Crossed Product Algebras for Piece-Wise Constant Functions

Abstract

In this paper we consider algebras of functions that are constant on the sets of a partition. We describe the crossed product algebras of the mentioned algebras with $\mathbb {Z}.$ We show that the function algebra is isomorphic to the algebra of all functions on some set. We also describe the commutant of the function algebra and finish by giving an example of piece-wise constant functions on a real line.

1 Introduction

An important direction of investigation for any class of non-commutative algebras and rings, is the description of commutative subalgebras and commutative subrings. This is because such a description allows one to relate representation theory, non-commutative properties, graded structures, ideals and subalgebras, homological and other properties of non-commutative algebras to spectral theory, duality, algebraic geometry and topology naturally associated with commutative algebras. In representation theory, for example, semi-direct products or crossed products play a central role in the construction and classification of representations using the method of induced representations. When a non-commutative algebra is given, one looks for a subalgebra such that its representations can be studied and classified more easily and such that the whole algebra can be decomposed as a crossed product of this subalgebra by a suitable action.

When one has found a way to present a non-commutative algebra as a crossed product of a commutative subalgebra by some action on it, then it is important to know whether the subalgebra is maximal commutative, or if not, to find a maximal commutative subalgebra containing the given subalgebra. This maximality of a commutative subalgebra and related properties of the action are intimately related to the description and classification of representations of the non-commutative algebra.

Some work has been done in this direction [2, 4, 6] where the interplay between topological dynamics of the action on one had and the algebraic property of the commutative subalgebra in the $C^*-$crossed product algebra $C(X)\rtimes \mathbb {Z}$ being maximal commutative on the other hand are considered. In [4], an explicit description of the (unique) maximal commutative subalgebra containing a subalgebra $\mathcal {A}$ of $\mathbb {C}^X$ is given. In [3], properties of commutative subrings and ideals in non-commutative algebraic crossed products by arbitrary groups are investigated and a description of the commutant of the base coefficient subring in the crossed product ring is given. More results on commutants in crossed products and dynamical systems can be found in [1, 5] and the references therein.

In this article, we take a slightly different approach. We consider algebras of functions that are constant on the sets of a partition, describe the crossed product algebras of the mentioned algebras with $\mathbb {Z}$ and show that the function algebra is isomorphic to the algebra of all functions on some set. We also describe the commutant of the function algebra and finish by giving an example of piece-wise constant functions on a real line.

2 Definitions and a Preliminary Result

Let $\mathcal {A}$ be any commutative algebra. Using the notation in [4], we let $\psi :\mathcal {A}\rightarrow \mathcal {A}$ be any algebra automorphism on $\mathcal {A}$ and define

$$\begin{aligned} \mathcal {A}\rtimes _{\psi }\mathbb {Z}:=\{f:\mathbb Z\rightarrow \mathcal {A}: f(n)=0 \text{ except } \text{ for } \text{ a } \text{ finite } \text{ number } \text{ of } n\}. \end{aligned}$$

It can be proved that $\mathcal {A}\rtimes _{\psi }\mathbb {Z}$ is an associative $\mathbb {C}-$algebra with respect to point-wise addition, scalar multiplication and multiplication defined by twisted convolution , $*$ as follows;

$$\begin{aligned} (f* g)(n)=\sum _{k\in \mathbb {Z}}f(k).\psi ^k(g(n-k)), \end{aligned}$$

where $\psi ^k$ denotes the $k-$fold composition of $\psi$ with itself for positive k and we use the obvious definition for $k\le 0$.

Definition 1

$\mathcal {A}\rtimes _{\psi }\mathbb {Z}$ as described above is called the crossed product algebra of $\mathcal {A}$ and $\mathbb {Z}$ under $\psi .$

A useful and convenient way of working with $\mathcal {A}\rtimes _{\psi }\mathbb {Z}$, is to write elements $f,g\in \mathcal {A}\rtimes _{\psi }\mathbb {Z}$ in the form $f=\sum _{n\in \mathbb Z}f_n\delta ^n$ and $g=\sum _{n\in \mathbb Z}g_m\delta ^m$ where $f_n=f(n),~g_m=g(m)$ and

$$\begin{aligned} \delta ^n(k)={\left\{ \begin{array}{ll}1, ~~\text{ if } k=n,\\ 0, ~~\text{ if } k\ne n . \end{array}\right. } \end{aligned}$$

Then addition and scalar multiplication are canonically defined and multiplication is determined by the relation

$$\begin{aligned} (f_n\delta ^n)*(g_m\delta ^m)=f_n\psi ^n(g_m)\delta ^{n+m}, \end{aligned}$$

(1)

where $m,n\in \mathbb {Z}$ and $f_n,g_m\in \mathcal {A}.$

Definition 2

By the commutant $\mathcal {A}'$ of $\mathcal {A}$ in $\mathcal {A}\rtimes _{\psi }\mathbb {Z}$ we mean

$$\begin{aligned} \mathcal {A}':=\{f\in \mathcal {A}\rtimes _{\psi }\mathbb {Z}:fg=gf \text{ for } \text{ every } g\in \mathcal {A}\}. \end{aligned}$$

It has been proven [4] that the commutant $\mathcal {A}'$ is commutative and thus, is the unique maximal commutative subalgebra containing $\mathcal {A}$. For any $f,g\in \mathcal {A}\rtimes _{\psi }\mathbb {Z},$ that is, $f=\sum _{n\in \mathbb Z}f_n\delta ^n$ and $g=\sum _{m\in \mathbb Z}g_m\delta ^m,$ then $fg=gf$ if and only if

$$\begin{aligned} \forall r:~\sum _{n\in \mathbb Z}f_n\phi ^n(g_{r-m})=\sum _{m\in \mathbb Z}g_m\phi ^m(f_{r-m}). \end{aligned}$$

Now let X be any set and $\mathcal {A}$ an algebra of complex valued functions on X. Let $\sigma :X\rightarrow X$ be any bijection such that $\mathcal {A}$ is invariant under $\sigma$ and $\sigma ^{-1},$ that is for every $h\in \mathcal {A},\ h\circ \sigma \in \mathcal {A}$ and $h\circ \sigma ^{-1}\in \mathcal {A}.$ Then $(X, \sigma )$ is a discrete dynamical system and $\sigma$ induces an automorphism $\tilde{\sigma }:\mathcal {A}\rightarrow \mathcal {A}$ defined by,

$$\begin{aligned} \tilde{\sigma }(f)=f\circ \sigma ^{-1}. \end{aligned}$$

Our goal is to describe the commutant of $\mathcal {A}$ in the crossed product algebra $\mathcal {A}\rtimes _{\tilde{\sigma }}\mathbb Z$ for the case where $\mathcal {A}$ is the algebra of functions that are constant on the sets of a partition. First we have the following results.

Definition 3

For any nonzero $n\in \mathbb Z,$ we set

$$\begin{aligned} Sep_{\mathcal {A}}^n(X)=\{x\in X~|~\exists h\in \mathcal {A}~:~h(x)\ne \tilde{\sigma }^n(h)(x)\}. \end{aligned}$$

(2)

The following theorem has been proven in [4].

Theorem 1

The unique maximal commutative subalgebra of $\mathcal {A}\rtimes _{\tilde{\sigma }}\mathbb Z$ that contains $\mathcal {A}$ is precisely the set of elements

$$\begin{aligned} {A}'=\left\{ \sum _{n\in \mathbb Z}f_n\delta ^n~|~\text{ for } \text{ all } n\in \mathbb Z:~f_n|_{Sep_{\mathcal {A}}^n(X)}\equiv 0\right\} . \end{aligned}$$

(3)

We observe that since $\tilde{\sigma }(f)=f\circ \sigma ^{-1},$ then

$$\begin{aligned} \tilde{\sigma }^2(f)=\tilde{\sigma }(f\circ \sigma ^{-1})=(f\circ \sigma ^{-1})\circ \sigma ^{-1}=f\circ \sigma ^{-2}, \end{aligned}$$

and hence for every $n\in \mathbb {Z}$, $\tilde{\sigma }^n(f)=f\circ \sigma ^{-n}.$ Therefore, by taking $X=\mathbb R$ and $\mathcal {A}$ as the algebra of constant functions on X we have: for every $x\in X$ and every $h\in \mathcal {A}$,

$$\begin{aligned} \tilde{\sigma }^n(h)(x):=h\circ \sigma ^{-n}(x)=h(\sigma ^{-n}(x))=h(x), \end{aligned}$$

since h is a constant function. It follows that in this case $Sep_{\mathcal {A}}^n(X)=\emptyset .$ Therefore in this case, $\mathcal {A}'=\mathcal {A}\rtimes _{\tilde{\sigma }}\mathbb Z.$

3 Algebra of Piece-Wise Constant Functions

Let X be any set, J a countable set and $\mathbb {P}=\{X_j~:~j\in J\}$ be a partition of X;  that is $X=\cup _{r\in J}X_r$ where $X_r \ne \emptyset$ and $X_r\cap X_{r^{\prime }}=\emptyset$ if $r\ne r^{\prime }.$

Let $\mathcal {A}$ be the algebra of piece-wise constant complex-valued functions on X. That is

$$\begin{aligned} \mathcal {A}=\{h\in \mathbb {C}^X: \text{ for } \text{ every } j\in J:~h(X_j)=\{c_j\}\}. \end{aligned}$$

Let $\sigma :X\rightarrow X$ be a bijection on X. The lemma below gives the necessary and sufficient conditions for $(X,\sigma )$ to be a dynamical system.

Lemma 1

The following are equivalent.

• 1. The algebra $\mathcal {A}$ is invariant under $\sigma$ and $\sigma ^{-1}$.

• 2. For every $i\in J$ there exists $j\in J$ such that $\sigma (X_i)=X_j.$

Proof

We recall that the algebra $\mathcal {A}$ is invariant under $\sigma$ if and only if for every $h\in \mathcal {A},~h\circ \sigma \in \mathcal {A}.$

Obviously, if for every $i\in J$ there exists a unique $j\in J$ such that $\sigma (X_i)=X_j,$ then

$$\begin{aligned} (h\circ \sigma )(X_i)=h(\sigma (X_i))=h(X_j)=\{c_j\}. \end{aligned}$$

Thus $h\circ \sigma \in \mathcal {A}$.

Conversely, suppose $\mathcal {A}$ is invariant under $\sigma$ but 2. does not hold. Let $x_1,x_2\in X_j$ and $X_r, X_{r^{\prime }}\in \mathbb {P}$ such that $\sigma (x_1)\in X_r$ and $\sigma (x_2)\in X_{r^{\prime }}$. Let $h:X\rightarrow \mathbb {C}$ be the function defined by

$$\begin{aligned} h(x)={\left\{ \begin{array}{ll} 1 ~~ \text{ if } x\in X_r,\\ 0~~ \text{ otherwise. } \end{array}\right. } \end{aligned}$$

Then $h\in \mathcal {A} .$ But $h\circ \sigma (x_1)=1$ and $h\circ \sigma (x_2)=0.$ Thus $h\notin \mathcal {A}$, which contradicts the assumption.    $\square$

The following lemma asserts that any bijection $\sigma _2:X\rightarrow X$ that preserves the structure of a partition essentially produces the same algebra of functions.

Lemma 2

Let $\mathbb {P}_1=\{X_j~:~j\in J\}$ and $\mathbb {P}_2=\{Y_j~:~j\in J\}$ be partitions of the sets X and Y respectively, and let

$$\begin{aligned} \mathcal {A}_X=\{h\in \mathbb {C}^X: \text{ for } \text{ every } j\in J:~h(X_j)=\{c_j\}\}, \end{aligned}$$

and

$$\begin{aligned} \mathcal {A}_Y=\{h\in \mathbb {C}^Y: \text{ for } \text{ every } j\in J:~h(Y_j)=\{d_j\}\}. \end{aligned}$$

Then $\mathcal {A}_X$ is isomorphic to $\mathcal {A}_Y.$

Proof

Choose points $x_i\in X$ and $y_i\in Y$ such that $x_i\in X_i$ if and only if $y_i\in Y_i ~ \forall ~i\in J$ and let $\mu :\mathcal {A}_X\rightarrow \mathcal {A}_Y$ be a function defined by

$$\begin{aligned} \mu (f)(y)=f(x_i) \text{ if } y\in Y_j,~\forall ~j\in J. \end{aligned}$$

(4)

It is enough to prove that $\mu$ is an algebra isomorphism.

• Let $f,g\in \mathcal {A}_X$ and let $\alpha ,\beta \in \mathbb {C}.$ Then if $y\in Y,$ then $y\in Y_i$ for some $i\in J,$ therefore,

  $$\begin{aligned} \mu (\alpha f+\beta g)(y)=(\alpha f+\beta g)(x_i)\\ =\alpha f(x_i)+\beta g(x_i)\\ = \alpha \mu (f)(y)+\beta \mu (g)(y)\\ =[\alpha \mu (f)+\beta \mu (g)](y). \end{aligned}$$

  Therefore $\mu$ is linear since y was arbitrary.

• For every $f,g\in \mathcal {A}_X$ and $y\in Y~(y\in Y_i),$

  $$\begin{aligned} \mu (fg)(y) = (fg)(x_i)\\ = f(x_i)g(x_i)\\ =\mu (f)(y)\mu (g)(y)\\ = [\mu (f)\mu (g)](y). \end{aligned}$$

  Thus $\mu$ is a multiplicative homomorphism.

• Now, suppose $f,g\in \mathcal {A}_X$ such that $f\ne g.$ Then there exists $i\in J$ such that $f(x_i)\ne g(x_i),~x_i\in X_i.$ Therefore, if $y\in Y_i,$

  $$\begin{aligned} \mu (f)(y)=f(x_i)\ne g(x_i)=\mu (g)y. \end{aligned}$$

  Therefore $\mu$ is injective.

• Finally, suppose $h\in \mathcal {A}_Y$ and let $f\in \mathcal {A}_X$ be defined by $f(x)=h(y_i)$. If $y\in Y,$ then $y\in Y_i$ for some $i\in J,$ and hence,

  $$\begin{aligned} h(y)=h(y_i)=f(x)=f(x_i)=\mu (f)(y). \end{aligned}$$

  It follows that $\mu$ is onto and hence an algebra isomorphism.

   $\square$

Theorem 2

Let $\mathbb {P}_1=\{X_j :j\in J\}$ and $\mathbb {P}_2=\{Y_j :j\in J\}$ be partitions of two sets X and Y and $\mathcal {A}_X$ and $\mathcal {A}_Y$ be algebras of functions that are constant on the sets of the partitions $\mathbb {P}_1$ and $\mathbb {P}_2$ respectively. Let $\sigma _1 :X \rightarrow X$ and $\sigma _2 :Y \rightarrow Y$ be bijections such that $\mathcal {A}_X$ is invariant under $\sigma _1~(and~\sigma _1^{-1})$ and $\mathcal {A}_Y$ is invariant under $\sigma _2~(and~\sigma _2^{-1})$ and that $\sigma _1(X_i)=X_j$ whenever $\sigma _2(Y_i)=Y_j$ for all $i, j \in J$. Suppose $\tilde{\sigma _1}:\mathcal {A}_X \rightarrow \mathcal {A}_X$ is the automorphism on $\mathcal {A}_X$ induced by $\sigma _1$, and $\tilde{\sigma _2}:\mathcal {A}_Y \rightarrow \mathcal {A}_Y$ is the automorphism on $\mathcal {A}_Y$ induced by $\sigma _2$. Then

$$\begin{aligned} \tilde{\sigma _2}\circ \mu =\mu \circ \tilde{\sigma _1}. \end{aligned}$$

(5)

where $\mu$ is given by (4). Moreover, for every $n\in \mathbb {Z},$

$$\begin{aligned} \tilde{\sigma _2}^n\circ \mu =\mu \circ \tilde{\sigma _1}^n. \end{aligned}$$

(6)

Proof

Let $y\in X$ such that $y\in Y_i$ for some $i\in J.$ Then for every $f\in \mathcal {A},$

$$\begin{aligned} (\tilde{\sigma _2}\circ \mu )(f)(y)=(\mu f)\circ \sigma _2^{-1}(y)\\ =(\mu f)(\sigma _{2}^{-1}(y))\\ =f(\sigma _1^{-1}(x_i))\\ =(f\circ \sigma _1^{-1})(x_i)\\ = \mu (f\circ \sigma _1^{-1})(y)\\ = \mu [\tilde{\sigma _1}(f)](y)\\ = [\mu \circ \tilde{\sigma _1}](f)(y). \end{aligned}$$

Since y is arbitrary, we have

$$\begin{aligned} (\tilde{\sigma _2}\circ )(f) \mu =\mu \circ \tilde{\sigma _1}(f) \end{aligned}$$

for every $f\in \mathcal {A}.$ And since f is arbitrary,

$$\begin{aligned} \tilde{\sigma _2}\circ \mu =\mu \circ \tilde{\sigma _1}. \end{aligned}$$

Now from (5), we have

$$\begin{aligned} \tilde{\sigma _2}^2\circ \mu =\tilde{\sigma _2}\circ (\tilde{\sigma _2}\circ \mu )=\tilde{\sigma _2}\circ (\mu \circ \tilde{\sigma _1})=(\tilde{\sigma _2}\circ \mu )\circ \tilde{\sigma _1}=(\mu \circ \tilde{\sigma _1})\circ \tilde{\sigma _1}=\mu \circ \tilde{\sigma _1}^2. \end{aligned}$$

Therefore the relation (6) holds for $n=2.$

Now suppose the relation (6) holds for k. Then:

$$\begin{aligned} \tilde{\sigma _2}^{k+1}\circ \mu =\tilde{\sigma _2}\circ (\tilde{\sigma _2}^k\circ \mu )=\tilde{\sigma _2}\circ (\mu \circ \tilde{\sigma _1}^k)=(\tilde{\sigma _2}\circ \mu )\circ \tilde{\sigma _1}^k=(\mu \circ \tilde{\sigma _1})\circ \tilde{\sigma _1}^k=\mu \circ \tilde{\sigma _1}^{k+1}. \end{aligned}$$

Therefore, from the induction principle,

$$\begin{aligned} \tilde{\sigma _2}^n\circ \mu =\mu \circ \tilde{\sigma _1}^n. \end{aligned}$$

   $\square$

Remark 1

From Theorem 2 above, we get two nice results. The first is that if $\mathbb {P}_1=\mathbb {P}_2$ are partitions of X and $\sigma _1,\sigma _2:X\rightarrow X$ are bijections on X which preserve the structure of the partition, they will give rise to the same automorphism. That is, suppose $\mathbb {P}_1=\{X_j~:~j\in J\}$ is a partition of X and $\sigma _1,\sigma _2:X\rightarrow X$ are bijections on X such that, if $\sigma _1 (X_i)=X_j,$ then $\sigma _2 (X_i)=X_j,$ for all $i,j\in J.$ Let $\tilde{\sigma }:\mathcal {A}\rightarrow \mathcal {A}$ be the automorphism on induced by $\sigma ,$ that is, for every $h\in \mathcal {A},$

$$\begin{aligned} \tilde{\sigma }(h)=h\circ \sigma ^{-1}. \end{aligned}$$

Then for every $f\in \mathcal {A},$

$$\begin{aligned} \tilde{\sigma _1}(f)=\tilde{\sigma _2}(f). \end{aligned}$$

This is given by the fact that if $\mathbb {P}_1=\mathbb {P}_2$, then in (5), we can take $\mu =id.$

The second is the following important theorem.

Theorem 3

Let $\mathbb {P}_1=\{X_j :j\in J\}$ and $\mathbb {P}_2=\{Y_j :j\in J\}$ be partitions of two sets X and Y and $\mathcal {A}_X$ and $\mathcal {A}_Y$ be algebras of functions that are constant on the sets of the partitions $\mathbb {P}_1$ and $\mathbb {P}_2$ respectively. Let $\sigma _1 :X \rightarrow X$ and $\sigma _2 :Y \rightarrow Y$ be bijections such that $\mathcal {A}_X$ is invariant under $\sigma _1~(and~\sigma _1^{-1})$ and $\mathcal {A}_Y$ is invariant under $\sigma _2~(and~\sigma _2^{-1})$ and that $\sigma _1(X_i)=X_j$ whenever $\sigma _2(Y_i)=Y_j$ for all $i, j \in J$. Suppose $\tilde{\sigma _1}:\mathcal {A}_X \rightarrow \mathcal {A}_X$ is the automorphism on $\mathcal {A}_X$ induced by $\sigma _1$, and $\tilde{\sigma _2}:\mathcal {A}_Y \rightarrow \mathcal {A}_Y$ is the automorphism on $\mathcal {A}_Y$ induced by $\sigma _2$. Then the crossed product algebras $\mathcal {A}\rtimes _{ \tilde{\sigma _1}}\mathbb {Z}$ and $\mathcal {A}\rtimes _{ \tilde{\sigma _2}}\mathbb {Z}$ are isomorphic.

Proof

We need to construct the an isomorphism between the crossed product algebras $\mathcal {A}_X\rtimes _{ \tilde{\sigma _1}}\mathbb {Z}$ and $\mathcal {A}_Y\rtimes _{ \tilde{\sigma _1}}\mathbb {Z}.$ Using the notation in [4], we let $f:=\sum _{n\in \mathbb {Z}}f_n \delta ^n$ be an element in $\mathcal {A}_X\rtimes _{\tilde{\sigma _1}}\mathbb {Z}.$ Define a function $\mu :\mathcal {A}_X\rtimes _{ \tilde{\sigma _1}}\mathbb {Z} \rightarrow \mathcal {A}_Y\rtimes _{ \tilde{\sigma _2}}\mathbb {Z}$ be defined by

$$\begin{aligned} \tilde{\mu }\left( \sum _{n\in \mathbb {Z}}f_n \delta _1^n\right) =\sum _{n\in \mathbb {Z}}\mu (f_n) \delta _2^n, \end{aligned}$$

(7)

where $\mu$ is defined in (4). Then, since $\mu$ is an algebra isomorphism, it is enough to prove that $\tilde{\mu }$ is multiplicative. To this end, we let $f:=\sum _{n\in \mathbb {Z}}f_n \delta _1^n$ and $g:=\sum _{m\in \mathbb {Z}}g_m \delta _1^m$ be arbitrary elements in $\mathcal {A}_X\rtimes _{\tilde{\sigma _1}}\mathbb {Z},$ then we prove that $\tilde{\mu }$ is multiplicative on the generators $f_n\delta _1^n$ and $g_m\delta _1^m$ respectively. Using (1) we have

$$\begin{aligned} \tilde{\mu }((f_n\delta _1^n)*(g_m\delta _1^m))=\tilde{\mu }(f_n\tilde{\sigma _1}^n(g_m)\delta _1^{n+m})\\ =\mu (f_n\tilde{\sigma _1}^n(g_m))\delta _2^{n+m}\\ =[\mu (f_n)\mu (\tilde{\sigma _1}^n(f_m))]\delta _2^{n+m}\\ =\mu (f_n)\tilde{\sigma _2}^n(\mu (f_m))\delta _2^{n+m} ~~~~ \text{ by } \text{(6) }\\ =\tilde{\mu }(f_n\delta _2^n) *\tilde{\mu }(f_m\delta _2^m). \end{aligned}$$

Therefore $\tilde{\mu }$ is multiplicative on the generators $f_n\delta ^n$ and since $\mu$ is linear, it is multiplicative on the elements $f=\sum _{n\in \mathbb {Z}}f_n\delta ^n \in \mathcal {A}_X\rtimes _{\tilde{\sigma }}\mathbb {Z}.$    $\square$

Remark 2

In Lemma 1 we proved the necessary and sufficient condition on a bijection $\sigma :X\rightarrow X$ such that the algebra $\mathcal {A}_X$ is invariant under $\sigma ,$ that is, for every $i\in J$ there exists $j\in J$ such that $\sigma (X_i)=X_j$ where the $X_i$ form a partition for X. From this, it can be shown that $\mathcal {A}$ is isomorphic to $\mathbb {C}^J,$ where by $\mathbb {C}^J$ we denote the space of complex sequences indexed by J. This can be done by constructing an isomorphism between $\mathcal {A}_X$ and $\mathbb {C}^J$ via $\sigma$ as follows.

Let $\tau : J\rightarrow J$ be a map such that $\tau (i)=j$ is equivalent to $\sigma (X_i)=X_j$ for all $i,j\in J.$ Then $\tau$ is a bijection that plays the same role as $\sigma _2$ in Lemma 2. Therefore, using the same Lemma, we deduce that the algebra $\mathcal {A}$ is isomorphic to $\mathbb {C}^J$. In Theorem 3, we have shown a method of constructing an isomorphism between the crossed product algebras $\mathcal {A}_X\rtimes _{ \tilde{\sigma _1}}\mathbb {Z}$ and $\mathcal {A}_Y\rtimes _{ \tilde{\sigma _2}}\mathbb {Z}$, when $\mathcal {A}_X$ and $\mathcal {A}_Y$ are isomorphic. It follows that the crossed product algebra $\mathcal {A}_X\rtimes _{ \tilde{\sigma _1}}\mathbb {Z}$ is isomorphic to $\mathbb {C}^J\rtimes _{ \tilde{\tau }}\mathbb {Z}$, where $\tilde{\tau }$ follows the same definition as $\tilde{\sigma }$.

In the next section we describe the commutant of our algebra $\mathcal {A}_X$ in the crossed product algebra $\mathcal {A}_X\rtimes _{ \tilde{\sigma }}\mathbb {Z}.$

3.1 Maximal Commutative Subalgebra

We take the same partition $\mathbb {P}=\cup _{j\in J}X_j$ and a bijection $\sigma :X\rightarrow X$ such that for all $i\in J,$ there exists $j\in J$ such that $\sigma (X_i)=X_j.$ For $k\in \mathbb {Z}_{> 0},$ let

$$\begin{aligned} C_k:=\left\{ x \in X ~|~k \text{ is } \text{ the } \text{ smallest } \text{ positive } \text{ integer } \text{ such } \text{ that } x, \sigma ^k(x) \in X_j \right. \\ \nonumber \left. \text { for some } j \in J \right\} . \end{aligned}$$

(8)

According to Theorem 1, the unique maximal commutative subalgebra of $\mathcal {A}\rtimes _{\tilde{\sigma }}\mathbb Z$ that contains $\mathcal {A}$ is precisely the set of elements

$$\begin{aligned} {A}'=\left\{ \sum _{n\in \mathbb Z}f_n\delta ^n~|~\text{ for } \text{ all } n\in \mathbb Z:~f_n|_{Sep_{\mathcal {A}}^n(X)}\equiv 0\right\} , \end{aligned}$$

(9)

where $Sep_{\mathcal {A}}^n(X)$ is given by (2). We have the following theorem which gives the description of $Sep_{\mathcal {A}}^n(X)$ in this case and is crucial in the description if the maximal commutative subalgebra.

Theorem 4

Let $\sigma :X \rightarrow X$ be a bijection on X as given above, $\tilde{\sigma }:\mathcal {A}_X\rightarrow \mathcal {A}_X$ be the automorphism on $\mathcal {A}_X$ induced by $\sigma$ and $C_k$ be given by (8). Then for every $n\in \mathbb Z,$

$$\begin{aligned} {Sep_{\mathcal {A}}^n}(X)=\left\{ \bigcup _{k\not \mid n}C_k \cup C_{\infty } \right\} , \end{aligned}$$

(10)

where

$$\begin{aligned} C_{\infty }=\{X_j\in \mathbb {P}~:~ \sigma ^k(X_j)\ne X_j ~\forall k~\ge 1\}. \end{aligned}$$

Proof

• 1. If $n\equiv 0\pmod {k}$ and $x\in X_j\in C_k$, the we can write $n=mk$ for some $m\in \mathbb Z.$ Then, since $\sigma ^k(X_j)=X_j$ it follows that $\sigma ^{-k}(X_j)=X_j$ and therefore for every $h\in \mathcal {A},$

  $$\begin{aligned} \tilde{\sigma }^n(h)(x)=\tilde{\sigma }^{mk}(h)(x)=(h\circ \sigma ^{-mk})(x)=h(\sigma ^{-mk}(x))=h(x), \end{aligned}$$

  since x and $\sigma ^{-mk}(x)\in X_j$ for all $m\in \mathbb Z.$

• 2. If $n\not \equiv 0\pmod {k},$ we can write $n=mk+j$ where $m,j\in \mathbb Z$ with $1\le j< k.$ It follows that for every $x\in X_j\in C_k,$

  $$\begin{aligned} \tilde{\sigma }^n(h)(x)=\tilde{\sigma }^{mk+j}(h)(x)\\ =(h\circ \sigma ^{-mk+j})(x)\\ =h(\sigma ^{-mk+j}(x))\\ =\tilde{\sigma }^j(h)(x). \end{aligned}$$

  But k is the smallest integer such that $\sigma ^k(X_j)=X_j.$ Therefore since $j<k$,

  $$\begin{aligned} \tilde{\sigma }^j(h)(x)\ne h(x). \end{aligned}$$

  Hence

  $$\begin{aligned} Sep_{\mathcal {A}}^n(X)= \{x\in X~|~\exists ~h\in \mathcal {A}:h(x)\ne \tilde{\sigma }^n(h)(x)\}\\ ={\left\{ \begin{array}{ll} \left\{ \cup _{j~:~X_j\notin C_k}X_j\right\} ~~~ \text{ if } n\equiv 0\pmod {k},\\ \left\{ \cup _{j~:~X_j\in C_k}X_j\right\} ~~~ \text{ if } n \not \equiv 0\pmod {k}, \end{array}\right. } \end{aligned}$$

  and if $x\in C_{\infty },$ then obviously $x\in Sep_{\mathcal {A}}^n$ for every $n\ge 1$, or simply

  $$\begin{aligned} Sep_{\mathcal {A}}^n(\mathbb R)=\bigcup _{k\not \mid n}C_k\cup C_{\infty }. \end{aligned}$$

From the above theorem, the description of the maximal commutative subalgebra in $\mathcal {A}\rtimes _{\tilde{\sigma }}\mathbb {Z}$ can be done as follows.

Theorem 5

Let $\mathcal {A}_X$ be the algebra of piece-wise constant functions $f:X\rightarrow \mathbb {C}$, $\sigma :X\rightarrow X$ any bijection on X, $\tilde{\sigma }:\mathcal {A}_X\rightarrow \mathcal {A}_X$ the automorphism on $\mathcal {A}_X$ induced by $\sigma$ and $C_k$ be as described above. Then the unique maximal commutative subalgebra of $\mathcal {A}_X\rtimes _{\tilde{\sigma }}\mathbb Z$ that contains $\mathcal {A}_X$ is given by

$$\begin{aligned} \mathcal {A}'=\left\{ \sum _{n\in \mathbb Z~:~k\mid n}\left( \sum _{j_n\in J}a_{j_n}\chi _{X_{j_n}}\right) \delta ^n \right\} .\end{aligned}$$

Proof

From (9) we have that the unique maximal commutative subalgebra of $\mathcal {A}_X\rtimes _{\tilde{\sigma }}\mathbb Z$ that contains $\mathcal {A}_X$ is precisely the set of elements

$$\begin{aligned} \mathcal {A}'=\left\{ \sum _{n\in \mathbb Z}f_n\delta ^n~|~\text{ for } \text{ all } n\in \mathbb Z:~f_n|_{Sep_{\mathcal {A}}^n(X)}\equiv 0\right\} , \end{aligned}$$

and from (2),

$$\begin{aligned} Sep_{\mathcal {A}}^n(\mathbb R)=\bigcup _{k\not \mid n}C_k. \end{aligned}$$

(11)

Combining the two results and using the definition of $h_n\in \mathcal {A}_X$ as

$$\begin{aligned} h_n=\sum _{j_n\in J}a_{j_n}\chi _{X_{j_n}}, \end{aligned}$$

we get

$$\begin{aligned} \mathcal {A}'=\left\{ \sum _{n\in \mathbb Z~:~k\mid n}\left( \sum _{j_n\in J}a_{j_n}\chi _{X_{j_n}}\right) \delta ^n \right\} .\end{aligned}$$

   $\square$

It can be observed from the results in Theorem 4 that it is possible to have $Sep_{\mathcal {A}}^n(X)=X$ for all $n\in \mathbb {Z}.$ For example, suppose J is infinite and let $\sigma :X \rightarrow X$ be a bijection such that $\sigma (X_j)=X_{j+1}$ for every $j\in J.$ Then it is easily seen that in in this case $Sep_{\mathcal {A}}^n(X)=X.$ However, this is not possible if J is finite since in this case $\sigma$ acts like a permutation on a finite group. In the following section, we treat one such a case. We let $X=\mathbb {R}$ and $\mathcal {A}_X$ be the algebra of piece-wise constant functions on $\mathbb {R}$ with N fixed jump points, where $N\ge 1$ is an integer. In order to work in the setting described before, we treat jump points as intervals of zero length. Then $\mathbb {R}$ is partitioned into $2N+1$ sub-intervals.

4 Algebra of Piece-Wise Constant Functions on the Real Line with N Fixed Jump Points

Let $\mathcal {A}$ be the algebra of piece-wise constant functions $f:\mathbb R\rightarrow \mathbb R$ with Nfixed jumps at points $t_1,t_2,\ldots ,t_N.$ Partition $\mathbb R$ into $N+1$ intervals $I_0,I_1,\ldots ,I_N$ where $I_{\alpha }=]t_{\alpha },t_{\alpha +1}[$ with $t_0=-\infty$ and $t_{N+1}=\infty .$ By looking at jump points as intervals of zero length, we can write $\mathbb {R}=\cup I_{\alpha }$ where $I_{\alpha }$ is as described above for $\alpha =0,1,\ldots N$ and $I_{\alpha }=\{t_{\alpha } \}$ if $\alpha >N.$ Then for every $h\in \mathcal {A}$ we have

$$\begin{aligned} h(x)=\sum _{\alpha =0}^{2N}a_{\alpha }\chi _{I_{\alpha }}(x), \end{aligned}$$

(12)

where $\chi _{I_{\alpha }}$ is the characteristic function of $I_{\alpha }.$ As in the preceding section, we let $\sigma :\mathbb R \rightarrow \mathbb R$ be any bijection on $\mathbb R$ and let $\tilde{\sigma }:\mathcal {A}\rightarrow \mathcal {A}$ be the automorphism on $\mathcal {A}$ induced by $\sigma$. Then we have the following lemma which gives the necessary and sufficient conditions for $(\mathbb R, \sigma )$ to be a discrete dynamical system.

Lemma 3

The algebra $\mathcal {A}$ is invariant under both $\sigma$ and $\sigma ^{-1}$ if and only if the following conditions hold.

• 1. $\sigma$ (and $\sigma ^{-1}$) maps the each jump point $t_k,~k=1,\ldots ,N$ onto another jump point.

• 2. $\sigma$ maps every interval $I_{\alpha },~\alpha =0,1,\ldots N$ bijectively onto any of the other intervals $I_0,I_1 \ldots I_N.$

Proof

Obviously, if the two conditions hold, then $\mathcal {A}$ is invariant under $\sigma$. So we suppose that $\mathcal {A}$ is invariant under $\sigma$ and prove that the two conditions must hold.

• 1. Suppose $\sigma (t_k)=t_0\notin \{t_1,t_2,\ldots ,t_N\}$ for some $k\in \{1,2,\ldots ,N\}.$ Then, since $\sigma$ is onto, there exists $x_0\in \mathbb R$ such that $\sigma (x_0)=t_k,$ that is, there exists a non jump point that is mapped onto a jump point. We show that this is not possible.

  Let

  $$\begin{aligned} h(x)={\left\{ \begin{array}{ll}1~~ \text{ if } ~~x=t_k,\\ 0~~ \text{ otherwise. } \end{array}\right. } \end{aligned}$$

  Then $h\in \mathcal {A}.$ But

  $$\begin{aligned} h\circ \sigma (x)={\left\{ \begin{array}{ll}1~~ \text{ if } ~~\sigma (x)=t_k,\\ 0~~ \text{ otherwise, } \end{array}\right. }={\left\{ \begin{array}{ll}1~~ \text{ if } ~~x=x_0,\\ 0~~ \text{ otherwise. } \end{array}\right. } \end{aligned}$$

  Therefore $h\circ \sigma \notin \mathcal {A}$ which is a contradiction, implying that $\sigma$ does not map a non jump point onto a jump point, proving the first condition.

• 2. Consider the bijection $\sigma :\mathbb R\rightarrow \mathbb R$ defined by

  $$\begin{aligned} \sigma (x)={\left\{ \begin{array}{ll}x~~ \text{ if } ~~x\ne t_k^{'} \text{ or } t_k^{''},\\ t_k^{''}~~ \text{ if } x=t_k^{'},\\ t_k^{'}~~ \text{ if } x=t_k^{''}, \end{array}\right. } \end{aligned}$$

  (13)

  where $t_k^{'}\in I_k$ and $t_k^{''}\in I_{k+1}$ for some $k\in \{1,2,\ldots , N\}.$ Then $\sigma$ is a bijection that permutes the jump points. Let $h\in \mathcal {A}.$ Then using (12) and for the $\sigma$ in Eq. (13) above, we have:

  $$\begin{aligned} \sigma (x)={\left\{ \begin{array}{ll}h(x)~~ \text{ if } ~~x\ne t_k^{'} \text{ or } t_k^{''},\\ a_{k+1}~~ \text{ if } x=t_k^{'},\\ a_k~~~~~ \text{ if } x=t_k^{''}. \end{array}\right. } \end{aligned}$$

  Therefore, $h\circ \sigma$ has jumps at points $t_1,\ldots , t_N,t_k^{'},t_k^{''}$ implying that $h\circ \sigma \notin \mathcal {A}$.

   $\square$

The following theorem gives the description of $Sep_{\mathcal {A}}^n(\mathbb R)$ for any $n\in \mathbb Z.$

Theorem 6

Let $\mathcal {A}$ be an algebra of piece-wise constant functions with N fixed jumps at points $t_1,\ldots , t_N$, $\sigma :\mathbb R \rightarrow \mathbb R$ be any bijection on $\mathbb R$ such that $\mathcal {A}$ is invariant under $\sigma$ and let $\tilde{\sigma }:\mathcal {A}\rightarrow \mathcal {A}$ be the automorphism on $\mathcal {A}$ induced by $\sigma$. Let

$$\begin{aligned} C_k:=\left\{ x \in \mathbb {R} ~|~k \text{ is } \text{ the } \text{ smallest } \text{ positive } \text{ integer } \text{ such } \text{ that } x, \sigma ^k(x) \in I_{\alpha } \right. \\ \nonumber \left. \text { for some } \alpha = 0, \ldots 2N \right\} . \end{aligned}$$

(14)

Then for every $n\in \mathbb Z,$

$$\begin{aligned} {Sep_{\mathcal {A}}^n}(\mathbb R)=\bigcup _{k\not \mid n}C_k. \end{aligned}$$

(15)

Proof

See Theorem 4 and observe that $C_{\infty }=\emptyset$ in this case.    $\square$

Example 1

Let $\mathcal {A}$ be the algebra of piece-wise constant functions with $4-$fixed jump points at $t_1,t_2,t_3,t_4.$ Partition $\mathbb R$ into five subintervals $I_0,\ldots , I_4$ where $I_{\alpha }=]t_{\alpha },t_{\alpha +1}[$ with $t_0=-\infty$ and $t_5=\infty .$

Let $\sigma :\mathbb R\rightarrow \mathbb R$ be a bijection such that $\sigma (I_0)=I_1,~\sigma (I_1)=I_2,~\sigma (I_2)=I_0, \sigma (I_3)=I_4$ and $\sigma (I_4)=I_3.$ It follows that $\sigma ^3 (I_0)=I_0,~\sigma ^3 (I_1)=I_1$ and $\sigma ^3 (I_2)=I_2.$ But $\sigma ^j (I_{\alpha })\ne I_{\alpha }$ for $\alpha =0,1,2$ and $1\le j<3.$

Also $\sigma ^2 (I_3)=I_3,~\sigma ^2 (I_4)=I_4$ but $\sigma ^j (I_{\alpha })\ne I_{\alpha }$ if $j\not \equiv 0\pmod {2}$ with $\alpha =3,4.$ Therefore:

$$\begin{aligned} Sep_{\mathcal {A}}^n(\mathbb R)= \{x\in \mathbb R~|~\exists ~h\in \mathcal {A}:h(x)\ne \tilde{\sigma }^n(h)(x)\}\\ =\mathbb R\setminus \{\{I_3\cup I_4\}\cup \{t_k:~\sigma ^2(t_k)=t_k,~~k=1,2,3,4\}\} \text{ if } n\equiv 0\pmod {2}\\ =\{I_0\cup I_1\cup I_2\}\cup \{t_k:~\sigma ^2(t_k)\ne t_k,~~k=1,2,3,4\} \text{ if } n\equiv 0\pmod {2}, \end{aligned}$$

and

$$\begin{aligned} Sep_{\mathcal {A}}^n(\mathbb R)=\{x\in \mathbb R~|~\exists ~h\in \mathcal {A}:h(x)\ne \tilde{\sigma }^n(h)(x)\}\\ =\mathbb R\setminus \{I_0\cup \{I_1\cup I_2\}\cup \{t_k:~\sigma ^3(t_k)=t_k,~~k=1,2,3,4\}\} \text{ if } n\equiv 0\pmod {3}\\ =\{I_3\cup I_4\}\cup \{t_k:~\sigma ^3(t_k)\ne t_k,~~k=1,2,3,4\} \text{ if } n\equiv 0\pmod {3}. \end{aligned}$$

From these results we have the following theorem.

Theorem 7

Let $\mathcal {A}$ be the algebra of piece-wise constant functions $f:\mathbb R\rightarrow \mathbb R$ with N fixed jumps at points $t_1,t_2,\ldots ,t_N.$ Partition $\mathbb R$ into $N+1$ intervals $I_0,I_1,\ldots ,I_N$ where $I_{\alpha }=]t_{\alpha },t_{\alpha +1}[$ with $t_0=-\infty$ and $t_{N+1}=\infty$ and $I_M=\{t_{\alpha } \}$ for $N+1\leqslant M \leqslant 2N$. Let $\sigma :\mathbb R \rightarrow \mathbb R$ be any bijection on $\mathbb R$ such that $\mathcal {A}$ is invariant under $\sigma$ and let $\tilde{\sigma }:\mathcal {A}\rightarrow \mathcal {A}$ be the automorphism on $\mathcal {A}$ induced by $\sigma$. Let

$$\begin{aligned} C_k:=\left\{ x \in \mathbb {R} ~|~k \text{ is } \text{ the } \text{ smallest } \text{ positive } \text{ integer } \text{ such } \text{ that } x, \sigma ^k(x) \in I_{\alpha } \right. \\ \nonumber \left. \text { for some } \alpha = 0, \ldots 2N \right\} . \end{aligned}$$

(16)

Then the unique maximal commutative subalgebra of $\mathcal {A}\rtimes _{\tilde{\sigma }}\mathbb Z$ that contains $\mathcal {A}$ is given by

$$\begin{aligned} \mathcal {A}'=\left\{ \sum _{n\in \mathbb Z~:~k\mid n}\left( \sum _{\alpha _n=0}^{2N}a_{\alpha _n}\chi _{I_{\alpha _n}}\right) \delta ^n \right\} . \end{aligned}$$

Proof

From (9) we have that the unique maximal commutative subalgebra of $\mathcal {A}\rtimes _{\tilde{\sigma }}\mathbb Z$ that contains $\mathcal {A}$ is precisely the set of elements

$$\begin{aligned} \mathcal {A}'=\left\{ \sum _{n\in \mathbb Z}f_n\delta ^n~|~\text{ for } \text{ all } n\in \mathbb Z:~f_n|_{Sep_{\mathcal {A}}^n(X)}\equiv 0\right\} , \end{aligned}$$

and from (11),

$$\begin{aligned} Sep_{\mathcal {A}}^n(\mathbb R)=\bigcup _{k\not \mid n}C_k. \end{aligned}$$

Combining the two results and using the definition of $h_n\in \mathcal {A}$ as

$$\begin{aligned} h_n=\sum _{\alpha _n=0}^{2N}a_{\alpha _n}\chi _{I_{\alpha _n}}, \end{aligned}$$

we get

$$\begin{aligned} \mathcal {A}'=\left\{ \sum _{n\in \mathbb Z~:~k\mid n}\left( \sum _{\alpha _n=0}^{2N}a_{\alpha _n}\chi _{I_{\alpha _n}}\right) \delta ^n \right\} . \end{aligned}$$

   $\square$

5 Some Examples

In this section we give some examples of how our results hold for well known simple cases. We treat two cases of piece-wise constant functions on the real line; those with one fixed jump point and those with two fixed jump points.

5.1 Piece-Wise Constant Functions with One Jump Point

Let $\mathcal {A}$ be the collection of all piece-wise constant functions on the real line with one fixed jump point $t_0.$ Following the methods in the previous section $\mathbb {R}$ is partitioned into three intervals $I_0=(-\infty , t_0),\quad I_1=(t_0,\infty )$ and $I_2=\{t_0\}.$ Then we can write $h\in \mathcal {A}$ as

$$\begin{aligned} h=\sum _{\alpha =0}^2 a_{\alpha }\chi _{I_{\alpha }}= a_0\chi _{I_0}+a_1\chi _{I_1}+a_2\chi _{I_2}. \end{aligned}$$

(17)

Let $\sigma : \mathbb {R}\rightarrow \mathbb {R}$ be any bijection on $\mathbb {R}$ and let $\tilde{\sigma }$ be the automorphism on $\mathcal {A}$ induced by $\sigma$. Note that by the first part of Lemma 3, invariance of the algebra $\mathcal {A}$ implies that $\sigma (t_0)=t_0.$ It follows therefore that $\sigma (I_0)=I_0$ or $\sigma (I_0)=I_1.$ We treat these two cases below.

5.1.1 $\sigma (I_0)=I_0$

In this case (and by bijectivity of $\sigma$), we have that $\sigma (I_1)=I_1$ and since $\sigma (t_0)=t_0,$ then for every $x\in \mathbb {R},~h\in \mathcal {A}$ and $n\in \mathbb {Z}$

$$\begin{aligned} \tilde{\sigma }^nh(x):=h\circ \sigma ^{-n}(x)=h(x), \end{aligned}$$

since x and $\sigma ^{-n}(x)$ will lie in the same interval. Therefore, all intervals $I_{\alpha },~\alpha =0,1,2$ belong to $C_1$ and hence

$$\begin{aligned} {Sep_{\mathcal {A}}^n}(\mathbb R)=\bigcup _{k\not \mid n}C_k =\emptyset . \end{aligned}$$

Therefore, the maximal commutative subalgebra will be given by

$$\begin{aligned} \mathcal {A}'=\left\{ \sum _{n\in \mathbb Z}f_n\delta ^n~|~\text{ for } \text{ all } n\in \mathbb Z:~f_n|_{Sep_{\mathcal {A}}^n(X)}\equiv 0\right\} \\ =\mathcal {A}\rtimes _{\tilde{\sigma }}\mathbb Z. \end{aligned}$$

5.1.2 $\sigma (I_0)=I_1$

In this case (and by bijectivity of $\sigma$), we have that $\sigma (I_1)=I_0$ and since $\sigma (t_0)=t_0,$ then for every $x\in \mathbb {R},~h\in \mathcal {A}$ and $n\in \mathbb {Z}\text { such that }2\mid n$ we have

$$\begin{aligned} \tilde{\sigma }^nh(x):=h\circ \sigma ^{-n}(x)=h(x), \end{aligned}$$

since x and $\sigma ^{-n}(x)$ will lie in the same interval. And for odd n, $\tilde{\sigma }^n(h)(x)=h(x)$ if and only if $x=t_0.$ Therefore, we have,

$$\begin{aligned} C_1=\{I_{\alpha }~|~\sigma (I_{\alpha })=I_{\alpha }\}=I_2, \end{aligned}$$

and

$$\begin{aligned} C_2=\{I_{\alpha }~|~ \sigma ^2(I_{\alpha })=I_{\alpha }\}=I_0\cup I_1. \end{aligned}$$

Therefore,

$$\begin{aligned} {Sep_{\mathcal {A}}^n}(\mathbb R)=\bigcup _{k\not \mid n}C_k ={\left\{ \begin{array}{ll} C_2&{} \text { if } k=1,\\ \emptyset &{} \text { if } k=2. \end{array}\right. } \end{aligned}$$

Therefore, the maximal commutative subalgebra will be given by

$$\begin{aligned} \mathcal {A}'=\left\{ \sum _{n\in \mathbb Z~:~k\mid n}\left( \sum _{\alpha _n=0}^{2}a_{\alpha _n}\chi _{I_{\alpha _n}}\right) \delta ^n \right\} \\ =\left\{ \sum _{n\in \mathbb Z~:~2\mid n}\left( \sum _{\alpha =0}^{2}a_{\alpha _n}\chi _{I_{\alpha }}\right) \delta ^n \right\} \\ =\left\{ \sum _{m\in \mathbb {Z}}(a_{0,m}\chi _{I_0}+a_{1,m}\chi _{I_1}+a_{2,m}\chi _{I_2})\delta ^{2m}+\sum _{m\in \mathbb {Z}}\left( a_{2,m}\chi _{I_2}\right) \delta ^{2m+1}\right\} . \end{aligned}$$

5.2 Piece-Wise Constant Functions with Two Jump Points

Let $\mathcal {A}$ be the collection of all piece-wise constant functions on the real line with two fixed jump points at $t_0$ and $t_1.$ Following the methods in the previous section $\mathbb {R}$ is partitioned into intervals $I_0=]-\infty , t_0[,\quad I_1=]t_0,t_1[\quad I_2=]t_1,\infty [,\quad I_3=\{t_0\}$ and $I_4=\{t_1\}.$ Then we can write $h\in \mathcal {A}$ as

$$\begin{aligned} h=\sum _{\alpha =0}^4 a_{\alpha }\chi _{I_{\alpha }}. \end{aligned}$$

(18)

Let $\sigma : \mathbb {R}\rightarrow \mathbb {R}$ be any bijection on $\mathbb {R}$ and let $\tilde{\sigma }$ be the automorphism on $\mathcal {A}$ induced by $\sigma$. Note that by the first part of Lemma 3, invariance of the algebra $\mathcal {A}$ implies that $\sigma (t_0)=t_0$ (and $\sigma (t_1)=t_1$) or $\sigma (t_0)=t_1$ (in which case $\sigma (t_1)=t_0$). Below we give a description for the maximal commutative subalgebra of $\mathcal {A}\rtimes _{\tilde{\sigma }}\mathbb {Z}$ for different types of $\sigma .$

5.2.1 $\sigma (I_{\alpha })=I_{\alpha }$ for all $\alpha =0,\ldots , 4$

This case is similar to the one in Sect. 5.1.1 in the sense that, for every $x\in \mathbb {R},~h\in \mathcal {A}$ and $n\in \mathbb {Z}$

$$\begin{aligned} \tilde{\sigma }^nh(x):=h\circ \sigma ^{-n}(x)=h(x), \end{aligned}$$

since x and $\sigma ^{-n}(x)$ will lie in the same interval. Therefore, all intervals $I_{\alpha },~\alpha =0,\ldots ,4$ belong to $C_1$ and hence

$$\begin{aligned} {Sep_{\mathcal {A}}^n}(\mathbb R)=\bigcup _{k\not \mid n}C_k =\emptyset . \end{aligned}$$

Therefore, the maximal commutative subalgebra will be given by

$$\begin{aligned} \mathcal {A}'=\left\{ \sum _{n\in \mathbb Z}f_n\delta ^n~|~\text{ for } \text{ all } n\in \mathbb Z:~f_n|_{Sep_{\mathcal {A}}^n(X)}\equiv 0\right\} \\ =\mathcal {A}\rtimes _{\tilde{\sigma }}\mathbb Z. \end{aligned}$$

5.2.2 $\sigma (I_0)=I_1,~\sigma (I_1)=I_0$ and $\sigma (I_{\alpha })=I_{\alpha },~\alpha =2,3,4$

In this case (and by bijectivity of $\sigma$), we have that $\sigma (I_1)=I_0$ and therefore for every $x\in \mathbb {R},~h\in \mathcal {A}$ and $n\in \mathbb {Z}\text { such that }2\mid n$ we have

$$\begin{aligned} \tilde{\sigma }^nh(x):=h\circ \sigma ^{-n}(x)=h(x), \end{aligned}$$

since x and $\sigma ^{-n}(x)$ will lie in the same interval. And for odd n,  $\tilde{\sigma }^n(h)(x)=h(x)$ if and only if $x\in I_2\cup I_3\cup I_4.$ Therefore, we have,

$$\begin{aligned} C_1=\{I_{\alpha }~|~\sigma (I_{\alpha })=I_{\alpha }\}=I_2\cup I_3\cup I_4, \end{aligned}$$

and

$$\begin{aligned} C_2=\{I_{\alpha }~|~\sigma ^2(I_{\alpha })=I_{\alpha }\}=I_0\cup I_1. \end{aligned}$$

Therefore,

$$\begin{aligned} Sep_{\mathcal {A}}^n(\mathbb R)=\bigcup _{k\not \mid n}C_k ={\left\{ \begin{array}{ll} C_2&{} \text { if } k=1,\\ \emptyset &{} \text { if } k=2. \end{array}\right. } \end{aligned}$$

It follows that for $n\in \mathbb {Z}$ such that $2\mid n$, the maximal commutative subalgebra will be given by

$$\begin{aligned} \mathcal {A}_1'=\left\{ \sum _{n\in \mathbb Z~:~k\mid n}\left( \sum _{\alpha _n=0}^{2N}a_{\alpha _n}\chi _{I_{\alpha _n}}\right) \delta ^n \right\} \\ =\left\{ \sum _{n\in \mathbb Z~:~2\mid n}\left( \sum _{\alpha _n=0}^{2N}a_{\alpha _n}\chi _{I_{\alpha _n}}\right) \delta ^n \right\} \\ =\sum _{m\in \mathbb {Z}}\left( \sum _{\alpha =0}^4a_{\alpha ,m}\chi _{I_{\alpha }}\right) \delta ^{2m}. \end{aligned}$$

And for odd n,  we have

$$\begin{aligned} \mathcal {A}_2'=\sum _n (a_{2,m}\chi _{I_2}+a_{3,m}\chi _{I_3}+a_{4,m}\chi _{I_4})\delta ^n. \end{aligned}$$

Therefore, the commutant $\mathcal {A}'$ is given by:

$$\begin{aligned} \mathcal {A}=\left\{ \sum _{m\in \mathbb {Z}}\left( \sum _{\alpha =0}^4a_{\alpha ,m}\chi _{I_{\alpha }}\right) \delta ^{2m}+\sum _{m\in \mathbb {Z}} (a_{2,m}\chi _{I_2}+a_{3,m}\chi _{I_3}+a_{4,m}\chi _{I_4})\delta ^{2m+1}\right\} . \end{aligned}$$

Similar results can be obtained for the following cases

• 1. $\sigma (I_0)=I_1,\quad \sigma (I_1)=I_0,\quad \sigma (I_3)=I_4,\quad \sigma (I_4)=I_3$ and $\sigma (I_2)=I_2.$

• 2. $\sigma (I_0)=I_2,\quad \sigma (I_2)=I_0$ and $\sigma (I_{\alpha })=I_{\alpha },\quad \alpha =1,3,4.$

• 3. $\sigma (I_0)=I_2,\quad \sigma (I_2)=I_0,\quad \sigma (I_3)=I_4\quad \sigma (I_4)=I_3$ and $\sigma (I_1)=I_1.$

• 4. $\sigma (I_1)=I_2,\quad \sigma (I_2)=I_1$ and $\sigma (I_{\alpha })=I_{\alpha }\quad \alpha =0,3,4.$

• 5. $\sigma (I_1)=I_2,\quad \sigma (I_2)=I_1,\quad \sigma (I_3)=I_4,\quad \sigma (I_4)=I_3$ and $\sigma (I_0)=I_0.$

Since in all these cases, $\sigma ^2(I_{\alpha })=I_{\alpha },~\alpha =0,\ldots ,4.$

5.2.3 $\sigma (I_0)=I_1,~\sigma (I_1)=I_2,~\sigma (I_2)=I_0$ and $\sigma (I_{\alpha })=I_{\alpha },~\alpha =3,4$

In this case, using similar methods we have,

$$\begin{aligned} C_1=\{I_{\alpha }~|~ \sigma (I_{\alpha })=I_{\alpha }\}=\cup I_3\cup I_4,\quad C_2=\emptyset , \end{aligned}$$

and

$$\begin{aligned} C_3=\{I_{\alpha }~|~ \sigma ^3(I_{\alpha })=I_{\alpha }\}=I_0\cup I_1\cup I_2. \end{aligned}$$

Therefore,

$$\begin{aligned} {Sep_{\mathcal {A}}^n}(\mathbb R)=\bigcup _{k\not \mid n}C_k ={\left\{ \begin{array}{ll} C_3&{} \text { if } k\ne 3,\\ \emptyset &{} \text { if } k=3. \end{array}\right. } \end{aligned}$$

It follows that for $n\in \mathbb {Z}$ such that $3\mid n$, the maximal commutative subalgebra will be given by

$$\begin{aligned} \mathcal {A}_1'=\left\{ \sum _{n\in \mathbb Z~:~k\mid n}\left( \sum _{\alpha _n=0}^{2N}a_{\alpha _n}\chi _{I_{\alpha _n}}\right) \delta ^n \right\} \\ =\left\{ \sum _{n\in \mathbb Z~:~3\mid n}\left( \sum _{\alpha =0}^{4}a_{\alpha _n}\chi _{I_{\alpha }}\right) \delta ^n \right\} \\ =\sum _{m\in \mathbb {Z}}\left( \sum _{\alpha =0}^4a_{\alpha ,m}\chi _{I_{\alpha }}\right) \delta ^{3m}. \end{aligned}$$

If $3\not \mid n,$ then

$$\begin{aligned} \mathcal {A}_2'=\sum _n (a_{3,n}\chi _{I_3}+a_{4,n}\chi _{I_4})\delta ^n. \end{aligned}$$

Therefore:

$$\begin{aligned} \mathcal {A}'=\left\{ \sum _{m\in \mathbb {Z}}\left( \sum _{\alpha =0}^4a_{\alpha ,m}\chi _{I_{\alpha }}\right) \delta ^{3m}+\sum _n (a_{3,n}\chi _{I_3}+a_{4,n}\chi _{I_4})\delta ^n\right\} . \end{aligned}$$

5.2.4 $\sigma (I_0)=I_1,~\sigma (I_1)=I_2,~\sigma (I_2)=I_0$ and $\sigma (I_{3})=I_4,~\sigma (I_4)=I_3$

In this case, using similar methods we have,

$$\begin{aligned} C_1=\emptyset ,\quad C_2=I_3\cup I_4, \end{aligned}$$

and

$$\begin{aligned} C_3=\{I_{\alpha }~|~ \sigma ^3(I_{\alpha })=I_{\alpha }\}=I_0\cup I_1\cup I_2. \end{aligned}$$

Therefore,

$$\begin{aligned} Sep_{\mathcal {A}}^n(\mathbb R)=\bigcup _{k\not \mid n}C_k ={\left\{ \begin{array}{ll} \mathbb {R}\setminus C_3&{} \text { if } k= 3,\\ \mathbb {R}\setminus C_2&{} \text { if } k=2,\\ \mathbb {R} &{} \text { if } k=1. \end{array}\right. } \end{aligned}$$

It follows that for $n\in \mathbb {Z}$ such that $3\mid n$, the maximal commutative subalgebra will be given by

$$\begin{aligned} \mathcal {A}_1'=\left\{ \sum _{n\in \mathbb Z~:~k\mid n}\left( \sum _{\alpha _n=0}^{2N}a_{\alpha _n}\chi _{I_{\alpha _n}}\right) \delta ^n \right\} \\ =\left\{ \sum _{n\in \mathbb Z~:~3\mid n}\left( \sum _{\alpha _n=0}^{2N}a_{\alpha _n}\chi _{I_{\alpha _n}}\right) \delta ^n \right\} \\ =\sum _{m\in \mathbb {Z}}\left( a_{0,m}\chi _{I_0}+a_{1,m}\chi _{I_1}+a_{2,m}\chi _{I_2}\right) \delta ^{3m}. \end{aligned}$$

If $2\mid n,$ then

$$\begin{aligned} \mathcal {A}_2'=\sum _{m\in \mathbb {Z}} (a_{3,m}\chi _{I_3}+a_{4,m}\chi _{I_4})\delta ^{2m}, \end{aligned}$$

and for all other values of $n,~\mathcal {A}'=\mathcal {A}.$ Hence:

$$\begin{aligned} \mathcal {A}'=\left\{ \sum _{m\in \mathbb {Z}}\left( a_{0,m}\chi _{I_0}+a_{1,m}\chi _{I_1}+a_{2,m}\chi _{I_2}\right) \delta ^{3m}+\sum _{m\in \mathbb {Z}} (a_{3,m}\chi _{I_3}+a_{4,m}\chi _{I_4})\delta ^{2m}\right\} . \end{aligned}$$