# Aspects of Noncommutative Geometry of Bunce-Deddens Algebras

Slawomir Klimek  Department of Mathematical Sciences, Indiana University-
Purdue University Indianapolis, 402 N. Blackford St., Indianapolis, IN 46202,
U.S.A<EMAIL_ADDRESS>,  Matt McBride  Department of Mathematics and
Statistics, Mississippi State University, 175 President’s Cir., Mississippi
State, MS 39762, U.S.A<EMAIL_ADDRESS>and  J. Wilson Peoples
Department of Mathematics, Pennsylvania State University, 107 McAllister Bld.,
University Park, State College, PA 16802, U.S.A<EMAIL_ADDRESS>
###### Abstract.

We define and study smooth subalgebras of Bunce-Deddens C∗-algebras. We
discuss various aspects of noncommutative geometry of Bunce-Deddens algebras
including derivations on smooth subalgebras, as well as K-Theory and
K-Homology.

##  1\. Introduction

The purpose of this paper is to study noncommutative geometry of Bunce-Deddens
(BD) algebras [3], [4] and their natural smooth subalgebras. Our previous,
related investigations of BD algebras are contained in [13] and [14] and are
mostly concerned with classifications of unbounded derivations.

The main objects of study in this paper are smooth subalgebras of Bunce-
Deddens algebras. The smooth subalgebras are dense *-subalgebras closed under
holomorphic functional calculus and complete in their own stronger locally
convex topology. As the name suggests, smooth subalgebras capture
differentiable structure in noncommutative geometry, in analogy with ordinary
differential geometry and thus are important objects to study.

The most common way to construct smooth subalgebras in C∗-algebras is as
smooth elements of an action of a Lie group, see [2]. There is a natural
circle action on BD algebras, however the resulting smooth subalgebras seem to
be too big and in particular don’t have a nice classification of derivations
on them.

The smooth-subalgebras studied in this paper are more closely related to the
work in [16] regarding smooth subalgebras of crossed products
$A\rtimes{\mathbb{Z}}$ of which Bunce-Deddens algebras are examples. A key
difference, however, is that the construction in [16] starts with a dense
subalgebra of $A$ with its own Frechet topology, while in our case a choice of
such a subalgebra is not obvious. Bunce-Deddens algebras are crossed-product
C∗-algebras obtained from odometers. Due to the topology of odometers, which
are Cantor sets with a minimal action of a homeomorphism, the smooth
subalgebras are naturally equipped with inductive limit Frechet (LF) topology.

Derivations naturally arise in differential geometry as vector fields and play
an analogous role in noncommutative geometry. We prove that the space of
continuous derivations on the smooth subalgebra of a BD algebra modulo inner
derivations is just one-dimensional and we exhibit its natural generator.

K-Theory and K-Homology are key invariants in the noncommutative topology of
quantum spaces. In addition to reviewing the known $K$-Theory of BD algebras,
we include two detailed calculations of $K$-Homology. One calculation uses
Rosenberg and Schochet’s Universal Coefficient Theorem, while the other uses
Pimsner and Voiculescu’s $6$-term exact sequence.

Smooth subalgebras of noncommutative spaces are also naturally present in
studying spectral triples and in cyclic cohomology which we do not address in
this paper. Examples of spectral triples for BD algebras are discussed in [8].

The paper is organized as follows. In section 2 we review supernatural numbers
and Bunce-Deddens algebras described as concrete C∗-algebras of operators in a
Hilbert space. In section 3 we discuss smooth subalgebras of Bunce-Deddens
algebras and their properties. In section 4 we classify continuous derivations
on the smooth subalgebras from section 3. We also discuss examples that show
the necessity of the assumption that the derivations are continuous. In
section 5 we discuss the $K$-Theory and $K$-Homology of Bunce-Deddens
algebras. Finally in section 6, the appendix, we provide an example based
guide for computations in homological algebra involving the Ext functor.

##  2\. Preliminaries

We begin this section by recalling some relevant information regarding
supernatural numbers.

###  2.1. Supernatural Numbers

A supernatural number $S$ is defined as the formal product:

$S=\prod_{p-\textrm{prime}}p^{\varepsilon_{p}},\;\;\;\varepsilon_{p}\in\\{0,1,\cdots,\infty\\}\,.$

When a supernatural number $S$ has the property that
$\sum\varepsilon_{p}<\infty$, we say that $S$ is a finite supernatural number.
Otherwise, $S$ is said to be an infinite supernatural number. Given another
supernatural number

$S^{\prime}=\prod_{p-\textrm{prime}}p^{\varepsilon_{p}^{\prime}},$

their product is defined by

$SS^{\prime}:=\prod_{p-\textrm{prime}}p^{\varepsilon_{p}+\varepsilon_{p}^{\prime}}\,.$

We say that a supernatural number $S$ divides $\mathcal{S}$ if there is some
supernatural number $S^{\prime}$ such that $\mathcal{S}=SS^{\prime}$. In terms
of the powers, this is equivalent to
$\varepsilon_{p}(S)\leq\varepsilon_{p}(\mathcal{S})$ for every prime $p$.

For the remainder of the paper we work with a fixed supernatural number $S$.
Consider the set of divisors of $S$:

$\mathcal{L}_{S}=\\{l:\;l|S,l<\infty\\}.$

Notice that $(\mathcal{L}_{S},\leq)$ is a directed set where $l_{1}\leq l_{2}$
if and only if $l_{1}|l_{2}|S$. We say that a sequence $\\{l_{n}\\}$ of
divisors of $S$ converges to $S$, $\lim_{n\to\infty}l_{n}=S$, if the exponents
converge:

$\lim_{n\to\infty}\varepsilon_{p}(l_{n})=\varepsilon_{p}(S).$

###  2.2. Odometers [6]

For each pair of divisors $l,k$ satisfying $l\geq k$, consider the ring
homomorphism
$\pi_{kl}:{\mathbb{Z}}/l{\mathbb{Z}}\rightarrow{\mathbb{Z}}/k{\mathbb{Z}}$
defined by

$\pi_{kl}(x)=x\ (\textrm{mod }k)$

which satisfy

$\pi_{kj}=\pi_{kl}\circ\pi_{lj}\textnormal{ for all }k\leq l\leq j\,.$

Then the inverse limit of the system can be denoted as:

${\mathbb{Z}}/S{\mathbb{Z}}:=\lim_{\underset{l\in\mathcal{L}_{S}}{\longleftarrow}}{\mathbb{Z}}/l{\mathbb{Z}}=\left\\{\\{x_{l}\\}\in\prod\limits_{l\in\mathcal{L}_{S}}{\mathbb{Z}}/l{\mathbb{Z}}:\pi_{kl}(x_{l})=x_{k}\right\\}\,,$

and let

$\pi_{l}:{\mathbb{Z}}/S{\mathbb{Z}}\ni\\{x_{l}\\}\mapsto
x_{l}\in{\mathbb{Z}}/l{\mathbb{Z}}$

be the corresponding homomorphisms. Notice that for finite supernatural number
$S$, the above notation is consistent with the usual meaning of the symbol
${\mathbb{Z}}/S{\mathbb{Z}}$. A more interesting observation is that if
$S=p^{\infty}$ for a prime p, then the above limit is equal to
${\mathbb{Z}}_{p}$, the ring of $p$-adic integers, see for example [18]. In
general we have the following simple consequence of the Chinese Reminder
Theorem.

######  Proposition 2.1.

If $S=\prod\limits_{\begin{subarray}{c}p-\textrm{prime}\\\
{\varepsilon_{p}\neq 0}\end{subarray}}p^{\varepsilon_{p}}$, then
${\mathbb{Z}}/S{\mathbb{Z}}\cong\prod\limits_{\begin{subarray}{c}p-\textrm{prime}\\\
{\varepsilon_{p}\neq
0}\end{subarray}}{\mathbb{Z}}/{p^{\varepsilon_{p}}}{\mathbb{Z}}$.

If the ring ${\mathbb{Z}}/S{\mathbb{Z}}$ is equipped with the Tychonoff
topology it forms a compact, Abelian topological group. Thus it has a unique
normalized Haar measure $d_{H}x$. In addition, if $S$ is an infinite
supernatural number then ${\mathbb{Z}}/S{\mathbb{Z}}$ is a Cantor set [24].

###  2.3. Dense Cyclic Subgroup

${\mathbb{Z}}/S{\mathbb{Z}}$ contains a dense copy of ${\mathbb{Z}}$, which we
now describe. Define a homomorphism
$q:{\mathbb{Z}}\rightarrow{\mathbb{Z}}/S{\mathbb{Z}}$ by:

$q(x)=\\{x\ (\textrm{mod
}l)\\}\in{\mathbb{Z}}/S{\mathbb{Z}}\subseteq\prod\limits_{l\in\mathcal{L}_{S}}{\mathbb{Z}}/l{\mathbb{Z}}.$
(2.1)

and let $q_{l}:{\mathbb{Z}}\rightarrow{\mathbb{Z}}/l{\mathbb{Z}}$ be the usual
quotient maps. We have the following simple property:

$\pi_{l}\circ q=q_{l}\,.$

From this fact, along with the structure of cylinder sets, the following
proposition can be deduced:

######  Proposition 2.2.

The range of $q$ is dense in ${\mathbb{Z}}/S{\mathbb{Z}}$.

###  2.4. Locally Constant Functions [9]

We denote by $\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$ the space of locally
constant functions on ${\mathbb{Z}}/S{\mathbb{Z}}$. This is a dense subspace
of the space of continuous functions on ${\mathbb{Z}}/S{\mathbb{Z}}$. Given an
$f\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$, consider the sequence:

$a_{f}(k)=f(q(k)),\ \ k\in{\mathbb{Z}}\,.$

We have the following observation:

######  Proposition 2.3.

If $f\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$, then there exists
$l\in\mathcal{L}_{S}$ such that

$a_{f}(k+l)=a_{f}(k)$

for every $k\in{\mathbb{Z}}$. Conversely, if $a(k)$ is an $l$-periodic
sequence for some $l\in\mathcal{L}_{S}$, then there is a unique
$f\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$ such that $a(k)=a_{f}(k)$.

The proof follows from the fact that $f$ is locally constant on
${\mathbb{Z}}/S{\mathbb{Z}}$ iff there is $l|S$ such that $f(x)=f(y)$ for all
$x=\\{x_{k}\\}$ and $y=\\{y_{k}\\}$ in ${\mathbb{Z}}/S{\mathbb{Z}}$ such that
$x_{k}=y_{k}$ if $l|k$.

###  2.5. Bunce-Dedden Algebras

To introduce Bunce-Deddens algebras we follow [14] which more generally
described algebras associated with infinite compact monothetic groups. Recall
that a monothetic group is a topological group that has a dense cyclic
subgroup. For an infinite supernatural number $S$, the group
${\mathbb{Z}}/S{\mathbb{Z}}$ is an infinite compact monothetic group with
${\mathbb{Z}}$, the image of the map $q$, being its dense cyclic subgroup.

Let $H$ be the $\ell^{2}$ Hilbert space of the cyclic subgroup, that is

$H=\ell^{2}(\\{q(l)\\}_{l\in{\mathbb{Z}}})\,,$

which is naturally isomorphic with $\ell^{2}({\mathbb{Z}})$. Let
$\\{E_{l}\\}_{l\in{\mathbb{Z}}}$ be the canonical basis in $H$ and let $U:H\to
H$ be the shift operator on $H$:

$UE_{l}=E_{l+1}\,.$

For a continuous function $f\in C({\mathbb{Z}}/S{\mathbb{Z}})$ we define an
operator $M_{f}:H\to H$ via:

$M_{f}E_{l}=f(q(l))E_{l}\,.$

Notice that $M_{f}$ is a diagonal multiplication operator on $H$. Due to the
density of the subgroup $\\{q(l)\\}_{l\in{\mathbb{Z}}}$, we immediately
obtain:

$\|M_{f}\|=\underset{l\in{\mathbb{Z}}}{\textrm{sup
}}|f(q(l))|=\underset{x\in{\mathbb{Z}}/S{\mathbb{Z}}}{\textrm{sup
}}|f(x)|=\|f\|_{\infty}\,.$

The algebra of operators generated by the $M_{f}$’s is thus isomorphic to
$C({\mathbb{Z}}/S{\mathbb{Z}})$. So, while the $M_{f}$’s carry all the
information about the space ${\mathbb{Z}}/S{\mathbb{Z}}$, the operator $U$
reflects the dynamics of the map $\beta$ on ${\mathbb{Z}}/S{\mathbb{Z}}$ given
by addition of the generator of the cyclic subgroup:

$\beta(x)=x+q(1).$ (2.2)

This relationship is summarized by:

$UM_{f}U^{-1}=M_{f\circ\beta}\,.$

We define the algebra $B_{S}$ to be the C∗-algebra generated by operators $U$
and $M_{f}$:

$B_{S}=C^{*}\\{U,M_{f}:f\in C({\mathbb{Z}}/S{\mathbb{Z}})\\}\,.$

$B_{S}$ is isomorphic with the crossed product algebra:

$B_{S}\cong C({\mathbb{Z}}/S{\mathbb{Z}})\rtimes_{\beta}{\mathbb{Z}}\,.$

Indeed, observe that ${\mathbb{Z}}$ is amenable, the action of ${\mathbb{Z}}$
on ${\mathbb{Z}}/S{\mathbb{Z}}$ given by $\beta$ is a free action, and $\beta$
is a minimal homeomorphism. Thus, the crossed product is simple and equal to
the reduced crossed product, see, for instance, [7]. Clearly, the operators
$U$ and $M_{f}$ define a representation of the crossed product
$C({\mathbb{Z}}/S{\mathbb{Z}})\rtimes_{\beta}{\mathbb{Z}}$, and the algebra
they generate must be isomorphic to it, by simplicity of the crossed product.
This crossed product is known as the Bunce-Deddens algebra.

###  2.6. Polynomial Subalgebra

Let $\chi$ be a character of the group ${\mathbb{Z}}/S{\mathbb{Z}}$. We have
$\chi\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$ and so $\chi$ comes from a
$l$ periodic function on ${\mathbb{Z}}$, by an abuse of notation also denoted
by $\chi$. Any $l$ periodic character is of the form $\chi_{l}^{k}$ where

$\chi_{l}(q(x))=e^{2\pi ix/l},$

and $k$ is an integer.

The algebra $B_{S}$ has a natural dense $*$-subalgebra $\mathcal{B}_{S}$ of
polynomials in $U$, $U^{-1}$, and the $M_{\chi}$’s, where $\chi$ is a
character of ${\mathbb{Z}}/S{\mathbb{Z}}$. We have:

$\mathcal{B}_{S}=\left\\{\sum_{n}U^{n}M_{f_{n}}:f_{n}\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}}),\textrm{
finite sums}\right\\}\,.$

Alternatively, any element $a$ in $\mathcal{B}_{S}$ can be written as

$a=\sum_{k=0}^{l-1}F_{k}(U)M_{\chi_{l}^{k}}\,,$ (2.3)

where $F_{k}$ are trigonometric polynomials.

###  2.7. Label Operator

It is often convenient to use the following diagonal label operator on $H$:

$\mathbb{L}E_{l}=lE_{l}\,.$

If $\\{b(l)\\}_{l\in{\mathbb{Z}}}$ is a bounded sequence, then by the
functional calculus, $b(\mathbb{L})$ is a bounded diagonal operator given by:

$b(\mathbb{L})E_{l}=b(l)E_{l}\,,$

and all bounded diagonal operators can be expressed this way. Moreover, we
have the following commutation relation:

$b(\mathbb{L})U=Ub(\mathbb{L}+I)\,.$ (2.4)

##  3\. Smooth Subalgebras

###  3.1. Definitions

To define the space of smooth elements of the Bunce-Deddens algebra we need
the following terminology. Let $\\{a_{n}\\}$ be a sequence of complex numbers.
We say $\\{a_{n}\\}$ is Rapid Decay, RD, if for every $k\geq 0$, there exists
a constant $C_{k}$ such that

$n^{k}|a_{n}|\leq C_{k}\,.$

We say a family of locally constant functions is Uniformly Locally Constant,
ULC, if there exists a divisor $l$ of $S$ such that for every $f$ in the
family we have

$f(x+l)=f(x)$

for all $x\in{\mathbb{Z}}/S{\mathbb{Z}}$.

We define the space of smooth elements of the Bunce-Deddens algebra,
$B_{S}^{\infty}$, to be the space of elements in $B_{S}$ whose Fourier
coefficients are ULC and norms are RD. Using Equation 2.4 the ULC condition on
Fourier coefficients can be written as:

$B_{S}^{\infty}=\left\\{b=\sum_{n\in{\mathbb{Z}}}U^{n}M_{f_{n}}:\\{\|f_{n}\|\\}\textrm{
is }RD,\textrm{ there is an }l|S,\,\,U^{l}bU^{-l}=b\right\\}\,.$

It’s immediate that $B_{S}^{\infty}$ is indeed a nonempty subset of $B_{S}$.

###  3.2. M-Norms

Our next goal is to prove that $B_{S}^{\infty}$ is a $*$-subalgebra of
$B_{S}$. This is achieved in several steps. The first is the definition of a
derivation [10].

Let $A$ be an algebra. A linear map $\delta:A\to A$ is called a derivation if
the Leibniz rule holds:

$\delta(ab)=a\delta(b)+\delta(a)b$

for all $a,b\in A$. An example of a derivation $B^{\infty}_{S}\to
B^{\infty}_{S}$ is given by

$\delta_{\mathbb{L}}(b)=[\mathbb{L},b].$

Since $[\mathbb{L},U^{n}]=nU^{n}$ and $[\mathbb{L},M_{f}]=0$, we see that

$\textrm{if}\quad
b=\sum_{n\in{\mathbb{Z}}}U^{n}M_{f_{n}}\,,\quad\textrm{then}\quad\delta_{\mathbb{L}}(b)=\sum_{n\in{\mathbb{Z}}}U^{n}nM_{f_{n}}\,.$

Thus if $\\{\|f_{n}\|\\}$ are RD, then so too are $\\{n\|f_{n}\|\\}$ and hence
$\delta_{\mathbb{L}}(b)$ is in $B_{S}^{\infty}$. This derivation is very
fundamental, and is used extensively in the sequel.

In the following lemma we describe a class of norms on $B_{S}$ that we call
$M$-norms. These $M$-norms are used to provide an alternative description of
the RD condition in the definition of $B_{S}^{\infty}$ (see Proposition 3.3
below).

######  Lemma 3.1.

Let $A$ be an algebra with a submultiplicative norm $\|\cdot\|$ and let
$\delta:A\to A$ be any derivation. For any $a\in A$ and $M$ a nonnegative
integer, define recursively:

$\|a\|_{M+1}=\|a\|_{M}+\|\delta(a)\|_{M}\,,$

with $\|a\|_{0}:=\|a\|$. Then, $\|\cdot\|_{M}$ is a submultiplicative norm on
$A$. Moreover

$\|a\|_{M}=\sum_{j=0}^{M}\ \begin{pmatrix}M\\\
j\end{pmatrix}\|\delta^{(j)}(a)\|\,.$

###### Proof.

Proving that $\|\cdot\|_{M}$ is a norm is on $A$ is an immediate consequence
of induction and the fact that $\|\cdot\|$ is a norm on $A$. To prove
submultiplicativeness, induction can also be used. The base case $M=0$ is
immediate as $\|\cdot\|$ is submultiplicative. Thus given an $M$, suppose
$\|\cdot\|_{M}$ is submultiplicative. We have the following calculation:

$\displaystyle\|ab\|_{M+1}$
$\displaystyle=\|ab\|_{M}+\|\delta(ab)\|_{M}=\|ab\|_{M}+\|\delta(a)b+a\delta(b)\|_{M}$
$\displaystyle\leq\|a\|_{M}\|b\|_{M}+\|\delta(a)\|_{M}\|b\|_{M}+\|a\|_{M}\|\delta(b)\|_{M}$
$\displaystyle\leq(\|a\|_{M}+\|\delta(a)\|_{M})(\|b\|_{M}+\|\delta(b)\|_{M})=\|a\|_{M+1}\|b\|_{M+1}\,.$

Thus $\|\cdot\|_{M}$ is submultiplicative.

It remains to prove the formula in the last part of the statement of the
lemma. Again, we use induction. The base case $M=0$ is trivial. Suppose the
formula is true for a given $M$ and consider

$\|a\|_{M+1}=\|a\|_{M}+\|\delta(a)\|_{M}=\sum_{j=0}^{M}\begin{pmatrix}M\\\
j\end{pmatrix}\|\delta^{(j)}(a)\|+\sum_{j=0}^{M}\begin{pmatrix}M\\\
j\end{pmatrix}\|\delta^{(j+1)}(a)\|\,.$

Resumming the second sum and using the binomial coefficient reduction formula,
we obtain

$\displaystyle\|a\|_{M+1}$
$\displaystyle=\|a\|_{0}+\sum_{j=1}^{M}\left[\begin{pmatrix}M\\\
j\end{pmatrix}+\begin{pmatrix}M\\\
j-1\end{pmatrix}\right]\|\delta^{(j)}(a)\|+\|a\|_{M+1}$
$\displaystyle=\sum_{j=0}^{M+1}\begin{pmatrix}M+1\\\
j\end{pmatrix}\|\delta^{(j)}(a)\|\,.$

This completes the proof. ∎

For the remainder of the paper we focus on the following particular $M$-norm
on $B_{S}$:

$\|b\|_{M}=\sum_{j=0}^{M}\ \begin{pmatrix}M\\\
j\end{pmatrix}\|\delta_{\mathbb{L}}^{(j)}(b)\|\,.$

In general, this norm is not well-defined on all of $B_{S}$ as $\mathbb{L}$ is
an unbounded operator. We must first properly define
$\|\delta_{\mathbb{L}}^{(j)}(b)\|$. We do this in the following discussion.
Let $\mathcal{D}\subseteq\ell^{2}({\mathbb{Z}})$ be the following subspace:

$\mathcal{D}=\left\\{x\in\ell^{2}({\mathbb{Z}}):x=\sum_{k\in{\mathbb{Z}}}x_{k}E_{k}\textrm{
with }\\{x_{k}\\}\textrm{-RD}\right\\}\,.$

Then pick $x\in\mathcal{D}$ and a $b\in B_{S}$, with $\\{b_{kl}\\}$ the matrix
coefficients of $b$ with respect to the canonical basis $\\{E_{l}\\}$, and
consider the following calculation:

$\displaystyle\|\delta_{\mathbb{L}}(b)x\|^{2}$
$\displaystyle=\left\|\sum_{k\in{\mathbb{Z}}}x_{k}\delta_{\mathbb{L}}(b)E_{k}\right\|^{2}=\left\|\sum_{k,l\in{\mathbb{Z}}}lx_{k}b_{kl}E_{l}-\sum_{k,l\in{\mathbb{Z}}}kx_{k}b_{kl}E_{l}\right\|^{2}$
$\displaystyle=\left\|\sum_{l\in{\mathbb{Z}}}\left(\sum_{k\in{\mathbb{Z}}}(l-k)b_{kl}x_{k}\right)E_{l}\right\|^{2}\,.$

The inner sum over $k$ is finite since $\\{b_{kl}\\}$ is a bounded sequence as
$b\in B_{S}$, $\\{x_{k}\\}$ is a RD sequence and $l-k$ is a polynomial in $k$.
Therefore we can define

$\|\delta_{\mathbb{L}}(b)x\|^{2}=\sum_{l\in{\mathbb{Z}}}\left|\sum_{k\in{\mathbb{Z}}}(l-k)b_{kl}x_{k}\right|^{2}$

which may or may not be finite, but is a series of nonnegative terms over $l$.
Thus for $b\in B_{S}$ we define $\|\delta_{\mathbb{L}}(b)\|$ by

$\|\delta_{\mathbb{L}}(b)\|=\underset{x\in\mathcal{D}}{\textrm{sup
}}\frac{\|\delta_{\mathbb{L}}(b)x\|}{\|x\|}\,.$

Using similar considerations, we can define
$\|\delta_{\mathbb{L}}^{(j)}(b)\|$. Now $\|b\|_{M}\in[0,\infty]$ makes sense
for every $b\in B_{S}$.

###  3.3. Basic Properties

Next we want to relate the RD condition for $b\in B_{S}^{\infty}$ to
finiteness of $M$-norms. This is done using a $1$-parameter group of
automorphisms of $B_{S}$ that is given by the following equation:

$\rho_{\theta}(b)=e^{2\pi i\theta\mathbb{L}}be^{-2\pi
i\theta\mathbb{L}}\quad\textrm{ for }b\in B_{S},$

where $\theta\in{\mathbb{R}}/2\pi{\mathbb{Z}}$. We have the following
formulas:

$\rho_{\theta}(U)=e^{2\pi i\theta}U\textrm{ and
}\rho_{\theta}(b(\mathbb{L}))=b(\mathbb{L})\,.$

It immediately follows that $\rho_{\theta}:\mathcal{B}_{S}\to\mathcal{B}_{S}$
and that $\rho_{\theta}:B_{S}^{\infty}\to B_{S}^{\infty}$.

Define $E:B_{S}\to C^{*}\\{M_{f}:f\in C({\mathbb{Z}}/S{\mathbb{Z}})\\}\cong
C({\mathbb{Z}}/S{\mathbb{Z}})$ via

$E(b)=\int_{0}^{1}\rho_{\theta}(b)\,d\theta\,.$

It’s easily checked that $E$ is an expectation on $B_{S}$. For a $b\in B_{S}$
we define the $n$-th Fourier coefficient, $b_{n}$ by the following:

$b_{n}=E(U^{-n}b)=\int_{0}^{1}\rho_{\theta}(U^{-n}b)\,d\theta\,=\int_{0}^{1}e^{-2\pi
in\theta}U^{-n}\rho_{\theta}(b)\,d\theta.$

From this definition, it’s clear that $b_{n}\in C^{*}\\{M_{f}:f\in
C({\mathbb{Z}}/S{\mathbb{Z}})\\}$. We have the following lemma:

######  Lemma 3.2.

Let $b$ and $b^{\prime}$ be elements of $B_{S}$. If $b_{n}=b_{n}^{\prime}$ for
every $n$, then $b=b^{\prime}$. Moreover, if $\\{\|b_{n}\|\\}$ is a RD
sequence, then

$b=\sum_{n\in{\mathbb{Z}}}U^{n}b_{n}$

where the sum is norm convergent.

###### Proof.

The first part follows from the standard techniques from Fourier analysis, see
for example [11].

If $\\{\|b_{n}\|\\}$ is a RD sequence, then, there exists a constant $C$ such
that $(|n|^{2}+1)\|b_{n}\|\leq C$. Thus

$\left\|\sum_{|n|\leq j}U^{n}b_{n}\right\|\leq\sum_{|n|\leq
j}\|b_{n}\|\leq\sum_{|n|\leq
j}\frac{1}{|n|^{2}+1}\leq\sum_{n\in{\mathbb{Z}}}\frac{1}{|n|^{2}+1}$

which is an absolutely convergent sum. Thus by the Weierstrass Test,
$\sum_{n}U^{n}b_{n}$ is a norm convergent sum. It follows that $b_{n}$ are the
Fourier coefficients of $b\in B_{S}$ and
$b=\sum_{n\in{\mathbb{Z}}}U^{n}b_{n}\,.$ ∎

With this lemma, we have the following equivalent definition of
$B_{S}^{\infty}$:

######  Proposition 3.3.

$B_{S}^{\infty}=\\{b\in B_{S}:\|b\|_{M}<\infty\,,\textrm{ for every
}M,\textrm{ there is an }l|S,\,\,U^{l}bU^{-l}=b\\}\,.$

###### Proof.

In order to prove the proposition, we only need to check that finite $M$-norms
is equivalent to RD Fourier coefficients. If $b\in B_{S}^{\infty}$, then
$\\{\|f_{n}\|\\}$ are RD. A straightforward calculation shows that:

$\textrm{if
}b=\sum_{n\in{\mathbb{Z}}}U^{n}M_{f_{n}}\,,\quad\textrm{then}\quad\delta_{\mathbb{L}}^{(j)}(b)=\sum_{n\in{\mathbb{Z}}}n^{j}U^{n}M_{f_{n}}\,,$

and consequently we have:

$||\delta_{\mathbb{L}}^{(j)}(b)||\leq\sum_{n\in{\mathbb{Z}}}|n|^{j}||f_{n}||\,.$

It follows that $\|b\|_{M}$ is finite for every $M$.

On the other hand, let $b\in B_{S}$ and assume that $\|b\|_{M}$ is finite for
every $M$. Then we get the following formula for the Fourier coefficients
$M_{f_{n}}$ of $b$:

$E\left(U^{-n}\delta_{\mathbb{L}}^{(j)}(b)\right)=n^{j}E(U^{-n}b)=n^{j}M_{f_{n}}\,,$

and hence

$n^{j}\|f_{n}\|=n^{j}\|M_{f_{n}}\|\leq\|\delta_{\mathbb{L}}^{(j)}(b)\|\leq\|b\|_{M}\,.$

Therefore if $\|b\|_{M}$ is finite for every $M$, it follows that
$\\{\|f_{n}\|\\}$ are RD. ∎

We now give a brief discussion of the natural topology on $B_{S}^{\infty}$ and
continuity of linear maps on it. Define the following spaces:

$B_{S,\,l}^{\infty}=\\{b\in B_{S}^{\infty}:U^{l}bU^{-l}=b\textrm{ for some
}l|S\\}\,.$

Notice that $B_{S,\,l}^{\infty}$ is a closed subspace of $B_{S}^{\infty}$.
Moreover if $l$ and $l^{\prime}$ are two different divisors of $S$ with
$l<l^{\prime}$, then it follows that $B_{S,\,l}^{\infty}\subsetneqq
B_{S,\,l^{\prime}}^{\infty}$. Also $B_{S,\,l}^{\infty}$ is a Fréchet space
with respect to $\|\cdot\|_{M}$ and

$B_{S}^{\infty}=\bigcup_{l|S}B_{S,\,l}^{\infty}$

as a strict inductive limit. To establish some of the results below it is also
useful to represent this inductive limit in the following way. Let
$1<l_{1}|\,l_{2}|\cdots$ be a strictly increasing sequence of divisors of $S$.
Notice that

$B_{S}^{\infty}=\bigcup_{n=1}^{\infty}B_{S,\,l_{n}}^{\infty}\,.$

Let $T:B_{S}^{\infty}\to B_{S}^{\infty}$ be a linear map. Then by the general
property of inductive limits, see [19], $T$ is continuous if and only if
$T:B_{S,\,l_{n}}^{\infty}\to B_{S}^{\infty}$ is continuous for every $n$.
Additionally we have the following useful result.

######  Lemma 3.4.

Let $T:B_{S}^{\infty}\to B_{S}^{\infty}$ be a continuous linear map. Then for
every $n$, there exists an $m$ such that $T(B_{S,\,l_{n}}^{\infty})\subseteq
B_{S,\,l_{m}}^{\infty}$.

###### Proof.

Since $B_{S,\,l_{n}}^{\infty}$ is separable, choose a sequence
$\\{g_{k}\\}_{k=1}^{\infty}$ with $g_{k}\in B_{S,\,l_{n}}^{\infty}$ for every
$k$, such that $\textrm{span}_{k}\\{g_{k}\\}$ is dense in
$B_{S,\,l_{n}}^{\infty}$ and $\|g_{k}\|_{1}=1$. Define $f_{k}=g_{k}/k$ and so
$f_{k}\to 0$ in $B_{S,\,l_{n}}^{\infty}$. Since $T$ is continuous, we have
$T(f_{k})\to 0$ in $B_{S}^{\infty}$. Thus by Theorem 5.17 in [19], it follows
that there exists an $m$ such that $T(f_{k})\subseteq B_{S,\,l_{m}}^{\infty}$
for every $k$. However,

$T(\textrm{span}_{k}\\{g_{k}\\})=T(\textrm{span}_{k}\\{f_{k}\\})\subseteq
B_{S,\,l_{m}}^{\infty}\,.$

But $\textrm{span}_{k}\\{g_{k}\\}$ is dense in $B_{S,\,l_{n}}^{\infty}$ and
$B_{S,\,l_{m}}^{\infty}$ is complete. Thus it follows that
$T(B_{S,\,l_{n}}^{\infty})\subseteq B_{S,\,l_{m}}^{\infty}$. ∎

The following lemma is a special case, adapted to our scenario, of Theorem 5.2
in [19].

######  Lemma 3.5.

Let $T:B_{S,\,l_{n}}^{\infty}\to B_{S,\,l_{m}}^{\infty}$ be a linear map. Then
$T$ is continuous if and only if for every $M$, there exist an $M^{\prime}$
and a constant $C=C(M)$ such that for every $b\in B_{S,\,l_{n}}^{\infty}$, the
following inequality holds:

$\|T(b)\|_{M}\leq C\|b\|_{M^{\prime}}\,.$

Together, Lemmas 3.4 and 3.5 imply the following theorem:

######  Theorem 3.6.

$T:B_{S}^{\infty}\to B_{S}^{\infty}$ be a linear map. $T$ is continuous if and
only if for every $n$, there exists an $m$ such that
$T(B_{S,\,l_{n}}^{\infty})\subseteq B_{S,\,l_{m}}^{\infty}$ and for every $M$,
there exist an $M^{\prime}$ and a constant $C=C(M)$ such that for every $b\in
B_{S}^{\infty}$, the following inequality holds:

$\|T(b)\|_{M}\leq C\|b\|_{M^{\prime}}\,.$

######  Corollary 3.7.

$\delta_{\mathbb{L}}:B_{S}^{\infty}\to B_{S}^{\infty}$ is a continuous
derivation.

To conclude this section, we prove that $B^{\infty}_{S}$ is a $*$-subalgebra
of $B_{S}$ closed under the holomorphic functional calculus [2].

######  Proposition 3.8.

$B_{S}^{\infty}$ is a $*$-subalgebra of $B_{S}$ with respect to
$\|\cdot\|_{M}$.

###### Proof.

Since $\|\cdot\|_{M}$ is a norm and the relation $a=U^{l}au^{-l}$ for some
divisor $l$ of $S$ is linear, it follows that $B_{S}^{\infty}$ is closed under
addition and scalar multiplication. Moreover since $\|\cdot\|_{M}$ is
$*$-preserving, it follows that $B_{S}^{\infty}$ is closed under taking $*$’s.
It only remains to show that $B_{S}^{\infty}$ is closed under multiplication.
Let $a$ and $b$ be two elements in $B_{S}^{\infty}$ and let $l_{1}$ and
$l_{2}$ be two divisors of $S$ such that

$a=U^{l_{1}}aU^{-l_{1}}\quad\textrm{and}\quad b=U^{l_{2}}bU^{-l_{2}}\,.$

Set $l=\textrm{lcm}(l_{1},l_{2})$, then

$U^{l}abU^{-l}=U^{l}aU^{-l}U^{l}bU^{-l}=ab\,.$

Finally, since $\|\cdot\|_{M}$ is submultiplicative by Lemma 3.1, it follows
that $ab\in B_{S}^{\infty}$. ∎

###  3.4. Holomorphic Calculus

We verify below that $B_{S}^{\infty}$ is closed under the holomorphic
functional calculus and hence has the same K-Theory as $B_{S}$.

######  Lemma 3.9.

Let $b\in B_{S}^{\infty}$ such that $b$ is invertible in $B_{S}$. Then,
$b^{-1}\in B_{S}^{\infty}$.

###### Proof.

Since $b\in B_{S}^{\infty}$, there exists a divisor $l$ of $S$ such that
$U^{l}bU^{-l}=b$. Since $b$ and $U$ are invertible in $B_{S}$, it immediately
follows that $U^{l}b^{-1}U^{-l}=b^{-1}$.

To check that the $\|b^{-1}\|_{M}$ is finite, notice that

$\displaystyle\|b^{-1}\|_{1}$
$\displaystyle=\|b^{-1}\|+\|\delta_{\mathbb{L}}(b^{-1})\|\leq\|b^{-1}\|+\|b^{-1}\|^{2}\|\delta_{\mathbb{L}}(b)\|$
$\displaystyle\leq\|b^{-1}\|^{2}\|b\|+\|b^{-1}\|^{2}\|\delta_{\mathbb{L}}(b)\|=\|b^{-1}\|^{2}\|b\|_{1}\,.$

By definition we know that

$\|b\|_{M+1}=\|b\|_{M}+\|\delta_{\mathbb{L}}(b)\|_{M}\,,$

and so by a similar argument we obtain

$\|b^{-1}\|_{2}\leq\|b^{-1}\|_{1}^{2}\|b\|_{2}\,.$

Proceeding inductively we get

$\|b^{-1}\|_{M+1}\leq\|b^{-1}\|_{M}^{2}\|b\|_{M+1}\,.$

Hence $\|b^{-1}\|_{M}$ is finite for all $M$ and thus $b^{-1}\in
B_{S}^{\infty}$. ∎

######  Proposition 3.10.

$B_{S}^{\infty}$ is closed under the holomorphic functional calculus. That is,
given $b\in B_{S}^{\infty}$ and a function $f$ that is holomorphic on an open
domain containing the spectrum of $b$, we have $f(b)\in B_{S}^{\infty}$.

###### Proof.

Let $b\in B_{S}^{\infty}$ and let $f$ be a holomorphic function on some open
set containing $\sigma(b)$ and $C$ a contour aroud the spectrum. Write $f$ in
Cauchy integral form:

$f(z)=\frac{1}{2\pi i}\int_{C}\frac{f(\zeta)}{\zeta-z}\,d\zeta$

which defines $f(b)$ as

$f(b)=\frac{1}{2\pi i}\int_{C}f(\zeta)(\zeta-b)^{-1}\,d\zeta\,.$

Since $b\in B_{S}^{\infty}$, there exists a divisor $l$ of $S$ such that
$b=U^{l}bU^{-l}$. It follows from the Cauchy integral form above for that
$f(b)=U^{l}f(b)U^{-l}$.

From Lemma 3.9, it follows that if $b\in B_{S,\,l}^{\infty}$ for some divisor
$l$ of $S$ and $b$ is invertible in $B_{S}$, then $b^{-1}\in
B_{S,\,l}^{\infty}$. Consequently, $(\zeta-b)^{-1}\in B_{S,\,l}^{\infty}$ for
$\zeta\notin\sigma(b)$. It follows from completeness of $B_{S,\,l}^{\infty}$
that $f(b)\in B_{S,\,l}^{\infty}\subseteq B_{S}^{\infty}$. This completes the
proof. ∎

##  4\. Classification of Derivations

The main goal in this section is to classify continuous derivations
$\delta:B_{S}^{\infty}\to B_{S}^{\infty}$. The results below are based on
ideas from [13], which described a classification of unbounded derivations
$\delta:\mathcal{B}_{S}\to B_{S}$. Here $\mathcal{B}_{S}$ is the polynomial
dense subalgebra of $B_{S}$. The reference [14] contains a generalization of
this classification to $C^{*}$-algebras associated to monothetic groups.

###  4.1. Preliminaries

We begin with recalling the basic concepts. Let $A$ be a Banach algebra and
$\mathcal{A}$ be a dense subalgebra of $A$. We say a derivation
$\delta:\mathcal{A}\to A$ is inner if there is a $x\in A$ such that

$\delta(a)=[x,a]\,$

for $a\in\mathcal{A}$. We say a derivation $\delta:\mathcal{A}\to A$ is
approximately inner if there are $x_{n}\in A$ such that

$\delta(a)=\lim_{n\to\infty}[x_{n},a]$

for $a\in\mathcal{A}$.

The main result of [13] is:

######  Theorem 4.1.

Suppose $\delta:\mathcal{B}_{S}\to B_{S}$ is a derivation in $B_{S}$ with $S$
infinite. Then there exists a unique constant $C$ such that

$\delta=C\delta_{\mathbb{L}}+\tilde{\delta}$

where $\tilde{\delta}$ is approximately inner.

Compared to [13], here we are studying derivations on a larger domain but with
smaller range and we need an added assumption of continuity of $\delta$. It
turns out that for continuous derivations $\delta:B_{S}^{\infty}\to
B_{S}^{\infty}$ there is a similar decomposition as above; the key difference
is that $\tilde{\delta}$ is inner, not just approximately inner.

Given $n\in{\mathbb{Z}}$, a continuous derivations $\delta:B_{S}^{\infty}\to
B_{S}^{\infty}$ is said to be a $n$-covariant derivation if the relation

$\rho_{\theta}^{-1}\delta\rho_{\theta}(a)=e^{-2\pi in\theta}\delta(a)$

holds. When $n=0$ we say the derivation is invariant. With this definition, we
point out that $\delta_{\mathbb{L}}:B_{S}^{\infty}\to B_{S}^{\infty}$ is an
invariant continuous derivation.

Definition: If $\delta$ is a continuous derivation on $B_{S}$, the $n$-th
Fourier component of $\delta$ is defined as:

$\delta_{n}(b)=\int_{0}^{1}e^{2\pi
in\theta}\rho_{\theta}^{-1}\delta\rho_{\theta}(b)\,d\theta\,.$

Notice that the integral in the above definition makes sense because of the
continuity of $\delta$ and the continuity of the map $\theta\to\rho_{\theta}$.

###  4.2. Covariant Derivations

The key step in classifying derivations is to analyze the properties of their
Fourier components. This will be done in steps leading to the formulation of
the main result. We will need below the standard observation that if
$\delta_{1}$ and $\delta_{2}$ are continuous derivations then their commutator
$[\delta_{1},\delta_{2}]$ is also a continuous derivation.

######  Lemma 4.2.

If $\delta:B_{S}^{\infty}\to B_{S}^{\infty}$ is a continuous derivation then
its $n$-th Fourier component $\delta_{n}$ is also a continuous derivation on
$B_{S}^{\infty}$. Moreover, for every $M\geq 0$ and every $k\geq 0$, there
exists $M^{\prime}\geq 0$ and a constant $C_{k}=C_{k}(M)$ such that, for every
$b\in B_{S}^{\infty}$:

$n^{k}\|\delta_{n}(b)\|_{M}\leq C_{k}\|b\|_{M^{\prime}}\,.$

###### Proof.

Since $\delta$ is a continuous derivation on $B_{S}^{\infty}$, there exists a
constant $C=C(M)$ and an $M^{\prime}\geq 0$ such that

$\|\delta(b)\|_{M}\leq C\|b\|_{M^{\prime}}\,.$

Using this and the fact that M-norms are $\rho_{\theta}$ invariant, we have

$\|\delta_{n}(b)\|_{M}=\left\|\int_{0}^{1}e^{2\pi
in\theta}\rho_{\theta}^{-1}\delta\rho_{\theta}(b)\,d\theta\right\|_{M}\leq\int_{0}^{1}\|\rho_{\theta}^{-1}\delta\rho_{\theta}(b)\|_{M}\,d\theta\leq
C\|b\|_{M^{\prime}}\,.$

Using integration by parts, the continuity of $[\delta_{\mathbb{L}},\delta]$,
and the continuous differentiability of
$\theta\mapsto\rho_{\theta}^{-1}\delta\rho_{\theta}$ on $B_{S}^{\infty}$, we
have the following calculation:

$\displaystyle 2\pi in\delta_{n}(b)$ $\displaystyle=\int_{0}^{1}2\pi ine^{2\pi
in\theta}\rho_{\theta}^{-1}\delta\rho_{\theta}(b)\,d\theta=\int_{0}^{1}\frac{d}{d\theta}\left(e^{2\pi
in\theta}\right)\rho_{\theta}^{-1}\delta\rho_{\theta}(b)\,d\theta$
$\displaystyle=-\int_{0}^{1}e^{2\pi
in\theta}\frac{d}{d\theta}\left(\rho_{\theta}^{-1}\delta\rho_{\theta}(b)\right)\,d\theta=-2\pi
i\int_{0}^{1}e^{2\pi
in\theta}\rho_{\theta}^{-1}[\delta_{\mathbb{L}},\delta]\rho_{\theta}(b)\,d\theta$
$\displaystyle=-2\pi i\left([\delta_{\mathbb{L}},\delta]\right)_{n}(b)\,,$

where $\left([\delta_{\mathbb{L}},\delta]\right)_{n}$ is the $n$-th Fourier
component of $[\delta_{\mathbb{L}},\delta]$. Using induction on the order of
derivatives in terms of $\theta$, the result follows. ∎

In general, the Fourier series for a derivation constructed as above does not
converge to the derivation in norm. We only have the usual Cesàro mean
convergence result for Fourier components of $\delta$: if $\delta$ is a
derivation on $B_{S}^{\infty}$ then

$\delta(b)=\lim_{M\rightarrow\infty}\frac{1}{M+1}\sum_{j=0}^{M}\left(\sum_{n=-j}^{j}\delta_{n}(b)\right)\,,$
(4.1)

for every $b\in B_{S}^{\infty}$, see Lemma 4.2 in [13] for more details. In
particular, the Fourier components $\delta_{n}$ completely determine the
derivation $\delta$. However, for continuous derivations on $B_{S}^{\infty}$,
the above lemma implies that Fourier components $\delta_{n}$ are RD. Hence we
have the following statement.

######  Proposition 4.3.

Let $\delta:B_{S}^{\infty}\to B_{S}^{\infty}$ be a continuous derivation, then

$\delta(b)=\sum_{n\in{\mathbb{Z}}}\delta_{n}(b)$

for every $b\in B_{S}^{\infty}$. Here the sum is norm convergent.

###### Proof.

We will use the following version of the estimate from Lemma 4.2: for every
$M\geq 0$ and every $k\geq 0$, there exists $M^{\prime}\geq 0$ and a constant
$C_{k}=C_{k}(M)$ such that, for every $b\in B_{S}^{\infty}$ we have

$\left(|n|^{k}+1\right)\|\delta_{n}(b)\|_{M}\leq C_{k}\|b\|_{M^{\prime}}\,.$

Thus, considering the tail end of the infinite series and using the above
inequality with $k=2$, we have

$\left\|\sum_{|j|\geq n}\delta_{j}(b)\right\|_{M}\leq\sum_{|j|\geq
n}\|\delta_{j}(b)\|_{M}\leq C_{2}\|b\|_{M^{\prime}}\sum_{|j|\geq
n}\frac{1}{|j|^{2}+1}\,,$

which clearly goes to zero as $n$ goes to infinity. Thus the series is norm
convergent. Since the series is norm convergent, the formula
$\sum_{n\in{\mathbb{Z}}}\delta_{n}(b)$ defines a continuous derivation on
$B_{S}^{\infty}$ with the same Fourier coefficients as $\delta$ hence it must
be equal to $\delta$. ∎

######  Proposition 4.4.

Let $\delta:B_{S}^{\infty}\to B_{S}^{\infty}$ be a continuous derivation. Then
$\delta_{n}:B_{S}^{\infty}\to B_{S}^{\infty}$ is a continuous $n$-covariant
derivation.

###### Proof.

We previously noticed that $\delta_{n}$ is a derivation and is well-defined on
$B_{S}^{\infty}$. Since $\delta$ is a continuous derivation and the
automorphism $\rho_{\theta}$ is continuous, it follows that $\delta_{n}$ is
also continuous. The following computation verifies that $\delta_{n}$ is
$n$-covariant:

$\rho_{\theta}^{-1}\delta_{n}\rho_{\theta}(b)=\int_{0}^{1}e^{2\pi
in\varphi}\rho_{\theta}^{-1}\rho_{\varphi}^{-1}\delta\rho_{\varphi}\rho_{\theta}(b)\,d\varphi=\int_{0}^{1}e^{2\pi
in\varphi}\rho_{\theta+\varphi}^{-1}\delta\rho_{\theta+\varphi}(b)\,d\varphi\,.$

Changing to new variable $\theta+\varphi$, and using the translation
invariance of the measure, it now follows that
$\rho_{\theta}^{-1}\delta_{n}\rho_{\theta}(b)=e^{-2\pi
in\theta}\delta_{n}(b)$.

Since $\delta:B_{S}^{\infty}\to B_{S}^{\infty}$, for any $b\in B_{S}^{\infty}$
there exists a divisor $l$ of $S$ such that

$\delta(b)=U^{l}\delta(b)U^{-l}.$

It follows from the definition of $\delta_{n}$ that

$\delta_{n}(b)=U^{l}\delta_{n}(b)U^{-l}\,.$

Hence $\delta_{n}(b)\in B_{S}^{\infty}$. ∎

It turns out that $n$-covariant derivations can be described explicitly. This
was done in [13] for unbounded $n$-covariant derivations
$\delta:\mathcal{B}_{S}\to B_{S}$. We state those results since they are
useful below.

######  Proposition 4.5.

[13] Let $\delta:\mathcal{B}_{S}\to B_{S}$ be an $n$-covariant derivation
where $n\neq 0$. There exists an $F\in C({\mathbb{Z}}/S{\mathbb{Z}})$ such
that

$\delta=[U^{n}M_{F},\cdot]\,,$

so $\delta$ is an inner derivation.

######  Proposition 4.6.

[13] Let $\delta:\mathcal{B}_{S}\to B_{S}$ be a $0$-covariant (invariant)
derivation. There exists a unique constant $C$ such that

$\delta=C\delta_{\mathbb{L}}+\tilde{\delta}\,,$

where $\tilde{\delta}$ is an approximately inner derivation.

Using Proposition 4.5, for $n\neq 0$, we classify all continuous $n$-covariant
derivations on $B_{S}^{\infty}$.

######  Proposition 4.7.

Let $\delta:B_{S}^{\infty}\to B_{S}^{\infty}$ be a continuous $n$-covariant
derivation on $B_{S}^{\infty}$. There exists a
$F\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$ such that

$\delta(b)=\left[U^{n}M_{F},b\right]$

for $b\in B_{S}^{\infty}$, so that $\delta$ is an inner derivation for each
$n\neq 0$.

###### Proof.

Notice that a derivation $B_{S}^{\infty}\to B_{S}^{\infty}$ defines (an
unbounded) derivation $\mathcal{B}_{S}\to B_{S}$, so we can use Proposition
4.5. In particular, there exists $F\in C({\mathbb{Z}}/S{\mathbb{Z}})$ such
that $\delta(b)=\left[U^{n}M_{F},b\right]$, at least for
$b\in\mathcal{B}_{S}$. We want to show that in fact
$F\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$. Then, by continuity, that
formula will work for any $b\in B_{S}^{\infty}$.

Let $\chi$ be a character on ${\mathbb{Z}}/S{\mathbb{Z}}$. Then, since
$M_{\chi}\in\mathcal{B}_{S}\subset B_{S}^{\infty}$, we have that
$\delta(M_{\chi})\in B_{S}^{\infty}$ is well-defined. Using the fact that
$\beta^{n}(\chi)(x)=\chi(x)\chi(q(n))$ for $x\in{\mathbb{Z}}/S{\mathbb{Z}}$,
we have

$\delta(M_{\chi})=\left[U^{n}M_{F},M\chi\right]=U^{n}\left(M_{F\chi}-M_{F\beta^{n}(\chi)}\right)=(1-\chi(q(n)))U^{n}M_{F}M_{\chi}.$

However, since $\chi$ is a character on ${\mathbb{Z}}/S{\mathbb{Z}}$, we have
$\chi\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$ and for each $n$ we can
choose $\chi$ such that $\chi(q(n))\neq 1$. Therefore we obtain:

$M_{F}=\frac{1}{1-\chi(q(n))}U^{-n}\delta(M_{\chi})M_{\chi}^{-1}\in
B_{S}^{\infty}\,.$

Thus indeed $F\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$, completing the
proof. ∎

In particular, the above proposition can be applied to the $n$-th Fourier
components of a continuous derivation $\delta:B_{S}^{\infty}\to
B_{S}^{\infty}$.

Next, utilizing Proposition 4.6 we classify all $0$-covariant (invariant)
derivations on $B_{S}^{\infty}$.

######  Proposition 4.8.

Let $\delta:B_{S}^{\infty}\to B_{S}^{\infty}$ be an invariant continuous
derivation. Then there exists a unique constant $C$ such that

$\delta=C\delta_{\mathbb{L}}+\tilde{\delta}\,,$

where $\tilde{\delta}$ is an inner derivation of the form:

$\tilde{\delta}(a)=[M_{G},a]$

for some $G\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$.

###### Proof.

Let $\delta:B_{S}^{\infty}\to B_{S}^{\infty}$ be a continuous invariant
derivation. Clearly $\delta$ defines a derivation on $\mathcal{B}_{S}$.
Following the ideas of the proof of Proposition 4.6, from [13], we have
$\delta(U)=UM_{F}\in B_{S}^{\infty}$ for some
$F\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$ and $\delta(M_{f})=0$ for
$f\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$ since there are no nonzero
derivations on $\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$ by invariance. We
decompose $F$ in the following way

$F=C+\tilde{F}\,,$

where $C=\int_{{\mathbb{Z}}/S{\mathbb{Z}}}F$ and
$\int_{{\mathbb{Z}}/S{\mathbb{Z}}}\tilde{F}=0$. Thus we obtain

$\delta=C[\mathbb{L},\cdot]+\tilde{\delta}\,,$

with $\tilde{\delta}(U)=UM_{\tilde{F}}$. To complete the proof, we must show
$\tilde{\delta}$ is an inner derivation. That is, we need to show there exists
a $G\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$ such that

$\tilde{\delta}(a)=[M_{G},a]$

for any $a\in B_{S}^{\infty}$. If that’s the case, then

$UM_{\tilde{F}}=\tilde{\delta}(U)=[M_{G},U]=U(M_{G\circ\beta}-G)$

which is true if

$\tilde{F}(x)=G(x+q(1))-G(x)$

for some $G$. However, such a $G$ exists since
$\tilde{F}\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$ and hence is a finite
linear combination of nontrivial characters on ${\mathbb{Z}}/S{\mathbb{Z}}$,
as $\int_{{\mathbb{Z}}/S{\mathbb{Z}}}\tilde{F}=0$. By continuity,
$\chi(q(1))=1$ if and only if $\chi$ is trivial. If $\tilde{F}(x)=\chi(x)$
with $\chi(q(1))\neq 1$, then we can choose

$G(x)=\frac{\chi(x)}{\chi(q(1))-1}$

which is clearly in $\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$. Thus in
general, such a $G$ will also be a finite linear combination of characters and
$G\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$. ∎

###  4.3. Classification

Finally, we classify all continuous derivations $\delta:B_{S}^{\infty}\to
B_{S}^{\infty}$.

######  Theorem 4.9.

Let $\delta:B_{S}^{\infty}\to B_{S}^{\infty}$ be a continuous derivation. Then
there exists a unique constant $C$ such that

$\delta=C\delta_{\mathbb{L}}+\tilde{\delta}$

where $\tilde{\delta}$ is inner.

###### Proof.

Let $\delta_{0}$ be the $0$-th Fourier component of $\delta$. It is an
invariant derivation, so by Proposition 4.6 we have the unique decomposition:

$\delta_{0}(b)=C[\mathbb{L},b]+\tilde{\delta_{0}}(b)$

for every $b\in B_{S}^{\infty}$, where $\tilde{\delta_{0}}$ is an inner
derivation. From Proposition 4.7 we have that the Fourier components
$\delta_{n}$, $n\neq 0$ are inner derivations. It follows from Propositions
4.7 and 4.3, by extracting $\delta_{0}$, that we have:

$\delta(b)=\delta_{0}(b)+\lim_{n\to\infty}\sum_{|j|\leq n,\,j\neq
0}\delta_{j}(b)=\delta_{0}(b)+\lim_{n\to\infty}\left[\sum_{|j|\leq n,\,j\neq
0}U^{j}M_{F_{j}},b\right]$

with $F_{j}\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}})$. We now need to
establish two things: that $\\{F_{n}\\}$ is ULC and that the series
$\sum_{j}U^{j}M_{F_{j}}$ is convergent in $B_{S}^{\infty}$.

Since $\delta:B_{S}^{\infty}\to B_{S}^{\infty}$ is continuous, from Lemma 3.4,
we know it’s a continuous map between the Fréchet spaces $B_{S,\,l}^{\infty}$
for some $l$. Thus, given $b\in B_{S}^{\infty}$, we have $\delta(b)\in
B_{S}^{\infty}$ and so there exists a divisor $l$ of $S$ such that
$U^{l}\delta(b)U^{-l}=\delta(b)$. But, since $\rho_{\theta}$ is a continuous
automorphism, we have $U^{l}\delta_{n}(b)U^{-l}=\delta_{n}(b)$ for any $n$.
Using this observation for $b=U$, and the formula:

$\delta_{n}(U)=U^{n+1}M_{\beta(F_{n})-F_{n}}\,,$

we see that for some $l|S$ and we have:

$\beta^{l}(\beta(F_{n})-F_{n})=\beta(F_{n})-F_{n}$

for every $n$. This means that

$F_{n}(x+(l+1)q(1))-F_{n}(x+lq(1))=F_{n}(x+q(1))-F_{n}(x),$

and so the difference $F_{n}(x+lq(1))-F_{n}(x)$ is $\beta$-invariant. By
minimality of the map $\beta$, there is a constant $c_{n}$ such that

$F_{n}(x+lq(1))=F_{n}(x)+c_{n}.$

Iterating this equation we get:

$F_{n}(x+klq(1))=F_{n}(x)+kc_{n}\,$

for $k=0,1,2,\ldots$. In the topology of ${\mathbb{Z}}/S{\mathbb{Z}}$ we have
$klq(1)\to 0$. By taking the limit $k\to\infty$ in the above formula we see
that $c_{n}=0$ for all $n$ and hence $F_{n}$ is ULC.

Finally, to see that

$\lim_{n\to\infty}\sum_{|j|\leq n,\,j\neq 0}U^{j}M_{F_{j}}\in B_{S}^{\infty},$

we check that $\\{\|F_{n}\|\\}$ are RD. From the proof of Proposition 4.7, for
$n\neq 0$ we have

$M_{F_{n}}=\frac{1}{1-\chi(q(n))}U^{-n}\delta_{n}(M_{\chi})M_{\chi}^{-1}\in
B_{S}^{\infty}\,$

for any character $\chi$ such that $\chi(q(n))\neq 1$. Therefore by Lemma 4.2,

$\|F_{n}\|=\|M_{F_{n}}\|=\|M_{F_{n}}\|_{0}\leq\frac{1}{|1-\chi(q(n))|}\cdot\frac{C_{k}}{n^{k}}$

for some constants $C_{k}$. The point now is to choose a particular character
to estimate the denominator in the above formula.

Let $g=\textrm{gcd}(n,S)$ and write $n=g\cdot n^{\prime}$ with $n^{\prime}$
and $S$ relatively prime. Then set $\chi(q(n))=e^{2\pi inj/l}$ for some
divisor $l$ of $S$ to be chosen below, $0\leq j<l$ and $j$ relatively prime to
$l$. Then $\chi(q(n))=1$ if and only if $l$ is a divisor of $nj=gn^{\prime}j$.
Take $l=gh$ so that $gh$ is a divisor of $S$. Then $\chi(q(n))=e^{2\pi
in^{\prime}j/h}$ and $n^{\prime}$ is relatively prime to $h$. Thus, there
exist integers $p$ and $q$ so that $pn^{\prime}+qh=1$. Take $j=p\gamma$ for
some integer $\gamma$ to be determined. With those choices we obtain:

$\chi(q(n))=e^{2\pi in^{\prime}p\gamma/h}=e^{2\pi i\gamma/h}\,.$

Choose $\gamma$ as follows:

$\gamma=\left\\{\begin{aligned} &\frac{h}{2}&&\textrm{if }h\textrm{ is
even.}\\\ &\frac{h+1}{2}&&\textrm{if }h\textrm{ is odd.}\end{aligned}\right.$

Thus, $\chi(q(n))=-1$ for even $h$ and $\chi(q(n))\approx-1$ for odd large
enough $h$. Thus

$|1-\chi(q(n))|>3/2$

for this choice of $\chi$ and therefore we get:

$\|F_{n}\|\leq\frac{2C_{k}}{3n^{k}}.$

Hence, $\\{\|F_{n}\|\\}$ are RD. ∎

###  4.4. Counterexamples

It is necessary that the derivation is continuous in order to conclude that
$\tilde{\delta}$ is in fact inner. Here we present a class of examples of
derivations $\delta:B_{S}^{\infty}\to B_{S}^{\infty}$ are not inner modulo
$C\delta_{\mathbb{L}}$, and consequently are not continuous.

We will study below examples of derivations on $B_{S}^{\infty}$ which are of
the form:

$\delta_{F}(a)=[F(U),a]\,,$

where the function $F(z)$ is not smooth, so that the derivations are not
inner. For $a\in B_{S}^{\infty}$ we will use the decomposition found in
Equation 2.3:

$a=\sum_{k=0}^{l-1}f_{k}(U)M_{\chi_{l}^{k}}\,.$

Then

$\delta_{F}(a)=[F(U),a]=\sum_{k=0}^{l-1}f_{k}(U)\left[F(U)-F\left(e^{2\pi
ik/l}U\right)\right]M_{\chi_{l}^{k}}\,.$

Notice that 0-th Fourier component of derivations $\delta_{F}$ is always zero.
To show that $\delta_{F}$ is not inner, we construct an $F(z)\notin
C^{\infty}(S^{1})$ such that for any divisor $l$ of $S$ and $0\leq k\leq l-1$,
$F(z)-F(e^{2\pi ik/l}z)\in C^{\infty}(S^{1})$. First consider the case when
$S=p^{\infty}$ for some prime $p$ and write $F$ in its Fourier decomposition:

$F(z)=\sum_{n\in{\mathbb{Z}}}a_{k}z^{k}$

for some coefficients $a_{k}$ to be determined. Notice that

$F(z)-F\left(e^{2\pi
ik/l}z\right)=\sum_{n\in{\mathbb{Z}}}a_{k}\left(1-e^{\frac{2\pi
ik}{p^{n}}}\right)z^{k}\,.$

Define the coefficients $a_{k}$ as follows:

$a_{k}=\left\\{\begin{aligned} &b_{l}&&k=p^{l}\,,\,l\geq 0\\\
&0&&\textrm{else}\,.\end{aligned}\right.$

Then we have

$F(z)-F\left(e^{2\pi
ik/l}z\right)=\sum_{l=0}^{\infty}b_{l}\left(1-e^{\frac{2\pi
ip^{l}}{p^{n}}}\right)z^{p^{l}}=\sum_{l=0}^{n-1}b_{l}\left(1-e^{\frac{2\pi
ip^{l}}{p^{n}}}\right)z^{p^{l}}\,,$

which is clearly in $C^{\infty}(S^{1})$ for every $b_{l}$. For example, if we
chose $b_{l}=1$ then $F(z)\notin C^{\infty}(S^{1})$ because its Fourier
coefficients are not rapidly decaying. To extend this to general supernatural
numbers $S$, let $1<l_{1}|\,l_{2}|\cdots$ be a sequence of increasing divisors
of $S$ that converge to $S$. If $l$ is any divisor of $S$, then there exists a
$n$ large enough, such that $l|\,l_{n}$. Repeating the above steps we get

$F(z)=\sum_{n=0}^{\infty}a_{k}z^{l_{n}}\quad\textrm{and}\quad
F(z)-F\left(e^{2\pi
ik/l}z\right)=\sum_{n=1}^{\infty}a_{n}\left(1-e^{\frac{2\pi
il_{n}}{l}}\right)z^{l_{n}}\,.$

The last sum in the above equation will also be finite for the same reasons as
the case $S=p^{\infty}$, and consequently such derivations are well-defined
for any sequence $\\{a_{n}\\}$.

##  5\. $K$-Theory and $K$-Homology

In this section we study the $K$-Theory and $K$-Homology of $B_{S}$. Though
the $K$-Theory of Bunce-Deddens algebras is known, we include an explicit
calculation of it for completeness. Using the known $K$-Theory, we invoke
Rosenberg and Schochet’s Universal Coefficient Theorem [21], as well as
Pimsner and Voiculescu’s $6$-term exact sequence [17], to compute the
$K$-Homology.

###  5.1. K-Theory

For the remainder of the paper, we assume that $S$ is an infinite supernatural
number. For a $C^{*}$-algebra $A$, we denote by $[\cdot]_{0}$ and
$[\cdot]_{1}$ the class of an element in $K_{0}(A)$ and $K_{1}(A)$,
respectively. Define the group

$G_{S}:=\left\\{\frac{k}{l}\in{\mathbb{Q}}:k\in{\mathbb{Z}},l|S\right\\}.$

The proof of the following proposition is based on lemmas which are stated and
proved directly after.

######  Proposition 5.1.

$K_{0}(B_{S})=G_{S}$, and $K_{1}(B_{S})={\mathbb{Z}}$.

###### Proof.

Let $\\{l_{i}\\}$ be a sequence of finite divisors of $S$ such that
$l_{i}|l_{i+1}$, $l_{i}<l_{i+1}$, and $\lim_{i\to\infty}l_{i}=S$. For each
$i$, consider

$B_{S}\supseteq B_{S,i}:=\left\\{a\in B_{S}:U^{l_{i}}aU^{-l_{i}}=a\right\\}.$

It is easy to see that $B_{S,i}$ can be identified with the Bunce-Deddens
algebra for finite number $l_{i}$. Therefore, from [13], there is an
isomorphism of $C^{*}$-algebras:

$B_{S,i}\cong C(S^{1})\otimes M_{l_{i}}(\mathbb{C}).$

Since $l_{i}|l_{i+1}$, we have that $B_{S,i}\subseteq B_{S,i+1}$. Moreover,
$\cup_{i=1}^{\infty}B_{S,i}$ is dense in $B_{S}$, and we can realize $B_{S}$
as an inductive limit of $C^{*}$-algberas

$B_{S}=\overline{\cup_{i=1}^{\infty}B_{S,i}}.$

By the stability of $K_{0}$ and $K_{1}$ we have:

$K_{0}(B_{S,i})\cong K_{0}(C(S^{1}))=\mathbb{Z},$

and

$K_{1}(B_{S,i})\cong K_{1}(C(S^{1}))=\mathbb{Z}.$

By the continuity of $K_{0}$ and $K_{1}$, it only remains to compute the
corresponding direct limits. By Lemma 5.2, for each $j$ we have that
$K_{0}(B_{S,i})$ is generated by the class of the projection
$M_{\kappa_{l_{i},j}}$, where

$\kappa_{l,j}(x):=\left\\{\begin{aligned} &1&&\textrm{if }l|(l-j)\\\
&0&&\textrm{else}\end{aligned}\right.$ (5.1)

Let $\iota_{i}:B_{S,i}\hookrightarrow B_{S,i+1}$ denote the inclusion. By
Lemma 5.3, the induced map $K_{0}(\iota_{i}):K_{0}(B_{S,i})\to
K_{0}(B_{S,i+1})$ on generators is given by

$K_{0}(\iota_{i})\left[M_{\kappa_{l_{i},0}}\right]_{0}=\frac{l_{i+1}}{l_{i}}\left[M_{\kappa_{l_{i+1},0}}\right]_{0}.$

From this, we see $K_{0}(B_{S})$ is isomorphic to the space of equivalence
classes on the disjoint union of copies of ${\mathbb{Z}}$:

$K_{0}(B_{S})\cong\left(\bigsqcup_{i=1}^{\infty}{\mathbb{Z}}\right)/\sim$

where $x_{i}\sim x_{j}\iff\textrm{ there exist }r\geq t,k\geq s\textrm{ such
that }\frac{l_{r}}{l_{t}}x_{i}=\frac{l_{k}}{l_{s}}x_{j}$.

The key trick to compute the inductive limit above is to identify
$K_{0}(B_{S,i})$ with the subset of rational numbers
$\\{k/l_{i}:k\in{\mathbb{Z}}\\}\cong{\mathbb{Z}}$. From the fact that

$\frac{k}{l_{i}}=\frac{k^{\prime}}{l_{i+1}}\iff\frac{kl_{i+1}}{l_{i}}=k^{\prime},$

we see that the inductive limit takes the form

$G^{\prime}_{S}=\left\\{\frac{k}{l_{i}}:i=1,2,\dots\right\\}\subseteq G_{S}.$

To see the reverse inclusion $G_{S}\subseteq G_{S}^{\prime}$, fix
$\frac{k}{l}\in G_{S}$. Since $\lim_{i\to\infty}l_{i}=S$, there exists $i$
such that $l|l_{i}$. For such an $i$ we have:

$\frac{k}{l}=\frac{k(l_{i}/l)}{l_{i}}\in G_{S}^{\prime}.$

Therefore, $K_{0}(B_{S})$ is isomorphic to $G_{S}$.

To compute $K_{1}(B_{S})$, the approach is similar. In this case, however, the
inclusion map $\iota_{i}$ induces the identity map between $K_{1}(B_{S,i})$
and $K_{1}(B_{S,i+1})$. This is verified in Lemma 5.4. Hence, the direct limit
is simply $\mathbb{Z}$. This completes the proof. ∎

We now proceed by proving the lemmas used in the above theorem.

######  Lemma 5.2.

For every $j$, the class $[M_{\kappa_{l_{i},j}}]_{0}$ generates
$K_{0}(B_{S,i})$.

###### Proof.

Notice that, under the isomorphism $B_{S,i}\cong C(S^{1})\otimes
M_{l_{i}}(\mathbb{C})$ mentioned above, $M_{\kappa_{l_{i},j}}$ corresponds to
an element of the form $1\otimes E_{j,j}$, where $E_{j,j}$ is a rank $1$
projection. Recall the well known facts that $[1]_{0}$ generates
$K_{0}(C(S^{1}))$, and that the map $f\mapsto f\otimes E_{j,j}$ induces an
isomorphism between $K_{0}(C(S^{1}))$ and $K_{0}(C(S^{1})\otimes
M_{l_{i}}({\mathbb{C}}))$ (see, for instance, [20] Exercise $11.2$ and
Proposition 4.3.8). From this it is clear that $[1\otimes E_{j,j}]_{0}$
generates $K_{0}(C(S^{1})\otimes M_{l_{i}}({\mathbb{C}}))$, and therefore that
$[M_{\kappa_{l_{i},j}}]_{0}$ generates $K_{0}(B_{S,i})$. ∎

######  Lemma 5.3.

The induced map $K_{0}(\iota_{i}):K_{0}(B_{S,i})\to K_{0}(B_{S,i+1})$ is given
by

$K_{0}(\iota_{i})\left[M_{\kappa_{l_{i},0}}\right]_{0}=\frac{l_{i+1}}{l_{i}}\left[M_{\kappa_{l_{i+1}},0}\right]_{0}$

###### Proof.

Notice that $l_{i}|x$ if and only if $l_{i+1}|x$ or $l_{i+1}|(x-l_{i})$ or
$l_{i+1}|(x-2l_{i})$ or …or, finally, $l_{i+1}|\left(x-l_{i+1}-l_{i}\right)$.
In other words, we have the formula

$M_{\kappa_{l_{i},0}}=M_{\kappa_{l_{i+1},0}}+M_{\kappa_{l_{i+1},l_{i}}}+M_{\kappa_{l_{i+1},2l_{i}}}+\ldots+M_{\kappa_{l_{i+1},\left(\frac{l_{i+1}}{l_{i}}-1\right)l_{i}}}.$

Since $\left[M_{\kappa_{l_{i},0}}\right]_{0}$ generates $K_{0}(B_{S,i})$ and
for any $j$, $\left[M_{\kappa_{l_{i+1},j}}\right]_{0}$ generates
$K_{0}(B_{S,i+1})$, we have the following calculation:

$K_{0}(\iota_{i})\left[M_{\kappa_{l_{i},0}}\right]_{0}=\left[M_{\kappa_{l_{i+1},0}}+M_{\kappa_{l_{i+1},l_{i}}}+M_{\kappa_{l_{i+1},2l_{i}}}+\ldots+M_{\kappa_{l_{i+1},\left(\frac{l_{i+1}}{l_{i}}-1\right)l_{i}}}\right]_{0}$
$=\left[M_{\kappa_{l_{i+1},0}}\right]_{0}+\left[M_{\kappa_{l_{i+1},l_{i}}}\right]_{0}+\left[M_{\kappa_{l_{i+1},2l_{i}}}\right]_{0}+\ldots+\left[M_{\kappa_{l_{i+1},l_{i+1}-l_{i}}}\right]_{0}=\frac{l_{i+1}}{l_{i}}\left[M_{\kappa_{l_{i+1},0}}\right]_{0}$

where we were able to separate the class of the sum into the sum of classes
since $\\{M_{\kappa_{l_{i+1},j}}\\}$ are mutually orthogonal. ∎

######  Lemma 5.4.

The inclusion map $\iota_{i}:B_{S,i}\to B_{S,i+1}$ induces the identity map
between $K_{1}(B_{S,i})$ and $K_{1}(B_{S,i+1})$.

###### Proof.

Denote by $z\in C(S^{1})$ the map $S^{1}\ni z\mapsto z\in{\mathbb{C}}$. Recall
the well known facts that $[z]_{1}$ generates $K_{1}(C(S^{1}))$, and that the
map $f\mapsto f\otimes E_{0,0}$ induces an isomorphism between
$K_{1}(C(S^{1}))$ and $K_{1}(C(S^{1})\otimes M_{l_{i}}({\mathbb{C}}))\cong
K_{1}(M_{l_{i}}(C(S^{1})))$ (see [20] Example 11.3.4 and [22] Lemma 7.1.8).
Since both algebras are unital, it is easily verified that under the above map
$f\mapsto f\otimes E_{0,0}$ we have the following correspondence:

$[z]_{1}\mapsto\left[\begin{pmatrix}z&0\\\
0&1_{l_{i}-1}\end{pmatrix}\right]_{1}$

(see [20] exercise $8.5$). We can easily adapt Lemma 2.1.5 in [20] to see that
by homotopy the above class is the same as:

$\left[\begin{pmatrix}z^{1/l_{i}}&0&0&\cdots&0\\\ 0&z^{1/l_{i}}&0&\cdots&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
0&0&0&\cdots&z^{1/l_{i}}\end{pmatrix}\right]_{1}.$

Under the identification between $B_{S,i}$ and $C(S^{1})\otimes
M_{l_{i}}({\mathbb{C}})$, we see that $z^{1/l_{i}}\otimes 1_{l_{i}}$
corresponds to $U$. Since $\iota_{i}(U)=U$, we have that $\iota_{i}$ induces
the identity map between $K_{1}(B_{S,i})$ and $K_{1}(B_{S,i+1})$. ∎

###  5.2. K-Homology and UCT

Before computing the $K$-Homology of $B_{S}$, we need the following lemma
describing homomorphisms from $K_{0}(B_{S})$ to ${\mathbb{Z}}$.

######  Lemma 5.5.

There are no nontrivial homomorphisms from $G_{S}$ to $\mathbb{Z}$.

###### Proof.

Let $\phi:G_{S}\to{\mathbb{Z}}$ be a homomorphism. For each divisor $l$ of
$S$, consider the subgroup

$G_{l}=\left\\{\frac{k}{l}:k\in{\mathbb{Z}}\right\\}\cong{\mathbb{Z}}.$

$\phi$ therefore must restrict to a homomorphism
$\phi_{l}:G_{l}\to{\mathbb{Z}}$. Such a homomorphism is clearly determined by
its action on $\frac{1}{l}$. Denote

$\phi\left(\frac{1}{l}\right)=\phi_{l}\left(\frac{1}{l}\right):=a_{l}.$

If $l^{\prime}$ is another divisor of $S$ such that $l|l^{\prime}$, clearly
$G_{l}\subseteq G_{l^{\prime}}$. Since
$\frac{k(l^{\prime}/l)}{l^{\prime}}=\frac{k}{l}$, we have that

$\left(l^{\prime}/l\right)a_{l^{\prime}}=a_{l}.$ (5.2)

With this in mind, let $l$ be any finite divisor of $S$, and let
$\\{l_{i}\\}_{i=1}^{\infty}$ be a sequence such that $l_{i}|l_{i+1}$,
$l_{i}<l_{i+1}$, $\lim_{i}l_{i}=S$, and $l_{1}=l$. Suppose that $a_{l}\neq 0$.
Then $a_{l}\in{\mathbb{Z}}$ is some finite integer such that $a_{l}$ is
divisible by $(l_{i}/l)$ for every $i$. Since $l$ is finite and
$\lim_{i}l_{i}$ is an infinite supernatural number, this is clearly a
contradiction. Hence, $a_{l}=0$ for every finite divisor $l$. Equation 5.2
shows this holds for any divisor. ∎

We are now ready to compute the $K$-homology of $B_{S}$. We will use the
identification $KK^{i}(A,\mathbb{C}):=K^{i}(A)$. Since $B_{S}$ is an inductive
limit of $C(S^{1})\otimes M_{l_{i}}(\mathbb{C})$, it is clear that $B_{S}$
falls into the bootstrap category (see [1], page 228), and therefore the
conclusion of the Universal Coefficient Theorem of Rosenberg and Schochet
holds. Using this fact, the calculation of $K^{0}(B_{S})$ is simple.

######  Proposition 5.6.

The K-Homology groups of $B_{S}$ are:

$K^{0}(B_{S})=0\textrm{ and
}K^{1}(B_{S})=\mathbb{Z}\oplus\left(\left(\mathbb{Z}/S\mathbb{Z}\right)/\mathbb{Z}\right).$

###### Proof.

The Universal Coefficient Theorem states that we have an exact sequence:

${0}$${\operatorname{Ext}^{1}_{\mathbb{Z}}(K_{1}(B_{S}),{\mathbb{Z}})}$${K^{0}(B_{S})}$${\operatorname{Hom}(K_{0}(B_{S}),{\mathbb{Z}})}$${0,}$

We direct readers unfamiliar with computations involving
$\operatorname{Ext}_{\mathbb{Z}}^{1}$ to the appendix. Since
$\operatorname{Ext}^{1}_{\mathbb{Z}}(K_{1}(B_{S}),{\mathbb{Z}})\cong 0$, we
have an isomorphism
$K^{0}(B_{S})\cong\operatorname{Hom}(K_{0}(B_{S}),{\mathbb{Z}})$. The latter
group is isomorphic to $0$ by the preceding Lemma. This proves the first
claim. For $K^{1}(B_{S})$, again from the Universal Coefficient Theorem, we
have an exact sequence

${0}$${\operatorname{Ext}^{1}_{\mathbb{Z}}(K_{0}(B_{S}),{\mathbb{Z}})}$${K^{1}(B_{S})}$${\operatorname{Hom}(K_{1}(B_{S}),{\mathbb{Z}})}$${0,}$

which splits unnaturally [19]. From the final example in the Appendix we have:

$\operatorname{Ext}^{1}_{\mathbb{Z}}(K_{0}(B_{S}),{\mathbb{Z}})\cong\left(\mathbb{Z}/S\mathbb{Z}\right)/\mathbb{Z}.$

Hence, we obtain
$K^{1}(B_{S})\cong{\mathbb{Z}}\oplus\left({\mathbb{Z}}/S{\mathbb{Z}}\right)/{\mathbb{Z}}$,
by the splitting of the sequence. ∎

###  5.3. K-Homology and PV

One pitfall of the above calculation is that we obtain no information
regarding which subgroup of $\mathbb{Z}/S\mathbb{Z}$ was quotiented out. We
amend this through the following explicit calculation that uses different
techniques that have merit on their own. It turns out that the subgroup is the
natural dense subgroup identified in Equation 2.1.

Recall that the Pimsner-Voiculescu $6$-term exact sequence for $K$-Homology of
the crossed product $C({\mathbb{Z}}/S{\mathbb{Z}})\rtimes_{\beta}{\mathbb{Z}}$
reads:

${K^{0}(C(\mathbb{Z}/S\mathbb{Z}))}$${K^{0}(C(\mathbb{Z}/S\mathbb{Z}))}$${K^{0}(B_{S})}$${K^{1}(B_{S})}$${K^{1}(C(\mathbb{Z}/S\mathbb{Z}))}$${K^{1}(C(\mathbb{Z}/S\mathbb{Z}))}$$\scriptstyle{\partial^{1}}$$\scriptstyle{1-\beta^{*}}$$\scriptstyle{\iota^{*}}$$\scriptstyle{\iota^{*}}$$\scriptstyle{1-\beta^{*}}$$\scriptstyle{\partial^{0}}$

The original formulation can be found in [17]. The notation $\beta^{*}$ we
used above is for the map induced in K-Homology by $\beta$ from Equation 2.2.

Since $\mathbb{Z}/S\mathbb{Z}$ is totally disconnected, it follows from
Exercise 3.4 in [20] that

$K_{0}(C(\mathbb{Z}/S\mathbb{Z}))\cong
C(\mathbb{Z}/S\mathbb{Z},\mathbb{Z})=\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}}).$

The last equality above follows because ${\mathbb{Z}}$ is discrete.
Additionally, from [5] Example III.2.5, $C({\mathbb{Z}}/S{\mathbb{Z}})$ is AF.
Therefore, by Exercise 8.7 in [20], we have:

$K_{1}(C({\mathbb{Z}}/S{\mathbb{Z}}))=0.$

Since $\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}})$ is free (see
exercise 7.7.5 in [9]), using the Universal Coefficient Theorem we obtain

$K^{1}(C({\mathbb{Z}}/S{\mathbb{Z}}))=0\textrm{ and
}K^{0}(C({\mathbb{Z}}/S{\mathbb{Z}}))=\operatorname{Hom}(\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}}),{\mathbb{Z}}).$

Putting together these identifications, we can rewrite the above exact
sequence as follows:

${0}$${\operatorname{Hom}(\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}}),{\mathbb{Z}})}$${\operatorname{Hom}(\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}}),{\mathbb{Z}})}$${K^{1}(B_{S})}$${0.}$$\scriptstyle{\iota^{*}}$$\scriptstyle{1-\beta^{*}}$$\scriptstyle{\partial^{1}}$

From exactness, we conclude the following identification:

$K^{1}(B_{S})\cong\operatorname{Hom}(\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}}),{\mathbb{Z}})/(\textrm{Im}(1-\beta^{*})).$

In what follows we compute the range of $1-\beta^{*}$. To do that we consider
a more geometrically motivated identification of
$\operatorname{Hom}(\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}}),{\mathbb{Z}})$,
outlined presently.

Borrowing from [12], given a supernatural number $S$, we consider a set:

$V_{S}=\\{(l,k):l|S\textrm{ and }0\leq k<l\\}.$

We define the following subset of functions on $V_{S}$ with values in
${\mathbb{Z}}$:

$\Phi:=\left\\{\phi:V_{S}\to\mathbb{Z}:\phi(l,k)=\sum_{j=0}^{(l^{\prime}/l)-1}\phi(l^{\prime},k+jl)\textrm{
whenever }l|l^{\prime}\right\\}.$

The set $\Phi$ is an Abelian group with respect to addition of functions. We
claim that $\Phi$ is naturally isomorphic to the group
$\operatorname{Hom}(\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}}),{\mathbb{Z}})$
through the following identification.

To each $\phi\in\Phi$, we associate a homomorphism
$\psi_{\phi}\in\operatorname{Hom}(\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}}),{\mathbb{Z}})$
defined on characteristic functions $\kappa_{l,k}$, defined in Equation 5.1,
by

$\psi_{\phi}(\kappa_{l,k})=\phi(l,k).$

Since for $f\in\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}})$, there
exists $l$ such that $f(x+l)=f(x)$ for all $x$, we can write

$f(x)=\sum_{0\leq k<l}f(k)\kappa_{l,k}(x).$

Therefore, any
$\psi\in\operatorname{Hom}(\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}}),{\mathbb{Z}})$
is completely determined by the values $\psi(\kappa_{l,k})$. Due to the
equality

$\kappa_{l,k}(x)=\sum_{j=0}^{(l^{\prime}/l)-1}\kappa_{l^{\prime},k+jl}(x)$

for any $l^{\prime}$ such that $l|l^{\prime}$, it is clear that any
homomorphism $\psi$ can be associated to a unique element in $\Phi$. Hence, we
have established the following proposition.

######  Proposition 5.7.

With above notation we have a group isomorphism:

$\Phi\cong\operatorname{Hom}(\mathcal{E}({\mathbb{Z}}/S{\mathbb{Z}},{\mathbb{Z}}),{\mathbb{Z}}).$

The goal now is to compute the range of $1-\beta^{*}$ under this
identification. Our method will be to define a surjective map
$\Phi\to{\mathbb{Z}}\oplus({\mathbb{Z}}/S{\mathbb{Z}})/{\mathbb{Z}}$ whose
kernel is precisely ${\operatorname{Im\,}}(1-\beta^{*})$. To define such a
map, we need the following observation.

For $l,l^{\prime}$ such that $l|l^{\prime}|S$ and $\phi\in\Phi$, we define an
integer

$R\phi(l,l^{\prime}):=\sum_{a=1}^{(l^{\prime}/l)-1}\sum_{j=0}^{al-1}\phi(l^{\prime},j).$

By convention, a summation from bigger to smaller index value is set to be
zero, so that $R\phi(l,l)=0$. The quantity $R\phi(l,l^{\prime})$ above has the
following key property.

######  Lemma 5.8.

With $l,l^{\prime},\phi$ as above we have:
$R\phi(1,l^{\prime})-R\phi(1,l)=lR\phi(l,l^{\prime}).$

###### Proof.

We begin with the left-hand-side. Note first that in general we have the
following summation decomposition for any function $f$:

$\sum_{i=0}^{l^{\prime}-1}f(i)=\sum_{b=0}^{(l^{\prime}/l)-1}\sum_{j=0}^{l-1}f(bl+j).$

Using this observation, we compute:

$R\phi(1,l^{\prime})-R\phi(1,l)=\sum_{i=1}^{l^{\prime}-1}\left(\sum_{j=0}^{i-1}\phi(l^{\prime},j)\right)-\sum_{a=1}^{l-1}\sum_{j=0}^{a-1}\phi(l,j)=$
$\sum_{b=0}^{(l^{\prime}/l)-1}\sum_{a=0}^{l-1}\sum_{j=0}^{bl+a-1}\phi(l^{\prime},j)-\sum_{b=0}^{(l^{\prime}/l)-1}\sum_{a=0}^{l-1}\sum_{j=bl}^{bl+a-1}\phi(l^{\prime},j)=\sum_{a=0}^{l-1}\sum_{b=0}^{(l^{\prime}/l)-1}\sum_{j=bl}^{bl+a-1}\phi(l^{\prime},j)$
$=l\left(\sum_{b=0}^{(l^{\prime}/l)-1}\sum_{j=0}^{bl-1}\phi(l^{\prime},j)\right)=lR\phi(l,l^{\prime}).$

∎

As a corollary, we obtain the following congruence relation.

######  Corollary 5.9.

$R\phi(1,l)\equiv R\phi(1,l^{\prime})\ \textrm{mod}(l)$

For convenience, we recall the definition of ${\mathbb{Z}}/S{\mathbb{Z}}$ as
an inverse limit:

${\mathbb{Z}}/S{\mathbb{Z}}=\left\\{(x_{l})_{l|S}:x_{l}\in{\mathbb{Z}}/l{\mathbb{Z}}\textrm{
and }x_{l}\equiv x_{l^{\prime}}\textrm{ mod}(l)\textrm{ when
}l|l^{\prime}\right\\}.$

With this in mind, we define the following two group homomorphisms:

$\tau:\Phi\to{\mathbb{Z}}\textrm{,}\quad\tau(\phi):=\phi(1,0)$

$\rho:\Phi\to{\mathbb{Z}}/S{\mathbb{Z}}\textrm{,}\quad\rho(\phi):=(R\phi(1,l))_{l|S}.$

Due to Corollary 5.9, the latter map is well defined.

For our final result, we need to check that $\rho$ is surjective. This is
verified in the following proposition.

######  Proposition 5.10.

The homomorphism $\rho:\Phi\to{\mathbb{Z}}/S{\mathbb{Z}}$ is surjective.

###### Proof.

We begin by rewriting $R\phi(1,l)$ more conveniently as

$R\phi(1,l)=\sum_{j=0}^{l-2}(j+1)\phi(l,j).$ (5.3)

We consider a sequence of divisors $\\{l_{i}\\}_{i=1}^{\infty}$ such that
$l_{i}|l_{i+1}$, $l_{i}<l_{i+1}$, and $\lim_{i}l_{i}=S$. Then, similarly to
the usual expansion of a p-adic integer, we can represent an arbitrary element
$x$ in $\mathbb{Z}/S{\mathbb{Z}}$ by

$x=\sum_{n=1}^{\infty}a_{n}l_{n-1}$

with $l_{0}=1$ and “digits” $0\leq a_{n}\leq l_{n}/l_{n-1}$. Given
$x\in\mathbb{Z}/S{\mathbb{Z}}$ we use the above expansion to define
$\phi:V_{S}\to{\mathbb{Z}}$ in the following way:

$\phi(l_{n},0)=a_{1}-a_{2}-\ldots-a_{n}.$ $\phi(l_{n},l_{k})=a_{k+1}\qquad
k=1,2,\ldots,n-1.$ $\phi(l_{n},j)=0\qquad\textrm{ otherwise. }$

It is not difficult to check that $\phi$ indeed defines a function in $\Phi$.
Moreover, we have:

$R\phi(1,l_{n})=\sum_{j=0}^{l_{n}-2}(j+1)\phi(l_{n},j)=a_{1}-a_{2}-\ldots-
a_{n}+\sum_{i=1}^{n-1}(l_{i}+1)a_{i+1}$ $=a_{1}+a_{2}l_{1}+\ldots
a_{n}l_{n-1}\equiv\left(\sum_{j=1}^{\infty}a_{j}l_{j-1}\right)\
\textrm{mod}(l_{n}).$

This completes the proof. ∎

We are now ready to explicitly describe the K-Homology group $K^{1}(B_{S})$.
Consider the natural dense subgroup
$\mathbb{Z}\subset{\mathbb{Z}}/S{\mathbb{Z}}$, described in Section 2.
Quotienting out by this subgroup, we obtain a map
$\tilde{\rho}:\Phi\to({\mathbb{Z}}/S{\mathbb{Z}})/{\mathbb{Z}}$ defined by

$\tilde{\rho}(\phi):=\rho(\phi)+{\mathbb{Z}}.$

This, along with the map $\tau:\Phi\to{\mathbb{Z}}$ defined above, leads us to
the final result of this section.

######  Theorem 5.11.

With the above notation we have:
$\textrm{Im}(1-\beta^{*})=\textrm{Ker}(\tau\oplus\tilde{\rho})$. Consequently,
we have isomorphisms:

$K^{1}(B_{S})\cong\Phi/\textrm{ Ker
}(\tau\oplus\tilde{\rho})\cong{\mathbb{Z}}\oplus\left((\mathbb{Z}/S\mathbb{Z})/{\mathbb{Z}}\right).$

###### Proof.

It is easily verified that

$(\beta^{*}\phi)(l,k)=\phi(l,k+1)\textrm{ mod}(l)).$

Note that, in addition, it is clear that $\phi\in\textrm{ Ker
}(\tau\oplus\tilde{\rho})$ if and only if $\tau(\phi)=0$ and there exists some
number $\psi(1,0)\in{\mathbb{Z}}$ with

$\rho(\phi)+\psi(1,0)=0$

in $\mathbb{Z}/S\mathbb{Z}$. The last equation is equivalent to the following
congruences for each divisor $l|S$:

$\rho(\phi)+\psi(1,0)\equiv 0\textrm{ mod}(l).$

Letting $\phi=(1-\beta^{*})\psi$, we have that

$\tau(\phi)=\psi(1,0)-\psi(1,1\textrm{ mod}(1))=0.$

Notice additionally that we have the following formula:

$R\phi(1,l)=\left(\sum_{a=1}^{l-1}\sum_{j=0}^{a-1}\phi(l,j)\right)=\left(\sum_{a=1}^{l-1}\sum_{j=0}^{a-1}\psi(l,j)-\psi(l,j+1)\right)=$
$(l-1)(\psi(l,0)-\psi(l,1))+(l-2)(\psi(l,1)-\psi(l,2))+\dots$
$\ldots+2(\psi(l,l-3)-\psi(l,l-2)+(\psi(l-2)-\psi(l,l-1))$
$=l\psi(l,0)-\sum_{j=0}^{l-1}\psi(l,j)=l\psi(l,0)-\psi(1,0).$

This establishes that
$\textrm{Im}(1-\beta^{*})\subseteq\textrm{Ker}(\tau\oplus\tilde{\rho})$.

For the other direction, let $(\tau\oplus\tilde{\rho})\phi=0$. We define
$\psi\in\Phi$ such that $(1-\beta^{*})\psi=\phi$ in the following way. First,
let $\psi(1,0)$ be the number in ${\mathbb{Z}}/l{\mathbb{Z}}$ such that

$\rho(\phi)+\psi(1,0)\equiv 0\textrm{ mod}(l).$

For each $l$, let

$\psi(l,0)=\frac{R\phi(1,l)+\psi(1,0)}{l}.$ (5.4)

We can then inductively define $\psi(l,k)$ for all other values of $k$ by

$\psi(l,k)=\psi(l,0)-\sum_{j=0}^{k-1}\phi(l,j).$

By construction, it is clear that we have:

$\phi(l,k)=\psi(l,k)-\psi(l,(k+1)\textrm{ mod}(l)).$

It therefore only remains to check that $\psi$ represents an element in
$\Phi$. i.e., we check that

$\psi(l,k)=\sum_{b=0}^{(l^{\prime}/l)-1}\psi(l^{\prime},k+bl).$ (5.5)

Notice that the desired equality holds if and only if

$\psi(l,0)-\sum_{j=0}^{k-1}\phi(l,j)=\frac{l^{\prime}}{l}\psi(l^{\prime},0)-\sum_{b=0}^{(l^{\prime}/l)-1}\sum_{j=0}^{k+bl-1}\phi(l^{\prime},j).$

Using that $\phi(l,j)=\sum_{b=0}^{(l^{\prime}/l)-1}\phi(l^{\prime},j+bl)$ and
cancelling, this is equivalent to

$\psi(l,0)=\frac{l^{\prime}}{l}\psi(l^{\prime},0)-R\phi(l,l^{\prime}).$

Rearranging, and using Equation 5.4 we see that Equation 5.5 holds if and only
if

$lR\phi(l,l^{\prime})=R\phi(1,l^{\prime})-R\phi(1,l)$

which was proved in Lemma 5.8. Hence, we obtain:

$\textrm{Ker}(\tau\oplus\tilde{\rho})=\textrm{Im}(1-\beta^{*}).$ (5.6)

Since $\rho$ is surjective, by Lemma 5.10, and by Equation 5.3, $R\phi(1,l)$
does not depend on $\phi(1,0)$, it is clear that $\tau\oplus\tilde{\rho}$ is
surjective. Hence, we have the following short exact sequence:

${0}$${\textrm{Ker}(\tau\oplus\tilde{\rho})}$${\Phi}$${{\mathbb{Z}}\oplus\left((\mathbb{Z}/S\mathbb{Z})/{\mathbb{Z}}\right)}$${0.}$$\scriptstyle{\iota}$$\scriptstyle{\tau\oplus\tilde{\rho}}$

Pairing this information with Equation 5.6, as well as the Pimsner-Voiculescu
$6$-term exact sequence, we obtain the following sequence of isomorphisms:

${\mathbb{Z}}\oplus\left((\mathbb{Z}/S\mathbb{Z})/{\mathbb{Z}}\right)\cong\Phi/\textrm{Ker}(\tau\oplus\tilde{\rho})\cong\Phi/\textrm{Im}(1-\beta^{*})\cong
K^{1}(B_{S}).$

This completes the proof. ∎

##  6\. Appendix

In this Appendix, we give a brief description of
$\textrm{Ext}^{i}_{\mathbb{Z}}(G,\mathbb{Z})$, where $G$ is any Abelian group.
There are two equivalent definitions of
$\textrm{Ext}^{i}_{\mathbb{Z}}(G,\mathbb{Z})$ that are useful to us. The first
is in terms of free resolutions of the group $G$.

Definition: Any Abelian group $G$ admits a resolution of the form of a short
exact sequence:

${0}$${G_{1}}$${G_{2}}$${G}$${0,}$

where $G_{1}$ and $G_{2}$ are free Abelian. We can then form the corresponding
co-chain complex

${0}$${\textrm{Hom}(G_{2},\mathbb{Z})}$${\textrm{Hom}(G_{1},\mathbb{Z})}$${0.}$$\scriptstyle{g}$

$\textrm{Ext}^{i}_{\mathbb{Z}}(G,\mathbb{Z})$ is defined to be the cohomology
in the $i$-th position of this complex. Namely,

$\textrm{Ext}_{\mathbb{Z}}^{0}(G,\mathbb{Z}):=\textrm{Ker}(g)=\textrm{Hom}(G_{2},\mathbb{Z})\quad\textrm{
and
}\quad\textrm{Ext}_{\mathbb{Z}}^{1}(G,\mathbb{Z})=\textrm{Hom}(G_{1},\mathbb{Z})/\textrm{Im}(g),$

while $\textrm{Ext}_{\mathbb{Z}}^{i}(G,\mathbb{Z})=0$ for all $i\geq 2$.

We also have the following equivalent definition.

Definition: Given an injective resolution of $\mathbb{Z}$ of the form

${0}$${\mathbb{Z}}$${I_{1}}$${I_{2}}$${0,}$

where $I_{1}$ and $I_{2}$ are divisible Abelian, we can form the following co-
chain complex

${0}$${\textrm{Hom}(G,I_{1})}$${\textrm{Hom}(G,I_{2})}$${0.}$$\scriptstyle{f}$

We define $\textrm{Ext}_{\mathbb{Z}}^{i}(G,\mathbb{Z})$ to be the i-th
cohomology of the above complex, that is,

$\textrm{Ext}_{\mathbb{Z}}^{0}(G,\mathbb{Z}):=\textrm{Ker}(f)=\textrm{Hom}(G,I_{1})\quad\textrm{
and
}\quad\textrm{Ext}_{\mathbb{Z}}^{1}(G,\mathbb{Z})=\textrm{Hom}(G,I_{2})/\textrm{Im}(f).$

For convenience we now state some basic properties of
$\textrm{Ext}^{1}_{\mathbb{Z}}(\cdot,\mathbb{Z})$. Proofs of the properties
contained in the following proposition can be found in [23].

######  Proposition 6.1.

1) Ext${}^{1}_{\mathbb{Z}}(G,\mathbb{Z})=0$ whenever $G$ is free Abelian.

2) Given a short exact sequence $0\to K\to G\to H\to 0$, one obtains the
following exact sequence:

${0}$${\operatorname{Hom}(H,\mathbb{Z})}$${\operatorname{Hom}(G,\mathbb{Z})}$${\operatorname{Hom}(K,\mathbb{Z})}$${\operatorname{Ext}^{1}_{\mathbb{Z}}(H,\mathbb{Z})}$${\operatorname{Ext}^{1}_{\mathbb{Z}}(G,\mathbb{Z})}$${\operatorname{Ext}^{1}_{\mathbb{Z}}(K,\mathbb{Z})}$${0}$

3)
$\textrm{Ext}^{1}_{\mathbb{Z}}\left(\bigoplus_{i}G_{i},\mathbb{Z}\right)\cong\prod_{i}\textrm{Ext}(G_{i},\mathbb{Z})$.

Now we present some example calculations of
$\textrm{Ext}^{1}_{\mathbb{Z}}(G,{\mathbb{Z}})$ for groups $G$ relevant for
this paper.

Example:
$\textrm{Ext}^{1}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z})\cong\mathbb{Z}/n\mathbb{Z}$.

###### Proof.

We have the following free resolution:

${0}$${\mathbb{Z}}$${\mathbb{Z}}$${\mathbb{Z}/n\mathbb{Z}}$${0.}$$\scriptstyle{n}$

Dualizing and identifying
$\textrm{Hom}(\mathbb{Z},\mathbb{Z})\cong\mathbb{Z}$, we obtain

${0}$${\mathbb{Z}}$${\mathbb{Z}}$${0,}$$\scriptstyle{\phi}$$\scriptstyle{n}$

and we see that we have:

$\textrm{Ext}^{1}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z})=\textrm{Ker}(\phi)/\textrm{Im}(n)\cong{\mathbb{Z}}/n{\mathbb{Z}}.$

∎

Example:
$\textrm{Ext}^{1}_{\mathbb{Z}}(\widehat{\mathbb{Z}/S\mathbb{Z}},\mathbb{Z})\cong\mathbb{Z}/S\mathbb{Z}$,
where $\widehat{\mathbb{Z}/S\mathbb{Z}}$ is the Pontryagin dual of
$\mathbb{Z}/S\mathbb{Z}$.

###### Proof.

It is easy to see [15] that we can make the identification:

$\widehat{\mathbb{Z}/S\mathbb{Z}}\cong\\{z\in S^{1}:z^{l}=1\textrm{ for some
}l|S\textrm{ }\\}.$

In particular, this is a pure torsion group and we therefore have that
$\textrm{Hom}(\widehat{\mathbb{Z}/S\mathbb{Z}},\mathbb{Z})\cong 0$. From the
second definition of $\textrm{Ext}^{1}_{\mathbb{Z}}(\cdot,\mathbb{Z})$ with
$I_{1}={\mathbb{Q}}$, $I_{2}={\mathbb{Q}}/\mathbb{Z}$,

$\textrm{Ext}^{1}_{\mathbb{Z}}(\widehat{{\mathbb{Z}}/S{\mathbb{Z}}},\mathbb{Z})=\textrm{Hom}(\widehat{{\mathbb{Z}}/S{\mathbb{Z}}},{\mathbb{Q}}/\mathbb{Z})/\textrm{Im}(0)\cong\textrm{Hom}(\widehat{Z/S{\mathbb{Z}}},{\mathbb{Q}}/\mathbb{Z})\cong\widehat{\widehat{{\mathbb{Z}}/S{\mathbb{Z}}}}.$

Since the double Pontryagin dual is isomorphic to the original group, we
obtain the result. ∎

Note that the above calculation holds if
$\widehat{{\mathbb{Z}}/S{\mathbb{Z}}}$ is replaced by any pure torsion group.

Example:
$\textrm{Ext}_{\mathbb{Z}}^{1}(G_{S},\mathbb{Z})\cong({\mathbb{Z}}/S{\mathbb{Z}})/{\mathbb{Z}}$.
Here $G_{S}:=\left\\{k/l\in{\mathbb{Q}}:k\in{\mathbb{Z}},l|S\right\\}$ is the
group from the previous section.

###### Proof.

Notice that we have the following short exact sequence:

${0}$${\mathbb{Z}}$${G_{S}}$${\widehat{{\mathbb{Z}}/S{\mathbb{Z}}}}$${0}$$\scriptstyle{\phi}$

where $\phi(k/l)=e^{2\pi ik/l}$. Hence, by 2) of Proposition 6.1, we can
obtain the exact sequence

$0\to\textrm{Hom}(G_{S},\mathbb{Z})\to{\mathbb{Z}}\to\textrm{Ext}^{1}_{\mathbb{Z}}(\widehat{{\mathbb{Z}}/S{\mathbb{Z}}},\mathbb{Z})\to\textrm{Ext}^{1}_{\mathbb{Z}}(G_{S},\mathbb{Z})\to
0$

where we used that
$\textrm{Ext}^{1}_{\mathbb{Z}}(\mathbb{Z},{\mathbb{Z}})\cong 0$ since
${\mathbb{Z}}$ is free. We have also seen in Lemma 5.5 that
$\textrm{Hom}(G_{S},{\mathbb{Z}})\cong 0$, so the above sequence is short
exact. Therefore, by the previous example:

$\textrm{Ext}^{1}_{\mathbb{Z}}(G_{S},\mathbb{Z})\cong\textrm{Ext}^{1}_{\mathbb{Z}}(\widehat{{\mathbb{Z}}/S{\mathbb{Z}}},\mathbb{Z})/{\mathbb{Z}}\cong({\mathbb{Z}}/S{\mathbb{Z}})/{\mathbb{Z}}.$

∎

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