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Edit File: CHAPT2A.TEX

\section {$*$-Wild algebras and relations}

Before passing to the comlexity problem of unitary description of
representations of $*$-algebras, we give definitions and some
results
concerning the ideology and methodology of $*$-wildness in Sections
\ref{sec:3.1.1} and~\ref{sec:3.1.2}. Then, in Sections
\ref{sec:3.1.3}--\ref{sec:3.1.6} we will give a number of examples of
$*$-wild algebras generated by projections and idempotents, generated
by quadratic, cubic and semi-linear relations, $*$-wild group algebras
etc.

\subsection{Majorization of $*$-algebras with respect to complexity
of their representations} \label{sec:3.1.1}

\noindent\textbf{1.} Let $\mathfrak{A}$ be a $*$-algebra. We recall (see
Section~\ref{sec:1.1.3}) that
a pair
$(\widetilde{\mathfrak{A}};\phi\colon\mathfrak{A}\mapsto\widetilde
{\mathfrak{A}})$, where $\widetilde{\mathfrak{A}}$ is a $*$-algebra
and
$\phi $ is a $*$-homomorphism, is called an enveloping $*$-algebra if 
for any $*$-representation $\pi\colon\mathfrak{A}\mapsto L(H)$ of the
algebra $\mathfrak{A}$ there exists a unique $*$-representation
$\tilde{\pi}\colon\widetilde{\mathfrak{A}}\mapsto L(H)$ such that the
diagram
\begin{center}
\resetparms
\btriangle[\mathfrak{A}`\widetilde{\mathfrak{A}}`L(H) ;\phi `\pi
`\widetilde{\pi}]
\end{center}
is commutative.

\medskip\noindent\textbf{2.}
To verify whether or not $(\mathfrak{A},
\phi\colon\mathfrak{A}\mapsto\widetilde{\mathfrak{A}})$ determines an
enveloping algebra, it is sufficient to consider the unital
representations of algebra $\mathfrak{A}$ only. Indeed, let for any
unital
representation $\pi$ of $\mathfrak{A}$ there exists the
representation
$\widetilde{\pi}$ of $\widetilde{\mathfrak{A}}$ making the following
diagram commutative

\begin{center}
\resetparms
\btriangle[\mathfrak{A}`\widetilde{\mathfrak{A}}`L(H) ;\phi `\pi
`\widetilde{\pi}]
\end{center}

Suppose that $\pi\colon\mathfrak{A}\mapsto L(H)$ is not unital
representation. Then $\pi (e)=E$ is a projection in $L(H)$ and for
any
$a\in\mathfrak{A}$ we have $\pi(a)=\pi(eae)=E\pi(a)E$. The space $H$
decomposes into the direct sum $H=\im E\oplus\ker E$ and the subspaces
$\im E$, $\ker E$ are invariant with respect to the operators
$\pi(a)$; moreover $\ker E\subset\ker \pi(a)$, and according to this
inclusion we have
\[
\pi(a)= 
\begin{pmatrix}
\pi^{(1)}(a) & 0\\
0 & 0
\end{pmatrix},
\]
where $\pi^{(1)}$ is unital representation of $\mathfrak{A}$. Since
the
homomorphism $\phi$ is unital, we have
$\pi^{(1)}(e)=\tilde{\pi}^{(1)}(\phi(e))$, and
\[
\widetilde{\pi}(a)=
\begin{pmatrix}
\tilde{\pi}^{(1)}(a) & 0\\
0 & 0
\end{pmatrix}
\]
is the needed homomorphism from $\mathfrak{A}$ to
$\tilde{\mathfrak{A}}$. Since
$\tilde{\pi}(e)=\tilde{\pi}(\phi(e))=E$, then any
representation of $\tilde{\mathfrak{A}}$, making the previous diagram
commutative, coincides with the presented above. Thus
$\tilde{\pi}$ is determined uniquely.

\medskip\noindent\textbf{3.}
If $(\widetilde{\mathfrak{A}},\ \phi)$ is an enveloping algebra for the
algebra $\mathfrak{A}$ and $(\widetilde{\mathfrak{A}}_1,\ \psi)$ is
an
enveloping algebra for $\widetilde{\mathfrak{A}}$, then
$(\widetilde{\mathfrak{A}}_1,\ \psi\circ\phi)$ is an enveloping algebra
for the
algebra $\mathfrak{A}$. The proof is evident.

\medskip\noindent\textbf{4.}
Now we define the relation of majorization for $*$-algebras.
Denote by $\mathfrak{K}$ the algebra of compact operators in a
separable Hilbert space $H_0$.
Let $\pi\colon\mathfrak{A}\mapsto L(H)$; it induces the
representation
\[
\hat\pi=id\otimes\pi\colon \mathfrak{K}\otimes\mathfrak{A}\mapsto
L(H_0\otimes H)
\]
of algebra $\mathfrak{K}\otimes\mathfrak{A}$. If
$(\tilde{\mathfrak{K}\otimes\mathfrak{A}}, \phi)$ is an
enveloping algebra
of $\mathfrak{K}\otimes\mathfrak{A}$ then, by the definition,
$\hat\pi$ determines
uniquely the representation $\tilde{\pi}$ of an enveloping
algebra $\tilde{\mathfrak{K}\otimes\mathfrak{A}}$
in the same Hilbert space $L(H_0\otimes H)$. 
Let now $\psi$ be a homomorphism of the algebra $\mathfrak{B}$ into
the algebra $\tilde{\mathfrak{K}\otimes\mathfrak{A}}$; then
\[
\tilde{\pi}\psi\colon\mathfrak{B}\mapsto
L(H_0\otimes H)
\]
is the representation of $\mathfrak{B}$.

\begin{lemma}
Let  $(\widetilde{\mathfrak{K}\otimes\mathfrak{A}}, \phi)$ be 
enveloping of
 $\mathfrak{K}\otimes\mathfrak{A}$ and $\pi_j\in\rep(\mathfrak{A})$,
$j=1$, $2$. Let  
$\tilde\pi_j$ denote its lifting to  enveloping $*$-algebra as 
described above. Then 
  $V\in B(H)$ intertwines $\pi_1$ and $\pi_2$ implies 
$E\otimes V$ intertwines $\tilde\pi_1$
and $\tilde\pi_2$ $(E$ is identity operator in $B(H))$. 
\end{lemma}

\begin{proof}
We will prove that for $V$ intertwining one representation $\pi$. 
If this is proved form 
$$
\pi=\left(
\begin{array}{cc}
\pi_1&0\\
0&\pi_2\\
\end{array}
\right)
$$
and 
$$
V=\left(
\begin{array}{cc}
0&U\\
U&0\\
\end{array}
\right)
$$
And this will give the result  for  $U$ intertwining $\pi_1$ and $\pi_2$.
It is obvious that $E\otimes V\in
\hat\pi(\mathfrak{K}\otimes\mathfrak{A})'$.
Let $\mathcal{A}$ denote $C^*$-algebra generated by
$\tilde{\pi}(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})$ 
and $\mathcal{B}$ denote its $C^*$-subalgebra generated by
$\hat\pi(\mathfrak{K}\otimes\mathfrak{A})$;   
then   $\mathcal{B}$ is $C^*$-subalgebra in  $\mathcal{A}$.
Let us show  that if $\pi_1$, $\pi_2\in\rep(\mathcal{A})$ and
$\pi_1\restriction\mathcal{B}=\pi_2\restriction\mathcal{B}$, then 
$\pi_1=\pi_2$. Indeed, representations $\pi_1\circ\tilde{\pi}$
and 
$\pi_2\circ\tilde{\pi}$ of
$\widetilde{\mathfrak{K}\otimes\mathfrak{A}}$ are lifting of  the 
representation
$\pi_1\circ\tilde{\pi}\circ\phi=\pi_2\circ\tilde{\pi}\circ\phi$,
since $\mathcal{B}$ is closure of
$\hat\pi(\mathfrak{K}\otimes\mathfrak{A})
=\tilde{\pi}(\phi(\mathfrak{K}\otimes\mathfrak{A}))$.      
By the definition of enveloping $*$-algebra such lifting is unique, so 
$\pi_1=\pi_2$. By Lemma~3.9. in~\cite{Woron}, $\mathcal{B}=\mathcal{A}$.  
By assumption  $E\otimes V\in
\hat\pi(\mathfrak{K}\otimes\mathfrak{A})'$, so 
$E\otimes V\in \mathcal{B}'$, hence $E\otimes V\in
\mathcal{A}'$, and consequently 
$
E\otimes
 V\in\tilde{\pi}(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})'$.
\end{proof}

\begin{definition}\label{majoriz}
We say that a $*$-algebra $\mathfrak{B}$ majorizes
a $*$-algebra $\mathfrak{A}$ \textup(and denote it by
$\mathfrak{B}\succ\mathfrak{A}$\textup)\textup, if there exists an
enveloping algebra $(\widetilde{\mathfrak{K}\otimes\mathfrak{A}},
\phi)$ of
 algebra $\mathfrak{K}\otimes\mathfrak{A}$, and the unital
 homomorphism
$\psi\colon\mathfrak{B}\mapsto\widetilde{\mathfrak{K}\otimes\mathfrak
{A}}$, such
that the functor $F\colon \rep \mathfrak{A}\mapsto \rep
\mathfrak{B}$ defined by the following rule\textup:
\begin{align*}
F(\pi)&=\tilde{\pi}\psi, \quad \text{for any $\pi\in \rep
\mathfrak{A}$},
\\
F(A)&=E\otimes A, \quad \text{for any operator
$A$ intertwining $\pi_1$ and $\pi_2$},
\end{align*}
is full.
\end{definition}

\begin{remark}
It follows from the proof of previous Lemma that 
to verify whether $F$ is full it is enough to verify that 
 for every $\pi\in\rep(\mathfrak{B})$ in $B(H)$ and any
operator $A$ in $B(H)$  
$E\otimes A\in F(\pi)(\mathfrak{A})'$ implies  $E\otimes
A\in \pi(\mathfrak{B})'$. 
\end{remark}

\medskip\noindent\textbf{5.}
For a class of $C^*$-algebras, it is possible
to consider only homomorphisms $\psi\colon \mathfrak{B} \to
\mathfrak{K}\otimes\mathfrak{A}$ in Definition~\ref{majoriz}, since for a
$C^*$-algebra
$\mathfrak{A}$,   
$\mathfrak{K}\hat\otimes\mathfrak{A}$  is also a $C^*$-algebra, and
the unique
enveloping
algebra of a $C^*$-algebra 
is the algebra itself ($\mathfrak{K}\hat\otimes\mathfrak{A}$ is completion 
of
algebraic tensor product in any $C^*$-prenorm. 
It is known to be independent of the choice of the prenorm).
	Moreover, we have the following fact.
	
\begin{lemma}
Let $\psi\colon \mathfrak{B} \to \mathfrak{A}$ be a morphism of\/ 
$C^*$-algebras.
If \linebreak$\pi(\psi(\mathfrak{B}))' =\pi(\mathfrak{A})'$
for any $\pi \in \rep{\mathfrak{A}}$,
then $\psi$ is a surjection. 
\end{lemma}

\begin{proof}
Denote $\mathfrak{B}_1=\psi(\mathfrak{B}) $. Assume that $
\mathfrak{B}_1 \neq \mathfrak{A}$ then 
there is non-zero continuous functional $f \in \mathfrak{A}^* $ such
that $f(b)=0$ for all $ b \in  \mathfrak{B}_1$.
 Then one can find (see proof of Lemma~3.9 in~\cite{Woron}) a
 representation $\pi\colon \mathfrak{A} \to B(H)$ and a 
finite dimensional  operator  $\rho \in B(H)$ such that
$f(a)=\tr(\rho\pi(a))$ for all $a\in \mathfrak{A}$. 
Since $ \pi(\psi(\mathfrak{B}))' =\pi(\mathfrak{A})' $
we have $\pi(\mathfrak{B}_1)'' =\pi(\mathfrak{A})''$. 
Then $\pi(\mathfrak{A}) \subseteq
\overline{\pi(\mathfrak{B}_1)}^{WOT}$
by the von~Neumann density theorem. 
Then for every $a  \in  \mathfrak{A}$ there is generalized sequence
$b_\alpha \in \mathfrak{B}_1$ such that 
$\pi(a)=\wlim \pi(b_\alpha) $.  So 
\begin{align*}
f(a)&=\tr(\rho \wlim
\pi(b_\alpha))
\\
&=\lim \tr(\rho\pi(b_\alpha) )
=\lim f(b_\alpha) =0.
\end{align*}
This contradiction proves that   $ \mathfrak{B}_1=\mathfrak{A}$.
\end{proof}

\begin{theorem}\label{bound}
A $C^*$-algebra $\mathfrak{B}$ majorizes a $C^*$-algebra
$\mathfrak{A}$ if and only if 
  $\mathfrak{B}$   contains  an ideal $\mathfrak{I}$ such that
  $\mathfrak{B}/ \mathfrak{I}\simeq \mathfrak{K}\hat\otimes\mathfrak{A}$.
\end{theorem}

\begin{proof}
Let us note  firstly that any  representation $\pi $ of
$\mathfrak{K}\otimes\mathfrak{A}$ has
 a form $ \pi=U \id \otimes \pi_0 U^* $ for some 
$\pi_0 \in \rep(\mathfrak{A}) $ and unitary $U$.  Indeed, since
$C^*$-algebra $\mathfrak{K}$ is of type $I$  and any
irreducible representation of  $\mathfrak{K}$ is unitary equivalent
to $\id$,  
the above statement is true for irreducible $\pi $ by theorem
\cite{Takesaki}.  Arbitrary representation can be decomposed as  
$\pi =\bigoplus_\alpha \pi_\alpha $, where $\pi_\alpha $ is
irreducible and so  $ \pi_\alpha=U_\alpha \id \otimes {\pi_\alpha}_0
U^*_\alpha $. 
Put $\pi_0=\bigoplus_\alpha {\pi_\alpha}_0   $, and
$U=\bigoplus_\alpha  U_\alpha $ then   $ \pi=U \id \otimes \pi_0 U^*
$. 
Let $F_\psi \colon \rep(\mathfrak{A}) \to \rep(\mathfrak{B})$ be full. 
We will show that the functor $F_\psi \colon
\rep(\mathfrak{K}\otimes\mathfrak{A}) \to
\rep(\mathfrak{B})$ is also full.
Indeed, take arbitrary  $\pi  \in \mathfrak{K}\otimes\mathfrak{A} $
then  $ \pi=U
\id \otimes \pi_0 U^* $. 
Let $V \in \pi(\psi(\mathfrak{B}))'$  then $U^*VU \in (\id
\otimes
\pi_0(\psi(\mathfrak{B})))'$
 so $U^*VU=E \otimes V_0$ where $V_0 \in
 \pi_0(\psi(\mathfrak{B}))'$. Then $V_0 \in \pi_0(\mathfrak{A})'$
 since 
$F_\psi \colon \rep(\mathfrak{A}) \to \rep(\mathfrak{B})$ is full. It
proves that  $U^*VU \in (\id \otimes \pi_0(\mathfrak{A}))'$ and so 
$V \in \pi(\mathfrak{A})'$. Now we can apply previous lemma to finish
the proof. 
\end{proof}

\medskip\noindent\textbf{6.}
Now consider general case of arbitrary $*$-algebras.
The following lemma is an analogue
of Theorem~6.3.5 in~\cite{murphi} about $C^*$-algebras.

\begin{lemma}
Let $\mathfrak{A}$ be unital $*$-algebra and let $\mathcal{B}$ be 
any $C^*$-algebra. 
Let $\pi\in\rep(\mathcal{B}\otimes\mathfrak{A})$ be representation in $B(H)$. 
Then there are representations $\phi\colon\mathfrak{A}\to B(H)$ and 
$\psi\colon\mathcal{B}\to B(H)$ such that for all $a\in\mathfrak{A}$ and 
$b\in\mathcal{B}$, 
\[
\pi(b\otimes a)=\psi(b)\phi(a)=\phi(a)\psi(b).
\]  
\end{lemma}

\begin{proof}
Let $a^\prime\in\mathfrak{A}$ and define $\psi_{a^\prime}\colon\mathcal{B}
\to B(H)\quad b\mapsto\pi(b\otimes a^\prime)$. Let us show that 
$\psi_{a^\prime}$  is continuous for all $a^\prime\in\mathfrak{A}$. 
Since $\psi_{a^\prime}$ is  mapping   between Banach spaces we can apply 
closed graph theorem. We need to show that if $b_{n}\rightarrow 0$ in $\mathcal{B}$ and 
$\psi_{a^\prime}(b_n)\rightarrow c$ in $B(H)$ then $c=0$. Replacing $b_n$ by 
$b_n^*b_n$ and $a^\prime$ by $(a^{\prime})^*a^\prime$ and $c$ by $c^*c$ we can consider $c\ge 0$. 
Let $\tau$ be arbitrary positive functional on $B(H)$. Then functional 
\[
\rho\colon\mathcal{B}\to\mathbb{C}\quad b\mapsto\tau(\pi(b\otimes a^\prime))
\]
 is positive and therefor continuous. Hence $\tau(c)=\lim\tau(b_n\otimes a^\prime)=\lim\rho(b_n)=0$ 
since $\lim b_n=0$. So $c=0$ which proves that $\psi_{a^\prime}$ is continuous. We can assume that 
 $\pi$ is non-degenerate. Define $H_0=\pi(\mathcal{B}\otimes\mathfrak{A})$. 
Every $z\in H_0$ can be written as $z=\sum_{i=1}^{n}\pi(b_i\otimes a_i)(x_i)$ 
where $b_i\in\mathcal{B},\ a_i\in\mathfrak{A},\ x_i\in H$. 
Let there $z=\sum_{j=1}^{m}\pi(b_i^\prime\otimes a_i^\prime)(x_i^\prime)$ be another presentation. 
Let $v_\mu$ be approximate identity for $\mathcal{B}$ and $a\in\mathfrak{A}$. 
Then 
\[
\pi(v_\mu\otimes a)(z)=\sum_{i=1}^{n}\pi(v_\mu b_i\otimes aa_i)(x_i)=
\sum_{j=1}^{m}\pi(v_\mu b_i^\prime\otimes aa_i^\prime)(x_i^\prime) 
\]
Since $\psi_a$ is continuous we can take limit 
\[
\lim_{\mu}\pi(v_\mu\otimes a)(z)=\sum_{i=1}^{n}\pi(b_i\otimes aa_i)(x_i)=
\sum_{j=1}^{m}\pi(b_i^\prime\otimes aa_i^\prime)(x_i^\prime) 
\]
So mapping $\phi(a)\colon H_0\to H_0,\quad z\mapsto\sum_{i=1}^{n}\pi(b_i\otimes aa_i)(x_i)$
is correctly defined. Since $\phi(a)=\lim_{\mu}\pi(v_\mu\otimes a)(z)$ it is 
obvious that $\phi(a)$ is linear. Since $\psi_a$ is continuous there  is 
$M=M(a)\in \mathbb{R}_{+}$ such that $\Vert \pi(b\otimes a)\Vert \le M\Vert b\Vert\quad (b\in \mathcal{B})$.  
So $\phi(a)$ is bounded operator. Since $\pi$ is non-degenerate, $H_0$ is 
dense in $H$ and $\phi(a)$ is uniquely extended to bounded operator on $H$ 
(which is also denoted by $\phi(a)$). Then $\phi\colon\mathfrak{A}\to B(H)$ 
and $\psi=\psi_{1}$ ($a^\prime=1$) are required  
representations. 
Below we use notations  $\phi=\pi_{\mathfrak{A}}$ and $\psi=\pi_{\mathcal{B}}$ 
\end{proof}

\begin{proposition}
Let $\mathfrak{A}$ be unital $*$-algebra then 
\begin{enumerate}
\item  for every $\pi\in\rep(\mathfrak{K}\otimes\mathfrak{A})$ in $B(H)$ there are 
unitary $U\in B(H)$ and representation $\pi_0\in\rep(\mathfrak{A})$ in $B(H_1)$ 
such that $H=H_0\otimes H_1$ and $U^*\pi U=id\otimes\pi_0$.
\item pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})\simeq\mathfrak{K}\hat\otimes$pro-$C^*(\mathfrak{A})$.
 Where $\mathfrak{K}\hat\otimes$pro-$C^*(\mathfrak{A})$ is unique pro-$C^*$-algebra 
which is completion of algebraic tensor product(see~\cite{Phil}) since  $\mathfrak{K}$ is nuclear.
\end{enumerate}
\end{proposition}
\begin{proof}
1. By previous lemma $\pi=\pi_{\mathfrak{K}}\pi_{\mathfrak{A}}$. 
Denote by $\mathcal{A}=\overline{\pi_{\mathfrak{A}}(\mathfrak{A})}$ the $C^*$-subalgebra 
in $B(H)$ generated by range of $\pi_{\mathfrak{A}}$. Let $j\colon\mathcal{A}\to  B(H)$ denote 
natural embedding. 
If $\pi\neq 0$ then  $\pi_{\mathfrak{K}}\ne 0$ and by simplicity of 
$\mathfrak{K}$ we have $\pi_{\mathfrak{K}}(\mathfrak{K})\simeq\mathfrak{K}$. 
The following diagram is commutative. 
\begin{center}
\resetparms
\qtriangle[\mathfrak{K}\otimes\mathfrak{A}`\mathfrak{K}\otimes\mathcal{A}`B(H);id\otimes\pi_{\mathfrak{A}}`\pi
`\pi_{\mathfrak{K}}\otimes j]
\end{center}
Since $\mathfrak{K}$ is of type $I$ representation  $ \pi_{\mathfrak{K}}\otimes j=U^*id\otimes\hat\pi U$
where $\hat\pi\in\rep(\mathcal{A})$  (see proof of theorem~\ref{bound}).
So $\pi=U^*id\otimes\pi_0 U$ where $\pi_0=\hat\pi|\mathfrak{A}$.

2. Take $\pi\in\rep(\mathfrak{K}\otimes\mathfrak{A})$ then
$H=H_0\otimes H_1$ and  $\pi=U^*id\otimes\pi_0 U$ . Denote by $\phi\colon\mathfrak{A}\to$pro-$C^*(\mathfrak{A})$ canonical homomorphism. 
By 
definition of 
enveloping algebra there is unique lifting $\widetilde{\pi_0}\in\rep($pro-$C^*(\mathfrak{A}))$
in $B(H_1)$ of 
representation $\pi_0$. Then define representation 
$\widetilde{\pi}\in\rep(\mathfrak{K}\otimes$pro-$C^*(\mathfrak{A}))$ via the rule  
$\widetilde{\pi}(k\otimes a)=U^*id\otimes \widetilde{\pi_0}U$. Since $\phi(\mathfrak{A})$ is quasi-dense  
in pro-$C^*(\mathfrak{A})$ 
algebra $\mathfrak{K}\otimes\phi(\mathfrak{A})$ is quasidense in $\mathfrak{K}\hat\otimes$pro-$C^*(\mathfrak{A})$. 
So lifting $\widetilde{\pi}$ is unique. 
\end{proof}
Now we are in a position to prove the main theorem about majorization.

\begin{proposition}
\begin{enumerate}
\item If $(\widetilde{\mathfrak{A}},\phi)$ is enveloping $*$-algebra
for 
$\mathfrak{A}$ then
pro-$C^*(\mathfrak{A})\simeq$pro-$C^*(\widetilde{\mathfrak{A}})$.
\item $*$-algebra $\mathfrak{B}$ majorizes a $*$-algebra
$\mathfrak{A}$ if 
and only if there exists continuous morphism 
$\psi :$
pro-$C^*(\mathfrak{B})\to$pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})$ 
with quasidense image. 
(such that for any representation $\pi\in
\rep($pro-$C^*(\mathfrak{K}\otimes\mathfrak{A}))$
the set $\pi(\psi($pro-$C^*(\mathfrak{B})))$ is dense in $\im\pi$)
\item If pro-$C^*(\mathfrak{A})$ is $\sigma$-$C^*$-algebra (for
example if 
$\mathfrak{A}$ is finitely generated) then
$\mathfrak{B}\succ\mathfrak{A}$ 
if and only if there exists continuous morphism 
 $\psi :$
 pro-$C^*(\mathfrak{B})\to$pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})$ 
with dense image. 
\end{enumerate}
\end{proposition}

\begin{proof}

$1$.  Let $\psi$ denote canonical morphism form
$\widetilde{\mathfrak{A}}$ to pro-$C^*(\widetilde{\mathfrak{A}})$.   
For every $\pi\in\rep(\mathfrak{A})$ there is unique lifting
$\widetilde{\pi}\in\rep(\widetilde{\mathfrak{A}})$
 and by definition of enveloping pro-$C^*$-algebra $\widetilde{\pi}$
 is uniquely  lifted to representation 
 $\hat\pi\in\rep($pro-$C^*(\widetilde{\mathfrak{A}}))$. Hence 
(pro-$C^*(\widetilde{\mathfrak{A}}),\psi\circ\phi)$ is enveloping
pro-$C^*$-algebra of $\mathfrak{A}$. 
Then  pro-$C^*(\mathfrak{A})\simeq$pro-$C^*(\widetilde{A})$ by
uniqueness of pro-$C^*$-enveloping algebra.

$2$.  Let $(\widetilde{\mathfrak{K}\otimes\mathfrak{A}}, \phi)$ be 
enveloping 
$*$-algebra of $\mathfrak{K}\otimes\mathfrak{A}$. 
If $\pi\in\rep(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})$ then 
$\pi=U\widetilde{\pi_0}U^*$ for some $\pi_0\in\rep(\mathfrak{A}) $ and unitary $U\in B(H)$. 
Indeed, since mapping 
$\rep(\mathfrak{K}\otimes\mathfrak{A})\ni\kappa\mapsto\widetilde{\kappa}\in \rep(\widetilde{\mathfrak{K}\otimes\mathfrak{A}}) $
is bijective $\pi=\widetilde{\kappa}$ for some $\kappa\in\rep(\mathfrak{K}\otimes\mathfrak{A})$. 
By previous proposition $\kappa=Uid\otimes\pi_0U^*$ where $id$ is identical representation of 
 $\mathfrak{K}$ in $B(H_0)$ and $\pi_0$ is representation of $\mathfrak{A}$ in $B(H_1)$ and $H=H_0\otimes H_1$. 
Notice that $U\widetilde{\pi_0}U^*$ is representation of $\widetilde{\mathfrak{K}\otimes\mathfrak{A}}$ which 
is lifting of $\kappa$ and by uniqueness of such lifting 
\[
U\widetilde{\pi_0}U^*=\widetilde{\kappa}=\pi
\]

$3$. Assume  that $\mathfrak{B}\succ\mathfrak{A}$, i.e. there is
$*$-homomorphism 
$\psi :\mathfrak{B}\to \widetilde{\mathfrak{K}\otimes\mathfrak{A}}$
such
 that functor $F_\psi\colon
 \rep(\mathfrak{A})\mapsto\rep(\mathfrak{B})$ is full. Denote 
$\mathcal{A}=$pro-$C^*(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})\simeq$
pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})$ and let
$\phi:\widetilde{\mathfrak{K}\otimes\mathfrak{A}} \to \mathcal{A}$ be
canonical  
morphism. Then functor $F:\rep(\mathcal{A})\to
\rep(\mathfrak{B})\quad \rep(\mathcal{A})\ni\pi\mapsto\pi\circ\phi\circ\psi\in\rep(\mathfrak{B})$ is also full. 
Indeed, $\pi\circ\phi\in\rep(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})$, so 
$\pi\circ\phi=U\widetilde{\pi_0}U^*$ for some $\pi_0\in\rep(\mathfrak{A})$
in $B(H)$ then 
\[
F(\pi)(\mathfrak{B})^\prime=F_{\psi}(\pi\circ\phi)(\mathfrak{B})^\prime=
F_{\psi}(U\widetilde{\pi_0}U^*)(\mathfrak{B})^\prime
\]
$F_{\psi}(U\widetilde{\pi_0}U^*)(\mathfrak{B})^\prime=(U\widetilde{\pi_0}(\mathfrak{B})U^*)^\prime=
(U\widetilde{\pi_0}(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})U^*)^\prime$
since functor $F_{\psi}$ is full. 
$(U\widetilde{\pi_0}(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})U^*)^\prime=\pi(\phi(\widetilde{\mathfrak{K}\otimes\mathfrak{A}}))^\prime=
\pi(\mathcal{A})^\prime$ since $\phi(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})$ is quasidense in $\mathcal{A}$

Morphism $\phi\circ\psi$ can be extended to continuous morphism
 $\tau$ of pro-$C^*(\mathfrak{B})$  
since if $\phi_1:\mathfrak{B}\to$pro-$C^*(\mathfrak{B})$ denotes
canonical morphism then $\phi_1(\mathfrak{B})$ is quasidense in
pro-$C^*(\mathfrak{B})$  
 and topology on $\phi_1(\mathfrak{B})$ induced from
 pro-$C^*(\mathfrak{B})$(projective topology induced by all
 representations of $\mathfrak{B}$) 
is stronger then topology induced by map $\phi\circ\psi$ (projective
topology induced by 
those representations of $\mathfrak{B}$ which come through
$\mathcal{A}$). Let us show that range of $\tau$ is dense in every
representation. 
Take arbitrary representation $\pi\in \rep(\mathcal{A})$ then natural
injection 
$j:\overline{\pi(\tau(\mathfrak{B}))}\to\pi(\mathcal{A})$ satisfies 
conditions of previous lemma. 
So  $\pi(\tau($pro-$C^*(\mathfrak{B})))$ is dense in $\im\pi$. 
The converse statement is obvious.

$3$.   If pro-$C^*(\mathfrak{A})$ is $\sigma$-$C^*$-algebra then 
pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})$ is $\sigma$-$C^*$-algebra
also. Therefor its topology can be determined by countably increasing
family $p_n(\cdot)$ of $C^*$-seminorms. We have proved that
$\tau($pro-$C^*(\mathfrak{B}))$ is dense  in
pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})$ in any $C^*$-semi-norm
(each 
semi-norm $p()$ define representation $\pi_{p}$ of
pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})$ and
$\pi_{p}(\tau($pro-$C^*(\mathfrak{B})))$ is dense in $\im(\pi)$).

Using Cantor's diagonal method we can prove that it also dense in
topology determined by family $p_n(\cdot)$. Which concludes the
proof.

\end{proof}

\begin{remark}

The relation $\succ$ is a quasi-order relation, i.e.
$\mathfrak{C}\succ\mathfrak{B}$ and $\mathfrak{B}\succ\mathfrak{A}$
imply that $\mathfrak{C}\succ\mathfrak{A}$.
\end{remark}
Indeed if
$\psi_1:$pro-$C^*(\mathfrak{C})\mapsto$pro-$C^*(\mathfrak{K}_1\otimes
\mathfrak{B})=\mathfrak{K}_1\hat\otimes$pro-$C^*(\mathfrak{A})$ 
 and 
$\psi_2:$pro-$C^*(\mathfrak{B})\mapsto$pro-$C^*(\mathfrak{K}_2\otimes
\mathfrak{A})=\mathfrak{K}_2\hat\otimes$pro-$C^*(\mathfrak{A})$  
have  dense range in every representation then composed homomorphism
$(id_{\mathfrak{K}_1}\otimes\psi_2)\circ\psi_1\colon$
pro-$C^*(\mathfrak{C})\mapsto\mathfrak{K}_1\hat\otimes\mathfrak
{K}_2\hat\otimes$pro-$C^*(\mathfrak{A})$
has also dense range in every representation and
$\mathfrak{K}_1\otimes\mathfrak{K}_2$ is also algebra of compact
operators. 
\begin{remark}
If $\mathfrak{B}\succ\mathfrak{A}$ then if $\pi_1$ and $\pi_2$ are 
distinct representation of $\mathfrak{A}$ in the same space $H$ then representation $F_\psi(\pi_1)$
and $F_\psi(\pi_1)$  are also distinct.
\end{remark}
Indeed, let $\psi:$pro-$C^*(\mathfrak{B})\to$pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})$ be 
morphism with quasidense range and let $\pi_1,\ \pi_2\in\rep(\mathfrak{A})$ be distinct representations in the same space $H$. Let $\hat\pi_i=id\otimes\pi_i,\ i=1,2$ be corresponding 
representations of $\mathfrak{K}\otimes\mathfrak{A}$ in $H_0\otimes H$. It is obvious that 
$\hat\pi_1\neq\hat\pi_2$. By definition 
of enveloping algebra they can be lifted to distinct representation $\widetilde{\pi_1}, \widetilde{\pi_2}$ of pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})$. 
Denote $\mathcal{A}=$pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})$. Define  for every $x\in\mathcal{A}:\  p(x)=max(\Vert\pi_1(x)\Vert, \Vert\pi_2(x)\Vert)$ 
then $p(\cdot)$ is a $C^*$-norm on $\mathcal{A}$ and we denote by $\mathcal{A}_p$ completion 
of   $\mathcal{A}$ in this norm. If $j$ denote natural morphism from 
$\mathcal{A}$ to $\mathcal{A}_p$ then there are representations $\pi_{i,p}\in\rep(\mathcal{A}_p),\ i=1,2$ and we have  commutative diagrams 
 \begin{center}
\resetparms
\qtriangle[\mathcal{A}`\mathcal{A}_p`B(H_0\otimes H);j`\widetilde{\pi}_i
`\pi_{i,p}]
\end{center}
Since $\psi$ has quasidense range and $\phi(\mathfrak{B})$ is 
quasidense in pro-$C^*(\mathfrak{B})$ (where $\phi:\mathfrak{B}\to$pro-$C^*(\mathfrak(B)))$ is natural morphism)
we have  that $*$-subalgebra $\mathcal{B}_p=j\circ\psi\circ\phi(\mathfrak{B})$ 
is quasidense  in $\mathcal{A}_p$ and 
since the last is $C^*$-algebra quasi-density implies  density.
If $F_\psi(\pi_1)=F_\psi(\pi_2)$ then $\pi_{1,p}=\pi_{2,p}$ on dense 
$*$-subalgebra $\mathcal{B}_p$ and as such must coincide. Then 
from commutative diagram $\widetilde{\pi_1}=\widetilde{\pi_2}$ which 
contradicts assertion.

\subsection{$*$-Wildness of $*$-algebras} \label{sec:3.1.2}

1. In the theory
of representations of algebras, it was suggested \cite{?} to consider the
representation problem to be wild if it contains a standard difficult
problem of representation theory, i.e. the problem to describe, up to
similarity, a pair of matrices without relations. To define an
analogue of wildness for *-algebras (*-wildness), it was suggested
[1] to
choose, for a standard difficult problem in the theory of
*-representations, the problem of describing pairs of self-adjoint
(or
unitary) operators up to a unitary equivalence (free *-algebra
$\mathfrak{S}_2$ (or $\mathfrak{U}_2$) generated by a pair of
self-adjoint (or unitary) generators ) and there where indications
that suggested to regard the problems which contain the standard
*-wild problems as *-wild, -- one can prove that these problems
contain as a subproblem the problem of describing *-representations
of
any affine *-algebra.

2. The following theorem holds.

\begin{theorem}\label{th_o}
	$\mathfrak{S}_2 \succ \mathfrak{S}_m$ for any $m=1$,
	$2$,\dots.
\end{theorem}

\begin{proof}
	For the algebra $\tilde \mathfrak{S}_m$, take the algebra
$\mathfrak{S}_m=\Bbb{C}\langle b_1, \dots, b_m \mid b_i=b_i^*,\,
i=1, \dots, m\rangle$ itself, $n=m+2$.  Define a homomorphism $\psi
\colon \mathfrak{S}_2 \to M_{m+2}(\mathfrak{S}_m)$ as follows:

\begin{align*}
	\psi(a_1) & =
	\begin{bmatrix}
	e           &&&&&&\\
		&\frac12 e  &&&&\rlap{\smash{\Large0}}&\\
 		&&\frac13e   &&&&\\
 		&&&\ddots     &&&\\
 		&&&&\frac1m e  &&\\
 		&&\llap{\smash{\Large0}}&&&\frac1{m+1}e&\\
 		&&&&&&\frac1{m+2}e
	\end{bmatrix}
	\\ \notag
	\psi(a_2) & =
	\begin{bmatrix}
		0   &e   &b_1 &      &  &   &\\
		e   &0   & e  &b_2   &  &\rlap{\smash{\Large0}}&\\
		b_1 &e   & 0  &e     &\ddots&&\\
    		&b_2 & e  &\ddots&  &b_{m-1}&\\
    		&    &\ddots&    &0 & e &b_m\\
		&\llap{\smash{\Large0}}&&b_{m-1}&e & 0 &e  \\
		&    &    &      &b_m& e&0
	\end{bmatrix}.
\end{align*}

One can directly check that the functor $F_\psi$ is full.
\end{proof}

Theorem~\ref{th_o} permits to say that the problem of unitary
classification of pairs of  self-adjoint operators contains, as a
subproblem, the problem of unitary classification of representation
of any $*$-algebra with a countable number of generators (because it
is always possible to  choose these generators to be self-adjoint).

\begin{corollary}
	$\mathfrak{S}_2 \succ \mathfrak{U}_m$ for any $m=1$, $2$,
	$3$, \dots.
\end{corollary}

\begin{theorem}\label{th_t}
	$\mathfrak{U}_2 \succ \mathfrak{S}_2$.
\end{theorem}

\begin{proof}
	Choose the enveloping algebra $\tilde \mathfrak{S}_2$ to  be
the quotient algebra of the algebra $\mathfrak{S}_2$ with respect to
the set
%\begin{gather*}
$$
\begin{array}{c}
	\Sigma = \{ a-ie, a+ie, b-ie, b+ie\}, \quad n=1,
	\\
	\psi(u_1) = (a-ie)(a+ie)^{-1}, \quad \psi(u_2) =
	(b-ie)(b+ie)^{-1}
\end{array}
$$
%\end{gather*}
(the Cayley transformation). The rest is obvious.
\end{proof}

{\bf 3} Theorems~\ref{th_o}, \ref{th_t} allow, as a model
of complexity for problems of unitary classification of
representations of $*$-algebras,  to choose the problem of unitary
classification of representations of the algebra $\mathfrak{U}_2$ or,
which is the same thing, enveloping $C^*$-algebra
$C^*(\mathcal{F}_2)$, where $\mathcal{F}_2$ is a free group with
two generators.

\begin{definition}
	A $*$-algebra $\mathfrak{A}$ is called $*$-wild if
	$\mathfrak{A} \succ C^*(\mathcal{F}_2)$.
\end{definition}

{\bf 4} Theorem~\ref{?} immediately implies the following statement.

\begin{proposition}
	A $C^*$-algebra $\mathfrak{A}$ is $*$-wild if and only if there
	exist $n\in \Bbb N\cup \{\infty \}$ and a $C^*$-ideal $\mathfrak
	I\subset \mathfrak{A}$ such that
	$\mathfrak{A}/\mathfrak{I}\approx
	M_n(C^*({\mathcal F}_2)$.
	 \end{proposition}

5.	Nuclear $*$-algebras have only hyper-finite factor
	representations and, consequently, they can not be $*$-wild.
	There also exist non-nuclear $C^*$-algebras which are not
	wild. For example, the group $C^*$-algebra $C^*(B(m,2))$ of
	the Burnside group with two generators and sufficiently
	large odd $m$ is not nuclear, but it is also not
	wild (see Section \ref{3.1.6}).

\subsection{$*$-Wild algebras generated by orthogonal
projections and idempotents} \label{sec:3.1.3}

{\bf 1}	For representations of the $*$-algebra $\mathfrak{P}_2
= \Bbb{C}\langle p_1, p_2 \mid p_1^* =p_1=p_1^2, p_2^*
=p_2=p_2^2\rangle$ (a pair of orthogonal projections $P_1$, $P_2$)
there is a structure theorem
that gives a decomposition of representations into a direct sum (or
integral) of irreducible representations which are either
one-dimensional or two dimensional, and their description, up to a unitary
equivalence (see Section~\ref{1.2.2}).

{\bf 2.} The problem to describe, up to a unitary
equivalence, a family of orthogonal projections $P_1$, $P_2$,
\dots,~$P_n$ for $n\ge3$ is $*$-wild. We give a fairly simple proof
that this problem is $*$-wild for $n=3$.

\begin{theorem}
	Let
	$$
	{\mathcal P}_3=\Bbb C\langle p_1,p_2, p_3\,|\, p_i^2=p_i^*=p_i\
	(i=1,2,3)\rangle.
	$$
	Then ${\mathcal P}_3\succ C^*({\mathcal F}_2)$, i.e. ${\mathcal
	P}_3$
	is
	$*$-wild.
\end{theorem}

\begin{proof}
	Let us define the homomorphism $\psi\colon {\mathcal P}_3 \to
	M_4({\mathcal F}_2)$ as follows:

\begin{align*}
	\psi(p_1)&=
		\begin{pmatrix}
			e & 0 & 0 & 0\\
			0 & e & 0 & 0\\
			0 & 0 & 0 & 0\\
			0 & 0 & 0 & 0
		\end{pmatrix},
		\\
	\psi(p_2)&=
		\begin{pmatrix}
			\frac 12 e & 0 & \frac 12 e & 0\\
			0 & \frac 12 e & 0 & \frac 12 e\\
			\frac 12 e & 0 & \frac 12 e & 0\\
			0 & \frac 12 e & 0 & \frac 12 e
		\end{pmatrix},
		\\
	\psi(p_3)&=
		\begin{pmatrix}
			\frac 38 e & \frac {\sqrt 3}8 uv^* & \frac
				{\sqrt3}8 u & \frac 38 e\\
			\frac {\sqrt3}8 vu^* & \frac 58 e & \frac  38
				v & -\frac {\sqrt3}8 vu^*\\
			\frac {\sqrt3}8 u^* & \frac 38 v^* & \frac 14
				e & 0\\
			\frac 38 e & -\frac {\sqrt3}8 uv^* & 0 &
				\frac 34 e
		\end{pmatrix}.
\end{align*}
It is easy to check that the corresponding functor $F_\psi\colon \rep
C^*({\mathcal F}_2) \to  \rep {\mathcal P}_3$ is full.
\end{proof}

{\bf 3.} Moreover, the following theorem holds [1].
\begin{theorem}\label{th_f}
	Let
	\[
		\mathcal{P}_{3,\perp2} = \mathbb{C}\langle p, p_1, p_2 \mid
p^* = p, p^2 = p, p_i^* = p_i, p_i^2 = p_i, p_1p_2 =p_2p_1=0\rangle.  \] Then
	$\mathcal{P}_{3,\perp2} \succ C^*(\mathcal{F}_2)$, i.e.
	$\mathcal{P}_{3,\perp2}$ is $*$-wild.  \end{theorem}

\begin{proof}

Let
\begin{gather*}
	E_k =\left[\smash[b]{\underbrace{%
	\begin{matrix}
		e&&0\\ &\ddots&\\ 0&&e
	\end{matrix}
	}_{\text{$k$ times}}}\right],
	\qquad \text{$e$ is the identity in the algebra
	$C^*(\mathcal{F}_2)$,}
	\\[5mm]
	J_1 = \begin{bmatrix}
		E_4\\0_{3\times4}\\0_{5\times4}
	\end{bmatrix}, \quad J_2 =
	\begin{bmatrix}0_{4\times3}\\E_3\\ 0_{5\times 3}
	\end{bmatrix}, \quad
	J_3= \begin{bmatrix}A_1\\A_2\\A_3 \end{bmatrix},
\end{gather*}
where
\[
	A_1 = \frac1N\begin{bmatrix}
	e & 0 & 0&0 & 0 \\
	0 &e  &0 &0 & 0\\
	0&0&2e &0&0\\
	0&0&0&3e&0
	\end{bmatrix}, \quad
	A_2 = \frac1N\begin{bmatrix}
	e& 0       & e&  e & e\\
	0        & 2e& e & u_1 & 0 \\
	0        & 0       & e &0         &  u_2
	\end{bmatrix},
\]
$u_1$, $u_2$ are generators of the algebra $C^*(\mathcal{F}_2)$, $A_3
= \sqrt{E_5 - A_1^*A_1 - A_2^* A_2}$. $N$ is chosen so that
$\|A_1^*A_1 + A_2^* A_2\| < 1$ in $M_5(C^*(\mathcal{F}_2))$. Then
$J_1^*J_1
= E_4$, $J_2^*J_2 = E_3$, and $J_3^*J_3 = E_4$. This implies that
$(J_iJ_i^*)^2 = J_iJ_i^*$, $i=1$, $2$, $3$. Moreover, since $J_1^*
J_2 =0$ and $J_2J_1^*=0$, we see that
\[
	(J_1J_1^*)(J_2J_2^*) = (J_2J_2^*)(J_1J_1^*) =0.
\]
Set $\psi(p_i) = J_iJ_i^*$, $i=1$, 2, $\psi(p) = J_3 J_3^*$ ;
$\psi$ defines a homomorphism of the
$*$-algebra $\mathcal{P}_{3,\perp2}$
into $M_{14}(C^*(\mathcal{F}_2))$. One can
directly check that the functor $F_\psi \colon \rep(\mathcal{P}_{3,\perp2})
\to\rep(C^*(\mathcal{F}_2))$ is full.
\end{proof}

\begin{corollary}
	The problem of a unitary classification of ``all but one''
	orthogonal projections $p$, $p_1$, \dots , $p_n$ ($p_i
	p_j=0$ for $i\neq j$) is $*$-wild if $n\ge2$:
\[
\mathcal{P}_{n+1,\perp n} =
\mathbb{C} \langle p, p_1, \dots, p_n | p, p_i \text{are orthoprojections},
p_i p_j=0, i \ne j\rangle \succ C^*(\mathcal{ F}_2).
\]
 \end{corollary}

{\bf 4.}	The problem of unitary classification of quadruples of
	orthogonal projections $p_1$, $p_2$, $p_3$, $p_4$ such that
	$$
		\alpha(p_1 + p_2 + p_3 + p_4)=I,\qquad 0<\alpha<1,
	$$
	for a fixed $\alpha\neq \frac 12$, has only a finite number
	of irreducible representations, the dimension of which
	depends on the parameter $\alpha$ (see Section \ref{2.2.1}). If
	$\alpha=\frac 12$, then irreducible representations are only
	in dimensions one and two.

It directly follows from Theorem~\ref{th_f} that the problem of
	unitary classification of five orthogonal projections $r_1$,
	$r_2$, $r_3$, $r_4$, $r_5$ such that $r_1 + r_2 + r_3 + r_4 +
	r_5=2I$ is $*$-wild: 
\[
\mathcal{R}_{5,2} =  \mathbb{C}\langle
r_1, \dots, r_5 \mid r_i^2 = r_i = r_i^*, i=1, \dots, 5, \sum
r_i = 2e \rangle \succ \mathcal{P}_{3, \perp 2}.
\]

Indeed, let $\psi \colon \mathcal{R}_{5,2} \to \mathcal{P}_{3, \perp 2}$ be
defined by formulas
\begin{gather*}
\psi(r_1) = p, \quad \psi(r_2) = e-p,
\\
\psi(r_3) = p_1, \quad \psi(r_4) = p_2, \quad \psi(r_5) = e - p_1 - p_2.
\end{gather*}
One can directly check that the functor $F_\psi \colon \rep(\mathcal{P}_{3,
\perp 2}) \to \rep (\mathcal{R}_{5,2})$ is full.

{\bf 5}	For a single idempotent, the situation is similar to the
situation for two orthogonal projections.  Consider the $*$-algebra
$\mathcal{Q}_1$ generated by a pair of an idempotent and its adjoint
$q_1$, $q_1^*$.  Let $q_1= a_1 + ib_1$, $q_1^* = a_1 - ib_1$, where
$a_1^* = a_1$, $b_1^* = b_1$. The $*$-algebra $\mathcal{Q}_1$
coincides with the algebra
	\[
		\mathbb{C} \langle a,b \mid a=a^*, b= b^*;
		\{a,b\} = ab + ba =0, a^2 - b^2 =e\rangle,
	\]
where $a= 2\bigl(a_1 -\frac12 e\bigr)$, $b= 2b_1$.

Irreducible representations of the algebra $\mathcal{Q}_1$ (see
Section~\ref{1.2.2}),
	up to a unitary equivalence, coincide with one of the
	following:

$1)$ two one-dimensional representations given by
	$\pi_0(q_1) = 0$ and $\pi_1(q_1) = 1;$

$2)$ a family, depending  on a parameter $\alpha>0$, of two
	dimensional representations:
	\begin{equation*}
	\pi_\alpha(q_1) =
		\begin{pmatrix}1 &\alpha \\ 0 & 0 \end{pmatrix}.
	\end{equation*}

By decomposing a representation of the algebra
$\mathcal{Q}_1$ into the direct sum of irreducible representations
on
a finite dimensional space $H$, we obtain a structure theorem
(see~[17,~18]) for the unitary description of idempotents in the
finite dimensional  case.

There is a structure theorem that gives a
description of any bounded idempotent on any separable Hilbert
space  in the form of an integral of irreducible
representations~[4].

{\bf 6} Consider the problem of a unitary description of pairs of
idempotents $Q_1$, $Q_2$ ($Q_1^2 = Q_1$, $Q_2^2= Q_2$). The fact that
the problem of a unitary description of pairs of idempotents is
difficult is just a mathematical folk. We will prove a
corresponding theorem and show that, even if an additional
restriction of self-adjointness is imposed on one of the idempotents
(one of the idempotents is an orthogonal projection), the problem
does not become easer.

\begin{theorem}\label{th_fv}
	Let $\mathcal{Q}_2 = \mathbb{C} \langle q_1, q_2 \mid
	q_1^2 = q_1, q_2^2 = q^2 \rangle$, $\mathcal{D}_{1,1} =
	\mathbb{C} \langle q, p \mid q^2 = q, p^2 = p = p^*
	\rangle$, $\mathfrak{S}_2 = \mathbb{C}\langle a_1, a_2 \mid
	a_1 = a_1^*, a_2 = a_2^* \rangle$. Then $\mathcal{Q}_2
	\succ \mathcal{D}_{1,1} \succ \mathfrak{S}_2$, so that the
	$*$-algebras $\mathcal{Q}_2$, $\mathcal{D}_{1,1}$ are $*$-wild.
\end{theorem}

\begin{proof}
	Because $\mathcal{D}_{1,1}$ is a factor algebra of the algebra
$\mathcal{Q}_2$, we have that $\mathcal{Q}_2\succ \mathcal{D}_{1,1}$ (we
choose an enveloping algebra for $\mathcal{D}_{1,1}$ to be the algebra
$\mathcal{D}_{1,1}$ itself, $n=1$, $\psi \colon \mathcal{Q}_2
\to \mathcal{D}_{1,1}$ is the natural epimorphism of the algebra onto the
factor algebra).

Let us show that $\mathcal{D}_{1,1}\succ \mathfrak{S}_2$.
Construct the homomorphism $\psi \colon \mathcal{D}_{1,1}\to
M_2(\mathfrak{S}_2)$:
\[
	\psi(q) = \begin{pmatrix} e& a_1 + ia_2\\ 0&0\end{pmatrix},
	\quad \psi(p) = \frac12 \begin{pmatrix}
	e&e\\e&e\end{pmatrix}.
\]
It is easy to check that the corresponding functor $F_\psi \colon
\rep \mathfrak{S}_2 \to \rep\mathcal{D}_{1,1}$ is full.
\end{proof}

\begin{corollary}
	The algebra $\mathcal{Q}_n$, for $n\ge2$ (the problem of
	unitary description of $n$ idempotents if $n\ge2$) is
	$*$-wild.
\end{corollary}

{\bf 7} Finally, we show that the $*$-algebra $\mathcal{Q}_{n,
\perp}$ (the problem of unitary classification of a family of
pairwise orthogonal idempotents $Q_1$, $Q_2$, \dots,~$Q_n$, $Q_iQ_j=
0$ for $i\ne j$) is $*$-wild for $n\ge2$.

\begin{theorem}
	Let
	$$
		\mathcal{Q}_{2, \perp} = \mathbb{C} \langle q_1,
		q_2 \mid q_1^2 = q_1, q_2^2 = q_2,\,   q_1q_2 =q_2q_1
		=0 \rangle.
	$$
	Then $\mathcal{Q}_{2, \perp} \succ \mathfrak{S}_2$, i.e.
	$\mathcal{Q}_{2, \perp}$  is a wild $*$-algebra.
\end{theorem}

\begin{proof}
	Let us define a homomorphism $\psi \colon \mathcal{Q}_{2,\perp}
\to M_3(\mathfrak{S}_2)$ as follows:
\[
	\psi(q_1) = \begin{bmatrix} e&e&a_1 + ia_2\\
	0&0&0\\0&0&0\end{bmatrix}, \quad \psi(q_2) =\begin{bmatrix}
	0&-e&-e\\0&e&e\\0&0&0\end{bmatrix}.
\]
One can directly check that $[\psi(q_k)]^2 = \psi(q_k)$, $k=1$, $2$,
$\psi(q_1) \, \psi(q_2) = \psi(q_2)\, \psi(q_1) =0$, and that the
functor $F_\psi \colon \rep \mathfrak{S}_2 \to \rep \mathcal{Q}_{2,
\perp}$ is full.
\end{proof}

\begin{corollary}
	The problem of unitary classification of pairs of commuting
	idempotents is  $*$-wild.
\end{corollary}

\begin{corollary}
	The $*$-algebra $\mathcal{Q}_{n,\perp} = \mathbb{C} \langle
	q_1, \dots, q_n \mid q_i^2 = q_i, i=1, \dots, n; \, q_iq_j
	=0 \text{ for }i\ne j \rangle$ (the problem of unitary
	classification of $n$ pairwise orthogonal idempotents) is
	$*$-wild for $n\ge2$.
\end{corollary}

\begin{corollary}
	The $*$-algebra $\mathbb{C}\langle q_1, \dots, q_n \mid q_i^2 =
	q_i, \, i=1,\dots, n; \, q_1 + q_2 + \dots + q_n =e\rangle$
	(the problem of unitary classification of $n$ idempotents
	$Q_1$, \dots,~ $Q_n$ such that $Q_1 + \dots + Q_n = I$) is
	$*$-wild for $n \ge3$.
\end{corollary}

\begin{proof}
	If $m=3$, the condition $q_1+q_2+q_3 =e$ implies that the
idempotents $q_1$, $q_2$, $q_3$ are pairwise orthogonal. Then the
algebra under consideration coincides with the
algebra $\mathcal{Q}_{2,\perp}$.
\end{proof}

\subsection{$*$-Wild semi-linear relations} \label{sec:3.1.4}

{\bf 1}   In Sections \ref{sec:1.3.2}-\ref{sec:1.3.5}
we studied the representations of the
semi-linear relations. In particular, it is was studied the
structure of the operators pair $A=A^*$, $B=B^*$ which satisfy to
the semi-linear relations
\begin{equation}\label{poly}
\sum_{k=1}^{n}f_k(A)Bg_k(A)=0
\end{equation}

For this relation corresponds the characteristic function
\begin{equation}
\Phi(t,s)= \sum_{k=1}^{n}f_k(t)g_k(s)=0
\end{equation}
(we suppose that $\Phi(t,s)=\overline{\Phi(s,t)}$), ($t,s \in
{\Bbb R}$). If the graph $({\Bbb R}^1, \Gamma
=\{(t,s)\in{\Bbb R}^2: \Phi(t,s)=0\})$ has only the connected
components of the form \onepointx, \earx or \edgex
then the irreducible representations of the
relation   (\ref{poly})  are one-dimensional and two-dimensional
and it was described in 1.3.5.

{\bf 2} We show that all other relations are $*$-wild.

\begin{proposition}
If the graph of semi-linear relation  (\ref{poly}) contains the
subgraphs   or   then  the relation is $*$-wild.
 \end{proposition}
 \begin{proof}
It is  supposed that the functions $f_k(\cdot)$ and $g_k(\cdot)$
are polynomial. We prove that $*$-algebra $\mathfrak A_{\Gamma}=
{\Bbb C}\langle a=a^*,b=b^* \mid \sum_{k=1}^{n}f_k(a)bg_k(a)=0
\rangle$ is $*$-wild if the $\Gamma \subset$ $(\lambda_1,
\lambda_2 \in {\Bbb R}$, $\lambda_1\ne \lambda_2$).

We give the $*$-homomorphism  $\psi \colon \mathfrak A_{\Gamma} \to
M_3(\mathfrak S_2)$ by the following way:
\begin{gather*}
\psi(a)=\begin{pmatrix} \lambda_1 e& 0 & 0\\
						 0 & \lambda_1 e&0\\
						 0&0&\lambda_2 e
		  \end{pmatrix}, \qquad
\psi(b) =\begin{pmatrix} a_1&e&e\\
						  e&a_2&0\\
						  e&0&0
		  \end{pmatrix}
\end{gather*}

It is easy to check that the functor $F_{\psi}$ is full.
\end{proof}

\begin{proposition}
If the graph $\Gamma$ of the semi-linear relation (\ref{poly})
contains the subgraph
($\lambda_1\ne \lambda_2 \ne\lambda_3\ne\lambda_1$; $ \lambda_1,
\lambda_2,\lambda_3\in {\Bbb R}$) than the relation is $*$-wild.
\end{proposition}

\begin{proof}
We prove that $*$-algebra $\mathfrak A_{\Gamma}=
{\Bbb C}\langle a=a^*,b=b^* \mid \sum_{k=1}^{n}f_k(a)bg_k(a)=0
\rangle$ is $*$-wild if the $\Gamma \subset$ $(\lambda_1,
\lambda_2 \in {\Bbb R}$, $\lambda_1\ne \lambda_2$).

We give the $*$-homomorphism  $\psi\colon \mathfrak A_{\Gamma} \to
M_7(\mathfrak S_2)$ by the following way:
\begin{gather*}
\psi(a)=\begin{pmatrix} \lambda_1 e& 0 & 0&0&0&0&0\\
						0&\lambda_1 e&0&0&0&0&0\\
						 0&0& \lambda_2 e&0&0&0&0\\
						 0&0&0&\lambda_2 e&0&0&0\\
						 0&0&0& 0&\lambda_2 e&0&0\\
						 0&0&0&0&0&\lambda_3 e&0\\
						 0&0&0&0&0&0&\lambda_3 e
\end{pmatrix}, \\
\psi(b) =\begin{pmatrix} 0& 0 & e&e&a_1+ia_2&0&0\\
						 0&0& 0&e&e&0&0\\
						 e&0&0&0&0&e&0\\
						 e&e&0& 0&0&0&2e\\
						 a_1-ia_2&e&0&0&0&0&0\\
						 0&0&e&0&0&0&0\\
						 0&0&0&2e&0&0&0
		  \end{pmatrix}
\end{gather*}

It is easy to check that the functor $F_{\psi}$ is full.
\end{proof}

\subsection{$*$-Wild quadratic and cubic relations}\label{sec:3.1.5}

{\bf 1.} The relation $(I_0)$ $0=0$  defines the standard
	wild $*$-algebra $\mathfrak S_2 =   \mathbb{C}\langle a,b \mid
a^* =a, b^* =b\rangle$.  By the theorem, the
$*$-representations theory of it contains $*$-representations of
every finitely generated $*$-algebra.

{\bf 2.} The relation $(I_1)$ $a^2=e$  defines the
	 $*$-algebra $ \mathbb{C}\langle a,b \mid
a^* =a, b^* =b, a^2=e\rangle$.

\begin{proposition}
The $*$-algebra $\mathfrak D= \mathbb{C}\langle a,b \mid
a^* =a, b^* =b, a^2=e\rangle$ is $*$-wild.
\end{proposition}
\begin{proof}
We will  show that    $\mathfrak D= \mathbb{C}\langle a,b
\mid a^* =a, b^* =b, a^2=e\rangle \succ \mathcal
P_{3,\perp2}=\mathbb{C}\langle p_1, p_2, p_3 \mid p_i^* = p_i, p_i^2 =
p_i, p_1p_2 =p_2p_1=0\rangle$.  To show it, we give  the
$*$-homomorphism $\psi:\mathfrak D \to \mathcal P_{3,\perp2}$ as
follows:
\begin{eqnarray*}
 \psi(a)=  p-(e-p)=2 p-e\\
\psi(b)=p_1+\frac {1}{2} p_2 + \frac{1}{3} (e-p_1-p_2)
\end{eqnarray*}
It is easy to check that the corresponding functor
$F_{\psi}: \rep \mathcal P_{3,\perp2} \to \rep \mathfrak D$ is
full.
\end{proof}

{\bf 3} Now we give a criterion
in terms of  coefficients, when a quadratic  $*$-algebra  $\frak A
=\Bbb C
\langle
a, b \mid a=a^*,\quad b=b^*,\quad
P_2(a,b)=  \alpha a^2 + \beta b^2 + \hbar /i\left [a,b\right ] +
  \gamma \{a,b\} +\delta a + \epsilon b  + \chi I = 0.
\rangle$ ($ \alpha,\beta , \hbar,\gamma,\delta  \epsilon, \chi \in
{\Bbb
R}^1$).
is $*$-wild.
    \begin{theorem}
  A $*$-algebra $\frak A$ is a $*$-wild if and only if there are
  one of the following conditions:
  \begin{enumerate}
   \item
$ \alpha=\beta=\gamma=\hbar=\delta=\epsilon=\chi=0$
\item
   $ (\chi-\frac {\delta^2}{4\alpha})\alpha<0$,
  $\beta=\gamma=\hbar=\epsilon=0$;
  \item
  $   (\chi-\frac {\epsilon^2}{4\beta})\beta<0$,
  $\alpha=\gamma=\hbar=\delta=0$;
   \item
   $\gamma^2=\beta\alpha \ne 0,
  \quad
  \alpha(\chi-\frac{\delta^2}{4\alpha})<0$, $
    \frac{\delta^2}{\alpha}=\frac{\epsilon^2}{\beta} $,
   $\hbar=0$.
   \end{enumerate}
    \end{theorem}

\begin{proof}.
  The $*$-algebra with two self-adjoint variables and quadratic
  relations is wild iff there exists a change of variables such that
  the algebra can be transform to the $*$-algebra $\mathfrak S_2 =
  \Bbb C
  \langle a, b \mid a=a^*,\quad b=b^*\rangle$ or $*$-algebra
  $\mathfrak D
  = \Bbb C \langle a, b \mid a=a^*,\quad b=b^*, \quad a^2=e \rangle$.
Such quadratic $*$-algebras one can define by imposing one of the
conditions 1-4.
   \end{proof}

{\bf 4} Now we consider the pair of self-adjoint operators which
satisfy to the cubic semi-linear relations:

\begin{equation}\label{cub}
\epsilon\{A^2,B\}+i\delta[A^2,B]+2\mu ABA
+i\gamma[A,B] +2\beta \{A,B\}+\alpha B=0
\end{equation}

For the relation (\ref{cub}) corresponds
a $*$-algebra $\mathfrak A_3 ={\Bbb C}\langle a,b \mid a=a^*,b=
b^*, \qquad \epsilon\{a^2,b\}+i\delta[a^2,b]+2\mu aba
+i\gamma[a,b] +2\beta \{a,b\}+\alpha b\rangle$.

Then the corresponding characteristic
function is following:  \begin{eqnarray}\label{qpline}
\Phi(t,s)&=\epsilon (t^2+s^2) + i\delta(t^2-s^2)+
 2\mu ts  +
i(\gamma)(t-s)+ 2\beta\{a,b\}+\alpha b
  \end{eqnarray}
($\alpha, \beta, \gamma, \delta \in {\Bbb R}^1 $);

It follows from general theory of semi-linear relations
that  the corresponding $*$-algebra is $*$-wild if and only
 the equation $\Phi(t,s)=0$ has  either two solutions of the next
form ($t_1,t_1$), ($t_1,t_2$), where $t_1 \ne t_2$ either two
solutions of the next form $(t_1,t_2)$, $(t_1,t_3)$, where
$t_1,t_2,t_3$ are different.

The equation $\Phi(t,s)=0$ decompose by natural way to the system
of two equations
\begin{gather*}
  \left \{ \begin{array}{lll} \Phi_1(t,s)&=&\epsilon t^2+2\mu ts +
\epsilon s^2 + 2\beta t + 2\beta s + \alpha=0 \\ \Phi_2 (t,s)& =&
\delta (t^2 - s^2)+ \gamma (t-s)=0 \end{array} \right.
\end{gather*}

At first we assume that  $\delta = \gamma =0$
($\Phi_2(t,s)\equiv 0$). The equation $\Phi_1(t,s)=0$
defines the curve of the second order. These curves have next
invariants:

\begin{gather*}
I_1=2\epsilon;\\
I_2= \begin{vmatrix} \epsilon &\mu \\
					 \mu & \epsilon
	   \end{vmatrix}=\epsilon^2 -\mu^2; \\
I_3=   \begin{vmatrix} \epsilon &\mu &\beta\\
					 \mu & \epsilon  &\beta \\
					  \beta & \beta &\alpha
	   \end{vmatrix}=(\epsilon^2 -\mu^2)\alpha
-2\beta^2(\epsilon-\mu)
 \end{gather*}
	   If $I_2\ne 0$ then the equation $\Phi_1(t,s)=0$ define  a
central curve. By affine transformation of variables the equation
one can bring to the following form:
\begin{equation}\label{tri}
(\epsilon +\mu){\hat t}^2+(\epsilon -\mu){\hat s}^2+
\frac{I_3}{I_2}=0
 \end{equation}

a) Let $I_2>0$

If $I_3=0$ then the solutions set of equation (\ref{tri}) (it
means that $\Phi(t,s)=0$ too) has only one point. Hence The
$*$-algebra $\mathfrak A_3$ is not wild.

If $I_3>0$ then the solutions set of (\ref{tri}) is empty.
Therefore the  $*$-algebra $\mathfrak A_3$ is not wild.

If $I_3<0$ then the equations (\ref{tri}) is an ellipse equation.
Therefore there are two solutions $(t_1, t_2)$, $(t_1,t_3)$ such that
$t_2\ne t_3$. Hence the $*$-algebra $\mathfrak A_3$ is $*$-wild.

b) Let $I_2<0$ then the equation  $\Phi(t,s)=0$ has a hyperbolic
type.

If $I_3=0$ then we have a pair of intersecting straight line. Hence
it is easy to see that the $*$-algebra $\mathfrak A_3$ is
$*$-wild.

If $I_3\ne 0$ then we have a hyperbola equation. This equation
 have not more then one solution only in case that the asymptotes
are orthogonal and parallel to the coordinate axis.  That is
case when $\epsilon =0$.  Hence, if the $\epsilon \ne 0$ then the
$*$-algebra $\mathfrak A_3 $ is wild.

c) Let $I_2=0$.  Then the equation $\Phi_1(t,s)=0$ has a parabolic
type (the curve is not central).

Let $I_3=0$ then $\epsilon=\mu$ and
\begin{gather}
\Phi_1(t,s)=\epsilon(t+s)^2+2\beta(t+s)+\alpha=0
\end{gather}

If $\beta^2-\alpha\epsilon<0$ then the equation $\Phi_1(t,s)=0$
have not solutions and the $*$-algebra $\frak A_3$ is not wild.

If  $\beta^2-\alpha\epsilon<0$  then $t+s=-frac{\beta}{\epsilon}$
 and it is obviously that the $*$-algebra $\frak A_3$ is not wild.

If $\beta^2-\alpha\epsilon>0$ then the equation $\Phi_1(t,s)=0$
describe a pair of parallel lines. Hence the  solutions set
contains a two solution $(t_1,t_2)$ , $(t_1,t_3)$ such that $t_2
\ne t_3)$. Therefore the $*$-algebra $\mathfrak A_3$ is $*$-wild.

If $I_2=0$, $I_3\ne0$ then the equation   $\Phi_1(t,s)=0$
is a parabola equation. In that case, if
$\beta^2-\epsilon\alpha-4\beta t >0$ we have two solutions of the
equation $\Phi_1(t,s)=0$ for one value $t$ and the $*$-algebra
$\mathfrak A_3$ is $*$-wild.

Now we consider the case  $\Phi_1(t,s)\equiv 0$, $\delta
\ne 0$.
\begin{equation}
\Phi_2(t,s)=(t-s)(\delta t+\delta s +\gamma)=0
\end{equation}
In that case the equation $\Phi_2(t,s)=0$ defines  the pairs of
intersecting and non coinciding lines. Therefore the $*$-algebra
$\mathfrak A_3$ is $*$-wild.

Now we consider the  case $\Phi_1(t,s)\Phi_2(t,s) \ne 0$.
It is obviously that if $\Phi_1(t,s)=0$ or $\Phi_2(t,s)=0$
generate the non $*$-wild $*$-algebras then
 \begin{equation}\label{sis}
   \left \{ \begin{array}{lll}
  \Phi_1(t,s)&=&0 \\
  \Phi_2 (t,s)& =& 0
\end{array}\right.
 \end{equation}
  corresponds a non $*$-wild $*$-algebra too.
Therefore  we should study the case $\Phi_1(t,s)=0$ and
  $\Phi_2(t,s)=0$ generate a $*$-wild $*$-algebras. It is
  following that $\delta(\epsilon^2+\mu^2)\ne 0$.

By affine change of variables the system (\ref{sis}) one can
  bring to the form
  \begin{gather*}
  \left \{ \begin{array}{lll}\tilde \Phi_1(\tilde
  t,\tilde s)&=&\epsilon {\tilde t}^2+2\mu{\tilde t}{\tilde s}
  +\epsilon {\tilde s}^2+2\tilde {\beta}\tilde s +\tilde {\alpha} =0
  \\
  \tilde \Phi_2(\tilde t, \tilde s) & =& {\tilde t}^2-{\tilde s}^2)=0
\end{array}\right.
\end{gather*}
  where $\tilde
  \beta = \beta - (\epsilon +\mu)\frac{\gamma}{2\delta}$, $\tilde
  \alpha = \alpha +(\epsilon +\mu)\frac{\gamma^2}{2\delta^2} -
  2\beta\frac{\gamma}{\delta}$)
   The solutions set is the intersection of $\tilde \Phi_1(\tilde
  t, \tilde s)=0$ and of the lines pair $\tilde s =\tilde p$ and
  $\tilde s= -\tilde t$.

Therefore,  it is necessary and sufficiently for the $\frak
  A_3$ to be $*$-wild, that the equation $\tilde \Phi(\tilde
  t,\tilde s)=0$ have two solutions of the next form $(\tilde
  t_0,\tilde t_0)$ and $(\tilde t_0,-\tilde t_0)$.
	It is easy to show  that such condition be fulfilled
	in one of the next case
\begin{itemize}
 \item
$\epsilon>0$, $\mu>0$, $2\delta\beta-\epsilon\gamma=0 $,
$\alpha\delta^2-\epsilon\gamma^2<0$;
\item
$\mu\ne 0$, $2\delta\beta-\epsilon\gamma\ne 0 $,
$\alpha\epsilon^2-I_3 +\frac{\epsilon\gamma}{\delta}\cdot
\frac{(\epsilon + \mu )\gamma-4\beta\delta}{2\delta}=0$
\end{itemize}
So, we prove the next theorem:

\begin{theorem} The $*$-algebra $\mathfrak A_3 = {\Bbb C}\langle a,b
\mid a=a^*,b=
b^*, \qquad \epsilon\{a^2,b\}+i\delta[a^2,b]+2\mu aba
+i\gamma[a,b] +2\beta \{a,b\}+\alpha b\rangle$ (here $\epsilon \geq
0$,
$\epsilon^2+\mu^2+\delta^2\ne 0$.
is $*$-wild if and
  only if:
 \begin{enumerate}
 \item if $\delta=\gamma=0$, then there are one
of the following conditions:
\begin{enumerate}
 \item $I_1>0$,
 $I_2>0$, $I_3<0$
\item
$I_2<0$, $I_3=0$
\item
$I_1>0$, $I_2<0$, $I_3\ne 0$
\item $I_2=0$, $I_3=0$, $ b^2-2ac>0 $
\item $I_2=0$,
 $I_3\ne 0$
\end{enumerate}
 \item if $\epsilon=\mu=\beta=\alpha=0$, then $\delta\ne 0$.
\item  If $\delta(\epsilon^2+\mu^2) \ne 0$ then there are one of the
following
conditions:
 \begin{enumerate}
  \item
 ether $\epsilon > 0$, $\mu=0$, $ 2\delta \beta-\epsilon\gamma =0$,
$\alpha\delta^2-\epsilon\gamma^2<0$
  \item
  $mu\ne 0$, $2\delta\beta-\epsilon\gamma \ne 0$, $\alpha
  \epsilon^2-I_3 +
\frac{\epsilon\gamma}{\delta}\cdot \frac{\epsilon+\mu)-\gamma -
4\beta\delta}{2\delta}=0$
 \end{enumerate}
\end{enumerate}
\end{theorem}

\textbf{5}. Passing to (non-semi-linear) cubic relations, we restrict ourselves
with the following proposition.

\begin{proposition}
$*$-Algebra 
\[
\mathcal{B}_2 = \mathbb{C} \langle a = a^*, b^* \mid aba = bab \rangle
\]
is $*$-wild.
\end{proposition}

\begin{proof}
Let homomorphism $\psi \colon \mathcal{B}_2 \to M_4(\mathfrak{S}_2)$ be defined
as follows
\[
\psi(a) = \begin{pmatrix}\begin{matrix}0&e\\e&0\end{matrix} & 0\\0 &0
\end{pmatrix}, \quad \psi(b) = \begin{pmatrix}0&\begin{matrix}e&0\\0&0
\end{matrix} \\ \begin{matrix}e&0\\0&0\end{matrix} & \begin{matrix}
a_1 & e \\ e & a_2\end{matrix}
\end{pmatrix}
\]
Then the functor $F_\psi$ is full.
\end{proof}

\begin{remark}
$*$-Wild $*$-algebra $\mathcal{B}_2 = \mathbb{C}\langle a= a^*, b= b^* \mid aba
= bab\rangle$ is obtained by introducing involution $x= x^*$, $y = y^*$ on the
algebra $\mathbb{C}\langle x, y \mid xyx = yxy \rangle$.

This involution is not unique. If one introduce involution on this algebra by
$x^\star = y$, the obtained $*$-algebra is $\mathfrak{C}_2
= \mathbb{C} \langle x,
x^\star \mid x x^\star x = x^\star x x^\star \rangle$. The $*$-algebra
$\mathfrak{C}_2$ is not $*$-wild (see Section~\ref{3.2.1}).

\end{remark}

\subsection{$*$-Wild groups. Periodic groups are not $*$-wild}
\label{sec:3.1.6}
In this section we study the complexity of description of 
unitary representations 
($*$-representations of group algebras) for some discrete countable 
groups $G$. For groups $G_1$ and $G_2$, for which $C^*(G_1) \succ 
C^*(G_2)$, we will write below $G_1 \succ G_2$. If $C^*(G)$ is 
$*$-wild, then the description of unitary representations of such group 
contains as a subproblem the description of representations of any 
countable group; in what follows, in the book we will call them 
$*$-wild (from the viewpoint of complexity of their unitary 
representations).

Below, we give a number of examples of both $*$-wild groups, and groups 
which are not $*$-wild.

1. Let us give a number of examples of $*$-wild groups.

\begin{example}
It follows directly from the given above list of $*$-wild $*$-algebras, 
that the groups $\mathcal{F}_n$, $n\ge 2$, and $\mathbb{Z}_n * 
\mathbb{Z}_m$, $n \ge 2$, $m \ge 3$ are $*$-wild.
\end{example}

2. The following simple statement holds.

\begin{proposition}
If a group $G$ contains a normal subgroup $N$ such that $G/N = G_1$, 
then $G \succ G_1$.
\end{proposition}

\begin{proof}
Denote by $\phi \colon G \to G_1$ the mapping which maps the element $g 
\in G$ its conjugacy class $\phi(g) in G_1 = G/N$. Then, introduce a 
unitary $*$-homomorphism $\psi \colon L(G) \ni f(\cdot) \to \psi(f) 
(g_1) = \sum_{g \colon \phi(g) = g_1} f(f) \in L(G_1) \subset C^* 
(G_1)$, and use the same notation $\psi$ for its unique  extension to 
unital $*$-homomorphism from $C^*(G)$ to $C^*(G_1)$. It is clear 
that the corresponding functor $F_\phi \colon \rep C^*(G_1) \to \rep 
C^*(G)$ is full.
\end{proof}

The proposition above implies the following statement.

\begin{corollary}
A group $G$ possessing a normal subgroup $N$ such that $G/N = 
\mathcal{F}_2$, is $*$-wild.
\end{corollary}

\begin{example}
Extension $G$ of any group $G_1$ by $\mathcal{F}_2$, is a $*$-wild 
group.
\end{example}

3. Since the majorization  $\succ$ is a quasi-order relation (see 
Section~\ref{3.1.1}), the following corollary from Proposition~\ref{??} 
holds.

\begin{corollary}
A group $G$ such that $G/N = G_1$ is $*$-wild, is $*$-wild itself.
\end{corollary}

\begin{example}
The group $SL(2,\mathbb{Z})$ is $*$-wild, since $SL(2,\mathbb{Z})/\{e, 
-e\} = PSL(2, \mathbb{Z}) = \mathbb{Z}_2 * \mathbb{Z}_3$.
\end{example}

\begin{example}
The braid group $B_2$ is $*$-wild, since its group $*$-algebra is
$\mathbb{C}[B_2] = \mathbb{C} \langle u,v \mid \text{$u$, $v$ are 
unitary}, uvu = vuv\rangle = \mathbb{C} \langle w = uvu, z = uv \mid 
\text{$w$, $z$ are unitary}, w^2 = z^3\rangle$, and its quotient 
algebra is $\mathbb{C} \langle w,z \mid \text{$w$, $z$ are unitary}, w^2 = 
z^3 = e\rangle = \mathbb{C} [\mathbb{Z}_2 * \mathbb{Z}_3]$.
\end{example}

\begin{remark}
It looks attractive to investigate, whether are $*$-wild the following 
groups:

a) Coxeter groups which are not affine;

b) $SL(n, \mathbb{Z})$, $n \ge 3$;

c) hyperbolic groups;

d) groups with a single generating relation,

\noindent
and many other known groups, which are not amenable.
\end{remark}

4. Consider some examples of groups which are not $*$-wild.

If $G$ is amenable, then $C^*(G)$ is nuclear, and cannot be $*$-wild. 
Thus, we have the following statement.

\begin{proposition}
If $G$ is an amenable group, then $G$ is not $*$-wild.
\end{proposition}

Also, the following theorem holds.

\begin{theorem}
Let $G$ be a periodic group, i.e., any element $g \in G$ is periodic. 
Then $G$ is not $*$-wild.
\end{theorem}

\begin{proof}
Suppose that $G$ is $*$-wild, i.e., there exists a homomorphism $\phi 
\colon G \to U_n (C^*(\mathcal{F}_2))$ such that the functor 
$F \colon \rep \mathcal{F}_2 \to \rep G$ is full. Consider a family of 
one-dimensional representations $h_t$ of the group $\mathcal{F}_2 = 
\langle u, v\rangle$ in the space $\mathbb{C}$ such that $h_t(u) =1$, 
$h_t(v) = e^{it}$, $t\in (0, 2\pi]$, and denote by $U_t(g) = \hat h_t 
\circ \phi(g)$, $g \in G$, the matrix with entries being continuous 
functions in $t$. Since the functor $F$ is full, the representations 
$\hat h_{t_1} \circ \phi$ and $\hat h_{t_2} \circ \phi$ of 
the group $G$, are unitarily inequivalent for all $t_1 \ne t_2$. Since 
the irreducible representations of the group $G$ in a 
finite-dimensional space  are uniquely defined, up to a unitary 
equivalence, by their characters (see, for example, \cite{kirillov}), 
there exists $g\in G$, such that $\tr U_{t_1}(g) \ne \tr U_{t_2}(g)$, 
Then $\tr U_t(g)$ is a continuous function in $t$, which is not a 
constant. Order the eigenvalues $k_i(t)$, $i=1$, \dots,~$n$, of the 
matrix $U_t(g)$ by the value of the argument. Then there exists such 
$i$, that $k_i(t) \ne \const$. Since for a unitary matrix the problem 
of finding eigenvalues is stable, the eigenvalue $k_i(t)$ is a 
nontrivial continuous function of $t$ on some interval $(t_1, t_2)$. 
But then there exists  $t_0$, such that $k_i(t_0)$ will not be a root 
of unity. Since $G$ is a periodic group, for each $g \in G$
there exists a power $N(g)$ such that $g^{N(g)} =e$; but on the other 
hand, $U_{t_0}^{N(g)} \ne 1$. The obtained contradiction completes the 
proof.
\end{proof}

\begin{corollary}
It follows from the theorem, that there are groups, which are not 
$*$-wild, and not amenable. Those are, for example, Burnside groups 
$B(m,n)$, which are not amenable for 
odd $n \ge 665$, and $m \ge 2$ (see \cite{adjan}).
\end{corollary}