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\begin{document}
\maketitle

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

The following Amitsur--Levitsky's theorem takes place~[5]:
{\em $M_n({\Bbb C})$ is
an ${{\mathcal F}}_{2n}$-algebra, but not an
${{\mathcal F}}_{2n-1}$-algebra\/}.

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

\begin{theorem}
Consider the following statements:
\begin{itemize}
\item[(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 $\pi \in {\mathcal L}$;
\item[(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
        $\pi \in  {\mathcal L}$;
\item[(iii)]
        $A$ is an ${{\mathcal F}}_{2n}$-algebra;
\item[(iv)]
	for any $\pi \in \Irrep A$,
	$\dim H_\pi \le n$.
\end{itemize}
We have the following implications: (i) $\Longrightarrow$
(ii) $\Longrightarrow$ (iii) $\Longrightarrow$ (iv).
\end{theorem}

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


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

$2$.
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
$$
\left(
	\begin{array}{ccc}0&&*\\&\ddots&\\0&&0\\
	\end{array}\right),
$$
which has only the trivial irreducible representation for
any $n\in {\Bbb N}$.

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

$3$.
If ${\frak A}$ is a $*$-algebra, and  one only considers its
$*$-rep\-re\-sen\-tations, then, evidently, $(i) \iff (ii)$.

(iii) does not imply (ii), since, for example, the algebra
$M_n({\Bbb C})$ with a nonstandard involution does not have nonzero
$*$-representations.

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

$4$.
If ${\frak A}$ is a $C^*$-algebra then
all conditions of the lemma are equivalent, since for a Banach semisimple
algebra
${\frak A}$, the set $\Irrep {\frak A}$ of its irreducible
representations is a residual family and, therefore, $(iv) \Longrightarrow
(i)$ (see [2]).

\section{Examples of ${\mathcal F}_n$-algebras
generated by idempotents and their
representations}\label{sec:2}

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

$1$.
Representations of the ${{\mathcal F}}_4$-algebra with unit generated by two
idempotents $q_1$, $q_2$,
\begin{eqnarray*}
Q_2&=&{\Bbb C}\langle q_1,q_2 \mid {q_1}^2=q_1,{q_2}^2=q_2 \rangle\\
&=&
{\Bbb C}\langle u=2q_1-e,\,v=2q_2-e \mid u^2=e,v^2=e \rangle
\end{eqnarray*}
are well known, nevertheless we present a description of
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 $z\in {\Bbb C}\backslash \{ 0,1
	 \}$, of
	 two-dimensional representations:
	 $$
	 \pi_z(q_1)=\left(
	 \begin{array}{cc}
					 1&0\\
					 0&0
	  \end{array}\right), \quad
 \pi_z(q_2)=\left(
	 \begin{array}{cc}
					 z&1\\
					 z-z^2&1-z
	  \end{array}\right).
$$

 Every $\pi\in \Irrep Q_2$ is one- or two-dimensional. Indeed, the space
$H={\Bbb C}\langle e_{\lambda},\pi(q_2)e_{\lambda} \rangle$
($e_{\lambda}$ is an eigen vector of $\pi_z(q_1)\cdot \pi_z(q_2)$,
$\|
e_{\lambda}\| =1$, $\lambda\ne 0$) is invariant for the  representation
$\pi$.
Representations $\pi_z$ and $\pi_{z'}$ $(z\ne z')$ are not equivalent, and
any irreducible 2-dimensional representation can be reduced to such a
form.

$2$.
In the algebra $Q_2$, one can introduce two natural structures
of an algebra with
involution: 1)
\begin{eqnarray*}
{{\mathcal P}}_2&=&{\Bbb C}\langle
p_1^{*_1}=p_1,p_2^{*_1}=p_2\mid p_1^2=p_1,p_2^2=p_2\rangle\\
&=&{\Bbb C}\langle
u^{*_1}=u,v^{*_1}=v\mid u^2=v^2=e\rangle\\
&=& {\Bbb C}\left[ {\Bbb Z}_2*{\Bbb
Z}_2\right]
\end{eqnarray*}
is a group $*_1$-algebra generated by two unitary
selfadjoint generators.
  Irreducible 2-dimensional $*$-representations of ${{\mathcal P}}_2$(up
to a
unitary equivalence) are:
	 $$
	 \pi_{\phi}(p_1)=\left(
	 \begin{array}{cc}
					 1&0\\
					 0&0
	  \end{array}\right), \quad
 \pi_{\phi}(p_2)=\left(
	 \begin{array}{cc}
					 \cos^2\phi&\cos\phi\sin\phi\\
					 cos\phi\sin\phi&\sin^2\phi
	  \end{array}\right),
\phi \in(0,\pi/2).$$
They are equivalent to the representations $\pi_z$, $z\in(0,1)\subset
{\Bbb
R}$.

2)
$$
{{\mathcal Q}}_2={\Bbb C}\langle q_1^{*_2}=q_2\mid q_1^2=q_1\rangle={\Bbb
C}\langle u^{*_2}=v\mid u^2=e\rangle $$ is a ${*_2}$-algebra generated by
an idempotent and its adjoint.
 Irreducible two-dimesional $*$-representations of ${{\mathcal Q}}_2$ are:
$$
\pi_{\alpha}=\left( \begin{array}{cc} 1&\alpha\\ 0&0 \end{array}
		     \right),\quad (\alpha>0).
$$
 They are equivalent to the representations $\pi_z$, $z\in(1,\infty)
\subset {\Bbb R}$.
\begin{proposition}
The 2-dimensional representations
$\pi_z$, $z\in {\Bbb C}\setminus \{
0,1 \}$, (as well as $\pi_{\phi},\phi\in(0,\frac{\pi}{2})$, and
$\pi_{\alpha}$, $\alpha>0$) form a residual family,
$Q_2$ is an ${{\mathcal F}}_4$-algebra.
\end{proposition}
\begin{proof}
Let us consider any $x\in Q_2$.
\begin{align*}
x = & \alpha_0 e +\sum_{i=1}^{N_1} a_i (pq)^i +\sum_{j=1}^{N_2}b_j
(qp)^j+\\
& + \sum_{k=0}^{N_3}c_k (pq)^k p +\sum_{l=0}^{N_4}d_l (qp)^l q,\quad
\alpha_0 , a_i , b_j , c_k\in\mathbb{C}
\end{align*}
Then one have
\[
\pi_{z}(x)= \left(
\begin{array}{ccc}
\alpha_0 & & 0 \\
& & \\
0 & & \alpha_0
\end{array}
\right) +
\left(
\begin{array}{ccc}
\sum_{i=1}^{N_1}a_i z^i & &\sum_{i=1}^{N_1}a_i z^{i-1}\\
& & \\
0 & & 0
\end{array}
\right) +
\]
\[
+
\left(
\begin{array}{ccc}
\sum_{j=1}^{N_2}b_j z^j & & 0 \\
& & \\
\sum_{j=1}^{N_2}b_j z^j (1-z) & & 0
\end{array}
\right)
+
\left(
\begin{array}{ccc}
\sum_{k=0}^{N_3}c_k z_k & & 0\\
& & \\
0 & & 0
\end{array}
\right) + 
\]
\[
+
\left(
\begin{array}{cccc}
\sum_{l=0}^{N_4}d_l z^{l+1} & & & \sum_{l=0}^{N_4}d_l z^l\\
& & & \\
& & & \\
\sum_{l=0}^{N_4}d_l z^{l+1}(1-z) & & & \sum_{l=0}^{N_4}d_l z^l (1-z)
\end{array}
\right) = 
\]
\[
= \left(
\begin{array}{ccccccc}
\alpha_0 + \sum_{i=1}^{N_1}a_i z^i +\sum_{j=1}^{N_2}b_j z^j +& & &
\sum_{i=1}^{N_1}a_i z^{i-1}+\sum_{l=0}^{N_4}d_l z^l & & &\\
+\sum_{k=0}^{N_3}c_k z^k +\sum_{l=0}^{N_4}d_l z^{l+1} & & & & & &\\
& & & & & & \\
(\sum_{j=1}^{N_2}b_j z_j +\sum_{l=0}^{N_4}d_l z^{l+1})(1-z) & & &
\alpha_0 + \sum_{l=1}^{N_4}(D_l - d_{l-1})z^l - & & &\\
& & & - d_{N_4} z^{N_4 + 1} & & &
\end{array}
\right)
\]
It is easy follows from the structure of the matrix $\pi_{z}(x)$ that
$\pi_{z}(x)=0$ if and only if then $x=0$
\end{proof}

$4$. The structure of indeomposable representations of this
algebra is more complicated than the structure of irreducible ones.
Let us present without proof the list of all indecomposable
representations of $Q_2$:\\
For $\dim H = 2 k$
\[
\pi_{\lambda}(u)=\left(
\begin{array}{ccccccc}
1 & & & & & &\\
& \ddots & &  & & 0 &\\
& & 1 &   & & & \\
& & & & & & \\
& & &  & -1  & & \\
& 0 & & & & \ddots  & \\
& & &  &  & & -1 
\end{array}
\right)
\]
\[
\pi_{\lambda}(v)=\left(
\begin{array}{ccc}
A & & B \\
& & \\
C & & D
\end{array}
\right)
\]
\[
A=\left(
\begin{array}{cccc}
2(1-\lambda)^{-1}-1 & 2(1-\lambda)^{-2} & \cdots & 2(1-\lambda)^{-k}\\
& 2(1-\lambda)^{-1}-1 &\ddots &\vdots\\
& & \ddots & 2(1-\lambda)^{-2}\\
& & & 2 (1-\lambda)^{-1}-1
\end{array}
\right)
\]
\[
C=\left(
\begin{array}{cccc}
2\lambda (1-\lambda)^{-1}& 2(1-\lambda)^{-2}&\cdots &
2(1-\lambda)^{-k}\\
& 2\lambda(1-\lambda)^{-1}& \ddots & \vdots\\
& & \ddots & 2(1-\lambda)^{-2} \\
& & & 2\lambda (1-\lambda)^{-1}
\end{array}
\right)
\]
\[
B=\left(
\begin{array}{cccc}
 -2(1-\lambda)^{-1}&-2(1-\lambda^{-2})&\cdots  & -2 (1-\lambda)^{-k}\\
 &-2(1-\lambda)^{-1} &\ddots &\vdots\\
& &\ddots & -2(1-\lambda)^{-2}\\
& & & -2(1-\lambda)^{-1}\\
\end{array}
\right)
\]
\[
D=\left(
\begin{array}{cccc}
-2\lambda(1-\lambda)^{-1}-1 & -2 (1-\lambda)^{-1}& \cdots &
-2(1-\lambda)^{-k}\\
&-2\lambda(1-\lambda)^{-1}& \ddots &\vdots \\
& &\ddots & -2(1-\lambda)^{-2}\\
 & & & -2\lambda (1-\lambda)^{-1}
\end{array}
\right)
\]
For $\dim H = 2k+1$
\[
\pi_{\lambda}(u)=\left(
\begin{array}{ccccccc}
1 & & & & & &\\
& \ddots & &  & & k+1 &\\
& & 1 &   & & & \\
& & & & & & \\
& & &  & -1  & & \\
& k+1 & & & & \ddots  & \\
& & &  &  & & -1 
\end{array}
\right)
\]
\[
\pi_{\lambda}(v)=\left(
\begin{array}{ccc}
A & & B \\
& & \\
C & & D
\end{array}
\right)
\]
\[
A=\left(
\begin{array}{cccc}
1 & 2 & \cdots & 2\\
& 1 &\ddots &\vdots\\
& & \ddots & 2\\
& & & 1
\end{array}
\right)
\]
\[
C=\left(
\begin{array}{ccccc}
0 & 2 &\cdots & 2 & 2\\
& 0 & \ddots & \vdots & \vdots\\
& & \ddots & 2 & 2\\
& & & 0 & 2\\
& & & &
\end{array}
\right)
\]
\[
B=\left(
\begin{array}{ccccc}
& -2&-2&\cdots  & -2 \\
& &-2 &\ddots &\vdots\\
& & &\ddots & -2\\
& & & & -2 \\
0 & 0 & 0 & 0 &
\end{array}
\right)
\]
\[
D=\left(
\begin{array}{cccc}
-1 & -2 & \cdots & -2\\
&-1& \ddots &\vdots \\
& &\ddots & -2\\
 & & & -1
\end{array}
\right)
\]

\bigskip
\noindent
\textbf{Example 1.2}
The
algebra $Q_2$ is the group algebra of the Coxeter group
$\bullet\buildrel\infty\over -\bullet$,
generated
by two flips without any relations.

Consider the Coxeter group $G_M$ with a matrix $M =
(m_{ij})_{i,j=1}^m$, ($m_{ij} \in {\Bbb N} \cup \{\infty\}$, $m_{ii} =1$;
$m_{ij} = m_{ji} >1$, $i \ne j$; $i$, $j=1$, \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 semidirect product of the lattice
${{\Bbb Z}}^{p(M)}$
by a finite group $G_f(M)$, $G_M = {{\Bbb Z}}^{p(M)} \ltimes G_f(M)$.

 Since the Coxeter group $G_M$ is generated by flips $w_i^2 =e$, $i=1$,
 \dots,
$m$, ${{\Bbb C}}[G_M]$ gives also an example of an algebra generated
by
$m$ projections (see, for example [7]).
 There is a natural involution in ${{\Bbb C}}[G_M]$ with which all of the
group
elements are unitary, $g^* = g^{-1}$. (Generally speaking, this is not
the unique
involution
that can be defined on ${{\Bbb C}}[G_M]$).

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


\begin{remark}
It is a every difficult problem
to describe indecomposable representations of ${\Bbb C}[G_M]$ (except
for the case where
the Coxeter group
$G_M$ is finite group or is ${\Bbb Z}\ltimes{\Bbb Z}_2$) [8].
\end{remark}

$6$
Let now ${{\frak A}}_k=
{\Bbb C}\langle
u_1^{(k)},\dots,u_{n_k}^{(k)}\mid (  )_k\rangle$
be  ${{\mathcal F}}_{2m_k}$-algebras
generated by flips $u_1^{(k)},\dots,u_{n_k}^{(k)}$ and relations $(
)_k$,
having a residual family of
$\pi^{(k)}$ with $\dim H_{\pi^{(k)}}\le m_k$ ($k=1,\dots,n$).
Of course, the algebra
$$
{\Bbb C}\langle u_1^{(1)},\dots,u_{n_n}^{(n)}\mid (  )_1,\dots,(  )_n,
[u_i^{(k)},u_j^{(l)}]=0(k\ne l)\rangle
$$
is an ${{\mathcal F}}_{2m_1\cdot\dots\cdot m_n}$-algebra that has the
residual family of
$\pi^{(1)}\otimes\dots\otimes\pi^{(n)}$.
 Examples of algebras that we will further consider
are also defined by generators
$u_1^{(1)},\dots,u_{n_n}^{(n)}$, but if the upper indeces are not equal
the generators pairwise
commute or anticommute.
In examples 2.1 and 2.2, 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
$j=1,\dots,n_l$.

$7$.
Let ${{\frak A}}_{n,\epsilon}$ be an algebra generated by $s_1$,
$\dots$, $s_n$:
$$
\begin{array}{c}
 {{\frak A}}_{n,\epsilon}={\Bbb C}\langle s_1, \dots,s_n \mid s_i^2=1,
s_is_j=\epsilon_{ij}s_js_i,(i,j=1,...n)\rangle,\\[2mm]
(\epsilon=(\epsilon_{ij}), \quad \epsilon_{ii}=1).
\end{array}
$$

The algebra ${{\frak A}}_{n,\epsilon}$ is finite dimensional and
semisimple, it has a finite residual family
of irreducible $*$-representations
$\pi_p$
($s_i^*=s_i (i=1,\dots,n)$) and is an ${{\mathcal F}}_m$-algebra, where $m\ge
2^n$.

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

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

$9$.
In Example 2.2, the generators  $u_i^{(k)}$ and $u_j^{(l)}$   commute
or
anti\-commute independently of $i$ and $j$.
In Example 2.3, whether the generators $u_i^{(k)}$ and $u_j^{(l)}$
commute or not
depends on $i$, $j$.

$10$. Let
 $A_k={\Bbb C}\langle u,v,s_1,\dots,s_k\mid u^2=v^2=s_i^2=e\
(i=1,\dots,k),\
us_i=\alpha_is_iu,\ vs_i=\beta_is_iv,\ s_is_j=\epsilon_{ij}s_js_i\
(i,j=1,\dots,k)\rangle$, 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.

\begin{remark}
The algebras

${\Bbb C}\langle u,v,s_1\mid us_1=-s_1u,\ vs_1=-s_1v\rangle$
 and
${\Bbb C}\langle u,v,s_1\mid us_1=-s_1u,\ vs_1=s_1v\rangle$ 
were considered in [3,4].
\end{remark}

\begin{lemma}
The algebra $A_k$ is an ${{\mathcal F}}_{2^{k+2}}$-algebra, and it has a
residual family
${{\mathcal L}}_k$ subject to the condition: $\forall \pi\in{{\mathcal L}}_k$,
$\dim
H_{\pi}\le 2^{k+1}$.
\end{lemma}
\emph{Proof.}
For $k=0$, $A_0$ is an ${{\mathcal F}}_4$-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 ${{\mathcal F}}_{2^{n+2}}$-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
$\hat{\pi}\in{{\mathcal L}}_{n+1}$,  $\hat{\pi}:H_{\pi}\oplus
H_{\pi}\longrightarrow H_{\pi}\oplus H_{\pi}$, by
$$
\hat{\pi}(u)=\left(
\begin{array}{cc} \pi(u)&0\\0&\alpha_{n+1}\pi(u) \end{array} \right),\quad
\hat{\pi}(v)=\left( \begin{array}{cc} \pi(v)&0\\0&\beta_{n+1}\pi(v)
\end{array} \right ),
$$
$$
\hat{\pi}(s_i)=\left(\begin{array}{cc}
\pi(s_i)&0\\0&\epsilon_{n+1i}\pi(s_i)
\end{array} \right ),\quad i=1,\dots,n,
$$
$\hat{\pi}(s_{n+1})=\left(
\begin{array}{cc} 0&I\\I&0
\end{array} \right )$,  $I:H_{\pi}\rightarrow H_{\pi}$ is the identity
operator.
Let us show that ${{\mathcal L}}_{n+1}$ is indeed a residual family.
For $A_{n+1}$,
$\forall x\in A_{n+1}$ there exists an expansion $x=b_1+s_{n+1}b_2$
($b_1,b_2\in
B$). If $b_2\ne 0$, then $\exists \pi\in {{\mathcal L}}_{n}$ such that
$\pi(b_2)\ne 0$  $\Rightarrow$
$$
 \hat{\pi}(x)=\left(
\begin{array}{cc} \pi(b_1)&*\\\pi(b_2)&* \end{array} \right)\ne 0
\qquad (\hat{\pi}\in{{\mathcal L}}_{n+1}).
$$

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


\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 selfadjoint, 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
selfajoint.
There is a residual family for $A_k$ consisting only of irreducible
\hbox{$*$-rep}\-re\-sen\-tations (see
Remark~1.3).

	For a description of irreducible \hbox{$*$-rep}\-re\-sen\-tations
of $A_k$, also see [11].
\end{remark}

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

Moreover we have the next theorem .

\begin{theorem}
 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 decompose the proof on four steps .

 1).Let us consider the algebra $A_k={\Bbb C}\langle u,v,s_1,\dots,s_k\mid
  \alpha_i,\beta_i,\epsilon_{ij}
\rangle$.
Suppose that there exists $i,j$ for which $\epsilon_{ij}=-1$,
for example,
$s_1s_2=-s_2s_1$.  Then, using the following substitution of generators
$s_1'=s_1, s_2'=s_2,$
$$ s_j'= \left
\{ \begin{array}{ccc} s_j,&\epsilon_{1j}=\epsilon_{2j}=1&\\
 s_1s_j,&\epsilon_{1j}=-\epsilon_{2j}=1&\\
 s_2s_j,&-\epsilon_{1j}=\epsilon_{2j}=1&\\
		   (i)s_1s_2s_j,&\epsilon_{1j}=\epsilon_{2j}=-1,&\quad
                   (i)=\sqrt{-1}--\mbox{imaging identity}
\end{array} \right.
$$
$
u'=
 \left \{
 \begin{array}{cc} u,&\alpha_1=\alpha_2=1\\
                   s_1u,&\alpha_1=-\alpha_2=1\\
		   s_2u,&\alpha_1=\alpha_2=1\\
		   (i)s_1s_2u,&\alpha_1=\alpha_2=-1
\end{array} \right.
 $
$
v'=
 \left \{
 \begin{array}{cc} v,&\beta_1=\beta_2=1\\
                   s_1v,&\beta_1=-\beta_2=1\\
		   s_2v,&-\beta_1=\beta_2=1\\
		   (i)s_1s_2v,&\beta_1=\beta_2=-1
\end{array} \right.
 $
we obtain that $A_k={\Bbb C}\langle u',v',s_1',\dots,s_k'\mid
 \alpha_1'=\alpha_2'=\beta_1'=\beta_2',\epsilon_{12}'=-1,
 \epsilon_{1j}'=\epsilon_{2j}'=1,j>2\rangle$.
Let us consider the subalgebra
$A_{k-2}=
 {\Bbb C}\langle s_3',\dots,s_k',u',v'\rangle$.
\begin{lemma}
The algebra $A_k$ is isomorphic to the $M_2(A_{k-2})$.
\end{lemma}
\begin{proof}
Let  $A_{k}$, $A_{k-2}$ are 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 $a_i\in A_{k-2}$.
which implies the following identity $
 x=1/2(1+s_1')s_2a_1'+1/2(1-s_1')s_2a_2'+1/2(1+s_1')a_3'+1/2(1-s_1')a_4'$.
It is easy to verify, that
 $\psi : A_{k}\rightarrow M_2(A_{k-2})$
$$
 \psi(s_1')=\left (
 \begin{array}{cc} e&0\\0&-e \end{array} \right)
 \psi(s_2')=\left (
 \begin{array}{cc} 0&e\\e&0 \end{array} \right)
 \psi(s_j')=\left (
 \begin{array}{cc} s_j'&0\\0&s_j' \end{array} \right)
$$
$$
 \psi(u')=\left (
 \begin{array}{cc} u'&0\\0&u' \end{array} \right)
 \psi(v')=\left (
 \begin{array}{cc} v'&0\\0&v' \end{array} \right)
$$
is the algebras isomorphism. The inverse mapping
 $\psi^{-1} : M_2(A_{k-2})\rightarrow A_{k}$ is determined by the
formula:
 $$
 \psi^{-1}\left(
 \begin{array}{cc} a_3'&a_1'\\a_2'&a_4' \end{array} \right)
 =1/2(1+s_1')s_2a_1'+1/2(1-s_1')s_2a_2'+1/2(1+s_1')a_3'+1/2(1-s_1')a_4'.
 $$
\end{proof}
Using this lemma one can obtain the following proposition:
\begin{proposition}
The algebra $A_k={\Bbb C}\langle u,v,s_{1},\dots,s_{k}\mid
 \alpha_i,\beta_i,\epsilon_{ij}\rangle$
is isomorphic to the $M_{2^m}(A_{k-2m})$, for some $m$, where 
\[
A_{k-2m}=
{\Bbb C}\langle u',v',s_{2m+1}',\dots,s_{k}'\mid
 \alpha_i',\beta_i',\epsilon_{ij}'=1\rangle .
\]
\end{proposition}
So, we must to study the structure of the algebra $A_k$ with condition
$\epsilon_{ij}=1$.

 2). Further we assume, without loss of generality, that for some
 $m\in{\Bbb N}$ the conditions $\alpha_i=\beta_i$,$1\le i<m$
 $\alpha_i\ne \beta_i$ $m\le i\le k $ hold.
Let us introduce the new generators $ u'=u,v'=v,$
$$ s_j'=
 \left \{
 \begin{array}{cc} s_j,&1\le j<m\\
		     s_js_k, & m\le j<k
\end{array} \right.
$$
Then $A_k={\Bbb 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 the
following two cases are possible
: $\alpha_k=\beta_k$ ore $\alpha_k\ne\beta_k$.
The first case we consider at item 3) and second at item  4).

3). Let us arrange the family $\{s_j\}$ , to hold 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 $ .
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 coefficients $\alpha_i'=\beta_i'=1$, $i<k$ .
If $\alpha_k'=\beta_k'=1$, then algebra $A_k$ is isomorphic to the $Q_{2k}$.
In the case $\alpha_k'=\beta_k'=-1$ we have the following proposition:

\begin{proposition}.
$A_k={\Bbb C}\langle u,v,s_1,\dots,s_k\mid ,
 \alpha_i=\beta_i=1,i<k,\alpha_k=\beta_k=-1\rangle \cong M_2(Z(A_k))$.
\end{proposition}
\begin{proof}
Let us denote  $f=(1+s_k)uv+(1-s_k)vu$,
   $B={\Bbb C}\langle s_1,\dots,s_{k-1},f,f^{-1} \rangle $ and present
any $x\in A_k$ in the form
\[
x=1/2(1+s_k)a_1+1/2(1-s_k)a_2+1/2(1+s_k)ua_3+1/2(1-s_k)ua_4,\ a_i\in B.
\]
Note, that this decomposition is unique. Indeed, if we have another one:
\begin{align*}
x=&1/2(1+s_k)b_1+1/2(1-s_k)b_2+1/2(1+s_k)ub_3+\\
&+1/2(1-s_k)ub_4,\quad b_i\in B.
\end{align*}
then we obtain the identity
\begin{align*}
0=&1/2(1+s_k)(b_1-a_1)+1/2(1-s_k)(b_2-a_2)+\\
&+1/2(1+s_k)u(b_3-a_3)+1/2(1-s_k)u(b_4-a_4).
\end{align*}
Multiplying this formula by $1+s_k$ we conclude that
\[
0=2(1+s_k)(b_1-a_1).
\]
Combinig it with identity
\[
u(1+s_k)(b_1-a_1)u=(1-s_k)(b_1-a_1)=0.
\]
we obtain that $(b_1-a_1)=0$. By 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
$\psi : A_k\rightarrow M_2(B)$,
$$
 \psi(x)=\left (
 \begin{array}{cc} a_1&a_3\\a_4&a_2 \end{array} \right)
$$
The direct verification shows, that $\psi$ is epimorphism and the
algebra $B$ is isomorphic to the $Z(A_k)$.
\end{proof}
4) Consider the second possibility. One can assume that
$\alpha_k=-\beta_k=-1$ (else we replace $u$ on $v$ and vice versa).
Using the results of item we have $\alpha_{j}=\beta_{j},j<k$.
Then by the method described in 3), we obtain the identities
\[
\alpha_{j}=\beta_{j}=1,\ j<k-1,\quad \alpha_{k-1}=\beta_{k-1}.
\]
So,
$\alpha_{j}=\beta_{j}=1$, $j<k-1$, $\alpha_k = -\beta_{k}$, and
$\alpha_{k-1}$ equals to $+1$ ore $-1$. These cases are considered in
the following propositions .
\begin{proposition}
\[
A_k={\Bbb C}\langle u,v,s_1,\dots,s_k\mid ,
 \alpha_i=\beta_i=1,i<k,\alpha_k=-\beta_k=-1\rangle \cong
M_2(Q_{2,k-1}).
\]
\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
\[A_k\cong {\Bbb C}\langle u,v_1,v_2,
s_1,\dots,s_{k}\rangle\cong M_2(A_{k-1})=\cong
M_2(Q_{2,k-1}),
\]
where $A_{k-1}={\Bbb 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 presented in the form
$x=1/2(1+s_k)a_1+1/2(1-s_k)a_2+1/2(1+s_k)ua_3+1/2(1-s_k)ua_4$, $a_i\in
 A_{k-1}$ and the following formula gives the needed isomorphism
$\psi : A_k\rightarrow M_2(A_{k-1})$ :
\[
\psi(x)=\left ( \begin{array}{cc}
 a_1&a_3\\a_4&a_2 \end{array} \right)
\]
\end{proof}

\begin{proposition}
\[
A_k={\Bbb C}\langle u,v,s_1,\dots,s_k\mid ,
  \alpha_i=\beta_i=1,i<k,\alpha_{k-1}=\beta_{k-1}=-1,
\alpha_k=-\beta_k=-1\rangle \cong M_4(Z(A_k)).
\]
\end{proposition}
\begin{proof}
It is easy to verify that the element
$1/2(1+s_k)(uv)^2+1/2(1-s_k)(vu)^2$ lies in the centre of algebra
$A_k$ and moreover this element and the fmily $\{s_i,i<k-1\}$ generate
the centre.
Then  the needed isomorhism is determined by the formulas:
$\psi : A_k\rightarrow M_4(Z(A_{k}))$ $$ \psi(x)=\left
(
 \begin{array}{cccc}x&0&0&0\\
                    0&x&0&0\\
                    0&0&x&0\\
                    0&0&0&x \end{array} \right),
x\in Z(A_{k})
$$
$$
 \psi(u)=\left (
 \begin{array}{cccc}0&0&1&0\\
                    0&0&0&1\\
                    1&0&0&0\\
                    0&1&0&0 \end{array} \right),\quad
 \psi(1/2(1+s_{k-1})v)=\left (
 \begin{array}{cccc}0&1&0&0\\
                    1&0&0&0\\
                    0&0&0&0\\
                    0&0&0&0 \end{array} \right),
$$
$$
 \psi(s_{k-1})=\left (
 \begin{array}{cccc}1&0&0&0\\
                    0&1&0&0\\
                    0&0&-1&0\\
                    0&0&0&-1 \end{array} \right),\quad
 \psi(s_k)=\left (
 \begin{array}{cccc}1&0&0&0\\
                    0&-1&0&0\\
                    0&0&1&0\\
                    0&0&0&-1 \end{array} \right)
$$
\end{proof}
The proof of the theorem is done
\end{proof}

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

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

$1$.
\begin{proposition}
The irreducible self-adjoint solutions $\{A,B\}$ of relations $(II_1),\
(III_0),\ (III_1)$ are the following:
\begin{enumerate}
\item
one-dimensional ($\dim H =1$), $A=\lambda_1\mathbf{1}$,
$B=\lambda_2\mathbf{1}$, where the pair $(\lambda_1,\lambda_2)$ belongs
to the circle $K_{(II_1)}^{(1)}=\{ (a,b)\in\mathbb{R}^2\mid\ a^2+b^2
=1\}$, to the pair of crossed lines
$K_{(III_0)}=\{(a,b)\in\mathbb{R}^2\mid\ a^2=b^2\}$, or to the
hiperbola
$K_{(III_1)}=\{(a,b)\in\mathbf{R}^2\mid\ a^2-b^2 =1\}$ respectivly.
\item
two-dimensional ($\dim H =2$)
\[
A= a \left(
\begin{array}{cc}
1& 0\\
0 & -1
\end{array}
\right)\quad
B= b\left(
\begin{array}{cc}
\cos\phi & \sin\phi \\
\sin\phi & -\cos\phi
\end{array}
\right),\quad 0<\phi<\pi,
\]
where the pair $(a,b)$ belongs respaectively to the set:
\begin{align*}
K_{(II_1)} &=\{(a,b)\mid\ a>0\ ,b>0,\ a^2+b^2 =1\},\\
K_{(III_0)} &=\{(a,b)\mid\ a>0\ ,b>0,\ a^2-b^2 =0\}, \\
K_{(III_1)}&=\{(a,b)\mid\ a>0,\ b>0,\ a^2-b^2 =1\}.
\end{align*}
\end{enumerate}
\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,
$A^2=a^2\mathbf{1}$, $B^2 =b^2\mathbf{1}$. If the representation is
not one-dimensional, then $a>0$, $b>0$ and $(A/a)^2
=(B/b)^2=\mathbf{1}$. Then the proposition follows from the
description of irreducible pairs of unitary operators $U=A/a$, $V=B/b$.
\end{proof}
\begin{remark}
For the ``circle'' $(II_1)$ $A^2+B^2=\mathbf{1}$ only bounded
solutions exist, since $\Vert A\Vert\le 1$, $\Vert B \Vert\le 1$. In
contrast, for the relations $(III_0)$ and $(III_1)$ a class of
``integarble'' representations by unbounded operators was dedefined
and investigated (see chapter $III$).

There is no need in using unbounded operators for the studying
irreducible representations of relations $(III_0)$ and $(III_1)$
either. Ireducibility here implies that the operators $A^2$ and $B^2$
commute with both $A$ and $B$, hence are scalar. Thus the operators
$A$, $B$ are bunded in all the ireeducible representations are given
by proposition above.

If we consider reducible representations of ($III_0$) and ($III_1$) in
unbounded operators, new representations appear (see chapter $III$).
\end{remark}

$2$.
\begin{proposition}
The following algebras without involution
\[
\mathbb{C}\bigl< a,b\mid\ a^2 + b^2 = e \bigr> 
\]
\[
\mathbb{C}\bigl< a,b\mid\ a^2 - b^2 = e \bigr> 
\]
\[
\mathbb{C}\bigl< a,b\mid\ a^2 = b^2  \bigr>=
=\mathbb{C}\bigl< a,b\mid\ \{a, b\} = 0 \bigr>
\]
are $\mathcal{F}_4$- algebras.
\end{proposition}
\begin{proof}
We present 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 polinomials of ywo 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 involuiton defined as follows:
\[
a=a^*,\quad b=b^*,\quad (a^{k_1}b^{k_2})^* = (-1)^{k_1\cdot k_2}
a^{k_1} b^{k_2}
\]
\begin{lemma}
The irreducible *-representations of a *-algebra
\[
\mathbb{C}\bigl< a,b\mid\ a=a^*,\ b=b^*,\ \{a, b\} = 0 \bigr>
\]
can be obtained by following formulas:
\begin{enumerate}
\item
one-dimensional ($\dim H = 1$), $A=\lambda_1\mathbf{1}$,
$B=\lambda_2\mathbf{1}$, 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
two-dimensional ($\dim H = 2$)
\[
A=\lambda_1\left(
\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}
\right),
\quad
B=\lambda_2\left(
\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}
\right)
\]
where the pair $(\lambda_1,\lambda_2)$ belongs to the
\[
K^{(2)}=\{(\lambda_1,\lambda_2)\in\mathbb{R}^2\mid\ \lambda_1>0,\
\lambda_2>0\}.
\]
Moreover, these representations form a resedual family.
\end{enumerate}
\end{lemma}
\begin{proof}
It follows directly from the proposition above, that presented list
gives all representations. It remains to show that these
representations separate the elements of our algebra. Let
\[
x=\alpha e + \beta a =\gamma b +\sum_{i,j} c_{ij}a^i b^j
\]
If $\pi (x)=0$ in any one-dimensional representation, then
$\alpha=\beta=\gamma =0$. If, further, $\pi (x)= 0$ in any
two-dimensional representation, then we have:
\begin{align*}
& \pi (x)=  \pi (\sum_{i,j}c_{ij}a^i b^j)=\\
=& (\sum_{i=2k,j=2l}c_{ij}\lambda_1^i\lambda_2^j)\left(
\begin{array}{cc}
1& 0\\
0 & 1
\end{array}
\right)
+ (\sum_{i=2k+1,j=2l}c_{ij}\lambda_1^i\lambda_2^j)\left(
\begin{array}{cc}
1 & 0\\
0 & -1
\end{array}
\right) + \\
+ &
(\sum_{i=2k,j=2l+1}c_{ij}\lambda_1^i\lambda_2^j)\left(
\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}
\right)
+ (\sum_{i=2k+1,j=2l+1}c_{ij}\lambda_1^i\lambda_2^j)\left(
\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}
\right) = 0
\end{align*}
which implies $c_{ij}= 0$, i.e. $ x=0$.
\end{proof}
Then by lemma (???) the algebra
$\mathbb{C}\bigl<a,b\mid\ a^2=b^2\bigr>$ is $\mathcal{F}_4$-algebra.
\end{proof}
\end{document}