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\section{Introduction to representations of $*$-algebras}

\subsection{$*$-Representations: key words.}

1. A representation of an algebra $\mathcal{A}$ in
finite-dimensional Hilbert (unitary) space $H$ is a homomorphism
$\pi$
of $\mathcal{A}$ into the algebra $L(H)$ of linear transformations
of
$H$.
A representation of $*$-algebra $\mathfrak{A}$ is its
$*$-homomorphism
$\pi$ into the $*$-algebra $L(H)$ of bounded operators in a separable
Hilbert space $H$. Dimension of representation is the dimension of
$H$.

We emphasize that in this chapter we restrict ourselves to
considering
only finite-dimensional representations of $\mathcal{A}$ if
$\mathcal{A}$ is an algebra without involution and
$*$-representations
by bounded operators in a separable Hilbert space ($\dim H
\le\infty$)
if $\mathfrak{A}$ is a $*$-algebra.

\smallskip
2. In the representations theory of
algebras the representations are studied up to some equivalence. We
call representations $\pi$ of $\mathcal{A}$  in $H$ and
$\tilde{\pi}$ in
$\tilde{H}$
equivalent if there exists invertible operator $C\colon
H\mapsto\tilde{H}$ which intertwines the representations $\pi$
and
$\tilde{\pi}$, i.e.,
\[
  C \pi (x) = \tilde{\pi}(x) C,\quad \forall x\in\mathcal{A}.
\]

In the representation theory of
$*$-algebras the representations are studied up to the unitary
equivalence. Representations $\pi$ of $\mathfrak{A}$  in $H$ and
$\tilde{\pi}$ in
$\tilde{H}$
are said to be unitarily equivalent if there exists a unitary
operator
$U\colon H\mapsto\tilde{H}$, such that
\[
  U \pi (x) = \tilde{\pi}(x) U,\quad \forall x\in\mathfrak{A}.
\]

To every $*$-algebra $\mathfrak{A}$ one can associate a category
$*$-$\rep\mathfrak{A}$, whose objects are $*$-representations of
$\mathfrak{A}$
considered up to  unitary equivalence and its morphisms are
intertwining
operators considered up to natural equivalence.

We have the following simple proposition.

\begin{proposition}
Any two finite-dimensional\/ $*$-representations of\/ $*$-algebra
$\mathfrak{A}$
are equivalent if and only if they are unitarily equivalent.
\end{proposition}

\begin{proof}
One have to prove only  that equivalence of\/ $*$-representations $\pi$
in
$H$
and $\tilde{\pi}$ in $\tilde{H}$ implies their unitary
equivalence.

Let $C\colon H\mapsto \tilde{H}$ be an invertible operator such
that
\begin{equation}\label{intertwine}
  C \pi (x) = \tilde{\pi}(x) C,\quad \forall x\in\mathfrak{A}. 
\end{equation}

Let us consider the polar decomposition of operator $C = U A$, where
$A=(CC ^*)^{1/2}$ is an invertible positive operator
in
$H$
and $U$ is a  unitary operator from $H$ to $\tilde{H}$
($U^{-1}=U^*$).
Then it follows from \eqref{intertwine} that
\begin{equation}
\pi (x) = A^{-1}U^{-1}\tilde{\pi}(x) U A,\quad \forall
x\in\mathfrak{A}. 
\end{equation}
Taking adjoint we obtain
\begin{equation}
\tilde{\pi} (x) = UA^{-1}{\pi}(x)A U^{-1},\quad \forall
x\in\mathfrak{A}. 
\end{equation}
Consequently,
\[
A^2\pi (x) = A U^{-1}\tilde{\pi}(x)U A =
AU^{-1}UA^{-1}\pi(x)AU^{-1}UA=
\pi(x)A^2.
\]
Since $A$ is positive operator we have $A^2\pi(x)=\pi(x)A^2$,
$x\in\mathcal{A}$, which implies  $A \pi(x)=\pi(x)A$ for any
$x\in\mathfrak{A}$. From this we obtain
\[
U\pi(x)A=UA\pi(x)=\tilde{\pi}(x)UA,
\]
and since $A$ is invertible,
\[
U\pi(x)=\tilde{\pi}(x)U,   \quad \forall x\in\mathfrak{A}.
\]
Hence we have unitary equivalence of the representations $\pi$ in $H$
and
$\tilde{\pi}$ in $\tilde{H}$.
\end{proof}

$3$.
In  general representation theory one distinguishes
irreducible and indecomposable representations
from the set of  all {\it finite-dimensional} representations.
A representation
$\pi\colon\mathcal{A}\mapsto L(H)$ is called irreducible if there
exists no nontrivial subspace of $H$ invariant with respect to 
all the  operators $\pi(x)\  (\ x\in\mathcal{A})$. A representation
$\pi\colon\mathcal{A}\mapsto L(H)$ is called indecomposable if there
exists no decomposition $H=H_1 + H_2$ in a sum of the two
nontrivial subspaces which are invariant with respect to the all
operators 
$\pi(x)\  ( \ x\in\mathcal{A})$ and $ H_1 \cap H_2=\{ 0\}$. It is
clear
that any irreducible
representation is indecomposable. Thus irreducible
representations is a subset  of the set of all indecomposable ones.
It depends on the structure of $\mathcal{A}$ how big this subset in
the whole set of indecomposable representations.

Description of  all indecomposable (particularly irreducible)
representations is one of the most important problems of the
representation theory.

In the case when $\mathfrak{A}$ is algebra with an involution one can
consider the irreducible $*$-representations and indecomposable
$*$-representations as well. However in this case these notions
coincide. Namely the following simple proposition holds.
\begin{proposition}
A $*$-representation $\pi$ is indecomposable if and only if it is
irreducible.
\end{proposition}
\begin{proof}
It is sufficient to prove that any indecomposable $*$-representation
is
irreducible. Assume the contrary that an indecomposable
representation is
reducible, i.e. , there exists a proper subspace $H_1$ in $H$
invariant
with
respect to  all $\pi(x)\   (\ x\in\mathfrak{A})$. The contradiction
follows from the lemma below.
\begin{lemma}
The subspace $H_1^{\perp} = \{y\in H\colon (y,f)=0\ \forall f\in
H_1\}$ is nontrivial and invariant with respect to  all $\pi(x)\ 
(x\in\mathfrak{A})$.
\end{lemma}
\begin{proof}
If $y\in H_1^{\perp}$ then
\[
\Bigl( \pi(x)y\ ,\ f \Bigr)=\Bigl( y\ ,\ \pi(x^*) f \Bigr) = 0
\]
for all $f$ belonging to the invariant subspace $H_1$, i.e.
$\pi(x)y\in H_1^{\perp}$.
\end{proof}
\end{proof}

$4$. Let $H$ be a  separable({\it generally speaking,
infinite-dimensional}) Hilbert space. Following the general
strategy of representation theory we  select also the ``simplest''
irreducible representations among all $*$-representations. A
$*$-representation $\pi\colon\mathfrak{A}\mapsto L(H)$ is called
irreducible, if there exists no nontrivial subspace in $H$ invariant
with respect to  all the operators $\pi(x)  \ (x\in\mathfrak{A})$.
The
proposition below gives an equivalent condition of irreducibility.

\begin{proposition}
A $*$-representation $\pi(\cdot)$
is irreducible if and only if any bounded operator $C\in L(H)$ such
that
\[
C\pi(x)=\pi(x)C\quad (\forall x\in\mathfrak{A})
\]
is a multiple of identity, i.e., $C=cI$, with $c\in\mathbb{C}$
\end{proposition}
\begin{proof}
If $A=A^*$ and $A$ commutes with $\pi(\cdot)$, i.e. ,
$A \pi(x)=\pi(x) A\ \ (\forall x\in\mathfrak{A})$ then
     \[
     E_A (\Delta)\pi(x)=\pi(x)E_A (\Delta)
     \]
for all $x\in\mathfrak{A}$ and Borel $\Delta\subset\mathbb{R}^1$
(here $E_A (\Delta)$ is a spectral projector the operator $A$).
 In this case $H_{\Delta}=E_A (\Delta) H$ is an
invariant subspace in $H$.

If representation $\pi$ is irreducible then all such $H_{\Delta}$ are
either $\{ 0\}$ or $H$, i.e. the spectral measure $E_A (\cdot)$ is
concentrated at one point $a\in\mathbb{R}^1$ and $A=aI$.

If $C=A+i B\ (A=A^*,\ B=B^*\in L(H))$ commutes with an irreducible
representation $\pi(\cdot)$ of the $*$-algebra $\mathfrak{A}$, then
the
operators $A,\ B$ commute with $\pi(\cdot)$ as well, and,
consequently,
$C=aI+ibI=(a+ib)I\ \ (a,b\in\mathbb{R})$.

Conversely, if a representation $\pi(\cdot)$ is reducible and
$H_1$ is a subspace invariant with respect to $\pi(x) \
 (x\in\mathfrak{A})$ then by lemma  $H_1^{\perp}$ is also
invariant. Then the operator
 \[
 C = \left(
 \begin{array}{cc}
 c_1 I_{H_1} & 0 \\
 0 & c_2 I_{H_1^{\perp}}
 \end{array}
 \right)
 \quad (c_1\neq c_2\ ,\ c_1,c_2\in\mathbb{C})
 \]
commutes with the representation and is not a multiple of identity.
\end{proof}
\begin{remark}
It is possible to determine the notion of indecomposable
representation
in the case when $H$ is a separable Hilbert space ($\dim H =\infty$) 
and to prove the analog of the previous propositions. However we are
not 
going to do it here. 
\end{remark}

Irreducible representations and their intertwining operators form a
full sub-category $*$-Irrep $\mathfrak{A}$ in the category $*$-Rep
$\mathfrak{A}$. The condition  for sub-category to be full means that
the embedding functor $F$ from  $*$-Irrep $\mathfrak{A}$ into
$*$-Rep $\mathfrak{A}$  is monomorphism of the objects and
isomorphism of
the
corresponding morphisms. In what follows, we will mainly deal with
$*$-algebras and 
their $*$-representations; thus we will sometimes omit the involution
sign 
in words algebra and representation when no ambiguity can arise.

\subsection{$C ^*$-representable $*$-algebras.}

$1$. An important class of $*$-algebras is a class of all
$*$-algebras
which have
``sufficiently many'' $*$-representations. The latter means that
there
exists a residual family (r.f.) of $*$-representations, i.e. for any
$x\in \mathfrak{A}\ ,\ x\neq 0$, there exists a $*$-representation
$\pi$
(it can be chosen irreducible) such that $\pi(x)\neq 0$. For any
$C ^*$-
algebra always there exists a r.f. of $*$-representations.

If a $*$-algebra is $C ^*$- representable, i.e. there exists
a
$*$-isomorphism of $\mathfrak{A}$ into a $C ^*$-algebra,
then it
is
clear that $\mathfrak{A}$ has a r.f. .
\begin{remark}

\begin{enumerate}

\item
A  $C ^*$-algebra  containing a dense $*$-subalgebra
which is $*$-isomorphic to a given one is not unique in general. For
example, the $*$-algebra $\mathbb{C}[a=a^*]$ is isomorphic to a dense
subalgebra in
any $C ^*$-algebra $C(K)$ of continuous functions on a
compact
$K\subset\mathbb{R}^1$, containing at least one internal point.
However,
$C(K_1)$ is isomorphic to $C(K_2)$ if
and only if $K_1$ and $K_2$ are homeomorphic.
\item
Not any $*$-algebra is $C ^*$-representable. For example, if
involution in
$\mathcal{A}$ is not proper \\
(involution is proper if $xx^*=0$ implies $x=0$
) then such $*$-algebra can not be
$C ^*$-representable, more then that it has no  r.f.  
\end{enumerate}
\end{remark}

Nevertheless if $\mathfrak{A}=\mathbb{C}[G]$ is a group $*$-algebra
of a countable discrete group $G$ with natural involution(given on
basis
vectors $g \in G$ via the rule $ g^*=g^{-1}$),
then the following proposition holds 
\begin{proposition}
The involutive algebra $\mathbb{C}[G]$ is
$C ^*$-representable.
\end{proposition}
\begin{proof}
Indeed, any operator $\pi_r (x)$ of the right regular representation
$\pi_r$ in $L_2 (G)$ is non-zero for any non-zero element
$x\in\mathbb{C}[G]$, and hence $\mathbb{C}[G]$ is isomorphic to
$\pi_r
(\mathbb{C}[G])$ which is a dense  $*$-subalgebra
in the
$C ^*$-algebra $C_r^* (G)$ generated by operators $\pi(x)
\quad
(x\in\mathbb{C}[G])$.
\end{proof}

$2$. The following  proposition holds.
\begin{proposition}
Consider the following properties of a $*$-algebra:\\
\ \ \ \ $(i)$ \ \ the algebra is $C ^*$-representable;\\
\ \ \ \ $(ii)$\ \ there exists a residual family of its
representations;\\
\ \ \ \ $(iii)$\ \ the involution on the algebra is completely proper
(i.e.  the equality $\sum_{k=1}^n x_k x_k^* = 0$ implies $x_k = 0$,
$k=1,\ldots ,n$, $n\in\mathbb{N}$);\\
\ \ \ \ $(iv)$\ \ the involution in the algebra is proper (i.e.
$xx^*=0$
implies $x=0$).\\
\ \ \ \ Then  $(i)\Rightarrow
(ii)\Rightarrow (iii)\Rightarrow (iv)$.\\
Neither of inverse implications holds.
\end{proposition}
\begin{proof}
To see that $(ii)$ does not imply $(i)$ one have to consider the
$*$-algebra $C ([0,\infty))$ of all continous functions
$f(\cdot)$ on $[0,\infty)$ with natural involution and pointwise
multiplication. It is obvious that this algebra has a r.f. of
one-dimensional representations, however it is not
$C ^*$-representable.
Indeed, assume that $C ([0,\infty))$ is embedded 
as a $*$-subalgebra in $C ^*$-algebra $A$
then there exists integer $N$, such that $\Vert f\Vert \le N$, where
$f(x)=
\begin{cases}
x-n,& \text{if $2n < x < 2n+1$};\\
n,& \text{if  $2n-1 \le  x \le  2n$}.
\end{cases}$
$(n \in \mathbb{N})$.

This implies that element $NI-f$ is invertible in $A$, but it
 is zero divizor in $C ([0,\infty))$, which is
 contradiction.

To see that $(iii)$ does not imply $(ii)$ let us consider the
$*$-algebra
\begin{align*}
&\mathfrak{a}= \lim_{\longrightarrow}\mathfrak{a}_n =\\
&=\lim_{\longrightarrow}\mathbb{C}\Bigl<a_1,\ldots , a_n
\mid\ a_i=a_i^*,\ a_1^2 +a_k^2 =\frac{1}{k^2}e,\ [a_i,a_j]=0
\ i,j,k=1,\ldots ,n\Bigr>,
\end{align*}
where $\mathfrak{a}_n\hookrightarrow\mathfrak{a}_{n+1}$ are the
natural injections. 
It is easy to see that $\mathfrak{a}_n=$lspan$(S_n)$, where $S_n= \{
a_1^ka_{i_1}\ldots a_{i_s}, k=0,1,2; 2 \le i_1 < i_2< \ldots <i_s \le
n
\}$. Let $P$ be non-trivial linear combination of elements of $S_n$,
such
that $\pi(P)=0$ for every  representation $ \pi$ of $\mathfrak{a}_n$.
To each $\alpha \in (0;\frac{1}{n})$ corresponds one-dimensional
representation $\pi_\alpha$, such that
$\pi_\alpha(a_1)=\alpha,\pi_\alpha(a_k)=\sqrt{\frac{1}{k^2}-\alpha^2}
$,
maps  $\pi_{\alpha,j}(a_s)=
\begin{cases}
\pi_\alpha(a_s),& \text{if $ s \neq j$,}\\
-\pi_\alpha(a_s), & \text{ $s=j$ ;}
\end{cases}$
extend to representations of  $\mathfrak{a}_n$, from this one can
deduce
that $P$ can be chosen to have a form $P=(\alpha_0I
+\alpha_1a_1^2)a_{i_1}
\ldots a_{i_s}$, where $\alpha_0,\alpha_1 \in \mathbb{C}$, then the 
function $\pi_{x}(P)=(\alpha_0I
+\alpha_1x^2)(\frac{1}{i_1^2}-x^2)^{1/2} \ldots
(\frac{1}{i_s^2}-x^2)^{1/2}$ has only finite number of roots, so
there
exists $\alpha \in (0;\frac{1}{n})$ that $\pi_\alpha(P) \neq 0$.
Obtained
contradiction proves that $\mathfrak{a}_n$ has r.f., $S_n$ is a
linear
basis
for $\mathfrak{a}_n$, and
$\mathfrak{a}_n\hookrightarrow\mathfrak{a}_{n+1}$.$\mathfrak{a}_n$ is 
completely proper, so is inductive limit.
It is easy to see that an involution on
$\mathfrak{a}$ is completely proper, however $\Vert\pi(a_1)\Vert =0$
in
any $*$-representation $\pi$ of $\mathfrak{a}$

To see that $(iv)$ does not imply (iii) we address a reader
to ~\cite{Belfi} and references given there.

\end{proof}
$3$. It is natural to consider among all $*$-algebras which have a
r.f.
those
which posses a  residual family of finite dimensional representations
(we call these algebras residually finite dimensional (r.f.d.))
\begin{proposition}
If $G$ is a residually finite group (i.e. $\forall g\neq e $ there
exists a normal subgroup $G_g\not\ni g$ such that $G\Big/G_g$ is
a finite group), then $\mathbb{C}[G]$ is residually finite
dimensional.
\end{proposition}
\begin{proof}
Let us recall firstly  the equivalent definition of residually
finiteness of a group $G$: for any finite set of non-identity
elements
of $G$ there exists normal subgroup, which does not contain any of
these elements, and such that the factor-group of $G$ by this
subgroup is finite.

Let $\alpha =\sum_{k}c_k g_k$ be an element of the group algebra
($c_k\neq
0\ ,\ k=1,\ldots , n$). Choosing a normal subgroup which does not
contain the elements $g_i g_j^{-1}\quad i\neq j\quad i,j=1,\ldots
,n$ we conclude that the image of $\alpha\in\mathbb{C}[G]$ under the
homomorphism
\[
\mathbb{C}[G]\mapsto\mathbb{C}\Bigl[G\Big/N(g_i g_j^{-1}\ ,\ i\neq j)
\Bigr]
\]
of $\mathbb{C}[G]$ on the group algebra of finite group is non-zero
and hence it is non-zero in
the finite-dimensional regular representation of\\
$\Bigl[ G\Big/ N(g_i g_j^{-1}\ ,\ i\neq j) \Bigr]$
\end{proof}

$4$. The following list of ${C}^*$-algebras is naturally
connected with a group $G$:
\begin{itemize}
\item $C_f^*(G)$. 
This algebra is completion of  $\mathbb{C}[G]$ by the following norm:
\[
\Vert\alpha\Vert_{C_{f}^{*}(g)} = \sup_{\pi\in
*-fdRep\mathbb{C}[G]}
\Vert\pi(\alpha)\Vert\quad (\alpha\in\mathbb{C}[G])
\]
(where $\sup$ is taken over all finite-dimensional
$*$-representations
of $\mathbb{C}[G]$).
\item  $C_r^* (G)$
This algebra is a completion of $\mathbb{C}[G]$ by the right regular
norm
\[
\Vert \alpha\Vert_{C_r(G)}= \Vert\pi_r
(\alpha)\Vert
\quad (\alpha\in\mathbb{C}[G])
\]
where $\pi_r (\alpha)$ is the operator of right regular
representation
of $G$ on  $L_2(G)$.
\item $C^*(G)$.
This algebra is completion  of $\mathbb{C}[G]$ by the following norm:
\[
\Vert\alpha\Vert_{ C^*(G)}=\sup_{\pi\in *-Rep\mathbb{C}[G]}
\Vert\pi(\alpha)\Vert\quad (\alpha\in\mathbb{C}[G])
\]
\end{itemize}
The $*$-algebra $\mathbb{C}[G]$ is $*$-bounded, i.e. $\Vert\pi
(\alpha)\Vert
\le C_{\alpha} < \infty $ for any
$\alpha\in\mathbb{C}[G]$ and $\pi\in *-Rep\ \mathbb{C}[G]$ (for more
detailed information about $*$-boundedness see $2.1.3$). Hence all
norms
defined above are finite for any $\alpha\in\mathbb{C}[G]$.
\begin{proposition}
If $G$ is residual finite group then
\[
\Vert\alpha\Vert_{L_1(G)} \ge 
\Vert\alpha\Vert_{ C^*(G)}\ge
\Vert\alpha\Vert_{{C}_f^* (G)}\ge
\Vert\alpha\Vert_{{C}_r^* (G)}\ge
\Vert\alpha\Vert_{L_2 (G)}
\]
for any $\alpha\in\mathbb{C}[G]$.
\end{proposition}
\begin{proof}
One has  to prove only the inequality
\[
\Vert\alpha\Vert_{{C}_f^* (G)}\ge
\Vert\alpha\Vert_{{C}_r^* (G)}\quad
(\forall\alpha\in\mathbb{C}[G])
\]
Let $\alpha=\sum_{k=1}^{m} c_kg_k \neq 0$, then
$\Vert\alpha\Vert_{{C}_r^* (G)}=d>0$
and there exists $\gamma\in L_2 (G)$ such that
$\Vert \gamma\Vert_{L_2 (G)}=1$ and
$\Vert\alpha\gamma\Vert_{L_2(G)}\ge d-\frac{\varepsilon}{2}$. Hence,
there exists $\delta =\sum_{1}^n \delta_k h_k$ such that
$\Vert \delta\Vert_{L_2 (G)}=1$ and
$\Vert \gamma -\delta\Vert_{L_2 (G)}\le\varepsilon\cdot\frac{1}{2d}$.
It
implies
\[
\Vert\alpha \gamma-\alpha\delta\Vert_{L_2
(G)}\le\frac{\varepsilon}{2}
\]
and
\[
\Vert\alpha\delta \Vert_{L_2 (G)}\ge d -\varepsilon.
\]
Further, let us choose a normal subgroup $\mathcal{N}$ of group $G$
which
does not contain the non-trivial  elements among $g_kg_l^{-1},\ 
h_th_s^{-1},\  g_kh_th_s^{-1}g_l^{-1};\  k,l=1, \ldots, m,
t,s=1,\ldots,
n$ and
such
that the quotient group of $G$ by this subgroup is
finite. Then in the regular representation of
$G\Big/\mathcal{N}$ we have
\[
\Vert\alpha\delta\Vert_{L_2 (G
/\mathcal{N})}=\Vert\alpha\delta\Vert_{L_2(G)}\ge
d-\varepsilon .
\]
Hence, for any $\varepsilon$ we have
$\Vert\alpha\Vert_{{C}_f^*(G)}\ge d-\varepsilon$. Therefore
\[
\Vert\alpha\Vert_{{C}_f^*(G)}\ge\Vert\alpha\Vert_{{C}
_r^*(
G)}
\]
\end{proof}
\begin{remark}
If $G$ is residually finite group then we have the following sequence
of $*$-homomorphisms
\[
C^*(G)\mapsto C_f^* (G)\mapsto C_r^* (G),
\]
which are identical on the dense $*$-subalgebra $\mathbb{C}[G]$. Let
us
note that these homomorphisms are epimorphisms. However, in general,
neither $C^*(G)$ nor $C_r^*(G)$ is r.f.d.
$*$-algebra. Indeed, $C^*(\mathcal{F}_2)$, where $\mathcal{F}_2$
is free group with two generators, is r.f.d. $*$-algebra, however,
$C_r^*(\mathcal{F}_2)$ is not r.f.d. $*$-algebra.
The residual finite group $SL(2,\mathbb{Z}[\frac{1}{p}])$ ($p$ is a 
prime number) gives an example of r.f. group $G$ for which 
$C^*(G)$ is not r.f.d. see~\cite{Harpe}.
\end{remark}

\subsection{Enveloping $*$-algebras and $C^*$-algebras.}

$1$. Sometimes it is possible to reduce the study of representations
of
$*$-algebra $\mathfrak{A}$ to the study of $*$-representations of its
enveloping $*$-algebra or $\sigma$-${C}^*$- algebra. Let us
recall the definition.
\begin{definition}
Let $\mathfrak{A}$ be  a $*$-algebra. The 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 enveloping $*$-algebra of algebra $\mathfrak{A}$ if for any
$*$-representation $\pi\colon\mathfrak{A}\mapsto\L(H)$ of algebra
$\mathfrak{A}$ there exists unique $*$-representation
$\widetilde{\pi}\colon\widetilde{\mathfrak{A}}\mapsto\L(H)$ such that
the
following diagram is commutative.
\begin{center}
\resetparms
\btriangle[\mathfrak{A}`\widetilde{\mathfrak{A}}`L(H);\phi`\pi
`\widetilde{\pi}]
\end{center}

\end{definition}
\begin{description}
\item[Example 1.] Let $\Sigma $ be any family of elements of
 $*$-algebra
$\mathfrak{A}$ which are invertible in any $*$-representation
$\pi\colon\mathfrak{A}\mapsto\L(H)$. Let us denote
$\widetilde{\mathfrak{U}}=\mathfrak{A}[\Sigma^{-1}]$ the algebra of
quotients
of $\mathfrak{A}$ by $\Sigma$ (see ~\cite{Gabriel}). Let 
$\phi\colon\mathfrak{A}\mapsto \widetilde{\mathfrak{U}}$ the
natural
homomorphism. Then we obtain the enveloping $*$-algebra for the
$*$-algebra
$\mathfrak{A}$. Let us note that in case when $\mathfrak{A}$ is
$C^*$-representable $\phi$ is injection. 
\end{description}

$2$. If $\widetilde{\mathfrak{A}}$ carries the structure of
${C}^*$-algebra then $\widetilde{\pi}$ is a continuous
$*$-homomorphism from the ${C}^*$-algebra
$\widetilde{\mathfrak{A}}$
to the ${C}^*$-algebra $L(H)$. In this case the pair
$(\widetilde{\mathfrak{A}}, \phi)$ is called enveloping
${C}^*$-algebra of algebra $\mathfrak{A}$. The enveloping
${C}^*$-algebra is unique in class of $C^*$-algebras (in the
case 
when it exists).
Indeed, in this case  $\widetilde{\mathfrak{A}}$ is an enveloping
$\sigma$-
$C^*$-algebra which is unique by theorem~\ref{env}. We will denote
the
enveloping $C ^*$-algebra of algebra $\mathfrak{A}$ by
$C^*(\mathfrak{A})$.

A $*$-algebra $\mathfrak{A}$ is called $*$-bounded if for  any
$x\in\mathfrak{A}$ there exists the number $C_x< \infty$ such that
$\Vert\pi(x)\Vert\le C_x$ for any its
$*$-representation $\pi\colon\mathfrak{A}\mapsto L(H)$. 
For a finitely-generated $*$-algebra $\mathfrak{A}=\mathbb{C}\big<
x_1,\ldots, x_n |S \big> $ to be $*$-bounded it is sufficient that
for any $*$-representation $\pi\colon\mathfrak{A}\mapsto L(H)$ and
any
$k=1,\ldots ,n$ there exists $C_k$ such that $\Vert\pi(x_k)\Vert\le
C_k$.

A $*$-algebra $\mathfrak{A}$ has the enveloping
$C ^*$-algebra $C^*(\mathfrak{A})$ if and only if it
is $*$-bounded. In this case the enveloping $C ^*$-algebra
is the completion  of $\mathfrak{A}$ by the
norm $\Vert x\Vert =\sup_{\pi \in
*-Rep(\mathfrak{A})}\Vert\pi(x)\Vert\le
C_{x}<\infty$ (sup
is taken under all $*$-representations $\pi$).
\begin{description}
\item[Example 2.]
The  group $*$-algebra $\mathbb{C}[G]$ is $*$-bounded since for any
generator $g\in G$ and any $\pi\in *-Rep\ \mathbb{C}[G]$ the
operator
$\pi(g)$ is unitary and $\Vert\pi(g)\Vert = 1$. Hence there exists
the
enveloping $C^*$-algebra
$C^*(G):=C^*(\mathbb{C}[G])$.

If $*$-algebra $\mathfrak{A}$, generated by  generators
$x_1,\ldots ,x_n$ and polynomial relations $P_k(x_1,\ldots ,x_n)=0$
, $k=1,\ldots ,m$, has the enveloping $C ^*$-algebra
$\widetilde{\mathfrak{A}}$, then we will denote it by
\[
C^*(\mathfrak{A})=C ^*(x_1,\ldots , x_n; P_k(\cdot)=0,
k=1,\ldots ,m)=C ^*(x_1,\ldots ,x_n; J)
\]
Where $J$ denotes $*$-ideal in free $*$-algebra generated by
$P_k(x_1,\ldots, x_n)\quad (k=1,\ldots, m)$  
\item[Example 3.]
The $*$-algebra $\mathcal{P}_n =\mathbb{C}\Bigl<p_1,\ldots ,p_n\mid
p_k^2=p_k=p_k^*\ ,\ k=1,\ldots ,n\Bigr>$ is $*$-bounded since $\Vert
P_k
\Vert\le 1\ (k=1,\ldots ,n)$. Then there exists unique\\
$C ^*(\mathcal{P}_n)=C ^*(p_1,\ldots ,p_n;
p_k^2=p_k=p_k^*\ ,\
 r=1,\ldots ,n)$. It is known that $C ^*$-algebra
 $C ^*(\mathcal{P}_2)=
\{f\in C([0,1],M_2(\mathbb{C}))\colon f(0),f(1)\  are\ diagonal\}$.
\end{description}

$3$. If $*$-algebra $\mathfrak{A}=\mathbb{C}\bigl< x_1,\ldots
,x_n\bigr>/J$
is not $*$-bounded it does not have enveloping
$C ^*$-algebra
$C ^*(x_1,\ldots ,x_n; J)$. However, we can always find
enveloping 
$\sigma$-$C^*$-algebra. Let us sketch this construction. 
If one replace in the
definition of the enveloping $C ^*$-algebra the condition of
commutativity of the diagram

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

for all $*$-representations of $\mathfrak{A}$ by the condition of its
commutativity for the representations with the restriction
\[
\Vert \pi(x_k)\Vert\le d_k\quad d_k>0\ ,\ k=1,\ldots ,n
\]
only, then there exists $C ^*$-algebra
$\widetilde{\mathfrak{A}}$ making this diagram commutative. We will
denote this $C ^*$-algebra by
\[
C ^*(x_1,\ldots ,x_n;\ \Vert x_k\Vert\le d_k, (k=1,\ldots
,n); J)
\]
\begin{description}
\item[Example 4.]
The algebra $\mathbb{C}[a=a^*]$ of complex polynomials in  one real
variable $(t \in \mathbb{R}^1)$
is not $*$-bounded since for any $\lambda\in\mathbb{R}$ there exists
representation $\pi_{\lambda}(a)=\lambda I$ with
$\Vert\pi_{\lambda}(a)\Vert=\mid\lambda\mid$.

However, there exists a $C ^*$-algebra $\mathfrak{A}_d=C[-d,
d]$
with one generator
$a(t)=t=t^*=a^*(t)$ and homomorphism
$\theta\colon\mathbb{C}[a=a^*]\ni a 
\mapsto a(\cdot) \in C[-d, d]$
which satisfies the following conditions:
\begin{enumerate}
\item $\Vert a(\cdot)\Vert\le d$
\item for any $*$-representation $\pi\colon\mathbb{C}[a=a^*]
\mapsto L(H)$ such that $\Vert\pi(a)\Vert\le d$ there exists a unique
representation $\widetilde{\pi}\colon =C^*(a;\ 
\Vert a\Vert\le d)
\mapsto L(H)$ such that  the following   diagram is commutative
\begin{center}
\resetparms
\btriangle[\mathbb{C}[a=a^*]`\mathfrak{A}_d`L(H);\theta`\pi`
\widetilde{\pi
}]
\end{center}
\end{enumerate}

\item[Example 5.]
The $*$-algebra $\mathcal{Q}_1=\bigl< q,\ q^*\ \mid\ q^2 =q\ ,\
(q^*)^2=q^*\bigr>$, generated by one idempotent is not $*$-bounded
since
for any $\lambda\in\mathbb{R}$ there exists the representation
\[
\pi_{\lambda}(q)=\left(
\begin{array}{cc}
1&\lambda\\
0 & 0
\end{array}
\right)
\]
with $\Vert\pi_{\lambda}(q)\Vert\rightarrow\infty\ ,\
\lambda\rightarrow\infty$. However there exists
\[
C ^*\bigl(q,q^*\ \mid\ \Vert q \Vert\le d\ ;\ q^2 =q\bigr)=
\{ f\in C([0,d], M_2(\mathbb{C}))\colon \ f(0)\ are\ diagonal\}.
\]
\end{description}

For every $*$-algebra $\mathfrak{A}= \mathbb{C}<x_1, \ldots, x_m,
\mathcal{J}>$ 
we can construct topological $*$-algebra $\widetilde{\mathfrak{A}}$
which
is also an 
enveloping $*$-algebra of $\mathfrak{A}$.
Indeed, in the previous section we have constructed algebras
$\mathfrak{A}_n=C^*(x_1, \ldots , x_m; \Vert x_j\Vert \le n, 1 \le j
\le
m;\mathcal{J})$ and 
$*$-homomorphisms $\phi_n :\mathfrak{A} \mapsto \mathfrak{A}_n $ with 
appropriate universal properties. Since $Ker \phi_n \supseteq Ker
\phi_{n+1}$ there 
exists  $*$-homomorphism $\psi_n^{n+1}:\mathfrak{A}_{n+1} \to
\mathfrak{A}_n$
 such that the following diagrams commute

\begin{center}
\resetparms
\Atriangle[\mathfrak{A}`\mathfrak{A}_{n+1}`\mathfrak{A}_{n};
\phi _{n+1}`\phi _{n}`\psi^{n+1}_{n}]
\end{center}
Consider subalgebra $\widetilde{\mathfrak{A}}$ in Cartesian product 
$\prod_{n \in \mathbb{N}} \mathfrak{A}_{n}$ consisting of elements 
$f: \mathbb{N} \to  \cup_{n \in \mathbb{N}}\mathfrak{A}_{n}$ such
that 
$\psi_n^{n+1}(f(n+1))=f(n) \quad (n \in \mathbb{N})$. Then 
$\widetilde{\mathfrak{A}}$ is topological $*$-algebra endowed with
the
weakest
topology  such that maps 
$\pi_n : \widetilde{\mathfrak{A}} \to \mathfrak{A}_{n}, \quad  f \to
f(n)$ are continuous.
We will denote $\widetilde{\mathfrak{A}}$ by $\varprojlim 
\mathfrak{A}_{n}$.
We need a lemma of Woronowicz lemma3.9.~\cite{Woron}.
\begin{lemma}\label{WSW}

Let $A$ be a $C^*$-algebra and $B$ be a $C^*$-subalgebra of $A$. 
Assume that for any two representations $\pi, \ \pi^\prime$ of $A$ if 
$\pi|B=\pi^\prime|B$ then $\pi=\pi^\prime$.
In this case $B=A$.

\end{lemma}

\begin{theorem}\label{env}

The pair $(\widetilde{\mathfrak{A}},  \phi)$ is unique enveloping 
$\sigma$-$C^*$-algebra for $\mathfrak{A}$. Homomorphism $\phi$ has
dense range. 
Moreover
$(\widetilde{\mathfrak{A}},  \phi)$    
 is also an enveloping $*$-algebra. 
\end{theorem}
\begin{proof}
 Let notice that homomorphisms $\pi_n : \widetilde{\mathfrak{A}} \to
 \mathfrak{A}_{n}$ have dense ranges.   
 The topology of $\sigma$-$C^*$-algebra $(\widetilde{\mathfrak{A}}$  
 can be determined by countably increasing
family $p_n(\cdot)$ of $C^*$-seminorms. Using Cantor's diagonal
method we can prove that it also dense in
topology determined by family $p_n(\cdot)$. Which proves that
Homomorphism $\phi$ has dense range.
 Let $\pi$ be a representation of $\mathfrak{A}$ in $B(\h)$.
 If $\pi \in \mathcal{R}_{n}$ define
 $\widetilde{\pi}=F(\pi)(\pi_{n})$
Since algebra $\phi(A)$ is dense in $\mathcal{A}$, $\widetilde{\pi}$
is
uniquely 
defined.
Let us prove that enveloping $\sigma$-$C^*$-algebra is unique.

\begin{enumerate}
\item 
  The set $Rep(\mathfrak{A})$ with ordering $ \pi_1 \le \pi_2$ iff $
  Ker(\pi_2) \subseteq Ker(\pi_1)$ is a net 
since $\pi_j \le \pi_1 \oplus \pi_2$. Choose any cofinite subnet
$\pi_n$ and define 
$\widetilde{\mathfrak{A}}_p=\varprojlim\overline{\pi_n(\mathfrak{A})}
$
\item Let $(\widetilde{\mathfrak{A}},\phi)$ be enveloping
$\sigma$-$\ca$-algebra of $\mathfrak{A}$.
We will prove that $\phi(\mathfrak{A})$ is dense in  $\mathfrak{A}$.
 Let $\mathcal{B}$ be a $\ca$-algebra and $j:\mathfrak{A} \to
 \mathcal{B}$ be 
a continuous $*$-surjection. If $\pi, \pi^\prime \in
Rep(\mathcal{B},\h) $ and 
$\pi\circ j\circ\phi = \pi^\prime\circ j\circ\phi$  then by
uniqueness 
of $\widetilde{\pi}$ in the definition of enveloping algebra we have 
$\pi\circ j = \pi^\prime\circ j$ then $\pi= \pi^\prime$ since $j$ is
surjective.
Then $j\circ\phi(A)$ is dense  in $\mathcal{B}$ by lemma~\ref{WSW}.
Present $\widetilde{\mathfrak{A}}$ as $\varprojlim \mathcal{B}_n$
(see prop.1.2.,~\cite{Phil}
 or prop.5.6.,~\cite{Arveson}), $\mathcal{B}_n$
 is a $\ca$-algebra  and $j_n : \widetilde{\mathfrak{A}} \to
 \mathcal{B}_n$ is canonical 
surjection. Previous arguments show that
$\overline{j_n(\phi(\mathfrak{A}))}=\mathcal{B}_n$.
 From which it follows that $\phi(\mathfrak{A})$ is dense $*$-algebra
 in $\widetilde{\mathfrak{A}}$
 since $ z_k \to z$ in $\widetilde{\mathfrak{A}}$ iff $j_n(z_k) \to
 j_n(z)$ for all $n$. 
\item 
We will prove that $\widetilde{\mathfrak{A}} \simeq
\widetilde{\mathfrak{A}}_p$
Passing if necessary to quotient by radical we can assume that $\phi$
is injection and so 
$\phi(\mathfrak{A}) \simeq \mathfrak{A}$.
 Since $\overline{\phi(\mathfrak{A})}= \mathfrak{A}$ we need only to
 prove that 
topology $\tau_1$ on $\phi(\mathfrak{A})$ induced from
$\widetilde{\mathfrak{A}}$ coincides with 
topology $\tau_2$ on $\phi(\mathfrak{A})$ defined by seminorms $\Vert
\widetilde\pi_n(\cdot) \Vert$.  
Note that topology $\tau_1$ defined by seminorms $\Vert \cdot
\Vert_n$ of 
faithful representations of $\mathcal{B}_n$, and since subnet $\pi_n$ 
is cofinite in the set of all representations topology $\tau_2$ is 
stronger then $\tau_1$. But since every representations $\pi_n$ can
be 
lifted to $\widetilde{\mathfrak{A}}$ then $\tau_1$ is stronger then 
$\tau_2$. It proves that $\tau_1=\tau_2$.
\end{enumerate}

Algebra $\widetilde{\mathfrak{A}}$ is a metrizable
locally-$\ca$-algebra then 
by (~\cite{FR}, corollary 4.7) every $*$-representation of
$\widetilde{\mathfrak{A}}$
is
 continuous. And so $\widetilde{\pi}$ is uniquely defined even
 without
 requirement
to be continuous. It proves that $*$-algebra
$\widetilde{\mathfrak{A}}$  is also enveloping $*$-algebra of
$\mathfrak{A}$.  
\end{proof}
Note that homomorphism $\phi$ is injection if and only if
$\mathfrak{A}$ 
has a r.f. of representations. In this case it is natural to call
$\mathfrak{A}$ $\sigma$-$C^*$-representable.
\begin{remark}
In case when $\mathfrak{A}$ is not finitely generated we can also
construct enveloping pro-$C^*$-algebra (index $n$ in above
construction  should be replaced by multi-index
$\alpha\in\mathbb{N}^{\mathbb{I}}$ where $\{x_\alpha;\
\alpha\in\mathbb{I}\}$ is the set of generators of  $\mathfrak{A}$)
with all the above statements fulfilled except that it is also
enveloping $*$-algebra and that range of homomorphism $\phi$ need not
be dense  but only quasi-dense, i.e. such that  
 for any representation $\pi\in
\rep(\mathfrak{A})$
the set $\pi(\phi(\widetilde{\mathfrak{A}}))$ is dense in $\im\pi$)

\end{remark}

It is convenient to adopt the following definitions :
\begin{definition}

We will call  a $*$-algebra of type I(nuclear) iff $\mathfrak{A}_n$
is of type I(nuclear)  
 for all $n \in \mathbb{N}$.

\end{definition}

\subsection{$*$-Representations of generators and relations}

1. To any $*$-representation of a finitely generated $*$-algebra 
$\mathfrak{B} = \mathbb{C} \langle x_1, \dots, x_n, x_1^*, \dots,
x_n^* 
\mid P_j(x_1, \dots, x_n, x_1^*, \dots, x_n^*) =0, j=1, \dots, 
m\rangle$ by bounded operators there corresponds a family of bounded 
operators $\{ X_i = \pi (x_i), X_i^* = \pi(x_i)^* = \pi 
(x_i^*)\}_{i=1}^n$, such that 
\begin{equation} \label{just_relations}
P_j(X_1,\dots, X_n, X_1^*, \dots, X_n^*) =0, \quad j=1, \dots, m.
\end{equation}
Conversely, a family of bounded operators $\{X_i, X_i^*\}_{i=1}^n$, 
satisfying \eqref{just_relations}, can be uniquely extended to a 
representation of the whole $*$-algebra $\mathfrak{B}$. For any 
finitely presented $*$-algebra, one can choose self-adjoint
generators 
$a_i = a_i^*$, $i=1$, \dots, $l$ (their number may be larger than
$n$), 
connected by self-adjoint relations $Q_j(a_1, \ldots, a_l) =
Q_j^*(a_1, 
\dots, a_l)$, $l=1$, \dots, $r$ (their number may grow, too); 
therefore, any representation $\pi$ of algebra $\mathfrak{B} = 
\mathbb{C} \langle a_1, \dots, a_l \mid a_i = a_i^*, i=1, \dots, l ; 
Q_j(a_1, \dots, a_l) = 0, j=1, \dots, r \rangle$ is uniquely
determined 
by a family of self-adjoint operators $A_i = A_i^* = \pi(a_i)$,
$i=1$, 
\dots, $l$, such that
\begin{equation}\label{self_relation}
Q_j(A_1, \dots, A_l) =0, \quad j=1, \dots, r.
\end{equation}

2. Since the properties of representation of an algebra
(irreducibility 
etc.) are completely determined by representation of its generators,
in 
the sequel we will use equivalent language of representations of 
relation \eqref{self_relation} by bounded self-adjoint operators.

Studying families of self-adjoint operators $A_1$, \dots, $A_n$, as
the 
simplest ones, we 
take,  as it is commonly accepted in the representation theory,
irreducible ones: we say that a family of self-adjoint 
operators $A_k = \int_{\mathbb{R}} \lambda_k \, dE_k (\lambda_k)$, $k
= 
1$, \dots, $n$, is irreducible, if there is no nontrivial (different 
from $H$ and $\{0\}$) subspace in $H$, which is invariant with
respect 
to all operators $E_k(\Delta)$, $k=1$, \dots, $n$; $\Delta \in 
\mathfrak{B}(\mathbb{R}^1)$. If the operators of the family are 
bounded, the irreducibility of the family means that there is no 
non-trivial subspace in $H$, invariant with respect to all operators
of 
the family $(A_k)_{k=1}^n$.

The following condition is equivalent to the irreducibility: a 
collection of self-adjoint operators $(A_k)_{k=1}^n$ is irreducible
if 
any bounded operator $C$ commuting with all $A_k$, $k=1$, \dots, $n$ 
(i.e., with all their spectral projections), is multiple of the 
identity operator.

3. For a single bounded self-adjoint operator $A= A^*$, its 
irreducibility means that $\dim H = 1$, and this operator is a 
multiplication by a constant; $A= \lambda$, $\lambda \in \mathbb{R}$, 
and the spectral theorem for a bounded self-adjoint operator gives
its 
decomposition into irreducible ones: $A = \int_{-\|A\|}^{\|A\|}
\lambda 
\, dE_A(\lambda) $, where $E_A(\cdot)$ is a spectral measure of the 
operator $A$.

4. Irreducible representations of a pair of bounded self-adjoint 
operators exist in a Hilbert space of arbitrary dimension.

Let $\dim H =n$, $e_1$, \dots, $e_n$, be orthonormal basis in $H$.
Take 
operators $A$ and $B$ such that in the basis $(e_k)_{k=1}^n$ they are 
defined by matrices
\[
A = \begin{pmatrix} \lambda_1 &&0 \\ & \ddots & \\0 && \lambda _n 
\end{pmatrix}, \quad B = (b_{ij}), \quad b_{ij} =\bar{b_{ji}},
\lambda_j
\in \mathbb{R},
\]
such that $\lambda_i \ne \lambda_j$, $i \ne j$, and for $i$ there 
exists $j$, $i\ne j$, such that $b_{ij} \ne 0$.

The pair of self-adjoint operators $A$, $B$, is irreducible. Indeed, 
if $C = (c_{ij})_{i,j=1}^n$ is a matrix commuting with the operators 
$A$ and $B$, then the condition $[A,C]=0$ gives
\[
C = \begin{pmatrix} c_{11}&&0 \\ & \ddots & \\ 0&&c_{nn} 
\end{pmatrix},
\]
and $[C,B]=0$ implies $c_{11} = \dots = c_{nn} = c$, i.e., $C = cI$, 
ant therefore, the pair $A$, $B$, is irreducible.

Let now $H$ be separable infinite-dimensional Hilbert space, and let 
$(e_k)_{k=1}^\infty$ be orthonormal basis in $H$. A pair of bounded 
self-adjoint operators 
having the following matrix representation
\[
A = \begin{pmatrix}\lambda_1 &&& 0 \\ & \ddots && \\ && \lambda _n &
\\ 
0 &&& \ddots \end{pmatrix}, \quad B = (b_{ij}),
\]
$\lambda_i \ne \lambda_j$, $i\ne j$; $|\lambda_k| \le C <\infty$, 
$k=1$, $2$,~\dots; $b_{ij} = \bar b_{ji}$, $i$, $j=1$, $2$,~\dots; 
$\forall i \ne j$ $\exists b_{ij} \ne 0$; $\sum_{j=1}^\infty
|b_{ij}|^2 
\le K < \infty$ $\forall i=1$, 2,~\dots, is irreducible.

5. Irreducible pairs connected by relation \eqref{self_relation}, 
exist, in general, not in any dimension.

Pairs of commuting bounded self-adjoint operators $A=A^*$, $B = B^*$, 
$AB=BA$, have only one-dimensional irreducible representations, $\dim
H 
= 1$, $A = \lambda_1$, $B = \lambda _2$, $(\lambda_1, \lambda_2) \in 
\mathbb{R}_2$. Joint spectral measure $E_{(A_1, A_2)}(\cdot, \cdot) = 
E_{A_1} (\cdot) \otimes E_{A_2}(\cdot)$ on the plane $\mathbb{R}^2$ 
gives a decomposition of a pair $A_1 = \int_{\mathbb{R}^2}
\lambda_1\, 
dE_{(A_1, A_2)}(\lambda_1, \lambda_2)$, $A_2 = \int_{\mathbb{R}^2} 
\lambda_2\, dE_{(A_1, A_2)}(\lambda_1, \lambda_2)$ into irreducible 
ones.

6. It may happen that there are no pairs of bounded self-adjoint 
operators $A$, $B$, connected by relation \eqref{self_relation} at
all.

For example, there are no bounded pairs of self-adjoint operators
$A$, 
$B$ (in particular, no irreducible pairs), connected by canonical 
commutation relations (CCR), $[A,B] = iI$. Indeed, otherwise we would 
have
\[
A^n B - B A^n = inA^{n-1},
\]
and 
\[
n \|A^{n-1}\| = n \|A\|^{n-1} \le 2 \|A\|^n \|B\|.
\]
Since $\|A \| \ne 0$, the latter imply $\|A\| \|B\| \ge n/2$ for all 
$n$, which 
contradicts the assumption that $A$ and $B$ are bounded.

The fact that pairs of operators, satisfying CCR, play crucial role 
in models of 
mathematical physics, stresses the need to study both bounded and 
unbounded families of operators satisfying relations.

7. As it is commonly accepted in representation theory, 
collections of operators are studied up to unitary equivalence. Two 
collections, $(A_k)_{k=1}^n$ in a Hilbert space $H$, and $(\tilde 
A_{k})_{k=1}^n$ in a Hilbert space $\tilde H$, are unitarily 
equivalent, if there exists a unitary operator $U \colon H \to
\tilde 
H$, such that the diagrams
\[
\begin{CD}
H @>{A_k}>> H \\ @V {U} VV @VV{U}V \\ {\tilde H} @>{\tilde A_k}>> 
{\tilde H}
\end{CD}
\]
are commutative for all $k=1$, \dots, $n$, i.e., $UA_k = \tilde A_k
U$.

The description of bounded representations of $*$-algebra 
$\mathfrak{B}$ up to unitary equivalence is the same thing as  the 
description of bounded representations of the generators up to
unitary 
equivalence.

8. Basic problem of $*$-representation theory is to describe for
given
relations 
\eqref{just_relations}, up to unitary equivalence, all irreducible 
families of bounded self-adjoint operators $A_1$, \dots, $A_n$, 
satisfying these relations; and this is the task we will deal with in 
the sequel.

\subsection{Pairs of self-adjoint operators satisfying quadratic 
relations}

In this chapter, we will study, in particular, pairs of self-adjoint
bounded
operators $A$, $B$, which satisfy the following
relation
\begin{gather}\label{oneone}
P_2(A,B)=
\alpha A^2+\beta\{ A,B\}+i\hbar[ B,A] + \gamma B^2 + \delta A
+ \epsilon B + \chi I =0
\\
(\alpha, \beta, \hbar, \gamma, \delta, \epsilon, \chi \in
\mathbb{R})\notag
\end{gather}

1. Let us start with the homogeneous quadratic relation
\begin{equation}\label{onetwo}
\frac q i [A,B]=\alpha A^2+\beta\{A,B\}+\gamma B^2,\quad
\alpha,\beta,\gamma, q\in\mathbb{R}.
\end{equation}

\begin{proposition}
By using a non-degenerate linear transformation, relation
\eqref{onetwo} can be reduced to one of the following forms:
\[
\begin{array}{|c|c|}
\hline
(0_0)\hfill                    & (IV_0)\hfill\\
0=0                            & [A,B]=0 \\ \hline
(I_0)\hfill                    & (V_0)\hfill  \\
A^2=0                          & \frac1i[A,B]=B^2\\ \hline
(II_0)\hfill                   & (VI_0)\hfill\\
A^2+B^2=0                      & \frac1i[A,B]=q(A^2+B^2)\\
                               & (q>0)            \\ \hline
(III_0)\hfill                  & (VII_0)\hfill     \\
A^2-B^2=0                      & \frac1i[A,B]=q(A^2-B^2) \\
                               & (q>0    )  \\ \hline
\end{array}
\]
\end{proposition}

\begin{proof}
By using a non-degenerate linear transformation, we can reduce
\[
\alpha A^2 + \beta \{A,B\} + \gamma B^2
\]
to a diagonal form. If $q=0$, then equation (\ref{onetwo})
will take one of the forms $(0_0)$--$(III_0)$. If $q\ne0$,
then by using the same transformation, we reduce the right hand
side of equality  \eqref{onetwo} to the corresponding form and
then if $\frac q i[A,B]=0$, we get $(IV_0)$, if $\frac q
i [A,B]=A^2$, replace $B$ by $q B$ to get $(V_0)$, if $\frac
q i[A,B]=A^2\pm B^2$, we get $(VI_0)$ and $(VII_0)$ with
$q=\frac1q>0$ by substituting $A$ with
$\frac{q}{|q|}A $. 
\end{proof}

2. By applying a similar arguments we can prove the following
statement.

\begin{proposition}
By using an affine change of variables, equation \eqref{oneone}
can be reduced to one of the following forms:
\[
\begin{array}{|c|c|c|}
\hline
(0_0)\hfill            & (0_1)\hfill          & (0_2) \hfill\\
0=0                    & \chi I=0\ (\chi\in\mathbb{R}, \chi\ne0)&
A=0\\
\hline
(I_0)\hfill            & (I_1)\hfill          & (I_2) \hfill\\
A^2=0                  & A^2=I                & A^2=B\\
                       & (I_1')\hfill          &             \\
                       & A^2=-I                &      \\ \hline
(II_0)\hfill           & (II_1)\hfill         &              \\
A^2+B^2=0              & A^2+B^2=I            &    \\
                       & (II_1')\hfill         &              \\
                       & A^2+B^2=-I           &    \\ \hline
(III_0)\hfill          & (III_1)\hfill        &               \\
A^2-B^2=0              & A^2-B^2=I            &    \\
\text{or } \{\tilde A,\tilde B\}=0                         & \text{or
}\{\tilde A,\tilde B\}=I 
		                              &      \\ \hline
(IV_0)\hfill           & (IV_1)\hfill         & (IV_2) \hfill \\
{}[A,B]=0                & \frac 1i[A,B]=I      & \frac1i[A,B]=A\\
\hline
(V_0)\hfill            & (V_1)\hfill          & (V_2) \hfill\\
\frac1i[A,B]=A^2       & \frac1i[A,B]=A^2+I   & \frac1i[A,B]=A^2+B\\
                       & (V_1')\hfill         &              \\
                       & \frac1i[A,B]=A^2-I   &              \\
                       \hline
(VI_0)\hfill           & (VI_1)\hfill         &              \\
\frac1i[A,B]=q(A^2+B^2)&\frac1i[A,B]=q(A^2+B^2)+I &    \\
(q>0)                  & (q\in\mathbb{R}, q\ne0)   &              \\
\hline
(VII_0)\hfill          & (VII_1)\hfill        &               \\
\frac1i[A,B]=q(A^2-B^2)&\frac1i[A,B]=q(A^2-B^2)+I &    \\
(q>0)                  & (q\in\mathbb{R}, q\ne0)   &           \\
\hline
\end{array}
\]
\end{proposition}

Our further aim is to describe for each relation
$(0_0)$--$(VII_0)$
pairs of (bounded or unbounded) self-adjoint
operators, which satisfy this relation.

Considering solutions of the equations we have:

$(0_1)$ $\chi I=0$ $(\chi\ne0)$. There are no pairs $A$, $B$
which satisfy $(0_1)$;

$(0_2)$ $A=0$. Since $B=B^*$ it is an arbitrary (bounded or
unbounded) self-adjoint operator, the only irreducible
representations are one-dimensional, $A=0$, $B=b$ and their
structure is given by the structure theorem for a single
operator $B$;

$(I_0)$ $A^2=0$. Because $A=A^*$, $A^2=A=0$ and case $(I_0)$ is
similar to $(0_2)$. The structure of any solution of equation
$(I_0)$ is the following: $A=0$,
$B=\int_{\mathbb{R}^1}\lambda\,dE_B(\lambda)$, where $E_B(\cdot)$ is
the resolution of identity for the operator $B$ concentrated on a
compact 
set $K \subset 
\mathbb{R}$;

$(I_1')$ $A^2=-I$. This equation doesn't have solutions.

$(I_2)$ $A^2=B$. The structure of any bounded
solution of equation $(I_2)$ has the form $A=\int_{\mathbb{R}^1}
\lambda\, dE_A(\lambda)$, $B=\int_{\mathbb{R}^1}\lambda^2\,
dE_A(\lambda)$, where $E_A(\cdot)$ is the resolution of identity
for the operator $A$, concentrated on a compact set $K \subset 
\mathbb{R}$;

$(II_0)$ $A^2+B^2=0$.  Here, $A=B=0$;

$(II_1')$ $A^2+B^2=-I$. There are no solutions.

The rest of the relations can be divided into four groups:
\[
\begin{array}{|c|c|c|}
\hline
%====================================================================
\multicolumn{3}{|c|}{\text {$*$-wild relations}}\\ \hline
%--------------------------------------------------------------------
(0_0)\hfill          &(I_1)\hfill            &                \\
0=0                  &A^2=I                  &          \\ \hline
%====================================================================
\multicolumn{3}{|c|}{\text {$\mathcal{F}_4$ relations}}\\ \hline
%--------------------------------------------------------------------
                     &(II_1)\hfill           &            \\
                     & A^2+B^2=I             &             \\ \hline
%--------------------------------------------------------------------
(III_0)\hfill        &(III_1)\hfill          &             \\
\{A,B\}=0            &\{A,B\}=I              &       \\ \hline
%====================================================================
\multicolumn{3}{|c|}{\text{Lie algebras and their non-linear
transformations}} \\ \hline
%--------------------------------------------------------------------
(IV_0)\hfill         &(IV_1)\hfill           &(IV_2)\hfill \\
{} [A,B]=0              &\frac1i[A,B]=I         &\frac1i[A,B]=A\\
\hline
%--------------------------------------------------------------------
(V_0)\hfill          &(V_1)\hfill            &(V_2)\hfill\\
\frac1i[A,B]=A^2     &\frac1i[A,B]=A^2+I     &\frac1i[A,B]=A^2+B\\
                     &(V_1')\hfill           &                \\
                     &\frac1i[A,B]=A^2-I     &         \\ \hline
%====================================================================
\multicolumn{3}{|c|}{\text{$q$-relations}} \\ \hline
%--------------------------------------------------------------------
(VI_0)\hfill         &(VI_1)\hfill          &          \\
\frac1i[A,B]=q(A^2+B^2)&\frac1i[A,B]=q(A^2+B^2)+I&\\
(q>0) & (q\in\mathbb{R}, q\ne0)&\\ \hline
%--------------------------------------------------------------------
(VII_0)\hfill         &(VII_1)\hfill          &          \\
\frac1i[A,B]=q(A^2-B^2)&\frac1i[A,B]=q(A^2-B^2)+I&\\
(q>0) & (q\in\mathbb{R}, q\ne0)&\\ \hline
%====================================================================
\end{array}
\]

In what follows, our aim is, in particular, to study irreducible 
representations for each of these
groups of relations.