\documentstyle{SibMatJ}

\topmatter
                                   %%% specific SMJ tags
\translator  Ya.~A. Kopylov\endtranslator
\opages      1199--1207\endopages
\iauthor     Avsyankin~O.G.\endiauthor
\UDclass     517.9\endUDclass
                                   %%% standard AMS tags
\title On the $C^*$-Algebra Generated by Multiplicative Discrete Convolution
Operators with Oscillating Coefficients
\endtitle
\author      O.~G. Avsyankin\endauthor
\date        January 24, 2014\enddate
\affil       O. G. Avsyankin\\
	         Southern Federal University,
	         Rostov-on-Don, Russia
\endaffil
\email       avsyanki\@math.rsu.ru\endemail
\address     Rostov-on-Don\endaddress
\keywords    multiplicative discrete convolution, $C^*$-algebra,
isomorphism, group, symbol, Noethericity, invertibility
\endkeywords
\abstract
We consider the $C^*$-algebra generated by multiplicative discrete convolution
operators and  multiplication operators by oscillating coefficients. For this algebra
we construct some symbol operator calculus in terms of which we obtain necessary
and sufficient conditions of the Noethericity for operators.
\endabstract
\endtopmatter

\noindent
{\bf Introduction.}
At present, there is a rather complete theory of integral operators with homogeneous
kernels (see, for example, [1--5] and the references therein). The discrete operators
with homogeneous kernels (they are called {\it multiplicative discrete convolution
operators}) are much less studied. In contrast to the one-dimensional continual
case, these operators are not reduced to convolution operators, and their study requires
absolutely different approaches. The operators were first introduced and studied
in~[6,\,7]. In~[8], the author constructed and investigated the Banach algebra
generated by multiplicative discrete convolution operators. A~symbol calculus
was constructed for this algebra, in terms of which some Noethericity criterion
for an operator and a~formula for computing its index were obtained.

This article continues ~[8]. It is devoted to the study
of the $C^*$-algebra $\frak{B}$ generated by multiplicative discrete convolution operators
and multiplication operators by oscillating coefficients of the form
$n^{i\alpha}\allowmathbreak (=e^{i\alpha \ln n})$. We stress that the coefficients $n^{i\alpha}$ play the same
role in the theory of multiplicative discrete convolutions as the coefficients of the form
$e^{i\alpha n}$ do in the theory of ordinary discrete convolutions. The above-mentioned
algebra $\frak{B}$ is essentially noncommutative and so~$\frak{B}$ has no scalar symbol.
For the study of this algebra, there is used Antonevich's approach, based on the theory
of $C^*$-algebras generated by dynamical systems [9, Chapter\,II]. This approach makes it
possible to construct a~ symbol operator calculus for the $C^*$-algebra $\frak{B}$, in terms
of which we obtain a~Noethericity criterion\footnote"$^{1)}$"{A ~ bounded linear operator $A$
is called {\it Noether\/} if the range of~$A$ is closed and the kernels of~$A$ and the adjoint
operator~$A^*$ are finite-dimensional.} for the operators of this algebra. In the final part
of the article, we distinguish a~class of operators in~$\frak{B}$ for which we point out
an~efficient scalar Noethericity condition for computing their indices.

\smallskip
Below we use the notations:
$\Bbb{N}$, $\Bbb{R}$, and $\Bbb{C}$ are the sets of natural, real, and complex numbers respectively;
$\dot{\Bbb{R}}$ is the compactification of~$\Bbb{R}$ by one infinitely remote point;
$\Cal{L}(X)$ is the Banach algebra of all  bounded linear operators acting in a~Banach space~$X$;
$n^{i\alpha}=e^{i\alpha \ln n}$; and
$\ell_2=\ell_2(\Bbb{N})$ is the Banach space of complex sequences
${\{\varphi_n\}}_{n=1}^\infty$ such that $\sum\nolimits_{n=1}^\infty
|\varphi_n|^2<\infty$.


\medskip
{\bf 1.~Prerequisites.}
The facts below are presented in detail in~[9, \S\,6--\S\,8;\,10].

\demo{Definition~1}
Suppose that $\frak{B}$ is a $C^*$-algebra and $\frak{A}$ is a $C^*$-subalgebra in~$\frak{B}$, while
$T$ is a~unitary representation of~$G$ in~$\frak{B}$, and the axioms are fulfilled:

(1) $T(g)aT^{-1}(g)\in\frak{A}$, $a\in\frak{A}$, $g\in G$;

(2) the set $\frak{B}^0$ of the finite sums of the form $\sum\nolimits_i a_iT(g_i)$, where
$a_i\in\frak{A}$ and $g_i\in G$, is norm-dense ~ in~$\frak{B}$.

Then the $C^*$-algebra $\frak{B}$ is said to be {\it generated\/} by ~
$\frak{A}$ {\it and the representation}~$T$ of~$G$, which is symbolized as
$\frak{B} = C^*(\frak{A}, G, T)$.
\enddemo

Recall that a~representation~$T$ of~$G$ in~$\frak{B}$ is called {\it unitary\/} if
 $T(g)\in \frak{B}$ is unitary for every $g\in G$.

Condition~1 means that each $T(g)$ generates the mapping
$$
  \widehat{T}_g: \frak{A}\rightarrow\frak{A}, \quad
  a\rightarrow T(g)aT^{-1}(g)
$$
which is an~automorphism of~$\frak{A}$.

In~[9,\,10], it was assumed that a~$C^*$-algebra $\frak{A}$ is $*$-isomorphic
to the endomorphism algebra $\operatorname{END}(\Cal{E})$ of an $N$-dimensional
complex vector bundle $\Cal{E}$ over a~compact topological space $\Cal{X}$. It was
supposed in addition that on~$\Cal{E}$ there was fixed a~Hermitian metric; i.e.,
each fiber~$\Cal{E}_x$, where $x\in \Cal{X}$, was endowed with an~inner product continuously
depending on~$X$. If~$\frak{A}$ is commutative (corresponds to the case of $N=1$) then
~$\frak{A}$ is $*$-isomorphic to~$C(\Cal{X})$.

Let $\frak{A}$ be a $C^*$-algebra of the above type. Each automorphism
$\widehat{T}_g$ of~$\frak{A}$ generates a~homeomorphism $\pi_g: \Cal{X}\rightarrow \Cal{X}$.
The~homeomorphisms $\pi_g$ define an~action of~$G$ on~$\Cal{X}$.

\demo{Definition~2}
The action of a~group~$G$ on an algebra~$\frak{A}$ by automorphisms is called
{\it topologically free\/} if, for every finite collection $F$ of elements in~$G$
and every open set $\Cal{W}\subset \Cal{X}$, there exists  $x_0\in \Cal{W}$
such that all points $\pi_g(x_0)$, $g\in F$, are pairwise disjoint.
\enddemo

This definition is equivalent to the fact that the set of fixed points of every mapping
$\pi_g$ with $g\ne e$ has empty interior.


\proclaim{Proposition 1 \rm (the isomorphism theorem [9, \S\,7])}
Given $C^*$-algebras $\frak{B} = C^*(\frak{A}, G, T)$ and
$\frak{B}_1 = C^*(\frak{A}_1, G, T_1)$ with the same group~$G$, suppose that there
exists a $*$-isomorphism $\varphi: \frak{A}\rightarrow\frak{A}_1$ such that
$$
  \varphi(T(g)aT^{-1}(g)) = T_1(g)\varphi(a)T_1^{-1}(g).
$$
If a $C^*$-algebra $\frak{A}$ is isomorphic to the endomorphism algebra $\operatorname{END}(\Cal{E})$
of an $N$-dimensional complex vector bundle~$\Cal{E}$ and the group~$G$
is admissible and acts on~$\frak{A}$ topologically freely by automorphisms then
the correspondence
$$
  \Phi: \sum_i a_i T(g_i)\rightarrow\sum_i\varphi(a_i)T_1(g_i)
$$
extends from the set $\frak{B}^0$ of finite sums to a~$*$-isomorphism of the $C^*$-algebras
$\frak{B}$ and~$\frak{B}_1$.
\endproclaim

The definition of admissible group can be found, for example, in~[9, \S\,7]. We observe
only that every abelian group is admissible.

\medskip
{\boldmath\bf 2. The~$C^*$-algebra $\frak{B}$ and its structure.}  In the ~$\ell_2$ space, define
the operator~$H$ by the formula
$$
  (H\varphi)_m =\sum\limits_{n=1}^\infty k(m,n) \varphi_n,\quad m\in \Bbb{N},
 \eqno(1)
$$
where $\varphi = \{\varphi_n\}_{n=1}^\infty$ and the function $k(x,y)$ is defined
on $(0,\infty) \times (0,\infty)$ and satisfies the conditions:

$(1^\circ)$~homogeneity of degree~($-1$), i.e.,
$$
  k(\alpha x, \alpha y)=\alpha^{-1} k(x,y), \quad \alpha>0;
$$

$(2^\circ)$~integrability, i.e.,
$$
  \kappa =\int\limits_0^\infty |k(1,y)| y^{-1/2}\,dy < \infty;
$$

$(3^\circ)$~the function $|k(1,y)|y^{1/2}$ has bounded variation on~$(0,\infty)$; i.e.,
$$
  v=\bigvee\limits_0^\infty [|k(1,y)| y^{1/2}]<\infty,
$$
where $\bigvee\nolimits_0^\infty [|k(1,y)| y^{1/2}]$ is the full variation of this function.

\goodbreak

The operator~$H$ is called a {\it multiplicative discrete convolution operator}.
It is known (see~[6]) that ~$H$ is bounded in~$\ell_2$ and $\|H\| \leqslant \kappa+v$.

Denote by $\frak{A}$ the least $C^*$-subalgebra in the $C^*$-algebra $\Cal{L}(\ell_2)$
containing all operators of the form $\lambda I+H+K$, where $\lambda \in \Bbb{C}$
and $K$ is a~compact operator in~$\ell_2$. In~[8], we constructed some symbol calculus
for~$\frak{A}$; i.e., to each operator $A \in \frak{A}$, we assigned the function
$\sigma_{A}(\xi)\in C(\dot{\Bbb{R}})$ called the {\it symbol\/} of~$A$. The symbol
of~$\lambda I+H$ is defined as follows:
$$
  \sigma_{\lambda I+H}(\xi)=\lambda
+ \int\limits_0^\infty k(1,y) y^{-1/2+i \xi}\,dy,
  \quad \xi\in \Bbb{R},
 \eqno(2)
$$
and $\sigma_{\lambda I+H}(\infty)=\lambda$.
Further, the set $\Cal{K}=\Cal{K}(\ell_2)$ of all compact operators
in~$\ell_2$ is a~closed two-sided ideal in~$\frak{A}$. Consider the quotient algebra
$\frak{A}/\Cal{K}$. Define the symbol $\sigma_{[A]}(\xi)$ of the coset $A+\Cal{K}$
by the equality $\sigma_{[A]}(\xi)=\sigma_{A}(\xi)$ (this is sound because
all operators in $A+\Cal{K}$ have the same symbol, see~[8]). It was proved in~[8]
that the $C^*$-algebra $\frak{A}/\Cal{K}$ is commutative and its maximal ideal space
with the Gelfand topology is homeomorphic to the compact space~$\dot{\Bbb{R}}$.
By the Gelfand--Naimark theorem the above implies (see, for example, [11, Section~2.1]):

\proclaim{Lemma 1}
The mapping
$$
  s: \frak{A}/\Cal{K} \rightarrow C(\dot{\Bbb{R}}),
  \quad A+\Cal{K} \rightarrow \sigma_{[A]}(\xi),
 \eqno(3)
$$
is a $*$-isomorphism.
\endproclaim

Define the operator $M_\alpha$, where $\alpha\in \Bbb{R}$, in~$\ell_2$
by the equality
$$
  (M_\alpha \varphi)_n = n^{i\alpha} \varphi_n, \quad n \in \Bbb{N}.
 \eqno(4)
$$

\proclaim{Lemma 2}
Let $A\in \frak{A}$. Then $A_\alpha=M_\alpha A M_\alpha^{-1}$ belongs
to~$\frak{A}$ for every $\alpha\in \Bbb{R}$.
\endproclaim

\demo{Proof}
Let $A=H$, where $H$ is an~operator of the form~(1). It is easy to see that
 $H_\alpha=M_\alpha H M_\alpha^{-1}$ is defined by the formula
$$
  (H_\alpha \varphi)_m= \sum\limits_{n=1}^\infty k_{\alpha}(m,n) \varphi_n,
\quad m\in \Bbb{N},
$$
where
$$
  k_\alpha(x,y) = k(x,y) (x/y)^{i\alpha}.
 \eqno(5)
$$
Since $k_\alpha(x,y)$ satisfies conditions $1^\circ$--$3^\circ$,
we have $H_\alpha\in \frak{A}$.

Consider
$$
  A=\sum\limits_i \prod_j (\lambda_{ij} I+H_{ij}+K_{ij}),
 \eqno(6)
$$
where $H_{ij}$ are operators of the form~(1), $K_{ij} \in \Cal{K}$,
and the sum and the product are finite. Then
$$
  A_\alpha= \sum\limits_i \prod_j M_\alpha (\lambda_{ij} I+H_{ij}+K_{ij})
M_\alpha^{-1}= \sum\limits_i \prod_j (\lambda_{ij} I+ (H_{ij})_\alpha
+(K_{ij})_\alpha),
$$
which implies that $A_\alpha \in \frak{A}$.

Let $A$ be an arbitrary operator in~$\frak{A}$. Since the set $\frak{A}_0$ consisting
of all operators of the form~(6) is everywhere dense in~$\frak{A}$, there is a~sequence
$\{A_s\} \subset \frak{A}_0$ with $\|A-A_s\| \to 0$ as $s\to \infty$. Then
$\|A_\alpha -(A_s)_\alpha\| \to 0$. Since $(A_s)_\alpha \in \frak{A}$, we have
$A_\alpha \in \frak{A}$.
\qed\enddemo

Denote by $\frak{B}$ the $C^*$-algebra generated by all operators~$A$ in~$\frak{A}$
and all operators~$M_\alpha$. Obviously, $\Cal{K}$ is a closed two-sided ideal in~$\frak{B}$.
Consider the quotient algebra $\frak{B}/\Cal{K}$ which is a $C^*$-subalgebra in
the $C^*$-algebra $\Cal{L}(\ell_2)/\Cal{K}$. It is the closure in the norm of
$\Cal{L}(\ell_2)/\Cal{K}$ of
$$
  (\frak{B}/\Cal{K})_0 = \Bigl\{ \sum\limits_i
\prod_j A_{ij} M_{\alpha_{ij}} +\Cal{K} \Bigr\},
$$
where the sums and the products are finite.

\goodbreak
Consider the set of all elements of the form $M_{\alpha}+\Cal{K}$, where
$\alpha \in \Bbb{R}$, belonging to $\Cal{L}(\ell_2)/\Cal{K}$. It is not
hard to see that an element $M_{\alpha}+\Cal{K}$ is unitary and
$(M_\alpha +\Cal{K}) (M_\beta +\Cal{K})= M_{\alpha+\beta} +\Cal{K}$
for all $\alpha,\beta\in \Bbb{R}$. Consequently, the homomorphism
$$
  T_M:  \Bbb{R}\to \Cal{L}(\ell_2)/\Cal{K},
  \quad \alpha\to M_\alpha +\Cal{K},
$$
is a~unitary representation of~$\Bbb{R}$.

\proclaim{Lemma 3}
The $C^*$-algebra $\frak{B}/\Cal{K}$ is generated by the $C^*$-algebra
$\frak{A}/\Cal{K}$ and the unitary representation~$T_M$ of~$\Bbb{R}$;
i.e., $\frak{B}/\Cal{K}=C^*(\frak{A}/\Cal{K}, \Bbb{R}, T_M)$.
\endproclaim

\demo{Proof}
Use Definition~1. Lemma~2 implies that, for all $A \in \frak{A}$ and
$\alpha \in \Bbb{R}$, the element
$(M_\alpha +\Cal{K}) (A+\Cal{K}) \big(M_\alpha^{-1} +\Cal{K}\big)$
belongs to $\frak{A}/\Cal{K}$, i.e., Axiom~1 holds.

Check Axiom~2, i.e., prove that the set
$$
  (\frak{B}/\Cal{K})^0=  \Bigl\{\sum\limits_i A_i M_{\alpha_i} +\Cal{K} \Bigr\},
$$
where the sums are finite, is everywhere dense in $\frak{B}/\Cal{K}$. To this end,
show that $(\frak{B}/\Cal{K})^0=(\frak{B}/\Cal{K})_0$. To verify that, check the equality
$$
  A_1 M_{\alpha_1} A_2 M_{\alpha_2}=A_3 M_{\alpha_1+\alpha_2}+K,
 \eqno(7)
$$
where $A_1, A_2, A_3 \in \frak{A}$ and $K\in \Cal{K}$. Indeed,
$A_1 M_{\alpha_1} A_2 M_{\alpha_2}=A_1(A_2)_{\alpha_1} M_{\alpha_1+\alpha_2}$;
moreover, by Lemma~2, $(A_2)_{\alpha_1}=M_{\alpha_1} A_2 M_{\alpha_1}^{-1}$
belongs to~$\frak{A}$. Since $A_1 (A_2)_{\alpha_1}=A_3+K$, where $A_3 \in \frak{A}$ and
$K \in \Cal{K}$ (see~[8]), we get~(7).
\qed\enddemo

Each operator $M_\alpha$ of the form~(4) generates a~mapping
$$
  \widehat{M}_\alpha: \frak{A}/\Cal{K} \to \frak{A}/\Cal{K},
  \quad A+\Cal{K} \to M_\alpha AM_\alpha^{-1}+\Cal{K},
$$
which is an~automorphism of the $C^*$-algebra $\frak{A}/\Cal{K}$.

 \proclaim{Lemma 4}
The action of the group~$\Bbb{R}$ on the $C^*$-algebra $\frak{A}/\Cal{K}$
by the automorphisms $\widehat{M}_\alpha$ is topologically free.
\endproclaim

\demo{Proof}
By Lemma~1, each automorphism $\widehat{M}_\alpha$ generates an~automorphism
$\widehat{F}_\alpha =s\circ \widehat{M}_\alpha \circ s^{-1}$ of~ $C(\dot{\Bbb{R}})$.
Find a~formula for $\widehat{F}_\alpha$. Let $\sigma(\xi)$ be an~arbitrary
function in $C(\dot{\Bbb{R}})$ and let $s^{-1}(\sigma)=A+\Cal{K}$, so that
$\sigma(\xi)=\sigma_{[A]}(\xi)$. Since $\widehat{M}_\alpha(A+\Cal{K})=A_{\alpha}+\Cal{K}$,
where $A_{\alpha}=M_\alpha A M_\alpha^{-1}$, it remains to find $s(A_{\alpha}+\Cal{K})$;
i.e., we have to calculate  the symbol $\sigma_{[A_{\alpha}]}(\xi)$ of the element
$A_{\alpha}+\Cal{K}$.

Consider the two cases:

1.~If $A=H$, where $H$ is as in ~(1), then from ~(2) and~(5) we infer
$$
  \sigma_{[H_\alpha]}(\xi)=\int\limits_0^\infty k_\alpha(1,y) y^{-1/2+i\xi}\,dy
=  \int\limits_0^\infty k(1,y) y^{-1/2+i(\xi-\alpha)}\,dy
= \sigma_{[H]}(\xi-\alpha).
$$

2.~Let $A$ be an~operator in~$\frak{A}$. Since the set $(\frak{A}/\Cal{K})_0$
of all elements of the form $\lambda I+H+\Cal{K}$ is everywhere dense in $\frak{A}/\Cal{K}$,
there is a~sequence $\{\lambda_j I+H_j+\Cal{K}\}$ such that
$$
\|A-(\lambda_j I+H_j)+\Cal{K}\|_{\frak{A}/\Cal{K}} \to 0.
$$
Then
$\|A_{\alpha}-(\lambda_j I+H_j)_{\alpha}+\Cal{K}\|_{\frak{A}/\Cal{K}} \to 0$.
By the above,
$$
  \sigma_{[\lambda_j I+(H_j)_\alpha]}(\xi)
=\sigma_{[\lambda_j I+H_j]}(\xi-\alpha),
  \quad \xi \in \dot{\Bbb{R}}.
\eqno(8)
$$
Since a $*$-isomorphism is an~isometry [11, Section~3.1], Lemma~1 implies that
$$
  \sup\limits_{\xi \in \dot{\Bbb{R}}} |\sigma_{[A]}(\xi)|
=  \|A+\Cal{K}\|_{\frak{A}/\Cal{K}}
$$
for every operator $A\in \frak{A}$. Taking this into account and passing to the limit as
$j \to \infty$ in~(8), we obtain
$$
  \sigma_{[A_\alpha]}(\xi)=\sigma_{[A]}(\xi-\alpha),
\quad \xi \in \dot{\Bbb{R}},
 \eqno(9)
$$
which  implies in particular that
$\sigma_{[A_\alpha]}(\infty)=\sigma_{[A]}(\infty)$.

It follows from~(9) that the automorphism $\widehat{F}_\alpha$ is defined by the formula
$$
  (\widehat{F}_\alpha \sigma)(\xi)
=\cases
    \sigma(\xi-\alpha), & \xi \in \Bbb{R}, \\
    \sigma(\infty),     & \xi=\infty,
  \endcases
$$
where $\sigma(\xi)$ is an~arbitrary function in~$C(\dot{\Bbb{R}})$. In turn,
$\widehat{F}_\alpha$ generates a~homeomorphism $\pi_\alpha$ of the compact topological
space $\dot{\Bbb{R}}$ which is defined as follows:
$$
  \pi_\alpha(\xi)=\cases
    \xi+\alpha, & \xi \in \Bbb{R}, \\
    \infty,     & \xi=\infty.
  \endcases
$$
Thus, if $\alpha\ne 0$ then the infinitely remote point is the only fixed point
of~$\pi_\alpha$. But this means that the group~$\Bbb{R}$ acts on the $C^*$-algebra
$\frak{A}/\Cal{K}$ by automorphisms topologically freely (see Section~1).
\qed

\medskip
{\bf 3.~A~Noethericity criterion in~$\frak{B}$.}
Let $\sigma(\xi)\in C(\dot{\Bbb{R}})$. Define the operator
$\Bbb{M}_\sigma$  in~$L_2(\Bbb{R})$ by
$$
  (\Bbb{M}_\sigma f)(\xi) = \sigma(\xi) f(\xi).
 \eqno(10)
$$
Denote by $\frak{D}$ the $C^*$-algebra generated by all operators $\Bbb{M}_\sigma$
of multiplication by functions in~$C(\dot{\Bbb{R}})$. Using the properties
of a~multiplication operator, it is easy to check that
$$
  \mu: C(\dot{\Bbb{R}}) \to \frak{D}, \quad \sigma(\xi)\to \Bbb{M}_\sigma
\eqno(11)
$$
is a $*$-isomorphism.

Next, consider the shift operator in~$L_2(\Bbb{R})$:
$$
  (U_\alpha f)(\xi) = f(\xi-\alpha), \quad \alpha \in \Bbb{R}.
  \eqno(12)
$$
It is easy to see that the homomorphism
$$
  T_U: \Bbb{R}\to \Cal{L}(L_2(\Bbb{R})), \quad \alpha\to U_\alpha,
$$
is a~unitary representation of~$\Bbb{R}$.

Denote the $C^*$-algebra generated by all operators $\Bbb{M}_\sigma$
of the form~(10) and all operators $U_\alpha$ of the form~(12) by~$\frak{B}_1$.
This algebra is the closure in the operator topology of the set
$$
  (\frak{B}_1)_0 = \Bigl\{ \sum\limits_i \prod_j
  \Bbb{M}_{\sigma_{ij}} U_{\alpha_{ij}}\Bigr\},
$$
where the sums and products are finite.

 \proclaim{Lemma 5}
The $C^*$-algebra $\frak{B}_1$ is generated by the $C^*$-algebra $\frak{D}$ and
the unitary representation $T_U$ of~$\Bbb{R}$; i.e.,
$\frak{B}_1=C^*(\frak{D},\Bbb{R},T_U)$.
\endproclaim

\demo{Proof}
For every function $\sigma(\xi)\in C(\dot{\Bbb{R}})$ and every $\alpha\in \Bbb{R}$,
we have $U_\alpha \Bbb{M}_\sigma U_\alpha^{-1} = \Bbb{M}_{\sigma_\alpha}$,
where $\sigma_\alpha(\xi)=\sigma(\xi-\alpha)$. Consequently,
$U_\alpha \Bbb{M}_\sigma U_{\alpha}^{-1} \in \frak{D}$; i.e., Axiom~1 of Definition~1
is fulfilled.

Check Axiom~2, i.e., prove that the set
$$
  \frak{B}_1^0= \Bigl\{ \sum\limits_i \Bbb{M}_{\sigma_i} U_{\alpha_i} \Bigr\},
$$
where the sums are finite, is everywhere dense in~$\frak{B}_1$. To this end, show
that $\frak{B}_1^0=(\frak{B}_1)_0$. This follows from the easy equality
$$
  \Bbb{M}_{\sigma_1} U_{\alpha_1} \Bbb{M}_{\sigma_2}
  U_{\alpha_2} = \Bbb{M}_\sigma U_{\alpha_1+\alpha_2},
$$
where $\sigma(\xi)=\sigma_1(\xi) \sigma_2(\xi-\alpha_1)$.
Thus, $\frak{B}_1=C^*(\frak{D}, \Bbb{R},T_U)$.
\qed\enddemo

Consider the mapping
$$
  s_1 : \frak{A}/\Cal{K} \to \frak{D},
  \quad  A+\Cal{K} \to \Bbb{M}_{\sigma_{[A]}},
$$
where $\sigma_{[A]}(\xi)$ is the symbol of  $A+\Cal{K}$. Since $s_1=\mu \circ s$,
where $s$ and $\mu$ are defined from~(3) and~(11) respectively, $s_1$ is a $*$-isomorphism.

 \proclaim{Theorem 1}
The correspondence
$$
  \gamma_0:  (\frak{B}/\Cal{K})^0 \to \frak{B}_1^0, \quad
  \sum\limits_i A_i M_{\alpha_i}+\Cal{K} \to
  \sum\limits_i s_1(A_i+\Cal{K}) U_{\alpha_i}
$$
on the set of finite sums $(\frak{B}/\Cal{K})^0$, extends to a~$*$-isomorphism
$\gamma: \frak{B}/\Cal{K} \to \frak{B}_1$.
\endproclaim

\demo{Proof}
Show that all conditions of the isomorphism theorem are fulfilled (Proposition~1).
By~(9), we have
$$
  s_1\bigl((M_\alpha+\Cal{K}) (A+\Cal{K}) \bigl(M_\alpha^{-1}+\Cal{K}\bigr)\bigr)
= s_1(A_\alpha+\Cal{K})=\Bbb{M}_{\sigma_{[A]}(\xi-\alpha)}.
$$
Since
$\Bbb{M}_{\sigma_{[A]}(\xi-\alpha)}=
U_\alpha \Bbb{M}_{\sigma_{[A]}(\xi)} U_\alpha^{-1}$,
we have
$$
  s_1\bigl((M_\alpha+\Cal{K}) (A+\Cal{K}) \bigl(M_\alpha^{-1}+\Cal{K}\bigr)\bigr)
= U_\alpha s_1(A+\Cal{K}) U_\alpha^{-1}.
$$

Now, the $C^*$-algebra $\frak{A}/\Cal{K}$ is $*$-isomorphic to the $C^*$-algebra
$C(\dot{\Bbb{R}})$ (Lemma~1), while the group~$\Bbb{R}$ is admissible and acts
on~$\frak{A}/\Cal{K}$ by automorphisms topologically freely (Lemma~4).
Applying Proposition~1, we infer that $\gamma_0$ extends to a $*$-isomorphism
$\gamma: \frak{B}/\Cal{K} \to \frak{B}_1$. The theorem is proved. \qed
\enddemo

Thus, in Theorem~1, we define a~$*$-isomorphism
$\gamma : \frak{B}/\Cal{K} \to \frak{B}_1$. Consider the $*$-homomorphism
$$
  p: \frak{B} \to \frak{B}/\Cal{K}, \quad B \to B+\Cal{K}.
$$
Then  $S=\gamma \circ p : \frak{B} \to \frak{B}_1$
is a ~$*$-homomorphism as well. Refer as the {\it symbol\/} of an~operator $B \in \frak{B}$
to its image $S(B)\in \frak{B}_1$. In~particular, if
$B=\sum\nolimits_{i=1}^r A_i M_{\alpha_i}$, where $A_i \in \frak{A}$, then the symbol of~$\frak{B}$
is the operator
$$
  S(B)=\gamma(B+\Cal{K})=\sum\limits_{i=1}^r s_1(A_i+\Cal{K})
U_{\alpha_i}=  \sum\limits_{i=1}^r \Bbb{M}_{\sigma_{A_i}} U_{\alpha_i}.
\eqno(13)
$$
(Recall that $\sigma_{A_i}(\xi)=\sigma_{[A_i]}(\xi)$.)

 \proclaim{Theorem 2}
Let $B\in \frak{B}$. The operator $B$ is Noether if and only if its symbol~$S(B)$
is invertible in~$\Cal{L}(L_2(\Bbb{R}))$.
\endproclaim

\demo{Proof}
An~operator $B\in \frak{B}$ is Noether if and only if the coset $B+\Cal{K}$ is
invertible in $\Cal{L}(\ell_2)/\Cal{K}$. Since a $C^*$-algebra $\frak{B}/\Cal{K}$
is a $C^*$-subalgebra in $\Cal{L}(\ell_2)/\Cal{K}$, the Noethericity of~$B$ is
equivalent to the invertibility of $B+\Cal{K}$ in the $C^*$-algebra $\frak{B}/\Cal{K}$.
Since the mapping $\gamma: \frak{B}/\Cal{K} \to \frak{B}_1$ is a~$*$-isomorphism,
the element $B+\Cal{K}$ is invertible in~$\frak{B}/\Cal{K}$ if and only if the operator
$\gamma(B+\Cal{K})=S(B)$ is invertible in~$\frak{B}_1$. The last is equivalent
to the invertibility of~$S(B)$ in~$\Cal{L}(L_2(\Bbb{R}))$.
\qed


\medskip
{\bf 4.~Particular cases.}
Let us distinguish the classes of operators for which it is possible to~obtain Noethericity
in the scalar (and not in the operator) form. In~$\ell_2$, consider the operator
$$
  B=I + A_1 + A_2 M_\alpha,
\eqno(14)
$$
where $M_\alpha$ is an~operator of the form (4), while $A_1$ and~ $A_2$ are operators in~$\frak{A}$
whose symbols satisfy the condition
$$
  \sigma_{A_1}(\infty)=\sigma_{A_2}(\infty)=0.
\eqno(15)
$$
In particular, as $A_j$, $j=1,2$, we can take the canonical multiplicative discrete
convolution operators, i.e., assume that $A_j=H_j$, where $H_j$ is an~operator
of the form~(1).

By~(13), the symbol of~$B$ is the following operator acting in~$L_2(\Bbb{R})$:
$$
  S(B)=I + \Bbb{M}_{\sigma_{A_1}} + \Bbb{M}_{\sigma_{A_2}} U_\alpha,
$$
where $U_{\alpha}$ is defined by~(12). If $f \in L_2(\Bbb{R})$ then
$$
  (S(B)f)(\xi)=(1+\sigma_{A_1}(\xi)) f(\xi)+ \sigma_{A_2}(\xi) f(\xi-\alpha).
$$
Conditions for the invertibility of~$S(B)$ are well known (see, for example,
Theorem~$4'$ in [12, Chapter~2,~ \S\,1]). Basing on them, we obtain

 \proclaim{Theorem 3}
Suppose that $B$ is an operator of the form~$(14)$, and~$(15)$ holds.
The~operator~$B$ is Noether if and only if
$$
  1+\sigma_{A_1}(\xi)\ne 0, \quad \xi \in \Bbb{R}.
 \eqno(16)
$$
If~$(16)$ is fulfilled then
$$
  \operatorname{ind}(B)= -\operatorname{ind} (1+\sigma_{A_1}(\xi))
  := -\frac{1}{2\pi} \Delta [ \arg (1+\sigma_{A_1}(\xi)) ]
  |_{-\infty}^\infty.
\eqno(17)
$$
\endproclaim

  \demo{Proof}
By Theorem~2, $B$ is Noether if and only if its symbol $S(B)$ is invertible
in~$\Cal{L}(L_2(\Bbb{R}))$. Since $|1+\sigma_{A_1}(\infty)|>|\sigma_{A_2}(\infty)|$,
(16) is a~necessary and sufficient condition for the invertibility of~$S(B)$
(see~[12,~ p.~54]).

Find the index of~$B$. Join~$B$ and~$I+A_1$ by the family $\{B_t\}_{t \in [0,1]}$
of the operators
$$
  B_t=I +A_1 +(1-t) A_2 M_\alpha.
$$
For every $t \in [0,1]$, \,\, $B_t$ is an operator of the form~(14).
Therefore, ~(16) guarantees the Noethericity of every operator~$B_t$.
It is not hard to see that the family $\{B_t\}$ is continuous in~$t$
in the uniform operator topology. Then, by the homotopy stability
of the index, $\operatorname{ind} B =\operatorname{ind}(I+A_1)$.
As is known (see~[8]), $\operatorname{ind}(I+A_1)= -\operatorname{ind}
(1+\sigma_{A_1}(\xi))$. This yields~(17).
\qed
%\enddemo

 \proclaim{Corollary 1}
An~operator $B_1=I+ A_1+ M_\alpha A_2$ is Noether if and only if $(16)$ holds.
In this case its index is calculated by~$(17)$.
\endproclaim

 \demo{Proof}
To verify this, it suffices to write down $B_1$ as $B_1=I+A_1+(A_2)_\alpha M_\alpha$,
where $(A_2)_\alpha=M_\alpha A_2 M_\alpha^{-1}$, and use Theorem~3.
\qed
%\enddemo

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\enddocument