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\section{Introduction to representations of $*$-algebras}
\label{sec:1.1}

\subsection{$*$-Representations: key words}\label{sec:1.1.1}
\markboth{Chapter 1. Pairs of self-adjoint operators}{1.1. Introduction
to representations of $*$-algebras}

\textbf{1.} A representation\index{representation} 
of an algebra $\mathcal{A}$ on a
finite-dimensional Hilbert (unitary) space $H$ is a homomorphism
$\pi$
of $\mathcal{A}$ into the algebra $L(H)$ of linear transformations
on
$H$.
A $*$-representation of a  $*$-algebra $\mathfrak{A}$ is a
$*$-homomorphism
$\pi$ from the algebra into the $*$-algebra $L(H)$ of bounded operators on a separable
Hilbert space $H$. The dimension of the 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 on a separable Hilbert space ($\dim H
\le\infty$)
if $\mathfrak{A}$ is a $*$-algebra.

\medskip\noindent\textbf{
2.} In the representation theory of
algebras,  representations are studied up to some equivalence. We
call representations of $\mathcal{A}$, $\pi$ on $H$ and
$\tilde{\pi}$ on
$\tilde{H}$,
equivalent\index{representations!equaivalent} 
if there exists an invertible operator $C\colon
H\mapsto\tilde{H}$ that intertwines the representations $\pi$
and
$\tilde{\pi}$, i.e.,
\[
  C \pi (x) = \tilde{\pi}(x) C,\qquad \forall x\in\mathcal{A}.
\]

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

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

We have the following simple proposition.

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

\begin{proof}
One has to prove only  that the equivalence of\/ $*$-representa\-tions
$\pi$
on
$H$
and $\tilde{\pi}$ on $\tilde{H}$ 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,\qquad \forall x\in\mathfrak{A}. 
\end{equation}

Let us consider the polar decomposition of the  operator $C$, $C = U A$, where
$A=(CC ^*)^{1/2}$ is an invertible positive operator
on
$H$
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,\qquad \forall
x\in\mathfrak{A}. 
\end{equation}
Taking adjoints of both sides, we obtain
\begin{equation}
\tilde{\pi} (x) = UA^{-1}{\pi}(x)A U^{-1},\qquad \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 a positive operator, we have $A^2\pi(x)=\pi(x)A^2$,
$x\in\mathcal{A}$, which implies that $A \pi(x)=\pi(x)A$ 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,
\[
U\pi(x)=\tilde{\pi}(x)U,   \qquad \forall x\in\mathfrak{A}.
\]
Hence we have a unitary equivalence of the representations $\pi$ on $H$
and
$\tilde{\pi}$ on $\tilde{H}$.
\end{proof}

\noindent\textbf{3.}
In the  general representation theory one distinguishes
irreducible and indecomposable representations
in the set of  all {\it finite-dimen\-sional} representations.
A representation
$\pi\colon\mathcal{A}\mapsto L(H)$ is called
irreducible\index{representation!irreducible} 
if there
exists no non-trivial subspace of $H$ invariant with respect to 
all operators $\pi(x)$, $x\in\mathcal{A}$. A representation
$\pi\colon\mathcal{A}\mapsto L(H)$ is called 
indecomposable\index{representation!indecomposable} 
if there
exists no decomposition $H=H_1 + H_2$ into a sum of the two
non-trivial subspaces that are invariant with respect to all the
operators 
$\pi(x)$  $x\in\mathcal{A}$, and $ H_1 \cap H_2=\{ 0\}$. It is
clear
that any irreducible
representation is indecomposable. Thus the set of irreducible
representations is a subset  of the set of all indecomposable 
representations.
 The size of this subset in
the whole set of indecomposable representations depends on the structure 
of~$\mathcal{A}$.

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

In the case where $\mathfrak{A}$ is an algebra with an involution, one
can
consider both irreducible $*$-representations and indecomposable
$*$\nobreakdash-repre\-sen\-ta\-tions.
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 $*$-repre\-sentation
is
irreducible. Assume the contrary, that is, let an indecomposable
representation be
reducible, i.e., there exists a proper subspace $H_1$ in $H$
invariant
with
respect to  all $\pi(x)$, $x\in\mathfrak{A}$.

\begin{lemma}
The subspace $H_1^{\perp} = \{y\in H\colon (y,f)=0,\ \forall f\in
H_1\}$ is non-trivial and invariant with respect to  all $\pi(x)$, 
$x\in\mathfrak{A}$.
\end{lemma}

\begin{proof}
If $y\in H_1^{\perp}$, then
\[
( \pi(x)y , f )=( y , \pi(x^*) f ) = 0
\]
for all $f$ from the invariant subspace $H_1$, i.e.
$\pi(x)y\in H_1^{\perp}$.
\end{proof}

 The contradiction immediately 
follows from the lemma.
\end{proof}

\noindent\textbf{4.}
 Let $H$ be a  separable (generally speaking,
infinite-dimensional) Hilbert space. Following the general
strategy of representation theory, we take irreducible representations
to be the ``simplest''
among all $*$-representations. A
$*$-representation $\pi\colon\mathfrak{A}\mapsto L(H)$ is called
irreducible, if there exists no non-trivial subspace in $H$ invariant
with respect to  all the operators $\pi(x)  \ (x\in\mathfrak{A})$.
The
following form of Schur's lemma, 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
\[
C\pi(x)=\pi(x)C,\qquad \forall x\in\mathfrak{A},
\]
is a multiple of the identity, i.e., $C=cI$, with $c\in\mathbb{C}$.
\end{proposition}

\begin{proof}
If $A=A^*$  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 sets $\Delta\subset\mathbb{R}^1$
(here $E_A (\Delta)$ is a spectral projector of the operator $A$).
 In this case, $H_{\Delta}=E_A (\Delta) H$ is an
invariant subspace in $H$.

If the 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$ also commute with $\pi(\cdot)$, 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 the lemma,  $H_1^{\perp}$ is also
invariant. Then the operator
 \[
 C = \left(
 \begin{array}{cc}
 c_1 I_{H_1} & 0 \\
 0 & c_2 I_{H_1^{\perp}}
 \end{array}
 \right),
 \qquad c_1\neq c_2 ,\qquad c_1,c_2\in\mathbb{C},
 \]
commutes with the representation and is not a multiple of the
identity.
\end{proof}

\begin{remark}
It is possible to define the notion of an indecomposable
representation
in the case where $H$ is a separable Hilbert space ($\dim H =\infty$) 
and to prove an 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,  $\sirrep\mathfrak{A}$, in the category 
$\srep\mathfrak{A}$.
The condition  for a sub-category to be full means that
the embedding functor $F$ from  $\sirrep\mathfrak{A}$ into
 $\srep\mathfrak{A}$  is an
isomorphism on
the
corresponding morphisms. In what follows, we will mainly deal with
$*$-algebras and 
their $*$-representations; thus we will sometimes omit the involution
sign 
with the words algebra, morphism, category, and representation, if
no ambiguity can arise.   

\subsection{$C^*$-representable $*$-algebras}\label{sec:1.1.2}

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

If a $*$-algebra $\mathfrak{A}$  is 
$C ^*$-representable\index{algebra@$*$-algebra!$C^*$-representatble}, 
i.e., there
exists
a
$*$-iso\-morphism of $\mathfrak{A}$ on  a $*$-subalgebra of 
a $C ^*$-algebra,
then it
is
clear that $\mathfrak{A}$ has a r.f. 

\begin{remark}
1).
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 an infinite 
compact set
$K\subset\mathbb{R}^1$.
However,
$C(K_1)$ is isomorphic to $C(K_2)$ if
and only if $K_1$ and $K_2$ are homeomorphic.

2).
Not any $*$-algebra is $C ^*$-representable. Moreover, not any
$*$-algebra possesses a r.f. of $*$-representations. For
example, if the
involution in
$\mathcal{A}$ is not proper 
(an involution is proper if $xx^*=0$ implies $x=0$),
 then such a $*$-algebra cannot be
$C ^*$-representable; moreover it does not possess  r.f. of
$*$-representations.
\end{remark}

Nevertheless, if $\mathfrak{A}=\mathbb{C}[G]= \mathbb{C} \langle g\mid
g\in G\rangle$
is a group $*$-algebra\index{algebra@$*$-algebra!group}
of a countable discrete group $G$ with a natural involution (given on
basis
vectors $g \in G$ by $ 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
$\pi_r$ on $L_2 (G)$ is non-zero for any non-zero element
$x\in\mathbb{C}[G]$, and hence the $*$-algebra $\mathbb{C}[G]$
is isomorphic to
$\pi_r
(\mathbb{C}[G])$ which is a dense  $*$-subalgebra
of the
$C ^*$-algebra $C_r^* (G)$ generated by the operators $\pi(x)$, 
$x\in\mathbb{C}[G]$.
\end{proof}

\noindent\textbf{2.}
 The following  proposition holds.

\begin{proposition}\label{prop:cstar}
Consider the following properties of a $*$-algebra\textup:

\begin{itemize}
\item[$(i)$]  the algebra is $C ^*$-representable\textup;

\item[$(ii)$] there exists a residual family of its
representations\textup;

\item[$(iii)$] the involution on the algebra is completely
  proper\index{involution!completely proper}
\textup(i.e.,
 the equality $\sum_{k=1}^n x_k x_k^* = 0$ implies $x_k = 0$,
$k=1$,\dots, $n$; $n\in\mathbb{N}$\textup)\textup;

\item[$(iv)$] the involution in the algebra is
  proper\index{involution!proper} 
\textup(i.e.,
$xx^*=0$
implies $x=0$\textup).
\end{itemize}

Then  $(i)\Rightarrow
(ii)\Rightarrow (iii)\Rightarrow (iv)$.

Neither of the inverse implications holds. 
\end{proposition}

\begin{proof}
The direct implications are easily verified.
To see that $(ii)$ does not imply $(i)$, one has to consider the
$*$-algebra $C ([0,\infty))$ of all continuous functions
$f(\cdot)$ on $[0,\infty)$ with the 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 a $C ^*$-algebra $\mathfrak A$,
then there exists an integer $N$ such that $\| f\| < N$, where
\[
f(x)=
\begin{cases}
x-n,& \text{if $2n < x < 2n+1$},\\
n,& \text{if  $2n-1 \le  x \le  2n$},
\end{cases}
\qquad n \in \mathbb{N}.
\]

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

In order to show that $(iii)$ does not imply $(ii)$, consider
the
$*$-algebra with unit $e$, 
$\mathfrak A= \mathbb{C}\langle a,x\mid a^*a=qaa^*,\, xx^*+aa^*=e\rangle$,
where $0<q<1$.
The set of words  which do not contain the sub-words
$a^*a$ and $xx^*$ forms
a linear basis for this algebra.
Take an arbitrary
 $z=\sum \alpha_iu_i\,x^{k_i}\in \mathfrak A$, where the words
$u_i$ do not
end with 
$x$, and $k_i \ge 0$. Denote by $F(z)$ the  sum  of coefficients
at those words $u_ix^{k_i}$ of minimal length that can also be written
in the form $ww^*$. Define $J=\{j\colon l(u_j) \le l(u_i), \,\forall
i \} $.
We will  show that $F(zz^*)=\sum_{j\in J}|\alpha_j|^2$.
Indeed, the canonical
 form of $u_ix^{k_i} x^{*k_j}u_j^*$
is $-u_i\bigl(\sum_{1\le s\le
\min(k_i,k_j)}x^{k_i-s}(x^*)^{k_j-s}\bigr)\,u_j^*
+ \mathop{\text{CF}}(u_iu_j^*)$,
where $\mathop{\text{CF}}(u_iu^*_i)$
denotes the canonical
form of $u_i u_i^*$.
Since $u_i$ and $u_j$ do not end with $x$,  $\mathop{\text{CF}}(u_iu_j^*)$
is a word
of length $|u_i|+|u_j|$. So if the unique shortest word, 
$\mathop{\text{CF}}(u_iu_j^*)$,
in $\mathop{\text{CF}}(zz^*)$ has minimal
length in $zz^*$, then $i$, $j\in J$ (hence $|u_i|=|u_j|$).
Let us show that if $u_iu_j^*=ww^*$,  then $u_i=u_j$.
Let $u_i$ be a word in $a$, $a^*$.
If $u_i$ ends with $a$, or $u_j$ ends with $a^*$,
then $\mathop{\text{CF}}(u_iu_j^*)$ is
$u_iu_j^*$ (as in the free
$*$-algebra), and  we conclude from $u_iu_j^*=ww^*$ 
that  $u_i=u_j$. In the opposite case,
write $u_i=v_ia^{*k}$ and $u_j=v_ja^{m}$, where $v_i$ ends with $a^*$
and  $v_j$ ends with $a$. Then
$\mathop{\text{CF}}(u_iu_j^*)=q^{km}v_ia^ma^{*k}v_j^*$.
If $m>k$, then, since $u_iu_j^*=ww^*$, we have that $v_ia^{m_1}=w$,
 $a^{m-m_1}a^{*k}v_j^*=w^*$, which is impossible, since
$w$ ends with
$a$ and $a^*$ simultaneously. So  $m=k$ and $w=v_ia^{k}=v_ja^{k}$. Hence,
$v_i=v_j$ and $u_i=u_j$.
Now let $u_k=u_{k,1}x^{\#}u_{k,2}$, $k=i$, $j$,
where $u_{k,1}$ does not contain
$x^\#$ ($x^{\#}$ stands for either $x$ or $x^*$).
Then it follows from  $u_iu_j^*=ww^*$  that $u_{1,1}=u_{2,1}$ and
$u_{1,2}u_{2,2}^*=w_1w_1^*$. By induction on $l(u_i)$,
we obtain the desired result.
This proves that, if  $u_iu_j^*=ww^*$,  then $u_i=u_j$.
From this it follows that $F(zz^*)=\sum_{j\in J}|\alpha_j|^2$.
Using the existence of $F(\cdot)$, it is easy to show that the
$*$-algebra is completely proper. 
But in every representation, we have that 
$\|aa^*\|=\|a^*a\|=q\,\|aa^*\|$, hence
$\|aa^*\|=0$, and so $A$ is not $C^*$-representable.



To see that $(iv)$ does not imply $(iii)$, we refer the reader
to~\cite{wich} (see also~\cite{dor_bel} and the references therein).
\end{proof}

\textbf{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.).\index{algebra!residually finite dimensional}

\begin{proposition}
If $G$ is a residually finite group\index{group!residually finite} 
\textup(i.e. $\forall g\neq e $ there
exists a normal subgroup $G_g\not\ni g$ such that $G/G_g$ is
a finite group\textup), then $\mathbb{C}[G]$ is residually finite
dimensional.
\end{proposition}

\begin{proof}
Let us first  recall an equivalent definition of the residual
finiteness of a group $G$: for any finite set $\{g_1, \ldots , g_n \}$
of non-identity
elements
of $G$ there exists a normal subgroup that does not contain any of
these elements and such that the quotient group of $G$ by this
subgroup is finite.

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

\noindent\textbf{4.}
The following list of ${C}^*$-algebras is naturally
connected with a group $G$:
\begin{itemize}
\item[1.] $C_f^*(G)$. 
This algebra is the completion of  $\mathbb{C}[G]$ in the following
norm:
\[
\Vert\alpha\Vert_{C_{f}^{*}(G)} = \sup_{\pi\in
\sfdrep\mathbb{C}[G]}
\Vert\pi(\alpha)\Vert,\qquad \alpha\in\mathbb{C}[G],
\]
where $\sup$ is taken over all finite-dimensional
$*$-representations
of $\mathbb{C}[G]$.

\item[2.]  $C_r^* (G)$.
This algebra is the completion of $\mathbb{C}[G]$ in the right
regular
norm
\[
\Vert \alpha\Vert_{C_r(G)}= \Vert\pi_r
(\alpha)\Vert,
\qquad \alpha\in\mathbb{C}[G],
\]
where $\pi_r (\alpha)$ is the operator of the right regular
representation
of $G$ on  $L_2(G)$.

\item[3.] $C^*(G)$.
This algebra is the  completion  of $\mathbb{C}[G]$ in the following
norm:
\[
\Vert\alpha\Vert_{ C^*(G)}=\sup_{\pi\in \srep\mathbb{C}[G]}
\Vert\pi(\alpha)\Vert, \qquad \alpha\in\mathbb{C}[G].
\]
\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 any $\pi\in \srep \mathbb{C}[G]$ (for more
detailed information about $*$-boundedness see \ref{sec:1.1.3}). Hence all
norms defined above are finite for any $\alpha\in\mathbb{C}[G]$.
The following proposition can be found in \cite{29}.

\begin{proposition}
If $G$ is a residually 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}
The only non-trivial part is the proof of  the inequality
\[
\Vert\alpha\Vert_{{C}_f^* (G)}\ge
\Vert\alpha\Vert_{{C}_r^* (G)},\qquad
\forall\alpha\in\mathbb{C}[G].
\]
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
$\Vert\alpha\gamma\Vert_{L_2(G)}\ge d-\varepsilon/2$. Hence,
there exists $\delta =\sum_{1}^n \delta_k h_k$ such that
$\Vert \delta\Vert_{L_2 (G)}=1$ and
$\Vert \gamma -\delta\Vert_{L_2 (G)}\le\varepsilon/{2d}$.
This
implies
$\Vert\alpha \gamma-\alpha\delta\Vert_{L_2
(G)}\le{\varepsilon}/{2}$,
and
$\Vert\alpha\delta \Vert_{L_2 (G)}\ge d -\varepsilon$.
Further, let us choose a normal subgroup ${N}$ of the group $G$
which
does not contain non-trivial  elements among $g_kg_l^{-1}$, 
$h_th_s^{-1}$,  $g_kh_th_s^{-1}g_l^{-1}$; $ k$, $l=1$, \dots, $m$,
$t$, $s=1$,\dots,
$n$, and
such
that the quotient group of $G$ by this subgroup is
finite. Then, in the regular representation of
$G/{N}$, we have
\[
\Vert\alpha\delta\Vert_{L_2 (G
/{N})}=\Vert\alpha\delta\Vert_{L_2(G)}\ge
d-\varepsilon.
\]
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}\label{rem:resid}
If $G$ is a 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 a r.f.d.\
$*$\nobreakdash-alge\-bra.
Indeed, $C^*(\mathcal{F}_2)$, where $\mathcal{F}_2$
is the free group with two generators, is a r.f.d.\ $*$-algebra~\cite{21};
however,
$C_r^*(\mathcal{F}_2)$ is not a r.f.d.\ $*$\nobreakdash-algebra,
since it is
simple~\cite{30}.
The residually finite group $SL(2,\mathbb{Z}[\frac{1}{p}])$ ($p$ is a 
prime number) is an example of a r.f.\ group $G$ for which 
$C^*(G)$ is not r.f.d.\ see~\cite{77}.
\end{remark}

\subsection{Enveloping $*$-algebras and $C^*$-algebras}\label{sec:1.1.3}

\textbf{1.}
Sometimes it is possible to reduce the study of representations
of a
$*$-algebra $\mathfrak{A}$ to the study of $*$-representations of its
enveloping $*$-algebra, $\sigma$-${C}^*$-algebra, or 
$C^*$-algebra.\index{algebra!enveloping $*$-algebra}
 Let us
recall the definition.

\begin{definition}
Let\/ $\mathfrak{A}$ be  a $*$-algebra. The pair
$(\tilde{\mathfrak{A}};
\phi\colon\mathfrak{A}\mapsto\tilde{\mathfrak{A}})$, where
$\tilde{\mathfrak{A}}$ is a $*$-algebra and $\phi$ is a
$*$-homomorphism, is
called an enveloping $*$-algebra of the algebra $\mathfrak{A}$ if for any
$*$-representation $\pi\colon\mathfrak{A}\mapsto L(H)$ of the algebra
$\mathfrak{A}$ there exists a unique $*$-representation
$\tilde{\pi}\colon\tilde{\mathfrak{A}}\mapsto L(H)$ such that
the
following diagram is commutative.
\begin{center}
\resetparms
\btriangle[\mathfrak{A}`\tilde{\mathfrak{A}}`L(H);\phi`\pi
`\tilde{\pi}]
\end{center}
\end{definition}

\begin{example} Let $\Sigma $ be any family of elements of
a  $*$-algebra
$\mathfrak{A}$ which are invertible in any $*$-representation
$\pi\colon\mathfrak{A}\mapsto L(H)$. Let us denote
 the algebra of
quotients
of $\mathfrak{A}$ with respect  to
$\Sigma$ by $\tilde{\mathfrak{U}}=\mathfrak{A}[\Sigma^{-1}]$  (see
~\cite{117}). Let 
$\phi\colon\mathfrak{A}\mapsto \tilde{\mathfrak{U}}$ be the
natural
homomorphism. Then we obtain an enveloping $*$-algebra for the
the $*$-algebra
$\mathfrak{A}$. Let us note that in the case where $\mathfrak{A}$ is
$C^*$-representable, $\phi$ is an injection. 
\end{example}

\noindent\textbf{2.}
If $\tilde{\mathfrak{A}}$ carries the structure of a
${C}^*$-algebra, then $\tilde{\pi}$ is a continuous
$*$-homomorphism from the ${C}^*$-algebra
$\tilde{\mathfrak{A}}$
to the ${C}^*$-algebra $L(H)$. In this case the pair
$(\tilde{\mathfrak{A}}, \phi)$ is called an enveloping
${C}^*$-algebra\index{algebra!enveloping $C^*$} 
of the algebra $\mathfrak{A}$. The enveloping
${C}^*$-algebra is unique in the class of $C^*$-algebras,
if it exists (see Theorem~\ref{env}).
Indeed, in this case  $\tilde{\mathfrak{A}}$ 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
$C^*(\mathfrak{A})$.

Let $\mathfrak{A}=\mathbb{C}\langle
x_1,\dots, x_n, x_1^*, \dots, x_n^*\mid P_k(\cdot)=0,\ k=1, \dots,
m \rangle=
\mathbb{C}\langle
x_1,\ldots, x_n,  x_1^*, \ldots, x_n^*\mid J \rangle $  denote
the $*$-algebra with generators\index{algebra@$*$-algebra!with
  generators and relations}  
$x_j$, $x_j^*$, $j=1$, \dots, $m$,  and relations  $P_k(x_1,\dots,
x_n, x_1^*, \dots, x_n^*)=0$,
 (where $P_k(\cdot)$ are non-commutative polynomials),  i.e.,
$\mathfrak{A}$ is the  quotient of
the free $*$-algebra $\mathbb{C}\langle
x_1,\dots, x_n, x_1^*, \dots, x_n^* \rangle$
with respect to the two-sided $*$-ideal $J$  generated by the relations.
 In the sequel, we will sometimes omit  the symbols
$x_1^*, \ldots, x_n^*$, if it is clear from the context, that the
considered algebra is a $*$-algebra. 
 
If a $*$-algebra $\mathfrak{A}$, generated by  generators
$x_1$, \dots , $x_n$ and polynomial relations $P_k(x_1,\dots ,x_n, x_1^*,
\dots, x_n^*)=0$, $k=1$,\dots , $m$, has the enveloping $C ^*$-algebra
$\tilde{\mathfrak{A}}$, then we will denote it by
\begin{align*}
C^*(\mathfrak{A})&=C ^*(x_1,\dots , x_n; P_k(\cdot)=0,
k=1,\dots ,m)
\\
&=C ^*(x_1,\dots ,x_n; J).
\end{align*}



A $*$-algebra $\mathfrak{A}$ is called 
$*$-bounded\index{algebra@$*$-algebra!$*$-bounded} if for  any
$x\in\mathfrak{A}$ there exists a number $C_x< \infty$ such that
$\Vert\pi(x)\Vert\le C_x$ for any of its
$*$\nobreakdash-representa\-tions $\pi\colon\mathfrak{A}\mapsto L(H)$. 
For a finitely-generated $*$-algebra 
\[
\mathfrak{A}=\mathbb{C}\langle
x_1,\dots, x_n \mid J \rangle 
\]
 to be $*$-bounded it is sufficient that
for any of its $*$-representations 
$\pi\colon\mathfrak{A}\mapsto L(H)$ and
any
$k=1$, \dots , $n$, 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 $\mathfrak{A}$ with respect to the
norm $\Vert x\Vert =\sup_{\pi \in
\srep(\mathfrak{A})}\Vert\pi(x)\Vert\le
C_{x}<\infty$ (sup
is taken over all $*$-representations $\pi$).

\begin{example}
The  group $*$-algebra $\mathbb{C}[G]$ is $*$-bounded, since for any
generator $g\in G$ and any $\pi\in\srep\mathbb{C}[G]$ the
operator
$\pi(g)$ is unitary and $\Vert\pi(g)\Vert = 1$. Hence there exists
the
enveloping $C^*$-algebra
$C^*(G)=C^*(\mathbb{C}[G])$.
\end{example}

\begin{example} \label{ex:proj}
The $*$-algebra 
\[
\mathcal{P}_n =\mathbb{C}\bigl<p_1,\dots ,p_n\mid
p_k^2=p_k=p_k^* ,\ k=1,\dots ,n\bigr>
\]
 is $*$-bounded, since $\Vert
P_k
\Vert\le 1$, $k=1$,\dots , $n$. Then there exists a unique
$C ^*(\mathcal{P}_n)=C ^*(p_1,\dots ,p_n;
p_k^2=p_k=p_k^* ,\
 k=1,\ldots ,n)$. It is known that the $C ^*$-algebra
 $C ^*(\mathcal{P}_2)$ can also be defined as $C ^*(\mathcal{P}_2)=
\{f\in C([0,1],M_2(\mathbb{C}))\mid f(0),f(1)\ \text{are diagonal}\}$
(see~\cite{114,115} and others).
\end{example}

\noindent\textbf{3.}
If a $*$-algebra $\mathfrak{A}=\mathbb{C}\langle x_1,\dots
,x_n\rangle/J$
is not $*$-bounded, it does not have an enveloping
$C ^*$-algebra,
$C ^*(x_1,\dots ,x_n; J)$. However, we can always find an
enveloping 
$\sigma$-$C^*$-algebra.\index{algebra!enveloping $\sigma$-$C^*$} 
Let us sketch this construction. 
If in the
definition of the enveloping $C ^*$\nobreakdash-algebra,
one replaces the condition that the diagram

\begin{center}
\resetparms
\btriangle[\mathfrak{A}`\widetilde{\mathfrak{A}}`L(H);\phi`\pi`
\widetilde{\pi}]
\end{center}
is commutative
for all $*$-representations of $\mathfrak{A}$ by the condition that it is
commutative only for the representations subject to the restriction
\[
\Vert \pi(x_k)\Vert\le d_k,\qquad d_k>0 ,\qquad k=1,\dots ,n,
\]
then there exists a $C ^*$-algebra
$\tilde{\mathfrak{A}}$ making this diagram commutative. We will
denote this $C ^*$-algebra by
\[
C ^*(x_1,\ldots ,x_n;\, \Vert x_k\Vert\le d_k,\, k=1,\dots
,n; J).
\]

\begin{example}
The
algebra $\mathbb{C}[a=a^*]$ of complex polynomials of  one real
variable $t \in \mathbb{R}^1$
is not $*$-bounded, since for any $\lambda\in\mathbb{R}$ there exists
a 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 self-adjoint generator
$a(t)=t$ and a 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 $\tilde{\pi}\colon C^*(a=a^*;\ 
\Vert a\Vert\le d)
\mapsto L(H)$ such that  the following   diagram is commutative
\begin{center}
\resetparms
\btriangle[\mathbb{C}[a=a^*]`\mathfrak{A}_d`L(H);\theta`\pi`
\tilde{\pi
}]
\end{center}
\end{enumerate}
\end{example}

\begin{example}
The
$*$-algebra $\mathcal{Q}_1=\langle q, q^* \mid q^2 =q ,\
(q^*)^2=q^*\rangle$, 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
\begin{gather*}
C ^*\bigl(q,q^* ; \Vert q \Vert\le d ;\ q^2 =q\bigr)
\\
=
\{ f\in C([0,d], M_2(\mathbb{C}))\colon  \text{$f(0)$ is diagonal}\}.
\end{gather*}
\end{example}

For every $*$-algebra $\mathfrak{A}= \mathbb{C}\langle x_1, \ldots, x_m;
J\rangle$ 
we can construct a topological $*$-algebra $\tilde{\mathfrak{A}}$
which
is also an 
enveloping $*$-algebra of $\mathfrak{A}$.
Indeed, in the previous section we have constructed the algebras
$\mathfrak{A}_n=C^*(x_1, \dots , x_m; \Vert x_j\Vert \le n,\, 1 \le j
\le
m;{J})$ and 
$*$\nobreakdash-homomor\-phisms $\phi_n \colon\mathfrak{A}
\to \mathfrak{A}_n $ with 
appropriate universal properties. Since $\ker \phi_n \supseteq \ker
\phi_{n+1}$, there 
exists a $*$-homomorphism $\psi_n^{n+1}\colon\mathfrak{A}_{n+1} \to
\mathfrak{A}_n$
 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 a subalgebra $\tilde{\mathfrak{A}}$ in the Cartesian
product 
$\prod_{n \in \mathbb{N}} \mathfrak{A}_{n}$ consisting of elements 
$f\colon \mathbb{N} \to  \cup_{n \in \mathbb{N}}\mathfrak{A}_{n}$ such
that 
$\psi_n^{n+1}(f(n+1))=f(n)$,  $n \in \mathbb{N}$. Then 
$\tilde{\mathfrak{A}}$ is a topological $*$-algebra endowed with
the
weakest
topology  such that the maps 
$\pi_n \colon \tilde{\mathfrak{A}} \to \mathfrak{A}_{n}$, $f \mapsto
f(n)$ are continuous.
We will denote $\tilde{\mathfrak{A}}$ by $\varprojlim 
\mathfrak{A}_{n}$.
We need the following statement~\cite[Lemma~3.9]{kru_wor}.

\begin{lemma}\label{WSW}
Let $A$ be a $C^*$-algebra and $B$ be a $C^*$-subalgebra of $A$. 
Assume that $\pi\restriction B=\pi'\restriction B$
for any two representations $\pi$, 
$\pi'$ of $A$ implies that 
$\pi=\pi^\prime$.
In this case $B=A$.  
\end{lemma}

The following theorem holds.

\begin{theorem}\label{env}
 The pair $(\tilde{\mathfrak{A}},  \phi)$ is a unique enveloping 
$\sigma$-$C^*$-algebra for $\mathfrak{A}$. The homomorphism $\phi$ has
dense range. 
Moreover
$(\tilde{\mathfrak{A}},  \phi)$    
 is also an enveloping $*$-algebra. 
\end{theorem}

\begin{proof}
 Let us notice that the homomorphisms $\pi_n \colon
\tilde{\mathfrak{A}} \to
 \mathfrak{A}_{n}$ have dense ranges.   
 The topology of the $\sigma$-$C^*$-algebra $\tilde{\mathfrak{A}}$  
 can be defined by a countable increasing
family $p_n(\cdot)$ of $C^*$-semi-norms. Using arguments similar
to Cantor's
diagonal
method, we can prove that it is also dense in
the topology defined by the family $p_n(\cdot)$, which proves that
the homomorphism $\phi$ has dense range.
 Let $\pi$ be a representation of $\mathfrak{A}$ in $L(H)$.
 If $\pi \in \mathcal{R}_{n}$, define
 $\tilde{\pi}=F(\pi)(\pi_{n})$.
Since the algebra $\phi(A)$ is dense in $\mathcal{A}$,
$\tilde{\pi}$
is
uniquely 
defined.
Let us prove that the enveloping $\sigma$-$C^*$-algebra is unique.

1.  We say that  $ \pi_1 \le \pi_2$ for $\pi_1$, $\pi_2 \in
  \rep(\mathfrak{A})$ iff $
  \ker \pi_2 \subseteq \ker \pi_1$. 
  Then the set $\rep \mathfrak{A}$  is a net 
since $\pi_j \le \pi_1 \oplus \pi_2$. Choose any co-finite subnet
$\pi_n$, and define 
$\tilde{\mathfrak{A}}_p=\varprojlim\overline{\pi_n(\mathfrak{A})}
$.

2. Let $(\tilde{\mathfrak{A}},\phi)$ be an enveloping
$\sigma$-$\ca$-algebra of $\mathfrak{A}$.
We will prove that $\phi(\mathfrak{A})$ is dense in  $\tilde{\mathfrak{A}}$.
 Let $\mathcal{B}$ be a $\ca$-algebra and $j\colon\mathfrak{A} \to
 \mathcal{B}$ be 
a continuous $*$-surjection. If $\pi$, $\pi^\prime \in
\rep\mathcal{B} $ and 
$\pi\circ j\circ\phi = \pi^\prime\circ j\circ\phi$,  then by
the uniqueness 
of $\tilde{\pi}$ in the definition of the enveloping algebra, we have 
$\pi\circ j = \pi^\prime\circ j$. Then $\pi= \pi^\prime$ since $j$ is
surjective.
Then Lemma~\ref{WSW} implies that $j\circ\phi(\mathfrak A)$ is dense  in
$\mathcal{B}$.   
Represent $\tilde{\mathfrak{A}}$ as $\varprojlim \mathcal{B}_n$
(see \cite[Proposition~1.2]{15}
 or \cite[Proposition~5.6]{13}), where $\mathcal{B}_n$
 is a $\ca$-algebra  and $j_n : \tilde{\mathfrak{A}} \to
 \mathcal{B}_n$ is the canonical 
surjection. Previous arguments show that
$\overline{j_n(\phi(\mathfrak{A}))}=\mathcal{B}_n$.
 From this it follows that $\phi(\mathfrak{A})$ is a dense
 $*$-algebra
 in $\tilde{\mathfrak{A}}$,
 since $ z_k \to z$ in $\tilde{\mathfrak{A}}$ iff $j_n(z_k) \to
 j_n(z)$ for all $n$. 

3.
We will prove that $\tilde{\mathfrak{A}} \simeq
\tilde{\mathfrak{A}}_p$.
Passing if necessary to the quotient with respect to the radical, we can assume that
$\phi$
is an injection, and so 
$\phi(\mathfrak{A}) \simeq \mathfrak{A}$.
 Since $\overline{\phi(\mathfrak{A})}= \mathfrak{A}$, we only need to
 prove that the 
topology $\tau_1$ on $\phi(\mathfrak{A})$ induced from
$\tilde{\mathfrak{A}}$ coincides with the 
topology $\tau_2$ on $\phi(\mathfrak{A})$ defined by the semi-norms $\Vert
\tilde\pi_n(\cdot) \Vert$.  
Note that the topology $\tau_1$ is defined by the semi-norms $\Vert \cdot
\Vert_n$ induced by 
faithful representations of $\mathcal{B}_n$, and since the subnet $\pi_n$ 
is co-finite in the set of all representations, the topology $\tau_2$ is 
stronger then $\tau_1$. But since every representations $\pi_n$ can
be 
lifted to $\tilde{\mathfrak{A}}$, $\tau_1$ is stronger then 
$\tau_2$. This proves that $\tau_1=\tau_2$.


Since the algebra $\tilde{\mathfrak{A}}$ is a metrizable
locally $C^*$-algebra,\index{algebra@$*$-algebra!locally $C^*$}
by \cite[Corollary 4.7]{16},  every $*$-representation of
$\tilde{\mathfrak{A}}$
is continuous. And thus $\tilde{\pi}$ is uniquely defined even
 without
the  requirement
of being continuous. It proves that the $*$-algebra
$\tilde{\mathfrak{A}}$  is also an enveloping $*$-algebra of
$\mathfrak{A}$.  
\end{proof}

Note that the homomorphism $\phi$ is an 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.\index{algebra@$*$-algebra!$\sigma$-$C^*$-representatble}

\begin{remark}
In the case where $\mathfrak{A}$ is not finitely generated, we can also
construct an enveloping pro-$C^*$-algebra\index{algebra!enveloping
  pro-$C^*$} 
(the index $n$ in the above
construction  should be replaced by the multi-index
$\alpha\in\mathbb{N}^{\mathbb{I}}$, where $\{x_\alpha;
\alpha\in\mathbb{I}\}$ is the set of generators of  $\mathfrak{A}$).
In such a case, 
all the above statements hold true except that the $*$-algebra is not
necessarily enveloping, and the range of the homomorphism $\phi$ need not
be dense  but only quasi-dense, i.e., such that,  
 for any representation $\pi\in
\rep(\mathfrak{A})$,
the set $\pi(\phi(\tilde{\mathfrak{A}}))$ is dense in $\im\pi$.
\end{remark}

\noindent\textbf{4.}
 It is convenient to adopt the following definition:

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

\end{definition}

\subsection{$*$-Representations of generators and relations}
\label{sec:1.1.4}

\noindent\textbf{1.}
To any $*$-representation of a finitely generated
$*$-algebra\index{algebra@$*$-algebra!finitely generated} 
\begin{align*}
\mathfrak{B}&
= \mathbb{C} \bigl< x_1, \dots, x_n, x_1^*, \dots,
x_n^* 
\mid
\\
& \quad\qquad P_j(x_1, \dots, x_n, x_1^*, \dots, x_n^*) =0,
\ j=1, \dots, m\bigr>
\end{align*}
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, \qquad 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 also increase); 
therefore, any representation $\pi$ of the 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, \qquad j=1, \dots, r.
\end{equation}

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

In studying families of self-adjoint operators $A_1$, \dots, $A_n$, 
as usual, the role of the simplest families  of  operators is played 
by irreducible ones. A family of self-adjoint 
operators $A_k = \int_{\mathbb{R}} \lambda_k \, dE_k (\lambda_k)$, $k
= 
1$, \dots, $n$, is irreducible,\index{operators!irreducible family} if there is no non-trivial (different 
from $H$ and $\{0\}$) subspace in $H$ 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 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 a multiple of the 
identity operator.

\medskip\noindent\textbf{
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 the spectral measure of the 
operator $A$.

\medskip\noindent\textbf{
4.} Irreducible representations of a pair of bounded self-adjoint 
operators\index{operators!bounded self-adjoint pair} 
exist in a Hilbert space of arbitrary dimension.

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

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

\medskip\noindent\textbf{5.}
In general, it is not necessary that irreducible pairs connected by
the relation
\eqref{self_relation} 
exist in every dimension. 

Pairs of commuting bounded self-adjoint operators $A=A^*$, $B = B^*$, 
$AB=BA$, have only one-dimensional irreducible representations, $\dim
H 
= 1$, $A = \lambda_1$, $B = \lambda _2$, $(\lambda_1, \lambda_2) \in 
\mathbb{R}_2$. The 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 the  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.

\medskip\noindent\textbf{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 the canonical 
commutation relations (CCR), $[A,B] = iI$. Indeed, otherwise, following,
e.g., \cite{reedsim}, we would have
\[
A^n B - B A^n = i\,nA^{n-1},
\]
and 
\[
n \,\|A^{n-1}\| = n\, \|A\|^{n-1} \le 2\, \|A\|^n \|B\|.
\]
Since $\|A \| \ne 0$, the latter implies $\|A\| \,\|B\| \ge n/2$ for all 
$n$, which 
contradicts the assumption that $A$ and $B$ are bounded. 

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

\medskip\noindent\textbf{7.}
As is commonly accepted in representation theory, 
collections of operators are studied up to unitary equivalence. Two 
collections, $(A_k)_{k=1}^n$ on a Hilbert space $H$, and $(\tilde 
A_{k})_{k=1}^n$ on 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 a $*$-algebra 
$\mathfrak{B}$ up to unitary equivalence is the same as 
the description of bounded representations of the generators up to
unitary 
equivalence.

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

\subsection{Pairs of self-adjoint operators satisfying quad\-ratic 
relations}\label{sec:1.1.5}

In this chapter, we will study, in particular, pairs of self-adjoint
bounded
operators $A$, $B$, which satisfy the following
relation\index{relations!quadratic}
\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,\notag
\\
\alpha, \beta, \hbar, \gamma, \delta, \epsilon, \chi \in
\mathbb{R}.
\end{gather}

\noindent\textbf{1.}
 Let us start with the homogeneous quadratic 
relation\index{relations!quadratic!homogeneous}
\begin{equation}\label{onetwo}
\frac q i [A,B]=\alpha A^2+\beta\{A,B\}+\gamma B^2,\qquad
\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\textup:
\[
\begin{array}{|c|c|}
\hline
(0_0)\hfill                    & (IV_0)\hfill\\
0=0                            & [A,B]=0 \\ \hline
(I_0)\hfill                    & (V_0)\hfill  \\
A^2=0                          & \frac1i[A,B]=A^2\\ \hline
(II_0)\hfill                   & (VI_0)\hfill\\
A^2+B^2=0                      & \frac1i[A,B]=q(A^2+B^2)\\
                               & (q>0)            \\ \hline
(III_0)\hfill                  & (VII_0)\hfill     \\
A^2-B^2=0                      & \frac1i[A,B]=q(A^2-B^2) \\
                               & (q>0    )  \\ \hline
\end{array}
\]
\end{proposition}

\begin{proof}
By using a non-degenerate linear transformation, we can reduce
the symmetric quadratic form $\alpha A^2 + \beta \{A,B\} + \gamma B^2$
to a diagonal form, i.e., we can assume $\beta=0$, $\alpha$, $\gamma \in
\{-1,0,1\}$.
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 $ q i[A,B]=0$, we get $(IV_0)$; if $- q
i [A,B]=A^2$, replace $B$ by $q B$ to get $(V_0)$; if $-
q i[A,B]=A^2\pm B^2$, we get $(VI_0)$ and $(VII_0)$ with
$q$ replaced with $q^{-1}$; by substituting $A$ with
$\sign qA $ we get $q>0$. 
\end{proof}

\noindent\textbf{2.}
By applying a similar argument, 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:
\[
\begin{array}{|l|l|l|}
\hline 
\tstr(0_0)\ 0=0         & (0_1)\ \chi I=0,\ \chi\ne0    & (0_2)\ A=0 \\
\hline
\tstr(I_0)\  A^2=0      & (I_1)\  A^2=I      & (I_2)\ A^2=B\\
\tstr                   & (I_1')\ A^2=-I     &             \\
\hline
\tstr(II_0)\  A^2+B^2=0 & (II_1)\ A^2+B^2=I  &              \\
\tstr                   & (II_1')\ A^2+B^2=-I&              \\
\hline
\tstr(III_0)\  A^2-B^2=0& (III_1)\ A^2-B^2=I &               \\
\tstr\text{or } \{\tilde A,\tilde B\}=0 
                        & \text{or }\{\tilde A,\tilde B\}=I & \\ 
\hline
\tstr(IV_0)\  {}[A,B]=0 & (IV_1)\ \frac1i[A,B]=I &(IV_2)\ \frac1i[A,B]=A\\
\hline
\tstr(V_0)\ \frac1i[A,B]=A^2  
                        & (V_1)\ \frac1i[A,B]=A^2+I 
                                             & (V_2) \ \frac1i[A,B]\\
\tstr                   &  (V_1')\ \frac1i[A,B]=A^2-I  &\hfill 
=A^2+B\\
\hline
\tstr(VI_0)\  \frac1i[A,B] 
                        & (VI_1)\   \frac1i[A,B]      &              \\
\hfill=q(A^2+B^2),            &\hfill=q(A^2+B^2)+I,        &    \\
\tstr\hfill q>0                    &\hfill q\ne0       &              \\
\hline
\tstr(VII_0)\  \frac1i[A,B]  
                        & (VII_1)\  \frac1i[A,B]     &           \\
\hfill{}=q(A^2-B^2),         &\hfill{}=q(A^2-B^2)+I,       &    \\
\tstr\hfill  q>0                  &\hfill   q\ne0              &           \\
\hline
\end{array}
\]
\end{proposition}

Our next goal is to describe for each of the relations
$(0_0)$--$(VII_0)$
pairs of bounded  self-adjoint
operators, which satisfy this relation.

Considering the 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
 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 $A=A^*$, we have $A^2=A=0$, and the 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 the identity for the operator $B$ concentrated on a
compact 
set $K \subset 
\mathbb{R}$;

$(I_1')$ $A^2=-I$. This equation does not 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 the 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: $*$-wild
relations,\index{relations!$*$-wild}
 $F_4$-relations,\index{relations@relations!$F_4$-relations} 
Lie algebras and their nonlinear
transformations,\index{Lie algebras!nonlinear transformations} 
and $q$-relations\index{realtions@relations!$q$-relations} 
(see the table below)
\[
\begin{array}{|l|l|l|}
\hline
%====================================================================
\multicolumn{3}{|c|}{\tstr\text {$*$-wild relations}}\\ \hline
%--------------------------------------------------------------------
\tstr(0_0)\ 0=0          &(I_1)\ A^2=I           &                \\
\hline
%====================================================================
\multicolumn{3}{|c|}{\tstr\text {${F}_4$-relations}}\\ \hline
%--------------------------------------------------------------------
\tstr                     &(II_1)\ A^2+B^2=I           &            \\
\hline
%--------------------------------------------------------------------
\tstr(III_0)\  \{A,B\}=0       &(III_1)\ \{A,B\}=I         &             \\
\hline
%====================================================================
\multicolumn{3}{|c|}{\tstr\text{Lie algebras and their nonlinear
transformations}} \\ \hline
%--------------------------------------------------------------------
\tstr(IV_0)\ [A,B]=0         &(IV_1)\ \frac1i[A,B]=I    &(IV_2)\ \frac1i[A,B]=A \\
\hline
%--------------------------------------------------------------------
\tstr(V_0)\ \frac1i[A,B]=A^2 &(V_1)\ \frac1i[A,B]=A^2+I  
                                        &(V_2)\frac1i[A,B]\\
\tstr
                    &(V_1')\ \frac1i[A,B]=A^2-I  & \hfill =A^2+B     \\
\hline
%====================================================================
\multicolumn{3}{|c|}{\tstr\text{$q$-relations}} \\ \hline
%--------------------------------------------------------------------
\tstr(VI_0)\ \frac1i[A,B]&(VI_1)\ \frac1i[A,B] &  \\
 \hfill=q(A^2+B^2),  &\hfill =q(A^2+B^2)+I,&\\
\tstr\hfill q>0           & \hfill q\ne0       &\\ 
\hline
%--------------------------------------------------------------------
\tstr(VII_0)\ \frac1i[A,B]&(VII_1)\ \frac1i[A,B]&  \\
\hfill =q(A^2-B^2),&\hfill=q(A^2-B^2)+I,&\\
\tstr\hfill q>0 &\hfill q\ne0&\\ 
\hline
%====================================================================
\end{array}
\]


In what follows, our aim is, in particular, to study irreducible 
representations for each of these
groups of relations.
Representations of $F_4$-relations are described in
Section~\ref{sec:1.2.3},
representations of Lie algebras and their nonlinear transformations 
by bounded operators are described in 
Section~\ref{sec:1.3.1}, representations 
of $q$-relations are described in
Sections~\ref{sec:1.4.1}--\ref{sec:1.4.3}.
About $*$-wild 
relations see Section~\ref{sec:3.1.5}.



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