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\section{$F_n$-algebras and their representations}
\subsection {About $*$-representations of
${F}_n$-algebras}\label{sec:1.2.1}
\markright{1.2. $F_n$-algebras and their representations}

\noindent\textbf{1.}
Let ${{F}}_n$ denote the standard polynomial of degree
$n$
in $n$
non-commuting variables:
$$
{{F}}_n(x_1,x_2,\dots,x_n)=
\sum_{\sigma \in S_n} (-1)^{p(\sigma)} x_{\sigma(1)}
\dots x_{\sigma(n)},
$$
where $p(\sigma)$ is the parity of the permutation $\sigma$,
$S_n$ is the symmetric
group. In the sequel, we say that $A$ is a
${{F}}_n$-algebra, if\/ $\forall x_1,\dots,x_n\in A$,
we have that ${{F}}_n(x_1,\dots,x_n)=0$.

The following Amitsur--Levitsky's theorem takes place~\cite{126}:
{\em the algebra $M_n({\mathbb C})$ is
a ${{F}}_{2n}$-algebra, but not a
${{F}}_{2n-1}$-algebra}.

${{F}}_n$-algebras
form one of the most simple class of algebras, if considered from
the standpoint of the structure of irreducible representations.

\begin{theorem}\label{th:r.f}
Consider the following statements:
\begin{itemize}
\item[\textup{(i)}]
        there exists a residual family ${\mathcal L}$
        of irreducible representations
        $\irrep A \supset {\mathcal L} \ni \pi$ such that $\dim H_\pi
        \le
        n$ for
        all $\pi \in {\mathcal L}$\textup;
\item[\textup{(ii)}]
        there exists a residual family ${\mathcal L}$ of
        representations $
        \rep
        A \supset {\mathcal L} \ni \pi$ such that $\dim H_\pi \le n$
        for
        all
        $\pi \in  {\mathcal L}$\textup;
\item[\textup{(iii)}]
        $A$ is a ${{F}}_{2n}$-algebra\textup;
\item[\textup{(iv)}]
        for any $\pi \in \irrep A$,
        $\dim H_\pi \le n$.
\end{itemize}
We have the following implications: \textup{(i)} $\Rightarrow$
\textup{(ii)}
$\Rightarrow$ \textup{(iii)} $\Rightarrow$ \textup{(iv)}. Neither of the
inverse implications hold.
\end{theorem}

\begin{proof}
Here we will only prove that (ii) $\Rightarrow$ (iii), since it
is
this
statement that will be used later  in examples to prove that
the corresponding algebra is a ${{F}}_{2n}$-algebra.


Assume that (ii) holds, but $A$ is not a
${{F}}_{2n}$-algebra. Then there exist $x_1$, \dots,
$x_{2n}\in
A$ such that ${F}_{2n}(x_1, \dots, x_{2n})=x \ne0$.
Let us choose $\pi \in{\mathcal L}$
such that $\pi(x) \ne0$. Then we get
\[
\pi({F}_{2n}(x_1, \dots, x_{2n})) =
{F}_{2n}(\pi(x_1), \dots, \pi(x_{2n})) = \pi(x)
\ne0.
\]
But since $\dim H_\pi \le n$, this contradicts the
Amitsur--Levitsky's theorem.

Neither of the inverse implications of Theorem~1 holds.
Indeed, to see that
(ii) does not imply (i), and (iv) does not imply (iii), consider
the nilpotent algebra of complex $n\times n$ matrices of the form
$$
X =\begin{pmatrix}0&&*\\&\ddots&\\0&&0
        \end{pmatrix},
$$
which only has the trivial irreducible representation for
any $n\in {\mathbb N}$.

Condition (iii) does not imply (ii). For example, the algebra of
matrices $X \in M_n({\mathbb C})$ of the form
$$
X =  \begin{pmatrix}
a_{11} &* &\dots&*\\
0 &\vdots&&\vdots \\
\vdots&*&\dots& *\\
0&\dots&0&a_{11}
\end{pmatrix} 
$$
is a ${F}_{2n-2}$-algebra. But for any representation
$\pi$,
$\dim H_\pi \le n-1$, and the nilpotent element
\[
S = \begin{pmatrix}0&1&&0 \\&\ddots&\ddots& \\&&0&1 \\
0&&&0
\end{pmatrix},
\]
we have that
$\pi(S^{n-1}) = (\pi (S))^{n-1} =0$, since $\pi(S)$ is a nilpotent
element in $M_{n-1}({\mathbb C})$. Hence such representations do not
separate $S^{n-1}$ and the zero element of the algebra.
\end{proof}

\noindent\textbf{2.}
If ${\mathfrak A}$ is a $*$-algebra, and  one only considers its
$*$-rep\-re\-sen\-tations, then, evidently, (i) $\Leftrightarrow$ (ii).

(iii) does not imply (ii), since, for example, the algebra
$M_n({\mathbb C})$ with a non-standard involution does not have non-zero
$*$-representa\-tions.

Condition (iv) does not imply (iii). For example, the Weyl
$*$\nobreakdash-alge\-bra
${{\mathbb C}}\langle P =P^*, Q = Q^* \mid [P, Q] = iI \rangle$ of
differential operators with the coefficients being polynomials in one
variable does not have $*$-representations in bounded operators, but
it is
not a ${{F}}_n$-algebra for any $n \in {\mathbb N}$.

\medskip\noindent\textbf{3.}
If ${\mathfrak A}$ is a $C^*$-algebra then
all conditions of the lemma are equivalent, since for a Banach
semi-simple
algebra
${\mathfrak A}$, the set $\irrep {\mathfrak A}$ of its irreducible
representations is a residual family (see \cite{142}) and,
therefore, $(iv)
\Rightarrow
(i)$.

 \subsection{Examples of ${F}_n$-algebras
generated by idempotents and their
representations}\label{sec:1.2.2}

Here, we give a number of examples of algebras and $*$-algebras
generated by idempotents, and 
 construct a residual family of representations or
 $*$-representations
 $\pi$ for each of them
such that $\dim H_\pi \le n$, and, therefore, show that these
algebras
are
${{F}}_{2n}$-algebras.

\medskip\noindent\textbf{1.}
Representations of the  ${{F}}_4$-algebra  generated
by
two
idempotents $q_1$, $q_2$, and the  unit element,
\begin{align*}
Q_2&={\mathbb C}\langle q_1,q_2 \mid {q_1}^2=q_1,\,{q_2}^2=q_2 \rangle
\\
&=
{\mathbb C}\langle u=2q_1-e,\,v=2q_2-e \mid u^2=e,\,v^2=e \rangle
\end{align*}
are well known; nevertheless we present a description of the
irreducible representation of the algebra to show the scheme of
investigations we will
follow in more complicated examples.

All finite dimensional irreducible representations of $Q_2$, up to
equivalence, are:

   a) four  one-dimensional representations:
   $\pi_{0,0}(q_1)=0$, $\pi_{0,0}(q_2)=0$;
          $\pi_{1,0}(q_1)=1$, $\pi_{1,0}(q_2)=0$;
        $  \pi_{0,1}(q_1)=0$, $\pi_{0,1}(q_2)=1$;
          $\pi_{1,1}(q_1)=1$, $\pi_{1,1}(q_2)=1$;

         b) the family, parameterized by $z\in {\mathbb C}\backslash \{ 0,1
         \}$, of
         two-dimensional representations:
         $$
         \pi_z(q_1)=
         \begin{pmatrix}
                                         1&0\\
                                         0&0
          \end{pmatrix}, \quad
 \pi_z(q_2)=
         \begin{pmatrix}
                                         z&1\\
                                         z-z^2&1-z
          \end{pmatrix}.
$$

Every $\pi\in \irrep Q_2$ is one- or two-dimensional. Indeed, the
 space
$H={\mathbb C}\langle e_{\lambda},\pi(v)e_{\lambda} \rangle$
($e_{\lambda}$ is an eigenvector of $\pi(u)\cdot \pi(v)$,
$\|
e_{\lambda}\| =1$, $\lambda\ne 0$) is invariant for the
representation
$\pi$. 

\medskip\noindent\textbf{2.}
In the algebra $Q_2$, one can introduce two natural structures
of an algebra with
involution: 

1) $*_1$-algebra
\begin{align*}
{{\mathcal P}}_2&={\mathbb C}\langle
p_1^{*_1}=p_1,p_2^{*_1}=p_2\mid p_1^2=p_1,p_2^2=p_2\rangle
\\
&={\mathbb C}\langle
u^{*_1}=u,v^{*_1}=v\mid u^2=v^2=e\rangle
\\
&= {\mathbb C}\left[ {\mathbb Z}_2*{\mathbb
Z}_2\right]
\end{align*}
is a group $*_1$-algebra generated by two unitary
self-adjoint generators.
  Irreducible two-dimensional $*$-representations of ${{\mathcal
  P}}_2$ (up
to a
unitary equivalence) are:
         $$
         \pi_{\phi}(p_1)=
         \begin{pmatrix}
                                         1&0\\
                                         0&0
          \end{pmatrix}, \quad
 \pi_{\phi}(p_2)=
         \begin{pmatrix}
                                         \cos^2\phi&\cos\phi\sin\phi\\
                                         \cos\phi\sin\phi&\sin^2\phi
          \end{pmatrix},
$$
$\phi \in(0,\pi/2)$. They
are equivalent to the representations $\pi_z$, $z\in(0,1)\subset
{\mathbb
R}$.

2) $*_2$-algebra
$$
{{\mathcal Q}}_1={\mathbb C}\langle q_1, q_1^{*_2}\mid
q_1^2=q_1\rangle={\mathbb
C}\langle u, u^{*_2}\mid u^2=e\rangle $$ is a ${*_2}$-algebra
generated by
an idempotent and its adjoint.
 Irreducible two-dimensional $*$-representations of ${{\mathcal Q}}_1$
 are:
$$
\pi_{\alpha}(q_1)=\begin{pmatrix} 1&\alpha\\ 0&0 \end{pmatrix}
                    ,\qquad \alpha>0.
$$
 They are equivalent to the representations $\pi_z$, $z\in(1,\infty)
\subset {\mathbb R}$.

\medskip\noindent\textbf{3.}
The algebra $Q_2$ and the both $*$-algebras constructed from it, has a
residual family of finite-dimensional representations.

\begin{proposition}
The two-dimensional representations
$\pi_z$, $z\in {\mathbb C}\setminus \{
0,1 \}$, \textup(as well as $\pi_{\phi}$, $\phi\in(0,{\pi}/{2})$, and
$\pi_{\alpha}$, $\alpha>0$\textup) form a residual family.
$Q_2$ is a ${{F}}_4$-algebra.
\end{proposition}

\begin{proof}
Let us consider any $x\in Q_2$.
\begin{align*}
x =  \alpha_0 & +\sum_{i=1}^{N_1} a_i (q_1q_2)^i
+\sum_{j=1}^{N_2}b_j
(q_2q_1)^j
\\ \displaybreak[0]
&+ \sum_{k=0}^{N_3}c_k (q_1q_2)^k q_1 +\sum_{l=0}^{N_4}d_l
(q_2q_1)^l q_2,
\end{align*}
$\alpha_0 $, $a_i $, $b_j $, $c_k$, $d_l\in\mathbb{C}$.
Then one has
\begin{align*}
\pi_{z}(x)&= \begin{pmatrix}
\alpha_0 &  0 \\[3pt]
0  & \alpha_0
\end{pmatrix}
+
\begin{pmatrix}
\sum_{i=1}^{N_1}a_i z^i & \sum_{i=1}^{N_1}a_i z^{i-1}\\[3pt]
0 &  0
\end{pmatrix}
\\
&+
\begin{pmatrix}
\sum_{j=1}^{N_2}b_j z^j  & 0 \\[3pt]
\sum_{j=1}^{N_2}b_j z^j (1-z)  & 0
\end{pmatrix}
+
\begin{pmatrix}
\sum_{k=0}^{N_3}c_k z^k  & 0\\[3pt]
0  & 0
\end{pmatrix}
\\
&+
\begin{pmatrix}
\sum_{l=0}^{N_4}d_l z^{l+1}  & \sum_{l=0}^{N_4}d_l z^l\\[3pt]
\sum_{l=0}^{N_4}d_l z^{l+1}(1-z)  & \sum_{l=0}^{N_4}d_l z^l (1-z)
\end{pmatrix}.
\end{align*}
It easily follows from the structure of the matrix $\pi_{z}(x)$ that
$\pi_{z}(x)=0$ for any $z\in \mathbb{C} \setminus \{0,1\}$ if and
only if  $x=0$.
\end{proof}

\noindent\textbf{4.}
The structure of indecomposable representations of the
algebra $Q_2$ is more complicated than the structure of irreducible ones.
Let us present  the list of all indecomposable
representations of $Q_2$ ~\cite{naz,127}.

For $\dim H = 2 k$,
\[
\pi_{\lambda}(u)=
\begin{pmatrix}
I_k & 0\\
0& -I_k 
\end{pmatrix},\quad
\pi_{\lambda}(v)=\pm
\begin{pmatrix}
A &  B \\
C  & D
\end{pmatrix},
\]
where $I_k$ is the identity  matrix of order $k$, and
\begin{gather*}
B=
\begin{pmatrix}
 -2(1-\lambda)^{-1}&-2(1-\lambda)^{-2}&\cdots  & -2
 (1-\lambda)^{-k}\\
 &-2(1-\lambda)^{-1} &\ddots &\vdots\\
\hfill\smash{\text{\Large 0}}& &\ddots & -2(1-\lambda)^{-2}\\
& & & -2(1-\lambda)^{-1}\\
\end{pmatrix},
\\
C=
\begin{pmatrix}
2\lambda (1-\lambda)^{-1}& 2(1-\lambda)^{-2}&\cdots &
2(1-\lambda)^{-k}\\
& 2\lambda(1-\lambda)^{-1}& \ddots & \vdots\\
\hfill\smash{\text{\Large 0}}& & \ddots & 2(1-\lambda)^{-2} \\
& & & 2\lambda (1-\lambda)^{-1}
\end{pmatrix},
\\*
A = -B - I_k, \quad D = -C - I_k, \qquad \lambda\in \mathbb{C} 
\setminus \{1\}.
\end{gather*}
For $\dim H = 2k+1$,
\[
\pi_{\lambda}(u)=
\begin{pmatrix}
I_{k+1} &0 \\
0 &- I_k
\end{pmatrix},
\quad\pi_{\lambda}(v)=\pm
\begin{pmatrix}
A &  B \\
C  & D
\end{pmatrix},
\]
where
\begin{gather*}
A=
\begin{pmatrix}
1 & 2 & \cdots & 2\\
& 1 &\ddots &\vdots\\
&& \ddots & 2\\
\smash{\text{\Large 0}} & & & 1
\end{pmatrix},
\quad
B=
\begin{pmatrix}
-2&\cdots  & -2 \\
0 &\ddots &\vdots\\
 &\ddots & -2\\
\smash{\text{\Large 0}}  &  & 0
\end{pmatrix},
\\[3pt]
C=
\begin{pmatrix}
0 & 2 &\cdots & 2 & 2\\
& 0 & \ddots & \vdots & \vdots\\
& & \ddots & 2 & 2\\
\smash{\text{\Large 0}}& & & 0 & 2
\end{pmatrix},
\quad
D=
\begin{pmatrix}
-1 & -2 & \cdots & -2\\
&-1& \ddots &\vdots \\
& &\ddots & -2\\
\smash{\text{\Large 0}} & & & -1
\end{pmatrix}.
\end{gather*}

The
algebra $Q_2$ is the group algebra of the Coxeter group
$\twopointinf$
generated
by two flips without any relations.

\medskip\noindent\textbf{5.}
Consider the Coxeter group $G_M$ with a matrix $M =
(m_{ij})_{i,j=1}^m$, $m_{ij} \in {\mathbb N} \cup \{\infty\}$; $m_{ii}
=1$
$m_{ij} = m_{ji} >1$, $i \ne j$; $i$, $j=1$, \dots, $m$, which is
defined in terms
of generators
$(w_i)_{i=1}^m$ and the relations $(w_i w_j)^{m_{ij}} =e$, $i$,
$j=1$,
\dots, $m$; if $m_{ij} =\infty$, then there is no relation between
the
generators $w_i$ and $w_j$.
 If the Cartan matrix $K
= \big( - \cos \pi /m_{ij}\bigr)_{i,j=1}^m$,
which corresponds to $M$, is positive definite (all
its principal minors are positive), then the group $G_M$ is finite;
if $\det K
=0$, but the other principal minors are positive, then the group
$G_M$ is
infinite, but $G_M$ is a semi-direct product of the lattice
${{\mathbb Z}}^{m-1}$
and a finite group $G_f(M)$, $G_M = {{\mathbb Z}}^{m-1} \rtimes G_f(M)$
see ~\cite{121} and others.

 Since the Coxeter group $G_M$ is generated by flips $w_i^2 =e$,
 $i=1$,
 \dots,
$m$, ${{\mathbb C}}[G_M]$ also gives an example of an algebra generated
by
$m$ projections.
 There is a natural involution in ${{\mathbb C}}[G_M]$ such that  all of
 the
group
elements are unitary, $g^* = g^{-1}$. (Generally speaking, this is
not
the unique
involution
that can be defined on ${{\mathbb C}}[G_M]$).

The dimensions of the irreducible $*$-representations
$\pi_{\alpha}$ of the group
\hbox{$*$-al}gebra of the Coxeter group $G_M={\mathbb Z}^{m-1}\rtimes
G_f$
are majorized by the number $|G_f|$.
These representations form a residual family, because irreducible
$*$-representations of
${\mathbb C}[G_M]$, with the involution $g^*=g^{-1}$, make a residual
family.
Hence ${\mathbb C}[G_M]$ is a ${{F}}_{2|
G_f|}$-algebra which is generated by flips.


\begin{remark}
It is a very difficult problem
to describe indecomposable representations of ${\mathbb C}[G_M]$ (except
for the case where
the Coxeter group
$G_M$ is a finite group or is ${\mathbb Z}\rtimes{\mathbb Z}_2$) \cite{bon_dr}.
\end{remark}

\noindent\textbf{6.}
Let now ${{\mathfrak A}}_k=
{\mathbb C}\langle
u_1^{(k)},\dots,u_{n_k}^{(k)}\mid (  )_k\rangle$
be  ${{F}}_{2m_k}$-algebras
generated by the flips $u_1^{(k)}$, \dots, $u_{n_k}^{(k)}$
and relations $(
)_k$,
such that
$\pi^{(k)}$ is a residual family
with $\dim H_{\pi^{(k)}}\le m_k$, $k=1$, \dots, $n$).
Of course, the algebra
\[
{\mathbb C}\langle u_1^{(1)},\dots,u_{n_n}^{(n)}\mid (  )_1,\dots,(
)_n,\,[u_i^{(k)},u_j^{(l)}]=0, \,k\ne l\rangle
\]
is a ${{F}}_{2m_1\cdot\dots\cdot m_n}$-algebra  having the
residual family
$\pi^{(1)}\otimes\dots\otimes\pi^{(n)}$.

\medskip\noindent\textbf{7.}
 Examples of algebras that we will  consider in the sequel
are also defined by generators
$u_1^{(1)}$, \dots, $u_{n_n}^{(n)}$, but if the upper indices are not
equal,
the generators pairwise
commute or anti-commute.
In items 7 and 8, these relations
are as follows:
$u_i^{(k)}u_j^{(l)}=\epsilon_{kl}\,u_j^{(l)}u_i^{(k)}$, $k\ne l$
($\epsilon_{kl}=+1$ or $-1$, $\epsilon_{kl}=\epsilon_{lk}$),
$k$, $l=1$, \dots, $n$, and do not depend on $i=1$,  \dots, $n_k$ and
$j=1$, \dots, $n_l$.


Let ${{\mathfrak A}}_{n,\epsilon}$ be an algebra generated by $s_1$,
$\dots$, $s_n$,
\begin{gather*}
 {{\mathfrak A}}_{n,\epsilon}={\mathbb C}\bigl< s_1, \dots,s_n \mid
 s_i^2=1,\,
s_is_j=\epsilon_{ij}s_js_i;\, i,j=1,\dots,n\bigr>,
\\
\epsilon=(\epsilon_{ij}), \qquad \epsilon_{ii}=1.
\end{gather*}

The algebra ${{\mathfrak A}}_{n,\epsilon}$ is finite dimensional and
semi-simple, it has a finite residual family
of irreducible $*$-representations
$\pi_p$ with $s_i^*=s_i$, $i=1$, \dots, $n$,
and is an ${{F}}_m$-algebra,
where
$m\ge
2^{{n}/{2}+1}$.

\medskip\noindent\textbf{8.}
 Let
${{\mathfrak B}}_{({{\mathfrak A}}_k),\epsilon}={\mathbb C}\langle u_1^{(1)},
\dots, u_{n_n}^{(n)}\mid ( )_1,\dots,(  )_n;\,
u_i^{(k)}u_j^{(l)}=\epsilon_{kl}u_j^{(l)}u_i^{(k)},\allowbreak\, k\ne l,\,
k,l=1\dots,n;
\, i=1,\dots,n_k,\, j=1,\dots,n_l\rangle$.

This is a ${{F}}_{2^{n+1}m_1\cdot\dots\cdot m_n}$-algebra
which
has a residual family of \hbox{$*$-rep}\-re\-sen\-tations
$\pi^{(1)}\otimes\dots\otimes\pi^{(n)}\otimes\pi_p$ with
$\dim{H_{\pi^{(1)}\otimes\dots\otimes\pi^{(n)}\otimes\pi_p}}\le2^n
\cdot
m_1\cdot\dots\cdot m_n$
(here, $\pi^{(1)}\otimes\dots\otimes\pi^{(n)}\otimes\pi_p(u_i^{(k)})=
1\otimes\dots\otimes\pi^{(k)}(u_i^{(k)})\otimes\dots\otimes
1\otimes\pi_p(s_k)$).

\medskip\noindent\textbf{9.}
In items 7
and 8 above, the generators  $u_i^{(k)}$ and $u_j^{(l)}$
commute
or
anti\-commute independently of $i$ and $j$.
In this item,
whether the generators $u_i^{(k)}$ and $u_j^{(l)}$
commute or not
depends on $i$, $j$.

 Let
 $A_k={\mathbb C}\langle u,v,s_1,\dots,s_k\mid u^2=v^2=s_i^2=e,\,
i=1,\dots,k, \allowbreak\,
us_i=\alpha_is_iu,\, vs_i=\beta_is_iv,\, s_is_j=\epsilon_{ij}s_js_i;\,
i,j=1,\dots,k\rangle$, where $\alpha_i=\pm 1$, $\beta_i=\pm 1$,
$ \epsilon_{ij}=\epsilon_{ji}=\pm 1$; $i\ne j$, $\epsilon_{ii}=1$.

        Of course, this algebra should have been denoted by $A_{k,
\alpha,\beta,\epsilon}$, but we leave out the symbols $\alpha$,
$\beta$, and $\epsilon$ for brevity.

Let us notice that the algebras
${\mathbb C}\langle u,v,s_1\mid us_1=-s_1u,\, vs_1=-s_1v\rangle$
 and
${\mathbb C}\langle u,v,s_1\mid us_1=-s_1u,\, vs_1=s_1v\rangle$
were considered in \cite{124,125}.


\begin{lemma}
The algebra $A_k$ is a ${{F}}_{2^{k+2}}$-algebra, and 
has a
residual family
${{\mathcal L}}_k$ subject to the condition\textup: $\forall
\pi\in{{\mathcal
L}}_k$,
$\dim
H_{\pi}\le 2^{k+1}$.
\end{lemma}

\begin{proof}
For $k=0$, $A_0$ is a ${{F}}_4$-algebra, and  has a
residual
family
${{\mathcal L}}_0$, since
the algebra $A_0=Q_2$.
Let $k=n$, and assume that all $A_n$ are ${{F}}_{2^{n+2}}$-algebras
and there
exists ${{\mathcal L}}_n$ with $\dim H_{\pi}\le 2^{n+1}$.
By induction, consider the algebra $A_{n+1}$.
It contains the subalgebra $B=C\langle u,v,s_1,\dots,s_n\mid
\alpha_i,\beta_i,\epsilon_{ij}\rangle$.
Clearly, $B$ is isomorphic to some $A_n$.
Therefore, the claim is true for $B$.
Let ${{\mathcal L}}_n$ be a residual family for $B$ with  $\dim
H_{\pi}\le
2^{n+1}$.
 We construct a residual family for $A_{n+1}$ by applying the
 following
procedure:
$\forall \pi \in {{\mathcal L}}_{n}$, $\pi:H_{\pi}\rightarrow
H_{\pi}$, introduce
$\hat{\pi}\in{{\mathcal L}}_{n+1}$,  $\hat{\pi}:H_{\pi}\oplus
H_{\pi}\rightarrow H_{\pi}\oplus H_{\pi}$, by
\begin{gather*}
\hat{\pi}(u)=
\begin{pmatrix} \pi(u)&0\\0&\alpha_{n+1}\pi(u) \end{pmatrix},\quad
\hat{\pi}(v)= \begin{pmatrix} \pi(v)&0\\0&\beta_{n+1}\pi(v)
\end{pmatrix},
\\
\hat{\pi}(s_i)=\begin{pmatrix}
\pi(s_i)&0\\0&\epsilon_{n+1i}\pi(s_i)
\end{pmatrix} ,\qquad i=1,\dots,n,
\\
\hat{\pi}(s_{n+1})=
\begin{pmatrix} 0&I\\I&0
\end{pmatrix},
\end{gather*}
  $I:H_{\pi}\rightarrow H_{\pi}$ is the
identity
operator.
Let us show that ${{\mathcal L}}_{n+1}$ is indeed a residual family.
For 
$\forall x\in A_{n+1}$ there exists an expansion $x=b_1+s_{n+1}b_2$,
with $b_1$, $b_2\in
B$. If $b_2\ne 0$, then $\exists \pi\in {{\mathcal L}}_{n}$ such
that
$\pi(b_2)\ne 0$  $\Rightarrow$
$$
 \hat{\pi}(x)=
\begin{pmatrix} \pi(b_1)&*\\\pi(b_2)&* \end{pmatrix} \ne 0,
\qquad \hat{\pi}\in{{\mathcal L}}_{n+1}.
$$

If $b_2=0$, $b_1\ne 0$, then $\exists \pi\in {{\mathcal L}}_n$:
$\pi(b_1)\ne
0$ $\Rightarrow$  $\hat{\pi}(b_1)\ne 0$.
Note that $\forall \hat{\pi}\in {{\mathcal L}}_{n+1}$,  $\dim
\hat{\pi}\le 2^{n+2}$.
\end{proof}


\begin{remark}
There is a natural involution in $A_k$ given by $u^*=u$, $v^*=v$,
$s_i^*=s_i$.
Since there exists a residual family for $Q_2$ such that $\forall \pi
\in
{{\mathcal L}}_0$ the operators $\pi(u)$, $\pi(v)$ are self-adjoint
(see \ref{sec:1.2.2}, item~2),
there
exists a residual family for $A_k$ satisfying the condition:
$\forall \pi \in
{{\mathcal L}}_k$ the operators $\pi(u)$, $\pi(v)$, $\pi(s_i)$ are
self-adjoint.
Therefore, there is a residual family for $A_k$ consisting only of
irreducible
\hbox{$*$-rep}\-re\-sen\-tations.

        For a description of irreducible \hbox{$*$-rep}\-re\-sen\-tations
of $A_k$, see also \cite{119}.
\end{remark}

\begin{remark}
$\forall\ k$, $\alpha_i$, $\beta_i$, $\epsilon_{ij}$, the algebra
$A_k$ is semi-simple.
\end{remark}

Moreover, the following theorem holds.

\begin{theorem}\label{th:q2}
 Let $Q_{2,m}=A_m$ with
 $\alpha_i=1$, $\beta_i=1$, $\epsilon_{ij}=1$, for all
$i$, $j$. Then every algebra $A_k$ is isomorphic to
$M_{2^n}(Q_{2,m})$ or
to
$M_{2^n}(Z(A_k))$, where $Z(A_k)$ is the center of $A_k$.
\end{theorem}

\begin{proof}
Let us split the proof into four steps.

 1). Let us define the algebra $A_k={\mathbb C}\langle
 u,v,s_1,\dots,s_k\mid
  \alpha_i,\beta_i,\epsilon_{ij}
\rangle$.
Suppose that there exist $i$, $j$ such that $\epsilon_{ij}=-1$,
for example,
$s_1s_2=-s_2s_1$.  Then, using the following substitution of
generators
$s_1'=s_1, s_2'=s_2,$
\begin{align*} 
s_j'&= \begin{cases}
 s_j,     &\epsilon_{1j}=\epsilon_{2j}=1,\\
 s_1s_j,  &\epsilon_{1j}=-\epsilon_{2j}=1,\\
 s_2s_j,  &\epsilon_{1j}=-\epsilon_{2j}=-1,\\ 
\sqrt{(-1)}\,s_1s_2s_j,&\epsilon_{1j}=\epsilon_{2j}=-1,
\end{cases}
\\
u'&=
 \begin{cases} u,&\alpha_1=\alpha_2=1,\\
                   s_1u,&\alpha_1=-\alpha_2=1,\\
                   s_2u,&\alpha_1=-\alpha_2=-1,\\
                   \sqrt{(-1)}\,s_1s_2u,&\alpha_1=\alpha_2=-1,
\end{cases}
\\
v' &=
 \begin{cases} v,&\beta_1=\beta_2=1,\\
                   s_1v,&\beta_1=-\beta_2=1,\\
                   s_2v,&\beta_1=-\beta_2=-1,\\
                   \sqrt{(-1)}\,s_1s_2v,&\beta_1=\beta_2=-1,
\end{cases} 
\end{align*}
we obtain that $A_k={\mathbb C}\langle u',v',s_1',\dots,s_k'\mid
 \alpha_1'=\alpha_2'=\beta_1'=\beta_2',\allowbreak\,\epsilon_{12}'=-1,
 \, \epsilon_{1j}'=\epsilon_{2j}'=1,\,j>2\rangle$.
Let
$A_{k-2}$ denote the subalgebra
$ {\mathbb C}\langle s_3',\dots,s_k',u',v'\rangle$.
\begin{lemma}
The algebra $A_k$ is isomorphic to $M_2(A_{k-2})$.
\end{lemma}

\begin{proof}
Let  $A_{k}$, $A_{k-2}$ be algebras  described above. Then $\forall
x\in
A_{k}$
there exists a unique decomposition
 $x=s_1'a_1+s_2'a_2+s_1's_2'a_3+a_4$,
where $a_i\in A_{k-2}$, $i=1$ \dots, 4.
This implies the following identity $
x=\bigl((1+s_1')\,s_2a_1'+(1-s_1')\,s_2a_2'+(1+s_1')\,a_3'+(1-s_1')\,a
 _4'\bigr)/2$.
It can be easily verified that
 $\psi \colon A_{k}\rightarrow M_2(A_{k-2})$,
\begin{gather*}
 \psi(s_1')=
 \begin{pmatrix} e&0\\0&-e \end{pmatrix} ,\quad
 \psi(s_2')=
 \begin{pmatrix} 0&e\\e&0 \end{pmatrix} ,\quad
 \psi(s_j')=
 \begin{pmatrix} s_j'&0\\0&s_j' \end{pmatrix} ,
\\
 \psi(u')=
 \begin{pmatrix} u'&0\\0&u' \end{pmatrix} ,\quad
 \psi(v')=
 \begin{pmatrix} v'&0\\0&v' \end{pmatrix} ,
\end{gather*}
is an  isomorphism. The inverse mapping
 $\psi^{-1} : M_2(A_{k-2})\rightarrow A_{k}$ is defined by the
formula:
 $$
 \psi^{-1}
 \begin{pmatrix} a_3'&a_1'\\a_2'&a_4' \end{pmatrix} 
=\frac12\bigl((1+s_1')s_2a_1'+(1-s_1')s_2a_2'+(1+s_1')a_3'+(1-s_1')a_
 4'\bigr).
 $$
which completes the proof of the lemma.
\end{proof}

Using this lemma one can obtain the following proposition:

\begin{proposition}
The algebra $A_k={\mathbb C}\langle u,v,s_{1},\dots,s_{k}\mid
 \alpha_i,\beta_i,\epsilon_{ij}\rangle$
is isomorphic to $M_{2^m}(A_{k-2m})$ for some $m$, where
\[
A_{k-2m}=
{\mathbb C}\langle u',v',s_{2m+1}',\dots,s_{k}'\mid
 \alpha_i',\beta_i',\epsilon_{ij}'=1\rangle .
\]
\end{proposition}
So, we must study the structure of the algebra $A_k$ with the
condition that
$\epsilon_{ij}=1$.

 2) We further assume, without loss of generality, that for some
 $m\in{\mathbb N}$, the relations $\alpha_i=\beta_i$, $1\le i<m$,
 $\alpha_i\ne \beta_i$, $m\le i\le k $, hold.
Let us introduce the new generators $ u'=u$, $v'=v$,
$$ s_j'=
 \begin{cases} s_j,&1\le j<m,\\
                     s_js_k, & m\le j<k.
\end{cases} 
$$
Then $A_k={\mathbb C}\langle u',v',s_1',\dots,s_k'\mid 
 \alpha_i=\beta_i,\,i<k,\,\alpha_k=\pm 1,\,\beta_k=\pm 1\rangle $, and 
there are two possibilities:
$\alpha_k=\beta_k$ or $\alpha_k\ne\beta_k$.
The first case is considered in  3), and the second in  4).

3) Let us arrange the family $\{s_j\}$ so that, for some $m$, the
conditions $\alpha_i=\beta_i=1$, $1\le i<m$, and
 $\alpha_i= \beta_i=-1$, $m\le i\le k $, hold.
Using the new generators $s_i'=s_i$, $1\le i<m$,
 $s_i'=s_is_{k}$, $m\le i<k-1$, $u'=u$, $v'=v$, we obtain the
algebra $A_k$ with the coefficients $\alpha_i'=\beta_i'=1$, $i<k$.
If $\alpha_k'=\beta_k'=1$, then the algebra $A_k$ is isomorphic to 
$Q_{2,k}$.
In the case where $\alpha_k'=\beta_k'=-1$, we have the following
proposition:

\begin{proposition}
$A_k={\mathbb C}\langle u,v,s_1,\dots,s_k\mid 
 \alpha_i=\beta_i=1,\, i<k,\allowbreak\,\alpha_k=\beta_k=-1\rangle \cong
 M_2(Z(A_k))$.
\end{proposition}
\begin{proof}
Let us denote  $B={\mathbb C}\langle s_1,\dots,s_{k-1},f,f^{-1} \rangle $,
$f=(1+s_k)uv+(1-s_k)vu$,
    and write
any $x\in A_k$ in the form
\[
x=\frac12\bigl((1+s_k)\,a_1+(1-s_k)\,a_2+(1+s_k)\,ua_3+(1-s_k)\,ua_4\bigr)
\]
$a_i\in
B$. Note, that this decomposition is unique. Indeed, if we have another
one:
$$
x=\frac12\bigl((1+s_k)\,b_1+(1-s_k)\,b_2+(1+s_k)\,ub_3
+(1-s_k)\,ub_4\bigr), 
$$
$b_i\in B$, then we obtain the identity
$0=1/2\bigl((1+s_k)(b_1-a_1)+(1-s_k)(b_2-a_2)
+(1+s_k)u(b_3-a_3)+(1-s_k)u(b_4-a_4)\bigr)$.
%\end{align*}
Multiplying this formula by $1+s_k$, we conclude that
\[
0=2(1+s_k)(b_1-a_1).
\]
Combining it with the identity
\[
u(1+s_k)(b_1-a_1)u=(1-s_k)(b_1-a_1)=0,
\]
we obtain that $(b_1-a_1)=0$. In the same way we show that $b_2=a_2$,
$b_3=a_3$, $b_4=a_4$.
Now we can write the formula for the isomorphism
$\psi \colon A_k\rightarrow M_2(B)$,
$$
 \psi(x)=
 \begin{pmatrix} a_1&a_3\\a_4&a_2 \end{pmatrix} .
$$
A direct verification shows that $\psi$ is an epimorphism and the
algebra $B$ is isomorphic to the $Z(A_k)$.
\end{proof}


4) Consider the second possibility. One can assume that
$\alpha_k=-\beta_k=-1$ (otherwise we replace $u$ with $v$ or vice
versa).
Using the previous results we have $\alpha_{j}=\beta_{j}$, $j<k$.
Then by the method described in 3), we obtain the identities
$
\alpha_{j}=\beta_{j}=1$, $j<k-1$, $\alpha_{k-1}=\beta_{k-1}$.
Thus,
$\alpha_{j}=\beta_{j}=1$, $j<k-1$, $\alpha_k = -\beta_{k}$, and
$\alpha_{k-1}$ equals $+1$ or $-1$. These cases are considered in
the following propositions.

\begin{proposition}
$A_k={\mathbb C}\langle u,v,s_1,\dots,s_k\mid 
 \alpha_i=\beta_i=1,\,i<k,\,\alpha_k=-\beta_k=-1\rangle \cong
M_2(Q_{2,k-1})$.
\end{proposition}

\begin{proof}
If we denote $v_1=1/2((1+s_k)v+(1-s_k)uvu)$,
 $v_2=1/2((1-s_k)v+(1+s_k)uvu)$, then
\[A_k\cong {\mathbb C}\langle u,v_1,v_2,
s_1,\dots,s_{k}\rangle\cong M_2(A_{k-1})\cong
M_2(Q_{2,k-1}),
\]
where $A_{k-1}={\mathbb C}\langle v_1,v_2,
s_1,\dots,s_{k-1}\rangle\cong Q_{2,k-1} $.
Indeed, any $x\in A_k$ can be represented in the form
$x=1/2\bigl((1+s_k)\,a_1+(1-s_k)\,a_2+(1+s_k)\,ua_3+(1-s_k)\,ua_4\bigr)$,
$a_i\in
 A_{k-1}$, and the  formula 
\[
\psi(x)=\begin{pmatrix}
 a_1&a_3\\a_4&a_2 \end{pmatrix}.
\]
gives the needed isomorphism
$\psi \colon A_k\rightarrow M_2(A_{k-1})$.
\end{proof}

\begin{proposition}
$
A_k={\mathbb C}\langle
u,v,s_1,\dots,s_k\mid
  \alpha_i=\beta_i=1,\,i<k,\allowbreak \, \alpha_{k-1}=\beta_{k-1}=-1,
\,\alpha_k=-\beta_k=-1
\rangle
\cong M_4(Z(A_k)).
$
\end{proposition}

\begin{proof}
It is easy to verify that the element
$1/2\bigl((1+s_k)(uv)^2+(1-s_k)(vu)^2\bigr)$ lies in the center of
the algebra
$A_k$, and, moreover, this element and the family $\{s_i,\,i<k-1\}$
generate
the center.
Then   the formulas
$\psi\colon A_k\rightarrow M_4(Z(A_{k}))$ 
\begin{gather*}
\psi(x)=
 \begin{pmatrix}x&0&0&0\\
                    0&x&0&0\\
                    0&0&x&0\\
                    0&0&0&x \end{pmatrix},\qquad
x\in Z(A_{k}),
\\[3pt]
 \psi(u)=
 \begin{pmatrix}0&0&1&0\\
                    0&0&0&1\\
                    1&0&0&0\\
                    0&1&0&0 \end{pmatrix},\quad
 \psi(1/2(1+s_{k-1})v)=
 \begin{pmatrix}0&1&0&0\\
                    1&0&0&0\\
                    0&0&0&0\\
                    0&0&0&0 \end{pmatrix} ,
\\[3pt]
 \psi(s_{k-1})=
 \begin{pmatrix}1&0&0&0\\
                    0&1&0&0\\
                    0&0&-1&0\\
                    0&0&0&-1 \end{pmatrix} ,\quad
 \psi(s_k)=
 \begin{pmatrix}1&0&0&0\\
                    0&-1&0&0\\
                    0&0&1&0\\
                    0&0&0&-1 \end{pmatrix} ,
\end{gather*}
determine the needed isomorphism.
\end{proof}
\nobreak
The proof of the theorem is completed.
\end{proof}


        By using Theorem~2.1, it is possible to obtain a description
of all irreducible representations of the algebras $A_k$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Non-commutative ``circle'', ``pair of intersecting
%\penalty-10000{} 
li\-nes''
 and
``hyperbola''. More examples of 
$F_4$-alge\-bras}\label{sec:1.2.3}

\noindent\textbf{1.}
Now we describe  bounded irreducible solutions of the relations
($II_1$)
$A^2+B^2=I$ (``non-commutative circle''), ($III_0$) $A^2-B^2
=0$
(``non-commutative pair of intersecting lines''), which is the same as the
relations $\{\widetilde{A},\widetilde{B}\}=0$, and  ($III_1$)
$A^2 -B^2 =I$ (``non-commutative hyperbola''), which is the
same
as the relation $\{\widetilde{A},\widetilde{B}\}=I$. We show
that these relations are $F_4$-relations, i.e., the
corresponding algebras are ${F}_4$-algebras.


\begin{proposition}
Irreducible self-adjoint solutions $A$, $B$ of the relations
$(II_1),\
(III_0),\ (III_1)$ are the following\textup:
\begin{itemize}
\item[$1)$]
one-dimensional \textup($\dim H =1$\textup),
$A=\lambda_1\mathbf{1}$,
$B=\lambda_2\mathbf{1}$, where the pair $(\lambda_1,\lambda_2)$
belongs
to the circle $K_{(II_1)}^{(1)}=\{
(\lambda_1,\lambda_2)\in\mathbb{R}^2\mid\
\lambda_1^2+\lambda_2^2
=1\}$, the pair of intersecting lines
$K_{(III_0)}^{(1)}=\{(\lambda_1,\lambda_2)\in\mathbb{R}^2\mid\
\lambda_1^2=\lambda_2^2\}$, or the
hyperbola
$K_{(III_1)}^{(1)}=\{(\lambda_1,\lambda_2)\in\mathbb{R}^2\mid\
\lambda_1^2-\lambda_2^2 =1\}$, respectively\textup;

\item[$2)$]
two-dimensional \textup($\dim H =2$\textup),
\[
A= \lambda_1 
\begin{pmatrix}
1& 0\\
0 & -1
\end{pmatrix},
\quad
B= \lambda_2
\begin{pmatrix}
\cos\phi & \sin\phi \\
\sin\phi & -\cos\phi
\end{pmatrix},
\]
$0<\phi<\pi$, where
the pair $(\lambda_1,\lambda_2)$ belongs respectively to the set\textup:
\begin{align*}
K_{(II_1)} &=\{(\lambda_1,\lambda_2)\mid \lambda_1>0
,\,\lambda_2>0,\, \lambda_1^2+\lambda_2^2 =1\},\\
K_{(III_0)} &=\{(\lambda_1,\lambda_2)\mid \lambda_1>0
,\,\lambda_2>0,\, \lambda_1^2-\lambda_2^2 =0\}, \\
K_{(III_1)}&=\{(\lambda_1,\lambda_2)\mid \lambda_1>0,
\,\lambda_2>0,\, \lambda_1^2-\lambda_2^2 =1\}.
\end{align*}
\end{itemize}
\end{proposition}

\begin{proof}
Since the operators $A^2$ and $B^2$ belong to the center of the
algebra, they are scalar in an irreducible representation,
$A^2=\lambda_1^2I$, $B^2 =\lambda_2^2 I$. If the representation is
not one-dimensional, then $\lambda_1>\nobreak0$, $\lambda_2>0$, and
$(A/\lambda_1)^2
=(B/\lambda_2)^2=I$. Then the proposition follows from the
description of irreducible pairs of unitary self-adjoint operators
$U=A/\lambda_1$,
$V=B/\lambda_2$.
\end{proof}

\begin{remark}
For the ``circle'' $(II_1)$ $A^2+B^2=I$ only bounded
solutions exist, since $\Vert A\Vert\le 1$, $\Vert B \Vert\le 1$. In
contrast with this, for relations $(III_0)$ and $(III_1)$, a class of
``integrable'' representations by unbounded operators was defined
and investigated (see~\cite{fa} and others).

There is no need to use unbounded operators for studying the
irreducible representations of the relations $(III_0)$ and $(III_1)$
either. Irreducibility here implies that the operators $A^2$ and $B^2$
commute with both $A$ and $B$, and hence are scalar. Thus the operators
$A$, $B$ are bounded in any irreducible representation which are given
in the proposition above.

If we consider reducible representations of ($III_0$) and ($III_1$)
by
unbounded operators, new representations appear (see~\cite{fa} and
others).
\end{remark}

\pagebreak[2]
\noindent\textbf{2.} We have the following proposition.
\nopagebreak

\begin{proposition}
The following algebras \textup(without involution\textup)
\[
\mathbb{C}\bigl< x,y\mid x^2 + y^2 = e \bigr> =\mathbb{C}\bigl<
a,b\mid a^2 - b^2 = e \bigr>,
\] and
\[
\mathbb{C}\bigl< x,y\mid x^2 = y^2  \bigr>=
\mathbb{C}\bigl< a,b\mid \{a, b\} = 0 \bigr>,
\]
are $F_4$-algebras.
\end{proposition}

\begin{proof}
We give the proof for the algebra
$\mathbb{C}\bigl< a,b\mid\{a, b\} = 0 \bigr>$ defined by two
generators $a$, $b$ and the relation $ab + ba =0$. As a vector space,
it is the same as the space of complex polynomials in two variables
but with the following multiplication of terms:
\[
a^{k_1}b^{k_2}\cdot a^{j_1}b^{j_2}=
(-1)^{k_2\cdot j_1}a^{k_1 +k_2}b^{j_1 +j_2}.
\]
Let us supply this algebra by involution defined as follows:
\[
a=a^*,\quad b=b^*,\quad (a^{k_1}b^{k_2})^* = (-1)^{k_1\cdot k_2}
a^{k_1} b^{k_2}.
\]

Irreducible $*$-representations of the $*$-algebra
\[
\mathbb{C}\bigl< a,b\mid a=a^*,\, b=b^*,\, \{a, b\} = 0 \bigr>
\]
can be obtained as follows:
\begin{enumerate}
\item[1)]
one-dimensional ($\dim H = 1$), $A=\lambda_1I$,
$B=\lambda_2I$, where the pair $(\lambda_1,\lambda_2)$
belongs
to the set
\[
K^{(1)}=\{(\lambda_1,\lambda_2)\in\mathbb{R}^2\mid\lambda_1\lambda_2
=
0\};
\]
\item[2)]
two-dimensional ($\dim H = 2$),
\[
A=\lambda_1
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
,
\quad
B=\lambda_2
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
,
\]
where the pair $(\lambda_1,\lambda_2)$ belongs to 
\[
K^{(2)}=\{(\lambda_1,\lambda_2)\in\mathbb{R}^2\mid \lambda_1>0,\,
\lambda_2>0\}.
\]
\end{enumerate}

Let us show that  these representations separate elements of the
algebra. Let
\[
x=\alpha e + \beta a +\gamma b +\sum_{i,j} c_{ij}a^i b^j.
\]
If $\pi (x)=0$ for any one-dimensional representation, then
$\alpha=\beta=\gamma =0$. If, further, $\pi (x)= 0$ for any
two-dimensional representation, then we have:
\begin{align*}
 \pi (x)&=  \pi \Bigl(\sum_{i,j}c_{ij}a^i b^j\Bigr)
\\
&= \Bigl(\sum_{i=2k,j=2l}c_{ij}\lambda_1^i\lambda_2^j\Bigr)
\begin{pmatrix}
1& 0\\
0 & 1
\end{pmatrix}
\\
&\quad{}+\Bigl(\sum_{i=2k+1,j=2l}c_{ij}\lambda_1^i\lambda_2^j\Bigr)
\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}
\\
&\quad{}+ 
\Bigl(\sum_{i=2k,j=2l+1}c_{ij}\lambda_1^i\lambda_2^j\Bigr)
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\\
&\quad{}+ \Bigl(\sum_{i=2k+1,j=2l+1}c_{ij}\lambda_1^i\lambda_2^j\Bigr)
\begin{pmatrix}
0 & 1\\
-1 & 0
\end{pmatrix}
= 0
\end{align*}
which implies $c_{ij}= 0$, $\forall i$, $j$, i.e., $ x=0$.

Then by Theorem~\ref{th:r.f}, the algebra
$\mathbb{C}\bigl<x,y\mid x^2=y^2\bigr> =\mathbb{C}\langle a,b \mid \{a,
b\}=0 \rangle$
is an ${F}_4$-algebra.
\end{proof}

\noindent\textbf{3.}
Let us notice that the algebra $Q_{4,2}=\mathbb{C}\bigl< q_1,
q_2, q_3, q_4\mid q_i^2=q_i,\, i=1,2,3,4;\, \sum_{i=1}^{4}q_i=2e \bigr>$
 is also a $F_4$-algebra.
Its representation will be
described below in Section~\ref{sec:2.2.1}.

\medskip\noindent\textbf{4.}
The algebra $\mathbb{C}\bigl< q_1, q_2, q_3, q_4\mid q_k^2=q_k ,\,
\bigl[q_k,\sum_{j=1}^{4}q_j\bigr]=0, \,
k=1,2,3,4 \bigr>$ is not a $F_n$-algebra for
all finite $n$, since it has irreducible representations in all finite
dimensions
(see Section~\ref{sec:2.2.1}).

\medskip\noindent\textbf{5.} The algebra 
\[
Q_{5,\lambda}=\mathbb{C}\Bigl<
q_1,\dots, q_5\mid q_k^2=q_k
,\, k=1,\dots,5;
\, \sum_{j=1}^{5}q_j=\lambda e  \Bigr>
\]
 is not a
$F_n$-algebra for all $\lambda\in \mathbb{C}$ and all finite
$n$ since it has an irreducible
infinite-dimensional representation (see ~\cite{rab_sam_fa}).
The  construction of this representation, which can be found in
~\cite{rab_sam_fa},
generalizes the one for five idempotents with the zero sum  given in 
~\cite{bart}.












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