\sum_{m=1}^n P_{i_1 \dots i_k m} & = \sum_{m=1}^n S_{i_1}\dots
S_{i_k} S_m S_m^* S_{i_k}^* \dots S_{i_1}^*
\\
&= S_{i_1}\dots S_{i_k} \biggl(\sum_{m=1}^n S_m S_m^*\biggr)
S_{i_k}^* \dots S_{i_1}^* = P_{i_1, \dots, i_k}.
\end{align*}
Then, according to \cite{ber1}, the spectral measure of the
commuting
family is concentrated on functions $\lambda(\cdot) \in
\{0,1\}^\Gamma$ such that 
\[
\sum_{m=1}^n \lambda(i_1,\dots, i_k,m) =
\lambda(i_1,\dots, i_k).
\]

Let $\lambda(\cdot)$ be such a function. Then $\lambda(\emptyset)$
is
either 0 or 1. If $\lambda(\emptyset) =0$, then $\lambda(1)+\dots
+\lambda(n)=0$ and $\lambda(1)=\dots=\lambda(n) =0$, etc., and
therefore,
$\lambda(\cdot)\equiv 0$. This implies that $E(\lambda(\cdot)) =0$,
since
the latter projection is in the kernels of all $P_i$, which is
impossible because  $P_1+\dots+P_m=I$.

To any function $\lambda(\cdot)$ with $\lambda(\emptyset)=1$ there
uniquely corresponds a sequence $(x_1, x_2,\dots) \in
\{1,\dots,n\}^\infty=\mathbb{Z}_n^\infty$
such that for any $k=1$, 2,\dots we have $\lambda(x_1,\dots,
x_k)=1$
and vice versa. To a cylinder set in $\mathbb{Z}_n^\infty$ there
corresponds a cylinder set in $\{0,1\}^\Gamma$, which enables us
to
define the image of the projection-valued measure $E(\cdot)$ under
the  mapping $\phi\colon \mathbb{Z}_n^\infty \to
\{0,1\}^\Gamma$ which was just defined; 
this image will also be denoted by $E(\cdot)$ as
long as this will not lead to  any confusion.

 Therefore, the spectral
decomposition of the commuting family of projections $P_\alpha$,
$\alpha \in \Gamma$, takes the form
\[
P_\alpha = \int_{\mathbb{Z}_n^\infty} \chi_{\alpha\times
\mathbb{Z}_n^\infty} (x) \, dE(x) = E(\alpha_1,\dots,\alpha_k
\times \mathbb{Z}_n^\infty).
\]

Now, Theorem~\ref{model-common} implies the statement.
\end{proof}

The established theorem provides a way to construct
representations of the Cuntz algebra. Indeed, one should take an
appropriate quasi-invariant measure on $\mathbb{Z}_n^\infty$,
choose an
invariant multiplicity function, and a unitary operator-valued
function.

\begin{remark}
The constructed commutative model can also be rewritten in the
space
of (vector-valued, in general) functions on the unit interval as
follows.

To each point $x$ of the unit interval $[0,1)$ we put into
correspondence
its $n$-ary representation $x=0.x_1x_2\dots$.
This correspondence is one-to-one except for the points with tails
consisting of $n$'s (they are identified with points having finite
decomposition). With such a correspondence, cylinder sets in
$\mathbb{Z}_n^\infty$ correspond to Borel sets from $[0,1)$, and
the representation \eqref{cuntz_model} takes the form
\begin{align*}
H &= \int_{[0,1)} H_\lambda\, dE(\lambda),
\\
(S_if)(\lambda) & = \chi_{\left[\frac{i-1}n, \frac
in\right)}(\lambda) \,
U_i(n\lambda -(i-1))
\\
& \quad{} \times
\biggl(\frac{d\mu(n\lambda - (i-1))}{d\mu(\lambda)}\biggr)^{1/2}
f(n\lambda-(i-1)),
\\
(S^*_i f)(\lambda)& = U_i^*(\lambda) \,
\biggl(\frac{d\mu((\lambda+i-1)/n)}{d\mu(\lambda)}\biggr)^{1/2}\,
f((\lambda + i-1)/n),
\end{align*}
which holds for all representations that do not contain a single
representation related to the orbit of the point $(n,n,\dots) \in
\mathbb{Z}_n^\infty$ (see example below).
\end{remark}

\begin{example} (Representations related to a single orbit).
The simplest class of irreducible representations can be obtained
as
follows. Take an arbitrary point $x=(x_1, x_2, \dots) \in
\mathbb{Z}_n^\infty$. The simplest quasi-invariant ergodic measure
is an atomic
measure $\mu$ concentrated on the orbit of a point $x$ with respect to the
action given in \eqref{cuntz_model}. The orbit consists of all
points having similar tails, i.e., the points $y = (y_1, y_2, \dots)$
such that there exist $l\ge1$, $m \in \mathbb{Z}$ such that $x_k =
y_{k+m}$ for all $k > l$. Denote such an orbit by $ O_x$.
In this case, any operator-valued function $U(x)$ is
equivalent to the identity, and the irreducibility implies that
$d(x)=1$
$\mu$-a.e.

Now, $\delta$-functions concentrated at points of the orbit form
a
basis of $H$, and the representation acts as follows:
\begin{align}
S_i \, \delta_{(x_1, x_2, \dots)}& = \delta_{(i, x_1, x_2, \dots)},
\notag
\\
S_i^* \, \delta_{(x_1, x_2, \dots)}& =
\delta_{i,x_1}\,\delta_{(x_2,x_3, \dots)}, \qquad i=1, \dots, n.
\end{align}
Two such representations are unitarily equivalent if and only if
they
correspond to the same orbit.

As one can easily see, all such representations fall into the class
of permutative ones \cite{brat-jorg}.


Depending on the orbit taken, the following situations may occur.

i. The point $x= (x_1, x_2, \dots)$  taken to construct
the
orbit has a constant tail of the form $(\dots, k, k, k, \dots)$.
Then
$ O_x$ is exactly the orbit of the point $(k, k, k, \dots)$. The
operator $S_k$ has a unitary part formed by its restriction to the subspace
spanned by the eigenvector
$\delta_{(k, k,\dots)}$, while the others are multiples of the
unilateral shift.

For such a representation, its restriction to UHF$_n$ is
irreducible, since the action of elements of the form $S_{i_1}\dots
S_{i_k}S_{j_1}^*\dots S_{j_k}^*$ act transitively on the orbit.

ii. Orbits of periodic points  $(x_1, x_2, \dots,
x_m, x_1, \dots)$ generate another class of irreducible
representations of $\mathcal{O}_n$. The action of UHF$_n$ splits the
orbit into $m$ different sub-orbits ($m$ is a minimal period)
corresponding to  $(x_1, x_2, \dots, x_m, \dots)$, $(x_2,
x_3, \dots, x_m, \dots)$, \dots, $(x_m, x_1, \dots, x_{m-1},
\dots)$;
therefore, such a representation of $\mathcal{O}_n$ splits into a direct
sum of $m$ inequivalent representations of UHF$_n$ corresponding
to these sub-orbits.

iii.  The non-periodic (irrational)  points $x=(x_1,x_2,
\dots)$ and $x'=(x_1',x_2', \dots)$ belong to the same orbit of
UHF$_n$ if and only if $x_i \ne x_i'$ for a finite number of $i$'s.
This implies that the corresponding representation of
$\mathcal{O}_n$
decomposes into an infinite direct sum of inequivalent irreducible
representations of UHF$_n$ corresponding to the sub-orbits.
\end{example}

\begin{example}
Another class of irreducible representations is related to product
measures. Any product measure $\mu$ on $\mathbb{Z}_n^\infty$ is
ergodic,
and choosing $H=L_2(\mathbb{Z}_n^\infty, d\mu)$ with $U(x)\equiv
1$, we
get an irreducible representation related to the measure $\mu$. Two
such
representations are unitarily equivalent if and only if the
corresponding measures are equivalent.

Since any product measure is ergodic with respect to the action of
UHF$_n$, the restriction of the corresponding representation to
UHF$_n$ remains irreducible.
\end{example}


\begin{example} (A family of irreducible representations).
For each $\alpha \in \mathbb{T}$, consider the following
representation
of the Cuntz algebra $\mathcal{O}_n$. Let $H =
L_2(\mathbb{Z}_n^\infty,
d\mu)$, and $d\mu(x) = \bigotimes_{k=1}^\infty d\mu_k(x_k)$,
$\mu_k(j)=
1/n$, $j=1$, \dots, $n$, and the operators be
\begin{align}\label{cuntz-alpha}
(S_j^{(\alpha)}f)(x_1, x_2, \dots) & = \alpha \delta_j(x_1) n^{1/2}
f(x_2,x_3, \dots),\notag
\\
({S_j^{(\alpha)}}^* f)(x_1, x_2, \dots) &= \alpha ^{-1}
n^{-1/2} f(j,x_1, x_2, \dots), \qquad j=1, \dots, n.
\end{align}
A direct computation shows that the expression for
${S_j^{(\alpha)}}^*$
gives indeed an operator adjoint to $S_j^{(\alpha)}$, and that
\eqref{cuntz-alpha} is
a representation of the Cuntz algebra.

Notice that the latter representations admit the following
realization
on $L_2([0,1], dt)$:
\begin{align*}
(S_j^{(\alpha)} f)(t) & = \alpha \chi_{[(j-1)/n, j/n]}(t) n^{1/2}
f(nt -
(j-1)),
\\
({S_j^{(\alpha)}}^* f)(t) & = \alpha ^{-1} n^{-1/2} f((t+(j-1)/n),
\qquad
j=1, \dots, n.
\end{align*}

\begin{proposition}
The representations $(S_j^{(\alpha)})$ are unitarily inequivalent
irreducible representations of the Cuntz algebra $\mathcal{O}_n$.
\end{proposition}

\begin{proof}
The irreducibility of representations of the form
\eqref{cuntz-alpha}
follows immediately from the ergodicity of the product measure
$\mu$. We
show that the representations are different for different $\alpha$.

Indeed, take $\alpha_1$ and $\alpha_2$, and form the corresponding
representations of $\mathcal{O}_n$. Then there exists a unitary
operator
$V \colon L_2(\mathbb{Z}_n^\infty, d\mu) \to
L_2(\mathbb{Z}_n^\infty,
d\mu)$ such that $V^* S_j^{(\alpha_1)}V = S_j^{(\alpha_2)}$,
$j=1$,
\dots, $n$. Since the
same
commutative family of projections on $L_2(\mathbb{Z}_n^\infty,
d\mu)$ corresponds to both representations,
the operator $V$ has the form $(Vf)(x) = v(x)\, f(x)$, where
$|v(x)|
=1$ $\mu$-a.e.

Since $V^* {S_j^{(\alpha_1)}}^* V = {S_j^{(\alpha_2)}}^*$, $j=1$,
\dots,
$n$, as well, we have
\begin{multline*}
v^{-1} (x_1,x_2, \dots) \alpha_1^{-1} n^{-1/2} v(j, x_1, x_2,
\dots)
f(j, x_1, x_2, \dots)
\\
= \alpha_2^{-1} n^{-1/2} f(j, x_1,x_2, \dots),
\qquad j=1, \dots, n,
\end{multline*}
which implies that for all $j =1$, \dots, $n$
\begin{equation}\label{cuntz-aux}
v(j, x_1, x_2, \dots)= \frac {\alpha_1}{\alpha_2} \, v(x_1, x_2,
\dots),
\qquad \text{$\mu$-a.e.}.
\end{equation}
This implies, in turn, that
\[
v(1, x_1, x_2, \dots)=v(2, x_1, x_2, \dots)=\dots= v(n, x_1, x_2,
\dots)
\]
 $\mu$ -a.e., i.e., the function $v(\cdot)$ is independent 
of
$x_1$. But then the right-hand side of \eqref{cuntz-aux} is
invariant
with respect to $x_1$, and the left-hand side is independent
of
$x_1$, $x_2$, etc. Therefore, $v(x)$ is independent of
any finite number of variables; since a product measure
is
ergodic with respect to such transformations, we get that $v(x)$ is
constant
$\mu$-a.e., and thus, $\alpha_1 = \alpha_2$.
\end{proof}
\end{example}


\begin{example} 
Now consider a slightly  more complicated example of representation
of $\mathcal{O}_n$ in the space of vector-valued functions. Let
\begin{gather*}
H= \int_{\mathbb{Z}_n^\infty} l_2(\mathbb{Z})\, d\mu(x) =
\bigoplus_{-\infty}^\infty L_2(\mathbb{Z}_n^\infty, d\mu) =
l_2(\mathbb{Z}) \otimes L_2(\mathbb{Z}_n^\infty, d\mu),
\\
d\mu(x)  = \bigotimes_{k=1}^\infty d\mu_k(x_k), \quad \mu_k(i) =
1/n,
\qquad i=1, \dots, n,
\\
u_1(x) =\dots= u_n(x) = u(x) = S \otimes 1(x), \quad Se_k =
e_{k+1}, \qquad k
\in \mathbb{Z}.
\end{gather*}

For vector-valued functions $\mathbf{f}(x) = (\dots, f_{-1}(x),
f_0(x), f_1(x), \dots)$, the representation acts as follows:
\begin{align*}
(S_j \mathbf{f})_k(x_1, x_2, \dots) & = \delta_j(x_1) \,n^{1/2}\,
\mathbf{f}(x_2,x_3,\dots)_{k-1}, \qquad k \in \mathbb{Z},
\\
(S_j^*\mathbf{f})_k(x_1, x_2,\dots) & = n^{-1/2}\,\mathbf{f}(j,
x_1, x_2,
\dots)_{k+1}, \qquad  k \in \mathbb{Z}.
\end{align*}

Choose a vector $\Omega = e_0 \otimes 1(x)$.
Then we have
\begin{align*}
(S_j \Omega)(x) & = e_1 \otimes \delta_i(x_1) \sqrt{n} \,
1(x_2,x_3,\dots),
\\
(S_j^*\Omega)(x) & = e_{-1} \otimes (\sqrt{n})^{-1} \, 1(j, x_1,
x_2, \dots),
\end{align*}
and writing $S_\alpha = S_{\alpha_1}\dots S_{\alpha_s}$ for a
multi-index $\alpha = (\alpha_1, \dots, \alpha_s)$, we get
\begin{equation}\label{cuntz-aux2}
(S_\alpha S_\beta^* \Omega)(x) = e_{|\alpha| - |\beta|} \otimes
\delta_{\alpha_1}(x_1) \dots \delta_{\alpha_s}(x_s) \,
n^{(|\alpha| - |\beta|)/2} \, 1(x_{s+1}, \dots),
\end{equation}
where $|\alpha|$ is the  length of the multi-index. Also, the vector
$\Omega$ is cyclic.

The corresponding
state on $\mathcal{O}_n$ is  defined by the following formula
\[
\omega(S_\alpha  S_\beta^*) = \delta_{\alpha,\beta} n^{-|\alpha|},
\]
where $\delta_{\alpha,\beta}$ is one if $|\alpha| = |\beta|$, and
zero otherwise.

Indeed, since $S_\alpha S^*_\beta \Omega \in H_{|\alpha| -
|\beta|}$,
we have $(S_\alpha S_\beta^* \Omega, \Omega)=0$, $|\alpha|\ne
|\beta|$,
and \eqref{cuntz-aux2} implies
\[
(S_\alpha S_\alpha^* \Omega, \Omega) = \int_{\mathbb{Z}_n^s}
\delta_{\alpha_1}
(x_1)\dots \delta_{\alpha_s}(x_s)\, d \mu_1(x_1)\dots d\mu_s(x_s) =
n^{-s},
\]
where $s=|\alpha|$, and the formula follows.

Passing to a unitarily equivalent realization, one has the following
formulas for the operators in 
$H = L_2(\mathbb{T} \times \mathbb{Z}_n^\infty, dz \otimes d\mu)$,
\[
(S_jf)(z,x_1, x_2, \dots)  = z n^{1/2}\delta_j(x_1) f(z, x_2, x_3,
\dots).
\]
The latter form gives an explicit decomposition of the
constructed
representation into a direct integral of irreducible inequivalent
representations $(S_j^\alpha)$ constructed above.
\end{example}

\begin{example}
(The KMS representation)
We give an explicit formula for the representation  of the  Cuntz
algebra $\mathcal{O}_n$, corresponding to the unique KMS state
related to the action of the gauge group.

Introduce some notations. Let $\mathbb{Z}_{n,0}^\infty$ be the set
of all finite sequences $\alpha = (\alpha_1,\dots,\alpha_n,0,0,\dots)$;
$\mathbb{Z}_{n,0}^\infty = \bigcup _k \mathbb{Z}_n^k$,
with $\mathbb{Z}_n^k \subset \mathbb{Z}_n^{k+1}$ given by
$(\alpha_1, \dots,\alpha_n) \mapsto (\alpha_1, \dots, \alpha_n, 0)$.
Let $|\alpha|$ be the number of the tailing non-zero $\alpha_j$ in $\alpha$.
Write $j_k= (0, \dots, 0,j,0,\dots)$ with $j$ in the $k$-th place,
$j\in \mathbb{Z}_n$, $k=1$, 2, \dots. The group operation in
$\mathbb{Z}_{n,0}^\infty$ will be denoted by the addition sign.

$\mathbb{Z}_{n,0}^\infty$ is a countable set, so we can consider the
separable Hilbert space $l_2(\mathbb{Z}_{n,0}^\infty)$.
For $x=(x_1, x_2, \dots) \in \mathbb{Z}_n^\infty$, put
$\sigma(x) = (x_2, x_3, \dots)$, $\sigma_j(x) = (j, x_1, x_2, \dots)$,
$j=0$, \dots, $n-1$.



The representation space in our example will have the form $H_0
\otimes L_2(\mathbb{Z}_n^\infty, d\mu(x))$, where $\mu$ is an
infinite product of Haar measures on $\mathbb{Z}_n$.

The space $H_0$ is generated by its orthogonal basis consisting of
vectors $e(i,k,\alpha)$, where $\alpha \in\Gamma$ is any
multi-index, $i\in \mathbb{Z}_n$, and $k=0$ if $\alpha\ne\emptyset$
and $i=0$, and $k=0$, 1, \dots, elsewhere, i.e., for $\alpha =
\emptyset$, or $i\ne 0$.

The operators $S_0$, \dots, $S_{n-1}$ act as follows
\begin{align*}
S_j e(i,k,\alpha) \otimes f(x) &= n^{1/2}\left \{\begin{array}{ll}  
e(j-\alpha_1, 0, \sigma(\alpha)),& i=0,
\alpha\ne \emptyset\\
 e(i, k+1, \alpha),&\text{otherwise} \end{array}\right \} 
\\
&\qquad\otimes
\delta_j(x_1) f(\sigma(x)),
\\
S_j^* e(i, k, \alpha) \otimes f(x) & = n^{-1/2} \left \{\begin{array}{ll} 
e(0,0,\sigma_{j-i}(\alpha)), &k=0\\  e(i, k-1, \alpha),& k \ne 0
\end{array}
\right \}
\\
&\qquad\otimes f(\sigma_j(x)).
\end{align*}

The vector $\Omega = e(0,0,\emptyset) \otimes 1(x)$ is cyclic, and the
corresponding state is
\[
\omega(S_\alpha S_\beta^*) = (S_\alpha S_\beta^* \Omega, \Omega) =
\delta_{\alpha, \beta} n^{-|\alpha|}.
\]

Notice that the vector $\Omega$ is not cyclic with respect to the UHF
subalgebra in $\mathcal{O}_n$, and the corresponding cyclic
subspace is a proper subspace of $H$. This means, in particular,
that the representation corresponding to the tracial state on UHF
cannot be extended to a representation of $\mathcal{O}_n$ in the same
space.

Also, choosing another product measure on $\mathbb{Z}_n^\infty$,
corresponding to weights $p_0$, \dots, $p_{n-1}$, $\sum p_i=1$, the
formula similar to that given above, but with $p_j^{1/2}$ replacing $n^{-1/2}$,
gives a
representation corresponding to states constructed as extension of
product states on UHF which have the form
\[
\omega(S_\alpha S^*_\beta) = \delta_{\alpha, \beta} p_{\alpha_1}
\dots p_{\alpha_s}, \qquad s = |\alpha|.
\]
\end{example}

\begin{example}
We give another example of a representation. The corresponding state
is
\[
\omega(S_\alpha S^*_\beta) = \begin{cases} n^{-|\alpha|}, &
|\alpha| = |\beta|,\, \sum \alpha_i = \sum \beta_i,\\ 0, &
\text{otherwise.} \end{cases}
\]
 The representation space is $\mathbb{Z}_n \otimes l_2(\mathbb{Z})
\otimes L_2(\mathbb{Z}_n^\infty, d\mu)$ with the standard product
measure $\mu$, and the formulas are
 \begin{align*}
 S_j e_l \otimes e_k \otimes f(x) & = n^{1/2} e_{l-j} \otimes
e_{k+1} \otimes  \delta_j(x_1) f(\sigma(x)),
 \\
 s_j^* e_l \otimes e_k \otimes f(x) & = n^{-1/2} e_{x_1+l} \otimes
e_{k-1} \otimes  f(\sigma_j(x)).
 \end{align*}
The state which is written above is obtained from the 
vector $e_0 \otimes e_0 \otimes
1(x)$, which is cyclic.
\end{example}

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