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\section {$*$-Wild algebras and relations}\label{sec:3.1} \markright{3.1. $*$-Wild algebras and relations} Before passing to the complexity problem of unitary description of representations for concrete classes of $*$-algebras, we give definitions and some results concerning the ideology and methodology of majorization of $*$-algebras and $*$-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 semilinear relations, $*$-wild group algebras, etc. We emphasize once again that the fact that some algebra is $*$-wild implies that the problem of unitary description of {\em all} representations is very complicated. However, the problem of construction and description of different classes of representations becomes even more attractive. Below in this chapter, for a $*$-algebra $\mathfrak{A}$ we will write $\rep \mathfrak{A}$, instead of $*$-$\rep\mathfrak{A}$, to simplify the notations. \subsection{Majorization of $*$-algebras with respect to the 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 $(\tilde{\mathfrak{A}},\phi\colon\mathfrak{A}\to\tilde {\mathfrak{A}})$, where $\tilde{\mathfrak{A}}$ is a $*$-algebra and $\phi $ is a $*$-homomorphism, is called an enveloping $*$-algebra if for any $*$-representation $\pi\colon\mathfrak{A}\to L(H)$ of the algebra $\mathfrak{A}$ there exists a unique $*$-representation $\tilde{\pi}\colon\tilde{\mathfrak{A}}\to L(H)$ such that the diagram \begin{center} \resetparms \btriangle[\mathfrak{A}`\tilde{\mathfrak{A}}`L(H) ;\phi `\pi `\tilde{\pi}] \end{center} is commutative. \medskip\noindent\textbf{2.} To verify whether or not a pair $(\mathfrak{A}, \phi\colon\mathfrak{A}\to\tilde{\mathfrak{A}})$ determines an enveloping algebra, it is only sufficient to consider unital representations of the algebra $\mathfrak{A}$. Indeed, let for any unital representation $\pi$ of $\mathfrak{A}$ there exist a representation $\tilde{\pi}$ of $\tilde{\mathfrak{A}}$ making the following diagram commutative \begin{center} \resetparms \btriangle[\mathfrak{A}`\tilde{\mathfrak{A}}`L(H) ;\phi `\pi `\tilde{\pi}] \end{center} Suppose that $\pi\colon\mathfrak{A}\to L(H)$ is not a 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 a unital representation of $\mathfrak{A}$. Since the homomorphism $\phi$ is unital, we have $\pi^{(1)}(e)=\tilde{\pi}^{(1)}(\phi(e))$, and \[ \tilde{\pi}(a)= \begin{pmatrix} \tilde{\pi}^{(1)}(a) & 0\\ 0 & 0 \end{pmatrix} \] is the needed homomorphism from $\mathfrak{A}$ to $\tilde{\mathfrak{A}}$. Since $\tilde{\pi}(e)=\tilde{\pi}(\phi(e))=E$, any representation of $\tilde{\mathfrak{A}}$ that makes the previous diagram commutative, coincides with the one presented above. Thus $\tilde{\pi}$ is determined uniquely. \medskip\noindent\textbf{3.} If $(\tilde{\mathfrak{A}}, \phi)$ is an enveloping algebra of the algebra $\mathfrak{A}$ and $(\tilde{\mathfrak{A}}_1, \psi)$ is an enveloping algebra of $\tilde{\mathfrak{A}}$, then $(\tilde{\mathfrak{A}}_1, \psi\circ\phi)$ is an enveloping algebra of the algebra $\mathfrak{A}$. The proof is evident. \medskip\noindent\textbf{4.} Now we introduce the relation of majorization for $*$-algebras. Denote by $\mathfrak{K}$ the algebra of compact operators in a separable (possibly finite-dimensional) Hilbert space $H_0$. The algebra $\mathfrak{K} \otimes \mathfrak{A}$ will occasionally be denoted by $M_n(\mathfrak{A})$, $n \in \mathbb{N} \cup \{\infty\}$. Let $\pi\colon\mathfrak{A}\to L(H)$ be a representation of $\mathfrak{A}$. It induces a representation, \[ \hat\pi=\id\otimes\pi\colon \mathfrak{K}\otimes\mathfrak{A}\to L(H_0\otimes H), \] of the algebra $\mathfrak{K}\otimes\mathfrak{A}$. If $(\widetilde{\mathfrak{K}\otimes\mathfrak{A}}, \phi)$ is an enveloping $*$-algebra of $\mathfrak{K}\otimes\mathfrak{A}$ then, by definition, $\hat\pi$ uniquely determines the representation $\tilde{\pi}$ of the enveloping algebra $\widetilde{\mathfrak{K}\otimes\mathfrak{A}}$ in the same Hilbert space $L(H_0\otimes H)$. Now let $\psi$ be a homomorphism of the algebra $\mathfrak{B}$ into the algebra $\widetilde{\mathfrak{K}\otimes\mathfrak{A}}$; then \[ \tilde{\pi}\circ\psi\colon\mathfrak{B}\mapsto L(H_0\otimes H) \] is a representation of $\mathfrak{B}$. \begin{lemma} Let $(\widetilde{\mathfrak{K}\otimes\mathfrak{A}}, \phi)$ be an enveloping $*$-algebra of $\mathfrak{K}\otimes\mathfrak{A}$ and $\pi_j\in\rep(\mathfrak{A})$, $j=1$, $2$. Let $\tilde\pi_j$ denote its lifting to the enveloping $*$\nobreakdash-algebra as described above. If $V\in L(H)$ intertwines $\pi_1$ and $\pi_2$, then $I\otimes V$ intertwines $\tilde\pi_1$ and $\tilde\pi_2$ $(I$ is the identity operator in $L(H))$. \end{lemma} \begin{proof} We will prove the lemma for $V$ intertwining one representation $\pi$, since, by setting $$ \pi= \begin{pmatrix} \pi_1&0\\ 0&\pi_2\\ \end{pmatrix}, $$ and $$ V= \begin{pmatrix} 0&U\\ U&0\\ \end{pmatrix} , $$ we obtain the result for $U$ intertwining $\pi_1$ and $\pi_2$. Let $\mathcal{A}$ denote the $C^*$-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 a $C^*$-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, the representations $\pi_1\circ\tilde{\pi}$ and $\pi_2\circ\tilde{\pi}$ of $\widetilde{\mathfrak{K}\otimes\mathfrak{A}}$ are liftings of the representation $\pi_1\circ\tilde{\pi}\circ\phi=\pi_2\circ\tilde{\pi}\circ\phi$, since $\mathcal{B}$ is the closure of $\hat\pi(\mathfrak{K}\otimes\mathfrak{A}) =\tilde{\pi}(\phi(\mathfrak{K}\otimes\mathfrak{A}))$. By the definition of enveloping $*$-algebra, such a lifting is unique, hence $\pi_1=\pi_2$. By Lemma~\ref{lemma2}. in Section~\ref{sec:1.1.3}, $\mathcal{B}=\mathcal{A}$. By the assumption, $I\otimes V\in \hat\pi(\mathfrak{K}\otimes\mathfrak{A})'$, so $I\otimes V\in \mathcal{B}'$, hence $I\otimes V\in \mathcal{A}'$, and consequently, $ I\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 the algebra $\mathfrak{K}\otimes\mathfrak{A}$, and a unital homomorphism $\psi\colon\mathfrak{B}\to\widetilde{\mathfrak{K}\otimes\mathfrak {A}}$ such that the functor $F\colon \rep \mathfrak{A}\to \rep \mathfrak{B}$ defined by the following rule\textup: \begin{align*} F(\pi)&=\tilde{\pi}\circ\psi, \qquad \text{for any $\pi\in \rep \mathfrak{A}$}, \\ F(A)&=I\otimes A, \qquad \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 the previous lemma that, in order to verify that $F$ is full, it is sufficient to verify that, for every $\pi\in\rep(\mathfrak{A})$ in $L(H)$ and any operator $A$ in $L(H)$, the inclusion $I\otimes A\in F(\pi)(\mathfrak{B})'$ implies $I\otimes A\in \pi(\mathfrak{A})'$. \end{remark} For some comments on Definition~\ref{majoriz}, see also Section~\ref{sec:3.2.3} below. \medskip\noindent\textbf{5.} For a class of\/ $C^*$-algebras, it is possible to only consider the homomorphisms $\psi\colon \mathfrak{B} \to \mathfrak{K}\mathbin{\hat\otimes}\mathfrak{A}$ in Definition~\ref{majoriz}, since for a $C^*$\nobreakdash-algebra $\mathfrak{A}$, $\mathfrak{K}\mathbin{\hat\otimes}\mathfrak{A}$ is also a $C^*$-algebra, and the unique enveloping algebra of a $C^*$-algebra is the algebra itself ($\mathfrak{K}\mathbin{\hat\otimes}\mathfrak{A}$ is the completion of the algebraic tensor product in the $C^*$-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}) $. Assuming that $ \mathfrak{B}_1 \neq \mathfrak{A}$, we have that there is a 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 the proof of Lemma~3.9 in \cite{kru_wor}) a representation $\pi\colon \mathfrak{A} \to L(H)$ and a finite-dimensional operator $\rho \in L(H)$ 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. Hence, for every $a \in \mathfrak{A}$ there is a 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\/ $\mathfrak{B}/ \mathfrak{I}\simeq \mathfrak{K}\otimes\mathfrak{A}$. \end{theorem} \begin{proof} Let us note that any representation $\pi $ of $\mathfrak{K}\mathbin{\hat\otimes}\mathfrak{A}$ has the 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 $ (see, e.g., \cite{take79}). 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 $. Setting $\pi_0=\bigoplus_\alpha {\pi_\alpha}_0 $ and $U=\bigoplus_\alpha U_\alpha $ we get $ \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 an arbitrary $\pi \in \rep(\mathfrak{K}\otimes\mathfrak{A}) $. Then $ \pi=U( \id \otimes \pi_0) U^* $. Let $V \in \pi(\psi(\mathfrak{B}))'$, then $U^*VU \in (\id \otimes \pi_0(\psi(\mathfrak{B})))'$, and so $U^*VU=I \otimes V_0$, 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. This proves that $U^*VU \in (\id \otimes \pi_0(\mathfrak{A}))'$, and so $V \in \pi(\mathfrak{A})'$. Now we can apply the previous lemma to finish the proof. \end{proof} In what follows, the $C^*$-algebra $\mathfrak{K} \otimes \mathfrak{A}$ will also be occasionally denoted by $M_n(\mathfrak{A})$, $n\in \mathbb{N} \cup \{\infty\}$. \medskip\noindent\textbf{6.} Now consider the general case of arbitrary $*$-algebras. The following lemma is a simple modification of Theorem~6.3.5 in~\cite{murphy}. \begin{lemma} Let\/ $\mathfrak{A}$ be a unital\/ $*$-algebra and let\/ $\mathcal{B}$ be any $C^*$-algebra. Let\/ $\pi\in\rep(\mathcal{B}\otimes\mathfrak{A})$ be a representation in $L(H)$. Then there are representations $\phi\colon\mathfrak{A}\to L(H)$ and $\psi\colon\mathcal{B}\to L(H)$ 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 $\psi_{a^\prime}\colon\mathcal{B} \to L(H)$, $ b\mapsto\pi(b\otimes a')$. Let us show that $\psi_{a^\prime}$ is continuous for all $a^\prime\in\mathfrak{A}$. Since $\psi_{a^\prime}$ is a mapping between Banach spaces, we can apply the 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 $L(H)$, then $c=0$. Replacing $b_n$ with $b_n^*b_n$, $a^\prime$ with $(a^{\prime})^*a^\prime$, and $c$ with $c^*c$, we can assume that $c\ge 0$. Let $\tau$ be an arbitrary positive functional on $L(H)$. Then the functional \[ \rho\colon\mathcal{B}\to\mathbb{C},\qquad b\mapsto\tau(\pi(b\otimes a^\prime)) \] is positive and, therefore, 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. Consider the linear subset $\mathcal{L}=\pi(\mathcal{B}\otimes\mathfrak{A})H$. Every $z\in \mathcal{L}$ 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 $z=\sum_{j=1}^{m}\pi(b_i^\prime\otimes a_i^\prime)(x_i^\prime)$ be another presentation. Let $v_\mu$ be an 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 pass to the 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 the mapping $\phi(a)\colon \mathcal{L}\to \mathcal{L}$, $ 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 $\Vert \pi(b\otimes a)\Vert \le M\Vert b\Vert$, $b\in \mathcal{B}$. So $\phi(a)$ is a bounded operator. Since $\pi$ is non-degenerate, $\mathcal{L}$ is dense in $H$ and $\phi(a)$ is uniquely extended to a bounded operator on $H$ (which is also denoted by $\phi(a)$). Then $\phi\colon\mathfrak{A}\to L(H)$ and $\psi=\psi_{1}$ ($a^\prime=1$) are the required representations. Below we use the notations $\phi=\pi_{\mathfrak{A}}$ and $\psi=\pi_{\mathcal{B}}$ \end{proof} \begin{proposition} Let $\mathfrak{A}$ be a unital $*$-algebra. Then \begin{itemize} \item[\textup{1.}] for every $\pi\in\rep(\mathfrak{K}\otimes\mathfrak{A})$ in $L(H)$ there are unitary $U\in L(H)$ and representation $\pi_0\in\rep(\mathfrak{A})$ in $L(H_1)$ such that $H=H_0\otimes H_1$ and $U^*\pi U=\id\otimes\pi_0$. \item[\textup{2.}] $\pro C^*(\mathfrak{K}\otimes\mathfrak{A}) \simeq\mathfrak{K}\mathbin{\hat\otimes} \pro C^*(\mathfrak{A})$, where $\mathfrak{K}\mathbin{\hat\otimes}\pro C^*(\mathfrak{A})$ is a unique,since $\mathfrak{K}$ is nuclear, $\pro C^*$-algebra which is the completion of the algebraic tensor product \textup(see~\textup{\cite{15})}. \end{itemize} \end{proposition} \begin{proof} 1. By the previous lemma, $\pi=\pi_{\mathfrak{K}}\otimes \pi_{\mathfrak{A}}$. Denote by $\mathcal{A}=\overline{\pi_{\mathfrak{A}}(\mathfrak{A})}$ the $C^*$-subalgebra in $L(H)$ generated by the range of $\pi_{\mathfrak{A}}$. Let $j\colon\mathcal{A}\to L(H)$ denote the 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 \Vtriangle[\mathfrak{K}\otimes\mathfrak{A}`\mathfrak{K}\otimes \mathcal{A}`L(H);\id\otimes\pi_{\mathfrak{A}}`\pi `\pi_{\mathfrak{K}}\otimes j] \end{center} Since $\mathfrak{K}$ is of type $I$, the representation $ \pi_{\mathfrak{K}}\otimes j=U^*\id\otimes\hat\pi U$, where $\hat\pi\in\rep(\mathcal{A})$ (see the proof of Theorem~\ref{bound}). So $\pi=U^*\id\otimes\pi_0 U$, where $\pi_0=\hat\pi\restriction\mathfrak{A}$. 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 $\phi\colon\mathfrak{A}\to\pro C^*(\mathfrak{A})$ the canonical homomorphism. By the definition of an enveloping algebra, the representation $\pi_0$ admits a unique lifting $\tilde\pi_0\in\rep(\pro C^*(\mathfrak{A}))$ in $L(H_1)$. Then define the representation $\tilde{\pi}\in\rep(\mathfrak{K}\mathbin{\hat\otimes}\pro C^*(\mathfrak{A}))$ via the rule $\tilde{\pi}(k\otimes a)=U^*\id\otimes \tilde\pi_0U$. Since $\phi(\mathfrak{A})$ is quasi-dense in pro-$C^*(\mathfrak{A})$, the algebra $\mathfrak{K}\otimes\phi(\mathfrak{A})$ is quasi-dense in $\mathfrak{K}\mathbin{\hat\otimes}\pro C^*(\mathfrak{A})$. Thus the lifting $\tilde{\pi}$ is unique. \end{proof} Now we are in a position to prove the main theorem about majorization. \begin{theorem}\label{th:main_major} \textup 1. If $(\tilde{\mathfrak{A}},\phi)$ is an enveloping $*$-algebra of \/ $\mathfrak{A}$, then $\pro C^*(\mathfrak{A})\simeq\pro C^*(\tilde{\mathfrak{A}})$. \textup 2. $*$-Algebra $\mathfrak{B}$ majorizes a $*$-algebra $\mathfrak{A}$ if and only if there exists a continuous morphism $\psi \colon \pro C^*(\mathfrak{B})\to\pro C^*(\mathfrak{K}\otimes\mathfrak{A})$ with quasi-dense image \textup(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 $\im\pi$\textup). \textup 3. If $\pro C^*(\mathfrak{A})$ is a $\sigma$-$C^*$-algebra \textup(for example if \/ $\mathfrak{A}$ is finitely generated\textup), then $\mathfrak{B}\succ\mathfrak{A}$ if and only if there exists a continuous morphism $\psi \colon \pro C^*(\mathfrak{B})\to \pro C^*(\mathfrak{K}\otimes\mathfrak{A})$ with dense image. \end{theorem} \begin{proof} $1$. Let $\psi$ denote the canonical morphism form $\tilde{\mathfrak{A}}$ to pro-$C^*(\tilde{\mathfrak{A}})$. For every $\pi\in\rep(\mathfrak{A})$ there is a unique lifting $\tilde{\pi}\in\rep(\tilde{\mathfrak{A}})$ and, by the definition of the enveloping pro-$C^*$-algebra, $\tilde{\pi}$ is uniquely lifted to a representation $\hat\pi\in\rep(\pro C^*(\tilde{\mathfrak{A}}))$. Hence (pro-$C^*(\tilde{\mathfrak{A}}),\psi\circ\phi)$ is an enveloping pro-$C^*$-algebra of $\mathfrak{A}$. Then pro-$C^*(\mathfrak{A})\simeq\pro C^*(\tilde\mathfrak{A})$ by the uniqueness of the pro-$C^*$-enveloping algebra. $2$. a) Let $(\widetilde{\mathfrak{K}\otimes\mathfrak{A}}, \phi)$ be an enveloping $*$-algebra of $\mathfrak{K}\otimes\mathfrak{A}$. If $\pi\in\rep(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})$, then $\pi=U\tilde\pi_0 U^*$ for some $\pi_0\in\rep(\mathfrak{A}) $ and unitary $U\in L(H)$. Indeed, since the mapping $\rep(\mathfrak{K}\otimes\mathfrak{A})\ni\kappa\mapsto\tilde{\kappa} \in \rep(\widetilde{\mathfrak{K}\otimes\mathfrak{A}}) $ is bijective, $\pi=\tilde{\kappa}$ for some $\kappa\in\rep(\mathfrak{K}\otimes\mathfrak{A})$. By the previous proposition, $\kappa=U\id\otimes\pi_0U^*$, where $\id$ is the identical representation of $\mathfrak{K}$ in $L(H_0)$ and $\pi_0$ is a representation of $\mathfrak{A}$ in $L(H_1)$, $H=H_0\otimes H_1$. Notice that $U\tilde{\pi_0}U^*$ is a representation of $\widetilde{\mathfrak{K}\otimes\mathfrak{A}}$ which is a lifting of $\kappa$, and by the uniqueness of such a lifting, \[ U\tilde\pi_0U^*=\tilde{\kappa}=\pi. \] b). Assume that $\mathfrak{B}\succ\mathfrak{A}$, i.e., there exists a $*$-homomorphism $\psi \colon\mathfrak{B}\to \widetilde{\mathfrak{K}\otimes\mathfrak{A}}$ such that functor $F_\psi\colon \rep(\mathfrak{A})\to\rep(\mathfrak{B})$ is full. Denote $\mathcal{A}=\pro C^*(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})\simeq \pro C^*(\mathfrak{K}\otimes\mathfrak{A})$, and let $\phi \colon \widetilde{\mathfrak{K}\otimes\mathfrak{A}} \to \mathcal{A}$ be the canonical morphism. Then the functor $F\colon\rep(\mathcal{A})\to \rep(\mathfrak{B})$, $\rep(\mathcal{A})\ni\pi\mapsto\pi\circ\phi\circ\psi\in\rep(\mathfrak{ B})$, is also full. Indeed, $\pi\circ\phi\in\rep(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})$, so $\pi\circ\phi=U\tilde\pi_0U^*$ for some $\pi_0\in\rep(\mathfrak{A})$ in $L(H)$. Thus, \[ F(\pi)(\mathfrak{B})^\prime=F_{\psi}(\pi\circ\phi)(\mathfrak{B})^\prime= F_{\psi}(U\tilde\pi_0U^*)(\mathfrak{B})^\prime, \] $F_{\psi}(U\tilde\pi_0 U^*)(\mathfrak{B})^\prime=(U\tilde \pi_0(\mathfrak{B})U^*)^\prime= (U\tilde\pi_0(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})U^*)^ \prime$ since the functor $F_{\psi}$ is full. $(U\tilde\pi_0(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})U^*) ^\prime=\pi(\phi(\widetilde{\mathfrak{K}\otimes\mathfrak{A}}))^\prime = \pi(\mathcal{A})^\prime$, since $\phi(\widetilde{\mathfrak{K}\otimes\mathfrak{A}})$ is quasi-dense in $\mathcal{A}$. The morphism $\phi\circ\psi$ can be extended to a continuous morphism $\tau$ of the pro-$C^*(\mathfrak{B})$, since if $\phi_1\colon\mathfrak{B}\to\pro C^*(\mathfrak{B})$ denotes the canonical morphism, then $\phi_1(\mathfrak{B})$ is quasi-dense in pro-$C^*(\mathfrak{B})$, and the topology on $\phi_1(\mathfrak{B})$ induced from pro-$C^*(\mathfrak{B})$ (the projective topology induced by all representations of $\mathfrak{B}$) is stronger than the topology induced by the map $\phi\circ\psi$ (projective topology induced by those representations of $\mathfrak{B}$ which come through $\mathcal{A}$). Let us show that the range of $\tau$ is dense in every representation. Take an arbitrary representation $\pi\in \rep(\mathcal{A})$. Then the natural injection $j\colon\overline{\pi(\tau(\mathfrak{B}))}\to\pi(\mathcal{A})$ satisfies the conditions of the 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 a $\sigma$-$C^*$-algebra, then pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})$ is also a $\sigma$\nobreakdash-$C^*$-algebra. Therefore its topology can be determined by a countable 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 $C^*$-seminorm (each seminorm $p(\cdot)$ defines a 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 arguments similar to Cantor's diagonal method we can prove that it is also dense in the topology determined by the family $p_n(\cdot)$. This completes the proof. \end{proof} \begin{corollary} \label{cor:q-order} The relation $\succ$ is a quasi-order relation, i.e., $\mathfrak{C}\succ\mathfrak{B}$ and\/ $\mathfrak{B}\succ\mathfrak{A}$ imply $\mathfrak{C}\succ\mathfrak{A}$. \end{corollary} Indeed, if $\psi_1\colon\pro C^*(\mathfrak{C})\to\pro C^*(\mathfrak{K}_1\otimes \mathfrak{B})=\mathfrak{K}_1\mathbin{\hat\otimes}\pro C^*(\mathfrak{A})$ and $\psi_2\colon\pro C^*(\mathfrak{B})\to\pro C^*(\mathfrak{K}_2\otimes \mathfrak{A})=\mathfrak{K}_2\mathbin{\hat\otimes}\pro C^*(\mathfrak{A})$ have dense range in every representation, then composed homomorphism \[ (\id_{\mathfrak{K}_1}\otimes\psi_2)\circ\psi_1\colon \pro C^*(\mathfrak{C})\to\mathfrak{K}_1\mathbin{\hat\otimes}\mathfrak {K}_2\mathbin{\hat\otimes}\pro C^*(\mathfrak{A}) \] also has dense range in every representation and $\mathfrak{K}_1\mathbin{\hat\otimes}\mathfrak{K}_2$ is also the 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 the representation $F_\psi(\pi_1)$ and $F_\psi(\pi_2)$ are also distinct. \end{remark} Indeed, let $\psi\colon\pro C^*(\mathfrak{B})\to\pro C^*(\mathfrak{K}\otimes \mathfrak{A})$ be a morphism with quasi-dense range, and let $\pi_1$, $\pi_2\in\rep(\mathfrak{A})$ be distinct representations on the same space $H$. Let $\hat\pi_i=\id\otimes\pi_i$, $ i=1$, $2$ be the corresponding representations of $\mathfrak{K}\otimes\mathfrak{A}$ on $H_0\otimes H$. It is obvious that $\hat\pi_1\neq\hat\pi_2$. By the definition of an enveloping algebra, they can be lifted to distinct representations $\tilde\pi_1$, $\tilde\pi_2$ of pro-$C^*(\mathfrak{K}\otimes\mathfrak{A})$. Denote $\mathcal{A}=\pro C^*(\mathfrak{K}\otimes\mathfrak{A})$. For every $x\in\mathcal{A}$ define \[ p(x)=\max(\Vert\pi_1(x)\Vert, \Vert\pi_2(x)\Vert). \] Then $p(\cdot)$ is a $C^*$-norm on $\mathcal{A}$, and we denote by $\mathcal{A}_p$ the completion of $\mathcal{A}$ in this norm. If $j$ denotes the 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 the commutative diagram \begin{center} \resetparms \qtriangle[\mathcal{A}`\mathcal{A}_p`L(H_0\otimes H);j`\tilde{\pi}_i `\pi_{i,p}] \end{center} Since $\psi$ has quasi-dense range and $\phi(\mathfrak{B})$ is quasi-dense in pro-$C^*(\mathfrak{B})$ (where $\phi\colon\mathfrak{B}\to\pro C^*(\mathfrak(B)))$ is the natural morphism), we have that the $*$-subalgebra $\mathcal{B}_p=j\circ\psi\circ\phi(\mathfrak{B})$ is quasi-dense in $\mathcal{A}_p$, and since the latter is a $C^*$-algebra, the quasi-density implies density. If $F_\psi(\pi_1)=F_\psi(\pi_2)$, then $\pi_{1,p}=\pi_{2,p}$ on the dense $*$-subalgebra $\mathcal{B}_p$ and as such must coincide. Then it follows from the commutativity of the diagram that $\tilde\pi_1=\tilde\pi_2$, which contradicts the assertion. \begin{remark} Below in this section, we give a number of examples of $*$-algebras and mappings $\psi\colon \mathfrak{A} \to M_n(\mathfrak {B})$ such that the functor $F_\psi \colon \rep \mathfrak{B} \to \rep \mathfrak{A}$ is full. But we do not discuss here methods of construction of such mappings. This is a separate topic to be discussed elsewhere. We also do not discuss the question what is the minimal number $n$, for which there exists a homomorphism $\psi \colon \mathfrak{A} \to M_n(\mathfrak{B})$ such that the corresponding functor $F_\psi$ is full. We only notice that in \cite{ols_zame}, it was shown that for the $C^*$-algebra $\mathcal{A}$ with $m$ self-adjoint generators $a_1$, \dots, $a_m$, for $n\ge -3+\sqrt{9+8m}$, $M_n(\mathcal{A})$ is generated by a pair of self-adjoint generators. It is shown in paper \cite{rab_umz}, that this statement holds for $n\ge\sqrt{m-1}$, and that this estimate is exact, i.e., there exists a commutative $C^*$-algebra $C(K)$ with $m$ self-adjoint generators such that $M_n(C(K))$ is not singly generated for $n<\sqrt{m-1}$, i.e., $M_n(C(K))$ is not generated by a pair of self-adjoint elements. \end{remark} \subsection{$*$-Wildness of $*$-algebras} \label{sec:3.1.2} \textbf{1.} In the theory of representations of algebras, it was suggested \cite{89} to consider the representation problem to be wild if it contains the classical unsolved 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 \cite{krusam} to choose, for a standard $*$-wild problem in the theory of $*$-representations, the problem of describing pairs of self-adjoint (or unitary) operators up to a unitary equivalence (representations of the $*$-algebra $\mathfrak{S}_2$ (or $\mathfrak{U}_2$) generated by a pair of free self-adjoint (or unitary) generators) and to regard the problems which contain the standard $*$-wild problem as $*$-wild. \medskip\noindent\textbf{2.} One can prove that the standard $*$-wild problem contains as a subproblem the problem of describing $*$-representations of any finitely generated $*$-algebra. The following theorem holds. \begin{theorem}\label{th_o} $\mathfrak{S}_2= \mathbb{C}\langle a,b\mid a=a^*, b=b^* \rangle \succ \mathfrak{S}_m= \Bbb{C}\langle a_1, \dots, a_m \mid a_i=a_i^*,\, i=1, \dots, m\rangle$ for any $m=1$, $2$,\dots. \end{theorem} \begin{proof} For the algebra $\tilde \mathfrak{S}_m$, take the algebra $\mathfrak{S}_m$ 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) & = \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(b) & = \begin{bmatrix} 0 &e &a_1 & & & &\\ e &0 & e &a_2 & &\rlap{\smash{\Large0}}&\\ a_1 &e & 0 &e &\ddots&&\\ &a_2 & e &\ddots& &a_{m-1}&\\ & &\ddots& &0 & e &a_m\\ &\llap{\smash{\Large0}}&&a_{m-1}&e & 0 &e \\ & & & &a_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} For any $m=1$, $2$,~\dots, $\mathfrak{S}_2 \succ \mathfrak{U}_m = \mathbb{C} \langle u_1, \dots, u_m,\allowbreak u_1^*, \dots, u_m^* \mid u_i u_i^* = u_i^*u_i =e, i=1, \dots, m \rangle$. \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 algebra of fractions of the algebra $\mathfrak{S}_2$ (see, e.g., \cite{117}) with respect to the set $\Sigma = \{ a-ie, a+ie, b-ie, b+ie\}$. Define the homomorphism $\psi \colon \mathfrak{U}_2 \to \tilde \mathfrak{S}_2$ as follows: $$ \psi(u_1) = (a-ie)(a+ie)^{-1}, \quad \psi(u_2) = (b-ie)(b+ie)^{-1} $$ (the Cayley transformation). The rest is obvious. \end{proof} \noindent{\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, the enveloping $C^*$-algebra $C^*(\mathcal{F}_2)$, where $\mathcal{F}_2$ is the free group with two generators, $u$, $v$. \begin{definition}\label{def:14} A $*$-algebra $\mathfrak{A}$ is called $*$-wild if\/ $\mathfrak{A} \succ C^*(\mathcal{F}_2)$. \end{definition} For comments on Definition~\ref{def:14}, also see Section~\ref{sec:3.2.3} below. \medskip\noindent {\bf 4.} Theorem~\ref{bound} 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\/ $\mathfrak{A}/\mathfrak{I}\simeq M_n(C^*({\mathcal F}_2))$. \end{proposition} \noindent\textbf{5.} Nuclear $*$-algebras have only hyperfinite 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{sec:3.1.6}). \subsection{$*$-Wild algebras generated by orthogonal projections and idempotents} \label{sec:3.1.3} {\bf 1.} For representations of the $*$-algebra $\mathcal{P}_2 = \Bbb{C}\langle p_1, p_2 \mid p_1^* =p_1=p_1^2,\, p_2^* =p_2=p_2^2\rangle$ (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- or two-dimensional, and their description, up to a unitary equivalence (see Section~\ref{sec:1.2.2}). \medskip\noindent{\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\\[2pt] \frac {\sqrt3}8 vu^* & \frac 58 e & \frac 38 v & -\frac {\sqrt3}8 vu^*\\[2pt] \frac {\sqrt3}8 u^* & \frac 38 v^* & \frac 14 e & 0\\[2pt] \frac 38 e & -\frac {\sqrt3}8 uv^* & 0 & \frac 34 e \end{pmatrix}. \end{align*} It is easy to see that the corresponding functor $F_\psi\colon \rep C^*({\mathcal F}_2) \to \rep {\mathcal P}_3$ is full. \end{proof} \begin{remark} The $*$-algebra $\mathcal{P}_3$ coincides with the group $*$-algebra of the Coxeter group $\coxetertriangle{\infty}{\infty}{\infty}$, see Section~\ref{sec:1.2.2}, i.e., \[ \mathcal{P}_3 = \mathbb{C} \langle w_i = w_i^* = 2p_i -e, i=1, 2, 3 \mid w_i^2 =e,\, i=1, 2, 3\rangle. \] Thus, the group $C^*$-algebra of this group is $*$-wild (see also Section~\ref{sec:3.1.6}). \end{remark} \noindent{\bf 3.} Even with additional conditions imposed on the triple of orthogonal projections, $p_1$, $p_2$, $p_3$ (the triple of flips $w_i= 2p_i -e$, $i=1$, 2, 3), the problem of their description may still be $*$-wild. \begin{theorem}\label{th:anti-wild} The $*$-algebra \begin{align*} \mathcal{P}_{3,2\text{\textup{anti}}} = \mathbb{C} \bigl\langle w_i = w_i^*, i=1,2,3&\mid w_i^2 = e, i=1, 2, 3; \\* &\{w_1, w_2\} = w_1 w_2 + w_2 w_1=0\}\bigr\rangle \end{align*} is $*$-wild. \end{theorem} \begin{proof} Define the homomorphism $\psi $ of the $*$-algebra $\mathcal{P}_{3, 2\text{anti}}$ into $M_4(\mathfrak{U}_2)$, where $\mathfrak{U}_2 = \mathbb{C} \langle u, u^*, v, v^* \mid uu^* = u^*u =vv^* = v^*v =e\rangle$, as follows: \begin{align*} \psi(w_1) &= \begin{pmatrix} e&0&&\smash[b]{\text{\lower7pt\hbox to 0pt{\kern-15pt\Large0\hss}}} \\0& e \\&&e&0\\ \smash{\text{\raise5pt\hbox to 0pt{\kern5pt\Large 0\hss}}} &&0&-e \end{pmatrix}, \\ \psi(w_2)& = \begin{pmatrix} &&e&0\\ \smash{\text{\raise5pt\hbox to 0pt{\kern5pt\Large 0\hss}}} &&0&e\\e&0\\0&e& \smash{\text{\raise5pt\hbox to 0pt{\kern5pt\Large 0\hss}}} \end{pmatrix}, \\ \psi(w_3)& = \frac1{\sqrt3}\begin{pmatrix} e & 0 & u^*&v^* \\ 0 &-e & e & -uv^*\\ u &e & 0 & -uv^*\\ v&-vu^*& -vu^* &0 \end{pmatrix}. \end{align*} It is easy to check that the corresponding functor $F_\psi \colon \rep \mathfrak{U}_2 \to \rep \mathcal{P}_{3,2\text{anti}}$ is full. \end{proof} \noindent\textbf{4.} It is a more complicated fact that the group $*$-algebra of the Coxeter group $\coxetertwoedge{\infty}{\infty}$ (i.e., the $*$-algebra $\mathcal{P}_{3,2\text{comm}} = \mathbb{C} \langle p_1, p_2, p_3 \mid p_i ^* = p_i^2 = p_i,\, i=1,2,3;\, [p_1, p_2] =0 \rangle = \mathbb{C} \langle w_i = w_i^* = 2p_i -e,\allowbreak i=1,2,3 \mid w_i ^2 =e,\, i=1, 2,3; \,[w_1, w_2]=0 \rangle$) is also $*$-wild. \begin{theorem} The $*$-algebra $\mathcal{P}_{3, 2 \text{\textup{comm}}}$ is $*$-wild. \end{theorem} Moreover, even a stronger statement of the following theorem holds \cite{krusam}. \begin{theorem}\label{th_f} Let \begin{align*} \mathcal{P}_{3,\perp2} = \mathbb{C}\langle p, p_1, p_2 & \mid p^* = p, \,p^2 = p, \\ &\quad p_i^* = p_i,\, p_i^2 = p_i,\, p_1p_2 =p_2p_1=0\rangle. \end{align*} 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 the 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 $M_{12}(C^*(\mathcal{F}_2))$. 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\/ $n\ge2$\textup: \begin{align*} \mathcal{P}_{n+1,\perp n} = \mathbb{C} \bigl\langle p, p_1, \dots, p_n &\mid p, p_i \text{ are orthogonal projections}, \\ &\quad p_i p_j=0, i \ne j\bigr\rangle \succ C^*(\mathcal{ F}_2). \end{align*} \end{corollary} \noindent{\bf 5.} 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{sec:2.2.1}). If $\alpha=\frac 12$, then there is an uncountable family of irreducible representations, and their dimension is one or two (see Section \ref{sec:2.2.1}). It directly follows from Theorem~\ref{th_f} that {\em 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}: $\mathcal{R}_{5,2}\succ \mathcal{P}_{3, \perp 2}$, where \[ \mathcal{R}_{5,2} = \mathbb{C}\Bigl\langle r_1, \dots, r_5 \mid r_i^2 = r_i = r_i^*, i=1, \dots, 5; \sum_{i=1}^5 r_i = 2e \Bigr\rangle . \] Indeed, let $\psi \colon \mathcal{R}_{5,2} \to \mathcal{P}_{3, \perp 2}$ be defined by the 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 directly checks that the functor $F_\psi \colon{} \rep(\mathcal{P}_{3, \perp 2}) \to \rep (\mathcal{R}_{5,2})$ is full. It directly follows from Theorem~\ref{th:anti-wild} that {\em the following $*$-algebra is $*$-wild}, \begin{align*} \mathcal{R}_{5,\frac52}& = \mathbb{C}\Bigl\langle r_1, \dots, r_5 \mid r_i^2 = r_i = r_i^*,\, i=1, \dots, 5; \sum_{i=1}^5 r_i = \frac52\,e \Bigr\rangle \\ & = \mathbb{C} \Bigr\langle w_i = 2r_i -e, i=1,\dots, 5 \mid w_i = w_i^*, w_i^2 =e, \\ & \qquad\qquad\qquad \qquad \qquad \qquad \qquad i=1, \dots, 5; \sum_{i=1}^5 w_i =0 \Bigr\rangle. \end{align*} Indeed, its quotient $*$-algebra $\mathcal{R}_{5, \frac32+1} = \mathbb{C} \bigl\langle r_1, \dots, r_5 \mid r_i^2 = r_i^* = r_i,\, i=1, \dots, 5; \sum_{i=1}^3 r_i =\frac32 e,\, r_4+ r_5 = e \bigr\rangle = \mathbb{C} \bigl\langle w_i = 2r_i -e, i=1, \dots, 5 \mid w_i = w_i^*, w_i^2 =e, i=1, \dots, 5; \sum_{i=1}^3 w_1 =0, w_4 + w_5 =0 \bigr\rangle = \mathbb{C} \bigl\langle z_1 = w_1 + w_2, z_2 = \frac1{\sqrt3} (w_1 - w_2), z_3 = w_4 \mid z_k ^* = z_k,\, z_k^2 = e,\, k=1, \dots, 3; \{z_1, z_2\} =0 \bigr\rangle = \mathcal{P}_{3, 2\text{anti}}$ is $*$-wild. \medskip\noindent {\bf6.} For a single idempotent, the situation is similar to the situation for two orthogonal projections. Consider the $*$-algebra $\mathcal{Q}_1$ generated by 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 \[ \mathbb{C} \bigl\langle a,b \mid a=a^*, b= b^*; \{a,b\} = ab + ba =0,\, a^2 - b^2 =e\bigr\rangle, \] where $a= 2\bigl(a_1 -\frac12 e\bigr)$, $b= 2b_1$. Irreducible representations of the algebra $\mathcal{Q}_1$ (see Section~\ref{sec: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 a direct sum of irreducible representations on a finite dimensional space $H$, we obtain the structure theorem (see~\cite{112,113}) 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. \medskip\noindent {\bf 7.} Further, 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 folklore. 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 $\mathcal{Q}_2 = \mathbb{C} \langle q_1, q_2, q_1^*, q_2^* \mid q_1^2 = q_1, q_2^2 = q^2 \rangle$, $\mathcal{D}_{1,1} = \mathbb{C} \langle q,q^*, p \mid q^2 = q, p^2 = p = p^* \rangle$, $\mathfrak{S}_2 = \mathbb{C}\langle a_1, a_2 \mid a_1 = a_1^*, a_2 = a_2^* \rangle$. Then $\mathcal{Q}_2 \succ \mathcal{D}_{1,1} \succ \mathfrak{S}_2$, so that the $*$-algebras $\mathcal{Q}_2$ and $\mathcal{D}_{1,1}$ are $*$-wild. \end{theorem} \begin{proof} Because $\mathcal{D}_{1,1}$ is a quotient 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 quotient 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 $n\ge2$ \textup(the problem of unitary description of $n$ idempotents if $n\ge2$\textup), is $*$-wild. \end{corollary} \noindent {\bf 8.} 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 $$ \mathcal{Q}_{2, \perp} = \mathbb{C} \bigl\langle q_1, q_2, q_1^*, q_2^* \mid q_1^2 = q_1,\, q_2^2 = q_2,\, q_1q_2 =q_2q_1 =0 \bigr\rangle. $$ 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 $\mathcal{Q}_{n,\perp} = \mathbb{C} \langle q_1, \dots, q_n \mid q_i^2 = q_i, i=1, \dots, n; \, q_iq_j =0 \text{ for }i\ne j \rangle$ \textup(the problem of unitary classification of $n$ pairwise orthogonal idempotents\textup) is $*$-wild for $n\ge2$. \end{corollary} \begin{corollary} The $*$-algebra $\mathbb{C}\langle q_1, \dots, q_n \mid q_i^2 = q_i, \, i=1,\dots, n; \allowbreak\, q_1 + \dots + q_n =e\rangle$ \textup(the problem of unitary classification of\/ $n$ idempotents $Q_1$, \dots,~ $Q_n$ such that $Q_1 + \dots + Q_n = I$\textup) 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 semilinear relations} \label{sec:3.1.4} {\bf 1.} In Sections \ref{sec:1.3.2}-\ref{sec:1.3.5} we studied representations of semilinear relations. In particular, the structure of pairs of operators $A=A^*$, $B=B^*$ which satisfy the semilinear relation \begin{equation}\label{poly} \sum_{k=1}^{n}f_k(A)\,B\,g_k(A)=0 \end{equation} was studied. This relation corresponds to 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 connected components of the form \onepointx, \earx, or \edgex, then irreducible representations of relation (\ref{poly}) are one- and two-dimensional and were described in 1.3.5. \medskip\noindent {\bf 2.} We show that all other relations are $*$-wild. \begin{proposition}\label{pro:wild1} If the graph of semilinear relation \eqref{poly} contains a subgraph \earedge or \twoedge{}{}{}, then the relation is $*$-wild. \end{proposition} \begin{proof} We assume that the functions $f_k(\cdot)$ and $g_k(\cdot)$ are polynomials and prove that the $*$-algebra $$ \mathfrak A_{\Gamma}= {\mathbb C}\Bigl\langle a=a^*,b=b^* \mid \sum_{k=1}^{n}f_k(a)\,b\,g_k(a)=0 \Bigr\rangle $$ is $*$-wild if $\Gamma \supset$ \earedge $(\lambda_1, \lambda_2 \in {\Bbb R}$, $\lambda_1\ne \lambda_2$). Define a $*$-homomorphism $\psi \colon \mathfrak A_{\Gamma} \to M_3(\mathfrak S_2)$ as follows: \begin{gather*} \psi(a)=\begin{pmatrix} \lambda_1 e& 0 & 0\\ 0 & \lambda_1 e&0\\ 0&0&\lambda_2 e \end{pmatrix}, \quad \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}\label{pro:wild2} If the graph $\Gamma$ of semilinear relation \eqref{poly} contains a subgraph \vrule width 0pt depth 8pt\twoedge{\lambda_1}{\lambda_2}{\lambda_3} , $\lambda_1\ne \lambda_2 \ne\lambda_3\ne\lambda_1$\textup; $ \lambda_1$, $\lambda_2$, $\lambda_3\in {\Bbb R}$, then the relation is $*$-wild. \end{proposition} \begin{proof} We prove that $\mathfrak A_{\Gamma}= {\Bbb C}\bigl\langle a=a^*,b=b^* \mid \sum_{k=1}^{n}f_k(a)\,b\,g_k(a)=0 \bigr\rangle$ is $*$-wild if \vrule width0pt depth 8pt$\Gamma \supset$\twoedge{\lambda_1}{\lambda_2}{\lambda_3} $(\lambda_1, \lambda_2,\lambda_3 \in {\Bbb R}$, $\lambda_1\ne \lambda_2 \ne \lambda_3\ne\lambda_1$). We construct the $*$-homomorphism $\psi\colon \mathfrak A_{\Gamma} \to M_7(\mathfrak S_2)$ as follows: \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 Theorem~\ref{th_o}, the theory of its $*$\nobreakdash-representations contains $*$-representations of every finitely generated $*$-algebra. \medskip\noindent {\bf 2.} The relation $(I_1)$ $a^2=e$ defines the $*$-algebra $\mathfrak{D}= \mathbb{C}\langle a,b \mid a^* =a, \,b^* =b,\, a^2=e\rangle$. \begin{proposition} The $*$-algebra $\mathfrak D$ is $*$-wild. \end{proposition} \begin{proof} We will show that $\mathfrak D \succ \mathcal P_{3,\perp2}=\mathbb{C}\langle p_1, p_2, p_3 \mid p_i^* = p_i,\, p_i^2 = p_i,\, p_1p_2 =p_2p_1=0\rangle$. To show this, we define a $*$-homomorphism $\psi:\mathfrak D \to \mathcal P_{3,\perp2}$ as follows: \begin{align*} \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{align*} It is easy to check that the corresponding functor $F_{\psi}: \rep \mathcal P_{3,\perp2} \to \rep \mathfrak D$ is full. \end{proof} \noindent{\bf 3.} Now we give a criterion, in terms of the coefficients, for the quadratic $*$-algebra $\mathfrak A =\mathbb C \langle a, b \mid a=a^*,\, b=b^*,\, P_2(a,b)= \alpha a^2 + \beta b^2 + q /i [a,b] + \gamma \{a,b\} +\delta a + \epsilon b + \chi I = 0 \rangle$, $ \alpha$, $\beta $, $q$, $\gamma$, $\delta$, $ \epsilon$, $\chi \in {\mathbb R}$ to be $*$-wild. \begin{theorem}\label{quadr1} A $*$-algebra $\frak A$ is $*$-wild if and only if one of the following conditions holds\textup: \begin{itemize} \item[$1$.] $ \alpha=\beta=\gamma=q=\delta=\epsilon=\chi=0$\textup; \item[$2$.] $ \bigl(\chi-\frac {\delta^2}{4\alpha}\bigr)\,\alpha<0$,\quad $\beta=\gamma=q=\epsilon=0$\textup; \item[$3$.] $ \bigl(\chi-\frac {\epsilon^2}{4\beta}\bigr)\,\beta<0$,\quad $\alpha=\gamma=q=\delta=0$\textup; \item[$4$.] $\gamma^2=\beta\alpha \ne 0, \quad \alpha(\chi-\frac{\delta^2}{4\alpha})<0$,\quad $ \frac{\delta^2}{\alpha}=\frac{\epsilon^2}{\beta} $,\quad $q=0$. \end{itemize} \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 transformed to the $*$-algebra $\mathfrak S_2 = \Bbb C \langle a, b \mid a=a^*,\, b=b^*\rangle$ or the $*$-algebra $\mathfrak D = \Bbb C \langle a, b \mid a=a^*,\, b=b^*, \, a^2=e \rangle$. One can define such a quadratic $*$-algebras by imposing one of the conditions 1--4. \end{proof} \noindent{\bf 4.} Now we consider a pair of self-adjoint operators which satisfy the cubic semilinear relation: \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} To relation (\ref{cub}) there corresponds the $*$-algebra $\mathfrak A_3 ={\Bbb C}\langle a,b \mid a=a^*,\,b= b^*, \, \epsilon\{a^2,b\}+i\delta[a^2,b]+2\mu\, aba +i\gamma[a,b] +2\beta \{a,b\}+\alpha b=0\rangle$. The corresponding characteristic function is the following: \begin{align*} \Phi(t,s)&=\epsilon (t^2+s^2) + i\delta(t^2-s^2) \\* &+ 2\mu\, ts + i\gamma(t-s)+ 2\beta(t+s)+\alpha, \end{align*} $\alpha$, $\beta$, $\gamma$, $\delta \in {\mathbb R}$. It follows from the general theory of semilinear relations that the corresponding $*$-algebra is $*$-wild if and only if the equation $\Phi(t,s)=0$ has either two solutions of the form ($t_1,t_1$), ($t_1,t_2$), where $t_1 \ne t_2$, or two solutions of the form $(t_1,t_2)$, $(t_1,t_3)$, where $t_1$, $t_2$, $t_3$ are distinct. The equation $\Phi(t,s)=0$ decomposes in a natural way into the system of two equations: \[ \left \{ \begin{array}{l} \Phi_1(t,s)=\epsilon\, t^2+2\mu\,ts + \epsilon s^2 + 2\beta t + 2\beta s + \alpha=0, \\ \Phi_2 (t,s) = \delta (t^2 - s^2)+ \gamma (t-s)=0 \end{array} \right. \] First we assume that $\delta = \gamma =0$ ($\Phi_2(t,s)\equiv 0$). The equation $\Phi_1(t,s)=0$ defines a curve of degree two. Such curves have the following invariants: \begin{gather*} I_1=2\epsilon, \quad 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 $\Phi_1(t,s)=0$ defines a central curve. By an affine transformation of variables, the equation can be reduced into 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 $*$-algebra $\mathfrak A_3$ is not wild. If $I_3>0$, then the set of solutions of (\ref{tri}) is empty. Therefore the $*$-algebra $\mathfrak A_3$ is not wild. If $I_3<0$, then equation (\ref{tri}) becomes an equation of an ellipse. 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 $\Phi(t,s)=0$ is of hyperbolic type. If $I_3=0$, then we have a pair of intersecting straight lines. Hence it is easy to see that the $*$-algebra $\mathfrak A_3$ is $*$-wild. If $I_3\ne 0$, then we have an equation of a hyperbola. This equation has no more than one solution only in case where $\epsilon =0$. Hence, if $\epsilon \ne 0$, then the $*$-algebra $\mathfrak A_3 $ is wild. c) Let $I_2=0$. Then the equation $\Phi_1(t,s)=0$ has parabolic type (the curve is not central). Let $I_3=0$. Then $\epsilon=\mu$ and \[ \Phi_1(t,s)=\epsilon(t+s)^2+2\beta(t+s)+\alpha=0. \] If $\beta^2-\alpha\epsilon<0$, the equation $\Phi_1(t,s)=0$ has no solutions, and the $*$-algebra $\frak A_3$ is not wild. If $\beta^2-\alpha\epsilon<0$, then $t+s=-{\beta}/{\epsilon}$, and it is obvious that the $*$-algebra $\frak A_3$ is not wild. For $\beta^2-\alpha\epsilon>0$, the equation $\Phi_1(t,s)=0$ describes a pair of parallel lines. Hence the set of solutions contains two solutions: $(t_1,t_2)$, $(t_1,t_3)$ 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 $\Phi_1(t,s)=0$ defines a parabola. In this 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 consider the case $\Phi_1(t,s)\equiv 0$, $\delta \ne 0$. We have \[ \Phi_2(t,s)=(t-s)(\delta t+\delta s +\gamma)=0. \] In that case the equation $\Phi_2(t,s)=0$ defines a pair of intersecting and non coinciding straight lines. Therefore the $*$-algebra $\mathfrak A_3$ is $*$\nobreakdash-wild. Consider the case $\Phi_1(t,s)\,\Phi_2(t,s) \ne 0$. It is obvious that if each equation $\Phi_1(t,s)=0$ and $\Phi_2(t,s)=0$ generates a non $*$-wild $*$-algebra, then \begin{equation}\label{sis} \left \{ \begin{array}{l} \Phi_1(t,s)=0, \\ \Phi_2 (t,s) = 0, \end{array}\right. \end{equation} corresponds a non $*$-wild $*$-algebra, too. Therefore we will consider the case when $\Phi_1(t,s)=0$ and $\Phi_2(t,s)=0$ generate $*$-wild $*$-algebras. Then we have that $\delta(\epsilon^2+\mu^2)\ne 0$. By an affine change of variables, system (\ref{sis}) one can reduced to the following form: \begin{gather*} \left \{ \begin{array}{l}\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 $\tilde \beta = \beta - (\epsilon +\mu){\gamma}/({2\delta})$, $\tilde \alpha = \alpha +(\epsilon +\mu){\gamma^2}/({2\delta^2}) - 2\beta{\gamma}/{\delta}$. Solutions of this system are solutions of the equation of $\tilde \Phi_1(\tilde t, \tilde s)=0$, where $\tilde s =\tilde p$ and $\tilde s= -\tilde t$. Therefore, for $\frak A_3$ to be $*$-wild it is necessary and sufficient that the equation $\tilde \Phi(\tilde t,\tilde s)=0$ have two solutions of the form: $(\tilde t_0,\tilde t_0)$ and $(\tilde t_0,-\tilde t_0)$. It is easy to show that this condition is fulfilled in one of the following cases. \begin{itemize} \item[i.] $\epsilon>0$,\quad $\mu>0$,\quad $2\delta\beta-\epsilon\gamma=0 $,\quad $\alpha\delta^2-\epsilon\gamma^2<0$; \item[ii.] $\mu\ne 0$,\quad $2\delta\beta-\epsilon\gamma\ne 0 $,\quad $\displaystyle\alpha\epsilon^2-I_3 +\frac{\epsilon\gamma}{\delta} \frac{(\epsilon + \mu )\gamma-4\beta\delta}{2\delta}=0$. \end{itemize} Thus we proved the following theorem. \begin{theorem}\label{th:cub1} The $*$-algebra $\mathfrak A_3 = {\mathbb C}\bigl\langle a,b \mid a=a^*,b= b^*,\,\epsilon\{a^2,b\}+i\delta[a^2,b]+2\mu\, aba +i\gamma[a,b] +2\beta \{a,b\}+\alpha b\bigr\rangle$, $\epsilon \geq 0$, $\epsilon^2+\mu^2+\delta^2\ne 0$, is $*$-wild if and only if one of the conditions \textup{1)}--\textup{3)} holds\textup: \begin{itemize} \item[$1)$] $\delta=\gamma=0$, and one of the following conditions holds\textup: \begin{itemize} \item[$(a)$] $I_1>0$, $I_2>0$, $I_3<0$, \item[$(b)$] $I_2<0$, $I_3=0$, \item[$(c)$] $I_1>0$, $I_2<0$, $I_3\ne 0$, \item[$(d)$] $I_2=0$, $I_3=0$, $ \beta^2-8\alpha \epsilon>0 $, \item[$(e)$] $I_2=0$. $I_3\ne 0$\textup; \end{itemize} \item[$2)$] $\epsilon=\mu=\beta=\alpha=0$, \textup(then $\delta\ne 0$\textup). \item[$3)$] $\delta(\epsilon^2+\mu^2) \ne 0$ and one of the following conditions holds\textup: \begin{itemize} \item[$(a)$] $\epsilon > 0$, $\mu=0$, $ 2\delta \beta-\epsilon\gamma =0$, $\alpha\delta^2-\epsilon\gamma^2<0$, \item[$(b)$] $\mu\ne 0$, $2\delta\beta-\epsilon\gamma \ne 0$, $\alpha \epsilon^2-I_3 + \displaystyle\frac{\epsilon\gamma}{\delta} \frac{\epsilon+\mu)-\gamma - 4\beta\delta}{2\delta}=0$. \end{itemize} \end{itemize} \end{theorem} \noindent\textbf{5}. For a non-semilinear cubic relation, we give only the following proposition. \begin{proposition} The $*$-algebra \[ \mathcal{B}_2 = \mathbb{C} \langle a = a^*, b=b^* \mid aba = bab \rangle \] is $*$-wild. Moreover, its quotient algebra $\mathbb{C}\langle a= a^*, b= b^* \mid aba = bab =0\rangle$ is $*$-wild. \end{proposition} \begin{proof} Let a homomorphism $\psi \colon \mathcal{B}_2 \to M_4(\mathfrak{S}_2)$ be defined as follows: \[ \psi(a) = \begin{pmatrix}\begin{matrix} 0&e\\e&0 \end{matrix} & \text{\Large0}\\\text{\Large0} &\text{\Large0} \end{pmatrix}, \quad \psi(b) = \begin{pmatrix}\text{\Large0}&\begin{matrix}e&0\\0&0 \end{matrix} \\ \begin{matrix}e&0\\0&0\end{matrix} & \begin{matrix} a_1 & e \\ e & a_2\end{matrix} \end{pmatrix}. \] One can check that $\psi(aba)=0$, and $\psi$ defines a homomorphism of the quotient algebra. The constructed functor $F_\psi$ is full. \end{proof} \begin{remark} The $*$-wild $*$-algebra $\mathcal{B}_2 = \mathbb{C}\langle a= a^*, b= b^* \mid aba = bab\rangle$ is obtained by introducing the involution in the algebra $\mathbb{C}\langle x, y \mid xyx = yxy \rangle$ by setting $x= x^*$, $y = y^*$. The $*$-structure on the algebra $\mathbb{C}\langle x,y\mid xyx=yxy\rangle$ is not unique. If one introduces an involution in 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{sec: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 a group contains, as a subproblem, the description of representations of any countable group; further 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 that are not $*$-wild. \medskip\noindent\textbf{1.} Let us give a number of examples of $*$-wild groups. \begin{example} It follows directly from the given above list of $*$-wild $*$\nobreakdash-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} \noindent\textbf{2.} The following simple statement holds. \begin{proposition}\label{pr:wildgroups} 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 an element $g \in G$ to its conjugacy class $\phi(g) \in G_1 = G/N$. Then, introduce a unitary $*$-homomorphism $$ \psi \colon L(G) \ni f(\cdot) \mapsto \psi(f) (g_1) = \sum_{g \colon \phi(g) = g_1} f(g) \in L(G_1) \subset C^* (G_1), $$ and use the same notation $\psi$ for its unique extension to a 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 $G$ that possesses 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} Notice that the proof of wildness of $\mathbb{Z}_2 * \mathbb{Z}_3$ given in Section~\ref{sec:3.1.3}, is not reduced to the check of the conditions of the corollary above. Since the majorization is a quasi-order relation (Section~\ref{sec:3.1.1}), the following corollary from Proposition~\ref{pr:wildgroups} holds. \begin{corollary} If a group $G$ possesses a normal subgroup $N$ such that $G/N = G_1$ is $*$-wild, then $G$ is $*$-wild. \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 $B_2/\mathbb{Z} = \mathbb{Z}_2 * \mathbb{Z}_3$, 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} \medskip\noindent\textbf{3.} It looks attractive to investigate whether the following groups are $*$-wild or not: a) Coxeter groups which are not affine (except for $\coxetertwoedge{\infty}{\infty}$ which is $*$-wild, see Section~\ref{sec:3.1.3}); b) non-elementary hyperbolic groups; \noindent and many other known groups, which are not amenable. The structure of quotient groups for listed above groups is the subject the authors are not familiar enough; thus we cannot estimate properly, how complicated the listed tasks are. \medskip\noindent\textbf{4.} Consider some examples of groups which are not $*$-wild. If $G$ is amenable, then $C^*(G)$ is nuclear. 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}\label{th:kal} 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 that are 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{kiril}), 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 $i$ such 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 $k_i(t_0)$ is not 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} There are groups which are not $*$-wild and not amen\-able. Those are, for example, Burnside groups $B(m,n)$ which are not amenable for odd $n \ge 665$ and $m \ge 2$ \textup(see \textup{\cite{olsh,adjan_book})}. \end{corollary} %%% Local Variables: %%% mode: latex %%% TeX-master: "the" %%% TeX-master: "the" %%% TeX-master: "the" %%% TeX-master: "the" %%% End: