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\section{On representations of some nuclear algebras}
\subsection{Commutative models}
Above, during the study of structure of operator families related to dynamical 
systems, we saw that depending on the properties of the dynamical 
system, the structure problem may be rather simple (in the case when 
each ergodic measure is concentrated on a single orbit), or more 
complicated, if there are non-trivial ergodic measures. In this 
section, we will study, what information about operator families related
to dynamical systems can be obtained in the general case. In this case 
complete description of structure of the family is problematic, 
however, we construct commutative model --- some standard form, in 
which any representation can be written. It should be noted that this 
form involves objects whose classification is still very complicated.

Consider a family of commuting bounded self-ajoint (or normal) operators 
$(A_k)$, where $k \in X$ ranges over 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}(\mathbb{A}),
\end{equation}
where $F_{kj}$ is a measurable function of the commuting family 
$\mathbb{A} = (A_k)$, $k \in X$, $j\in  Y$.

\begin{theorem}
Let families  $(A_k)$, $A_k= A_k^*$, $k \in X$, $(B_j)$, $j \in  Y$, 
in a Hilbert space $H$ satisfy relations \eqref{models-relations}. Then 
$H$ can be decomposed into direct sum $H = H_1 \oplus H_2 \oplus 
\dots\oplus H_\infty$  in such a way that
 for all $m=1$, $2$, \dots,$\infty$, the space 
$H_m$ is invariant with respect to  $(A_k)$, $k \in X$, $(B_j)$, 
$(B_j^*)$, $j \in Y$, and each $H_m$ is unitarily equivalent to 
$\bigoplus_{i=1}^m L_2(\mathbb{R}^X, d\mu_m)$; the operators act as 
follows:
\begin{align}
(A_k f)(x) & = x_k f(x), \quad k \in X,
\\
(B_j f)(x) & = b_j(x) \chi_{\delta_j}(x) \sqrt{\rho(x)} 
f(F_{j}^{-1}(x)), \quad j \in Y.
\end{align} 
\end{theorem}

\begin{proof}
The decomposition of the commuting self-adjoint family  
\end{proof}

\subsection{Centered operators}

\subsubsection{Introduction}


In \cite{camp,morr}, a bounded operator $T$ was called centered 
if the operators $T^k(T^*)^k$, $(T^*)^kT^k$ form a commuting family i.e.,
for all $k,j\in \Bbb N$
\begin{equation}\label{cent}
[T^k(T^*)^k,T^j(T^*)^j]= [T^k(T^*)^k,(T^*)^jT^j]= [(T^*)^kT^k,(T^*)^jT^j]=0.
\end{equation}

In \cite{camp,krupir} the structure of operators from a more wide 
class of  operators, such that only $T^*T$ and $TT^*$ commute,
was studied. It was shown in  \cite{krupir} that the problem of the 
unitary classification of such bounded operators is wild, i.e.,
it ``contains  as a subproblem'' the same problem for a pair 
of arbitrary self-adjoint operators (the precise definition of the words 
``contains as a subproblem'' is explained in the paper being cited).
The same is true if one assume in addition that the operator $T$ is
a partial isometry.

In this section we study the structure of centered
(bounded or unbounded) operators. We show that the problem
of the unitary classification of the centered operators is 
not wild in the sense mentioned above (is not $p$-wild \cite{pirsam}),
but one can construct a centered operator $T$ which together with
$T^*$ generates a factor which is not of type $I$ (is $f$-wild in
the sense of \cite{pirsam}).  

The form in which we rewrite the relations enables us to apply
the technique of dynamical systems \cite{bos,romp}. In particular,
one can construct the commutative model for centered operators.
Similarly to \cite{vai}, we describe the centered operators
for which $\ker T\ne\{0\}$ or $\ker T^*\ne \{0\}$. In particular, we
show that any irreducible centered partial isometry is either unitary,
or isometry, or co-isometry, or finite-dimensional.

We also describe up to the unitary equivalence all the finite-dimensional 
irreducible centered operators. The set of $n$-dimensional irreducible
representations is parameterized by $n+1$ parameter i.e., the number
of parameters is unbounded, which is one more indication on the complexity
of the classification problem.

\subsubsection{Dynamical system associated with centered operators.}

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

Let $T=UC$ be the polar decomposition of the operator $T$. 
We also denote by $E_{\bold A,\bold B}(\cdot,\cdot)$ the joint 
resolution of the identity
of the commuting self-adjoint family $(A_k,B_k)_{k\in\Bbb N}$.

\begin{proposition} 
The relations \eqref{cent} is equivalent to the following
relations
\begin{equation}\label{im}
E_{\bold A,\bold B}(\Delta)U=UE(F^{-1}(\Delta)),\qquad 
\Delta \in \frak B(\Bbb R^{\Bbb N}\times\Bbb R^{\Bbb N}),
\end{equation}
where the mapping $F^{-1}(\cdot)$ is defined as follows:
\begin{multline*}
F^{-1}(\bold x,\bold 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}
Since the operators $A_k$ and $B_k$ form a commuting family, 
there exists a dense in $H$ set $\cal D$  consisting of entire vector for 
these operators. Note that $\cal D\subset D(T^k)\cap D((T^*)^k)$ for any $k$.
Thus, for any $f,g\in \cal D$, $k\in \Bbb N$
\begin{align*}
(Cf,U^*A_kg)&= (UCf, A_kg)=(Tf,A_kg) =(Tf, T^k(T^*)^kg)\\
  &=(T^*Tf,A_{k-1}T^*g)=(B_1f, A_{k-1}T^*g)\\
  &=(UCA_{k-1}B_1f,g)=(UA_{k-1}B_1Cf,g),
\end{align*}
which implies $(Cf,U^*A_kg)=(Cf,B_1A_{k-1} U^* g)$. Denote by $P$ the 
projection on $(\ker C)^\perp$, then since $\ker C=\ker U$ and $UP=U$
we have $A_kUf=UA_{k-1}B_1f$. Similarly one can obtain the equalities
$A_1Uf=UB_1f$ and $B_kA_1Uf=UB_{k+1}f$. 

Applying the results of \cite{romp} to $A_k$, $A_{k-1}$ and $B_1$
one can show that for any Borel set  $\Delta\subset \Bbb R^{\Bbb N}$ 
the following relation hold:
\begin{equation}\label{f1}
E_{B_1,\bold A}(\Bbb R\times\Delta)U=
UE_{B_1,\bold A}(F_1^{-1}(\Bbb 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 $E_{B_1,\bold A}(\{0\}\times\Bbb R
\times\Bbb R\times\dotsb )$ is the projection on $\ker B_1=\ker U$,
the whole expression is zero. Thus, for convenience, we can put 
$$ 
F_1^{-1}(y_1,0,x_2,x_3,\dotsc)=(0,0,0,\dotsc).
$$

In a similar manner, from the equation $B_kA_1Uf=UB_{k+1}f$ one 
can easily derive the equality
\begin{equation}\label{f2}
E_{A_1,\bold B}(\Bbb R\times \Delta)U^*=
U^*E_{A_1,\bold B}(F_2^{-1}(\Bbb 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 the needed 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 $U$ (being a partial isometry) satisfying the
relations \eqref{im} generates a centered operator, is a direct
calculation.
\end{proof}

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

\begin{proof}
Indeed, the relations \eqref{im} imply that the corresponding
von~Neumann algebra is a crossed product of a commutative one 
by the group $\Bbb Z$. By \cite{dye} we get the assertion.
\end{proof}

\begin{corollary}
Since for a pair of arbitrary selfadjoint operators there 
exists a non-hyperfinite factor representations, the
description problem for centered operators by the previous
statement is not $p$-wild in the sense of \cite{pirsam}. 
\end{corollary}

\begin{proposition}
Let $T$ be a centered (in general setting, closed unbounded) 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 in $H_1$ \ $\ker T\cup \ker T^*=\{0\}$ and 
$H_0$ is generated by $\ker T\cup \ker T^*$.
\end{proposition}

\begin{proof} By the previous Proposition we can only refer to \cite{vai}. 
\end{proof}

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

\begin{example}
Consider the operator $T$ satisfying the relation 
$$
TT^*=F(T^*T),
$$
where $F(\cdot)\colon \Bbb R\to \Bbb R$ is a continuous function
(such operators are always centered).
If the dynamical system $x\to F(x)$ is ergodic, then there exist
a representation of this relation which generates a factor which is
not of type I \cite{romp,vai}.
\end{example}

Although the structure of non-degenerated centered operators is 
complex, we can give a realization of such operators
as ``operator-valued shift'' operators.

\begin{theorem}\label{thmodel}
Let $T$ be a centered (unbounded) operator with zero kernel and dense image.
Then it can be realized in the space $L_2(\Bbb R_+^\infty\times\Bbb R_+^\infty,
\cal H, d\mu(\cdot,\cdot))$ of vector-valued square-integrable functions
having their values in a certain Hilbert space $\cal H$ by the formula
\begin{equation}\label{model}
(Tf)(\bold x,\bold y) =x_1^{1/2} u(\bold x,\bold y)(d\mu(F(\bold x,\bold y))/
d\mu(\bold x,\bold y))^{1/2}f)F(\bold x,\bold 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 $\cal H$, $\mu(\cdot,\cdot)$ and $u(\cdot,\cdot)$
having the mentioned properties generates by the formula above a centered
operator.
\end{theorem}

\begin{proof}
Directly follows from the theorem on commutative models \cite{bos,romp}
applied to the relations \eqref{im}.
\end{proof}

\subsubsection{Partial isometries and degenerate representations}

In this section we describe up to unitary equivalence all the
irreducible representations of \eqref{cent} for which $\ker T
\cup \ker T^*\ne \{0\}$. We start with a classification of
centered partial isometries.

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

\begin{theorem}\label{thiso}
Any centered irreducible representation of \eqref{iso1} is
one of the following:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

To apply this theorem to the description of the irreducible
solutions of \eqref{cent}, we prove an auxiliary statement which
may be of independent use.

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

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

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

We divide the proof into several steps.

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

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

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

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

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

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

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

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

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

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


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

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


Now we apply the results on partial isometries to the description
of centered operators (note that the same result can be obtained 
using \cite{vai}).

\begin{theorem}\label{thdeg}
All the irreducible representations of the relations \eqref{cent}
for which $\ker T\cup\ker T^*\ne\{0\}$ fall into the following
classes:
\begin{itemize}
\item[(i)] finite-dimensional representations on $\Bbb C^n$
\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 on $l_2$
$$
Te_j=\lambda_je_{j+1},\qquad j=1,2,\dotsc,\ \lambda_j>0;
$$

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

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

\subsubsection{Finite-dimensional representations}

We conclude the paper with the complete list of the 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 the irreducible finite-dimensional representations of \eqref{cent}
are either the finite-dimensional representations described by 
Theorem~\ref{thdeg} or the representations of the form:
\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 the 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}

The set of $n$-dimensional irreducible representations is naturally
parameterized by $n+1$ parameters $\lambda_j$ and $\alpha$ (the degenerate
representations also fit in this family for $\lambda_n=0$). This
gives an example of finite-generated $*$-algebra for which the number
of parameters defining the finite-dimensional irreducible representations
grows with a dimension, but this algebra is not $p$-wild in the sense 
of \cite{pirsam}.


\subsection{Representations of Cuntz algebras}

We consider representations of Cuntz algebra $\mathcal{O}_n$. 
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, \quad i=1,\dots,n, \quad \sum_{i=1}^n S_iS_i^*
 =I.
\end{equation}
Notice that the relations imply $S_i^*S_j=0$ for $i\ne j$.

Our purpose is to construct a commutative model for representations
of the Cuntz algebra and 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:
\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)
\,\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),
\label{cuntz_model} 
\\
(S_i^*f)(x_1, x_2, \dots)& = U_i^*(x_1, x_2, \dots) \, 
\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) \notag
\\
& i = 1,\dots, n.\notag
\end{align}
Here, $\mu(\cdot)$ is a probability measure defined on the cylinder
$\sigma$-algebra, quasi-invariant with respect to transforms
$\mu(x_1,
x_2,\dots)\mapsto \delta_i(x_1)\otimes \mu(x_2, x_3,\dots)$, $i=1$,
$2$,~\dots; $H_x$ is a measurable field of Hilbert spaces, such 
that
$d(x) = \dim H_x$ is invariant $\mu$-a.e. with respect to 
transforms
$d(x_1, x_2, \dots) \mapsto d(i, x_1, x_2, \dots)$; $U_1(x)$,
\dots, $U_n(x)$ are measurable unitary operator-valued functions.

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

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

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

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

iii. collections ofunitary operator functions $(U_i(x))$, and 
$(\tilde U_i(x))$ are equivalent in the following
sense: there exists a measurable unitary operator-valued function
$V(x)$, such that
\[
V^*(x_1, x_2,\dots)\,U_i^*(x_1, x_2, \dots)\, V(1,x_1,x_2,\dots) = 
\tilde U_i^*(x_1,x_2,\dots), \quad i=1,\dots, n.
\]

Conversely, a family consisting of a quasi-invariant measure
$\mu(\cdot)$, dimension function $d(\cdot)$, and a collection of
unitary operator-valued functions $(U_i(\cdot))$,
possessing the listed properties, determine 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 muti-indices 
$\alpha$ will be denoted by $\Gamma$. The following statement
is an easy direct calculation.

\begin{proposition}
The operators $P_\alpha$,
where $\alpha$ changes 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^*,
\label{cuntz_ds}
\\
i,\alpha_1,\dots, \alpha_s=1,\dots,n, \quad
s=1, 2,\dots. \notag
\end{gather}
\end{proposition}

According to \cite{bos}, 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. Out further task is to describe the joint spectrum of the
mentioned commuting family of projections, and to
study the corresponding dynamical system on it.

According to the spectral theorem for 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 a joint resolution
of the identity of 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{ber0}, 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 $E(\lambda(\cdot)) =0$, 
since
the latter projection is in 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 one 
to
define the image of the projection-valued measure $E(\cdot)$ under
the just defined mapping $\phi\colon \mathbb{Z}_n^\infty \to
\{0,1\}^\Gamma$; this image will also be denoted by $E(\cdot)$ as 
far as this will not lead to  a 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 on commutative models (see \cite{bos},
\cite{umz88}, \cite{berkon}), implies the statement.
\end{proof}

The established theorem provides a way to construct
representations of the Cuntz algebra. Indeed, one should take
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 decomposition into infinite sequence $x=0.x_1x_2\dots$.
This correspondence is one-to-one except for the points with tails
consisting of $n$'s (they are identified with points having finite
decomposition). Under such correspondence, cylinder sets in
$\mathbb{Z}_n^\infty$ correspond to Borel sets from $[0,1)$, and
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))
\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, which do not contain single
representation, related to the orbit of 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 atomic
measure $\mu$ concentrated on an orbit of the point $x$ under the
action described in \eqref{cuntz_model}. The orbit consists of all
points having similar tails, i.e., 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 orbit by $O_x$.
In this case, any operator-valued function $U(x)$ is
equivalent to identical identity, and irreducibility implies 
$d(x)=1$
$\mu$-a.e.

Now, $\delta$-functions concentrated at points of the orbit, form 
the
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)}, \quad 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 $x= (x_1, x_2, \dots)$ which was taken to construct 
the
orbit, has 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 unitary part formed by its eigenvector
$\delta_{(k, k,\dots)}$, while the others are multiples of the
unilateral shift.

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

iii. For non-periodic (irrational) sequences, 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 finite number of $i$'s.
This implies that the corresponding representation of 
$\mathcal{O}_n$
decomposes into infinite direct sum of non-equivalent irreduible
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 irreducible representation related to the measure $\mu$. Two 
such
representations are unitary equivalent if and only if the
corresponding measures are equivalent.

Since any product measure is ergodic with respect to 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 are
\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), \quad j=1, \dots, n.
\end{align}
It is a direct calculation that the expression for 
${S_j^{(\alpha)}}^*$
is indeed the adjoint to $S_j^{(\alpha)}$ operator, and that 
\eqref{cuntz-alpha} is
a representation of the Cuntz algebra.

Notice that the latter representations admit the following 
realization
in $L_2([0,1], dt)$:
\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), 
\quad
j=1, \dots, n.
\end{align*}

\begin{proposition}
The representations $(S_j^{(\alpha)})$ are unitary 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 immideately from the ergodicity of the product measure 
$\mu$. We
show that the representatoins 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 to the both representations there corresponds the 
same
commutative family of projections in $L_2(\mathbb{Z}_n^\infty, 
d\mu)$,
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),
\quad j=1, \dots, n,
\end{multline*}
which implies
\begin{equation}\label{cuntz-aux}
v(j, x_1, x_2, \dots)= \frac {\alpha_1}{\alpha_2} \, v(x_1, x_2, 
\dots),
\quad \text{$\mu$-a.e.}, \quad j =1, \dots, n.
\end{equation}
This implies, in turn, that
\[
v(1, x_1, x_2, \dots)=v(2, x_1, x_2, \dots)=\dots= v(n, x_1, x_2, 
\dots)
\quad \text{$\mu$ -a.e.},
\]
i.e., the function $v(\cdot)$ is invariant with respect to changes 
of
$x_1$. But then the right-hand side in \eqref{cuntz-aux} is 
invariant
with respect to $x_1$, and the left-hand side is invariant with 
respect
to $x_1$, $x_2$, etc. Therefore, $v(x)$ is invariant with respect 
to
change of any finite number of variables; since a product measure 
is
ergodic with respect to such transforms, we get that $v(x)$ is 
constant
$\mu$-a.e., and thus, $\alpha_1 = \alpha_2$.
\end{proof}
\end{example}


\begin{example} (KMS representation).
Consider now a bit more complicated example of representation 
of $\mathcal{O}_n$ in the space of vector-valued functions. Let
\begin{align*}
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,
\quad i=1, \dots, n,
\\
u_1(x)& =\dots= u_n(x) = u(x) = S \otimes 1(x), \quad Se_k = 
e_{k+1}, \quad k
\in \mathbb{Z}.
\end{align*}

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}, \quad 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}, \quad  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 a multi-index. Also, the vector
$\Omega$ is cyclic. Indeed, from the latter formula we see that the
operator $n^{1/2} S_j^*$ maps $\Omega =\Omega_0$ into $\Omega_{-1} 
=
e_{-1} \otimes 1(x)$, and each $\Omega_k = e_k \otimes 1(x)$ into
$\Omega_{k-1}$; also 
\begin{multline*}
n^{-1/2}(S_1+\dots + S_n) \Omega_k 
\\
= e_{k+1}(\delta_1(x_1) + \dots + \delta_n(x_1)) 1(x_2, x_3, \dots) 
\\
=  e_{k+1} \otimes 1(x_1, x_2, \dots) = \Omega_{k+1}.
\end{multline*}
Each $\Omega _k$ is cyclic in the corresponding subspace $H_k = e_k
\otimes L_2(\mathbb{Z}_n^\infty, d\mu)$ with respect to the 
commuting
family of projections $S_\alpha S^* _\alpha$, where $\alpha$ runs 
over 
the set of  all finite multi-indices.

The corresponding
state on $\mathcal{O}_n$ is the unique KMS state $\omega$
on the Cuntz algebra $\mathcal{O}_n$ with respect to gauge group 
(see \cite{BrRob}) 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 unitary equivalent realization, one have the following
formulas for the operators
\begin{align*}
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), \quad j=1, \dots, n,
\end{align*}
or in yet another form, in $H = L_2(\mathbb{T} \times [0,1], dz 
\otimes
dt)$:
\[
(S_j f)(z, t) = z n^{1/2} \chi_{[(j-1)/n, j/n]} f(z, nt-(j-1)), 
\quad
j=1, \dots, n.
\]
Any of the latter forms 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}
(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 action of 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 $\mathbb{Z}_n^k \subset \mathbb{Z}_n^{k+1}$ by 
$(\alpha_1, \dots,\alpha_n) \mapsto (\alpha_1, \dots, \alpha_n, 0)$. 
Let $|\alpha|$ be the naumer of the last nonzero $\alpha_j$ in $\alpha$. 
Write $j_k= (0, \dots, 0,j,0,\dots)$ with $j$ at 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 
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 
infinte 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}\begin{cases} i=0, 
\alpha\ne \emptyset& e(j-\alpha_1, 0, \sigma(\alpha))\\ 
\text{otehrwise} & e(i, k+1, \alpha) \end{cases} \otimes 
\delta_j(x_1) f(\sigma(x)),
\\
S_j^* e(i, k, \alpha) \otimes f(x) & = n^{-1/2} \begin{cases} k=0 & 
e(0,0,\sigma_{j-i}(\alpha))\\ k\ne 0& e(i, k-1, \alpha) \end{cases} 
\otimes f(\sigma_j(x))
\end{align*}

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 UHF 
subalgebra in $\mathcal{O}_n$, and the corresponding cyclic 
subspace is a proper subspace of $H$. This means, in particular, 
that representation corresponding to the tracial state on UHF 
cannot be extended to 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 like above, but with $p_j^{1/2}$ istead of $n^{-1/2}$ gives 
representation corresponding to states constructed as extention of 
product states on UHF, which have the form
\[
\omega(S_\alpha S^*_\beta) = \delta_{\alpha, \beta} p_{\alpha_1} 
\dots p_{\alpha_s}, \quad s = |\alpha|.
\]
\end{example}

\begin{example}
We give another example of 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 needed state is obtained from vector $e_0 \otimes e_0 \otimes 
1(x)$, which is cyclic.
\end{example}