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%\volume{228}
%\Copyrightyear{2004}
%\DOI{003-0001}
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%\DOI{003-xxxx-y}
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%\commby{inhouse}
%\submitted{March 14, 2003}
%\received{March 16, 2000}
%\revised{June 1, 2000}
%\accepted{July 22, 2000}
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%Insert here the title, affiliations and abstract:
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\title[On configurations of subspaces $\dots$] {On Coxeter graph
  related configurations of subspaces of a Hilbert space}


%----------Author 1
\author{Popova~N.~D.}

\address{%
Institute of Mathematics,\\
Tereschenkivska, 3,\\
01601, Kyiv,\\
Ukraine}

\email{popova@imath.kiev.ua}

%----------Author 2
\author{Samo{\u\i}lenko~Yu.~S.}
\address{Institute of Mathematics,\\
Tereschenkivska, 3,\\
01601, Kyiv,\\
Ukraine}
\email{yurii\_sam@imath.kiev.ua}

%----------Author 3
\author{Strelets~A.~V.}
\address{Institute of Mathematics,\\
Tereschenkivska, 3,\\
01601, Kyiv,\\
Ukraine}
\email{sav@imath.kiev.ua}
%----------classification, keywords, date
\subjclass{Primary *46C07; Secondary 46K10}

\keywords{Family of subspaces, Coxeter graph, Temperley--Lieb
  algebras, $*$-representation}

\date{January 1, 2004}
%----------additions
%\dedicatory{To my boss}
%%% ----------------------------------------------------------------------

\begin{abstract}
  For a Hilbert space $\sH$, we study configurations of its
  subspaces related to Coxeter graphs $\mathbb{G}_{4,4}$, which are
  arbitrary trees such that two edges have type $4$ and the rest are
  of type $3$. We prove that such irreducible configurations exist
  only in a finite dimensional $\sH$, where the dimension of $\sH$
  does not exceed the number of vertices of the graph plus the
  number of vertices of the subgraph that lies between the edges
  of type $4$. We give a description of all irreducible
  nonequivalent configurations; they are indexed with a continuous
  parameter. As an example, we study all irreducible configurations
  related to a graph that consists of three vertices and two type
  $4$ edges.
\end{abstract}


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\maketitle
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%\tableofcontents

\section*{Introduction}

One of important and fruitful directions of M.~G.~Krein's original
mathematical work was a development of algebraic methods in
functional analysis and operator theory.

The main object we study in this paper is a family $S=(\sH; \sH_1,
\dots, \sH_n)$ of subspaces $\sH_1,\dots,\sH_n$ of a Hilbert space
$\sH$. For any family of subspaces $S$ there is a unique
collection of orthogonal projections $P_k=P_{\sH_k}$, which are
orthogonal projections of $\sH$ onto $\sH_k$, $k=1,\dots, n$.

\sloppy A family $S=(\sH; \sH_1, \dots, \sH_n)$ is \emph{unitarily
equivalent} to a family $\tilde{S}=(\tilde\sH; \tilde\sH_1, \dots,
\tilde\sH_n)$ of subspaces $\tilde\sH_1,\dots,\tilde\sH_n$ of a
Hilbert space $\tilde\sH$ if there exists a unitary operator $U$
acting from $\sH$ onto $\tilde\sH$ such that $U$ maps $\sH_k$ onto
$\tilde\sH_k$, $k=1,\dots,n$, that is, the corresponding
collections of the orthogonal projections $P_k$ and $\tilde P_k$,
$k=1,\dots,n$, are unitarily equivalent.

\fussy A family $S=(\sH; \sH_1, \dots, \sH_n)$ is \emph{irreducible}
if the identity $[A,P_k]=0$ satisfied for all $k=1,\dots, n$, where
$A\in\sBH$, implies that $A=\lambda I$, where $\lambda\in\sC$, $I$
is the identity operator on $\sH$, that is, the collection of
orthogonal projections $P_k, k=1,\dots,n$, is irreducible.

Irreducible pairs of subspaces exist only in one- or two-dimensional
Hilbert spaces $\sH$. A list of such subspaces, up to unitary
equivalence, is the following:
\begin{enumerate}
\item\label{list:I} $\sH=\sC^1$,
  \begin{equation*}
    (\sC^1;0,0),\qquad (\sC^1;\sC^1,0),\qquad
    (\sC^1;0, \sC^1),\qquad (\sC^1;\sC^1,\sC^1);
  \end{equation*}
\item\label{list:II} $\sH=\sC^2=\slg e_1,e_2 \srg$ ($\|e_k\|=1$,
  $k=1,2$; $e_1\bot e_2$),
  \begin{equation*}
    (\sC^2;\slg e_1 \srg, \slg \cos\varphi e_1+\sin\varphi e_2\srg),
  \end{equation*}
  ($\varphi\in(0,\frac\pi2)$ is the angle between the subspaces
  $\slg e_1 \srg$ and $\slg \cos\varphi e_1+\sin\varphi
  e_2\srg$).\\
  Denoting $P_1=P_{\slg e_1\srg}$,
  $P_2=P_{\slg\cos\varphi{}e_1+\sin\varphi{}e_2\srg}$ we have
  \begin{align*}
    P_1P_2P_1&=\tau{}P_1,\\
    P_2P_1P_2&=\tau{}P_2,
  \end{align*}
  where $\tau=\cos^2\varphi\in(0,1)$.
\end{enumerate}

For any pair of subspaces $S=(\sH;\sH_1,\sH_2)$ there is a spectral
theorem, --- the space $\sH$ can be decomposed into a direct sum, or
an integral, of irreducible pairs of subspaces that are equivalent
to pairs of subspaces listed in~\ref{list:I}
and~\ref{list:II} above, see~\cite{Halmos:TAMS_en_1969}.%???


The problem of describing irreducible $n\text{-}$tuples of
subspaces,
$$
S=(\sH;\sH_1,\dots,\sH_n),
$$
and a similar problem of representing $n\text{-}$tuples as a
direct sum, or
an integral, of irreducible ones for $n\geqslant3$ is a $*\text{-}$wild problem%???
. The problem of describing triples of subspaces,
$S=(\sH;\sH_1,\sH_2,\sH_3)$, such that $\sH_2\bot\sH_3$ is also
$*-$wild~\cite{KrugSam:FAP_ru_1980,KrugSam:PAMS_en_2000}.

It is thus natural to study $n-$tuples of subspaces $\sH_1,
\dots, \sH_n$, with a condition on the angles between the subspaces %???
$\sH_i, \sH_j$ for $ i,j=\overline{1,n}, i\neq j.$

In~\cite{Popova:PIM_en_2004,VlasPop:UMZ_2004} and others, the
authors study configurations of subspaces, $S_{\Gamma,\tau,\bot}$,
related to a simple graph $\Gamma$ with edges labeled with
$\tau=\{\tau_{ij}\}$, where $\tau_{ij}\in (0;1)$.  (By a simple
graph we mean a finite non-oriented graph without multiple edges and
loops.)  In the configuration $S_{\Gamma,\tau,\bot}$, the number of
subspaces coincides with the number of vertices in the graph, and if
vertices $i$ and $j$ are not connected with an edge, the
corresponding subspaces are assumed to be orthogonal,
$$
P_iP_j=P_jP_i=0,
$$
and if there is an edge, then we fix an angle between $\sH_i$ and
$\sH_j$, $\phi_{ij}\in(0,\frac{\pi}{2})$,
($\cos^2\phi_{ij}=\tau_{ij}$, where $\tau_{ij}\in(0,1)$ is the
number located above the edge connecting the vertices $i$ and
$j$), that is, we have
$$
P_iP_jP_i=\tau_{ij}P_i \text{ and } P_jP_iP_j=\tau_{ij}P_j.
$$

Since the subspaces corresponding to vertices in different connected
components are orthogonal, we always assume that $\Gamma$ is a
connected graph.

The problem of describing the configurations $S_{\Gamma,\tau,\bot}$
can be reformulated in terms of $*-$representations of corresponding
$*$-algebras $TL_{\Gamma,\tau,\bot}$ that are defined as follows:
\begin{equation*}
  \begin{aligned}
    TL_{\Gamma,\tau,\bot} = \sC\slg p_1,\dots, p_n\sei
    &p_i^2=p_i^{*}=p_i; \\
    &p_ip_j=p_jp_i=0, \text{ if the vertices } i,j
    \text{ are not connected}\\
      &\qquad \text{with an edge};\\
    &p_ip_jp_i=\tau_{ij} p_i, p_jp_ip_j=\tau_{ij} p_j, \text{
      otherwise } \srg.
  \end{aligned}
\end{equation*}

Note that $TL_{\Gamma,\tau,\bot}$ are the quotient algebras of
introduced in~\cite{Graham:thesis} generalized Temperley--Lieb
algebras (which are, in their turn, quotient algebras of the
Hecke algebras related to $\Gamma$%???????????
).

The dimension and the growth of these algebras do not depend on
the numbers $\tau_{ij}$, see~\cite{Vlas:MFAT_2004}, namely, the
algebra $TL_{\Gamma,\tau,\bot}$
\begin{itemize}
\item has a finite dimension equal to $n^2+1$ if $\Gamma$ is a tree,
  where $n$ is the number of the vertices;
\item has linear growth if $\Gamma$ has exactly one cycle;
\item contains a free subalgebra with two generators if $\Gamma$
  has more than one cycle.
\end{itemize}

If we fix a graph $\Gamma$, then for each collection of
the parameters %???
$\tau_{ij}$ there is a finite number of unitarily nonequivalent
configurations $S_{\Gamma,\tau,\bot}$ if and only if the graph
$\Gamma$ is a tree, see~\cite{VlasPop:UMZ_2004,
  PopSamStr:UMG_ru_2008}. The dimension of the space $\sH$ does not
exceed the number of vertices, and the dimensions of the subspaces
are not greater than $1$.

If the graph $\Gamma$ has one cycle, i.e., it is unicyclic, then
there is an arrangement $\tau$ of numbers at its edges such that
there exist infinitely many irreducible nonequivalent configurations
$S_{\Gamma,\tau,\bot}$. The dimensions of the subspaces do not
exceed $1$, and the dimension of the whole space is not greater than
the number of vertices in the graph~\cite{VlasPop:UMZ_2004}.

%%%%%%% Kokster
A more general case is obtained by not fixing the angles between
the subspaces but allowing them to assume values from a fixed
finite set.  Hence, we come to configurations
$S_{\mathbb{G},g,\bot}$, where $\mathbb{G}$ is a Coxeter graph
with the vertices corresponding to subspaces, and $g$ is a
collection of polynomials that define permissible angles between
the subspaces.

By a \textit{Coxeter graph} $\mathbb{G}$ we mean a finite
non-oriented marked graph without multiple edges and with no
loops. We will write ${\mathbb{G}} = (V, R)$, where
$V=\{1,\dots,n\}$ is the set of vertices, $R$ is the set of edges.
The edge between $i$ and $j$ will be denoted by $\gamma_{ij}$ (we
take $\gamma_{ij}=\gamma_{ji}$). All edges of a Coxeter graph
$\mathbb{G}$ can be subdivided into different types,
\begin{equation*}
  R=\bigsqcup_{s=3}^\infty R_s.
\end{equation*}
The corresponding edges will be called $R_3\text{-}$edges,
$R_4\text{-}$edges, etc., or we will say that the edge has type $3$,
$4$, etc.

Let $g$ be a mapping that assigns a polynomial $g_{ij}$ to each
edge $\gamma_{ij}\in R_s$, $s=2k+\sigma\geqslant3$, $k\in\sN$,
$\sigma\in\{0,1\}$, such that $\deg g_{ij}\leqslant k-1$ and
$g_{ij}(0)=0$ if $\sigma =0$,
\begin{equation*}
  g:R\to\sR[x]:\gamma_{ij}\mapsto g_{ij}(x)=
  \sum_{m=1-\sigma}^{k-1}\tau_{ij}^{(m)}x^m\in \sR[x].
\end{equation*}

The number of subspaces in a configuration $S_{\mathbb{G},g,\bot}$,
as in the case of a simple graph, coincides with the number of
vertices in the graph. If vertices $i$, $j$ are not connected with
an edge, the corresponding subspaces $\sH_i$, $\sH_j$ are considered
to be orthogonal,
$$
P_i P_j=P_j P_i=0,
$$
and if there is an edge of type $s=2k+\sigma\geqslant3$,
$k\in\sN$, $\sigma\in\{0,1\}$, then the following relations hold:
\begin{equation}\label{eq:angels}
  (P_iP_j)^kP_i^\sigma=g_{ij}(P_iP_j)P_i^\sigma \quad \text{ and }
  \quad (P_jP_i)^kP_j^\sigma=g_{ij}(P_jP_i)P_j^\sigma.
\end{equation}

It is convenient to reformulate the problem about the configurations
$S_{\mathbb{G},g,\bot}$ in terms of finding $*$-representations of
the corresponding $*$-algebras $TL_{\mathbb{G},g,\bot}$.

\begin{Def}\label{Def_algebra}
  $TL_{{\mathbb{G}}, g,\bot}$ is an associative $*$-algebra over
  $\mathbb{C}$, with an identity $e$, and is defined by generators
  and relations determined by the graph $\mathbb{G}$ and the mapping
  $g$,
  \begin{equation}\label{algebra}
    \begin{aligned}
      TL_{{\mathbb{G}}, g,\bot} = \sC\slg p_1,\dots, p_n\sei
      &p_i^2=p_i^*=p_i; \quad
      p_ip_j=p_jp_i=0, \text{ if }  \gamma_{ij}\not\in R;\\
      &(p_ip_j)^k p_i^\sigma=
      g_{ij} (p_ip_j)p_i^\sigma, \;\; (p_jp_i)^k p_j^\sigma=
      g_{ij} (p_jp_i)p_j^\sigma,\\
      &\text{if } \gamma_{ij}\in R_s,\quad s=2k+\sigma\geqslant3,
      \sigma\in\{0,1\} \srg.
    \end{aligned}
\end{equation}
\end{Def}

For an edge $\gamma_{ij}\in R_{s}$, $s=2k+\sigma$, and a polynomial
$g_{ij}$, we will also consider the polynomial
$f_{ij}(x)=x^k-g_{ij}(x)$. Then relations~\eqref{algebra} can be
rewritten in the form
$f_{ij}(p_ip_j)p_i^\sigma=f_{ij}(p_jp_i)p_j^\sigma=0$.

If an edge $\gamma_{ij}$ has type $3$, then the corresponding
polynomial is $g_{ij}(x)=\tau_{ij}$, and the relations satisfied by
$p_i$ and $p_j$ will be
$$
p_ip_jp_i=\tau_{ij} p_i, \quad p_jp_ip_j=\tau_{ij} p_j.
$$

If an edge $\gamma_{ij}$ has type $4$, then $g_{ij}(x)=\tau_{ij}x$
and the relations satisfied by $p_i$ and $p_j$ will be
$$
(p_ip_j)^2=\tau_{ij} p_ip_j, \quad (p_jp_i)^2=\tau_{ij} p_jp_i.
$$

The dimension and the growth of the algebras $TL_{{\mathbb{G}},
  g,\bot}$ were studied in~\cite{PopSamStr:UMG_ru_2007}. The
dimension and the growth of the algebra $TL_{{\mathbb{G}},
g,\bot}$ depend on the type of the graph $\mathbb{G}$, which can
be a tree, an unicyclic graph, a graph with two or more cycles,
and on the type of its edges, but do not depend on the way the
polynomials $g_{ij}$ are positioned at its edges. Let us give here
brief formulations of the results. To this end, we introduce the
following notations: $\hat{R}_r=\bigsqcup_{s=r}^\infty R_s$, and
$s_{\mathbb{G}}\geqslant3$ denotes the index $s$ such that
$R_{s_{\mathbb{G}}}\ne\varnothing$ and $R_s=\varnothing$ if $s>
s_{\mathbb{G}}$.

\begin{Theorem}\label{th:dim}
  Let ${\mathbb{G}}$ be a tree. Then we have the following.
  \begin{itemize}
  \item [0)] If $|\hat{R}_4|=0$, then $\dim
    TL_{{\mathbb{G}},g,\bot}=
    |{\mathbb{G}}|^2+1$.
  \item [1)] If $|\hat{R}_4|=|{R}_{s_{\mathbb{G}}}|=1$, then
    $\mathbb{G}$ consists of two $R_3\text{-}$connected components
    $\mathbb{G}_1$ and $\mathbb{G}_2$, and
    \begin{equation*}
      \dim TL_{{\mathbb{G}},g,\bot}=
      \begin{cases}
        m|{\mathbb{G}}|^2+1,&
        \text{ if } s_{\mathbb{G}}=2m+1,\\
        (m-1)|{\mathbb{G}}|^2+|\mathbb{G}_1|^2+|\mathbb{G}_2|^2+1,&
        \text{ if } s_{\mathbb{G}}=2m.
      \end{cases}
    \end{equation*}
  \item [2)] If $|\hat{R}_4|\geqslant2$, then $\dim
    TL_{{\mathbb{G}}, g,\bot}=\infty$. In the case where
    $|\hat{R}_4|=2$, the algebra has polynomial growth if
    $|\hat{R}_6|=0$, and contains a free subalgebra with two
    generators if $|\hat{R}_6|\geqslant1$.
  \item [3)] If $|\hat{R}_4|\geqslant3$, then
    $TL_{{\mathbb{G}},g,\bot}$ contains a free algebra with two
    generators.
  \end{itemize}
\end{Theorem}

\begin{Theorem}
  Let ${\mathbb{G}}$ be a connected Coxeter graph.
  \begin{itemize}
  \item [0)] If ${\mathbb{G}}$ contains a cycle and $|\hat{R}_4|=0$,
    then $\dim TL_{{\mathbb{G}},g,\bot}=\infty$. If also there is
    precisely one cycle, then the algebra has linear growth, and if
    ${\mathbb{G}}$ has two cycles, then the algebra has a free
    algebra with two generators.
  \item [2)] If ${\mathbb{G}}$ contains at least one cycle and
    $|\hat{R}_4|\geqslant1$, then $TL_{{\mathbb{G}},g,\bot}$
    contains a free algebra with two generators.
  \end{itemize}
\end{Theorem}

The authors in~\cite{PopSamStr:UMG_ru_2008} studied
$*$-representations of the algebras $TL_{{\mathbb{G}},g,\bot}$. They
gave a description of all irreducible $*$-representations of the
finite dimensional algebras $TL_{{\mathbb{G}},g,\bot}$ on a Hilbert
space. In particular, it was shown that, if an irreducible pair of
projections $P_1$ and $P_2$ satisfies relations~\eqref{eq:angels},
then at least one of the roots $\lambda\in[0,1)$ of the polynomial
$f_{12}(x)$ satisfies the relations
\begin{equation*}
  \begin{cases}
    P_1P_2P_1=\lambda P_1\, \text{ and }\,   P_2P_1P_2=\lambda
    P_2,&\quad \lambda\ne0,\\
    P_1P_2=P_2P_1=0, &\quad\lambda=0.
  \end{cases}
\end{equation*}
Hence, using roots of the polynomial $f_{12}(x)$ we can define
admissible angles between the subspaces. There was also proved a
theorem stating that \emph{only graphs that are trees with no more
  than one edge of type $s>3$ define algebras of finite Hilbert
  type, i.e., the algebras that have a finite number of irreducible
  $*\text{-}$representations on a Hilbert space for all values of
  the parameters.}

In this paper, we use $*$-representations of the algebra
$TL_{\mathbb{G}_{4,4},g,\bot}$ to study configurations of the
subspaces, $S_{\mathbb{G}_{4,4},g,\bot}$, that are related to the
Coxeter graph $\mathbb{G}_{4,4}$, which is an arbitrary tree with
exactly two edges of type $4$ and others are of type $3$. As
Theorem~\ref{th:dim} shows, this is the simplest case where the
algebra is already not finite dimensional but still has polynomial
growth.

In Section~\ref{sec:1}, we give necessary definitions and introduce
notations; we also introduce the notion of a proper
$*$-representation of the algebra $TL_{\mathbb{G}_{4,4},g,\bot}$.

Section~\ref{sec:2} gives a construction of a family of irreducible
proper $*$-repre\-sentations $\pi_{\nu}$ of the algebra
$TL_{\mathbb{G}_{4,4},g,\bot}$. This family is parametrized with a
parameter $\nu$ running over a subset $\Sigma_{\mathbb{G}_{4,4},g}$
of the interval $(0,1)$. It is shown that $*$-representations
corresponding to different values of the parameter are not unitarily
equivalent.

In Section~\ref{sec:3}, we show that any irreducible proper
$*$-representation $\pi$ of the algebra
$TL_{\mathbb{G}_{4,4},g,\bot}$ is unitarily equivalent to one of the
$*$-representations $\pi_{\nu}$ constructed in Section~\ref{sec:2},
which finishes the description of all irreducible proper
\hbox{$*$-re}pre\-sentations.

As an example, in Section~\ref{sec:4}, we study all irreducible
$*$-representations of the algebra $TL_{\mathbb{G}_{4,4},g,\bot}$,
where the graph $\mathbb{G}_{4,4}$ consists of three vertices and
two edges of type~$4$.


\section{Definitions and notations}\label{sec:1}

A path of length $m$ in a Coxeter graph $\mathbb{G}$,
\begin{equation*}
  l=l(i_0)=(i_0,i_1,\dots,i_m), \quad \gamma_{i_{k-1},i_{k}}\in R,
\end{equation*}
will be called a \emph{path without repetitions} if $i_k\ne i_j$ for
$k,j=0,\dots,m$, $k\ne j$. The path $l=(i_0)$ is considered as a path
of length $0$ without repetitions, and it is convenient to consider
the path $l=()$ as ``empty''. For a path $l=(i_0,i_1,\dots,i_m)$,
define $l^*=(i_m,i_{m-1},\dots,i_0)$. A \textit{union of paths}
$l_1=(i_0,\dots,i_{k-1},i_k)$ and $l_2=(i_k,i_{k+1}\dots,i_t)$ is
defined to be the path $l_1\cup
l_2=(i_0,\dots,i_{k-1},i_k,i_{k+1},\dots,i_t)$.  To any path
$l=(i_0,i_1,\dots,i_m)$, we make correspond the product
$\Pi_{l}=p_{i_{0}}\dots p_{i_m}$ in the algebra, to the ``empty''
path, we set $\Pi_l=e.$

By $\mathbb{G}_{4,4}$, we will denote a Coxeter graph, which is a
tree having two edges of type $4$ and the rest of edges are of type
$3$.  We index the vertices in such a way that the edges
$\gamma_{0,1}$, $\gamma_{m-1,m}$ are of type $4$ and the vertices
$1$,~$m-1$ are connected with the path $(1,2,\dots, m-1)$,
\begin{center}
\includegraphics{fig2.eps}
\end{center}

It is natural to split the graph into three parts,
\begin{equation*}
  V=V_0\cup V_{in}\cup V_{m},
\end{equation*}
namely, any two different vertices in each part are connected with a
path consisting of type $3$ edges.

Denote by $\hat{l}$ the path $(m,m-1,\dots,1,0)$, and by $\mathcal
P$ the set of all paths $l=(i_0,i_1,\dots, 0)$ such that $\Pi_{l}$
is a normal word that does not have $\Pi_{\hat{l}^*\cup\hat{l}}$
as a subword. For normal words, Groebner bases, the composition
lemma, we refer to e.g.~\cite{Ufn:SPM_1990}. For the algebra
$TL_{\mathbb{G},g,\bot}$, normal words are precisely the words
that do not contain, as subwords, the leading words of the
defining relations~\eqref{algebra} of the algebra
$TL_{\mathbb{G},g,\bot}$, see~\cite{PopSamStr:UMG_ru_2007}. That
is, a normal word should not contain, as subwords, the following
words:
\begin{equation*}
  \begin{aligned}
    &p_i^2, i\in V; \\
    &p_ip_j, \;\; p_jp_i, \text{ if }  \gamma_{ij}\not\in R;\\
    &(p_ip_j)^k p_i^\sigma, \;\; (p_jp_i)^k p_j^\sigma, \text{ if } \gamma_{ij}\in R_s,\quad s=2k+\sigma\geqslant3,
    \sigma\in\{0,1\}.
  \end{aligned}
\end{equation*}


It is clear that the set $\mathcal{P}$ consists of two parts
containing paths $l\in\mathcal P$ without and with repetitions,
denoted by $\mathcal{S}$ and $\mathcal{L}'$, respectively. The set
$\mathcal{L}'$, in its turn, can be slitted into two more parts
containing the the paths $l$ that do or do not have
$\Pi_{\hat{l}}$ as a subword of $\Pi_l$, and are denoted by
$\mathcal L$ and $\mathcal L_{0}$, respectively. We set
$\hat{\mathcal{P}}=\mathcal{S}\cup \mathcal{L}$.

\begin{Prop}\label{prop:paths}
  \begin{itemize}
  \item[1)] For each path $l\in\mathcal{L}$ there exists a unique
    vertex $j\in\{1,2\dots,m-1\}$ and a unique collection of paths
    without repetitions,
    \begin{align*}
      l_s&=(i_0,i_1,\dots, j) , \\
      l_e&=(j,j-1,\dots, 0),\\
      \tilde{l}&=(j,j+1,\dots, m),
    \end{align*}
    such that $l=l_s\cup\tilde{l}\cup\hat{l}$ and $\tilde{l}^*\cup
    l_e=\hat{l}$. The identity
    \begin{equation*}
      \omega(l)=l_s\cup l_e
    \end{equation*}
    defines an injective mapping $\omega:\mathcal{L}\to
    \mathcal{S}$.
  \item[2)] There are naturally defined bijective mappings $\psi_*:
    V\to\mathcal{S}$ and $\varphi_*:V_{in}\to\mathcal{L}$. We also
    have $\omega(\varphi_*(j))=\psi_*(j)$ for any $j\in V_{in}$.
  \end{itemize}
\end{Prop}

For any \begin{equation*}
  l=(i_0,i_1,\dots,0)\in\hat{\mathcal{P}}\backslash\{\psi_*(0)\},
\end{equation*}
define a truncation operation, $\eta$, of the path $l$ by
\begin{equation*}
  \eta(l)=(i_1,\dots,0).
\end{equation*}

For a $*$-representation $\pi$ of the algebra
$TL_{\mathbb{G}_{4,4},g,\bot}$, denote $P_i=\pi(p_i)$, $i\in V$.

\begin{Def}
  A $*$-representation $\pi$ of the algebra
  $TL_{\mathbb{G}_{4,4},g,\bot}$ will be called \emph{proper} if
  none of the following relations is satisfied:
  \begin{align}
    &P_0P_1=0, \label{eq:ort_0}\\
    &P_{m-1}P_{m}=0, \label{eq:ort_m}\\
    &P_1P_0P_1-\tau_{0,1}P_1= 0, \label{eq:three_0}\\
    &P_{m-1}P_{m}P_{m-1}-\tau_{m-1,m}P_{m-1}= 0. \label{eq:three_m}
\end{align}
\end{Def}

If some of relations~\eqref{eq:ort_0}--\eqref{eq:three_m} are
satisfied, this would mean that the \hbox{$*$-re}pre\-sentation
$\pi$ is a lift of a certain $*$-representation of the algebra
obtained by taking the quotient with respect to an ideal
generated by a corresponding relation. All such quotient algebras
are finite dimensional, and their $*$-representations were studied
before, see~\cite{VlasPop:UMZ_2004}, \cite{PopSamStr:UMG_ru_2008}.
Thus, a complete description of all irreducible $*$-representations
of the algebra $TL_{\mathbb{G}_{4,4},g,\bot}$ require a description
of all irreducible proper $*$-representations.

For each $\nu\in\Sigma_{\mathbb{G}_{4,4},g}$, we will construct an
irreducible proper $*$-representation $\pi_{\nu}$, where
$\Sigma_{\mathbb{G}_{4,4},g}$ is a subset of the interval $(0,1)$.
An exact definition of this set will be given later. Let us note
that, in general, this set could be empty. This means that the
algebra $TL_{\mathbb{G}_{4,4},g,\bot}$ does not have proper
$*$-representations. On the other hand, for any Coxeter graph
$\mathbb{G}_{4,4}$ there is $g$ such that
$\Sigma_{\mathbb{G}_{4,4},g}$ contains an entire
interval~\cite{PopSamStr:UMG_ru_2008}. It will be shown that
$*$-representations $\pi_{\nu_1}$ and $\pi_{\nu_2}$ are not
unitarily equivalent if $\nu_1\ne\nu_2$. The fact that for every
irreducible proper $*$-representation $\pi$ there exists
$\nu\in\Sigma_{\mathbb{G}_{4,4},g}$ such that $\pi$ is unitarily
equivalent to $\pi_{\nu}$ finishes the description of all
irreducible proper $*$-representations.


\section{The set $\Sigma_{\mathbb{G}_{4,4},g}$,
  proper $*$-representations $\pi_\nu$,
  $\nu\in\Sigma_{\mathbb{G}_{4,4},g}$}\label{sec:2}

Let $L$ be a linear complex space obtained as a set of all formal
linear combinations of paths taken from the set $\hat{\mathcal{P}}$.

For each $\nu\in(0,1)$, introduce a Hermitian sesquilinear form
$B_{\mathbb{G}_{4,4},g}^{\nu}$ on the linear space $L$ by defining
it on the formal linear basis $\hat{\mathcal{P}}$ to be
\begin{align*}
  B_{\mathbb{G}_{4,4},g}^{\nu}(l,l)&=1,
  \qquad
  l\in\hat{\mathcal{P}},\\
  B_{\mathbb{G}_{4,4},g}^{\nu}(l,\eta(l))
  &=B_{\mathbb{G}_{4,4},g}^{\nu}(\eta(l),l)\\
  &=\begin{cases}
    \sqrt{\tau_{ij}},
                                %    \qquad
    &l\in\hat{\mathcal{P}}\backslash\{\psi_*(0), \psi_*(m),
    \varphi_*(m-1)\},\\
    \sqrt{\nu\tau_{m-1,m}},
                                %    \qquad
    &l=\psi_*(m),\\
    \sqrt{(1-\nu)\tau_{m-1,m}},
                                %    \qquad
    &l=\varphi_*(m-1),
  \end{cases}\\
  B_{\mathbb{G}_{4,4},g}^{\nu}(l_1,l_2)&=0,
  \qquad
  l_1,l_2\in\hat{\mathcal{P}}, l_1\ne l_2, l_1\ne\eta(l_2), l_2\ne\eta(l_1),
\end{align*}
where the paths $l$ and $\eta(l)$ start at $i$ and $j$,
correspondingly.

Let $\Sigma_{\mathbb{G}_{4,4},g}$ be the set of all $\nu\in(0,1)$
such that the sesquilinear form $B_{\mathbb{G}_{4,4},g}^{\nu}$ is
nonnegative definite. For $\nu\in\Sigma_{\mathbb{G}_{4,4},g}$,
denote by $\sH_{\nu}$ the Hilbert space obtained by equipping the
linear space $L/L_\nu$, where $L_\nu=\{x\sei
B_{\mathbb{G}_{4,4},g}^{\nu}(x,x)=0\}$, with the scalar product
$\slg y_1+L_\nu,y_2+L_\nu\srg_{\nu}=
B_{\mathbb{G}_{4,4},g}^{\nu}(y_1,y_2)$.  Let $\rho_\nu$ be the
linear mapping
\begin{equation*}
  \rho_\nu:L\to\sH_\nu:y\mapsto y+L_\nu .
\end{equation*}
By the definition of the form $B_{\mathbb{G}_{4,4},g}^{\nu}$, any
$l\in\hat{\mathcal{P}}$ does not belong to $L_\nu$. Hence, there
is a bijection between the set
$\rho_\nu(\hat{\mathcal{P}})=\{l+L_\nu\sei
l\in\hat{\mathcal{P}}\}\subset\sH_\nu$ and the set
$\hat{\mathcal{P}}$, where the former generates a linear space
$\sH_\nu$, however, if the form is not positive definite, the set
$\rho_\nu(\hat{\mathcal{P}})$ is not a set of linearly independent
vectors. Set
\begin{align*}
  \psi=\psi_\nu&=\rho_\nu\circ\psi_*:\mathcal{S}\to\sH_\nu,\\
  \varphi=\varphi_\nu&=\rho_\nu\circ\varphi_*:\mathcal{L}\to\sH_\nu
  .
\end{align*}

For an arbitrary vertex $i\in{V}_{in}$, define an operator
$P_{\nu,i}$ to be the orthogonal projection onto the linear span of
the pair of vectors $\psi(i), \varphi(i)\in\sH_\nu$, and, for an
arbitrary vertex $i\in{V}\backslash{V}_{in}$, the operator
$P_{\nu,i}$ is defined to be an orthogonal projection onto the
linear span of the vector $\psi(i)\in\sH_\nu$.

\begin{Prop}
  For any $x\in\sH_\nu$, we have the formula
  \begin{equation*}
    P_{\nu,i}x =\begin{cases}
      \slg x, \psi(i)\srg_\nu\psi(i)+
      \slg x, \varphi(i)\srg_\nu\varphi(i),&
      \qquad i\in{V}_{in},\\
      \slg x, \psi(i)\srg_\nu\psi(i),&
      \qquad i\in{V}\backslash{V}_{in}.
    \end{cases}
  \end{equation*}
\end{Prop}

\begin{proof}
  It is sufficient to notice that $\slg P_{\nu,i}x,
  \psi(i)\srg_\nu=\slg x, \psi(i)\srg_\nu$ for any $i\in{V}$, and
  $\slg P_{\nu,i}x, \varphi(i)\srg_\nu=\slg x, \varphi(i)\srg_\nu$
  for any $i\in{V}_{in}$.
\end{proof}

\begin{Lemma}\label{lemma:family}
  For each $\nu\in\Sigma_{\mathbb{G}_{4,4},g}$, the mapping
  \begin{equation*}
    \pi_\nu: TL_{\mathbb{G}_{4,4},g,\bot}\to \sB{\sH_\nu}:p_i\mapsto P_{\nu,i}
  \end{equation*}
  is an irreducible proper $*\text{-}$representation.
\end{Lemma}

\begin{proof}
  For the sake of brevity, set $\varphi(i)=0$ for
  $i\in{}V\backslash{V}_{in}$. Then, for any $i\in{V}$,
  $x\in\sH_\nu$, we have
  \begin{equation*}
    P_{\nu,i}x =
    \slg x, \psi(i)\srg_\nu\psi(i)+
    \slg x, \varphi(i)\srg_\nu\varphi(i).
  \end{equation*}


  Let us show that $\pi_{\nu}$ is a $*\text{-}$representation.

  It is clear that for any $x\in\sH_\nu$, $P_{\nu,i}^2x=P_{\nu,i}x$,
  since $\slg \psi(i),\varphi(i) \srg_\nu=0$.

  Now, if the vertices $i$ and $j$ are not connected with an edge,
  then for any vector $x\in\sH_\nu$,
  \begin{align*}
    P_{\nu,i}P_{\nu,j}x&=P_{\nu,i}(\slg x, \psi(j)\srg_\nu\psi(j)+
    \slg x, \varphi(j)\srg_\nu\varphi(j))\\
    &=\slg x, \psi(j)\srg_\nu(\slg \psi(j), \psi(i)\srg_\nu\psi(i)+\slg \psi(j),
    \varphi(i)\srg_\nu\varphi(i))\\
    &+\slg x,
    \varphi(j)\srg_\nu(\slg\varphi(j),\psi(i)\srg_\nu\psi(i)
    +\slg\varphi(j),\varphi(i)\srg_\nu\varphi(i))=0 .
  \end{align*}

  Let now the vertices $i$ and $j$ be joined with a type $3$ edge.
  Then we can assume that $\psi_*(i)=\eta(\psi_*(j))$. Moreover, we
  have that either $i,j\in{V}\backslash{V}_{in}$, then
  $\varphi(i)=\varphi(j)=0$, or $i,j\in{V}_{in}$ implying that one
  of the following two identities is verified: if
  $j\in\{2,\dots,m-1\}$ then $\varphi_*(j)=\eta(\varphi_*(i))$,
  otherwise $\varphi_*(i)=\eta(\varphi_*(j))$. Hence, for an
  arbitrary vector $x\in\sH_\nu$, we have
  \begin{align*}
    P_{\nu,j}P_{\nu,i}P_{\nu,j}x&=P_{\nu,j}P_{\nu,i}(\slg x,
    \psi(j)\srg_\nu\psi(j)+
    \slg x, \varphi(j)\srg_\nu\varphi(j))\\
    &=\sqrt{\tau_{ij}}P_{\nu,j}(\slg x, \psi(j)\srg_\nu \psi(i)+
    \slg x, \varphi(j)\srg_\nu\varphi(i))\\
    &=\tau_{ij}(\slg x, \psi(j)\srg_\nu \psi(j)+
    \slg x, \varphi(j)\srg_\nu\varphi(j))\\
    &=\tau_{ij} P_{\nu,j}x .
  \end{align*}

  It remains to check the relations for the orthogonal projections
  corresponding to the vertices joined with edges of type $4$. For
  an arbitrary $x\in\sH_\nu$, we have
  \begin{align*}
    P_{\nu,0}P_{\nu,1}P_{\nu,0}x&=P_{\nu,0}P_{\nu,1}\slg x, \psi(0)\srg_\nu\psi(0)\\
    &=\sqrt{\tau_{0,1}}P_{\nu,0}\slg x, \psi(0)\srg_\nu \psi(1)\\
    &=\tau_{0,1}\slg x, \psi(0)\srg_\nu \psi(0)\\
    &=\tau_{0,1} P_{\nu,0}x,
  \end{align*}
  \begin{align*}
    P_{\nu,m}P_{\nu,m-1}P_{\nu,m}x&=P_{\nu,m}P_{\nu,m-1}\slg x, \psi(m)\srg_\nu\psi(m)\\
    &=\sqrt{\tau_{m-1,m}}\slg x,\psi(m)\srg_\nu
    P_{\nu,m}(\sqrt{\nu}\psi(m-1)+\sqrt{1-\nu}\varphi(m-1))\\
    &=\tau_{m-1,m}\slg x, \psi(m)\srg_\nu (\nu+(1-\nu))\psi(m)\\
%    &=\tau_{m-1,m}\slg x, \psi(m)\srg_\nu \psi(m)\\
    &=\tau_{m-1,m} P_{\nu,m}x .
  \end{align*}
  This implies that
  \begin{align*}
    &P_{\nu,0}P_{\nu,1}P_{\nu,0}P_{\nu,1}=\tau_{0,1} P_{\nu,0}P_{\nu,1},\\
    &P_{\nu,1}P_{\nu,0}P_{\nu,1}P_{\nu,0}=\tau_{0,1} P_{\nu,1}P_{\nu,0},\\
    &P_{\nu,m}P_{\nu,m-1}P_{\nu,m}P_{\nu,m-1}=\tau_{m-1,m} P_{\nu,m}P_{\nu,m-1},\\
    &P_{\nu,m-1}P_{\nu,m}P_{\nu,m-1}P_{\nu,m}=\tau_{m-1,m} P_{\nu,m-1}P_{\nu,m}.
  \end{align*}

  It is easy to see that $*\text{-}$representation
  $\pi_{\nu}$ is proper.

  Let us show that the constructed $*\text{-}$representation
  $\pi_{\nu}$ is irreducible.

  Assume that an operator $A\in\sB{\sH_\nu}$ commutes with all
  $P_{\nu,i}$, $i\in{V}$. Then, $A(\mathrm{Im} P_{\nu,i})\subset
  \mathrm{Im} P_{\nu,i}$. Consequently, there exists a number
  $\lambda\in\sC$ such that $A\psi(0)=\lambda\psi(0)$. Then $\lambda
  \sqrt{\tau_{0,1}}\psi(1)=\lambda P_{\nu,1}\psi(0)=P_{\nu,1}
  A\psi(0)=A \sqrt{\tau_{0,1}}\psi(1)$. In a similar way, it is easy
  to show that if $A\psi(i)=\lambda\psi(i)$ and the vertices $j$ and
  $i$ are joined with an edge, then $A\psi(j)=\lambda\psi(j)$.

  Let us show that $A\varphi(m-1)=\lambda\varphi(m-1)$. It was
  proved before that $A\psi(m)=\lambda\psi(m)$. Then, on the one
  hand,
  \begin{align*}
    P_{\nu,m-1}A\psi(m)
    &=\lambda P_{\nu,m-1}\psi(m)\\
    &=\lambda(\sqrt{\nu\tau_{m-1,m}}\psi(m-1)+\sqrt{(1-\nu)\tau_{m-1,m}}\varphi(m-1)).
  \end{align*}
  On the other hand, we have also shown that
  $A\psi(m-1)=\lambda\psi(m-1)$ and, consequently,
  \begin{align*}
    AP_{\nu,m-1}\psi(m)&=A(\sqrt{\nu\tau_{m-1,m}}\psi(m-1)+\sqrt{(1-\nu)\tau_{m-1,m}}\varphi(m-1))\\
    &=\lambda\sqrt{\nu\tau_{m-1,m}}\psi(m-1)+\sqrt{(1-\nu)\tau_{m-1,m}}A\varphi(m-1).
  \end{align*}
  So, $A\varphi(m-1)=\lambda\varphi(m-1)$.

  Now, for any vertices $i,j \in V_{in}$, which are connected with
  an edge, the identity $A\varphi(i)=\lambda\varphi(i)$ implies that
  $A\varphi(j)=\lambda\varphi(j)$. Indeed,
  $P_{\nu,j}A\varphi(i)=\lambda P_{\nu,j}\varphi(i)=\lambda
  \sqrt{\tau_{ij}}\varphi(j)=AP_{\nu,j}\varphi(i)=
  \sqrt{\tau_{ij}}A\varphi(j)$.

  Thus, $A=\lambda I$ and, consequently, the
  $*\text{-}$representation $\pi_{\nu}$ is irreducible.
\end{proof}


\begin{Lemma}
  The $*\text{-}$representations $\pi_{\nu_1}$ and $\pi_{\nu_2}$,
  $\nu_1,\nu_2\in\Sigma_{\mathbb{G}_{4,4},g}$, are unitarily
  equivalent if and only if $\nu_1=\nu_2$.
\end{Lemma}

\begin{proof}
  Let us consider the operator
  \begin{equation*}
    W_\nu=U^\nu_{0,1}U^\nu_{1,2}\dots U^\nu_{m-1,m}U^\nu_{m,m-1}\dots U^\nu_{2,1}U^\nu_{1,0},\qquad
    U^\nu_{i,j}=\frac{P_{\nu,i}P_{\nu,j}}{\sqrt{\tau_{i,j}}} .
  \end{equation*}
  It is easy to see that if the vertices $i,j\in V_{in}$ are joined
  with an edge, then
  \begin{equation*}
    U^\nu_{i,j}(\mu_1\psi(j)+\mu_2\varphi(j))=(\mu_1\psi(i)+\mu_2\varphi(i)),
  \end{equation*}
  and, consequently,
  \begin{align*}
    W_\nu\psi(0)&=U^\nu_{0,1}U^\nu_{1,2}\dots{}U^\nu_{m-1,m}U^\nu_{m,m-1}\dots{}U^\nu_{2,1}\psi(1)\\
    &=U^\nu_{0,1}U^\nu_{1,2}\dots{}U^\nu_{m-1,m}U^\nu_{m,m-1}\psi(m-1)\\
    &=\sqrt{\nu}\,U^\nu_{0,1}U^\nu_{1,2}\dots{}U^\nu_{m-1,m}\psi(m)\\
    &=\sqrt{\nu}\,U^\nu_{0,1}U^\nu_{1,2}\dots{}U^\nu_{m-2,m-1}
    (\sqrt{\nu}\psi(m-1)+\sqrt{1-\nu}\varphi(m-1))\\
    &=\sqrt{\nu}\,U^\nu_{0,1}(\sqrt{\nu}\psi(1)+\sqrt{1-\nu}\varphi(1))\\
    &=\nu\psi(0).
  \end{align*}

  Assume that the $*$-representations $\pi_{\nu_1}$ and
  $\pi_{\nu_2}$ are unitarily equivalent, that is, there exists a
  unitary operator
  \begin{equation*}
    V:\sH_{\nu_1}\to\sH_{\nu_2}
  \end{equation*}
  such that $V\pi_{\nu_1}(a)=\pi_{\nu_2}(a)V$ for any $a\in
  TL_{\mathbb{G}_{4,4},g,\bot}$. Then
  \begin{equation*}
    W_{\nu_2}V\psi_{\nu_1}(0)=VW_{\nu_1}\psi_{\nu_1}(0)=\nu_1{}V\psi_{\nu_1}(0),
  \end{equation*}
  so that $V\psi_{\nu_1}(0)$ is an eigenvector of the operator
  $W_{\nu_2}$ with an eigenvalue $\nu_1$. Consequently, this vector
  belongs to $\mathrm{Im} P_{\nu_2,0}$, so that there is a number
  $\beta\in\sC\setminus\{0\}$ such that $V\psi_{\nu_1}(0)=\beta
  \psi_{\nu_2}(0)$. But, in this case,
  \begin{equation*}
    W_{\nu_2}V\psi_{\nu_1}(0)=\beta
    W_{\nu_2}\psi_{\nu_2}(0)=\beta\nu_2\psi_{\nu_2}(0)=\beta\nu_1\psi_{\nu_2}(0),
  \end{equation*}
  whence we get that $\nu_1=\nu_2$.

  We have thus shown that if
  $\nu_1,\nu_2\in\Sigma_{\mathbb{G}_{4,4},g}$ are different,
  the $*$-representations $\pi_{\nu_1}$ and $\pi_{\nu_2}$ are not
  unitarily equivalent.
\end{proof}



\section{A description of all proper $*$-representations of
  the algebra $TL_{\mathbb{G}_{4,4},g,\bot}$}\label{sec:3}

Let now $\pi$ be some $*\text{-}$representation of the algebra
$TL_{\mathbb{G}_{4,4},g,\bot}$ on a Hilbert space $\sH$,
$P_i=\pi(p_i)$, $i\in V$. Denote $w=\Pi_{\hat{l}^{*}\cup\hat{l}}$,
$W=\pi(w)$. Note that
$W=(\pi(\Pi_{\hat{l}}))^{*}\pi(\Pi_{\hat{l}})$, where
$\pi(\Pi_{\hat{l}})=P_mP_{m-1}\dots P_0$, thus $W$ is a positive
operator.

\begin{Prop}
  Let $\pi$ be an irreducible $*$-representation, $P_0P_1\ne0$, and
  $P_{m-1}P_{m}\ne0$. Then $W\ne0$.
\end{Prop}

\begin{proof}
  Assume that $W=0$. Since $P_0P_1\ne0$, there exists $x_0$ such
  that $P_0x_0=x_0$, $P_1P_0x_0\ne0$.

  Consider the linear span $\sH'$ of a finite set of the vectors
  $\{\pi(\Pi_{l})x_0\}_{l\in\mathcal{P}}$. It is easy to see that
  $\sH'$ is invariant with respect to $\pi$ and, consequently, it
  coincides with $\sH$.

  It is clear that $\pi(\Pi_{\hat{l}})=0$. Moreover, $\pi(\Pi_{l}) =
  0$ for any $l\in\mathcal{P}$ such that $\Pi_{l}$ contains
  $\Pi_{\hat{l}}$ as a subword. Let us show that
  $P_mP_{m-1}\pi(\Pi_l)x_0=0$ for any $l\in \mathcal{P}$. Indeed,
  $P_{m-1}\pi(\Pi_l)\ne 0$ only if the initial point of $l$
  coincides with the vertex $m-1$ or is joined to it with an edge,
  and $\Pi_{l}$ does not contain $\Pi_{\hat{l}}$ as a subword. There
  are three cases possible,
  \begin{itemize}
  \item [1)] $l=(m-1,m-2,\dots,0)$,
  \item [2)] $l=(m-2,\dots,0)$,
  \item [3)] $l=(j,m-1,m-2,\dots,0)$,
  \end{itemize}
  where $j$ is any vertex distinct from $m-2$ and jointed to the
  vertex $m-1$ with an edge of type $3$. In the first two cases,
  $P_mP_{m-1}\pi(\Pi_l)=\pi(\Pi_{\hat{l}})=0$, and in the third one,
  $P_mP_{m-1}\pi(\Pi_l)=\tau_{j,m-1}\pi(\Pi_{\hat{l}})=0$. Hence,
  $P_{m}P_{m-1}\sH=\{0\}$, and this contradicts to
  $P_{m}P_{m-1}\ne0$.
\end{proof}

Since we are interested in proper $*\text{-}$representations of the
algebra $TL_{\mathbb{G}_{4,4},g,\bot}$, in the sequel we will
consider only $*\text{-}$representations $\pi$ such that $W\ne0$.

\begin{Prop}
  Let $\pi$ be an irreducible $*\text{-}$representation of the
  algebra $TL_{\mathbb{G}_{4,4},g,\bot}$, such that $W\ne0$. Then
  there exist $\xi\in(0,1]$ and $0\ne x_0\in\sH$ such that
  $P_0x_0=x_0$ and $Wx_0=\xi x_0$.
\end{Prop}

\begin{proof}
  Consider the linear subspace $L=\mathrm{Im}\,W\ne\{0\}$ invariant
  with respect to the operator $W$. If some $x\in\sH$ satisfies
  $W^2x=0$, then
  \begin{equation*}
    0=\slg W^2x,x\srg=\slg Wx,Wx\srg= \|Wx\|^2,
  \end{equation*}
  that is, we have shown that $\mathrm{Im}\,W\cap\ker{W}=\{0\}$.

  Denote by $\tilde{\pi}$ the $*\text{-}$representation of the generated by an element
  $w$ commutative subalgebra of the algebra
  $TL_{\mathbb{G}_{4,4},g,\bot}$ on the Hilbert space $\tilde{\sH}=\overline{L}$, which is
  defined by
  \begin{equation*}
    \tilde{\pi}(w)=\tilde{W}=W\arrowvert_{\tilde{\sH}}.
  \end{equation*}

  If $\tilde{\pi}$ is irreducible, then $\dim\tilde{\sH}=1$ and
  $\tilde{W}=\xi\in\sC$. Moreover, since $W$ is a nonzero positive
  operator with the norm being less or equal to one, we see that
  $\xi\in(0,1]$. Take $0\ne{x_0}\in\tilde\sH$. Then $Wx_0=\xi{x_0}$,
  $P_0x_0=x_0$.

  Assume that $\tilde{\pi}$ is reducible. Then there are nonzero
  subspaces $\tilde{\sH}_1$ and $\tilde{\sH}_2$ of the space
  $\tilde{\sH}$ that are invariant with respect to the action of
  $\tilde{\pi}$ and such that
  $\tilde\sH=\tilde\sH_1\oplus\tilde\sH_2$. Since $\tilde{W}\ne0$,
  there exist $0\ne x\in\tilde{\sH}_1\cap L$ and $0\ne
  y\in\tilde{\sH}_2\cap L$.

  Let $\sH_0$ be the closure of the linear span of the vectors
  $\{\pi(\Pi_l)x\}_{l\in\mathcal{N}}$, where $\mathcal{N}$ is a set
  of paths $l=(i_0,i_1,\dots 0)$ such that $\Pi_l$ is a normal word.
  It is clear that $\sH_0$ is invariant with respect to $\pi$ and,
  hence, $\sH=\sH_0$. Then, for any $l\in\mathcal{N}$,
  \begin{equation*}
    \slg y, \pi(\Pi_l) x\srg = \sum\limits_{l'\in{\mathcal{N}}_0}\lambda_{l,l'}
    \slg y, \pi(\Pi_{l'}) x\srg,
  \end{equation*}
  where $\mathcal{N}_0=\{l\in\mathcal{N}\sei i_0=0\}$.

  For any $l'\in \mathcal{N}_0$, the monomial $\Pi_{l'}$ can be
  written in one of the following forms:
  \begin{align*}
    p_0,\quad p_0p_1p_0,\quad w^n,\, n\in\sN.
  \end{align*}
  To show that $\slg y, \pi(\Pi_{l'}) x\srg=0$, it is sufficient to
  prove that $\pi(\Pi_{l'})x\in\tilde\sH_1$. It is clear that
  $P_0x=x\in\tilde\sH_1$, $W^nx\in\tilde\sH_1$. Now,
  $P_0P_1P_0x=\tau_{0,1}x\in\tilde\sH_1$, since
  $x\in\mathrm{Im}\,W$.

  Hence, $\slg y, \pi(\Pi_l) x\srg=0$, that is,
  $y\in\sH_0^\bot=\sH^\bot$ and, consequently, $y=0$, which
  contradicts to $y\ne0$.
\end{proof}

Fix $x_0\in\sH$ and $\xi\in(0,1]$, which exist by the previous
proposition. Everywhere in the sequel, we consider $\|x_0\|=1$.
Denote
\begin{equation}\label{eq:expr nu}
  \nu = \xi\big(\prod_{i=1}^{m}\tau_{i-1,i}\big)^{-1} > 0,
\end{equation}
and introduce
\begin{equation*}
  \tau_{i,j}'=\begin{cases}
    \sqrt{\tau_{i,j}},\quad &\gamma_{i,j}\ne\gamma_{m-1,m},\\
    \sqrt{\nu\tau_{m-1,m}},\quad &\gamma_{i,j}=\gamma_{m-1,m}.
  \end{cases}
\end{equation*}

For a path $l=(i_0,i_1,\dots,i_k)$, let us introduce
\begin{equation*}
  W_l=\begin{cases}
    P_{i_0},\quad & \text{if } k=0,\\
    \frac{P_{i_0}P_{i_1}}{\tau'_{i_0,i_1}}
    \cdot
    \frac{P_{i_1}P_{i_2}}{\tau'_{i_1,i_2}}
    \cdot\ldots\cdot
    \frac{P_{i_{k-1}}P_{i_{k}}}{\tau'_{i_{k-1},i_{k}}},
    \quad &\text{otherwise}.
  \end{cases}
\end{equation*}

\begin{Prop}
  The following identity holds:
  \begin{equation*}
    W_{_{\hat{l}^*\cup\hat{l}}}x_0=x_0.
  \end{equation*}
\end{Prop}

\begin{proof}
  We have $W_{_{\hat{l}^*\cup\hat{l}}}x_0
  =\big(\prod_{i=1}^{m}\tau'_{i-1,i}\big)^{-2}Wx_0
  =\big(\nu\prod_{i=1}^{m}\tau_{i-1,i}\big)^{-1}\xi x_0=x_0$.
\end{proof}

\begin{Prop}\label{prop:inv}
  The linear span $\sH'$ of the vectors
  $\{W_lx_0\}_{l\in\hat{\mathcal{P}}}$ is invariant with respect to
  an irreducible $*\text{-}$representation $\pi$.
\end{Prop}

\begin{proof}
  For any $l$ such that $\Pi_l$ is a normal word, $W_l$ coincides
  with $\pi(\Pi_{l})$ up to a numeric coefficient. Let us show that,
  for any $a\in TL_{\mathbb{G}_{4,4},g,\bot}$ and any
  $l_0\in\hat{\mathcal{P}}$, the following identities are verified:
  \begin{equation*}
    \pi(a)W_{l_0}x_0=\sum_{l\in\mathcal{N}}\lambda_{l}\pi(\Pi_l)x_0=
    \sum_{l\in\mathcal{N}}\lambda'_{l}W_lx_0=\sum_{l\in\mathcal{P}}\lambda''_{l}W_lx_0
    =\sum_{l\in\hat{\mathcal{P}}}\lambda'''_{l}W_lx_0,
  \end{equation*}
  where $\mathcal{N}$ is a set of paths $l$ that end in the vertex
  $0$ and such that $\Pi_l$ is a normal word. To prove the identity
  before the last one, it will suffice to note that any $l\in
  \mathcal{N}$ can be represented as
  \begin{equation*}
    l=l'\cup  \underbrace{(\hat{l}^*\cup\hat{l}) \cup\dots\cup (\hat{l}^*\cup\hat{l})}_{k \text{ times
      }},\quad k\geq0,
  \end{equation*}
  where $l'\in\mathcal{P}$. To prove the last identity, we note
  that, for $l\in\mathcal{L}_{0}$,
  \begin{equation*}
    W_{l}x_0=W_{l'}W_{(0,1,0)}W_{\hat{l}^*\cup\hat{l}}x_0
    =W_{l'}W_{\hat{l}^*\cup\hat{l}}x_0=W_{l'}x_0,
  \end{equation*}
  where $l'\in \mathcal{S}$.
\end{proof}

\begin{Corollary}\label{cor:dimH}
  If $\pi$ is an irreducible $*\text{-}$representation of the
  algebra $TL_{\mathbb{G}_{4,4},g,\bot}$, then $H=H'$ and $\dim
  H\leqslant |V|+|V_{in}|$.
\end{Corollary}

\begin{Prop}\label{product}
  \begin{itemize}
  \item[1)] For any $l\in\mathcal{S}$,
    \begin{equation*}
      \slg W_lx_0, W_lx_0 \srg =1.
    \end{equation*}
  \item [2)] For any $l\in\mathcal{L}$,
    \begin{align*}
      &\slg W_lx_0, W_{\omega(l)}x_0 \srg =1,\\
      &\slg W_lx_0, W_lx_0 \srg = \nu^{-1}.
    \end{align*}
  \end{itemize}
\end{Prop}

\begin{proof}
  1) It is clear that $\slg W_{\psi_*(0)}x_0,
  W_{\psi_*(0)}x_0\srg=\slg x_0,x_0\srg=1$. Now,
  \begin{align*}
    \slg W_{\psi_*(1)}x_0, W_{\psi_*(1)}x_0\srg&=
    \sblg\frac{P_{1}P_{0}}{\tau'_{0,1}}x_0,
    \frac{P_{1}P_{0}}{\tau'_{0,1}}x_0\sbrg
    =\sblg\frac{P_{0}P_{1}}{\tau'_{0,1}}
    \cdot \frac{P_{1}P_{0}}{\tau'_{0,1}}
    W_{\hat{l}^*\cup\hat{l}}x_0,x_0\sbrg\\
    &=\sblg\frac{P_{0}P_{1}}{\tau'_{0,1}}
    \cdot \frac{P_{1}P_{0}}{\tau'_{0,1}}
    \cdot \frac{P_{0}P_{1}}{\tau'_{0,1}}W_{\eta(\hat{l}^*\cup\hat{l})}x_0,x_0\sbrg\\
    &=\sblg\frac{P_{0}P_{1}}{\tau'_{0,1}}W_{\eta(\hat{l}^*\cup\hat{l})}x_0,x_0\sbrg
    =\slg W_{\hat{l}^*\cup\hat{l}}x_0,x_0\srg=1,\\
    \slg W_{\psi_*(m)} x_0, W_{\psi_*(m)} x_0\srg&=
    \slg W_{\hat{l}}x_0, W_{\hat{l}}x_0\srg=
    \slg W_{\hat{l}^*\cup\hat{l}}x_0,x_0\srg=1.
  \end{align*}

  Let $l\in\mathcal{S}\backslash\{\psi_*(0),\psi_*(1), \psi_*(m)\}$,
  and $j$, $j'$ be the initial points of the paths $l$ and
  $\eta(l)$, correspondingly. Assuming that $\slg W_{\eta(l)}
  x_0,W_{\eta(l)} x_0\srg=1$ has been proved, we have
  \begin{align*}
    \slg W_{l}x_0, W_{l}x_0\srg&=
    \sblg\frac{P_{j}P_{j'}}{\tau'_{j,j'}}W_{\eta(l)}x_0,
    \frac{P_{j}P_{j'}}{\tau'_{j,j'}}W_{\eta(l)}x_0\sbrg\\
    &=\slg W_{\eta(l)}x_0, W_{\eta(l)}x_0\srg= 1.
  \end{align*}

  2) Let $l=l_s\cup\tilde{l}\cup\hat{l}\in \mathcal{L}$,
  $\omega(l)=l_s\cup l_e$, see Proposition~\ref{prop:paths}. Then
  \begin{align*}
    \slg W_{l}x_0,W_{\omega(l)}x_0\srg
    &=\slg W_{l_s}W_{\tilde{l}}W_{\hat{l}}x_0,W_{l_s}W_{l_e}x_0\srg\\
    &=\slg W_{\hat{l}}x_0,W_{\tilde{l}}^*W_{l_e}x_0\srg\\
    &=\slg W_{\hat{l}}x_0,W_{\hat{l}}x_0\srg=1.
  \end{align*}

  Now, consider $l=\varphi_*(m-1)$. Then $\eta(l)=\hat{l}$ and
  \begin{align*}
    \slg W_{l}x_0, W_{l}x_0\srg&=
    \sblg\frac{P_{m-1}P_m}{\tau'_{m-1,m}}W_{\hat{l}}x_0,
    \frac{P_{m-1}P_m}{\tau'_{m-1,m}}W_{\hat{l}}x_0\sbrg=\\
    &=\sblg\frac{P_mP_{m-1}}{\tau'_{m-1,m}}\cdot\frac{P_{m-1}P_m}{\tau'_{m-1,m}}
    \cdot\frac{P_mP_{m-1}}{\tau'_{m-1,m}}W_{\eta(\hat{l})}x_0,
    W_{\hat{l}}x_0\sbrg=\\
    &=\frac{1}{\nu}\sblg\frac{P_mP_{m-1}}{\tau'_{m-1,m}}W_{\eta(\hat{l})}x_0,
    W_{\hat{l}}x_0\sbrg=\\
    &=\nu^{-1}\slg W_{\hat{l}}x_0,W_{\hat{l}}x_0\srg=\nu^{-1}.
  \end{align*}
  Let now $l\in \mathcal{L}\backslash\{\varphi_*(m-1)\}$, and $j$,
  $j'$ be initial points of the paths $l$ and $\eta(l)$,
  correspondingly. Assuming that $\slg W_{\eta(l)}x_0,
  W_{\eta(l)}x_0\srg= \nu^{-1}$, we have
  \begin{align*}
    \slg W_{l}x_0, W_{l}x_0\srg=
    \sblg\frac{P_{j}P_{j'}}{\tau'_{j,j'}}W_{\eta(l)}x_0,
    \frac{P_{j}P_{j'}}{\tau'_{j,j'}}W_{\eta(l)}x_0\sbrg=\slg
    W_{\eta(l)}x_0, W_{\eta(l)}x_0\srg= \nu^{-1}.
  \end{align*}
  This finishes the proof.
\end{proof}

Let us recall that $\pi$ is an irreducible $*$-representation of
the algebra $TL_{\mathbb{G}_{4,4},g,\bot}$, and the number $\nu$
is given by the formula~\eqref{eq:expr nu}.

\begin{Prop}
  The parameter $\nu$ belongs to the interval $(0,1]$. Here, if
  $\nu=1$, then $P_{m-1}P_{m}P_{m-1}=\tau_{m-1,m}P_{m-1}$ and
  $\dim\sH\leqslant|V|$.
\end{Prop}

\begin{proof}
  It follows from~\eqref{eq:expr nu} that $\nu >0$.

  Consider $l\in\mathcal{L}$ and calculate the norm of
  $(W_{l}-W_{\omega(l)})x_0$,
  \begin{align*}
    \|(W_{l}-W_{\omega(l)})x_0\|^2=
    \slg W_{l}x_0-W_{\omega(l)}x_0,
    W_{l}x_0-W_{\omega(l)}x_0 \srg=\nu^{-1}-1 \geqslant 0.
  \end{align*}
  Hence, $\nu\leqslant1$.

  If $\nu=1$, then $(W_{l}-W_{\omega(l)})x_0=0$, that is,
  $W_{l}x_0=W_{\omega(l)}x_0$. This shows that the linear span of
  $\{W_lx_0\}_{l\in\mathcal{S}}$ coincides with the linear span of
  $\{W_lx_0\}_{l\in\hat{\mathcal{P}}}$ and, consequently, $\dim
  H\leqslant|V|$, see Proposition~\ref{prop:inv}.

  Let us show that in this case,
  $P_{m-1}P_{m}P_{m-1}-\tau_{m-1,m}P_{m-1}=0$.

  Indeed, $P_{m-1}W_{l}\ne0$ only if the initial point of
  $l\in\mathcal{S}$ coincides with the vertex $m-1$ or is joined to
  it with an edge. There are four cases to consider,
  \begin{itemize}
  \item [1)] $l=\psi_*(m-1)=(m-1,m-2,\dots,0)$,
  \item [2)] $l=\psi_*(m-2)=(m-2,\dots,0)$,
  \item [3)] $l=\hat{l}=\psi_*(m)=(m,m-1,m-2,\dots,0)$,
  \item [4)] $l=\psi_*(j)=(j,m-1,m-2,\dots,0)$,
  \end{itemize}
  where $j$ is any vertex, distinct from $m-2$, joined to the vertex
  $m-1$ with an edge of type $3$. For the first case, we get
  \begin{equation*}
    (P_{m-1}P_{m}P_{m-1}-\tau_{m-1,m}P_{m-1})W_lx_0= \tau_{m-1,m}
    (W_{\varphi_*(m-1)}-W_{\psi_*(m-1)})x_0=0.
  \end{equation*}
  For the second case,
  \begin{equation*}
    (P_{m-1}P_{m}P_{m-1}-\tau_{m-1,m}P_{m-1})W_lx_0= \tau_{m-1,m}
    \tau_{m-2,m-1}' (W_{\varphi_*(m-1)}-W_{\psi_*(m-1)})x_0=0.
  \end{equation*}
  In the third case,
  \begin{equation*}
    (P_{m-1}P_{m}P_{m-1}-\tau_{m-1,m}P_{m-1})W_{\hat{l}}x_0\\=
    \frac{1}{\tau_{m-1,m}'}P_{m-1}P_{m}(P_{m-1}P_m-\tau_{m-1,m})W_{\psi_*(m-1)}x_0=0.
  \end{equation*}
  For the fourth case,
  \begin{equation*}
    (P_{m-1}P_{m}P_{m-1}-\tau_{m-1,m}P_{m-1})W_{l}x_0=
    \tau_{m-1,j}'(P_{m-1}P_{m}P_{m-1}-\tau_{m-1,m}P_{m-1})W_{\psi_*(m-1)}x_0=0.
  \end{equation*}

  Hence, we have shown that
  $(P_{m-1}P_{m}P_{m-1}-\tau_{m-1,m}P_{m-1})\sH=\{0\}$.
\end{proof}

\begin{Corollary}
  If $\pi$ is a proper $*\text{-}$representation, then
  $\nu\in(0,1)$.
\end{Corollary}

\begin{Def}
  For any $l\in\hat{\mathcal{P}}$, define
  \begin{align*}
    y_l&=\begin{cases}
      W_lx_0,\qquad  &l\in\mathcal{S},\\
      \sqrt{\frac{\nu}{1-\nu}}\big(W_l-W_{\omega(l)}\big)x_0,
      \qquad  &l\in\mathcal{L}.
    \end{cases}
  \end{align*}
\end{Def}

Recall that $\sH'$ denotes the linear span of the vectors
$\{W_lx_0\}_{l\in\hat{\mathcal{P}}}$.

\begin{Prop}
  The linear span of the vectors $\{y_l\}_{l\in\hat{\mathcal{P}}}$
  coincides with $\sH'$.
\end{Prop}

\begin{Prop}
  The following identities are satisfied:
  \begin{itemize}
  \item [1.] for any $l,l'\in\hat{\mathcal{P}}$ such that their
    initial points do not coincide and are not joined with an edge,
    \begin{equation*}
      \slg y_{l}, y_{l'}\srg=0;
    \end{equation*}
    \item [2.]
      \begin{itemize}
      \item[(a)] for any $l\in\hat{\mathcal{P}}$,
        \begin{equation*}
          \slg y_{l}, y_{l}\srg=1;
        \end{equation*}
      \item [(b)] for any $l\in \mathcal{L}$,
        \begin{equation*}
          \slg y_{l}, y_{\omega(l)}\srg=0;
        \end{equation*}
      \end{itemize}
      \item[3.]
        \begin{itemize}
        \item [(a)] for any
          $l\in\hat{\mathcal{P}}\backslash\{\psi_*(0)\}$,
          \begin{equation*}
            \slg y_{l}, y_{\eta(l)}\srg=\begin{cases}
              \sqrt{\nu}\sqrt{\tau_{m-1,m}}, & l=\psi_*(m),\\
              \sqrt{1-\nu}\sqrt{\tau_{m-1,m}}, & l=\varphi_*(m-1),\\
              \sqrt{\tau_{j,j'}}, & \text{otherwise},
            \end{cases}
          \end{equation*}
          where $j$ and $j'$ are initial points of $l$ and
          $\eta(l)$, correspondingly;
        \item[(b)] for any $l\in
          \mathcal{L}\backslash\{\varphi_*(m-1)\}$,
          \begin{equation*}
            \slg y_{\eta(l)}, y_{\omega(l)}\srg=0,\qquad \slg y_{l}, y_{\omega(\eta(l))}\srg=0.
          \end{equation*}
        \end{itemize}
      \end{itemize}
    \end{Prop}

\begin{proof}
  1. \ In this case, the proof is clear.\smallskip

  2. \ Let $l\in\mathcal{S}$. Then $\slg y_l, y_l\srg = \slg W_lx_0,
  W_lx_0\srg=1$.

  Now, for any $l\in \mathcal{L}$,
  \begin{align*}
    \slg y_{l}, y_{l}\srg&=\frac{\nu}{1-\nu}\slg
    W_{l}x_0-W_{\omega(l)}x_0, W_{l}x_0-W_{\omega(l)}x_0\srg \\
    &=\frac{\nu}{1-\nu}\big(\nu^{-1}-2+1\big)=1.
  \end{align*}

  Hence, (a) is proved for $l\in\hat{\mathcal{P}}$. Let us now prove
  (b) for $l\in\mathcal{L}$,
  \begin{equation*}
    \slg y_{l}, y_{\omega(l)}\srg=\sqrt{\frac{\nu}{1-\nu}}\slg
    W_{l}x_0-W_{\omega(l)}x_0, W_{\omega(l)}x_0\srg=0.
  \end{equation*}
  \smallskip

  3. \ Let $l\in \mathcal{S}\backslash\{\psi_*(0)\}$, and $j$, $j'$
  be initial points of the paths $l$ and $\eta(l)$, respectively.
  Then
  \begin{align*}
    \slg y_{l}, y_{\eta(l)}\srg&=
    \slg P_{j}W_{l}x_0, P_{j'}W_{\eta(l)}x_0\srg=
    \slg W_{l}x_0, P_{j}P_{j'}W_{\eta(l)}x_0\srg\\
    &=\tau'_{j,j'}\slg W_{l}x_0, W_{l}x_0\srg=\tau'_{j,j'}.
  \end{align*}


  Before proving~(a) for $l\in \mathcal{L}$, let us prove~(b).

  Let $l\in \mathcal{L}\backslash\{\varphi_*(m-1)\}$, and $j$, $j'$
  be initial points of the paths $l$ and $\eta(l)$,
  respectively. Then
  \begin{align*}
    \slg y_{l}, y_{\omega(\eta(l))}\srg&= \sqrt{\frac{\nu}{1-\nu}}\slg
    W_{l}x_0 - y_{\omega(l)} , y_{\omega(\eta(l))}\srg \\
    &=
    \sqrt{\frac{\nu}{1-\nu}}\bigg(\slg W_{l}x_0,
    y_{\omega(\eta(l))}\srg - \tau'_{j,j'}\bigg)\\
    &=\sqrt{\frac{\nu}{1-\nu}}\bigg(\sblg P_{j'}P_{j}\cdot
    \frac{P_{j}P_{j'}}{\tau'_{j,j'}} W_{\eta(l)}x_0,
    y_{\omega(\eta(l))}\sbrg - \tau'_{j,j'}\bigg)\\
    &=\tau'_{j,j'}\cdot\sqrt{\frac{\nu}{1-\nu}}\bigg(\slg  W_{\eta(l)}x_0,
    y_{\omega(\eta(l))}\srg - 1\bigg)=0;\\
%%%%
    \slg y_{\eta(l)}, y_{\omega(l)}\srg&= \sqrt{\frac{\nu}{1-\nu}}\slg
    W_{\eta(l)}x_0 - y_{\omega(\eta(l))} , y_{\omega(l)}\srg\\
    &=
    \sqrt{\frac{\nu}{1-\nu}}\bigg(\slg W_{\eta(l)}x_0,
    y_{\omega(l)}\srg - \tau'_{j,j'}\bigg)\\
    &=\sqrt{\frac{\nu}{1-\nu}}\bigg(\slg P_{j}P_{j'} W_{\eta(l)}x_0,
    y_{\omega(l)}\srg - \tau'_{j,j'}\bigg)\\
    &=\tau'_{j,j'}\cdot\sqrt{\frac{\nu}{1-\nu}}\bigg(\slg  W_{l}x_0,
    y_{\omega(l)}\srg - 1\bigg)=0.
  \end{align*}

  Now we prove (a) for $l\in \mathcal{L}$.

  Consider $l=\varphi_*(m-1)$. Then $\eta(l)=\psi_*(m)=\hat{l}$,
  $\eta(\hat{l})=\omega(l)=\psi_*(m-1)$, and
  \begin{align*}
    \slg y_{l}, y_{\hat{l}}\srg&= \sqrt{\frac{\nu}{1-\nu}}\slg
    W_{l}x_0 - y_{\omega(l)} , y_{\hat{l}}\srg\\
    &=\sqrt{\frac{\nu}{1-\nu}}\bigg(\slg W_{l}x_0,
    W_{\hat{l}}x_0\srg - \tau'_{m-1,m}\bigg)\\
    &=\tau'_{m-1,m}\cdot\sqrt{\frac{\nu}{1-\nu}}
    \bigg(\sblg\frac{P_{m-1}P_m}{\tau'_{m-1,m}}W_{\hat{l}}x_0,
    \frac{P_{m-1}P_m}{\tau'_{m-1,m}}W_{\hat{l}}x_0\sbrg -
    1\bigg)\\
    &=\tau'_{m-1,m}\sqrt{\frac{1-\nu}{\nu}}=\sqrt{(1-\nu)\tau_{m-1,m}}.
  \end{align*}

  Let $l\in \mathcal{L}\backslash\{\varphi_*(m-1)\}$, and $j$, $j'$
  be initial points of the paths $l$ and $\eta(l)$,
  correspondingly. Then
  \begin{align*}
    \slg y_{l}, y_{\eta(l)}\srg&=
    \sqrt{\frac{\nu}{1-\nu}}\slg y_{l} ,
    W_{\eta(l)}x_0 - y_{\omega(\eta(l))}\srg\\
    &=
    \frac{\nu}{1-\nu}\slg W_{l}x_0 - y_{\omega(l)} ,
    W_{\eta(l)}x_0\srg\\
    &=\frac{\nu}{1-\nu}\slg W_{l}x_0 - y_{\omega(l)} ,
    P_jP_{j'}W_{\eta(l)}x_0\srg\\
    &=\tau'_{j,j'}\frac{\nu}{1-\nu}\slg W_{l}x_0 - y_{\omega(l)} ,
    W_{l}x_0\srg= \tau'_{j,j'}.
  \end{align*}
\end{proof}

This implies the following.
\begin{Lemma}
  If $\pi$ is a proper irreducible $*\text{-}$representation, then
  $\nu\in\Sigma_{\mathbb{G}_{4,4},g}$. Moreover, the unitary operator
  $V:\sH_\nu\to\sH:\rho_\nu(l)\mapsto{}y_l$,
  $l\in\hat{\mathcal{P}}$, defines a unitary equivalence between the
  $*\text{-}$representations $\pi_\nu$ and $\pi$.
\end{Lemma}

\begin{proof}
  It is clear that for any $l,l'\in\hat{\mathcal{P}}$, we have
  $B_{\mathbb{G}_{4,4},g}^{\nu}(l,l')=\slg y_l, y_{l'}\srg$. Hence,
  $V$ is a unitary operator and the form
  $B_{\mathbb{G}_{4,4},g}^{\nu}(\cdot,\cdot)$ is nonnegative definite.

  The operator $V$ intertwines the $*\text{-}$representations
  $\pi_\nu$ and $\pi$. Indeed, for any $j\in V_{in}$ and any
  $l\in\hat{\mathcal{P}}$,
  \begin{align*}
    VP_{\nu,j}\rho_\nu(l)&=\slg \rho_\nu(l),\psi(j)\srg_{\nu} V\psi(j)+\slg
    \rho_\nu(l),\varphi(j)\srg_{\nu} V\varphi(j)\\
    &= \slg \rho_\nu(l),\psi(j)\srg_{\nu} y_{\psi_*(j)}+\slg
    \rho_\nu(l),\varphi(j)\srg_{\nu} y_{\varphi_*(j)},\\
    P_{j}V\rho_\nu(l)&
    =P_jy_l=\slg y_l,y_{\psi_*(j)}\srg y_{\psi_*(j)}+\slg y_l,y_{\varphi_*(j)}\srg y_{\varphi_*(j)}.
  \end{align*}
  For any $i\in V\backslash V_{in}$ and any $l\in\hat{\mathcal{P}}$,
  we similarly have
  \begin{align*}
    VP_{\nu,i}\rho_\nu(l)&=\slg \rho_\nu(l),\psi(i)\srg_{\nu} V\psi(i)
    = \slg \rho_\nu(l),\psi(i)\srg_{\nu} y_{\psi_*(i)},\\
    P_{i}V\rho_\nu(l)&
    =P_iy_l=\slg y_l,y_{\psi_*(i)}\srg y_{\psi_*(i)}.
  \end{align*}
\end{proof}

This proves the following theorem.

\begin{Theorem}
  There is a one-to-one correspondence between the set
  $\Sigma_{\mathbb{G}_{4,4},g}$ and the set of classes of unitarily
  equivalent irreducible proper $*\text{-}$representations of the
  algebra $TL_{\mathbb{G}_{4,4},g,\bot}$.
\end{Theorem}

\section{A description of all irreducible $*$-representations
  of the algebra $TL_{\mathbb{G}_{4,4},g,\bot}$
  generated by three projections}\label{sec:4}

As an example, we consider the case where the graph
$\mathbb{G}_{4,4}$ consists of precisely three vertices and two
edges of type $4$, that is, $V=\{0,1,2\}$,
$R=R_4=\{\gamma_{01},\gamma_{12}\}$, $g_{01}(x)=\tau_{1}x$,
$g_{12}(x)=\tau_{2}x$, $\tau_{j}\in(0,1)$, $j\in\{1,2\}$. It is
clear that $V_{in}=\{1\}$ and, consequently, the dimension of proper
$*\text{-}$representations does not exceed $4$.

\begin{center}
  \includegraphics{fig1.eps}
\end{center}


\begin{Theorem}
  All irreducible $*\text{-}$representations of
  $TL_{\mathbb{G}_{4,4},g,\bot}$, up to unitary equivalence, are as
  follows:
  \begin{itemize}
  \item [a)] four improper one-dimensional
    $*\text{-}$representations,
    \begin{align*}
      &P_0=0,\quad P_1=0,\quad P_2=0; \\
      &P_0=1,\quad P_1=0,\quad P_2=0; \\
      &P_0=0,\quad P_1=1,\quad P_2=0; \\
      &P_0=0,\quad P_1=0,\quad P_2=1;
    \end{align*}
  \item[b)] two improper two-dimensional
    $*\text{-}$representations,
    \begin{align*}
      &P_0=\begin{pmatrix}
        1&0\\
        0&0
      \end{pmatrix},\:
      P_1=\begin{pmatrix}
        \tau_1&\sqrt{\tau_{1}(1-\tau_{1})}\\
        \sqrt{\tau_{1}(1-\tau_{1})}&1-\tau_{1}
      \end{pmatrix},\:
      P_2=\begin{pmatrix}
        0&0\\
        0&0
      \end{pmatrix};\\
      &P_0=\begin{pmatrix}
        0&0\\
        0&0
      \end{pmatrix},\:
      P_1=\begin{pmatrix}
        \tau_2&\sqrt{\tau_{2}(1-\tau_{2})}\\
        \sqrt{\tau_{2}(1-\tau_{2})}&1-\tau_{2}
      \end{pmatrix},\:
      P_2=\begin{pmatrix}
        1&0\\
        0&0
      \end{pmatrix};\\
    \end{align*}
    in the case where $\tau_1+\tau_2=1$, there is a third
    two-dimensional improper $*\text{-}$repre\-sentation,
    \begin{align*}
      &P_0=\begin{pmatrix}
        1&0\\
        0&0
      \end{pmatrix},\:
      P_1=\begin{pmatrix}
        \tau_1&\sqrt{\tau_{1}\tau_{2}}\\
        \sqrt{\tau_{1}\tau_{2}}&\tau_{2}
      \end{pmatrix},\:
      P_2=\begin{pmatrix}
        0&0\\
        0&1
      \end{pmatrix};
    \end{align*}
  \item[c)] one improper three-dimensional
    $*\text{-}$representation, if $\tau_1+\tau_2<1$,
    \begin{align*}
      &P_0=\begin{pmatrix}
        1&0&0\\
        0&0&0\\
        0&0&0
      \end{pmatrix},\:
      P_1=\begin{pmatrix}
        \tau_1&\sqrt{c\tau_1}&\sqrt{\tau_1\tau_2}\\
        \sqrt{c\tau_1}&c&\sqrt{c\tau_2}\\
        \sqrt{\tau_1\tau_2}&\sqrt{c\tau_2}&\tau_2
      \end{pmatrix},\:
      P_2=\begin{pmatrix}
        0&0&0\\
        0&0&0\\
        0&0&1
      \end{pmatrix},
    \end{align*}
    where $c = 1-\tau_1-\tau_2$;

    one proper three-dimensional $*\text{-}$representation
    $\pi_{\nu_0}$, if $\tau_1+\tau_2>1$,
    \begin{align*}
      &P_0=\begin{pmatrix}
        1&0&0\\
        0&0&0\\
        0&0&0
      \end{pmatrix},\:
      P_1=\begin{pmatrix}
        \tau_1&{r(c,\tau_1)}&{r(\tau_1,\tau_2)}\\
        {r(c,\tau_1)}&c&-{r(c,\tau_2)}\\
        {r(\tau_1,\tau_2)}&-{r(c,\tau_2)}&\tau_2
      \end{pmatrix},\:
      P_2=\begin{pmatrix}
        0&0&0\\
        0&0&0\\
        0&0&1
      \end{pmatrix},
    \end{align*}
    where $c = 2-\tau_1-\tau_2$, $r(x,y)=\sqrt{(1-x)(1-y)}$,
    \begin{equation*}
      \nu_0=\frac{r^2(\tau_1,\tau_2)}{\tau_1\tau_2}=
      \left(\frac{1}{\tau_{1}}-1\right)\left(\frac{1}{\tau_{2}}-1\right)
      \in\Sigma_{\mathbb{G}_{4,4},g};
    \end{equation*}
  \item [d)] a family of four-dimensional proper $*\text{-}$representations
    $\pi_\nu(\cdot)$, where $\nu\in(0,1)$, if
    $\tau_1+\tau_2\leqslant1$, and $\nu\in(0,\nu_0)$, if
    $\tau_1+\tau_2>1$,
    \begin{align*}
      &P_0=\begin{pmatrix}
        1&0&0&0\\
        0&0&0&0\\
        0&0&0&0\\
        0&0&0&0
      \end{pmatrix},\:
      P_2=\begin{pmatrix}
        0&0&0&0\\
        0&0&0&0\\
        0&0&0&0\\
        0&0&0&1
      \end{pmatrix},\\
      &P_1=\begin{pmatrix}
        \tau_1&\sqrt{c\tau_1}&0&\sqrt{\nu\tau_1\tau_2}\\
        \sqrt{c\tau_1}&c+\frac{\nu(1-\nu)\tau_2^2}{c}&
        -\frac{\sqrt{d(1-\nu)\nu}\tau_2}{c}&\frac{b\sqrt{\nu\tau_2}}{\sqrt{c}}\\
        0&-\frac{\sqrt{d(1-\nu)\nu}\tau_2}{c}&\frac{d}{c}&\frac{\sqrt{d(1-\nu)\tau_2}}{\sqrt{c}}\\
        \sqrt{\nu\tau_1\tau_2}&\frac{b\sqrt{\nu\tau_2}}{\sqrt{c}}&\frac{\sqrt{d(1-\nu)\tau_2}}{\sqrt{c}}&\tau_2
      \end{pmatrix},\:
    \end{align*}
    where $b=1-\tau_1-\tau_2$, $c=b+(1-\nu)\tau_2$,
    $d=b+(1-\nu)\tau_1\tau_2$.
  \end{itemize}
\end{Theorem}

\begin{proof}
  With respect to the formal linear basis
  \begin{equation*}
    \hat{\mathcal{P}}=\big\{\psi(0)=(0),\, \psi(1)=(1,0),
    \, \psi(2)=(2,1,0),\, \varphi(1)=(1,2,1,0)\big\},
  \end{equation*}
  the matrix of the sesquilinear form $B^\nu_{\mathbb{G}_{4,4}, g}$
  will be
  \begin{equation*}
    \begin{pmatrix}
      1&\sqrt{\tau_{1}}&0&0\\
      \sqrt{\tau_{1}}&1&\sqrt{\nu\tau_{2}}&0\\
      0&\sqrt{\nu\tau_{2}}&1&\sqrt{(1-\nu)\tau_{2}}\\
      0&0&\sqrt{(1-\nu)\tau_{2}}&1\\
    \end{pmatrix}.
  \end{equation*}
  For the form to be positive definite, it is necessary and
  sufficient that
  \begin{align*}
    &1-\tau_{1}> 0,\\
    &1-\tau_{1}-\tau_{2}+(1-\nu)\tau_{2}> 0,\\
    &1-\tau_{1}-\tau_{2}+(1-\nu)\tau_{1}\tau_{2}> 0.
  \end{align*}

  It is clear that the first inequality holds, since
  $\tau_{1}\in(0,1)$. Moreover, if $\tau_{1}+\tau_{2}\leqslant1$,
  then the second and the third inequalities hold for any
  $\nu\in(0,1)$. Hence, if the condition
  $\tau_{1}+\tau_{2}\leqslant1$ is satisfied, then
  $\Sigma_{\mathbb{G}_{4,4},g}=(0,1)$ and $\dim\sH_\nu=4$ for any
  $\nu\in(0,1)$.

  Let now $\tau_{1}+\tau_{2}>1$. It is clear that
  \begin{equation*}
    1-\tau_{1}>
    1-\tau_{1}-\tau_{2}+(1-\nu)\tau_{2}>
    1-\tau_{1}-\tau_{2}+(1-\nu)\tau_{1}\tau_{2}.
  \end{equation*}
  Now, the third inequality can be rewritten as
  \begin{equation*}
    \nu<\left(\frac{1}{\tau_{1}}-1\right)\left(\frac{1}{\tau_{2}}-1\right)=\nu_0.
  \end{equation*}
  This shows that $(0,\nu_0)\subset\Sigma_{\mathbb{G}_{4,4},g}$, and
  $\dim\sH_\nu=4$ for any $\nu\in(0,\nu_0)$.

  It is also clear that
  $1-\tau_{1}-\tau_{2}+(1-\nu)\tau_{1}\tau_{2}<0$ for $\nu>\nu_0$.
  If $\nu=\nu_0$, we get
  \begin{align*}
    &1-\tau_{1}> 0,\\
    &1-\tau_{1}-\tau_{2}+(1-\nu_0)\tau_{2}> 0,\\
    &1-\tau_{1}-\tau_{2}+(1-\nu_0)\tau_{1}\tau_{2}= 0.
  \end{align*}
  Hence, $\Sigma_{\mathbb{G}_{4,4},g}=(0,\nu_0]$ and
  $\dim\sH_{\nu_0}=3$.

  Let $\nu\in\Sigma_{\mathbb{G}_{4,4},g}$ be such that
  $\dim\sH_\nu=4$. Then, using the system of vectors
  $\psi(0),\psi(2), \psi(1),\varphi(1)$ we get an orthonormal basis
  $y_0,y_3,y_1,y_2$ such that
  \begin{align*}
    \psi(0)&=y_0,\\
    \psi(1)&=\sqrt{\tau_1}y_0+\sqrt{c}y_1+\sqrt{\nu\tau_2}y_3,\\
    \varphi(1)&=-\frac{\sqrt{\nu(1-\nu)}\tau_2}{\sqrt{c}}y_1
    +\frac{\sqrt{d}}{\sqrt{c}}y_2+\sqrt{(1-\nu)\tau_2}y_3,\\
    \psi(2)&=y_3,
  \end{align*}
  where $b=1-\tau_1-\tau_2$, $c=b+(1-\nu)\tau_2$,
  $d=b+(1-\nu)\tau_1\tau_2$.

  It is clear that $P_{\nu,0}$ and $P_{\nu,2}$ are orthogonal
  projections onto the subspaces generated by the vectors $y_0$ and
  $y_3$, correspondingly. Let us calculate $P_{\nu,1}$ on the basis
  vectors by the formula $P_{\nu,1}y_i=\slg
  y_i,\psi(1)\srg\psi(1)+\slg y_i,\varphi(1)\srg\varphi(1)$,
  \begin{align*}
    P_{\nu,1} y_0&=\tau_1y_0+\sqrt{c\tau_1}y_1+\sqrt{\nu\tau_1\tau_2}y_3,\\
    P_{\nu,1} y_1&=\sqrt{c\tau_1}y_0+(c+\frac{\nu(1-\nu)\tau_2^2}{c})y_1
    -\frac{\tau_2\sqrt{d(1-\nu)\nu}}{c}y_2
    +\frac{b\sqrt{\nu\tau_2}}{\sqrt{c}}y_3,\\
    P_{\nu,1} y_2&=-\frac{\tau_2\sqrt{d(1-\nu)\nu}}{c}y_1
    +\frac{d}{c}y_2+\frac{\sqrt{d(1-\nu)\tau_2}}{\sqrt{c}}y_3,\\
    P_{\nu,1} y_3&=
    \sqrt{\nu\tau_1\tau_2}y_0+\frac{b\sqrt{\nu\tau_2}}{\sqrt{c}}y_1+
    \frac{\sqrt{d(1-\nu)\tau_2}}{\sqrt{c}}y_2\
    +\nu\tau_2y_3.
  \end{align*}

  Similar calculations give the tree-dimensional proper
  $*\text{-}$representation for $\tau_1+\tau_2>1$, $\nu=\nu_0$.

  By writing all irreducible $*\text{-}$representations
  for the quotient algebras $TL_{\mathbb{G}_{4,4},g,\bot}/\slg
  p_1p_0p_1-\tau_1p_1 \srg$, $TL_{\mathbb{G}_{4,4},g,\bot}/\slg
  p_1p_2p_1-\tau_1p_1 \srg$ and $TL_{\mathbb{G}_{4,4},g,\bot}/\slg
  p_0p_1, p_1p_0, p_1p_2, p_2p_1 \srg$, we obtain all improper
  $*\text{-}$representations of $TL_{\mathbb{G}_{4,4},g,\bot}$.
\end{proof}

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\end{thebibliography}


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